1

SOLUTION TO COMBINED TEST-10 FOR CLASS 11TH ALL (HELD ON 12-JAN-20) PHYSICS (PART I)

SECTION I 1. Two forces with equal magnitudes F act on a body and the magnitude of the resultant force is

F/3. The angle between the two forces is

(A) 1 17cos

18

(B) 1 1cos

3

(C) 1 2cos

3

(D) 1 8cos

9

Solution (A):

net

2 2 21 2 1 2F F F 2F F cos

2F

3

2 2 2F F 2F cos 17

cos18

2. The force-time (F – t) curve of a particle executing linear motion is as shown in the figure. The momentum acquired by the particle in time interval from zero to 8 second will be

(A) – 2 N-s (B) + 4 N-s (C) 6 N-s (D) Zero Solution (D):

Momentum acquired by the particle is numerically equal to area enclosed between the F-t curve and time axis. For the given diagram area in upper half is positive and in lower half is negative (and equal to upper half), so net area is zero. Hence the momentum acquired by the particle will be zero.

3. A man is standing at a spring platform. Reading of spring balance is 60 kg wt. If man jumps outside platform, then reading of spring balance

(A) First increases then decreases to zero (B)Decreases (C) Increases (D)Remains same Solution (A):

For jumping he presses the spring platform, so the reading of spring balance increases first and finally it becomes zero.

4. A block can slide on a smooth inclined plane of inclination kept on the floor of a lift. When the

lift is descending with a retardation a, the acceleration of the block relative to the incline is (A) (g a)sin (B) (g a) (C) gsin (D) (g a)sin

Solution (A):

Acceleration of block in a stationary lift = gsin

If lift is descending with acc. then it will be (g–a)sin. but in the problem acceleration = – a (retardation) Acceleration of block = [g ( a)]sin = (g a)sin

5. A given object takes n times as much time to slide down a 45° rough incline as it takes to slide down a perfectly smooth 45° incline. The coefficient of kinetic friction between the object and the incline is given by

(A) 2

11

n

(B)

2

1

1 n (C)

2

11

n

(D)

2

1

1 n

Solution (A):

2

1tan 1

n

2

11

n [As 45 ]

2

6. Pulling force making an angle to the horizontal is applied on a block of weight W placed on a

horizontal table. If the angle of friction is , then the magnitude of force required to move the

body is equal to

(A) Wsin

gtan( )

(B)

Wcos

cos( )

(C)

Wsin

cos( )

(D)

Wtan

sin( )

Solution (C):

Conceptual

7. A block of mass m lying on a rough horizontal plane is acted upon by a horizontal force P and another force Q inclined at an angle to the vertical. The block will remain in equilibrium, if the

coefficient of friction between it and the surface is

(A) (P Qsin )

(mg Qcos )

(B)

(Pcos Q)

(mg Qsin )

(C) (P Qcos )

(mg Qsin )

(D)

(Psin Q)

(mg Qcos )

Solution (A):

By drawing the free body diagram of the block for critical condition

F R P Qsin (mg Qcos )

P Qsin

mg Qcos

8. A particle moves under the effect of a force F = Cx from x = 0 to 1x x . The work done in the

process is

(A) 21Cx (B) 2

1

1Cx

2 (C) 1Cx (D) Zero

Solution (B): 11 1

xx x 221

0 0 0

x 1W F.dx Cx dx C Cx

2 2

9. Two bodies of masses 1m and 2m have equal kinetic energies. If 1p and 2p are their respective

momentum, then ratio 1 2p : p is equal to

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

(C) 1 2m : m (D) 2 21 2m : m

Solution (C):

P 2mE P m (if E const.) 1 1

2 2

P m

P m

10. A sphere of mass m moving with a constant velocity u hits another stationary sphere of the same mass. If e is the coefficient of restitution, then the ratio of the velocity of two spheres after collision will be

3

(A) 1 e

1 e

(B)

1 e

1 e

(C)

e 1

e 1

(D) 2e 1

te 1

Solution (A):

