JEE Physics Circular Motion MCQs Set 10

Answer: (b) 0.50

Explanation:
1. The tension is largest at the lowest point of the bob's swing: \( v^2=2gl(1-\cos60^\circ)=gl \).
2. \( T_{max}=mg+\frac{mv^2}{l}=mg(3-2\cos60^\circ)=2mg=40\ \text{N} \).
3. \( \mu\times80\ge40\Rightarrow\mu\ge0.5 \).

Teacher's Note:
1. The block is most likely to slip when the tension is greatest.
2. T = mg(3 − 2cosθ0) at the lowest point for amplitude θ0.

Answer: (b) \( 2l \)

Explanation:
1. Minimum speed at the top \( v=\sqrt{gl} \) (horizontal). The top is at height 2l above the lowest point.
2. Time of fall \( t=\sqrt{\frac{2(2l)}{g}}=2\sqrt{\frac lg} \).
3. \( x=vt=\sqrt{gl}\times2\sqrt{\frac lg}=2l \).

Teacher's Note:
1. After the break the body is a horizontal projectile.
2. The height dropped is the diameter of the circle.

Answer: (d) \( g\sqrt{3\sin^2\theta+1} \)

Explanation:
1. Speed after falling a height \( l\sin\theta \): \( v^2=2gl\sin\theta \), so the radial acceleration is \( \frac{v^2}{l}=2g\sin\theta \).
2. Tangential acceleration \( =g\cos\theta \) (the component of g along the arc).
3. \( a=g\sqrt{4\sin^2\theta+\cos^2\theta}=g\sqrt{3\sin^2\theta+1} \).

Teacher's Note:
1. The angle is measured from the horizontal, so the tangential component of gravity is g cosθ.
2. Combine radial and tangential parts as perpendicular vectors.

Answer: (a) 1, 2 and 3

Explanation:
1. In a vertical circle the speed changes with height, and so do the centripetal force \( \frac{mv^2}{r} \) and the tension.
2. The weight (earth's pull) is constant.

Teacher's Note:
1. Gravity does work, so speed varies.
2. The weight mg stays the same everywhere.

Answer: (b) 1, 3 and 4

Explanation:
1. Constant speed means no tangential acceleration (1).
2. The normal acceleration is \( \frac{v^2}{r} \) (3), and since it is the only component, the total acceleration is also \( \frac{v^2}{r} \) (4).

Teacher's Note:
1. Statement 2 contradicts statement 3.
2. Total acceleration equals the normal one when the tangential one is zero.

Answer: (b) 2 and 4

Explanation:
1. Top: \( T+mg=\frac{mv^2}{l}\Rightarrow v^2=20 \), so the speed is √20 m/s (2 is correct).
2. Bottom: \( v_b^2=v^2+4gl=60 \), so the speed is √60 m/s (3 is wrong).
3. \( T_b=mg+\frac{mv_b^2}{l}=10+60=70\ \text{N} \) (4 is correct); tension varies with position (1 is wrong).

Teacher's Note:
1. Use energy conservation for the speed change of 4gl between top and bottom.
2. Tension is greatest at the bottom.

Answer: (b) 2 and 3

Explanation:
1. Along the string: \( T-mg\cos\theta=\frac{mv^2}{L} \) (2).
2. Perpendicular to the string the only force is \( mg\sin\theta \), so the tangential acceleration is \( g\sin\theta \) (3).
3. Statements 1 and 4 hold only if v = 0.

Teacher's Note:
1. Resolve the weight along and perpendicular to the string.
2. T = mg cosθ only at the extreme positions.

Answer: (b) 2, 3 and 4

Explanation:
1. At θ = 0 the only acceleration is centripetal, directed upward toward the pivot (2 is correct, 1 is wrong).
2. At θ = 90° the speed is zero, so the acceleration is g downward (3).
3. At angle θ from the vertical, \( v^2=2gl\cos\theta \), so the radial acceleration is \( 2g\cos\theta \) (towards the pivot) and the tangential acceleration is \( g\sin\theta \). The downward component of the total acceleration is \( g\sin^2\theta-2g\cos^2\theta=g(1-3\cos^2\theta) \), which is zero when cos θ = 1/√3, so the acceleration is horizontal (4).

Teacher's Note:
1. The pivot is above the bob, so the centripetal acceleration at the bottom points up.
2. Check the vertical component of the total acceleration for statement 4.

Answer: (c) A is true, but R is false.

Explanation:
1. Bending inward makes the torque of the weight and normal reaction about the centre of gravity balance the torque of friction, so A is true.
2. Lowering of the centre of gravity is not the reason for bending, so R is false.

Teacher's Note:
1. Bending provides the torque balance, not a lower centre of gravity.
2. Lean angle is given by tanθ = v²/(rg).

Answer: (d) A is false, but R is true.

Explanation:
1. The separation depends on the difference in density of the components in the rotating frame; gravity (buoyancy) also plays a role, so the phrase 'only' makes A false.
2. R is true.

