JEE Physics Circular Motion MCQs Set 06

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

Explanation:
1. At the minimum speed the normal reaction at the top is zero, so \( mg = \dfrac{mv^2}{r} \).
2. \( v = \sqrt{gr} \).

Teacher's Note:
1. The condition at the top is that gravity alone gives the centripetal force.
2. This is the critical speed.

Answer: (a) \( \sqrt{2gL} \) and 3mg

Explanation:
1. Energy conservation over a fall of L: \( \tfrac{1}{2}mv^2 = mgL \Rightarrow v = \sqrt{2gL} \).
2. At the bottom \( T - mg = \dfrac{mv^2}{L} = 2mg \), so \( T = 3mg \).

Teacher's Note:
1. Remember the result: speed \( \sqrt{2gL} \) and tension 3mg from the horizontal.
2. Check the dimension of each quantity in the options.

Answer: (b) \( T_1 \gt T_2 \)

Explanation:
1. Tension at angle \( \theta \) from the lowest point: \( T = \dfrac{mu^2}{r} + mg(3\cos\theta - 2) \), where u is the speed at the bottom.
2. \( \cos\theta \) is larger at 15° than at 45°, so \( T_1 \gt T_2 \).

Teacher's Note:
1. Tension falls steadily as the particle climbs from the bottom to the top.
2. A smaller angle from the vertical means a larger tension.

Answer: (b) 1 : 3

Explanation:
1. Bottom: \( T_b = mg + \dfrac{m(8gR)}{R} = 9mg \).
2. Top speed: \( v^2 = 8gR - 4gR = 4gR \), so \( T_t = \dfrac{m(4gR)}{R} - mg = 3mg \).
3. Ratio \( T_t : T_b = 3 : 9 = 1 : 3 \).

Teacher's Note:
1. Use energy conservation to get the speed at the top.
2. Subtract weight at the top, add weight at the bottom.

Answer: (a) 3 m/s

Explanation:
1. At the lowest point, \( T - mg = \dfrac{mv^2}{r} \).
2. \( 30 - 3 = \dfrac{0.3\,v^2}{0.1} = 3v^2 \), so \( v^2 = 9 \) and \( v = 3 \) m/s.

Teacher's Note:
1. Net upward force equals the centripetal force at the bottom.
2. Weight is 0.3 \( \times \) 10 = 3 N.

Answer: (d) 1.8 m

Explanation:
1. After the string breaks, the mass leaves along the tangent, which is vertical, with speed 6 m/s.
2. It rises \( h = \dfrac{v^2}{2g} = \dfrac{36}{20} = 1.8 \) m.

Teacher's Note:
1. A body released from a circle flies off along the tangent.
2. At the level of the centre the tangent is vertical.

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

Explanation:
1. The height dropped is \( \ell(1-\cos\theta) \).
2. \( \tfrac{1}{2}mv^2 = mg\ell(1-\cos\theta) \Rightarrow v = \sqrt{2g\ell(1-\cos\theta)} \).

Teacher's Note:
1. Height is measured from the lowest point, not from the support.
2. At \( \theta = 90^\circ \) this gives \( \sqrt{2g\ell} \).

Answer: (b) \( \dfrac{1}{2} \) kg

Explanation:
1. The difference between the tension at the bottom and at the top is always 6 mg.
2. \( 6m = 3 \) kg wt, so \( m = \dfrac{1}{2} \) kg.

Teacher's Note:
1. The 6mg difference does not depend on the radius or speed.
2. Remember it for vertical circle problems.

Answer: (b) x = 2L

Explanation:
1. Minimum speed at the top: \( v = \sqrt{gL} \), horizontal.
2. The fall is 2L, so \( t = \sqrt{\dfrac{4L}{g}} \).
3. \( x = vt = \sqrt{gL}\cdot\sqrt{\dfrac{4L}{g}} = 2L \).

Teacher's Note:
1. A horizontal projectile covers range = speed times time of fall.
2. The time of fall comes from \( 2L = \tfrac{1}{2}gt^2 \).

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

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

Teacher's Note:
1. Treat the liquid as a point mass at its centre of mass.
2. Check dimensions: force is mass times length times \( \omega^2 \).

Answer: (b) 0, 10 m/s

Explanation:
1. In one full lap the displacement is zero, so the average velocity is 0.
2. Average speed \( = \dfrac{2\pi\times 100}{62.8} = 10 \) m/s.

Teacher's Note:
1. Average velocity uses displacement and average speed uses distance.
2. They differ for a closed path.

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

Explanation:
1. Weightlessness means the normal reaction is zero, so \( mg = \dfrac{mv^2}{r} \).
2. \( v = \sqrt{gr} = \sqrt{200} \approx 14.1 \) m/s, which lies between 14 and 15 m/s.

Teacher's Note:
1. Zero normal force means gravity alone gives the centripetal force.
2. \( \sqrt{200} \) is just above 14.

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

Explanation:
1. The coin stays if the required centripetal force does not exceed the maximum friction: \( m\omega^2 r \le \mu mg \).
2. So \( r \le \dfrac{\mu g}{\omega^2} \).

Teacher's Note:
1. Coins near the centre stay on, and those near the edge slide off.
2. The mass cancels.

