FIITJEE REVIEW FULL TEST – I

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ANSWERS, HINTS & SOLUTIONS

PAPER–1 (20.4.2017)

Q. No. PPhhyyssiiccss CChheemmiissttrryy MMaatthheemmaattiiccss

1. A D C

2. B D C

3. C C A

4. A D C

5. B A B

6. B A B

7. C B C

8. D C C

9. B C B

10. C C D

11. A, C A, B, C B, C

12. A, B, C, D A, B, D A, C

13. B, D B, C, D A

14. A, D A, B, C D

15. A, B, D A, C, D C

1. 2 6 1

2. 4 3 0

3. 4 2 6

4. 2 8 7

5. 2 1 0

FIITJEE REVIEW FULL TEST – I

Review Full Test-I-(Paper-1)-SOL


2

PPhhyyssiiccss PART – I

SECTION – A

Single Correct Choice Type

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

1. A particle is projected under gravity velocity 2Hg from a poll at a

height H above the level plane. The maximum area on ground in which it can hit the target

(A) 8 H (B) 4 H (C) 2 H (D) None

u

H

Sol. A

For max. Range 2uR tan

g

R tan 2H

2

22

gRH Rtan 1 tan2u

2RH 2H H

4H

R 8H A = 2R 8 H 2. One mole of an ideal monoatomic gas is taken along the process in

which xPV = constant. The graph shown represents the variation of molar heat capacity of such a gas with respect to X. The value of A

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

C

A X

Sol. B

R RC1 1 x

3. If the thread AB is the following system is cut, the acceleration of mass

m2 just after the cutting is (m2 > m1)

(A) 1 2

2

m m gm

(B) 1 2

1

m m gm

(C) 2 1

2

m m gm

(D) 2

1

mg

m

m1 m2

B

A



3

Sol. C

2

2 1m

2

m g m gam

4. A particle of charge q and mass m is projected from the origin with velocity 0

ˆv v i

in a non-

uniform magnetic field 0ˆB B xk

. Here v0 and B0 are positive constants of proper dimensions.

Find the maximum positive x coordinate of the particle during its motion.

(A) 0

0

2mvB q

(B) 0

0

mvB q

(C) 0

0

mv2B q

(D) 0

0

mv2

B q

Sol. A 5. For an equilateral prism, it is observed that when a ray strikes grazing at one face it emerges

grazing at the other. It’s refractive index is

(A) 32

(B) 2

(C) 23

(D) Data not sufficient

Sol. B sin90 sin30 2

30o 30o 120o

60o

90o

6. An unknown particle is bombarded on N14, as a result O17 nucleus is formed and a proton is

emitted. The unknown particle is (A) Neutron (B) particle

(C) Electron (D) Positron Sol. B 4 14 17 1

2 7 1He N O P 7. If 1 and 2 (> 1) are the lengths of an air column for the first and second resonance when a

tunning fork of frequency n is sounded on a resonance tube, then the minimum distance of the anti-node from the top end of the resonance tube is

(A) 2 (2 - 1) (B) ½ (21 - 2)

(C) 23 12

(D) 2 1

2

Sol. C 1 + x = /4

2 + x = 4

3

x = 23 12

2

1

x



4

8. The potential difference between point A & B in a section of the circuit shown is

(A) 5 volt (B) 1 volt (C) zero volt (D) 8 volt

A B 3A

1A 3A

2 2 2 2

1 3V 2V

Sol. D A BV 2 3 3 2 1 V B AV V 8 volt 9. A transverse wave traveling in air is given by y 4sin 2t 4z . This wave now passes through

a medium of x fraction index n = 2. Find the of wave in that medium.

(A) 9 (B)

4

C) 2 (D)

16

Sol. B 10. The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of 108.5

nm. The ground state energy of an electron of this ion will be (A) 3.4 eV (B) 13.6 eV (C) 54.4 eV (D) 122.4 eV

Sol. C

Multiple Correct Answer(s) Type This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. 11. A spool of wire rests on a horizontal surface as shown in figure.

As the wire is pulled, the spool does not stop at contact point P. On separate trials, each one of the forces F1, F2, F3 and F4 is applied to the spool. For each one of these forces the spool

(A) will rotate anti clockwise, if F1 is applied

(B) will not rotate, if F2 is applied (C) will rotate anticlockwise, if F3 is applied (D) will rotate clockwise, if F4 is applied

F1

F2

F3

F4

Sol. A, C 12. A constant force F is applied on a spring block system as shown

in figure. The mass of the block is m and spring constant is k. The block is placed over a smooth surface. Initially the spring was unstretched. Choose the correct alternative:-

(A) block will execute S.H.M.

(B) amplitude of oscillation is Fk

(C) Time period of oscillation is m2k

(D) sum of P.E. and K.E. is maximum at right extreme.

m F

Sol. A, B, C, D



5

13. A uniform rope of mass m and length L hangs from the ceiling. A small wave pulse is created at bottom, then which of following is correct regarding wave pulse.

(A) Rate of change of speed with respect to time varies. (B) Rate of change of speed with respect to time is constant. (C) Rate of change of speed with respect to time is zero. (D) Wave speed varies

Sol. B, D 14. A first L-C circuit has capacitance C1 = C and inductance L1 = L. A second L-C circuit has

C2 = C3

, L2 = 3L and third L-C circuit has C3 = 3C, L3 = L/3, all the three capacitors are charged to

the same potential V & then made to oscillate. Then (A) angular frequency of oscillation is same for all the three circuit (B) Maximum current is greatest in first circuit (C) Maximum current is greatest in second circuit (D) Maximum current is greatest in third circuit Sol. A, D

1wLC

and 2 2m

1 1 CLi CV i V2 2 L

15. An ideal gas undergoes a thermodynamic cycle as shown in

figure. Which of the following statements is/are correct? (A) Straight line AB cannot pass through O. (B) During process AB, temperature decreases while that

during process BC increases. (C) During process BC, work is done by the gas against

external pressure, and temperature of gas increases. (D) During process CA, work is done by the gas against

external pressure and heat supplied to the gas is exactly equal to this work.

O

P

A B

C

1v

Sol. A, B, D

SECTION – C

Integer Answer Type

This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). 1. When an object is placed at a distance of 25 cm from a mirror, the magnification is m1. The

object is moved 15 cm farther away from mirror. Its magnification is m2. If 1

2

m4

m the focal

length of mirror is 10 x cm then x will be? Sol. 2

1

1 1 1 2 1 1v 25 f 25m 25 f __________(1)

2

1 1 1 1 1 1v 40 f 40m 40 f __________(2)

Comprising from (1) & (2)



6

1

1 1 1 125m 40m 25 40

1 2m 4m

2 1

2 2

40m m 1540m 25m 25 40

2 2 1 240m 25m 15 m m = 2

2 240m 100m 15 um

m2 = 1

1 1 1 1f 40 40 20

f = 20 x = 2 2. In an L R series circuit a sinusoidal voltage v = 0v sinwt is applied. It is given that L = 35 mH,

R = 11 , Vrms = 220 V, 50Hz2

and 227

the amplitude of current is 5n A. Find n.

Sol. 4

00 2 2

L

VI

X R

=

2

23

220 2

2235 10 50 2 117

= 220 2 20 5h11 2

n 4

3. Oil enters the bend of a pipe in the horizontal plane

with velocity 4 m/s & pressure 280 3 210 N / m A1/4=A2 as shown. If pressure at Q in (KN/m2) is N 43 . Find N (specific gravity 0.9 & cos37 0.8 )

Sol. 4 1 1 2 2A v A V V2 =16 m/s Applying Bernoulli’s equation

2 21 1 2 2

1 1P v P v2 2

( = 0.9 103 kg/m3)

P

Q

4 m/s

37o

A1

A1/4 = A2

4. In the figure shown A and B are two thin transparent slabs of

refractive indices 1 and 2 (2 > 1) and there is no gap between A and B. A beam of light of wavelength falls on A normally. A part of it reflects at upper face and a part reflects at the interface of A and B and then come back to the previous medium. The phase difference between two reflected waves at a particular

point at any time t is 1 12K t

. Find the value of K.

