Aiats Iitjee 2012 Test-2 P-II Solution

8/18/2019 Aiats Iitjee 2012 Test-2 P-II Solution

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Test - 2 (Paper - II) (Answers & Hints) All India Aakash Test Series for IIT-JEE 2012

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TEST - 2 (Paper - II)

A N S W E R S

1. (B)

2. (A)

3. (B)

4. (A)

5. (A)

6. (A)

7. (D)

8. (C)

9. (A, B, D)

10. (A, C)

11. (A)

12. (A, B, C)

13. (3)

14. (6)

15. (9)

16. (8)

17. (0)

18. (9)

19. A → (p, q, s)

B → (q, s)

C → (q, t)

D → (q, r)

20. A → (p, q)

B → (q)

C → (q, s, t)

D → (p, r)

CHEMISTRY MATHEMATICS PHYSICS

21. (A)

22. (B)

23. (B)

24. (D)

25. (D)

26. (D)

27. (C)

28. (B)

29. (A, B, C, D)

30. (A, B, C, D)

31. (A, B)

32. (B, D)

33. (1)

34. (9)

35. (4)

36. (2)

37. (3)

38. (3)

39. A → (p, r)

B → (t)

C → (p, q, s)

D → (p, q, r, s)

40. A → (s)

B → (q)

C → (p)

D → (q)

41. (C)

42. (B)

43. (C)

44. (C)

45. (D)

46. (B)

47. (B)

48. (B)

49. (A, B, C)

50. (A, C)

51. (B, C, D)

52. (B, C)

53. (7)

54. (2)

55. (2)

56. (2)

57. (6)

58. (4)

59. A → (t)

B → (r)

C → (s)

D → (q)

60. A → (t)

B → (t)

C → (r)

D → (p)


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All India Aakash Test Series for IIT-JEE 2012 Test - 2 (Paper - II) (Answers & Hints)

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1. Answer (B)

Decreases in denser medium.

2. Answer (A)

Wave in same phase add-up together.

3. Answer (B)

π∗p y has two nodal planes

4. Answer (A)

7 22 2

1 1 11.09677 10 (1)

(4) (7)

= × −

λ

⇒ λ = 2.17 × 10 –9 m/s

5. Answer (A)

1 atomst : 5 4 3 2 1 4 lines (different)

2 atomnd

3 atomrd

4 atomth

5 atomth

6 atomth

7 atomth

:

::

:

:

:

5

55

5

5

5

4

43

3

2

1

11

1

1

1

1 line

1 line1 line

1 line

1 line

1 line

10 lines

4

2

5 → 3 → 2 → 1 has repeated lines

6. Answer (A)

Factual.7. Answer (D)

3 21

2 32

T (4) (2) 4T 9(3) (4)

= × =

8. Answer (C)

0 01

h h KE (KE)h

ν = ν + ∴ ν = ν + .

9. Answer (A, B, D)

O N O

O

–

ANSWERS & HINTS

PART - I (CHEMISTRY)


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10. Answer (A, C)

Bond order of O 2 – = 1.5

Bond order of O 22– = 1

Bond order of O 2 = 2

Bond order of O 3 = 1.5

11. Answer (A)

Cp – C v = R ⇒ (1.25 M – 0.75 M) = 2

0.5 M = 2 ∴ Molar mass = 4 g

12. Answer (A, B, C)

Factual

13. Answer (3)

5 mole steam absorbs 660 kJ.

To maintain temperature constant CO(g) must evolve.

660 kJ ∴ n CO = 6 ⇒ 2On = 3

14. Answer (6)

Wrev =3.99V

2.303 nRT logV

−

= –2.303 × 1 × 8.314 × 522.27 log(3.99)

= –6000 J = –6 kJ

15. Answer (9)

∆G = ∆H – T ∆S for minimum ∆H, ∆G = 0

∆H = T ∆S = 298 × 30.2 = 9000 = 9 kJ

16. Answer (8)

Equation (i) × 2 + Equation (ii) × 2 – Equation (iii)

= –65 × 2 + (–97) × 2 – (–340) = 340 – 194 – 130 = 16 kcal

= (2 kcal) × 8

17. Answer (0)

H2O( ) H2O(g)

Molar heat capacity = ∞ =10

18. Answer (9)

∆G° = –2.303 × 8.314 × 298 log K = 55.6 × 10 3

∴

355.6 10log K 9.744 ( x 0.744)

2.303 8.314 298×= − = − = − −

× ×

∴ x = 9

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

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


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PART - II (MATHEMATICS)

