physics - dpp - Resonance

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E ST INFORMA TIO

Course : VIJETA (JP)

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DPPDPPDPPDAILY PRACTICE PROBLEMS

NO. B19 TO B20

2. DPP Syllabus :

DPP No. : B19 (JEE-ADVANCED) Total Marks : 41 Max. Time : 24 min. Single choice Objective ('–1' negative marking) Q.1 to Q.3 (3 marks 2 min.) [09, 06] One or more than one options correct type ('–1' negative marking) Q.4 to Q.9 (4 marks 2 min.) [24, 12] Match the Following (no negative marking) Q.10 (8 marks 06 min.) [08, 06]

ANSWER KEY OF DPP No. : B19 1. (D) 2. (A) 3. (C) 4. (A,B,D) 5. (A,D) 6. (A,B,C) 7. (B*) 8. (C*) 9. (C*) 10. (A) p,r (B) q (C) p,r,s (D) p,r,s

1. The circuit was in the shown state for a long time. Now if the switch S is closed then the net charge that

flows through the switch S, will be

ifjiFk yEcs le; ls iznf'kZr voLFkk esa FkkA vc ;fn fLop S cUn ¼pkyw½ dj fn;k tkrk gS rks fLop S ls izokfgr dqy vkos'k gksxk &

S

4 F

50V

2 F 4 F

2 F

(A) 3

400C (B) 100 C (C)

3

100 C (D*) 50 C

Sol. S

4 F

50V

2 F

4 F

2 Fq =100 Cf

q =50 Cf

q =100/3 C1

q =50/3 C2

1

2

q=200/3 Ci

q=200/3 Ci

Initial and final charges are marked on 4f

and 2f capacitors as shown. Hence charge passing through segment 1 and 2 are

q1 = 3

100c q1 =

3

50c

charge through switch = q1 + q2 = 50 C.




Sol. S

4 F

50V

2 F

4 F

2 Fq =100 Cf

q =50 Cf

q =100/3 C1

q =50/3 C2

1

2

q=200/3 Ci

q=200/3 Ci

izkjfEHkd o vfUre vkos'k 4f o 2f la/kkfj=kksa ij n'kkZ;s vuqlkj vafdr gSA vr% Hkkx 1 o 2 ls izokfgr vkos'k gSa :-

q1 = 3

100c q1 =

3

50c

fLop ls izokfgr vkos'k = q1 + q2 = 50 C

2. A cyclic process ABCD is shown on V-T diagram. The P-T and P-V diagram for this cyclic process is

shown in the figure(given below). Select the correct choices.

,d pØh; izØe ABCD dk V-T oØ fp=k esa iznf'kZr gSA pØh; izØe ds fy;s uhps iznf'kZr P-T rFkk P-V oØksa esa ls lgh oØ dk p;u dhft,A

(A*) (B) (C) (D)

3. The figure shows the tungsten filament of a bulb with constant diameter except a piece of it which has

half the diameter of the rest of the wire. Assume that the temperature is constant within each part and changes suddenly between the parts. If the temperature of thick part is 2000 K, the temperature of the thin part of the filament is (Assume that no radiations are incident on filament)

iznf'kZr fp=k fu;r O;kl ds VaxLVu fQykesaV ds cYc dks iznf'kZr djrk gSa ftudks vk/ks O;kl ds VaxLVu rkj }kjk tksM+k x;k gSA izR;sd Hkkx esa rkieku fu;r ekusa rFkk Hkkxksa ds e/; rkieku vpkud ifjofrZr gksrk gSA ;fn eksVs Hkkx dk rkieu 2000 K gks rks irys Hkkx ds rUrq dk rkieku gksxk % (;g ekfu, fd fQykesaV ij dksbZ fofdj.k vkifrr ugha gksrh gS)

(A) 2000 K (B) 21/4(2000) K (C*) 81/4|(2000) K (D) 41/2(2000) K Sol. In steady state

2R = T4 × surface area

22r

= T42r

T4 3r

1

4/3

1

2

2

1

r

r

T

T




4. The vessel shown in the figure has two sections. The lower part is a rectangular vessel with area of cross-section A and height h. The upper part is a conical vessel of height h with base area ‘A’ and top

area ‘a’ and the walls of the vessel are inclined at an angle 30° with the vertical. A liquid of density fills both the sections upto a height 2h. Neglecting atmospheric pressure.

fp=k esa n'kkZ;s vuqlkj ,d uyh ds nks Hkkx gSA fupyk Hkkx ,d vk;rkdkj uyh gSa ftldk vuqizLFk dkV dk {ks=kQy A rFkk ÅapkbZ h gSA Åijh Hkkx ,d 'kaDokdkj uyh gS ftlds vk/kkj dk {ks=kQy A rFkk ÅapkbZ h gS vkSj 'kadq dk Åijh

{ks=kQy a gSA 'kaDokdkj uyh dh nhokj m/okZ/kj ls 30° ds dks.k ij gSA ,d nzo ftldk /kuRo gS] nksuksa Hkkxksa esa 2h

ÅapkbZ rd Hkjrs gSA rc ¼ok;qe.Myh; nkc dks ux.; ekfu;s½

(A*) The force F exerted by the liquid on the base of the vessel is 2hgA

(B*) the pressure P at the base of the vessel is 2hg (C) the weight of the liquid W is greater than the force exerted by the liquid on the base (D*) the walls of the vessel exert a downward force (F–W) on the liquid.

(A*) uyh ds vk/kkj ij æo }kjk vkjksfir cy F = 2hgA gSA

(B*) uyh ds vk/kkj ij nkc P = 2hg gSA

(C) æo dk Hkkj W, vk/kkj ij æo }kjk vkjksfir cy ls T;knk gSA (D*) uyh dh nhokjsa æo ij uhps dh rjQ ,d cy (F–W) vkjksfir djrh gSA 5. A wooden block, with a coin placed on its top, floats in water as shown in figure. The distances l and h

are shown there. After some time the coin falls into the water. Then

fp=kkuqlkj ikuh esa rSj jgs ydM+h ds CykWd dh Åijh lrg ij ,d flDdk j[kk gqvk gSA nwfj;k¡ l rFkk h fp=k esa n'kkZbZ xbZ gSaA dqN le; i'pkr~ flDdk ikuh esa fxj tkrk gS rc %

(A*) decreases (B) h increases (C) increases (D*) h decreases

(A*) ?kVsxh (B) h c<+sxh (C) c<+sxh (D*) h ?kVsxh Solution :

6. An electron is at P at t = 0.It is circulating in anticlockwise direction with a constant angular speed along the shown circular path.Magnetic field at Q(CQ = 2R,where R is radius of the circle) will be recorded as zero at times:

t = 0 ij ,d bysDVªkWu P ij fLFkr gSA bysDVªkWu okekorZ fn'kk esa fu;r dks.kh; pky ls fp=k esa iznf'kZr o`Ùkh; iFk ij xfreku gSA fdl le; ij Q ij pqEcdh; {ks=k 'kwU; gksxk & (fn;k gS CQ = 2R,tgk¡ R o`Ùk dh f=kT;k gS)

P QC

(A*)

3 (B*)

3

5 (C*)

3

7 (D)

3

8

Sol. Zero magnetic field will be recorded at Q when particle is at A and B such that AQ and BQ are tangent to circle.

Q ij pqEcdh; {ks=k 'kwU; gksxkA tc d.k A rFkk B ij gksaxs rc AQ rFkk BQ o`Ùk ij Li'kZ js[kk gksxhA




Comprehension

vuqPNsn

Two thin similar spherical glass pieces are joined together front to front with rear portion of one piece silvered and the combination of glass pieces is placed at a distance 60 cm from a screen. A point object is placed on the optical axis of the combination at the mid point of the line joining this combination and screen. Initially there is air between glass pieces as well as between combination and screen.

nks irys le:i xksyh; dkWp ds VqdM+ks dks lkeus ls tksM+dj] rFkk ihNs ds Hkkx dks jtrhr (silverd) djds] mijksDr la;kstu dks insZ ls 60 cm dh nwjh ij j[kk tkrk gSA ,d fcUnw oLrq O la;kstu dh izdkf'kd v{k ij] la;kstu o insZ ds e/; fcUnw ij] j[kh xbZ gSA izkajHk esa dk¡p ds VqdM+ks ds e/; gok Hkjh gS rFkk insZ ,o la;kstu ds e/; Hkh gok gSA

7. If image of the object is formed on the screen, then the radius of curvature of the either glass piece is :

;fn oLrq dk izfrfcEc insZ ij curk gS rks dkWp ds VqdM+s dh oØrk f=kT;k D;k gksxh A (A) 60 cm (B*) 40 cm (C) 30 cm (D) 15 cm

8. If air between the glass pieces is replaced by water

3

4, the new system (combination of glass

pieces and water) behaves like :

;fn dk¡p ds VqdM+ks ds e/; ok;q ds LFkku ij ty

3

4 Hkj fn;k tk;s rks u;s la;kstu ¼dkWp VqdM+ks rFkk ty

dk la;kstu½ dk O;ogkj fuEu gksxk : (A) Convex mirror of focal length 12 cm (B) Convex mirror of focal length 24 cm (C*) Concave mirror of focal length 12 cm (D) Concave mirror of focal length 24 cm

(A) 12 cm Qksdl nwjh dk mÙky niZ.k (B) 24 cm Qksdl nwjh dk mÙky niZ.k (C*) 12 cm Qksdl nwjh dk vory niZ.k (D) 24 cm Qksdl nwjh dk vory niZ.k 9. In the previous question, the distance through which the object must be displaced so that the sharp

image is again formed on screen is :

mijksDr iz'u esa oLrq dks fdruh nwjh vksj foLFkkfir fd;k tk;s rkfd oLrq dk Li"V izfrfcEc iqu% insZ ij izkIr gks tk,A

(A) 20 cm towards the combination (B) 20 cm away from the combination (C*) 15 cm towards the combination (D) 15 cm away from the combination

(A) 20 cm la;kstu dh rjQ (B) 20 cm la;kstu ls nwj dh rjQ

(C*) 15 cm la;kstu dh rjQ (D) 15 cm la;kstu ls nwj dh rjQ

Sol.(7 to 9) u = 30 v = – 60

f = –2

r = – 20 r = 40 cm

LM f

2–

f

1

F

1

Gives ¼fn;k gS½ F = – 12 cm

From ls F = – 12 cm

& V = – 60 cm u = – 15 cm

So, shift = 15 cm towards mirrors. (vr%, f'k¶V = 15 cm gS tksfd niZ.k dh vkSj)




10. A uniform solid cylinder of mass m and radius R is placed on a rough horizontal surface where friction is sufficient to provide pure rolling. A horizontal force of magnitude F is applied on cylinder at different positions with respect to its centre O in each of four situations of column-I, due to which magnitude of acceleration of centre of mass of cylinder is a. Match the appropriate results in column-II for conditions of column-I.

m nzO;eku o R f=kT;k ds ,d leku Bksl csyu dks {kSfrt [kqjnjh lrg ij j[kk tkrk gS tgk¡ lEidZ lrg ij 'kq) ykSVuh xfr djkus ds fy, i;kZIr ?k"kZ.k mifLFkr gSA F ifjek.k dk ,d {kSfrt cy csyu ij LrEHk-I esa n'kkZbZ xbZ pkjksa fHkUu fLFkfr;ksa ¼dsUnz ds lkis{k½ esa vkjksfir fd;k tkrk gS ftlds dkj.k csyu ds nzO;eku dsUnz dk Roj.k dk ifjek.k a gSA LrEHk-I esa nh xbZ izR;sd fØ;k ds lgh ifj.kke dk p;u LrEHk-II ls dhft;sA

Column-I Column-II

(A)

F

RO

(p) Friction force on cylinder will not be zero.

(p) csyu ij ?k"kZ.k cy 'kwU; ugh gksxkA

(B)

F

OR2

(q) a =m

F

(C) F O

(r) a m

F

(D) O

R2

F

(s) the direction of friction force acting on

cylinder is towards left

csyu ij ?k"kZ.k cy dh fn'kk cka;h vksj gSA Ans. (A) p,r (B) q (C) p,r,s (D) p,r,s

NCERT Questions 6.1 to 6.17





E ST INFORMA TIO


PPHHYYSSIICCSS


NO. B19 TO B20

DPP No. : B20 (JEE-ADVANCED) Total Marks : 41 Max. Time : 27 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.3 (3 marks 2 min.) [09, 06] One or more than one options correct type ('–1' negative marking) Q.4 to Q.8 (4 marks 2 min.) [20, 10] Subjective Questions ('–1' negative marking) Q.9 (4 marks 5 min.) [04, 05] Match the Following (no negative marking) Q.10 (8 marks 06 min.) [08, 06]

ANSWER KEY OF DPP No. : B20 1. (B) 2. (A) 3. (A) 4. (B,C) 5. (A,C,D) 6. (A) 7. (C) 8. (B) 9. 64 10. (A) – p,r ; (B) – q ; (C) – s ; (D) – p,r,t

1. The capacitors A and B are connected in series with a battery as shown in the figure. When the switch

S is closed and the two capacitors get charged fully, then-

fp=kkuqlkj la/kkfj=kksa A rFkk B dks ,d cSVjh ls la;ksftr fd;k x;k gSA fLop S dks pkyw djus ij nksuksa la/kkfj=k iw.kZ:i ls vkosf'kr gks tkrs gS] rks&

S

10V

A B

2 F 3 F

(A) The potential difference across the plates of A is 4V and across the plates of B is 6V (B*) The potential difference across the plates of A is 6V and across the plates of B is 4V (C) The ratio of electric energies stored in A and B is 2 : 3 (D) The ratio of charges on A and B is 3 : 2

(A) A dh IysVksa ds e/; foHkokUrj 4V rFkk B dh IysVksa ds e/; 6V gSA (B) A dh IysVksa ds e/; foHkokUrj 6V rFkk B dh IysVksa ds e/; 4V gSA (C) A rFkk B esa lafpr fo|qr ÅtkZvksa dk vuqikr 2 : 3 gSA (D) A rFkk B ij vkos'kksa dk vuqikr 3 : 2 gSA

Sol. In serues Js.kh Øe esa qA = qB = q = 10 × 5

32 × 10–6 = 12 C

VA =AC

q= 6V VB=

BC

q=4V

In series Js.kh Øe esa U C

1

2

3

C

C

U

U

A

B

B

A

2. Imagine a short dipole is at the centre of a spherical surface. If magnitude of electric field at a certain point on the surface of sphere is 10 N/C, then which of the following cannot be the magnitude of electric field anywhere on the surface of sphere

,d xksykdkj lrg ds dsUnz ij ,d NksVk oS|qr f}/kqzo ekfu;sA ;fn xksys dh lrg ds fdlh fcUnq ij oS|qr {ks=k dk ifjek.k 10 N/C gS, rc fuEu esa ls dkSulk xksys dh lrg ds fdlh Hkh fcUnq ij oS|qr {ks=k dk ifjek.k ugha gks ldrk gSA

(A*) 4 N/C (B) 8 N/C (C) 16 N/C (D) 20 N/C




Sol. If Eaxial = 10 N/C then Eeq = 5 N/C If Eeq = 10 N/C then Eaxial = 20 N/C So magnitude of electric field at any point

5N/C E 20 N/C

Hindi ;fn Ev{k

= 10 N/C

rc Efuj{k

= 5 N/C

;fn Efuj{k

= 10 N/C

rc Ev{k

= 20 N/C

vr% xksys ds fdlh fcUnq ij oS|qr {ks=k dk ifjek.k 5N/C E 20 N/C 3. A sphere of radius R and relative density 4 is hanging with the help of a string such that it remains just

inside water. The ratio of force exerted by the liquid on upper and lower half of the sphere is (take Patm

= 0)

R f=kT;k ,oa lkis{k ?kuRo 4 dk ,d xksyk Mksjh dh lgk;rk ls bl izdkj yVdk gqvk gS fd ;g ikuh ds Bhd vUnj

fLFkr gksrk gSA nzo }kjk xksys ds Åijh v)Z Hkkx ,oa fupys v)Z Hkkx ij vkjksfir cy dk vuqikr gksxk (Patm = 0 ysaa)

(A*) 1 : 5 (B) 1 : 4 (C) 2 : 5 (D) 1 : 3

Sol. Consider the upper half sphere only, let F1 & F2 be forces exerted by the water on curved and flat

surface respectively

dsoy Åijh v)Z xksys dks yhft,, ekukfd ikuh }kjk oØ ,oa lery lrg ij vkjksfir cy Øe'k% cy F1 o F2 gSA

F2 – F1 = FB

F1

F2

2 3

1

2Rg( R ) F R g

3

3

1

1F R g

3

Similarly for lower half blhizdkj ls fupys v)ZHkkx ds fy,

3 2 B

F F F

2 3

3

2F Rg( R ) R g

3

3

3

5F R g

3




F2

F3

1

3

F 1

F 5

4. A light ray enters into a medium whose refractive index varies along the x-axis as 0

xn(x) n 1

4

where n0 = 1. The medium is bounded by the planes x = 0, x = 1 & y = 0. If the ray enters at the origin

at an angle 30º with x–axis.

(A) equation of trajectory of the light ray is y = [ 3 x – 3 ]

(B*) equation of trajectory of the light ray is y = 2[ 3 x – 3 ]

(C*) the coordinate the point at which light ray comes out from the medium is [1, 2(2 – 3 )]

(D) the coordinate the point at which light ray comes out from the medium is [0, 2(2 – 3 )]

,d izdk'k fdj.k ,d ,sls ek/;e esa izos'k djrh gS ftldk viorZukad x-v{k ds vuqfn'k lEcU/k 0

xn(x) n 1

4

ds vuqlkj ifjofrZr gksrk gSA ;gk¡ n0 = 1 gSA ;g ek/;e x = 0, x = 1 rFkk y = 0 ry ls f?kjk gSA ;fn fdj.k x–v{k

ds lkFk 30º dks.k cukrs gq;s ewy fcUnq ij izos'k djrh gS rks

(A) izdk'k fdj.k ds iFk dh lehdj.k y = [ 3 x – 3 ] gSA

(B*) izdk'k fdj.k ds iFk dh lehdj.k y = 2[ 3 x – 3 ] gSA

(C*) fdj.k ek/;e ds ftl fcUnq ls ckgj fudysxh mlds funsZ'kkad [1, 2(2 – 3 )] gSA

(D) fdj.k ek/;e ds ftl fcUnq ls ckgj fudysxh mlds funsZ'kkad [0, 2(2 – 3 )] gSA




Sol.

(a)

1 × sin 30º = n sin i

sin i = 1

2n

tan i = 2

1

4n 1

dy

dx=

1

x 3

y x

1/2

0 0

dy (x 3) dx

y = 2 x 3 3

(b) when tc x = 1

y = 2( 1 3 3 )

y = 2 2 3

Position at which ray comes out of the medium is (1, 2(2 3)) .

fLFkfr tgk¡ fdj.k ek/;e ls fuxZr gksrh gS (1, 2(2 3))

5. Four point masses each of mass m are placed on vertices of a regular tetrahedron. Distance between

any two masses is r. (A*) Gravitation field at centre is zero

(B) Gravitation potential at centre is 4Gm

r

(C*) Gravitation potential energy of system in 26Gm

r

(D*) Gravitation force on one of the point mass is 2

2

6Gm

r

pkj fcUnq nzO;eku] izR;sd dk nzO;eku m gS] tks ,d leprq"Qyd ds 'kh"kkZsa ij j[ks gq, gSA fdUgh nks nzO;ekuksa ds e/; nwjh r gS %

(A*) dsUnz ij xq:Roh; {kS=k 'kwU; gSA

(B) dsUnz ij xq:Roh; foHko 4Gm

r

gSA

(C*) fudk; dh xq:Roh; fLFkfrt ÅtkZ 26Gm

r

gSA

(D*) fdlh Hkh ,d fcUnq nzO;eku ij xq:Roh; cy 2

2

6Gm

r gSA

Sol. Eg = 0 By symmetry

Potential energy of system = 26Gm

r

Force between two masses F = 2

2

Gm

r




Angle between any two force is 60°

1 2 3F F F F

2

netF = 2 2 2

1 2 2F F F + 1 22F F

+ 2 32F F

+ 3 12F F

2

netF = 3F2 + 3F2

Fnet = 6F

Eg = 0 leferh ls

fudk; dh fLFkfrt ÅtkZ = 26Gm

r

nks nzO;ekuksa ds e/; cy F = 2

2

Gm

r

fdUgh nks cyksa ds e/; dks.k 60° gSA

1 2 3F F F F

2

netF = 2 2 2

1 2 2F F F + 1 22F F

+ 2 32F F

+ 3 12F F

2

netF = 3F2 + 3F2

Fnet = 6F

Paragraph for Question Nos. 6 to 8

iz'u 6 ls 8 ds fy, vuqPNsn

A tank of height 'H' and base area 'A' is half filled with water and there is a very small orifice at the

bottom and there is a heavy solid cylinder having base area3

A.The water is flowing out of the orifice.

Here cylinder is put into the tank to increase the speed of water flowing out. It is given that height of the cylinder is same as that of the tank.

,d Vsad ftldh Å¡pkbZ 'H' gS rFkk vk/kkj dk {ks=kQy 'A' gS] dks ty ls vk/kk Hkjk x;k gSA bl VSad ds iSUns esa ,d

vYi vuqizLFk dkV dk cgqr NksVk fNnz gS rFkk ,d3

A vk/kkj dk ,d Bksl Hkkjh csyu gSA fNnz ls ty ckgj vkrk gSA

ckgj fudyrs ty dh pky esa of) djus ds fy, mijksDr csyu dks Vsad esa j[kk x;k gS rFkk ;g fn;k x;k gS dh csyu dh Å¡pkbZ VSd dh Å¡pkbZ ds leku gSA

HH2

V

6. The speed of water flowing out of the orifice before the cylinder kept inside the tank

csyu dks Vsad esa j[kus ds iwoZ fNnz ls ty ds ckgj vkus dh pky gksxh &

(A*) gH (B) 1.414 gH (C) 2

gh (D)

2

gh




Sol.

P0

HH2

V P0

V = gH2

Hg2gh2

7. The speed of water flowing out of orifice after the cylinder is kept inside it (height of water in tank was

2

Hjust before inserting cyllinder).

csyu dks Vsad esa j[kus ds ckn fNnz ls ty ds ckgj vkus dh pky gksxh ¼csyu dks Vsad esa j[kus ds Bhd igys ikuh

dh Å¡pkbZ 2

H gS½ &

(A) gH3 (B) gH2 (C*) 2

gH3 (D)

2

gH

Sol.

A/3

Let h be height of water column just after putting cylinder,

ekuk csyu dks j[kus ds rqjar ckn ty LrEHk dh Å¡pkbZ h gSA

A2

H

3

A–Ah

h´ = 4

3H

V´ = gH2

3gh2

8. After long time, when the height of water inside the tank again becomes equal to2

H . The solid cylinder

is taken out. Then the velocity of liquid flowing out of orifice will be

yEcs le; ckn] tc nksckjk VSad ds Hkhrj ty dh Å¡pkbZ 2

H gks tkrh gS rc Bksl csyu dks ckgj fudky fy;k tkrk

rks fNnz ls ckgj fudyus okys ty dh osx gksxh &

(A)

2

Hg2 (B*)

3

Hg2 (C)

3

gH (D)

2

gH3

Sol.

A''h3

A–A

2

H h =

3

H

v = gH3

2''gh2




9. A ball of mass m is connected by a light inextensible cord and is rotated in a circle of radius R on a

smooth fixed horizontal table. Initially the angular velocity of the ball was 0 and pulling force applied

was T1. Now the pulling force is increased to T2, until the radius of rotation of the ball becomesR

4. Then

ratio 2

1

T

Tis :

nzO;eku m dh ,d xsan gYdh vforkU; Mksjh ls tqM+h gS rFkk fpduh {kSfrt est ij f=kT;k R ds o`Ùk esa ?kqekbZ tkrh gSA çkjEHk esa xsan dk dks.kh; osx 0 rFkk f[kapko cy T1 FkkA vc f[kapko cy c<+kdj T2 dj fn;k tkrk gS] tc rd

fd ?kw.kZu dh f=kT;k R

4 ugh gks tkrh gS rks vuqikr 2

1

T

T gksxk :

Ans. 64 Sol. No external torque is acting on the ball. So applying angular momentum conservation about point O.

xsan ij cká cyk?kw.kZ ugha gS] vr% fcUnq O ds ifjr% dks.kh; laosx laj{k.k ls

m(r) (r) = constant fu;rkad 2

1

r So vr%

2

1

(1/ 4) = 16 times xquk

T = m2r = (16)2 × 1

4 = 64 times xquk

10. Match the coluumn : fuEu dks lqesfyr dhft, &

Column–I Column–II

(A) (p) may be zero

m1 and m2 are the masses connected to a spring which is initially in natural length and whole system is placed on a smooth horizontal surface. Now m2 is imparted a velocity u as shown in figure. The minimum kinetic energy of mass m2 in resulting motion

(B) (q) must be zero

A rod of mass m1 and length is hinged about a fixed, smooth, horizontal axis passing through its one end. A block

of mass m2 is placed on rod at a distance5

4 as shown in

figure. System is released from rest then normal reaction on mass m2 by rod just after release

(C)

V V

(r) may be positive




One mole of a gas of unknown atomicity undergoes

the process A B (isochoric) and then B C (isothermal)

as shown in P–V diagram. Q for the path ABC

(D) (s) must be positive

Friction coefficient at both the contacts is and tan < .

If system remains at rest then 1

12

m

m–m (t) may be negative

dkWye–I dkWye–II

(A) (p) 'kwU; gks ldrk gSA

m1 rFkk m2 nzO;eku lkekU; yEckbZ dh fLçax ls tqM+s gq, gS rFkk lEiw.kZ fudk; fpduh {kSfrt lrg ij j[kk gqvk gSA vc m2 dks fp=kkuqlkj u osx fn;k tkrk gSA ifj.kkeh xfr ds nkSjku m2 nzO;eku dh U;wure xfrt ÅtkZ

(B) (q) 'kwU; gksxkA

m1 nzO;eku rFkk yEckbZ dh NM+ fLFkj fpdus fcUnq ls fuyfEcr gSA fuyfEcr fcUnq ls {kSfrt v{k xqtj jgh gSA m2 nzO;eku dk CykWd

fuyfEcr fcUnq ls fp=kkuqlkj NM+ ij5

4 nwjh ij j[kk gqvk gSA

fudk; dks fojkekoLFkk ls NksM+k tkrk gS rks NksM+us ds rqjUr i'pkr~ m2 nzO;eku ij NM+ }kjk vkjksfir vfHkyEc izfrfØ;k cy

(C) (r) /kukRed gks ldrk gSA

,d eksy rFkk vKkr ijek.kqdrk okyh xSl ds fy, çØe A B (le&vk;rfu;) rFkkB C (lerkih;)

P–V xzkQ esa çnf'kZr gS rks ABC iFk ds fy,Q gksxkA

(D) (s) /kukRed gksxkA

nksauks lEidksZ ds fy, ?k"kZ.k xq.kkad gS rFkk tan < gSA

;fn fudk; fojkekoLFkk esa gks rks1

12

m

m–m dk eku gksxkA

(t) _.kkRed gks ldrk gSA

Ans. (A) – p,r ; (B) – q ; (C) – s ; (D) – p,r,t

Sol. (C) QAB = 2

f (2P0V0 – P0V0)

=2

f P0V0

QBC = –2 P0V0 n2

–1.4 P0V0 QAB + QBC > 0 For any value of f




(D) Figure shows the forces acting on the two blocks . As we are looking for the maximum value of M / m, the equilibrium is limiting . Hence , the frictional forces are equal to times the corresponding normal forces.

(D) çnf'kZr fp=k esa CykWdksa ij vkjksfir cy fn[kk;s x;s gSA M / m dk vf/kdre eku Kkr djus ds fy, lkE;koLFkk dh fLFkfr ds fy, (nksauks CykWdksa ij ?k"kZ.k cy rFkk muds vfHkyEc cyksa ds xq.kuQy ds cjkcj gksxk)

Equilibrium of the block m gives m nzO;eku dh lkE;koLFkk esa T = N1 and rFkk N1 = mg

which gives gy djus ij T = mg ......(i) Next, consider the equilibrium of the block M. Taking components parallel to the incline

vc M nzO;eku dh lkE;koLFkk esa cyksa ds ?kVd urry ds vuqfn'k ysus ij T + N2 = Mg sin

Taking components normal to the incline urry ds yEcor~ ?kVd ysus ij = Mg cos

These give gy djus ij T = Mg(sin – cos) .......(ii)

From (i) and (ii) ,mg = Mg(sin –cos)

(i) rFkk (ii) ls , mg = Mg(sin – cos)

or, ;k M/m =

cos–sin

If tan< , (sin – cos) < 0 and the system will not slide for any value of M/m.

;fn tan< gks rks (sin –cos) < 0 gksxk rFkk fudk; M/m ds fdlh Hkh eku ds fy, ugha fQlysxkA





E ST INFORMA TIO


PPHHYYSSIICCSS


NO. B21 TO B23

2. DPP Syllabus :

DPP No. : B21 (JEE-MAIN)

Total Marks : 60 Max. Time : 40 min. Single choice Objective ('–1' negative marking) Q.1 to Q.20 (3 marks 2 min.) [60, 40]

ANSWER KEY OF DPP No. : B21 1. (A) 2. (D) 3. (C) 4. (A) 5. (B) 6. (D) 7. (A) 8. (A) 9. (D) 10. (C) 11. (D) 12. (C) 13. (A) 14. (B) 15. (C) 16. (B) 17. (A) 18. (B) 19. (D) 20. (B)

1. A hemispherical shell is uniformly charged positively. The electric field at a point on a diameter away from the centre (inside the boundary of hemispherical shell) is directed

dksbZ v/kZxksyk ,dleku /kukosf'kr gSA xksys ds dsUnz ls ijs blds fdlh O;kl ij ¼v)Z xksys dh lhek ds Hkhrj½ fLFkr fcUnq ij tks dsUnz ls nwj gS] fo|qr {ks=k dh fn'kk

(A*) perpendicular to the diameter bl O;kl ds yEcor~ gSA (B) parallel to the diameter bl O;kl ds lekUrj gSA (C) at an angle tilted towards the diameter bl O;kl dh vksj fdlh dks.k ij >qdh gSA (D) at an angle tilted away from the diameter bl O;kl ls nwj fdlh dks.k ij >qdh gSA

2. A uniform magnetic field exists in region given by ˆ ˆ ˆB 3i 4 j 5k

. A rod of length 5 m is placed along

y-axis is moved along x-axis with constant speed 2 m/sec. Then induced e.m.f. in the rod will be:

fdlh LFkku esa ,d leku pqEcdh; {ks=k] ˆ ˆ ˆB 3i 4 j 5k

}kjk iznf'kZr djrs gSaA y-v{k ds vuqfn'k j[kh 5 eh- yEch

NM+ dks x-fn'kk esa 2 eh-@ls- dh fu;r pky ls xfreku djrs gSaA rks NM+ ij izsfjr fo- ok- c- dk eku gS&

(A) zero 'kwU; (B) 25 V (C) 20 V (D*) 50 V

Sol. = B.(V )

= ˆ ˆ ˆ ˆ ˆ(3i 4 j 5k).[2i 5 j]

= 50 volt. 3. A parallel plate capacitor is made of two dielectric blocks in series, One of the blocks has thickness d1

and dielectric constant k1 and the other has thickness d2 and dielectric constant k2 as shown in fig.. This arrangement can be thought as a dielectric slab of thickness d (= d1 + d2) and effective dielectric constant k. The k is :

dksbZ lekUrj ifV~Vdk la/kkfj=k nks Js.khc) ijkoS|qr xqVdksa ls cuk gSA buesa fp=k esa n'kkZ, vuqlkj ,d xqVds dh eksVkbZ d1 rFkk ijkoS|qrkad k1 rFkk nwljs xqVds dh eksVkbZ d2 rFkk ijkoS|qrkad k2 gSA bl O;oLFkk dks ,d ,slk ijkoS|qr xqVdk ekuk tk ldrk gS ftldh eksVkbZ d (= d1 + d2) rFkk izHkkoh ijkoS|qrkad k gSA rc k dk eku gS %

K1

K2

(A) 21

2211

dd

dkdk

(B) 21

2211

kk

dkdk

(C*) 1 2 1 2

1 2 2 1

k k (d d )

(k d k d )

(D) 21

21

kk

kk2




4. Two batteries of emf 1 and 2 (2 > 1) and internal resistances r1 and r2 respectively are connected in parallel as shown in Fig.

nks cSVfj;kW ftuds emf 1 rFkk 2 (2 > 1) rFkk vkUrfjd izfrjks/k Øe'k% r1 rFkk r2 gSa] fp=k 3.1 esa n'kkZ, vuqlkj lekUrj Øe esa la;ksftr gSaA

(A*) The equivalent emf eq of the two cells is between 1 and 2, i.e. 1 < eq < 2.

nksuksa lsyksa dk rqY; emf rqY;, 1 rFkk 2 ds chp vFkkZr~, 1 < rqY; < 2 gSA.

(B) The equivalent emf eq is smaller than 1

rqY; emf rqY;, 1 ls de gSA

(C) The eq is given by eq = 1 + 2 always

lnSo rqY; = 1 + 2 gksrk gSA

(D) eq is independent of internal resistance r1 and r2.

rqY; vkUrfjd izfrjks/kksa r1 rFkk r2 ij fuHkZj ugha gSA

5. Two cells of emf’s approximately 5V and 10V are to be accurately compared using a potentiometer of length 400 cm.

5V rFkk 10V lfUudV emf ds nks lsyksa dh rqyuk ifj'kq) :i ls 400 cm yEckbZ ds foHkoekih }kjk dh tkuh gSA (A) The battery that runs the potentiometer should have voltage of 8V.

foHkoekih esa mi;ksx gksus okyh cSVjh dh oksYVrk 8V gksuh pkfg,A

(B*) The battery of potentiometer can have a voltage of 15V and R adjusted so that the potential drop across the wire slightly exceeds 10V.

foHkoekih dh oksYVrk 15V gks ldrh gS rFkk R dks bl izdkj lek;ksftr dj ldrs gSaa fd rkj ds fljksa ij foHkoikr 10V ls FkksM+k vf/kd gksA

(C) The first portion of 50 cm of wire itself should have a potential drop of 10V

Lo;a rkj ds igys 50 cm Hkkx ij foHkoikr 10V gksuk pkfg,A

(D) Potentiometer is usually used for comparing resistances and not voltages.

foHkoekih dk mi;ksx izk;% izfrjks/kksa dh rqyuk ds fy, fd;k tkrk gS] foHkoksa ds fy, ughaA 6. An electron is projected with uniform velocity along the axis of a current carrying long solenoid. Which

of the following is true ?

,d bysDVªkWu dks fdlh yEch /kkjkokgh ifjukfydk ds v{k ds vuqfn'k ,dleku osx ls iz{ksfir fd;k tkrk gSA fuEufyf[kr esa dkSu lk izdFku lR; gS \

(A) The electron will be accelerated along the axis

bysDVªkWu v{k ds vuqfn'k Rofjr gksxkA (B) The electron path will be circular about the axis

v{k ds ifjr% bysDVªkWu dk iFk o`Ùkkdkj gksxkA (C) The electron will experience a force at 45º to the axis and hence execute a helical path

bysDVªkWu v{k ls 45º ij cy vuqHko djsxk vkSj blizdkj daqMfyuh iFk ij xeu djsxkA (D*) The electron will continue to move with uniform velocity along the axis of the solenoid.

bysDVªkWu ifjukfydk ds v{k ds vuqfn'k ,dleku osx ls xfr djrk jgsxkA 7. Biot-Savart law indicates that the moving electrons (velocity v) produce a magnetic field B such that

ck;ks lkoVZ fu;r bafxr djrk gS fd v osx ls xfreku bysDVªkWuksa }kjk mRiUu pqEcdh; {ks=k B bl izdkj dk gksrk gS fd

(A*) B v (B) B || v (C) it obeys inverse cube law

;g O;qRØe ?ku fu;e dk ikyu djrk gSA

(D) it is along the line joining the electron and point of observation

;g isz{k.k fcUnq bysDVªkWu dks feykus okyh js[kk ds vuqfn'k gksrk gSA




8. A ray of light incident at an angle on a refracting face of a prism emerges from the other face normally. If the angle of the prism is 5° and the prism is made of a material of refractive index 1.5, the angle of incidence is:

fdlh fizTe ds ,d viorZd Qyd ij dks.k cukrs gq, vkifrr gksus okyh ,d izdk'k fdj.k nwljs Qyd ls vfHkyEcr% fuxZr gksrh gSA ;fn fizTe dk dks.k 5° gS rFkk fizTe 1.5 viorZukad ds inkFkZ dk cuk gS, rks vkiru dks.k gS :

(A*) 7.5 (B) 5° (C) 15° (D) 2.5° 9. A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling

through the slab, the first colour to emerge is : (A) blue (B) green (C) violet (D*) red

'osr izdk'k dk ,d y?kq Lian ok;q ls dk¡p ds ,d LySc ij yEcor~ vkifrr gksrk gSA LySc ls xqtjus ds i'pkr~ lcls igys fuxZr gksus okyk o.kZ gksxk :

(A) uhyk (B) gjk (C) cSaxuh (D) yky 10. Which of the following pairs of physical quantities does not have same dimensional formula? (A) Work and torque (B) Angular momentum and Planck’s constant. (C*) Tension and surface tension. (D) Impulse and linear momentum.

HkkSfrd jkf'k;ksa ds fuEufyf[kr tksM+ksa esa ls fdl tksM+s dk foeh; lw=k leku ugha gSa \ (A) dk;Z vkSj cy&vk?kw.kZ (B) dks.kh; laosx vkSj Iyk¡d fu;rkad

(C) ruko vkSj i`"B ruko (D) vkosx vkSj js[kh; laosx 11. You measure two quantities as A = 1.0 m ± 0.2 m, B = 2.0 m ± 0.2 m. We should report correct value

for AB as:

nks jkf'k;ksa dks eki dj vki mudk eku A = 1.0 m ± 0.2 m, B = 2.0 m ± 0.2 m çkIr djrs gSaA AB dk lgh eku gksxk&

(A) 1.4 m ± 0.4 m (B) 1.41m ± 0.15 m (C) 1.4m ± 0.3 m (D*) 1.4m ± 0.2 m 12. A lift is coming from 8th floor and is just about to stop on 4th floor. Taking ground floor as origin and

positive direction upwards for all quantities, which one of the following is correct? Here x- position, v - velocity, a -acceleration.

,d fy¶V vkBoha eafty ls uhps vk jgh gS vkSj pkSFkh eafty ij :dus okyh gSA ;fn lHkh jkf';ksa ds fy, Hkwry dks ewy fcanq rFkk Åij dh vksj /kukRed fn'kk ysa rks fuEufyf[kr esa dkSu lgh gS \ ;gk¡ x- fLFkfr,

v -osx, a -Roj.k gSA (A) x < 0, v < 0, a > 0 (B) x > 0, v < 0, a < 0 (C*) x > 0, v < 0, a > 0 (D) x > 0, v > 0, a < 0 13. A body of mass 2kg travels according to the law x(t) = pt + qt2 + rt3 where p = 3ms–1 , q = 4ms–2 and

r = 5 ms–3 . The force acting on the body at t = 2 seconds is (A*) 136 N (B) 134 N (C) 158 N (D) 68 N

2kg nzO;eku dk dksbZ fiaM lehdj.k x(t) = pt + qt2 + rt3 ds vuqlkj xfr djrs gS] tgk¡ p = 3ms–1, q = 4ms–2 vkSj r = 5ms–3 gSA

t =2s ij fiaM ij yxus okyk cy gS& – – (A) 136 N (B) 134 N (C) 158 N (D) 68 N




14. A circular loop of mass m and radius r is kept in a horizontal position (X – Y plane) on a rough table as

shown in figure. A uniform magnetic field B is applied parallel to x-axis. The current in the loop, so that

its one edge just lifts from the table, is :

m nzO;eku rFkk r f=kT;k dk ,d o`Ùkkdkj ywi {kSfrt fLFkfr esa (X–Y ry esa) fp=k esa n'kkZ;suqlkj ,d [kqjnjh Vscy ij

j[kk gqvk gSA ,dleku pqEcdh; {kS=k B, x-v{k ds lekUrj vkjksfir fd;k tkrk gSA ywi esa /kkjk D;k gks] rkfd ywi

dk ,d fljk Bhd Vscy ls mB tk;sA

x

y Top view

(A) Br

mg2

(B*) rB

mg

(C)

rB2

mg

(D)

mg

rB

(E) None of these buesa ls dksbZ ugha

Sol. mag = iR2B = mgR

15. A bicyclist comes to a skidding stop in 10 m. During this process, the force on the bicycle due to the road is 200N and is directly opposed to the motion. The work done by the cycle on the road is

,d lkbfdy lokj czssd yxkus ds ckn 10m dh nwjh fQlyrs gq, vkrk gSA bl izfØ;k es lM+d }kjk lkbfdy ij yxk;k x;k cy 200 N gS vkSj xfr dk Bhd fojks/k djrk gSA lkbfdy }kjk lM+d ij fd;k x;k dk;Z gSA

(A) + 2000J (B) – 200J (C*) zero 'kwU; (D) – 20,000J

16. The potential energy function for a particle executing linear SHM is given by 2kx2

1)x(V where k is the

force constant of the oscillator (Fig). For k = 0.5N/m, the graph of V(x) versus x is shown in the figure. A particle of total energy E turns back when it reaches x = ± xm . If V and K indicate the P.E. and K.E., respectively of the particle at x = +xm, then which of the following is correct?

ljy vkorZ xfr (SHM) djrs fdlh d.k dk fLFkfrt ÅtkZ Qyu gS & 2kx2

1)x(V fn;k tkrk gS tgk¡ k nksfy=k

dk cy fu;rkad gSA k = 0.5N/m, ds fy, V(x) vkSj x dk xzkQ fp=k esa n'kkZ;k x;k gSA E ÅtkZ dk dksbZ x = ± xm

igq¡p dj okil ykSVrk gSA ;fn x = +xm, ij V ,oa K Øe'k% d.k dh fLFkfrt ÅtkZ (P.E) ,oa xfrt ÅtkZ (K.E)

fu:fir djrs gksa rks fuEufyf[kr esa dkSu&lk dFku lgh gS \

(A) V = O, K = E (B*) V = E, K = O (C) V < E, K = O (D) V = O, K < E.

