Rotational Motion

Vectors are categorized into two types based on their symmetry :

• 1. Polar Vectors : These represent actual physical displacements. They flip their sign (direction) when the coordinate system is inverted ( $\vec{A} \rightarrow-\vec{A}$ ).

• 2. Axial Vectors (Pseudo-vectors) : These usually represent rotations or are the result of a cross product of two polar vectors. They do not flip their sign under coordinate inversion $(\overrightarrow{\mathrm{B}} \rightarrow \overrightarrow{\mathrm{B}})$.

Velocity is the rate of change of position: $\vec{v}=\frac{d \vec{r}}{d t}$

Since $\vec{r}$ changes to $-\vec{r}$, then $\frac{d(-\vec{r})}{d t}=-\vec{v}$.

So, when position vector changes sign then velocity vector also get flipped.

Momentum is defined as $\overrightarrow{\mathrm{p}}=\mathrm{m} \overrightarrow{\mathrm{v}}$.

Since $\overrightarrow{\mathrm{v}}$ flips to $-\overrightarrow{\mathrm{v}}$, then $\overrightarrow{\mathrm{p}}$ becomes $\mathrm{m}(-\overrightarrow{\mathrm{v}})=-\overrightarrow{\mathrm{p}}$.

So the linear momentum also flips if position vector changes sign.

Acceleration is the rate of change of velocity: $\vec{a}=\frac{d \vec{v}}{d t}$.

Since $\overrightarrow{\mathrm{v}}$ flips to $-\overrightarrow{\mathrm{v}}$, then $\frac{\mathrm{d}(-\overrightarrow{\mathrm{v}})}{\mathrm{dt}}=-\overrightarrow{\mathrm{a}}$.

So, the acceleration also gets flipped when position vector changes sign.

Angular momentum is defined by the cross product :

$\overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}}$.

If we apply the sign change :

$ \overrightarrow{\mathrm{L}^{\prime}}=(-\overrightarrow{\mathrm{r}}) \times(-\overrightarrow{\mathrm{p}}) $

$\Rightarrow $ $ \overrightarrow{\mathrm{L}^{\prime}}=(\overrightarrow{\mathrm{r}}) \times(\overrightarrow{\mathrm{p}}) $

So, it does not flip. It is an axial vector. Hence, the correct option is (D).

To find the angular acceleration $(\alpha)$, we use the formula :

$ \tau=\mathrm{I}_{\text {total }} \cdot \alpha \Rightarrow \alpha=\tau / \mathrm{I}_{\text {total }} $

where $\mathrm{I}_{\text {total }}$ is the total moment of inertia of the assembly about the axis AB .

The rod has mass $\mathrm{M}=600 \mathrm{~g}$ and length $\mathrm{L}=30 \mathrm{~cm}$. The axis AB is perpendicular to the rod at a distance of 10 cm from one end.

The centre of mass (CM) of the rod is at 15 cm from either end.

The distance (d) between the axis AB and the CM of the rod is: $\mathrm{d}=15 \mathrm{~cm}-10 \mathrm{~cm}=5 \mathrm{~cm}$.

Using the Parallel Axis Theorem :

$ I_{\text {rod }}=\frac{M^2}{12}+{M d}^2 $

$\Rightarrow $ $I_{\mathrm{rod}}=\frac{600 \times 30^2}{12}+600 \times 5^2$

$\Rightarrow $ $ I_{\mathrm{rod}}=(50 \times 900)+(600 \times 25)=60,000 \mathrm{~g} \cdot \mathrm{~cm}^2 $

The axis $A B$ is parallel to the diameter of the discs.

The moment of inertia of a disc about its diameter is $I_{\text {dia }}= \frac{1}{4} \mathrm{MR}^2$.

For the left disc, distance from axis $A B$ is $r_1=10 \mathrm{~cm}$.

$ I_1=\frac{1}{4} M R^2+M r_1^2=\frac{1}{4}(600)(10)^2+600(10)^2 $

$\Rightarrow $ $ \mathrm{I}_1=75000 \mathrm{~g} \cdot \mathrm{~cm}^2 $

For the right disc, distance from axis AB is $\mathrm{r}_2=20 \mathrm{~cm}$.

$ \mathrm{I}_2=\frac{1}{4} \mathrm{MR}^2+\mathrm{Mr}_2^2=\frac{1}{4}(600)(10)^2+600(20)^2 $

$ \mathrm{I}_2=255000 \mathrm{~g} \cdot \mathrm{~cm}^2 $

So, total moment of inertia is,

$ \mathrm{I}_{\text {total }}=\mathrm{I}_{\mathrm{rod}}+\mathrm{I}_1+\mathrm{I}_2 $

$\Rightarrow $ $I_{\text {total }}=60000+75000+255000=390000 \mathrm{~g} \cdot \mathrm{~cm}^2$

$\Rightarrow $ $ \mathrm{I}_{\text {total }}=3.9 \times 10^5 \mathrm{~g} \cdot \mathrm{~cm}^2 $

Given Torque $\tau=43 \times 10^5$ dyne $\cdot \mathrm{cm}$.

$ \alpha=\frac{\tau}{\mathrm{I}_{\text {total }}}=\frac{43 \times 10^5}{3.9 \times 10^5} $

$\Rightarrow $ $ \alpha=\approx 11.025 \mathrm{rad} / \mathrm{s}^2 $

The nearest integer value is 11 . Hence, the correct option is (D).

The rod is pivoted at a distance $\mathrm{L} / 3$ from one end. Using the parallel axis theorem, the moment of inertia about this pivot point: $\mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Md}^2$

Where, moment of inertia about the centre of mass $\mathrm{I}_{\mathrm{cm}}=\frac{\mathrm{ML}^2}{12}$ and distance between the center of mass and the pivot $\mathrm{d}=\frac{\mathrm{L}}{2}-\frac{\mathrm{L}}{3}=\frac{\mathrm{L}}{6}$.

$ \mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Md}^2=\frac{\mathrm{ML}^2}{12}+\mathrm{M}\left(\frac{\mathrm{~L}}{6}\right)^2 $

$ \mathrm{I}=\frac{\mathrm{ML}^2}{12}+\frac{\mathrm{ML}^2}{36}=\frac{3 \mathrm{ML}^2+\mathrm{ML}^2}{36}=\frac{4 \mathrm{ML}^2}{36}=\frac{\mathrm{ML}^2}{9} $

When the rod falls from a vertical to a horizontal position,

Initial height of COM (relative to the pivot) $\mathrm{h}_1=\frac{\mathrm{L}}{6}$ above the pivot.

Final height of COM, $\mathrm{h}_2=0$ (it is at the same horizontal level as the pivot when lying on the table).

Loss in Potential Energy:

$ \Delta U=M g \Delta h=M g\left(\frac{L}{6}\right) $

The loss in potential energy is converted entirely into rotational kinetic energy ( $\mathrm{K}_{\mathrm{rot}}$ ) at the moment of impact.

$ \Delta U=K_{\mathrm{rot}} $

$\Rightarrow $ $\operatorname{Mg}\left(\frac{L}{6}\right)=\frac{1}{2} I \omega^2$

$\Rightarrow \frac{\mathrm{MgL}}{6}=\frac{1}{2} \times\left(\frac{\mathrm{ML}^2}{9}\right) \omega^2$

$\Rightarrow \frac{\mathrm{g}}{6}=\frac{\mathrm{L} \omega^2}{18}$

$\begin{aligned} & \Rightarrow \omega^2=\frac{3 \mathrm{~g}}{\mathrm{~L}} \\ & \Rightarrow \omega=\sqrt{\frac{3 \mathrm{~g}}{\mathrm{~L}}}\end{aligned}$

The heavier mass starts from rest ( $u=0$ ) and falls a certain distance ( $h$ ) in time ( $t$ ).

Using the kinematic equation :

$ \mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2 $

$\Rightarrow $ $ \mathrm{h}=\frac{1}{2} \mathrm{at}^2 \Rightarrow \mathrm{a}=\frac{2 \mathrm{~h}}{\mathrm{t}^2} $

The given distance $\mathrm{h}=81 \mathrm{~cm}=0.81 \mathrm{~m}$

The time taken is $\mathrm{t}=9 \mathrm{~s}$

$ a=\frac{2 \times 0.81}{9^2}=\frac{1.62}{81} $

$\Rightarrow $ $ \mathrm{a}=0.02 \mathrm{~m} / \mathrm{s}^2 $

Since the pulley has mass (and inertia), the tension is not the same on both sides. Let $\mathrm{T}_1$ be the tension on the heavier mass side ( $\mathrm{m}_1$ ) and $\mathrm{T}_2$ be the tension on the lighter mass side ( $\mathrm{m}_2$ ).

For heavier mass ( $\mathrm{m}_1=0.4 \mathrm{~kg}$ ) which is moving down.

$ \mathrm{m}_1 \mathrm{~g}-\mathrm{T}_1=\mathrm{m}_1 \mathrm{a} $

$\Rightarrow $ $ \mathrm{T}_1=\mathrm{m}_1(\mathrm{~g}-\mathrm{a}) \ldots(\mathrm{i}) $

For lighter mass ( $\mathrm{m}_2=0.35 \mathrm{~kg}$ ) which is moving up.

$ \mathrm{T}_2-\mathrm{m}_2 \mathrm{~g}=\mathrm{m}_2 \mathrm{a} $

$\Rightarrow $ $ \mathrm{T}_2=\mathrm{m}_2(\mathrm{~g}+\mathrm{a}) \ldots \text { (ii) } $

The pulley rotates due to the difference in tensions. The net torque $(\tau)$ causes angular acceleration $(\alpha)$.

$ \tau=\left(\mathrm{T}_1-\mathrm{T}_2\right) \mathrm{R} $

Also, from Newton's Second Law for rotation:

$ \tau=\mathrm{I} \alpha $

Therefore :

$ \left(\mathrm{T}_1-\mathrm{T}_2\right) \mathrm{R}=\mathrm{I} \alpha $

Assuming the string does not slip, the linear acceleration of the string

$ \mathrm{a}=\mathrm{R} \alpha \Rightarrow \alpha=\frac{\mathrm{a}}{\mathrm{R}} $

Substituting in the torque equation :

$ \left(\mathrm{T}_1-\mathrm{T}_2\right) \mathrm{R}=\mathrm{I}\left(\frac{\mathrm{a}}{\mathrm{R}}\right) $

$ \mathrm{I}=\frac{\left(\mathrm{T}_1-\mathrm{T}_2\right) \mathrm{R}^2}{\mathrm{a}} \ldots (iii)$

On subtracting (ii) from (i)

$ \mathrm{T}_1-\mathrm{T}_2=\left[\mathrm{m}_1(\mathrm{~g}-\mathrm{a})\right]-\left[\mathrm{m}_2(\mathrm{~g}+\mathrm{a})\right] $

$\Rightarrow $ $\mathrm{T}_1-\mathrm{T}_2=\left(\mathrm{m}_1 \mathrm{~g}-\mathrm{m}_1 \mathrm{a}\right)-\left(\mathrm{m}_2 \mathrm{~g}+\mathrm{m}_2 \mathrm{a}\right)$

$\Rightarrow $ $ \mathrm{T}_1-\mathrm{T}_2=\left(\mathrm{m}_1-\mathrm{m}_2\right) \mathrm{g}-\left(\mathrm{m}_1+\mathrm{m}_2\right) \mathrm{a} $

Putting the values,

$ \mathrm{T}_1-\mathrm{T}_2=(0.400-0.350) \times 9.8-(0.400+0.350) \times 0.02 $

$\Rightarrow $ $\mathrm{T}_1-\mathrm{T}_2=(0.05 \times 9.8)-(0.75 \times 0.02)$

$\Rightarrow $ $ \mathrm{T}_1-\mathrm{T}_2=0.475 \mathrm{~N} $

So, putting in equation (iii),

$ \mathrm{I}=\frac{(0.475) \times \mathrm{R}^2}{\mathrm{a}} $

Given radius $\mathrm{R}=2 \mathrm{~cm}=0.02 \mathrm{~m}$.

