NEET Angular Momentum and Conservation of Angular Momentum MCQs & PYQs

Practice NEET Angular Momentum and Conservation of Angular Momentum Questions

Question 21:

easy

A force $\vec{F} = \alpha\hat{i} + 3\hat{j} + 9\hat{k}$ is acting at a point $\vec{r} = 2\hat{i} – 6\hat{j} – 12\hat{k}$. The value of $\alpha$ for which angular momentum about origin is conserved is: (2015 Re)

For angular momentum to be conserved, torque $\vec{\tau} = \vec{r} \times \vec{F}$ must be zero, meaning $\vec{r}$ and $\vec{F}$ are collinear. Taking the ratio of their components: $\frac{2}{\alpha} = \frac{-6}{3} = \frac{-12}{9}$, which simplifies to $\frac{2}{\alpha} = -2$, giving $\alpha = -1$.

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Question 22:

easy

When a mass is rotating in a plane about a fixed point, its angular momentum is directed along

(2012 Pre)

Angular momentum is defined as $\vec{L} = \vec{r} \times \vec{p}$. According to the properties of the cross product, the vector $\vec{L}$ is directed perpendicular to the plane containing the position vector $\vec{r}$ and momentum vector $\vec{p}$.

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Question 23:

easy

A thin circular ring of mass $M$ and radius $r$ is rotating about its axis with constant angular velocity $\omega$. The objects each of mass $m$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with angular velocity given by:

(2010 Mains)

By conservation of angular momentum, $I_{1}\omega_{1} = I_{2}\omega_{2}$. Initially, $I_{1} = Mr^{2}$. Finally, the moment of inertia is $I_{2} = Mr^{2} + 2mr^{2} = (M+2m)r^{2}$. Equating the two yields $Mr^{2}\omega = (M+2m)r^{2}\omega_{2}$, so $\omega_{2} = \frac{M\omega}{M+2m}$.

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Question 24:

easy

A thin circular ring of mass $M$ and radius ‘$r$’ is rotating about its axis with a constant angular velocity $\omega$. Four objects each of mass $m$, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be:

(2003)

By conservation of angular momentum, $I_{initial}\omega_{initial} = I_{final}\omega_{final}$.
Initial moment of inertia $I_i = Mr^2$. Final moment of inertia $I_f = Mr^2 + 4mr^2$.
Thus, $$Mr^2 \omega = (M + 4m)r^2 \omega' \implies \omega' = \frac{M\omega}{M + 4m}$$.

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Question 25:

easy

A disc is rotating with angular speed $\omega$. If a child sits on it, what is conserved:

(2002)

When the child sits on the rotating disc gently, no external torque acts on the system.
According to Newton's second law for rotation, if net external torque is zero, the total angular momentum of the system remains conserved.

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Question 26:

easy

A circular ring of mass $M$ and radius $R$ is rotating about its axis with constant angular velocity $\omega$. Two particles each of mass $m$ are attached gently to the opposite ends of a diameter of the ring. The angular velocity of the ring will now become:

(1998)

Since the particles are attached gently, external torque is zero, meaning angular momentum is conserved.
$I_1\omega_1 = I_2\omega_2 \implies (MR^2)\omega = (MR^2 + 2mR^2)\omega'$.
Solving for the new angular velocity yields $\omega' = \frac{M\omega}{M + 2m}$.

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