PRE-SERIES-OLT-2021-T1-FT-I-KVPY-CLASS-XII FULL TEST – I

PART – I MATHEMATICS

1. Number of solution of the equation ( )6log x

2 6log x 3 log x+ = is

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

Sol. ( )6log x

2 6log x 3 log x+ =

Let u

6u log x x 6= =

u u u6 3 2+ =

u

u 21 2

3

+ =

LHS = RHS happens for only one value of u, which is –1

( LHS is increasing whereas RHS is decreasing)

6

1log x 1 x

6= − =

2. Give that 2 2S x 4x 5 x 2x 5= + + − + + for all real values of x, then the maximum

value of 4S is

(A) 1 (B) 2 (C) 4 (D) 3 Ans. C Sol. This problem can be viewed as

( ) ( )2 2 2S x 2 1 x 1 2= + + − + +

S PA PB= −

But PA PB AB−

max.

PA PB AB 2 S − = = =

4S 4 =

B (–1, 2)

A (–2, 1)

y

P (x, 0) –2 –1

3. If

=+ 30

lnxJ dx

1 x and

=+ 30

x ln xK dx,

1 x then

(A) J K 0+ = (B) J K 0+

(C) J K 0+ (D) nothing can be said about +J K

Ans. A

Sol. ( ) +

+ =+ 30

1 x lnxJ K dx

1 x

Put =1

x .t

+ + = −

+

02

3 2

t 1 1 1J K t ln dt

tt 1 t

( )

( ) +

= − = − ++ 30

t 1 lntdt J K

1 t

4. There is only one way to choose real numbers M and N such that when the polynomial

4 3 25x 4x 3x Mx N+ + + + is divided by the polynomial 2x 1+ , the remainder is 0. If M and

N assume these unique values, then M N− is: (A) –6 (B) –2 (C) 6 (D) 2 Ans. C

Sol. Let ( ) 4 3 2P x 5x 4x 3x Mx N= + + + +

Let ( ) 2Q x x 1= +

If the quotient is Q

Then ( ) ( )2P x Q x 1= +

If x i,= then ( )P i 0=

If x i,= − then ( )P i 0− =

Hence, 5 4i 3 Mi N 0− − + + =

Hence, N Mi 2 4i+ = − + N 2; M 4= − =

M N 6− =

5. If A and B are two square matrixes such that 2A B BA= and if ( )10 k 10AB A B= then k is

(A) 1022 (B) 1023 (C) 1024 (D) 1025

Ans. B

Sol. ( ) ( ) 3 2AB . AB A B=

( )3 7 3AB A B=

( )nn 2 1 nAB A B−=

6. Let n

r

nC ,

r

=

then the value of

2000 2000 2000 2000.................

2 5 8 2000

+ + + +

equals

(A) 20012 1

3

+ (B)

20002 1

3

−

(C) 20002 1

3

+ (D)

20012 1

3

−

Ans. B

Sol. Let ( ) ( )2000

2000 k

k 0

2000f x 1 x x

k=

= + =

( ) ( ) ( )2 2 2000 2000 2000f 1 f f 3 ........

2 5 2000

+ + = + + +

( ) ( )200020002000 2 22 1 1 + + + +

7. The minimum distance between the circle 2 2x y 9+ = and the curve 2 22x 10y 6xy 1+ + =

is:

(A) 2 2 (B) 2

(C) 3 2− (D) 1

311

−

Ans. B

Sol. ( )2 2 2r 2cos 10sin 6sin cos 1 + + =

2 1 1r 1

3sin2 4cos2 6 6 5= =

− + −

Minimum distance between curves 3 1 2= − =

( )r cos , r sin

8. Let ABC be a triangle. Let A be the point (1, 2), y = x is the perpendicular bisector of AB

and x 2y 1 0− + = is the angle bisector of angle C. If the equation of BC is given by

ax by 5 0+ − = , then the value of a b+ is:

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

Ans. B Sol. Image of A say A’ w.r.t x 2y 1 0− + = lies

on BC.

( )

2

1 4 1x 1 y 2 42

1 2 51 2

− +− −= = − =

− +

9 2

A ' ,5 5

=

Equation of BC joining 9 2

A ' ,5 5

and B

(2, 1) is ( ) ( )

21

35y 1 x 2 x 29 1

25

−

− = − = −

−

3x y 5 0 a b 3 1 2− − = + = − =

C

A

(1, 2)

(2, 1)

B

x – 2y + 1 = 0 9. Two tangents drawn to parabola at points (3, 9) and (9, 7) intersect at (2, 2) then slope

of directrix is

(A) 3

2 (B)

2

3−

(C) 5

2− (D)

2

5

Ans. B Sol. Use the property : “The line segment joining the point of intersection tangents drawn at

the points A and B on the parabola to the midpoint of the corresponding chord of contact is always parallel to the axis of the parabola”.

Here, If ( ) ( )A 3, 9 , B 9, 7= = and ( )C 2, 2= and also let M ( )6, 8 is the mid point of AB.

Then slope of CM = 8 2 6 3

6 2 4 2

−= =

−

Slope of directrix = 2

3

−

10. A pair of dice is rolled till a sum of either 5 or 7 is obtained. Then the probability that 5

come before 7 is

(A) 1

5 (B)

2

5

(C) 3

5 (D)

4

5

Ans. B Sol. A = getting five P(A) = 1/4 B = getting 7 P(B) = 1/6

n (AUB) = 10

Prob. that sum of numbers is 5 or 7 10 5

36 18= =

Prob. That sum is neither 5 or 7 13

18=

Required prob. 1 13 1 13 13 7 2

........9 18 9 18 18 9 5+ + + =

11. Let f : 0, 4 0, → be defined by ( ) ( )1f x cos cosx−= . The number of points

x 0, 4 satisfying the equation ( )10 x

f x10

−= is

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

Sol. 0,4 0, →

( ) 1f x cos cosx−=

( )10 x x

f x 110 10

−= = −

2 3 4

3 solutions

12. If y1 and y2 are two different solutions of the equation dy

P(x)y Q(x)dx

+ = and y = 3y1 +

2y2 is also the solution of the equation, (where - = 4), then which of the following is

not correct. (, R)