1 2mu mv mv

1 2u v v

2u ue 2v

2

u(1 e)v

2

1

u(1 e)u v

2

1

1 eu v

2

1

2

v 1 e

v 1 e

11. A particle of mass m moving with a velocity V makes a head on elastic collision with another particle of same mass initially at rest. The velocity of the first particle after the collision will be

(A) V (B) V (C) 2V (D) Zero

Solution (D): In perfectly elastic head on collision of equal masses velocities gets interchanged

12. A bullet of mass a and velocity b is fired into a large block of mass c. The final velocity of the system is

(A) c

ba b

(B) a

ba c

(C) a b

.ac

(D)

a cb

a

Solution (B):

By the conservation of momentum

a b + 0 = (a + c) V ab

Va c

13. A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is (A) MgL (B) MgL/3 (C) MgL/9 (D) MgL/18

Solution (D):

2 2

MgL MgL MgLW

182n 2(3) (n = 3 given)

14. An open knife edge of mass 'm' is dropped from a height 'h' on a wooden floor. If the blade penetrates upto the depth 'd' into the wood, the average resistance offered by the wood to the knife edge is

(A) mg (B) h

mg 1d

(C)

hmg 1

d

(D)

2h

mg 1d

4

Solution (C):

Let the blade stops at depth d into the wood. 2 2v u 2aS 20 ( 2gh) 2(g a)d

by solving h

a 1 gd

So the resistance offered by the wood h

mg 1d

15. A bird flies for 4 s with a velocity of |t 2|m/s in a straight line, where t is time in seconds. It

covers a distance of (A) 2 m (B) 4 m (C) 6 m (D) 8 m

Solution (B): The velocity time graph for given problem is shown in the figure.

Distance travelled S = Area under curve= 2+2=4m 16. A body of mass m kg is rotating in a vertical circle at the end of a string of length r metre. The

difference in the kinetic energy at the top and the bottom of the circle is

(A) mg

r (B)

2mg

r (C) 2mgr (D) mgr

Solution (C):

Difference in K.E. = Difference in P.E. = 2mgr

17. If cmL is the angular momentum about COM of a system of particles , cmr is the position vector

of the centre of mass relative to any point ‗P‘ and p is the total momentum the angular

momentum L of the system relative to the point P is given by

(A) cmL L (B) cm cmL p r L

(C) cm cmL r p L (D) cm cmL L r p Solution (C):

cm cmL L r P

18. If the rotational inertia parameter of a body rolling down a rough inclined plane (of inclination

and height ‗h‘) without slipping is given by cm2

I

MR then the time taken by the body to teach

the bottom of the incline plane is given by

(A) 1 2h

t (1 )sin g

(B) 1 2h

t (1 )cos g

(C) 1 2h

t (1 )sin g

(D) 1 h

t (1 )sin g

Solution (A):

cm

gsina

1

Hence

2h 1t

sin gsin

5

19. A particle of mass 1 kg is projected at an angle q with horizontal. Its co-ordinates at any instant are (5m, 5m) and it is having velocity components along X-axis and Y-axis as 8 m/s and 4 m/s respectively. Its angular momentum about origin is

(A) – 20 k̂ (B)+ 20 k̂ (C)– 60 k̂ (D)+ 60 k̂ Solution (A): Conceptual

20. A uniform disc of mass m and radius R is rolling up a rough inclined plane which makes an

angle of 300with the horizontal. If only gravitational and frictional forces are acting on the disc, the magnitude of the frictional force acting along the inclined plane is

(A)mg

6upwards (B)

mg

6downwards (C)

mg

4upwards (D)

mg

4 downwards

Solution (A): Conceptual

SECTION-II 21. A block of mass 5 kg is on a rough horizontal surface and is at rest. Now a force of 24 N is

imparted to it with negligible impulse. If the coefficient of kinetic friction is 0.4 and 2g 9.8m/s , then the acceleration of the block (in m/s2) is

Solution (0.88):

