KVPY – CLASS-XII PART TEST – 4

(OLTS-1819-T4-PT-4-KVPY-XII)

PART – I

MATHEMATICS

1. If A and B are two matrices such that AB B= and BA A,= then 2 2A B+ is equal to

(A) 2AB (B) 2BA (C) A B+ (D) AB Ans. C

Sol. We have ( ) ( )2 22 2A B BA AB+ = +

( )( ) ( )( )BA BA AB AB= +

( ) ( )B AB A A BA B= +

( ) ( )B BA A AB BA AB A B= + = + = + .

2.

2 2

2 2

sin x cos x 1

cos x sin x 1

10 12 2−

is equal to

(A) 0 (B) 2 212 cos 10sin x−

(C) 2 212sin x 10cos x 2− − (D) 210sin x

Ans. A

Sol. Apply 1 1 2 3R R R R ,→ + − then the given determinant is equal to

2

2

0 cos x 1

0 sin x 1 0

0 12 2

= .

3. If the function ( )( ) ( )

cx df x

x 1 x 4

+=

− −has a turning point at the point (2, –1) then

(A) c 2, d 0= = (B) c 1, d 0= =

(C) c 1, d 1= = − (D) c 1, d 1= =

Ans. B

Sol. ( )( ) ( ) ( ) ( )

( ) ( )2 2

c x 1 x 4 cx d 2x 5f ' x

x 1 x 4

− − − + −=

− −

So, ( )( )2c 2c d d

0 f ' 2 d 04 4

− + += = = =

Also ( )2c d

1 f 2 c2

+− = = = −

− c 1 =

4. For the parabola 2y 16x= , the ratio of the length of the subtangent to the abscissa is

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

(C) x : y (D) 2x : y

Ans. A

Sol. Differentiating, dy

2y 16dx

= So dy 8

dx y= .

Thus the length of the subtangent is 2dx y 16x

y. 2xdy 8 8

= = = .

Hence length of the subtangent : abscissa 2x:x 2 :1= = .

5. The trigonometric equation 1 1sin x 2sin a− −= has a solution for

(A) all real values of a (B) 1

a2

(C) 1

a2

(D) 1 1

a2 2

Ans. B

Sol. 1 1 2sin x sin 2a 1 a− −= − if 1

a2

2x 2a 1 a= − which is possible

If ( )2 2 2x 4a 1 a 1= − 11 sin x 1− −

or if 4 24a 4a 1 0− + if ( )2

22a 1 0−

Which is true, so 1

a2

6. Let

b c q r y z

c a r p z x

a b p q x y

+ + +

= + + +

+ + +

and 1

x a p

y b q

z c r

= then

(A) 12 = (B) 12 = −

(C) 14 = (D) 14 = −

Ans. A

Sol. Using 1 1 2 3R R R R ,→ + + we get

( ) ( ) ( )2 a b c 2 p q r 2 x y z

c a r p z x

a b p q x y

+ + + + + +

= + + +

+ + +

Taking 2 common from 1R and applying 2 2 1R R R→ − , 3 3 1R R R ,→ − we obtain

a b c p q r x y z

2 b q y

c r z

+ + + + + +

= − − −

− − −

Applying 1 1 2 3R R R R ,→ + + we get ( ) ( )

a p x a x p

2 1 1 b q y 2 b y q

c r z c z r

= − − = −

1

x a p

2 y b q 2

z c r

= =

7. If 1

x b b

a x b

a a x

= and 2

x b,

a x = then

(A) ( )2

1 23 = (B) ( )1 2

d3

dx =

(C) ( ) 2

1 2

d3

dx = (D) 3/2

1 23 =

Ans. B

Sol. We have 1

1 b b x 0 b x b 0d

0 x b a 1 b a x 0dx

0 a x a 0 x a a 1

= + +

x b x b x b

a x a x a x= + +

23=

8. If sinx cosx tanx cot x sec x cosec x 7+ + + + + = and sin2x a b 7,= − then ordered pair

(a, b) can be: (A) (6, 2) (B) (8, 3) (C) (22, 8) (D) (11, 4) Ans. C Sol. sinx cosx tanx cot x sec x cosec x 7+ + + + + =

( )( )sinx cos x1

sinx cos x 7sinxcos x sinx.cos x

+ + + + =

( )2 2

sin x cos x 1 7sin2x sin2x

+ + = −

( ) 2

4 41 sin2x 1

sin2xsin 2x

+ + +

2

4 2849

sin2xsin 2x= + − (squaring both sides)

3 2sin 2x 44sin 2x 36sin2x 0− + =

sin2x 22 8 7= −

9. Which of the following is the solution set of the equation ( )2

1 1

2

2x 12cos x cot

2x 1 x

− − −

= −

?

(A) (0, 1) (B) ( ) 1, 1 0− −

(C) ( )1, 0− (D) 1, 1−

Ans. A

Sol. 2

1 1

2

2x 12cos x cot

2x 1 x

− − −

= −

Put x cos ;= LHS 2 ; 0= and 1 x 1− ……(i)

RHS = ( )1 1cos2cot cot cot 2 2

2cos sin

− −

= =

……(ii)

If 0 2 or 02

From equation (i) and (ii), 02

( )x 0, 1

10. There is an equilateral triangle with side 4 and a circle with the centre on one of the vertex of

that triangle. The arc of that circle divides the triangle into two parts of equal area. How long is the radius of the circle?