(A) 4 + = 5 (B) + 4 = 7

(C) 5 + 5 + 2 = 0 (D) 5 + 10 + 13 = 0 Ans. B

Sol. 11

dyP(x)y Q(x)

dx+ = , 2

2

dyP(x)y Q(x)

dx+ =

( )1 21 2

d(3 y y )P(x) 3 y 2 y Qx

dx

+ + + =

3Q(x) + 2(Q(x)) = Q(x) 3 + 2 = 1 and - = 4

9

5 = ,

11

5

− =

13. If roots of quadratic equation (x 3) (x p) 7− − = have integral root then sum of all

possible integral values of p is (A) 6 (B) 9 (C) 7 (D) none of these Ans. A Sol. Consider the following case

x 3− x p−

1 7 7 1 -1 -7 -7 -1

14. If in a triangle ABC, cosAcosB sinA sinBsinC 1,+ = then the sides are proportional to

(A) 1:1: 2 (B) 1: 2:1

(C) 2 : 1:1 (D) none of these

Ans. A

Sol. cosAcosB sinAsinBsinC 1+ =

1 cos A cosB

sinCsin A sinB

−= …..(i)

1 cosAcosB sinAsinB +

or ( )cos A B 1−

( )cos A B 1− =

A B=

From Eq. (i)

2

2

1 cossinC

sin A

−=

= 1

C 90 =

A B 45 = =

a : b : c sin A : sin B : sin C=

1 1

: :12 2

=

or a : b : c 1:1: 2=

15. Let r, s and t be the roots of the equation, 38x 1001x 2008 0+ + = . The value of

( ) ( ) ( )3 3 3

r s s t t r+ + + + + is:

(A) 251 (B) 751 (C) 735 (D) 753 Ans. D

Sol. 38x 1001x 2008 0+ + =

2008

r s t 0, rst 2518

+ + = = − = −

So, ( ) ( ) ( )3 3 3

r s s t t r+ + + + +

( ) ( )3 3 3t s r 3rst 3 251= − + + = − = − −

= 753

As r s t 0,+ + = so 3 3 3r s t 3rst+ + =

16. If ( )

e 21 1 e

2ee

elog

3 2log xxy tan tan ,

1 6log xlog ex

− −

+ = +

−

then 2

2

d y

dx is

(A) 2 (B) 1 (C) 0 (D) –1 Ans. C

Sol. ( )

e 21 1 e

2ee

elog

3 2log xxy tan tan

1 6log xlog ex

− −

+ = +

−

1 1e e

e e

1 2 log x 3 2 log xtan tan

1 2 log x 1 3.2log x

− − − +

= + + −

( ) ( ) ( ) ( )1 1 1 1

e etan 1 tan 2 log x tan 3 tan 2log x− − − −= − + +

( ) ( )1 1tan 1 tan 3− −= +

dy

0dx

=

17. The number of distinct solutions of the equation

2 4 4 6 65cos 2x cos x sin x cos x sin x 2

4+ + + + = in the interval 0, 2 is

(A) 5 (B) 2 (C) 6 (D) 8 Ans. D

Sol. On simplification, w get n

cos4x 0 4x 2n x2 2 8

= = =

Number of distinct solution on 0, 2 is 8.

18. Let a, b, c are complex numbers and roots of 3 2z az bz c 0+ + + = are unimodular then

a b− equal to

(A) 0 (B) 1 (C) –1 (D) 2 Ans. A

Sol. a ++ =

b + + =

1 1 1

b + + =

b a + + = =

19. Let ( )

1

n2

n1

n

k lim n 2019 sinx 2020cosx x dx→

−

= + . The value of is equal to

(A) 2021 (B) 2022 (C) 2020 (D) 2019 Ans. C

Sol. ( )1

n1

n

1 1 12019sinx 2020cosx x dx 4040 sin cos 1

n n n−

+ = + −

21 1 1

4040 sin 2sinn n 2n

= −

Hence,

2

n

1 1sin sin

n 2nk 4040 lim 20201 1

2n 2n

→

= − =

20. If then the value of x is equal to

(A) 90cot1 cos 1ec (B) 90sec1

(C) 90cot1 (D) none of these

Ans. C

Sol. S 1sin2 2sin4 3sin6 ... 89sin178= + + + +

S 89sin178 88sin176 87sin174 ... 1sin2= + + + +

Adding the two, we get

( )sin89

S 90 sin2 sin4 ... sin178 90 sin90 90cot1sin1

= + + + = =

PHYSICS

21. A thin prism 1P with angle o4 and made from glass ( = 1.54) is combined with another

prism 2P made of another glass of = 1.72 to produce dispersion without deviation. The

angle of prism 2P is

(A) o53.3 (B) o4

(C) o3 (D) o2.6

Ans. C Sol. For no deviation

(1 – 1)A1 = (2 – 1)A2 4o(1.54 – 1) = (1.72 – 1)A2

A2 = 4 0.54

0.72

= 3°.

22. By applying a force ( ) ˆ ˆF 3xy 5z j 4zk= − + a particle is moved

along the path y = x2 from point (0, 0, 0) to (2, 4, 0). The work done by the force F on the particle is

(A) 280

5 unit (B)

140

5unit

(C) 232

5unit (D)

192

5unit

(2, 4, 0)

y = x2

y

x (0,0,0)

Ans. D

Sol. w F.dr=

( )3xy 5z dy= − ; 4

0

1923 y ydy

5=

23. A boat which has a speed of 6 km/h in still water crosses a river of width 1 km along the shortest possible path in 20 min. The velocity of the river water in km/h is (A) 1 (B) 3

(C) 4 (D) 3 3

Ans. D

Sol. t = 20 min = 1

hr3

1 km = 2 2 1(6) V hr

3

−

V = 3 3 km/hr.