Net force = Applied force – Friction force ma 24 mg 24 0.4 5 9.8 24 19.6

24.4a 0.88 m/s

5

22. When a spring is stretched by 2 cm, it stores 100 J of energy. If it is stretched further by 2 cm, the stored energy will be increased by (in J) Solution (300):

21100 kx

2 (given)

2 2 2 22 1

1 1W k(x x ) k[(2x) x ]

2 2 21

3 kx 3 100 300J2

23. A weight lifter lifts 300 kg from the ground to a height of 2 meter in 3 second. The average power (in Watt) generated by him is

Solution (1960):

PWorkdone

Time =

mgh

t

300 9.8 21960W

3

24. A thief is running away on a straight road in jeep moving with a speed of 91ms .

A police man

chases him on a motor cycle moving at a speed of 101ms .

If the instantaneous separation of the

jeep from the motorcycle is 100 m, how long (in s) will it take for the police to catch the thief Solution (100):

The relative velocity of policeman w.r.t. thief 10 9 1m/s .

Time taken by police to catch the thief100

1 =100 sec

25. A particle does uniform circular motion in a horizontal plane. The radius of the circle is 20 cm. The centripetal force acting on the particle is 10 N. It's kinetic energy (in J) is Solution (1.0):

2mv10

r 21 r

mv 10 1J2 2

6

CHEMISTRY (PART II)

SECTION I 26. You are provided two aqueous solutions A (500 mL of 5 M) and B (500 mL of 2M) of NaOH. If

solutions of A and B are to be used in appropriate amount to prepare maximum volume V mL of 3M solution of NaOH, what is the value of V?

(A) 500 mL (B) 1000 mL (C) 800 mL (D) 750 mL Solution (D):

To prepare maximum volume of 3 M solution from A and B, 2M solution should be completely used and some 5M solution should be mixed in order to increase molarity.

No let ‗a‘ mL of 5M be used, then 500 2 (5 a) 3 (500 a) or a 250mL

Thus, total volume 500 250 750mL

27. The mass of but-1,3-diene which contains 1 mole of double bonds: (A) 54 g (B) 27 g (C) 108 g (D) 9 g

Solution (B):

2 2 4 6CH CH CH CH or C H

(Molar mass 54 g)

54g 4 6C H has 2N double bonds or 2 mole double bonds.

28. The number of radial nodes of 3s and 2p-orbital are respectively: (A) 2, 0 (B) 0, 2 (C) 1, 2 (D) 2, 1

Solution (A):

No. of radial node n 1 l .

29. Photo-electric emission is observed from a surface for frequencies 1 2v and v of the incident

radiation 1 2(v v ) . If the maximum kinetic energies of the photoelectrons in the two cases are in

the ratio 1:k then the threshold frequency 0v is given by:

(A) 2 1v v

k 1

(B) 1 2kv v

k 1

(C) 2 1kv v

k 1

(D) 2 1v v

k

Solution (B):

21 0 1

1hv hv mu

2

22 0 2

1hv hv mu

2

2 21 2

1 1 1mu mu

2 k 2

from eq. (i) 2 21 0 2 2 1 0

1 1hv hv mu or mu khv khv

2k 2

By Eqs. (ii) and (iv) 2 0 0 1hv hv khv khv

1 20 2 1 0

kv vor v (1 k) v kv or v

(k 1)

30. The electron identified by quantum numbers n and l, (i) n 4, 1 l (ii) n 4, 0 (iii) l n 3, 2 l

(iv) n 3, 1 l can be placed in order of increasing energy from the lowest to highest:

(A) (iv)<(ii)<(iii)<(i) (B) (ii)<(iv)<(i)<(iii) (C) (i)<(iii)<(ii)<(iv) (D) (iii)<(i)<(iv)<(ii)

Solution (A):

Lower is (n ) l , lower is energy level; Also, if (n ) l is same, lower in n, lower is energy level.

7

31. A hydrocarbon gas diffuses through a porous plug at th1/6 rate of diffusion of 2H gas under

similar conditions of P and T. The gas is:

(A) 2 6C H (B) 4CH (C) 2 2C H (D) 5 12C H

Solution (D):

Hg 2

H g g2

Mr 1 2

r 6 M M gM 72 Gas is 5 12C H

32. The average speed of a helium atom at o25 C is 3 11.23 10 ms . What is the average wavelength

(in meter) of helium atom at o25 C ?

