JEE Physics Laws Of Motion MCQs Set 31

Answer: (b) 25 N

Explanation:
1. Along the smooth incline the balance supports \( mg\sin30^\circ = 5\times10\times\frac12 = 25 \) N.

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (d) 4 N

Explanation:
1. \( mg-N = ma \Rightarrow N = m(g-a) = 0.5\times8 = 4 \) N, and by Newton's third law A presses B with 4 N.

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (c) 13 N

Explanation:
1. The wall exerts normal force 12 N and friction 5 N (balancing the weight; limiting friction is 7.2 N). Resultant \( = \sqrt{12^2+5^2} = 13 \) N.

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (c) 0 m/s²

Explanation:
1. Limiting friction \( = 0.3\times2\times10 = 6 \) N, which exceeds the applied 5 N, so the body stays at rest and \( a = 0 \).

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (d) x = 2.5 m

Explanation:
1. Speed is maximum when the acceleration is zero: \( g\sin37^\circ = \mu g\cos37^\circ \Rightarrow \tan37^\circ = 0.3x \).
2. \( \frac34 = 0.3x \Rightarrow x = 2.5 \) m.

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (b) \( \tan\theta \)

Explanation:
1. At the point of contact the board lies along a tangent that is inclined at θ to the horizontal; the contact force (normal + friction) must be vertical to balance the weights.
2. Hence \( \frac fN = \tan\theta = \mu \).

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (c) The spring will be in its natural length under all conditions

Explanation:
1. If \( a\le\mu g \) both blocks move with the plank, needing only \( ma \) of friction each, so the spring stays natural.
2. If \( a>\mu g \) both slip and each has acceleration \( \mu g \) (kinetic friction \( \mu mg \) divided by its own mass), so their separation does not change and the spring is still unstretched.

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (a) 2.5 s

Explanation:
1. The block's acceleration is limited to \( \mu g = 2 \) m/s², while the truck reaches 5 m/s at t = 1 s and then moves uniformly.
2. The block reaches 5 m/s when \( 2t = 5 \), i.e. at t = 2.5 s.

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (b) kinetic, up the incline, 13.5 N

Explanation:
1. Normal reaction \( N = mg\cos37^\circ-F\sin30^\circ = 32-5 = 27 \) N, so limiting friction \( = 0.5\times27 = 13.5 \) N.
2. Net push down the plane without friction \( = 24-F\cos30^\circ = 24-8.7 = 15.3 \) N, which exceeds 13.5 N, so the block slides down and kinetic friction of 13.5 N acts up the incline.

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (d) \( \frac{\mu}{\sin\theta-\mu\cos\theta} \)

Explanation:
1. At the limit \( T = \mu mg \) (block m about to slide towards the pulley) and for M: \( Mg\sin\theta-\mu Mg\cos\theta = T \).
2. \( M(\sin\theta-\mu\cos\theta) = \mu m \Rightarrow \frac Mm = \frac{\mu}{\sin\theta-\mu\cos\theta} \).

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (c) 3/2

Explanation:
1. Accelerations: \( a_1 = \frac{(m_1-m_2\sin30^\circ)g}{m_1+m_2} \), \( a_2 = \frac{(m_2-m_1\sin30^\circ)g}{m_1+m_2} \).
2. Half the time means four times the acceleration: \( a_1 = 4a_2 \Rightarrow m_1-\frac{m_2}2 = 4m_2-2m_1 \Rightarrow \frac{m_1}{m_2} = \frac32 \).

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (c) \( \frac{m_1m_2g}{m_1+m_2}(\sin\alpha+\sin\beta) \)

Explanation:
1. \( a = \frac{(m_1\sin\alpha-m_2\sin\beta)g}{m_1+m_2} \). For B: \( T-m_2g\sin\beta = m_2a \).
2. \( T = m_2g\left[\sin\beta+\frac{m_1\sin\alpha-m_2\sin\beta}{m_1+m_2}\right] = \frac{m_1m_2g}{m_1+m_2}(\sin\alpha+\sin\beta) \).

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (d) 0.8 m/s²

Explanation:
1. If they moved together, \( a = \frac{100}{42} = 2.4 \) m/s², but A can be accelerated by static friction at most \( \frac{0.5\times70}{35} = 1 \) m/s². So B slips on A.
2. Kinetic friction \( = 0.4\times70 = 28 \) N, so \( a_A = \frac{28}{35} = 0.8 \) m/s².

