Elite2011 Pet1 Solns p1

7/28/2019 Elite2011 Pet1 Solns p1

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1

BRILLIANTS

MOCK ALL-INDIA TESTS (B.MAT)

NATIONAL SIT-DOWN

PROGRESSIVE EVALUATION TEST 1FOR STUDENTS OF

OUR TWO-YEAR ELITE POSTAL COURSE

TOWARDS

IIT-JOINT ENTRANCE EXAMINATION, 2011

PART A : CHEMISTRY

SECTION I

1. (D) Radius of nucleus = 1.25 1013 A1/3 cm

= 1.25 1013 641/3 cm

= 5.0 1013 cm

Radius of atom = 1 = 1 108 cm

Volume of nucleusVolume of atoms

=

4

3 5

10

13 3

4

3 1 10

8 3

= 125 10 39

1 10 24

= 125 1015

= 1.25 1013

Brilliant Tutorials Pvt. Ltd. EELP/I PET/CMP/P(I)/Solns

ELITE 2011

I PET/CMP/P(I)/SOLNS

PAPER I SOLUTIONS

CHEMISTRY MATHEMATICS PHYSICS


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2

2. (A) First line in Lyman series of Hydrogen atom is for n2

= 2 to n1

= 1

1

= R

HZ

2 1

n1

2

1

n2

2 Z = 1 (for H)

1

= 109678 3

1 4= 82259 cm

1

3. (C) In OF2 molecule, two bond pair of electrons and two lone pair of electrons

are present in oxygen. Hence hybridisation of oxygen in this molecule is sp3.

4. (C) The d-orbital involved in sp3d hybridisation is dz

2.

5. (D) 100 P =V240

100P

V2

= 100 100

40= 250 cm

3

V2 = volume of bulb A + volume of bulb B

Hence volume of bulb B = 250 100 = 150 cm3

6. (D)r.m.s. velocity =

3RT

M

r.m.s. velocity for CH4

r.m.s. velocity for SO2

=M of SO

2

M of CH4

at the same temperature

100

V= 64

16= 4 = 2

V = 100

2= 50 msec

7. (D) Z2O3 + 3H2 2 Z + 3H2O1 mole 3 mole

If x is the atomic weight of metal, molecular weight of Z2O3 will be (2x + 48)

No. of moles of Z2O3 =0.1596

2x 48

No. of moles of Hydrogen =0.006

2= 0.003



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3

According to the balanced equation,

0.1596

2x 48: 0.003 = 1 : 3

or3 0.1596

2x 48= 0.003

0.1596

2x 48= 0.001 or 2x 48 = 0.1596

0.001

2x = 159.6 48 = 111.6

x =111.6

2 = 55.8

Atomic weight of the metal = 55.8

8. (B) In the presence of phenolphthalein indicator, whole of NaOH and half ofNa2CO3 are neutralised by HCl. The remaining half of Na2CO3 is neutralised

by further addition of HCl using methyl orange indicator.

Half of Na2CO3 present in solution = 25 0.2 milliequivalents

= 5.0 milliequivalents

NaOH + half of Na2CO3 present in solution

= 300 0.1 milliequivalents= 30 milliequivalents

Amount of NaOH present in solution

= 30 5 = 25 milliequivalents

= 25 103 40

= 1.0 g

SECTION II

9. (B), (C), (D)

P vs V is a curve. But others form linear plots.10. (A), (B), (C)

PH3 has one lone pair and three bond pairs of electrons and P is in sp3 hybrid

state. SO3

2 ion as well as SO4

2 ion have tetrahedral shape. XeF4 has four

bond pair and two lone pair of electrons. It is square planar but the Xe atomis in sp3d2 hybrid state.



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11. (B), (D)Isotones have same no. of neutrons.

32

76Ge,

33

77As and

34

78Se all have 44 neutrons in their nuclei.

