Past NEET Papers

Hint

As we know in equilibrium net moment of force is equals to zero.

\(
\begin{aligned}
&2 g \times 20=0.5 \mathrm{~g} \times 60+\mathrm{mg} \times 120 \\
&\mathrm{~m}=\frac{0.5}{6} \mathrm{~kg}=\frac{1}{12} \mathrm{~kg}
\end{aligned}
\)

Hint

The moment of inertia of a ring about an axis passing through the centre and perpendicular to the ring is \(I=M R^2\)
Mass of the remaining portion of the ring as shown in figure \((ii)\) is \(M-\frac{M}{4}=\frac{3 M}{4}\)
Moment of inertia of the remaining portion of the ring about the given axis is \(I^{\prime}=\frac{3}{4} M R^2\) But \(I^{\prime}=k M R^2\) (given)
\(
\therefore k=\frac{3}{4}
\)

Hint

The given situation is shown in the figure.

Position vector of centre of mass,
\(
\begin{aligned}
\boldsymbol{R}_{\mathrm{CM}} &=\frac{m_1 \mathbf{r}_1+m_2 \mathbf{r}_2}{M_1+M_2} \\
&=\frac{M O \mathbf{A}+M O B}{M+M} \\
&=\frac{M \times 2 \hat{\mathbf{i}}+M \times 2 \hat{\mathbf{j}}}{2 M}=\hat{\mathbf{i}}+\hat{\mathbf{j}}
\end{aligned}
\)

Hint

Given, position vector, \(\mathbf{r}=2 \hat{\mathbf{k}} \mathrm{m}\)
Force, \(\mathbf{F}=3 \hat{\mathbf{j}} \mathrm{N}\)
As, torque, \(\tau=\mathbf{r} \times \mathbf{F}=2 \hat{\mathbf{k}} \times 3 \hat{\mathbf{j}}=6(-\hat{\mathbf{i}})[latex] [latex]=-6 \hat{\mathbf{i}} \mathrm{N}\mathrm{m}\)
Hence, the correct option is (b).

Hint

Given, \(m_1=5 \mathrm{~kg}, m_2=10 \mathrm{~kg}\) and \(r=1 \mathrm{~m}=100 \mathrm{~cm}\)
Let the centre of mass lies at origin O.
\(
\begin{aligned}
&\therefore \quad \frac{m_1 r_1-m_2 r_2}{m_1+m_2}=0\\
&\Rightarrow \quad 5 r_1-10 r_2=0\\
&\Rightarrow \quad r_2=\frac{r_1}{2}\\
&\text { Also, } \quad r_1+r_2=100\\
&\Rightarrow \quad r_1+\frac{r_1}{2}=100\\
&\Rightarrow \quad 3 r_1=200\\
&\Rightarrow \quad r_1=\frac{200}{3}=67 \mathrm{~cm}\\
&\text { Hence, correct option is (b). }
\end{aligned}
\)

Hint

Initial angular speed of wheel,
\(
\begin{aligned}
\omega_0 &=2 \pi f_0=2 \pi \times \frac{360}{60} \mathrm{rad} / \mathrm{s} \\
&=12 \pi \mathrm{rad} / \mathrm{s}
\end{aligned}
\)
Final angular speed of wheel,
\(
\begin{aligned}
\omega &=2 \pi \mathrm{f} \\
&=2 \pi \times \frac{1200}{60} \mathrm{rad} / \mathrm{s}=40 \pi \mathrm{rad} / \mathrm{s} \\
t &=14 \mathrm{~s}
\end{aligned}
\)
From the equation of rotational motion,
\(\omega=\omega_0+\alpha t\)
\(\Rightarrow \quad \alpha=\frac{\omega-\omega_0}{t}\)
\(=\frac{40 \pi-12 \pi}{14}=\frac{28 \pi}{14}=2 \pi \mathrm{rad} / \mathrm{s}^2\)

The correct answer is (a).