Teacher's Note:
1. Words like 'only' usually make an assertion false.
2. Heavier particles move outward, lighter ones inward.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. On an unbanked road \( v_{max}=\sqrt{\mu rg} \), which rises with the friction coefficient, so A is true.
2. Banking also increases the safe speed, so R is true, but it is not the explanation of A.

Teacher's Note:
1. Both statements are true but independent.
2. Banking adds a component of the normal reaction towards the centre.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Speed and radius are constant, so \( a=\frac{v^2}{r} \) has constant magnitude, and the particle does experience a centripetal force.
2. Both are true, but the existence of the force does not by itself explain why its magnitude is constant.

Teacher's Note:
1. Acceleration is constant in magnitude but changes in direction.
2. Constancy of v and r explains the constant magnitude.

Answer: (d) A is false, but R is true.

Explanation:
1. A change of direction requires a force (centripetal) even if the speed is constant, so A is false.
2. R is a true statement.

Teacher's Note:
1. Constant speed does not mean zero force.
2. Velocity changes whenever its direction changes.

Answer: (d) 80 N

Explanation:
1. Energy conservation to the horizontal position (height l): \( v^2=u^2-2gl=100-20=80 \).
2. At this position the weight is tangential, so \( T=\frac{mv^2}{l}=80\ \text{N} \).

Teacher's Note:
1. At the horizontal position only the tension acts along the radius.
2. Use energy conservation to find v.

Answer: (d) 50 N

Explanation:
1. At the top, \( v^2=u^2-4gl=60 \).
2. \( T+mg=\frac{mv^2}{l}\Rightarrow T=60-10=50\ \text{N} \).

Teacher's Note:
1. Weight points toward the centre at the top.
2. The speed is enough to keep the string taut (v² > gl).

Answer: (b) 8.02 m/s

Explanation:
1. On the smooth plane the effective gravity along the plane is \( g'=g\sin37^\circ=5.9\ \text{m/s}^2 \).
2. \( T-mg'=\frac{mv^2}{l}\Rightarrow 110-7.08=\frac{1.2\,v^2}{0.75} \).
3. \( v^2=64.3\Rightarrow v\approx8.02\ \text{m/s} \).

Teacher's Note:
1. Only the component of gravity in the plane acts in the circle's plane.
2. Replace g by g sin θ in the vertical-circle equations.

Answer: (a) 6.83 m/s

Explanation:
1. From the lowest point to the highest, the bob rises 2l along the plane's effective gravity: \( v_t^2=v_b^2-4g'l=64.3-4(5.9)(0.75) \).
2. \( v_t^2\approx46.6\Rightarrow v_t\approx6.83\ \text{m/s} \).

Teacher's Note:
1. Use energy conservation with g′ in place of g.
2. The height gain along the effective gravity is the diameter 2l.

Answer: (d) 67.5 N

Explanation:
1. At the top the effective weight points toward the centre: \( T+mg'=\frac{mv_t^2}{l} \).
2. \( T=\frac{1.2\times46.6}{0.75}-1.2\times5.9\approx74.6-7.1\approx67.5\ \text{N} \).

Teacher's Note:
1. Both the tension and the effective weight point towards the centre at the top.
2. Use the speed found at the top.

Answer: (a) 14 m/s²

Explanation:
1. Speed \( v=\frac{ds}{dt}=3t^2=12\ \text{m/s} \) and tangential acceleration \( a_t=6t=12\ \text{m/s}^2 \).
2. Radial acceleration \( \frac{v^2}{R}=\frac{144}{20}=7.2\ \text{m/s}^2 \).
3. \( a=\sqrt{12^2+7.2^2}\approx14\ \text{m/s}^2 \).

Teacher's Note:
1. Find the tangential and radial components separately.
2. Combine them by Pythagoras.

Answer: (d) \( -\frac{v^2}{R}\cos\theta\,\hat i-\frac{v^2}{R}\sin\theta\,\hat j \)

Explanation:
1. The position is \( R(\cos\theta\,\hat i+\sin\theta\,\hat j) \) and the acceleration points to the centre, opposite to the position vector.
2. \( \vec a=-\frac{v^2}{R}(\cos\theta\,\hat i+\sin\theta\,\hat j) \).

Teacher's Note:
1. Centripetal acceleration is in the direction of −r.
2. Check at θ = 0: the acceleration must be along −i.

Answer: (c) angular momentum changes in direction but not in magnitude

Explanation:
1. The bob moves in a horizontal circle below the suspension point, so its angular momentum about that point has constant magnitude and constant angle to the vertical, but its horizontal component rotates.
2. So the direction changes, the magnitude does not (the gravity and tension torques are non-zero).

Teacher's Note:
1. Torque of the weight about the suspension point is non-zero, so L is not conserved.
2. L has a fixed vertical component and a rotating horizontal component.