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7340010333

COMPUTER BASED TEST (CBT)

Questions & Solutions

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JEE (Main)

2021

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http://www.resonance.ac.in/reso/results/jee-main-2014.aspx

| JEE MAIN-2021 | DATE : 25-02-2021 (SHIFT-1) | PAPER-1 | OFFICAL | MATHEMATICS

Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005


This solution was download from Resonance JEE (MAIN) 2021 Solution portal

PAGE # 1

7340010333

PART : MATHEMATICS

1. Let be the angle between the lines whose direction cosines satisfy the equations

l + m – n = 0 and l 2 + m2 – n2 = 0. Then the value of sin4 + cos4 is :

ekuk nks js[kk,sa ftudh fnDdksT;k;sa lehdj.kksa l + m – n = 0 rFkk l 2 + m2 – n2 = 0 dks lUrq"V djrh gSa] ds chp ,d

dks.k gSA rks sin4 + cos4dk eku gS %

(1) 8

3 (2)

4

3 (3)

8

5 (4)

2

1

Ans. (3)

Sol. + m = n…….(i)

2 + m2 = n2…….(ii) put n = (+m) in (2)

2 + m2 = (+m)2

m = 0

= 0 m = 0

m = n n=

2 + m2 + n2 = 1 2 + m2 + n2 = 1

0 + m2 + n2 = 1 2 + 0 + 2 = 1

m = ±2

1 2 = ±

2

1

direction cosine are direction cosine are

2

1,

2

1,0

2

1,0,

2

1

2

1,

2

1,0

2

1,0,

2

1

Now if we take direction cosines of lines as

2

1,0,

2

1and

2

1,

2

1,0

2

1|

2

100|cos

8

5

16

10

16

1

16

9

2

1

2

3cossin

44

44

2. If Rolle’s theorem holds for the function f(x) = x3 – ax2 + bx – 4, x [1, 2] with 03

4f

, then ordered

pair (a, b) is equal to :

;fn 03

4f

ds lkFk Qyu f(x) = x3 – ax2 + bx – 4, x [1, 2] ds fy, jksys dk izes; ykxw gksrk gS] rks Øfer

;qXe (a, b) cjkcj gS :

(1) (5, 8) (2) (–5 , 8) (3) (5, –8) (4) (–5, –8)







PAGE # 2

7340010333

Ans. (1) Sol. f(1) = f(2)

1 – a + b – 4 = 8 – 4a + 2b – 4 3a – b = 7

f'(x) = 3x2 – 2ax + b 03

4'f

0ba

3

8–

9

163

–8a + 3b = – 16 ………….(2) a = 5, b = 8

3. The coefficients a, b and c of the quadratic equation, ax2 + bx + c = 0 are obtained by throwing a dice three times. The probability that this equation has equal roots is :

f}?kkrh; lehdj.k ax2 + bx + x = 0 ds xq.kkad a, b rFkk c, ,d ikls dks rhu ckj mNky dj izkIr fd, tkrs gSaA

bl lehdj.k ds ewy cjkcj gksus dh izkf;drk gS %

(1) 36

1 (2)

72

1 (3)

54

1 (4)

216

5

Ans. (4)

Sol. ax2 + bx + c = 0

for equal roots = 0

b2 – 4ac = 0

b2 = 4ac

possible value of ac is 1,4 or 9

case -I if ac = 1 b = 2

(a, b, c ) (1,2,1) 1 way case -II

if ac = 4 b = 4

a

4

1

2

c

1

4

2

3ways

case-3 if ac = 9 b = 6

a

3

c

3

one possible way

total possible ways = 1 + 3 + 1 = 5 Required probability = 216

5

666

5







PAGE # 3

7340010333

4. A tangent is drawn to the parabola y2 = 6x which is perpendicular to the line 2x + y = 1. Which of the

following, points does NOT lie on it ?

Ikjoy; y2 = 6x ij ,d Li'kZ js[kk [khaph xbZ gS tks js[kk 2x + y = 1 ds yacor gSA rks fuEu esa ls dkSu lk fcnq bl

ij fLFkr ugha gS \

(1) (–6, 0) (2) (0, 3) (3) (4, 5) (4) (5, 4)

Ans. (4)

Sol. Equation of tangent : m2

3mxy

2

1mT = ( perpendicular to line 2x + y = 1)

tangent is 32

xy x – 2y + 6 = 0

clearly (5, 4) does not satisfies the above equation

5. All possible values of [0, 2] for which sin 2 + tan 2 > 0 lie in :

[0, 2] ds lHkh laEko eku] ftuds sin 2 + tan 2 > 0 gS] fuEu esa ls fdl esa gaSA

(1)

6

7,

4

3,

22,0 (2)

6

11,

2

3

4

3,

24,0

(3)

2

3,

2,0 (4)

4

7,

2

3

4

5,

4

3,

24,0

Ans. (4)

Sol. sin2 + tan2 > 0

0tan–1

tan2

tan1

tan222

tan = t

0t–1

t2

t1

t222

0)t1)(t–1(

t2t2t2–t222

33

0)t1)(t–1(

t422

0)1t)(1–t(

t22

0)1t)(1–t(

t

–1

+

0

+

1

t (–, –1) (0,1)







PAGE # 4

7340010333

tan (–, –1) (0,1)

1

0

–1

/4 /2 3/2 2 3/4 5/4 7/4

4

7,

2

3

4

5,

4

3,

24,0

6. The statement A (B A) is equivalent to :

dFku A (B A) fuEu esa ls fdlds rqY; gS \

(1) A (A B) (2) A (A B) (3) A (A B) (4) A (A B)

Ans. (2)

Sol. A (BA) ~A (BA)

A (~BA) A~B) (~AA) ~B t ~B t

option (2)

A (AB) ~A (~AB) (~AA) B tB t

7. Let f, g : N N such that f(n + 1) = f(n) + f(1) n N and g be any arbitrary function. Which of the

following statements is NOT true ?

(1) If f is onto, then f(n) = n nN (2) If g is onto, then fog is one-one

(3) f is one-one (4) If fog is one–one, then g is one–one

Ekkuk f, g : N N gSa] ftuds fy, f(n + 1) = f(n) + f(1) nN gS rFkk g ,d LosPN Qyu gSA fuEu esa ls dkSu lk

dFku lR; ugha gS \

(1);fn f vkPNknd gS] rks f(n) = n nN gS (2) ;fn g vkPNknd gS] rks fog ,dSdh gS

(3) f ,dSdh gS (4) ;fn fog ,dSdh gS] rks g ,dSdh gS

Ans. (2)

Sol. f(2) = 2f(1)

f(3) = 3f(1)

f(x) = xf(1)

f(g(x)) = g(x).f(1) fog(x) is one-one

g(x) will also be one-one

8. The value of 1

1–

]x[2 dxex3

, where [t] denotes the greatest integer t, is :

1

1–

]x[2 dxex3

] tgk¡ [t] e“gÙke iw.kk±d t gS] dk eku gS %

(1) e3

1–e (2)

e3

1e (3)

e3

1 (4)

3

1e

Ans. (2)







PAGE # 5

7340010333

Sol. I = dxe.x

1

1

]x[2 3

Let x3 = t

I = dte3

1 1

1

]t[

dxedxe3

1 1

0

00

1

1–

e3

e1

3

1

e3

1

9. The integer ‘k’, for which the inequality x2 – 2(3k – 1)x + 8k2 – 7 > 0 is valid for every x in R, is :

iw.kk±d ‘k’ ftlds fy, vlfedk x2 – 2(3k – 1)x + 8k2 – 7 > 0, R esa izR;sd x ds fy,] ekU; gS] gS %