(A) 12 3

(B)

24 3

(C) 30 3

(D)

6 3

Ans. A

Sol. 23 1.16 2. r .

4 2 3

=

2r 12 3 =

12 3

r =

B

A

r 60o

C

11. Let f be a twice differentiable function such that ( ) ( ) ( )f 1 3, f 2 2, f 3 1,= = = then identify the

correct statement:

(A) ( )f ' x 1= − for atleast two ( )x 1, 2 (B) ( )f " x 0= for atleast one ( )x 1, 3

(C) ( )f " x 0= for atleast one ( )x 1,2 (D) ( )f " x 0= for atleast two ( )x 2, 3

Ans. B

Sol. Consider ( ) ( )H x f x x 4= + −

( )H x is continuous and differentiability in (1, 3)

( ) ( ) ( )H 1 H 2 H' x 0= = for atleast one

( )x 1, 2

( ) ( ) ( )H 2 H 3 H' x 0= = for atleast one

( )x 2,3

( )H" x 0= for atleast one ( )x 1, 3

12. The exact value of o o o

o o o

96sin80 sin65 sin35

sin20 sin50 sin110+ + is equal to:

(A) 12 (B) 24 (C) –12 (D) 48

Ans. B

Sol. A

sin A 4 cos2

= in Dr. as A + B + C =

13. Consider a rectangular sheet of perimeter 6 meters. A circular sector of radius equal to

smaller side of sheet and centre at one corner is removed such that the area of remaining portion is maximum, then area of removed portion is:

(A) ( )

24

+ (B)

( )2

9

4 +

(C) ( )

2

36

4

+ (D)

( )2

9

4

+

Ans. D

Sol. Area of remaining portion 21A ab a

4= −

( ) 21a 3 a a

4= − −

dA a 6

3 2a 0 ada 2 4

= − − = =

+

and 2

2

d A2 0

2da

= − −

6

a4

=+

at 6

a ,4

=+

A is maximum

Removed portion 21.a

4=

( ) ( )

2 2

1 36 9.

4 4 4

=

+ +

a

2( a+b )= 6

a

b

14. A car is to be driven 200kms on a highway at an uniform speed of x km/hr (speed rules of

the highway require 40 x 70 ). The cost of diesel is Rs 30/ litre and is consumed at the

rate of 2x

10060

+ litres per hour. If the wage of the driver is Rs200 per hour then the most

economical speed to drive the car is (A) 55.5 (B) 70 (C) 40 (D) 80 Ans. B Sol. Let cost incurred to travel 200kms be c(x)

( )2x 200 200 640000

c x 100 30 200 100x60 x x x

= + + = +

( ) c ' x 0 in x 40,70

c(x) is minimum for x=70 in x 40,70

15. If the functions f( )x and g( )x are defined on R → R such that

3, rational

f( )4 , irrational

+ =

x xx

x x and 5, irrational

g( ), rational

+ =

−

x xx

x x

Then (f g) ( )− x is

(A) one-one and onto (B) neither one-one nor onto (C) one-one but not onto (D) onto but not one-one Ans. B

Sol. We have, ( ) ( ) ( ) ( )( )f g x f x g x− = −

2x 3, x rational

3x 5, x irrational

+ =

−

As, 3 5

f 0 f2 3

− = =

or ( )5 1

f 1 1 f3

+− = =

and so on

( )f x is many one function

Also, 5 does not belong to the range of ( )f x , because of 3x 5 5− = −

x 0 Q=

( )f x is into function.

16. If f(x) = sin-1 cosec(sin-1 x) + cos-1(sec(cos-1 x)) then range of f(x) has (A) exactly two values (B) one value (C) undefined (D) infinite values Ans. B

Sol. f(x) is defined only for x = 1

and ( ) ( )f 1 f 12

= − =

17. If A and B are two square matrices of order 3 x 3 which satisfy AB = A, BA = B then ( )7

A B+

is

(A) ( )64 A B+ (B) ( )16 A B+

(C) ( )4 A B+ (D) none of these

Ans. A Sol. A AB= and ( ) ( )

2 2 2A B A AB BA B 2 A B+ = + + + = +

2A ABA AB A= = = ( ) ( ) ( )4 2

A B 4 A B 8 A B+ = + = +

2 3 2A A A A A = = = ( ) ( ) ( )6 2

A B 16 A B 32 A B+ = + = +

2 3 4 5 6 7A A A A A A A = = = = = = ( ) ( )7

A B 64 A B+ = +

Similarly 2 3 4 5 6 7B B B B B B B= = = + + =

18. Vertices of a variable triangle are (3,4), ( ) sin5,cos5 and ( )− cos5,sin5 , where ,R . Locus

of its orthocenter is

(A) ( ) ( ) 1007yx1yx22=−−+−+ (B) ( ) ( ) 1001yx7yx

22=−−+−+

(C) ( ) ( ) 1001yx7yx22=−++−+ (D) ( ) ( ) 1001yx7yx

22=+−+−+

Ans. D Sol. Origin is circumcentre centre of the variable triangle.

