JEE Physics Circular Motion MCQs Set 02

Answer: (d) No work gets done on it

Explanation:
1. The speed is constant but the direction of the velocity changes, so there is a centripetal acceleration and a net force.
2. The net force is perpendicular to the velocity, so the work done on the body is zero.

Teacher's Note:
1. Constant speed in a circle still needs a net force towards the centre. That force does no work.

Answer: (d) Away from the centre of the path

Explanation:
1. In the car's frame the bob experiences a pseudo (centrifugal) force away from the centre.
2. The string tilts until the tension's horizontal component supplies the centripetal force; the bob therefore swings outwards, away from the centre.

Teacher's Note:
1. The bob lags behind the car's inward turn, so it appears to move outwards. In the ground frame the tension's horizontal part is the centripetal force.

Answer: (a) 100 rad/s

Explanation:
1. \( T = m\omega^2 r \Rightarrow 100 = 0.1\times\omega^2\times 0.1 \).
2. \( \omega^2 = 10^4 \Rightarrow \omega = 100 \) rad/s.

Teacher's Note:
1. Convert the mass to kilograms. The breaking tension gives the maximum angular speed.

Answer: (d) 2F

Explanation:
1. \( F = m\omega^2 r \) with constant ω (constant frequency).
2. Doubling r doubles F.

Teacher's Note:
1. At constant frequency the force is proportional to the radius. At constant speed it would be inversely proportional.

Answer: (b) 20 N

Explanation:
1. \( F = \dfrac{mv^2}{r} = \dfrac{0.5\times 16}{0.4} = 20 \) N.

Teacher's Note:
1. Substitute directly into F = mv²/r. Check units: kg m²/s² per m gives newton.

Answer: (d) a, c

Explanation:
1. The centripetal force \( mv^2/r \) depends on \( v^2 \), so its magnitude is unchanged.
2. It still points towards the centre, so its direction is also unchanged.
3. Statements (a) and (c) are correct.

Teacher's Note:
1. Reversing the sense of rotation reverses the velocity but not the radial direction of the force. Always compare vectors with care.

Answer: (c) a_r ≠ 0 but a_t = 0

Explanation:
1. Circular motion needs a radial acceleration v²/r, so a_r ≠ 0.
2. Constant speed means there is no tangential acceleration, a_t = 0.

Teacher's Note:
1. Uniform circular motion has only the centripetal component. Any tangential part would change the speed.

Answer: (a) Perpendicular to each other

Explanation:
1. The velocity is tangential and the acceleration is radial (towards the centre).
2. A tangent is perpendicular to the radius, so the two vectors are perpendicular.

Teacher's Note:
1. This is why the speed does not change: acceleration has no component along the velocity.

Answer: (a) 20 rad/s

Explanation:
1. \( T_{max} = m\omega^2 r \Rightarrow 10 = 0.25\times\omega^2\times 0.1 \).
2. \( \omega^2 = 400 \Rightarrow \omega = 20 \) rad/s.

Teacher's Note:
1. Convert grams to kilograms and centimetres to metres. The maximum tension limits ω.

Answer: (a) \( 36\times 10^{21} \) N

Explanation:
1. \( r = 1.5\times 10^{11} \) m.
2. \( F = m\omega^2 r = 6\times 10^{24}\times 4\times 10^{-14}\times 1.5\times 10^{11} = 36\times 10^{21} \) N.

Teacher's Note:
1. Convert kilometres to metres before computing. The gravitational pull of the sun supplies the centripetal force.

Answer: (b) 1000 N

Explanation:
1. \( v = 36\times\tfrac{5}{18} = 10 \) m/s.
2. \( F = \dfrac{mv^2}{r} = \dfrac{500\times 100}{50} = 1000 \) N.

Teacher's Note:
1. Convert km/h to m/s by multiplying by 5/18 before using F = mv²/r.

Answer: (c) 4 times

Explanation:
1. \( F = mv^2/r \propto v^2 \).
2. Doubling the velocity makes the force 4 times.

Teacher's Note:
1. Because of the v², a small increase in speed has a large effect on the force needed on a curve.

Answer: (a) Centripetal force

Explanation:
1. A net force directed towards the centre is needed to change the direction of the velocity.
2. This is the centripetal force.

Teacher's Note:
1. Centrifugal force is a pseudo force that appears only in the rotating frame. It is not needed in the ground frame.

Answer: (b) Due to lack of proper centripetal force

Explanation:
1. On a flat road the centripetal force is provided by static friction.
2. If the speed is too high, the required force mv²/r exceeds the maximum friction and the car skids out.