Hint

Let \(\mathrm{T}_{\mathrm{A}}\) and \(\mathrm{T}_{\mathrm{B}}\) are the time periods of particle \(A\) and \(B\) respectively. According to question,
\(\mathrm{T}_{\mathrm{A}}=\mathrm{T}_{\mathrm{B}}=\mathrm{T}\)
If \(\omega_A\) and \(\omega_B\) are their angular speeds, then
\(\omega_A=\frac{2 \pi}{T_A}\) and \(\omega_B=\frac{2 \pi}{T_B}\)
\(
\therefore \frac{\omega_{\mathrm{A}}}{\omega_{\mathrm{B}}}=\frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{TA}}=\frac{\mathrm{T}}{\mathrm{T}}= 1: 1
\)

Hint

From third equation of motion for circular motion
\(
\begin{aligned}
&\qquad \omega^2-\omega_0^2=2 \alpha \theta \dots(i)\\
&\text { where, } \omega=\text { final angular velocity of particle } \\
&\omega_0=\text { initial angular velocity } \\
&\alpha=\text { angular acceleration and } \\
&\theta=\text { angular displacement } \\
&\text { Here, } \omega=\frac{v_0}{r} \mathrm{rad} / \mathrm{s} \text { (where, } r \text { radius of } \\
&\begin{array}{l}
\omega_0=0 \text { (initially particle is at rest) } \\
\theta=2 \pi n \text { (for } n \text { rounds) }
\end{array} \\
&\begin{array}{l}
\text { Substituting these values in Eq. (i), } \\
\text { we get } \\
\Rightarrow \quad\left(\frac{v_0}{r}\right)^2-0=2 \alpha(2 \pi n)
\end{array} \\
&\quad \alpha=\frac{v_0^2}{4 \pi n r^2} \mathrm{rad} / \mathrm{s}^2
\end{aligned}
\)

Hint

Key Idea According to the work-energy theorem, the change in kinetic energy of a particle is the amount of work done on the particle to move, i.e.
\(
W=-\Delta K E=K E_f-K E_i
\)
Given, mass of cylinder, \(m=2 \mathrm{~kg}\)
radius of cylinder, \(r=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}\)
rotational velocity, \(\omega=3 \mathrm{rpm}\)
\(=3 \times \frac{2 \pi}{60}=\frac{\pi}{10} \mathrm{rad} / \mathrm{s}\) and \(\theta=2 \pi\) revolution \(=2 \pi \times 2 \pi=4 \pi^2 \mathrm{rad}\).
The work done in rotating an object by an angle \(\theta\) from rest is given by \(W=\tau \theta\)
As the cylinder is brought to rest, so the work done will be negative.
According to work-energy theorem,
Work done \(=\) Change in rotational kinetic energy
\(
\begin{aligned}
&-\tau \theta=\frac{1}{2} I \omega_{\mathrm{f}}^2-\frac{1}{2} I \omega_1^2=\frac{1}{2} I\left(\omega_{\mathrm{f}}^2-\omega_i^2\right) \\
&\Rightarrow \quad \tau \quad \frac{I\left(-\omega_i^2\right)}{2 \theta} \quad\left[\because \omega_{\mathrm{f}}=0\right] \\
&\quad=\frac{1}{2}\left(\frac{1}{2} m r^2\right) \frac{\omega_i^2}{\theta}
\end{aligned}
\)
\(
\begin{aligned}
&\quad\left[I=\frac{1}{2} M R^2 \text { (for cylinder) }\right] \\
&=\frac{1}{4} m r^2 \frac{\omega^2}{\theta} \quad \quad\left[\because \omega_i=\omega\right] \\
&=\frac{1}{4} \times 2 \times\left(4 \times 10^{-2}\right)^2 \times\left(\frac{\pi}{10}\right)^2 \times \frac{1}{4 \pi^2} \\
&=\frac{1}{4} \times 2 \times 16 \times 10^{-4} \times \frac{\pi^2}{100} \times \frac{1}{4 \pi^2} \\
&=\frac{2}{100} \times 10^{-4}=2 \times 10^{-6} \mathrm{~N}\mathrm{m}
\end{aligned}
\)

Hint

(a) Work done to stop the disc \(=\) change in total kinetic energy of disc
Final \(\mathrm{KE}=0\)
Initial KE = Translational K.E. \(+\) Rotational K.E.
\(
\begin{aligned}
&=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} l \omega^2 \\
&=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \times \frac{\mathrm{mR}^2}{2} \times\left(\frac{\mathrm{v}}{\mathrm{R}}\right)^2 \\
&=\frac{1}{2} \mathrm{mv}^2+\frac{1}{4} \mathrm{mv}^2=\frac{3}{4} \mathrm{mv}^2 \\
&=\frac{3}{4} \times 100 \times\left(20 \times 10^{-2}\right)^2=3 \mathrm{~J} \\
&|\Delta \mathrm{KE}|=3 \mathrm{~J}
\end{aligned}
\)

