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Section - C : Assertion - Reason Type 1.
Answer (4) Direction of accelerations of the blocks might be different from the direction of motion. Hence, tension in any of sections of the strings may be greater.
2.
Answer (1)
N
mgsinθ
f mgcosθ
Speed at point of contact is zero (at every moment) wfriction is zero. 3.
Answer (1) As the bottommost point either moves or has tendency to move leftwards, friction on it acts rightwards. Thus, net external force acts rightwards and the centre of mass moves rightwards.
4.
Answer (1)
A
I about axis perpendicular to the plane and passing through O is
mA 2 + m(OC )2 12
C A /2
2
mA 2 mA 2 ⎛A⎞ + m⎜ ⎟ = = = constant 12 3 ⎝2⎠
5.
θ
B
O
Answer (4)
For a rigid cylinder rolling on a rough horizontal surface, as the bottommost point is at rest, no friction acts. Therefore V and ω are maintained. 6.
Answer (3) As the surface is smooth the rod does not experience any force in horizontal direction and its centre accelerates in vertical direction. As shown in the figure it moves vertically down from A to A′.
A A′
The rod is also experiencing a normal contact force so acceleration of centre of mass =
mg − N and m
not g. Hence, it does not fall freely. 7.
Answer (4)
Acceleration of a solid sphere rolling down an incline =
=
g sin θ , I 1+ MR 2 g sin θ 5g sin θ = . 2 7 1+ 5
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Mechanics-II
137
Answer (2) L for a particle with linear momentum p about a point in space = r × p , where r = position vector of the particle with respect to the point in space. Second statement is also correct. But these two statements are independent.
9.
Answer (1) As bottommost point moves rightwards / has tendency to move rightwards, friction acts leftwards.
10. Answer (4) 11. Answer (1) A body a be treated to be executing pure rotation about a point instantly at rest (or instantaneous centre of rotation). 12. Answer (3) At point C, although the ball has translational kinetic energy equal to zero, but it has same rotational kinetic as that at the bottom. ⇒ By conservation of energy h′ < h. Second statement is incorrect. In rolling, work done by friction is zero and therefore, energy is conserved. 13. Answer (3) Angular momentum of the ice skater = Iω = constant. As ice skater folds his limbs I decreases, ω increases. 14. Answer (4) Any point on the rim of a rolling wheel moves on cycloid and covers distance equal to 8r in one revolution. But angular displacement of wheel in one revolution, about its axis, is 2π. 15. Answer (3) a c.m. =
⇒
m1a1 + m2 a2 m1 g + m2 g = =g m1 + m2 m1 + m2
a of particles relative to c.m. = g − g = 0
⇒ Particles move uniformly. 16. Answer (2) p with respect to centre of mass
= Σmi (v i − v c.m. ) = Σmi v i − (Σmi )v c.m. ⎛ Σm v i i = Σmi v i − (Σmi )⎜ ⎜ Σmi ⎝
⎞ ⎟ ⎟ ⎠
= 0 Also, F
external
= ma c.m.
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17. Answer (4) F external =
⇒
dp dt
p = constant if F
external
=0
⇒ Internal forces can’t change momentum of the system. But internal forces can change individual momentum of the particles and hence can change kinetic energy of the system. 18. Answer (2) 19. Answer (4) Angular momentum = mvr sinφ = constant
φ
v
r S
20. Answer (4) 21. Answer (1) g′ = g – ω2 Rcos2λ for poles λ = 90° So, g ′ = g (Hence no effect of rotation) 22. Answer (3) Wext = change in total energy Since Wext < 0. So, the total energy will decrease. The kinetic energy is negative of total energy so the speed will increase. 23. Answer (2) The rms speed of the gas molecules on earth is less than the escape speed so they cannot escape from the gravitation field of the earth. 24. Answer (2) The absolute escape velocity does not depend on the direction of projection. But the velocity w.r.t. ground may depend on the direction of projection and so does the resultant velocity acquired by the projected satellite. 25. Answer (2) Total energy = −
GMm 2a
where a = semi major axis 26. Answer (3) Total energy remains constant. In elliptical path the gravitation force is perpendicular to the velocity of the satellite at two points only, so at rest of the points power of gravitational force is non-zero. Aakash Educational Services Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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27. Answer (3) Time period of geostationary satellite is 24 hours. The polar satellite are used for weather forecasting. 28. Answer (1) To shield a region from gravity we need both attractive and repulsive force. 29. Answer (4) 1
g∝
so the acceleration of the satellite will be less than the acceleration due to gravity.