17. The density of a non-uniform rod of length 1m is given by (x) = a(1+bx2)

where a and b are constants and o x 1 . The centre of mass of the rod will be at

1m yach fdlh NM+ dk ?kuRo fn;k x;k gS (x) = a(1+bx2)

;gk¡ a rFkk b fLFkjkad gSa rFkk o x = 1 .

bl NM+ dk nzO;eku dk dsUnz gksxk

(A*) )b3(4

)b2(3

(B) )b3(3

)b2(4

(C) )b2(3

)b3(3

(D) )b2(3

)b3(4




Sol. When b 0, the density becomes uniform and hence the centre of mass is at x = 0.5. Only option (A)

tends to 0.5 as b 0

tc b 0, ?kuRo ,d leku gks tkrk gS vkSj blfy, nzO;eku dsUnz x = 0.5 ij gksxkA tc b 0 rks dsoy fodYi (A) dh vksj izo`r gksrk gSA

18. A Merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed

? A person of mass M is standing on it. At one instant, the person jumps off the round,radially away from the centre of the round (as seen from the round). The speed of the round afterwards is

f=kT;k R nzO;eku M ds NYys tSl IysVQkWeZ dk cuk dksbZ esjh & xks&jkmaM dks.kh; pky ls ifjØe.k dj jgk gSA M nzO;eku dk dksbZ O;fDr bl >wys ij [kM+k gSA fdlh {k.k fo'ks"k ij ;g O;fDr bl >qys ls] bl >wys ds dsUnz ls ijs f=kT;k ¼>wys ls ns[kus ij½ dwnrk gSA blds i'pkr~ >wys dh pky gSA

(A) 2 (B*) (C) 2

(D) 0

19. Two identical long conducting wires AOB and COD are placed at right angle to each other, with one

above other such that 'O' is their common point for the two. The wires carry 1 and 2 currents, respectively. Point P is lying at distance 'd' from 'O' along a direction perpendicular to the plane containing the wires. The magnetic field at the point 'P' will be :

nks loZle (,d ls) yEcs pkyd rkj AOB rFkk COD ,d nwljs ds Åij vkil esa yEcor~ j[ks x;s gSa] vkSj 'O' fcUnq ij ,d nwljs dks dkVrs gSaA buls Øe'k% 1 rFkk 2 /kkjk;sa izokfgr gks jgh gSaA fcUnq 'O' rFkk 'd' nwjh ij nksuks rkjksa ds ry ds yEcor~ fn'kk ds vuqfn'k fLFkr fdlh fcUnq 'P' ij pqEcdh; {ks=k dk eku gksxk :

(A)

2

10

d2 (B) )(

d221

0

(C) )(d2

22

21

0

(D*) 2/122

21

0 )(d2

Ans. (D)

Sol.

B

due to wire (A) 1B

= id2

i10

rkj (A) ds dkj.k pqEcdh; {ks=k = 1B

= id2

i10

B

due to wire (B) = 2B

= )j(d2

i20

rkj (B) ds dkj.k pqEcdh; {ks=k B

= )j(d2

i20

|Bnet| = 2

221

0 iid2




20. A capacitor of 7 F is connected as shown in the circuit (Fig.) The internal resistance of the battery is

0.5 . The amount of charge on the capacitor plates will be :

fp=k n'kkZ, vuqlkj ifjiFk 7 F dk la/kkfj=k tqM+k gSA cSVjh dk vkUrfjd izfrjks/k 0.5 gSA la/kkfj=k dh IysVksa ij vkos'k dh ek=kk gksxh &

(A) 0 (B*) 14C (C) 16 C (D*) 8 C






E ST INFORMA TIO


PPHHYYSSIICCSS


NO. B21 TO B23

DPP No. : B22 (JEE-ADVANCED)

Total Marks : 42 Max. Time : 27 min. Single choice Objective ('–1' negative marking) Q.1 to Q.2 (3 marks 2 min.) [06, 04] One or more than one options correct type ('–1' negative marking) Q.3 to Q.8 (4 marks 2 min.) [24, 12] Subjective Questions ('–1' negative marking) Q.9 (4 marks 5 min.) [04, 05] Match the Following (no negative marking) Q.10 (8 marks 6 min.) [08, 06]

ANSWER KEY OF DPP No. : B22 1. (B) 2. (A) 3. (A,D) 4. (B,C) 5. (A,C) 6. (C) 7. (A)

8. (C) 9. R =3a

2bs, = a

22

3

4bs1

a

10. (A) q,s (B) p,s (C) q,s (D) p,s

1. Three particles each of mass m can slide on fixed frictionless horizontal circular tracks in the same

horizontal plane as shown in the figure. The particle m1 moving with speed V0 collides with m2. The coefficient of restitution being e = 0.5. Assuming that m2 and m3 are at rest initially and lie along a radial line before impact and the spring is initially unstretched, then maximum extension in spring in subsequent motion.

m nzO;eku ds rhu d.k fLFkj ?k"kZ.kjfgr {kSfrt o`Ùkkdkj iFk ij leku {kSfrt ry esa fp=kkuqlkj fQly ldrs gSA m1

æO;eku dk d.k V0 pky ls xfr djrk gqvk m2 æO;eku ds d.k ls Vdjkrk gSA izR;koLFkku xq.kkad e = 0.5 gSA ;g ekfu;s fd m2 rFkk m3 izkjEHk esa fojke ij gS rFkk VDdj ds igys f=kT;h; js[kk ds vuqfn'k fLFkr gS] rFkk fLiazx izkjEHk eas ewy yEckbZ esa gSA rc vxyh xfr esa fLiazx esa vf/kdre~ foLrkj gksxkA

(A) 0

3 mV

4 k (B*) 0

3 mV

4 5 k (C) 0

3 2mV

4 5 k (D) 0

3 mV

5 k




Sol. Velocity just after collision

VDdj ds Bhd i'pkr~ osx

V0 = V1 + V2 ....(A)

1

2 = 1 2

0

V V

V

–V1 + V2 = 0V

2 ....(B)

from (A) & (B)

(A) rFkk (B) ls

V1 = 0V

4 & V2 = 03V

4

When maximum extension occur’s then angular speed with respect to centre for mass m2 & m3 are same. Using angular momentum conservation about centre of circle.

tc vf/kdre~ foLRkkj gksxk rc nzO;eku m2 rFkk m3 ds dsUnz ds lkis{k dks.kh; pky leku gksxhA o`Ùk ds dsUnz ds lkis{k dks.kh; laosx laj{k.k ls

m 0

3V

4 (2R) + 0 = mR2 + m (2R)2

= 0V3

10 R

So velcoity of m2 = 3

5 V0

vr% m2 dk osx = 3

5 V0

& velcoity of m3 = 3

10 V0

rFkk m3 dk osx = 3

10 V0

When extension is maximum using energy conservation.

tc foLrkj vf/kdre~ gks rc ÅtkZ laj{k.k ds mi;ksx ls

2

0

1 3m V

2 4

=

2

0

1 3m V

2 5

+

2

0

1 3m V

2 10

+ 2max.

1kx

2 xmax =

3

4 V0

m

5 k




2. Figure shows the roller coaster track. Each car will start from rest at point A and will roll with negligible friction. It is important that there should be at least some small positive normal force exerted by the track on the car at all points, otherwise the car would leave the track. With the above fact, the minimum safe value for the radius of curvature at point B is (g = 10 m/s2) :

fp=k esa jkWyj dksLVj (>wyk) dk iFk çnf'kZr gSA çR;sd dkj fcUnq A ls fojke ls 'kq: gksrh gS rFkk ux.; ?k"kZ.k ls yq<+drh gSA iFk ds lHkh fcUnqvksa }kjk dkj ij /kukRed vfHkyEc cy yxuk vko';d gSA vU;Fkk dkj iFk dks NksM+ nsxhA bl rF; dks /;ku j[krs gq, B fcUnq ij oØrk f=kT;k dk U;wure lqjf{kr eku gS & (g = 10 m/s2) :

(A*) 20 m (B) 10 m (C) 40 m (D) 25 m

Sol. VB = 2 10 10 ; 2Bm v

R < mg ; R >

2Bv

g R > 20 m

3. There is a square loop of side = 10 cm. The resistance of loop is 0.5 /cm. The loop is moved with

constant velocity 2 i m/s in two uniform magnetic field as shown in the figure. The magnitude of magnetic field is 10T in both regions.

;gk¡ = 10 cm Hkqtk dk ,d oxkZdkj ywi gSA ywi dk izfrjks/k 0.5 /cm gSA ywi fp=k esa n'kkZ;s x;s nks ,dleku

pqEcdh; {ks=k esa fu;r osx 2 i m/s ls xfr'khy gSA nksuksa ifj{ks=k esa pqEcdh; {ks=k dk ifjek.k 10 T gSA

x x

x x

x x

D

A

C

B x

y

2 m/s

ˆB 10Tk

ˆB 10Tk

The potential difference across the points : (A*) A & B is 1V (B) B & C is zero (C) C & D is zero (D*) D & A is 1V

fcUnqvksa ds lkis{k foHkokUrj gksxk (A*) A o B ds e/; 1V gSA (B) B o C ds e/; 'kwU; gSA (C) C o D ds e/; 'kwU; gSA (D*) D o A ds e/; 1V gSA

4. A smooth track is in the form of a quarter circle of radius 10 m lies in the vertical plane. A bead moves

from A to B along the circular track under the action of forces 1F

, 2F

and 3F

. Force 1F

acts always

towards point B and is always 20N in magnitude. Force 2F

always acts horizontally and is always 25 N

in magnitude. Force 3F

always acts tangentially to the track and is of constant magnitude 10 N. Select

the correct answer.

,d fpduk iFk 10 m f=kT;k ds ,d or ds pkSFkkbZ Hkkx ds :i esa m/okZ/kj ry esa fLFkr gSA ,d eudk A ls B rd

o`Ùkh; iFk ds vuqfn'k xfr'khy gSA xfr ds nkSjku bl ij cy 1F

, 2F

o 3F

vkjksfir gSA cy 1F

ges'kk fcUnq B dh

rjQ rFkk bldk ifjek.k ges'kk 20N gSA cy 2F

ges'kk {kSfrt fn'kk esa vkjksfir gS rFkk bldk ifjek.k ges'kk 25 N

gS cy 3F

ges'kk o`Ùkh; iFk ds Li'kZ js[kh; fn'kk esa vkjksfir gS rFkk bldk fu;r ifjek.k 10 N gSA lgh mÙkj dk

p;u djksA




(A) Work done by 1F

is 750 J on the bead euds ij 1F

}kjk fd;k x;k dk;Z 750 J gSSA

(B*) Work done by 2F

is 250 J on the bead euds ij 2F

}kjk fd;k x;k dk;Z 250 J gSSA

(C*) Work done by 3F

is 50J on the bead euds ij 3F

}kjk fd;k x;k dk;Z 50J gSSA

(D) Work done by 3F

is 100 2 J on the bead euds ij 3F

}kjk fd;k x;k dk;Z 100 2 J gSSA

Sol. Work done by F1 = 20 × 210 = 2200 J

F1 }kjk fd;k x;k dk;Z = 20 × 10 2 = 200 2 J

Work done F2 = 25 × 10 = 250 J

2F

}kjk fd;k x;k dk;Z = 25 × 10 = 250 J

Work done by F3 = 10 × 2 R

4

= 50 J.

3F

}kjk fd;k x;k dk;Z = 10 ×2 R

4

= 50 J.

5. A point charge of specific charge q

m = 0.1 C/kg is projected in uniform magnetic field. The particle

moves in magnetic field such that its position vector at any instant is given by ˆ ˆ ˆr 3sin t i 3cos t j 4tk

.

Select correct statements from following :

(A*) Magnetic field in space is 10T

(B) The distance traveled by the particle in 5s is 20m

(C*) Power of magnetic force is zero

(D) The radius of curvature of the path is 3m

fof'k"V vkos'k q

m = 0.1 C/kg dk ,d fcUnq vkos'k ,d leku pqEcdh; {kS=k esa iz{kSfir fd;k tkrk gSA d.k pqEcdh;

{kS=k esa bl izdkj xfr djrk gS fd fdlh {k.k ij bldk fLFkfr lfn'k ˆ ˆ ˆr 3sin t i 3cos t j 4tk

}kjk fn;k x;k

gSA fuEu esa ls lgh dFku@dFkuksa dk p;u dhft,A

(A*) ifj{kS=k esa pqEcdh; {kS=k 10T gSA (B) 5s esa d.k }kjk r; dh x;h nwjh 20m gSA

(C*) pqEcdh; cy dh 'kfDr 'kwU; gSA (D) iFk dh oØrk f=kT;k 3m gSA

Sol. = 1 qB

1m

B = 10T

Speed (pky) = 5 m/s




COMPREHENSION

One mole of a diatomic gas is heated under a "Kibolinsky Process" in which gas pressure is

temperature controlled according to law P=5

4CT3/2. Where C is Kibolinsky constant. In this process,

gas is heated by 300 K temperature. Answer the following questions.

,d f}ijek.kqd xSl ds ,d eksy dks ^^fdcksfyUldh^^ (Kibolinsky Process) izfØ;k }kjk Å"ek iznku dh tkrh gS

ftlesa xSl dk nkc fuEu fu;e ds vuqlkj rki ls fu;af=kr gksrk gSA P = 5

4CT3/2 tgk¡ C fdcksfyUldh fu;rkad gS

bl izfØ;k esa xSl dks 300 K rki of) gksus rd Å"ek iznku dh tkrh gSA fuEu iz'uksa dk mÙkj nhft;sA 6. The amount of workdone by gas under above heating is :

nh xbZ Å"ek ds vUrZxr xSl }kjk fd;s x;s dk;Z dh ek=kk gksxh &

(A) 600 R (B) 750 R (C*) – 150 R (D) None of these buesa ls dksbZ ugh 7. The amount of heat supplied to the gas under above heating is :

nh xbZ Å"ek ds vUrZxr xSl dks iznku dh xbZ Å"ek dh ek=kk gksxh &

(A*) 600 R (B) – 750 R (C) – 150 R (D) None of these buesa ls dksbZ ugh 8. Under above process the molar specific heat of gas is :

mijksä izfØ;k ds vUrZxr xSl dh eksyj fof'k"V Å"ek gksxh &

(A) 2

R (B) R (C*) 2 R (D) None of these buesa ls dksbZ ugh

Sol. For P = 5

4CT3/2

We have PV3 = Constant Thus molar specific heat of gas is

C = CV +

31

R

= C

V –

2

R = 2 R [as C

V =

2

R5]

Heat supplied to gas in temperature increment by T = 300 K in this process is

Q = nCT = n (2R) (300) = 600 R Charge in internal energy of gas in this process is

U = nCVT =

2

R5× 300 = 750 R.

Thus work done by the gas is

W = Q – U = – 150 R.

gy% ds fy, or P = 5

4 CT3/2

ge tkurs gS PV3 = fu;rkad vr% xSl dh eksyj fo'kf"V Å"ek gksxh

C = CV +

31

R

= C

V –

2

R = 2 R [D;ksafd C

V =

2

R5]

bl izfØ;k esa T = 300 K rki o`)h esa xSl dks iznku dh xkbZ Å"ek Q = nCT = n (2R) (300) = 600 R

bl izfØ;k esa xSl dh vkUrfjd ÅtkZ esa ifjorZu

U = nCVT =

2

R5× 300 = 750 R.

vr% xSl }kjk fd;k x;k dk;Z W = Q – U = – 150 R.




9. A point moves in the plane so that its tangential acceleration wt = a & its normal acceleration wn = bt4, where a & b are positive constants & t is time. At the moment

t = 0 the point was at rest. Find how the curvature radius R of the point's trajectory & the total acceleration w depend on the distance covered s.

,d fcUnq ,d ry esa xfreku gS mldk Li'khZ; Roj.k wt = a rFkk vfHkyEcor~ Roj.k wn = bt4 gS, tgk¡ a rFkk b

/kukRed fu;rkad vkSj t le; gSaA t = 0 ij fcUnq fojke voLFkk esa FkkA Kkr dhft, fd s nwjh r; djus ij bldh oØrk f=kT;k R ,oa dqy Roj.k W fdl izdkj ls fuHkZj djsaxs \

[ Ans: R =3a

2bs, = a

22

3

4bs1

a

]

10. Three wires are carrying same constant current i in different directions. Four loops enclosing the wires

in different manners are shown. The direction of d is shown in the figure :

rhu rkjksa esa fu;r /kkjk i fHkUu fn'kkvksa esa izokfgr gSA rkjksa ds ?ksjs esa fHkUu izdkj ls pkj ywi n'kkZ;s x;s gSAd dh

fn'kk fp=k esa fn[kk;s vuqlkj gS &

Loop-1

Loop-2

Loop-3

Loop-4

i i

i

Column I Column II

(A) Along closed Loop-1 (p) B .d

= µ0 i

cUn ywi 1 ds vuqfn'k

(B) Along closed Loop-2 (q) B .d

= –µ0 i

cUn ywi 2 ds vuqfn'k

(C) Along closed Loop-3 (r) B .d

= 0

cUn ywi 3 ds vuqfn'k (D) Along closed Loop-4 (s) net work done by the magnetic force to move a unit charge

along the loop is zero.

cUn ywi 4 ds vuqfn'k ,dkad vkos'k dks cUn ywi ds vuqfn'k xfr djkus esa ifj.kkeh pqEcdh; cy }kjk dk;Z 'kwU; gS &

Ans. (A) q,s (B) p,s (C) q,s (D) p,s Sol. Work done by magnetic force on a charge = 0 in any part of its motion.

'S' is matching for all parts (i), (ii), (iii), (iv)

For loop 1 in = – i + i – i = – i B.d

= 0 ( i)

For loop 2 in = i – i + i = i B.d

= 0(i)

For loop 3 in = –i – i +i = –i B.d

= 0 ( i)

For loop 4 in = + i + i – i = +i B.d

= 0(i)

(Note : That current will be taken as positive which produces lines of magnetic field in the same sense

in which d is taken)




Sol. xfr ds fdlh Hkh Hkkx ds fy, pqEcdh; cy }kjk ,d vkos'k ij fd;k x;k dk;Z = 0

'S' izR;sd Hkkx (i), (ii), (iii), (iv) ls esy [kkrk gSA

ywi 1 ds fy, in = – i + i – i = – i B.d

= 0 ( i)

ywi 2 ds fy, in = i – i + i = i B.d

= 0(i)

ywi 3 ds fy, in = –i – i +i = –i B.d

= 0 ( i)

ywi 4 ds fy, in = + i + i – i = +i B.d

= 0(i)

(Note : og /kkjk /kukRed yh tk,xh tks mlh fn'kk esa pqEcdh; {ks=k dh js[kk,a mRiUu djsxh ftlesa d fy;k x;k gSA)

NCERT Questions

7.1 to 7.15

Board level Questions 1. Why is the use of a.c. voltage preferred over d.c. voltage? Give two reasons.

[Alternating current]

d.c. oksYVrk dh rqyuk esa a.c. oksYVrk ds mi;ksx dks izkFkfedrk D;ksa nh tkrh gS\ nks dkj.k nhft,A

Sol.: The main reason for preferring use of ac voltage over dc voltage is that ac voltages can be easily and

efficiently converted from one voltage to the other by means of transformer. Further, electrical energy

can be transmitted economically over long distances using AC.

d.c. oksYVrk dh rqyuk esa ac oksYVrk ds mi;ksx dks izkFkfedrk nsus dk eq[; dkj.k gS fd ,slh oksYVrk dks ljyrk rFkk n{krk ls ,d oksYVrk ls vU; oksYVrk esa Vªk¡lQkWjej dh lgk;rk ls ifjofrZr dj ldrs gSA blds vfrfjDr

oS/kqr ÅtkZ dks ac ds :Ik esa cgqr de [kpZ esa ,d LFkku ls nwljsa LFkku ij igqpka;k tk ldrk gSA 2. (a) state Faraday’s law of electromagnetic induction (b) Explsin, with the help of a suitable example, how we can show that lenz’s law is a consequence of

the principle of conservation of energy. (c) Use the expression for Lorentz force acting on the charge carries of a conductor to obtain the

expression for the induced emf across the conductor of length l moivng with velocity v through magnetic field B acting perpendicular to its length.

(a) QSjkMs dk fo|qr&pqEcdh; izsj.k dk fu;e fyf[k,A (b) ysat dk fu;e ÅtkZ laj{k.k fl)kUr dk fu"d"kZ gS] bldks dSls n'kkZ,xs \ mfpr mnkgj.k lfgr O;k[;k dhft,A

(c) fdlh pkyd ds vkos'k okgdksa ij dk;Zjr yfjUV~t cy ds O;atd dk mi;ksx ml izsfjr fo|qr&okgd cy (emf)

tks fdlh pqEcdh; {ks=k B ds yEcor~ l yEckbZ ds pkyd dks osx v ls xfr djus ij pkyd esa mRiUu gksrk gS] ds fy, O;atd izkIRk djus esa dhft,A





E ST INFORMA TIO


PPHHYYSSIICCSS


NO. B21 TO B23

DPP No. : B23 (Special DPP) Total Marks : 118 Max. Time : 86 min. Single choice Objective ('–1' negative marking) Q.1 to Q.2 (3 marks 2 min.) [06, 04] One or more than one options correct type ('–1' negative marking) Q.3 to Q.20 (4 marks 2 min.) [72, 36] Subjective Questions ('–1' negative marking) Q.21 to Q.28 (4 marks 5 min.) [32, 40] Match the Following (no negative marking) Q.10 (8 marks 06 min.) [08, 06]

ANSWER KEY OF DPP No. : B23 (Special DPP) 1. (B) 2. (B) 3. (A,B,C) 4. (A,B,C) 5. (B,C) 6. (A,C,D) 7. (B,C,D) 8. (B,C) 9. (A,B,C,D) 10. (B) 11. (A,B,C) 12. (B) 13. (A,D) 14. (B) 15. (C) 16. (B) 17. (C) 18. (A) 19. (D) 20. (B) 21. 8 22. 21/4 23. 1 24. 3 25. 2 26. 6 27. 8 28. 2 29. (A) p,r (B) p,r (C) q,s (D) q,s 30. (A) p,s (B) p,s (C) q,r (D) p,s

1. A wire frame is in the shape of a parabola y = 5x2. It is being rotated about y-axis with an angular

velocity . A small bead of mass (m = 1 kg) is at point P and is in equilibrium with respect to the frame.

Then the angular velocity is (g = 10 m/s2)

,d rkj dks ijoy;kdkj Ýse y = 5x2 dh vkÑfr esa eksM+k x;k gSA ;g y-v{k ds ifjr% dks.kh; osx ls ?kw.kZu dj

jgk gSA ,d NksVk eksrh (nzO;eku m = 1 kg) fcUnq P ij fLFkr gS rFkk Ýse ds lkis{k lkE;koLFkk esa gSA dks.kh; osx

dk eku gksxk (g = 10 m/s2)

P(1, 5)

Y

X

(A) = 5 rad/sec (B*) = 10 rad/sec (C) = 20 rad/sec (D) = 15 rad/sec




Sol.

m2

x

mg

x

N

Y

X

(1) m2x cos = mg sin tan = 2x

g

; tan =

dy

dx = 10x

(2) N = m2x sin + mg cos 10 rad/ sec

2. The particle of mass 1 kg is acted upon by a force

x yˆ ˆF F i F j for time t = 4 sec starting from t = 0.

Find the velocity of the particle at t = 4 sec., if its initial velocity is ˆ ˆu i j and the variation of applied

force with time is as shown in the graphs :

1 kg nzO;eku ds d.k ij izkjEHk esa t = 0 ls t = 4 sec rd cy

x yˆ ˆF F i F j dk;Zjr gSA t = 4 sec i'pkr~ d.k dk

osx Kkr dhft,A ;fn bldk izkjfEHkd osx ˆ ˆu i j rFkk le; ds lkFk cy esa ifjorZu xzkQ esa iznf'kZr gSA

2 0

5

–5

T (sec)

4

FX (N)

2 0

5

T (sec) 4

FY (N)

(A) ˆ ˆv i j (B*)

ˆ ˆv i 11j (C) ˆv 10 j (D)

ˆ ˆv 2 i 10 j

Sol.

4

x x x x

0

F dt P m(v u )

0 = m (vx – ux)

vx = ux = – 1

4

y y y y

0

F dt P m(v u )

y y

14 5 1(v 1) v 11

2

ˆ ˆv i 11j




3. A small block of mass m is released from rest from point P in a smooth fixed hemispherical bowl of

radius R, as shown. Choose the correct alternative(s) :

m nzO;eku ds ,d NksVs CykWd dks R f=kT;k ds fpdus tM+or~ v)Z xksykdkj I;kys ds vUnj fcUnq P ls fojkekoLFkk ls

NksM+k tkrk gS tSlk fd fp=k esa iznf'kZr gSA lgh fodYiksa dk p;u dhft,

60°

B

O R m

P

A

(A*) The speed of block at A is 2gR

A ij CykWd dh pky 2gR gSA

(B*) The speed of block at B is gR

B ij CykWd dh pky gR gSA

(C*) The normal reaction at B is 3mg

2

B ij vfHkyEc cy 3mg

2gS

(D) The net force on the block at B is 3mg

2

B ij CykWd ij dqy cy 3mg

2gSA

Sol. At point A

fcUnq A ij

mgR = 2A A

1mv V 2gR

2

m

mg

N1

ar

N1 – mg = mar = 2AmV

R

N1 = mg + 2mg = 3mg

At point B

fcUnq B ij

mg(R cos 60) = 2B B

1mV V gR

2




2Bv

R

mg

mgcos60

N2

mgsin60

N2 –mg cos 60 = 2BmV

R

N2 = 3mg

2

Fnet = 2BmV

mgR

4. Two smooth spheres A and B of masses 4 kg and 8 kg move with velocities 9 m/s and 3 m/s in opposite directions. If A rebounds on its path with velocity 1 m/s, then

(A*) The velocity of sphere B after the impact 2 m/s

(B*) The coefficient of impact e = 1

4

(C*) The loss of kinetic energy = 180 J (D) Impulse imparted by A on B is 24 Ns.

nks fpdus xksys A o B ds nzO;eku 4 kg o 8 kg gS] tks foifjr fn'kk esa Øe'k% 9 m/s o 3 m/s ds osx ls xfr'khy gSA ;fn A blds iFk ij 1 m/s ds osx ls okil yksVrk gSA rc :

(A*) VDdj ds i'pkr~ xksys B dk osx 2 m/s gSA

(B*) VDdj dk izR;koLFkku xq.kkad e = 1

4 gSA

(C*) xfrt ÅtkZ esa gkfu 180 J gSA (D) A }kjk B ij vkjksfir vkosx 24 Ns gSA Sol. Fext = 0

i fP P

36 – 24 = – 4 + 8v2

v2 = +2ms–1 & v1 = – 1ms–1

e = sep

app

v

v =

3

12 =

1

4

K = 2 21 1(4)(9) 8(3)

2 2

– 2 21 1

4(1) 8(2)2 2

K = 180 J

B

= B f im (v v )

= 8(2 – (–3)) = 40 N-S

5. Consider a rope of mass 4m and length 4R on a fixed rough pulley of radius R as shown in the figure. The rope is in equilibrium. Length of vertical hanging parts is shown in the figure.

R

2R

O

(A) Torque of tension force about O on pulley is mgR (B*) Torque of normal force between rope and pulley on pulley about O is zero (C*) Torque of friction force between rope and pulley on pulley about O is mgR (D) Torque of friction force between rope and pulley on pulley about O is zero




ekfu;s dh 4m nzO;eku ,oae~ 4R yEckbZ dh ,d jLlh fp=kkuqlkj R f=kT;k dh [kqjnjh tM+or~ f?kjuh ij fLFkr gSA jLLkh lkE; voLFkk esa gSA Å/okZ/kj yVdh gqbZ jLlh dh yEckbZ fp=k esa n'kkZ;s vuqlkj gS %

R

2R

O

(A) f?kjuh ij O ds lkis{k ruko cy dk cy vk?kw.kZ mgR gSA (B*) f?kjuh ij O ds lkis{k jLlh o f?kjuh ds e/; vkjksfir vfHkyEc cy dk cy vk?kw.kZ 'kwU; gSA (C*) f?kjuh ij O ds lkis{k jLlh ,oae~ f?kjuh ds e/; ?k"kZ.k cy dk cy vk?kw.kZ mgR gSA

(D) f?kjuh ij O ds lkis{k jLlh ,oae~ f?kjuh ds e/; ?k"kZ.k cy dk cy vk?kw.kZ 'kwU; gSA

Sol. FBD of rope in contact with pulley is shown here we can see torque of dN about O is zero Torque of friction is balanced by torque T1 and T2.

f?kjuh ds lEidZ esa jLlh dk FBD fp=k esa n'kkZ;k x;k gSA ge ns[k ldrs gS] fd dN dk cy vk?kw.kZ O ds lkis{k 'kwU; gSA

?k"kZ.k dk cy vk?kw.kZ T1 o T2 ds cy vk?kw.kZ ls lUrqfyr gSA

dN

df

mgT2 = 2mgT1 = mg

O

6. A collision takes place between two particles each moving with speed v, the collision is oblique. After

collision two blocks move together with velocity v as shown in figure. Then which of the following option is/ are true :

nks d.k tks izR;sd v pky ls xfr'khy gS, ds e/; fr;Zd VDdj gksrh gSA VDdj ds i'pkr~ d.k dk mHkfu"B osx v

fp=kkuqlkj xfr'khy gSA rc fuEu esa ls dkSulk@dkSSuls fodYi lgh gS :

(A*) 2sin

tan1 2cos

(B)

2costan

1 2sin

(C*) Final kinetic energy of system 2mv

6 (5 + 4cos)

fudk; dh vfUre xfrt ÅtkZ 2mv

6 (5 + 4cos)

(D*) loss of kinetic energy = 2

3mv2 (1 – cos)

xfrt ÅtkZ esa gkfu = 2

3mv2 (1 – cos)




Sol. ˆ ˆ3mv ' (mv 2mv cos ) i 2mv sin j

v 2ˆ ˆv ' (1 2cos )i sin j3 3

tan = 2sin

1 2cos

K = ki – kf

22 2 23 1 3 mv

mv (3m)v ' mv (5 4cos )2 2 2 6

7. A particle moves along a horizontal circle such that the radial force acting on it is directly proportional to

square of time. Then choose the correct option :

,d d.k {kSfrt o`Ùk esa bl çdkj ls xfr dj jgk gS fd bl ij dk;Zjr f=kT;h; cy (radial force) le; ds oxZ ds lh/ks lekuqikrh gS rks lgh fodYi pqfu, &

(A) tangential force acting on it is directly proportional to time (B*) power developed by total force is directly proportional to time (C*) average power developed by the total force over first t second from rest is directly proportional to

time (D*) angle between total force and normal decreases with time

(A) bl ij dk;Zjr Li'kZ js[kh; cy le; ds lekuqikrh gSA (B*) dqy cy }kjk iznku dh xbZ dqy 'kfä le; ds lekuqikrh gSA (C*) dqy cy }kjk iznku dh xbZ dqy 'kfDr izkjEHk ls] le; t ds lekuqikrh gksxhA (D*) dqy cy rFkk vHkhyEc fn'kk (normal) e/; dks.k le; ds lkFk ?kVsxkA Sol. an = kt2

2v

R = kt2

v = kR t

at =dv

dt = kR constant fu;rkad

tan = t2

n

a 1

a t

p = Ft v

t

<p> t. 8. The figure shows a block of mass M=2m having a spherical smooth cavity of radius R placed on a

smooth horizontal surface .There is a small ball of mass m moving at an instant vertically downward with a velocity v with respect to the block .At this instant : [tough]

M = 2m nzO;eku dk ,d CykWd ftlesa R f=kT;k dh fpduh xksyh; xqfgdk cuh gqbZ gS] fpduh {kSfrt lrg ij j[kk gSA m nzO;eku dh ,d NksVh xsan CykWd ds lkis{k v osx ls Å/okZ/kj uhps dh vksj xfr dj jgh gSA bl {k.k ij :

M

m

horizontal smooth surface

(A) The normal reaction on the ball by the block is 2mv

R

(B*) The normal reaction on the ball by the block is 2

3

2mv

R

(C*) The acceleration of the block with respect to the ground is 2v

3R

(D) The acceleration of the block with respect to the ground is 2v

2R




(A) CykWd }kjk xsan ij vfHkyEc çfrfØ;k 2mv

R gSA (B*) CykWd }kjk xsan ij vfHkyEc çfrfØ;k

2

3

2mv

R gksxhA

(C*) tehu ds lkis{k CykWd dk Roj.k 2v

3R gSA (D) tehu ds lkis{k CykWd dk Roj.k

2v

2R gksxkA

Sol. Let the normal force between the block and the ball be N.

ekuk CykWd rFkk xsan ds e/; vfHkyEc cy N gSA

M

N

a

mg

N ma(pseudo force)

For the block, from Newton’s IInd law , we have N = Ma = 2ma

For ball (with respect to the block), from Newton’s IInd law , we have N + ma = 2mv

R

Solve the two equations.

CykWd ds fy, U;wVu ds f}Ùkh; fu;e ls xsan , N = Ma = 2ma

xsan (CykWd ds lkis{k) ds fy,] U;wVu ds f}rh; fu;e ls N + ma = 2mv

R

bu lehdj.kksa dks gy djus ij 9. A double conical pendulum consists of two masses, m and M, connected by a massless string passing

over a frictionless, massless pulley. The entire apparatus rotates freely at constant angular speed (rad/s) about the vertical axis (dashed line) passing through centre of pulley as shown. After the system

comes in steady state, the length of string on either sides of pulley are and L. Pick up the correct option(s).

,d f} (double) 'kaDokdkj yksyd fp=kkuqlkj nks nzO;eku m rFkk M j[krk gS tks fd ,d Hkkjghu jLlh }kjk ?k"kZ.kghu o Hkkjghu f?kjuh ls fp=kkuqlkj tqM+s gSA iwjh O;oLFkk fu;r dks.kh; pky (rad/s) ls fcuk fdlh ck/kk ds Å/okZ/kj v{k (fp=kkqulkj iqyh ds dsUnz ls xqtjus okys dashed line) ds ifjr% ?kw.kZu dj jgh gSA tc ;g fudk; LFkk;h voLFkk esa vkrk gS rks fp=kkuqlkj f?kjuh ds izR;sd rjQ dh yEckbZ fp=kkuqlkj yEckbZ o L gSA lgh fodYi gS&

(A*) cos m

cos M

(B*) cos =

2

g

L (C*) m = ML (D*) cos =

2

g

Sol. For conical pendulum

dksfudy is.Mqye

2 = g g

cos Lcos

cos = 2

g

.... (1), cos =

2

g

L .... (2)

and cos = L cos .... (3)

Also T cos = Mg and T cos = mg

cos m

cos M

.... (4)

from (3) and (4)

ML = m




Comprehension –1

Four particles of masses m, 2m, 3m, and 4m are placed on vertices of a square ABCD as shown in the figure. Now the particles at A, B, C and D are displaced to their new positions B, C, D and A respectively.

m, 2m, 3m rFkk 4m nzO;eku ds pkj d.k fp=k esa n'kkZ;suqlkj ,d oxZ ABCD ds 'kh"kksZ ij j[ks gq, gSA vc A, B, C

rFkk D ij j[ks d.kksa dks mudh Øe'k% u;h fLFkfr;ksa B, C, D rFkk A rd foLFkkfir fd;k tkrk gSA

3m

4m

2m

m

P R

S

Q

A

B C

D

10. Distance of new position to of center of mass from of it’s old position is –

u;s nzO;eku dsUnz dh fLFkfr dh iqjkus nzO;eku dsUnz dh fLFkfr ds e/; nwjh gksxh –

(A) (B*) 2

5

(C)

5 2

(D)

2 2

Sol. cm

ˆ ˆ ˆ ˆm j 2m i 3m j 4m id

10m

ˆ ˆ2 i 2 j

10

ˆ ˆ( i j)5

cm

2d

5

11. AB, AC, PQ, PR are moment of inertia about axis in plane and passing through respective points. For which axes moment of inertia will change after above displacements :

ry esa fLFkr lEcfU/kr fcUnqvksa ls xqtjus okyh v{kksa ds lkis{k tM+Ro vk?kw.kZ AB, AC, PQ, PR gSA buesa ls fdruh v{kksa ds lkis{k tM+Ro vk?kw.kZ mijksDr foLFkkiu ds i'pkr~ ifjofrZr gksrk gS &

(A*) AB (B*) AC (C*) PQ, (D) PR

Sol. For PR axis all particles are at distance /2 in both situations

PR v{k ds fy, nksuksa fLFkfr;ksa esa lHkh fcUnq /2 nwjh ij gSA Comprehension-2

A solid cylinder of mass m is rotating with angular velocity about its fixed axis of symmetry. It is brought in contact with a plank of mass m placed on a smooth surface. There exist friction between cylinder and plank because of which plank starts moving. After some time plank moves such that the there is no slipping between cylinder and plank.

m nzO;eku dk ,d Bksl csyu bldh fLFkj lefer v{k ds ifjr% dks.kh; osx ls ?kw.kZu dj jgk gSA ;g {kSfrt lrg ij j[ks gq, m nzO;eku ds r[rs ds lkFk lEidZ esa yk;k tkrk gSA csyu rFkk r[rs ds e/; ?k"kZ.k gksus ds dkj.k r[rk xfr djuk izkjEHk djrk gSA dqN le; i'pkr~ r[rk bl izdkj xfr djrk gS fd csyu rFkk r[rs ds e/; dksbZ fQlyu ugha gksrh gSA

0R

m

m

Smooth surface

Rough surface

0R

m

m[kqjnjh lrg

fpduh lrg 12. The final velocity of the plank when there is no slipping between cylinder and plank is :

tc csyu rFkk r[rs ds e/; dksbZ fQlyu ugha gksrh gS] rc r[rs dk vfUre osx gksxk :

(A) 03R

4

(B*) 0R

3

(C) 0R

2

(D) 02R

3




Sol. On applying conservation of angular momentum about axis of rotation.

20

1mR

2 = 21

mR2

+ mR2 ( is final angular velocity of cylinder)

= 0

3

v = 0R

3

13. Select correct statements :

lR; dFkuksa dk p;u dhft;s

(A*) magnitude of angular impulse imparted to the cylinder during this process 2

0mR

3

(B) magnitude of angular impulse imparted to the cylinder during this process 2

02mR

3

(C) the total work done by friction force in this process – 2 20

1mR

9

(D*) the total work done by friction force in this process – 2 20

1mR

6

(A*) bl izØe ds nkSjku csyu dks iznku fd;s x;s dks.kh; vkosx dk ifjek.k 2

0mR

3

(B) bl izØe ds nkSjku csyu dks iznku fd;s x;s dks.kh; vkosx dk ifjek.k 2

02mR

3

(C) bl izØe esa ?k"kZ.k cy }kjk fd;k x;k dqy dk;Z – 2 20

1mR

9

(D*) bl izØe esa ?k"kZ.k cy }kjk fd;k x;k dqy dk;Z – 2 20

1mR

6

Sol. (0 – ) = 2mR

2(0 – ) =

20mR

3

Comprehension–3 A solid non uniform sphere of mass m and radius R is released on smooth horizontal surface from the

situation shown in the figure. In given situation centre of mass (C) and geometric centre (O) are on

same horizontal line. The distance OC is 2

R. The moment of inertia about axis passing through C and

perpendicular to the plane of the paper is = 2

MR2

. Acceleration of centre of mass is a and angular

acceleration of the sphere is .

,d vle:i nzO;eku forj.k ds ,d Bksl xksys dk nzO;eku m o f=kT;k R gS] ftls fpduh {kSfrt lrg ij fp=k esa n'kkZ;h xbZ fLFkfr ls NksM+k tkrk gSA nh xbZ O;oLFkk esa izkjEHk esa nzO;eku dsUnz (C) o T;kferh dsUnz (O) leku

{kSfrt js[kk esa fLFkr gSA nwjh OC dk eku 2

R

gSA C ls xqtjus okyh o dkxt ry ds yEcor~ v{k ds lkis{k tM+Ro

vk?kw.kZ = 2

MR2

gSA nzO;eku dsUnz dk Roj.k a rFkk xksys dk dks.kh; Roj.k gSA

O C

R

2




14. Just after release relation between a & is

NksM+us ds rqjUr ckn a o esa lEcU/k gksxk &

(A) a = R (B*) a = 2

R

(C) a = 2R (D) a =

2

R5

Sol. Just after release acceleration of O must be zero.

NksM+us ds Bhd ckn O dk Roj.k 'kwU; gh gksxk

So vr% a = R

2

15. At the moment when centre of mass is at the lowest point, angular velocity of the sphere is –

nzO;eku dsUnz fuEure fcUnq ij gS] rc bl {k.k ij xksys dk dks.kh; osx gksxk &

(A) g

R (B)

g

2R

(C*) 2g

R (D) None of these buesa ls dksbZ ugha

Sol. mgR 1

2 2

22mR

2

= 2g

R

Comprehension–4

The following set of figures shows components of a mechanical system in figure-1 and assembled system in figure-2. There are two small balls of mass m and 2m, a massless tube of length L, a spring

of natural length (> L) and an inextensible, massless rope of length L. In the assembly, the string holds the two balls in position, keeping the spring compressed.

The arrangement lies on a smooth horizontal table (x-y plane) as shown. At this moment, the velocity of

centre of mass of system is v0 i and angular velocity is 0k

. Take this moment as t = 0.

fp=kksa ds fuEu leqPp;ksa esa fp=k-1 esa ,d ;kaf=kd fudk; ds ?kVd n'kkZ;s x;s gS rFkk fp=k-2 esa mudksa la;qDr djds cuk fudk; n'kkZ;k x;k gSA ;gk¡ m o 2m nzO;eku dh nks NksVh xsnsa, ,d L yEckbZ dh nzO;ekughu [kks[kyh ufydk]

izkÑfrd yEckbZ (> L) dh ,d fLiazx rFkk L yEckbZ dh ,d vforkU; nzO;ekughu Mksjh gSA la;qDr fudk; esa Mksjh nksuksa xsanksa dks ck/ksa j[krh gS] tks fLiazx dks lEihM+hr j[krh gSA

lEiw.kZ O;oLFkk fp=k esa n'kkZ;suqlkj {kSfrt Vscy (x-y ry) ij fLFkr gSA bl {k.k ij fudk; ds nzO;eku dsUnz dk osx

v0 i rFkk dks.kh; osx 0k

gSA bl {k.k dks t = 0 ysosaA

2m

Figure fp=k-1

m

L

hollow tube

spring

fLiazx

string

2m

m

x-axis v{k

y-axis v{k

(0,L)

(0,0)

Figure fp=k -2

Mksjh

[kks[kyh ufydk

16. The kinetic energy of the system in centre of mass frame is

nzO;eku dsUnz ds rU=k esa fudk; dh xfrt ÅtkZ gksxh &

(A) 2 20

3m L

2 (B*) 2 2

0

1m L

3 (C) 2 2

0

2m L

3 (D) 2 2

0

4m L

3

Sol. KE in frame of CM

nzO;eku dsUnz ds rU=k esa xfrt ÅtkZ

= 2cm

1

2 =

1

2 2 21 2

1 2

m mL

m m




17. Angular momentum of the system about origin of the table frame (where the point 2m is located at

origin at t = 0) is

Vscy funsZ'k ra=k esa ewyfcUnq ds lkis{k fudk; dk dks.kh; laosx gksxk (ewy fcUnq ij t = 0 ij 2m nzO;eku fLFkr gS)

(A) 3 mL 00

L ˆ2v k2

(B) – mL 0 0

2 ˆv L k3

(C*) – mL 0 0

2 ˆv L k3

(D) 0 0

2 ˆmL v L k9

Sol. L

= (–mVcmrcm + cm0) k = – mL 0 0

2 ˆv L k3

Comprehension–5 The cross-section of a fixed cylinder (not allowed to rotate and translate) with horizontal axis is as

shown. One end of a light inelastic string is fixed at top of cylinder of radius R and a small block of mass m is tied to the other end of string. Initially the block is at rest with the portion of string not in contact with cylinder being vertical and having length L as shown. At the lowest position the block is given initial

horizontal velocity u = 2g L and the block moves in vertical plane. When the block reaches the

highest point of its trajectory, the length of string not in contact with cylinder is L + R

3

. (where g is

acceleration due to gravity).

