Solutions

AITS-CRT-III-PCM(Sol)-JEE(Main)/15

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1

ANSWERS, HINTS & SOLUTIONS

CRT – III (Main)

PHYSICS CHEMISTRY MATHEMATICS 1. B B A 2. B B D 3. B B B 4. D D B 5. D D D 6. A A B 7. A B B 8. B C C 9. A A B 10. A B C 11. B B B 12. C B D 13. B B B 14. C A B 15. B B C 16. B B C 17. A B D 18. B B C 19. D D C 20. A B A 21. C B A 22. A D C 23. A D A 24. D C A 25. A D C 26. C A D 27. D D B 28. C C B 29. D D A 30. C B A

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2

PPhhyyssiiccss PART – I

SECTION – A 1. Tdt Mv Mu and T d t Mv

uv2

2 2

A newv u 3MgT Mg M Mg Ml 4l 2

A oldT Mg

M

TA

Tdt Tdt

v

v u

2. Let equivalent resistance between A and B is R.

ViR V, i R3

ii3

2V 2iR3 3

R2

B

A R1 R1

R i i – i i

2R 2R

12ViR3

1

2

R 4R 3

3. Conserving angular momentum

2

23mRmvR 3mR2

2v 3R6R

4. Let at any time a charge q flows through the circuit. The

circuit is as shown

Time constant 2RC( )3

+ –

2C

4V

C

R + –

q

3t

2RC4CVq 1 e3

5. 1

MM 2

21

1

VI

1R LC

Similarly, 2

MM 2

22

2

VI

1R LC

Since, both 1MI and

2MI are n times less than resonance current amplitude,



3

M1 M2I I

M M2 2

2 21 2

1 2

V V

1 1R L R LC C

or 1 21 2

1 1L LC C

or 2 21 2

1 2

LC 1 1 LCC C

or 1 2 1 2 1 2LC( ) ( ) 0 or 1 21 LC 0

21 2 0

1LC

0 1 2

6. 01 Tf2l

0

0

df 1 dTf 2 T

0Tf 2 15 100 15021

Hz.

7. W PdV and P V 9. The object distance for the mirror

2015 304 / 3

cm

For mirror,

1 1 1v u f

1 1 1v 30 20

v 60 cm

The distance from water surface equals 60 – 15 = 45 cm

12. Least count = PitchNumber of divisions on circular scale

= 1mm 0.01mm 0.001cm100

Zero error = – 0.03 mm Measurement = 3.76 – (– 0.03) = 3.79 mm

13. cma aˆ ˆr i j2 2

(considering A to be origin AD to x-axis and AB to be y-axis)

i cm cm 0a aˆ ˆ ˆ ˆL mr v m i j v (2i 4 j)2 2

= 0 0ˆ ˆ ˆmav (2k k) mav k



4

2 2 2

2 2f

m a ma maL (a a ) m12 2 6 2

From conservation of angular momentum about point A f iL L

20

2 ma mav3

03v2a

0 0cm

3v 3va av2a2 2 2 2

14. If 0V be volume of the gas at C, 0 0 0P V 3RT W = 0 0 03R(2T )ln2 P V = 0 06RT ln2 3RT = 03RT [2 0.693 1] = 03RT [1.386 1] = 01.16RT 15. 0mv R = I + mvR

= 2 vmR mvRR

v = 0v2

iK = 20

1 mv2

fK = 2 2

2 2 20 02

v v1 1 1 1mv I m mR2 2 2 2 2 4R

= 20

1 mv4

So fW = 20

1 mv4

16. Initial current in 3R , i1 3

EiR R

Final current 2f

2 3 2 31

2 3

REiR R R RR

R R

= 2

1 2 3 2 3

R ER (R R ) R R

f1 3

1 32

EiR R R RR

As f ii i , so current in the resistor R3 will decrease.