Hint

When a body rolls i.e. have rotational motion, the total kinetic energy of the system will be
\(
K E=\frac{1}{2} m v^2\left(1+\frac{k^2}{R^2}\right)
\)
where, \(m=\) mass of body, \(v=\) velocity and \(k=\) radius of gyration
Given, \(m=2 \mathrm{~kg}, \theta=30^{\circ}, \mathrm{v}=4 \mathrm{~ms}^{-1}\)
Let \(h\) be the height of the inclined plane, then from law of conservation of energy,
(1V) \(\begin{aligned} K E &=P E \\ \frac{1}{2} m v^2\left(1+\frac{K^2}{R^2}\right) &=m g h \end{aligned}\)
Substituting the given values in the above equation, we get
\(
\begin{aligned}
\frac{1}{2} \times 2 \times 16\left(1+\frac{1}{2}\right) &=2 \times 10 \times \mathrm{h} \\
& {\left[\because \text { For cylinder } \frac{k^2}{R^2}=\frac{1}{2}\right] } \\
\Rightarrow \quad & 8 \times \frac{3}{2}=10 \mathrm{~h} \Rightarrow h=1.2 \mathrm{~m}
\end{aligned}
\)
From the above diagram
\(
\begin{aligned}
\sin \theta &=\frac{h}{x} \\
\Rightarrow x=\frac{h}{\sin \theta}=\frac{1.2}{\sin 30^{\circ}}=12 \times 2=2.4 \mathrm{~m} \\
{\left[\because \sin 30^{\circ}=\frac{1}{2}\right] }
\end{aligned}
\)

Hint

The moment of inertia of a rotating solid sphere about its symmetrical (diametric) axis is given as, \(I=\frac{2}{5} m r^2\)

The rotational kinetic energy of a solid sphere is
\(
\begin{aligned}
K_t &=\frac{1}{2} I \omega^2 \\
&=\frac{1}{2} \times \frac{2}{5} m r^2 \omega^2=\frac{1}{5} m r^2 \omega^2
\end{aligned}
\)
Angular velocity, \(\omega=v r\)
As, we know that external torque,
\(
\tau_{\text {ext }}=\frac{d L}{d t}
\)
where, \(L\) is the angular momentum.
Since, in the given condition, \(\tau_{\text {ext }}=0\) \(\Rightarrow \frac{d L}{d t}=0\) or \(L=\) constant
Hence, when the radius of the sphere is increased keeping its mass the same, only the angular momentum remains constant. But other quantities like the moment of inertia, rotational kinetic energy, and angular velocity changes.

Hint

In rolling motion, rotational translational kinetic energy.
\(
\mathrm{K}_{\mathrm{t}}=\frac{1}{2} m v^2
\)
\(
\begin{aligned}
&\text { And, } \mathrm{K}_{\mathrm{t}}+\mathrm{K}_{\mathrm{r}}=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2\\
&=\frac{1}{2} m v^2+\frac{1}{2}\left(\frac{2}{5} m r^2\right)\left(\frac{v}{r}\right)^2=\frac{7}{10} m v^2\\
&\therefore \quad \frac{K_t}{K_t+K_r}=\frac{\frac{1}{2} m v^2}{\frac{7}{10} m v^2}=\frac{5}{7}
\end{aligned}
\)

Hint

Work done required to bring an object to rest is given as
\(
W=\frac{1}{2} I \omega^2
\)
where, \(I\) is the moment of inertia and \(\omega\) is the angular velocity.
Since, here all the objects spin with the same \(\omega\), this means,
\(
\begin{aligned}
&\text { As, } I_A \text { (for a solid sphere) }=\frac{2}{5} M R^2 \\
&I_B \text { (for a thin circular disk) }=\frac{1}{2} M R^2 \\
&\begin{aligned}
I_C(\text { for a circular ring) }&=M R^2 \\
\therefore \quad W_A: W_B: W_C &=I_A: I_B: I_C \\
&=\frac{2}{5} M R^2: \frac{1}{2} M R^2: M R^2 \\
&=\frac{2}{5}: \frac{1}{2}: 1 \\
&=4: 5: 10 \\
\Rightarrow \quad W_A &<W_B<W_C
\end{aligned}
\end{aligned}
\)