,d fLFkj csyu dk vuqizLFk dkV {kSfrt v{k ds lkFk fp=k esa iznf'kZr gSA csyu ?kw.kZu ,oa LFkkukUrj.k xfr ds fy, Lora=k gSA gYdh vizR;kLFk Mksjh dk ,d fljk R f=kT;k ds csyu ds Åijh fljs ij tqM+k gqvk gS rFkk m nzO;eku dk ,d NksVk CykWd Mksjh ds nwljs fljs ij ca/kk gqvk gSA izkjEHk esa CykWd Å/okZ/kj Mksjh ¼tks csyu ds lEidZ esa ugh gS½ dss lkFk fojkekoLFkk esa gS ,oa Mksjh ds Å/oZ Hkkx dh dqy yEckbZ L gSA fp=k esa iznf'kZr fuEure fLFkfr ij CykWd dks

izkjEHk esa u = 2g L {kSfrt osx fn;k tkrk gS rFkk CykWd Å/okZ/kj ry esa xfr djrk gSA tc CykWd blds iFk ds

mPpre fcUnq ij igq¡prk gS rks csyu ls vlaifdZr Mksjh dh yEckbZ ¼tks Mksjh csyu ds lEidZ esa ugh gS½ L + R

3

gks

tkrh gS ¼tgk¡ g xq:Roh; Roj.k gS½

m

Rfixed cylinder

u= 2gL

L

m

RfLFkj csy u

u= 2gL

L

18. The distance between block and centre of cylinder when block is at highest position will be:

tc CykWd mPpre fcUnq ij igq¡prk gS rks CykWd rFkk csyu ds dsUnz ds e/; nwjh gksxh &

(A*) 2R (B) 5 R (C) 3R (D) 2 R

3

19. The least tension in string is :

Mksjh esa U;wure ruko gksxk &

(A) mg

6 (B)

mg

5 (C)

3

2 mg (D*)

mg

2

20. Tangential acceleration of block at highest position is :

mPpre fcUnq ij CykWd dk Li'kZ js[kh; Roj.k gksxk &

(A) g

2 (B*)

3

2g (C)

g

3 (D)

g

6




Sol. Since u = Lg2 , the highest point to which the block shall reach isg2

u2

= L distance above its initial

position. Hence at highest point, the thread has rotated by 3

= 60° and the block is at same horizontal

level as centre of cylinder as shown.

u = 2g L , vr% CykWd mPpre fcUnq] ftl rd CykWd igq¡psxk izkjfEHkd fLFkfr ls 2u

2g= L Åij gksxkA blfy;s

mPpre fcUnq ij /kkxk 3

= 60° ds dks.k ls ?kwe tkosxk vkSj CykWd fp=kkuqlkj cSyu ds dsUnz ds leku {kSfrt ry esa

gksxkA

R

60

O x

L+R 3

30

60

mg

x = R

sin30 = 2R

T = mg cos 60° = mg

2

tangential acceleration (Li'kZ js[kh; Roj.k) at = g sin 60° = 3g

2

21. A cubical block of mass 1 Kg is pulled by horizontal force F as shown in figure. The Coefficient of

friction of surface is = 0.2. The value of force F for which block will topple is xN. Find the value of x.

1 Kg nzO;eku ds ?kukdkj CykWd dks {kSfrt cy F }kjk fp=kkuqlkj [khapk tkrk gSA lrg ds fy, ?k"kZ.k xq.kkad

= 0.2 gSA CykWd ds iyVus ds fy, cy F dk eku xN gks rks] x dk eku Kkr djksA

F

Ans. 8

Sol. F f N.2 2 2

F = N – f

= 10 – 0.2 × 10

= 8 N Ans.

22. Two particles of mass M = 3 kg each are kept on a horizontal circular platform on two mutually

perpendicular radii at equal distance R = 1 m from the centre of the table. The particles are connected with a string, which is just taught when the platform is not rotating. Coefficient of friction between the

platform and block is = 0.1. Then the maximum angular speed ( in rad/sec) of platform about its centre so that the blocks remain stationary relative to platform. (take g = 10 m/s2)

nks d.k izR;sd dk nzO;eku M = 3 kg est ds dsUnz ls R = 1 m dh nwjh ij ,d {kSfrt o`Ùkh; IysVQkeZ ij nks ijLij yEocr~ f=kT;kvksa ij j[ks x;s gSA nksuksa d.k ,d jLlh ls tqM+s gS] tks Bhd ruh gqbZ gS tc IysVQkeZ ugha ?kwe jgk gSA CykWd o IysVQkeZ ds e/; ?k"kZ.k xq.kkad = 0.1 gSA rc blds dsUnz ds ijhr% IysVQkeZ dh vf/kdre dks.kh; pky (rad/sec esa) D;k gksxh rkfd IysVQkeZ ds lkis{k d.k fLFkj jgsaaA (g = 10 m/s2 yhft,A)

Ans. 21/4




23. The atwood machine has a third mass attached to it by a slacked and light string. After being released,

the 2m mass falls a distance y = 8

3 m before the slack string becomes taut. Just after the slack string

becomes taut the speed of masses is X g . Find X. (Assume that pulley is ideal and take acceleration

due to gravity = g)

,VoqM e'khu }kjk rhljk nzO;eku <hyh ¼gYdh½ jLlh }kjk tksM+k tkrk gSA NksM+us ds i'pkr~ 2m nzO;eku }kjk <hyh

jLlh ruus rd uhps dh rjQ pyh xbZ nwjh y = 8

3m gSA <hyh jLlh ruus ds rqjUr i'pkr~ nzO;ekuksa dh pky

X g gks rks X Kkr djksA (ekuk f?kjuh vkn'kZ gS rFkk xq:Roh; Roj.k g gSA)

Ans. 1

Sol. a = (2m m)g

3m

=

g

3

V2 = 0 + g

23

(y)

From conservation of mometum. ¼laosx laj{k.k ls½ 3mV = (4m) v’

v’ = 3V 3gy

4 8

24. Two particles of masses m1 and m2 are connected as shown. The strings are massless and

inextensible and gravity is absent. If the system is rotated about O with uniform angular velocity , the

ratio of tensions in the strings is 5

,2

then find the ratio 1

2

m.

m (Both the strings remain in the straight line

during rotation).

m1 rFkk m2 nzO;eku ds nks d.k fp=kkuqlkj tqM+s gq, gSA jLlh;ka nzO;ekughu ,oa vforkU; gS] ,oe~ xq:Ro vuqifLFkr

gSA ;fn fudk; O ds ifjr% le:i dksf.k; osx ls ?kw.kZu dj jgk gks rks jLLkh;ksa esa ruko dk vuqikr 5

2 gS] rks

vuqikr 1

2

m

m Kkr djksA (?kw.kZu xfr ds nkSjku nksuksa jLlh;ka ,d ljy js[kk esa jgrh gSA)

Ans. 3




Sol. 22 2T m (2l)

21 2 1T T m

2 21 1 2T m 2m

given (fn;k gS)

1

2

T 5.

T 2

1 2

2

m 2m 5

2m 2

1

2

m3.

m

25. A masonry column of density has height h, top cross sectional area A1 and bottom cross sectional area A2. Top cross sectional of the column is uniformly loaded with load F. Stress at any cross section

of the column is same and ghA1 = 2F. If 2

1

A

A= ex , then find the value of x -

,d fpukbZ (masonry) LrEHk dk ?kuRo rFkk Å¡pkbZ h gS, 'kh"kZ vuqizLFk dkV {ks=kQy A1 rFkk ryh dk vuqizLFk dkV {ks=kQy A2 gSA LrEHk dk 'kh"kZ vuqizLFk dkV Hkkx ,d leku Hkkj F ls Hkkfjr gSA LrEHk ds fdlh Hkh vuqizLFk dkV {ks=k

ij izfrcy leku gS ,oa ghA1 = 2F gSA ;fn 2

1

A

A= ex rks x dk eku Kkr djksA

Ans. 2

Sol. Since stress at any cross section is same

pwafd fdlh Hkh Hkkx dk izfrcy leku gSA

Stress at top 'kh"kZ ij izfrcy = stress at general cross sectional fdlh Hkh vuqizLFk Hkkx ij izfrcy

1

F

A =

W F

A

Here ;gk¡ W = weight of column above the general cross section fdlh Hkh Hkkx ds Åij ds LrEHk dk Hkkj gSA

1

F

A =

Agdz F

A

1

FA

A – F = g dzA

Differentiating w.r.t. z on both sides nksuksa vksj z ds lkis{k vodyu djus ij

gAdz

dA

A

F

1

2

1

A

A

dAA

1 =

h

0

1 dzF

gA

n 1

2

A

A =

F

hgA1 =

F

F2 = 2

1

2

A

A = e2 = 7.38

Closest integer fudVre iw.kk±d = 7




26. A shell is stationary in the free space split in to two fragments of masses 4 kg and 2 kg by it's explosion adds on an energy of 24 J to fragments. The magnitude of relative velocity (in m/s) between the two fragments after explosion is : (Assume that energy is released in form of kinetic energy only)

eqDr {ks=k esa ,d fLFkj dks'k vkUrfjd foLQksV ds dkj.k 4 kg o 2 kg nzO;ekuksa ds nks VqdM+ksa esa foHkDr gks tkrk gSA ;g foLQksV blds VqdM+ksa dks 24 J ÅtkZ nsrk gSA foLQksV ds i'pkr~ VqdM+ksa ds e/; lkis{k osx dk ifjek.k (m/s esa) gksxk : (ekfu,asa fd foLQksV esa izkIr ÅtkZ dsoy xfrt ÅtkZ ds :i esa mi;ksx gksrh gSA)

Ans. 6

Sol. 2 21 1

1 2

m1m V E

2 m m

V1 = 2

1 2 1

m2E

m m m

V2 = 1

1 2 2

m2E

m m m

Vrel = 1 2

1 2 1 2

m m2E

m m m m

= 1 2

1 2

2E(m m ) 24 62

m m 8

27. A particle is projected from ground with an initial velocity 20 m/sec making an angle 60° with horizontal.

If R1 and R2 are radius of curvatures of the particle at point of projection and highest point respectively,

then find the value of 1

2

R

R.

,d d.k izkjfEHkd osx 20 m/sec ls {kSfrt ls 60° ds dks.k ij iz{ksfir fd;k tkrk gSA ;fn R1 rFkk R2 Øe'k% iz{ksi.k

fcUnq rFkk mPPre fcUnq ij d.k dh oØrk f=kT;k gS rks 1

2

R

R dk eku Kkr djksA

Ans. 8

Sol. R1 = 2

0v

gcos

R2 = 2

0(v cos )

g

v0v cos0

g

g

1

2

R

R=

3

18

(cos )




28. A block of mass m is released on a smooth inclined wedge of inclination 30º and mass M. Height of the block (from ground) varies with time as h = 1.5 – 1.5t2. (h in metre and t = time in second). The

acceleration of M is 3

xm/s2. Find the value of x :

M nzO;eku rFkk 30º dks.k dh fpduh ur lrg okys ost ij m nzO;eku ds CykWd dks NksM+k tkrk gSA CykWd dh tehu ls Å¡pkbZ le; ds lkFk h = 1.5 – 1.5t2 }kjk nh tkrh gSA (h ehVj esa gS rFkk t = lsd.M esa gSA) M dk Roj.k

3

x m/s2 gks rks x dk eku Kkr djks :

Sol. Ans. 2 ax and ay component of acceleration of m and a acceleration of M.

ax rFkk ay, m nzO;eku ds Roj.k ds ?kVd gS rFkk a, nzO;eku M dk Roj.k gSA

y

x

a

a a=

1

3

a + ax = 3 ay = 3 3

mg – N cos30º = may N sin 30º = max

ay =

2

23

d h

dt

on solving gy djus ij

a = 2

3

x = 2




29. A small block lies on a rough horizontal platform above its centre C as shown figure. The plank is moved in vertical plane such that it always remains horizontal and its centre C moves in a vertical circle

of centre O with constant angular velocity . There is no relative motion between block and the plank and the block does not loose contact with the plank anywhere. P, Q, R and S are four points on circular trajectory of centre C of platform. P and R lie on same horizontal level as O. Q is the highest point on the circle and S is the lowest point on the shown circle. Match the statements in column-I with points in column-II.

,d NksVk CykWd ,d [kqjnjs {kSfrt IysVQkeZ ds Åij blds dsUnz C ij fp=kkuqlkj j[kk gS vc r[rs ¼IysVQkeZ½ dks bl Å/okZ/kj ry esa o`Ùk esa bl rjg ls xfr djkbZ tkrh gS fd ;g ges'kk {kSfrt jgrk gS ftldk dsUnz C Å/okZ/kj

o`Ùkh; iFk ¼dsUnz O) ij xfr'khy gS rFkk bldh fu;r dks.kh; pky gSA ;gka CykWd o r[rs ds e/; dksbZ lkisf{kd xfr ugha gS rFkk CykWd r[rs ds lkFk dHkh Hkh lEidZ ugh NksM+rk gSA P, Q, R rFkk S pkj fcUnq IysVQkeZ ds dsUnz C

ds o`Ùkkdkj iFk fLFkr fcUnq gSA LrEHk-I esa fn;s x;s oDrO;ksa dks LrEHk-II esa fn;s x;s fcUnqvksa ls lqesfyr dfj, &

(fcUnq O, P o R ds ,d gh {kSfrt js[kk ij fLFkr fcUnq gSA o`Ùk dk mPpre o fuEure fcUnq Øe'k% Q o S gS)

Column-I Column-II (A) Magnitude of frictional force on block is maximum (p) when block is at position P (B) Magnitude of normal reaction on block is equal to mg (q) when block is at position Q (C) Magnitude of frictional force is zero (r) when block is at position R (D) Net contact force on the block is directed toward centre (s) when block is at position S LrEHk-I LrEHk-II

(A) CykWd ij ?k"kZ.k cy dk ifjek.k vf/kdre gS (p) tc CykWd fLFkfr P ij gSA

(B) CykWd ij vfHkyEc cy dk ifjek.k mg ds cjkcj gS (q) tc CykWd fLFkfr Q ij gSA

(C) ?k"kZ.k cy dk ifjek.k 'kwU; gS (r) tc CykWd fLFkfr R ij gSA

(D) CykWd ij ifj.kkeh lEidZ cy dsUnz dh rjQ (s) tc CykWd fLFkfr S ij gSA funsZf'kr gksrk gS

Ans. (A) p,r (B) p,r (C) q,s (D) q,s Sol. At position P and R frictional force is maximum and normal reaction is mg. At position S and Q frictional force is zero and normal reaction points towards centre. fLFkfr P rFkk R ij ?k"kZ.k cy vf/kdre gS rFkk vfHkyEc izfrfØ;k mg gSA

fLFkfr S rFkk Q ij ?k"kZ.k cy 'kwU; gS rFkk bu fcUnqvksa ij vfHkyEc cy dsUnz dh rjQ gSA

30. A block of mass m is placed on wedge also of mass m. The wedge is placed on smooth horizontal fixed

surface. One end of a light spring is connected to block and the other end to a light support S rigidly fixed to wedge as shown. Friction is absent every where. The system is initially released from rest with spring unstressed. Match statements in column-I with corresponding statements in column-II.

m nzO;eku dk CykWd m nzO;eku ds ost ij fLFkr gS rFkk ost fpduh {kSfrt rFkk fLFkj lrg ij fLFkr gSA ,d gYdh fLizax dk ,d fljk

CykWd ls tqM+k gS rFkk nwljk fljk ,d gYds vk/kkj S ds lkFk fp=kkuqlkj ost ij fLFkj fd;k x;k gSA ?k"kZ.k gj txg vuqifLFkr gSA nh xbZ

fLizax izkÑfrd yEckbZ dh voLFkk esa gS vc iwjs fudk; dks fojke ls NksM+k tkrk gSA dkWye-I esa fn;s x;s oDrO;ksa dks dkWye-II esa laxr

oDrO;ksa ls lqesfyr dhft,A [DPP JF 2011-2012]

S

m

m

smooth horizontal surface

S

m

m

fpd uh {kSfr t l r g




Column-I Column-II (A) At the instant compression in spring is maximum (p) speed of block is zero (B) At the instant spring has natural length, that is, (q) speed of block is non zero it is unstressed. (C) At the instant net force on wedge is zero (r) speed of block is maximum (D) At the instant elastic potential energy stored in (s) speed of block is minimum spring is least dkWye-I dkWye-II

(A) fLizax esa laihMu vf/kdre gksus ds {k.k ij (p) CykWd dh pky 'kwU; gS

(B) fLizax ds izkÑfrd yEckbZ dh voLFkk es gksus ds {k.k (q) CykWd dh pky v'kwU; gS ij vFkkZr~ fcuk f[kaph gqbZ voLFkk

(C) ost ij dqy cy 'kwU; gksus ds {k.k ij (r) CykWd dh pky vf/kdre gS

(D) fLizax esa lafpr izR;kLFk fLFkfrt ÅtkZ ds U;wure gksus ds {k.k ij (s) CykWd dh pky U;wure gS

Ans : (A) p,s (B) p,s (C) q,r (D) p,s Sol. The velocity of centre of mass of system of block and wedge is always zero. (A) At maximum compression velocities of block and wedge must i.e. same and equal to velocity of

centre of mass. Hence velocity of both is zero and also minimum (B) At natural length of spring the velocity of system is zero as given. (C) At the instant acceleration of wedge is zero its velocity must be maximum. Hence velocity of block

must be maximum. (D) When elastic energy stored in spring is least it must be in state of natural length. CykWd rFkk ost ds fudk; dk nzO;eku dsUnz dk osx lnSo 'kwU; gksrk gSA

(A) vf/kdre laihMu ij CykWd rFkk ost dk osx leku vkSj nzO;eku dsUnz ds osx ds cjkcj gksxkA vr% nksuks dk osx 'kwU; rFkk U;wure gSA

(B) fLizax dh LokHkkfod yEckbZ ij fudk; dk osx 'kwU; gksrk gSA

(C) ml {k.k tc ost dk Roj.k 'kwU; gksrk gS bldk osx vf/kdre gksxk vr% CykWd dk osx vf/kdre gksxkA

(D) tc fLizax esa lafpr izR;kLFk fLFkfrt ÅtkZ U;wure gksrh gS rks fLizax LokHkkfod yEckbZ dh voLFkk esa gksxhA





E ST INFORMA TIO


PPHHYYSSIICCSS


NO. B24 TO B25

2. DPP Syllabus :

DPP No. B24 (JEE–Main) Total Marks : 94 Max. Time : 72 min. Single choice Objective ('–1' negative marking) Q.1 to Q.20 (3 marks 2 min.) [60, 40]

ANSWER KEY OF DPP No. : B24 1. (B) 2. (C) 3. (B) 4. (C) 5. (B) 6. (D) 7. (A) 8. (C) 9. (A) 10. (C) 11. (B) 12. (B) 13. (D) 14. (C) 15. (D) 16. (B) 17. (B) 18. (A) 19. (B) 20. (B)

1. A string of mass 2.5 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the

transverse jerk is struck at one end of the string, the disturbance will reach the other end in (A) one second (B*) 0.5 second (C) 2 seconds (D) data given is insufficient.

2.5 kg æO;eku dh ,d Mksjh esa 200 N dk ruko gSA rfur Mksjh dh yackbZ 20.0 m gSA ;fn Mksjh ds ,d fljs ij ,d vuqizLFk Lian mRiUu fd;k tk, rks fo{kksHk blds nwljs fljs ij igq¡psxk %

(A) ,d lsd.M esa (B*) 0.5 lsd.M esa (C) 2 lsd.M esa (D) fn, x, vkadM+s vi;kZIr gSa 2. A mercury drop of water has potential 'V' on its surface. 1000 such drops combine to form a new drop.

Find the potential on the surface of the new drop.

ikjs dh ,d cwan ds i`"B ij foHko V gSA bl izdkj dh 1000 cwans feydj ,d ubZ cwan cukrh gSA ubZ cwan ds i`"B ij foHko gksxk&

(A) V (B) 10V (C*) 100V (D) 1000V Sol. (C) Let the radius of each mercury drop be r. If q is charge on each drop

The potential of drop V = 0

q

4 r or q = 4 0 rV ...(1)

Let R be the radius of the new drop formed by combination of 1000 drops of radius r.

4

3

R3 = 1000

4

3

r3 R = 10 r ....(2)

Potential of new drop 0

(1000q)

4 R=

0

(100q)

4 r= 100 V




3. If the switch is closed then current will flow :

;fn fLop dks cUn dj fn;k tk;s rks /kkjk izokfgr gksxh %

(A) from A to B (A ls B)

(B*) from B to A (B ls A)

(C) 0 (D) cannot say unless resistance of galvanometer is given.

dqN ugh dgk tk ldrk tc rd fd /kkjkekih dk izfrjks/k u fn;k gksaA

4. All straight wires are very long. Both AB and CD are arcs of the same circle, both subtending right angles at the centre O. Then the magnetic field at O is :

lHkh lh/ks rkj cgqr yEcs gSaA AB rFkk CD nksuksa leku o`Rr ds pki gSaA nksuks dsUnz O ij ledks.k varfjr djrs gS ] rks O ij pqEcdh; {ks=k gS &

(A) 0i

4 R

(B) 0i2

4 R

(C*) 0i

2 R

(D) 0i( 1)

2 R

Sol. field due to AA' = 0i

4 R

= field due to BB' field due to CC' = field due to DD' = 0

field due to BA = 0i

8R

field due to CD = 0– i

8R

net field at O = 0i

2 R

.

gy% AA' ds dkj.k {ks=k = 0i

4 R

= BB' ds dkj.k {ks=k CC' ds dkj.k {ks=k = DD' ds dkj.k {ks=k = 0

BA ds dkj.k {ks=k = 0i

8R

CD ds dkj.k {ks=k = 0– i

8R

O ij dqy {ks=k = 0i

2 R




5. If a charged particle goes unaccelerated in a region containing electric and magnetic fields :

(A) E

must be parallel to B

(B*) v

must be perpendicular to Electric field

(C) v

must be perpendicular to B

(D) E must be equal to vB.

;fn fo|qr rFkk pqEcdh; {ks=k okys LFkku ls ,d vkosf'kr d.k fcuk Rofjr gq, xqtjrk gks rks_

(A) E

,B

ds lekUrj gksuk pkfg,A (B*) v

fo|qr {ks=k ds yEcor gksuk pkfg,A

(C) v

, B

ds yEcor gksuk pkfg,A (D) E, vB ds cjkcj gksuk pkfg,A.

Sol. qE = – q(V B)

E

= – (V B)

E V

and E B

6. The maximum load a wire can withstand without breaking, when its length is reduced to half of its

original length, will (A) be double. (B) be half. (C) be four times. (D*) remain same.

;fn fdlh rkj dh viuh ewy yackbZ ?kVdj vk/kh jg tkrh gS] rks og vf/kdre Hkkj] tks ;g rkj fcuk VwVs lgu dj ldrk gS] gksxkA

(A) nksxquk (B) vk/kk (C) pkj xquk (D) mruk gh ¼leku½ 7. A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit,

horizontally between two pillars (Fig.). A mass m is suspended from the mid point of the wire. Strain in the wire is :

yackbZ 2L rFkk vuqizLFk dkV {ks=kQy A ds fdlh e`nw bLikr ds rkj dks bldh izR;kLFkrk lhek ds Hkhrj nks LraHkksa ds chp fp=kkuqlkj {kSfrtr% rfur fd;k tkrk gSA dksbZ nzO;eku m blds e/; fcUnq ls fuyafcr fd;k tkrk gSA rkj es fodfr gS &

(A*) 2

2

L2

x (B)

L

x (C)

L

x2

(D) L2

x2

8. A rectangular frame is to be suspended symmetrically by two strings of equal length on two supports

(Fig.). It can be done in one of the following three ways;

fdlh vk;rkdkj Ýse dks nks leku yackbZ dh Mksfj;ksa }kjk nks voyacks ls lefer :i ls fuyafcr fd;k tkrk gS A bls uhps fn, rhu izdkj ls O;ofLFkr fd;k tk ldrk gS &

(a) (b) (c)

The tension in the strings will be : (A) the same in all cases. (B) least in (a). (C*) least in (b). (D) least in (c).

Mksfj;ksa essa ruko

(A) lc izdj.kksa eas leku gksxkA (B) (a) esa lcls de gksxkA (C) (b) esa lcls de gksxkA (D) (c) esa lcls de gksxkA




9. An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters 2.5 cm and 3.75 cm. The ratio of the velocities in the two sections is

dksbZ vkn'kZ rjy] o`Ùkh; vuqizLFk dkV ds vleku ikbi ls izokfgr gksrk gS ftlds nks vuqHkkxksa ds O;kl Øe'k% 2.5

cm rFkk 3.75 cm gSA bu nks [k.Mksa ls izokfgr rjy ds osxksa dk vuqikr gSA

(A*) 9:4 (B) 3:2 (C) 3: 2 (D) 2: 3 10. The angle of contact at the interface of water-glass is 0°, Ethylalcohol-glass is 0°, Mercury-glass is 140°

and Methyliodide-glass is 30°. A glass capillary is put in a trough containing one of these four liquids. It is observed that the meniscus is convex. The liquid in the trough is

(A) water (B) ethylalcohol (C*) mercury (D) methyliodide.

Li'kZ dks.k dk eku ty & dk¡p] varjki`"B ij 0°, ,sfFky vYdksgy&dk¡p varjki`"B ij 0°, ikjk&dk¡p varjki`"B ij 140°vkSj feFkkby vk;ksMkbM&dk¡p varjki`"B ij 30° gSA fdlh ik=k esa Hkjs bu pkjksa esa ls fdlh ,d nzo esa dk¡p dh ds'kuyh dks j[kk x;k A ;g ik;k tkrk gS fd uopUnzd mÙky gSA ik=k esa Hkjk nzo gS &

(A) ty (B) ,sfFky vYdksgy (C) ikjk (D) esfFky vk;ksMkbM 11. A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is

heated uniformly to raise its temperature slightly (A) its speed of rotation increases. (B*) its speed of rotation decreases. (C) its speed of rotation remains same. (D) its speed increases because its moment of inertia increases.

dksbZ ,d leku /kkrq dh NM+ yacor~ f}Hkktd ds ifjr% ,d leku dks.kh; pky ls ?kw.kZu djrh gSA ;fn bldk FkksM+k rki c<+kus ds fy, bls leku xeZ djsa rks bldh

(A) ?kw.kZu&pky c<+ tkrh gSA (B) ?kw.kZu&pky ?kV tkrh gSA (C) ?kw.kZu pky vifjofrZr jgrh gSA (D) bldh pky c<+rh gS D;kasfd bldk tM+Ro vk?kw.kZ c<+rk gS 12. The graph between two temperature scales A and B is shown in Fig. Between upper fixed point and

lower fixed point there are 150 equal division on scale A and 100 on scale B. The relationship for conversion between the two scales is given by :

fp=k esa nks rkiØe iSekuksa A rFkk B ds chp xzkQ n'kkZ;k x;k gSA Ldsy A rFkk Ldsy B ij fu;r fuEu rkikad rFkk fu;r mPp rkikad ds chp Øe'k% 150 rFkk 100 leku Hkkx gS rks iSekuksa ds vkil esa ifjorZu ds fy, lgh laca/k fn;k x;k gSA

(A) 150

t

100

180t BA

(B*) 100

t

150

30t BA

(C) 100

t

150

180t AB

(D) 180

t

100

40t AB




13. The radius of a metal sphere at room temperature T is R, and the coefficient of linear expansion of the

metal is . The sphere is heated a little by a temperature T so that its new temperature is T +T . The increase in the volume of the sphere is approximately

dejs ds rki T ij fdlh /kkrq ds xksys dh f=kT;k R gS rFkk /kkrq dk jSf[kd izlkj xq.kkad gSA xksys dks FkksM+k xeZ djds bldk rki T ls T+T fd;k tkrk gSa xksys ds vk;ru esa gqbZ yxHkx o`f) gS &

(A) 2 RT (B) R2T (C) 4R3T/3 (D*) 4R3T

Sol. original volume ewy vk;ru Vo = 3R3

4

Coeff of linear expansion jSf[kd o`f) xq.kkad =

Coeff of volume expansion vk;ru of) xq.kkad = 3

3dT

dV

V

1

dV = 3VdT 4R3T 14. A sphere, a cube and a thin circular plate, all of same material and same mass are initially heated to

same high temperature. (A) Plate will cool fastest and cube the slowest (B) Sphere will cool fastest and cube the slowest (C*) Plate will cool fastest and sphere the slowest (D) Cube will cool fastest and plate the slowest.

leku nzO;eku rFkk leku inkFkZ ds cus ,d xksys] ,d ?ku ,oa iryh o`Ùkkdkj IysV dks leku mPPk rki rd xeZ fd;k x;k gSA

(A) IysV lcls vf/kd rsth ls vkSj ?ku lcls /khjs BaM+k gksxkA (B) xksyk lcls vf/kd rsth ls vkSj ?ku lcls /khjs BaMk gksxkA (C) IysV lcls vf/kd rsth ls vkSj xksyk lcls /khjs BaMk gksxkA (D) ?ku lcls vf/kd rsth ls vkSj IysV lcls /khjs BaMh gksxhA 15. If an average person jogs, hse produces 14.5 × 103 cal/min. This is removed by the evaporation of

sweat. The amount of sweat evaporated per minute (assuming 1 kg requires 580 × 103 cal for evaparation) is:

;fn dksbZ lkekU; O;fDr eaFkj xfr ls pyrk gS rks og 14.5 × 103 cal/min Å"ek mRiUu djrk gSA ;g Å"ek ilhus ds ok"iu ls 'kjhj ls ckgj fudy tkrh gSA ¼;g ekurs gq, fd 1 kg ilhus ds ok"iu ds fy, 580 × 103 cal pkfg, )

rc izfr feuV okf"ir ilhus dh ek=kk gS : (A) 0.25 kg (B) 2.25 kg (C) 0.05 kg (D*) 0.025 kg

16. The displacement of a particle is represented by the equation y = 3 cos

t24

.

The motion of the particle is :

(A) simple harmonic with period 2/. (B*) simple harmonic with period / (C) periodic but not simple harmonic. (D) non-periodic.

,d d.k dk foLFkkiu fuEufyf[kr lehdj.k }kjk O;Dr gksrk gS y = 3 cos

t24

.

d.k dh xfr : (A) ljy vkorZ xfr gS ftldk nksyu dky 2/. (B*) ljy vkorZ xfr gS ftldk nksyu dky /gSA (C) vkorhZ gS ijarq ;g ljy vkorZ xfr ugha gSA (D) vkorhZ xfr ugha gSA




17. Four pendulums A, B, C and D are suspended from the same elastic support as shown in Fig.. A and C are of the same length, while B is smaller than A and D is larger than A. If A is given a transverse displacement (perpendicular to the plane of the paper),

A

B

C D

(A) D will vibrate with maximum amplitude. (B*) C will vibrate with maximum amplitude. (C) B will vibrate with maximum amplitude. (D) All the four will oscillate with equal amplitude.

pkj yksyd A, B, C ,oa D ,d gh izR;kLFk vk/kkj ls fp=k ds vuqlkj yVdk;s x;s gSaA

A

B

C D

A ,oa C dh yEckbZ cjkcj gS] B dh yEckbZ A ls de gS tcfd D dh yEckbZ A ls vf/kd gSA ;fn A dks ,d vuqizLFk

foLFkkiu fn;k tk;s rks ¼dkxt ds ry ds yEcor~½ : (A) D vf/kd vk;ke ds lkFk nksyu djsxk (B*) C vf/kd vk;ke ds lkFk nksyu djsxk (C) B vf/kd vk;ke ds lkFk nksyu djsxk (D) lHkh ¼pkjksa½ yksyd leku vk;ke ds lkFk nksyu djsaxs 18. A particle executing S.H.M. has a maximum speed of 30 cm/s and a maximum acceleration of 60

cm/s2. The period of oscillation is

ljy vkorZ xfr djrs gq, d.k dh vf/kdre pky 30 cm/s rFkk vf/kdre Roj.k 60 cm/s2 gSA bldk vkorZ&dky gS :

(A*) s. (B) 2

s (C) 2s (D)

t

s




19. When a mass m is connected individually to two springs S1 and S2, the oscillation frequencies are

1 and 2. If the same mass is attached to the two springs as shown in Fig. the oscillation frequency would be

,d æO;eku m dks tc nks fLizaxksa S1 ,oa S2 ls iFkd&iFkd tksM+dj nksyu djk;k tkrk gS rks nksyu vko`fÙk Øe'k% 1 ,oa 2 ikbZ tkrh gSA ;fn ml æO;eku dks mu fLizaxksa ds lkFk fp=k esa fn[kk;s x;s vuqlkj tksM+dj nksyu djk;k tk, rks nksyu vkofÙk gksxh :

(A) 1 + 2 (B*) 22

21 (C)

1

21

11

(D) 22

21

20. Equation of a plane progressive wave is given by y = 0.6sin2x

t2

. On reflection from a denser

medium its amplitude becomes 2/3 of the amplitude of the incident wave. The equation of the reflected wave is

,d lery izxkeh rjax dk lehdj.k y = 0.6sin2x

t2

fn;k x;k gSA ,d l?ku ek/;e ls ijkorZu gksus ij

bldk vk;ke vkifrr rjax ds vk;ke dk 2/3 gks tkrk gSA ijkofrZr rjax dk lehdj.k gS :

(A) y = 0.6sin2 x

t2

(B*) y = –0.4sin2x

t2

(C) y = 0.4sin2 x

t2

(D) y = –0.4sin2x

t2


Board Level questions 1. (a) Draw a labelled diagram of a moving coil galvanometer. Describe briefly its principle and

working. (b) Answer the following: (i) Why is it necessary to introduce a cylindrical soft iron core inside the coil of a galvanometer? (ii) Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity. Explain, giving reason.

(a) py dq.Myh xSYosuksehVj dk ukekafdr vkjs[k [khafp,A blds fl)kar vkSj dk;Ziz.kkyh dk la{ksi esa o.kZu dhft,A (b) fuEufyf[kr ds mÙkj nhft,%

(i) xSYosuksehVj dh dq.Myh ds chp ueZ yksgs ds csyukdkj ØksM dks j[kuk D;ksa vko';d gS\

(ii) fdlh xSYosuksehVj dh /kkjk lqxzkfgrk esa of) djus dk rkRi;Z ;g ugha gS fd mldh oksYVrk lqxzkfgrk esa Hkh vfuok;Zr% of) gks tk,xhA dkj.k nsrs gq, O;k[;k dhft,A

Sol. (i) The soft iron cylindrical core makes the magnetic field stronger and radial, such that into whatever position the coil rotates, the magnetic field is always parallel to its Plane.

uje yksgs dh csyukdkj ØksM izcy rFkk f=kT;h; pqEcdh; {ks=k bl izdkj cukrh gS fd dq.Myh dh fLFkfr fdlh Hkh

izdkj gks pqEcdh; {ks=k lnSo blds ry ds lekUrj gksxk

(ii) C

= BINA (i)

= BNA

C ------------ (ii).

From lehdj.k ls eqn V = I R

V = RBNA

C ------- (ii) .




From eqn (i) ) lehdj.k (i) ls I

=

C

BNA

Current Sensitivity /kkjk lqxzkghrk KL = I

=

C

BNA --------- (iii)

Voltage permittivity oksYVst fo|qr'khyrk Kv = V

=

CR

BNA ------ (iv)

If we increase the Number of turns N then current sensitivity also increases

;fn ge ?ksjks dh la[;k c<krs gS rc /kkjk lqxzkghrk Hkh c<rh gSA

When N increase, length of wire also increase, R

N remain constant. So voltage sensitivity doesn’t

increase.

tc N c<+krs gS rks rkj dh yEckbZ Hkh c<rh gS vr% R

N fu;r jgrk gS vr% oksYVst lqxzkghrk ugha c<sxhA

2. (a) Using phasor diagram, derive the expression for the current flowing in an ideal inductor connected

to an a.c. source of voltage, v = v0 sin t. Hence plot graphs showing variation of (i) applied voltage and (ii) the current as a function of wt.

(b) Derive an expression for the average power dissipated in a series LCR circuit.

(a) Qzstj vkjs[k dk mi;ksx djds a.c. L=kksr] ftldh oksYVrk v = v0 sin t. gS] ls lEc) fdlh vkn'kZ izsjd esa izokfgr /kkjk ds fy, O;atd O;qRiUu dhft,A bl izdkj wt ds Qyu ds :i esa (i) vuqiz;qä oksYVrk vkSj (ii) /kkjk ds fopj.k ds xzkQ [khfp,A

(b) fdlh Js.kh LCR ifjiFk esa vkSlr 'kfä&{k; ds fy, O;atd O;qRiUUk dhft,A





E ST INFORMA TIO


PPHHYYSSIICCSS


NO. B24 TO B25


ANSWER KEY OF DPP No. : B25

1. (A) 2. (B) 3. (D) 4. (C,D) 5. (A,C) 6. (B,C) 7. (A,D) 8. (D) 9. (C) 10. (a) R (b) S (c) Q (d) P

1. Calculate ratio of i1 and (i2 + i3).

i1 rFkk (i2 + i3) ds vuqikr dh x.kuk dhft, &

(A*) 4

3 (B)

4

9 (C)

2

3 (D) zero 'kwU;

Sol.

i1 = 4A

i2 = 6A i3 = –3A.




2. A small circular wire loop of radius a is located at the centre of a much larger circular wire loop of radius b as shown above(b>>a). Both loops are coaxial and coplanar. The larger loop carries a time (t) varying

current I = I0 cos t, where 0 and are constants. The magnetic field generated by the current in the large loop induces in the small loop an emf that is approximately equal to which of the following.

,d a f=kT;k dk NksVk o`Ùkkdkj ywi tks fd ,d cM+h f=kT;k b ds o`Ùkkdkj ywi ds dsUnz ij gS tSlk fd fp=k esa crk;k x;k gS (b > > a)A nksuksa ywi lek{kh; rFkk leryh; gSA cM+s okys ywi esa le; ifjorhZ /kkjk I = I0 cos t gSA tgk¡ 0 rFkk fu;r gSA cM+s okys ywi dh /kkjk ds dkj.k mRiUu pqEcdh; {ks=k ds dkj.k NksVs ywi esa fo- ok- cy (emf)

izsfjr gksrk gS rks og buesa ls yxHkx fdlds cjkcj gksxk &

(A) 2

0 0 acos t

2 b

(B*)

20 0 a

sin t2 b

(C) 0 0

2

asin t

2 b

(D) 0 0

2

acos t

2 b

Sol. The magnetic flux in inner loop due to current in outer loop is

ckgjh ywi ds dkj.k vkUrfjd ywi esa pqEcdh; ¶yDl gSA

= BA = 20 2 i. a

4 b

=

2

0 0

ai cos t

2 b

e = d

dt

= 0

20a i

2b

sint

3. An isolated smooth ring of mass M = 2m with two small beads each of mass m is as shown in the

figure. Initially both the beads are at diametrically opposite points and have velocity v0 (for each) in same direction. The speed of the beads just before they collide for the first time is (complete system is placed on a smooth horizontal surface and assume each point of ring is touching the surface)

M = 2m nzO;eku dh ,d foyfxr fpduh oy; ds lkFk nks NkssVs euds] çR;sd dk nzO;eku m fp=k esa n'kkZ;s vuqlkj gSA çkjEHk esa euds O;kl vfHkeq[k fcUnqvksa ij gS rFkk çR;sd leku fn'kk esa osx v0 j[krs gSA eudksa dh çFke ckj Vdjkus ds Bhd igys pky gksxhA (lEiw.kZ fudk; fpduh {kSfrt lrg ij j[kk gqvk gS rFkk ekfu, fd oy; dk çR;sd fcUnq lrg dks Li'kZ djrk gS)

(A) v0 (B) 0

2v

3 (C) 0v

2 (D*) 0

3v

2




Sol.