21. Answer (A)

Let,1 2 2 3 1 3

1 3 5(say)

| | | | | |

= = =

− − −

k

z z z z z z

∴ 1 2 2 3 1 31 3 5| | , | | and | |− = − = − =z z z z z z k k k

or, 1 2 1 2 2 3 2 3 1 3 1 32 2 21 9 25

( )( ) , ( )( ) and ( )( )− − = − − = − − =z z z z z z z z z z z z k k k

So, 2 1 21 2

1( )= −

−k z z

z z

22 3

2 3

9( )= −

−k z z

z z

23 1

3 1

25 ( )= −−

k z z z z

∴1 2 2 3 3 1

1 9 250+ + =

− − −z z z z z z

22. Answer (B)

From question

2a 2α 2 + b α + c = 0

∴ b α + c = –2 a 2α 2 …(i)

also, 3 a 2β2 – 2 b β – 2 c = 0

∴2 23

2β+ = βb c a …(ii)

Now, let f (x ) = 4 a 2x 2 + 3 bx + 3 c

Then f (α ) = 4 a 2α 2 + 3( b α + c )

= 4 a 2α 2 – 6 a 2α 2 …[from (i)]

= –2 a 2α 2 < 0

f (β) = 4 a 2β2 + 3 b β + 3 c

= 4 a 2β2 + 3( b β + c )

= 2 2 2 29

42

β + βa a …[from (ii)]α

βx -axisγ

= 2 217

02

β >a

f (α ) f (β) < 0, then one root of the equation f (x ) = 0 must lie between α and β.

23. Answer (B)

Let, the seating arrangements of the persons are as shown below,

S N S N N S N S N N

where, S → persons may be selected

N → persons may not be selected


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As there are 6N’s and 4S’s.

So, the problem is similar to that of selecting any 4 places out of 7 gaps (shown below)

X N X N X N X N X N X N X

which can be done is 7C 4 ways

= 35 ways

24. Answer (D)

6 = 1 + 5 Required number of ways

= 2 + 43

161 61 61

2!51 2! 4! 3! 3! 2!

C

= + + ×

= 3 + 3

25. Answer (D)

Let, f (x ) = ax 2 + bx + c ; a , b , c ∈ R ; a ≠ 0

Given, α and β are the roots of the equation f (x ) = 0

For (1) :

2

2 21 − + − = − + =

a b a bx cx f c

x x x x

∴ cx 2 – bx + a = 0 has roots1 1

and− −α β

For (2) :

Equation with roots ( α + β) and αβ is

Given by x 2

– ( α + β + αβ )x + (α + β)αβ = 0

or, 2 2 0

− − + + − = b c bc

x x a a a

or, a 2x 2 + (b – c )ax – bc = 0

For (3) :

22 2 4 ( ) (say)

2 4 2

= + + = + + = x ax bx f c ax bx c g x

Now, g (x + 1) = a (x + 1) 2 + 2 b (x + 1) + 4 c

= a (x 2 + 1 + 2 x ) + 2 bx + 2 b + 4 c

= ax 2 + 2( a + b )x + a + 2 b + 4 c

∴ Roots of the equation ax 2 + 2( a + b )x + a + 2 b + 4 c = 0 are 2 α – 1 and 2 β – 1.

26. Answer (D)

5 5| | = − +z z

z z

5 5| |

≤ − +z z z

or, 5| | 4 | |≤ +z z


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or,5

| | 4| |

− ≤z z

or, | z |2 – 4| z | – 5 ≤ 0

or, | z |2 – 4| z | + 4 – 9 ≤ 0

or, ( | z | – 2) 2 ≤ 9

∴ | z | – 2 ≤ 3

or | z | ≤ 5

Obviously | z | max = 5

27. Answer (C)

Let, n 1, n 2, n 3 and n 4 be the number of white, red, yellow and brown balls that are selected.

∴ n 1 + n 2 + n 3 + n 4 = 8

So, the required number of ways

= coefficient of x 8 in (1 + x + x 2 + ........) 4

= coefficient of x 8 in (1 – x ) –4

= coefficient of x 8 in (1 + 4C 1x +5C 2x

2 + 6C 3x 3 + ........)