24. The Young’s modulus of a wire of length L and radius r is Y newton per square metre. If the length is reduced to L/2 and radius to r/2, its Young’s modulus will be

(A) Y/2 (B) Y (C) 2Y (D) 4Y Ans. B Sol. Young’s modulus does not depend on the geometry of the wire but its materials. 25. A copper wire of diameter 1.02 mm carries a current of 1.7 amp. The drift velocity (vd) of

electrons in the wire. Given n, number density of electrons in copper = 8.5 1028 /m3. (A) 3.0 mm/s (B) 2.0 mm/s (C) 1.5 mm/s (D) 1 mm/s Ans. C Sol. I = 1.7 A J = current density

= 2 3 2

I 1.7

r (0.51 10 )−=

= nevd

= 8.5 1028 (1.6 10-19 ) vd

vd = 3 2 28 19

1.7

(0.51 10 ) 8.5 10 1.6 10− −

= 1.5 10-3 m/sec. = 1.5 mm/sec.

26. A cylindrical conductor of length and inner radius

R1 and outer radius R2 has specific resistance . A

cell of emf is connected across the two lateral faces of the conductor. Find the current drown from the cell.

(A) 2

1

2

Rln

R

(B) 2 2

2 1

2

(R R )

−

dx x

R2

R1

(C) 2 2

2 1

2

(R R )

−

(D) None of thee Ans. A Sol. Consider the differential element of the cylinder as shown in the figure.

2

1

R

R

dxR ( R )

2 x a = =

l

l

R = 2

1

Rln

R2

l

I =R

; I =

2

1

2

Rln

R

l

27. Two capillary tubes of radii 0.2 cm and 0.4 cm are dipped in the same liquid. The ratio of heights through which liquid will rise in the tube is

(A) 1:2 (B) 2:1 (C) 1:4 (D) 4:1 Ans. B Sol. capillary rise

h = ( )2Tcos

r g

1 2

2 1

h r 0.42 :1

h r 0.2= = =

28. A particle executes S.H.M. with a time period of 4 s. Find the time taken by the particle

to go directly from its mean position to half of its amplitude.

(A) 1 s (B) 1

s2

(C) 1

S3

(D) 1

S4

Ans. C

Sol. x = A sin ( )ot +

At t = 0, x = 0

oAsin 0 = or o = 0

Hence, x = A sin ( )t

or A/2 = A sin (t)

or 1/2 = sin (t)

t = sin-11

2 6

=

t = ( ).T

6 6 2

=

as 2 / T=

t T/12 = 1/3 s

29. In hydrogen like atoms the ratio of difference of energies 4n 2nE E− and 2n nE E− varies

with atomic number z and principle quantum number n as

(A) 2

2

z

n (B)

4

4

z

n

(C) z

n (D) none of these

Ans. D

Sol.

1 1

2 24n 2n

1 12n n2 2

E EE E 16n 4n

E EE E

4n n

−−

=−

−

= 1

4 = constant

30. A small sphere with radius r and density falls at a speed v through a fluid of density

and coefficient of viscosity . The dimensions of ‘’ are ML-1T-1. If the friction force F opposing the motion is proportional to V, it could also be proportional to

(A)

but independent of r (B) r but independent of

(C) 2 but independent of (D) r but independent of Ans. B

Sol. F 6x rv= −

31. A liquid is filled in a spherical container of radius R till a height

h. At this position the liquid surface at the edges is also horizontal. The contact angle is

(A) 0 (B) 1 R hcos

R

− −

(C) 1 h Rcos

R

− −

(D) 1 R hsin

R

− −

h

R R

Ans. B

Sol. ( )R h

sin 90R

−− =

90-

R R-h

32. An ideal gas initially at P1, V1 is expanded isothermally to P2, V2 and then compressed adiabatically to the same volume V1 and pressure P3. If W is the net work done by the gas in complete process, which of the following is true

(A) W > 0; P3 > P1 (B) W < 0; P3 > P1 (C) W > 0; P3 < P1 (D) W < 0; P3 < P1

Ans. B Sol.

P1

P2

P3

v1 v2 Area under curve is work done

33. A train of mass M is moving on a circular track of radius R with constant speed v. The

length of train is half the perimeter of track. The magnitude of linear momentum of the train will be

(A) 0 (B) 2M/ (C) MvR (D) Mv Ans. B

Sol. Total momentum =

90

90

(dm)v cos

+

=−

dm = M

RdR

Total momentum = M 2Mv

vRd cosR

=

34. A point object O is placed at a distance of 0.3 m from a convex

lens (focal length 0.2 m) cut into two halves each of which is displaced by 0.0005 m as shown in figure. Find the distance between the images (in mm)

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

cm05.02

30cm

O

L2

L1

• •

f=20cm Ans. C Sol. V = 0.6 m

I I

o

h h

h 0.05= = 2 hI = 0.1 cm.

Distance between them = 0.3 cm = 3 mm.

35. In a vertical plane inside a smooth hollow thin tube, a block of same mass as that of tube is released as shown. When it is slightly disturbed it moves towards right. By the time the block reaches the right end of the tube, the displacement of the tube will be (where ‘R’ is the mean radius of tube the assume that the tube remains in vertical plane) towards left

(A) 2R

(B)

4R

(C) R

2 (D) R

R

m

m

Ans. C

Sol. Initial position of block xcm = R Final position of block

xcm = m(R x) m(2R x)

2m

− + −

Now, m(R x) m(2R x)

R2m

− + −=

m

m

x

x

y

x

x x

36. In the shown circuit involving a resistor of resistance R , capacitor of capacitance C farad and an ideal cell of emf E volt, the capacitor is initially uncharged and the key is in position 1. At t = 0 second the key is pushed to position 2 for t0 = RC seconds and then key is pushed back to position 1 for t0 = RC seconds. This process is repeated again and again. Assume the time taken to push key from position 1 to 2 and vice versa to be negligible.