12. (B), (C), (D)

0.1 mole of O2 gas = 0.1 32 = 3.2 g

0.1 mole of SO2 gas = 0.1 64 = 6.4 g

6.02 1022 molecules of SO2 gas =6.02 10

22

6.02 1023

64 = 6.4 g

1.204 1023 molecules of O2 gas =1.204 10

23

6.02 1023

32

=32

5= 6.4 g

SECTION III

13. (B) Molecular weight of the gas = Mx

Molecular weight of helium = MHe = 4

According to Grahams law of diffusion

Rate of diffusion of gas

Rate of diffusion of He=

MHe

Mx

1

4= 4

Mx

1

16= 4

Mx

or Mx = 16 4 = 64

14. (D) Let the volume of gas diffused = V mL

Rate of diffusion of gas x =V

2m

2sec

Rate of diffusion of O2 =V

5.65m

2sec



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Rate of diffusion of gas

Rate of diffusion of O2

=MO2M

x

V

2

V

5.65

= 32

x

5.65

2= 2.825 = 32

x

2.825 2.825 =32

x

x = 32

2.825 2.825= 4

Molecular weight of the gas = 4

15. (B) Rate of diffusion of H2 gas =50

20= 2.5 m

2minute

Rate of diffusion of O2 gas =40

t

mL minute

Rate of diffusion of H2

Rate of diffusion of O2

=M

O2

MH

2

2.5t

40= 32

2= 4

t = 40 4

2.5= 64 minutes

16. (A) The correct order of strengths of H bonds is H ... F > H ... O > H ... N, sincethe electronegativity ofF is greater than O which is greater than nitrogenatom.

17. (C) Boiling point is proportional to the size of the molecules. Hence the correctorder isGeH4 > SiH4 > CH4.

18. (B) BeF2 molecule being symmetrical and linear has zero dipole moment. H2O isbent and has net dipole moment.



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SECTION IV

19. (A) (p), (r); (B) (t); (C) (p), (q); (D) (p), (s)

(A) O2 is paramagnetic; Bond order is 2.0

(B) N2 is diamagnetic; Bond order is 3.0

(C) O2

is paramagnetic; Bond order is 1.5

(D) N2

is paramagnetic; Bond order is 2.5

20. (A) (s); (B) (q), (s); (C) (r), (t); (D) (p)

(A) SF6 d2 sp3 hybridisation for sulphur atom.

(B) XeF4 sp3d2 hybridisation for xenon atom and has square planar shape.

(C) CCl4 sp3 hybridisation for carbon atom and has tetrahedral shape.

(D) NO2

linear and nitrogen atom has sp hybridisation.



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7

PART B : MATHEMATICS

SECTION I

21. (D) Using the numbers in AP, we get 3 log 3b = 3 log 5c

3b = 5c

Also a, b, c in GP b2 = ac

5

3

2

c2

= ac

Also b = 35 a, c = 925 a

a is the largest side

A is the largest angle

cos A = b2

c2

a2

2bc=

9

25a

2

81

625a

2 a

2

2 3

5a 9

25a

= ve

triangle is obtuse angled.

22. (D) E =x 4y

2x y

z 4y

2z y=

x 4 2xzx z

2x 2xz

x z

z 4 2xzx z

2z 2xz

x z

[Q x, y, z are in HP]

= x2

9xz

2x2

z

2 9xz

2z2

= 1

2

9

2

z

x

x

z

1

2

= 1 9

2

z

x

x

z

which is > 1 +9

2 2 since x, y, z are positive

= 10



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23. (B) rth term = r1 3 5 ... 2r 1

= 1

2

2r 1 1

1 3 5 2r 1

= 1

2

1

1 3 5 2r 1

1

1 3 5 ... 2r 1

T1

= 1

21 1

1 3

T2 = 12 11 3 11 3 5

...........................

Tn

= 1

2

1

1 3 5 2n 1

1

1 3 5 2n 1

Sn = T1 + T2 + ... + Tn

=1

21 1

1 3 5 2n 1

Ltn

Sn = 12 1 0 = 12

24. (B) 1 < 2 < 3 < 4 ... (1)

sin 1 = sin 2 = sin 3 = sin 4 = (positive) ... (2)

we can choose 2, 3, 4 as follows

2 = 1 3 = 2 + 1 4 = 31

E = 4 sin

1

2 3 sin

2

1

2 2 sin

1

2 sin

3

2

1

2

= 4 sin

1

2 3 cos

1

2 2 sin

1

2 cos

1

2

= 2 sin

1

2 cos

1

2

= 2 1 sin 1

= 2 1



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25. (D) 2 + kx2

+ sin

1

(x2

2x + 2) + cos

1

(x2

2x + 2) = 0

2 + kx2 + sin1 ((x 1)2 + 1) + cos1 ((x 1)2 + 1) = 0

2 + k +

2= 0

( Q sin1 ((x 1)2 + 1) etc. is valid only when x = 1)

k =

2 2

26. (C) rth term tr = sin1 r r 1

r r 1

= sin1 r

r r 1

r 1

r r 1

= sin1 1

r1 1

r 1

1

r 11 1

r

= sin

1

x 1 y

2

y 1 x

2

where x =1

ry = 1

r 1

= sin1 x sin1 y

tr

= sin1 1

r sin

1 1

r 1

t1

= sin1

1 sin1 1

2

t2

= sin1 1

2 sin

1 1

3

tn

= sin1 1

n sin

1 1

n 1



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Adding Sn = sin1 1 sin1 1n 1