Hint

Key Concept Moment of force is defined as the cross product of the force and the force arm.
\(
\text { Given, } \quad \begin{aligned}
\quad & \mathbf{F}=4 \hat{\mathbf{j}}+5 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} \\
\mathbf{r}_1=2 \hat{\mathbf{i}}+0 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\
\quad \mathbf{r}_2=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}
\end{aligned}
\)
Moment of force \(=\mathbf{r} \times \mathbf{F}\)
\(=\left(\mathbf{r}_1-\mathbf{r}_2\right) \times \mathbf{F}\)
\(=[-(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})+(2 \hat{\mathbf{i}}+0 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})]\)
\(\times[4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}]\)
\(=[0 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-1 \hat{\mathbf{k}}] \times[4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}]\)
\(=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & 2 & -1 \\ 4 & 5 & -6\end{array}\right|\)
\(
\begin{aligned}
&=\hat{\mathbf{i}}[(-6 \times 2)-(-1 \times 5)] \\
&-\hat{\mathbf{j}}[(-6 \times 0)-(-1 \times 4)]+\hat{\mathbf{k}}[(0 \times 5)-2 \times 4] \\
&=-7 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}
\end{aligned}
\)

Hint

Mechanical advantage, M. A. \(=\frac{\text { Load }}{\text { Effort }}\) If M.A. \(>1 \Rightarrow\) Load \(>\) Effort
Centre of mass may or may not coincide with centre of gravity. Net torque of gravitational pull is zero about centre of mass.
\(\tau_g=\Sigma \tau_i=\Sigma r_i \times m_{i g}=0\)
The correct answer is option d

Hint

Given, mass of cylinder \(\mathrm{m}=3 \mathrm{~kg}\), \(\mathrm{R}=40 \mathrm{~cm}=0.4 \mathrm{~m}\), \(\mathrm{F}=30 \mathrm{~N} ; \alpha=\)?
As we know, torque \(\tau=I \alpha\) \(\mathrm{F} \times \mathrm{R}=\mathrm{MR}^2 \alpha\)
\(
\begin{aligned}
&\alpha=\frac{F \times R}{\mathrm{MR}^2} \\
&\alpha=\frac{30 \times(0.4)}{3 \times(0.4)^2} \text { or, } \alpha=25 \mathrm{rad} / \mathrm{s}^2
\end{aligned}
\)

The correct answer is c.

Hint

When no external torque acts on system then, angular momentum of system remains constant. Angular momentum before contact \(\mathrm{I} \omega_1+\mathrm{I} \omega_2=2 \mathrm{I} \omega\)

\(\omega=\frac{\omega_1+\omega_2}{2}\)

Angular momentum after the discs brought into contact
\(
=I_{\text {net }} \omega=\left(I_1+I_2\right) \omega
\)
So, final angular speed of system \(=\omega\)
\(
=\frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2}
\)
\(
\begin{aligned}
&(\text { K.E. })_i=\frac{1}{2} I \omega_1^2+\frac{1}{2} I \omega_2^2 \\
&(\text { K.E. })_f=\frac{1}{2} \times 2 I \omega^2=I\left(\frac{\omega_1+\omega_2}{2}\right)^2
\end{aligned}
\)
Loss in K.E. \(=(\mathrm{K} . \mathrm{E})_{\mathbf{f}}-(\mathrm{K} . \mathrm{E})_{\mathrm{i}}=\frac{1}{4} \mathrm{I}\left(\omega_1-\omega_2\right)^2\)