R

2 E

1

C

K

The charge on capacitor at t = 2RC second is

(A) CE (B) 1

CE 1e

−

(C) 2

1 1CE

e e

−

(D)

2

1 1CE 1

e e

− +

Ans. C

Sol. For charging q = t /RCCE(1 e )−−

Charge at t = RC qo = CE(1 – e–1) At t = RC discharging starts

q = qo(e–t/RC) = CE(1 – e–1) × 2

1 1 1CE

e e e

= −

37. Two small balls A and B, each of mass m, are joined rigidly at the ends of a light rod of length L. They are placed on a frictionless horizontal surface. Another ball of mass 2 m moving with speed u towards one of the ball and perpendicular to the length of the rod on the horizontal frictionless surface as shown in the figure. If the coefficient of restitution is 1/2 then the angular speed of the rod after the collision will be

m

L

m

u

2m

A

B

(A) 4 u

3 (B)

u

(C) 2 u

3 (D) None of these

Ans. C

Sol. Velocity of ball ‘A’ just after the collision 1 2

V 1 u u2 3

= + =

Velocity of ball ‘B’ just after the collision = 0.

u 0 u−

= =

38. A cubical frame is made by connecting 12 identical uniform conducting rods as shown in the figure. In the steady state the temperature of junction A is 1000 C

while that of the G is 00 C . Then,

(A) H will be Hotter than B

(B) Temperature of F is 040 C

(C) Temperature of D is 066.67 C

(D) Temperature of E is 050 C

E F

B A

H

D a C

G

a

00C

Ans. B

Sol. 100°C – Ir – I

r Ir2

− = 0°C

Ir = 40°C ; tF = 0° + Ir = 40°C 39. An open organ pipe of length L vibrates in second harmonic mode. The pressure

vibration is maximum (A) at the two ends (B) at a distance L/4 from either end inside the tube (C) at the mid-point of the tube (D) none of these Ans. B Sol. Pressure node is formed at the both the ends and in the

middle.

40. In order to shift a body of mass m from a circular orbit of radius 2R to a higher orbit of radius 4R around the earth, (Where R is radius and M is mass of earth) the work done on the body is

(A) GMm

8R (B)

GMm

2R

(C) GMm

4R (D)

GMm

R

Ans. A

Sol. ( ) ( )

GMm GMmW

2 4R 2 2R

= − − −

=

GMm

8R

CHEMISTRY

41. Molecule AB has a bond length if 1.61 Å and a dipole moment of 0.380 D. The fractional charge on each atom (absolute magnitude) is (e0 = 4.802 × 10-10 esu)

(A) 0.5 (B) 0.05 (C) 0 (D) 1.0 Ans. B Sol. q d =

10 300.380 q 1.61 10 3.33 10− −=

20q 0.786 10−=

Fraction charge 20

19

0.786 10

1.6 10

−

−

10.5 10−

0.05

42. Electromagnetic radiations having 310Å = are subjected to a metal sheet having work

function = 12.8 eV. What will be the velocity of photo electrons with maximum kinetic energy

(A) 0, no emission will occurs (B) 2.18 × 106 m/s

(C) 62.18 2 10 m/s (D) 8.72 × 106 m/s

Ans. C

Sol. 12500

E 40.32310

=

19 31 2127.3 1.6 10 9.1 10 v

2

− − =

2 10v 6 1.6 10=

12 69.6 10 2.18 2 10 m/s

43. The number of lone pair on central atom in 4 6BrF , XeF−

and 4SF are:

(A) 2, 0, 1 (B) 1, 0, 0 (C) 2, 1, 1 (D) 2, 1, 0 Ans. C Sol.

Br

F

F

F

F

Xe

F

F

F

F

F

F

44. Which of the following arrangements of molecules is correct on the basis of their dipole

moments? (A) 3 3 3BF NF NH (B) 3 3 3NF BF NH

(C) 3 3 3NH BF NF (D) 3 3 3NH NF BF

Ans. D Sol.

N

H

H H

N

F

F F

B

F F

F

0 =

45. What are A and B in the following reaction?

Cl

Br

Mg/T.H.F

(ii) aq. NH4Cl

CH3 CHO(i) A B

(A)

MgCl

Br

CHOHCH3

Br

and

(B)

Cl

MgBr

Cl

CHOHCH3

and

(C)

Cl

MgBr

CHOHCH3

Br

and

(D) None of these

Ans. B

Sol. Cl

MgBr

CH3

CH CH3

OHCH3 CHO

46. 2 2 45 10 5 9

Br /hv Mg/Ether (i)Acetone H SO

(ii) HC H C H Br P . P is

(only one isomer)

(A)

(B)

(C)

(D)

Ans. B Sol.

Br2/hv

Br

Br2/Ether

MgBr

CH3 C CH3

O

+

(ii) N+

CH OHCH3

CH3

H2SO4/CH OHCH3CH3CH3

+

47. Which of the following compound is not formed at all? (A) 3IF (B) 3BrCl

(C) 3FCl (D) 3ClF

Ans. C Sol. Fluorine combines through one bond only. 48. Which of the following species exhibit the diamagnetic behaviour?

(A) NO (B) 2

2O −

(C) 2O + (D) 2O

Ans. B Sol. Refer MOT

49. Cl

SH

Acetone/NaSHReaction Mechanism

(A) Only one product will form (B) Two products are formed and they will be enantiomer of each other (C) Mechanism of reaction is SN2 (D) Mechanism of the reaction is SN1 Ans. B Sol. Based on SNi mechanism. 50. Which of the following is most reactive towards SN2?