19. Let ( ) ( )( )( )1/3 7f x ln log log sinx a= + be defined for every real value of x, then the possible

value of a is (A) 2 (B) 4 (C) 8 (D) 6 Ans. B Sol. 1 < sinx + a < 7

20. a, b, c are the length of sides BC, CA, AB respectively of ABC satisfying

clog 1 loga logb log 2

a

+ + − =

Also, the quadratic equation ( ) ( )2 2a 1 x 2bx c 1 x 0− + + + = has two equal roots.

The value of ( )sinA sinB sinC+ + is equal to

(A) 5

2 (B)

12

5

(C) 8

3 (D) 2

Ans. B

Sol. Equal roots oC 90 =

and sinA 1 2sinB cosB 2sinB 1+ = = − ( )oA B 90+ =

2 21 sin B 4sin B 4sinB 1 − = − +

4

sinB5

=

12

sinA sinB sinC 3sinB5

+ + = =

PHYSICS

21. In a photodiode the conductivity increases when light of wavelength less than 620 nm is incident. The bandgap is

(A) 1.12 eV (B) 1.8 eV (C) 2.0 eV (D) 1.62 eV Ans. C

Sol. ( )( )g

1242 1242E eV 2eV

nm 620= = =

22. If x and y be the distances of the object and image formed by a concave mirror from its focus and f be the focal length then

(A) xf = y2 (B) xy = f2 (C) x/y = f (D) x/y = f2 Ans. B Sol. According to Newton's formula xy = f2 .

Note that m = f f v

f u f

−=

−

or f y

x x= xy = f2

23. An electron (mass = me) and proton (mass = 1836 me) are moving with the same speed. The

ratio of their de-Broglie wavelength electron

proton

will be

(A) 1 (B) 1836 (C) 1/1836 (D) 918 Ans. B

Sol. h h

p mv = =

When speed is same

1

m

electron

proton

= e

e

1836 m1836

m=

24. A real image of magnification m1 is formed on a screen by a convex lens. If the lens is

moved through a distance x and the object is the moved until a new image of magnification m2 is formed on the screen. The focal length of the lens is

(A)2 1

x

m m- (B)

1 2

x

m - m

(C)1 2

x

m m (D) None of these

Ans. A

Sol. In first case, 1 1 1 q

andp q f p+ = =m1

1+m1 =q

f …(1)

In the second case 1 1 1

q x p f+ =

+

And2

q xm

p

+=

m2 =

q x

f

+ …(2)

(1) and (2)

m2 – m1 = x/f

f = 2 1

x

m m−

25. Imagine a Young’s double slit interference experiment performed with wave associated with fast moving electrons produced from an electron gun. The distance between successive maxima will decrease maximum if

(A) the accelerating potential in the electron gun is decreased. (B) the accelerating potential is increased and the distance of screen from slit is decreased. (C) the distance of the screen from the slit is increased. (D) the distance between the slits is decreased. Ans. B

Sol. If V is increased will decrease and hence fringe width will decrease.

26. The dimensions of Stefan-Boltzmann constant can be written in terms of Planck’s constant

h, Boltzmann constant kB and the speed of light c as = Bh k c . Here

(A) 3, 4and 3 = = = − (B) 3, 4 and 2 = = − =

(C) 3, 4and 2 = − = = − (D) 2, 3and 1 = = − = −

Ans. C

Sol. h k c

=

( ) ( ) ( )3 4 2 1 2 2 1 1MT K ML T ML T K LT

− − − − − − =

1 4

2 2 0 3

4 2

+ = =

+ + = = −− = − = −

27. Write the Boolean expression of the block diagram

represented in figure (A) 1 (B) 0

(C) A A+ (D) ( )A A+

• y

•

•

A

A

Ans. A Sol. Fact based 28. A spectral line results from the transition n =2 to n =1, in the atoms/species given below

which one of these will produce the shortest wavelength emission? (A) Hydrogen atom (B) Singly ionised Helium (C) Doubly ionized Lithium (D) Deuterium atoms Ans. C Sol. According to Bohr theory

2

2 2

1 2

1 1 1RZ

n n

= −

Thus 2

1

Z

More the atomic number, smaller is the wavelength obtained in the transition of electron (for identical transition).

29. When an electron jumps from an orbit with n = 1 to n = 2, the change in orbital angular momentum is

(A)h

2 (B)

h

4

(C)3h

4 (D) Zero

Ans. A Sol. According to Bohr postulates for hydrogen like atoms, electron can orbit in those orbits in

which it’s orbital angular momentum is integral multiple of h/2. Thus change in orbital angular momentum

= 2h 1 h h

2 2 2− =

30. When ultraviolet radiation of a certain frequency falls on a potassium target, the

photoelectrons released can be stopped completely by a retarding potential of 0.6 V. If the frequency of the radiation is increased by 10%, this stopping potential rises to 0.9 V. The work function of potassium is

(A) 2.0 eV (B) 2.4 eV (C) 3.0 eV (D) 2.8 eV Ans. B

Sol. h – 0 = 0.6 ev. 1.1 h – 1.1 = 0.66

1.1 h - = 0.9 ev. 1.1h 0.9−

+− − =

0.1 = 0.24

= 2.4 ev.

