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).

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}$

To determine the new angular velocity of the system, we use the principle of conservation of angular momentum. When no external torque acts on a system, its angular momentum remains constant. Let's denote the initial angular momentum and the final angular momentum, respectively, as $L_{\text{initial}}$ and $L_{\text{final}}$.

The initial angular momentum of the system is given by:

$ L_{\text{initial}} = I_{\text{initial}} \cdot \omega $

Here, $I_{\text{initial}}$ is the moment of inertia of the first disc. For a thin circular disc, the moment of inertia about its center is:

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

Thus,

$ L_{\text{initial}} = \left( \frac{1}{2} M R^2 \right) \omega $

When the second disc is placed gently on the first disc, the two discs rotate together with a common angular velocity $\omega'$. The moment of inertia of the second disc is:

$ I_{\text{second}} = \frac{1}{2} \left( \frac{M}{2} \right) R^2 = \frac{1}{4} M R^2 $

The combined moment of inertia of the system after placing the second disc is:

$ I_{\text{final}} = I_{\text{initial}} + I_{\text{second}} = \frac{1}{2} M R^2 + \frac{1}{4} M R^2 = \frac{3}{4} M R^2 $

Thus, the final angular momentum of the system is:

$ L_{\text{final}} = I_{\text{final}} \cdot \omega' = \left( \frac{3}{4} M R^2 \right) \omega' $

By the conservation of angular momentum:

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

This simplifies to:

$ \left( \frac{1}{2} M R^2 \right) \omega = \left( \frac{3}{4} M R^2 \right) \omega' $

Solving for $\omega'$:

$ \omega' = \frac{\left( \frac{1}{2} M R^2 \right) \omega}{\left( \frac{3}{4} M R^2 \right)} = \frac{\omega}{\frac{3}{2}} = \frac{2}{3} \omega $

So, the new angular velocity of the system is:

$\omega' = \frac{2}{3} \omega$

Thus, the correct answer is:

Option D: $\frac{2}{3} \omega$

For hollow sphere

$\begin{aligned} & I_1=\frac{2}{3} m R^2=\left(\sqrt{\frac{2}{3}} R\right)^2 m \\ & \therefore \quad K_1=\sqrt{\frac{2}{3}} R \end{aligned}$

For solid cylinder

$\begin{aligned} & I_2=\frac{1}{4} m R^2+\frac{m}{3}(4 R)^2 \\ & =\left(\sqrt{\frac{67}{12}} R\right)^2 m \\ & \therefore \quad K_2=\sqrt{\frac{67}{12}} R \\ & \Rightarrow \frac{K_1}{K_2}=\sqrt{\frac{2}{3} \times \frac{12}{67}} \\ & =\sqrt{\frac{8}{67}} \end{aligned}$

$\Rightarrow \quad x=67$

Initial Conditions:

Consider a solid disc of mass $ M $ and radius $ R $ rolling without slipping on a horizontal surface. The center of mass of the disc has an initial speed $ v $.

Kinetic Energy of the Rolling Disc:

For an object that rolls without slipping, its total kinetic energy $ K $ is the sum of its translational kinetic energy and its rotational kinetic energy.

Translational Kinetic Energy:

$ K_{\text{trans}} = \frac{1}{2} M v^2. $

Rotational Kinetic Energy:

The moment of inertia $ I $ of a solid disc about its center is:

$ I = \frac{1}{2} M R^2. $

Since the disc rolls without slipping, the angular velocity $\omega$ is related to the linear speed by:

$ v = \omega R \implies \omega = \frac{v}{R}. $

Therefore, the rotational kinetic energy is:

$ K_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{v^2}{R^2}\right) = \frac{1}{4} M v^2. $

Total Initial Kinetic Energy:

Summing these gives:

$ K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2. $

Energy Conservation as the Disc Climbs the Incline:

As the disc moves up the incline, there are no dissipative forces (assuming a smooth incline and no slipping), so energy is conserved. The total kinetic energy at the bottom will be converted entirely into gravitational potential energy at the maximum height $ h $:

$ K_{\text{total}} = M g h. $

Substitute $ K_{\text{total}} = \frac{3}{4} M v^2 $:

$ \frac{3}{4} M v^2 = M g h. $

Canceling $ M $ from both sides:

$ \frac{3}{4} v^2 = g h. $

Thus,

$ h = \frac{3}{4} \frac{v^2}{g}. $

Final Answer:

$ \boxed{\frac{3}{4} \frac{v^2}{g}} $

This corresponds to Option A.