1

2 t2

t1 A

B

air

air

Sol. 2 5. Three resistors of resistances 2 , 3 and 6 are available to be connected across a battery of

internal resistance 1 and emf. 4V by means of conducting wires. The maximum current (in amp) that can be drown through the cell by using all the given resistances is

Sol. 2 Req = 1

max4vI 2 Amp

1 1



7

CChheemmiissttrryy PART – II

SECTION – A


This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 1. Which of the following is a cyclic oxoacid? (A) H4P2O7 (B) H2P2O6 (C) H5P5O15 (D) H3P3O9 Sol. D H3P3O9 is a cyclic trimetaphosphoric acid.

O

P

P

O

O

PO

OHOH

O

O OH

2. Which quantum number is not related with Schrodinger equation? (A) Principal quantum number (B) Azimuthal quantum number (C) Magnetic quantum number (D) Spin quantum number Sol. D 3. 200°C 120°C

4 2 HeatGypsum

B CaSO .2H O A

strongly heated

C A, B and C are respectively: (A) Plaster of Paris, dead burnt plaster, calcium sulphide (B) Dead burnt plaster, plaster of Paris, lime (C) Plaster of Paris, dead burnt plaster, lime (D) Anhydrous calcium sulphate, plaster of Paris, calcium oxide Sol. C . 200°C 120°C 1

4 4 2 4 22Heat(B) Gypsum (A)

CaSO Dead burnt plaster CaSO .2H O CaSO H OPlaster of Paris

strongly heated

12

CaO+ SO + O2 2Lime(C)

4. Which of the following cell can produce more electrical work?

(consider same electrode pressure of H2 gas in all cases) (A) Pt, H2 |0.1 M NH4Cl||0.1 M CH3COOH | H2,Pt (B) Pt, H2 |0.1 M HCl || 0.1 M NaOH | H2, Pt (C) Pt, H2 |0.1 M HCl || 0.1 M CH3COOK | H2,Pt (D) Pt, H2 |0.1 M CH3COOK ||0.1 M HCl | H2,Pt 4. D

a

cell

c

H0.059E log1 H



8

5. Hybridisation of carbon in C3O2 is (A) sp (B) sp2 (C) sp3 (D)sp3d2. Sol. A O=C=C=C=O 6. Calculate the mass of 2

3SeO in a solution on the basis of following data: 20 ml of M/60 solution of KBrO3 was added to definite volume of -2

3SeO solution. The Bromine evolved was removed by boiling and excess of KBrO3 was back titrated with 5.1 ml of N/25 solution of NaAsO2. The reactions involved are given below (At mass of Se = 79)

SeO3–2 + BrO3

– + H+ SeO4–2 + Br2 + H2O

BrO3– + AsO2

– + H2O Br– + AsO4–3 + H+

(A) 0.084 g (B) 0.042 g (C) 0.84 g (D) 0.42g Sol. A By observing given reactions, we find n factor for KBrO3 is 5 and 6 respectively.

Initial moles of KBrO3 = 20 × -31 ×1060

Equivalent of KBrO3 =

-3120× ×10 ×560

= 1.66 × 10–3

Equivalents of AsO2– reacted with excess of KBrO3 = -315.1× ×10 ×2

25 (n factor of AsO2

– is 2)

= 4.08 × 10–4 Equivalents of KBrO3 in excess = 4.08 × 10–4

moles of excess of KBrO3 = -4104.08×

6

= 6.8 × 10–5 Equivalents of KBrO3 in excess of n – factor 5 = 5 × 6.8 × 10–5 = 3.4 × 10–4 Equivalents of KBrO3 reacted with SeO3

–2 = (1.66 – 0.34) × 10–3 = 1.32 × 10–3 Equivalents of SeO3

–2 = 1.32 × 10–3

Mass of SeO3–2 = 1.32 × 10–3 × 127 = 0.084

2 g

7. The complexes given below are:

M

A

A

en en and M

en

A

A

en

(A) Position Isomers (B) Identical (C) Geometrical Isomers (D) Optical Isomers Sol. B



9

8. Which of the following is most acidic?

List –I (compound)

(A) CH3

O COOC2H5

(B) CH3

O

CH3

O

(C) O O

(D)

CH3O

Sol. C

Compound (pKa)

(A) CH3

O COOC2H5

(1) 10.7

(B) CH3

O

CH3

O

(2) 8.9

(C) O O

(3) 5

(D) CH3

O (4) 17

9.

To distinguish between the above isomeric hydrocarbon molecules, which will work best: (A) Bromine water test (B) O3/H2O, heating and passing evolved gas through lime water (C) O3/H2O, , I2/NaOH (D) Tollen’s test Sol. C Both molecules will respond to A & B; neither will respond to D. PhCH2COOH and PhCOCH3 can be distinguished by iodoform reaction. 10. fΔG for the formation of n–butane is –15.69 kJmol–1 and for isobutane is –21.39 kJmol–1.

Starting with 1 mole of n–butane, calculate the moles of isobutane in the mixture after heating (in the presence of catalyst, e.g. AlCl3) until equilibrium is obtained. (Given that 2.303 RT = 5.7 at 298 K)

(A) 0.045 (B) 0.45 (C) 0.91 (D) 0.091 Sol. C



10

CH3

CH3

CH3

CH3

CH3

1 mole 01-x x

0 0ΔG ΔG iso-butane ΔG n butanef f

= –21.39 + 15.69 = – 5.7 = 2.303 RT logK 5.7 = 5.7 log K log K = 1, K = 10

10

x1x

x = 10 – 10x 11x = 10

x = 10 0.9111

Multiple Correct Answer(s) Type

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. 11. When an inorganic compound (X) having 3c–2e as well as 2c–2e bonds reacts with ammonia gas

at a certain temperature, gives a compound (Y), isostructural with benzene. Compound (X) with ammonia at a high temperature produces a substance (Z). Which of the following is/are correct?

(A) (X) is B2H6 (B) (Z) is known as inorganic graphite (C) (Y) is B3N3H6 (D) (Z) is soft like graphite Sol. A, B, C Diborane, B2H6, is a compound consisting 3c–2e and 2c–2e bonds.

B2H6 + 2NH3 low temp. 200°C2 6 3 3 3 6

(Y)B H .2NH B N H . (Y) has structure similar to benzene, it is

called inorganic benzene. hightemp.2 6 3 x

(Z)B H +NH (BN) . (Z) is a hard substance, and is called

inorganic graphite. 12. A compound X of formula C2H5NO2 on treatment with HNO2 (NaNO2 + HCl) gives off an inactive

gas Y and forms Z. When heated with soda lime, X gives A(CH5N) soluble in acid. Find the true statements among the following:

(A) A gives carbylamine test (B) X is an amino acid (C) Y is CO2 (D) Z is a hydroxyl acid Sol. A, B, D X = H2NCH2COOH Y = N2 Z = HOCH2 – COOH A = H3CNH2 13. Which of the following will show geometrical isomerism?

