JEE Physics Circular Motion MCQs Set 19

Answer: (c) \( 20\pi\text{ rad s}^{-1} \)

Explanation:
1. 600 rev/min = 10 rev/s.
2. \( \omega = 2\pi f = 2\pi\times10 = 20\pi \) rad/s.

Teacher's Note:
1. Convert rpm to rev/s by dividing by 60 before using \( \omega = 2\pi f \).
2. One revolution is \( 2\pi \) rad.

Answer: (d) \( 2.99\times10^{4}\text{ m s}^{-1} \)

Explanation:
1. \( T = 365\times24\times3600 \) s \( = 3.15\times10^{7} \) s.
2. \( v = 2\pi r/T = 2\pi\times1.5\times10^{11}/3.15\times10^{7} \approx 2.99\times10^{4} \) m/s.

Teacher's Note:
1. Use v = 2 pi r / T for uniform circular motion.
2. Earth's orbital speed of about 30 km/s is a useful number to remember.

Answer: (c) 3.07 s

Explanation:
1. \( a = 4\pi^2 r/T^2 = 3g \).
2. \( T = 2\pi\sqrt{r/3g} = 2\pi\sqrt{7/29.4} = 3.07 \) s.

Teacher's Note:
1. Centripetal acceleration can be written as v²/r, ω²r or 4π²r/T².
2. Choose the form that uses the quantity given in the problem.

Answer: (c) 3.5 m s⁻² downwards

Explanation:
1. \( a = v^2/r = 49/14 = 3.5 \) m/s².
2. At the topmost point the centre of the wheel lies below the passenger, so the acceleration is directed downwards.

Teacher's Note:
1. Uniform circular motion has only a centripetal acceleration, always pointing to the centre.
2. Check the direction as well as the magnitude.

Answer: (b) \( 5.7\times10^{-2}\text{ N} \)

Explanation:
1. \( \omega = 2\pi\times1 = 6.283 \) rad/s.
2. \( F = m\omega^2 r = 0.018\times39.48\times0.08 = 5.7\times10^{-2} \) N.

Teacher's Note:
1. Convert grams to kilograms and centimetres to metres first.
2. The friction between coin and disc supplies this force.

Answer: (d) 0.8

Explanation:
1. Friction supplies the centripetal force: \( \mu mg = mv^2/r \).
2. \( \mu = v^2/(rg) = 156.25/196 = 0.80 \).

Teacher's Note:
1. The mass cancels in the expression μ = v²/(rg).
2. Larger speed or smaller radius needs more friction.

Answer: (b) \( \tan^{-1}(9/40) \)

Explanation:
1. 54 km/h = 15 m/s.
2. \( \tan\theta = v^2/(rg) = 225/1000 = 9/40 \), so \( \theta = \tan^{-1}(9/40) \).

Teacher's Note:
1. For a frictionless banked road, tan θ = v²/(rg).
2. Always convert km/h to m/s by multiplying by 5/18.

Answer: (c) \( 2.0625\times10^{5}\text{ N upward} \)

Explanation:
1. 900 km/h = 250 m/s.
2. At the lowest point \( F - mg = mv^2/r \), so \( F = 5000\times250^2/2000 + 5000\times10 = 1.5625\times10^5 + 0.5\times10^5 = 2.0625\times10^5 \) N, directed upwards.

Teacher's Note:
1. At the lowest point the net force points to the centre, i.e. upwards, so the air force exceeds the weight.
2. Do not forget to add mg.

Answer: (c) \( mg\cos\theta_0 \)

Explanation:
1. At the extreme position the speed is zero, so the centripetal force required is zero.
2. Along the string: \( T - mg\cos\theta_0 = mv^2/r = 0 \), so \( T = mg\cos\theta_0 \).

Teacher's Note:
1. At the turning point the tension only balances the radial component of weight.
2. The tension is largest at the mean position, not at the extreme.

Answer: (d) 8.6 m s⁻¹

Explanation:
1. Minimum speed at the lowest point is \( \sqrt{5gr} \).
2. \( v = \sqrt{5\times9.8\times1.5} = \sqrt{73.5} \approx 8.6 \) m/s.

Teacher's Note:
1. Minimum speed at the top is √(gr); at the bottom it is √(5gr).
2. Energy conservation links the two values.