(A) 101.35 10 (B) 341.35 10 (C) 103.97 10 (D) 323.97 10

Solution (C):

34 23

av 3 3av

h 6.6 10 6.023 10

mu 4 10 1.23 10

(de Bronglie concept) 103.97 10 m

33. For one mole of a van der Waals‘ gas when b 0 and T 300K, the

PV vs. 1/V plot is shown below. The value of the van der Waals‘

constant a (atm. 2 2litre mol ) is:

(A) 1.0 (B) 4.5 (C) 1.5 (D) 3.0 Solution (C):

Van der Waal‘s equation for 1 mol of real gas is 2

aP [V b] RT

V

Given that b 0

2

aP (V) RT

V

aPV RT

V

Following y mx c for the curve 1

PV vsV

Slope a

a 1.5

34. The value of pK for 2 4 3 2NH COONH (s) 2NH (g) CO (g), if total pressure is P:

(A) 32P

27 (B) 327

P2

(C) 34P

27 (D) 327

P4

Solution (C):

' 'NH CO3 2

2X XP P; P P

3X 3X

21.6 20 1Slope 1.5

2 3

2 4 3 22X X

NH COONH (s) 2NH (g) CO (g)

22 3

1 NH CO3 2

2PK (P ) (P ) P/3 4p /27

3

0 2.0 3.0

11(mol litre )

V

(Graph not to scale)

20.1

21.6

23.1

24.6

PV

(litr

e-atm

mol

-1

0 2.0 3.0

11(mol litre )

V

(Graph not to scale)

20.1

21.6

23.1

24.6

PV

(litr

e-atm

mol

-1

8

N

N NH

35. At constant temperature, the equilibrium constant p(K ) for the decomposition reaction.

2 4 2N O 2NO is expressed by, 2

p 2

(4x P)K

(1 x )

Where, P pressure, x extent of decomposition. Which of the following statement is true?

(A) pK increases with increase of P

(B) pK increases with increase of x

(C) pK increase with decrease of x

(D) pK remains constant with change in P and x decreases with pressure

Solution (D):

pK is a characteristic constant for a given reaction and changes only with temperature.

36. Arrange the following in order of their bpK value:

(a) (b) 3 3CH NH CH (c) 2H N CH NH

(A) a b c (B) b a c (C) c b a (D) b c a

Solution (D):

Option ―a‖ represent Guanadine type compound while (c) represent the imidine class of compound and (b) represent aliphalic amines. The conjugate acid of Guanadine is most resonance stabilized followed by imidine. Hence a is more basic than c and b.

37. Stability order of following alkoxide ions is:

(A) c b a (B) a c b (C) b a c (D) c a b

Solution (A):

When negative charge is delocalized with electron withdrawing group like 2(NO ) then stability

increase.

(a) Negative charge is delocalized with 2NO group

(b) Negative charge is delocalized with carbon of alkene

(c) Negative charge is localized 38. Which one the following among each pair will release maximum energy on gaining one electron

(a F, Cl), (b S, Se), (c Li, Na)

(A) (a) Cl, (b) S, (c) Li (B) (a) S, (b), Cl, (c) Li

(C) (a) Li, (b) Cl, (c) S (D) (a) Cl, (b) Li, (c) S

Solution (A):

Theoretical

39. Correct bond energy order of following is: (A) C–Cl>C–Br>C–I>C–F (B) C–F<C–Cl<C–Br<C–I (C) C–F>C–Cl>C–Br>C–I (D) C–I<C–Br<C–F<C–Cl

Solution (C):

Bond energy 1

Bond length

2NO

O

2NO

O 2O NO

(a) (b) (c)

9

NO O

..+

O

NO O

..+

O

>

..

N

..

