iit study circle paper2

8/8/2019 IIT Study Circle Paper2

http://slidepdf.com/reader/full/iit-study-circle-paper2 1/17

x0 L

SOLUTIONSMOCK TEST PAPER II

PHYSICS

1. (D)

d A+q0

Potential at A,VA = Potential due to 0

q + Potential due to 0−ε on inner surface

+ Potential due to 0+ε on outer sur face

0 0 0 01

0

kq kq kq q 1 1V

d R / 2 R 4 d R

= + − + = − π ∈

After grounding charge on outer surface is zero.

0 0 02

0

kq kq q 1 2V

d R / 2 4 d R

= + − = − π ∈ hence.

2. (B) Motional emf e = Blv induced for motion from

for red x dsin t= ω x = 0 to x = d, VQ > VP

v d cos t= ω ω x = d to x = 0, VP > VQ

hence e Bld cos t= ω ω x = 0 to x = – d, VQ > VP

x = –d to x = 0, VP > VQ

3. (B)F. B. D at instant given

k x1 1 k x2 2

m

x

k x1 1 k x2 2

2 2k x−

4. (B) When blocks both of them rest on similar surface, T is independent of µ hence T

for x < 0 and x > L are same, in betweenT decrease

5. (C) Since there are no external horizontal forces displacement of the centre of mass

along horizontal is zero.

Equ. of state for



2c

c

s

c

r

left chamber( ) ( )

( )0P AL P 'A L x

................ 1T T

−=

right chamber ( ) ( ) ( )0P AL P' L x ........... 2T 2T

+=

from (1) and (2)

0

Lx

3=

No. displacement of centre of mass

hence 0 0x x

MX m X m X2 2

= − + −

displacement of box to right( )

mLX

3 M 2m=

+.

6. (A) Centripetal force is provided by friction for small ω, Tension appears beyondlimiting case, coin B required greates centripetal force.

If 1 2f 0 and f 0≠ = then tension on B > 0 and force on A towards centre in greater

than B which is not possible.

7. (D) Pressure decrease along the direction of acceleration.hence dip is to the left. air bubble should have a component of force to the left as

well as upward

hence ar

is as shown in (D)

8. 1. All objects above OK exit radiation with varying intensity over the entirespectrum.

2. ( )0

dT KAT T

dt ms− = − fall in temperatures depends on other factors.

3. I.e., cannot exits vibrational energy levels in metals

9. (A) Radius r = H Tanc

1sinc

= µ

2

H

1= µ −

(C)( ) 22 1 cosc 1Intensity transmitted

1Intensity incident 2

π − µ − = = − π µ

2 21 1Intensity reflected1 1

Intensity incident

µ − µ −= − − =

µ µ



10. (A), (B), (D)

(A) for the section ( )22Tsin n R. sinθ = ∆ ω θ = θ

2 2T R= µω

θ

ω

T

θθ

T

θ

(B) Speed of propagation of kink w.r.t medium

Tv R= = ω

µ

(D) kink,obs kinkmedium mediumobsv v v 0= + =r r r

(clockwise vr

and anticlockwise vr

)

11. (B), (D)

At steady state, Heat currect across every section is equal.(B) for any section

heat currentdT

Hdx

kA

=

x

R

dx

(D)( )

22

dT H H

dx k R x b ak a

l

= =π −

π +

12. (A) Applying Bernoulli’s theoremsBetween A and B

2 2

0 0 0

F 1 1P V P gH V

A 2 2

+ + ρ = + ρ + ρ

............(I)



F

V

A

B

V0

and continuity equation at A and B

0AV av= .............(2) F = 5500N

(D) Pressure at bottom =5

0

FP 6.5 10 pas cals

A+ = ×

13. (A)

To achieve the given condition

( )2Rsin l ............. 1θ ≤

and ( ) ( )R 1 cos b ........... 2− θ ≤

from (1) and (2) 45ºθ ≤

14. (C)

( )

( ) ( )

Rsin b ........... 1

x R 1 cos ............ 2

θ =

= − θ

from (1) and (2)b

x3

= .