31. A transistor having equal to 80 has a change in base current of 250 A, then the change in collector current is

(A) ( )250 8 A+ (B) 250/80 A

(C) ( )250 80 A− (D) 80 250 A

Ans. D

Sol. c BI I=

80 250 A=

Transistor M, A 32. Two plane mirrors A and B are aligned parallel to each

other as shown in figure. A light ray is incident at an angle of 30° at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is

(A) 28 (B) 30 (C) 32 (D) 34

0.2 m

32 m B

30°

A Ans. B

Sol. From the law of refection

tan30° = BC BC

AB 0.2= ;

1BC 0.2 0.115

3= =

Total no. of reflection = 30

C

30° 30°

A

B

33. In a pure semiconductor the number of conduction electrons is 196 10 /m3. The number of

holes in a sample of 1 cm 1 cm 1 mm are

(A) 136 10 (B) 126 10

(C) 156 10 (D) 166 10

Ans. B

Sol. Since 19 2 2 3

i ih n 6 10 10 10 10− − −= =

= 126 10

34. Two identical P-N junctions may be connected in series with a battery in three ways, as

shown in figure. The potential drops across the P-N junctions are equal in (A) circuit 1 and circuit 2. (B) circuit 2 and circuit 3. (C) circuit 3 and circuit 1. (D) circuit 1 only.

P n n P

+ −

P n n P

+ −

n P

+ −

n P

(3) (2) (1) Ans. B Sol. Factual

35. 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°.

36. Two coherent point sources of frequency 10v

(fd

=

where v is speed of light) are placed at a distance d apart as shown in figure. The receiver is free to move along the dotted line shown in the figure. Find total number of maxima observed by receiver.

(A) 6 (B) 7 (C) 5 (D) 8

S1

2d

3

d

3 S2

Ans. A

Sol. f = v = 10 v d

.d 10

=

Maximum path diff. can be d

.3

Hence, number of maximum will be 6.

37. Critical angle of light passing from glass to air is maximum for (A) red colour. (B) green colour (C) yellow colour. (D) blue colour.

Ans. A

Sol. 1

v r

1C Sin and u u

u

−=

38. A glass prism of refractive index 1.5 is immersed in water (R.I. =

4/3). The beam of light incident normally on the face AB is totally reflected to reach the face BC, if

(A) sin 8/9

(B) sin < 2/3

(C) 2/3 < sin < 8/9 (D) None of these

C

B A

Ans. A Sol. The light ray is totally reflected internally at the point P (say).

Therefore the critical angle C . Also,

c w

g

sin n

sin 90 n

=

sin c =

4 / 3 8

3 / 2 9=

sin 8/9

A

B

p

C

39. The angles of incidence and refraction of a monochromatic ray of light of wavelength at an

air-glass interface are I and r, respectively. A parallel beam of light with a small spread in

wavelength about a mean wavelength is refracted at the same air-glass interface. The

refractive index of glass depends on the wavelength as () = a + b/2 where a and b are constants. Then the angular spread in the angle of refraction of the beam is

(A) 3

sini

cosr

(B)

3

2b

(C) 3

2btanr

a b

+ (D)

2

3

2b(a b / )sini+

Ans. C

Sol. sin i = sin r

d

...(i)dr tanr

=

2 3

a b d 2b...(ii)

dx

+ − = =

From (i) and (ii)

dr = 3

2btan rd

a b

+

40. In Young’s Double slit Experiment, the wavelength of the red light is 7800 Å and that of blue

light is 5200 Å. The value of n for which nth bright band due to red light coincides with (n + 1)th bright band due to blue light, is

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

Sol. 1n(red)

n Dy

d

=

2nt(blue)

(n 1) Dy

d

+ =

Apply yn (red) = yn+1 (blue)

n(r) = (n + 1)b n(7800) = (n + 1) (5200)

n = 2

CHEMISTRY

41. ZnS reacts with dilute H2SO4 to produce a colorless gas, which on reaction with Cl2 yields (A) HCl (B) SCl4 (C) HOCl (D) S2Cl2 Ans. A Sol. The colorless gas is H2S. 42. When copper ore is mixed with silica, in a reverberatory furnace, copper matte is produced.

The copper matte contains ____________. (A) sulphides of copper (II) and iron (II) (B) sulphides of copper (II) and iron (III) (C) sulphides of copper (I) and iron (II) (D) sulphides of copper (I) and iron (III)

Ans. C Sol. Cu2S and FeS 43. Which of the following are peroxoacids of sulphur?

(A) H2SO5 and H2S2O8 (B) H2SO5 and H2S2O7 (C) H2S2O7 and H2S2O8 (D) H2S2O6 and H2S2O7

Ans. A Sol. H2SO5 (Peroxomonosulphuric acid) and H2S2O8 (Peroxodisulphuric acid) 44. On addition of small amount of KMnO4 to concentrated H2SO4, a green oily compound is

obtained which is highly explosive in nature. Identify the compound from the following. (A) Mn2O7 (B) MnO2 (C) MnSO4 (D) Mn2O3

Ans. A

Sol. 2HMnO4 – H2O → Mn2O7 (covalent green oil)

45. Which of the following complex does NOT exist at room temperature? (A) [BiI4]– (B) [CuI4]–2 (C) [PbI4]–2 (D) [HgI4]–2 Ans. B Sol. Cu(II) oxidizes I– to I2. 46. Which of the following compounds liberate CO2 on heating? (A) Li2CO3 (B) Na2CO3 (C) K2CO3 (D) All of these Ans. A

Sol. 2 3 2 2Li CO Li O CO⎯⎯→ +

47. Which of the following is NOT a carbonate ore? (A) Siderite (B) Malachite (C) Zincite (D) Calamine Ans. C Sol. Siderite-FeCO3, Malachite-CuCO3.Cu(OH)2, Zincite-ZnO, Calamine-ZnCO3. 48. Among the given properties, which is correct for both borazine and benzene? (A) They are aromatic compounds (B) They are isoelectronic having total 42 electrons each (C) The B and N-atoms in borazine and C-atoms in benzene are sp2-hybridized (D) All of these Ans. D Sol. H

H

H

H

H

HB

N

N

B

B

N

H

H

H

H

H

H

Benzene Borazine 49. Which of the following molecules is NOT aromatic?