(1) 4 (2) 2 (3) 0 (4) 3

Ans. (4)

Sol. Discriminant D<0 [a>0]

4(3k–1)2 – 4(8k2–7) < 0

9k2+1–6k –8k2 + 7 < 0

k2–6k +8 < 0

(k–2) (k –4) < 0

k(2,4) integral value of k is 3

10. The value of the integral

d

2cos–1

6sin3sin2)sinsin(sin2sin.sin 24246

is :

(where c is a constant of integration)

lekdyu

d

2cos–1

6sin3sin2)sinsin(sin2sin.sin 24246

cjkcj gS %

¼tgk¡ c ,d lekdyu vpj gS½

(1) csin2–sin9sin18–1118

12

342 (2) ccos2–cos9cos18–11

18

12

342

(3) csin6–sin3sin2–918

12

3246 (4) ccos6–cos3–cos2–9

18

12

3246

Ans. (2)

Sol.

d

sin2

6sin3sin2sinsinsin)cossin2(sin2

24246

Let sin= t

cos . d= dt

6t3t2ttt 24246 dt

dtt6t3t2ttt 24635

Let 2t6 + 3t4 + 6t2 =z

12(t5+t3+t) dt = dz







PAGE # 6

7340010333

dz12

Z2

1

c18

Z 2

3

c18

sin6sin3sin2c

18

t6t3t2 2

32462

3246

c

18

) cos–6(1 + ) cos–3(1 + cos– (122

3

22232

c

18

) cos–6(1 + ) 2cos – cos+3(1 + 3cos + 3cos – cos– 2(1 2

3224426

c

18

11 + 18cos – 9cos + 2cos– 2

3246

11. A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A,

with uniform speed. At that point, angle of depression of the boat with the man’s eye is 30° (Ignore

man’s height). After sailing for 20 seconds, towards the base of the tower (which is at the level of

water), the boat has reached a point B, where the angle of depression is 45°. Then the time taken (in

seconds) by the boat from B to reach the base of the tower is :

,d lraHk ds 'kh"kZ ls ,d iq:"k ns[k jgk gS fd ,d fuf'pr fcanq A ls ,d uko ,d leku xfr ls LraHk dh vksj vk

jgh gSA ml le; Ïiq:"k dh vk¡[k ls uko dk voueu dks.k 30º gS ¼ Ïiq:"k dh Åpk¡bZ dk /;ku u nsa½ LraHk ds vk/kkj

¼tks ikuh dh lrg ij gS½ dh rjQ uko 20 lsd.M pyus ds i'pkr~~ ,d fcanq B ij igq¡prh gS] tgk¡ voueu dks.k

45º gSA uko ds B ls LraHk ds vk/kkj rd igq¡pus esa fy;k x;k le; ¼lsd.M esa½ gS %

(1) 310 (2) 1310 (3) 10 (4) 1–310

Ans. (2)

Sol.

30° 45°

B A

h

45° 30°

P

C

y h

(t=0 s) (t = 20 s)

Let speed of boat is v m/s

Let height of tower = 'h' m

BC = h

y = h cot30º– h cot45º







PAGE # 7

7340010333

y = h 1–3

s/m20

1–3h

20

yV

Time taken from B to C

t1 =

sec1310

20

1–3h

h

v

h

12. Let the lines (2 – i)z = (2 +i) z and (2 + i)z + (i – 2) z – 4i = 0, (here i2 = – 1) be normal to a circle C. If the

line iz + z + 1 + i = 0 is tangent to this circle C, then its radius is :

Ekkuk js[kk,sa (2 – i)z = (2 +i) z rFkk (2 + i)z + (i – 2) z – 4i = 0 (;gk¡ i2 = – 1) ,d o``Ùk C ij vfHkyEc gSaA ;fn

js[kk iz + z + 1 + i = 0 ] o``Ùk C dh Li'kZ js[kk gS] rks bldh f=kT;k gS %

(1) 22

1 (2)

22

3 (3) 23 (4)

2

3

Ans. (2)

Sol. 2(x + iy) – i(x + iy) = (2 + i) (x – iy)

2 x + y + i (2y – x) = 2x – 2iy + ix + y

2i (2y – x) = 0 y = 2

x ……………(1)

(2 + i) (x + iy) + (i – 2 ) (x – iy) – 4i = 0

(2 +i) (x + iy) + (i – 2) (x – iy) – 4i = 0

2x + 2iy + ix –y + ix + y – 2x + 2iy – 4i = 0

i (2y + x + x + 2y – 4) = 0

4y + 2x = 4 2y + x = 2 ………………(2)

from (1) and (2) x = 1, y = 2

1

(x + iy) i + x – iy + 1 + i = 0 ix – y + x– iy + 1 + i = 0

i(x – y + 1) + (x – y + 1) = 0 x – y + 1 = 0

radius = 22

3

2

12

1–1

13. If the curves, 1b

y

a

x 22

and 1d

y

c

x 22

intersect each other at an angle of 90°, then which of the

following relations is TRUE ?

;fn oØ 1b

y

a

x 22

rFkk 1d

y

c

x 22

,d nwljs dks 90° ds dks.k ij dkVrs gSa] rks fuEu esa ls dkSu lk lac/k

lR; gS\

(1) a – c = b + d (2) ba

dcab

(3) a + b = c + d (4) a – b = c – d

Ans. (4)







PAGE # 8

7340010333

Sol. 1b

y

a

x 22

…….(i) Ellipse

1)d(

y

c

x 22

……(ii) Hyperbola

(i) and (ii) are orthogonal

these will be confocal

)d(cba

(a–b)= c–d.

14. If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is

2–x

8yx4–x2 , then this curve also passes through the point :

;fn ,d oØ ewyfcanq ls gksdj tkrk gS rFkk blds fdlh fcanq (x, y) ij Li'kZ js[kk dh izo.krk 2–x

8yx4–x2 gS]

rks ;g oØ fuEu eas ls fdl fcanq ls Hkh gksdj tkrk gS \

(1) (4, 4) (2) (5, 5) (3) (5, 4) (4) (4, 5)

Ans. (2) Sol. Let (y+4)= t(x–2)

t)2x(dx

dt

dx

dy

given differential equation between

)2x(

t)2x()2x(t)2x(

dx

dt 2

t)2x(t)2x(dx

dt

1dx

dt

t = x + c 2x

4yt

cx2x

4y

it passes through (0, 0)

–2=0+c c = –2

2–x2x

4y

y + 4 = (x–2)2

clearly it will pass through (2, –4)

Check the option curve also passes through the point (5, 5)







PAGE # 9

7340010333

15.

n

2

limn

n

n

1........

2

11

1

is equal to :

n

2

limn

n

n

1........

2

11

1

cjkcj gS %

(1) 2

1 (2)

e

1 (3) 0 (4) 1

Ans. (4) Sol. Let limit be L

So, L =

n

n

1...

3

1

2

11

limn

e = ek (say)

Now assume n = 2p + , {0,1,2,….,2p – 1}

Now assume 1 +

.....

7

1

6

1

5

1

4

1

3

1

2

1

12

1...

12

1

2

1p1p1p

+ S2

1...

12

1

2

1ppp

So, S < 1 +

times)1(

ppp 2

1...

2

1

2

1....