>

N

OH

+>

O–H+

>

NH2

OCH3

NH2

+

+

OCH3

40. Correct order of first ionization energy of the following metals Na, Mg, Al, Si in KJ mol–1 respectively are: (A) 497, 737, 577, 786 (B) 497, 577, 737, 786 (C)786, 739, 577, 497 (D) 739, 577, 786, 487 Solution (A):

Correct order of ionisation energy will be : Na < Al < Mg < Si 41. Among the following least 3rd ionization energy is for:

(A) Mn (B) Co (C) Fe (D) Ni Solution (C):

6 226Fe [Ar]3d rs

42. Arrange the order of C—OH bond length of the following compounds. Methanol Phenol p-Ethoxyphenol

(a) (b) (c) (A) a>b>c (B) a>c>b (C) c>b>a (D) b>c>a

Solution (B):

There is no resonance in 3CH –OH . Resonance is poor in p-Ethoxyphenol than phenol.

43. Which of the following correctly represent decreasing order of stability of resonating structure

(A) (B) (C) (D) Solution (B):

44. Which species is planner?

(A) (B) (C) (D)

Solution (A):

Planer

Pyramidal

Pyramidal

Tetrahedral

SECTION II

45. N2 and O2 are converted to mono cations N2+ and O2

+ respectively, which is wrong?

(A) In N2+, N–N bond become weaker

(B) In O2+, the O–O bond order increases

(C) In O2+, the paramagnetism decreases

(D) N2+ becomes diamagnetic

Solution (D):

46. A H-like atom radiates out energy in form of X-rays of wavelength 83.055 10 m during its

transition from 2nd to 1st orbit. What is the atomic number of H-like atom? Solution (2):

OH OH

2 3O CH CH

3CH OH

(a) (b) (c)

23CO 2

3SO 3ClO

4BF

23CO 2sp

23SO 3sp

3ClO 3sp

4BF 3sp

10

NH2

..NH2

..NH2

..

N..

N.. ON

H

..

N..

N.. ON

2(sp )

2

(Bridgehead carbon

Cannot besp

hybridized)

O

NH2

..

O

O

H

2NH

3PH

2SR

2OH

3NH

N

HHO

O

H

2NH

3PH

2SR

2OH

3NH

N

HH

O P O

O3sp

O

2spS

O

O O

2spN

O

O O

47. Calculate p d bonds present in 24 3SO , NO & 3SO . (Report your answer as (x+y+z))

Solution (3): p d

2

0

1 48. In how many species positive charge is delocalized.

Solution (4):

In 2nd, 6th, 7th and 8th structure +ve charge cannot be resonance delocalized because of octet violation in the resonating structure.

49. In how many compound lone pair of nitrogen is involved in resonance?

(I) (II) (III) (IV)

(V) (VI) (VII) Solution (4):

(I) No conjugation

(IV) Lone pair is not in proper orientation to overlap with the p-orbital of bond.

(VI)

So In (I), (II), (V) and (VII) lone pair is involved in resonance.

50. The mass per cent of oxygen in 2 4 3 2Al (SO ) 18H O (molar mass 666 g 1mol is) :

Solution (72.07):

666g 2 4 3 2Al (SO ) 18H O 30 mol of O atom 16 30 g O.

CH—C—NH3 2

O..

11

MATHEMATICS (PART III)

SECTION I

51. If sec2 p tan2 , then the value of 2sin in terms of p is given by:

(A)

2

2

p 1

2 p 1

(B)

21 p 1

2 p 1

(C)

2

2

p 1

2 p 1

(D)

2

2

1 p 1

2 p 1

Solution (A):

sec2 p tan2 sec2 tan2 p

1

sec2 tan2p

1p

psec2

2

2

2pcos2

p 1

2

2

2p1 2sin

p 1

22

2

p 1sin

2 p 1

52. If 1

sin cos5

& 0 , then tan is

(A) 4

3 (B)

3

4 (C)

3

4 (D)

4

3

Solution (A): Squaring both the sides:

1

1 sin225

24

sin225

Let t tan

2

2t 24

251 t

250t 24 24t 0 212t 25t 12 0 4t 3 3t 4 0

4 3

t or t3 4

(rejected) as if 3

tan4

, then ,2

then

1sin cos

5 .