(A)

O

(B)

(C)

(D)

Ans. D

Sol.

HH

contains sp3 – carbon and molecule does not obey Huckel’s rule.

50. The correct structure for Gammaxane is

(A)

Cl

Cl

Cl

Cl

Cl

Cl

(B)

Cl

Cl

Cl

Cl

Cl

Cl

(C)

Cl

Cl

Cl

Cl

Cl

Cl

(D) None of these

Ans. B Sol. Gammaxane is benzene hexachloride C6H6Cl6. 51. Which of the following reagents can be used to distinguish But-1-yne and But-2-yne? (A) H2/Ni (B) dil.H2SO4/HgSO4(aq) (C) HBr/acetone (D) NaNH2 Ans. D

Sol. 2Na NH

3 2 3 2 3CH CH C CH CH CH C C Na NH+ −

− +− − ⎯⎯⎯⎯→ − − +

52. The increasing order of basicity of the following compounds is:

(a)

(b)

(c)

(d)

(A) (a) < (b) < (c) < (d) (B) (b) < (a) < (c) < (d) (C) (b) < (a) < (d) < (c) (D) (d) < (b) < (a) < (c) Ans. C Sol. Higher the electron density on N, more its basic strength. 53. In graphite and diamond, the percentage of p-characters of the hybrid orbitals in

hybridization are respectively: (A) 33 and 25 (B) 33 and 75 (C) 67 and 75 (D) 50 and 75 Ans. C Sol. Graphite (sp2), Diamond (sp3) 54. Xenon hexafluoride on partial hydrolysis produces X and Y. Compounds X and Y and

oxidation state of Xe, respectively are: (A) XeOF4 (+6) and XeO3 (+6) (B) XeOF4 (+6) and XeO2F2 (+6)

(C) XeO2F2 (+6) and XeO2 (+4) (D) XeO2 (+4) and XeO3 (+6)

Ans. B Sol. XeF6 + H2O → XeOF4 + 2HF XeF6 + 2H2O → XeO2F2 + 4HF 55. The number of P – O bonds in P4O6 is (A) 9 (B) 18 (C) 12 (D) 6 Ans. C Sol.

56. The increasing order of the acidity of the following carboxylic acids is :

(A) III < II < IV < I (B) I < III < II < IV (C) IV < II < III < I (D) II < IV < III < I Ans. A Sol. Acidity of the acid (HA) depends on stability of its conjugate anion (A-). 57. In KO2, the nature of oxygen species and oxidation state of oxygen atom are respectively (A) Superoxide and -1/2 (B) Peroxide and -1/2 (C) Oxide and -2 (D) Superoxide and -1 Ans. A Sol. Potassium superoxide, K+ O2

- 58. In Wilkinson’s catalyst, the hybridization of central metal ion and its shape are respectively: (A) d2sp3, octahedral (B) sp3d, trigonal bipyramidal (C) sp3, tetrahedral (D) dsp2, square planar Ans. D Sol. Wilkinson’s catalyst, [(Ph3P)3RhCl]

59. Mg3N2 + H2O → X(solution) + Y(gas) Gas (Y) in the above reaction is (A) N2 (B) NH3 (C) NO (D) NO2

Ans. B

Sol. Mg3N2 + H2O → Mg(OH)2 + NH3 60. What is the oxidation number of hydrogen in CaH2?

(A) +1 (B) − 1

(C) − 2 (D) + 2 Ans. B Sol. The oxidation numbers of calcium and hydrogen in CaH2 are, respectively, +2 and -1

PART – II

MATHEMATICS

61. If ( )3sin sin 2 , = + then ( )tan 2tan + − is

(A) independent of (B) independent of

(C) independent of both and (D) independent of none of these

Ans. C

Sol. ( )sin 2 3sin + =

( )

( )

sin 2 sin 3 1

sin 2 sin 3 1

+ + + =

+ − −

( )

( )( )

2sin cos2 tan 2 tan 0

2cos sin

+ = + − =

+

62. Let

3 1

1 12 2P ,A

0 11 3

2 2

= = −

and Q TPAP ,= then T 2015P Q P is

(A) 1 2015

0 1

−

(B) 2015 1

0 2015

(C) 1 2015

0 1

(D) 2015 2015

0 2015

Ans. C

Sol.

cos sin6 6

P

sin cos6 6

=

−

T

cos sin6 6

P

sin cos6 6

−

=

Since T 1 0PP

0 1

=

T 1P P−=

We have T 1Q PAP PAP−= =

( )2015

2015 1 2015 1Q PAP PA P− −= =

Thus, ( )T 2015 1 2015 1P Q P P PA P P− −=

( ) ( )1 2015 1P P A P P− −=

Now, A I B= + where 0 1

B0 0

=

Since, 2B O,= we get rB O r 2= .