15. (B)

16. (D) Particles with 2v 0> should not reach A hence ( )z

v 2g H Z< −

Particles with zv 0> should not rebound with z components which can take

them to A.

Hence, 2 2Zv 2gz v 2gH+ = <

( )zv 2g H z< −

hence ( ) ( )z2g H z v 2g H z− − < < −



17. (D) Neutrons which enter cavity at the highest position close to A with zv 0= take

maximum time to reach the plate A after reflection during which duration all

other neutrons with speeds above the range ( ) ( )22g H z v 2g H z− − < < −

will get absorbed hence minimum length of cavity required

( ) 2min x xmax

2HL v t v

g= ∆ = hence (D)

18. (C) ( ) ( )H

z

0

P z dz 2 M 2g H z dz nh= - =ò ò Ñ

3/ 2

2 / 3

3/2 2/3

42gM H nh

3

3nh 3hH or H n

4 2gM 4 2gM

=

= =

122 33

2

9hH n

32gM

=

19. AB → isobaric process dT > 0 hance dQ > 0, dU > 0, dW > 0

BC → isothermal process U 0, P 0, v 0, dW 0,dQ 0∆ = ∆ < ∆ > ∆ > >

CD → isobaric process. dT < 0, dU < 0 dW < 0 dQ < 0

DE → P V∝ hence isochoric process V 0; W 0,dU 0 dQ 0∆ = ∆ = < <

EA → Isothermal process dU = 0, dW < 0, dQ < 0

20. mF id B= ×r r ur

, induced emf e B v= l

induced current is zero in frame C hence Fm = 0

for A and B frames induced current is clockwise, D anticlockwise.but of varying magnitude.



MOCK TEST PAPER II

CHEMISTRY

21. (A)3 2

2 3X O X O

+ +

→

(vf = 2)2 electron –––– 1 molecule of X2O3

3 mole ←–– 1.5 mole

∴ 3 equals required of reductant

22. (D)2H

996n 0.4446 mole

22.4= =

2O

2.2n 0.0982 mole

22.4

= =

eq 2 eq 2 2 8 eq 2

eq 2 2 8

n O n H S O n H

0.0982 4 n H S O 0.4446 2

+ =

× + = ×

2 2 8

eq 2 2 8

H S O

n H S O 0.4964

0.4964m 194 48g

2

=

∴ = × =

23. (A)2 4

2 2 2

0

1 Z e mTE

8 n h= −

∈

24. (A) Using ( )2

H 2

1

1 11 R

1 x

1 = = λ

( )2

H 2

2

1 1 1and 2 R

2 32

= − λ

2

9x

5⇒ λ = by division

25. (C)3 2 2 2

C O 2O 3CO+ →

Since gn 0∆ = so volume change will be absent in this case.

26. (A)

27. (A)



1EN 12

6

2

= ×

=

28. (C) The half life period is1

2min

hence the 3rd

half life in the run will be1 1 1 1

ie., min2 2 2 8

× ×

29. (A) From unit of k we infer that it is a Oth order reaction

so[ ] [ ]

0A A

k t

−= indicates that

a decrease in molarity of 0.1 will occur every second and hence decrease of 6

every minute

30. (A) Since the % ionic character will decrease

So, the radius of the metal will increase

31. (A)133 17 S

181 19 S

−=

−

Solving we get S = 11.458

32. (A) The ( )Li

68 78 76 82R ang

4+

+ + +=

= 76

Li 0d 76 140

216 pm

−∴ = +=

33. (B) Factual question

34. (A) Since the particles are truly changed

35. (D) Factual question

36. (A)0.693 1 100

ln

13 432 10

=

0.0533 = 0.0533

hence statement – 1 is true.and statement – 2 is a correct explanation of statement – 1 .