$ I=\frac{0.475 \times(0.02)^2}{0.02}=0.475 \times 0.02=9.5 \times 10^{-3} \mathrm{kgm}^2 $

Therefore, the rotational inertia of the pulley is $9.5 \times 10^{-3} \mathrm{kgm}^2$.

Hence, the correct option is (D).

Let the mass m be at position $\mathrm{x}=0$.

The mass 2 m is at position $\mathrm{x}=\mathrm{d}$.

The position of the centre of mass $\left(\mathrm{X}_{\mathrm{CM}}\right)$ is given by :

$ \mathrm{x}_{\mathrm{CM}}=\frac{\mathrm{m}_1 \mathrm{x}_1+\mathrm{m}_2 \mathrm{x}_2}{\mathrm{~m}_1+\mathrm{m}_2} $

$ X_{C M}=\frac{m(0)+2 m(d)}{m+2 m}=\frac{2 m d}{3 m}=\frac{2 d}{3} $

So, the distances of the masses from the centre of mass are :

Distance of mass m from $\mathrm{CM}\left(\mathrm{r}_1\right)$ :

$ r_1=\frac{2 d}{3} $

Distance of mass 2 m from CM ( $\mathrm{r}_2$ ) :

$ \mathrm{r}_2=\mathrm{d}-\frac{2 \mathrm{~d}}{3}=\frac{\mathrm{d}}{3} $

The moment of inertia of the system about the axis passing through the CM is the sum of the moments of inertia of the individual masses :

$ \mathrm{I}=\mathrm{m}_1 \mathrm{r}_1^2+\mathrm{m}_2 \mathrm{r}_2^2 $

$\Rightarrow $ $I=m\left(\frac{2 d}{3}\right)^2+2 m\left(\frac{d}{3}\right)^2$

$\Rightarrow $ $I=m\left(\frac{4 d^2}{9}\right)+2 m\left(\frac{d^2}{9}\right)$

$\Rightarrow $ $I=\frac{4 \mathrm{md}^2}{9}+\frac{2 \mathrm{md}^2}{9}$

$\Rightarrow $ $ \mathrm{I}=\frac{6 \mathrm{md}^2}{9}=\frac{2}{3} \mathrm{md}^2 $

The angular momentum L is related to the moment of inertia I and angular velocity $\omega$ by the formula:

$ \mathrm{L}=\mathrm{I} \omega $

$\Rightarrow \omega=\frac{L}{I}=\frac{L}{\left(\frac{2}{3} \mathrm{md}^2\right)} $

$ \Rightarrow \omega=\frac{3}{2} \frac{\mathrm{~L}}{\mathrm{md}^2} $

Therefore, the angular velocity of the system is $\frac{3}{2} \frac{\mathrm{~L}}{\mathrm{md}^2}$.

Hence, the correct option is (B).

Total mass of the square loop is M which is made of 4 identical solid cylinders.

So, mass of each cylinder $=m=\frac{M}{4}$

The length of each cylinder is L and radius is R .

Axis of rotation passes through the midpoints of two opposite sides.

Let's assume the square lies in the xy -plane and the axis bisects the square vertically.

For the cylinder B and D

The axis passes through the center of these cylinders and is perpendicular to their length.

The Moment of Inertia ( $\mathrm{I}_{\mathrm{B}}, \mathrm{I}_{\mathrm{D}}$ ) of a solid cylinder about a transverse axis passing through its center is:

$ \mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{D}}=\frac{\mathrm{mL}^2}{12}+\frac{\mathrm{mR}^2}{4} $

$\Rightarrow $ $ \mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{D}}=2\left(\frac{\mathrm{~mL}^2}{12}+\frac{\mathrm{mR}^2}{4}\right)=\left(\frac{\mathrm{mL}^2}{6}+\frac{\mathrm{mR}^2}{2}\right) $

For the cylinders A and C :

The axis is parallel to the length of these cylinders.

The axis is at a distance $\mathrm{d}=\frac{\mathrm{L}}{2}$ from the center of each cylinder.

The Moment of Inertia ( $\mathrm{I}_{\mathrm{CM}}$ ) of a solid cylinder about its own longitudinal axis is $\frac{1}{2} \mathrm{mR}^2$.

Using the Parallel Axis Theorem ( $\mathrm{I}=\mathrm{I}_{\mathrm{CM}}+\mathrm{md}^2$ ), the inertia of one such cylinder about the central axis is:

$ \mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{C}}=\frac{1}{2} \mathrm{mR}^2+\mathrm{m}\left(\frac{\mathrm{~L}}{2}\right)^2=\frac{\mathrm{mR}^2}{2}+\frac{\mathrm{mL}^2}{4} $

$\Rightarrow $ $ \mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{C}}=2 \times\left(\frac{\mathrm{mR}^2}{2}+\frac{\mathrm{mL}^2}{4}\right)=\left(\mathrm{mR}^2+\frac{\mathrm{mL}^2}{2}\right) $

Hence, the total moment of inertia of system about the given axis of rotation is,

$ \mathrm{I}_{\text {total }}=\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{D}} $

$\Rightarrow $ $\mathrm{I}_{\text {total }}=\left(\frac{\mathrm{mL}^2}{6}+\frac{\mathrm{mR}^2}{2}\right)+\left(\mathrm{mR}^2+\frac{\mathrm{mL}^2}{2}\right)$

$\Rightarrow $ $\mathrm{I}_{\text {total }}=\left(\frac{1}{6}+\frac{1}{2}\right) \mathrm{mL}^2+\left(\frac{1}{2}+1\right) \mathrm{mR}^2$

$\Rightarrow $ $\mathrm{I}_{\text {total }}=\frac{4}{6} \mathrm{~mL}^2+\frac{3}{2} \mathrm{mR}^2$

$\Rightarrow $ $\mathrm{I}_{\text {total }}=\frac{2}{3} \mathrm{~mL}^2+\frac{3}{2} \mathrm{mR}^2$

$\Rightarrow $ $\mathrm{I}_{\text {total }}=\frac{2}{3}\left(\frac{\mathrm{M}}{4}\right) \mathrm{L}^2+\frac{3}{2}\left(\frac{\mathrm{M}}{4}\right) \mathrm{R}^2$

$\Rightarrow $ $\mathrm{I}_{\text {total }}=\frac{2}{12} \mathrm{ML}^2+\frac{3}{8} \mathrm{MR}^2$

$\Rightarrow $ $ \mathrm{I}_{\text {total }}=\frac{1}{6} \mathrm{ML}^2+\frac{3}{8} \mathrm{MR}^2 $

Hence, the correct option is (D).

The mass of the rod is $M=20 m$

The length of the rod is $\mathrm{L}=12 \mathrm{~cm}$

Distance of mass m from center, $\mathrm{r}_1=4 \mathrm{~cm}$

Distance of mass 2 m from center, $\mathrm{r}_2=2 \mathrm{~cm}$

Let the geometrical center of the rod be the origin $(0,0)$.

The position of the new center of mass $\left(\mathrm{x}_{\mathrm{cm}}\right)$ after the collision :

$ \mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{M}(0)+\mathrm{m}\left(\mathrm{r}_1\right)-2 \mathrm{~m}\left(\mathrm{r}_2\right)}{\mathrm{M}+\mathrm{m}+2 \mathrm{~m}} $

$\Rightarrow $ $\mathrm{x}_{\mathrm{cm}}=\frac{20 \mathrm{~m}(0)+\mathrm{m}(4)-2 \mathrm{~m}(2)}{23 \mathrm{~m}}$

$\Rightarrow $ $ \mathrm{x}_{\mathrm{cm}}=\frac{4 \mathrm{~m}-4 \mathrm{~m}}{23 \mathrm{~m}}=0 $

Since $\mathrm{x}_{\mathrm{cm}}=0$, the center of mass of the system does not shift. The entire system will only rotate about the geometrical center of the rod.

The initial angular momentum $\left(\mathrm{L}_{\mathrm{i}}\right)$ of the system about the center must equal the final angular momentum $\left(\mathrm{L}_{\mathrm{f}}\right)$.

The initial angular momentum of the system is,

$ \mathrm{L}_{\mathrm{i}}=\left(\mathrm{m} \cdot \mathrm{v} \cdot \mathrm{r}_1\right)+\left(2 \mathrm{~m} \cdot \mathrm{v} \cdot \mathrm{r}_2\right) $

$\Rightarrow $ $\mathrm{L}_{\mathrm{i}}=\mathrm{m} \cdot \mathrm{v} \cdot 4+2 \mathrm{~m} \cdot \mathrm{v} \cdot 2$

$\Rightarrow $ $ \mathrm{L}_{\mathrm{i}}=4 \mathrm{mv}+4 \mathrm{mv}=8 \mathrm{mv} $

The final moment of inertia ( $\mathrm{I}_{\mathrm{f}}$ ) about the center is the sum of the moment of inertia of the rod and the two point masses.

$ \mathrm{I}_{\mathrm{f}}=\mathrm{I}_{\mathrm{rod}}+\mathrm{I}_{\mathrm{m}}+\mathrm{I}_{2 \mathrm{~m}} $

$\Rightarrow $ $I_f=\frac{M L^2}{12}+m r_1^2+2 m r_2^2$

$\Rightarrow $ $\mathrm{I}_{\mathrm{f}}=\frac{(20 \mathrm{~m})(12)^2}{12}+\mathrm{m}(4)^2+2 \mathrm{~m}(2)^2$

$\Rightarrow $ $ \mathrm{I}_{\mathrm{f}}=20 \mathrm{~m}(12)+16 \mathrm{~m}+2 \mathrm{~m}(4)=264 \mathrm{~m} $

The angular momentum of the system is given as $\mathrm{L}=\mathrm{I} \omega$, where $\omega$ is the angular speed and I is the moment of inertia.

Using conservation of angular momentum,

$ \mathrm{L}_{\mathrm{i}}=\mathrm{L}_{\mathrm{f}} \Rightarrow 8 \mathrm{mv}=264 \mathrm{~m} \omega $

$\Rightarrow \omega=\frac{8 \mathrm{mv}}{264 \mathrm{~m}}$

$\Rightarrow $ $\frac{\omega}{v}=\frac{1}{33} \Rightarrow \frac{v}{\omega}=33$

As the tube rotates, the gas molecules are pushed outward due to the rotation. This outward push (centrifugal effect) creates a pressure gradient; the pressure must increase as we move from the axis of rotation A towards the outer end B to provide the necessary centripetal force to keep the gas in circular motion.

Let the cross-sectional area of the tube be S . Assume a small element of thickness dx be at a distance x from the axis of rotation A .

Let the pressure on the inner face be P and the pressure on the outer face be $\mathrm{P}+\mathrm{dP}$.

