JEE Physics Circular Motion MCQs Set 03

Answer: (c) 1/3 kg

Explanation:
1. \( T_{max} - T_{min} = 6mg \) for a complete loop.
2. \( 6m\,g = 2g \) (2 kg wt) \( \Rightarrow m = \tfrac{1}{3} \) kg.

Teacher's Note:
1. The difference of 6mg between the bottom and top tension is independent of the radius and speed. It is a very common result.

Answer: (a) \( mg + \dfrac{mv^2}{r} \)

Explanation:
1. At the lowest point of a concave road the net force is upward: \( N - mg = \dfrac{mv^2}{r} \).
2. \( N = mg + \dfrac{mv^2}{r} \).

Teacher's Note:
1. On a concave bridge the road pushes harder than the weight, on a convex bridge it pushes less. Draw the free-body diagram first.

Answer: (b) 2 cm

Explanation:
1. To complete the loop, the speed at the bottom must satisfy \( v^2 \ge 5gR \).
2. \( gh = \tfrac{1}{2}v^2 \ge \tfrac{5}{2}gR \Rightarrow R \le \tfrac{2}{5}h = 2 \) cm.

Teacher's Note:
1. The starting height must be at least 2.5R. Equivalent statement: R can be at most 0.4h.

Answer: (c) \( \sqrt{gr} \)

Explanation:
1. At the top the minimum speed corresponds to zero normal force from the seat/air: \( mg = \dfrac{mv^2}{r} \).
2. \( v_{min} = \sqrt{gr} \).

Teacher's Note:
1. Critical speed at the top is √(gr); at the bottom it must be at least √(5gr) for a string.

Answer: (b) 5 m/s

Explanation:
1. \( T - mg = \dfrac{mv^2}{r} \Rightarrow 52 - 2 = \dfrac{0.2v^2}{0.1} \).
2. \( v^2 = 25 \Rightarrow v = 5 \) m/s.

Teacher's Note:
1. At the bottom the net upward force T - mg supplies the centripetal force. Take care to convert all quantities to SI units.

Answer: (c) \( \sqrt{2g\ell(1-\cos\theta)} \)

Explanation:
1. The bob falls a height \( h = \ell(1 - \cos\theta) \).
2. \( \tfrac{1}{2}mv^2 = mgh \Rightarrow v = \sqrt{2g\ell(1-\cos\theta)} \).

Teacher's Note:
1. Find the vertical drop from the geometry, then use energy conservation.

Answer: (c) At the extreme positions

Explanation:
1. \( T - mg\cos\theta = mv^2/\ell \).
2. T = mg cos θ only when v = 0, which happens at the extreme positions.

Teacher's Note:
1. At the extremes the bob is momentarily at rest, so the radial equation has no centripetal term.

Answer: (a) \( \dfrac{ML\omega^2}{2} \)

Explanation:
1. The liquid's centre of mass is at \( L/2 \) from the axis.
2. Force at the far end \( = M\omega^2\dfrac{L}{2} = \dfrac{ML\omega^2}{2} \).

Teacher's Note:
1. For a uniform distribution, use the centre of mass as the radius. The end wall supplies the centripetal force.

Answer: (b) 0, 10 m/s

Explanation:
1. After a full lap the displacement is zero, so the average velocity is zero.
2. Average speed \( = \dfrac{2\pi\times 100}{62.8} = 10 \) m/s.

Teacher's Note:
1. Average velocity uses displacement; average speed uses path length. They differ for a complete circle.

Answer: (c) 14 m/s and 15 m/s

Explanation:
1. Weightlessness needs \( mg = \dfrac{mv^2}{r} \Rightarrow v = \sqrt{gr} = \sqrt{200} \).
2. \( v \approx 14.1 \) m/s, which lies between 14 and 15 m/s.

Teacher's Note:
1. Weightlessness means the normal force is zero, so gravity alone provides the centripetal force.

Answer: (d) \( r \le \dfrac{\mu g}{\omega^2} \)

Explanation:
1. Friction supplies the centripetal force: \( m\omega^2 r \le \mu mg \).
2. \( r \le \dfrac{\mu g}{\omega^2} \).

Teacher's Note:
1. The coin stays only if the needed centripetal force does not exceed maximum static friction. Coins near the edge slip first.

Answer: (d) 5 m/s²

Explanation:
1. \( \omega = \dfrac{2\pi}{0.2\pi} = 10 \) rad/s.
2. \( a = \omega^2 r = 100\times 0.05 = 5 \) m/s².

Teacher's Note:
1. Convert 5 cm to 0.05 m. The π cancels neatly in the angular velocity.

Answer: (d) 30 m/s

Explanation:
1. \( v = \sqrt{rg\tan\theta} = \sqrt{90\times 10\times 1} = 30 \) m/s.

Teacher's Note:
1. On a frictionless banked road tan θ = v²/rg. The mass of the car is not needed.

Answer: (b) \( \sqrt{\mu_s Rg} \)

Explanation:
1. \( \dfrac{mv^2}{R} \le \mu_s mg \).
2. \( v_{max} = \sqrt{\mu_s Rg} \).