5

17. y

x

vdy 1dx v3

So x yv v 3

2 2 2x y 0v v v

2 2 2y y 03v v v

0y

vv

2

0x

3vv

2

So 0

0

3v223vy e

1y e 18. If the string is pulled down by distance x, it is equivalent that a mass of x is removed from the

upper end of the string and is connected to the lower end So g i fW U U U

gGM( x) GM xWR L R

= 1 1GM xR R L

19. amax = 2A = Km

A

If mass and amplitude both are doubled maximum acceleration will not change. 20. Resultant wave amplitude is given by A = 2 2

1 2 1 2 1 2A A 2A A cos( ) Value of cos(1 – 2) lies between 1 and – 1 Hence A1 – A2 A A, + A2 24. If three waves meet in same phase then the resultant amplitude will be algebraic addition of all

three amplitudes R = 3A A Amplitude of superimposed wave R Amplitude of resultant wave than I = 9I0 as [I R2] 25. If process starts and ends on same isotherm then product of P and V must be constant hence

process a, b, d will be on same isotherm.

26. 1 1 2 2mix

1 1 2 2

n Cp n Cpn Cv n Cv

=

7 52 R 1 R 192 25 3 132 R 1 R2 2



6

27. 22 2

K

l 1 1R(Z 1)1 2

.....(i)

22 2

K

l 1 1R(Z 1)1 2

....(ii)

Solving (i) and (ii) Z = 9 28. = A( – 1) Slope of line will be A.

30. x = 2 (SS1 + S1O) – (SO) =

2

2 22 D d 2D2

2

2d2D 1 1

22D

d = D2



7

CChheemmiissttrryy PART – II

SECTION – A 1. In gas phase inductive effect play a key role. 2. Geometrical isomers = 8 = 2n Optical isomer = 8 = 23 = 8 = 16 3. Due to bulky effect. 4.

O

O

H

O

OH

O

OH

O

OH

H

O

OH

Tautomerise

O

O

5. Birch reduction. 6. 3o carbocation is more stable. 7. 2: CCl intermediate. 8. 3 3 2

Red ppt.CH CHO 2CuO CH COOH Cu O

9.

D

H

D

H

D

H

D3D O

D

D

D

D

D

D



8

10.

N OH

H

NC

H

O

Polymerisation NH (CH2)5 C

O

nCaprolactum Nylon-6

11. It is belong to f-block – Lu element. 12. IP4 – IP5 required more energy, So IVA group elements. 15.

Mg

H

H

H

H

Mg

16. 3NaOH + 2O3 2NaO3 + NaOH.H2O(s) + 21 O2

17. 2 2 2 2B.A 110 118 105 112

Cl O, ClO , F O, Br O

19. (II) and (IV) have unpaired electron. 20. 2 4 Yellow ppt

H S CdSO CdS

21. Let 1 mole of mixture has x mole N2O4 2 27.6 = x(92) + (1 – x) 46 x = 0.2

22. 2p 3p 3d

m 1 1 1s 1/ 2 1/ 2 1/ 2

23. % by weight 10 d 36.5 10 1.2M 12 MM.w 36.5

36.5 1000 1000m 15.7 m36.5 100 36.5 63.5

24. 3 4ClO ClO 2e x 2 oxidation x = 5, x = 7 3 2e ClO ClO x 1 reduction x = 5, x = 4

WME M2

25. Ideal gas has no force of attraction and has negligible volume. Hence it can’t be liquefied at any T

and P.



9

26. H = nxCp T Cp = Cv + K

PVn 0.05RT

Cp = 20.794 JK–1

For reversible adiable r 1TV constant

r 1 r 12 2 1 1T V T V r = 1.66 for argon

T2 = 189.85 k T = – 110.15 Thus H = 0.05 20.794 –110.15) = – 114.52 J 27.

3 3 3 3CH COCH CH CH g CO g

At t 0 150 0 0At t t 150 x x x

x 1150 x 3

x = 75 mm. 28. 2

2 2ssMg OH Mg 2OH

4s3 = 8.9 10–12, s = 1.3 10–4 m [OH] = 2 1.305 10–4 pOH = 3.58832 pH = 10.4168 29. 1 mole of A will form 2 moles of B and C after completion of reaction when 75%. A converted into

B and C then total no. of moles 0.5 2 1.5 3.5 moles 30. Along one body diagonal 2X atoms from 2 corners, one Y particle (at the centre of cube) will be

removed.