37. (A)

38. (A)

39. (A)



40. (A) → (p), (r), (s)

(B) → (r)

(C) → (s)

(D) → (p), (r)

41. (A) → (r), (s)

(B) → (p)

(C) → (p), (p), (r)

(D) → (q), (r)

42. (A) → (p), (q), (s)

(B) → (r)

(C) → (p), (q), (s)

(D) → (p) , (q), (s)

MOCK TEST PAPER II

MATHEMATICS

43.(C) ( ) ( )g ' x f x=

g is not monotonic

If [ ]x 2nT k, k 0,2T= + ∈

{ }n N 0 and∈ ∪

( ) ( ) ( ) ( )x 2T k

0 0 0g x f t dt n f t dt f t dt 0= = + >∫ ∫ ∫

because of the condition given in C.

C is the correct answer.

D is not correct as g(x) > 0 for ( )x 0,T∈

44.(A) Centre of S1 is (2, 4)

Centre of S2 is (4, 2)

Radius of circle S1 = radius of circle S2 = 4Equation of circle S2

( ) ( )2 2

x 4 y 2 16− + − =

( )2 2x y 8x 4y 4 0 .............. 1⇒ + − − + =

Equation of circle touching y = x at (1, 1)can be taken as

( ) ( ) ( )2 2

x 1 y 1 x y 0− + − + λ − =

( ) ( ) ( )2 2x y x 2 y 2 2 0 .......... 2⇒ + + λ − + − λ− + =



As this is orthogonal to S2

( ) ( )2 2

2 . 4 2 2 4 2

2 23

λ − − λ− ⇒ − + − = +

⇒ λ =

Required equation of the circle is2 2x y x 5y 2 0+ + − + =

45.(D) Equation of the line PR is

x 3 y 7 z 1r

1 2 641 41 41

− − −= = =−

Any point on the line say Q is

r 2r 6r3 , 7,141 41 41

+ + −

If it lies on the plane,25 41

r59

=

46.(B) Coefficient of x4

in

( ) ( ) ( ) ( )404 303 202 1014 4 4 4 4

0 1 2 3 4C 1 x C 1 x C 1 x C 1 x C+ − + + + − + +

i.e., in ( )4

1011 x 1 + −

i.e., in ( )4

101 101 2 101 101

1 2 101C x C x ...... C x+ + +

i.e., ( ) ( )4 4101

1C 101=

47.(A),(B),(C),(D)T TA A AA I= = , also

T 2A A, so, A I= =

⇒ A is involutary matrix22A A 1 or A 1= = = ±

But ( )( )2 2 2

a b c

A b c a a b c a b c ab bc cac a b

= = + + + + − − −

2 2 2A a b c ab bc ca 0= + + − − − ≥

A 1,∴ = − Hence 3 3 3a b c 3abc 1+ + − =

Again,2 2 2a b c ab bc ca 1+ + − − − =

( )1 b ab bc ca 1

ab bc ca 0

⇒ − + + =

⇒ + + =

⇒ at least one of a, b, c is negative



48.(A),(C),(D)

Let a, b, ∈ R such that

z a ib, b 0= + ≠ 5 4 2 3 5Imz 5a b 10a b b= − +

Solving ( )2

x a / b yields=

( )5

22

5

Im z5x 10x 1 5 x 1 4

Im z= − + = − −

The minimum value = – 4 for x = 1

( )

2a

xb

a b

z a 1 i

=

⇒ = ±∴ = ±

49.(C),(D)

( )dy dx

y x, y 2 1dx dy

+ = − =

Solution, It is a clairaut’s equation, 2y xp p= −

general solution: 2y cx c= −

x y 1 0 for c 1∴ + + = =

Singular solution is2x

y4

=

50.(A),(B),(C),(D)

( ) ( )2f ' x 6 x 2 x3

λ = − λ + +

It has two distinct roots if ( )

2 42 0

3

λ⇒ λ + − >

2 84 0

3

λ⇒ λ + + > which is true for ∀ real λ

51.(A),(B),(C),(D)

( )1 f " x 1− ≤ ≤

On integrating it twice in the limits 0 to x we get



( )2

xf x

2

1 1f 2 8

≤

± ≤

and ( )f 2 2± ≤

MATCH THE FOLLOWING

52. (A) → (q), (B) → (p), (C) → (s), (D) → (s)

[ )

[ )