(A) C C

Cl

Cl C

H

C

H

H

H

HH

(B) C C

Cl

H CH

Cl

CH2



11

(C) H2C C

C

Cl

CH2

Cl

(D) CH CH

C CH2

Cl

Cl

Sol. B, C, D 14. Which of the following statement(s) is/are correct? (A) A triplet carbene is more stable than a singlet carbene

(B) The carbocation, F3C – +

2CH is less stable than the carbocation, +

3F C (C) Methyl carbanion is both isostructural and isoelectronic with ammonia (D) A singlet carbene is a paramagnetic species. Sol. A, B, C

15. Among the following statements which is/are correct (A) Calamine and Siderite are carbonate ores. (B) Argenite and Cuprites are oxide ores. (C) Malachite and Azurites are ores of Cu. (D) Carnallite and Sylvite are chloride ores. Sol. A, C, D Calamine ZnCO3 , Siderite FeCO3 , Argentite Ag2S , Cuprite Cu2O , Sylvite KCl Malachite CuCO3. Cu(OH)2 , Azurite Cu(OH)2. 2CuCO3 , Carnallite KCl. MgCl2. 6H2O

SECTION – C

Integer Answer Type

This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). 1. One mole of nitrogen gas at 0.8 atm takes 38 s to diffuse through a pinhole, whereas one mole of

an unknown compound of xenon with fluorine (XeFn) at 1.6 atm takes 57 s to diffuse through the same hole. Determine the value of ‘n’. [Xe = 131].

Sol. 6 The rate of diffusion depends on the following factors.

r p and r 1M

taking these together, we get

1/2

2 2 1

1 1 2

r p M=r p M

since r 1t

, we can write

1/2

1 2 1

2 1 2

t p M=t p M

or

2

2 22 1

1 1

p tM = Mp t

Identify the script 1 with nitrogen and 2 with unknown gas, we get

2

21.6 56M = ×0.8 37.9

(28 g mol–1) = 244.52 g mol–1

Let the molecular formula of the unknown compound be XeFn. We will have MXe + nMF = 244.52 g mol–1 i.e. [131 + n(19)] g mol–1 = 244.52 g mol–1



12

244.52 -131n = = 5.975 619

2. An ion (atomic number Z) isoelectronic with hydrogen, is in nth excited state. This ion emits two

photons of energies 10.20 eV and 17.00 eV successively to return to first excited state. It can also emit two photons of energies 4.25 and 5.95 eV successively to return to second excited state. Find the value of Z. (Ionisation energy of H atom is 13.6 eV).

Sol. 3 Let excited state be n. Case I: There is a transition to first excited state i.e. 2nd level from nth level.

22 2

1 110.20+17.00 =13.6Z -2 n

…(1)

Case II : There is a transition to second excited state i.e. 3rd level from nth level.

22 2

1 14.25+5.95 =13.6Z -3 n

…(2)

on dividing (1) by (2), we have 2

2

1 14 n1 19 n

-27.20 =10.20 -

n = 6. On putting values of n in (1) or (2) we have Z = 3. 3. RH2 (cation exchange resin) can replace Ca2+ in a sample of hard water as

RH2 + Ca2+ RCa + 2H+. 1L of such a hard water sample is passed through a cation exchange resin bed and the resultant solution requires 20 ml of 0.5 (M) NaOH solution for complete neutralization. Hardness of water is α x 102 ppm w.r.t. Ca2+. What is the value of α?

Sol. 2 RH2 + Ca2+ → RCa + 2H+

Moles of 2+ 1Ca =2

× moles of H+ = 21

× moles of NaOH

= 21

× 20 × .5 × 10–3

wt of Ca2+ = 21

× 20 × .5 × 10–3 × 40

Hardness = 3 6

3

40 20 .5 10 102 10

ppm = 200 ppm

4. How many of Chiral carbons are present in the following compound?

CHO

CH(OH)CH3

Cl

Cl

Sol. 8



13

3

21

CHO

6

4

5

7

8

Cl

Cl

CH3H

OH

5. The relative lowering of vapour pressure of an aqueous solution of a non–volatile solute of

molecular weight 60 (which neither dissociates nor associates in the solution) is 0.018. If Kf of water is 1.86 K mol–1 kg, then what will be the ratio of depression in freezing point to cryoscopic constant?

Sol. 1 2

21

nx = = 0.018n

or, 2 1

1 2

w M = 0.018w M

or, 2

1 2 1

1000w 0.018×1000 0.018×1000= = =1w M M 18

or, m = 1

f

f

ΔT = m = 1,K



14

MMaatthheemmaattiiccss PART – III

SECTION – A


This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 1. If log10x + log10y 2, then the smallest possible value of x + y is (A) 10 (B) 30 (C) 20 (D) none of these 1. C 1. log10x + log10y 2 xy 100 (x + y)2 (x – y)2 + 4xy 400.

2. Let f be a positive function. Let k

1I k

I xf x 1 x dx

, k

2I k

I f x 1 x dx

where 2k – 1 > 0, then

1

2

II

is

(A) 2 (B) k

(C) 12

(D) 1

Sol. C

Given k

1I k

I xf x 1 x dx

….. (1)

k

2I k

I f x 1 x dx

….. (2)

Here, k k

1I k I k

I xf x 1 x dx k 1 k x f k 1 k x 1 k 1 k x dx

= k k k

I k I k I k

1 x f 1 x x dx f 1 x x dx xf 1 x x dx

I1 = I2 – I1

1

2

I 1I 2

3. If a, b, c, d are the roots of the equation x4 + 5x3 + 4x2 + 5x + 3 = 0 then value of (1 + a2)(1 + b2)(1 + c2)(1 + d2) is (A) 0 (B) 1 (C) 4 (D) 16 Sol. A x4 + 5x3 + 4x2 + 5x + 3 = (x – a)(x – b)(x – c)(x – d) Let x = i 1 – 5i – 4 + 5i + 3 = (i – a)(i – b)(i – c)(i – d) (1 + a2)(1 + b2)(1 + c2)(1 + d2) = 0



15

4. If normals to parabola y2 = 4ax at three points, P, Q, R meet at A and S be the focus, then

2SP SQ SR

SA

is (where SP, SQ, SR, SA are distances)

(A) 4a (B) 2a (C) a (D) none of these Sol. C Normal y + tx = at3 + 2at Let A = (h, k) at3 + 2at – ht – k = 0 SP = 2

1a 1 t , SQ = 22a 1 t , SR = 2

3a 1 t

22

1 1 1 22t t 2 t t 2a ha

222 2

1 2 1 2 1 2 2 3 22a ht t t t 2 t t t t

a

SP.SQ.SR = 2 22a h a k a SA 5. Let f(x) satisfies the requirements of Lagrange’s Mean value theorem in [0, 2]. If f(0) = 0 and

1f ' x2

for all x in [0, 2], then

(A) f(x) 2 (B) |f(x)| 1 (C) f(x) = 2x (D) f(x) = 3 for at least one x in [0, 2] Sol. B Let x [0, 2]. Since f(x) satisfies all the conditions of L. M. V. theorem on [0, 2] it also satisfies on [0, x] [0, 2]

1f x f 0 f ' x

x 0

where 0 < x1 < x < 2 i.e., 0 < x1 < 2 f(x) = xf(x1) [f(0) = 0]

|f(x)| = |xf(x1)| = |x| |f(x1)| 12 12 1

1x 2 and f' x2

f x 1

6. If 1x2

then 3 3 7

2 2 4 4 81 2x 2x 4x 4x 8x .....

