JEE Physics Laws Of Motion MCQs Set 30

Answer: (d) At t = 1 s the velocity of A is 6 m/s towards the left

Explanation:
1. Retardation \( a = \frac{m_Bg}{m_A+m_B} = \frac{20}{5} = 4 \) m/s².
2. \( v(1) = 10-4 = 6 \) m/s. A stops at \( t = 2.5 \) s and returns to 100 m at \( t = 5 \) s (\( x = 100+10t-2t^2 \)); at 5 s the distance is 100 m, not 108 m, and at 10 s it is far inside the table edge, so the other statements are false.

Teacher's Note:
1. Hanging mass decelerates A until it stops, then pulls it back.
2. Position x = 100 + 10t - 2t² describes the whole motion.

Answer: (b) \( \frac{3g}{2\sqrt2} \)

Explanation:
1. Retardation \( = g\sin45^\circ + \mu g\cos45^\circ \).
2. \( = \frac{g}{\sqrt2}(1+0.5) = \frac{3g}{2\sqrt2} \).

Teacher's Note:
1. Both gravity and friction act down the plane while moving up.
2. Add the two retarding terms.

Answer: (c) \( \frac{M_1M_2(\sin\beta+\sin\alpha)g}{M_1+M_2} \)

Explanation:
1. Let \( M_2 \) go down: \( M_2g\sin\beta - T = M_2a \) and \( T - M_1g\sin\alpha = M_1a \).
2. Eliminating a: \( T = \frac{M_1M_2g(\sin\alpha+\sin\beta)}{M_1+M_2} \).

Teacher's Note:
1. Write the two equations along the planes.
2. The tension is symmetric in α and β.

Answer: (a) zero

Explanation:
1. Component of \( M_1g \) along the plane \( = 10\times10\times0.5 = 50 \) N.
2. Weight of \( M_2 = 50 \) N. The forces balance, so the acceleration is zero.

Teacher's Note:
1. Compare M₁ sin θ with M₂.
2. Here 10 sin30° = 5 kg.

Answer: (b) 36.75 N

Explanation:
1. \( a = \frac{5g-0.5\times5g}{10} = \frac{24.5}{10} = 2.45 \) m/s².
2. For the hanging block \( T = 5(g-a) = 5\times7.35 = 36.75 \) N.

Teacher's Note:
1. Find a first from the system.
2. Then use one block alone to find T.

Answer: (d) 3.3

Explanation:
1. Constant speed: \( T = mg\sin30^\circ - \mu mg\cos30^\circ = 50 - 17.3 = 32.7 \) N.
2. \( m_B = T/g = 3.27 \approx 3.3 \) kg.

Teacher's Note:
1. Zero acceleration for both blocks.
2. Friction acts up the slope since A moves down.

Answer: (c) 168 N

Explanation:
1. \( a = \frac{(40-30)g\sin30^\circ}{70} = 0.7 \) m/s².
2. For the 30 kg block: \( T = 30(g\sin30^\circ + a) = 30(4.9+0.7) = 168 \) N.

Teacher's Note:
1. Heavier block goes down, lighter goes up.
2. Use one block to find T.

Answer: (c) \( \frac{\sqrt3-1}{\sqrt3+1} \)

Explanation:
1. At 30°: \( F = mg(\sin30^\circ+\mu\cos30^\circ) \). At 60°: \( F = mg(\sin60^\circ-\mu\cos60^\circ) \).
2. Equate: \( 0.5+0.866\mu = 0.866-0.5\mu \), \( \mu = \frac{0.366}{1.366} = \frac{\sqrt3-1}{\sqrt3+1} \).

Teacher's Note:
1. Friction reverses direction between the two cases.
2. Equate the two expressions for F.

Answer: (d) 1/2

Explanation:
1. Along the plane: friction \( f = mg\sin45^\circ - T\cos45^\circ = \frac{150-50}{\sqrt2} = \frac{100}{\sqrt2} \) N.
2. Perpendicular: \( N = mg\cos45^\circ + T\sin45^\circ = \frac{200}{\sqrt2} \) N, so \( \mu = f/N = \frac12 \).

Teacher's Note:
1. The horizontal string presses the block into the plane.
2. Limiting friction gives μ = f/N.

Answer: (b) (a) and (d)

Explanation:
1. Total driving force \( 6g\sin37^\circ = 36 \) N; friction \( 0.75\times4\times10\times0.8 + 0.25\times2\times10\times0.8 = 24+4 = 28 \) N.
2. \( a = \frac{36-28}{6} = 1.33 \) m/s². For \( M_2 \): \( 12 - 4 - T = 2a \), so \( T = 5.3 \) N.

Teacher's Note:
1. The lower block would slide faster alone, so the string is taut.
2. Apply F = ma to each block.

Answer: (c) 1 : 2

Explanation:
1. Case 1 (F on 2m, pushing m): \( a = \frac{F}{3m} \), contact force \( = ma = \frac F3 \).
2. Case 2 (F on m, pushing 2m): contact force \( = 2m\cdot\frac{F}{3m} = \frac{2F}{3} \). Ratio 1 : 2.

Teacher's Note:
1. The contact force accelerates only the block ahead of it.
2. Same acceleration in both cases.

Answer: (a) due to inertia of motion, the upper part of the body continues to move forward while the feet come to rest on touching the road

Explanation:
1. Feet stop on touching the road (friction), but the upper body keeps its forward velocity.
2. This is inertia of motion.

Teacher's Note:
1. Inertia of motion keeps the upper part moving.
2. Habit is not a physical explanation.