(A) Cl (B)

Cl

(C) Cl (D)

Ph C

O

CH2 Cl Ans. D Sol. e— w.g. stabilize T.S. so answer D.

51. To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid ( )3 3H PO ,

the volume of 0.1 M aqueous KOH solution required is (A) 60 mL (B) 20 mL (C) 40 mL (D) 10 mL Ans. C Sol. milli equivalent of 3 3H PO = milli equivalent of KOH

2 20 0.1 V 0.1

V 40ml

=

=

52. A mixture of 2N and 2H is caused to react in a closed container to form 3NH . The

reaction ceases before reactant has totally consumed. At this stage, 2.0 moles each of

2 2N , H and 3NH are present. The moles of 2N and 2H present originally were

respectively. (A) 4 and 4 moles (B) 3 and 5 moles (C) 3 and 4 moles (D) 4 and 5 moles

Ans. B

Sol. 2 2 3

x y

N 3H 2NH+ ⎯⎯→ 2Z 2, x Z 2, y 3Z 2

Z 1 x 3 y 5

= − = − =

= = =

x z y 3z 2z

2 2 2

− −

53. Which is incorrect about Hunsdiecker reaction? (A) Cl2 can give alkyl halide (B) I2 will give ester as major product when treated with RCOOAg (C) The reaction proceeds through free radical Intermediate

(D) The R-COO is formed during reaction Ans. D Sol.

RCOOAgBr2

R C O

O

Br

R C O

O

BrR R + CO2

.

.Br.

54. O

CN

OHKCN/H+

OH

(A) Forward reaction is nucleophilic addition and backward reaction is E1 (B) Forward reaction is nucleophilic addition and backward reaction is E1CB (C) Forward reaction is nucleophilic addition and backward reaction is E2 (D) Forward reaction is nucleophilic addition and backward reaction is SN1 Ans. B Sol. Anion is formed so answer B. 55. O

OH

C2H5

(i) NaOH +CHCl3Major Product

(A)

OC2H5

OH

OCH

(B)

OC2H5

O Na

OCH

+

(C)

OC2H5

O Na

CHO

+

(D)

OC2H5

OHCHO

Ans. B

Sol. O

OH

C2H5

NaOH

O

O N O

C2H5

+ CCl2

O

O N O

C2H5

OCH

+(i) (ii) :

(iii) NaOH

O— having more +m effect more – O – C2H5 .

56. 100 ml of tap water containing ( )3 2Ca HCO was titrated with N 50HCl with methyl orange

as indicator. It 30 ml of HCl were required, what is the degree of temporary hardness as ppm.

(A) 300 ppm (B) 500 pm (C) 30 ppm (D) 50 ppm Ans. A

Sol. mili equivalent of HCl = ( )3 32Ca HCO CaCO= = 330

1050

−

Mass of 3

milieq. molar massCaCO

nf

=

= 330 10 100

50 2

− = 0.03

Hardness in ppm = 60.03 10

300ppm100

=

57. 2g of aluminium is treated separately with excess of dilute 2 4H SO and excess of NaOH

solution. The ratio of volumes of hydrogen evolved under similar conditions is (A) 1: 2 (B) 1: 1 (C) 2: 1 (D) 2: 3 Ans. B

Sol. ( )2 4 2 4 232A 3H SO A SO 3H+ ⎯⎯→ +

2 2 22A 2NaOH 2H O 2NaA O 3H+ + ⎯⎯→ +

58. The difference in the wavelength of the 1st line of Lyman series and 2nd line of Balmer

series in a hydrogen atom is

(A) 9

2R (B)

4

R

(C) 88

15R (D) None of these

Ans. B

Sol. 1st line in Lyman, 2 1⎯⎯→

1

1 1 1 3RR

12 22 4

= = − =

IInd line in Balmer 4 R⎯⎯→

2 2

2

1 1 1 3R R

162 4

= = − =

Difference 1 2

4 16 4

3R 3R R − = − =

59. O

L.D.A

HCHO

HCHO/NaOH(dil.)A B (Product)

(product)

A and B are respectively are

(A)

O

CH2OHand

O

CH2OH

(B)

O

CH2OHBoth

(C) CH2OH

O

Both

(D) CH2OH

O O

CH2OHand

Ans. A Sol.

L.D.A CH N

CH3

CH3

HC

CH3

CH3

Li+

is Buky Base then NaOH.

60. Cyclohexene on ozonolysis followed by reaction will zinc dust and water gives

compound E. Compound E on further treatment with aqueous KOH yield compound F. Compound F is

(A) CHO

(B) CHO

(C) COOH

(D) COOH

COOH

Ans. A Sol.

(i) O3

(ii) Zn/H2OCHO CH2 CH2 CH2 CH2 CHO

aq. KOH

CHO

PART – II

MATHEMATICS

61. The value of c such that areas of shaded region are equal is

(A) 32

27 (B)

23

27

(C) 34

27 (D)

25

274

y =8x – 27 x3

y = c

Ans. A

Sol. ( )( ) ( )( )a b

3 3

0 a

c 8x 27x dx 8x 27x c dx− − = − −

2 4O 4b 27b bc= − −

( )2 327O 4b b 8b 27b

4= − − −

4 281b b 4 0

4

− =

2

2 4b 0 b

81 =

4b

81=

3C 8b 27b= −

32

27=

y = 8x– 27 x3

(a, c) (b, c)

62. Value of Definite integration( )

( ) ( )( )

22 1

1

2 2 10

2x 1 x cot xdx

1 x 1 1 x cot x

−

−

− +

+ − + is

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

Ans. A

Sol. ( )

1

221

102

2xcot x

1 xI dx

1cot x

1 x

−

−

−+

=

−+

( )

1

2 2 221

102

2x 1 1cot x

1 x 1 x1 xdx

1cot x

1 x

−

−

− + −+ ++

=

−+

( )