1 x x 1 x x 1 x x

is

(A) 1 (B) 87

(C) 47

(D) 716

Sol. B

(1 + x + x2)(1 – x + x2)(1 – x2 + x4) + ….. n 1 n2 21 x x

= n n 12 21 x x

ln(1 + x + x2) + ln(1 – x + x2) + ln(1 – x2 + x4) + ….. = n n 12 2ln 1 x x

On differentiation both side

n n 1

n n 1

3 n 2 1 n 1 2 1

2 2 4 2 2 2

1 2x 2x 4x 1 2x 2 x 2 x.....1 x x 1 x x 1 x x 1 x x

as n and 1x2



16

RHS = 1 1 81 1 712 4

7. Let f(x) = sin3 x + sin2 x, where x2 2

. The values of for which f(x) has exactly one

minima and exactly one maxima is/are

(A) 3 3, 2 2

(B) 3 3,

2 2

(C) 3 3, {0}2 2

(D) (1, 1)

Sol. C

f ' x 0 2sin x3

or sin x = 0

21 13

and 0

8. x = f(t) cos t + f(t) sin t, y = –f(t) sin t + f(t) cos t, then

12 2 2dx dy dt

dt dt

equals

(A) f(t) + f(t) + c (B) f(t) + f(t) + c (C) f(t) + f(t) + c (D) f(t) – f(t) + c Sol. C

dx f t cos t f t sin t f t sin t f ' t cos t f t f ' t cos tdt

dy f t sin t f t cos t f t cos t f ' t sin t f t f ' t sin tdt

2 2dx dy

dt dt

= 1

22 22 2f t f ' t cos t sin t f t f ' t

12 2 2dx dy dt f " t f t c

dt dt

9. The sides AB, BC and CA of a triangle are c, a, b respectively if a2 + b2 = 1993c2, then

cot Ccot A cotB

is equal to

(A) 1996 (B) 996 (C) 1990 (D) none of these Sol. B a2 + b2 – c2 = 1992c2

cot Ccot A cotB

= 2

cosC sin A sinBsin C

= 2 2 2

2a b c

2c = 996.

10. A test paper has four section A, B, C, D section A, B, C, D have 7, 4, 5, 7 question respectively.

Each question in section C has alternate number of ways in which student can attempt the question paper, taking atleast two question from each section is

(A) 222 (B) 27 4 52 1 2 1 2 1

(C) 27 4 52 8 2 5 3 1 (D) 27 4 52 8 2 5 3 11 Sol. D



17

Selecting atleast two question from section A is 7 7 7

0 12 C C

from section B is 4 4 40 12 C C from section C is 5 10

13 1 C

Multiple Correct Answer(s) Type This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. 11. A player throws an ordinary die with faces numbered 1 to 6, whenever he throws 1, he gets a

further throw. The probability of obtaining a total score of exactly r is

(A) r 11 115 6

if 1 < r 6 (B) r1 115 6

if 1 < r 6

(C) r 6 r 11 1 15 6 6

if r > 6 (D) r 6 r1 1 15 6 6

if r > 6

Sol. B, C If 1 < r 6 it can be occur by following ways

r1 1 1 1 1 1 1.....6 6 6 6 6 6 6 r

1 115 6

if r > 6 then 1 1 1 ….. 1 6 + 1 1 ….. 1 5 + ….. + 1 1 ….. 1 2

= r 6 r 11 1 15 6 6

12. Values of R for which A–1 – I is singular, where 6 2 2

A 2 3 12 1 3

and I is third order unit

matrix, are

(A) 12

(B) 14

(C) 18

(D) 12

Sol. A, C 1A I 0 I A 0

1A I 0

Let 1k

A kI 0 k = 2, 8 13. Area of the region bounded by the curves y = 2x, y = 2x – x2, x = 0 and x = 2 is given by

(A) 3 4log2 3

(B) 3 4log2 3

(C) 43log23

(D) none of these

Sol. A



18

Required area = 2

x 2

0

2 2x x dx

= 2x 3

2

0

2 xxlog2 3

= 4 8 14log2 3 log2

= 3 4log2 3

(0, 1)

(0, 0) (2, 0)

(1, 1)

(2, 4)

14. In how many ways can 99 books be distributed in 3 students so that the third student gets books

in multiple of 3 (A) 1700 (B) 1919 (C) 5151 (D) 1717 Sol. D Let, the students receive x, y and 3z books (since, third student gets books in multiple of 3). Thus, x + y + 3z = 99 x + y = 99 – 3z where 0 z 33 Thus, number of ways = 99 3z 2 1 100 3z

2 1 1C C 100 3z

Total number of ways = 33

z 0

3 33 34100 3z 34002

= 3400 – 1683 = 1717

15.

sin xf x3 x

is

(A) continuous for all x R (B) continuous and differentiable for all x R (C) continuous and differentiable everywhere except for x [–3, –2) (D) f(x) is twice differentiable for all x R Sol. C

0 ; x R [ 3, 2)f x

N.D. ; x [ 3, 2)

f(x) is continuous and differentiable for all x R – [–3, –2)

SECTION – C

Integer Answer Type

This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive).

1. If x sin14

then value of 4x(1 + x – 2x2) is _____

Sol. 1

14

sin 4 = cos 3

8 sin3 – 4 sin2 – 4 sin + 1 = 0 2. If tangent and normal to a rectangular hyperbola xy = 42 at a point cut off intercepts a1 and a2 on

the x–axis and b1, b2 on the y–axis then a1a2 + b1b2 is equal to _____ Sol. 0



19

OA ODOB OC

a1a2 + b1b2 = 0

B D

C O A

3. The medians of a triangle ABC make angles , , with each other then value of

cot cot cot6cot A cotB cotC

is _____

Sol. 6 BGC =

3AD AG2

; 3BE BG2

; 3CF CG2

BC2 + AB2 = 2(BE2 + CE2)

BG2 = 2 2 21 2a 2c b9

F

A

B D C

E G

2 2 2 21CG 2a 2b c9

2 2 2BG CG BCcos2BG CG

2 2 21 b c 5acot9 4 ar BGC

= 2 2 2b c 5a

12

= ar(ABC)

cot + cot + cot = 2 2 2a b c

4

2 2 2a b ccot A cotB cotC

4

4. A and B are two persons standing in the frame as shown in figure. A can take a step eastward or

northward with equal probabilities while B can take a step westward or southward with equal probabilities. In each move they can step only to an adjoining square (having a common side). Both A and B take 3 step at random. The probability that they will meet in a shaded square is p/q where p and q are co–prime then (p + q)/3 is _____

Sol. 7

3 3 3 31 1 1 1

23

1 1 C C C C 1 1

2

= 516

N

S

W E

B

A 5. ABCD is a cyclic quadrilateral with AC BD and ‘O’ is center of its circumcircle, then

OA OB OB OC OC OD OD OA

is equal to (radius of circle is 5) _____ Sol. 0



20

OA OC BD

OB OD AC

OA OC OB OD 0

A

B D

C

O


ANSWERS, HINTS & SOLUTIONS

PAPER–2 (20.42017)

Q. No. PPhhyyssiiccss CChheemmiissttrryy MMaatthheemmaattiiccss

1. C D A

2. B A B

3. B C D

4. D B A

5. C A A

6. C C B

7. A D C

8. C B A

9. D C B

10. A B C

11. A D D

12. C A B

13. A B A

14. C A A

15. A, B, D A, B, C, D A, D

16. A, B A, B, C, D A, B

17. B A, B, D D

18. A, B, C A, B, D B, D

19. A, B, C, D B, C A, B, C

20. A, B, C, D A, B A, D

FIITJEE REVIEW FULL TEST – II

Review Full Test-II-(Paper-2)-SOL


2

PPhhyyssiiccss PART – I

SECTION – A

Straight Objective Type

This section contains 8 multiple choice questions numbered 1 to 8. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. Electromagnetic radiations fall on a metallic body whose work function is 2eV. For a particular

radiation of frequency , the maximum kinetic energy of the photoelectrons is found to be 4 eV.

What would be the maximum kinetic energy of photoelectrons for the radiation of frequency 3

5?