Thus, 2015 1 2015A I 2015B

0 1

= + =

63. If the tangent to the curve 3 2x y 0− = at ( )2 3m , m− is parallel to 31y x 2m

m= − − , then the

value of 2m is

(A) 1

3 (B)

1

6

(C) 2

3 (D)

2

3

−

Ans. C

Sol. Differentiating 3 2x y 0,− = we have ( )2 3

2 4

3

m , m

dy 3x dy 3m 3m

dx 2y dx 22m−

= = − = −

According to the given condition 23 1 2m m

2 m 3− = − =

64. The value of ( ) ( )1 1 1 31tan tan2A tan cot A tan cot A

2

− − − + +

for 0 A

4

is:

(A) ( )14tan 1− (B) ( )12tan 2−

(C) 0 (D) none of these Ans. A

Sol. ( ) ( )1 1 1 31tan tan2A tan cot A tan cot A

2

− − − + +

3

1 1

4

1 cot A cot Atan tan2A tan

2 1 cot A

− − + = + +

−

0 A cot A 14

( )

( ) ( )

2

1 1

2 2 2

cot A 1 cot Atan Atan tan

1 tan A 1 cot A 1 cot A

− −+

= + + − − +

1 1

2 2

tanA cot Atan tan

1 tan A 1 cot A

− − = + +

− −

( )14tan 1 A−= =

65. If ( )1 1A x ,y and ( )2 2B x ,y are two points on curve ( )2 x x

1 2y e x 0 and x 0+

= such that

tangents at points A and B are perpendicular, then number of such pair of points A and B, is: (A) 0 (B) 1 (C) 2 (D) more than 2 Ans. A

Sol. If 13x3x dyx 0, y e 3e

dx = =

2xx dyx 0, y e e

dx

−− = = −

1 2 1 23x x 3x x 13e .e 1 e 1

3

− −= =

1 23x x 0 − which is impossible

66. A spherical balloon is expanding. If at any instant rate of increase of its volume is 16 times of

rate of increase of its radius, then its radius at that instant, is

(A) 1

(B)

2

(C) 2

(D)

4

3

Ans. B

Sol. 34V r

3=

2 2dV dr 24 r 16 4 r r

dt dt= = =

67. If ( )x be a differentiable function such that ( )5 0 and ( )3 0 − and Rolle’s theorem is

not applicable to ( )( )

2x 2x 15f x

x

− −=

in 3, 5− , then:

(A) ( )x has atleast one root in 3, 5−

(B) ( )x has exactly one root in 3, 5−

(C) ( )x has exactly two root in 3, 5−

(D) ( )x has no root in 3, 5−

Ans. A

Sol. ( ) ( )f 3 f 5 0− = =

As rolle’s theorem is not applicable for ( )f x in 3, 5− so either ( )f x is not continuous or

not differentiable. So, ( )c 0 = for at least one ( )c 3, 5 − .

68. Let nA and nB be square matrices of order 3, which are defined as:

n ijA a = and n ijB b = where ij 2n

2i ja

3

+= and ij 2n

3i jb

2

−= for all i and j, 1 i, j 3 .

If ( )2 3 n

1 2 3 nn

l LimTr. 3A 3 A 3 A ...... 3 A→

= + + + + and

( )2 3 n

1 2 3 nn

m LimTr. 2B 2 B 2 B .......... 2 B→

= + + + + , then find value of ( )l m

3

+

[Note : Tr. (P) denotes the trace of matrix P.] (A) 3 (B) 5

(C) 7 (D) none of these Ans. C

Sol. ( )r 1 11 22 33T A a a a= + +

( )1

3 6 9 29

= + + =

( ) ( )r 2 4 4

1 18T A 3 6 9

3 3= + + =

l = 9

( )r 1 11 22 33T B b b b= + + ( )1

2 4 64

= + +

( ) ( )r 2 4

1T B 2 4 6

2= + +

m = 12

69. Let ( ) 3100 99f x x x= − − then set of values of x satisfying the inequality

( ) ( )( ) ( )( )3

100 99 100 1f x f x f x− − − is

(A) (–99, 0) (99, ) (B) (– , –1) (0, 1)

(C) (–1, 0) (1, ) (D) (–100, 0) (100, ) Ans. C

Sol. f(x) is everywhere decreasing. f(f(x))>f(100(1–x)) f(x)<100(1–x). 70. Vertices of a variable acute angle triangle ABC lies on a fixed circle. If a, b, c are length of

sides and angles of triangle ABC and x1,x2, x3 are distances of orthocenter from A, B, C

respectively then find the maximum value of 1 2 3dx dx dx3 11

da db dc

+ + +

(A) 2 (B) 3 (C) 4 (D) 5 Ans. A

Sol. 1 2 3x 2Rcos A, x 2RcosB,x 2RcosC= = =

1dx da2RsinA also a 2RsinA 2RcosA

dA dA= − = =

so, 1 2 3dx dx dxtanA, tanB, tanC

da db dc= − = − = −

tan A tanB tanC 3 3+ +

1 2 3dx dx dxSo, 3 11 2

da db dc

+ + +

PHYSICS

71. If I = 0.1 (mA) [eV/VT – 1] is valid for a pn junction. Then find the resistance when V = 0.5 volt and VT = 0.025 volt.