Force pushing outward on the inner face $=\mathrm{P} \cdot \mathrm{S}$

Force pushing inward on the outer face $=(\mathrm{P}+\mathrm{dP}) \cdot \mathrm{S}$

The net inward force provided by the pressure difference is:

$ \mathrm{F}_{\mathrm{net}}=(\mathrm{P}+\mathrm{dP}) \cdot \mathrm{S}-\mathrm{P} \cdot \mathrm{~S}=\mathrm{S} \cdot \mathrm{dP} $

This net inward force acts as the centripetal force required to keep this small mass of gas moving in a circle of radius x with angular velocity $\omega$.

The mass of this small element (dm) is its volume (S ⋅ dx) multiplied by the local density ( $\rho$ ) of the gas :

$ \mathrm{dm}=\rho \cdot \mathrm{S} \cdot \mathrm{dx} $

At equilibrium,

$ S \cdot d P=(\rho \cdot S \cdot d x) \cdot x \cdot \omega^2 $

$\Rightarrow \mathrm{dP}=\rho \cdot \omega^2 \cdot \mathrm{x} \cdot \mathrm{dx} \ldots(\mathrm{i})$

It is given that it's an ideal gas at a constant temperature T .

From the ideal gas equation:

$ \mathrm{PV}=\mathrm{nRT} $

The number of moles $n=\frac{m}{M}$, where $m$ is the mass and $M$ is the molar mass (molecular weight).

$ \mathrm{PV}=\left(\frac{\mathrm{m}}{\mathrm{M}}\right) \mathrm{RT} $

$\Rightarrow $ $ \mathrm{PM}=\left(\frac{\mathrm{m}}{\mathrm{~V}}\right) \mathrm{RT} $

Since density $\rho=\frac{\mathrm{m}}{\mathrm{V}}$, we get :

$ \mathrm{PM}=\rho \mathrm{RT} $

$\Rightarrow $ $ \rho=\frac{\mathrm{PM}}{\mathrm{RT}} $

Now, substitute this expression for $\rho$ into equation (i) :

$ \mathrm{dP}=\left(\frac{\mathrm{PM}}{\mathrm{RT}}\right) \omega^2 \mathrm{xdx} $

$ \frac{\mathrm{dP}}{\mathrm{P}}=\left(\frac{\mathrm{M} \omega^2}{\mathrm{RT}}\right) \mathrm{xdx} $

At end $\mathrm{A}(\mathrm{x}=0)$, the pressure is $\mathrm{P}_{\mathrm{A}}$.

At end $\mathrm{B}(\mathrm{x}=\mathrm{l})$, the pressure is $\mathrm{P}_{\mathrm{B}}$.

$ \int\limits_{\mathrm{P}_{\mathrm{A}}}^{\mathrm{P}_{\mathrm{B}}} \frac{1}{\mathrm{P}} \mathrm{dP}=\left(\frac{\mathrm{M} \omega^2}{\mathrm{RT}}\right) \int\limits_0^1 \mathrm{x} \mathrm{dx} $

$\Rightarrow $ $[\ln (\mathrm{P})]_{\mathrm{P}_{\mathrm{A}}}^{\mathrm{P}_{\mathrm{B}}}=\left(\frac{\mathrm{M} \omega^2}{\mathrm{RT}}\right)\left[\frac{\mathrm{x}^2}{2}\right]_0^1$

$\Rightarrow $ $\ln \left(\mathrm{P}_{\mathrm{B}}\right)-\ln \left(\mathrm{P}_{\mathrm{A}}\right)=\left(\frac{\mathrm{M} \omega^2}{\mathrm{RT}}\right)\left(\frac{\mathrm{l}^2}{2}-0\right)$

$\Rightarrow $ $\ln \left(\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\mathrm{A}}}\right)=\frac{\mathrm{M} \omega^2 \mathrm{l}^2}{2 \mathrm{RT}}$

$\Rightarrow $ $\frac{P_B}{P_A}=e^{\left(\frac{M \omega^2 l^2}{2 R T}\right)} $

$\Rightarrow P_B=P_A e^{\left(\frac{M \omega^2 l^2}{2 R T}\right)}$

Hence, the correct option is (C).

We have two solid spheres touching each other. The axis of rotation is the common tangent passing exactly through their point of contact.

Using the Parallel Axis Theorem.

$ \mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Md}^2 $

Where $I_{c m}$ is the moment of inertia about the centre of mass, $M$ is the mass of object whose centre of mass is at distance d from the axis of rotation.

For a uniform solid sphere the moment of inertia about its diametric centre axis is,

$ \mathrm{I}_{\mathrm{cm}}=\frac{2}{5} \mathrm{Mr}^2 $

The distance $d$ from the center to the tangent is the radius $r$.

$ \mathrm{I}_{\text {tangent }}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Mr}^2 $

$\Rightarrow $ $ \mathrm{I}_{\text {tangent }}=\frac{2}{5} \mathrm{Mr}^2+\mathrm{Mr}^2=\frac{7}{5} \mathrm{Mr}^2 $

For sphere $\mathrm{s}_1$

• Mass $\mathrm{m}_1=5 \mathrm{~kg}$

• Radius $\mathrm{r}_1=10 \mathrm{~cm}=0.1 \mathrm{~m}$

So, the moment of inertia of sphere $\mathrm{s}_1, \mathrm{I}_1=\frac{7}{5} \mathrm{~m}_1 \mathrm{r}_1^2$

$ \mathrm{I}_1=\frac{7}{5} \times 5 \times(0.1)^2 $

$\Rightarrow $ $ \mathrm{I}_1=7 \times 0.01=0.07 \mathrm{~kg} \cdot \mathrm{~m}^2 $

For Sphere $2 \mathrm{~s}_2$ :

• Mass $\mathrm{m}_2=10 \mathrm{~kg}$

• Radius $\mathrm{r}_2=20 \mathrm{~cm}=0.2 \mathrm{~m}$

So, the moment of inertia of sphere $\mathrm{s}_2, \mathrm{I}_2=\frac{7}{5} \mathrm{~m}_2 \mathrm{r}_2^2$

$ I_2=\frac{7}{5} \times 10 \times(0.2)^2 $

$\Rightarrow $ $I_2=7 \times 2 \times 0.04$

$\Rightarrow $ $ \mathrm{I}_2=\frac{1}{4} \times 0.04=0.56 \mathrm{~kg} \cdot \mathrm{~m}^2 $

So, the total moment of inertia of two sphere system is,

$ \mathrm{I}_{\text {total }}=\mathrm{I}_1+\mathrm{I}_2 $

$ \begin{aligned} & \Rightarrow \mathrm{I}_{\text {total }}=0.07+0.56 \\ & \Rightarrow \mathrm{I}_{\text {total }}=0.63 \mathrm{~kg} \cdot \mathrm{~m}^2 \end{aligned} $

Therefore, the moment of inertia of given system of spheres about their common tangent is $0.63 \mathrm{~kg} \cdot \mathrm{~m}^2$.

Hence, the correct option is (D).

For a hoop of mass $M$ and radius $R$, the moment of inertia is given as,

$ \mathrm{I}_{\mathrm{rim}}=\mathrm{MR}^2 $

Each rod has mass $M$ and length $L=2 R$. The moment of inertia of a rod rotating about its center is $\frac{1}{12} M L^2$.

$ \mathrm{I}_{\text {rod }}=2 \times \frac{1}{12} \mathrm{M}(2 \mathrm{R})^2=\frac{8 \mathrm{MR}^2}{12}=\frac{2}{3} \mathrm{MR}^2 $

Total moment of inertia of the system is,

$ \mathrm{I}=\mathrm{I}_{\text {rim }}+\mathrm{I}_{\text {rod }}=\mathrm{MR}^2+\frac{2}{3} \mathrm{MR}^2=\frac{5}{3} \mathrm{MR}^2 $

Let a be the linear acceleration of the blocks and $\alpha$ be the angular acceleration of the pulley. Assuming no slipping, $\mathrm{a}=\mathrm{R} \alpha$.

For Mass M,

$ \mathrm{Mg}-\mathrm{T}_1=\mathrm{Ma} \ldots \text { (i) } $

For Mass m,

$ \mathrm{T}_2-\mathrm{mg}=\mathrm{ma} . $

The net torque $\tau$ about the centre of pulley is provided by the difference in tensions:

$ \tau=\left(\mathrm{T}_1-\mathrm{T}_2\right) \mathrm{R}=\mathrm{I} \alpha $

Substituting $\mathrm{I}=\frac{5}{3} \mathrm{MR}^2$ and $\alpha=\frac{\mathrm{a}}{\mathrm{R}}$ :

$ \begin{aligned} \left(\mathrm{T}_1-\mathrm{T}_2\right) \mathrm{R} & =\left(\frac{5}{3} \mathrm{MR}^2\right) \frac{\mathrm{a}}{\mathrm{R}} \\ \Rightarrow \mathrm{T}_1-\mathrm{T}_2 & =\frac{5}{3} \mathrm{Ma} \ldots(\text { iii }) \end{aligned} $

Adding equations (i) and (ii):

$ \begin{aligned} & \left(M g-T_1\right)+\left(T_2-m g\right)=M a+m a \\\\ & \Rightarrow (M-m) g-\left(T_1-T_2\right)=(M+m) a \end{aligned} $

Substituting the value of ( $\mathrm{T}_1-\mathrm{T}_2$ ) from equation (iii) :

$ (M-m) g-\frac{5}{3} M a=(M+m) a $

$\Rightarrow $ $(M-m) g=(M+m) a+\frac{5}{3} M a$

$\Rightarrow $ $(M-m) g=\left(M+m+\frac{5}{3} M\right) a$

$\Rightarrow $ $(M-m) g=\left(\frac{8}{3} M+m\right) a$

$\Rightarrow $ $ a=\frac{(M-m) g}{\left(\frac{8}{3} M+m\right)} $

Therefore, the acceleration of the blocks is $\frac{(M-m) g}{\left(\frac{8}{3} M+m\right)}$.

Hence, the correct option is (A).

Immediately after one string is cut, the only remaining upward force is the Tension (T) at end A and the downward force is the Weight $(\mathrm{mg})$ acting at the centre of mass (distance $1 / 2$ from each end).

Let a be the downward linear acceleration of the center of mass and $\alpha$ be the angular acceleration of the rod about its centre of mass.

For translational motion

Using Newton's Second Law ( $\mathrm{F}_{\text {net }}=\mathrm{ma}$ ):

$ \mathrm{mg}-\mathrm{T}=\mathrm{ma} $ .....(i)

For rotational motion

The torque $(\tau)$ about the centre of mass is provided by the tension T at distance $\frac{1}{2}$ :

$ \tau=\mathrm{T} \cdot \frac{1}{2} $

Also, $\tau=\mathrm{I} \alpha$, where I is the moment of inertia of a rod about its center $\left(\mathrm{I}=\frac{1}{12} \mathrm{ml}^2\right)$ :

$ \mathrm{T} \cdot \frac{\mathrm{l}}{2}=\left(\frac{1}{12} \mathrm{ml}^2\right) \alpha $

Immediately after the cut, end A is effectively the pivot point. The relationship between linear acceleration (a) of the center and angular acceleration ( $\alpha$ ) is:

$ a=\alpha \cdot \frac{l}{2} \Rightarrow \alpha=\frac{2 a}{l} \ldots (ii)$

From relations (ii) and (iii)

$ \mathrm{T} \cdot \frac{\mathrm{l}}{2}=\frac{1}{12} \mathrm{ml}^2\left(\frac{2 \mathrm{a}}{\mathrm{l}}\right) $

$\Rightarrow $ $\mathrm{T}=\frac{4 \mathrm{ma}}{12}=\frac{\mathrm{ma}}{3}$

$ \Rightarrow \mathrm{ma}=3 \mathrm{~T} $

From relation (i),

$ \mathrm{mg}-\mathrm{T}=3 \mathrm{~T} $

$\Rightarrow $ $\mathrm{mg}=4 \mathrm{~T}$

$\Rightarrow $ $ \mathrm{T}=\frac{\mathrm{mg}}{4} $

Therefore, immediately after one string is cut, the tension in the remaining string $\frac{\mathrm{mg}}{4}$.