As n the above = sin1 1 sin1 0

=

2 0 =

2

27. (A)

1

2= BD

DC=

area of ABD

area of ADC

=

1

2AB AD sin

1

2AD AC sin

Hencesin

sin = 2 AB

AC... (1)

Similarlysin

sin = 2 AC

AB... (2)

The product gives the result = 4

28. (A) The equation is

cosx

4

sin x sin x

4

cos x 2 cos2

x sin2

x cos x = 0

sin5x

4 cos x = 2

Since both the terms of LHS are 1 the above is valid if

sin 5x

4= 1 and cos x = 1, simultaneously



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5x4 = 2n 2 and x = 2m

x =8n 2

5 and x = 2m

2 4n 1

5= 2m 4n 1 = 5m

This will be true if n = 1, 6, 11, .... 14, 9, 4, ....

All these will be exhausted if we take n = 5k + 1.

x =2 4 5k 1 1

5

= 2

520k 5

= 2 (4k + 1)

= 8k + 2

Finally k can be replaced by n

x = 8n + 2

SECTION II

29. (B), (D)

Given equation is sin x + 2 cos a cos x = 2

LHS lies between 4cos2

a 1 and 4cos2

a 1

4cos2

a 1 2 4cos2

a 1

4 cos2 a + 1 4

4 cos2 a 3

cos2 a 3

4... (D) is correct

1 sin2 a 3

4

sin2 a 1

4... (B) is correct

sin2 a 1

4



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12

30. (B), (C)

Case I: Let ADC = 45

then by (m n) formula

2 cot 45 = 1 cot B 1 cot 30

2 = cot B 3

cot B = 2 + 3

= cot 15

B = 15

Case II: Let ADB = 45 then ADC = 135

2 cot 135 = cot B 3

2 = cot B 3

cot B = 2 + 3

B = 105

31. (A), (B), (C)

sin2x sin2y =3

2

sin (x + y) sin (x y) = 3

2

sin (x + y) sin 3

= 32

( Q x y = 13

)

sin (x + y) = 1 (x + y) = (4n + 1)

2

x + y = 2n +1

2, n I



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13

Put n = 0 x + y = 12

x y =1

3solving (x, y) =

5

12, 1

12.... (A)

Put n = 1 x + y =5

2

x y =1

3solving (x, y) =

17

12, 13

12... (B)

n = 1 gives (C)

n = 12

gives (D)

but n is an integer. Hence D is not valid.

32. (A), (B)

We can take the 4 terms as a

r3

, a

r, ar, ar

3

Product a4 = 9

a2 = 3; a = 3

Also

a

r =

1

ar3

a2r2 = 1 ar = 1

r = 1

3

There are 4 cases

Sum to infinity =

a

r3

1 r2

=

a

r

1

r2

1 r2

ar

= 3 taking a and r as 3, 13

or 3, 13

= 3 3

1 1

3

=9 3

2= 27

2

Similarly S

= 27

2in the other cases.



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14

SECTION III

33. (C) tn = Sn Sn 1

= a 3n a (a 3n 1 a)

= a 3n 1 2 = 2a 3n 1

The sequence is 2a, 2a 3 2a 32 ....

It is a GP with common ratio 3

34. (A) tk = Sk Sk 1 = a 4k a (a 4k 1 a)

= a 4

k 1

3= 3a 4k 1

1

tk

= 1

3a

1

4

k 1

Hence 1

1

tk

= 1

3a1 1

4

1

4

2

...