53. If O be the origin and if the co-ordinates of any two points 1Q & 2Q be 1 1x ,y & 2 2x ,y

respectively, then 1 2 1 2OQ .OQ .cos Q OQ

(A) 1 2 1 2x x y y (B) 1 1 2 2x y x y (C) 1 2 1 2x x y y (D) 1 1 2 2x y x y

Solution (C):

2

2 2 2 22 2 2 21 1 1 1 1 2 1 2

1 21 2

x y x y x x y x

cos Q OQ2 OQ OQ

1 2 1 2 1 2 1 2OQ OQ cos Q OQ x x y y

54. The values of x satisfying 5log xx 5 lie in the interval

(A) 0, (B) 1

0, 5,5

(C) 1, (D) 1, 2

Solution (B):

Q (x y )2 2 2

Q (x y )1 1 1

O

12

Let y5log x y 5 x

y

y 1 25 55 5 y 1 y 1 or y 1 log x 1 or log x 1

5 5 5 5

1 1log x log or log x log 5 0 x orx 5

5 5

10 x 5,

5

55. The solution set of 2log 4 5x 2 is

(A) 8

,5

(B)

4 8,

5 5

(C) 8

, 0 ,5

(D) none of these

Solution (C):

22 2log 5x 4 log 2 & 5x 4 0

4 4

5x 4 4 & x R ~ 5x 4 4 or5x 4 4 & x R ~5 5

8 4

x 0 or x & x R5 5

8

x .0 ,5

56. The coordinates of the foot of perpendicular from a,0 on the line a

y mxm

are

(A) a

0,m

(B) a

0,m

(C)

a,0

m

(D) a

,0m

Solution (A):

a

y mxm

a

mx y 0m

Let coordinates of the foot of perpendicular be x,y .

2

ama 0

x a y 0 mm 1 m 1

2

2

x a y 0 m a a

m 1 m m 1

x a a

ym m

x a a a

& ym m m

ax 0 & y

m

Hence coordinates of foot of perpendicular are a

0,m

.

57. If the points 2,0 , 0,1 , 4,5 and 0,c are concyclic, where c 1 , then the value of c is

(A) 1 (B) 14

3 (C)

11

2

(D) none of these

Solution (B):

Let A 2,0 ,B 0,1 & C 4,5 .

Equation of the straight line AB is:

1 0

y 0 x 20 2

x 2y 2 0

Hence the equation of family of circles passing through A & B is:

x 2 x 0 y 0 y 1 x 2y 2 0

2 2x y 2x y x 2y 2 0

13

But the above circle is passing through C 4,5 .

16 25 8 5 4 10 2 0

12 28 7

3

Hence the equation of the circle passing through 2,0 , 0,1 and 4,5 is:

2 2 7x y 2x y x 2y 2 0

3

As per question the above circle passes through 0,c .

2 7c c 2c 2 0

3

14c c 1 c 1 0

3

14

c 1 c 03

14c 1,

3

14c .

3

58. In a ABC, the perimeter = 2s and the ex-radii are 1 2r ,r and 3r . Then 1 2 2 3 3 1r r r r r r is equal to

(A) 2s (B) 22s (C) 23s (D) 24s

Solution (B):

1 2 3r ,r & rs a s b s c

1 2 2 3 3 1r r r r r r . . .s a s b s b s c s c s a

2 1 1 1

s a s b s b s c s c s a

2 s c s a s b

s a s b s c

2 23s a b c 3s 2s

s a s b s c s a s b s c

2 s

s a s b s c

2 22 2 2

2

s ss

s s a s b s c

59. The difference of the values of x satisfying the equation 2x x2 8 2 12 is

(A) log3

1log2

(B) 3

log2

(C) log3

log2 (D) 1

Solution (C):

Let x2 y 2y 8y 12 2y 8y 12 0 y 2 y 6 0 y 2,6

x x2 2 or 2 6 2x 1or x log 6

2 2 2x 1 or x log 2 3 log 2 log 3 log3

1log2

log3

x 1,1log2

Hence difference of the values of x satisfying the given equation is log3

log2

60. The value of ‗a‘ for which one root of the quadratic equation 2 2a 5a 3 x 3a 1 x 2 0 is

twice as large as the other, is: (A) 2/3 (B) 2/3 (C)1/3 (D) 1/3

Solution (A):

2 2a 5a 3 x 3a 1 x 2 0

14

Let and 2 be the roots of this equation.