Hence, the correct option is (B).

The cylinder is slipping, which means its translational motion and rotational motion are not in order $(\mathrm{v}=\mathrm{r} \omega)$ for pure rolling.

Initial translational velocity is $\mathrm{v}_0=49 \mathrm{~m} / \mathrm{s}$.

Initial angular velocity is $\omega_0=\frac{v_0}{4 R}$, where $R$ is the radius of solid cylinder of length $L$.

For pure rolling to start, the condition $\mathrm{v}=\omega \mathrm{R}$ must be satisfied.

Initially, $\omega_0 \mathrm{R}=\frac{\mathrm{v}_0}{4}$, which is less than $\mathrm{v}_0$, so the cylinder is sliding forward.

Since the cylinder is sliding forward, kinetic friction $\left(\mathrm{f}_{\mathrm{k}}\right)$ acts backward at the point of contact.

$ \mathrm{f}_{\mathrm{k}}=\mu_{\mathrm{k}} \mathrm{~N}=\mu_{\mathrm{k}} \mathrm{mg} . $

Using Newton's second law of motion ( $\mathrm{F}=\mathrm{ma}$ ) linear retardation of cylinder is,

$ \mathrm{F}=\mathrm{ma} \Rightarrow-\mu_{\mathrm{k}} \mathrm{mg}=\mathrm{ma} \Rightarrow \mathrm{a}=-\mu_{\mathrm{k}} \mathrm{~g} $

Friction provides a torque $(\tau)$ about the center of mass.

$ \tau=\mathrm{f}_{\mathrm{k}} \cdot \mathrm{R}=\left(\mu_{\mathrm{k}} \mathrm{mg}\right) \mathrm{R} $

We also know $\tau=\mathrm{I} \alpha$.

For a solid cylinder, the moment of inertia $\mathrm{I}=\frac{1}{2} \mathrm{mR}^2$.

$ \left(\mu_{\mathrm{k}} \mathrm{mg}\right) \mathrm{R}=\left(\frac{1}{2} \mathrm{mR}^2\right) \alpha \Rightarrow \alpha=\frac{2 \mu_{\mathrm{k}} \mathrm{~g}}{\mathrm{R}} $

We need to find the time t when the final translational velocity v and final angular velocity $\omega$ satisfy $\mathrm{v}=\omega \mathrm{R}$.

Using equation of linear motion, final linear velocity is $\mathrm{v}=\mathrm{v}_0+\mathrm{at}=\mathrm{v}_0-\mu_{\mathrm{k}} \mathrm{gt}$

Using equation of rotational motion, final angular velocity is,

$ \omega=\omega_0+\alpha \mathrm{t}=\frac{\mathrm{v}_0}{4 \mathrm{R}}+\frac{2 \mu_{\mathrm{k}} \mathrm{~g}}{\mathrm{R}} \mathrm{t} $

Applying the rolling condition $\mathrm{v}=\omega \mathrm{R}$ :

$\mathrm{v}_0-\mu_{\mathrm{k}} \mathrm{gt}=\left(\frac{\mathrm{v}_0}{4 \mathrm{R}}+\frac{2 \mu_{\mathrm{k}} \mathrm{g}}{\mathrm{R}} \mathrm{t}\right) \mathrm{R}$

$\Rightarrow $ $\mathrm{v}_0-\mu_{\mathrm{k}} \mathrm{gt}=\frac{\mathrm{v}_0}{4}+2 \mu_{\mathrm{k}} \mathrm{gt}$

$\Rightarrow $ $\mathrm{v}_0-\frac{\mathrm{v}_0}{4}=2 \mu_{\mathrm{k}} \mathrm{gt}+\mu_{\mathrm{k}} \mathrm{gt}$

$\Rightarrow $ $\frac{3 \mathrm{v}_0}{4}=3 \mu_{\mathrm{k}} \mathrm{gt}$

$\Rightarrow $ $ \mathrm{t}=\frac{\mathrm{v}_0}{4 \mu_{\mathrm{k}} \mathrm{~g}} $

Substituting the given values, $\mathrm{v}_0=49 \mathrm{~m} / \mathrm{s}, \mu_{\mathrm{k}}=0.25$ and $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$

$ \mathrm{t}=\frac{49}{4 \times 0.25 \times 9.8} $

$\Rightarrow $ $ t=\frac{49}{9.8} s=5 s $

Therefore, the time taken for the cylinder to start rolling is 5 seconds.

Hence, the correct option is (B).

Let a sphere of mass M and radius R be acted upon by a force F at its highest point.

The external force F and the friction force f (acting at the contact point) determine the linear acceleration a :

$ \mathrm{F}+\mathrm{f}=\mathrm{Ma} \ldots \text { (i) } $

The torque $\tau$ about the center of mass is produced by both F and f :

$ \tau=\mathrm{F} \cdot \mathrm{R}-\mathrm{f} \cdot \mathrm{R}=\mathrm{I} \alpha $

Where I is the moment of inertia and $\alpha$ is the angular acceleration.

For rolling without slipping, $\mathrm{a}=\mathrm{R} \alpha$.

Substituting $\alpha=\mathrm{a} / \mathrm{R}$ into the torque equation :

$ \mathrm{FR}-\mathrm{fR}=\mathrm{I}\left(\frac{\mathrm{a}}{\mathrm{R}}\right) $

$\Rightarrow $ $\mathrm{F}-\mathrm{f}=\frac{\mathrm{Ia}}{\mathrm{R}^2} \ldots (ii) $

Adding equation (i) and equation (ii) to eliminate friction f :

$ (F+f)+(F-f)=M a+\frac{I a}{R^2} $

$\Rightarrow $ $2 \mathrm{~F}=\mathrm{a}\left(\mathrm{M}+\frac{\mathrm{I}}{\mathrm{R}^2}\right)$

$\Rightarrow $ $ \mathrm{a}=\frac{2 \mathrm{~F}}{\mathrm{M}+\mathrm{IR}^2} $

For solid sphere (A)

The mass of solid sphere is $\mathrm{M}_{\mathrm{A}}=5 \mathrm{~m}$

The moment of inertia of solid sphere is given as $I_A=\frac{2}{5} M_A R^2=\frac{2}{5}(5 m) R^2=2 m R^2$

So, the linear acceleration of solid sphere A is,

$ a_A=\frac{2 F}{5 m+\frac{2 m R^2}{R^2}}=\frac{2 F}{5 m+2 m}=\frac{2 F}{7 m} $

For spherical shell (B)

The mass of spherical shell is $M_B=m$

The moment of inertia of spherical shell is given as $I_B=\frac{2}{3} M_B R^2=\frac{2}{3}(m) R^2=\frac{2}{3} m R^2$

So, the linear acceleration of spherical shell $B$ is,

$ a_B=\frac{2 F}{m+\frac{2 m R^2}{3 R^2}}=\frac{2 F}{m+\frac{2}{3} m}=\frac{2 F}{5 / 3 m}=\frac{6 F}{5 m} $

Hence, the ratio of linear acceleration of A and B is,

$ \frac{\mathrm{a}_{\mathrm{A}}}{\mathrm{a}_{\mathrm{B}}}=\frac{(2 \mathrm{~F} / 7 \mathrm{~m})}{6 \mathrm{~F} / 5 \mathrm{~m}} $

$\Rightarrow $ $ \frac{\mathrm{a}_{\mathrm{A}}}{\mathrm{a}_{\mathrm{B}}}=\frac{10}{42}=\frac{5}{21} $

Therefore, the correct option is (A).

The mass of the solid sphere is $M=5 \mathrm{~kg}$ and its radius is $R=4 \mathrm{~cm}=0.04 \mathrm{~m}$

For a solid sphere rotating about its centre, the moment of inertia is $I=\frac{2}{5} M R^2$

$ I=\frac{2}{5} \times 5 \times(0.04)^2 $

$\Rightarrow $ $ \mathrm{I}=2 \times 0.0016=0.0032 \mathrm{~kg} \cdot \mathrm{~m}^2 $

The initial angular velocity is $\omega_0=1200 \mathrm{rpm}$ (rotations per minute).

The rate of rotation, usually measured in radians per second $\left(\frac{\mathrm{rad}}{\mathrm{s}}\right)$.

$ \omega_0=2 \pi \times \frac{1200}{60}=40 \pi \mathrm{rad} / \mathrm{s} $

The final angular velocity is $\omega=0$

The time taken before it comes to rest is $\mathrm{t}=10 \mathrm{~s}$.

Using the rotational kinematic equation,

$ \omega=\omega_0+\alpha t $

$\Rightarrow $ $0=40 \pi+\alpha(10)$

$\Rightarrow $ $ \alpha=-\frac{40 \pi}{10}=-4 \pi \mathrm{rad} / \mathrm{s}^2 $

The negative sign shows the solid sphere is slowing down.

Using the law of rotational motion, $\tau=\mathrm{I} \alpha$, where $\tau$ is the torque acting and $\alpha$ is the resultant angular acceleration.

$ \tau=I \times|\alpha| $

$\Rightarrow $ $ \tau=0.0032 \times 4 \pi=0.0128 \pi \mathrm{Nm} $

Using kinematic equation for rotational motion $\theta=\omega_0 \mathrm{t}+\frac{1}{2} \alpha \mathrm{t}^2$, the total angle covered is,

$\theta=40 \pi \times 10+\frac{1}{2} \times(-4 \pi) \times 10^2$

$\Rightarrow $ $ \theta=400 \pi-2 \pi \times 100=200 \pi \text { radians } $

Since one full rotation is $2 \pi$ radians, the number of rotations is:

$ \mathrm{N}=\frac{\theta}{2 \pi}=\frac{200 \pi}{2 \pi}=100 \text { rotations } $

Therefore, the magnitude of applied torque is $0.0128 \pi \mathrm{Nm}$ and the number of rotations is 100 . Thus, the correct option is (D).

For rotational motion with uniform angular acceleration, the angular displacement in time $t$ is

$ \theta = \omega_0 t + \frac{1}{2}\alpha t^2 $

Since the wheel is initially at rest,

$ \omega_0 = 0 $

So,

$ \theta = \frac{1}{2}\alpha t^2 $

Now let us find $\theta_1$ and $\theta_2$.