= 11C 8

= 11C 3

28. Answer (B)

210 = 2 × 3 × 5 × 7

x 1 x 2 x 3 x 4 = 210∴ Each of x i may be equal to 2 or 3 or 5 or 7

So, the possible number of solutions

= 4 × 4 × 4 × 4

= 256

29. Answer (A, B, C, D)

Let, z 1 = e i α

and z 2 = re i θ

∴( ) – ( )

1 2α−θ θ−α= =i i z z r e r e

OP = | z 1 | = 1 = OA

Also, 1 2 1 2 2= = = =OR z z z z z OQ

α –( – )αθ

θ

P z ( )1

A x O

R z z ( )1 2

y

(1, 0)

Q z ( )2

and ∠QOP = ( θ – α ) = ∠ROA

Obviously ∆POQ and ∆AOR are congruent

∴ PQ = AR

or, 2 1 1 2 1− = −z z z z

Also, Area ∆POQ = Area ∆AOR


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Now, 1 1 1 21 2 1 2 2 1

1arg arg arg arg

= = =

z z z z

z z z z z z

∆POQ and ∆AOR are congruent

∴ ∠PQO = ∠ARO

⇒2 1 1 2

2 1 2arg arg

1

−=

− z z z z

z z z

or, 12 1 2

1arg 1 arg 1

1

− = +

− z z z z

or, ( )1 11 2 1 2

1 1arg 1 arg 1 11

z z z z z z

− = − =

− ∵

So, all the options are correct.

30. Answer (A, B, C, D)

Number of straight lines = 7C 2 – 3C 2 + 1 = 19Number of triangles = 7C 3 –

3C 3 = 34

For quadrilateral ; it can be formed in following cases :Selected collinear

points (3)Selected non-collinear

points (4)Number of

quadrilaterals formed

(i) 1 3 = × = 123 4C C 1 3

(ii)

(iii)

2

0

2

4

= × = 183 4C C 2 2

= × = 13 4C C 0 4

So, number of quadrilaterals = 12 + 18 + 1 = 31

Similarly, number of pentagons = 3C 1 × 4C 4 + 3C 2 × 4C 3= 3 + 12 = 15

31. Answer (A, B)

21 1= + + −z t i t

⇒ Re( z ) = x = 1 + t

2Im( ) 1= = −z y t

So, ( x – 1) 2 + y 2 = t 2 + 1 – t 2O

y

P z ( )

c Q (2, 0)

x (0, 0) (1, 0) 1

1

or, ( x – 1) 2 + y 2 = 1

i.e. , | z – 1| = 1

⇒ z lies on a circle with radius 1 and centered at (1, 0). Obviously the circle passes throgh origin.

Also, | z |max = 2

⇒ | z | ≤ 2 ∀ t ∈ R

From figure, its clear that ∠CPQ ≠ 2π

2π

∠ = ∵ OPQ

⇒1

arg2 2

− π ≠ −

z z

∆CPQ is an isosceles ∆, so its quite possible that if PQ also becomes 1 (for some P on the circle),

Then ∆CPQ may become equilateralSo, it is possible that | z – 1| = | z – 2| for some z or for some values of t .


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32. Answer (B, D)

The given equation is x 2 + 2( m – 1) x + (m + 5) = 0

Here, D r = (2( m – 1)) 2 – 4( m + 5)

= 4[m 2 + 1 – 2 m – m – 5]

= 4[ m 2 – 3 m – 4]

= 4( m + 1) ( m – 4)

For, the equation to have at least one negative root

(i) Either both the roots are real and negative.

OR

(ii) Exactly one root is negative.

Case I : D ≥ 0 and α + β < 0 and αβ > 0

⇒ m ≤ –1 or m ≥ 4 …(i)

and m > 1 …(ii)and m > –5 …(iii)

(i) ∩ (ii) ∩ (iii) : –5 –1 1 4

⇒ m ≥ 4 …(a)

Case II : αβ < 0

m < –5 …(b)

From (a) and (b)

m ∈ (– ∞, –5) ∪ [4, ∞]33. Answer (1)

This is possible when

k 2 – 4 = 0 ⇒ k = ±2

k 2 – 5 k + 6 = 0 ⇒ k = 3, 2

and k 2 – 3 k + 2 = 0 ⇒ k = 1, 2

The common value of k for which all the conditions are satisfied is k = 2

So, there is only one value of such ‘ k ’ is possible.