(A) 6 eV (B) 3

14eV

(C) 8eV (D) 3

10eV

Sol. C T1 T2 = h(v1 v2) T2 = T1 + h(v2 v1) 2. Two charges q1 and q2 are kept on x axis and the variation of

electric field strength at different point on x axis is described in the adjacent figure. Choose correct alternative

(A) q1 negative, q2 negative 1 2q q

(B) q1 positive, q2 negative 1 2q q

(C) q1 positive, q2 negative 1 2q q

(D) q1 negative, q2 positive 1 2q q

q1 q2

Sol. B 2q is less than 1q since in middle So q1 + w & q2 – w 3. Two atoms interact with each other according to the following force F and potential k diagram.

What is their equilibrium separation.

u w

F

x (Separation) y z

U

x (Separation)

(A) the separation u which is equal to y. (B) the separation u which is equal to z (C) the separation w which is equal to y (D) the separation w which is equal to z Sol. B



3

4. A short linear object of length b lies along the axis of a concave mirror of focal length f at a distance u from the pole of the mirror, the size of the image is approximately equal to:

A) 1/2u fb

f

(B) u f

u

C) 1/2fb

u f

(D) 2fb

u f

Sol. D 5. Two parallel ray are traveling in a medium of refractive index 1 4 / 3 one of the ray passes

through a parallel glass slab of thickness ‘2t’ & refractive index 2 3 / 2 the optical path difference with respect to given medium between the two rays due to the glass slab will be

(A) t8

(B) 3t

C) t4

(D) 2t9

Sol. C

3 9 t22t 1 2t 14 8 4

3

6. Along cylindrical conductor of radius R contains current of uniform

density J. Two small holes of radius r r d are drilled symmetrically throughout the length of the cylinder. The axes of the holes are parallel to the axes of cylinder and at distance d from it in the same plane. Magnetic field on the axes of one of the hole will be :

(A) 0J d2

(B) 0J rd 22

C) 2

0J rd2 2d

(D) None of these

R

d d

Sol. C

due to full cylinder 0J d2

Due to absent

0 20i Jr

2 2d 4d

So, 2

0J rd2 2d

7. A conducting wire is bent to form a quarter of circle of radius r

and moved in a uniform magnetic field of induction B as shown in the figure. The emf induced across the ends of the wire will be

(A) Bv0r (B) 2

Bv0r

(C) 4

Bv0r (D) Bv04r

×

×

×

× × × ×

×

×

× ×

× ×

×

× ×

× ×

× ×

× × ×

× ×

× ×

× v0

Sol. A Draw projection of the length of the arc perpendicular to the velocity V = Brv0



4

8. A battery of internal resistance 4 is connected to the network of resistance as shown. In order to give the maximum power to the network, the value of R should be

(A) 49 (B) 8

9

(C) 397 (D) 11

7

4

E R

R

R R

R 4R

6R

Sol. C

Comprehension Type

This section contains 3 paragraphs. Based upon paragraph 2 multiple choice questions have to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Questions 9 and 10 Read the following write up carefully and answer the following questions: In the standard Young’s double slit experiment shown, the source is monochromatic and a plate of thickness ‘p’ is placed in front of the upper slit whose refractive index is varying with time t as n = 0n kt , where k<<1, Taking O as

the origin and the co-ordinate axis as shown. Find

D

d S1

S2

Screen n = n0+kt

O

p

x

y

S

9. The Co-ordinates of central maxima at any time t is

(A) 02DD, n kt 1 pd

(B) 0

DD, n kt 1 p2d

(C) 0DD, n kt 1 pd

(D) 0

DD, n kt 1 pd

Sol. D

y = D n 1 pd

= 0D n kt 1 pd

Co-ordinates of central maxima are

0DD, n kt 1 pd

10. Velocity and acceleration of central maxima at any time t.

(A) ˆDkpjv

d

(B) ˆDkpjv 2

d

(C) ˆDkpjv

2d

(D) ˆ4Dkpjv

d

Sol. A

central max ima

ˆDkpjvd

Central maxima moves along positive y-axis with a constant velocity.



5

Paragraph for Questions 11 and 12 Read the following write up carefully and answer the following questions: A small sphere of mass 1 kg is rolling without slipping with linear

speed 200v7

m/s at the horizontal plane. Now it ascends to a

sufficiently rough inclined plane BC such that slipping does not occur. It leaves the inclined plane at point C.

A B

C

30o 1 m

11. Find its linear speed at point C.

(A) 1007

m/s (B) 507

m/s

(C) 10035

m/s (D) 20035

m/s

Sol. A 12. Find the ratio of rotational and translational kinetic energy of the sphere when it strikes the

ground after leaving from point C.

(A) 25

(B) 23

(C) 16

(D) 12

Sol. C

Paragraph for Questions 13 and 14 Read the following write up carefully and answer the following questions: A small sphere is charged uniformly and placed in air (k = 1) at point A so that at point B (8, 7) electric field strength is ˆ ˆ54i 72 j NC–1 and potential is

+900 V.

x

y

z 13. The magnitude of charge over sphere is

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

Sol. A 14. Coordinates of point A where sphere is placed, are

(A) (0, 0) (B) (1, 2) (C) (2, –1) (D) None

Sol. C (13 - 14)Let charge is q and placed at (x, y)

Distance of B and centre C is 2 2r 8 x y 7

2

K.Q 90r

……….. (1)



6

KQ 900r

………. (2) 2 2

2

2

K Q900 900r

KQ 90r

Q = 1000 x 10–9 From (1) and (2) x = 2, y = –1

Multiple Correct Answer(s) Type

This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. 15. Figure shows the variation of energy with the orbit radius r of

a satellite in a circular motion. Mark the incorrect statements: (A) C shows the total energy, B the kinetic energy and A the

potential energy of satellite. (B) A shows K.E. & B the total energy and C the P.E. of

satellite. (C) As B are the K.E and P.E. and C the total energy of

satellite. (D) C and A are the K.E. and P.E. respectively and B the total

energy of satellite.

A

C B r

Energy

Sol. A, B, D

16. Suppose the potential energy between electron & proton at a distance r is given by 2

3

ke3r .

Application of Bohr’s theory of hydrogen atom in this case shown that : (A) Energy in the nth orbit is proportional to n6 (B) Energy is proportional to 3m (m : mass of electron) (C) Energy in the nth orbit is proportional to 2n (D) Energy is propotional to 4m (m = mass of electron) Sol. A, B

KE = 12

mv2 = 2

3

3Ker

nhmvr2

___________(1)

U = 2

3

Ke3r

F = 2 2

4

dU 3Ke mvdr rr

3r mv2 = 3 Ke2__________(2)

2

2

2 rm1

2

2

3kvn

2

mrn

22

2 1Vr1 n



7

22

1Vr4n

8

4 2

nm n

6

23

1 nmv2 m

VE 6

3

nm

17. In the given figure, brv

= velocity of boat with respect to still water

rv

= velocity of water W = width of river Now, if | brv

| = 2| rv

|

vbr

vr

W

(A) the minimum drift would be zero & corresponding value of is 90. (B) the minimum drift would be zero & corresponding value of is 120.