4

1

4

1

4

1

4

1

2

1

2

1

S < 1pp2

1....111

timesp

Hence k 02

1plim

pn

Also, S >

timesn

1n

1....

n

1

n

1

n

1

Hence k 0n

1limn

So, L = 1

16. If 0 < , <2

,

0n

n2cosx ,

0n

n2siny and

n2

0n

n2 sin.cosz then :

;fn 0 < , <2

,

0n

n2cosx ,

0n

n2siny rFkk n2

0n

n2 sin.cosz

gSa rks :

(1) xy – z = (x + y)z (2) xy + yz + zx = z (3) xyz = 4 (4) xy + z = (x + y)z

Ans. (4)

Sol. x = 1+cos2 + cos4

2cos1

1x







PAGE # 10

7340010333

cos2 = x

1x

2sin1

1y

sin2 = y

1y

22 cossin1

1z

1yx

xyz

x

)1x(

y

)1y(1

1z

z(x+y) = xy + z

17. The image of the point (3, 5) in the line x – y + 1 = 0, lies on :

;fn x – y + 1 = 0 esa fcanq (3, 5) dk izfrfcac fuEu esa ls fdl ij fLFkr gS \

(1) (x – 2)2 + (y – 2)2 = 12 (2) (x – 4)2 + (y – 4)2 = 8

(3) (x – 4)2 + (y + 2)2 = 16 (4) (x – 2)2 + (y – 4)2 = 4

Ans. (4)

Sol.

11

15–32–

1–

5–y

1

3–x

11–

5–y

1

3–x

x = 4, y = 4

p'(4,4) which satisfies (x – 2)2 + (y – 4)2 = 4

18. The equation of the line through the point (0, 1, 2) and perpendicular to the line

2–

1–z

3

1y

2

1–x

is :

;fn (0, 1, 2) ls gksdj tkus okyh rFkk js[kk 2–

1–z

3

1y

x

1–x

ds yEcor js[kk dk lehdj.k gS

(1) 3

2–z

4–

1–y

3

x (2)

3

2–z

4

1–y

3

x (3)

3

2–z

4

1–y

3–

x (4)

3–

2–z

4

1–y

3

x

Ans. (3)

P(3,5)

x–y + 1 = 0

P'(x,y)







PAGE # 11

7340010333

Sol. A(0,1,2)

B

r2

1z

3

1y

2

1x

B(2r + 1, 3r – 1, –2r + 1)

direction ratios of AB are : 2r + 1, 3r – 2, –2r – 1

AB is perpendicular to given line

(2r + 1) (2) + (3r–2) (3) + (–2r–1) (–2) 4r + 2 + 9r – 6 + 4r + 2 = 0

17r –2 = 0 17

2r

Direction ratios of AB : 17

21,

17

28,

17

21

Simplified direction ratios of AB 3, –4, –3 OR –3, 4,3

equation of AB : 3

2z

4

1y

3

0x

19. When a missile is fired from a ship, the probability that it is intercepted is 3

1 and the probability that the

missile hits the target, given that it is not intercepted, is 4

3. If three missiles are fired independently from

the ship, then the probability that all three hit the target, is :

tc ,d iz{ksikL+=k fdlh tgkt ls nkxk tkrk gS] rks blds vo:) gksus dh izkf;drk 3

1 gS rFkk ;g fn, gksus ij fd

;g vo:) ugha gksrk] blds fu'kkus ij yxus dks izkf;drk 4

3 gSA ;fn tgkt ls rhu iz{ksikL=k Lora=k :i ls nkxs

tkrs gSa] rks rhuksa ds fu'kkus ij yxus dh izkFkfedrk gS %

(1) 8

1 (2)

4

3 (3)

8

3 (4)

27

1

Ans. (1)

Sol. Required probability = 8

1

4

3

3

2

4

3

3

2

4

3

3

2

20. The total number of positive integral solutions (x, y, z) such that xyz = 24 is :

xyz = 24 ds /ku iw.kk±d gyksa (x, y, z) dh dqy la[;k gS %

(1) 36 (2) 24 (3) 30 (4) 45

Ans. (3)

Sol. x y z = 24

xyz = 23 × 3

Number of positive integral solution = (3 + 3 – 1C3–1) ×(1+3–1C3–1)

= 5C2 × 3C2 = 10 × 3 = 30







PAGE # 12

7340010333

21. The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5, is ________ .

vadksa 1, 2, 3, 4, 5 ls 100 rFkk 1000 ds chp dh cukbZ tk ldus okyh la[;kvksa] ;fn la[;kvksa] ;fn dksbZ Hkh vad

nksgjk;k ugha tkrk gS rFkk la[;k,sa ;k rks 3 ls 5 ;k ls foHkkT; gS] dh dqy la[;k gS ________ .

Ans. 32

Sol. Case-1 1,2,3 3! = 6

Case-2 1,3,5 3! = 6

2,3,4 3!= 6

Case-3 3,4,5 3! = 6

________

24

three digit number divisible by 5

431 = 12

common Numbers 1,3,5 3,1, 5

so total number = 24 +12 –4 = 32

22. The number of points, at which the function f(x) = |2x + 1| –3|x + 2| +|x2 + x – 2|, x R is not

differentiable, is ________ .

mu fcUnqvksa dh la[;k] ftu ij Qyu f(x) = |2x + 1| –3|x + 2| +|x2 + x – 2|, x R vodyuh; ugha gS] gS

________ A

Ans. 2

Sol.

1x;7x

1x2

1;3x2x

2

1x2;5x6x

2–x;3x2x

)x(f

2

2

2

2

1x;x2

1x2

12x2

2

1x26x2

2–x;2x2

)x(f

at x = –2

f'(x) =

2

1x26x2

2x:2x2

=

2

1x2264

2x2

at x = –2, f(x) is differentiable







PAGE # 13

7340010333

at x = 2

1

1x2

1:2x2

2

1x2:6x2

)x('f

1x2

1:121

2

1x2:561

at x = 2

1, f(x) is not differentiable

at x = 1

f(x) =

1x,2

1x2

1–,4–

f(x) is not differentiable at x = 1

f(x) is not differentiable at two points x = 2

1– and x = 1

23. If

02

tan

2tan–0

A and (I2 + A) (I2 – A)–1 =

ab

b–a, then 13(a2 + b2) is equal to ________ .

;fn

02

0tan

2

0tan–0

A rFkk (I2 + A) (I2 – A)–1 =

ab

b–a gSa] rks 13(a2 + b2) cjkcj gS________ A

Ans. 13

Sol.

12

tan

2tan–1

1

12

tan

2tan1

L

12

tan

2tan–1

12

tan

2tan–1

2tan1

1

2

12

tan2

tan2

2tan2–

2tan1

2sec

1

2

2

2

so

2sec

2tan2

band

2sec

2tan1

a22

2

13(a2+b2)= 132

tan42

tan1

2sec

13 22

2

4







PAGE # 14

7340010333

24. The graphs of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area A. The A4 is equal to ________ .

Lkkbu rFkk dkslkbu ds xzkQ ,d nwljs dks cgqr ls fcUnqvksa ij dkVrs gSa] buds nks Øekxr izfrPNsu fcUnqvksa dsa chp

esaa ls nks xzkQ ,d leku {ks=kQy A ?ksjrs gSaA rks A4 cjkcj gS ________ A

Ans. 64

Sol.

A = 4

5

4

4

5

4

)xsinxcos(dx)xcosx(sin

2

1

2

1

2

1

2

1

A = 22

A4 = 64

25. Let k–j2ia

, j–ib

and k–j–ic

be three given vectors. If r is a vector such that acar

and 0b.r

, the a.r

is equal to _______ .

Ekkuk rhu lfn'k k–j2ia

, j–ib

rFkk k–j–ic

fn, x, gSaA ;fn r ,d lfn'k gS] ftlds fy,

acar

rFkk 0b.r

rks a.r

cjkcj gS _______ A

Ans. 12

Sol. acar

0ac–r

ac–r

b.ab.cb.r

0 = 2 + (–1)

= 2

ca2r

k–j–ik2–j4i2r

k3–j3i3r

a.r

= 3 + 6 + 3 = 12







PAGE # 15

7340010333

26. If the system of equations

kx + y + 2z = 1

3x – y – 2z = 2

– 2x – 2y – 4z = 3

has infinitely many solutions, then k is equal to _________ .