So effective no. of X particle in a unit cell = 1 154 28 4

And effective no. of Y particle in a unit cell = 4 – 1 = 3

15X : Y : 34

5 : 4



10

MMaatthheemmaattiiccss PART – III

SECTION – A 1. n + n = –pn ….. (1)

Again since

is a root of equation

xn + 1 + (x + 1)n = 0

n n

1 1 0

n + n + ( + )n = 0 n + n = –( + )n = –(–p)n ….. (2) From equation (1) and (2), we get –pn = –(–p)n pn = (–p)n pn = (–1)n pn (–1)n = 1 n is an even integer 2. From the figure, OM is bisector of angle BOP of triangle OBP

250 OB50 OP

2 2300 x5x

25x2 x2 = (300)2 24x2 = (300)2

2 300 300x 3250 25 25 624

x 25 6.

P

M

B A

300 m

250 m

50 m

O (observer)

x m

3. 2 31 11

1 2 2 2P 10 10 10 10 …..

2 31 1 11 .....2 2 2P 10

=

111 2210 10 100

log0.01P = log0.01(100) = –1

4. 2

2ax b ax b x ab2

Consequently, for all x > 0

2x x 1y

ax b 2x ab 2 ab

5. x = n n n n

1 2 3 nn n n

1 2 n

0.1 1 C 2 C 3 C n C1 C C C

= n

nr

r 0r C

=

n 1n

r 1r 1

nn

rr 0

nr Cr

C

=

nn 1

r 1r 1

nn

rr 0

n C

C

=

n 1

nn2

2

= n2



11

6. Greatest coefficient is r 3m r

n! 15!3! 4!q! q 1 !

ln = 15 and m = 4, q = 3 and r = 37 15 = 4 3 – 13 7. Two triangles may have equal areas f is not one-one Since each positive real number can represent area of a triangle f is onto

8. Area of PAB = 32sin2

(PAB)min when 22

4

h = OC cosec = 4 2

C

P(0, h)

B A O

9. Equation of chord joining and is x ycos sin cosa 2 b 2 2

+ = 3

x ycos 0a 2 b

It passes through the centre (0, 0)

10. Ordinate of the point of intersection of the line x y ma b and the hyperbola, given by

x y x y 2y 1a b a b b

i.e. 2ym m 1b

i.e. 2b 1 m

y2m

Similarly ordinate of the point of intersection of the line x y ma b and the hyperbola is given by

2b m 1y

2m

Sum of the ordinates is zero 11. Let 1 1 2 2P(x ,y ) , Q(x , y ) be the given points on x 2y 2

1x 21y 2 and 2x 2

2y 2

ÔP i

= projection of OP

on x-axis = x1 = 1 y1 = 2 and ÔQ i

= projection of OQ

on x-axis = x2 = 2 y2 = 16

ˆ ˆ ˆ ÔP i 2 j , OQ 2i 16j

ˆ ÔQ 4OP 6i 8i

ˆ ÔQ 4OP 6i 8 j 10



12

12. The point B is (–q, –p) and the point C is (–p, –q) mid-point of AC is (0, 0) 13. The point Q is (2, 1)

Image of P(1, 2) in the line x – 2y + 1 = 0 is given by x 1 y 2 41 2 5

Ordinate of the point are 9 2,5 5

Since this point lies on QR Equation of QR is 3x – y – 5 = 0 + = 2 14. Let (, 3 – ) be any point on x + y = 3 Equation of chord of contact is x + (3 – )y = 9 i.e. (x – y) + 3y – 9 = 0 The chord passes through the point (3, 3) for all values of 15. Equation of the two circles be (x – r)2 + (y – r)2 = r2 i.e. x2 + y2 – 2rx – 2ry + r2 = 0 where r = r1 and

r2 condition of orthogonally gives 2r1r2 + 2r1r2 = 2 21 2r r

4r1r2 = 2 21 2r r , circle passes through (a, b)

a2 + b2 – 2ra – 2rb + r2 = 0 i.e. r2 – 2r(a + b) + a2 + b2 = 0 4(a2 + b2) = 4(a + b)2 – 2(a2 + b2) a2 – 4ab + b2 = 0