(C) the minimum drift would be 3W & corresponding value of = 120 (D) the minimum drift would be 2W & corresponding value of = 90 Sol. B 18. When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have

maximum kinetic energy TA eV and de-Broglie wavelength A . The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB = (TA – 1.50) eV. If the de-Broglie wavelength of these photoelectrons is B A2 , then (A) the work function of A is 2.25 eV (B) the work function of B is 4.20 eV (C) TA = 2 eV (D) TB = 2.75 eV

Sol. A, B, C 19. The energy, the magnitude of linear momentum and orbital radius of an electron in a hydrogen

atom corresponding to the quantum number n are E, P and r respectively. Then according to Bohr’s theory of hydrogen atom

(A) EPr is proportional to 1n

(B) P/E is proportional to n

(C) Er is constant for all orbits (D) Pr is proportional to n Sol. A, B, C, D



8

20. Three infinitely long thin wires, each carrying current I in the same direction are in the x – y plane in a gravity free space (see Fig.). The central wire is at x = 0 while the other two wires are at x = d . The net magnetic field

(A) cannot be zero beyond B (i.e. x d ) (B) cannot be zero beyond A (I,e. x < –d) (C) will be zero between x = 0 and x = d (D) will be zero between x = 0 and x = –d

Sol. A, B, C, D



9

CChheemmiissttrryy PART – II

SECTION – A


This section contains 8 multiple choice questions numbered 1 to 8. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. In the reaction

O

Cl

[X](i) C2H5O

(ii) H3O+

What is the IUPAC name of product [X]? (A) Hemiacetal (B) Cyclic Ester (C) 2 – chloro cyclobutan – 1 - ol (D) ethyl -1-cyclopropylmethanoate Sol. D ethyl -1-cyclopropylmethanoate

OC2H5

O

2. Two gases in adjoining vessels were brought into contact, by opening a stop cock between them,

one vessel contain 0.25 L volume & NO at 800 torr at 220 K. The other measured 0.100 L and contained O2 at 600 torr at 220 K. The reaction to form N2O4 (s) exhausts the limiting reactant completely. Neglecting the vapour pressure of N2O4, What mass of N2O4 is formed?

(A) 0.402 g (B) 4.02 g (C) 6.716 g (D) 0.6716 g Sol. A 2NO + O2 N2O4(s)

n(NO) = PVRT

= 0.0146 mol NO at 220 K

n(O2) = 0.00437 mol O2 at 220 K in the above reaction O2 is limiting reagent. So pressure will be due to NO gas left = 0.0059 mol NO

nRTp =V

= 0.304 atm = 231 torr

Limiting reagent will decide the wt of N2O4 = 0.00437 × 92 = 0.402 g N2O4 3. x% oleum in water is treated with 0.5L of 2.75 M Ca(OH)2 solution. The resulting solution required

15.7 g of H3PO3 solution for complete neutralization. Calculate the amount of free SO3 in oleum. (A) 6.88 (B) 31.2 (C) 68.8 (D) 3.12 Sol. C Total Meq of Ca(OH)2 taken initially = 2.75 × 2 × 0.5 × 1000 = 2750 Meq

Meq of Ca(OH)2 left behind = 15.7 ×100041

= 383 Meq

Meq of H2SO4 (Total) = 2750 – 383 = 2367 Meq



10

Wt of H2SO4 = 49×2367 =115.481000

gm = w

(100 – w) + 1.225 w = 115.48 0.225 w = 15.48 w = 15.48/0.225 = 68.8 gm Amount of free SO3 in oleum = 68.8 gm 4. Two hydrogen atoms collide head on and end up with zero kinetic energy. Each atom then emits

a photon of wavelength 121.6 nm. What is the principal quantum number of excited level. (A) 3 (B) 2 (C) 4 (D) 5 Sol. B Wavelength emitted in UV region and thus n1 = 1; For H atom

H 2 2

1 1 1= R -λ 1 n

7–9 2 2

1 1 1=1.097×10 -121.6×10 1 n

n = 2 5. A tetra-atomic molecule (A) on reaction with nitrogen (I) oxide, produces two substances (B) and

(C). (B) is a dehydrating agent while substance (C) is a diatomic gas which shows almost inert behaviour. The substances (A), (B) and (C) are:

(a) P4, P4O10, N2 (b) P4, N2O5, N2 (c) P4, P2O3, Ar (d) P4, P2O3, O2 Sol. A 4 2 4 10 2

(A) (c)(B)P +10N O P O +10N

6. Which of the following is not known? (A) KHF2 (B) PbCl2 (C) PbI4 (D) BiF5 Sol. C Pb4+ is an oxidizing agent and readily changes to Pb2+ I– is a reducing agent. 7.

C6H5

OH

OH 2NH OHH

HA B C

HC E

HB D

Which of the following is correct? (A) B and C are geometrical isomers of each other. (B) B and C both have one chiral centre each. (C) D and E both have two chiral centres, each. (D) All above are correct. Sol. D



11

C6H5

OH

OHC6H5

OH

+H+

C6H5

OH+

C6H5

O

NH2OH

H+

(Pinacole-Pinacolone rearrangement)(A)

N

C6H5

OH

+

N

C6H5

OH(C)(B)

N

C6H5

OHN

C6H5

OH2

H+

+

N

C6H5

+

N

C6H5

OH

N

C6H5O

H

H2O

(Beckmann Rearrangement)(D)

C6H5

N

OH(C)

H+

N

C6H5

OH

N

C6H5

O

H

(E) 8. Below is a figure showing splitting of degenerate d-orbitals in an octahedral field.

Which of the following statement is true? (A) Stronger the ligand, smaller is the splitting.

d-orbital in free ion.

average energy of d-orbitals.

0

dxy, dyz, dzx Splitting of d-orbitals

in an octahedral crystal field.

t 2 g

dx²y², dz² eg.



12

(B) Higher the oxidation state, greater is the magnitude of ∆o. (C) Larger Δo, absorb light of larger λ, and thus smaller ν (D) All of above are correct. Sol. B

Comprehension Type


Paragraph for Question Nos. 9 to 10 Read the following write up carefully and answer the following questions: Carbonyl compounds having or without H do undergo many named organic reaction which find lots of use in synthesis of many organic compounds; an application of those have been asked below: 9. The major product of the following reaction after acidification is:

OHCH C

H3C

H3C

O

H H2C CH C

O

CH3

(A) O

(B) O

(C) O

(D) O

Sol. C

OHCH C

H3C

H3C

O

H H2C CH C

O

CH3C C

H3C

H3C

O

H

1, 4 addition

C C

H3C

H3C

O

H

H2C CH C

O

CH3

2

OH aldolH O

O

10.

C6H5CHO CH3 C

O

O C

O

CH3+ Base

forms a 'x' membered cyclic intermediate. 'x' will be

(A) 4 (B) 6 (C) 5 (D) 8



13

Sol. B

Ph C

O

H CH2 C

O

O C

O

CH3+ Base_

O

C

C

CH2

O

CO

O CH3

Ph

O

C

C

CH2

O

CO

O Me

Ph

_

_

Paragraph for Question Nos. 11 to 12 Read the following write up carefully and answer the following questions: A yellow inorganic solid (P) when treated with concentrated HNO3 gives a brown gas (Q). (P) upon boiling with Na2SO3 gives a clear solution (R) which upon acidification gives a colourless gas (S) and a white turbidity (T). 11. The yellow solid (P) can show which of the following properties? (A) It is rhombic in its most stable allotropic forms. (B) It burns in O2 to give a gas (S) which can react with acidified K2Cr2O7 to give green

colouration. (C) It reacts with concentrated H2SO4 to give the gas (S) (D) All of the above. Sol. D 12. Compounds Q, R, S and T respectively are: (A) NO2, Na2S2O3, SO2 and S (B) NO2, Na2SO4, SO3 and NaOCl (C) NO2, Na2S2O3, SO2 and Na2SO4 (D) Br2, NaBr, HOBr and NaOBr.



14

Sol. A

Sconcentrated HNO3

NO2(P) (Q)

boiling withNa2SO3 Na2S2O3

(R)

H+

SO2 + S(S) (T)

Most stable allotrope of S is rhombic S. S + O2 → SO2

SO2 + K2Cr2O7 H 2

4SO + Cr+3(green) S + H2SO4 → SO2 + H2O.