$\begin{aligned} & \mathrm{L}=m u \cos \theta H \\ & =m u \cos \theta \times \frac{u^2 \sin ^2 \theta}{2 g} \\ & =\frac{m u^3}{2 g} \times \frac{\sqrt{3}}{2} \times\left(\frac{1}{2}\right)^2=\frac{\sqrt{3} m u^3}{16 g} \end{aligned}$

First, let us restate the problem in our own words:

We have a uniform iron bar of weight $12\,\mathrm{kgf}$ (meaning its mass is $12\,\mathrm{kg}$, so its weight is $12g\,\mathrm{N}$ in SI units).

One end of the bar is on the ground (assume a frictionless contact), and the other end is on a man’s shoulder (also assume frictionless contact).

The bar makes an angle of $60^\circ$ with the horizontal.

We want to find how much load (“effective weight”) the man feels on his shoulder.

Under these (common) assumptions of frictionless contacts at both ends, each contact force is perpendicular to the bar. We can solve the problem by taking torques about the ground contact point.

1. Labeling and geometry

Let the length of the bar be $L$.

The bar is inclined at $60^\circ$ from the horizontal.

Because the bar is uniform, its center of gravity (C.O.G.) lies at its midpoint, i.e., at a distance $L/2$ from either end.

Let the weight of the bar be $W = 12g$.

Let the reaction force from the man’s shoulder be $R$. (Magnitude unknown.)

Let the reaction force at the ground be $R_g$. (We do not actually need its exact value to answer the question.)

We will place our origin (for torque calculation) at the ground contact point.

2. Torques about the ground contact

Torque due to the bar’s weight

The weight $W = 12g$ acts downward at the midpoint of the bar.

The position vector from the pivot (ground) to the bar’s midpoint is $\tfrac{L}{2}$ long and makes $60^\circ$ with the horizontal.

The weight is vertical downward, i.e., $90^\circ$ from the horizontal.

The angle between the position vector ($60^\circ$ from horizontal) and the weight ($90^\circ$ from horizontal) is

$ 90^\circ - 60^\circ \;=\; 30^\circ. $

Hence, the torque magnitude from the weight is

$ \tau_W \;=\; \bigl(12g\bigr)\,\bigl(\tfrac{L}{2}\bigr)\,\sin(30^\circ) \;=\; 12g \cdot \tfrac{L}{2}\cdot \tfrac12 \;=\; 12g \cdot \tfrac{L}{4}. $

Torque due to the man’s reaction force $R$

The man’s shoulder is at the other end of the bar, i.e., $L$ from the pivot.

For frictionless contact, $R$ is perpendicular to the bar. Since the bar is at $60^\circ$ from horizontal, a line perpendicular to the bar is $90^\circ$ away from the bar’s direction, so the angle between the bar’s position vector ($\vec{r}$) and $R$ is $90^\circ$.

Thus, the torque from $R$ about the ground contact is

$ \tau_R \;=\; R \,\times\, L \,\times\, \sin(90^\circ) \;=\; R \cdot L. $

Because the bar is in static equilibrium, the net torque about the ground contact must be zero. Taking the clockwise direction (by weight) as positive, we have

$ \tau_W \;-\; \tau_R \;=\; 0 \quad\Longrightarrow\quad 12g \cdot \frac{L}{4} \;=\; R \cdot L. $

Solving for $R$,

$ R \;=\; \frac{12g}{4} \;=\; 3g. $

Hence, the force on the man’s shoulder has magnitude $3g$ in newtons.

3. Converting to “kg weight”

Often, such problems phrase the answer in “kg of weight” rather than newtons. Since $W = mg$ for a mass $m$ in Earth’s gravity, a force of $3g\,\mathrm{N}$ corresponds to a “weight” of $3\,\mathrm{kgf}$.

So the man effectively feels 3 kg of load on his shoulder.

The correct answer is Both A and R are correct and R is the correct explanation of A.

Explanation:

Assertion A: An electric fan continues to rotate for some time after the current is switched off. This is a correct statement. When you switch off the fan, it doesn't stop immediately but continues to rotate for some time.

Reason R: Fan continues to rotate due to inertia of motion. This is also a correct statement. Inertia is the resistance of any physical object to any change in its state of motion. This includes changes to the object's speed or direction of motion. An object will stay in its state of motion unless a force acts on it. In the case of the fan, after the current is switched off, the fan blades have inertia and continue to move due to this inertia until the frictional forces (like air resistance and friction in the fan's bearings) cause it to stop. Therefore, inertia of motion is the correct reason for the fan's continued motion after the current is switched off.

