Q.When a torque acting on a system is zero, which of the following is constant?

The Reason states the wrong conserved quantity: it's angular momentum $L=I\omega$ that stays constant with no external torque, not $\omega$ itself.

Assertion (A): Tornado wind speeds genuinely are alarmingly high — this is an observed, true fact.

Reason (R): It claims that with no external torque, angular velocity $\omega$ remains constant. This is FALSE. The actual conservation law is for angular momentum, $L=I\omega$: with no external torque, $L$ stays constant, but $\omega$ itself is free to change if the moment of inertia $I$ changes.

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If earth's radius halves (mass unchanged), the day shortens from 24 hours to 6 hours by conservation of angular momentum.

Treating the earth as a uniform sphere, moment of inertia $I = \dfrac{2}{5}MR^2$. If $R \to R/2$ while $M$ stays the same:

$I_{new} = \dfrac{2}{5}M\left(\dfrac{R}{2}\right)^2 = \dfrac{1}{4}\cdot\dfrac{2}{5}MR^2 = \dfrac{I_{old}}{4}$

With no external torque, angular momentum $L = I\omega$ is conserved:

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Just as zero net force keeps linear momentum constant, zero net external torque keeps angular momentum constant -- this is the law of conservation of angular momentum.

The rotational analogue of Newton's second law is:

tau (net torque) = dL/dt (rate of change of angular momentum)

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The stone's speed, and therefore its kinetic energy, remains constant because the string tension is always perpendicular to the velocity and does no work; angular momentum about the fixed axis is NOT conserved here since the effective radius keeps shrinking while the speed stays the same.

As the stone winds around the peg, at every instant its path is a curve (an involute of the peg's circle) and the string is tangent to the peg at the point where it currently leaves the peg. The stone's instantaneous velocity is always perpendicular to the (locally straight) string, because the string is the only thing constraining its motion and it can only pull, never push.

A force that is always perpendicular to the velocity of a particle does zero work on it (dW = F.v dt = 0, since F is perpendicular to v). So the tension does no work on the stone as it winds in — the stone's speed does not change, and hence its kinetic energy remains constant throughout the process.

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Torque plays the same role for angular momentum that force plays for linear momentum: zero net torque means angular momentum is conserved.

Newton's second law for rotation states:

tau (net torque) = dL/dt

where L is the angular momentum of the system. If the net external torque tau = 0, then dL/dt = 0, which means L is constant in time - this is the law of conservation of angular momentum.

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By the law of conservation of angular momentum, L = I x omega stays constant exactly when the net external torque on the system is zero, since (rate of change of L) = (net external torque) -- this is analogous to how linear momentum is conserved when net external force is zero.

Just as Newton's second law in linear form gives F = dp/dt, its rotational analogue is:

tau_ext = dL/dt

where tau_ext is the net external torque and L is the angular momentum.

If the net external torque acting on a system is zero (tau_ext = 0), then dL/dt = 0, which means L is constant -- angular momentum is conserved. This is the principle behind phenomena like a spinning ice-skater speeding up when pulling her arms in (no external torque, so L = I x omega stays fixed even as I decreases and omega increases).

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The rotational analogue of Newton's first law says angular momentum is conserved when net external torque is zero; since angular momentum is not listed among the choices, the correct pick here is 'none of these'.

Just as linear momentum is conserved when net force is zero, angular momentum L is conserved when net torque tau is zero: if tau = dL/dt = 0, then L is constant.

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