(A) 50 (B) 25

(C) 10 (D) zero Ans. D

Sol. 3 20

V/V

T

dI 0.1 0.1 10 ee T

dV V 0.025

− = =

R = 8

dV 2500

dI 5 10= →

72. A nuclear reactor is operating at 600 M Watt per hour. If energy released per fission is

3 10–11 J, the number of nuclei consumed per minute will be

(A) 2 1019 (B) 1.2 1021

(C) 2 1013 (D) 3.3 1017

Ans. D

Sol. Energy per Hour = 600 106 Joules

Energy per minute = 6

7600 1010

60

= Joule

Number of nuclei consumed per minute

7

17

11

103.3 10

3 10−=

73. Two coherent monochromatic light beams of intensities I and 4I are superimposed. The

maximum and minimum possible intensities in the resulting beam are: (A) 5 I and I (B) 5 I and 3I

(C) 9 I and I (D) 9 I and 3 I Ans. C

Sol. Intensity (Amplitude)2

I A2 When two waves (beams) of amplitude A1 and A2 superimpose, at maxima and minima, the

amplitude of the resulting wave are (A1 + A2) and (A1 - A2) respectively. If the maximum and minimum possible intensities are Imax and Imin respectively, then

Imax (A1 +A2 ) 2

And Imin (A1 – A2 ) 2

2

12

max 1 2 2

1min 1 2

2

A1

I A A A

AI A A1

A

+ +

= = − −

where 1

2

A I 1

A 24I= =

max

min

I 9

I 1= Imax = 9I, Imin = I

74. A thin rod of length f/3 is placed along the optic axis of a concave mirror of focal length f such that its image, which is real and elongated, just touches the rod. The magnification is

(A) 2 (B) 4 (C) 2.4 (D) 1.5 Ans. D Sol. For B

1 1 1

fV f(2f )

3

+ =−

− −

, 1 3 1 3 5

V 5f f 5f

−= − =

V = – 5f

2 , |AB | =

5f f2f

2 2− =

Magnification =

f

2 1.5f

3

=

A B

f/3

B

75. 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) b

1/2u f

f

−

(B) bu f

f

−

(C) b

1/2f

u f

− (D) b

2f

u f

−

Ans. D

Sol. 1 1 1

u v f+ =

1 1 1

v f u= − + v =

uf

(f u)−

Differentiating 1 1 1

u v f+ =

2 2

1 1du

u v− = − dv

f is constant

2 2

2 2

dv v f

du u (u f)= − = −

− m =

I dv

O du= − =

2

2

f

(u f )−

I = 2

2

bf

(u f )−

76. Two thin convex lens of focal lengths f1

and f2 are separated by a horizontal

distance d (where 1 2d f , d f ), and their

centres are displaced by a vertical

separation as shown in the figure. Taking the origin of coordinates, O at the centre of left lens, the x and y coordinates of the focal point of this lens system, for a parallel beam of rays coming from the left, are given by

y

O

d

x

(A) 1 2

1 2

f fx

f f=

+ y = 0 (B)

( )1 2

1 2

f f dx

f f d

+=

+ −

2

1 2

yf f

=

+

(C) ( )1 2 1

1 2

f f d f dx

f f d

+ −=

+ −

( )1

1 2

f dy

f f d

−=

+ − (D)

( )1 2 1

1 2

f f d f dx

f f d

+ −=

+ − y = 0

Ans. C

Sol. For the second lens ; ( )1 2u f d , f f= − =

1 1 1

v u f− =

( )( )

2 1

1 2

f f fv

f f d

−=

+ −

x d v= + 2

1 2 1

1 2

f f f d dx

f f

+ −=

+

Height from principal axis = (y – )

( )y v

mu

− − = =

( )( )

1

1 2

f dy

f f d

−=

+ −

77. The refractive index n of a medium within a certain region x > 0,

y >0 changes with y till it aquires a value nmax. After it aquires the value nmax it remains constant. A light ray travelling in air along the x-axis, strikes the medium at a grazing angle and moves through the medium along a circular arc as shown in the figure. If angular deviation of the ray before it starts moving on a straight line is 600, then nmax is

(A) 2 (B) 3

(C) 3

2 (D) 2

Air

Air

z

x

y

Ans. A Sol. (1) sin 900 = (nmax) sin 300 nmax = 2

Air x

y

300

600 60

0

78. Choose the incorrect option regarding photoelectric effect supports quantum nature of light because

(A) these is a minimum frequency of light below which no photoelectron are emitted. (B) the maximum kinetic energy of photo electrons depends only on the frequency of light

and not on it’s intensity. (C) even when the metal surface is faintly illuminated when the photoelectron leave the

surface immediately. (D) electric charge of the photoelectrons in quantized. Ans. D Sol. Factual

79. A prism ( = 1.5) has a refracting angle of 30°. The deviation of a monochromatic ray

incident normally on its one surface will be (sin48°36 = 0.75)

(A) 18°36 (B) 22°38 (C) 18° (D) 22°1

Ans. A

Sol. 1.5 sin 30° = 1 sin r

3

sinr4

= = 0.75 = sin48°36

r = 48°36

= r – i = 48°36 – 30° = 18°36

r

30°

30°

=1.5

80. A ray incident at a point as an angle of incidence of 60° enters a glass sphere of R.I. n = 3

and is reflected and refracted at the farther surface of the sphere. The angle between the reflected and refracted rays at this surface is (A) 50° (B) 60° (C) 90° (D) 40°

Ans. C Sol. Refraction at P.