$a = {{g\sin \theta } \over {1 + {{{K^2}} \over {{R^2}}}}}$

$v = \sqrt {{{2Sg\sin \theta } \over {1 + {{{K^2}} \over {{R^2}}}}}} $

$ \Rightarrow {{{v_c}} \over {{v_{ss}}}}\sqrt {{{1 + {{K_{ss}^2} \over {{R^2}}}} \over {1 + {{K_c^2} \over {{R^2}}}}}} = \sqrt {{{1 + {2 \over 5}} \over {1 + {1 \over 2}}}} $

$ \Rightarrow \sqrt {{{{7 \over 5}} \over {{3 \over 2}}}} = \sqrt {{{14} \over {15}}} $

${\overrightarrow L _0} = {\overrightarrow L _{of\,cm}} + {\overrightarrow L _{about\,cm}}$

$ \Rightarrow {a \over 3}{R^2}\omega = mvR + {2 \over 3}m{R^2}\omega = {5 \over 3}m{R^2}\omega $

$ \Rightarrow a = 5$

$\alpha = {{d\omega } \over {dt}} = 6{t^2} - 2t$

$\int_0^\omega {d\omega = \int_0^t {(6{t^2} - 2t)dt} } $

so $\omega = 2{t^3} - {t^2} + 10$

and ${{d\theta } \over {dt}} = 2{t^3} - {t^2} + 10$

so $\int_4^\theta {d\theta = \int_0^t {(2{t^3} - {t^2} + 10)dt} } $

$\theta = {{{t^4}} \over 2} - {{{t^3}} \over 3} + 10t + 4$

Taking torque from B

${F_w} \times 5 = {3 \over 2}mg$

$ \Rightarrow {F_w} = {3 \over {10}} \times 10 \times 10$

$ = 30\,N$

$N = mg = 100\,N$

and ${f_r} = {F_w} = 30\,N$

so ${F_f} = \sqrt {{N^2} + f_r^2} = \sqrt {10900} = 10\sqrt {109} \,N$

so ${{{F_w}} \over {{F_f}}} = {3 \over {\sqrt {109} }}$

(A) Moment of inertia of solid sphere of radius R about a tangent $ = {2 \over 5}M{R^2} + M{R^2} = {7 \over 5}M{R^2}$

$\Rightarrow$ A $-$ (II)

(B) Moment of inertia of hollow sphere of radius R about a tangent $ = {2 \over 3}M{R^2} + M{R^2} = {5 \over 3}M{R^2}$

$\Rightarrow$ B $-$ (I)

(C) Moment of inertia of circular ring of radius (R) about its diameter = ${{\left( {M{R^2}} \right)} \over 2}$

$\Rightarrow$ C $-$ (IV)

(D) Moment of inertia of circular disc of radius (R) about any diameter

$ = {{M{R^2}/2} \over 2} = {{M{R^2}} \over 4}$

$\Rightarrow$ D $-$ (III)

$m{\omega ^2}({l_0} + x) = kx$

$ \Rightarrow m{\omega ^2}{l_0} = (k - m{\omega ^2}) \times x$

$ \Rightarrow x = {{m{\omega ^2}{l_0}} \over {(k - m{\omega ^2})}}$

$K{E_R} = {1 \over 2}l{w^2}$

$ = {1 \over 2} \times {2 \over 5} \times {\omega ^2} \times (m{R^2})$

$K{E_{total}} = {1 \over 2} \times {7 \over 5} \times m{R^2} \times {\omega ^2}$

$\therefore$ ${{K{E_R}} \over {K{E_{total}}}} = {2 \over 7}$

${I_1}{\omega _1} = {I_2}{\omega _2}$

$M{R^2}{\omega _1} = (M{R^2} + 2m{R^2}){\omega _2}$

${\omega _2} = \left( {{M \over {M + 2m}}} \right){\omega _1}$

${\omega _2} = 2\left( {{M \over {M + 2m}}} \right)$

$\overrightarrow \tau = \overrightarrow r \times \overrightarrow F $

$ = (2\widehat i + \widehat j + 2\widehat k) \times (3\widehat i + 4\widehat j - 2\widehat k)$

$ = - 10\widehat i + 10\widehat j + 5\widehat k$