Angle rotated in first 2 s

$ \theta_1 = \frac{1}{2}\alpha (2)^2 = 2\alpha $

Total angle rotated in first 4 s

$ \theta_{0\to4} = \frac{1}{2}\alpha (4)^2 = 8\alpha $

Angle rotated in next 2 s

This means angle rotated from $t=2$ s to $t=4$ s:

$ \theta_2 = \theta_{0\to4} - \theta_1 = 8\alpha - 2\alpha = 6\alpha $

Now the ratio is

$ \frac{\theta_2}{\theta_1} = \frac{6\alpha}{2\alpha} = 3 $

So, the correct answer is:

$ \boxed{3} $

Option B

For a body rolling without slipping, the initial kinetic energy has two parts:

$ K = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega^2 $

Since it rolls without slipping,

$ \omega = \frac{v_0}{R} $

So,

$ K = \frac{1}{2}mv_0^2 + \frac{1}{2}I\frac{v_0^2}{R^2} $

When the object reaches the maximum height, it momentarily comes to rest, so all the initial kinetic energy converts into gravitational potential energy:

$ mgh = \frac{1}{2}mv_0^2 + \frac{1}{2}I\frac{v_0^2}{R^2} $

Divide by $m$:

$ gh = \frac{1}{2}v_0^2 + \frac{1}{2}\frac{I}{mR^2}v_0^2 $

Given,

$ h = \frac{7v_0^2}{10g} $

So,

$ g\left(\frac{7v_0^2}{10g}\right)=\frac{1}{2}v_0^2+\frac{1}{2}\frac{I}{mR^2}v_0^2 $

$ \frac{7v_0^2}{10}=\frac{1}{2}v_0^2+\frac{1}{2}\frac{I}{mR^2}v_0^2 $

Divide by $v_0^2$:

$ \frac{7}{10}=\frac{1}{2}+\frac{1}{2}\frac{I}{mR^2} $

$ \frac{1}{2}\frac{I}{mR^2}=\frac{7}{10}-\frac{1}{2}=\frac{2}{10}=\frac{1}{5} $

$ \frac{I}{mR^2}=\frac{2}{5} $

Now compare with standard moments of inertia:

• Ring: $\frac{I}{mR^2}=1$

• Solid cylinder/disc: $\frac{I}{mR^2}=\frac{1}{2}$

• Solid sphere: $\frac{I}{mR^2}=\frac{2}{5}$

Hence, the object is a

$ \boxed{\text{solid sphere}} $

So, the correct option is D.

Let the large stationary disk have its centre at the origin 0 . The two smaller disks Disk 1 and Disk 2 each have a radius $\mathrm{r}=\frac{\mathrm{R}}{50}$.

When the two smaller disks are in contact with each other while resting on the circumference of the large disk, the distance between the centre of the large disk and the centre of either smaller disk is :

$ \mathrm{R}_{\mathrm{c}}=\mathrm{R}+\mathrm{r} $

Since the two smaller disks are in contact, the distance between their centres is exactly 2 r .

Let $\Delta \theta$ be the angular separation between their centres at centre of the large disk $O$.

Using the properties of an isosceles triangle formed by O and the two small disk centres :

$ \sin \alpha=\frac{r}{R+r} $

Here, $\alpha=\frac{\Delta \theta}{4} \Rightarrow \sin \left(\frac{\Delta \theta}{4}\right)=\frac{\mathrm{r}}{\mathrm{R}+\mathrm{r}}$

Given the small-angle approximation $\sin (\Delta \theta) \approx \Delta \theta$, we can also approximate

$ \sin \left(\frac{\Delta \theta}{4}\right) \approx \frac{\Delta \theta}{4} $

Therefore $\frac{\Delta \theta}{4}=\frac{\mathrm{r}}{\mathrm{R}+\mathrm{r}}$

$ \Delta \theta=\frac{4 r}{R+r} $

Substituting $\mathrm{R}=50 \mathrm{r}$ into this equation:

$ \Delta \theta=\frac{4 \mathrm{r}}{50 \mathrm{r}+\mathrm{r}}=\frac{4}{51} \text { radians } $

Let the small disk rotate about its own centre with an angular velocity $\omega^{\prime}$. The point of contact with the stationary large disk must have a linear velocity of zero to satisfy the no-slip condition.

From rotation about its own centre relative to the contact point $\mathrm{v}_{\mathrm{cm}}=\omega^{\prime} \cdot \mathrm{r}$

From orbital path around the large disk's centre $\mathrm{v}_{\mathrm{cm}}=\Omega \cdot(\mathrm{R}+\mathrm{r})$

$\Omega (R + r) = \omega' \cdot r$

$\Rightarrow $ $\Omega = \omega' \left(\frac{r}{R + r}\right)$

Substituting $R = 50r$:

$\Omega = \omega' \left(\frac{r}{51r}\right) = \frac{\omega'}{51}$

It is given that the two smaller disks spin with constant angular velocities $\omega$ and $2\omega$ in opposite directions along the circumference.

For Disk 1 ($\omega'_1 = \omega$):

$\Omega_1 = \frac{\omega}{51}$

For Disk 2 ($\omega'_2 = 2\omega$):

$\Omega_2 = \frac{2\omega}{51}$

Since they travel in opposite directions around the large disk, their relative orbital angular velocity

$\Omega_{\mathrm{rel}} = \Omega_1 + \Omega_2 = \frac{\omega}{51} + \frac{2\omega}{51} = \frac{3\omega}{51}$

Initially, the two disks are in contact with an angular separation between their centres equal to

$\Delta \theta = \frac{2}{51}$

For them to come into contact again, the total angular path covered by the two centres relative to each other must equal to $\Phi = 2\pi - \Delta \theta$

Since each disk has a radius $r$, when they meet again, their centres will once again be separated by $\Delta \theta$. Thus, the net angular displacement between the two centres around the large disk is :

$\theta_{\mathrm{rel}} = 2\pi - 2\Delta \theta$

Substituting the value of $\Delta \theta = \frac{2}{51}$:

$\theta_{\mathrm{rel}} = 2\pi - \frac{4}{51}$

The time $\tau$ taken to cover this relative angular displacement at a relative angular speed $\Omega_{\mathrm{rel}}$ is :

$\tau = \frac{\theta_{\mathrm{rel}}}{\Omega_{\mathrm{rel}}}$

$\tau = \frac{2\pi - \frac{4}{51}}{\frac{3\omega}{51}}$

$\tau = 51 \times \left(2\pi - \frac{4}{51}\right) / 3\omega$

Therefore, the correct option is (C).

When the uniform solid cylinder rolls along the horizontal table and strikes the sharp corner edge, its smooth rolling motion changes. The corner point P acts as a temporary pivot.

The cylinder begins to pivot or rotate purely about this fixed corner point P . As long as it stays in contact with this corner, the centre of mass ( $C$ ) executes circular motion of radius $R$ around $P$.

The moment of inertia of a uniform solid cylinder of mass M and radius R about its central longitudinal axis is :

$ \mathrm{I}_{\mathrm{cm}}=\frac{1}{2} \mathrm{MR}^2 $

According to the parallel axis theorem, the moment of inertia about an axis at a distance $\mathrm{d}=\mathrm{R}$ from the centre is :

$ \mathrm{I}_{\mathrm{p}}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Md}^2=\frac{1}{2} \mathrm{MR}^2+\mathrm{MR}^2=\frac{3}{2} \mathrm{MR}^2 $

The centre of mass immediately after the cylinder pivots onto the corner gravity is a non-impulsive force, so its effect over the infinitesimally short duration is negligible.

The reaction forces at the corner P are impulsive, but they pass directly through P . Therefore, the net torque about the pivot P during the collision is zero, which means angular momentum about P is conserved.

The cylinder is rolling without slipping with a centre of mass velocity $\mathrm{v}_0$ and angular velocity $\omega_0=\mathrm{v}_0 \mathrm{R}$.

The total angular momentum about the level of the corner P which is at a height R below the centre of mass path is:

$L_{\text{initial}} = Mv_0R + I_{\text{cm}}\omega_0$

$L_{\text{initial}} = Mv_0R + \left(\frac{1}{2}MR^2\right)\left(\frac{v_0}{R}\right) = \frac{3}{2}Mv_0R$

After impact the cylinder is rotating about the pivot P with a new angular velocity $\omega_1$. Its angular momentum about P is:

$L_{\text{final}} = I_p\,\omega_1 = \frac{3}{2}MR^2\omega_1$

Using conservation of angular momentum

$L_{\text{initial}} = L_{\text{final}}$

$\frac{3}{2}Mv_0R = \frac{3}{2}MR^2\omega_1 \rightarrow \omega_1 = \frac{v_0}{R}$

As the cylinder tilts forward around the corner, the angle $\theta$ with the vertical increases from $0$. The centre of mass descends vertically. When the line PC has tilted by an angle $\theta$.

Initial height of centre of mass relative to pivot $= R$

Height of centre of mass at angle $\theta$ $= R\cos\theta$

Loss in potential energy

$= Mg(R - R\cos\theta) = MgR(1 - \cos\theta)$

By conservation of total mechanical energy between the initial vertical position of PC ($\theta = 0$, angular speed $\omega_1$) and a tilted position $\theta$ (angular speed $\omega$):

$\Delta K = -\Delta U$

$\frac{1}{2}I_p\omega^2 - \frac{1}{2}I_p\omega_1^2 = MgR(1 - \cos\theta)$

Substituting $I_p = \frac{3}{2}MR^2$ and $\omega_1 = \frac{v_0}{R}$:

$\frac{1}{2}\left(\frac{3}{2}MR^2\right)\omega^2 - \frac{1}{2}\left(\frac{3}{2}MR^2\right)\left(\frac{v_0}{R}\right)^2 = MgR(1 - \cos\theta)$

$\frac{3}{4}MR^2\omega^2 - \frac{3}{4}Mv_0^2 = MgR(1 - \cos\theta)$

Since the linear velocity of the centre of mass at any instant during this circular path is $v = \omega R$,

So, $R^2\omega^2 = v^2$:

$\frac{3}{4}Mv^2 = \frac{3}{4}Mv_0^2 + MgR(1 - \cos\theta)$

$v^2=v_0^2+\frac{4}{3} \operatorname{gR}(1-\cos \theta) \ldots (i)$

The centre of mass travels along a circular arc of radius R around the pivot P .

For motion along the radial direction pointing toward the centre of rotation P .

The component of gravity pulling toward the pivot is $\mathrm{Mg} \cos \theta$

The normal contact force exerted by the corner pulling outward away from the pivot is N

The net inward centripetal force is $Mg\\cos\theta - N = \frac{Mv^2}{R}$

The cylinder loses contact with the corner when the normal contact force drops to zero $N = 0$:

$Mg\cos\theta = \frac{Mv^2}{R} \rightarrow v^2 = gR\cos\theta \ldots \text{(ii)}$

So, from energy conservation $v^2 = v_0^2 + \frac{4}{3}gR - \frac{4}{3}gR\cos\theta \ldots \text{(i)}$

From the radial force equilibrium $\cos\theta = \frac{v^2}{gR} \rightarrow \frac{4}{3}gR\cos\theta = \frac{4}{3}v^2 \ldots \text{(i)}$

$v^2 = v_0^2 + \frac{4}{3}gR - \frac{4}{3}v^2$

$v^2 + \frac{4}{3}v^2 = v_0^2 + \frac{4}{3}gR$

$\frac{7}{3}v^2 = v_0^2 + \frac{4}{3}gR$

The initial velocity is $v_0 = \sqrt{\frac{gR}{3}}$, $\rightarrow v_0^2 = \frac{gR}{3}$.

$\frac{7}{3}v^2 = \frac{gR}{3} + \frac{4}{3}gR$

$\frac{7}{3}v^2 = \frac{5}{3}gR$

$7v^2 = 5gR$

$v^2 = \frac{5gR}{7}$

$v = \sqrt{\frac{5gR}{7}}$

Therefore, the correct option is (B).