Teacher's Note:
1. The mass cancels. Remember v_max = √(μRg) for a flat curve.

Answer: (b) 2

Explanation:
1. Equal forces: \( \dfrac{m v_1^2}{r} = \dfrac{2m v_2^2}{r/2} \Rightarrow v_1^2 = 4v_2^2 \).
2. \( v_1 = 2v_2 \), so n = 2.

Teacher's Note:
1. Write the centripetal force for each stone and equate. Be careful to put the lighter stone's speed on the left.

Answer: (d) The magnitude of the velocity of the particle is 8 m/s

Explanation:
1. \( |\vec R| = 4 \) m for all t: circular path. The angular speed is \( 2\pi \) rad/s.
2. \( v = \omega R = 8\pi \) m/s, not 8 m/s. Acceleration \( = -\omega^2\vec R \) (towards the centre).
3. Hence the statement about 8 m/s is wrong.

Teacher's Note:
1. Differentiate to get speed: v = ωR = 8π m/s. The factor π is often forgotten.

Answer: (a) 0.1 m/s²

Explanation:
1. \( \tfrac{1}{2}mv^2 = 8\times 10^{-4} \Rightarrow v^2 = 0.16 \) m²/s².
2. Distance in two revolutions \( s = 2\times 2\pi\times 0.064 \approx 0.804 \) m.
3. \( a_t = \dfrac{v^2}{2s} = \dfrac{0.16}{1.608} \approx 0.1 \) m/s².

Teacher's Note:
1. Use v² = 2a_t s where s is the arc length. The kinetic energy gives v² directly.

Answer: (d) \( \sqrt{5gR} \)

Explanation:
1. At the top, \( v_{top}^2 \ge gR \).
2. Energy: \( \tfrac{1}{2}mu^2 = \tfrac{1}{2}mv_{top}^2 + mg(2R) \Rightarrow u^2 \ge 5gR \).

Teacher's Note:
1. Memorise √(5gR) at the bottom, √(gR) at the top. A rod instead of a string needs only √(4gR) at the bottom.

Answer: (b) \( \sqrt{gR\dfrac{\mu_s + \tan\theta}{1 - \mu_s\tan\theta}} \)

Explanation:
1. At maximum speed friction acts down the slope: \( N\cos\theta = mg + \mu N\sin\theta \) and \( N\sin\theta + \mu N\cos\theta = \dfrac{mv^2}{R} \).
2. Dividing: \( v_{max}^2 = Rg\,\dfrac{\mu_s + \tan\theta}{1 - \mu_s\tan\theta} \).

Teacher's Note:
1. Check dimensions: v² must have the dimension of Rg, which rules out the other options. Setting μ = 0 gives the frictionless result.

Answer: (a) 5.7 m/s

Explanation:
1. The radial (centripetal) component is \( a\cos30^\circ = 15\times 0.866 = 12.99 \) m/s².
2. \( v^2 = a_c R = 12.99\times 2.5 = 32.5 \Rightarrow v \approx 5.7 \) m/s.

Teacher's Note:
1. Only the radial component of the total acceleration equals v²/R. The tangential component changes the speed.

Answer: (c) Velocity is perpendicular to r and acceleration is directed towards the origin

Explanation:
1. \( \vec v = \omega(-\sin\omega t\,\hat x + \cos\omega t\,\hat y) \), so \( \vec v\cdot\vec r = 0 \).
2. \( \vec a = -\omega^2\vec r \), directed towards the origin.

Teacher's Note:
1. Position vectors of this form describe uniform circular motion. Differentiate twice to see that a = -ω²r.

Answer: (a) 7.5 s

Explanation:
1. Let the interval start at \( t_0 \), when \( \omega_0 = 2t_0 \).
2. \( 100 = \omega_0(5) + \tfrac{1}{2}(2)(5)^2 = 10t_0 + 25 \).
3. \( t_0 = 7.5 \) s.

Teacher's Note:
1. The speed at the start of the interval is not zero because the particle has been accelerating. Use θ = ω₀t + ½αt² for the interval.

Answer: (d) \( \tfrac{5}{4}D \)

Explanation:
1. For a body just completing a loop of radius R = D/2: \( h = \tfrac{5}{2}R \).
2. \( h = \tfrac{5}{2}\times\tfrac{D}{2} = \tfrac{5}{4}D \).

Teacher's Note:
1. Use h = 2.5R for the minimum starting height. Convert radius to diameter carefully.

Answer: (a) 7 s

Explanation:
1. Speed at the start of the interval \( \omega_1 = 3 + 3t_0 \).
2. \( 120 = 4\omega_1 + \tfrac{1}{2}(3)(16) = 12 + 12t_0 + 24 \).
3. \( t_0 = 7 \) s.

Teacher's Note:
1. First find the angular speed at the start of the interval. Then use the angle formula for the 4 s window.

Answer: (c) 24.20 m

Explanation:
1. With \( v^2 = rg\tan\theta \) and tan θ fixed, \( r \propto v^2 \).
2. \( r' = 20\times(1.1)^2 = 24.2 \) m.

Teacher's Note:
1. Radius varies as the square of the speed at a fixed banking angle. A 10% speed increase means a 21% radius increase.