Just before the collision, situation is as shown

VDdj ds Bhd igys] fLFkfr n'kkZ;s vuqlkj gS Let the beads have velocity V1 w.r.t. the ring and the ring has velocity V2.

ekuk euds oy; ds lkis{k osx V1 j[krs gS o oy; dk osx V

2 gS

Then, by momentum conservation rc laosx laj{k.k ls 2mv0 = (M + 2m)V2

02

VV

2 ..........(1)

By mechanical energy conservation

;kaf=kd ÅtkZ laj{k.k ls

2

2 2 00

V1 1 1(2m)V 2mv 2m

2 2 2 2

where v = velocity of beads w.r.t. ground

tgk¡ v = euds dk /kjkry ds lkis{k osx gS

So, vr% 0

3V V

2

4. In the figure shown radius of curvature of either surface of equi convex half lens is 40 cm and refractive

index 1.5. Its one side is silvered. A plane mirror is also placed. A small object O is placed such that there is no parallax between final image formed by lens and mirror. If transverse length of final image formed by silvered lens system is twice that formed by mirror. Choose the correct options :

iznf'kZr fp=k esa ,d vk/ks f}mÙky ySUl dh nksauks lrg dh oØrk f=kT;k 40 cm gS ,oa bldk viorZukad 1.5 gSA blds ,d Hkkx dks jtrh; fd;k x;k gS rFkk ,d lery niZ.k ds lkFk bldks j[kk x;k gS] fp=k nsf[k,A ,d NksVh oLrq O dks bl çdkj j[kk x;k gS fd ySUl rFkk niZ.k ls cus gq, çfrfcEcksa ds e/; lkis{k ik'oZ foLFkkiu (parallax)

çkIr ugha gksrk gSA ;fn ySUl (silvered) fudk; }kjk cus vfUre çfrfcEc dh vuqçLFk yEckbZ (transverse length)

niZ.k }kjk cus çfrfcEc dh vuqçLFk yEckbZ ls nqxquh gS rks lgh dFku pqfu, &

(A) a = 2.5 cm (B) b = 5 cm (C*) Distance of image from object is 15 cm (D*) Image formed by silvered lens system is virtual (A) a = 2.5 cm (B) b = 5 cm

(C*) oLrq ls çfrfcEc dh nwjh 15 cm gSA (D*) jtrh; ySUl fudk; ls cuk çfrfcEc dkYifud gSA




Sol. 2(a b) a

2a

a + 2b = 2a 2b = a

2

1 2(1.5 1)

f 40

f2 = 40

fm = –20

m 2

1 1 2

F f f f = –10 cm

1 1 1

V u F a = 5

1 1 1

2a a 10

b = 2.5

5. A particle of mass m is attached to an end of a uniform rod of mass M=2m and length , which is suspended through its mid point by an inextensible string as shown.Initially the rod is in horizontal position and at rest. The system is released from this position.Just after the release

m nzO;eku dk ,d d.k M = 2m nzO;eku rFkk yEckbZ dh ,dleku NM+ ds ,d fljs ls tqM+k gqvk gS rFkk ;g NM+ blds e/; ls vforkU; Mksjh }kjk fp=kkuqlkj yVdh gSA çkjEHk esa NM+ {kSfrt fLFkfr esa fojke ij gSA fudk; dks bl fLFkfr ls NksM+k tkrk gSA NksM+us ds Bhd i'pkr~&

m

(A*) The angular acceleration of the system is 6g

5

(B) The angular acceleration of the system is 2g

5

(C*) The Tension in the string is 12

5mg

(D) The Tension in the string is 2

5mg

(A*) fudk; dk dks.kh; Roj.k 6g

5 gSA (B) fudk; dk dks.kh; Roj.k

2g

5 gSA

(C*) Mksjh esa ruko 12

5mg gSA (D) Mksjh esa ruko

2

5mg gSA

Sol. mg

2

=

2 22m m

12 4

= 6g

5

acm = 6

=

g

5

3mg – T = 3mg

5

T = 12mg

5




Alternate :

N

F

mg

The FBD of the rod and the ball are shown. Applying = about the C.M of the rod, we have,

NM+ rFkk xsan dk FBD fp=k esa çnf'kZr gSA NM+ ds nzO;eku dsUnz ds lkis{k = yxkus ij

N2

=

2M

12

Writing newtons IInd law in the vertical direction on the CM of the rod we have

NM+ ds nzO;eku dsUnz ij Å/okZ/kj fn'kk esa U;wVu dk f}Ùkh; fu;e T– N – 2mg = 0 and writing newtons II law in the vertical direction on the ball we have

rFkk xsan ij Å/okZ/kj fn'kk esa U;wVu dk f}Ùkh; fu;e fy[kus ij

m g – N= m 2

Solve these equations. bu lehdj.kksa dks gy djus ij 6. A capacitor of capacitance C and an ideal inductor of inductance L are connected in series with an ideal

battery of emf E. The resistance of circuit is negligible. If initially capacitor is uncharged and key is closed at t = 0 then select the correct statement(s) :

C /kkfjrk dk ,d la/kkfj=k rFkk L izsjdRo dh ,d vkn'kZ izsjd dq.Myh Js.khØe esa E fo|qr okgd cy dh vkn'kZ cSVjh ls tqM+s gq, gSA ifjiFk dk izfrjks/k ux.; gSA ;fn izkjEHk esa la/kkfj=k vukosf'kr gS ,oa t = 0 ij dqath can dh tkrh gS rc lgh dFku@dFkuksa dk p;u fdft,A

(A) Maximum charge on capacitor is CE

la/kkfj=k ij vf/kdre~ vkos'k CE gksxkA

(B*) Maximum charge on capacitor is 2CE

la/kkfj=k ij vf/kdre~ vkos'k 2CE gksxkA

(C*) Maximum potential difference across inductor is E

izsjd dq.Myh ds fljks ij vf/kdre~ foHkokUrj E gksxkA (D) Maximum potential difference across inductor is 2E

izsjd dq.Myh ds fljks ij vf/kdre~ foHkokUrj 2E gksxkA Sol. WB = UC + UL

when q is maximum i = 0 UL = 0

tc q vf/kdre~ gS i = 0 UL = 0

qE = 2q

2C qmax = 2CE

Charge on the capacitor will oscillates between 0 & 2CE

la/kkfj=k ij vkos'k 0 o 2CE ds e/; nksyu djsxkA

By KVL ls

E = q

C + VL

Minimum value of q is zero so maximum value of VL = E

q dk U;wure~ eku 'kwU; gS vr% VL dk vf/kdre~ eku = E gksxkA




7. A rod of uniform cross–section but non–uniform thermal conductivity which vary as k = k0

2

2

x

L (1 x L)

(as shown in figure) is kept between fixed temperature difference for a long time. Select the correct option(s) :

,d le:i vuqçLFk dkV {ks=kQy ysfdu vle:i rkih; pkydrk dh ,d NM+ ftldh pkydrk k = 2

2

x

Lk0 (1 x

L) ds vuqlkj ifjofrZr gksrh gSA fp=k eas n'kkZ;s vuqlkj fuf'pr rkikUrj ds e/; yEcs le; ds fy, j[kh tkrh gSA lgh fodYi@fodYiksa dk p;u dhft, &

(A*) (B) (C) (D*)

Ans. (A,D)

Sol. i = 2

0

2

k x dTA

dxL

H

T x2

20T 1

iL dxdT

k A x

T – TH =

x2

0 1

iL 1

k A x

Comprehension

vuqPNsn

An elastic string of unit cross–sectional area and natural length (a + b) where a > b and modulii of elasticity Y has a particle of mass m attached to it at a distance a from one end , which is fixed to a point A of a smooth horizontal plane. The other end of the string is fixed to a point B so that string is just unstretched. If particle is displaced towards right by distance x0 and then released,

,dkad vuqçLFk dkV {ks=kQy rFkk çkd`frd yEckbZ (a + b) tgk¡ a > b gS rFkk çR;kLFkrk xq.kkad Y dh ,d çR;kLFk Mksjh ls ,d m nzO;eku dk d.k blds ,d fljs ls a nwjh ij tqM+k gSA bl jLlh dk ,d fljk fcUnq A ij tM+or~ gS tks fpdus {kSfrt ry ij fLFkr gSA jLlh dk nwljk fljk B ij tM+or~ gS rkfd Mksjh Bhd fcuk [khaph gqbZ fLFkfr esa jgrh gSA ;fn izkjaHk esa d.k dks x0 nka;h vksj f[kapdj NksM+k tk, rks

A

O

a b B

8. The time period of the oscillation will be : nksyu dk vkorZdky gksxk &

(A) (a b)m

2Y

(B)

m2 ( a b)

Y (C)

(a b)m

Y

(D*)

m( a b)

Y

9. The separation between two extreme positions will be : nksauks pje fLFkfr;ksa ds e/; nwjh gksxh &

(A) a + b (B) 2 a (C*) 0

a b x

a

(D) 0

a b x

b




Sol. A

O

a b

from B 0t ls

T = y x

a

2

2

d x yx

amdt

B 0

amt

2 y

simply B 0t other ext. = bm

2 y

blh çdkj B 0t vU; çlkj ds fy, = bm

2 y

Time period vkorZdky = m

( a b )y

Energy conservation ÅtkZ laj{k.k ls

2 2

0y x yc

2a 2b

c = x0 b

a

Distance nwjh (ext. to ext.) (çlkj ls çlkj)

x0 + c = x0 b a

a

10. A beam consisting of four types of ions A, B, C and D enters a region that contains a uniform magnetic

field as shown. The field is perpendicular to the plane of the paper, but its precise direction is not given. All ions in the beam travel with the same speed. The table below gives the masses and charges of the ions.

Ion beam

r 1r 2r 4 r 3

Region containing Magnetic field

2 4 3 1

ION MASS CHARGE

A 2m e

B 4m – e

C 2m – e

D m + e

r4 > r3 = r2 > r1. The ions fall at different positions 1, 2, 3 and 4, as shown. Correctly match the ions with respective

falling positions.

,d iqat ftlesa A, B, C o D pkj izdkj ds vk;u gSA fp=k esa n'kkZ;sa vuqlkj le:i pqEcdh; {ks=k esa izos'k djrk gSA {kS=k dkxt ds ry ds yEcor gS ijUrq bldh mfpr fn'kk ugha nh xbZ gSA lHkh vk;u iqat esa leku pky ls pyrs gSA uhps nh xbZ lkj.kh vk;uksa ds nzO;eku o vkos'k dks crkrh gSA




vk;u iaqt

r 1 r 2r 4 r3

pqEcdh; {kS=k dh mifLFkfr okyk LFkku

2 4 3 1

ION MASS CHARGE

A 2m e

B 4m – e

C 2m – e

D m + e r4 > r3 = r2 > r1.

vk;u fofHkUu fLFkfr;ksa 1, 2, 3 o 4 ij n'kkZ;sa vuqlkj fxjrs gSaA vk;uksa dks muds fxjus dh lgh fLFkfr;ksa ds lkFk lqesfyr djokb;saA

Column I Column II (a) A (P) 1 (b) B (Q) 2 (c) C (R) 3 (d) D (S) 4 Ans. (a) R (b) S (c) Q (d) P

Sol. r = mV

qB

rA = 2mV

eB

rB = 4mV

eB

rC = 2mV

eB;

rD = mV

eB





E ST INFORMA TIO


PPHHYYSSIICCSS


NO. B26 TO B28

2. DPP Syllabus :



ANSWER KEY OF DPP No. : B26 1. (A) 2. (B) 3. (B) 4. (A) 5. (B) 6. (D) 7. (A) 8. (C) 9. (A) 10. (A) 11. (D) 12. (A) 13. (B) 14. (A) 15. (B) 16. (C) 17. (A) 18. (B) 19. (C) 20. (B)

1. Which of the following statement(s) is/are wrong-

[a] A body can have constant speed and varying acceleration

[b] A body can have zero velocity and non zero acceleration at a moment.

[c] A body can have constant velocity and uniform acceleration

(A*) Only c (B) Both b and c (C) Both a and c (D) All a, b, c

fuEu esa ls vlR; dFku gS&

[a] ,d oLrq dh vpj pky rFkk ifjorhZ Roj.k gks ldrk gSA

[b] ,d oLrq dk fdlh {k.k ij osx 'kwU; ,oa v'kwU; Roj.k gks ldrk gSA

[c] ,d oLrq dk vpj osx rFkk ,d leku Roj.k gks ldrk gSA

(A) dsoy c (B) b rFkk c nksuksa (C) a rFkk c nksuksa (D) lHkh a, b, c

Sol. When velocity is constant and acceleration is zero.

tks osx fu;r gS rFkk Roj.k 'kwU; gSaaA

2. A body travelling with uniform acceleration crosses two points A and B with velocities 20 m/sec and

30 m/sec respectively. Then the speed of the body at mid-point of A and B is-

(A) 25 m/sec (B*) 25.5 m/sec (C) 24 m/sec (D) 610 m/sec

,d oLrq leku Roj.k ls xfr djrh gqbZ nks fcUnq A rFkk B dks Øe'k% 20 eh-@ls- rFkk 30 eh-@ls- ds osx ls ikj djrh

gSA rc A o B ds e/; fcUnq ij oLrq dk osx gS&

(A) 25 eh-@ls- (B) 25.5 eh-@ls- (C) 24 eh-@ls- (D) 610 eh-@ls-

Sol. v = 2

vv 22

21

= 25.5 m/s




3. A block of mass m1 = 2 kg on a smooth inclined plane at angle 30° is connected to a second block of mass m2 = 3 kg by a cord passing over a frictionless pulley as shown in fig. The acceleration of each

block is- (assume g = 10 m/sec2)

,d ur ry ftldk {ksfrt ls >qdko 30° gS] ij fLFkr ,d m1 = 2 fdxzk nzO;eku dk CykWd] m2 = 3 fdxzk nzO;eku ds nwljs CykWd ls Mksjh dh lgk;rk ls tqM+k gqvk gS rFkk Mksjh fp=kkuqlkj ,d ?k"kZ.k jfgr f?kjuh ls xqtjrh gS] izR;sd nzO;eku dk Roj.k gS& (g = 10 eh@ls2)

30°

m 1 m2

(A) 2 m/sec

2 (B*) 4 m/sec

2 (C) 6 m/sec

2 (D) 8 m/sec

2

Sol. a = 2 1

1 2

m g m gsin30

m m

a = 2

g5

= 4 m/s2

4. If kinetic energy is doubled, then its momentum becomes n times, here n is :

;fn xfrt ÅtkZ nqxuh dj nh tk;s rks] laosx n xquk gks tkrk gS n rks dk eku Kkr djksA

(A*) 2 (B) 2 2 (C) 1

2 (D)

1

2 2

Sol. KE = 2p

2m KE p2

2

1

p

p = 2

1

KE

KE = 2

5. A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. The work done by the force

of gravity during the time the particle goes up is :

100 g nzO;eku dk d.k Å/oZ Åij dh vksj 5 m/s dh pky ls Qsadk tkrk gSA Å/oZ xeu ds nkSjku xq:Roh; cy }kjk d.k ij fd;k x;k dk;Z gksxk&

(A) – 0.5 J (B*) – 1.25 J (C) 1.25 J (D) 0.5 J Sol. The height (h) tranversed by particle while going up is :

Åij tkus ij d.k }kjk r; dh xbZ ÅpkbZ (h)

h = 2u

2g =

25

2 9.8

Work done by gravity force = mg.h

xq:Roh; cy }kjk fd;k x;k dk;Z = mg.h

= 0.1 × g × 25

cos1802 9.8

(Angle between force and displacement is 180°)

(cy rFkk foLFkkiu ds e/; dks.k 180° gS)

W = – 0.1 ×25

2 = – 1.25 J




6. A particle of mass m is executing a uniform motion along a circular path of radius r. If the magnitude of its linear momentum is p, the radial force acting on the particle will be.

m nzO;eku dk ,d d.k r f=kT;k ds iFk ij ,d leku o`Ùkh; xfr dj jgk gSA ;fn blds js[kh; laosx dk ifjek.k P

gks rks d.k ij dk;Zdkjh f=kT; cy gksxk &

(A) pmr (B) rm/p (C) mp2/r (D*) p2/mr

Sol. F =2mv

r = p2/mr

7. A shell is fired from a cannon with a velocity v at an angle with the horizontal direction. At the highest point in its path, it explodes into two equal pieces, one retraces its path to the cannon and the speed of the other piece immediately after the explosion is :

,d xksys dks rksi ls {kSfrt ls dks.k ij v osx ls nkxk tkrk gSA blds iFk ds mPpre fcUnq ij ;g nks cjkcj Hkkxksa esa fo[kf.Mr gks tkrk gSA ,d Hkkx blh iFk dks rksi dh rjQ vuqlj.k djrk gS rks nwljs Hkkx dh pky VDdj ds rqjUr ckn gksxh &

(A*) 3 cos (B) 2 cos (C) 3

2

cos (D) 3

2 cos

Sol. P1 = Pf

mV cos = m

2 (–V cos +

m

2 V)

V = 3V cos

8. A 1.0kg ball drops vertically into a floor from a height of 25 cm. It rebounds to a height of 4cm. The coefficient of restitution for the collision is -

1.0kg nzO;eku okyh xsan fdlh Q'kZ ij 25 cm dh ÅapkbZ ls m/okZ/kj fxjrh gSA og 4cm dh ÅapkbZ rd okil mNyrh gSA VDdj dk izR;koLFkku xq.kkad gS&

(A) 0.16 (B) 0.32 (C*) 0.40 (D) 0.56

Sol. Velocity just before collision = u = 2gh = 5 m/s

VDDj ds Bhd igys osx = u = 2gh = 5 m/s

velocity just after collision = v = 2gh' = 0.8 m/s

VDdj ds Bhd i'pkr~ osx = v = 2gh' = 0.8 m/s

e = v

0.4u

9. If the radius of the earth suddenly contracts to half of its present value, then the duration of day will be of (mass of earth remain same)

(A*) 6 hours (B) 12 hours (C) 18 hours (D) 24 hours

;fn i`Foh dh f=kT;k {k.k Hkj esa fldqM+dj orZeku f=kT;k dh vk/kh gks tk;s] rks ,d fnu dh vof/k gksaxsA ¼i`Foh dk æO;eku fu;r jgrk gS½

(A) 6 ?kUVs (B) 12 ?kUVs (C) 18 ?kUVs (D) 24 ?kUVs Sol. Using energy conservation

ÅtkZ laj{k.k ls 1 = 2

1 2

2 1

R T

R T

T2 = 6 hours




10. A solid sphere and a solid cylinder having the same mass and radius, roll down the same incline. The ratio of their acceleration will be

leku æO;eku o f=kT;k ds Bksl xksyk o Bksl csyu fdlh ur ry ij yq<+drs gSaA buds Roj.kksa dk vuqikr gksxkA

(A*) 15 : 14 (B) 14 : 15 (C) 5 : 3 (D) 3 : 5

Sol. a =

2

gsin

1MR

1

2

11

a 22a

15

= 15 : 14

11. Two simple pendulums of lengths 1 meter and 16 meteres respectively are both given small

displacements in the same direction at the same instant. They will again be in phase at same position after the shorter pendulum has completed n oscillations where n is

nks ljy yksyd] ftudh yEckbZ;k¡ Øe'k% 1 ehVj 16 ehVj gSa] nksuksa dks ,d lkFk ,d gh fn'kk esa vYi foLFkkiu nsrs gSaA ;s nksuksa mlh fcUnq ij fQj ,d gh dyk esa gksrs gSa] tc NksVk yksyd n nksyu iw.kZ dj ysrk gSA tgk¡ n gSaA

(A) 1

4 (B) 2 (C) 5 (D*) 4

Sol. 1

2

2

1

n

n

41

16

n

n

2

1

n = 4 12. If you set up the seventh harmonic on a string fixed at both ends, how many nodes and antinodes are

set up in it -

;fn vki nksuksa fljksa ij fLFkj ¼tMor~½ Mksjh esa lIre lukafn mRiUUk djrs gS rks blesa fdrus fuLiUn rFkk fdrus izLiUn mRiUUk gksxsA

(A*) 8, 7 (B) 7, 7 (C) 8, 9 (D) 9, 8 Sol. 8, 7 13. Two identical flutes produce fundamental notes of frequency 300 Hz at 27°C. If the temperature of air in

one flute is increased to 31°C, the number of the beats heard per second will be

nks ,dleku cklqfj;k¡ 27°C ij 300 Hz vkofÙk ds ewy Loj mRiUu djrh gSA ;fn ,d ck¡lqjh esa ok;q rki c<+kdj 31°C, dj fn;k tk;s rks izfr lSd.M lqukbZ nsus okys foLianksa dh la[;k gks tk;sxhA

(A) 1 (B*) 2 (C) 3 (D) 4

Sol. f

f

=

1

2

f = 2 14. The gases carbon-monoxide (CO) and nitrogen at the same temperature have kinetic energies E1 and

E2 respectively. Then : (A*) E1 = E2 (B) E1 > E2 (C) E1 < E2 (D) E1 and E2 cannot be compared

leku rki ij dkcZu eksuksvkWDlkbM (CO) rFkk ukbVªkstu N2 xSlksa dh xfrt ÅtkZ;s Øe'k% E1 rFkk E2 gS] rks (A*) E1 = E2 (B) E1 > E2

(C) E1 < E2 (D) E1 rFkk E2 dh rqyuk ugha dh tk ldrh




Sol. The gases carbon-monoxide (CO) and nitrogen (N2) are diatomic, so both have equal kinetic energy

5

KT2

, i.e. E1 = E2.

gy xSls dkcZu eksuksvkWDlkbZM (CO) rFkk ukbVªkWtu (N2) f}ijek.kfod xSl gS vr% nksuks dh xfrt ÅtkZ leku gksxhA

5

KT2

, vFkkZr E1 = E2.

(CO) rFkk ukbVªkstu (N2) nksuksa xSl f}v.kqd gS] rFkk nksuks dh xfrt ÅtkZ cjkcj 5

KT2

vFkkZr E1 = E2.

15. For a black body at temperature 727°C, its net radiating power is 60 watt and temperature of

surrounding is 227°C. If temperature of black body is changed to 1227°C then its net radiating power will be-

(A) 304 W (B*) 320 W (C) 240 W (D) 120 W

727°C ij fLFkr ij fLFkr ,d Ñ".k fiaM ds fy, ifj.kkeh fodj.k 'kfDr 60 okWV rFkk cká; okrkoj.k dk rki

227°C gSA ;fn Ñ".k fiaM dk rki 1227°C rd ifjofrZr fd;k tkrk gS rks bldh ifj.kkeh fofdj.k 'kfDr gksxh&

(A) 304 okWV (B*) 320 okWV (C) 240 okWV (D) 120 okWV

Sol. 60 = K(10004 – 500

4) ...(i)

E = K(15004 – 500

4) ...(ii)

from (i) and (ii) 4 4

4 4

E 1500 500

60 1000 500

E = 320

lehdj.k (i) rFkk (ii) 4 4

4 4

E 1500 500

60 1000 500

E = 320

16. A body takes 10 minutes to cool down from 62°C to 50°C. If the temperature of surrounding is 26°C

then in the next 10 minutes temperature of the body will be :

;fn okrkoj.k dk rki 26°C gks rks ,d oLrq 62°C ls 50°C rd B.Mk gksus esa 10 feuV dk le; ysrh gSA vxys 10 feuV ds ckn oLrq dk rki gksxk :

(A) 38°C (B) 40°C (C*) 42°C (D) 44°C Sol. From Newton's cooling law

U;wVu ds 'khryu ds fu;e ls

1 2 1 2K 26t 2

Case (izFke fLFkfr): 12

10 = K × 30

K = 12

10 30 =

1

25

Case f}rh; fLFkfr : 2 250 50K 26

10 2

2 250 21

10 25 2

250 – 52 = 2 – 2 62 = 252

2 =252

6 = 42°C

17. A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y plane

and the other in x-z plane. If the current in the loop is i. The resultant magnetic field due to the two semicircular parts at their common centre is

,d /kkjkokgh dq.Myh dks nks ,dleku R f=kT;k okys v)Zo`Ùkh; Hkkxksa dks tksM+dj cuk;k tkrk gS] ftlesa ,d Hkkx x-

y ry ds vuqfn'k rFkk vU; nwljk Hkkx x-z ry ds vuqfn'k gSA ;fn dq.Myh esa izokfgr /kkjk i gks rks nksuksa v)Zo`Ùkh; Hkkx ds dkj.k] muds }kjk mHk;fu"B dsUæ ij ifj.kkeh pqEcdh; {ks=k gksxk

(A*) 0i

2 2R

(B) 0i

2R

(C) 0i

4R

(D) 0i

2R




Sol. The loop mentioned in the question must look like one as shown in the figure.

Magnetic field at the centre due to semicircular loop lying in x-y plane, 0xy

i1B

2 2R

negative z

direction

Similarly field due to loop in x–z plane, Bxz =0i1

2 2R

in negative y direction.

Magnitude of resultant magnetic field,

B =

2 22 2 0 0xy xz

i iB B

4R 4R

= 0 0i i2

4R 2 2R

Sol. iz'u esa fn, x, ywi fp=k esa n'kkZ;s vuqlkj gksaxs

x-y ry esa fLFkr v)Zo`Ùkkdkj ywi ds dkj.k dsUæ ij pqEcdh; {ks=k 0xy

i1B

2 2R

, _.kkRed z fn'kk

blh izdkj x–z ry esa ywi ds dkj.k {ks=k , Bxz =0i1

2 2R

_.kkRed y fn'kk

ifj.kkeh pqEcdh; {ks=k dk ifjek.k

B =

2 22 2 0 0xy xz

i iB B

4R 4R

= 0 0i i2

4R 2 2R

18. In a cyclotron, the angular frequency of the charged particle is independent of (A) mass (B*) speed (C) charge (D) magnetic field

,d lkbDyksVªkWu esa vkosf'kr d.k dh dks.kh; vko`fÙk fuHkZj ugha djrh &

(A) nzO;eku ij (B) pky ij (C) vkos'k ij (D) pqEcdh; {ks=k ij

SOL. osc

qBf

2 m

19. A particle of charge per unit mass is released from origin with a velocity 0ˆv v i

in a uniform

magnetic field 0ˆB B k

. If the particle passes through (0, y, 0) then y is equal to

,d leku pqEcdh; {ks=k 0ˆB B k

esa vkos'k izfr bdkbZ nzO;eku dk ,d d.k ewy fcUnq ls 0

ˆv v i

ds osx ls NksM+k tkrk gSA ;fn d.k (0, y, 0) ls xqtjrk gS rks y cjkcj gksxk&

(A) 0

0

2v

B

(B) 0

0

v

B (C*) 0

0

2 v

B (D) 0

0

v

B




Sol. Directin of force along +y direction . so partical passes through the+y axis at a distance 2r form origin.

pqEcdh; cy dh fn'kk +y fn'kk ds vuqfn'k gSA vr% d.k /kukRed y v{k ls ewy fcUnq ls 2r nwjh ij xqtjrk gSA

y = 2r = 2mv

qB = 0

0

2 v

B

20. Two identical capacitors A and B shown in the given circuit are joined in series with a battery. If a dielectric slab of dielectric constant K is slipped between the plates of capacitor B and battery remains connected, then the energy of capacitor A will-

fn;s x;s ifjiFk esa ,d tSls nks la/kkfj=k A o B ,d cSVjh ls Js.khØe esa tqM+s gq;s gSA K ijkoS|qrkad dh ,d ijkoS|qr ifêdk dks B la/kkfj=k dh IysVksa ds e/; j[krs gS] rFkk ogh cSVjh tqM+h gqbZ jgrh gS] rks la/kkfj=k A dh ÅtkZ&

(A) Decrease (B*) Increase (C) Remain the same (D) Be zero since circuit will not work

(A) de gksxhA (B) c<sxhA

(C) leku jgsxhA (D) ifjiFk dk;Z ugha djrk gS] vr% 'kwU; jgsxhA Sol. qA = qB = q q =

q' = q' > q





E ST INFORMA TIO


PPHHYYSSIICCSS


NO. B26 TO B28



1. (A) 2. (C) 3. (C) 4. (A,C) 5. (B,C,D) 6. (A,B,D) 7. (B,C) 8. (A,C) 9. (a) P,S (B) P,R (C) Q,S (D) P,R

1. The spectra of a black body at temperatures 273°C and 546°C are shown in the figure. If A1 and A2 be

the areas under the two curves respectively, the value of 1

2

A

A is

rkieku 273°C rFkk 546°C ij d`".k oLrq ds fy, LizsDVªe fp=k esa iznf'kZr gSA ;fn bu oØks ls f?kjk gqvk {ks=kQy

Øe'k% A1 rFkk A2 gksrks 1

2

A

A dk eku gksxk A

546°CE

273°C

(A*) 16

81 (B)

1

16 (C)

8

27 (D)

81

16

2. The time constant of the circuit shown in the figure is

iznf'kZr fp=k ds fy, ifjiFk dk le; fu;rkad gksxkA R

CR

R

R

(A) 3RC

2 (B) RC (C*)

5RC

3 (D)

2RC

3




3. In a arrangement, 3n cells of emf and internal resistance r are connected in series. Out of 3n cells, polarity of n cells is reversed. Current in the circuit is

iznf'kZr O;oLFkk esa fo|qr okgd cy rFkk vkUrfjd izfrjks/k r okys 3n lSy Js.kh Øe esa tqMs gq, gSA 3n lSyksa esa ls n lSy foifjr /kzqork ds lkFk tqMs gq, gSA rks ifjiFk esa /kkjk gksxhA

A

r r

(A) 2

r

(B)

2

3r

(C*)

3r

(D)

r

4. Electron jumps to some higher energy state where its angular momentum becomes twice. In final state (A*) radius of orbit becomes 4 times (B) radius of orbit becomes 2 times (C*) speed of electron will half (D) speed of electron remain constant

bysDVªkWUk mPp d{kk esa dwnrk gS tgk¡ mldk dks.kh; laosx nqxuk gks tkrk gSA vfUre voLFkk esa (A*) d{kh; f=kT;k 4 Xkquk gks tk;sxh (B) d{kh; f=kT;k 2 Xkquk gks tk;sxh (C*) bysDVªkWu dh pky vk/kh gksxh (D) bysDVªkWUk dh pky vifjofrZr jgsxh

Sol. r z

n2

rFkk and L = 2

nh

r = 4r1

v n

z

5. A long, straight wire carries a current along the Z-axis. One can find two points in the X-Y plane such

that (A) the magnetic fields are equal (B*) the directions of the magnetic fields are the same (C*) the magnitudes of the magnetic fields are equal (D*) the field at one point is opposite to that at the other point ,d yEcs ,oa lh/ks rkj esa Z-v{k ds vuqfn'k /kkjk izokfgr gks jgh gSA x - y ry esa fLFkr nks fcUnqvksa ds fy;s dksbZ

O;fDr izsf{kr dj ldrk gS] fd - (A) pqEcdh; {ks=k ,d leku gSA (B*) pqEcdh; {ks=kksa dh fn'kk,¡ ,d leku gSA (C*) pqEcdh; {ks=kksa ds ifjek.k ,d leku gSA (D*) ,d fcUnq ij pqEcdh; {ks=k dh fn'kk nwljs fcUnq ij {ks=k dh fn'kk ls foifjr gSA 6. A particle starts moving from the origin & moves along positive x-direction. Its rate of change of kinetic

energy with time shown on y-axis varies with time t as shown in the graph. If position, velocity, acceleration & kinetic energy of the particle at any time t are x, v, a & k respectively then which of the option (s) may be correct ?

,d d.k ewy fcUnq ls /kukRed x-fn'kk esa xfr djuk izkjEHk djrk gSA le; t ds lkFk bldh xfrt ÅtkZ esa ifjorZu dh nj xzkQ esa iznf'kZr gSA xfrt ÅtkZ esa ifjorZu dks y-v{k ij fy;k x;k gSA le; t ij d.k ds fLFkfr] osx ] Roj.k o xfrt ÅtkZ Øe'k% x, v, a ,oa k ls iznf'kZr gS rks dkSulk@dkSuls fodYi lgh gks ldrs gS \

(A*) (B*) (C) (D*)

x

k




Sol. dk

dt t

k t2 ...........(i)

v t .............(ii) v vs t : st. line

x t2 ...........(iii) x vs t : parabola

(i) & (iii) k x ...........(iv) k vs x : st. line

(ii) a = constant

(ii) & (iii) v2 x..............(v) v2 vs x : st. line

gy. dk

dt t

k t2 ...........(i)

v t .............(ii) v o t : ljy js[kk x t2 ...........(iii) x o t : ijoy;

(i) o (iii) k x ...........(iv) k o x : ljy js[kk (ii) a = fu;rkad (ii) o (iii) v2 x..............(v) v2 o x : ljy js[kk

Comprehension

Water is filled to a height h in a fixed vertical cylinder placed on horizontal surface. At time t=0 a small hole is drilled at a height h/4 from bottom of cylinder as shown. The cross section area of hole is a and the cross-section area of cylinder is A such that A > > a.

{kSfrt lrg ij j[ks gq;s fLFkj Å/okZ/kj csyu esa h Å¡pkbZ rd ty Hkjk gqvk gSA t = 0 le; ij csyu dh ryh ls h/4

Å¡pkbZ ij ,d NksVk fNnz fp=kkuqlkj fd;k tkrk gSA fNnz dk vuqizLFk dkV {ks=kQy a gS rFkk csyu dk vuqizLFk dkV {ks=kQy A bl izdkj gS fd A > > a.

h/4

Horizontal surfacex

h/4

{kSfrt lrg x

7. Let the value of horizontal distance of point where the water fall on horizontal surface from bottom of cylinder be x as shown.Then from time t = 0 till water comes out of hole, pick the correct statement:

ekuk csyu dh ryh ls ml fcUnq dh {kSfrt nwjh dk eku tgka ty fxjrk gS fp=kkuqlkj] x gS rc t = 0 ls izkjEHk djrs gq;s tc rd ikuh fNnz ls ckgj dh vksj vkrk gSA lgh dFku dk p;u dhft;sA

(A) x increases with time. (B*) x decreases with time. (C*) After coming out from hole water falls in parabolic path (D) Velocity of efflux increases with time

(A) le; ds lkFk x c<+rk gSA (B*) le; ds lkFk x ?kVrk gSA (C) fNæ ls ckgj fudyus ij ikuh ijoy;kdkj iFk ij fxjrk gSA (D) le; ds lkFk cká L=kko osx c<+rk gSA




8. For the above question (A*) As long as water comes out of hole, the time taken by a water particle starting from hole to reach

the horizontal surface is h

2gand is always constant

(B) As long as water comes out of hole, the time taken by a water particle starting from hole to reach

the horizontal surface is h

2gand is decrease with time.

(C*) The duration of time for which water flows out of hole is g2

h3

a

A

(D) The duration of time for which water flows out of hole is A h

a g

mijksDr iz'u gsr&

(A*) tc rd ikuh fNnz ls ckgj vkrk gS] nzo d.kksa }kjk fNnz ls {ksfrt lrg rd igqpusa esa fy;k x;k le;

h

2ggksxk rFkk ;g lnso fu;r jgsxkA

(B) tc rd ikuh fNnz ls ckgj vkrk gS] nzo d.kksa }kjk fNnz ls {ksfrt lrg rd igqpusa esa fy;k x;k le;

h

2ggksxk rFkk ;g le; ds lkFk ?kVsxkA

(C*) le;kof/k g2

h3

a

Ard ikuh fNnz ls ckgj fudyrk jgsxkA

(D) le;kof/k A h

a grd ikuh fNnz ls ckgj fudyrk jgsxkA

Sol. The initial velocity of water coming out of hole is horizontal and hole is at a height 4

h from ground.

Hence time taken by water to reach ground is t = g

)4/h(2 which remains constant.

x = vt where v is velocity of effux.... Since v decreases wth time x will decrease.

v

v

y

Let y be the height of water surface above hole

–dt

dy=

A

av =

A

gy2a

t

0

0

4/h3

dtA

a

gy2

dy t =

g2

h3

a

A

fNnz ds ckgj vkus ij ty dk izkjfEHkd osx {kSfrt gS rFkk fNnz {kSfrt lrg ls 4

h ÅpkbZ ij gSA vr% ty }kjk

{kSfrt lrg rd igqpus es yxk; le; t = g

)4/h(2 gksxk tks fu;r jgrk gSA

x = vt tgk”v ckgj vkus dk osx ....

v

v

y

pawfd v le; ds lkFk ?kV jgk gSa vr% x c<+sxkA ekuk fNnz ls ty lrg dh ÅapkbZ y gS




– dt

dy=

A

av =

A

gy2a

t

0

0

4/h3

dtA

a

gy2

dy t =

g2

h3

a

A

9. Match the statements in Column with the results in Column

LrEHk esa fn;s x;s dFkuksa dks LrEHk esa fn;s x;s ifj.kkeksa ls lqesfyr dhft;s Column – I Column – II (a) A thin uniform spherical shell of surface area 'S' has an (P) is independent of 'S' initial temperature more than its surrounding atmosphere. Then magnitude of rate of change of its temperature with time (assuming density of material and thickness of shell remains constant) (b) A soap bubble initially in equilibrium is given a charge Q, (Q) depends on 'S' which distributes uniformly over its surface. The centre of the bubble is always fixed. For the duration the bubble having surface area 'S' expands, the magnitude of electric potential at a fixed point always lying outside the bubble (c) A container with open top and filled with ideal liquid is (R) remains constant placed at rest on a smooth horizontal table. A small hole of area 'S' is drilled at the bottom of a side wall of container. The magnitude of force exerted by escaping liquid on the container (d) An infinitely long straight current carrying wire lies along the (S) decreases with time axis of a closed cylindrical surface of total surface area 'S' in space. As the magnitude of current in the wire is continuously increased, the magnitude of the magnetic flux through the surface of this cylinder LrEHk – I LrEHk – II

(A) ,d iryk ,dleku xksykdkj dks'k ftldk i`"B {ks=kQy 'S' (p) 'S' ls LorU=k gSA gS rFkk bldk izkjfEHkd rki] blds ckgjh okrkoj.k ds rki ls T;knk gS rc blds rkieku ds ifjorZu dh nj dk ifjek.k le; ds lkFk (ekuk inkFkZ dk ?kuRo rFkk dks'k dh eksVkbZ fu;r jgrk gSA)

(B) ,d lkcqu dk cqycqyk izkjEHk esa lkE;oLFkk esa gS rFkk bldks vkos'k (q) 'S' ij fuHkZj djrk gSA Q fn;k tkrk gS tks fd bldh lrg ij ,dleku :i ls forfjr gks tkrk gSA cqycqys dk dsUnz ges'kk fLFkj (fixed) gSA ml nkSjku ds fy, tc cqycqys dk i`"B {ks=kQy S QSyrk gS] rks cqycqys ds ckgj fLFkr ,d tM+or (fixed) fcUnq ij fo|qr foHko dk ifjek.k

(C) ,d ik=k ¼crZu½ ftldk Åijh Hkkx [kqyk gS rFkk bldks vkn'kZ nzo (r) fu;r jgrk gSA ls Hkjk tkrk gS o bldks ,d fpduh {kSfrt Vscy ij fojke esa j[kk tkrk gSA crZu dh ik'oZ nhokj (side wall) ds fupys ry (bottom) ij 'S'

{ks=kQy dk ,d NksVk Nsn fd;k tkrk gSA rks ckgj fudyus okys nzo }kjk ik=k (crZu) ij yxk;s x;s cy dk ifjek.k

(D) ,d cUn (closed) csyukdkj lrg ftldk dqy {ks=kQy 'S' gSA ;gk¡ (s) le; ds lkFk ?kVrk gSA vkdk'k (space) esa gS bldh v{k ds vuqfn'k ,d vuUr yEck /kkjkokgh lh/kk rkj fo|eku gSA tSls tSls bl rkj esa /kkjk dk ifjek.k yxkrkj c<+k;k tkrk gS] rks bl csyu dh lrg ls ikfjr pqEcdh; ¶yDl dk ifjek.k Ans. (a) P,S (B) P,R (C) Q,S (D) P,R




Sol. (a) The rate of change of heat energy of thin shell of mass m, specific heat C, radius R and width dR is given by

mCdt

dT = sA (T4 – T0

4)

or (r 4pR2 dR) C

dt

dT= s (4 p R2) (T4 – T0

4)

Hence the magnitude of rate of change of temperatureC

)TT(

dt

dT 40

4

is independent of S = 4 pR2 and decreases with time as temperature of shell T decreases with time.

(B) The potential outside bubble can be found by simply concentrating net charge on bubble at its centre. Hence potential at fixed point outside is constant and independent of S.

(C) The magnitude of force exerted by liquid of density r escaping with speed v is rSv2. Hence this force depends on S and decreases with time as amount of liquid in container decreases resulting in continuous decrease of v.

(D) The magnitude flux through surface S of closed cylinder is zero. Hence it is constant and independent of S.

gy% (A) m nzO;eku, C fof'k"V Å"ek, R f=kT;k rFkk dR eksVkbZ ds irys dks'k dh Å"ek ÅtkZ ds ifjorZu dh nj nh tkrh

gSA mC dt

dT= sA (T4 – T0

4)

;k (r 4pR2 dR) C

dt

dT= s (4 p R2) (T4 – T0

4)

blfy;s rki ifjorZu dh nj dk ifjek.k C

)TT(

dt

dT40

4

;g S = 4 px2 ls LorU=k gS rFkk le; ds lkFk ?kVrk gS D;ksafd dks'k dk rki le; ds lkFk ?kVrk gSA

(B) cqycqys ds ckgj foHko usV vkos'k dks blds dsUnz ij fLFkr ekudj fudkyk tk ldrk gSA blfy;s ckgj ds fLFkj fcUnq ij foHko fu;r gS rFkk S ls LorU=k gSA

(C) v pky ls fudyrs gq;s r ?kuRo ds nzo }kjk vkjksfir cy dk ifjek.k r S v2 gSA blfy;s ;g cy S ij fuHkZj djrk gS rFkk le; ds lkFk ?kVrk gS tc crZu esa nzo dh ek=kk ?kV tkrh gS ftlds ifj.kkeLo:i v lrr~ :i ls ?kVrk gSA

(D) cUn csyu dh lrg S ls gksdj tkus okyk pqEcdh; ¶yDl 'kwU; gSA blfy;s ;g fu;r gS rFkk S ls LorU=k gSA





E ST INFORMA TIO


PPHHYYSSIICCSS


NO. B26 TO B28


Single choice Objective ('–1' negative marking) Q.1 to Q.2 (3 marks 2 min.) [06, 04] One or more than one options correct type ('–1' negative marking) Q.3 to Q.9 (4 marks 2 min.) [28, 14] Match the Following (no negative marking) Q.10 (8 marks 06 min.) [08, 06]

ANSWER KEY OF DPP No. : B28 1. (D) 2. (A) 3. (A,C,D) 4. (B,C) 5. (A,C) 6. (A,B,C,D)

7. (A,C,D) 8. (B) 9. (D) 10. (A) p,q,r (B) p,q,r, (C) p,s (D) p,q,r

1. A body of mass 6 kg is acted upon by a force which causes a displacement in it given by x = 4

t 2

metre

where t is the time in second. The work done by the force is 2 seconds is:

,d oLrq dk æO;eku 6 kg gS bl ij ,d cy dk;Zjr gS ftlds dkj.k bldk foLFkkiu x = 4

t 2

ehVj gksrk gS] tgk¡

t lSd.M esa le; gSA 2 lsd.M esa fd;k x;k dk;Z gSA (A) 12 J (B) 9 J (C) 6 J (D*) 3 J Sol. The velocity of the body a time t is given by

2

t

4

t

dt

d

dt

dx 2

At t = 0 , = u = 0 and t = 2 s, = 1ms-1, Now, work done = increase in KE

0m2

1mu

2

1m

2

1 222

22 )1(62

1m

2

1 = 3J, Hence the correct choice is (d).