0

1

Sin 603

Sin r=

1

1Sinr

2= r1 = 30°

Since r2 = r1

r2 = 30°

Refraction at Q 2

2

Sin r 1

Sin i 3=

Putting R2= 30° we obtain i2 = 60° Reflection at Q

o

2 2r r 30 = =

0

2 2180 (r i ) = − +

= 180°- (30°+60°) = 90°

600

i2r’2

r2

r1

P

Q

CHEMISTRY

81. NaCl + MnO2 + H2SO4 → MnCl2 + NaHSO4 + H2O + [X]; Which of the following is/are correct about [X]?

(A) Its aqueous solution loses its color on standing. (B) [X] on reaction with white phosphorus gives a compound which fumes in moisture. (C) It turns acidified light green solution of FeSO4 into yellow (D) All of these. Ans. D Sol. [X]: Cl2 82.

CH3

NO2

( ) ( )

( ) ( ) ( )( )

o2

3 2 4 3

i Sn/HCl ii NaNO /HCl,0 5 C

iii H PO iv KMnO /H v NHX Major product+

−⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→

The molecular formula of (X) is (A) C7H6N2O3 (B) C7H9NO2 (C) C6H6N2O2

(D) C6H5NO3 Ans. B

Sol

CH3

NO2

( ) oi Sn/HCl ,0 5 C−⎯⎯⎯⎯⎯⎯→

( ) 3v NH⎯⎯⎯⎯

( ) 3 2iii H PO⎯⎯⎯⎯→

( ) 2

o

ii NaNO /HCl

0 5 C−⎯⎯⎯⎯⎯⎯→

CH3

NH2

CH3

N2 Cl

CH3

COOH

(iv) KMnO /H4

COO NH 4 83. The process of producing ‘syn-gas’ from coal at 1270 K is called ‘coal gasification’. Here,

production to dihydrogen can be increased by (A) carrying out the reaction at high pressures. (B) adding carbon monoxide to the reaction mixture. (C) reacting syn-gas mixture with carbon dioxide in presence of iron chromate as catalyst. (D) reacting syn-gas mixture with steam. Ans. D Sol. Coal gasification

( ) ( ) ( ) ( )2 2

Syn gas

C s H O g CO g H g+ +

Water-gas shift reaction

( ) ( ) ( ) ( )2 2 2CO g H O g CO g H g+ +

84. The major product of the reaction is

2H /H O+⎯⎯⎯⎯⎯→

(A)

OH

(B) OH

(C)

OH

(D)

OH Ans. C Sol.

2H O

H−⎯⎯⎯→H

⎯⎯⎯→Rearrangement

⎯⎯⎯⎯⎯⎯⎯⎯⎯→

OH 85. Which of the following species is paramagnetic? (A) BaO2 (B) KO2 (C) N2O (D) Li2O Ans. B

Sol. K+ 1s22s22p63s23p6

z

2 *2 2 *2 2

2 1s 1s 2s 2s 2PO− x x

2 *2

2P 2P

y y

2 *1

2P 2P

KO2 contains one unpaired electron hence paramagnetic. 86. ( )

( )

( )( )

( )3 2 4

o2

i Conc.HNO /H SO

ii Sn/HCl

iii NaNO /HCl, 0 5 C

iv KI/acetone

X Major product ; X is−

⎯⎯⎯⎯⎯⎯⎯⎯→

(A)

NO2

I

(B)

NH2

I

(C)

NH2

NO2

I

(D)

I

Ans. D Sol.

NO2NH2

3

2 4

Conc.HNO

H SO⎯⎯⎯⎯⎯⎯⎯→ o

2NaNO /HCl

0 5 C⎯⎯⎯⎯⎯⎯⎯→

−

Sn/HCl⎯⎯⎯⎯⎯→

N2 Cl

KI/acetone

I

87. The correct combination is: (A) [NiCl4]2– – square-planar; [Ni(CN)4]2– – paramagnetic (B) [NiCl4]2– – dimagnetic; [Ni(CO)4] – square-planar (C) Ni(CN)4]2– – tetrahedral; [Ni(CO)4] – paramagnetic (D) [NiCl4]2– - paramagnetic; [Ni(CO)4] – tetrahedral Ans. D Sol. As per Valance Bond theory, in [NiCl4]2–, Ni+2 has 3d8 configuration with 2 unpaired

electrons.

88. When XO2 is fused with an alkali metal hydroxide in presence of an oxidizing agent such as KNO3; a dark green product is formed which disproportionates in acidic solution to afford a dark purple solution. X is

(A) Ti (B) Mn (C) Cr (D) V Ans. B Sol. XO2: MnO2 ; Dark green product: MnO4

2- ; Dark purple solution: MnO4-

89. Which of the following complexes will show geometrical isomerism? (A) potassium tris(oxalato)chromate(III) (B) pentaaquachlorochromium(III) chloride (C) potassium amminetrichloroplatinate(II) (D) aquachlorobis(ethylenediamine)cobalt(II) chloride Ans. D Sol.

90. The IUPAC name of the following compound is:

(A) (B)

(C) (D) Ans. D Sol. As per IUPAC Nomenclature system