34. Answer (9)Since, ω and ω2 are the roots of the equation

1 1 1 3+ + =+ + +a x b x c x x

or [x 2 + (b + c )x + bc + x 2 + ( a + c )x + ac + x 2 + ( a + b )x + ab ]x = 3( a + x )(b + x )(c + x )

or 3 x 3 + 2( a + b + c )x 2 + x (ab + bc + ca ) = 3[ x 3 + ( a + b + c )x 2 + ( ab + bc + ca )x + abc ]

or (a + b + c )x 2 + 2( ab + bc + ca )x + 3 abc = 0

Obviously, 22( ) 1+ +ω+ω = − = −

+ +ab bc ca

a b c …(i)

and 2 3· 1ω ω = =+ +abc

a b c …(ii)


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From (i) and (ii), we have

2(ab + bc + ca ) = 3 abc

or,1 1 1 3

2+ + =

c a b

or, 1 1 1 3 932 2

+ + ++ + = + =c a b c a b

or,1 1 1

2 9+ + +

+ + = a b c

a b c 35. Answer (4)

Let, 2 n – 1 and 2 n + 1 be two consecutive odd integers and roots of the equation ax 2 + bx + c = 0.

Then, 2 n – 1 + 2 n + 1 = −b a

…(i)

and (2 n – 1)(2 n + 1) =c a

…(ii)

Therefore,2

4

− b c a a

= (4 n )2 – 4 × (4 n 2 – 1)

= 16 n 2 – 16 n 2 + 4

= 4

∴

2

24

4− =b c

a a

or,2

24 4− =b ac

a

36. Answer (2)1 Row

st

2 Rownd

3 Rowrd

As there are 1, 3 and 4 boxes in the 1 st, 2 nd and 3 rd row respectively.

Then the required number of ways of selecting the boxes, so that none of the row may remain empty is

= Coefficient of x 6 in the expansion of ( 1C 1x )(3C 1x +

3C 2x 2 + 3C 3x

3)(4C 1x +4C 2x

2 + 4C 3x 3 + 4C 4x

4)

= Coefficient of x 6 in x (3x + 3 x 2 + x 3)(4x + 6 x 2 + 4 x 3 + x 4)

= Coefficient of x 6 in x 3(3 + 3 x + x 2)(4 + 6 x + 4 x 2 + x 3)= Coefficient of x 3 in (3 + 3 x + x 2)(4 + 6 x + 4 x 2 + x 3)

= 21

Now,

The word LOKPAL may be arranged in6!2

ways in the boxes. Therefore, total number of required ways

=6!

212

×

=2(5 2 1) (2 2 2)!

2

× + × × +

∴ n = 2


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37. Answer (3)

9x 2 – 6 x + y 2 = 24

or, (3 x – 1) 2 + y 2 = 25

The possible solutions are

3 – 1x

y

±5

0

0

±5

±3

±4

±4

±3

(2, 0), (–1, 3) and (–1, –3) only

∴ Number of ordered pair = 3

38. Answer (3)

N = 5880 = 2 3 × 3 × 5 × 7 2

= 2 × (2 2 × 3 × 5 × 7 2)

Number of even divisors = (2 + 1)(1 + 1)(1 + 1)(2 + 1)

= 3 × 2 × 2 × 3

= 36

= 4 × (3) 2

∴ n = 3

39. Answer → A(p, r), B(t), C(p, q, s), D(p, q, r, s)

(A) If we select 4 vertices, two diagonals can be formed and we get one point of intersection.

So, total number of points of intersection of diagonals = 6C 4 × 1 =6C 4 = 15

(B) One line divides the plane in 2 regions

Two lines divides the plane in 4 regions

Three lines divides the plane in 7 regions

Four lines divides the plane in 11 regions

So, there is a series like

2, 4, 7, 11, ........

and we have to find the eighth term of the series

S T n n = 2 + 4 + 7 + 11 + ........ +S T T n n n = 2 + 4 + 7 + ........ + + – 1or, (–) (–)

0 = 2 + 2 + 3 + 4 + ........ + – n T n

Let,

∴ T n = 2 + 2 + 3 + 4 + ........ + n

=( 1)

12+

+ n n

∴ T 8 = 1 + 8 ×92

= 1 + 36

= 37


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(C) f : {1, 2, 3, 4} → {p , q , r , s }

1 may be related to any of p , q , r , s i.e. , 4 ways

f has to be one-one onto

∴ 2 may be related in 3 ways

Now, 3 may be related in 2 ways

and 4 may be related in only 1 way

So, the number of one-one onto function = 4 × 3 × 2 × 1

= 24

∴ The possible number of invertible function is = 24

(D) Let number consider the side A1A2.

Now to form a ∆ whose one side is common with that of then -gon, available vertices are,

A4, A

5, ........, A

n – 1.

i.e. , (n – 4) vertices are available.