Sol. fdlh le; t ij oLrq dk osx fn;k tkrk gS %

2

t

4

t

dt

d

dt

dx 2

t = 0, = u = 0 o t = 2 s, = 1ms-1, vc fd;k x;k dk;Z = xfrt ÅtkZ esa o`f)

0m2

1mu

2

1m

2

1 222

22 )1(62

1m

2

1 = 3J

vr% lgh fodYi (d). gS A




2. In the figure shown, a small ball of mass 'm' can move without sliding in a fixed semicircular track of radius R in vertical plane. It is released from the top. The resultant force on the ball at the lowest point of the track is

fp=k esa ,d 'm' nzO;eku dh ,d NksVh xsan fp=kkuqlkj R f=kT;k ds ,d fLFkj v/kZo`Ùkkdkj iFk ij fcuk fQlys Å/okZ/kj ry esa xfr dj ldrh gSA bldks 'kh"kZ ls eqDr fd;k tkrk gSA xsan ij vius iFk ds lcls fuEure fcUnq ij ifj.kkeh cy gS &

(A*) 7

mg10 (B)

7

mg17 (C)

7

mg3 (D) zero 'kwU;

Sol. From conservation of energy, the kinetic energy of ball at lowest portion is (vc = speed of centre of ball)

ÅtkZ laj{k.k ls] lcls fuEure fLFkfr ij xsan dh xfrt ÅtkZ (vc = xsUn ds dsUnz dh pky)

2cmv

2

1 + 2

cmv5

2

2

1 = mgR

or 2cmv

10

7 = mgR

Since net tangential force on sphere at lowest point is zero, net force on sphere at lowest position is

pwafd ifj.kkeh Li'kZjs[kh; cy xksys ij lcls fuEure fcUnq ij 'kwU; gS vr% xksys ij lcls fuEure fcUnq ij ifj.kkeh cy

mg7

10

R

mv2c = upwards. Åij dh rjQA

3. An elastic rod will change its length, if (A*) the rod is suspended at one end (B) The rod is allowed to fall freely under gravity (C*) the rod is rotated about one end on a frictionless horizontal table (D*) the rod is given a horizontal acceleration by a force applied at one end

,d izR;kLFk NM+ dh yEckbZ esa ifjorZu gksxk ;fn

(A*) NM+ ,d fljs ls vkyfEcr gSA (B) NM+ dks xq:Ro ds v/khu eqDr :i ls fxjus fn;k tkrk gS (C*) NM+ ,d fljs ds lkis{k ?k"kZ.kghu {kSfrt est ij ?kw.kZu djrh gSA (D*) NM+ dks ,d fljs ij cy yxkdj {kSfrt Roj.k fn;k tkrk gS 4. Figure shows a solid resistance wire of uniform material but non–uniform cross–section area. A,B and

C are the three sections. Current flowing is i, electric field at a section is E, potential is V and flux of

electric field cut by a cross–section is then,

fp=k esa vleku vuqizLFk dkV {ks=kQy dk rFkk ,d le:i inkFkZ dk Bksl izfrjks/k rkj n'kkZ;k x;k gSA A,B rFkk C

rhu Hkkx gSA izokfgr /kkjk i, gSA Hkkx esa fo|qr {ks=k E gS] foHko V gS rFkk vuqizLFk dkV ls xqtjus okyk fo|qr {ks=k dk QyDl gS rks

(A) iB > iA > iC (B*) EB > EA > EC (C*) A = B = C (D) VB > VA > VC




Sol. (A) izR;sd Hkkx ds fy, /kkjk leku gksaxhA Current will be same for each section (B) E a Vd, so EB > EA > EC

J E 1

EA

(C) = J

E.A . A

vr% So, = i

as 'i' is same for each section, so is also same

pawdh 'i' izR;sd Hkkx ds fy, leku gSA vr% Hkh leku gkssxkA (D) current flows from high potential to low potential so VA > VB > VC .

(D) /kkjk mPp foHko ls fuEu foHko dh vksj izokfgr gkszxhA vr% VA > VB > VC .

5. Two identical balls each of mass m moving on straight track approaching towards each other with same speed. Final kinetic energy of the two ball system is equal to the total energy loss during collision. E is the total kinetic energy of the balls before collision and E' is after collision and coefficient of restitution is e. Then choose the correct options

m nzO;eku dh nks ,d leku xsans lh/ks iFk ij leku pky ls ,d&nwljs dh vksj xfr dj jgh gSA nksuksa xsanksa ds fudk; dh vfUre xfrt ÅtkZ VDdj ds nkSjku dqy ÅtkZ gkfu ds cjkcj gksrh gSA VDdj ls igys xsankas dh dqy xfrt ÅtkZ E gS rFkk VDdj ds i'pkr~ xsanksa dh dqy xfrt ÅtkZ E' gS o izR;koLFkku xq.kkad e gS rks lgh fodYiksa dk p;u dhft,A

(A*) E

2E'

(B) e = 1

2 (C*) e =

1

2 (D)

E2

E'

Sol. mm u u

E = mu2

E = 21 m(2u)

2 2

(1 – e2) = mu2(1 – e2)

E' = E – E = mu2(1 – 1+ e2) = mu2e2

since E' = E mu2e2 = mu2(1 – e2)

e2 = 1 – e2 e2 = 1

2

1e

2 and

2

E 12

E' e

6. Select the correct alternative/alternatives : lgh fodYi@fodYiksa dks pqfu;sa

(A*) Current in circuit = 2 sin [100 (sec–1) t – 37°] Ampere.

ifjiFk esa /kkjk = 2 sin [100 (sec–1) t – 37°] ,fEi;j gSA (B*) Average power consumption in the circuit is 8 Watt.

ifjiFk esa vkSlr 'kfDr dk miHkksx 8 okWV gSA

(C*) Power factor of circuit is 4

5.

ifjiFk dk 'kfDr xq.kkad 4

5 gSA

(D*) Resonance frequency of circuit is less than 50 Hz.

ifjiFk dh vuquknh vkofr 50 Hz ls de gSA




Sol. |z| = 5, = 37°

= 0V

z sin (100 t – )

= 10

5sin (100 t – 37°)

Pav = 2( 2) .4 = 8 watt

Power factor 'kfDr xq.kkad = cos 37° = 4

5

Since (pawfd) XL > XC So (vr%) 50Hz > fr

COMPREHENSION

A uniform and constant magnetic field ˆ ˆ ˆB (20 i 30 j 50k)

Tesla exists in space. A charged particle

with charge to mass ratio 3q 10

m 19

C/kg enters this region at time t = 0 with a velocity

ˆ ˆ ˆV (20 i 50 j 30k)

m/s. Assume that the charged particle always remains in space having the given

magnetic field. (Use 2 = 1.4)

,d ,dleku o fu;r pqEcdh; {ks=k ˆ ˆ ˆB (20 i 30 j 50k)

Vslyk vkdk'k (space) esa fo|eku gSA ,d vkosf'kr

d.k ftldk vkos'k o nzO;eku dk vuqikr 3q 10

m 19

C/kg gS og bl {ks=k esa osx ˆ ˆ ˆV (20 i 50 j 30k)

m/s ds

lkFk t = 0 ij izos'k djrk gSA ;g ekfu, fd vkosf'kr d.k ges'kk bl fn;s x;s pqEcdh; {ks=k esa jgrk gSA

( 2 = 1.4 dk iz;ksx djsaA)

7. During the further motion of the particle in the magnetic field B

,

pqEcdh; {ks=k B esa d.k dh vkxs dh xfr ds nkSjku &

(A*) the angle between the magnetic field B

and velocity of the particle remains constant

(B) the angle between the magnetic field B

and velocity of the particle increases (C*) Radius of curvature remain constant (D*) the angle between the acceleration and velocity of the particle does not depend on mass of the

particle.

(A*) pqEcdh; {ks=k B

rFkk d.k ds osx ds chp dk dks.k fu;r jgrk gS

(B) pqEcdh; {ks=k B

rFkk d.k ds osx ds chp dk dks.k c<+rk gS (C*) oØrk f=kT;k fu;r jgrh gS (D*) d.k ds Roj.k rFkk osx ds e/; dk dks.k d.k ds æO;eku ij fuHkZj ugha djrk gSA Sol. The component of velocity of charged particle along the magnetic field does not change. The

component of velocity of charged particle normal to magnetic field only changes in direction but always remains normal to magnetic field. Hence angle between velocity and magnetic field remains same.

pqEcdh; {ks=k ds vuqfn'k osx dk ?kVd ifjofrZr ugh gksrk gSA pqEcdh; {ks=k ds yEcor~ osx dk ?kVd dsoy fn'kk esa ifjofrZr gksrk gS ysfdu ges'kk {ks=k ds yEcor~ jgrk gSA blfy;s osx rFkk pqEcdh; {ks=k ds chp dk dks.k leku jgrk gSA

8. The frequency of the revolution of the particle in cycles per second will be

d.k ds ifjØe.k dh vkofÙk pØ izfr lsd.M (cycle per second) esa gksxh &

(A) 310

19 (B*)

410

38 (C)

410

19 (D)

410

2 19

Sol. f = qB

2 m =

310 3800

19 2

=

410

38




9. The pitch of the helical path of the motion of the particle will be

d.k dh dq.Mfyuhdkj (helical) xfr ds iFk dk pwM+h vUrjky ¼fip½ gksxk &

(A) 100

m (B)

125

m (C)

215

m (D*)

250

m

Sol. Pitch = T.V||

pwM+h vUrjky = T.V||

= 4

1 V.B 38.400.

f | B | 10 . 3800

= 3

4

10

m =

250

m

10. A circular loop of conducting wire, carrying a current I in anticlockwise sense, lies in x-y plane as shown

in figure. In each statement of column-I, a external constant magnetic field is given, which exist in space containing the loop. The term Bo given in column-I is positive constant. With the given magnetic field in each situation of column-I, match the results given in column-II.

pkyd rkj dk ,d o`Ùkkdkj ywi ftlesa /kkjk I okekorZ fn'kk esa izokfgr gS] fp=kkuqlkj x-y ry esa j[kk gSA LrEHk-I ds izR;sd dFku esa ,d ckº; fu;r pqEcdh; {ks=k fn;k gS] tks ywi ds {ks=k esa fo|eku gSA LrEHk-I esa fn;k in Bo /kukRed fu;rkad gSA LrEHk-I esa fn;s izR;sd pqEcdh; {ks=k dks LrEHk-II esa fn;s ifjek.k ls lqesfyr dhft;sA

Column-I Column-II

(A) iBB 0

(p) Resultant magnetic force due to external magnetic

field on the loop is zero

(B) jBB 0

(q) Magnitude of torque due to external magnetic field

on loop is non zero

(C) kBB 0

(r) Magnetic potential energy of current carrying loop

in presence of external magnetic field is nonnegative.

(D) )ji(BB 0

(s) Magnetic force on the loop has tendency to expand

the loop

LrEHk-I LrEHk-II

(A) iBB 0

(p) ckº; pqEcdh; {ks=k ds dkj.k ywi ij ifj.kkeh pqEcdh;

cy 'kwU; gS

(B) jBB 0

(q) ckº; pqEcdh; {ks=k ds dkj.k ywi ij cyk?kw.kZ dk ifjek.k v'kwU; gSA

(C) kBB 0

(r) ckº; pqEcdh; {ks=k dh mifLFkfr esa /kkjkokgh ywi dh pqEcdh;

fLFkfrt ÅtkZ v_.kkRed (non negative) gSA

(D) )ji(BB 0

(s) ywi ij pqEcdh; cy dh izo`fÙk ywi dks QSykus (expand) dh gSA Ans. (A) p,q,r (B) p,q,r, (C) p,s (D) p,q,r




Sol. (A), (B) and (D): In all these three cases, uniform and constant magnetic field is in plane of the loop, Hence angle between magnetic field and dipole moment of loop is 90o. Therefore magnetic potential energy is zero, that is, nonnegative. Torque is maximum in magnitude. Force on a current carrying loop in uniform magnetic field is always zero. The magnetic field only causes the the loop to rotate, as it acts at each point on the loop in direction perpendicular to plane of the loop. Hence magnetic force on the loop has no tendency to expand the loop.

(C) In this option, dipole moment of loop and external magnetic field are in same direction, hence magnetic potential energy is negative and net torque on loop is zero. Magnetic field exert zero net zero force on loop in such a way that, on each point on loop magnetic force is radially outwards leading to tendency of expansion of loop.

(A), (B) rFkk (D): lHkh rhuks fLFkfr;ksa esa ywi ds ry esa le:i rFkk fu;r pqEcdh; {ks=k gSA vr% pqEcdh; {ks=k rFkk ywi ds f}/kzqo vk?kw.kZ ds chp dks.k 90o gSA vr% pqEcdh; fLFkfrt ÅtkZ 'kwU; gSA tks fd v_.kkRed gSA cyk?kw.kZ ifj.kke esa vf/kdre gSA le:i pqEcdh; {ks=k ywi esa dsoy ?kw.kZu iSnk djrk gSA tks fd ywi ds izR;sd fcUnq ij ywi ds ry ds yEcor~ fn'kk esa gSA vr% ywi ij dk;Zjr pqEcdh; cy ywi dks QSykus dh ps"Vk ugh djrkA

(C) bl fodYi esa] ywi dk f}/kzqo vk?kw.kZ rFkk ckº; pqEcdh; {ks=k leku fn'kk esa gSA vr% pqEcdh; fLFkfrt ÅtkZ _.kkRed gksxh rFkk ywi ij dqy cyk?kw.kZ 'kwU; gksxkA pqEcdh; {ks=k ywi dqy v'kwU; cy bl izdkj vkjksfir djrk gS fd ywi dqy v'kwU; cy bl izdkj vkjksfir djrk gS fd ywi ds izR;sd fcUnq ij cy f=kT;h; ckgj dh rjQ nks tks ywi dks QSykus dh ps"Vk djrk gSA

NCERT Questions: 11.20 to 11.37





E ST INFORMA TIO


PPHHYYSSIICCSS


NO. B29 TO B31

2. DPP Syllabus :




1. (D) 2. (B) 3. (A) 4. (A) 5. (D) 6. (B) 7. (C) 8. (C) 9. (B) 10. (B) 11. (A) 12. (A) 13. (A) 14. (D) 15. (C) 16. (C) 17. (A) 18. (C) 19. (D) 20. (B)

1. A rectangular loop of sides of length and b is placed in xy plane. A uniform but time varying magnetic

field of strength B

= 20 t i + 10 t2 j + 50 k where t is time elapsed. The magnitude of induced e.m.f. at

time t is:

,d vk;rkdkj ywi ftldh Hkqtkvksa dh yEckbZ rFkk b gS] dks xy ry es j[krs gSA ,d le:i ijUrq le; ds lkFk

ifjorhZ pqEcdh; {ks=k B

= 20 t i + 10 t2 j + 50k vkjksfir djrs gSA tgka t le; gSA le; t ij çsfjr fo|qr okgd

cy gS&

(A) 20 + 20 t (B) 20 (C) 20 t (D*) zero 'kwU;

Sol. The area vector of loop A

= ± b k

& B

= 20t i + 10 t2 j + 50 k

Magnetic flux is = B

. A

= ± 50 lb

emf = dt

d= 0

2. The dimensional formula of magnetic flux density is :

pqEcdh; ¶yLd ?kUkRo dk foeh; lw=k gS %

(A) M1 L2 T1 1 (B*) M1 T 2 1 (C) M1 L2 T 2 1 (D) M1 L1 T 2 1

3. A circular coil of radius R and a current , which can rotate about a fixed axis passing through its diameter is initially placed such that its plane lies along magnetic field B. Kinetic energy of loop when it

rotates through an angle 90° is : (Assume that remains constant) ,d o`Ùkkdkj dq.Myh dh f=kT;k R rFkk blesa /kkjk gS, ;g blds O;kl ls xqtjus okys fLFkj v{k ds lkis{k ?kw.kZu dj

ldrh gS] izkjEHk es bl izdkj j[kh gqbZ gS fd bldk ry] pqEcdh; {ks=k B ds vuqfn'k gSA dq.Myh ;k ywi dh xfrt mtkZ D;k gksxh tc ;g 90° dks.k ls ?kwe tk,A : (I dks fu;r ekurs gq,)

(A*) R2B (B) 2

BR2 (C) 2R2B (D) 2R

2

3

Sol Loss in potential energy = gain kinetic energy (–MB cos900) – (–MBcos00) = KE




4. A semi–circular current carrying wire having radius R is placed in x–y plane with its centre at origin O.

There is a position x dependent non–uniform magnetic field kR2

xBB 0

(here B0 is positive constant)

existing in the region. The force due to magnetic field acting on the semi–circular wire will be along :

,d v)Zo`Ùkkdkj /kkjkokgh rkj] ftldh f=kT;k R gS] x–y ry esa fLFkr gSA v/kZo`Ùk dk dsUnz ewy fcUnq O ij fLFkr gSA

;gk¡ fLFkfr x ij fuHkZj vle:i pqEcdh; {ks=k kR2

xBB 0

(;gk¡ B0 /kukRed fu;rkad gS) {ks=k esa fo|eku gSA pqEcdh;

{ks=k }kjk v/kZo`Ùkkdkj rkj ij cy dh fn'kk gksxhA

(A*) negative x-axis (B) positive x-axis (C) negative y-axis (D) positive y-axis

(A*) _.kkRed x-v{k (B) /kukRed x-v{k (C) _.kkRed y-v{k (D) /kukRed y-v{k Sol. The direction of forces on the two elements taken symmetrical on two sides of the y–axis are shown.

Clearly the net force will be on negative x-axis.

y–v{k ds nksauks vksj lefer vo;oksa ij cyksa dh fn'kk n'kkZbZ xbZ gSA Li"Vr% dqy cy _.kkRed x-v{k dh vksj gksxkA

5. A spring of spring constant ‘K’ is fixed at one end has a small block of mass m and charge q is attached at the other end. The block rests over a smooth horizontal surface. A uniform and constant magnetic

field B exists normal to the plane of paper as shown in figure. An electric field E

= E0 i (E0 is a positive constant) is switched on at t = 0 sec. The block moves on horizontal surface without ever lifting off the surface. Then the normal reaction acting on the block is :

‘K’ fLizax fu;rkad okyh ,d fLizax dk ,d fljk fLFkj gS vkSj nwljs fljs ij m nzO;eku vkSj q vkos'k dk ,d CykWd tqM+k gqvk gSA CykWd {kSfrt lrg ij fojkekoLFkk esa j[kk gqvk gSA dkxt ds ry ds yEcor~ ,d le:i rFkk fu;r

pqEcdh; {ks=k B fp=kkuqlkj yxk;k x;k gSA t = 0 ij ,d fo|qr {ks=k E

= E0 i (;gk¡ E0 ,d /kukRed fu;rkad gS) 'kq: dj fn;k tkrk gSA CykWd {kSfrt lrg ij fcuk lEidZ [kks, xfr dj jgk gSA CykWd ij yxus okyk vfHkyEc izfrfØ;k cy gksxk :

(A) Maximum at extreme position and minimum at mean position. (B) Maximum at mean position and minimum at extreme position. (C) Is uniform throughout the motion. (D*) Is both maximum and minimum at mean position.

(A) pje fLFkfr;ksa ij vf/kdre vkSj ek/; fLFkfr ij U;wure (B) ek/; fLFkfr ija vf/kdre vkSj pje fLFkfr ij U;wure (C) lEiw.kZ xfr ij fu;r (D*) ek/; fLFkfr ij gh vf/kdre vkSj U;wure




Sol. The velocity is maximum at mean position. Hence the magnetic force on block is maximum, at its mean position. The magnetic force on the block while it crosses the mean position towards right and left is as shown

ek/; fLFkfr esa osx vf/kdre gksrk gSA vr% ek/; fLFkfr ij block ij yxus okyk pqEcdh; cy Hkh vf/kdre gksxkA

tc block nk;ha ;k ck;ha fn'kk esa xfr djrk gqvk] ek/; fLFkfr dks ikj djrk gS] rc bl ij yxus oky pqEcdh; cy fp=k esa fn[kk;k x;k gSA

Vmax

qV Bmax

Vmax

N1

mg+qV Bmax

Case - 1 Case - 2

mg

N2

Hence normal reaction is maximum in case 1 & minimum in case 2.

vr% case : 1 esa vfHkyEc izfrfØ;k cy vf/kdre vkSj case 2 esa U;wure gSA

6. An air capacitor is completely charged upto the energy U and removed from battery.Now distance

between plates is increased slowly by an external agent. If work done by external agent is 3U then ratio of final sepration between the plates to the intial sepration :

,d ok;q la/kkfj=k dks cSVjh dh lgk;rk ls U ÅtkZ rd vkosf'kr fd;k x;k gSA blds ckn bldks cSVjh ls gVk fy;k x;k gSA vc cká; dkjd }kjk IysVksa ds e/; nwjh esa o`f) vR;Ur /khjs&/khjs dh tkrh gSA ;fn cká dkjd 3U dk;Z djrk gS rks la/kkfj=k dh IysVksa ds e/;] vfUre nwjh rFkk çkjfEHkd nwjh dk vuqikr gksxk:

(A) 5 (B*) 4 (C) 3 (D) 1.5

Sol. Initially, çkjEHk esa U = C2

q2

= 0

2

KC2

q ............(i)

When di–electric is pulled out, then U + 3U = 0

2

C2

q

tc ijkoS|qr [khapk tkrk gS rc U + 3U = 0

2

C2

q

4U = 0

2

C2

q ............(ii)

Then, rc (ii) & (i) K = 4.

7. A capacitor of capacitance C carrying charge Q is connected to a source of emf E. Finally the charge

on capacitor would be: C /kkfjrk dk ,d la/kkfj=k ftl ij vkos'k Q gS] dks E fo-ok-cy ds lzksr ls tksM+k tkrk gSA vUr es la/kkfj=k ij

vkos'k gksxk% (A) Q (B) Q + CE (C*) CE (D) none 8. The galvanometer shown in the figure reads 3A, while the ideal voltmeter reads 24volt.The value of

R = 7 The galvanometer resistance is:

fp=k esa n'kkZ;k xsYoksehVj 3A ikB~;kad ns jgk gS] tcfd vkn'kZ oksYVehVj 24 oksYV ikB~;kad iznf'kZr dj jgk gSA R = 7 gS] rks xsYouksehVj dk izfrjks/k gksxk :

(A) 10 (B) 5 (C*) 1 (D) 20Sol. V = i(G+R) or 24 = 3(G+7) or G = 1




9. In the potentiometer experiment the balancing with a cell is at length 240 cm. On shunting the cell with

a resistance of 2 , the balancing length becomes 120 cm. The internal resistance of the cell is

,d foHkoekih ds iz;ksx esa lSy ds lkFk larqyu voLFkk 240 cm yEckbZ ij izkIr gksrh gSA lSy dks ;fn 2 izfrjks/k ds lkFk 'kaV ¼lekUrj Øe esa yxkuk) fd;k tk, rks larqyu yEckbZ 120 cm gks tkrh gSA rks lSy dk vkUrfjd izfrjks/k gksxkA

(A) 4 (B*) 2 (C) 1 (D) 0.5 10. Figure shows a hemisphere of charge Q and radius R and a sphere of charge 2Q and radius R. The

total potential energy of hemisphere is UH and that of the sphere is US. Then.

fp=k esa R f=kT;k rFkk Q vkos'k dk ,d v)Zxksyk ,oa R f=kT;k rFkk 2Q vkos'k dk ,d xksyk n'kkZ;k x;k gSA v)Zxksys dh dqy fLFkfrt ÅtkZ UH gS] rFkk xksys dh US gSA rc

(A) 2UH = US (B*) 2UH < US

(C) 2UH > US (D) UH = US

Sol. Us = 2Uh + interection energy vUr% ÅtkZ

11. A block of silver of mass 4 kg hanging from a string is immersed in a liquid of relative density 0.72. If relative density of silver is 10, then tension in the string will be:[ take g = 10 m/s2 ]

4 kg nzO;eku dk pk¡nh dk ,d xqVdk ,d Mksjh ls yVdk gS] ,oa vkisf{kd ?kuRo 0.72 ds nzo esa Mwck gSA ;fn pk¡nh dk vkisf{kd ?kuRo 10 gS, rks Mksjh esa ruko gksxk & :[ take g = 10 m/s2 ]

(A*) 37.12 N (B) 42 N (C) 73 N (D) 21 N

Sol. Let S, L be the density of silver and liquid. Also m and V be the mass and volume of silver block.

Tension in string = mg – bouyant force

(A) ekuk S, L pk¡nh o nzo ds ?kuRo gS rFkk m o V pk¡nh ds xqVds ds nzO;eku o vk;ru gSA Mksjh es ruko = mg – mRIykod cy

T = SVg – L Vg = (S – L) Vg

Also o V = s

m

T =

s

sSmg =

3

3

1010

10)72.010(

× 4 × 10 = 37.12 N.

12. A small solid ball is dropped from a height above the free surface of a liquid. It strikes the surface of

the liquid at t = 0. The density of the material of the ball is 500 kg/m3 and that of liquid is 1000 kg/m3. If the ball comes momentarily at rest at t = 2 sec then initial height of the ball from surface of liquid was (neglect viscosity) :

,d NksVh Bksl xsan dks dqN Å¡pkbZ ls nzo dh eqDr lrg ij fxjk;k tkrk gSA ;g nzo dh lrg ij t = 0 le; ij Vdjkrh gSA xasn ds inkFkZ dk ?kuRo 500kg / m3 rFkk nzo dk ?kuRo 1000 kg/m3 gSA vxj t = 2 sec. ij xsan {kf.kd :i ls :d tkrh gks rks xsan dh nzo dh lrg ls izkjfEHkd Å¡pkbZ gksxh ¼';kUkrk dks ux.; ekus½ &

(A*) 20 m (B) 10 m (C) 15 m (D) 25 m




Sol.

Velocity of ball when it reaches to surface of liquid

a = V500

gV500gV1000 ; where V is the volume of the ball.

a = 10 m/sec2

apply v = u + at 0 = gh2 – 10t

gh2 = 10 × (2)

2 × 10 × h = 400 h = 20 m Ans 13. The displacement x of a particle as a function of time t is shown in following figure. The figure indicates: fdlh d.k dk le; t ds lkFk foLFkkiu x fp=k esa iznf'kZr gSA xzkQ iznf'kZr djrk gS fd

(A*) the particle starts with a certain velocity, but the motion is retarded and finally particle stops (B) the velocity of particle is constant throughout (C) the acceleration of the particle is constant throughout (D) the particle starts with a constant velocity, the motion is accelerated. (A*) d.k fuf'pr izkjfEHkd osx ls xfr djrk gqvk eafnr gksdj :d tkrk gSA (B) d.k dk osx ges'kk fu;r jgrk gSA (C) d.k dk Roj.k ges'kk fu;r jgrk gSA (D) d.k fu;r osx pyuk 'kq: djuk gS] ,oa xfr Rofjr gSA Sol. From the graph ; we observe that slope is non–zero positive at t = 0 & slope is continuously decreasing

with time and finally becomes zero. Hence we can say that the particle starts with a certain velocity, but the motion is retarded (decreasing velocity)

14. A hollow cubical box P is moving on a smooth horizontal surface in the x–y plane with constant

acceleration of j4i3a

m/s2 . A block Q of mass 2kg is at rest inside the cubical box as shown in

figure. lf the coefficient of fricition between the surface of the cube P and the block Q is 0.6. Then the force of fricition between P and Q is :

,d [kks[kyk ?kukdkj ckWDl P, fu;r Roj.k j4i3a

m/s2 ls x–y ry esa ?k"kZ.kjfgr {kSfrt lrg ij xfr dj jgk

gSA 2kg nzO;eku dk ,d CykWd Q ?kukdkj ckWDl ds vUnj fp=kkuqlkj fojke ij j[kk gqvk gSA ;fn ?ku P rFkk CykWd Q dh lrg ds e/; ?k"kZ.k xq.kkad 0.6 gS rks P rFkk Q ds e/; ?k"kZ.k cy gksxk :

Z

P

Y

X

Q

(A) 5 N (B) 8 N (C) 12 N (D*) 10 N




Sol. (D)

22 43a =5 m/s

fricition force ?k"kZ.k cy f = m a

= 2 × 5 = 10 N

15. Three waves of amplitude 10m, 4m and 7 m arrive at a point with successive phase difference of

/ 2. The amplitude of the resultant wave is

10m, 4m rFkk 7 m vk;ke dh rhu rjaxsa fdlh fcUnq ij ,dkarj / 2 dykarj ij vkifrr gksrh gSA ifj.kkeh rjax dk vk;ke gksxk %

(A) 2m (B) 7m (C*) 5m (D) 1

16. In the resonance tube experiment, first resonant length is 1 and the second resonant length is 2 , then the third resonant length will be ?

vuqukn uyh iz;ksx esa izFke vuquknh yEckbZ 1 rFkk f}rh; vuquknh yEckbZ 2 gks, rks rrh; vuquknh yEckbZ gksxh ?

(A) 5 (B) 2(2 – 1) (C*) 22 – 1 (D) 32 – 21

Sol. for first resonance

izFke vuqukn ds fy,

1 + = 0f4

v

for second resonance

f}rh; vuqukn ds fy,

2 + = 0f4

v3

for the third resonance

r`rh; vuqukn ds fy,

3 + = 0f4

v5

Solving get 3 = 22 – 1 17. Magnetisation (M) of a material depends on magnetising field (H) according to following graph :

fdlh inkFkZ dk pqEcdRo (M), pqEcdu {ks=k (H) ij fp=k esa iznf'kZr xzkQ ds vuqlkj ifjofrZr gksrk gS

H

M

If same variation is studied at higher temperature then slope of the above graph will:

;fn mPp rki ij izsf{kr ifjorZu leku gks rks xzkQ dk <+ky % (A*) Remain same (B) increases (C) Decreases (D) may increase or decrease

(A*) leku jgsxk (B) c<+sxk

(C) ?kVsxk (D) ?kV ;k c<+ ldrk gS Sol. Susceptibility of diamagnetic substance does not depend on temperature so slope of above graph will

remain unchanged




18. Magnetisation (M) of a material depends on magnetising field (H) according to following graph :

fdlh inkFkZ dk pqEcdRo (M), pqEcdu {ks=k (H) ij fp=k esa iznf'kZr xzkQ ds vuqlkj ifjofrZr gksrk gS

H

M

If same variation is studied at higher temperature then slope of the above graph will:

;fn mPp rki ij izsf{kr ifjorZu leku gks rks xzkQ dk <+ky % (A) Remain same (B) increases (C*) Decreases (D) may increase or decrease

(A) leku jgsxk (B) c<+sxk

(C*) ?kVsxk (D) ?kV ;k c<+ ldrk gS Sol. Susceptibility of paramagnetic substance inversely proportional to temperature so slope of above graph

will decrease. 19. Two magnetic bars are suspended in horizontal plane and oscillates in earth magnetic field. Their time

period of oscillations are T1 and T2. The ratio of their magnetic dipole moments is :

{kSfrt ry esa yVdh gqbZ nks NM+ pqEcd iFoh ds pqEcdh; {ks=k esa nksyu dky T1 rFk T2 ls nksyu dj jgh gSA buds pqEcdh; f}/kzqo vk?kw.kks± dk vuqikr gksxk:

(A) 2

1

T

T (B)

1

2

T

T (C)

2

2

1

T

T

(D*)

2

1

2

T

T

20. A body initially at rest starts moving in x–direction. Its acceleration a is plotted against x as shown in

figure. Find the maximum velocity (in m/s) of the body ?

,d oLrq izkjEHk esa fojke ij gS] x–fn'kk esa pyuk izkjEHk djrh gSA blds Roj.k a dk x ds lkFk xzkQ fp=k esa iznf'kZr gSA oLrq dk vf/kdre osx (m/s eas) Kkr djksA

(A) 2 (B*) 4 (C) 6 (D) 8

Sol. Area = dx a = dxdx

vdx

=

maxv

0

dv v = 2

v max2

= 2

1× (1 + 3) × 4 vmax = 4 m/s





E ST INFORMA TIO


PPHHYYSSIICCSS


NO. B29 TO B31


Single choice Objective ('–1' negative marking) Q.1 (3 marks 2 min.) [03, 02] One or more than one options correct type ('–1' negative marking) Q.2 to Q.8 (4 marks 2 min.) [28, 14] Subjective Questions ('–1' negative marking) Q.9 (4 marks 5 min.) [04, 05] Match the Following (no negative marking) Q.10 (8 marks 10 min.) [08, 06]


1. (C) 2. (A,B,C,,D) 3. (B,D) 4. (A,D) 5. (A,C,D) 6. (A,B,D)

7. (A,C,D) 8. (B*) 9. g

2

10. A(p, r), B(p, q, s, t), C(p, r), D(p)

1. A particle P collides with 15 ms–1 at an angle 60° from vertical on a smooth horizontal surface. The

coefficient of restitution (e) is 3

1 . The range R, as shown in diagram will be : [Take g = 10 m/s2]

,d d.k P {kSfrt fpduh lrg ls Å/okZ/kj ls 60° dks.k ij 15 m/sec osx ls Vdjkrk gSA izR;koLFkku xq.kkad (e)

3

1 gks rks fp=k esa iznf'kZr ijkl R dk eku gksxk \ [g = 10 m/s2]

(A) 12.5 m (B) 12 m (C*) 11.25 m (D) 12.75 m

Sol.(C)]

x

Given fn;k gS 3

1e

tan e

1tan 3tan sin =

10

3

Along horizontal, m 15 sin 60° = m v sin {kSfrt esa, m 15 sin 60° = m v sin




32

10515v

=13.7 ms–1

x=

g

2sinv

g

22sinv

22

=10

10

32)7.13( 2

= 11.25 m

2. A proton and an electron are moving with the same de–Broglie wavelength (consider the non–

relativistic case). Then : (A*) in a uniform magnetic field both the particles may describe circles of same radius (B*) both the particles may have the same momentum (C*) the speed of the proton and the electron are in the ratio me / mp, where me is the electron mass and

mp, the proton mass (D*) the product of mass and kinetic energy is the same for both particles

,d çksVkWu rFkk ,d bysDVªkWu leku Mh&czksxyh rjaxnS/;Z ls xfr dj jgs gS (lkis{kokn ds fl)kUr ij fopkj u dhft,) rks: (A*) ,dleku pqEcdh; {ks=k esa nksauks d.k leku f=kT;k dk o`Ùk fu:fir dj ldrs gSaA

(B*) nksauks d.kksa dk laosx leku gks ldrk gSA (C*) çksVkWu rFkk bysDVªkWu dh pky dk vuqikr me / mp ds vuqikr esa gS tgk¡ me bysDVªkWu dk nzO;eku gS rFkk mp

çksVkWu dk nzO;eku gSA (D*) nzO;eku rFkk xfrt ÅtkZ dk xq.kuQy nksauks d.kksa ds fy, leku gSA

Sol. d = h

p p is the same leku gS

p = mv p e

e p

v m

v m

p = 2mK mK = 2p

2

Ans. (A), (B), (C) AND (D) 3. Which of the following statements is/are correct for an x-ray tube ?

x-fdj.k ufydk ds fy, fuEu es ls dkSulk@dkSuls dFku lR; gS ?

(A) On increasing potential difference between filament and target, photon flux of x-rays decreases (B*) On increasing potential difference between filament and target, cut-off frequency of x-rays

increases (C) On increasing filament current, cut–off wavelength increases (D*) On increasing filament current intensity of x-rays increases

(A) ;fn y{; rFkk fQykes.V ds e/; foHkokUrj c<+krs gS rks x-fdj.k QksVksu dk ¶yDl ?kV tkrk gSA

(B*) ;fn y{; rFkk fQykes.V ds e/; foHkokUrj c<+krs gS rks x-fdj.k dh nsgyh (cut-off) vkofÙk c<+ tkrh gSA (C) fQykes.V /kkjk c<+kus ls nsgyh rjaxnS/;Z (cut-off wavelength) c<+ tkrh gSA (D*) fQykes.V /kkjk c<kus ij x-fdj.k rhozrk c<+ tkrh gSA

Sol. eVa=cutoff

hc




4. In the following figures which one corresponds to which type of equilibrium :

fuEu fp=k dkSulh lkE;koLFkk dks n'kkZrk gS &

(A*) figure (i) corresponds to unstable equilibrium (B) figure (ii) corresponds to unstable equilibrium (C) figure (iii) corresponds to stable equilibrium (D*) figure (iv) corresponds to stable equilibrium

(A*) fp=k (i) vLFkk;h lkE;koLFkk ds laxr gSA (B) fp=k (ii) vLFkk;h lkE;koLFkk ds laxr gSA (C) fp=k (iii) LFkk;h lkE;koLFkk ds laxr gSA (D*) fp=k (iv) LFkk;h lkE;koLFkk ds laxr gSA 5. A U tube contains water so that air column in each tube 2.20 m. A tuning fork of frequency 345 Hz

Vibrate above one of the limbs, say limb A. An immiscible liquid C of specific gravity 0.6 is now slowly poured into the other limb B till air column in limb A resonates. Speed of sound in air = 345 ms–1, then at this position, (Neglect the end correction)

(A*) Air column in limb B will not resonate with the same tuning fork (B) Air column in limb B will resonate with the same tuning fork, in 2nd overtone. (C*) The difference between liquid levels in the two limbs is 0.6 m (D*) The height of liquid column C is 1.5 m

,d U uyh ikuh ls bl izdkj Hkjh gS fd bldh izR;sd uyh esa ok;q LrEHk dh yEckbZ 2.20 m gSA ,d 345 Hz

vko`fÙk dk Lofj=k fdlh ,d uyh ¼ekuk A½ ij dfEir fd;k tkrk gSA ,d vfefJr æo C ftldk fof'k"V xq:Ro 0.6 gS uyh ds nwljs fljs B esa /khjs ls Hkjk tkrk gS] tc rd fd uyh A esa ok;q LrEHk nqckjk vuqukn ugha djus yx tk;sA ok;q esa /ofu dh pky 345 ms–1 gks rks bl fLFkfr esa (fljk la'kks/ku dks ux.; ekfu,½

(A*) uyh B esa fLFkr ok;q LrEHk leku Lofj=k ds }kjk vuqukfnr ugha gskxk (B) uyh B esa fLFkr ok;q LrEHk leku Lofj=k ds }kjk f}rh; vf/kLojd esa vuqukfnr gskxk (C*) nksuksa ufy;ksa esa æo Lrj dh Å¡pkbZ;ksa esa varj 0.6 m gSA (D*) æo C ds LrEHk dh Å¡pkbZ 1.5 m gSA

Sol. For 345 Hz, = v/f = 1m, initially column is not resonance, since 9 × /4 = 9 × 0.25 = 2.25 m and 7 × 0.25 = 1.75 m. When resonance takes place.

( odd multiples of /4, i.e., 7 × 0.25)

1.75

0.45

x

0.450.45

220

AABA

x × 0.6 = (0.90) × 1

x = 1.5 m d is correct

Air column in B is 2.20 + 0.45 – 1.50 = 1.15 m

Difference in level is 0.6 C is correct

Will not resonate with 345 Hz, since for 2nd overtone. Air column required is 0.25 × 5 = 1.25 m




6. A conducting rod of length is moved at constant velocity ‘v0’ on two parallel, conducting, smooth, fixed rails, that are placed in a uniform constant magnetic field B perpendicular to the plane of the rails as shown in figure. A resistance R is connected between the two ends of the rail. Then which of the following is/are correct :

,d pkyd NM+ ftldh yEckbZ gS nks lekUrj ?k"kZ.k fofgu pkyd NM+ksa ds chp ‘v0’ vpj osx ls py jgh gSA fp=kkuqlkj tks fd vpj pqEcdh; {ks=k B esa fLFkr gS rFkk pqEcdh; {ks=k NM+ksa ds ry ds vfHkyEcor~ gSA NM+ksa dks ,d izfrjks/k R ls tksM+k x;k gSA rc fuEu esa ls dkSu lgh gS@gSa :

X B

x

R V0

X

(A*) The thermal power dissipated in the resistor is equal to rate of work done by external person

pulling the rod. (B*) If applied external force is doubled than a part of external power increases the velocity of rod.

(C) Lenz’s Law is not satisfied if the rod is accelerated by external force (D*) If resistance R is doubled then power required to maintain the constant velocity v0 becomes

half.

(A*) izfrjks/k esa Å"eh; 'kfDr cká O;fDr }kjk NM+ dks [khapus esa fd;s x;s dk;Z dh nj ds cjkcj gksxhA (B*) vxj cká cy dks nqxuk fd;k tk, rc] cká 'kfDr dk dqN Hkkx NM+ dh pky dks c<+krk gSA

(C) vxj cká cy }kjk NM+ dks Rofjr fd;k tk, rc ySat fu;e oS| ugha gSA (D*) vxj izfrjks/k R dks nqxuk fd;k tk, rc NM+ dks vpj osx v0 ij cuk;s j[kus ds fy;s vko';d 'kfDr vk/kh

gks tk;sxhA Sol. Rate of work done by external agent is :

dt

dw =

dt

dx.LB = BLv & thermal power dissipated in the resistor = e = (BvL)

clearly both are equal, hence (A) . If applied external force is doubled, the rod will experience a net force and hence acceleration. As a

result velocity increases, hence (B)

Since ; = R

e

On doubling ‘R’, current and hence required power becomes half.

Since, P = BLv Hence (D).

COMPREHENSION A horizontal thin tube of length 2L completely filled with a liquid of density rotates about a vertical axis

passing through one of its end and along the diameter of its side face with an constant angular velocity

. Two vertical thin long tubes are fitted with the horizontal tube to measure the pressure difference between points A and B which are at same horizontal level as shown. Neglect viscous forces and surface tension of liquid.

2L yEckbZ dh iryh {kSfrt uyh ?kuRo ds æo ls iw.kZr% Hkjh gqbZ gS rFkk ;g uyh vius ,d fdukjs ls xqtjus okyh Å/okZ/kj v{k ifjr% ?kw.kZu dj jgh gS rFkk fp=kkuqlkj csyu ds fljs ds Å/okZ/kj O;kl ds ifjr% bldk fu;r dks.kh; osx gSA {kSfrt uyh ds nks fcUnqvksa A rFkk B ds e/; nkckUrj ekius ds fy, leku {kSfrt Lrj ij fp=kkuqlkj fcUnq A rFkk B ij nks Å/okZ/kj iryh yEch ufydk,W yxkbZ tkrh gSA ';kucy rFkk nzo dk i`"B ruko ux.; ekusA




7. The correct variation of the pressure gradient

dx

dPat a distance x from the axis of rotation with

various quantities:

nkc izo.krk

dx

dP dk ?kw.kZu v{k ls nwjh x ds lkFk mfpr xzkQ crkvks :

(A*) (B)

(C*)

(D*)

8. The difference in the heights (h) of the liquid column in the two vertical thin tubes is :

nksuksa Å/okZ/kj iryh ufy;ksa ds nzo LrEHk dh Å¡pkbZ esa vUrj D;k gksxk :

(A) g2

L22 (B*)

g2

L3 22 (C)

g2

L5 22 (D) None of these buesa ls dksbZ ughaA

Sol.

dm = Adx

dPA = (Adx)a

dx

dP = 2x

The difference in heights

ÅWpkbZ esa vUrj h = y2 – y1 = g2

L

g2

)L2( 2222

=

g2

L3 22

9. A wire is bend at an angle . A rod of mass m can slide along the bended wire without friction as shown in figure. If a soap film is maintained in the frame and frame is kept in a vertical position and rod is in equilibrium. If rod is displaced slightly in vertical direction. Find the time period of oscillation.