2

3a 12

a 5a 3

2

3a 1

3 a 5a 3

Also 2

2.2

a 5a 3

2

2

22

a 5a 3

2

2 2

3a 1 22

3 a 5a 3 a 5a 3

2

2 22

3a 1 1

a 5a 39 a 5a 3

2

2

3a 11

9 a 5a 3

, (since coefficient of 2x cannot be zero, hence 2a 5a 3 0 )

2 23a 1 9 a 5a 3 2 29a 1 6a 9a 45a 27 45a 6a 27 1 39a 26 2

a3

61. Let A, B and C are the angles of a triangle and A 1 B 2

tan ,tan2 3 2 3 . Then

Ctan

2 is equal to

(A) 7

9 (B)

2

9 (C)

1

3 (D)

2

3

Solution (A):

A B C A B C

2 2 2 2

A B Ctan tan

2 2 2 2

A Btan tan

C2 2 cotA B 21 tan .tan2 2

1 213 3

1 2 C1 . tan

3 3 2

9/9 1

C7/9 tan2

C 7

tan2 9

62. 2 2 2 2

1 1 1 11 1 1 .... 1

2 3 4 n

equals

(A) n 1

2n

(B)

n 1

2n

(C)

n 1

n

(D)

n 1

n

Solution (B):

2 2 2 2

1 1 1 11 1 1 .... 1

2 3 4 n

2 2 2

2 2 2

2 1 3 1 4 1...

2 3 4

2 2 2

2 2 2

n 2 1 n 1 1 n 1...

nn 2 n 1

2 2

2 1 2 1 3 1 3 1...

2 3

2 2

n 1 1 n 1 1 n 1 n 1...

nn 1

1 3 2 4 3 5...

2 2 3 3 4 4

n 2 n n 1 n 1...

n 1 n 1 n n

1 2 3 n 2 n 1....

2 3 4 n 1 n

3 4 5 n n 1....

2 3 4 n 1 n

1 n 1

n 2

n 1

2n

.

15

63. The value of o o o ocos1 cos2 cos3 ....cos179 is

(A) 1/ 2 (B) 0 (C) 1 (D) None of these

Solution (B):

For, one of the factors of the given expression is ocos90 which has the value 0.

64. Given that tanA and tanB are the roots of 2x px q 0, then the value of 2sin A B is

(A)

2

22

p

p 1 q

(B)

2

2 2

p

p q

(C)

2

22

q

p 1 q

(D)

2

2

p

p q

Solution (A): tanA tanB p and tanAtanB q

tanA tanB p

tan A B1 tanAtanB 1 q

sin A B

22

p

p 1 q

65. If and are the roots of the equation 2x x 11 0, then the value of

3 2 3 23 3 2 2 11 is

(A) 33 (B) -33 (C) 22 (D) -22 Solution (D):

2 3 211 0 3 3 33 0 ... (1)

2 3 211 0 2 2 22 0 ... (2)

(1) + (2)

3 2 3 23 3 33 2 2 22 0

3 2 3 23 3 2 2 11 22 22

66. If the sum of n terms of an A.P., Sn is given by n

1S np n n 1 q,

2 then the common difference

of the A.P. is: (A) 2p + 3q (B) p + q (C) 2q (D) q

Solution (D):

n n n 1T S S nq p q Obviously common difference is q.