2 221 1

10 02

1 2x

1 x 1 xdx 1dx

1cot x

1 x

−

−+ +

= − +

−+

1

1

e 2

0

1log cot x 1

1 x

− = − − +

+

e e

1log log 1 1

2 4 2

= − − + − +

e

12log 1

1

2 4

−

= + −

63. In a ABC the maximum value of

2 Aacos

2

a b c

+ +

is

(A) 1

4 (B)

1

2

(C) 3

4 (D) 1

Ans. C

Sol. ( ) ( ) ( )

( )

a 1 cos A b 1 cosB c 1 cosC

2 a b c

+ + + + +

+ +

( )1 R

sin2A sin2B sin2C2 4s

= + + +

3 2

1 R abc 1 abc4

2 4s 28R 8R s

= + = +

2

1 4R 1 r 1 r 1 1 31 1

2 2 2R 2 R 2 2 48R s

= + = + = + +

64. If the line + − =x y 1 0 is a tangent to a parabola with focus (1, 2) at A and intersects the

directrix at B and tangent at vertex at C respectively, then AC. BC is equal to: (A) 2 (B) 1

(C) 1

2 (D)

1

4

Ans. A

Sol. Using power of C

( )( ) ( )2

2 1 2 1BC AC CS 2

2

+ − = = =

C A

B

S (1, 2) x + y – 1 = 0

65. For x 0 , let

2x 10 0

x

A 0 x 0

0 0 16

+

=

and

2

5x0 0

x 1

3B 0 0

x

10 0

4

+ =

be two matrices.

Three other matrices X, Y, and Z are defined as

( ) ( ) ( ) ( )1 2 3 n

nX AB AB AB ............ AB , Y LimX

− − − −

→= + + + = and 1Z Y 2I,−= −

Then the value of ( )1det. adj 5 Y− is equal to

(A) ( )2

5! (B) ( )235 5!

(C) ( )2

5 5! (D) ( )225 5!

Ans. C

Sol.

2

2

5x0 0x 1

0 0 x 1 5 0 0x3

AB 0 x 0 0 0 0 3 0x

0 0 16 0 0 41

0 04

+ + = =

( ) ( )

2

1 2

2

2

110 00 0

5 5

1 1AB 0 0 , AB 0 0

3 3

1 10 0 0 0

4 4

− −

=

and so on

2 n

2 n

2 3 n

1 1 1..... 0 0

5 5 5

1 1 1x 0 ..... 0

3 3 3

1 1 1 10 0 .....

4 4 4 4

+ +

= + + + + + +

n

1

5 0 01

115 0 041

13Y Lim 0 0 Y 0 01 2

13 1

0 01 3

40 01

14

→

−

= =

−

−

1

4 0 0

Y 0 2 0

0 0 3

−

=

66. Let a, b, c be the sides of a triangle. No two of them are equal and R . If the roots of

the equation ( ) ( )2x 2 a b c x 3 ab bc ca 0+ + + + + + = are real and distinct, then

(A) 4

3 (B)

5

3

(C) 1 1

,3 5

(D)

4 5,

3 3

Ans. A Sol. Roots are real and distinct. Δ 0

( ) ( )2

4 a b c 12 ab bc ca 0+ + − + +

( ) ( ) ( )2 2 2a b c 2 ab bc ca 3 ab bc ca 0+ + + + + − + +

( )2a

3 2ab

−

.…(i)

Now, in a triangle Difference of two sides < third side

i.e. a b c, b c a− − and c a b−

( ) ( ) ( )2 2 2 2 2 2a b b c c a a b c− + − + − + +

( )2 2 2a b c 2 ab bc ca+ + + +

or

2a2

ab

….(ii)

From Eqs. (i) and (ii), we get

2a3 2 2

ab −

4

3 2 23

−

67. f is an odd function. It is also known that ( )f x is continuous for all values of x and is

periodic with period 2. If ( ) ( )x

0g x f t dt,= then:

(A) ( )g x is odd (B) ( )g n 0, n N=

(C) ( )g 2n 0, n N= (D) ( )g x is non – periodic

Ans. C

Sol. ( ) ( )x

0g x f t dt,= so, ( ) ( ) ( )

x x

0 0g x f t dt f t dt

−

− = = − −

( ) ( )x

0g x f t dt,− = as ( ) ( )f t f t ,− = − so, ( ) ( )g x g x− =

Also ( ) ( ) ( ) ( )x 2 2 2 x

0 0 2g x 2 f t dt f t dt f t dt

+ +

+ = = +

( ) ( ) ( )x

0g x 2 g 2 f t 2 dt+ = + +

( ) ( ) ( ) ( )x

0g 2 f t dt g 2 g x= + = +

Now, ( ) ( ) ( ) ( ) ( ) ( )2 1 2 1 0

0 0 1 0 1

g 2 f t dt f t dt f t dt f t dt f t 2 dt−

= = + = + +

( ) ( ) ( ) ( )1 0 1

0 1 1

g 2 f t dt f t dt f t dt 0− −

= + = = as ( )f t is odd

( ) ( ) ( )g 2 0 g x 2 g x= + =

( )g x is periodic with period 2.