;fn lehdj.k fudk;

kx + y + 2z = 1

3x – y – 2z = 2

– 2x – 2y – 4z = 3

ds vuUr gy gSa] rks k cjkcj

Ans. 21 Sol. kx + y + 2z = 1 ……(1) 3x – y – 2z = 2 ……(2) 2x + 2y + 4z = –3 ……(3) 5(2) + 3(3), we get 21x + y + 2z = 1 ……(4) Now for getting infinite solution (1) and (4) should be identical

k = 21

27. Let

yxz

xzy

zyx

A , where x, y and z are real numbers such that x + y + z > 0 and xyz = 2.

If A2 = I3 then the value of x3 + y3 + z3 is ________ .

Ekuk

yxz

xzy

zyx

A gS] tgk¡ x, y rFkk z okLrfod la[;k,sa gSa] ftuds fy, x + y + z > 0 rFkk xyz = 2 gSaA ;fn

A2 – I3 gS] rks x3 + y3 + z3 dk eku gS ________ .

Ans. 7

Sol. A2 = 3

take determinant both sides

|A|2 = 1 |A| = –1 (|A| 1 as |A| = – (x3 + y3 + z3– 3xyz) < 0)

3xyz – x3 – y3 – z3 = –1

x3 + y3 + z3 = 7

28. Let A1, A2, A3 …….. be squares such that for each n 1, the length of the side of An equals the length

of diagonal of An + 1. If the length of A1 is 12 cm, then the smallest value of n for which area of An is less

than one, is _________ .

ekuk A1, A2, A3 …….. oxZ gS tc fd izR;sd n 1 ds fy,] An dh Hkqtk dh yEckbZ An + 1 ds fod.kZ dh yEckbZ ds

cjkcj gSA ;fn A1 dh Hkqtk dh yEckbZ 12 cm gS rks n dk U;wure eku] ftlds ds fy, An dk {ks=kQy ,d ls de

gS] gS _________ A

Ans. 9

Sol. Let a1 , a2, a3,…………..an be the side length of square A1, A2, A3, ………….An respectively

a1 = 12 , a2 = 2

12, a3 =

22

12, a4 =

32

12







PAGE # 16

7340010333

similarly an = 1–n

2

12

area of An =

2

1–n2

12

= 1

2

1441–n (Given)

2n–1 > 144 n – 1 > 7 n > 8

smallest value of n is 9

29. Let f(x) be a polynomial of degree 6 in x, in which the coefficient of x6 is unity and it has extrema at x =

– 1 and x = 1. If 1x

)x(f3

lim0x , then 5f(2) is equal to ________ .

Ekkuk x esa ,d cgqin f(x) dh ?kkr 6 gS] rFkk in x6 dk xq.kkad ,d gS vkSj x = – 1 rFkk x = 1 blds pje fcanq gSaA

;fn 1x

)x(f3

lim0x gS] rks 5.f(2) cjkcj gS ________ .

Ans. 144 Sol. Let f(x) = x6 + bx5 + cx4 + dx3 + ex2 + fx + g

1x

)x(flim

30x

d = 1; e = f = g = 0

f(x) = x6 + bx5 + cx4 + x3

f(x) has extremum values at x = – 1 and x = 1

f '(–1) = f '(1) = 0

f '(x) = 6x5 + 5bx4 + 4cx3 + 3x2

f '(1) = 0 6 + 5b + 4c + 3 = 0 5b + 4c + 9 = 0 ….(1)

f '(–1) = 0 – 6 + 5b – 4c + 3 = 0 5b – 4c – 3 = 0 ….(2)

Solving (1) and (2), we get

b = – 5

3 and c = –

2

3

f(x) = x6 – 5

3.x5 –

2

3x4 + x3

f(2) = 64 – 5

96 – 24 + 8 = 48 –

5

96 =

5

144

5f(2) = 144

30. The locus of the point of intersection of the lines 034–kykx3 and 0k34–y–x3 is a

conic, whose eccentricity is ________ .

js[kkvksa 034–kykx3 rFkk 0k34–y–x3 ds izfrPNsnu fcanq dk fcnqiFk ,d 'kkado gS] ftldh

mRdsUnzrk gS ________ A

Ans. 2

Sol. 34

y–x3

x3y

34k

3x2 – y2 = 48

148

y–

16

x 22

; it is a hyperbola 216

481e Ans.



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7340010333





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JEE (Main) 2021




| JEE MAIN-2021 | DATE : 25-02-2021 (SHIFT-1) | PAPER-1 | OFFICIAL | CHEMISTRY

Resonance Eduventures Ltd.


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7340010333

PART : CHEMISTRY

Single Choice Type

This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its answer,

out of which Only One is correct.

1. Which of the following equation depicts the oxidizing nature of H2O2 ?

(1) I2 + H2O2 + 2OH– 2I– + 2H2O + O2 (2) KIO4 + H2O2 KIO3 + H2O + O2

(3) 2I– + H2O2 + 2H+ I2 + 2H2O (4) Cl2 + H2O2 2HCl + O2

Ans. (3)

Sol. 2I– + H2O2 + 2H+ I2 + 2H2O

(–1) (0)

Hence, H2O2 at as an oxidising agent.

2. Which statement is correct ?

(1) Synthesis of Buna-S nascent oxygen

(2) Neoprene is an addition copolymer used in plastic bucket manufacturing

(3) Buna-N is a natural polymer

(4) Buna-S is a synthetic and linear thermosetting polymer.

Ans. (1)

Sol. Theoretical

3. Complete combustion of 1.8 g of an oxygen containing compound (CxHyOz) give 2.64 g of CO2 and 1.08

g H2O. The percentage of oxygen in the organic compound is :

(1) 50.33 (2) 51.63 (3) 63.53 (4) 53.33

Ans. (4)

Sol. CxHyOz +

2

z

4

yx O2(g) xCO2(g) +

2

y H2O

z16yx12

8.1

z16yx12

8.1x +

z16yx12

8.1

2

y

44z16yx12

8.1xW

2CO

= 2.64 …(i)

18z16yx12

8.1

2

yW OH2

= 1.08 …(ii)

2

x2

W

W

OH

CO

2

2 = 1 so y = 2x






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PAGE # 2

7340010333

on putting y = 2x in equation (i)

we get x = z

so formula of compound

CxH2xOx CH2O

so mass % of oxygen =

10016212

16 = 53.33 %

4. Ellingham diagram is a graphical representation of :

(1) G vs T (2) G vs P (3) H vs T (4) (G – TS) vs T

Ans. (1)

Sol. Ellingham diagram is the plot of ΔƒG vs T for formation of oxides of elements i.e., for the reaction,

2xM(s) + O2(g) → 2MxO(s)

5. The major product of the following chemical reaction is :

CH3CH2CN 24

2

3

H,BaSO/Pd)3

SOCl)2,OH)1

?