Teacher's Note:
1. The car does not get 'thrown out' by a new force. It simply moves straight along the tangent when friction cannot supply enough centripetal force.

Answer: (a) \( \dfrac{v^2b}{Rg} \)

Explanation:
1. For safe banking without friction, \( \tan\theta = \dfrac{v^2}{Rg} \).
2. For a small angle, \( \tan\theta \approx \dfrac{h}{b} \), so \( h = \dfrac{v^2b}{Rg} \).

Teacher's Note:
1. The banking angle is found from h/b. This is the standard formula for the height of the outer edge.

Answer: (c) \( \tan\theta = \dfrac{r\omega^2}{g} \)

Explanation:
1. Horizontal component of tension gives the centripetal force: \( T\sin\theta = m\omega^2 r \).
2. Vertical component balances the weight: \( T\cos\theta = mg \).
3. Dividing, \( \tan\theta = \dfrac{\omega^2 r}{g} \).

Teacher's Note:
1. Resolve the tension into horizontal and vertical parts and divide the two equations. This is the same idea as the conical pendulum.

Answer: (a) \( \sqrt{rg\tan\theta} \)

Explanation:
1. \( T\sin\theta = \dfrac{mv^2}{r} \) and \( T\cos\theta = mg \).
2. Dividing: \( \tan\theta = \dfrac{v^2}{rg} \Rightarrow v = \sqrt{rg\tan\theta} \).

Teacher's Note:
1. Horizontal component gives the centripetal force and vertical component balances the weight. Dividing eliminates the tension.

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

Explanation:
1. Energy conservation: \( mgL = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2gL} \).
2. At the lowest point \( T - mg = \dfrac{mv^2}{L} = 2mg \Rightarrow T = 3mg \).

Teacher's Note:
1. Learn this result: tension 3mg at the bottom when released from the horizontal. It appears often in JEE problems.

Answer: (a) 6 N

Explanation:
1. At the top, \( T + mg = \dfrac{mv^2}{r} \).
2. \( T = \dfrac{1\times 16}{1} - 10 = 6 \) N.

Teacher's Note:
1. At the top both the tension and the weight point towards the centre. Subtract mg from mv²/r.

Answer: (a) At the highest point

Explanation:
1. At the top, \( T + mg = mv^2/r \). If \( v^2 = gr \), then T = 0.
2. At lower points the tension must also support part of the weight, so it cannot be zero.

Teacher's Note:
1. The tension is smallest at the top, so that is where it can first vanish. This gives the critical speed √(gr) there.

Answer: (c) The string is vertical with the stone at the lowest position

Explanation:
1. \( T = \dfrac{mv^2}{r} + mg\cos\theta \) (θ from the lowest point), and v is largest at the bottom.
2. Both terms are largest at the lowest point.

Teacher's Note:
1. The stone moves fastest at the bottom and the weight also adds to the required tension there.

Answer: (a) 5 rad/s

Explanation:
1. The tension is largest at the lowest point: \( T - mg = m\omega^2 r \).
2. \( 30 - 5 = 0.5\times\omega^2\times 2 \Rightarrow \omega^2 = 25 \Rightarrow \omega = 5 \) rad/s.

Teacher's Note:
1. Always apply the breaking condition at the point where the tension is greatest.

Answer: (b) x = 2L

Explanation:
1. At the top the minimum speed is \( v = \sqrt{gL} \), horizontal.
2. The body falls 2L: \( 2L = \tfrac{1}{2}gt^2 \Rightarrow t = 2\sqrt{L/g} \).
3. \( x = vt = \sqrt{gL}\times 2\sqrt{L/g} = 2L \).

Teacher's Note:
1. After the string breaks, treat the motion as a horizontal projectile from height 2L. The minimum speed at the top is √(gL).

Answer: (b) T₁ > T₂

Explanation:
1. Tension \( T = \dfrac{mv^2}{r} + mg\cos\theta \) with θ measured from the lowest point.
2. At 30° the speed is greater and cos θ is larger than at 60°, so T₁ > T₂.

Teacher's Note:
1. Tension decreases steadily as the particle rises from the bottom. A smaller angle from the vertical means a greater tension.

Answer: (c) 3g

Explanation:
1. At the top \( v^2 = gr \). When the string is horizontal, the body has dropped through a height r: \( v'^2 = gr + 2gr = 3gr \).
2. \( a_c = \dfrac{v'^2}{r} = 3g \).

Teacher's Note:
1. Use energy conservation from the top to the horizontal position. The centripetal acceleration is v²/r.