The moment of inertia of single rod about the axis passing through its center and making angle $\theta$ along the length of rod is :

$ \mathrm{I}_{\text {center }}=\frac{1}{12} \mathrm{ml}^2 \sin ^2 \theta $

If the axis instead passes through one end of the rod at an angle $\theta$,

$ \mathrm{I}_{\mathrm{end}}=\frac{1}{3} \mathrm{ml}^2 \sin ^2 \theta $

Structure (P) consists of two uniform rods, each of mass m and length l, joined at a right angle ($90^\circ$) at point C.

The axis OCO' passes through the junction C and bisects the angle, making an angle $\theta = 45^\circ$ with both rods.

Since the axis passes through the end of both rods,

$I_1 = \frac{1}{3}ml^2\sin^2(45^\circ) = \frac{1}{3}ml^2\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{6}ml^2$

$I_2 = \frac{1}{3}ml^2\sin^2(45^\circ) = \frac{1}{3}ml^2\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{6}ml^2$

The total moment of inertia of structure (P) is the sum of the two parts:

$I_P = I_1 + I_2 = \frac{1}{6}ml^2 + \frac{1}{6}ml^2 = \frac{1}{3}ml^2$

$P \rightarrow 5$

Rods CA and CB have one end attached at vertex C, which lies directly on the axis of rotation. Each rod is inclined at an angle $\theta = 60^\circ$ to the axis.

$I_{CA} = \frac{1}{3} m l^2 \sin^2(60^\circ) = \frac{1}{3} m l^2 \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{3} m l^2 \left(\frac{3}{4}\right) = \frac{1}{4} m l^2$

$I_{CB} = \frac{1}{3} m l^2 \sin^2(60^\circ) = \frac{1}{3} m l^2 \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{3} m l^2 \left(\frac{3}{4}\right) = \frac{1}{4} m l^2$

Base rod AB is horizontal and completely parallel to the axis OCO'.

This distance is equal to the altitude (height h) of the equilateral triangle.

$h = l \sin(60^\circ) = \frac{\sqrt{3}}{2} l$

Since the entire mass m of rod AB is distributed uniformly at a constant distance h from the axis, its moment of inertia is:

$I_{AB} = m h^2 = m \left(\frac{\sqrt{3}}{2} l\right)^2 = \frac{3}{4} m l^2$

So, the total moment of inertia for structure Q

$I_Q = I_{CA} + I_{CB} + I_{AB}$

$I_Q = \frac{1}{4} m l^2 + \frac{1}{4} m l^2 + \frac{3}{4} m l^2 = \frac{5}{4} m l^2$

$Q \rightarrow 1$

The rods $A B$ and $B C$ meet at vertex $B$.

The axis passing through A and C , it forms a right-angled triangle ABC where the diagonal AC is the axis. The angle made by axis with every rod is $45^{\circ}$. As all four sides are symmetric about axis of rotation therefore, the total moment of inertia of the square loop about its diagonal axis is:

$ \mathrm{I}=\mathrm{I}_{\mathrm{AB}}+\mathrm{I}_{\mathrm{BC}}+\mathrm{I}_{\mathrm{CD}}+\mathrm{I}_{\mathrm{DA}}=4 \times \frac{1}{3} \mathrm{ml}^2 \sin ^2 45^{\circ}=4 \times \frac{1}{3} \mathrm{ml}^2 \times \frac{1}{2}=\frac{2}{3} \mathrm{ml}^2 $

R→ 4

Rods CA and CB have one end attached at vertex C, which lies directly on the axis of rotation. Each rod is inclined at an angle $\theta = 30^{\circ}$ to the axis.

$I_{CA} = \frac{1}{3}ml^2 \sin^2(30^{\circ}) = \frac{1}{3}ml^2\left(\frac{1}{2}\right)^2 = \frac{1}{3}ml^2\left(\frac{1}{4}\right) = \frac{1}{12}ml^2$

$I_{CB} = \frac{1}{3}ml^2 \sin^2(30^{\circ}) = \frac{1}{3}ml^2\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{3}ml^2\left(\frac{1}{4}\right) = \frac{1}{12}ml^2$

So, the total moment of inertia for structure S

$I_S = I_{CA} + I_{CB}$

$I_S = \frac{1}{12}ml^2 + \frac{1}{12}ml^2 = \frac{2}{12}ml^2 = \frac{1}{6}ml^2$

$S \rightarrow 2$

So, the correct matches are $\mathrm{P} \rightarrow 5, \mathrm{Q} \rightarrow 1, \mathrm{R} \rightarrow 4, \mathrm{~S} \rightarrow 2$.

Therefore, the correct option is (A).

First, relate the length of the rod to the circumference of the ring:

$ L = 2\pi R $

The mass of the ring, denoted by $ M $, is the product of the linear mass density and the length of the rod:

$ M = \lambda \times L $

The moment of inertia of a ring about a diameter is given by the formula:

$ I = \frac{MR^2}{2} $

Substituting the expression for mass $ M $ and using the relation for $ R $ derived from the circumference:

$ I = \frac{\lambda \times L}{2} \times \left(\frac{L}{2\pi}\right)^2 $

Simplifying the expression, we arrive at:

$ I = \frac{\lambda L^3}{8\pi^2} $

This is the moment of inertia of the ring about any of its diameters.

Let's examine each expression step by step:

$\vec{\tau} = \vec{r} \times \vec{L}$

  Here, $\vec{L}$ is the angular momentum. However, torque is defined as the time derivative of angular momentum:

  $\vec{\tau} = \frac{d\vec{L}}{dt},$

  not as the cross product of the position vector with the angular momentum. In fact, if you write

  $\vec{r} \times \vec{L} = \vec{r} \times (\vec{r} \times \vec{p}),$

  you don't obtain the standard expression for torque. Thus, this expression is not correct.

$\vec{\tau} = \frac{d}{dt}(\vec{r} \times \vec{p})$

  For a particle, the angular momentum is defined as

  $\vec{L} = \vec{r} \times \vec{p}.$

  Taking the time derivative gives

  $\frac{d}{dt}(\vec{r} \times \vec{p}) = \frac{d\vec{r}}{dt} \times \vec{p} + \vec{r} \times \frac{d\vec{p}}{dt}.$

  Since $\frac{d\vec{r}}{dt} = \vec{v}$ and $\vec{p} = m\vec{v},$ the term

  $\vec{v} \times m\vec{v}$

  is zero. This simplifies to

  $\vec{\tau} = \vec{r} \times \frac{d\vec{p}}{dt},$

  which is a standard expression for torque. So, this expression is correct.

$\vec{\tau} = \vec{r} \times \frac{d \vec{p}}{d t}$

  This is the standard definition of torque, as $\frac{d \vec{p}}{d t}$ is the net force $\vec{F}.$ Hence, we can also write

  $\vec{\tau} = \vec{r} \times \vec{F}.$

  This expression is correct.

$\vec{\tau} = I \vec{\alpha}$

  This relation applies to rigid bodies rotating about a fixed axis (where the moment of inertia $I$ is constant and can be treated as a scalar). It is a common form used in rotational dynamics, although one must be cautious since it is a special case. In the context of this problem, it is acceptable as a correct expression.

$\vec{\tau} = \vec{r} \times \vec{F}$

  This is the fundamental definition of torque in physics. It directly relates the force applied to a particle and its lever arm. This expression is clearly correct.

To summarize:

Expression A is not a standard or generally valid expression for torque.

Expressions B, C, D, and E are acceptable under the usual assumptions in mechanics.

Looking at the provided options, the correct answer is:

  Option C: B, C, D and E Only.

Let's analyze the situation step by step.

The particle’s position vector is given by

$\vec{r} = a(\hat{i}\cos\omega t + \hat{j}\sin\omega t),$

which represents uniform circular motion in the xy-plane.

Its velocity is the time derivative of the position:

$\vec{v} = \frac{d\vec{r}}{dt} = -a\omega\sin\omega t\,\hat{i} + a\omega\cos\omega t\,\hat{j}.$

The acceleration, obtained by differentiating the velocity, is:

$\vec{a} = \frac{d\vec{v}}{dt} = -a\omega^2\cos\omega t\,\hat{i} - a\omega^2\sin\omega t\,\hat{j}.$

Notice that this can be rewritten as:

$\vec{a} = -\omega^2 \vec{r}.$

This indicates that the acceleration (and hence the force, since $\vec{F}=m\vec{a}$) is directed opposite to the position vector $\vec{r}$—that is, radially inward.

To check against the given options:

Option A (Opposite to $\vec{L}$): The angular momentum $\vec{L}=m\vec{r}\times\vec{v}$ is perpendicular to the plane of motion (along $\pm\hat{k}$) and does not indicate a direction in the plane.

Option B (Opposite to $\vec{L}\times\vec{P}$): For a particle in circular motion, when you compute $\vec{L}\times\vec{P}$ (with $\vec{P} = m\vec{v}$), you’ll find that it is proportional to $-\vec{r}.$ Thus, the direction opposite to $\vec{L}\times\vec{P}$ would be the same as $\vec{r}$, which is the outward radial direction—not matching the inward (centripetal) force.

Option D (Opposite to $\vec{P}$): The linear momentum is tangential; the force (centripetal) is not tangential but directed radially inward.

Option C (Opposite to $\vec{r}$): As we found, the acceleration (and hence force) is given by $\vec{F}= m\vec{a} = -m\omega^2 \vec{r},$ which is clearly opposite to $\vec{r}.$

Therefore, the force is directed opposite to $\vec{r},$ making Option C the correct answer.

Torque about bottom point

$\begin{aligned} & \mathrm{F} \times 2 \mathrm{r}=\mathrm{I} \alpha \\ & 49 \times 2 \mathrm{r}=\frac{7}{5} \mathrm{mr}^2 \alpha \\ & 14=4 \mathrm{r} \alpha \end{aligned}$

As sphere rolls without slipping

$\begin{aligned} & \mathrm{a}=\mathrm{r} \alpha \\ & \mathrm{a}=\frac{14}{4}=\frac{7}{2}=3.5 \mathrm{~m} / \mathrm{s}^2 \end{aligned}$

The moment of inertia of a circular ring with mass $ M $ and diameter $ r $ about a tangential axis lying in the plane of the ring is calculated as follows:

Given the diameter $ r $ of the ring, the radius $ R $ is $ \frac{r}{2} $.

The formula for the moment of inertia about a tangential axis in the plane of the ring is:

$ I_{\text{tangential}} = \frac{3}{2} M \left(\frac{R}{2}\right)^2 $

Substitute the value of $ R = \frac{r}{2} $ into the equation:

$ I_{\text{tangential}} = \frac{3}{2} M \left(\frac{r}{2}\right)^2 = \frac{3}{8} Mr^2 $

Therefore, the moment of inertia of the ring about the specified axis is $\frac{3}{8} Mr^2$.

$\alpha=\frac{\mathrm{M} \ell^2}{12}\quad\text{..... (1)}$

$\begin{aligned} &\begin{aligned} & \alpha^{\prime}=2\left[\frac{\frac{\mathrm{M}}{2}\left(\frac{\ell}{2}\right)^2}{12}\right] \\ & \alpha^{\prime}=\frac{\mathrm{M} \ell^2}{48}=\frac{\alpha}{4} \end{aligned}\\ &\text { Correct option is (2) } \end{aligned}$

$\begin{aligned} &\text { Since the lamina is equilibrium. }\\ &\therefore \mathrm{F}_{\text {net }}=0 \& \tau_{\text {net }}=0 \end{aligned}$

$\mathrm{T}_{\mathrm{o}}=10 \ell-\mathrm{F} \ell \Rightarrow \mathrm{F}=10 \mathrm{~N}$

To determine the angular velocity of the wheel after the cord is unwound by 1 meter, we start by calculating the work done by the force applied on the cord.