(1) CH3CH2CHO (2) (CH3CH2CO)2O (3) CH3CH2CH3 (4) CH3CH2CH2OH

Ans. (1)

Sol. CH3–CH2–CN OH3 CH3–CH2–COOH SOCl 2

CH3–CH2–C–Cl

O

BaSO Pd, ,H 42

CH3–CH2–C–H

O

6. The hybridization and magnetic nature of [Mn(CN)6]4– and [Fe(CN)6]3–, respectively are :

(1) sp3d2 and diamagnetic (2) sp3d2 and paramagnetic

(3) d2sp3 and paramagnetic (4) d2sp3 and diamagnetic

Ans. (3)

Sol. In [Mn(CN)6]4– 25Mn+2 in presence of SFL, (CN–) =

[Ar]18

In [Fe(CN)6]3– Fe+3, due to SFL (CN–) = 1,2,2g2t , eg0,0 paramagnetic nature

(1 unpaired e–)






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7340010333

7. Which of the glycosidic linkage between galactose and glucose is present in lactose ?

(1) C–1 of galactose and C–4 of glucose (2) C–1 of galactose and C–6 of glucose

(3) C–1 of glucose and C–4 of galactose (4) C–1 of glucose and C–6 of galactose

Ans. (1)

Sol.

O

CH2OH

6

5

H

OH

3

H

OH H

HO

H

1 4

2

O

CH2OH

6

5

H

OH

3

H

OH H

H

H

1

2

O 4

OH

-D-Galactose -D-Glucose

Lactose

H

-D(+) Glucose + -D (+) galactose (-1,4- glycosidic linkage)

8. The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is :

[Assume : No cyano complex is formed ; Ksp(AgCN) = 2.2 × 10–16 and Ka(HCN) = 6.2 × 10–10]

(1) 1.6 × 10–6 (2) 0.625 × 10–6 (3) 2.2 × 10–16 (4) 1.9 × 10–5

Ans. (4)

Sol. AgCN Ag+ + CN–

S’ [CN–]

Ksp = 2.2 × 10–16 = S’ [CN–] ….(1)

H+ + CN– HCN

10–3 [CN–] S’

aK

1 =

]CN[10

'S3 ….(2)

(1) × (2)

2.2 × 10–16 × 10102.6

1

= 310

)'S(

S’ = 10

316

102.6

10102.2

= 10

19

102.6

102.2

= 2.6

102.2 9

= 810035.0 = 0.187 × 10–4 1.9 × 10–5

= 1.9 × 10–5 M






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9. The plots of radial distribution functions for various orbitals of hydrogen atom against 'r' are given below:

(A)

(B)

(C)

(D)

The correct plot for 3s orbital is :

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

Ans. (1)

Sol. From NCERT

10. Which one of the following reactions will not form acetaldehyde ?

(1) CH3CN OH)ii

HDIBAl)i

2

(2) CH2=CH2 + O2

OH

)II(Cu/)II(Pd

2

(3) CH3CH2OH K573

Cu (4) CH3CH2OH 423 SOHCrO

Ans. (4)






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Sol. (1) CH3CN + DIBAL–H CH3CHO (acetaldehyde)

(2) CH2=CH2 + O2 waterin)II(Cu),II(Pd

Catalyst CH3CHO (acetaldehyde)

(3) CH3CH2OH + Cu, heat CH3CHO (acetaldehyde)

(4) CH3CH2OH + CrO3, H2SO4 CH3COOH (acetic acid)

11. According to molecular orbital theory, the species among the following that does not exist is :

(1) He2– (2) Be2 (3) O2

2– (4) He2+

Ans. (2)

Sol. (1) He2– = 1s2, 1s2, 2s1 BO =

2

1(3 – 2) =

2

1

(2) Be2 = 1s2, 1s2, 2s2, 2s2 BO = 2

1(4 – 4) = 0

(3) O22– = 1s2, 1s2, 2s2, 2s2, 2pz

2, (2px2 = 2py

2), (2px2 = 2py

2) BO = 2

1(10 – 8) = 1

(4) He2+ = 1s2, 1s1 BO =

2

1(2 – 1) =

2

1

12. Identify A in the given chemical reaction :

CH2–CH2–CH2–CH–CH3

CH3 CH3

atm2010K773

OMo 32

productmajor

'A'

(1)

(2) CH3

(3)

(4) CH3

Ans. (2)

Sol. n–Alkanes having six or more carbon atoms on heating to 773 K at 10–20 atompheric pressure in the

presence of oxides of vanadium, molybdenum or chromium supported over alumina get dehydrogenated

and cyclised to benzene and its homologues.

13. The correct statement about B2H6 is :

(1) Its fragment, BH3, behaves as a Lewis base.

(2) All B–H–B angles are of 120º

(3) Terminal B–H bonds have less p-character when compared to bridging bonds

(4) The two B–H–B bonds are not of same length

Ans. (3)






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7340010333

Sol. B2H6 have 4 2c-2e bonds and 2 3c-2e bonds. Bridging bonds have larger bond length than terminal

bonds. Angle between terminal bonds is more than angle between bridging bonds if all 4 terminal bonds

are in one plane then bridging bonds are in perpendicular plane.

As terminal H–B–H > bridge H–B–H

% s-character B.A.

As external bond has more % s-character or less % p-character.

14. Compound(s) which will liberate carbon dioxide with sodium bicarbonate solution is/are :

A =

NH2

NH2 NH2

OH

B =

COOH

C =

OH

NO2

NO2

NO2

(1) A and B only (2) B only (3) B and C only (4) C only

Ans. (3)

Sol. Compounds which are more acidic than H2CO3 release CO2 gas with NaHCO3

15. In which of the following pairs, other outer most electronic configuration will be the same ?

(1) Cr+ and Mn2+ (2) Ni2+ and Cu+ (3) Fe2+ and Co+ (4) V2+ and Cr+

Ans. (1)

Sol. 25Mn+2 = 3d5 4s0

Cr+1+ = 3d5 4s0

Cu+2 = 3d9 4s0

Ni+2 = 3d8 4s0

Co+2 = 3d7 4s0

B B

H

H

H

Ht

H

Hb

2 1

3C – 2e bond–

x

3C – 2e bond–

y






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7340010333

16. Given below are two statements :

Statement I : An allotrope of oxygen is an important intermediate in the formation of reducing smog.

Statement II : Gases such as oxides of nitrogen and sulphur present in troposphere contribute to the

formation of photochemical smog.

(1) Both Statement I and Statement II are false (2) Statement I is false Statement II are true

(3) Both Statement I and Statement II are true (4) Statement I is true but Statement II is false

Ans. (1)

Sol. From NCERT

17. In Freundlich adsorption isotherm at moderate pressure, the extent of adsorption

m

xdirectly

proportional to px. The value of x is :

(1) Zero (2) 1 (3) (4) n

1

Ans. (4)

Sol. Freundlich adsorption isotherm

m

x = K(P)1/n

18. Identify A and B in the chemical reaction.

OCH3

NO2

HCl

)major(

]A[acetonedry

Na

)major(

]B[

(1) A =

OCH3

NO2

Cl B =

OCH3

NO2

Cl

(2) A =

OCH3

NO2

Cl B =

OCH3

NO2

(3) A =

OCH3

NO2

Cl

B =

Cl

NO2

(4) A =

OCH3

NO2

Cl

B =

Cl

NO2

Ans. (2)






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7340010333

Sol.

OCH3

NO2

HCl NaI

OCH3

NO2

Cl

Acetone

OCH3

NO2

I

19. In which of the following reaction/s will not give p-aminoazobenzene ? [OC-II_XII_Amines_M]

A -

NO2

(i) Sn/HCl

(ii) HNO2

(iii) Aniline

B -

(i) NaBH4

(ii) NaOH

(iii) Aniline

NO2

C -

(i) HNO2

(ii) Aniline, HCl

NH2

(1) A only (2) B only (3) C only (4) A and B

Ans. (2)

Sol.

NO2

(1) Sn, HCl NO2

(2) NaNO2, HCl

N2+

(3) HCl, Aniline

NH2

N=N


Statement I : CeO2 can be used oxidation of aldehydes and ketones.

Statement II : Aqueous solution of EuSO4 is a strong reducing agent.