( )g 4 0=

( ) ( )f 6 0, g 2n 0, n N= =

68. A line 1L parallel to ˆ ˆ ˆ3i 2j 4k+ + passes through ( ) ˆ ˆ ˆA a 7i 6j 2k= + + and another line 2L

parallel to ˆ ˆ ˆ2i j 3k+ + passes through ( ) ˆ ˆ ˆB b 5i 3j 4k= + + . If line 3L perpendicular to

ˆ ˆ ˆ2i 2j k− − intersects 1L and 2L at ( )A a and C respectively, then find the value of AC

[Note : [y] denotes greatest integer less than or equal to y] (A) 3 (B) 4 (C) 5 (D) 6

Ans. B

Sol. 1 1

x 7 y 6 z 2L r

3 2 4

− − −= = = =

2 2

x 5 y 3 z 4L : r

2 1 3

− − −= = =

Any point C on 2L is ( )2 2 25 2r , 3 r ,4 3r+ + +

Direction ratio of 3L line is

( )2 2 25 2r 7, 3 r 6, 4 3r 2+ − + − + −

( )2 2 22r 2, r 3,3r 2 − − + and it is perpendicular to

vector ˆ ˆ ˆ2i 2j k− −

( ) ( ) ( ) ( ) ( )2 2 22r 2 2 r 3 2 3r 2 1 − + − − + + −

20 r 0= =

C is (5, 3, 4)

( ) ( ) ( )2 2 2

AC 7 5 6 3 2 4 17 = − + − + − =

AC 4 =

L1

L2

L3

A (7, 6, 2)

B (5, 3, 4)

C

69. If p is the product of the sines of angles of a triangle and q the product of their cosines,

the tangents of the angle are roots of the equation

(A) ( )3 2qx px 1 q x p 0− + + − = (B) ( )3 2px qx 1 p x q 0− + + − =

(C) ( ) 3 21 q x px qx q 0+ − + − = (D) none of the above

Ans. A

Sol. Here, p sin A sinBsinC=

and q cos A cosBcosC=

p

tanA tanBtanCq

=

And tan A tanB tanB tanC tanCtan A+ +

sin A sinBcosC sinBsinCcos A+

sinCsinA cosB

cos A cosB cosC

+=

( ) ( )sinB sin A cosC cos A sinC sinC sin A cosB

cos A cosBcosC

+ +=

( )sinBsin A C sinCsin A cosB

cos A cosBcosC

+ +=

2sin B sinCsinA cosB

cos A cosBcosC

+=

( )( )1 cosB cosB sinC sin A

q

− −=

( )( )1 cosB cos A C sin A sinC

q

− − + −=

and tan A tanB tanC+ +

( )tanA tanBtanC in a ABC=

p

q=

Required equation is 3 2p 1 q px x x 0

q q q

+− + − =

( )3 2qx px 1 q x p 0− + + − =

70. If the domain of f(x) is ( )x R 1,1 − − , then the domain of the function

f sin x cosx 1

−

(where [.] denotes greatest function) is

(A) x R (B) ( )x R 1,1 − −

(C) x (D) ( )x 1,1 −

Ans. C

Sol

1 sin x cos 1x 1

−

And domain of function is ( )x R 1,1 − −

sinx cos 1x 1

= −

sinx cos 1x 1

= −

either sin x 1,cos 1x 1

= = − −

( )sin x 1 x 4n 12

= = +

cos 1 x 1 1x 1

x 0,1 2,3

x

= − − = −

Or [sinx]=-1,

cos 1 but cos 1x 1 x 1

= − −

hence x

PHYSICS

71. If the kinetic energy of a body is directly proportional to time ‘t’, the magnitude of the force acting on the body is (Body is moving on straight line path)

(A) directly proportional to t

(B) inversely proportional to t

(C) directly proportional to the speed of the body (D) inversely proportional to the speed of the body Ans. D

Sol. 21mV t

2 V2 t

V t dV 1

dt t

F 1

t F

1

V

72. Two blocks of masses m and nm are connected by a massless string

passing over a frictionless pulley. The value of n for which both the blocks moves with an acceleration of g/10 is

(A) 9/11 (B) 11/9 (C) Both (A) and (B) (D) None

m nm Ans. C Sol. Case (1) Assume that mass m is accelerating upward.

T − mg = mg

10 T =

11mg

10

nmg − T = nmg

10 nmg −

11mg nmg

10 10=

m nm

nmg

T T

mg

Figure 1

9 11

n10 10

= and n = 11

9

Case (2) Assume If m mass is moving downward

mg − T = mg

10 T =

9mg

10

T − nmg = nmg/10

9 nmg

mg nmg10 10

− =

9 n 11n

n10 10 10

= + =

n = 9/11

73. A particle of mass 2 kg is performing SHM, given by equation x = 10 sin 2t (x is in m and t in s). The work done on the particle in time 0.25 sec to 0.75 sec, will be

(A) 2002J (B) 1002J

(C) 50J (D) zero Ans. D

Sol. F = 2

2

2

d xm 2x (10 4 sin 2 t)

dt= −

= − 80 2 sin 2t

dx = 20 cos 2 t dt

W = 0.75 0.75

2

0.25 0.25

F dx 80 20 sin2 t cos2 t dt= −

= − 40 2 20 0.75

0.25

sin 4 t dt = 3

0.75

0.25

800cos 4 t 0

4

=

74. A cylindrical hall has a horizontal smooth floor. A ball is projected along the floor from A

point on the wall in a direction making an angle with the radius through the point the ball returns back to the initial point after two impacts with the wall. If the coefficient of

restitution is e then tan2 will be

(A) 2

3

1 e e

e

+ + (B)

2

1 e

e

+

(C) 2e

1 e+ (D)

3

2

e

1 e e+ +

Ans. D

Sol. tan = usin tan

eucos e

=

tan = 1

tane

tan = 1

tane

2 + 2 + 2 = 180°

+ = (90 – )

tan( + ) = cot

tan tan 1

1 tan tan tan

+ =

−

2

2

3

1 1tan tan

1e e1 tan

1 tane

+

=

−

2 2

2 3

1 1 1tan 1 tan

e e e

+ = −

2

2 3

1 1 1tan 1

e e e

+ + =

3

2

2

2 3

1 etan

1 1 1 1 e e

e e e

= + + + +

75. There is a horizontal film of soap solution. On it a thread is placed in the form of a loop. The film is pierced inside the loop and thread becomes a circular loop of radius R. If the surface tension of the solution be T, then what will be the tension in the thread?

(A)2R

T

(B) 2R T

(C) 2 RT (D) RT

Ans. D

Sol.