Answer: (c) 3

Explanation:
1. \( \theta_1 = \tfrac12\alpha(2)^2 = 2\alpha \).
2. Angle in 4 s is \( \tfrac12\alpha(4)^2 = 8\alpha \), so \( \theta_2 = 6\alpha \) and \( \theta_2/\theta_1 = 3 \).

Teacher's Note:
1. Distances in successive equal intervals from rest are in the ratio 1 : 3 : 5.
2. The same rule holds for angles under constant angular acceleration.

Answer: (a) 1.5 m s⁻¹

Explanation:
1. \( \sin\theta = 0.3/0.5 = 0.6 \), so \( \tan\theta = 0.75 \).
2. \( v^2 = rg\tan\theta = 0.3\times9.8\times0.75 = 2.205 \), so \( v \approx 1.5 \) m/s.

Teacher's Note:
1. For a conical pendulum, T sin θ = mv²/r and T cos θ = mg.
2. Dividing the two gives tan θ = v²/(rg).

Answer: (d) 50 rpm

Explanation:
1. Slipping condition: \( \mu g = \omega^2 r \).
2. This does not involve the mass, so two coins slip at the same speed, 50 rpm.

Teacher's Note:
1. The limiting speed depends on μ and r only.
2. Adding mass adds equal amounts to both friction and required centripetal force.

Answer: (a) \( \sqrt{98}\text{ m s}^{-1} \)

Explanation:
1. \( T_{max} = mv^2/r + mg \), \( T_{min} = mv^2/r - mg \).
2. \( 3(v^2/r + g) = 5(v^2/r - g) \), so \( v^2 = 4gr = 4\times9.8\times2.5 = 98 \) and \( v = \sqrt{98} \) m/s.

Teacher's Note:
1. Maximum tension occurs at the lowest point and minimum at the highest point.
2. Set up the ratio and solve for v²/r in terms of g.

Answer: (d) \( \pi\sqrt2/30\text{ cm/s} \)

Explanation:
1. \( v = \omega r = (2\pi/60)(1) = \pi/30 \) cm/s.
2. In 15 s the hand turns \( 90^\circ \), so \( |\Delta v| = \sqrt2\, v = \pi\sqrt2/30 \) cm/s.

Teacher's Note:
1. For a turn of angle θ, |Δv| = 2v sin(θ/2).
2. Here θ = 90°, so |Δv| = √2 v.

Answer: (c) \( 10\pi\text{ rad} \)

Explanation:
1. 300 rpm = 5 rev/s.
2. Angle in 1 s \( = 5\times2\pi = 10\pi \) rad.

Teacher's Note:
1. Angle = ωt, with ω in rad/s.
2. Each revolution is 2π rad.

Answer: (d) The change in kinetic energy is zero

Explanation:
1. The speed is constant, so the kinetic energy does not change.
2. The change in momentum is \( 2mv \), not \( mvR \).

Teacher's Note:
1. Kinetic energy is a scalar and depends only on speed.
2. Momentum changes because its direction reverses.

Answer: (c) 3mg

Explanation:
1. Speed at mean position: \( v^2 = 2gl \).
2. \( T = mg + mv^2/l = mg + 2mg = 3mg \).

Teacher's Note:
1. Tension at the bottom for a release from angle θ0 is mg(3 - 2cos θ0).
2. For θ0 = 90° this gives 3mg.

Answer: (a) 60.075 kg-wt

Explanation:
1. The balance supplies both weight support and centripetal force: \( F = m\sqrt{g^2 + (v^2/r)^2} \).
2. \( v^2/r = 0.49 \), so \( F/\,\text{weight} = \sqrt{1+(0.49/9.8)^2} = 1.00125 \), giving 60.075 kg-wt.

Teacher's Note:
1. The two perpendicular components, mg and mv²/r, add vectorially.
2. The centripetal part is tiny, so the reading is only slightly above 60.

Answer: (d) \( m\left(g+\dfrac{\pi^2n^2r}{900}\right) \)

Explanation:
1. \( \omega = 2\pi n/60 = \pi n/30 \) rad/s.
2. At the lowest point \( T = mg + m\omega^2 r = m\left(g + \pi^2n^2 r/900\right) \).

Teacher's Note:
1. Convert revolutions per minute into rad/s before substituting.
2. At the bottom the tension must exceed the weight.