In the light of the above statements, choose the correct answer from the options given below :

(1) Statement I is false but Statement II is true

(2) Both Statement I and Statement II are false

(3) Both Statement I and Statement II are true

(4) Statement I is true Statement II is false

Ans. (3)

Sol. In Ce, Eu +3 is more stable oxidation state.






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PAGE # 9

7340010333

Numerical Value Type

This section contains 10 Numerical value type questions.

21. In basic medium CrO42– oxidises S2O3

2– to SO42– and itself changes into Cr(OH)4

–.

The volume of 0.154 M CrO42– required to react with 40 mL of 0.25 M S2O3

2– is …….. mL. (Rounded-off

to the nearest integer)

Ans. (173)

Sol. –24

6

OCr

+ –23

2

2 OS

–4

3

(OH)Cr

+

6–2

4SO

vf = 3 vf = 4 × 2 = 8

Mili eq. of CrO42– = Mili eq. of S2O3

2–

3 [0.154 × V] = 8 [0.25 × 40]

V = 173.16 mL

22. A car tyre is filled with nitrogen gas at 35 psi at 27ºC. If will burst if pressure exceeds 40 psi.

The temperature in ºC at which the carb tyre will burst is ………… (Rounded-off to the nearest integer)

Ans. (70)

Sol. 2

2

1

1

T

P

T

P

2T

40

300

35

T2 = 35

30040 = 342.85 K = 69.85ºC 70ºC

23. Using the provided information in the following paper chromatogram :

B

A

5 cm

2 cm

2 cm

Base line Spot

Solvent front

Fig : Paper chromatography for compounds A and B.

the calculated Rf value of A …………… × 10–1.

Ans. (4)






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Sol. Retardation factor (Rf) = linebasefromsolventthebymovedcetanDis

linebasefromcetansubsthebymovedcetanDis

= 5

2 = 0.4

= 4 × 10–1

24. Among the following, the number of halides(s) which is/are inert to hydrolysis is …………… .

(1) BF3 (2) SiCl4 (3) PCl5 (4) SF6

Ans. (1)

Sol. SF6 (Steric crowding)

SiCl4 + 4H2O H4SiO4 + 4HCl

PCl5 + 2H2O POCl3 + 2HCl

BF3 + 3H2O H3BO3 + 3HF

SF6 + H2O no reaction

25. The ionization enthalpy Na+ formation from Na(g) is 495.8 kJ mol–1, while the electron gain enthalpy of Br

is –325.0 kJ mol–1. Given the lattice enthalpy of NaBr is –728.4 kJ mol–1. The energy for the formation of

NaBr ionic solid is (–) ………. × 10–1 kJ mol–1.

Ans. (5576)

Sol. Na(g) + Br(g) NaBr(s)

Na+(g) + Br–(g)

HI.E. Heg

HLE

Hºƒ(NaBr) = HIE(Na) + Heg(Br) + HLE

Hºƒ(NaBr) = 495.8 – 325 –728.4 = –557.6

= –5576 × 10–1

Note : In question Hsub(Na, S), Hvap(Br2, ) and HBDE(Br2, g) must be given, we solve this question

with the help of given information which is not correct.

26. CH CH 3AlClHCl, CO,(2)

K 873Tube, Fe hot Red(1) Product

The number of sp2 hybridized carbon atoms(s) present in the product is …………… .

Ans. (7)






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Sol. CHCH

Tube Fe hot Red AlClAnhyd.HClCO, 3

CHO

Benzaldehyde

It is Gattermann Koch reaction

No. of sp2 c–atom in Benzaldehyde is 7.

27. The reaction of cyanamide, NH2CN(s) with oxygen was run in a bomb calorimeter and U was found to

be –742.24 kJ mol–1. The magnitude of H298 for the reaction

NH2CN(s) + 2

3O2(g) N2(g) + O2(g) + H2O(l) is________ kJ. (Rounded off to be nearest

integer) (Assume ideal) gases and R = 8.314 J mol–1 k–1] Given :

Ans. (741)

Sol. Hºreaction = Uºreaction + ngRT

= –742.14 + 2981000

314.8

2

1

= –742 + 1.24

= |–740.9| kJ/mole.

= 740.9 741 kJ/mole.

28. 0.4 g mixture of NaOH, Na2CO3 and some inert impurities was first titrated with 10

NHCl using

phenolphthalein as an indicator, 17.5 mL of HCl was required at the end point. After this methyl orange

was added and titrated. 1.5 mL of same HCl was required for the next end point. The weight percentage

of Na2CO3 in the mixture is _______ (Rounded-off to the nearest integer) Given :

Ans. (4)

Sol. MeOH end point

NaHCO3 + HCl H2CO3 + NaCl

1106

x–4.0

= 3105.1

10

1

(0.4 – x) = 106 × 1.5 × 10–4

0.4 – x = 0.0159

x = 0.3841 gram

% of NaOH = 4.0

3841.0 × 100 = 96.025

% of Na2CO3 = 4 %






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29. 1 molal aqueous solution of an electrolyte A2B3 is 60% ionised. The boiling point of the solution at 1 atm

is _______ L. (Rounded-off to the nearest integer) [Given Kb for (H2O) = 0.52 K kg mol–1] Given 375

Ans. (375)

Sol. Tb = i kbm

i = 1 + (n – 1) = 1 + [5 – 1]0.6 = 3.4

Tb = 3.4 × 0.52 × 1 = 1.768 K = 1.77 K

T’b – 373 = 1.77

T’b = 374.77

Ans. 375

30. For the reaction, aA + bB cC + dD, the plot of log k us T

1is given below :

log K

1/T

slope = – 10000 K

The temperature at which the rate constant of the reaction is 10–4 s–1 is _______ K.

(Rounded-off to the nearest integer)

[Given : The rate constant of the reaction is 10–5 s–1 at 500 K]

Ans. (526)

Sol. k = RT

Ea

Ae

log k = log A – T

1

R303.2

Ea

Slope = –R303.2

Ea = – 10,000 = –104

At T = 500 K log10–5 = log A –

500

104

–5 = log A – 20

log A = +15

at temperature T log10–4 = 15 –

T

104

–4 = 15 –T

104

T

104

= 19

So, T = 19

000,10

T = 526.3 K



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7340010333





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JEE (Main) 2021




| JEE MAIN-2021 | DATE : 25-02-2021 (SHIFT-1) | PAPER-1 | OFFICAL PAPER | PHYSICS


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PAGE # 1

7340010333

PART : PHYSICS

Single Choice Type

This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its answer,

out of which Only One is correct.


Statement I : A speech signal of 2 kHz is used to modulate a carrier signal of 1MHz. The bandwidth

requirement for the signal is 4 kHz.

Statement II : The side band frequencies are 1002 kHz and 998 kHz.

In the light of the above statements, choose the correct answer from the options given below:

(1) Both Statement I and Statement II are false

(2) Statement I is true but Statement II is false

(3) Both Statement I and Statement Ii are true

(4) Statement I is false but Statement II is true

Ans. (3)

Sol. Maximum frequency of modulated signal = carrier wave frequency + signal frequency

Minimum frequency of modulated signal = carrier wave frequency – signal frequency

2. In an octagon ABCDEFGH of equal side, what is the sum of AHAGAEADACAB

If k4j3i2AO .

O

C

D

E F

G

H

A B

(1) k32j24i16 (2) k32j24i16 (3) k32j24i16 (4) k32j24i16

Ans. (2)

Sol.