=1

3a 1

1 1

4

=4

9a

35. (D) tk =k k 1 k 2

12

k 1 k k 1

12

=1

12k k 1 3

=k k 1

4

1t

k

= 4k k 1

Let it be uk = 41

k

1

k 1

u1 = 4 1 1

2



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15

u2 = 4 12 13

un = 41

n

1

n 1

Adding Sn = 4 1 1

n 1

Sum to = 4 [1 0] = 4

36. (A) B1 + B2 =

given expression issin C

1

sin B1

sin C

2

sin B1

=sin C

1 sin C

2

sin B1

=

c1

c2

b

Now a2 = b2 + c2 2 bc cos A

c2 2 bc cos A + (b2 a2) = 0

The two roots are c1 and c2

c1 + c2 = 2b cos A

expression =2b cos A

b= 2 cos A

37. (B) Required sum of the areas

= 1

2ab sin C

1 sin C

2

As per Qn. 36 c1 + c2 = 2b cos A

sin C1 + sin C2 = 2 sin B cos A



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16

sum of the areas = 12 ab 2 sin B cos A

=a sin B

bb

2cos A

= sin A b2 cos A

=1

2b

2sin 2A

38. (C) a2 = b2 + c2 2bc cos A

72

2

= 16 c2

2 4 c 12

c2 4c + 16 49

4= 0

c2 4c + 15

4= 0

Roots are c1 and c2

c1 + c2 = 4

c1

c2

= 15

4

(c1 c2)2 = 16 4

15

4

= 16 15

c1 c2 = 1

SECTION IV

39. (A) (q); (B) (r); (C) (s); (D) (p)

(A) tk = tan1

k2

k 1 k2

k 1

1 k

2

k 1 k

2

k 1= tan1 (k2 + k + 1) tan1 (k2 k + 1)

t1 = tan1 3 tan1 1

t2 = tan1 5 tan1 3

tn = tan1 (n2 + n + 1) tan1 (n2 n + 1)



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17

Sn = tan

1

(n2

+ n + 1) tan

1

1

Ltn

Sn

=

2

4=

4

(B) 81cos

2x

81

81cos

2x

= 30

y + 81

y= 30 where y = 81

cos2

x

y

2

30 y + 81 = 0y = 27 or y = 3

34cos

2x

= 33

34cos

2x

= 31

4 cos2 x = 3

cos x = 3

2cos2 x =

1

4

cos x = 1

2

A pair of solutions is 6

and 3

Sum =

2

(C) Given equation is

a b c

c

b c a

b= 3

(b + c)2 a2 = 3 bc

b2

+ c2

a2

= bc

b2 + c2 a2 = 2 bc 1

2

cos A =1

2

A =

3



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18

(D) E = sin x 6

cos x 6

= 2 sin x

6

4

Maximum E is attained

when x

6

4=

2

x =

1240. (A) (q); (B) (p); (C) (s); (D) (r)

(A) LHS k = 1

n

j = 1

k2j

= k = 1

n2 k k 1

2

= k = 1

nk

2 k

= n n 1 2n 1

6

n n 1

2

= 1

6n n 1 2n 1 3

= 1

3n n 1 n 2

= 1

3n

3 3n

2 2n

= 1

3n

3 n

2

2

3n

A =1

3B = 1 C =

2

3

3A + 3C = 1 + 2 = 3

3A + 3C + B = 4



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(B) Let A and R be the first term and common ratio of the GP

A Rp 1 = 64 ... (1)

A Rq 1 = 27 ... (2)

A Rr 1 = 36 ... (3)

1

2R

p q=

64

27= 4

3

3

3

2

Rr q

= 36

27

= 4

3

R3r 3q = Rp q

3r 3q = p q

3r p = 2q

3r p

q= 2

(C) cot B = cos B

sin B= c

2 a

2 b

2

2casin B= c

2 a

2 b

2

4

cot C = a2

b2

c2

4and cot A = b

2 c

2 a

2

4

cot B cot C

cot A = 2a

2

b2

c2

a2

= 2a2

3a2

a2

= 1

(D) The equation is

1 + sin x sin2x

2= 0

(sin x) sin2 x

2= 1

sin x = 1 and sin2x

2= 1

which is not possible. Hence no solution.



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PART C : PHYSICS

SECTION I

41. (A) r = ct bt2

2

Velocityv =dr

dt= c bt

v v = c bt c bt

v2 = c c bbt2 2bct

v2 = c2 + b2 t2 + 2 bct, since b and c are collinear

= (c + bt)2

v = c + bt

Aliter: Since c and b are constant and collinear we can write the givenequation as

r = ct + b t2

2

Differentiating with respect to t

v = dr

dt= c bt

v = c + bt

42. (B) v = 2s

Squaring we get v2 = 2s2

Differentiating with respect to t we get

2v

dv

dt = 2 2s

ds

dt

dv

dt= 2s

i.e., a = 2s

a

s= 2



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43. (C) U = A xx

2 B

2

[B] = [x2] = L2 (since x is distance)

A =U x

2 B

2

x

= M1L

2T

2L

2

L12

= M1

L7/2

T2

[A] [B] = M1 L7/2 T2 L2

= M1 L11/2 T2

44. (B) Circumference = 2.40 cm

But being a number has unlimited significant figures.