A1 A2

A3An

Now, as there is n -gon. So, we may take any of the n sides of n -gon to be the common side.

Therefore, total number of such triangles = ( n – 4) × n

As we have n = 10

∴ Number of such triangles = (10 – 4 ) × 10

= 6 × 10

= 60

40. Answer → A(s), B(q), C(p), D(q)

(A) Obviously,

| z 1 – z 2 |min = 0

and | z 1 – z 2 |max = 8

∴ sum = 8

(B) (1 + αω 5)n = (α + ω4)n

or, (1 + αω 2)n = (α + ω)n

or, ω2n (ω + α )n = (α + ω)n

B

A z z ( ) ( )2 1≡

2

2

x

y

( )z 1 | | = 4z 1

O

| – 1 + 3| = 2z i 2

(1, – 3 )

4

or, ( ω + α )n [ω2n – 1] = 0

⇒ ω 2n = 1

∴ Least positive value of n = 3

(C)

1

1 2 2

11 2

2

3 31 13 4 4 4333 4 1144

+ ++= =

− −−

z ik z z z

z z z ik z

314 1314

+= =

− +

ik

ik


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Since, 1 2 1 2 0+ =z z z z

or, 1 12 2

= −z z z z

or, 1 12 2

= −

z z z z

⇒1

2=

z ik

z (i.e. , 1

2

z z

is purely imaginary)

(D) Let, z 1 = cos α + i sin α , z 2 = cos β + i sin β and z 3 = cos γ + i sin γ

∴ z 1 + z 2 + z 3 = (cos α + cos β + cos γ ) + i (sin α + sin β + sin γ )

= 0 + i 0 (from question)

= 0

Now, (z 1 + z 2 + z 3)2 = z 12 + z 22 + z 32 + 2 ( z 1z 2 + z 2z 3 + z 1z 3)

or, 0 = z 12 + z 2

2 + z 32 + 2 ∑z 1z 2 …(i)

Now, 1 1 11 2 31 2 3

1 1 1 − − −+ + = + +z z z z z z

= (cos α + cos β + cos γ ) – i (sin α + sin β + sin γ )

= 0

i.e. , 2 3 1 3 1 21 2 3

0+ +

=z z z z z z

z z z

⇒ z 1z 2 + z 2z 3 + z 1z 3 = 0 …(ii)

From (i) and (ii), we get

z 12 + z 2

2 + z 32 = 0

⇒ cos2 α + cos2 β + cos2 γ = 0 = sin2 α + sin2 β + sin2 γ

Now, cos2 α + cos2 β + cos2 γ = 0

⇒ 2(cos 2α + cos 2β + cos 2γ ) – 3 = 0 ( cos2 θ = cos 2θ – sin 2θ)

or, 2(cos 2α + cos 2β + cos 2γ ) = 3

PART - III (PHYSICS)41. Answer (C)

2

(cos sin ) mv T mg − θ+ θ =

21 (1 cos ) sin2

mv mg F = − − θ + θ

3 2sin 24

T mg π = θ+ −

max 3 2 2T mg = −


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42. Answer (B)

2

2cos mv mg

R β =

2

1 2

1cos cos2mgR mv mgR β = + β

v 2 = 2 gR (cos β1 – cos β2)

Rg cos β2 = 2 Rg (cos β1 – cos β2)

12

2coscos

3β

β =

43. Answer (C)

0du dx

=

50,2

x =

x = 0 is local maxima

at52

x = ± u = – 6.25 J

K.E. = 42.25 J

v max = 6.5 m/s

44. Answer (C)

Applying energy conservation

2 21 12 (1 cos ) 2 (1 sin ) 22 2

mgr mgr mgr mv mv = − θ + − θ + +

[ ]2 2 cos 2sin 13

v gr = θ+ θ−

for maximum v 2

tan θ = 2

45. Answer (D)

By work energy theorem2

gravity1 02 F

mv W W − = +

212 2

r mv Fr mg = −

21 10 (200 50) 102

v × = − ×

v = 300

= 10 3

= 17.3 m/s


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46. Answer (B)

2 cos2

T mr f ω = θ+

sin ;2T

f f mg θ = = µ

2 2 sin(45 )g r µω = + θ

max2 g r µ

ω =

47. Answer (B)

v = 4 t + 5

a = 4

m = 2

∆P = 2[8 – 0]