O

C

D

E F

G

H

A B





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PAGE # 2

7340010333

k4j3i2AO

AHAGAFAEADACAB

BCABAC

)DECDBCAB()CDBCAB()BCAB(AB + )EFDECDBCAB(

+ )FGEFDECDBCAB( + )GHFGEFDECDBCAB(

]DEHA,CDGH,BCFG,ABEF[

)DE()DECD()DECDBC()DECDBCAB()CDBCAB()BCAB(AB

4 × AE4)DECDBCAB(

AO24

)k4j3i2(8

k32j24i16

3. A student is performing the experiment of resonance column. The diameter of the column tube is 6 cm.

The frequency of the tuning fork is 504 Hz. Speed of the sound at the given temperature is 336 m/s. The

zero of the metre scale coincides with the top end of the resonance column tube. The reading of the water

of the water level in the column when the first resonance occurs is :

(1) 14.8 cm (2) 13 cm (3) 16.6 cm (4) 18.4 cm

Ans. (1)

Sol. V = f

= V/f

=504

336

4

e

5044

3361063.0 2

= 14.87 cm





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PAGE # 3

7340010333

4. Magnetic fields at two points on the axis of a circular coil at a distance of 0.05 m and 0.2m from the centre

are in the ratio 8 : 1 The radius of coil is___.

(1) 0.2 m (2) 0.15 m (3) 1.0 m (4) 0.1 m

Ans. (4)

Sol. 2/322

20

)RX(2

NIRB

8B

B

2

1

8

)RX(2

NIR

)RX(2

NIR

2/3222

20

2/3221

20

4RX

)RX(22

1

222

(0.2)2 + R2 = 22 R)05.0(4

22 R100100

254R

100

4

22 R4100

1R

100

4

2R3100

3

10

1R = 0.1 m

5. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : When a rod lying freely is heated, no thermal stress is developed in it.

Reason R : On heating the length of the rod increases.

In the light of the above statements, choose the correct answer from the options given below:

(1) A is false but R is true

(2) A is true but R is false

(3) Both A and R true and R is the correct explanation of A

(4) Both A and R are true but R is NOT the correct explanation of A

Ans. (4)





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PAGE # 4

7340010333

6. An particle and a proton are accelerated from rest by a potential difference of 200 V. After this , their

de Broglie wavelengths are and p respectively. The ratio

p is :

(1) 3.8 (2) 7.8 (3) 2.8 (4) 8

Ans. (3)

Sol. = P

h =

mqv2

h

em

e2m4P

= 22

7. The angular frequency of alternating current in L-C-R Circuits is 100 rad/s. The components connected

are shown in figure. Find the value of inductance of the coil and capacity of condenser.

(1) 0.8 H and 250 f (2) 0.8 H and 150 f (3) 1.33 H and 150 f (4) 1.33 H and 250 f

Ans. (1)

Sol.

Current through resistance is 20/40 = 1/2A

Voltage across capacitor is 10Volt and current is 1/4A

104

1

C

1

F250C

Voltage across inductor is 20Volt and current is 1/4A

204

1L

L = 0.8H





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PAGE # 5

7340010333

8. The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is

put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a

wire is placed between the jaws, the first linear scale division is clearly visible while 72nd division on

circular scale coincides with the reference line. The radius of the wire is :

(1) 1.64 mm (2) 1.80 mm (3) 0.90 mm (4) 0.82 mm

Ans. (4)

Sol. LC = 100

1

division of number

pitch

error = 8 × 100

1

Reading (2R) = 1 + 72 × 100

1 – 8 ×

100

1

2R = 1.64

R = 0.82 mm

9. A proton, a deuteron and an particle are moving with same momentum in a uniform magnetic field. The

ratio of magnetic forces acting on them is ____and their speed is

(1) 2 : 1 : 1 and 4 : 2 : 1

(2) 1 : 2 : 4 and 1 : 1 : 2

(3) 1 : 2 : 4 and 2 : 1 : 1

(4) 4 : 2 : 1 and 2 : 1 : 1

Ans. (1)

Sol. F = q v B

= m

B)mv(q

F m

q

F1 : F2 : F3 : = 4

2:

2

1:

1

1

= 4 : 2 : 2

= 2 : 1 : 1

P = mv v m

1

v1 : v2 : v3 : = 4

1:

2

1:

1

1= 4 : 2 : 1





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PAGE # 6

7340010333

10. A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force F1.

Now a spherical cavity of radius

2

R is made in the sphere (as shown in figure) and the force becomes

F2. The value of F1 : F2 is :

(1) 41 : 50 (2) 50 : 41 (3) 36 : 25 (4) 25 : 36

Ans. (2)

Sol. F1 = 2R9

GmM

F2 = F1 – 2

R2

5

m8

MG

F2 =

50

1

9

1

R

GMm2

2R

GMm

509

41

41

50

F

F

2

1

11. Two radioactive substances X and Y originally have N1 and N2 nuclei respectively. Half life of X is half of

the half life of Y. After three half lives of Y, number of nuclei of both are equal. The ratio 2

1

N

N will be equal

to :

(1) 1

3 (2)

1

8 (3)

8

1 (4)

3

1

Ans. (2)

Sol. We know n

0

T/t

0

)2(

N

)2(

NN

Given that N = 1n

1

)2(

N=

2n

2

)2(

N

Also given that n1 = 2n2 = 6

So 36nn

2

1 2)2(N

N21 = 8





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PAGE # 7

7340010333

12. The current (i) at time t = 0 and t = respectively for the given circuit is :

(1) 18

E5,

55

E18 (2)

33

E10,

18

E5 (3)

18

E5,

33

E10 (4)

55

E18,

18

E5

Ans. (2)

Sol. at t = 0

inductor offers '' resistance

5 5

1

E

I1

9 E

Req =

5

18

15

96

I1 =

5

18

E =

18

E5

at t =

inductor offers 0 resistance

5 5

4

5

5

4

E

I2

Req = 10

825

2

4

2

5

5

14

2

5

=

10

33

l2 = 33

E10

10

33

E





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PAGE # 8

7340010333

13. Match List-I with List-II :

List-I List-II

(a) h (Plank's constant) (i) [M L T–1]

(b) E (kinetic energy) (ii) [M L2 T–1]

(c) V (electric potential) (iii) [ML2T–2]

(d) P (Linear momentum) (iv) [M L2I–1T–3]

Choose the correct answer from the options given below:

(1) (a) (ii), (b) (iii), (c) (iv), (d) (i) (2) (a) (iii), (b) (ii), (c) (iv), (d) (i)

(3) (a) (iii), (b) (iv), (c) (ii), (d) (i) (4) (a) (i), (b) (ii), (c) (iv), (d) (iii)

Ans. (1)

14. Two satellites A and B of masses 200 kg and 400 kg are revolving round the earth at height of 600 km

and 1600 km respectively. If TA and TB are the time periods of A and B respectively then the value of

TB – TA :

[Given : radius of earth = 6400 km, mass of earth = 6 × 1024 kg]

(1) 3.33 × 102s (2) 4.214 ×103 s (3) 4.24 ×102 s (4) 1.33 × 103 s

Ans. (4)

Sol.

600 km 6400 km

1600 km

B 400 kg

m=6×1023kg

Earth 200 kg

A





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PAGE # 9

7340010333

T = Gm

)R(2T

2/

Thus,

2/332/33AB 107000108000

GM

2TT

= 1.287 × 103 sec

15. Two coherent light sources having intensity in the ratio 2x produce an interference pattern. The ratio

minmax

minmax

II

II

will be:

(1) 1x2

x22

(2)

1x

x2

(3)

1x2

x2

(4)

1x

x22

Ans. (1)

Sol. 2

122

12

212

212

minmax

minmax

)II()II(

)II()II(

II

II

= 2

1

2

2

1

2

2

1

2

2

1

2

1I

I1

I

I

1I

I1

I

I

= 22

22

)1x2()1x2(

)1x2()1x2(

=

)x221x2()x221x2(

)x221x2()x221x2(

=

1x2

x22

2x4

x24

16. Given below are two statement : one labelled as Assertion A and the other is labelled as Reason R.

Assertion A : The escape velocities of planet A and B are same. But A and B are of unequal mass.