Answer: (d) 30 m/s

Explanation:
1. \( v^2 = rg\tan\theta = 90\times 10\times 1 = 900 \).
2. \( v = 30 \) m/s.

Teacher's Note:
1. The mass of the car does not matter.
2. At 45° the safe speed is simply \( \sqrt{rg} \).

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

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

Teacher's Note:
1. Maximum friction provides the centripetal force.
2. The mass cancels.

Answer: (b) 2

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

Teacher's Note:
1. Equate the two centripetal forces and cancel common factors.
2. Take the square root at the end.

Answer: (a) 0.1 m/s\( ^2 \)

Explanation:
1. \( \tfrac{1}{2}mv^2 = 8\times10^{-4} \Rightarrow v^2 = \dfrac{2\times 8\times10^{-4}}{0.01} = 0.16 \) m\( ^2 \)/s\( ^2 \).
2. Distance \( s = 2\times 2\pi r = 4\pi\times 0.064 \approx 0.804 \) m.
3. \( a_t = \dfrac{v^2}{2s} = \dfrac{0.16}{1.608} \approx 0.1 \) m/s\( ^2 \).

Teacher's Note:
1. Use \( v^2 = 2a_t s \) along the arc.
2. Convert grams and centimetres to SI units first.

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

Explanation:
1. At the top the minimum speed is \( \sqrt{gR} \).
2. Energy conservation from the bottom: \( \tfrac{1}{2}mu^2 = \tfrac{1}{2}m(gR) + mg(2R) \Rightarrow u^2 = 5gR \).

Teacher's Note:
1. Remember: top speed \( \sqrt{gR} \), bottom speed \( \sqrt{5gR} \).
2. This holds for a string or the inside of a loop.

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

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

Teacher's Note:
1. The position vector traces a unit circle.
2. Differentiate twice and compare with \( \vec{r} \).

Answer: (c) 10 rad/s

Explanation:
1. The normal force provides the centripetal force: \( N = m\omega^2 r \).
2. Friction balances the weight: \( \mu N = mg \Rightarrow \mu m\omega^2 r = mg \).
3. \( \omega^2 = \dfrac{g}{\mu r} = \dfrac{10}{0.1\times 1} = 100 \), so \( \omega = 10 \) rad/s.

Teacher's Note:
1. The wall pushes inward to supply circular motion and friction carries the weight.
2. The mass cancels.

Answer: (c) \( \dfrac{V_0^2}{4\pi n r^2} \) rad/s\( ^2 \)

Explanation:
1. Angle turned in n rounds \( \theta = 2\pi n \), and \( \omega = V_0/r \).
2. \( \omega^2 = 2\alpha\theta \Rightarrow \dfrac{V_0^2}{r^2} = 2\alpha(2\pi n) \).
3. \( \alpha = \dfrac{V_0^2}{4\pi n r^2} \).

Teacher's Note:
1. Use the rotational equation \( \omega^2 = 2\alpha\theta \).
2. Convert the linear speed to angular speed with \( \omega = V_0/r \).

Answer: (c) 4 s

Explanation:
1. Let \( \omega_1 \) be the angular velocity at the start of the interval.
2. \( 72 = \omega_1(4) + \tfrac{1}{2}(2)(4)^2 = 4\omega_1 + 16 \Rightarrow \omega_1 = 14 \) rad/s.
3. \( \omega_1 = \omega_0 + \alpha t \Rightarrow 14 = 6 + 2t \Rightarrow t = 4 \) s.

Teacher's Note:
1. First find the angular velocity at the start of the interval.
2. Then use \( \omega = \omega_0 + \alpha t \).

Answer: (d) 43.2 m

Explanation:
1. \( v^2 = rg\tan\theta \Rightarrow r \propto v^2 \) for a fixed angle.
2. New \( r = 30\times (1.2)^2 = 30\times 1.44 = 43.2 \) m.

Teacher's Note:
1. Radius varies as the square of the safe speed.
2. A 20% rise in speed means a 44% rise in radius.

Answer: (b) 6 : 9 : 10

Explanation:
1. C is at distance \( 6\ell \) from O, so \( \omega = \dfrac{v_0}{6\ell} \).
2. Tension in BC supplies the centripetal force of C: \( T_{BC} = m\omega^2(6\ell) = 6m\omega^2\ell \).
3. \( T_{AB} = m\omega^2(3\ell + 6\ell) = 9m\omega^2\ell \) and \( T_{OA} = m\omega^2(\ell + 3\ell + 6\ell) = 10m\omega^2\ell \).
4. Ratio \( T_{BC} : T_{AB} : T_{OA} = 6 : 9 : 10 \).

Teacher's Note:
1. Each section of thread pulls all the particles beyond it.
2. Add the centripetal forces from the outer end inwards.

Answer: (d) \( \sqrt{2(u^2 - g\ell)} \)

Explanation:
1. At the horizontal position, \( v^2 = u^2 - 2g\ell \), and the velocity is vertical.
2. Initially the velocity u is horizontal, so the two velocities are perpendicular.
3. \( |\Delta\vec{v}| = \sqrt{u^2 + v^2} = \sqrt{2u^2 - 2g\ell} = \sqrt{2(u^2 - g\ell)} \).

Teacher's Note:
1. Velocity change is a vector difference, not a difference of speeds.
2. The two velocities are at right angles here.