Paragraph for Question Nos. 13 to 14

Read the following write up carefully and answer the following questions: Vapour pressure of a solvent is the pressure exerted by vapour when they are in equilibrium with its solvent at that temperature. The vapour pressure of solvent is dependent of nature of solvent, temperature, addition of non-volatile solute as well as nature of solute to dissociate or associate. The vapour pressure of a mixture obtained by mixing two volatile liquids is given by PM = P0

A A + P0B B

where P0A and P0

B are vapour pressures of pure components A and B and A , B are their mole fractions in mixture. For solute solvent system, the relation becomes. PM = P0

A A where B is non-volatile solute. 13. The amount of solute (mol. Wt. 60) required to dissolve in 180 g water to reduce the vapour

pressure to 4/5 of the pure water. (A) 120 g (B) 150 g (C) 200 g (D) 60 g Sol. B

Let 2H OP = Po Psolution =

54 PO,

PoPs = 15

Po

o ssolute

o

P PP

15

= n180n+18

n 1=n+10 5

, n = 2.5

weight of solute = 2.560 = 150gm 14. A mixture of two volatile liquids ‘A’ and ‘B’ of 1 and 3 moles respectively has a V.P. of 300 mm at

270C. If one mole of ‘A’ is further added to this solution, the vapour pressure becomes 290 mm at 270C. The vapour pressure of ‘A’ is:

(A) 250 mm (B) 316 mm (C) 220 mm (D) 270 mm Sol. A



15

Initially A = 41

; B = 43

.

Ps = A PAo

+ B PBo

300 = 41

PAo +

43

PBo

... (1)

In new mixture

A = 52

, B = 53

PS = 290 = 52

PAo +

53

PBo ... (2)

From equation 1 & 2 PA

o = 250mm Hg

Multiple Correct Answer(s) Type This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. 15. Which of the following statement(s) is/are correct? (A) The sugar which contain free aldehydic group can reduce Tollen’s reagent or Fehling’s

reagent. (B) Nucleotide is a phosphate ester of nucleoside. (C) Enzymes are the biological catalyst which increases the rate of metabolism and they are

highly specific in nature. (D) Proline and glycine are sometimes known as "helix breakers" because they disrupt the

regularity of the α helical backbone conformation Sol. A, B, C, D 16. Which of the following is/are correct order? (A) I2 < F2 < Br2 < Cl2 [Increasing Bond Energy] (B) HF < HCl < HBr < HI [Increasing acid strength] (C) BiH3 < AsH3 < PH3 < NH3 [Increasing base strength]

(D) H2O < H2S < H2Se < H2Te [Increasing acid strength]

Sol. A, B, C, D 17. Which of the following is correctly matched.

Formula Cryoscopic measurement

Total unit charges Structure

(A) CoCl3.6NH3 4 particles 6 [Co(NH3)6]3+.3Cl- (B) CoCl3.5NH3 3 particles 4 [Co(NH3)5Cl]2+.2Cl¯ (C) Co(NO2)3.3KNO2 4 particles 4 K3[Co(NO2)6] (D) Co(NO2)3.2KNO2.NH3 3 particles 4 K2[Co(NH3)(NO2)5]

Sol. A, B, D 18. Given below is an ester which can give following reactions:

R C

O

O (A) It reacts with sodium ethoxide to give a -keto ester (P) provided atleast two -H are present.



16

(B) (P) upon saponification and treatment with H+ followed by heating undergoes decarboxylation and a ketone(Q) is produced.

(C) Q responds to Tollen’s reagent test.

(D) Q upon treatment with ||

2

O

Br ZnCH C OEt , H+ and heat can give -unsaturated ester. Sol. A, B, D

R CH2 C

O

OEtEtO

-

(claisen condensation)R CH2 C

O

C

R

H

C

O

OEt(P)

EtO

O

CC

O

CCH2

R R

OEt

-

(i) H2O/OH-(ii)H

+

(P) R CH2 C

O

CH

R

C

O

OH + EtOH

R CH2 C

O

CH2 R + CO2

(Q)

R CH2 C

O

CH2 RBrZnH2C C

O

OEt(i)

(ii) H+

R CH2 C

OH

CH2 C

H2C R

O

OEt

Reformatsky reaction

R CH2 C CH C

H2C

O

OEt

R 19. For the dissociation

A2B3(g) 2AB(g) + )g(B21

2

If, M = Molecular mass of A2B3(g) D = vapour density of equilibrium mixture P0 = Initial pressure of A2B3(g) then, identify the correct statment(s) (A) Equilibrium pressure can be expressed as 02P D

M

(B) Equilibrium pressure can be expressed as 0P M

2D



17

(C) Degree of dissociation of A2B3(g) can be expressed as M- 2D3D

(D) Increase in temperature will increase the magnitude of D. Sol. B, C 20. Quinhydrone is charge transfer complex of quinone and hydroquinone. Which of the following

statement is/are incorrect with respect to quinhydron? (A) Quinone is e acceptor (B) Hydroquinone is e donor (C) Quinone is e donor (D) Hydroquinone is e acceptor Sol. A, B O

O

OH

OHQuinhydron



18

MMaatthheemmaattiiccss PART – III

SECTION – A


This section contains 8 multiple choice questions numbered 1 to 8. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. Let A(x1, y1) and B(x2, y2) be any two points on the parabola y = ax2 + bx + c and let C(x3, y3) be

the point on the arc AB where tangent is parallel to the chord AB then (A) 2x3 = x1 + x2 (B) x3 = |x1 – x2| (C) x1 + x2 + x3 = 0 (D) none of these Sol. A

2 13

2 1

f x f xf ' x

x x

2 22 2 1 1

3 32 1

ax bx c ax bx cf ' x 2ax b

x x

2x3 = x1 + x2 2. If x = cot + tan ; y = sec – cos , then 4 / 3 2 / 3 2 / 3 4 / 3x y x y is equal to (A) 0 (B) 1 (C) 2 (D) depends on Sol. B As x tan = sec2 ; y sec = tan2 x2 y = sec3 and xy2 = tan3

Now 2 / 3 2 / 3 2 / 3 2 / 34 / 3 2 / 3 2 / 3 4 / 3 2 2 3 3x y x y x y xy sec tan 1

3. The value of 1

1 2 1 2

1

1 1sin x cos x dx2 2

, (where [.] denotes greatest integer function)

is

(A) 0 (B) 2

(C) (D) 2 Sol. D

2 1x2

is either 0 or 1

if 2 1x 02

then 1 2 1 21 1sin x cos x

2 2

if 2 1x 12

then 1 2 1 21 1sin x cos x

2 2

4. If 2f(sin x) + f(cos x) = x; then range of f(x) is

(A) 2 , 3 3

(B) 2, 3 3

(C) R (D) none of these Sol. A



19

x x2

2f cos x f sinx x2

f(cos x) = x3

1f x cos x3

. Hence range of f (x) is 2 ,

3 3

.

5. The radius of circle passing through the vertices of the triangle formed by a pair of lines 7x2 + 8xy – y2 = 0 and 2x + y = 1 is given by

(A) 4 1013

(B) 6 1013

(C) 8 1013

(D) none of these

Sol. A x2 + y2 + 2gx + 2fy = 0 will be circumcircle of given x2 + y2 + (2gx + 2fy) (2x + y) = 0 will represent 7x2 + 8xy – y2 = 0

1 4g 2f g 1 2f7 4 1

4 10r13

6. Suppose f(x) = x3 + ax2 + bx + c, where a, b, c are chosen respectively by throwing a die three

times then the probability that f(x) is an increasing function is

(A) 29

(B) 49

(C) 59

(D) 13

Sol. B f(x) > 0 x R a2 3b a and b can be selected by 16 ways

p = 6

13

16 C6

7. A variable plane lx + my + nz = p (p > 0) l, m, n are D.C. of normal to plane, intersects the

coordinate axes at A, B, C then area of ABC may be

(A) 23 p2

(B) 2p

2

(C) 23 3 p2

(D) p2

Sol. C

2 2 2 1/ 32 2 2l m n l m n

3

lmn 13 3



20

= 2p

2lmn.