Hint

Given: Radius of disc, \(R=50 \mathrm{~cm}\) Angular acceleration, \(\alpha=2.0\) rads \(^{-2}\); time \(t=2 \mathrm{~s}\) Particle at periphery (assume) will have both radial (one) and tangential acceleration \(a_t=R \alpha=0.5 \times 2=1 \mathrm{~m} / \mathrm{s}^2\)
Angular speed
\(\omega=\omega_0+\alpha t\) \(\omega=0+2 \times 2=4 \mathrm{rad} / \mathrm{sec}\) Centripetal acceleration
\(\mathrm{a}_{\mathrm{c}}=\omega^2 \mathrm{R}=(4)^2 \times 0.5=16 \times 0.5=8 \mathrm{~m} / \mathrm{s}^2\), Linear acceleration, \(\mathrm{a}_{\mathrm{t}}=\alpha . \mathrm{R}=1 \mathrm{~m} / \mathrm{sec}^2\)
Net acceleration,
\(\mathrm{a}_{\text {total }}=\sqrt{\mathrm{a}_{\mathrm{t}}^2+\mathrm{a}_{\mathrm{c}}^2}=\sqrt{1^2+8^2} \approx 8 \mathrm{~m} / \mathrm{s}^2\)
The acceleration of the particle moving in a circle has two components; tangential acceleration (\(a_t\)) which is along the tangent and radial acceleration (\(a_r\)) which 15 towards the centre of circle. As the two components are mutually perpendicular therefore, net acceleration of the particle will be \(a=\sqrt{a_t^2+a_r^2}\)

Option a is the correct answer.

Hint

Considering the information given in the question, let us draw the figure

If the above figure is considered, then moment of inertia of disc will be given as
\(
I=I_{\text {remain }}+I_{[R / 2]} \Rightarrow I_{\text {remain }}=I-I_{[R / 2]}
\)
Putting the values, we get
\(
\begin{aligned}
&=\frac{M R^2}{2}-\left[\frac{\frac{M}{4}\left(\frac{R}{2}\right)^2}{2}+\frac{M}{4}\left(\frac{R}{2}\right)^2\right] \\
&=\frac{M R^2}{2}-\left[\frac{M R^2}{32}+\frac{M R^2}{16}\right] \\
&=\frac{M R^2}{2}-\left[\frac{M R^2+2 M R^2}{32}\right] \\
&=\frac{M R^2}{2}-\frac{3 M R^2}{32} \\
&=\frac{16 M R^2}{32}-3 M R^2 \\
\text I_{\text {remain }} &=\frac{13 M R^2}{32}
\end{aligned}
\)

Hint

Time of descent \(\propto \frac{\mathrm{K}^2}{\mathrm{R}^2}\)
Order of value of \(\frac{\mathrm{K}^2}{\mathrm{R}^2}\)
for disc; \(\frac{\mathrm{K}^2}{\mathrm{R}^2}=\frac{1}{2}=0.5\)
for sphere; \(\frac{\mathrm{K}^2}{\mathrm{R}^2}=\frac{2}{5}=0.4\)
(sphere) \(<\) (disc)
\(\therefore \quad\) Sphere reaches first
The value \(\frac{\mathrm{K}^2}{\mathrm{R}^2}\) is a measure of moment of inertia. Its. value is fixed for a particular shape of the body and independent of mass and radius of the body.

The correct option is (a).

Hint

COM(Centre Of Mass) of \(m_1\) and \(m_2\) masses lies at
\(
r=\frac{m_1 r_1+m_2 r_2}{m_1+m_2}
\)
\(\therefore\) Moment of inertia of the point masses about the given axis is
\(
\begin{aligned}
I &=\sum m_i r_i^2 \Rightarrow I=m_1 r_1^2+m_2 r_2^2 \\
&=m_1\left(\frac{m_2 I}{m_1+m_2}\right)^2+m_2\left(\frac{m_1 I}{m_1+m_2}\right)^2 \\
&=\frac{m_1 m_2 I^2}{\left(m_1+m_2\right)^2}\left(m_2+m_1\right)=\frac{m_1 m_2 I^2}{\left(m_1+m_2\right)}
\end{aligned}
\)

Hint

As we know, the kinetic energy of a rotating body,
\(
K E=\frac{1}{2} I \omega^2
\)
\(
=\frac{1I^2 \omega^2}{2 I}
\)
\(
=\frac{L^2}{2 I}
\)
Also, angular momentum, \(L=I \omega\)
Thus, \(\quad K_A=K_B\)
\(\Rightarrow \quad \frac{1}{2} \frac{L_A^2}{I_A}=\frac{1}{2} \frac{L_B^2}{I_B}\)
\(
\begin{aligned}
&\Rightarrow \quad\left(\frac{L_A}{L_B}\right)^2=\frac{I_A}{I_B} \Rightarrow \frac{L_A}{L_B}=\sqrt{\frac{I_A}{I_B}}\\
&\therefore \quad L \propto \sqrt{I}\\
&\therefore \quad L_A<L_B \quad\left[\because I_B>I_A\right]
\end{aligned}
\)