16. 2 3y 6 x2

Equation of directrix x = 0

Let coordinate P be 23 3 t , 3t2 2

Coordinate of M are (0, 3t) 2MS 9 9t

23 3MP t2 2

9 + 9t2 = 2

22 23 3 9t 1 t2 2 4

4 = 1 + t2 Length of side = 6 17. yy1 = 2a(x + x1)

x2 = 4by = 11

2a4b x xy

y1x2 – 8ab x – 8ab x1 = 0 D = 0 gives xy = –2ab



13

18. Equation of the tangent to the ellipse at P(5 cos , 4 sin ) is xcos y sin 15 4

It meets the line x = 0 at Q(0, 4 cosec ) Image of in the line y = x is R(4 cosec , 0) Equation of the circle is x(x – 4 cosec ) + y(y – 4 cosec ) = 0 i.e. x2 + y2 – 4(x + y) cosec = 0 Each member of the family passes through the intersection of x2 + y2 = 0 and x + y = 0 i.e. The point (0, 0)

19. Equation of normal at (13 cos , 5 sin ) of 2 2x y 1

169 25

13siny 5 sin x 13cos5cos

it passes through (0, 6) then cos (15 + 72 sin ) = 0

cos = 0, 5sin24

1 5sin24

21. Given, f(x) = (x – 1)(x – 2)(x – 3)(x – 4) Consider the interval [1, 2] Since, f(x) is a polynomial function, it is continuous in [1, 2] and differentiable in (1, 2) Also, f(1) = 0 = f(2) By Rolle’s theorem these exists a point x = c in (1, 2) such that f(c) = 0 i.e. these exists a root of

f(x) = 0 in the interval (1, 2) which is positive Similarly, two positive roots of f(x) = 0 exist in the open interval (2, 3) and (3, 4) Hence, the three roots of f(x) are positive 22. f(x) g(y) + f(x) g(y) = g(y) (f(x) + f(x)) g(y) = g(y)

f(x) + f(x) =

g' yg y

= constant ….. (1)

Put y = 0

g' 0 1g 0 1

Now from equation (1), we get f(x) + f(x) = 1 and

g' y1

g y

g(y) = ey 23. 10 10 10 10

1 2 3 4C C C C 10 45 120 210 385 24. Total number of numbers without restriction two numbers have all the digits equal Hence, the required number of numbers = 25 – 2 = 30

25. Given 7a – 9b = 0 b = 7 a9

So, the number of pairs (a, b) can be (9, 7), (18, 14) (27, 21) and (36, 28)

Required probability = 392

4 4741C



14

26. Two numbers are selected from {1, 2, 3, 4, 5, 6} n(S) = 6 5 (as one by one with out replacement) favourable events = minimum of the two

numbers is less than 4 n(E) = 6 4 As for the minimum of the two is less then 4, we can select one from {1, 2, 3, 4} and other from

{1, 2, 3, 4, 5, 6}

Required probability =

n E 24 4n S 30 5

27. 4

dx

sin xcot x = 2

dxsin x cot x =

2cosec x dxcot x

Put cot x = t2 cosec2 x dx = –2t dt

= 2t dtt

= –2t + c = 2 cot x c = 2 ctanx

28. Let a = 2x, b = –3x – 1, c = –1 a3 + b3 + c3 = 3abc

Now, a3 + b3 + c3 – 3abc = 21 a b c a b2

factorizing and get a + b + c = 0 as a b c

Hence, the given equation is; 2x – 3x – 1 – 1 = 0 By trial x = 1 and 2 are solution of this Now, 2x = 3x – 1 + 1 has no other solution Number of real solution exactly two 29. 2 sin2 – 5 sin + 2 > 0

1sin2

( –1 sin 1)

50, , 26 6

30. Let h(x) = f2(x) + (f(x))2 h(x) = –2x(f(x))2 g(x) x = 0 is the point of maximum

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