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (b) 6 N

Explanation:
1. The string is taut because the lower block would otherwise slide faster. System: \( a = g\sin60^\circ-\frac{(0.1\times6+0.6\times4)g\cos60^\circ}{10} = 8.66-1.5 = 7.16 \) m/s².
2. For the 6 kg block: \( 60\sin60^\circ-0.1\times30-T = 6a \Rightarrow T = 51.96-3-42.96 = 6 \) N.

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (c) Only (1) and (3)

Explanation:
1. The hanging block moves \( \cos37^\circ \) times as fast: \( a_2 = 0.8a_1 \Rightarrow 4a_1 = 5a_2 \).
2. \( T\cos37^\circ = 5a_1 \Rightarrow a_1 = 0.16T \), and \( 100-T = 10a_2 = 8a_1 = 1.28T \), so \( T = 43.86 \) N.

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (a) Only (1) and (3)

Explanation:
1. Horizontal: \( N\sin\theta = ma = 5 \); vertical: \( N\cos\theta = mg = 10 \).
2. \( \tan\theta = \frac12 \) and \( N = \sqrt{25+100} = 5\sqrt5 \) N.

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (b) Only (1), (2) and (4)

Explanation:
1. Effective gravity \( g' = 12 \) m/s². Normal reaction \( N = 12\cos30^\circ = 6\sqrt3 \) N.
2. The contact force balances \( mg' \) in the elevator frame, i.e. 12 N. The block accelerates with the elevator so the net force is \( ma = 2 \) N, not zero.

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (d) Only (1) and (3)

Explanation:
1. Below the slit \( T_1 = 4(10-a) \), above the slit \( T_2 = T_1-10 = 2(10+a) \). So \( 40-4a-10 = 20+2a \Rightarrow a = \frac53 \) m/s².
2. \( T_2 = 2(10+\frac53) = \frac{70}{3} \) N. The tension changes at the slit, and after interchange the same equation gives \( a = \frac53 \) again, not \( \frac{10}3 \).

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (c) All four statements (1), (2), (3) and (4)

Explanation:
1. (1) \( \tan\theta = \frac ag = 1 \). (2) A spring along the line of the net apparent gravity gives the same equilibrium direction.
2. (3) At constant velocity the bob hangs vertically, which is 45° from the perpendicular to a 45° roof. (4) In free descent along a smooth incline the apparent gravity is perpendicular to the roof, so the string is perpendicular to the roof (0°).

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (c) Only (3) and (4)

Explanation:
1. Static friction can be anything from zero up to its limit, so it need not exceed kinetic friction (1 false); friction accelerates a body in the ground frame in many cases such as a box on an accelerating truck (2 false).
2. (3) and (4) are the standard definitions.

Teacher's Note:
1. Static friction adjusts itself up to a limit.
2. Contact force = normal + friction.

Answer: (d) Only (2) and (4)

Explanation:
1. The contact force is the vector sum of normal force and friction, so it equals the normal force only if friction is zero (2).
2. Friction can be zero between rough surfaces if there is no tendency to slip (4); the surface need not be frictionless (1 false).

Teacher's Note:
1. Contact force = √(N² + f²).
2. Friction is self-adjusting.

Answer: (b) Only (1) and (2)

Explanation:
1. Total pull on the system \( = 125+125 = 250 \) N. Ground friction limit \( = 0.2\times1000 = 200 \) N, so it moves with \( a = \frac{250-200}{100} = 0.5 \) m/s².
2. For the man: \( 125-f = 60\times0.5 \Rightarrow f = 95 \) N (below the limit 180 N), so statements 3 and 4 are false.

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (c) Only (1), (2) and (4)

Explanation:
1. Momentum: \( \frac m2u = \frac{3m}2v \Rightarrow v = \frac u3 \). Energy lost \( = \frac14mu^2-\frac1{12}mu^2 = \frac16mu^2 = \frac23 \) of the initial \( \frac14mu^2 \).
2. \( a_B = \mu g \) (retarding), \( a_A = \frac{\mu(m/2)g}{m} = \frac{\mu g}2 \); the relative acceleration is \( \mu g+\frac{\mu g}{2} = \frac{3\mu g}{2} \).

Teacher's Note:
1. Draw a free-body diagram and write F = ma along each axis.
2. Check limiting friction before assuming a block moves.