Hint

The correct answer is option (b)
Key Idea KE of a rotating rigid body,
\(K E=\frac{1}{2} I \omega^2\)
\(\therefore\) KE of sphere, \(K_s=\frac{1}{2} I \omega_1^2=\frac{1}{2} \frac{2}{5} m R^2 \omega_1^2\)
\(=\frac{1}{5} m R^2 \omega_1^2\)
KE of cylinder, \(K_c=\frac{1}{2} \frac{1}{2} m R^2 \omega_2^2\)
\(=\frac{1}{4} m R^2 \omega_2^2\)
\(\therefore \quad \frac{K_s}{K_c}=\frac{\frac{m R^2 \omega_1^2}{5}}{\frac{m R^2 \omega_2^2}{4}}\)
\(=\frac{4}{5} \frac{\omega_1^2}{\omega_2^2}=\frac{4}{5} \frac{\omega_1^2}{\left(2 \omega_1\right)^2}\)
\(=\frac{1}{5} \quad\) (given, \(\omega_2=2 \omega_1\) )

Hint

The correct option is (b)
Applying angular momentum conservation
\(
\mathrm{mv}_0 \mathrm{R}_0=(\mathrm{m})(\mathrm{v})\left(\frac{\mathrm{R}_0}{2}\right)
\)
\(
\therefore \quad \mathrm{v}^{\prime}=2 \mathrm{v}_0
\)
Therefore, final \(\mathrm{KE}=\frac{1}{2} \mathrm{~m}\left(2 \mathrm{v}_0\right)^2=2 \mathrm{mv}_0^2\)

Hint

As the weight w balances the normal reactions.

So, \(\quad w=N_1+N_2 \quad\)…(i)
Now balancing torque about the COM,
i.e. anti-clockwise momentum
\(=\) clockwise momentum
\(\Rightarrow \quad N_1 x=N_2(d-x)\) Putting the value of \(N_2\) from Eq. (i), we get
\(
\begin{array}{ll}
& N_1 x=\left(w-N_1\right)(d-x) \\
\Rightarrow & N_1 x=w d-w x-N_1 d+N_1 x \\
\Rightarrow & N_1 d=w(d-x) \\
\Rightarrow & N_1=\frac{w(d-x)}{d}
\end{array}
\)

Hint

As the velocity of an automobile vehicle,
\(
v=54 \mathrm{~km} / \mathrm{h}=54 \times \frac{5}{18}=15 \mathrm{~m} / \mathrm{s}
\)
Angular velocity of a vehicle, \(v=\omega_0 r\) \(\Rightarrow \quad \omega_0=\frac{v}{R}=\frac{15}{0.45}=\frac{100}{3} \mathrm{rad} / \mathrm{s}\)
So, angular acceleration of an automobile,
\(
\begin{aligned}
\alpha &=\frac{\Delta \omega}{t}=\frac{\omega_f-\omega_0}{t}=\frac{0-\frac{100}{3}}{15} \\
&=\frac{-100}{45} \mathrm{rad} / \mathrm{s}^2
\end{aligned}
\)
Thus, the magnitude of average torque transmitted by its brakes to the wheel
\(
\tau=I \alpha \Rightarrow 3 \times \frac{100}{45}=6.66 \mathrm{kgm}^2 \mathrm{~s}^{-2}
\)