Answer: (b) the combination of forces acting on it balance each other

Explanation:
1. Rest means zero acceleration, so the net force is zero.
2. Individual forces (gravity, normal) may still act.

Teacher's Note:
1. Zero net force, not zero force.
2. Newton's first law.

Answer: (d) \( \frac W2\cot\theta \)

Explanation:
1. At each support, vertical component of tension \( = \frac W2 \), so \( T\sin\theta = \frac W2 \) at the support.
2. Horizontal component \( T\cos\theta = \frac W2\cot\theta \) is the same everywhere, and at the lowest point the chain is horizontal, so the mid-point tension is \( \frac W2\cot\theta \).

Teacher's Note:
1. Horizontal component of tension is constant along the chain.
2. At the lowest point the tension is horizontal.

Answer: (a) (2) only

Explanation:
1. Horizontal component \( 40\cos30^\circ = 34.6 \) N, not 20 N, so (1) is false.
2. Weight is vertical (true). Vertical forces: \( N + 40\sin30^\circ = mg \), so the net vertical force is zero, not 30 N.

Teacher's Note:
1. The horizontal component uses cos 30°.
2. The body does not leave the table, so vertical forces balance.

Answer: (c) \( g\tan\alpha \)

Explanation:
1. Block at rest relative to the incline: normal N has vertical part \( N\cos\alpha = mg \) and horizontal part \( N\sin\alpha = ma \).
2. So \( a = g\tan\alpha \).

Teacher's Note:
1. Only gravity and the normal force act on the block.
2. The horizontal part of N gives the acceleration.

Answer: (c) N(AB) remains constant and N(BC) increases

Explanation:
1. Vertical balance: \( N_{AB}\cos30^\circ = mg \), so \( N_{AB} \) is fixed.
2. Horizontal: \( N_{BC} - N_{AB}\sin30^\circ = ma \) (acceleration directed to the side of BC), so \( N_{BC} \) grows with a.

Teacher's Note:
1. The vertical equation involves only N(AB).
2. Horizontal acceleration is supplied by the difference of horizontal components.

Answer: (a) \( \frac{4}{m_1} = \frac1{m_2}+\frac1{m_3} \)

Explanation:
1. Tension in the string over the fixed pulley \( T = m_1g \) when m₁ is at rest.
2. The massless movable pulley needs \( T = 2T' \), where \( T' = \frac{2m_2m_3g}{m_2+m_3} \) is the tension in the inner string. So \( m_1 = \frac{4m_2m_3}{m_2+m_3} \), i.e. \( \frac4{m_1} = \frac1{m_2}+\frac1{m_3} \).

Teacher's Note:
1. A massless pulley has zero net force on it.
2. Tension of an Atwood string is 2m₂m₃g/(m₂+m₃).

Answer: (d) 10 m/s

Explanation:
1. \( a = \frac{(2m-m)g}{3m} = \frac{10}{3} \) m/s².
2. \( v^2 = 2as = 2\times\frac{10}{3}\times15 = 100 \), so \( v = 10 \) m/s.

Teacher's Note:
1. Atwood acceleration is g/3.
2. Use v² = 2as from rest.

Answer: (b) a₁ > a₂ > a₃

Explanation:
1. A: \( a_1 = \frac{2mg-mg}{m} = g \). B: \( a_2 = \frac{2m-m}{3m}g = \frac g3 \). C: \( a_3 = \frac{mg-mg}{m} = 0 \).
2. So a₁ > a₂ > a₃.

Teacher's Note:
1. A pulled string is not the same as a hanging mass.
2. A hanging mass has to be accelerated as well.

Answer: (b) 4 m/s²

Explanation:
1. Spring force on the 10 kg mass \( = 10\times12 = 120 \) N.
2. On the 20 kg mass the spring pulls back with 120 N: \( a = \frac{200-120}{20} = 4 \) m/s².

Teacher's Note:
1. Spring forces on the two masses are equal and opposite.
2. Apply F = ma to each separately.

Answer: (c) 3g/2

Explanation:
1. Spring is stretched by the lower block: spring force \( = mg \), pulling the upper block downward.
2. Upper block: \( 2ma = 2mg + mg \), so \( a = \frac{3g}{2} \).

Teacher's Note:
1. The spring force is unchanged at the instant of cutting.
2. The upper block now has weight and spring force both downward.

Answer: (a) 120 N

Explanation:
1. \( T = m(g+a) = 10\times12 = 120 \) N.

Teacher's Note:
1. Upward acceleration increases the apparent weight.
2. Reading equals the tension.

Answer: (d) 80 N

Explanation:
1. \( T = m(g-a) = 10\times8 = 80 \) N.

Teacher's Note:
1. Only acceleration matters, not velocity.
2. Downward acceleration reduces the apparent weight.

Answer: (b) 720 N

Explanation:
1. \( N = m(g+a) = 60\times12 = 720 \) N.

Teacher's Note:
1. Velocity direction is irrelevant.
2. Apparent weight is the normal reaction.

Answer: (d) 600 N

Explanation:
1. Zero acceleration, so the normal reaction equals the weight: 60 × 10 = 600 N.

Teacher's Note:
1. Constant velocity means equilibrium.
2. Weight is mg.

Answer: (a) 480 N

Explanation:
1. \( N = m(g-a) = 60\times8 = 480 \) N.

Teacher's Note:
1. Downward acceleration reduces the reading.
2. Equivalent to a smaller effective g.

Answer: (c) 15 kg

Explanation:
1. The system is in equilibrium; tension in each string is 15 kg-wt.
2. The balance reads the tension in it, 15 kg.

Teacher's Note:
1. Compare with fixing one end to a wall.
2. Reading is not the sum of the two pulls.