10 i n 0,tan1

tan x

1in tan1,

− =

∞

[ )

[ )1

1 in 0,cot1cot x

0 in cot1,

− = ∞

( )

[ )

[ )

[ )

1 1

1 in 0,cot1

f x tan x cot x 0 in cot1,tan1

1 in tan1,

− −

−

= − = ∞

( )

[ ]

( )

[ ]

1 in 0,cos1

g x 0 i n cos1,sin1

1 in sin1,1

=

Total no of points at which f(x) and g(x) are not differentiable is one

The least value of f(x) is –1

The greatest value of g(x) 1The greatest value of f(x) – the least value of g(x) is (1 – 0) = 1

53.(A) → (r), (B) → (s), (C) → (q), (D) → (p)

(i)124 r r

124 2 4r 1 rt C 5 11

−

+ =

r 0,4,8,12,........,124.=

No of terms which are integers = 32

(ii) 4 lines intersect each other in 42

C 6= points

4 circles intersect in 42

P 12= points.

Each line cuts 4 circles into 8 points

4 lines cut 4 circles into 32 points

required no of points = 6 + 12 + 32 = 50



(iii) 2

0

1 sin3xI dx

1 2sinx

π +=

+∫

( )

2

0

22 2

0 0

1 sin3 x2

dx

1 2sin x2

1 cos3xdx 2cos x cosx 1 dx

1 2cosx

1

π

π π

π + − =π + −

−

= = − + ++

=

∫

∫ ∫

(iv) ( ) ( )sinx cosx 2cosx 1 0+ + =

1tanx 1, cosx

2

= − = −

There are 4 solutions.

SUBJECTIVE:-

54. Let ( )2

1S f x dx= ∫

( )( )1 1

1 1

2 2

f 2xf x dx dx

3=∫ ∫

Let u 2x=

( ) ( )1 2

11

2

f 2x f u 5dx du

3 6 6= =∫ ∫

Similarly, ( )2

1

21 2

2

5f x dx 6=∫

( )n 1

n 1

1

21 n

2

5f x dx

6

−

−

=∫

( )n

1

1 2 3 n

2

5 5 5 5f x dx .........

6 6 6 6= + + + +∫

n

11

65

5

− =



An ( )1

0

5n , f x dx 1

5→ ∞ = =∫

S 5⇒ =

55. Let i1

z re 2x= i

2z re2 y=

( ) ( )

( )

( ) ( )

( )

1 2

21 2

1 2

2

1 2

r z z cos x y

r z z cos x y

r z z sin y x

r z z sin y x

+ −=

+ +

− −=

− +

( ) ( ) ( )

( )

( )

( ) ( ) ( )

2 2 2 2

1 2 1 2 2 2

2 2 2 21 2 1 2

r z z r z z cos x y sin x y

cos x y sin x y 1r z z r z z cos x y sin x y

+ − − −

+ = + ≥ − + − = + − + +

56. Let a, b, c denote the numbers of –1’s, 1’s, and 2’s is the sequence respectively. We neednot consider the zeroes. Then, a, b, c are non-negative integers satisfying.

a b 2c 19, a b 4c 99− + + = + + =

a 40 c, b 59 3c∴ = − = −

0 c 19, Since b 0≤ ≤ ≥

So, 3 3 31 2 n

x x ............. x a b 8c+ + + = − + +

19 6c= +

The lower bonds is achieved when

c = 0, (a = 40, b = 59)The upper bonds is achieved when

a = 21, b = 2. Thus m = 19

and M = 133

So,M 133

7m 19

= =

57. xyz xy yz zx+ + +

( )( )( ) ( )x 1 y 1 z 1 x y z 1

abc 13

= + + + − + + −

= −

where a=x+1, b=y+1, c=z+1 are positive integerswhose sum is 15.

abc is largest when a=b=c=5

∴ Thus the answer is

5 × 5 × 5 – 13 = 112

xyz xy yz zx 1127

16 16

+ + +∴ = =

58. Multiplying by 7! on both the sideswe get



2 3 4 5 6 73600 2520a 840a 210a 42a 7a a= + + + + +

73600 a− is a multiple of 7, which implies that 7

a 2=

Thus2 3 4 5 6

3598 514 360a 120a 30a 6a a7

= = + + + +

6514 a⇒ − is a multiple of 6,

6a 4=

Thus2 3 4 5

51085 60a 20a 5a a

6= = + + +

Then it follows that 585 a− is a multiple of 5, where 5

a 0= , Continuing in this process,

we obtain 4 3 2a 1, a 1, a 1.= = = Thus the desired sum is

= 1 + 1 + 1 + 0 + 4 + 2 = 9

59.