= 16 N-s

48. Answer (B)

a 1 = a (1 + cos θ)

ma 1 = mg sin α + ma cos α – T

N = mg cos α – ma sin α

For horizontal motion of wedge

ma = T + T cos θ – T cos α + N sin θa 1 a

2 2

(1 cos )sin

(1 cos cos ) sin

mg a M m

+ θ α= + + θ− α + α

here m = M

θ = 60°

α = 30°

49. Answer (A, B, C)

Maximum acceleration of 7 kg block = 2 215

m/s 2.1 m/s7

=

Maximum acceleration of 7 kg and 3 kg block as one system = 0.4 m/s 2

Acceleration of whole system if it moves together = 210 5

m/s12 6

=

Then 2 kg block will slip on 3 kg block and there will be no slipping between 3 kg and 7 kg block.

Acceleration of 2 kg = 210 4 3 m/s

2− =

Acceleration of 3 kg and 7 kg block = 0.4 m/s 2


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50. Answer (A, C)

25 kg 15 kg F

Wall

25 kg

0.4 F 0.4 F

F F

2T

250 N

⇒ 2T = 0.8 F + 250 …(i)

T + 0.4 F = 150 …(ii)

0.8 F + 125 = 150

0.4 N

F

T

15 kg

150 N

F

25 250 31.25 N0.8 8

F = = =

T = 150 – 0.4 × 31.25

T = 137.5 N

51. Answer (B, C, D)

Work done by friction on plank in plank frame = 0

Work done by friction on block in plank frame = 212

mv −

Total work done = heat produced = 212

mv

Work done on block in ground frame = 21

K.E.2

mv = ∆

Work done on plank in ground frame = 2 2 21 12 2

mv mv mv − − =

52. Answer (B, C)

2

a g g

−=

2 1

2a a

g −

=

g

g a

a 1

a 23g

a 2a

1

g 2

13

2g a

a +

=

Solving the three equations

a = 3 g

a 2 = 7 g

a 1 = 5 g


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53. Answer (7)

2 20

1 12

2 2mv mg R mv + = …(i)

20

B

mv mg N R − = …(ii)

2

Amv

N mg R

− = …(iii)

54. Answer (2)

45

T Ma m a a

= + +

95

R M m a

= +

930 4·5 5

ma a m Ma ma − − − =

37°

M a a

a m

T ma

1730

5ma Ma = +

= (18 + 42) a

55. Answer (2)

T sin θx 1 – Tx 2 = 0

1 1 2sin cos 0d v x v

dt

θθ+ θ − =

as x 1 is very small

v 1sin θ = v 2

1 1 2sin cos d a v v

dt θθ+ θ =

m

m

30°x 1

T

T m

x 2

T 0

m

m T 2

T

mg

as v 1 = 0

a 1sinθ = a 2

a 1 = 2 a 2

⇒ 1 222T ma ma = =

mg – T = ma 2

mg = 5 ma 2

22 2 m/s5

g a = =

56. Answer (2)

Applying conservation of energy

21 1 1500 0.4 0.4 15 500 (0.2)(0.2)2 2 2

v × × × = × + × ×


17/18


17/18

57. Answer (6)

·d F dr ω=

ˆ ˆ ˆ ˆ( )·( )yi xj dxi dyj = + +

ydx xdy = +∫( )d xy = ∫4,3

2,1/2

2 312 9 J

2xy

× = = − = ∫Work done equals change in kinetic energy

219 82

v = × ×

3 m/s2v =

6 m/s4

v =

58. Answer (4)

26dy F r dx

= − = −

2mv F r

=

mv 2 = 6 r 3

Total energy = P.E. + K.E.

= 2 r 3 + 3 r 3

= 5 r 3

r = 5

⇒ 54 J

59. Answer → A(t), B(r), C(s), D(q)

For reacting point C C v gR ≥

52R

h ≥

For reacting point D 16D gR

v ≥

6932

h R ≥

h should be greater of the two

2

16

C C

mv N mg

R − =

N C = 17 mg


18/18


18/18

BP = v D × t

21 5 22 2 8D

R R mv mg mg R

= − +

175 4P v gR gR = −

34gR =

2 178

t g

=

5116

BP R =

60. Answer → A(t), B(t), C(r), D(p)2a 1 + a 2 = a 3

48 – 2 T = 4 a 1

24 – T = 2 a 2

T – 12 = a 3

a 1 = a 2 =212 m/s

5

a 1

a 2a 3

a 3 = 236

m/s5

[here a 1, a 2 and a 3 are acceleration w.r.t. elevator]

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Aiats Iitjee 2012 Test-2 P-II Solution