,d rkj dks dks.k ij eksM+k x;k gSA ,d m æO;eku dh NM+ bl eqM+s gq, rkj ds vuqfn'k fp=kkuqlkj fcuk ?k"kZ.k ds fQly ldrh gSA NM+ dks Å/okZ/kj ry esa lkE; fLFkfr esa j[kus ds fy;s bl Ýse ij lkcqu dh fQYe cukbZ tkrh gSA ;fn NM+ gYdh lh Å/okZ/kj fn'kk esa LFkkfir gksrh gS rks nksyuksa dk vkorZdky Kkr djks:

lkcqu dh fQYe NM+ dk æO;eku m

Ans. g

2




Sol.

mg2

tan2T2

,

if x be the displacement in vertical direction of the rod from equilibrium position

Fres = – 2x tan 2T2

, a = xm

2tanT4

,

T = g

2

4

mg4

m2

2tanT4

m2

10. Column I lists the field in a region and Column II lists the path of a charge q and mass m on which the

particle can move. Match the appropriate entries of Column II with entries of Column I. [Consider all fields to be uniform]

Column I Column II

(A) Only electric field E is present (p) The particle can move on a straight line

(B) Only magnetic field B is present (q) The particle can move on a circle

(C) Only gravitational field g is present (r) The particle can move on parabolic path

(D) Both electric field E and magnetic field (s) The particle can remain in rest

B are present

(t) The particle can move in a helical path of constant pitch

LrEHk I esa dqN {kS=k iznf'kZr gS rFkk LrEHk II esa q vkos'k rFkk m æO;eku ds d.k dk iFk iznf'kZr gS ftl ij d.k xfr dj ldrk gSA LrEHk I esa nh xbZ laxr fLFkfr;ksa dks LrEHk II ls lqesfyr dhft,A [lHkh {kS=kksa dks le:i ekfu,]

Column I Column II

(A) dsoy fo|qr {ks=k E mifLFkr gS (p) d.k lh/kh js[kk ij xfr dj ldrk gSA

(B) dsoy pqEcdh; {ks=k B mifLFkr gS (q) d.k o`Ùk ij xfr dj ldrk gSA

(C) dsoy xq:Roh; {ks=k g mifLFkr gS (r) d.k ijoy;kdkj iFk ij xfr dj ldrk gSA

(D) fo|qr {ks=k E rFkk pqEcdh; {ks=k

B (s) d.k fLFkj jg ldrk gSA

nksuksa mifLFkr gSa

(t) d.k dq.Mfyuhdkj iFk ij xfr dj ldrk gSA ftldk pwM+h varjky fu;r gS

Ans. A(p, r), B(p, q, s, t), C(p, r), D(p)

NCERT Questions

12.1 to 12.17




Board Level questions 1. When the electron orbiting in hydrogen atom in its ground state moves to the third excited state, show

how the de Broglie wavelength associated with it would be affected.

gkbMªkstu ijek.kq esa viuh fuEure voLFkk esa ifjØe.k djus okyk bysDVªkWu tc rrh; mÙksftr voLFkk esa xeu djrk gS] rc blls lEc) ns czkWXyh rajxnS/;Z fdl izdkj izHkkfor gksrh gS\

2. The graph shows variation of stopping potential V0 versus frequency of incident radiation v for two photosensitive metals A and B. Which of the two metals has higher threshold frequency and why?

Stopping Potential

(V0) v0

e

W0

Metal B

Frequency of incident Radiation (v)

v0'

Metal A

e

W0

O

xzkQ esa nks izdk'k&lqxzkgh /kkrqvksa A vkSj B ds fy,] vkifrr fofdj.k dh vko`fÙk v ds lkFk fujks/kh foHko V0 dk fopj.k n'kkZ;k x;k gSA bu nksuksa /kkrqvksa esa ls fdl /kkrq dh nsgyh vko`fÙk vf/kd gS vkSj D;ksa\

fujks/kh foHko(V0)

v0

e

W0

/kkrq B

vkifrr fofdj.k dh vkofÙk (v)

v0'

/kkrq A

e

W0

O

Sol.: K max = h

V = ee

h

0 = h

A B (From figure fp=kls)

00 BA

3. Using Rutherford model of the atom, derive the expression for the total energy of the electron in

hydrogen atom. What is the significance of total negative energy possessed by the electron?

ijek.kq ds jnQksMZ ekWMy dk mi;ksx djrs gq, gkbMªkstu ijek.kq esa bysDVªkWu dh dqy ÅtkZ ds fy, O;atd O;qRié dhft,A bysDVªkWu dh dqy ÅtkZ _.kkRed gksus dk D;k vFkZ gS\




Sol : The electrostatic force of attraction Fe between the revolving electrons and the nucleus provide the

requisite centripetal force (Fe) to keep them in their orbits ?k wers g q, bysDVª k Wu rFkk ukfH kd ds e/; dk;Zjr fLFkj oS| qr vkd"k Z .k cy bysDVª k Wu dk s bldh d{kk e s a cuk; s j[ku s

d s fy, vfH kd s Unz h; cy iznku djrk g SA

Fc = Fe

2

2

0

2

r

e

4

1

r

mV

Thus the relation between the orbit radius and electron velocity is vr% d{kh; f= kT;k rFkk bysDVª k Wu osx ds e/; lEcU/k gk sxkA

R = 2

0

2

mv4

e

K.E = 2

1 mv2 =

R

e

0

2

U = R4

e

0

2

The – ve sign in U signifies that the electrostatics Force is in the – r direction U es a _.kkRed fpUg dk vFk Z g S fd fLFkj oS| qr vkd"k Z .k cy -r dh fn'kk e s a g SA

Total energy dqy Åtk Z E = K + U = r4

e

r8

e

0

2

0

2

= r8

e

0

2

Total energy is negative . This implies that electron is bound to the atom. dqy Åtk Z _.kkRed g SAbldk vFk Z g S fd bysDVª k Wu ijek.k q e s c) voLFkk e s a g SA





E ST INFORMA TIO


PPHHYYSSIICCSS


NO. B29 TO B31


Single choice Objective ('–1' negative marking) Q.1 to Q.2 (3 marks 2 min.) [09, 06] One or more than one options correct type ('–1' negative marking) Q.3 to Q.5 (4 marks 2 min.) [12, 06] Subjective Questions ('–1' negative marking) Q.6 (4 marks 5 min.) [04, 05] Match the Following (no negative marking) Q.7 (8 marks 6 min.) [08, 06]

ANSWER KEY OF DPP No. : B31 1. (D) 2. (B) 3. (B,D) 4. (B,D) 5. (B) 6. 5 7. (A)

1. A cylinder of mass m and radius R is spined to a clockwise angular velocity o and then gently placed

on an inclined plane for which coefficient of friction = tan is the angle of inclined plane with

horizontal. The centre of mass of the cylinder will remain stationary for time:

m nzO;eku rFkk R f=kT;k dk ,d csyu tks o dks.kh; osx ls nf{k.kkorZ ?kw.kZu dj jgk gSA bls /khjs ls = tan

?k"kZ.k xq.kkad rFkk {kSfrt ls dks.k okys urry ij j[kk tkrk gSA csyu dk nzO;eku dsUnz fdrus le; rd fojke esa jgsxkA

(A) oR/gsin (B) 2oR/3gsin (C) 2oR/5gsin (D*) oR/2gsin2. A particle of mass 10 gm is placed in a potential field given by V = (50 x2 + 100) J/kg. The frequency of

oscillation in cycle/sec is :

10 xzke nzO;eku ds d.k dks foHko {ks=k V = (50 x2 + 100) twy@fdxzk- es j[kk x;k gSA nksyu dh vko`fÙk pDdj@ls- gS :

(A)

10 (B*)

5

(C)

100 (D)

50

Sol. Potential energy U = mV

U = (50x2 + 100) 10–2

F = dx

dU

= – (100x) 10–2

m2x = – (100 × 10–2 ) x

10 × 10–3 2x = 100 × 10–2 x

2 = 100, = 10

f = 2

=2

10 =

5




3. An electric dipole is placed in an electric field generated by an infinitely long wire having uniform charge density.

(A) The net electric force on the dipole must be zero (B*) The net electric force on the dipole may be zero (C) The torque on the dipole due to the field must be zero (D*) The torque on the dipole due to the field may be zero

le:i vkos'k ?kuRo okys vuUr yEckbZ ds rkj ds }kjk mRiUu fo|qr {ks=k esa ,d fo|qr f}/kzqo fLFkr gS %.

(A) f}/kzqo ij dqy fo|qr cy 'kwU; gksuk pkfg;s

(B*) f}/kzqo ij dqy fo|qr cy 'kwU; gks ldrk gS (C) f}/kzqo ij {ks=k ds dkj.k dqy cy vk?kw.kZ 'kwU; gksuk pkfg;s (D*) f}/kzqo ij {ks=k ds dkj.k dqy cy vk?kw.kZ 'kwU; gks ldrk gSA

Paragraph for Question Nos. 4 to 5

iz'u 4 ls 5 ds fy, vuqPNsn

A uniform magnetic field B0, exists perpendicular to plane of paper in the region ABCD as shown in the figure. A positive charge q of mass m enters in the region at mid-point of line AB perpendicular to line

AB with velocity v0 (which is in the plane of paper)as shown in the figure. AB = BC = CD = 2 fp=kkuqlkj {ks=k ABCD esa le pEcdh; {ks=k B0 i`"B ry ds yEcor~ gSA fp=kkuqlkj m nzO;eku o /kukosf'kr q dk d.k

js[kk AB ds yEcor~ o blds e/; fcUnq ls v0 osx ls izos'k (i`"B ry esa) djrk gSA AB = BC = CD = 2

It is given that mv0 = qB0. Based on the given information answer the following questions :

fn;k x;k gS fd mv0 = qB0.

nh xbZ tkudkfj;ksa ds vk/kkj ij fuEu iz'uksa ds mÙkj nhft, % 4. Select correct statement(s) :

(A) Distance travelled by charged particle in the region ABCD is 4

3

.

(B*) Distance travelled by charged particle in the region ABCD is 2

3

.

(C) Time spent in magnetic field is 0

v

.

(D*) Change in velocity till it comes out of magnetic field is 0

v 3

lgh dFku@dFkuksa dk p;u dhft,A

(A) {ks=k ABCD esa vkosf'kr d.k }kjk r; nwjh 4

3

gSA

(B*) {ks=k ABCD esa vkosf'kr d.k }kjk r; nwjh 2

3

gSA

(C) pqEcdh; {ks=k esa O;rhr le; 0

v

gSA

(D*) pqEcdh; {ks=k ls ckgj vkus rd blds osx esa ifjorZu 0

v 3 gSA




5. If same particle enters with speed n v0 such that it comes out via mid point of CD then select correct statement(s) : (A) n = 2 (B*) n = 3

(C) time that charged particle will spend in region ABCD is 0

v

.

(D) time that charged particle will spend in region ABCD is 0

4

v

.

;fn leku d.k pky nv0 ls pqEcdh; {ks=k esa bl izdkj izos'k djrk gS fd ;g CD ds e/; fcUnq ls ckgj fudyrk gS] rc lgh dFku@dFkuksa dk p;u dhft,A (A) n = 2 (B*) n = 3

(C) {ks=k ABCD esa vkosf'kr d.k }kjk O;frr le; 0

v

gSA

(D) {ks=k ABCD esa vkosf'kr d.k }kjk O;frr le; 0

4

v

gSA

6. Atoms of a hydrogen like gas are in a particular excited energy level. When these atoms de–excite they

emits photons of different energies. Maximum and minimum energies of emitted photons are Emax =

52.224 eV and Emin = 1.224 eV respectively. Calculate the principal quantum number of initially excited

level. (Ionisation energy of hydrogen atom = 13.6 eV)

,d gkbMªkstu rqY; xSl dk ijek.kq] fo'ks"k mÙksftr ÅtkZ voLFkk okys mtkZLrj esa gS ;s fuEu voLFkk esa vkrs gq, vyx&vyx mtkZvksa ds QksVksu mRlftZr djrs gS ;fn QksVksuks dh vf/kdre rFkk U;wure mRlftZr ÅtkZ Øe'k% Emax

= 52.224 eV rFkk Emin = 1.224 eV gS rks izkajHk es mÙksftr ÅtkZ voLFkk ds fy, eq[; DokaVe la[;k dh x.kuk djksA (gkbMªkstu ijek.kq dh vk;uu ÅtkZ = 13.6 eV)

Ans. 5 Sol. Let principal quantum number of initially excited energy level be n. Photon of maximum energy is

emitted during transitions n 1 Hence, Emax = [En – E1] ..........(i)

Since, difference between energy of two consecutive levels decreases when n increases therefore

photon of minimum energy is emitted during transition n (n – 1) Hence, Emin = En – E(n–1) .........(ii)

But energy of nth level is given by

En = 1

2

E

n

From equation (i), 1

12

EE

n = 52.224 eV .........(iii)

and from equation (iii), 1 1

2 2

E E

n (n 1)

= 1.224 eV .........(iv)

Solving equation (iii) and (iv), E1 = – 54.4 eV

and n = 5 Ans. Let atomic number of the gas be z E1 = –13.6 z2

z = 2 Hence, the gas is singly ionised helium Ans.




gy. ekuk eq[;k Dok.Ve la[;k n gS ftlesa izkjEHk esa mRrsftr byDVªkWu mifLFkr gS vf/kre ÅtkZ ds mRltZu ds fy, ;fn bysDVªkWu dks n 1 rd vkuk gksxk

vr% Emax = [En – E1] ..........(i)

pqafd nks fudVe ÅtkZ Lrjks dk ÅtkZ vUrj de gksrk tkrk gS] tc ge n dks c<krs gS rFkk U;wure ÅtkZ dk QksVkWu mRlftZr gks blls bySDVªkWu dks n (n – 1) rd vkuk gksxkA

vr% Emin = En – E(n–1) .........(ii)

pqafd nth Lrj dh ÅtkZ fuEu gksxhA

En = 1

2

E

n

lehdj.k (i) ls, 1

12

EE

n = 52.224 eV .........(iii)

rFkk lehdj.k (iii) ls, 1 1

2 2

E E

n (n 1)

= 1.224 eV .........(iv)

lehdj.k (iii) rFkk (iv) ls gy djus ij E1 = – 54.4 eV

rFkk n = 5 Ans.

ekuk ijek.kq Øekad z gSA E1 = –13.6 z2

z = 2

vr% xSl dsoy vkosf'kr gkbMªkstu vk;u gS Ans.

7. In column-1 a system is shown and a quantity called x is defined for that system. In column-2 value of x

is given. Match the proper entries from column-2 to column-1 using the codes given below the columns.

(P) x = 22 h

g

(1) x = 4

A uniform solid cylinder of height h, floating in an ideal liquid of density .

Area of cross section of cylinder is much leser than area of cross

section of container. is the angular frequency of small vertical oscillations.

(Q) x = 29 h

4 g

(2) x = 2

A uniform solid cylinder of height h, floating in an ideal liquid of density .

Area of cross section of container is four time the area of cross section of cylinder.

is the angular frequency of small vertical oscillations.

(R) x = 2 2R

gh

(3) x = 3




Ideal liquid is filled in a cylindrical container upto (3/4)th height. Now the liquid is being rotated about

vertical axis passing through its axis of symmetry with constant angular velocity , such that liquid is on

the verge of falling out of the container

(S) x =g2

h3 2 (4) x = 1

A U tube of uniform cross section is rotated with respect to one of its limb with constant angular velocity

, such that it is half filled.

(P) (Q) (R) (S)

(A*) 2 3 4 1

(B) 3 2 1 4

(C) 3 1 4 2

(D) 2 4 1 3

LrEHk-1 esa ,d fudk; rFkk ,d jk'kh x n'kkZ;h x;h gS] tks ml fudk; ds fy, ifjHkkf"kr gSA LrEHk&2 esa x ds eku

fn;s x;s gSA LrEHk ds uhps fn;s x;s dksM dk mi;ksx djrs gq, LrEHk&1 dks LrEHk&2 ls feyk;sA

(P) x =22 h

g

(1) x = 4

,d le:i csyu ftldh Å¡pkbZ h gS, ,d ?kuRo ds vkn'kZ nzo esa rSj jgk gSA csyu dk vuqizLFk dkV dk {ks=kQy]

ik=k ds vuqizLFk dkV ds {ks=kQy dh rqyuk esa ux.; gSA Å/okZ/kj NksVs nkSyuksa dh dks.kh; vko`fÙk gSA

(Q) x = 29 h

4 g

(2) x = 2

,d le:i csyu ftldh Å¡pkbZ h gS, ,d ?kuRo ds vkn'kZ nzo esa rSj jgk gSA ik=k dk vuqizLFk dkV dk {ks=kQy]

csyu ds vuqizLFk dkV dk {ks=kQy dk pkj xquk gSA Å/okZ/kj NksVs nkSyuksa dh dks.kh; vko`fÙk gSA

(R) x = 2 2R

gh

(3) x = 3

,d csyukdkj ik=k esa (3/4)th Å¡pkbZ rd ,d vkn'kZ nzo Hkjk x;kA ftldk ?kuRo gSA vc nzo dks csyu dh

Å/okZ/kj lefer v{k ds lkis{k dks.kh; osx ls ?kqek;k tkrk gS, rkfd nzo ik=k ls ckgj fudy ldsA




(S) x =g2

h3 2 (4) x = 1

,d le:i U ufydk dks mldh ,d Hkqtk ds lkis{k dks.kh; osx ls ?kqek;k tkrk gS, rkfd ;g nzo ls vk/kh Hkjh

jgsA

Sol. (P) Restoring force = Ayg = Ah 2y

2pg

h2 2

(Q) Ay = 3Ay’

y’ = y

3

Restoring force = 4

3 Agy = Ah2y

3pg4

h9 2

(R) y = g2

R22

h

2=

g2

R22

gh

R22 = 1

(S) P2 = P1 + 1

2 2 ( 2 2

2 1r r )

P2 – P1 = 1

2gh = 2

22 h

h4

g2

h3 2 = 4

BOARD LEVEL QUESTIONS

1. A parallel plate capacitor of capacitance C is charged to a potaential V. It is then connected to another uncharged capacitor having the same capacitace. Find out the ratio of the energy stroed in the combined system to the that stored initially in the single capacitor.

/kkfjrk C ds fdlh lekURkj ifV`dk la/kkfj=k dks foHko V rd vkosf'kr fd;k x;k A blds i'pkr~ bls leku /kkfjrk ds chp vU; la/kkfj=k] tks vukosf'kr gS] ls la;ksftr fd;k tkrk gSA la;qDr fudk; esa lafpr ÅtkZ vkSj vkjHHk esa ,dy la/kkfj=k eas lafpr ÅtkZ dk vuqikr Kkr dhft,A




2. Show diagrammatically the behaviour of magnetic field lines in the presence of (i) paramagnetic and (ii) diamagnetic substances. How does one explain this distinguishing feature ?

(i) vuqpqECkdh; inkFkZ ,oa (ii) izfrpqEcdh; inkFkZ dh mifLFkfr esa pqEcdh; {ks=k js[kkvksa dk O;ogkj vkjs[k [khapdj n'kkZb, A bl foHksnudkjh y{k.k dh O;k[;k fdl izdkj dh tkrh gSaA

3. Deduce the expression for the torque acting on a dipole of dipole moment P in the presence of a

uniform electric field E .

,d leku fo|qr {ks=k E dh mifLFkfr esa ,d f}/kzqo ftldk f}/kzqo vk?kw.kZ

P gS bl ij yxus okys cy vk?kw.kZ ds

fy;s O;atd O;qRiUu dhft,A





E ST INFORMA TIO


PPHHYYSSIICCSS


NO. B32 TO B33

2. DPP Syllabus :



ANSWER KEY OF DPP No. : B32 1. (B) 2. (B) 3. (C) 4. (C) 5. (A) 6. (C) 7. (A)

8. (D) 9. (D) 10. (B) 11. (A) 12. (D) 13. (C) 14. (D) 15. (A) 16. (C) 17. (C) 18. (A)

1. A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle the volume of the gas increases from V to 32 V, the efficiency of the engine is :

,d dkuksZa batu esa vkn'kZ f} ijek.kqd xSl dks dk;Zdkjh inkFkZ ds :i esa iz;qDr fd;k tkrk gSA pØ ds :nks"e izlkj ds nkSjku xSl dk vk;ru V ls 32 V rd c<+rk gS] batu dh n{krk gksxhA

(A) 0.5 (B*) 0.75 (C) 0.99 (D) 0.25

Sol. TV – 1 = constant

T1 1

5

7

V

= T2

15

7

)V32(

1

2

T

T =

4

1

)32(

15/2 = 1 –

1

2

T

T = 1 –

4

3

4

1

2. A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h. What is the speed of its centre of mass when the cylinder reaches its bottom

M æO;eku ,oa R f=kT;k dk ,d csyu L yEckbZ rFkk h Å¡pkbZ ds ur ry ij fcuk fQlys uhps dh vksj yq<+d jgk gSaA tc csyu blds ry ij igq¡prk gS rc blds æO;eku&dsUæ dh pky gksxhA

(A) 4

3gh (B*) gh

3

4 (C) gh4 (D) gh2

Sol. Mgh = 22 mv2

1

2

1

Mgh = 2mv2

3

2

1

v = gh3

4

3. A man of mass M stands at one end of a plank of length L which lies at rest on a frictionless horizontal

surface . The man walks to the other end of the plank. If the mass of the plank is M/3, the distance that the man moves relative to the ground is

?k"kZ.k jfgr {kSfrt lrg ij j[ks L yEckbZ ds r[rs ds ,d fljs ij M nzO;eku dk euq"; [kM+k gSA euq"; r[rs ds nwljs fljs rd pyrk gSA ;fn r[rs dk nzO;eku M/3 gS] rks euq"; }kjk tehu ds lkis{k pyh nwjh gS &

(A) 3 L/4 (B) 4 L/5 (C*) L/4 (D) none of these buesa ls dksbZ ugha




Ans. 4

L

Sol.

Mx + 3

M (x + L) = 0

3

M4x = –

3

ML

x = – 4

L

4. Two identical long, solid cylinders are used to conduct heat from temp T1 to temp T2. Originally the

cylinder are connected in series and the rate of heat transfer is H. If the cylinders are connected in parallel then the rate of heat transfer would be :

nks le:i yEcs Bksl csyuksa dk mi;ksx ] rki T1 ls rki T2 rd Å"ek lapj.k esa fd;k tkrk gSA izkjEHk esa csyu Js.khØe esa tqM+s gSa rFkk Å"ek lapj.k dh nj H gSA ;fn csyuksa dks lekUrj Øe esa tksM+k tk;s rks Å"ek lapj.k dh nj gksxh &

(A) H /4 (B) 2H (C*) 4H (D) 8H

Sol. (B) Initially effective resistance = 2R. In parallel effective resistance = 2

R . It has reduced by a factor of

1/4 so rate of heat transfer would be increased by a factor of 4, keeping other parameters same.

5. In a metre bridge experiment null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. If X < Y, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y ?

ehVj lsrq dh iz;ksfxd O;oLFkk esa larqyu fcUnq ,d fljs ls 20 cm dh nwjh ij izkIr gksrk gS] tc izfrjks/k X, vU; izfrjks/k Y ds fo:) larqfyr voLFkk esa gSA ;fn X < Y gks rks leku lUrqyu fcUnq dh ubZ fLFkfr D;k gksxh] tc izfrjks/k 4X dks Y ds fo:) larqfyr fd;k tk;sA

(A*) 50 cm (B) 80 cm (C) 40 cm (D) 70 cm

6. A ferromagnetic material is placed in an external magnetic field. The magnetic domains (A) increase in size (B) decrease in size (C*) may increase or decrease is size (D) have no relation with the field

,d yksg pqEcdh; inkFkZ dks ckg~; pqEcdh; {ks=k esa j[kk tkrk gS] rks pqEcdh; Mksesu

(A) dk vkdkj c<+sxk (B) dk vkdkj ?kVsxk

(C*) dk vkdkj c<+ ;k ?kV ldrk gS (D) ds vkdkj dk {ks=k ds lkFk dksbZ laca/k ugha gSA

7. The desirable properties for making permanent magnets are (A*) high retentivity and high coercive force (B) high retentivity and low coercive force (C) low retentivity and high coercive force (D) low retentivity and low coercive force

LFkk;h pqEcd ds fuekZ.k ds fy, vko';d xq.k gS

(A*) mPp /kkj.k'khyrk (retentivity) rFkk mPp fuxzkfgrk (coercive force)

(B) mPp /kkj.k'khyrk (retentivity) rFkk fuEu fuxzkfgrk (coercive force)

(C) fuEu /kkj.k'khyrk (retentivity) rFkk mPp fuxzkfgrk (coercive force)

(D) fuEu /kkj.k'khyrk (retentivity) rFkk fuEu fuxzkfgrk (coercive force)




8. Electromagnets are made of soft iron because soft iron has (A) high retentivity and high coercive force (B) high retentivity and low coercive force (C) low retentivity and high corecive force (D*) low retentivity and low coercive force

fo|qr pqEcd ueZ yksgs ds inkFkkZs }kjk cuk;h tkrh gS] D;ksfd ueZ yksgk

(A) mPp /kkj.k'khyrk (retentivity) rFkk mPp fuxzkfgrk (coercive force)

(B) mPp /kkj.k'khyrk (retentivity) rFkk fuEu fuxzkfgrk (coercive force)

(C) fuEu /kkj.k'khyrk (retentivity) rFkk mPp fuxzkfgrk (coercive force)

(D*) fuEu /kkj.k'khyrk (retentivity) rFkk fuEu fuxzkfgrk (coercive force)

9. A compass needle which is allowed to move in a horizontal plane is taken to a gemoagnetic pole. It (A) will stay in north-south direction only (B) will stay in east-west direction only (C) will stay exactly mid way between east–west and north–south (D*) will stay in any position.

{ksfrt ry esa ?kw.kZu ds fy, LorU=k fn'kk lwpd ;a=k dh lwbZ dks HkwpqEcdrRo /kzqo ekuk tk, rks ;g (A) dsoy mÙkj nf{k.k fn'kk esa LFkk;h gks tkrh gSA

(B) dsoy iwoZ if'pe fn'kk esa LFkk;h gks tkrh gSA (C) ;g tM+or~ gks tk,sxh rFkk dksbZ fopyu iznf'kZr ugha djsxh

(D*) fdlh Hkh fLFkfr esa O;ofLFkr gks tk,sxh

10. A semicircular wire of radius R is rotated with constant angular velocity about an axis passing through one end and perpendicular to the plane of the wire. There is a uniform magnetic field of strength B. The induced e.m.f. between the ends is :

,d v/kZo`Rrh; 'R' f=kT;k dk ywi vpj dks.kh; osx ls] ,d v{k ds lkis{k ?kqek;k tkrk gS] tks fd rkj ds ry ds yEcor~ gS vkSj rkj ds ,d fljs ls xqtjrh gS ;fn ;gk¡ ij le:i pqEcdh; {ks=k B fp=kkuqlkj mifLFkr gks rks rkj ds fljksa ds e/; çsfjr fo-ok-cy gksxk&

(A) B R2/2 (B*) 2 B R2 (C) is variable ifjorZu'khy gS (D) none of these

buesa ls dksbZ ugha Sol. We connect a conducting wire from A to C and complete the semicircular loop.

ge pkyd rkj dks A ls C rd tksM+rs gSa vkSj v/kZo`Ùkh; ywi iwjk dj ysrs gSaA

The loop emf in the semicircular loop is zero because its magnetic flux does not change.

emf of section APC + emf of section CQA = 0

or emf of section APC = emf of section AQC = 2Ba2 ywi fo0ok0c0 ywi esa 'kwU; gksxk] D;ksafd bldk pqEcdh; ¶yDl ifjofrZr ughs gks jgk gSA APC [k.M dk fo-ok-c- + CQA [k.M dk fo-ok-c- = 0

;k APC [k.M dk fo-ok-c- = AQC [k.M dk fo-ok-c- = 2Ba2

Ans. 2Ba2




11. A particle moves so that its position as a function of time in MKS unit is r

= ktjt4i 2 . The trajectory

of the particle is - (A*) A parabola (B) An ellipse (C) A circle (D) None Sol. x = ct and y = 4t2

y = 2

2x

c

4

Hence parabola

,d d.k bl izdkj xfr'khy gS fd MKS i)fr esa le; ds Qyu ds :i esa bldh fLFkfr r

= ktjt4i 2 }kjk

iznf'kZr dh tkrh gS d.k dk iFk gksxkA

(A*) ijoy;dkj (B) nh?kZo`Ùkkdkj (C) o`Ùkkdkj (D) buesa ls dksbZ ugha Sol. x = ct and y = 4t2

y = 2

2x

c

4

Hence parabola 12. Two solid spheres A and B of equal volumes but of different densities dA and dB are connected by a

string. They are fully immersed in a fluid of density dF. They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if

leku vk;ru ds nks Bksl xksyks ¼A rFkk B) ds fHkUu&fHkUu ?kuRo Øe'k% dA rFkk dB gSA budks jLlh ls cka/kk tkrk gSA budks dF ?kuRo ds nzo esa lEiw.kZ :i ls Mqcks;k tkrk gS ;fn iznf'kZr fp=k ds vuqlkj jLlh esa mRiUu ruko dh lgk;rk ls ;g lkE;oLFkk dh fLFkfr esa O;ofLFkr gks rks ;g O;oLFkk lEHko gksxh ;fn

(A) dA < dF (B) dB dF (C) dA = dF (D*) dA + dB = 2dF

13. A uniform rope of length lies on a table. If the coefficient of friction is then the maximum length 1 of the part of this rope which can overhang from the edge of the table without sliding down is

(A)

(B) 1

(C*)

1

(D)

1

yEckbZ o ,d leku eksVkbZ dh jLlh Vscy ij j[kh gSA ;fn ?k"kZ.k xq.kkad µ gS rks bl jLlh dh og vf/kdre

yEckbZ 1 D;k gksxh ftlds Vscy ls uhps yVdus ij Hkh jLlh uhps u fQlys &

(A)

(B) 1

(C*)

1

(D)

1

Sol. The length of the rope which can overhang from the edge of the table without sliding down is given by

jLlh dh yackbZ tks Vscy dh fdukjh ls yVdk, tkus ij ugh fQlys gksxh

1 =

1

14. The ratio of magnetic inductions at the centre of a circular coil of radius a and on its axis at a distance

equal to its radius, will be -

a f=kT;k dh o`Ùkkdkj dq.Myh ds dsUæ vkSj mlds v{k ij dsUæ ls f=kT;k dh cjkcj nwjh ij mRiUu pqEcdh; {ks=kksa dks vuqikr gksxkA

(A) 2

1 (B)

1

2 (C)

22

1 (D*)

1

22




Sol. Bc = a2

i0

BR = 2/322

20

)aa(2

ia

=

a24

i0

R

C

B

B=

1

22

15. A metallic rod completes its circuit as shown in the figure. The circuit is normal to a magnetic field of B

= 0.15 tesla. If the resistances of the rod is 3the force required to move the rod with a constant velocity of 2 m/sec is -

fn[kk;h gqbZ vkd`fr esa ,d /kkrq dh NM+ bldk ifjiFk iw.kZ djrh gSA ifjiFk pqEcdh; {ks=k B = 0.15 tesla ds lkFk yEcor~ gS] vxj NM+ ds izfrjks/k dk eku 3gS] rks NM+ dks 2 m/sec fu;r osx ls xfr esa ykus ds fy, vko';d cy gksxk&

v=2m/s50cm

B=0.15T (into page)

(A*) 3.75 × 10–3 N (B) 3.75 × 10–2 N (C) 3.75 × 102 N (D) 3.75 × 10–4 N

Sol. e = Bv

i = R

vB

R

e

F – iB = 0

F = iB= 3.75 × 10–3 N 16. The amplitude of a particle due to superposition of following S.H.Ms. Along the same line is

leku fn'kk essa fuEu ljy vkorZ xfr;ksa ds v/;kjksi.k ds dkj.k ,d d.k dk vk;ke gksxk

X1 = 2 sin 50 t ; X2 = 10 sin (50 t + 37º)

X3 = 4 sin 50 t ; X4 = 12 cos 50 t

(A) 4 2 (B) 4 (C*) 6 2 (D) none of these Hint : Amplitude phasor diagram :

resultant amplitude = 26




17. A uniform rod AB of mass m and length at rest on a smooth horizontal surface. An impulse P is applied to the end B. The time taken by the rod to turn through a right angle is:

m nzO;eku o yEckbZ dh ,d ,dleku NM+ A B ,d {ksfrt fpdus ry ij fLFkj j[kh gq;h gSA blds fljs B ij ,d vkosx P vkjksfir fd;k tkrk gSA NM+ }kjk ledks.k ls ?kwe tkus esa yxk le; gS &

(A) P

m2 (B)

P

m

3

(C*)

P

m

12

(D)

P

m

3

2

Sol. (C) Impulse = change in momentum

vkosx = laosx ifjorZu

P. 2

=

12

m 2. (about centre of AB) (AB ds dsUnz ds ifjr%)

= m

P6

For = 2

ds fy,;

2

= t t =

2 =

p62

m

t = p12

m Ans. .

18. At 450 to the magnetic meridian, the apparent dip is 300. Find the true dip.

pqEcdh; ;keksÙkj ls 450 dks.k ij vkHkklh ufr dks.k 300 gks rks okLrfod ufr dks.k Kkr djksA.

(A*) 1tan 1/ 6 (B) 1tan 1/3 (C) 1tan 1/ 2 (D) 1tan 2 /3

Sol. At 450 to the magnetic meridian, the effective horizontal component of the earth’s magnetic field is

B’H = BH cos 450 = 2

1BH. The apparent dip is given by

tan2B

B2

'B

B'tan

HH

where is the true dip. Thus,

tan 300 = tan2

or, 6/1tan 1

Board Questions 1. A carbon resistance is coded with following colours of four rings : Red, Brown, Yellow, Golden [1 Marks]

dkcZu izfrjks/k fuEu jax dh pkj fjaxksa ds lkFk lkadsfrd fd;k x;k gS: yky, Hkwjk, ihyk, lqugjk [1 Marks]

Ans. 21 × 104 ± 5% 2. What are the factors which affect current sensitivity of a moving coil galvanometer ? [1 Marks]

py dq.Myh /kkjkekih dh /kkjk lqxzkfgrk fdu dkjdksa ls izHkkfor gksrh gS \

Ans. Current sensitivity is reciprocal of figure of merit

3. Prove that electric lines of forces are perpendicular to equipotential surface.

fl) djks fd oS|qr cy js[kk,sa lefoHko i`"B ds yEcor~ gksrh gS \ [2 Marks]

4. Why does radial magnetic field is used in moving coil galvanometer. [2 Marks]

py dq.Myh /kkjkekih esa f=kT;h; pqEcdh; {ks=k D;ksa dke esa fy;k tkrk gS \





E ST INFORMA TIO


PPHHYYSSIICCSS


NO. B32 TO B33


Single choice Objective ('–1' negative marking) Q.1 (3 marks 2 min.) [03, 02] One or more than one options correct type ('–1' negative marking) Q.2 to Q.7 (4 marks 2 min.) [24, 12] Subjective Questions ('–1' negative marking) Q.8 to Q.9 (4 marks 5 min.) [08, 10] Match the Following (no negative marking) Q.10 (8 marks 6 min.) [08, 06]


1. (D) 2. (A,B) 3. (A,C) 4. (A,C,D) 5. (A,D) 6. (A) 7. (C)

8. 3 9. 9 10. A P, Q ; B P, R, S ; C P, S

1. There exists a uniform magnetic and electric field of magnitude 1 T and 1 V/m respectively along

positive y-axis. A charged particle of mass 1 kg and of charge 1 C is having velocity 1 m/sec along x-

axis and is at origin at t = 0. Then the co-ordinates of particle at time seconds will be:

/kukRed y-v{k ds vuqfn'k 1 VSlyk rFkk 1 oksYV@eh- ds ,d le:i pqEcdh; rFkk fo|qr {ks=k vfLrRo es gSaA 1 kg

nzO;eku rFkk 1 C vkos'k ds vkosf'kr d.k dk x-v{k ds vuqfn'k osx 1 eh-@ls- gS rFkk t = 0 ij ;g ewyfcUnq ij gSA

rc le; lsd.M ij d.k ds funsZ'kkad gksaxs &

(A) (0, 1, 2) (B) (0, 2/2, 2) (C) (2, 2/2, 2) (D*) (0, 2/2, 2) Sol. The particle will move in a non-uniform helical path with increasing pitch as shown below:

d.k vle:i dq.Mfyuh iFk esa c<+rs gq, pwM+h vUrjky ds lkFk uhps n'kkZ;s vuqlkj xfr djsxk&

Its time period will be : bldk vkorZ dky gksxk &

T = qB

m2 = 2seconds

Changing the view, the particle is seemed to move in a circular path in (x – z) plane as below

n`'; cnyus ij] d.k (x – z) ry esa o`Ùkkdkj iFk esa xfr djrk gqvk izrhr gksrk gSA

After -seconds the particle will be at point 'P', hence x coordinate will be 0

lSd.M i'pkr~ d.k fcUnq P ij gksxk] bl izdkj x funsZ'kkad 0 gksxk For linear motion along y-direction.

y-fn'kk ds vuqfn'k js[kh; xfr ds fy, &




y() = 2)(m

Eq

2

1)(0

y() = 2

2 and OP = 2 Hence the coordinate

2,

2,0

2

.

y() = 2

2 vkSj OP = 2 blfy, funZs'kkad

2,

2,0

2

gSA

2. Consider a non viscous & incompressible liquid filled in a tank upto a height h. A narrow pipe is attached to the bottom of the tank as shown. The cross sectional area A1 at the point 1 is very large as compared to the areas at the points 2 and 3. The relation between cross sectional areas is A2 = 2A3. If P1, P2, P3 are the pressures at cross-section 1, 2 & 3 respectively and P0 is atmospheric pressure then choose the correct options. ekfu, fd ,d v';ku rFkk vlEihfM+r nzo dks VSad esa h ÅWpkbZ rd Hkjk x;k gSA ,d iryh uyh bl VSad ds isans esa fp=kkuqlkj tksM+h xbZ gSA fcUnq 1 dk vuqizLFk dkV {ks=kQy A1, fcUnq 2 rFkk fcUnq 3 ds {ks=kQy dh rqyuk esa cgqr vf/kd cM+k gSA vuqizLFk dkV {ks=kQy A2 = 2A3 dk lEcU/k j[krs gSA ;fn P1, P2, P3 Øe'k% vuqizLFk dkV 1, 2 rFkk 3

ds nkc gks ,oa P0 ok;qe.Myh; nkc dks iznf'kZr djrk gks rks lgh fodYiksa dks pqfu;sA

(A*) 3V 2gh (B*) P2 = P0 + 3

4 gh (C) P2 = P0 +

gh

2

(D) 2V 2gh

Ans. (A) & (B) Sol. By Equation of continuity

lkUrR;rk lehdj.k ls A1v1 = A2v2 = A3v3

Since pwafd A1 >> A2, A3, so vr% v1 0

A3 = 2A

2 v3 = 2v2

By Bernoulli theorem at 1 and 3

1 rFkk 3 ij cjuksyh izes; ls

P0 + gh = P0 + 23

1V

2 v3 = 2gh , v2 =

2gh

2

At 1 and 2 ; 1 rFkk 2 ij

P0 + gh = P2 + 22v

2

P2 = P0 + 3

4 gh




3. Figure shows a LCR 'circuit connected with a dc battery of emf ‘’ and internal resistance ‘R’. After a long time (initially the capacitor was uncharged):

fp=k esa '' fo|qr okgd cy rFkk 'R' vkUrfjd çfrjks/k dh dc cSVjh ls tqM+k LCR ' ifjiFk çnf'kZr gSA yEcs le; i'pkr~: (izkjEHk esa la/kkfj=k vukosf'kr gS):

(A*) Current through the inductor is R8

(B) Charge stored in the capacitor is 4

C

(C*) Charge stored in the capacitor is 2

C

(D) Potential difference across the terminals of battery is 4

(A*) çsjd dq.Myh ls çokfgr /kkjk R8

gSA (B) la/kkfj=k esa laxzfgr vkos'k

4

C gSA

(C*) la/kkfj=k esa laxzfgr vkos'k 2

C gSA (D) cSVjh ds VfeZuy ds fljksa ij foHkokUrj

4

gSA

4. A metallic V shaped rod OAC is rotated with respect to one of its end in uniform magnetic field, such that axis of rotation is parallel to the direction of magnetic field. Length of each arm of rod is L and angle between the arms is 60º. P is the mid–point of section AC. Magnitude of magnetic field is B.

Then choose the correct relations.

V vkdkj dh ,d /kkfRod NM+ OAC blds ,d fljs ds lkis{k le:i pqEcdh; {ks=k esa bl çdkj ?kwerh gS] dh ?kw.kZu v{k pqEcdh; {ks=k dh fn'kk ds lekUrj gSA NM+ dh çR;sd Hkqtk dh yEckbZ L gS rFkk Hkqtkvksa ds e/; dks.k 60º gSA P

[k.M AC dk e/; fcUnq gSA pqEcdh; {ks=k dk ifjek.k B gSA

rc lgh lEcU/kksa dk p;u dhft,A

(A*) VA – V0 = 2

BL2 (B) VA – VC =

2

BL2 (C*) VC – VP =

8

BL2 (D*) VA – VP =

8

BL2




Sol.