67. The incentre of the triangle with vertices 1, 3 , 0,0 and 2,0 is:

(A) 3

1,2

(B) 2 1

,3 3

(C) 2 3

,3 2

(D) 1

1,3

Solution (D):

Let A 0,0 , B 1, 3 and C 2,0

AB 2, BC 2 and CA 2 ABC is an equilateral triangle therefore incentre is same as centroid.

16

Incentre is 3 1

1, i.e., 1,3 3

.

68. Locus of centroid of the triangle whose vertices are acost,asint , bsint, bcost and 1,0

where ‗t‘ is the parameter, is:

(A) 3 2 2 23x 1 3y a b (B)

2 2 2 23x 1 9y a b

(C) 2 2 2 23x 1 9y a b (D)

2 2 2 23x 1 9y a b

Solution (B): Let the centroid be (x, y)

3x acost bsint 1

and 3y asint bcost

2 2 2 23x 1 9y a b

69. The equation of the smallest circle passing through the intersection of 2 2x y 2x 4y 4 0

and the line x y 4 0 is:

(A) 2 2x y 3x 5y 8 8 (B) 2 2x y x 3y 0

(C) 2 2x y 3x 5y 0 (D) 2 2x y x 3y 8 0

Solution (C): The required circle must have the common chord as its diameter. The family of circles with

x y 4 0 as common chord is 2 2x y x 2 y 4 4 4 0 .

The centre of this circle, namely, 2 4

,2 2

should lie on x + y – 4 = 0.

1 2 4 02 2

= -1

70. If the straight line x – 2y + 1 = 0 intersects the circle x2 + y2 = 25 in the points P and Q, then the

coordinates of the point of intersection of tangents drawn at P and Q to the circle, are: (A) (25, 50) (B) (-25, -50) (C) (-25, 50) (D) (25, -50) Solution (C):

Equation of chord of contact is, T = 0 Hx + ky – 25 = 0 Compare with x – 2y + 1 = 0.

h k 25

1 2 1

h = -25, k = 50

SECTION II

71. For the circle 2 2x y 2x 6y 1 0 the maximum length of the chord which is drawn through

1,2 is

Solution (6):

Since 1,2 lies inside the circle, therefore the maximum length of the chord is the diameter = 6.

72. The value of the expression:

2 o 2 o 2 o 2 osin 1 sin 2 sin 3 ... sin 90 , is

Solution (45.5):

P

Q

x-2y+1M(h,k)

17

Regrouping the terms and using the fact that o o osin cos 90

2 o 2 2 o 2 osin 1 sin sin 3 ... sin 90

2 o 2 o 2 o 2 osin 1 sin 89 ... sin 44 sin 46 2 o 2 osin 45 sin 90

2 o 2 o 2 o 2 ocos 89 sin 89 ... cos 46 sin 46 2 o 2 osin 45 sin 90 1

44 1 45.52

73. In triangle ABC, if 2 2 2a c 2011b then cotA cotC

cotB

is equal to 1/. Then = ------.

Solution (1005):

cotA cotC

cotB

cos A cosC

sinA sinCcosB

sinB

2 2 2 2 2 2

2 2 2

b c a 2R a b c 2R

2bc a 2ab c

c a b 2R

2ca b

,

a b cusing sine rule 2R

sinA sinB sinC

2 2 2 2 2 2

2 2 2

b c a a b c

c a b

2 2

2 2 2 2 2

2b 2b

c a b 2011b b

, 2 2 2a c 2011b

2

2

2b 1

10052010b

74. The AM of 2 numbers is 75

4 and GM is 15. The value of larger number is

Solution (30):

Let the numbers be a & b .

As per question a b 75

& ab 152 4

75a b & ab 225

2

Therefore a & b are roots of the quadratic 2x a b x ab 0 2 75x x 225 0

2

22x 75 x 450 0 75 5625 3600 75 45

x4 4

15x 30,

2

Hence the larger number is 30.

75. If , are roots of 02x25x375 2 and n nn r

r 1

s then s

is /µ, where and µ are co-

prime numbers. Then + µ = ------. Solution (13):

02x25x375 2 750

301025x

1,1 &

375

2,

15

1

1r

22r .......S

11

1

2

12

1

348

29