Therefore = 22

7= 3.14159 given

circumference = 3.141 2.40 cm

= 7.5384

= 7.54 cm

45. (D) Because the particle has zero displacement at t = 0, the slope at t = 0, is zero.

Between 4 and 6s, the displacement versus time curve is a parabola pointingup.

Between 2 and 4s the displacement versus time curve is a straight line.

At t = 8 sec, v = 0. Hence the slope at t = 8 sec is zero.

46. (A) Distance travelled by the particle projected vertically up, in one second is

y = ut 12

gt2

= 10 1 1

2 10 1 1

y = 5 m

Let the particle projected at an angle have a velocity of projection u. Thehorizontal component of velocity is evidently u cos . Horizontal distance



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22

covered in one second is u cos 1 = 1 m (the distance of separation betweenthe two particles). The equation of trajectory for the projectile is given by

y = x tan gx

2

2u2

cos2

In our case y = 5 m

u cos = 1 m/s

x = 1 m

5 = 1 tan 10 1 1

2 1 1

5 = tan 5

tan = 10

= tan1 (10)

47. (C) Since speed is constant for both the particles let us assume that A collideswith B after travelling a distance x along the circumference of the circle.Then

x

vA

=2R x

vB

x (vA+ vB) = 2 RvA

x =2R v

A

vA

vB

=2 5 0.7

2.2

=21.99

2.2= 9.996

t = xvA

= 9.9960.7

= 14.3 s

48. (C) Since the blocks have the same acceleration they can be considered togetheras a single system. Then the tension in the spring becomes an internal force.Therefore, net external force on the system is

F = F1 F2 = (72 32)N

= 40 N



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23

Total mass = (7 + 3) kg = 10 kg

acceleration of the system =40 N

10 kg= 4 ms

2

Let us now consider the 3 kg block.

Force on the 3 kg block is

72 T = 3 4

T = 60 N

Stiffness of the spring K = 1000 N/m = 10 N/cm

Extension =T

K= 60 N

10 Ncm= 6 cm

SECTION II

49. (A), (D)

When the speed of the man is x, we have

vm

= x i

Letvr

= a i b j for the rain

Then vrm

= vr

vm

(by definition)

vrm

= a i b j x i ... (1)

= a x i b j

Butvrm = b j (given) ... (2)

a x = 0

a = x

In the second case

vrm

= 3 2 cos 45 i 3 2 sin 45 j

vr

vm

= a i b j 2x i

= a 2x i b j



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24

3 2 cos 45 i 3 2 sin 45 j = a 2x i b j a 2x = 3

b = 3

Since a = x

x 2x = 3

x = 3

vm

= 3 i

Speed is vm

= 3 i = 3 kmhr

Speed of rain is b = 3 km/hr

Vector diagram

(Before doubling the speed) (After doubling the speed)

50. (C), (D)

Force = Fg newton ( Q 1 kg f = g newton)

But F = ma (by Newtons II Law)

ma = Fg ... (1)

Also v = at (Q u = 0)

=

Fgt

m from 1

Now v2 = 2ax ( Q u = 0)

x = v2

2a= v

2m

2Fgusing (1)

Also x =1

2at

2= 1

2

Fg

m t

2



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25

51. (A), (D)

F.B.D for the mass m1 is

T1 = m1 g where T1 is the tension.