OA = OC = L

OP = 2

L3

VA – V0 = L

BL2 VA – VC = 0

VC – V0 = 2

BL2 VP – V0 =

2

B

4

L3 2

= 8

BL3 2

VC – VP = (VC – V0) – (VP – V0) = 2

BL2 –

8

BL3 2 =

8

BL2

VA – VP = (VA – V0) – (VP – V0) = 2

BL2 –

8

BL3 2 =

8

BL2

Comprehension

vuqPNsn

Two fixed and horizontal cylinders A and B having pistons (both massless) of cross sectional area 100 cm2 and 200 cm2 respectively, are connected by massless rod. The piston can move freely without

friction. The cylinder A contains 100 gms of an ideal gas ( = 1.5) at pressure 105N/m2 and temperature T0. The cylinder B contains identical gas at same temperature T0 but has different mass. The piston are held at the state such that volume of gas in cylinder A and cylinder B are same and is equal to 10–2m3. The walls and piston of cylinder A are thermally insulated where as gas in cylinder B is maintained at constant temperature T0. The whole system is in vacuum. Now the pistons are slowly released and they move towards left and mechanical equilibrium is reached at the state when the volume of gas in cylinder A becomes 25 × 10–4 m3.

nks fLFkj rFkk {kSfrt csyu A rFkk B ftuds fiLVu (nksuksa nzO;ekughu) ds vuqizLFk dkV {ks=kQy Øe'k% 100 cm2 rFkk 200 cm2 gS, ,d nzO;ekughu NM+ }kjk tqM+s gq, gSA fiLVu fcuk ?k"kZ.k ds Lora=krkiwoZd xfr dj ldrk gSA csyu A esa T0 rki rFkk 105N/m2 nkc ij 100 gms dh ,d vkn'kZ xSl ( = 1.5) Hkjh gqbZ gSA csyu B esa Hkh leku rki T0 ij ,dleku fdUrq fHkUu nzO;eku dh vkn'kZ xSl Hkjh gqbZ gSA fiLVu bl izdkj yxs gq, gSa fd csyu A rFkk B esa xSl dk vk;ru leku gS rFkk 10–2m3 ds cjkcj gSaA csyu A dh nhokjs rFkk fiLVu Å"ek ds dqpkyd gS tcfd csyu B esa xSl fu;r rki T0 ij cuk;s j[krh gSA lEiw.kZ fudk; fuokZr esa gSA vc fiLVu dks /khjs ls NksM+k tkrk gS rFkk ;s cka;h vksj xfr djrs gS rFkk tc ;kaf=kd lkE;koLFkk esa igq¡prs gS rc csyu A esa xSl dk vk;ru 25 × 10–4 m3 gks tkrk gSA

A B




5. Select correct statement :

lgh dFku pqfu;s % (A*) Number of moles in B is 10 times in number of moles in A (B) The mass of gas in cylinder A is 600 gms (C) Number of moles in B is 5 times in number of moles in A (D*) The mass of gas in cylinder B is 1 kg

(A*) B esa eksyksa dh la[;k A esa eksyksa dh la[;k ls 10 xquk gS (B) csyu A esa xSl dk æO;eku 600 gms gSA (C) B esa eksyksa dh la[;k A esa eksyksa dh la[;k dh 5 xquk gS (D*) csyu B esa xSl dk æO;eku 1 kg gSA

6. The change in internal energy of gas in cylinder A is :

csyu A esa xSl dh vkUrfjd ÅtkZ esa ifjorZu gksxk (A*) 2000 J (B) 1000 J (C) 500 J (D) 3000 J

7. The compressive force in the connecting rod at equilibrium is :

lkE;koLFkk ij la;ksftr NM+ksa esa lEihM+u cy gksxk (A) 2000 N (B) 4000 N (C*) 8000 N (D) 10000 N Sol. Let the initial temperature, pressure and volume of gas in ‘A’ be T0, P0, V0 and the area of the position A

and B be a, 2a. Now gas in chamber ‘A’ undergoes adiabatic compression whereas gas in chamber ‘B’ undergoes

isothermal expansion. Now solving for gas in ‘A’

A2A2A1A1 VPVP 00VP = P2A (0.25V0) P2A = 8P0

Solving for gas is chambers ‘B’

P1BV1B = P2BV2B P1BV0 = P2B x 2.5V0 P2B = 0.4 P1B

Also (P2A x a) = (P2B x 2a) for gas in ‘B’

8P0a = P2 x 2a P2 = 4P0

0.4 P1 = 4P0 P1 = 10P0

Now comparing the moles of gas in A and B

nA = 0

00

RT

VP ,

0

00B

RT

VP10n = 10nA

mB 10 mA = 10 x 100gm = 1kg.

Again for gas is A, Q = U + W

0 = U + W

0 = U +

)1(

4

VP8VP 0

000

U = 2P0V0

= 2 x 105 x 102 = 2000J

F = 8P0a = 8 x 105 x 102 = 8000 N

Sol. ekuk ‘A’ easa xSl dk izkjfEHkd rki] nkc rFkk xSl dk vk;ru T0, P0, V0 gS rFkk fLFkfr A rFkk B esa {ks=kQy a, 2a gSA .

vc d{k ‘A’ esa xSl :)ks"e laihMu ds varxZr gS tcfd d{k ‘B’ esa xSl lerkih; izlkj ds varxZr gSA vc ‘A’ esa xSl ds fy;s gy djus ij

A2A2A1A1 VPVP 00VP = P2A (0.25V0) P2A = 8P0

d{k ‘B’ esa xSl ds fy, gy djus ij

P1BV1B = P2BV2B P1BV0 = P2B x 2.5V0 P2B = 0.4 P1B

rFkk (P2A x a) = (P2B x 2a) ‘B’ esa xSl ds fy, 8P0a = P2 x 2a P2 = 4P0

0.4 P1 = 4P0 P1 = 10P0




vc A rFkk B esa xSl ds eksyksa dh rqyuk djus ij

nA = 0

00

RT

VP ,

0

00B

RT

VP10n = 10nA

mB 10 mA = 10 x 100gm = 1kg.

A esa xSl ds fy, iqu% , Q = U + W

0 = U + W

0 = U +

)1(

4

VP8VP 0

000

U = 2P0V0

= 2 x 105 x 102 = 2000J

F = 8P0a = 8 x 105 x 102 = 8000 N 8. In the shown figure a conducting ring of mass m = 2kg and radius R = 0.5 m. lies on a smooth

horizontal plane with its plane vertical. The ring carries a current of = A1

. A horizontal uniform

magnetic field of B = 12T is switched on at t = 0. The initial angular acceleration in rad./sec2 of the ring will be 4x if x is :

iznf'kZr fp=k esa m = 2kg æO;eku rFkk R = 0.5 m f=kT;k dh ,d pkyd oy; ?k"kZ.kjfgr {kSfrt ry ij fLFkr gS

rFkk bl oy; dk ry Å/okZ/kj gS] oy; esa izokfgr /kkjk = A1

gSA B = 12T dk ,d leku {kSfrt pqEcdh; {ks=k

t = 0 ij pkyw fd;k tkrk gS oy; dk izkjfEHkd dks.kh; Roj.k rad./sec2 esa 4x gS rc x gksxk :

Ans. 3

Sol. y-axis = Iy-axis

(I.r2) B = 1/2 mr2 = 12 rad/sec2

X = 3

9. A current carrying ring, carrying a constant current 2

Amp., radius 1m, mass 3

2kg and having 10

windings is free to rotate about its tangential vertical axis. A uniform magnetic field of 1 tesla is applied perpendicular to its plane. How much minimum angular velocity (in rad/sec.) should be given to the ring in the direction shown, so that it can rotate 270º in that direction. Write your answer in nearest single digit in rad/sec.

,d /kkjkokgh oy;] ftlesa çokfgr fu;r /kkjk 2

Amp., f=kT;k 1m, nzO;eku 3

2kg o 10 Qsjs gS] blds Li'kZ js[kh;

Å/oZ v{k ds ifjr% ?kw.kZu ds fy, LorU=k gSA blds ry ds vfHkyEcor~ 1 Vslyk dk le:i pqEcdh; {ks=k yxk;k x;kA n'kkZbZ fn'kk esa oy; dks fdruk U;wure dks.kh; osx (rad/sec. esa) fn;k tk,] ftlls ;g mlh fn'kk esa 270º

dks.k ls ?kwe tk;s A viuk mÙkj jsfM;[email protected] ek=kd esa fudVre ,d vadh; eku esa O;Dr dhft,A

Ans. 9




Sol.

to reach = 270º, it has to cross the potential energy barrier at = 180º and to cross = 180º angular

velocity at = 180º should be 0+

= 270º, igq¡pus ds fy, blh = 180º ij fLFkfrt ÅtkZ vojks/k ikj djuk gksxk rFkk = 180º ikj djus ds fy, = 180º ij dks.kh; osx 0+ gksuk pkfg,

ki + Ui = kf + Uy

22MR2

3

2

1

+ (–Mi AB cos 0º) = 0 + (–NiAB cos180º)

= 80 9 rad/sec.

10. Figure shows snap shot of a progressive wave and standing wave along a string. Match the column I and II.

fuEu fp=kksa esa fdlh Mksjh ds vuqfn'k] izxkeh rFkk vcizxkeh rjaxksa ds LiUn fp=k (snap shot) n'kkZ;s x;s gSA rks dkWye-I rFkk dkWye-II dk feyku dhft,A

Progressive wave

Standing wave

y

y

x

x

D

H

A

E

B

F

C

G

izxkeh rjax

vizzxkeh rjax

Column-I Column-II

dkWye-I dkWye-II

(A) Particles in same phase (P) A & rFkk D

d.k leku dyk esa gS (B) Particles with same amplitude (Q) E & rFkk F

of oscillations

d.kksa dk nksyu vk;ke leku gS (R) B & rFkk C

(C) Particles always having same speed (S) G & rFkk H

d.kksa dh pky lnSo leku gS Ans. A P, Q ; B P, R, S ; C P, S






E ST INFORMA TIO


PPHHYYSSIICCSS


NO. B34 TO B35

2. DPP Syllabus :


Total Marks : 62 Max. Time : 46 min. Single choice Objective ('–1' negative marking) Q.1 to Q.18 (3 marks 2 min.) [54, 36] Subjective Questions ('–1' negative marking) Q.19 to Q.20 (4 marks 5 min.) [08, 10]

ANSWER KEY OF DPP No. : B34 1. (D) 2. (C) 3. (B) 4. (A) 5. (B) 6. (B) 7. (D) 8. (B) 9. (C) 10. (C) 11. (C) 12. (C) 13. (A) 14. (B) 15. (C) 16. (A) 17. (C) 18. (B) 19. 0.05 20. 0.1 mm

1. The S unit of inductance, the Henry can not be written as :

çsjdRo dk S ek=kd gsujh dks ugha fy[k ldrs gS :

(A) weber/ampere (B) volt-second/ampere (C) joule/(ampere)2 (D*) ohm / second

(A) oscj/,Eih;j (B) oksYV-lSd.M/,Eih;j (C) twy/(,Eih;j)2(D*) vkse / lSd.M

Sol. (A) L = i

or Henry gSujh =

Ampere

Weber oscj/,Eih;j

(B) e = – L

dt

di

L = dt/di

e or Henry gSujh=

Ampere

ondsecVolt oksYV-lSd.M/,Eih;j

(C) U = 2

1Li2

L = 2i

U2 or Henry gSujh =

2)Ampere(

Joule twy/(,Eih;j)2

(D) U = 2

1Li2 = i2 Rt

[L] = [Rt]. or Henry gSujh = ohm-second vkse-lSd.M 2. The acceleration time graph of a particle moving along a straight line is as shown in the figure. At what

time the particle acquires its velocity equal to initial velocity ?

ljy js[kk es xfr djrs gq, d.k dk Roj.k le;&xzkQ fn[kk;k x;k gSA d.k fdl le; ij viuk osx çkjfEHkd osx ds cjkcj çkIr dj ysxkA

(A) 12 sec (B) 5 sec (C*) 8 sec (D) none of these bues ls dksbZ ugha




3. A particle of positive charge q and mass m enters with velocity jV at the origin in a magnetic field

)k(B which is present in the whole space. The charge makes a perfectly inelastic collision with free

identical particle at rest at its maximum y-coordinate. After collision the combined charge will move on

trajectory :(where qB

mVr )

m nzO;eku ,oa q /kukRed vkos'k dk ,d d.k ewy fcUnq ij jV osx ls pqEcdh; {ks=k )k(B tks fd lEiw.kZ {ks=k esa

mifLFkr gS] esa izos'k djrk gSA vkos'k vius le:i vU; fLFkj d.k ls iw.kZ vçR;kLFk VDdj djrk gS] tks fd vius egÙke y-funsZ'kkad rd xfr djus ds fy, LorU=k gSA VDdj ds i'pkr~ la;qDr vkos'k fuEu iFk esa xfr djsxk : (tgka

qB

mVr )

(A) y = )i(qB

mv (B*) (x + r)2 + (y – r/2)2 = r2/4

(C) (x – r)2 + (y – r)2 = r2 (D) (x – r)2 + (y + r/2)2 = r2/4

Sol. Hence (B). 4. When a person throws a meter stick it is found that the centre of the stick is moving with speed 10 m/s

and left end of stick with speed 20m/s. Both points move vertically upwards at that moment. Then angular speed of the stick is:

tc ,d vkneh us ,d ehVj NM+ dks QSadk rks ;g ns[kk x;k fd mlds dsUnz dh pky 10 m/s gS vkSj ckb± Nksj dh pky 20 m/s gSA nksuksa fcUnq ml {k.k Åij dh vksj tk jgs Fks] rks NM+ dh dks.kh; pky gksxh :

(A*) 20 rad/ sec(B) 10 rad/sec (C) 30 rad/sec (D) none of these buesa ls dksbZ ugh

Sol. Angular velocity (dks.kh; osx) w = 5.0

1020 = 20 rad/sec.

5. A car fitted with a device which transmits sound 60 times per minute. There is no wind and speed of

sound in still air is 345 m/s. If you hear the sound 68 times per minute when you are moving towards the car with a speed of 12 m/s, the speed of the car must be nearly.

,d dkj ftlesa ,d ;a=k tks vkokt dks ,d feuV esa 60 ckj izlkfjr djrk gS LFkkfir fd;k gqvk gSA ogk¡ ij dksbZ gok izokfgr ugha gS vkSj /ofu dh pky gok esa vHkh 345 m/s gSA tc rqe 12 m/s dh pky ls dkj dh rjQ ls xqtjrs gks rks rqe ,d feuV esa 68 ckj /ofu lqurs gksA rc dkj dh pky gksuh pkfg,:

(A) 20.0 m/s towards you (B*) 30.0 m/s towards you (C) 10.0 m/s away from you (D) 10.0 m/s towards you

(A) 20.0 m/s vkidh rjQ (B*) 30.0 m/s vkidh rjQ

(C) 10.0 m/s vkils nwj (D) 10.0 m/s vkidh rjQ




Sol. (B) C = 345 m/s fob = 68 time/1min

f’ = 2VC

12C

f0

60

68=

2V345

12345

345 – V2 = 357 × 68

60

V2 = 345 – 68

60357 = 345 – 315 = 30 m/s

6. An electron orbiting around a nucleus has angular momentum L. The magnetic field produced by the

electron at the centre of the orbit can be expressed as :

,d bysDVªkWu ,d ukfHkd ds pkjksa vksj L dks.kh; laosx ls d{kk esa ?kwe jgk gSA bysDVªkWu ds }kjk mRiUu pqEcdh; {ks=k d{k ds dsUæ ij bl izdkj n'kkZ;k tk ldrk gS :

(A) L)mr8/e(B 30 (B*) L)mr4/e(B 3

0

(C) L)mr/e(B 30 (D) L)mr4/e(B 3

0

Sol. (B)

L

B=

mVr

r

eV

4

20

L = m V r

L

B=

mVr

1

r

eV

4 2

0

B =

4

0

r

L =

3

0

mr4

eL

7. A particle of mass 0.2 kg moves along a path given by the relation : jtsin3itcos2r

. Then the

torque on the particle about the origin is :

,d 0.2 kg æO;eku dk d.k ,d iFk ij lEcU/k: jtsin3itcos2r

ls pyrk gSA rc d.k ij ewy fcUnq ds

lkis{k cy vk?kw.kZ gS:

(A) Nmk13 (B) Nmk3

2 (C) Nmk

2

3 (D*) 0

Sol. (D)

L = m Vr

V = jtcos3itsin2

ttanconsk60tcos3tsin2

tsin6tcos6R0tsin3tcos2

kji22

L = 0




8. Two equal masses are connected by a spring satisfying Hooke's law and are placed on a frictionless

table. The spring is elongated a little and allowed to go. Let the angular frequency of oscillations be . Now one of the masses is stopped. The square of the new angular frequency is :

nks leku æO;eku ,d jLlh ls gqd ds fu;e dk ikyu djrs gq, ca/ks gq, gSa vkSj ?k"kZ.k jfgr est ij j[ks gSaA jLlh dqN foLrkfjr gS vkSj xfr ds fy, NksM+ nh tkrh gSA ekuk nksyu dh dks.kh; vkofÙk gSA vc muesa ls ,d æO;eku dks jksd fn;k tk;s rks ubZ dks.kh; vkofÙk dk oxZ gS:

(A) 2 (B*) 2

2 (C)

3

2 (D) 22

Sol.

k

=

= 21

21

mm

mm

= m

k2

2 = m

k

22 =

m

k =

2

2 . Ans. (B)

9. In a steel factory it is found that to maintain M kg of iron in the molten state at its melting point an input

power P watt is required. When the power source is turned off, the sample completely solidifies in time t second. The latent heat of fusion of iron is

,d LVhy QSDVªh esa ;g ik;k x;k fd M kg ds yksgs dks xyukad fcUnq ij bldh xyu voLFkk esa cuk;s j[kus ds fy;s ,d 'kfDr P okWV dh vko';drk gksrh gSA tc 'kfDr L=kksr dks cUn djrs gSa rks t lsd.M esa uewuk iwjh rjg ls Bkslh; gks tkrk gS rks yksgs dh xyu dh xqIr Å"ek gS

(A) M

Pt2 (B)

M2

Pt (C*)

M

Pt (D)

t

PM

Sol. (c)

Pt = ML L = M

Pt

10. The Schrodinger equation for a free electron of mass m and energy E written in terms of the wave

function is 0h

mE8

dx

d2

2

2

2

. The dimensions of the coefficient of in the second term must be:

m æO;eku o E ÅtkZ ds ,d eqDr bysDVªkWu ds fy;s JksfMaxj lehdj.k rjax Qyu ds :i esa fuEu izdkj fy[kh xbZ

gS 0h

mE8

dx

d2

2

2

2

. nwljs in esa ds xq.kkad dh foek gksuh pkfg,:

(A) [M1 L1] (B) [L2] (C*) [L–2] (D) [M1 L–1 T1]

Sol. (C) h = C

E

Dimension of from first term of expression = [L2]

So dimension of coefficient of in second term = [L–2]




11. Let r be the distance of a point on the axis of a short bar magnet from its centre. The magnetic field at such point is proportional to

ekuk ,d NksVh NM+ pqEcd dh v{k ij fLFkr ,d fcUnq dh nwjh blds dsUæ ls r gSA bl fcUnq ij pqEcdh; {ks=k lekuqikrh gSA

(A) r

1 (B)

2r

1 (C*)

3r

1 (D) none of these

12. A paramagnetic material is kept in a magnetic field. The field is increased till the magnetization

becomes constant. If the temperature is now decreased, the magnetization (A) will increase (B) decrease (C*) ramain constant (D) may increase of decrease

,d vuqpqEcdh; inkFkZ ,d pqEcdh; {ks=k esa j[kk gSA {ks=k rc rd c<+rk gS tc rd fd pqEcdRo fu;r ugha gks tkrk gSA ;fn vc rkieku ?kVrk gS rks pqEcdRo %

(A) c<+sxk (B) ?kVsxk (C*) fu;r jgsxk (D) c<+ ;k ?kV ldrk gS 13. A very long bar magnet is placed along the axis of a circular loop carrying an electric current i. The

north pole of bar magnet is concilding with the centre of the circular loop. The magnetic field due to the magnet at a point on the periphery of the wire is B. The radius of the loop is a. The force on the wire is

,d cgqr yEch NM+ pqEcd ,d o`Ùkh; ywi ds dsUæ ds lkFk ftlesa ,d fo|qr /kkjk i izokfgr gS bl izdkj fLFkr gS fd bldk mÙkjh /kzqo o`Ùkkdkj ywi ds dsUæ ds lkFk lEikrh gSA rkj dh ifj/kh ds fdlh fcUnq ij pqEcd ds }kjk mRiUu pqEcdh; {ks=k B gSA ywi dh f=kT;k a gS rks ywi ij vkjksfir cy gksxk:

(A*) very nearly 2aiB perpendicular to the plane of the wire (B) 2aiB in the plane of the wire

(D) aiB along the magnet (D) zero

(A*) 2aiB ds cgqr djhc rkj ds ry ds yEcor~ (B) 2aiB rkj ds ry esa (D) aiB pqEcd ds vuqfn'k (D) 'kwU;

14. r.m.s. value of current i = 3 + 4 sin ( t + /3) is:

/kkjk i = 3 + 4 sin ( t + /3) dk oxZ ek/;ewy eku gksxk&

(A) 5 A (B*) 17 A (C) 2

5A (D)

2

7 A

Sol. i = 3 + 4 sin (t + /3)

i2 = 9 + 16 sin2 (t + /3) + 24 sin (t + /3) < i2 > = 9 + 16 [1/2] + 24 [0]

= 9 + 16

2

1

= 17

r.m.s. value = 17 15. S1 and S2 are two coherent sources of sound of frequency 110Hz each. They have no initial phase

difference. The intensity at a point P due to S1 is 0 and due to S2 is 40. If the velocity of sound is 330 m/s then the resultant intensity at P is

S1 rFkk S2 izR;sd 110Hz vko`fr ds nks dyklEc) /ofu L=kksr gSA muesa izkjfEHkd dykUrj ugha gSA fcUnq P ij S1

ds dkj.k rhozrk 0 rFkk S2 ds dkj.k rhozrk 40 gSA ;fn /ofu dk osx 330 m/s gS rks fcUnq P ij ifj.kkeh rhozrk gS &

(A) I0 (B) 9I0 (C*) 3I0 (D) 8I0




Sol. The wavelength of sound source = 110

330= 3 metre.

The phase difference betwen interfering waves at P is

= = 2

(S2P – S1P) = 3

2 (5 – 4) =

3

2

Resultant intensity at P = I0 + 4I0 + 3

2cos42 00

= 3 I0

L=kksr dh /ofu dh rjaxnS/;Z = 110

330 = 3 metre.

fcUnq P ij v/;kjksfir gksus okyh rjaxksa ij dykUrj

= = 2

(S2P – S1P) = 3

2 (5 – 4) =

3

2

ifj.kkeh rhozrk fcUnq P ij = I0 + 4I0 + 3

2cos42 00

= 3 I0

16. A ball of mass m moving with velocity v, collide at the centre of the fixed frictionless wall of height d elastically as shown in figure. After the collision the change in angular momentum of the ball about point P is -

v osx ls xfr dj jgh m nzO;eku dh ,d xsan d Å¡pkbZ dh fLFkj ?k"kZ.kjfgr nhokj ds dsUnz ij çR;kLFk :i ls fp=kkuqlkj Vdjkrh gSA VDdj ds i'pkr~ fcUnq P ds lkis{k xsan ds dks.kh; laosx esa ifjorZu gksxk -

(A*) 2

3mvd (B) mvd (C) 3 mvd (D) None of these

Sol.

Change in Angular momentum, dks.kh; laosx esa ifjorZu L = (P) 2

d = 2mv cos 30

2

d =

2

3mvd

17. A man of 80 kg attempts to jump from the small boat of mass 40 kg on to the shore. He can generate a

relative velocity of 6 m/s between him and boat. His velocity towards shore is:

80 kg nzO;eku dk euq";] 40 kg fdxzk nzO;eku dh ,d NksVh uko ls fdukjs ij dwnus dk iz;kl djrk gSA og vius rFkk uko ds chp 6 eh/ls0 dk vkisf{kd osx mRiUu dj ldrk gSA mldk fdukjs dh rjQ osx gSA

(A) 4 m/s (B) 8 m/s (C*) 2 m/s (D) 3 m/s

18. The vernier constant of a vernier callipers is 0.1 mm. There are 10 divisions on Vernier scale. These 10 divisions coincides with N-divisions of main scale. Then N should be :

ofuZ;j dsyhij ds ofuZ;j fu;rkad dk eku 0.1 mm. gSA ofuZ;j iSekus ij 10 Hkkx gSA ;g 10 Hkkx] eq[; iSekus ds N-Hkkxks ls lEikrh gS N dk eku gksxkA

(A) 11 (B*) 9 (C) 19 (D) 10




19. 19-divisions on the main scale of a vernier callipers coincide with 20 divisions on the vernier scale. If each division on the main scale is of 1 mm units, determine the least count of the instrument.

;fn ofuZ;j dsfyij ds eq[; iSekus dk 19 ok¡ Hkkx 20 osa ofuZ;j iSekus ds lEikrh gS rFkk eq[; iSekus dk ,d Hkkx 1mm bdkbZ ds cjkcj gS rks ofuZ;j dk vYirekad Kkr dhft;sA

Ans. 0.05 Sol. (N + 1) divisions on the vernier scale = N divisions on main scale

1 division on vernier scale = 1N

N

divisions on main scale

Each division on the main scale is of a units.

1 division on vernier scale = 1N

N

a units = a’ (say)

Least count = 1 main scale division – 1 vernier scale division

= a – a’ = a –1N

N

a =

1N

a

Put a = 1 mm and N = 19 20. In screw gauge screw moves 5 mm on main scale in 50 revolutions what is pitch of screw gauge.

50 pØks ds nkSjku LØw xSt] eq[; iSekus ij 5 mm foLFkkfir gksrk gSA LØw xSt dk pwM+h vUrjky D;k gksxkA Ans. 0.1 mm

BOARD PROBLEMS 1. Define the term ‘magnetic declination’.

pqEcdh; >qdko dks iznf'kZr dhft,A Ans. Magnetic declination : The small angle between magnetic axis and geographic axis at a place is

called the magnetic declination. It is denoted by .

pqEcdh; fnDikr~ dks.k ¼>qdko dks.k½: fdlh LFkku ij pqEcdh; v{k rFkk HkkSxksfyd v{k ds e/; vYi dks.k dks fnDikr~ dks.k dgrs gSaA bls ls iznf'kZr djrs gSaA

2. Two charges +Q and –Q are kep tat (–x2, 0) and (x1, 0) respectively in the x – y plane. Find the

magnitude and direction of the net electric field at the origin (0, 0)

nks vkos'k +Q rFkk –Q Øe'k% (–x2, 0) rFkk (x1, 0) fcUnqvksa ij x – y ry esa fLFkr gSA ewy fcUnq (0, 0) ij dqy fo|qr {ks=k dk ifjek.k rFkk fn'kk crkvksaA

Ans. Total charge, q = Q + (–Q) = 0

They form an electric dipole.

21 EEE

ix

KQi

x

KQ21

22

3. Draw the shapes of the suitable Gaussian surfaces, while applying Gaus’s law to calculate the electric field due to :

(a) a uniformly charged long straight wire, (b) a uniformly charged infinite plane sheet.

fuEufyf[kr ds fy;s xkml ds fu;e }kjk fo|qr {ks=k Kkr djus gsrq mi;qDr xkmlh; i`"B dks iznf'kZr djks ,oa fo|qr {ks=k Kkr djks :

(a) ,dleku vkosf'kr lh/ks rkj ds fy;s (b) ,dleku vkosf'kr vuUr ifêdk ds fy;s





E ST INFORMA TIO


PPHHYYSSIICCSS


NO. B34 TO B35


Single choice Objective ('–1' negative marking) Q.1 (3 marks 2 min.) [03, 02] One or more than one options correct type ('–1' negative marking) Q.2 to Q.7 (4 marks 2 min.) [24, 12] Subjective Questions ('–1' negative marking) Q.8 to 9 (4 marks 5 min.) [08, 10] Match the Following (no negative marking) Q.10 (8 marks 6 min.) [08, 06]

ANSWER KEY OF DPP No. : B35 1. (C) 2. (A,B,D) 3. (A,B,C)4. (B,C) 5. (A,B,C) 6. (B) 7. (A) 8. 8 9. 2 10. (A) – p, r, t ; (B) – q, s ; (C) – q ; (D) – q, r

1. During an experiment with a metre bridge, the galvanometer shows a null point when the joceky is

pressed at 40.0 cm using a standard resistance of 90 , as shown in the figure. The least count of the scale used in the meter bridge is 1 mm. The unknown resistance is :

,d ehVj czht ls 90 ds ekud izfrjks/k ds lkFk ,d iz;ksx djrs le;] tc tkWdh dks rkj ds ck;sa fljs ls 40.0 cm ij nck;k tkrk gS] rc xSYouksehVj ij 'kwU; fo{ksi iznf'kZr gksrk gS] tSlk fp=k esa fn[kk;k x;k gSA ehVj czht esa iz;qä iSekus dk vYirekad(least count) 1 m.m. gSA vKkr izfrjks/k dk eku gS %

[JEE (Advanced)-2014, 3/60, –1]

(A) 60 ± 0.15 (B) 135 ± 0.56 (C*) 60 ± 0.25 (D) 135 ± 0.23 Ans. (C)

Sol. For balanced meter bridge lsrq larqyu ds fy;s

R

X =

)100(

40

X =

60

90 X = 60

X = R )100(

60

1.0

40

1.0

100X

X

X= 0.25

so vr% X = (60 + 0.25)




2. A source emit sound waves of frequency 1000 Hz. The source moves to the right with a speed of 32 m/s relative to ground. On the right a reflecting surface moves towards left with a speed of 64 m/s relative to ground. The speed of sound in air is 332 m/s :

,d L=kksr 1000 Hz vko`fÙk dh /ofu rjax mRiUu djrk gSA L=kksr /kjkry ds lkis{k 32 m/s dh pky ls nka;h vksj xfr djrk gSA nwljh rjQ ijkofrZr lrg /kjkry ds lkis{k 64 m/s dh pky ls cka;h vksj xfr djrh gSA gok esa /ofu dh pky 332 m/s gSA

(A*) wavelength of sound in ahead of source is 0.3 m (B*) number of waves arriving per second which meets the reflected surface is 1320 (C) speed of reflected wave is 268 m/s (D*) wavelength of reflected waves is nearly 0.2 m

(A*) L=kksr ds vkxs /ofu dh rjaxnS/;Z 0.3 m gSA (B*) çfr lSd.M igq¡pus okyh rjaxks dh la[;k tks ijkofrZr lrg ls feyrh gS] 1320 gSA (C) ijkofrZr rjax dh pky 268 m/s gSA (D*) ijkofrZr rjax dh rjaxnS/;Z yxHkx 0.2 m gSA

Sol. ' = f

VV s=

1000

32332 = 0.3 m

f' = s

0

VV

)VV(f

= 1000 × 32332

64332

= 1320 Hz

'' = 'f

VV 0= 0.2 m.

3. A real 1 mm long object is kept perpendicular to the principal axis of a concave mirror. The size of the image formed is 1/4 mm. Now the mirror starts moving, away from the object with a velocity 2 cm/s along the principal axis.

,d okLrfod oLrq ftldh yEckbZ 1 mm gS] dks vory niZ.k dh eq[; v{k ds yEcor~ j[kk tkrk gSA blls izkIr izfrfcEc dk vkdkj 1/4 mm gSA vc ;fn niZ.k dks oLrq ls nwj dh rjQ 2 cm/s ds osx ls eq[; v{k ds vuqfn'k pykrs gS %

(A*) the velocity of the image along the principle axis at the given instant will be 17

8 cm/s towards the

mirror.

fn;s x;s {k.k ij] eq[; v{k ds vuqfn'k izfrfcEc dk osx 17

8 cm/s, niZ.k dh vksj gksxkA

(B*) the length of the image will decrease as the mirror moves

niZ.k ds xfr djus ij izfrfcEc dh yEckbZ ?kVsxhA (C*) the speed of the image will always be more than 2 cm/s.

izfrfcEc dk osx lnSo 2 cm/s ls vf/kd gksxkA (D) the rate with which the length of the image will change is a constant

ftl nj ls izfrfcEc dh yEckbZ esa ifjorZu gksxk og nj fu;r gksxhA

Sol. Lateral magnification ik'oZ vko/kZu m = 1

4

With respect to mirror : niZ.k ds lkis{k

velocity of object = 2 cm/s towards left

oLrq dk osx = 2 cm/s cka;h vksj

velocity of image =

21

2cm s4

towards right

izfrfcEc dk osx =

21

2cm s4

nka;h vksj 2 1

16 8

With respect to the ground tehu ds lkis{k

Velocity of image = 1 cm

2 towardsright8 s

izfrfcEc dk osx = s

cm2

8

1

nka;h vksj = 17 cm

8 s towards right nka;h vksj




4. Let y > 0 be the region of space with a uniform and constant magnetic field ˆBk . A particle with charge q and mass m travels along the y-axis and enters in magnetic field at origin with speed v0. In region the

particle is subjected to an additional friction force F kv

. Assume that particle remains in region y > 0 at all times. The coordinates of the particle where it will finally stop are (x, y) then.

Ekkuk ,d le:i rFkk fu;r pqEcdh; {ks=k ˆBk , LFkku y > 0 esa mifLFkr gSA ,d vkosf'kr d.k ftldk vkos'k q rFkk nzO;eku m gS] y-v{k ds vuqfn'k pyrk gqvk v0 pky ls] ewy fcUnq ij pqEcdh; {ks=k esa izos'k djrk gSA bl {ks=k esa

d.k ij] ,d vfrfjDr ?k"kZ.k cy F kv

Hkh mifLFkr gSA ;g ekfu;s fd d.k iw.kZ le; esa y > 0 {ks=k esa gh mifLFkr gS rks ml fcUnq ds funsZ'kkad] ftl ij d.k vUr esa :dsxk ] (x, y) gS rks

(A) x = 0

2 2

kmv

k (qB) (B*) 0

2 2

qBmvx

k (qB)

(C*) 0

2 2

kmvy

k (qB)

(D) 0

2 2

qBmvy

k (qB)

Sol. F ma

= x yˆ ˆk(v i v j) + x y

ˆ ˆq(v i v j) × ˆBk

max = – kvx + qvyB may = – kvy – qvxB At t = 0, vx = 0 vy = v0 x = 0 y = 0 finally vx = 0 vy = 0 x = x1 y = y1 mxo = – kx1 + qy1B

–mv0 = – ky1 – qx1B x1 = 0

2 2

qBmv

k (qB) y1 = 0

2 2

kmv

k (qB)

COMPREHENSION vuqPNsn

A thin superconducting (zero resistance) ring is held above a vertical long solenoid, as shown in the figure. The axis of symmetry of the ring is same to that of the solenoid. The cylindrically symmetric magnetic field around the ring can be described approximately in terms of the vertical and radial

component of the magnetic field vector as Bz = B0(1 – z) and Br = B0r, where B0, and are positive constants, and z & r are vertical and radial position coordinates, respectively. Initially plane of the ring is horizontal, has no current flowing in it. When released, it starts to move downwards with its axis still vertical. Initial coordinates of the centre of the ring ‘O’ is z = 0 and r = 0.

In the given diagram point O is on the axis and slightly above the solenoid having vertical and radial position coordinates as (0, 0). Ring has mass m, radius r0 and self inductance L. Assume the acceleration due to gravity as g.

,d iryh vfr pkyd ('kwU; izfrjks/k) dh ,d oy; Å/okZ/kj yEch ifjukfydk ds Åij fp=kkuqlkj j[kh gqbZ gSA oy; dh lefer v{k ifjukfydk dh v{k ds leku gSA oy; ds pkjksa vksj csyukdkj lefer pqEcdh; {ks=k dks pqEcdh; {ks=k lfn'k ds Å/okZ/kj rFkk f=kT;h; ?kVd ds inksa esa Bz = B0(1 – z) rFkk Br = B0r ds vuqlkj n'kkZ;k tk ldrk gS, tgk¡ B0, rFkk /kukRed fu;rkad gS rFkk z o r Øe'k% Å/okZ/kj rFkk f=kT;h; fLFkfr;ksa ds funsZ'kkad gSaA izkjEHk esa oy; dk ry {kSfrt gS rFkk blesa dksbZ /kkjk izokfgr ugha gSA tc bls NksM+k tkrk gSa rks ;g bldh Å/okZ/kj v{k ds vuqfn'k uhps dh vksj xfr izkjEHk djrh gSA oy; ds dsUnz ‘O’ dk izkjfEHkd funsZ'kkad z = 0 rFkk r = 0 gSA

fn;s x;s fp=k esa fcUnq O v{k ij fLFkr gSa rFkk ;g ifjukfydk ds FkksM+k lk Åij gSA ‘O’ ds Å/okZ/kj rFkk f=kT;h; fLFkfr funsZ'kkad (0, 0) gS] oy; dk nzO;eku m, f=kT;k r0 rFkk LoizsjdRo L gSA xq:Roh; Roj.k g ekusaA

z

B BO

g




5. When the ring is at a vertical position z.

tc oy; ,d Å/okZ/kj fLFkfr z ij gSA

(A*) Magnetic flux 200 rB

(B*) The current in the ring is zrBL

1 200

(C*) Magnitude of Lorentz force on the ring is 2 2 40 02B r z

L

(D) Magnitude of net force on the ring is 2 2 40 02B r z

L

(A*) pqEcdh; ¶yDl 200 rB gSA

(B*) oy; esa /kkjk zrBL

1 200 gSA

(C*) oy; ij yksjsUt cy dk ifjek.k 2 2 40 02B r z

L

gSA

(D) oy; ij dqy cy dk ifjek.k 2 2 40 02B r z

L

gSA

6. Find the vertical coordinates z for equilibrium position of the ring.

oy; dh lkE;koLFkk dh fLFkfr ds fy, Å/okZ/kj funsZ'kkad z Kkr dhft,

(A) 40

220 rB2

mgL

(B*)

40

220 rB2

mgL

(C) zero 'kwU; (D) None of these buesa ls dksbZ ugh

7. Find the time period of SHM (for small displacement along z–axis) of the ring.

oy; ds ljy vkorZ xfr (z-v{k ds vuqfn'k vYi foLFkkiu ds fy,) dk vkorZdky Kkr dhft,A

(A*) mL2

rB

1200

(B) mL2

rB2

1200

(C) 3

mL

rB

1200

(D) ring will not perform SHM oy; SHM ugha djrh

Sol. 5-7.

Total magnetic flux at any position LrB 20z

Since, R = 0, so Lr)z1(B 200 = constant

From initial condition (z = 0, I = 0), the value of constant is 200 rB

Using the above equation the current in the ring I = zrBL

1 200

The lorentz force acting on the ring (which can only be vertical, because of the symmetry of the assembly) can be expressed as

Fz = –Br(z)2r0 = L

zrB2 40

220 = – kz

Equation of motion of the ring is maz = Fz – mg = –kz – mg Equilibrium position z0 = –mg/k

m

k0




Sol. 5-7.

fdlh fLFkfr ij dqy pqEcdh; ¶yDl LrB 20z

pwafd , R = 0, vr% Lr)z1(B 200 = fu;rkad

izkjfEHkd 'krZ ls (z = 0, I = 0), fu;rkad dk eku 200 rB

oy; esa /kkjk dh mijksDr lehdj.k ds mi;ksx ls I = zrBL

1 200

oy; ij dk;Zjr yksjsat cy (tks leferh ds dkj.k dsoy Å/okZ/kj gks ldrk gS) dks fuEu izdkj iznf'kZr dj ldrs gSa

Fz = –Br(z)2r0 = L

zrB2 40

220 = – kz

oy; dh xfr ds lehdj.k ls maz = Fz – mg = –kz – mg

lkE;koLFkk fLFkfr z0 = –mg/k

m

k0

8. Durring Searle’s experiment, zero of the Vernier scale lies between 3.20 × 10–2 m and 3.25 × 10–2 m of

the main scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 × 10–2 m and 3.25 × 10–2 m of the main scale but now the 45th division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is 2m. and its cross-sectional area is 8 × 10–7 m2. The least count of the Vernier scale is 1.0 × 10–5 m. The maximum percentage error in the Young’s modulus of the wire is

lyZ ds iz;ksx esas ofuZ;j iSekus dk 'kwU; eq[; iSekus ij 3.20 × 10–2 m rFkk 3.25 × 10–2 m ds chp gSA ofuZ;j iSekus dk chlok¡ Hkkx (20th division) eq[; iSekus ds fdlh ,d Hkkx ds fcydqy lh/k esa gSA rkj ij 2 kg dk vfrfjDr Hkkj yxkus ij] ;g ns[kk x;k fd ofuZ;j iSekus dk 'kwU; vHkh Hkh eq[; iSekus ij 3.20 × 10–2 m rFkk 3.25 × 10–2 m ds chp gS] ijUrq vc ofuZ;j iSekus dk iSarkfylok¡ Hkkx (45th division) eq[; iSekus ds fdlh vU; Hkkx ds fcydqy lh/k esa gSA /kkrq ds irys rkj dh yEckbZ 2m rFkk bldk vuqizLFk dkV dk {ks=kQy 8 × 10–7 m2 gSA iSekus dk vYirekad (least count)1.0 × 10–5 m gSA Rkkj ds ;ax izR;kLFkrk xq.kkad (Young’s modulus) esa vf/kdre izfr'kr =kqfV gSA

Elasticity Ans. 8

Sol. Observation izs{k.k – 1

Let weight used is W1, extension 1

ekuk mi;ksx fy;k x;k Hkkj W1 gS] izlkj 1

y = L/

A/W

1

1

W1 = L

yA 1 1 = 3.2×10–2 + 20×10–5

Observation izs{k.k – 2

Let weight used is W2 extension 2

ekuk mi;ksx fy;k x;k Hkkj W2 gS] izlkj 2

y = L/

A/W

2

2

W1 = L

yA 2

1 = 3.2×10–2 + 45×10–5

W2–W1 = )(L

yA12 y =

)(yA

L/WW

12

12

max

y

y

=

12

12

= 5

5

1025

102

max

y

y

× 100% =

25

2× 100% = 8%




9. A particle P is moving on a circle under the action of only one force, which always acts towards a fixed

point O lying on the circumference. Find ratio of 2

2

dt

d to

2

dt

d

at the moment when = 45º. (C is

centre of circle)

,d d.k P lnSo dsoy ,d cy ds vUrxZr xfr dj jgk gS ;g cy lnSo ifjf/k ij fLFkr fLFkj fcUnq O dh rjQ

dk;Zjr~ gksrk gSA 2

2

dt

d dk

2

dt

d

ds lkFk ml {k.k ij vuqikr Kkr dhft, tc = 45º gks tkrk gSA (C o`Ùk dk

dsUnz gS)

F

P

OC

Ans. 2

Sol. at = m

Fsin

2

2

dt

)2(Rd =

m

Fsin

2

2

dt

d =

mR2

sinF ...(1)

ac = m

Fcos

Rm

F)2(

dt

d2

cos ...(ii)

2

2

2

dt

d

dt

d

= 2 tan = 2

10. To create uniform electric field, a capacitor containing two infinitely large plates are used. Particles in

column-I are projected horizontally from the middle with same kinetic energy. Neglect force acting between the particle and also neglect gravity. Match the column.

le fo|qr {ks=k mRiUu djus ds fy, nks vuUr {ks=kQy dh cM+h IysVksa ls cus la/kkfj=kksa dk iz;ksx djrs gSA LrEHk-I esa fn;s x;s d.kksa dks IysVksa ds e/; ls leku xfrt ÅtkZ ds lkFk {ksfrt fn'kk esa iz{ksfir fd;k x;k gSA d.kksa ds e/; vkil esa yxus okys cy vkSj xq:Roh; cyksa dks ux.; ekfu;sA LrEHkksa dks lqesfyr dhft,A




Column- Column- (A) -particle (p) Particles which will move along path–(1) (B) 23Na+1 (q) Particles which will move along path–(2) (C) 6Li+1 (r) Particles which will strike the negative plate in

minimum time (D) 2D+1 (s) Particles which will strike the negative plate in

maximum time (t) Particles which will strike the plate with maximum kinetic energy

dkWye- dkWye- (A) –d.k (p) d.k iFk–(1) ds vuqfn'k xfr djsaxsA

(B) 23Na+1 (q) d.k iFk–(2) ds vuqfn'k xfr djsaxsA (C) 6Li+1 (r) d.k tks U;wure le; esa _.kkRed IysV ls Vdjk,xkA (D) 2D+1 (s) d.k tks vf/kdre le; esa _.kkRed IysV ls Vdjk,xkA (t) d.k tks IysV ij vf/kdre xfrt ÅtkZ ls Vdjk,xkA Ans. (A) – p, r, t ; (B) – q, s ; (C) – q ; (D) – q, r

Sol. x = ut, y = 2tm

qE

2

1

y =

2

u

x

m

qE

2

1

KE = same for all lHkh ds fy, leku y q

So particle with greater q, will have greater y.

vr% d.k ftldk q vf/kd gS] dk y vf/kd gksxkA

So -particles will move along path 1 to strike the plate.

vr% -d.k ikFk 1 ds vuqfn'k xfr djrs gq, IysV ls Vdjk,xkA (for R,S) (R,S ds fy,)

2tm

qE

2

1

2

d

t =

q

m

E

d

Particle having more q

m will strike later.

d.k ftudk q

m vf/kd gS] nsjh ls Vdjk,xkA


BOARD PROBLEMS1. Two concentric circular coils, one of small radius r1 and the other of large radius r2, such that r1 <<n r2

are placed coaxially with centres coinciding. Obtain the mutual inductance of the arrangement.

nks ldsUæh; dq.Mfy;ksa esa ls ,d NksVh dq.Myh dh f=kT;k r1 rFkk vU; cM+h dq.Myh dh f=kT;k r2 bl izdkj gS fd r1 <<n r2 gSA ;g lek{kh; :i ls bl izdkj fLFkr gS fd buds dsUæ lEikrh gSaA bl O;oLFkk ds fy;s vU;ksU; izsj.k Kkr djksA

2. A bar magnet falls from a height ‘h’ through a metal ring. Will its acceleration be equal to g ? Give reason for your answer.