A

B

C

M

O

P(h, k)

Q

θ

N

y

x

Let OC = h, BC = h tanθ

h tan θ = 2

In OAQ OA 2sin and∆ ⇒ = θ in 2AON, AN 2sin Q∆ =

( )2 2

2

2 2

k 2sin , P is 2cot ,2sin

2 8k 2sin

1 cot 4 h

= θ θ θ

= θ = =+ θ +

locus of P is2

8y

4 x=

+

(0, 2) y = 84 + x

2

–2 20

(–2,1) (2,1)

Area =4 × 1 = 4 sq unit

60. The direction ratios of the line OP are 1, 2, 2.



Also, the equation of any plane through P(1, 2, 2) is

( ) ( ) ( ) ( )A x 1 B y 2 C z 2 0 ............ 1− + − + − =

If it is perpendicular to OP, then the normal to this plane is parallel to OP and soA B C

1 2 2= =

From (1) the equation of the plane through P(1, 2, 2) and perpendicular to OP is given by

( ) ( ) ( )1 x 1 2 y 2 2 z 2 0− + − + − =

⇒ x 2y 2z 9+ + =

The intercepts made by this plane on coordinate axes9 9

9, ,2 2

The coordinates of A (9, 0, 0),9 9

B 0, ,0 , C 0,0,2 2

respectively.

The coordinates of the projections of ∆ABC on the yz-planeare (0, 0, 0),

9 90, ,0 , 0,0,

2 2

respectively.

x∆ = area of projection of ABC∆ on yz-plane

0 0 1

1 9 810 1

2 2 8

90 1

2

= =

y∆ = area of projection of ABC∆ on zx- plane

=81

4

z∆ = area of projection if ABC∆ on xy-plane

=81

4

The required area = 2 2 2

x y z∆ + ∆ + ∆

2 2 281 81 81

8 4 4

81 11 1

4 4

81 3 243

4 2 8

= + +

= + +

= × =

According to the questions

Area243 243

72 8

λ= =



9λ =

61.A

B CD

E

F k

1–k

k

k

1–k

1–k

(a)→

(b)→ (c)

→

→

The P. V of ( )D 1 k b kc= − +ur r r

The P. V of ( )E 1 k c= −ur r

The P. V of F kb=r r

( )

( )

AD 1 k b kc

BE 1 k c b

CF kb c

= − +

= − −

= −

uuur r r

uuur r r

uuur r r

AD BE CF 0+ + =uuur uuur uuur

AD,BE,CFuuur uuur uuur

for a triangle

Area of the triangle =

1

AD BE2 ×

uuur uuur

( ) ( )1

1 k b kc 1 k c b2

= − + × − − r r r r

( ) ( )

( ) ( )

21A k k k 1 b c

2

1A ' k 2k 1 b c 0

2

1k

2

= − + ×

= − × =

=

r r

r r

( )1

A " k b c 02

= × >r r

A(k) is minimum for1

k 2

=

When1

k 2

= , D, E, F because the midpoint of BC, CA and AB respectively.

D( l , 0, 0), E(0, m, 0), F(0, 0, n)

( )2

2 2 2 2BCEF BC 4 m n

4= ⇒ = +



( ) ( )

( )

( )

( )

2 2 22 2 2 2

2 2 2 2 2 2

1 18 m n 1AB BC CA k k 1 4 2

2 m n 2 m n

+ + + + + + + + =

+ + + +

l

l l

∴ 8 7

72 4

= × =

iit study circle paper2

Documents