r2
30. Answer (4) Breaking stress of a wire depends only on the material of the wire. Breaking stress =
Breaking load Area
Breaking load depends on area but not the breaking stress. 31. Answer (1) Stress =
Force Area
Strain =
Change in length Original length
If force and area of cross-section are same, then stress will be same. If change in length is different but original length is same then strain will be different. 32. Answer (2) Height of liquid in capillary is given by h =
As θ =
If θ <
2T cos θ rρ g
π ,h=0 2
π , h will be positive. 2
33. Answer (3) A strong force can be exerted by creating pressure difference Force = Pressure difference × Area Pressure below the wings is greater than at the top. ∴
Force = Pressure × Area acts on the wings which lift the aeroplane.
For an ideal fluid in a horizontal pipe, flow may be there while there is no pressure difference. 34. Answer (1) When the scrap-iron is in boat then the volume of displaced water is greater than the water displaced when the scrap-iron is droped in water. Aakash Educational Services Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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35. Answer (1) When the system starts accelerating up, then the value of geff will be same for both block and water. ∴ The block will not go further inside the block. 36. Answer (1) As the temperature of water falls from 4°C to 0°C the density decreases and volume increases. So, the level of water will rise. 37. Answer (1) P – P0 =
4T as r-increase, P - will decrease. r
38. Answer (3) The pressure at hole is less than the atmospheric pressure. Therefore the Hg will not flow. 39. Answer (4) Liquid has only Bulk modulus and no Young’s modulus and shear modulus because liquid cannot be stretched as a wire and its free surface cannot sustain the shear stress. 40. Answer (1) As geff on the artificial satellite is zero. ∴ Due to surface tension liquid will rise upto full length of the tube. 41. Answer (3)
p2
r1
r2
p1
42. Answer (1) When an open vessel containing water accelerates then in the horizontal direction then the resultant force G R = (ma)2 + (mg )2 and the surface of water will be perpendicular to R . 43. Answer (1) 44. Answer (2) For SHM, acceleration = –ω2 (displacement) or
a = –ω2x.
45. Answer (4) If F = kx2 then for x > 0 the force F > 0 and for x < 0 the force F > 0 so the particle will not oscillate. 46. Answer (1) For SHM the potential energy must be minimum at the mean position. It is not necessarily that the potential energy is zero. Total energy = K + U It U is taken zero at the extreme point then the total energy at extreme position will become zero. Aakash Educational Services Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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47. Answer (1) For simple pendulum G G G ma = T + mg
⇒
G G G T = ma − mg
G G At mean position a and g are opposite. 48. Answer (2) l
T ′ = 2π
2
g +a 2
2
2
g + a > g so time period decreases.
Since
49. Answer (4) The phase difference between the acceleration and the displacement is 180°. 50. Answer (1) For the coin N – mg = ma ⇒
N
N = ma + mg
Since a = ω2A sin ωt N = ω2A sin ωt + mg
mg
Nav = 0 + mg = mg. 51. Answer (2) Work done by the spring is alternatively positive and negative. 52. Answer (2) If x = A sin ωt, y = B cos ωt then
x
2
A
2
y
+
B
2 2
= 1 (equation of ellipse)
G 2 2 |v |= ω A . 53. Answer (3) K=
1 2 2 2 mω ( A − x ) 2
Kmax =
1 2 2 mω A 2
So if A is doubled the Kmax becomes four times. While the minimum potential energy does not depend on the amplitude. Aakash Educational Services Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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54. Answer (4) G Displaceme nt v av = is the sin2ωt definition of the average velocity so the formula is always applicable. time
55. Answer (2) U=
1 2 2 2 mω A sin ωt 2
=
1 2 2 mω A . 4
56. Answer (1) Both statements are ture and statement-2 explains statement-1 57. Answer (1) Both statements are ture and statement-2 explains statement-1
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