,d /kkrq dh oy; ij ‘h’ Å¡pkbZ ls NM+ pqEcd dks NksM+k tkrk gS D;k bldk Roj.k g ds cjkcj gksxk ? viuk mÙkj dkj.k lfgr le>kb;sA

3. If the number of turns in the solenoid is doubled, keeping other factors constant, how does the self-

inductance of the coil change ?

;fn ifjukfydk esa vU; izkapyksa dks fu;r j[krs gq, ;fn blds ?ksjksa dh la[;k nksxquh dj nh tk;s rks bldk Loiszj.k

fdl izdkj ifjofrZr gksxk?

4. Which physical quantity has the unit Wb/m2. Is it a scalar or a vector quantity ?

fdl HkkSfrd jkf'k dk ek=kd Wb/m2 gSA ;g jkf'k lfn'k gS ;k vfn'kA





E ST INFORMA TIO


PPHHYYSSIICCSS


NO. B36 TO B37

2. DPP Syllabus :


One or more than one options correct type ('–1' negative marking) Q.1 to Q.5 (4 marks 2 min.) [20, 10] Subjective Questions ('–1' negative marking) Q.6 to Q.7 (4 marks 5 min.) [08, 10] Match the Following (no negative marking) Q.8 (8 marks 6 min.) [08, 06] Assertion and Reason (no negative marking) Q.9 (3 marks 2 min.) [03, 02]


1. (A,B,C) 2. (A,D) 3. (A,D) 4. (B,C) 5. (B,C) 6. v = a/g

7. m = 300 gm. 8. (A) p,s (B) q,s (C) p,s (D) s 9. (D) 1. Wavelength and intensity of monocromatic light falling on a photoelectric cell as a function of time is

given by = 10000 e–t/10 Å and = 1000 (1 – e–t/10). The threshold wave length of metal used in cell is 3700 Å.

(A*) Work function of metal plate is equal to 3.35 eV. (B*) At t = 0 sec photo current is zero. (C*) At t = 5 sec photo current is zero. (D) At t = 20 sec photo current is zero. izdk'k oS|qr lsy ij vkifrr ,d o.khZ; izdk'k dk rjaxnS/;Z ,oa rhozrk le; ds Qyu esa = 10000 e–t/10 Å rFkk

= 1000 (1 – e–t/10) }kjk fn;k x;k gSA lsy esa iz;qDr /kkrq dk nSgyh rjaxnS/;Z 3700 Å gSA (A*) /kkfRod IysV dk dk;Z Qyu 3.35 eV gSA (B*) t = 0 sec ij izdk'k /kkjk 'kwU; gSA

(C*) t = 5 sec ij izdk'k /kkjk 'kwU; gSA (D) t = 20 sec ij izdk'k /kkjk 'kwU; gSA

2. The electron in a hydrogen atom makes a transition n1 n2 where n1 and n2 are the principal quantum numbers of two states. Assume the Bohr model to be valid. The time period of the electron in the initial state is eight times that is in the final stage. The possible values of n1 & n2 are

gkbMªkstu ijek.kq esa ,d bySDVªkWu n1 n2 laØe.k djrk gS] tgk¡ n1 rFkk n2 nks voLFkkvksa dh eq[; Dok.Ve la[;k gSA ekfu; fd ckSgj ijek.kq izfr:i ekU; gSA izkjfEHkd voLFkk esa bySDVªkWu dk vkoZr dky vfUre voLFkk esa vkorZ dky dk 8 xquk gSA n1 o n2 ds lEHko eku gksaxs &

(A*) n1 = 4, n2 = 2 (B) n1 = 8, n2 = 2 (C) n1 = 8, n2 = 1 (D*) n1 = 6, n2 = 3

Sol. T2 r3 & r n2 so vr% r3 T2 3i3f

r

r =

2i2f

T

T = 82 i

f

r

r = 4

i

f

r

r =

2i2f

n

n = 4 i

f

n

n = 2




3. 'Identical dielectric slabs are inserted into two identical capacitors A and B. These capacitors and a battery are connected as shown in figure. Now the slab of capacitor B is pulled out with battery remaining connected :

,dleku ijkoS|qr ifêdk] nks ,dleku la/kkfj=kksa A rFkk B esa fLFkr gSA ;s la/kkfj=k fp=kkuqlkj cSVjh ls tqM+s gq, gSaA vc la/kkfj=k B ls ijkoS|qr ifêdk dks ckgj [khapk tkrk gS ,oa cSVjh tqM+h gqbZ jgrh gSA

a b+ –

A B

(A*) During the process positive charge flows from a to b. (B) Finally charge on capacitor B will be less than that on capacitor A. (C) During the process, a work is done by the external force F, which completely appears as heat in the

circuit. (D*) During the process, internal energy of the battery increases. (A*) bl izØe ds nkSjku /kukRed vkos'k a ls b dh rjQ izokfgr gksxkA (B) la/kkfj=k B ij vafre vkos'k la/kkfj=k A ds vkos'k ls de gksxkA (C) izØe ds nkSjku cká cy F }kjk dk;Z fd;k tkrk gS tks fd ifjiFk esa Å"ek ds :i esa mRiUu gksrk gSA (D*) bl izØe ds nkSjku cSVjh dh vkarfjd ÅtkZ c<+rh gSA Sol. (A) Charge on capacitor B decreases as dielectric slab is taken out. Charge from positive plate of B

flows towards battery. Charge on A and B can not be different, as being connected in series. During the process, battery is being charged. COMPREHENSION

vuqPNsn

There are two blocks A and B placed on a smooth surface. Block A has mass 10 kg and it is moving with velocity 0.8 m/s towards stationary B of unknown mass. At the time of collision, their velocities are given by the following graph :

;gk¡ fpduh lrg ij j[ks nks CykWd A o B gSA CykWd A dk nzO;eku 10 kg rFkk ;g 0.8 m/s ds osx ls vKkr nzO;eku ds fLFkj CykWd B dh vksj xfr djrk gSA VDdj ds le; muds osx fuEu vkjs[k }kjk fn, x, gSA

0.8

1

4. Select correct statements : lgh dFku@dFkuksa dks pqfu;s % (A) Impulse of deformation is 1.5 Ns (B*) Coefficient of restitution of the collision is 1 (C*) Impulse of Reformation is 3 Ns (D) mB = mA (A) fo:i.k dk vkosx 1.5 Ns gSA

(B*) VDdj dk–izR;koLFkku xq.kkad 1 gSA

(C*) la:i.k dk vkosx 3 Ns gSA (D) mB = mA




5. Choose the coorect option(s) : (A) Maximum deformation potential energy is 1.6J (B*) Maximum deformation potential energy is 1.2J

(C*) Minimum kinetic energy of the system is 2J (D) Minimum kinetic energy of the system is 1.8J

lgh dFkuksa dk pquko dfj;s : (A) vf/kdre fo:i.k fLFkfrt ÅtkZ 1.6J gSA (B*) vf/kdre fo:i.k fLFkfrt ÅtkZ 1.2J gSA

(C*) fudk; dh U;wure xfrt ÅtkZ 2J gSA (D) fudk; dh U;wure xfrt ÅtkZ 1.8J gSA

Sol. (4 to 5) mA × 0.8 = mA × 0.2 + mB × 1.0 mA × 0.6 mB × 1.0 mB = 0.6 mA

e = 8.0

2.0–1= 1 = 1.5

d = 6 × 0.5 – 6 × 0 = 3N – 5 = 10 × {0.8 – 0.5} = 10 × 0.3 = 3 NS

U = 2

1 × 10 × (0.8)2 –

2

1× 16 × (0.5)2 = 5 × 0.64 8 × 0.25 = 3.2 – 2.0 = 1.2 J

6. A car moves uniformly along a horizontal sine curve y = a sin (x/), where a and are certain constants. The coefficient of friction between the wheels and the road is equal to . At what maximum velocity will the car ride without sliding?

,d dkj {kSfrt T;koØ y = a sin (x/) ds vuqfn'k ,dleku :i ls xfr djrh gS, tgk¡ a rFkk ,d fuf'pr fu;rkad gSA dkj ds ifg;ksa rFkk lM+d ds e/; ?k"kZ.k xq.kkad .ds cjkcj gSA dkj fcuk fQlys fdrus osx ls xfr dj ldrh gS ?

[Ans : v = a/g

7. A wooden cube (density 0.5 gm/cc) of side 10 cm is floating in water kept in a cylindrical beaker of base

area 1500 cm2. When a mass m is kept on the wooden block the level of water rises in the beaker by 2mm. Find the mass m.

,d ydM+h dk ?ku (?kuRo 0.5 gm/cc) ftldh çR;sd Hkqtk 10 cm gS] csyukdkj chdj esa Hkjs ikuh esa rSj jgk gSA chdj dk vk/kkj {ks=kQy 1500 cm2 gSA tc nzO;eku m dks ydM+h ds ?ku ij j[kk tkrk gS rks chdj esa ikuh dk Lrj 2mm ls c<+ tkrk gSA nzO;eku m Kkr dhft,A

Sol. Let the cube dips further by y cm and water level rises by 2 mm.

Then equating the volumes (/// volume = \\\ volume in figure) volume of water raised = volume of extra depth of wood

100 y = (1500 – 100) 10

2 = 1400 ×

10

2 = 280

y = 2.8 cm Extra upthrust water × (2.8 + 0.2) × 100 g = mg m = 300 gm. m = 300 gm. ....Ans.




8. Match the statements in Column with the results in Column LrEHk esa fn;s x;s dFkuksa dks LrEHk esa fn;s x;s ifj.kkeksa ls lqesfyr dhft;s Column – I Column – II

(A) A variable resistor is connected across a non-ideal cell . (p) first increases for As the resistance of the variable resistor is continuously some time and then increased from zero to a very large value, the electric decreases. power consumed by the variable resistor (B) A circular ring lies in space having uniform and constant (q) first decreases for

magnetic field. Initially the direction of magnetic field is some time and then parallel to plane of the ring. Keeping the centre of ring fixed, increases. the ring is rotated by 1800 about one of its diameter with constant angular speed. For the duration the ring rotates, the magnitude of induced emf in the ring (C) A thin rod of length 1 cm lies along principal axis of a (r) is always constant convex lens of focal length 5 cm. One end of rod is at a distance 10 cm from optical centre of the lens. The convex lens is moved (without rotation) perpendicular to initial principal axis by 5 mm and brought back to its initial position.The length of the image of the rod (D) A bulb (of negligible inductance) and a capacitor in series (s) increases or may are connected across an ideal ac source of constant peak increase over voltage and variable frequency. As frequency of ac source is some time interval continuously increased, the brightness of bulb LrEHk – I LrEHk – II

(A) ,d ifjorhZ izfrjks/k dks ,d vukn'kZ lsy ds fljksa ij tksM+k (p) igys dqN le; ds fy,

tkrk gSA tSls tSls ifjorhZ izfrjks/k dk izfrjks/k eku lrr~ c<+rh@c<+rk gS rFkk fQj :i ls 'kwU; ls ,d cgqr cM+s eku rd c<+k;k tkrk gS] rks ?kVrh@?kVrk gSA ifjorhZ izfrjks/k }kjk miHkksx dh xbZ fo|qr 'kfä

(B) ,d o`Ùkkdkj oy; ,d le:i rFkk fu;r pqEcdh; {ks=k okys (q) igys dqN le; ds fy,

LFkku esa j[kh gqbZ gSA izkjEHk esa pqEcdh; {ks=k dh fn'kk oy; ?kVrh@?kVrk gS rFkk ds ry ds lekUrj gSA oy; ds dsUnz dks fLFkj j[krs gq;s] fQj c<+rh@c<+rk gSA oy; dks vius ,d O;kl ds ifjr% fu;r dks.kh; pky ls 180° ds dks.k ls ?kqek;k tkrk gSA oy; ds ?kweus ds nkSjku] oy; esa izsfjr fo-ok- cy dk ifjek.k (C) 1 cm yEckbZ dh ,d iryh NM+ 5 cm Qksdl nwjh ds mRry (r) ges'kk fu;r gSA ySUl dh eq[; v{k ds vuqfn'k fLFkr gSA NM+ dk ,d fljk ysUl ds izdkf'kd dsUnz ls 10 cm dh nwjh ij gSA mRry ysUl ds izkjfEHkd eq[; v{k ds yEcor~ 5 mm }kjk ¼fcuk ?kw.kZu ds½ xfr djk;h tkrh gS rFkk okil bldh izkjfEHkd fLFkfr esa yk;k tkrk gSA NM++ ds izfrfcEc dh yEckbZ

(D) (ux.; izsjdRo dk) ,d cYc rFkk ,d la/kkfj=k fu;r f'k[kj (s) c<+rh@c<+rk gS ;k dqN

oksYVst rFkk ifjorhZ vko`fÙk ds ,d vkn'kZ izR;korhZ lzksr ds le; vUrjky ds fy,

fljksa ij Js.khØe esa tksM+s tkrs gSA tSls tSls izR;korhZ lzksr dh c<+ ldrh@ldrk gSA vko`fÙk lrr~ :i ls c<+k;h tkrh gS] rks cYc dh ped Ans. (A) p,s (B) q,s (C) p,s (D) s




Sol. (A) The plot of power P consumed by variable resistor of resistance R w.r.t. R when connected across a non-ideal cell is given by

(B) At the instant ring has rotated by q = wt (w = angular speed) in uniform and constant magnetic field

B, the magnetic flux through the ring is f = BA sinwt

or emf = dt

d = Baw coswt

Hence for first half cycle, magnitude of emf first decreases and then increases. (C) As the lens is moved up, the projection of length of image on principal axis remains same, but its

slant with principal axis increases. This increase is due to the fact that lateral magnification of each point of image of rod is different. Hence as lens moves up the size of image increases and when it moves down the size of image decreases.

(D) As frequency of ac source increases, reactance of capacitor decreases. Therefore the bulb glows more brighter.

gy% (A) R izfrjks/k ds ifjorhZ izfrjks/k R }kjk mi;ksx dh xbZ 'kfä P dk R ds lkis{k xzkQ fp=kkuqlkj gksxk tc ;g vukn'kZ lsy ds fljksa ij tksM+k tkrk gSA

(B) ml {k.k] tc oy; le:i rFkk fu;r pqEcdh; {ks=k B esa q = wt (w = dks.kh; pky) ls ?kwe tkrh gS] oy; ls

tkus okyk pqEcdh; ¶yDl gSA f = BA sinwt

;k emf = dt

d = Baw coswt

blfy;s igys v/kZpØ ds fy,] fo-ok- cy dk ifjek.k igys ?kVrk gS rFkk fQj c<+rk gSA (C) tc ysUl Åij dh vksj mBrk gS izfrfcEc dh yEckbZ dk eq[; v{k ij iz{ksi leku jgrk gS] ysfdu eq[; v{k ls

bldk >qdko c<+rk gSA o`f) bl rF; ds dkj.k gS fd NM+ ds izR;sd fcUnq dk ik'ohZ; vko/kZu fHkUu gSA blfy;s tc ysUl Åij mBrk gS rks izfrfcEc dk vkdkj c<+rk gS vkSj tc ;g uhps vkrk gS rks izfrfcEc dk

vkdkj ?kVrk gSA (D) tc ac lzksr dh vko`fÙk c<+rh gS] la/kkfj=k dk izfr?kkr ?kVrk gSA blfy;s cYc vf/kd ped ls tyrk gSA 9. There exist uniform time varing magnetic field in cylindrical region of radius R as shown in figure R f=kT;k ds csyukdkj {ks=k esa le; ds lkFk ifjorhZ le:i pqEcdh; {ks=k iznf'kZr gS

x

x x x

x x x

x x x

A

B

Statement – 1 : Work done by induced electric field in moving a point charge from A to B is different for

two paths Shown Statement – 2 : Induced electric field is a non conservative electric field oDrO; – 1 : nks fHkUu fHkUu iFkksa }kjk ,d vkosf'kr d.k dks A ls B rd foLFkkfir djus esa izsfjr fo|qr {ks=k }kjk fd;s

x;s dk;Z fHkUu&fHkUu gksxk oDrO; – 2 : izsfjr fo|qr {ks=k ,d vlaj{kh fo|qr {ks=k gSA




(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True;Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D*) Statement-1 is False, Statement-2 is True. (A) oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA (B) oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k ugha gSA (C) oDrO;-1 lR; gS, oDrO;-2 vlR; gS, (D*) oDrO;-1 vlR; gS,, oDrO;-2 lR; gS Solution : By faraday law

x

x x x

x x x

x x x

A

B

O

– d.E = –

dt

d

q d.E = q

dt

d

d.F = q

dt

d

WAB + WAO + WOB = q dt

d

WAO = WOB = 0 (Since electrici field and d are perpendicular)

WAB = q dt

d

is same for both closed path.


BOARD PROBLEMS

1. (a) Show, giving a suitable diagram, how unpolarized light can be polarized by reflection. (b) Two polaroids P1 and P2 are placed with their pass axes perpendicular to each other. Unpolarised

light of intensity 0 is incident on P1. A third polaroid P3 is kept in between P1 and P2 such that its pass axis makes an angle of 60° with that of P1. Determine the intensity of light transmitted through P1, P2 and P3. (a) mi;qä vkjs[k nsrs gq,] ;g n'kkZb, fd ijkorZu }kjk v/kzqfor izdk'k dks fdl izdkj /kqfor fd;k tk ldrk gSA

(b) nks iksysjkWbMksa P1 rFkk P2 dks bl izdkj j[kk x;k gS fd buds ikfjr&v{k ,d&nwljs ds yEcor~ gksaA rhozrk I0 dk v/kqfor izdk'k P1 ij vkifr gSA fdlh rhljs iksysjkWbM P3 dks P1 vkSj P2 ds chp bl izdkj j[kk x;k gS fd bllss ikfjr&v{k P1 ds ikfjr&v{k ls 60° dks.k cukb,A P1, P2 vkSj P3 ls xqtjus okys izdk'k dh rhozrk fu/kkZfjr dhft,A




Sol. (a)

Incident

Medium

Refracted

Reflected

vkifrr

ek/;e

viofrZSr

ijkofrZr

In the incident light and refracted light both dots and arrows imply both polarization are present. The

oscillating electrons in the water produce the reflected wave. Consider the case when reflected and refracted rays are perpendicular to each other In that case reflected light is linearly polarized perpendicular to plane of figure

vkifrr izdk'k rFkk viofrZr izdk'k nksuksa esa fcUnq rFkk rhj dk vFkZ ;g gS fd nksuks /kqoz.k mifLFkr gSA ikuh esa nksfyr bysDVªkWu ijkofrZr

rjax mRiUUk djrs gSA ml fLFkfr ij fopkj dhft, tc ijkofrZr rFkk viofrZr fdj.k ,d nwljs ds yEcor~ gksrs gSA ml fLFkfr esa ijkofrZr

izdk'k fp=k ds yEcor~ js[kh; :i ls /kqzfor gksrk gSA

Thus reflected light can be polarized completely with E to plane of incidence, when reflected and

refracted are to each other

vr% vkiru ry ijkofrZr izdk'k ds yEcor~ E ds lkFk iw.kZ :i ls /kqzfor gks ldrk gSA tc ijkofrZr rFkk viofrZr izdk'k ,d nwljs ds

yEcor~ gksrs gS

= rSin

iSin B = B

B

icos

isin = tan iB

(b) I through P1 = I0 /2 (P1 ls xqtjus ij I = I0 /2)

I through P2 = 20I

cos2 60° P2 ls xqtjus ij I = 20I

cos2 60°

P1 P2

P3

= 80I

I through P3 = 80I

cos230°

P3 ls xqtjus ij I = 80I

cos230° = 32

3 0I

2. Explain with the help of diagram the terms magnetic declination at a given place.

,d fn;s x;s LFkku ij pqEcdh; >qdko dks mi;qDr fp=k }kjk le>kb;sA




3. A beam of -particles, projected along (+) X-axis, experiences a force due to a magnetic field along the

(+) Y axis.

(a)What is the direction of the magnetic field?

-d.kksa dk ,d iqat /kukRed (+) X-v{k ds vuqfn'k iz{ksfir fd;k tkrk gS, ;g (+) Y fn'kk esa] pqEcdh; {ks=k ds dkj.k

cy vuqHko djrk gS?

(a) pqEcdh; {ks=k dh fn'kk D;k gksxh?

(b) In place of -particle, if the beam is of electrons; then what will be the direction of the magnetic

field?

(b) ;fn -d.kksa ds LFkku ij bysDVªkWuksa dk iqat iz{ksfir fd;k tk;s rks pqEcdh; {ks=k dh fn'kk D;k gksxh? x

y

z

v





E ST INFORMA TIO


PPHHYYSSIICCSS


NO. B36 TO B37

DPP No. : B37 (JEE-MAIN) Total Marks : 58 Max. Time : 41 min. Single choice Objective ('–1' negative marking) Q.1 to Q.18 (3 marks 2 min.) [54, 36] Subjective Questions ('–1' negative marking) Q.19 (4 marks 5 min.) [04, 05]

ANSWER KEY OF DPP No. : B37 1. (A) 2. (D) 3. (D) 4. (D) 5. (C) 6. (C) 7. (C) 8. (A) 9. (A) 10. (B) 11. (A) 12. (C) 13. (C) 14. (B)

15. (C) 16. (A) 17. (A) 18. (B) 19. U = 2GM

4R

1. Figure shows an arrangement of three point charges. The total potential energy of this arrangement is

zero. The ratio q

Q is :

iznf'kZr fp=k esa rhu fcUnqor vkos'k n'kkZ;s x;s gSa bl fudk; dh lEiw.kZ fLFkfrt ÅtkZ 'kwU; gks rks q

Q dk eku gksxk:

-Q+q+q

2r

r

(A*) 4 : 1 (B) 2 : 1 (C) 1 : 1 (D) 2 : 1

Sol. (A)

Usys = 0

1

4qQ ( q)( q) Q( q)

r 2r r

= 0

– Q + q

2 – Q = 0 or 2Q =

q

2 or

q

Q =

4

1.

2. Four point charges +q, +q, –q and –q are placed at the corners of a square having side 2 a as shown in figure.

pkj fcUnqor vkos'k +q, +q, –q rFkk –q ,d oxZ ds dksuksa ij fLFkr gSa ,oa oxZ dh Hkqtk fp=kkuqlkj 2 a gSA

–q

+q

–q

+q• •

••

•O

2 a

Electric potential at point O is fcUnq O ij fo|qr foHko gksxk

(A) Kq

a (B)

2Kq

a (C)

4 2 Kq

a (D*) zero 'kwU;

Sol. (D)




3. Which one of the following equation is correct for the circuit shown. The currents in each branch of circuit are indicated and all three cells are ideal.

fn[kk;s x;s ifjiFk ds fy, fuEu lehdj.kksa esa ls dkSu lk ,d lgh gSA ifjiFk dh izR;sd 'kk[kk esa /kkjk iznf'kZr dh xbZ gS rFkk lHkh rhu lsy vkn'kZ gSaA

(A) 2 – I1 – 2I2 = 0 (B) 2 – 2I1 – 2I2 – 4I3 = 0 (C) 4 – I1 + 4I3 = 0 (D*) –2 – I1 – 2I2 = 0 Sol. Applying Kirchoff voltage law in left loop. cka;s ywi esa fdjpkQ oksYVst fu;e yxkus ij – 2 + I1 + 4 + 2I2 = 0 \ 2 + I1 + 2I2 = 0 4. Suppose a smooth tunnel is dug along a straight line joining two points on the surface of the earth and

a particle is dropped from rest at its one end. Assume that mass of earth is uniformly distributed over its Volume. Then

i`Foh ds ,d fljs ls nwljs fljs rd ,d ?k"kZ.k jfgr lh/kh lwjax [kksnh tkrh gSA ,d d.k lwjax ds ,d fljs ls fojkekoLFkk ls NksM+k tkrk gsA ekuk iFoh dk nzO;eku ,d leku :i ls blds vk;ru esa forfjr gSA rc

(A) the particle will emerge from the other end with velocity e

e

GM

2Rwhere Me and Re are earth’s mass

and radius respectively, (B) the particle will come to rest at centre of the tunnel because at this position, particle is closest

to earth centre. (C) potential energy of the particle will be equal to zero at centre of tunnel if it is along a diametre. (D*) acceleration of the particle will be proportional to its distance from midpoint of the tunnel.

(A) d.k nwljs fljs ls e

e

GM

2R osx ls lwjax ls ckgj fudyrk gSA tgk¡ M

e rFkk R

e Øe'k% i`Foh dk nzO;eku rFkk

f=kT;k gSA (B) d.k lwjax ds e/; fcUnq ij fojke esa gksxk] D;ksafd bl fLFkfr ij d.k i`Foh ds dsUnz ds fudV gSA (C) lwjax ds e/; fcUnq ij fLFkfrt ÅtkZ 'kwU; gksxh] ;fn d.k O;kl ds vuqfn'k gksA (D*)d.k dk Roj.k lwjax ds e/; fcUnq ls bldh nwjh ds lekuqikrh (proportional) gksxkA 5. A bead of mass m is located on a parabolic wire with its axis vertical and vertex at the origin as shown

in figure and whose equation is x2 = 8ay. The wire frame is fixed and the bead can slide on it without friction. The bead is released from the point y = 2a on the wire frame from rest. Then the magnitude of acceleration of the bead at y = 2a (that is, just after the release of bead is) :

fp=kkuqlkj ijoy; dh lehdj.k x2 = 8ay gSA bldh v{k Å/okZ/kj rFkk 'kh"kZ ewy fcUnq ij gS] bl ijoy; rkj ij m nzO;eku dk eudk fLFkr gSA rkj dk Ýse fLFkj gS rFkk eudk ¼eksrh½ fcuk ?k"kZ.k ds ijoy; ij fQly ldrk gSA rkj Ýse ij y = 2a fcUnq ls eudk fLFkjkoLFkk ls NksM+k tkrk gSA tc eudk y = 2a fLFkfr ij gS rks blds Roj.k dk ifjek.k gksxkA ¼vFkkZr~ euds dks NksM+us ds Bhd i'pkr~½ :

(A) g

2 (B)

3g

2 (C*)

g

2 (D)

g

5




Sol.

x2 = 8ay Differentiating w.r.t. y, we get y ds lkis{k vodyu

dy

dx =

x

4a

At (4a, 2a), = 1 hence vr% = 45°

the component of weight along tangential direction is mg sin . Hkkj dk Li'kZ js[kh; fn'kk esa ?kVd mg sin . 6. A particle starts moving from rest along a straight line at t = 0. Its acceleration-time graph is shown.

Correct relation between magnitudes of displacements between time durations t = 2s to t = 3s and t = 8s to t = 9s represented by S23 and S89 respectively is :

,d d.k t = 0 ij ljy js[kk ds vuqfn'k fojkekoLFkk ls xfr çkjEHk djrk gSA bldk Roj.k le; vkjs[k n'kkZ;k x;k gSA t = 2s ls t =3s rFkk t = 8s ls t = 9s le;kUrjkyks es d.k ds foLFkkiuks Øe'k% S23 o S89 ds ifjek.kks ds fy;s lgh lEcU/k gS %

(A) S23 > S89 (B) S89 > S23 (C*) S23 = S89 (D) insufficient information to decide (lwpuk vi;kZIr gSA)

7. Consider the circuit consisting of two capacitors having capacitances 3F and 6F and an ideal battery of emf e = 4 volts as shown. The switch S is open for a long time and then closed. After the switch is closed, the work done by the battery is :

fp=k esa 3F rFkk 6F ds la/kkfj=k] e = 4 oksYV fo|qr okgd cy dh vkn'kZ cSVjh ls tqM+s gSA dqath S yEcs le; rd [kqyh jgrh gS ,oa fQj cUn dj nh tkrh gSA dqUth cUn djus ds ckn cSVjh }kjk fd;k x;k dk;Z Kkr dhft,A

S

6 F3 F

=4v (A) 16 J (B) 32 J (C*) 64 J (D) 128 J

Sol. Initial charge on 6 mF capacitor = 3 6

43 6

= 8 C

final charge on 6 mF capacitor = 6 × 4 = 24 C charge passing through cell (in direction aided by cell) = qf – qi = q = 24 – 8 = 16 C work done by cell = q e = 16 × 4 = 64 J

6 mF ds la/kkfj=k ij izkjfEHkd vkos'k = 3 6

43 6

= 8 C

6 mF ds la/kkfj=k ij vfUre vkos'k = 6 × 4 = 24 C

lsy esa izokfgr vkos'k (lsy }kjk iznÙk fn'kk esa) = qf – qi = Dq = 24 – 8 = 16 C

lsy }kjk fd;k x;k dk;Z = q e = 16 × 4 = 64 J




8. Six identical uniform conducting rods are fixed to make a regular hexagon as shown. The ends A and C are maintained at constant temperature 2T0 and 6T0. After the steady state is reached, pick up the correct relation between temperatures TB, TD, TE and TF of ends B, D, E and F respectively.

N% ,d leku pkyd NM+ksa dks fp=kkuqlkj O;ofLFkr djds le"kVQyd cuk;k x;k gSA A rFkk C fljs dks fu;r rki 2T0 rFkk 6T0 ij O;ofLFkr fd;k x;k gSA LFkk;h voLFkk esa igq¡pus ds i'pkr~ fljs B, D, E rFkk F ds rki Øe'k% TB,

TD, TE rFkk TF ds e/; lgh lEcU/k dk p;u dhft;sA 2T0

A B

C

DE

F 6T0

(A*) TD > TE = TB > TF (B) TD = TB > TE > TF (C) TD > TE > TF = TB (D) TB > TF = TE > TD

Sol.

2T0

A B

C

DE

F 6T0

4T0

5T04T0

3T0

At steady temperature of ends B, D, E and F are TB = 4T0, TD = 5T0 , TE = 4T0 and TF = 3T0.

TD > TE = TB > TF 9. The loop shown is placed in uniform constant magnetic field B. A condcuting loop PQR has current I

and length of side PQ is L. The direction and magnitude of the magetic force on the loop PQR is : fp=k esa iznf'kZr ywi ,d leku fu;r pqEcdh; {ks=k B esa j[kk gSA ,d pkyd ywi PQR esa izokfgr /kkjk I gS rFkk

Hkqtk PQ dh yEckbZ L gSA ywi PQR ij pqEcdh; cy dh fn'kk rFkk ifjek.k gS : Q

P R

I60º

BL

(A*) zero (B) ILB out of the page

(C) 2

1ILB into the page (D) ILB into the page

(A*) 'kwU; (B) ILB dkxt ls ckgj dh vksj

(C) 2

1ILB dkxt esa vanj dh vksj (D) ILB dkxt esa vanj dh vksj

Sol. Vector sum PQ + QR +RP = 0 Thus force on PQR = 0.

Sol. lfn'k ;ksx PQ + QR + RP = 0

vr% PQR ij cy = 0 . 10. A thin circular ring of mass 2 kg and radius 1 m is rotating about its axis (without friction) with a constant

angular velocity 12 rad/s. Two objects each of mass 0.5 kg, are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity :

2 kg nzO;eku o 1 m f=kT;k dh ,d iryh oy; bldh v{k ds ifjr% ¼fcuk ?k"kZ.k ds½ fu;r dks.kh; osx 12 Rad/s

ls ?kw.kZu djrh gSA nks oLrq,sa izR;sd 0.5 kg nzO;eku dh bl oy; ds O;kl ds foifjr fdukjksa ij /khjs ls j[k nh tkrh gSA vc oy; dk dks.kh; osx gksxkA

(A) 9.6 rad/s (B*) 8 rad/s (C) 12 rad/s (D) 21 rad/s




11. A long plank of mass M is initially at rest on a frictionless surface. A small block with mass m and initial speed u0 slides on top of the larger plank. The coefficient of friction between the block and plank is .

M æO;eku dk ,d yEck r[rk izkjEHk esa ?k"kZ.kjfgr lrg ij fojkekoLFkk esa fLFkr gSA ,d m æO;eku dk NksVk CykWd bl r[rs dh Åijh lrg ij u0 izkjfEHkd pky ls fQlyrk gSA CykWd rFkk r[rs ds e/; ?k"kZ.k xq.kkad gSA.

M

m

u0

L Net work done by friction if the top block falls off the plank after sliding over its length L is ;fn CykWd r[rs ij L nwjh pyrk gqvk fxj tk;s rks bl nkSjku ?k"kZ.k }kjk dqy dk;Z Kkr djksA

(A*) –mgL (B) gLM

mM

(C) gL

mM

M

(D) –MgL

12. The figure below shows the forces that three charged particles exert on each other. Which of the four

situations shown can be correct. uhps iznf'kZr fp=k esa rhu vkos'kksa }kjk ,d nwljs ij vkjksfir cy iznf'kZr gSA uhps nh xbZ pkjksa fLFkfr;ksa esa ls dkSulh

fLFkfr lgh gks ldrh gSA

(I) (II) (III) (IV)

(A) all of the above (B) none of the above (C*) II, III (D) II, III & IV (A) mijksDr lHkh (B) mijksDr esa ls dksbZ ugha (C*) II, III (D) II, III & IV 13. A slender rod of mass M and length L is hinged about an end to swing freely in a vertical plane.

However, its density is nonuniform and varies linearly from hinged end to the free end, doubling its value. The moment of inertia of the rod, about the rotation axis passing through the hinge point is:

M æO;eku rFkk L yEckbZ dh ,d iryh NM+ blds ,d fdukjs ls yVdh gqbZ gS vkSj Å/okZ/kj ry esa nksyu ds fy;s Lora=k gSA bldk ?kuRo vle:i gS rFkk fuyEcu fcUnq ls eqDr fljs rd js[kh; :i ls ifjofrZr gksrk gqvk nqxquk gks tkrk gSA fuyEcu fcUnq ls xqtjus okyh ?kw.kZu v{k ds lkis{k NM+ dk tM+Ro vk?kw.kZ D;k gksxk \

(A) 9

LM2 2

(B) 16

LM3 2

(C*) 18

LM7 2

(D) None buesa ls dksbZ ugha

14. The secondary coil of an ideal step down transformer is delivering 500 watt power at 12.5 A current. If

the ratio of turns in the primary to the secondary is 5 : 1, then the current flowing in the primary coil will be :

,d vipk;h (stepdown) Vªk¡lQkeZj dh f}rh;d dq.Myh 12.5 ,Eih;j /kkjk ij 500 okWV ÅtkZ iznku djrh gSA vxj izkFkfed rFkk f}rh;d dq.Myh esa Qsjksaa dk vuqikr 5 : 1 gS rks izkFkfed dq.Myh esa /kkjk dk eku gksxkA

(A) 62.5 A (B*) 2.5 A (C) 6 A (D) 0.4 A Sol. P = V For secondary :

V2 = 2

2P

=

5.12

500 = 40 volts

For an ideal transformer (100% efficient) Pinput = Poutput V1 1 = V22

1 = 1

22

V

V =

540

)5.12(40

= 2.5 A

[ 2

1

2

1

V

V

n

n

40

V

1

5 1 ]




15. An inductor

H100

1L , a capacitor

F500

1C and a resistance (3) is connected in series with

an AC voltage source as shown in the figure. The voltage of the AC source is given as V = 10 cos(100 t) volt. What will be the potential difference between A and B ?

,d çsjdRo

H100

1L , ,d la/kkfj=k

F500

1C rFkk ,d çfrjks/k (3) fp=kkuqlkj izR;korhZ foHko lzksr ls

Js.khØe esa tqM+s gSA izR;korhZ lzksr dk foHko V = 10 cos(100 t) oksYV gSA A rFkk B ds e/; foHkokUrj gksxk ?

(A) 8 cos(100 t – 127º) volt (B) 8 cos(100 t – 53º) volt (C*) 8 cos(100 t – 37º) volt (D) 8 cos(100 t + 37º) volt

Sol. z = 22 )15(3 = 5

i = 2 cos(100t + 53º) vAB = 8 cos(100t + 53º – 90º) volt vAB = 8 cos(100t – 37º) volt Ans. 16.. Consider a L – R circuit shown in figure. There is no current in circuit. Switch S is closed at t = 0, time

instant when current in inductor is equal to current in resistor 2R will be :

fp=k esa n'kkZ;s vuqlkj ,d L – R ifjiFk ysosaA ifjiFk esa dksbZ /kkjk izokfgr ugha gSA t = 0 ij dqath S can dh tkrh gS, fdl {k.k ij izsjdRo esa /kkjk 2R izfrjks/k esa /kkjk ds cjkcj gksxh :

(A*) 2nR

L (B) 2n

R

L2 (C) 2n

R2

L (D)

R2

L

Sol.

ii = R

V (1 – e–t/)

i2 = R2

V

i2 = i2

R

V (1 – e–t/) =

R2

V

1 – e–t/ = 2

1 e–t/ =

2

1

et/ = 2 t

= n2 t = n2 = 2nR

L




17. Consider a L – C oscillation circuit. Circuit elements has zero resistance. Initially at t = 0 all the energy is stored in the form of electric field and plate-1 is having positive charge :

,d L–C nksfy=k ifjiFk gSA ifjiFk ds vo;o 'kwU; çfrjks/k j[krs gSA çkjEHk esa t = 0 ij lHkh ÅtkZ oS|qr {ks=k ds :i esa laxzfgr gS rFkk IysV-1 /kukRed vkos'k j[krh gSA

at time t = t1 plate-2 attains half of the maximum +ve change for the first time. Value of t1 is : le; t = t1 ij IysV-2 çFke ckj vf/kdre /kukRed vkos'k dk vk/kk vkos'k çkIr djrh gSA t1 dk eku gksxk –

(A*) LC3

2 (B) LC

3

(C) LC

3

4 (D) LC

Sol.

q1 = q0 sin (t + /2)

at t = t1 ij q1 = – 2

q0

t1 =

3 =

3

2 = LC

3

2

18. The diameter of a cylinder is measured using a vernier callipers with no zero error. It is found that the

zero of the vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The vernier scale has 50 division equivalent to 2.45 cm. The 24th division of the vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is :

,d csyu dk O;kl ekius ds fy, 'kwU; =kqfV jfgr ,d ofuZ;j dSfyilZ dk mi;ksx gksrk gSA ekius ds nkSjku iSekus dk 'kwU;] eq[; iSekus ds 5.10 cm vkSj 5.15 cm ds chp esa ik;k tkrk gSA ofuZ;j iSekus ds 50 Hkkx 2.45 cm ds rqY; gSA bl ofuZ;j iSekus dk pkSchlok¡ (24th) Hkkx eq[; iSekus ds ,d Hkkx ls lVhd lEikrh gksrk gSA csyu dk O;kl gS %

(A) 5.112 cm (B*) 5.124 cm (C) 5.136 cm (D) 5.148 cm Ans. (B) Sol. 50 VSD = 2.45 cm

1 VSD = 50

45.2 cm = 0.049 cm

Least count of vernier = 1MSD – 1 VSD = 0.05 cm – 0.049 cm = 0.001 cm Thickness of the object = Main scale reading + vernier scale reading × least count = 5.10 + (24) (0.001) = 5.124 cm. Hindi. 50 VSD = 2.45 cm

1 VSD = 50

45.2cm = 0.049 cm

ouhZ;j dk vYirekad = 1MSD – 1 VSD = 0.05 cm – 0.049 cm = 0.001 cm oLrq dh eksVkbZ = eq[; iSekus dk ikB~;kad + ofuZ;j iSekus dk ikB~;kad × vYirekad = 5.10 + (24) (0.001) = 5.124 cm.




19. Two identical stars of mass M each orbit around their centre of mass which is at rest. Each orbit is circular and has radius R, so that the two stars are always on opposite sides on a diameter.Find the minimum energy that would be required to separate the two stars to infinity.

nks ,d leku rkjs ftudk nzO;eku M gS ] muds fLFkj nzO;eku dsUnz ds ifjr% pDdj yxk jgs gSA çR;sd d{k o`Ùkh; gS rFkk f=kT;k R gS rFkk nksuksa rkjs lnSo O;kl ds foijhr fljks ij jgrs gSA nksuks rkjksa dks vuUr rd vyx vyx djus ds fy, vko';d U;wure&ÅtkZ Kkr dhft;sA

Ans U = 2GM

4R

Sol. The necessary centripetal force to both stars is provided by grativational attraction.

2

2

Mv GMM

R 4R .... (A)

From equation (A)

orbital speed of each star is v = GM

4R Ans.

Also from equation (A)

MMR RO

2

2 3

v GM

R 4R

2

3

2 GM

T 4R

or, time period T = 4 3R

GM Ans.

(b) Net energy of two star system = kinetic energy + gravitational potential energy.

= 1

2Mv2 +

1

2 Mv2 –

GMM

2R =

2 2GM GM

4R 2R =

2GM

4R

Hence the minimum energy to dismantle the system = – (Net energy) = 2GM

4R

Sol. M

MR RO

nksuksa rkjksa dks vko';d vfHkdsUnzh; cy xq:Rokd"kZ.k ls izkIr gksrk gSA

2

2

Mv GMM

R 4R .... (A)

lehdj.k (A) ls

izR;sd rkjs dh d{kh; pky v = GM

4R Ans.

fQj lehdj.k (A) ls

2

2 3

v GM

R 4R

2

3

2 GM

T 4R

vkorZdky T = 43R

GM Ans.

(b) nksuksa rkjksa ds fudk; dh dqy ÅtkZ = xfrt ÅtkZ $ xq:Roh; fLFkfrt ÅtkZ

= 1

2Mv2 +

1

2 Mv2 –

GMM

2R =

2 2GM GM

4R 2R =

2GM

4R

vr% nksuksa rkjksa dks vyx vyx djus ds fy, U;wure ÅtkZ = – (dqy ÅtkZ) = 2GM

4R.

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