F F

2F

Fnet = T.2R

76. If along a uniform rod of length carrying current I,

the voltage V changes with position x along the length of the rod such that dV/dx = – k, where k is a positive number, then the resistance of the rod is

x

x=0

(A) k/2I (B) k/I

(C) I/k (D) kI

Ans. B

Sol. dV = – IdR =dx

IA

= −

dV

I kdx A

= − = −

R = k/I

II Method: 2 1

dVV V d kL IR

d

− = = =

77. The magnetic field at the centre O of the arc in figure is (r is the radius of circular arc)

(A) 0l 24 r

+

(B) ( )0l 2 12 r 4

+ −

(C) 0 l2 r

4 r

+

(D) 0 l2

4 r 4

+

O

l

l

l

900 90

0

450 r

Ans. B

Sol. o o0 oI IB sin 90 sin 45

r 4 r 22

2

= − +

= oI 2 2 24 r 2

− +

= oI ( 2 1)2 r 4

+ −

78. A disc of mass m and radius R is lying on a smooth horizontal

surface. A particle of mass m moving horizontally with a

velocity 0v , collides with the disc at B and sticks to it. Speed of

the point A on the disc just after impact will be

(A) 0

31v

8 (B)

0

5v

16

(C) 05v

16 (D) 0v

2

A m

0v B

R/2

Ans. A

Sol. COM, 00 cm cm

vˆ ˆmv i 2mv v i2

= =

COAM about CM 2 2

20

R R 1 mRˆmv sin30º ( k) m mR2 4 2 4

− = + +

0v ˆ( k)4R

= − , A cmv v r= +

0 0v v R 3 3Rˆ ˆ ˆ ˆi ( k) . i j2 4R 2 2 4

= + − +

0 011v 3vˆ ˆi j16 16

= − , A 0

31| v | v

8=

A m 0v B

R/2 cm

cmv

79. A mirror ( = 3/2) is 10 cm thick. An object is placed 15 cm in front of it. The position of image from the front surface is

(A) 15 cm (B) 21.67 cm (C) 28.34 cm (D) 35 cm Ans. C

Sol. Xapp = 10 10 15

3 / 2 3 / 2 1+ +

= 40

153

+ = 28.33

10 cm

15 cm

3/2

80. In a radioactive reaction an unstable nucleus A

dis-integrates into a stable nucleus B. But A is generated at a constant rate of q nucleus per second. Then at steady state number of nucleus of A will be

(A) q (B) q

(C) q – (D) q

q (sec

–1) A

B

(Parent Nucleus)

(Daughter Nucleus)

Ans. B Sol. At steady state. Rate of generation of A = Rate of decay of A.

q = NA

A

qN =

CHEMISTRY

81. Which of the following will not under go Cannizaro reaction?

(A) CH CHO

CH3

CH3

(B)

CHO

Cl

(C)

CHO

(D) Cl3C CHO

Ans. D

Sol.

C C

Cl

Cl

Cl O

H

C C

Cl

Cl

Cl O

H

O H

Na+ OHC Na

+

Cl

Cl

Cl H C OH

O

+

CHCl3 + HCOO Na

82. The de broglie wavelength of neutron at 27ºC is . The wavelength at 927ºC will be

(A) /9 (B) / 4

(C) /2 (D) /3

Ans. C

Sol. 1 2

2 1 2

Th 12004

T 30003m KT

= = = =

83. If the uncertainity in velocity and position is same, the uncertainity in momentum will be

(A) hm

4 (B)

4m

4

(C) h

4 m (D)

1 h

m 4

Ans. A

Sol. h

x. V4 m

=

( )x V =

h

X4 m

=

hm

P m V4

= =

84.

CN CHO

XX can be

(A) LiAlH4/Ether/H2O (B) NaBH4/C2H5OH (C) DIBAl –H (D) Na/C2H5OH Ans. C Sol. Information Based.

85. Which of the following represents correct order of decreasing reducing nature in aqueous solution?

(A) Li Na K Rb (B) Rb K Na Li

(C) Rb Li Na K (D) Li Rb K Na

Ans. D Sol. On moving down the group, reducing nature increases. Li behave exceptionally due to

its small size. 86. Which of the following is correct?

(A) Radius of 2 2Ca Cl S+ − − (B) Radius of 2 2Cl S Ca− − +

(C) Radius of 2 2S Cl Ca− − += = (D) Radius of 2 2S Cl Ca− − +

Ans. A Sol. 2 2Ca Cl S

Z e 20 18 17 18 16 18

+ − −

Size decrease as z/e increases 87.

OO OH NH2

OH.

CH3.MgBr

(excess)CH4 + Other Product

The number of moles of CH4 liberated is (A) 3 (B) 4 (C) 5 (D) 6 Ans. C Sol. 5 Acidic H are present in reactant.

88. The total volume of 0.1 M KMnO4 solution that is needed to oxidize 100 mg each of

ferrous oxalate and ferrous sulphate in a mixture in acidic medium (A) 1.096 ml (B) 1.32 ml (C) 5.48 ml (D) none of these Ans. C Sol. Equivalent of KMnO4 = Equivalent of FeC2O4 + Equivalent of FeSO4

100 100

0.1 v 5 3 1144 152

= +

v 5.48 ml=

89. The dehydration step in aldol condensation in presence of dilute NaOH follow (A) E1 mechanism (B) E2 mechanism (C) Ei mechanism (D) E1 CB mechanism

Ans. D Sol. Anion is formed as intermediate. 90. Which of the following reaction will not produce alcohol?

(A) CH3 MgBrO

(i)

(ii) H+

P

(B) Ph MgBr(i) O2

(ii) H+ P

(C) MgBr(i) O2

(ii) H+ P

(D) CH3 CH CH2 MgBr

CH3

(i) O2

(ii) H+

Ans. B Sol. O H

In B product is which is not alcohol