Reason R : The product of their mass and radius must be same. M1R1 = M2R2.

In the light of the above statements, choose the most appropriate answer from the options given below:

(1) A is not correct but R is correct

(2) Both A and R are correct but R is NOT the correct explanation of A

(3) A is correct but R is not correct

(4) Both A and R are correct and R is the correct explanation of A

Ans. (3)

Sol. R

GM2VE

2

2

1

1

R

M

R

M





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PAGE # 10

7340010333

17. A diatomic gas, having Cp = R2

7 and CV = R

2

5, is heated at constant pressure. The ratio dU : dQ : dW

(1) 5 : 7 : 2 (2) 5 : 7 : 3 (3) 3 : 5 : 2 (4) 3 : 7 : 2

Ans. (1)

Sol. U = nCvT = T2

R5n

W = nRT

Q = nCpT = T2

R7n

U : Q : W = 5 : 7 : 2

18. An engine of a train, moving with uniform acceleration, passes the signal-post with velocity u and the last

compartment with velocity v. The velocity with which middle point the train passes the signal post is

(1) 2

uv (2)

2

uv 22 (3)

2

uv 22 (4)

2

vu

Ans. (2)

Sol. v2 = 2cv +2a [length of train = 2]

2cv = u2 + 2a

v2 – 2cv = 2

cv – u2

2

vuv

22

c

19. A 5 V battery is connected across the points X and Y. Assume D1 and D2 to be normal silicon diodes.

Find the current supplied by the battery if the +ve terminal of the battery is connected to point X.

(1) ~ 1.5 A (2) ~ 0.86 A (3) ~ 0.5 A (4) ~ 0.43 A

Ans. (4)

Sol. i = A43.010

7.05





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PAGE # 11

7340010333

20. If the time period of a two meter long simple pendulum is 2 s, the acceleration due to gravity at the place

where pendulum is executing S.H.M. is :

(1) 16 m/s2 (2) 9.8 m/s–2 (3) 22 m/s–2 (4) 2 m/s–2

Ans. (3)

Sol. planetg

2T

2

2

planetT

4g

gplanet = 22

Numerical Value Type (la[;kRed izdkj)

This section contains 10 Numerical value type questions.

bl [k.M esa 10 la[;kRed izdkj ds iz'u gSaA

1. In the given circuit of potentiometer, the potential difference E across AB (10 m length) is larger than

E1 and E2 as well. For key K1 (closed), the jockey is adjusted to touch the wire at point J1 so that there is

no deflection in the galvanometer. Now the first battery (E1) is replaced by second battery (E2) for working

by making K1 open and K2 closed. The galvanometer gives then null deflection at J2. The value of 2

1

E

E is

b

a, where a = _______.

Ans. 1

Sol. We know E l

E1 = K(380) ; E2 = K(760)

b

a

2

1

E

E

2

1





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PAGE # 12

7340010333

2. A coil of inductance 2 H having negligible resistance is connected to a source of supply whose voltage is

given by V = 3t volt (where t is in second). If the voltage is applied when t = 0. Then the energy stored in

the coil after 4 s is____ J.

Ans. 144

Sol. v = dt

diL

i = dtL

v4

0

i = dt2

t34

0

=

4

0

2

4

t3

i = 12

E = 2Li2

1

= 2)12(x2x2

1 = 144 J

3. The same size images are formed by convex lens when the object is placed is 20 cm or at 10 cm form

the lens. The focal length of convex lens is _____cm.

Ans. 15

Sol. As size of image is same

So |m||m| 21

21 uf

f

uf

f

21 uf

f–uf

f

as one image is real & other is virtual

f + u2 = – f – u1

2f = – u2 – u1

2f = – (– 10) – (–20)

2f = 10 + 20

f = 15 cm





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PAGE # 13

7340010333

4. A transmitting station releases waves of wavelength 960 m. A capacitor of 256 F is used in the resonant

circuit. The self inductance of coil necessary for resonance is ______×10–8 H.

Ans. 10

Sol. v = n

960

103Vn

8

At resonance C

1L

L = 61622 102561094

960960

C

1

L = 10×10–8 Hz

5. A small bob tied at one end of a thin string of length 1 m is describing a vertical circle so that the maximum

and minimum tension in the string are in the ratio 5 : 1. The velocity of the bob at the highest position is

______m/s. (Take g = 10 m/s2)

Ans. 5

Sol.

v

B

u A

T2

T1

r

r

mg

mg

We know that Tmax. – Tmin = 6mg

and given that Tmax/Tmin. = 5

Solving these Tmax = mg2

15 and Tmin = mg

2

3

T2 = mgr

mvT

2

min ; r

mvmg

2

5 2

v = 5 m/s





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PAGE # 14

7340010333

6. The potential energy (U) of a diatomic molecule is a function dependent on r (interatomic distance) as

3rr

U510

where, and are positive constants. The equilibrium distance between two atoms will

beb

a

2

, where a = _____.

Ans. 1

Sol. U = 10r

–

5r

–3

F = –dr

dU =

11r

10 –

6r

5

At Equilibrium F = 0

11r

10 =

6r

5

r5 =

5

10

r = 5

1

2

= b

a

2

b

a =

5

1

a = 1

b = 5

7. A monoatomic gas of mass 4.0 u is kept in an insulated container. Container is moving with velocity

30 m/s. If container is suddenly stopped then change in temperature of the gas (R = gas constant) is R3

x

. Value of x is _____.

Ans. 3600

Sol. 2

1× n × Mv2 = n . Cv T

= TR2

3.n

T = R3

Mv 2

= R3

30304

= R3

3600





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7340010333

8. The electric field in a region is given byC

NjE

5

4iE

5

3E 00

. The ratio of flux of reported field through

the rectangular surface of area 0.2 m2 (parallel to y – z plane) to that of the surface of area 0.3 m2 (parallel

to x – z plane) is a : b, where a = _____.

[Here j,i , and k are unit vectors along x, y and z-axes respectively]

Ans. 1

Sol. Flux = A.E

For Y-Z plane, A1 = 0.2 i

Then

)i2.0.(jE5

4iE

5

3001

= 0E5

6.0

For X-Z plane, A2 = 0.3 i

j3.0.jE5

4iE

5

3002

= 0E5

2.1

so b

a

2

1

2.1

6.0

1

2

9. In a certain thermodynamical process, the pressure of a gas depends on its volume as kV3. The work

done when the temperature changes from 100°C to 300°C will be ______nR, where n denotes number

of moles of gas.

Ans. 50

Sol. PdV

= dVkV3

=

2

1

V

V

4

4

kv

= 4

kVkV 41

42

232 PkV , 1

31 PkV , 22

42 VPkV , 11

41 VPkV

=4

)TT(nR 12 )TT(nRkVkV 1241

42

= 4

)100300(nR = 50 nR





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7340010333

10. 512 identical drops of mercury are charged to potential of 2v each. The drops are joined to from a single

drop. The potential of this drop is ______V.

Ans. 128

Sol. R radius of bigger drop

r radius of small drop

Volumebigger drop = 512 Volumesmaller drop

3R3

4 = 512 3r

3

4

R = 8r

V = r

Kq

2 = r

kq q =

k

r2

Q = 512 q : charge on bigger drop

Q = k

r1024

V ' = r

KQ=

r8k

r1024k

V' = 128 Volt



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