Work Done by Force (W_F):

$ W_F = 20 \, \text{N} \times 1 \, \text{m} = 20 \, \text{J} $

Relationship with Kinetic Energy:

The work done (W_F) is converted into the change in kinetic energy of the wheel. Therefore, we have:

$ \Delta \text{KE} = 20 \, \text{J} = \frac{1}{2} I \omega^2 $

where $ I $ is the moment of inertia of the wheel, and $ \omega $ is the angular velocity.

Determine the Moment of Inertia (I):

Since the wheel has mass $ M = 10 \, \text{kg} $ and radius $ R = 10 \, \text{cm} = 0.1 \, \text{m} $, and knowing it's a solid disk:

$ I = MR^2 = 10 \times 0.1^2 = 0.1 \, \text{kg} \cdot \text{m}^2 $

Solve for Angular Velocity ($\omega$):

Apply the equation for kinetic energy:

$ 20 = \frac{1}{2} \times 0.1 \times \omega^2 $

Solving for $\omega$, we get:

$ 20 = 0.05 \omega^2 $

$ \omega^2 = \frac{20}{0.05} = 400 $

$ \omega = \sqrt{400} = 20 \, \text{rad/s} $

Therefore, the angular velocity of the wheel after the cord is unwound by 1 meter is $ 20 \, \text{rad/s} $.

$\begin{aligned} & \tau_{\text {Net }}=0 \Rightarrow(400 \mathrm{~g} \times 30)=(250 \mathrm{~g} \times 10)(\mathrm{mg} \times 50) \\ & \mathrm{m}=\frac{12000-2500}{50}=\frac{9500}{50} \\ & \mathrm{M}=190 \mathrm{~g} \end{aligned}$

$\begin{aligned} & \mathrm{t}=\sqrt{\frac{2 \ell}{\mathrm{a}_{\mathrm{cm}}}} \\ & \mathrm{a}_{\mathrm{cm}}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}_{\mathrm{cm}}}{\mathrm{MR}^2}} \\ & \mathrm{a}_1=\mathrm{a}_{\mathrm{cm}_1}=\frac{5 \mathrm{~g} \sin \theta}{7} \ldots . . \text { Solid } \\ & \mathrm{a}_2=\mathrm{a}_{\mathrm{cm}_2}=\frac{3 \mathrm{~g} \sin \theta}{5} \ldots . \text { Hollow } \\ & \mathrm{a}_1>\mathrm{a}_2 \\ & \mathrm{t}_1<\mathrm{t}_2 \end{aligned}$

$\begin{aligned} & \frac{\text { Linear KE }}{\text { Rotational K.E }}=\frac{\frac{1}{2} \mathrm{mv}_{\mathrm{cm}}^2}{\frac{1}{2} \mathrm{I} \omega^2} \\ & \frac{\mathrm{mv}_{\mathrm{cm}}^2}{\frac{2}{5} \mathrm{mR}^2 \omega^2}=\frac{5}{2} \quad(\mathrm{~V}=\omega \mathrm{R}) \end{aligned}$

The acceleration of the center of mass for a rolling object down an incline is given by:

$ a = \frac{g \sin\theta}{1 + \frac{I}{m r^2}} $

For a uniform solid cylinder, the moment of inertia about its central axis is:

$ I = \frac{1}{2} m r^2 $

Substituting this into the expression for acceleration:

$ a = \frac{g \sin\theta}{1 + \frac{\frac{1}{2} m r^2}{m r^2}} = \frac{g \sin\theta}{1 + \frac{1}{2}} = \frac{g \sin\theta}{\frac{3}{2}} = \frac{2}{3} g \sin\theta $

For an incline of $45^\circ$, we have:

$ \sin 45^\circ = \frac{\sqrt{2}}{2} $

Thus, the acceleration becomes:

$ a = \frac{2}{3} g \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2} g}{3} $

Therefore, the linear acceleration of the cylinder's axis is:

$ \frac{\sqrt{2} g}{3} $

Moment of Inertia and Angular Motion

For a solid circular disk, the moment of inertia is given by

$ I = \frac{1}{2}MR^2. $

The angular position is defined as

$ \theta(t) = 5t^2 - 8t. $

Angular Velocity and Acceleration

Differentiate with respect to time to obtain the angular velocity:

$ \omega(t) = \frac{d\theta}{dt} = 10t - 8. $

Differentiating again, the angular acceleration is:

$ \alpha(t) = \frac{d\omega}{dt} = 10. $

At time $ t = 2 \, \text{s} $:

Angular velocity:

$ \omega(2) = 10(2) - 8 = 12 \, \text{rad/s}. $

Angular acceleration:

$ \alpha(2) = 10 \, \text{rad/s}^2. $

Torque Calculation

The torque applied by the external force is related to the moment of inertia and angular acceleration:

$ \tau = I \alpha = \frac{1}{2}MR^2 \times 10 = 5MR^2. $

Power Delivered

Power delivered by a torque is given by:

$ P = \tau \omega. $

At $ t = 2 \, \text{s} $, substitute the values:

$ P = 5MR^2 \times 12 = 60MR^2. $

Thus, the power delivered by the applied torque at $ t = 2 \, \text{s} $ is:

$ \boxed{60MR^2}. $

Let's analyze the problem step-by-step.

Energy conservation:

When the sphere rolls without slipping, its gravitational potential energy converts into both translational and rotational kinetic energy. The energy conservation equation is given by:

$mgL\sin\theta = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

where:

$L\sin\theta$ is the vertical height,

$I$ is the moment of inertia, and

$\omega$ is the angular speed.

Moment of Inertia and Rolling Condition:

For a solid sphere, the moment of inertia about its center is:

$I = \frac{2}{5}mr^2$

Since the sphere rolls without slipping, the linear speed $v$ and the angular speed $\omega$ are related by:

$v = r\omega \quad \Rightarrow \quad \omega = \frac{v}{r}$

Total Kinetic Energy:

Substituting the rolling condition into the rotational kinetic energy gives:

$ \begin{aligned} K &= \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right) \left(\frac{v}{r}\right)^2 \\ &= \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right) \frac{v^2}{r^2} \\ &= \frac{1}{2}mv^2 + \frac{1}{5}mv^2 \\ &= \frac{7}{10}mv^2 \end{aligned} $

Finding $v^2$ in Terms of $L$ and $\sin\theta$:

Equate the initial potential energy to the final kinetic energy:

$mgL\sin\theta = \frac{7}{10}mv^2$

Solving for $v^2$:

$v^2 = \frac{10}{7}gL\sin\theta$

Apply to the Two Cases:

For $\theta = 30^\circ$, since $\sin(30^\circ) = \frac{1}{2}$:

$v_1^2 = \frac{10}{7}gL\left(\frac{1}{2}\right) = \frac{5}{7}gL$

For $\theta = 45^\circ$, since $\sin(45^\circ) = \frac{1}{\sqrt{2}}$:

$v_2^2 = \frac{10}{7}gL\left(\frac{1}{\sqrt{2}}\right)$

Calculate the Ratio $\frac{v_1^2}{v_2^2}$:

$ \begin{aligned} \frac{v_1^2}{v_2^2} &= \frac{\left(\frac{5}{7}gL\right)}{\left(\frac{10}{7}gL\frac{1}{\sqrt{2}}\right)} \\ &= \frac{5}{10}\sqrt{2} \\ &= \frac{\sqrt{2}}{2} \end{aligned} $

Expressing this ratio as $v_1^2 : v_2^2$, we have:

$v_1^2 : v_2^2 = 1 : \sqrt{2}$

Conclusion:

According to the options given, the correct answer is:

Option C: $1:\sqrt{2}$.

First, torque ($\vec{\tau}$) is found by taking the cross product of the position vector ($\vec{r}$) and the force vector ($\vec{F}$):

$\vec{\tau} = \vec{r} \times \vec{F}$

Here, $\vec{r} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{F} = 2\hat{i} + \hat{j} + 2\hat{k}$.

We use a determinant to find the cross product: $ \vec{\tau} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 1 & 2 \\ \end{array} \right| $

Expanding the determinant:

$\vec{\tau} = \hat{i}[(1)(2)-(1)(1)] -\hat{j}[(1)(2)-(1)(2)] + \hat{k}[(1)(1)-(1)(2)]$

This gives: $\vec{\tau} = \hat{i}(2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 - 2)$

So, $\vec{\tau} = \hat{i} - 0\hat{j} - \hat{k}$

Mass of removed disc, $m = {M \over {\pi {R^2}}} \times \pi {\left( {{R \over 2}} \right)^2}$

$ \Rightarrow m = {M \over 4}$

$I = {I_1} - {I_2}$

where, I$_1$ = MOI of whole disc about perpendicular axis through O

I$_2$ = MOI of removed dis about $ \bot $ axis through O

$ \Rightarrow I = {{M{R^2}} \over 2} - \left[ {{{{M \over 4}{{\left( {{R \over 2}} \right)}^2}} \over 2} + {M \over 4}{{\left( {{R \over 2}} \right)}^2}} \right]$ (using parallel axis theorem)

$ \Rightarrow I = {{M{R^2}} \over 2} - \left( {{{M{R^2}} \over {32}} + {{M{R^2}} \over {16}}} \right)$

$ \Rightarrow I = M{R^2}\left( {{1 \over 2} - {1 \over {32}} - {1 \over {16}}} \right) = {{13} \over {32}}M{R^2}$

$ \Rightarrow I = {{13} \over {32}}M{R^2}$

Let's disc rolls to $\phi$ angular displacement about its centre:

As the disc rolls without slipping. we can write,

$ x=r \phi=(R-r) \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Now, Let's apply $\tau=I \alpha$ about contact

$ \begin{aligned} & \text { point } \\ & \tau=m g \sin \theta \cdot r+k x r=\left(\frac{m r^2}{2}+m r^2\right) \alpha^{\prime} \\ & \Rightarrow m g \theta \cdot r+k(R-r) \theta r=\frac{3 m r^2}{2} \alpha^{\prime} \\ & \Rightarrow[m g+k(R-r)] \theta r=\frac{3}{2} m r^{\prime} \alpha^{\prime} \\ & \Rightarrow[m g+k(R-r)] \theta=\frac{3}{2} m r \alpha^{\prime}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

here, $\alpha^{\prime}=\frac{d^2 \phi}{d t^2}$

from (1), $\phi=\left(\frac{R-r}{r}\right) 0$

$ \begin{aligned} & \Rightarrow \quad \alpha^{\prime}=\frac{d^2 \phi}{d t^2}=\left(\frac{R-r}{r}\right) \frac{d^2 \theta}{d t^2} \\ & \Rightarrow \quad \alpha^{\prime \prime}=\left(\frac{R-r}{r}\right) \alpha\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $

Now, from (2) and (3),

$ \begin{aligned} & {[m g+k(R-r)] \theta=\frac{3}{2} m r\left(\frac{R-r}{r}\right) \alpha} \\ & \Rightarrow[m g+k(R-r)] \theta=\frac{3}{2} m(R-r) \alpha \end{aligned} $