8. The value of

3x2 1 x

2x

3x 1lim4x 1

is

(A) 0 (B) 1 (C) 2 (D) 3 Sol. A

3 3

3 3

x x1 xx 1 x2 x 2

2 1 x 1 x2x x x 22

1 1x x3x 1 3 33 3L lim lim lim114 44x 1 xx44

=

3

3

x1 xx

1 x

x x 2

73 12lim lim 1

14 x4

Now,

3x1 x

x

3 3lim 04 4

and

33

2 2

xx 1 1x x x 11 x 4 4

x x2 2

7 712 12lim 1 lim 1

1 1x x4 4

= 7

12e

L = 0 × 7

12e = 0

Comprehension Type


Paragraph for Question Nos. 9 to 10 Read the following write up carefully and answer the following questions: Limiting points of a system of co–axial circles are the centres of the point circles belonging to the family (point circle means radius of circle is zero). Co–axial circles have same radical axis. 9. Co–ordinates of the limiting points of the system of circles determined by the two circles x2 + y2 + 5x + y + 4 = 0 and x2 + y2 + 10x – 4y – 1 = 0 are (a, b); (c, d) then |a + b + c + d| is (A) 3 (B) 6 (C) 0 (D) 1 Sol. B 10. Equation of circle orthogonal to co–axial circle given by question (1) can be (A) x2 + y2 + 2x + 4y + 4 = 0 (B) x2 + y2 + 2x + 4y + 2 = 0 (C) 2x2 + 2y2 + x + 5y – 3 = 0 (D) 2x2 + 2y2 + x + 5y + 3 = 0 Sol. C x2 + y2 + 5x + y + 4 + (x2 + y2 + 10x – 4y – 1) = 0 ( –1)

2 2 5 10 1 4 4x y x y 0

1 1 1



21

radius = 0 = 12

, 16

Limiting points are (–2, –1) and (0, –3) Circle passing through limiting point will be orthogonal to co–axial circle.

Paragraph for Question Nos. 11 to 12

Read the following write up carefully and answer the following questions: Let w be a non real cube root of unity then w3 = 1 and 1 + w + w2 = 0 11. Number of distinct elements in the set {(1 + w + w2 + ….. + wn)m; m, n N} is (A) 3 (B) 5 (C) 6 (D) 7 Sol. D 12. The least possible degree of a polynomial with real coefficients having roots 2w, (2 + 3w), (2 + 3w2), (2 – w – w2) is (A) 3 (B) 5 (C) 6 (D) 7 Sol. B (1 + w + w2 + ….. + wn)m can be –1, 0, w, w2, 1, –w, –w2 therefore total number in given set will be

7 (2 + 3w) and (2 + 3w2) are conjugate of each other therefore least degree will be 5

Paragraph for Question Nos. 13 to 14 Read the following write up carefully and answer the following questions: The number of circular permutations of n distinct objects is (n –1)! In this case, anticlockwise and clockwise orders of arrangement are regarded as distinct permutations. But if anticlockwise and clockwise order of arrangement are not distinct, the number of circular

permutations of n distinct items = 1 n 1 !2

e.g. arrangement of beads in a necklace, arrangement of

flowers in a garland are arrangements of the above type. 13. The chief ministers of 11 states of India meet to discuss food problem. The number of ways they

can seat themselves at a round table so that Punjab and Bihar chief ministers sit together is (A) 9! × 2! (B) 9! (C) 10! (D) none of these Sol. A Consider the Punjab and Bihar chief ministers as one PB + 9 others. These can be arranged in

(10 – 1)! = 9! Ways. Corresponding to each of these, both Punjab and Bihar chief ministers can interchange their places in 2! Ways. Thus, total number of ways = 9! · 2!

14. 7 men and 4 women are to be seated at a round table. If all the women sit together, then the

number of ways are (A) 7! · 4! (B) 6! · 3! (C) 10! – 7! · 4! (D) none of these Sol. A



22

Multiple Correct Answer(s) Type This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.

15. If the matrix a b c

A b c ac a b

where a, b, c are positive real number such that abc = 1 and ATA = 1.

Then find the value of a3 + b3 + c3

(A) 2

2x 2

x 4limx 3x 2

(B) x 0 e

sin2xlimlog 1 x

(C) x 0

xlim1 x 1

(D) x 0

x xcos sin2 24 limcosx

Sol. A, D Here, AT = A ATA = A2 = 1 (given)

We know, 3 3 3a b cb c a a b c 3abcc a b

2A 1 23 3 3a b c 3abc 1 23 3 3a b c 3 1 abc 1

3 3 3

1/ 3 3 3 3a b c a b c abc3

a3 + b3 + c3 3 a3 + b3 + c3 – 3 –1 a3 + b3 + c3 – 3 = 1 a3 + b3 + c3 = 4 16. Solution of (1 – xy – x5y5)dx = x2(x4y4 + 1)dy is

(A) loge x = xy + 5xy5

+ c (B) e xyxlog

e

= 5xy5

+ c

(C) loge y = xy + 4xy4

+ c (D) e xyylog

e

= 5xy5

+ c

Sol. A, B (1 – xy – x5y5) dx = (x6y4 + x2) dy (1 – xy) dx – x2 dy = x6y4dy + x5y5 dx (1 – xy) dx – x2 dy = x5y4 (x dy + y dx) dx – x(y dx + x dy) = x5 y4 [d(xy)]

4 4dx x y 1 d xy cx

5e

xylog x xy c

5

17. A(3, 0) and B(6, 0) are two fixed points and U(x1, y1) is a variable point of the plane. AU and BU

meet the y axis at C and D. AD meets OU at V, then CV passes through (o is origin)

(A) (2, –1) (B) 9 , 02

(C) (2, 1) (D) (2, 0)



23

Sol. D

AU y – y1 = 11

1

yx x

x 3

1

1

3yC 0,

3 x

and 1

1

6yD 0,

6 x

1

1

y 6 xxAD 13 6y

; OU 1

1

y xy

x

1 1

1 1

6x 6yV ;

6 x 6 x

CV 1

1

3y xy 13 x 2

18. The positive real valued function f satisfies

2x yf x 7

f y , y belong to its domain, 3f

2 4

then

(A) 3f x x x (B) 2 2sin x cos x

1 1

0 0

f x sin tdt cos tdt

(C) x

3

0

f x 2t dt (D) f ' 7 0

Sol. B, D f(x) is a constant function because f(x) 0 x R 19. If equations a2x + b2y + c2 = 0, a4x + b4y + c4 = 0 and x + y + 1 = 0 are consistent then (A) a = –b (B) |b| = |c| (C) c = a (D) none of these Sol. A, B, C

2 2 2

4 4 4

a b c

a b c 01 1 1

(a2 – b2)(b2 – c2)(c2 – a2) = 0 20. If S1, S2, S3 are the sums to n, 2n, 3n terms respectively for a G.P. If S2 – S1 = 141 and S3 – S2 = 168 then (A) [S1] = 105 (where [.] is greatest integer function) (B) [S1] = 107 (where [.] is greatest integer function)

(C) 1 2 2

1 3 1

S S SS S S

(D) 1 2 1

1 3 2

S S SS S S

Sol. A, D S1, S2 – S1, S3 – S2 G.P.

16627S56

FIITJEE REVIEW FULL TEST – I

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