F.B.D for the movable pulley

T1 = 2T2 (since the pulley is at rest)

If we consider the masses m2 and m3 as the system, we have,

net force is (m2 m3)g assuming m2 > m3Total mass is (m2 + m3)

acceleration of the system ism

2 m

3g

m2

m3

For the mass m2 we have

m2g T2 = m2a

T2 = m2g m2m

2 m

3

m2 m3

g

= gm

2

2 m

2m

3 m

2

2 m

2m

3

m2

m3

T2

=2m

2m

3

m2

m3

g



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26

52. (A), (B), (D)

For the particle at O, ao = 8 cm s2

For the particle at O, ao = 4 cm s2

OO = 36 cm

With respect to the particle starting from O, the particle starting from O has

u = (70 40) cm s1

a = ( 4 8) cm s2

Now s = ut +1

2at

2

36 = 30t 12

2t

2

t2 5t + 6 = 0

t2 3t 2t + 6 = 0

(t 2) (t 3) = 0 t = 2 sec and t = 3 sec

Hence the particles meet twice, at time 2 sec and 3 sec

x(2) = 40 2 + 4 22 = 96 cm

x(3) = 40 3 + 4 32 = 156 cm

Aliter

Using equation

x(t) = x(0) + vx(0)t +1

2

ax t

For the particle starting from O

x (t) = 0 + 40 t +1

2 8 t2

For the particle starting from O

x(t) = 36 + 70 t 1

2 4 t2



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27

If they meet x(t) = x(t)

0 + 40t + 4t2 = 36 + 70 t 2t2

i.e., 6t2 30t + 36 = 0

t2 5t + 6 = 0

t = 2s or 3s

x(2) = 40 2 + 4 22 = 96 cm

x(3) = 40 3 + 4 32 = 156 cm

SECTION III

53. (A) m = 102 kg, motion is along positive x-axis

F x = K

2x2

, K = 102 Nm2 (given)

At t = 0, x = 1.0 m and v = 0 (given)

F x = K

2x2

i.e., m dvdt

= K2x

2

m dv

dx

dx

dt= K

2x2

Here dx

dtis ve

i.e., m dv

dxv = K

2x2

or m

0

v

vdv =

1

x

K

2x2dx

i.e., mv2

2= K

2x 1

x

i.e., mv2

2= K

2

1

x 1



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28

or v2 = Km 1x 1

v = K

m

1

x 1

When x = 0.5

v = K

m

1

0.5 1 = 1 ms

Since retardation is along + ve x-axis and velocity is decreasing along positive

x-axiswe have v = 1 m/s

54. (C) We have v =dx

dt= 1 x

xQ

K

m= 1 ms

i.e., x

1 xdx = dt

1

0.25

x

1 xdx =

0

t

dt

Put x = sin2

dx = 2 sin cos d

2

6

2 sin2

d = t

cos 2 = 1 2 sin2

2 sin2 = 1 cos 2

2

6

1 cos 2 d = t

sin 2

2 2

6

= t



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29

i.e., 6 12 sin 3

2 12 sin

= t

6

2

1

2

3

2= t

3

3

4= t

t =

3

3

4

55. (C) Time taken to go from x = 1 m to x = 0. This is

1

0.5x

1 xdx =

0

t

dt

i.e., 2

4

1 cos 2 d = 0

t

dt (by virtue of solution above)

sin 2

2 2

4

= t

4

1

2sin

2

2

1

2sin = t

4

1

2

2= t

4

1

2= t

t =

4

1

2

average acceleration =v

2 v

1

t2

t1

= 1 0

4

1

2 0

= 4

2ms

2



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30

56. (B) Suppose the string moves from the position A to A and the block from C to Cin a small time interval t. In the Figure, AP represents the perpendiculardrawn from A on line AB. When the displacement is small, AB = PB. Now

AB + BC = AB + PB.

When the displacement is small, AB = PB. Therefore AB + BC = A B + BC(as length of string remains constant)

AP + PB + BC = AB + BC

i.e., AP = BC BC = CC

or AA cos = CC

or

AA cos

t =

CC

t

CC

t speed of the block

AA

t speed of the ring

speed of the block = v cos

57. (A) If the initial acceleration of the ring be a, then the acceleration of the blockwill be acos .

Suppose T is the tension in the string at this instant. Force acting on theblock are

Mg T = Ma cos ... (1)



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Forces acting on the ring are

Taking component along the rod, we get

T cos = ma or T =ma

cos ... (2)

Substituting for T in (1), we get

Mg ma

cos = Ma cos

a =Mg cos

m M cos2

=2m 9.8 cos 30

m 2m cos2

30

=

2 9.8 3

2

1 2 3

4

= 6.8 ms2

58. (D) From equation (1), we get

T = 1 a

cos = 6.8

cos 30

= 6.8 2

3= 7.8 N

SECTION IV

59. (A) (p), (q); (B) (r); (C) (r); (D) (q), (s)

60. (A) (r); (B) (q); (C) (s), (t); (D) (p)


Elite2011 Pet1 Solns p1

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