Now, we know, for SHM,

$ \begin{gathered} \alpha=\omega^2 \theta \\ \Rightarrow \quad\left[\frac{m g+k(R-r)}{\frac{3}{2} m(R-r)}\right] \phi=\omega^2 \phi \\ \Rightarrow \frac{2}{3} \frac{g}{(R-r)}+\frac{2}{3} \frac{k}{m}=\omega^2 \\ \Rightarrow \omega=\sqrt{\frac{2}{3}\left[\frac{g}{(R-r)}+\frac{k}{m}\right]} \end{gathered} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \frac{K_{\mathrm{rot}}}{K_{\mathrm{trans}}}=\frac{\frac{1}{2} I \omega^2}{\frac{1}{2} m v^2} \\ & \quad=\frac{\frac{1}{2} \times \frac{2}{5} m r^2 \times\left(\frac{v}{r}\right)^2}{\frac{1}{2} m v^2}=\frac{\frac{1}{5}}{\frac{1}{2}} \\ & \quad=\frac{2}{5} \\ & \therefore K_{\mathrm{rot}}: K_{\text {trans }}=2: 5 \end{aligned} \end{aligned} $

Moment of inertia of thin rod about an axis perpendicular to its length and passing through one end

$ \begin{aligned} & I=\frac{1}{3} M L^2 \\ \text { But } I= & M K^2 \\ \therefore \quad & M K^2=\frac{1}{3} M L^2 \\ \Rightarrow \quad & K=\frac{L}{\sqrt{3}} \Rightarrow \frac{K}{L}=\frac{1}{\sqrt{3}} \\ \Rightarrow \quad & K: L=1: \sqrt{3} \end{aligned} $

Density $\rho$, the length of the wire is

$ L=\frac{m}{\rho} $

Since, the wire is bent into a circular ring, its length is equal to the circumference

$ L=2 \pi R $

Therefore, $R=\frac{L}{(2 \pi)}=\frac{\left(\frac{m}{\rho}\right)}{(2 \pi)}=\frac{m}{(2 \pi \rho)}$

The moment of inertia of a circular ring about its diameter is given by

$ I_D=\frac{1}{2} m R^2 $

Using the parallel axis theorem, the moment of inertia about an axis parallel to the diameter at a distance $R$ (the radius) from the centre is

$ \begin{aligned} I_T & =I_D+m R^2 \\ & =\frac{1}{2} m R^2+m R^2=\frac{3}{2} m R^2 \\ & =\frac{3}{2} m\left(\frac{m}{2 \pi \rho}\right)^2=\frac{3 m^3}{8 \pi^2 \rho^2} \end{aligned} $

The acceleration (a) of a body rolling down on an inclined plane without slipping

$ a=\frac{g \sin \theta}{1+\frac{I}{M R^2}} $

For solid sphere, $I=\frac{2}{5} M R^2$

$ \therefore \quad a_s=\frac{g \sin \theta}{1+\frac{\frac{2}{5}}{M R^2} M R^2} $

$ \begin{aligned} &\begin{aligned} a_s & =\frac{5}{7} g \sin \theta \\ \text { but } \quad a_s & =3 \mathrm{~m} / \mathrm{s}^2 \\ \therefore \quad 3 & =\frac{5}{7} g \sin \theta \Rightarrow \sin \theta=\frac{21}{5 \mathrm{~g}} \end{aligned}\\ &\text { Thus, for disc, } I=\frac{1}{2} M R^2\\ &\begin{aligned} \therefore \quad a_d & =\frac{g \sin \theta}{\frac{1}{2} M R^2}=\frac{2}{3} g \sin \theta \\ & 1+\frac{M R^2}{M} \\ & =\frac{2}{3} g \times \frac{21}{5 g}=\frac{14}{5}=2.8 \mathrm{~m} / \mathrm{s}^2 \end{aligned} \end{aligned} $

Step 1: Understanding the Setup

We have a uniform meter scale that weighs 100 g. Each end of the scale has a plate that weighs 200 g. The scale is balanced on a support at the 45 cm mark. We put a 300 g weight on the plate at the 0 cm end and put vegetables (unknown weight $x$) on the plate at the 100 cm end.

Step 2: Calculating Clockwise Torque

The things making a clockwise turning effect (torque) around the pivot at 45 cm are:

• The 200 g plate at 100 cm, which is 55 cm from the pivot ($100 - 45 = 55$),
• The vegetable weight ($x$) on this plate, also 55 cm from pivot,
• Part of the scale's own mass to the right of the pivot (100 g is the total mass, half of it is on each side, but for 5 cm past pivot, that's $5$ g at 5 cm, but original calculation uses $100 \times 5$). Here we use $100$ g at a distance of $5$ cm ($50 - 45 = 5$ cm, but the calculation uses 5 cm for the rest mass).

Step 3: Calculating Anticlockwise Torque

The things making an anticlockwise torque are:

• The 200 g plate at 0 cm, 45 cm away from pivot,
• The 300 g weight at 0 cm, 45 cm away from pivot.

Step 4: Setting Torques Equal (Balance)

For balance, clockwise torque = anticlockwise torque.

$ \begin{aligned} 11,500 + 55x &= 22,500 \\ 55x &= 22,500 - 11,500 = 11,000 \\ x &= \frac{11,000}{55} = 200~\text{g} \\ \end{aligned} $

So, the actual mass of the vegetables is $200$ g.

Step 5: Calculate the Error

The balance shows $300$ g (apparent weight), but the real weight is $200$ g (actual weight).

Error $= 300 - 200 = 100~\text{g}$

Moment of inertia of ring about a tangential axis in its plane

$ I=I_{\mathrm{CM}}+M R^2=\frac{M R^2}{2}+M R^2 $

[The moment of inertia of a ring about its diameter $I_{\mathrm{CM}}=\frac{M R^2}{2}$ ]

$ I=\frac{3}{2} M R^2 $

If $K_R$ be the radius of gyration, then

$ M K_R^2=\frac{3}{2} M R^2 \Rightarrow K_R=\sqrt{\frac{3}{2}} R $

Moment of inertia of circular disc about tangential axis in its plane

$ \begin{aligned} I & =I_{C M}+M R^2 \\ & =\frac{M R^2}{4}+M R^2=\frac{5}{4} M R^2 \end{aligned} $

[The moment of inertia of a disc about its diameter is $I_{\mathrm{CM}}=\frac{1}{4} M R^2$ ]

$ \begin{aligned} &\text { If } K_C \text { be the radius of gyration, then }\\ &\begin{aligned} M K_C^2 & =\frac{5}{4} M R^2 \\ K_C & =\sqrt{\frac{5}{4}} R \\ \text { Since, } \frac{K_R}{K_C} & =\frac{\sqrt{12}}{\sqrt{K}} \\ \frac{\sqrt{\frac{3}{2}} R}{\sqrt{\frac{5}{4}} R} & =\frac{\sqrt{12}}{\sqrt{K}} \Rightarrow \frac{3}{2} \times \frac{4}{5}=\frac{12}{K} \\ \Rightarrow \quad K & =10 \end{aligned} \end{aligned} $

Step 1: Find the moment of inertia of the disc.

The moment of inertia $(I)$ for a disc about its center is $I = \frac{1}{2} M R^2$

Here, the mass $M = \frac{10}{\pi^2}\ \mathrm{kg}$, and the radius $R = 2\ \mathrm{m}$.

So, $I = \frac{1}{2} \times \frac{10}{\pi^2} \times (2)^2 = \frac{1}{2} \times \frac{10}{\pi^2} \times 4 = \frac{20}{\pi^2}\ \mathrm{kg\,m}^2$

Step 2: Change revolutions per minute (rev/min) to angular speed (rad/s).

The first speed is $90\ \mathrm{rev}/\mathrm{min}$. To convert to radians per second:

$w_1 = 90 \times \frac{2\pi}{60} = 3\pi\ \mathrm{rad/s}$

The second speed is $120\ \mathrm{rev}/\mathrm{min}$: $w_2 = 120 \times \frac{2\pi}{60} = 4\pi\ \mathrm{rad/s}$

Step 3: Use the rotational kinetic energy formula to find work done.

The work done to increase speed equals the change in rotational kinetic energy: $\text{Work Done} = \Delta KE = \frac{1}{2}I(w_2^2 - w_1^2)$

Plug in the values: $\text{Work} = \frac{1}{2} \times \frac{20}{\pi^2} \left[(4\pi)^2 - (3\pi)^2\right]$

Calculate:

$(4\pi)^2 = 16\pi^2$

$(3\pi)^2 = 9\pi^2$

So, $16\pi^2 - 9\pi^2 = 7\pi^2$

Therefore, $\text{Work} = \frac{1}{2} \times \frac{20}{\pi^2} \times 7\pi^2 = 10 \times 7 = 70\ \mathrm{J}$

When ice from the polar regions melts and flows towards the equator, the mass distribution of the Earth changes. This shift in mass causes in increases in the Earth's moment of inertia. According to the principle of conservation of angular momentum, if the angular momentum ( $L=I \omega$ ) remains constant and the moment of inertia ( $I$ ) increases, the angular velocity $(\omega)$ must decrease. A decrease in angular velocity means the Earth rotates more slowly, leading to an increase in the duration of a day.

Moment of inertia of ring about its center:

$ I_{\mathrm{CM}}=M R^2 $

Using the parallel axis theorem:

The parallel axis theorem says: If you know the moment of inertia of an object about its center ($I_{\mathrm{CM}}$), then the moment of inertia about a parallel axis a distance $R$ away is:

$ I = I_{\mathrm{CM}} + M R^2 $

For a thin circular ring, this axis through the edge is a distance $R$ from the center, so:

$ I = M R^2 + M R^2 = 2 M R^2 $

So, $ M R^2 = \frac{I}{2} \qquad \cdots (i) $

Finding the moment of inertia about the diameter:

The moment of inertia of a ring about its diameter is:

$ I_d = \frac{M R^2}{2} $

Substituting from equation (i):

Replace $M R^2$ with $\frac{I}{2}$: $ I_d = \frac{\frac{I}{2}}{2} = \frac{I}{4} $

As we know,

Moment of inertia about axis of cylinder,

$ I_{\text {long }}=\frac{1}{2} M R^2 $

Moment of inertia about an axis through centre and perpendicular to length.

$ I_{\text {transverse }}=\frac{1}{12} M L^2+\frac{1}{4} M R^2 $

By using, given ratio,

$ \begin{aligned} & \frac{1}{2} M R^2=\frac{1}{n}\left(\frac{1}{12} M L^2+\frac{1}{4} M R^2\right) \\ & \left(\frac{2 n-1}{4}\right) R^2=\frac{1}{12} L^2 \\ & 3(2 n-1)=\frac{L^2}{R^2} \Rightarrow \frac{L}{R}=\sqrt{3(2 n-1)} \end{aligned} $

$ \begin{aligned} &\text { Moment of inertia of solid cylinder }\\ &\begin{aligned} I & =\frac{M R^2}{2}=\frac{2.5 \times(0.1)^2}{2} \\ & =1.25 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^2 \\ & =0.0125 \mathrm{~kg}-\mathrm{m}^2 \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} \omega & =\omega_0+\alpha t \\ \Rightarrow \quad \alpha & =\frac{\omega-\omega_0}{t}=\frac{21-6}{1.5} \\ & =10 \mathrm{rad} / \mathrm{s}^2 \\ \text { Since, } L & =I \omega \\ \Rightarrow \quad \frac{d L}{d t} & =I \frac{d \omega}{d t}=I \alpha \\ & =100 \times 10^{-3} \times 10=1 \mathrm{Nm} \end{aligned} $