Hint

The correct option is (d)
From the figure, Total moment of inertia of the rod,
\(
\begin{aligned}
&\mathrm{I}=\mathrm{m}_1 \mathrm{x}^2+\mathrm{m}_2(\mathrm{~L}-\mathrm{x})^2 \\
&\mathrm{I}=\mathrm{m}_1 \mathrm{x}^2+\mathrm{m}_2\left[\mathrm{~L}^2+\mathrm{x}^2-2 \mathrm{Lx}\right] \\
&\mathrm{I}=\mathrm{m}_1 \mathrm{x}^2+\mathrm{m}_2 \mathrm{~L}^2+\mathrm{m}_2 \mathrm{x}^2-2 \mathrm{~m}_2 \mathrm{Lx} \\
&\text { As I is minimum, so, } \frac{\mathrm{dI}}{\mathrm{dx}}=0 \\
&\Rightarrow \frac{\mathrm{dI}}{\mathrm{dx}}=2 \mathrm{~m}_1 \mathrm{x}+2 \mathrm{~m}_2 \mathrm{x}-2 \mathrm{~m}_2 \mathrm{~L}
\end{aligned}
\)
Using equation (i)
\(
\begin{aligned}
&\Rightarrow \quad 2 \mathrm{~m}_1 \mathrm{x}+2 \mathrm{~m}_2 \mathrm{x}-2 \mathrm{~m}_2 \mathrm{~L}=0 \\
&\Rightarrow \quad \mathrm{x}=\frac{\mathrm{m}_2 \mathrm{~L}}{\left(\mathrm{~m}_1+\mathrm{m}_2\right)}
\end{aligned}
\)
When \(\mathrm{I}\) is minimum, the work done by rotating a rod will be minimum.

Alternate way:

Work required to set the rod rotating with angular velocity \(\omega_0\).
K.E. \(=\frac{1}{2} \mathrm{I} \omega_0^2\)
Work is minimum, when \(I\) is minimum. I is minimum about the centre of mass.
\(
\begin{aligned}
&\therefore m_1 x_1=m_2(L-x) \\
& m_1 x-m_2 L-m_2 x \\
&\therefore x=\frac{m_2 L}{m_1+m_2}
\end{aligned}
\)

Hint

(d) From Newton’s second law for rotational motion,
\(\vec{\tau}=\frac{\mathrm{d} \overrightarrow{\mathrm{L}}}{\mathrm{dt}}\), if \(\overrightarrow{\mathrm{L}}=\) constant then \(\vec{\tau}=0\)
So, \(\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}=0\)
\((2 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}) \times(\alpha \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})=0\)
Solving we get \(\alpha=-1\)

Hint

(d) Moment of inertia of shell 1 along the diameter
\(
\mathrm{I}_{\text {diameter }}=\frac{2}{3} \mathrm{Mr}^2
\)
Moment of inertia of shell \(2=\) Moment of Inertia of shell 3
Usning the parallel axis theorem
\(
=\mathrm{I}_{\text {tangential }}=\frac{2}{3} \mathrm{Mr}^2+\mathrm{Mr}^2=\frac{5}{3} \mathrm{Mr}^2
\)
\(
\begin{aligned}
&\text { So, I of the system along } X X^{\prime}\\
&=\mathrm{I}_{\text {diameter }}+\left(\mathrm{I}_{\text {tangential }}\right) \times 2\\
&\text { or, } \mathrm{I}_{\text {total }}=\frac{2}{3} \mathrm{Mr}^2+\left(\frac{5}{3} \mathrm{Mr}^2\right) \times 2\\
&=\frac{12}{3} \mathrm{Mr}^2=4 \mathrm{Mr}^2
\end{aligned}
\)

Hint

Given, \(m=50 \mathrm{~kg}, r=0.5 \mathrm{~m}\),
\(\alpha=2 \mathrm{rev} / \mathrm{s}^2 = =4 \pi \mathrm{rad} / \mathrm{s}^2 \text { (given) }
\)
Torque produced by the tension in the string.
\(
=T \times r=T \times 0.5=\frac{T}{2} \mathrm{~N}\mathrm{m}
\)
\(
\mathrm{I}_{\text {cylinder }}=\frac{1}{2} \mathrm{Mr}^2=\frac{1}{2}(50)(0.5)^2=\frac{25}{4} \mathrm{Kg}-\mathrm{m}^2
\)
As \(\tau=\mathrm{I} \alpha\) so \(\mathrm{Tr}=\mathrm{I} \alpha\)
\(
\Rightarrow \mathrm{T}=\frac{\mathrm{I \alpha}}{\mathrm{r}}=\frac{\left(\frac{25}{4}\right)(4 \pi)}{(0.5)} \mathrm{N}=50 \pi \mathrm{N}=157 \mathrm{~N}
\)

Hint

In this case, \(a_1=\frac{g \sin \theta}{1+\frac{k^2}{R^2}}=\frac{g \sin \theta}{1+\frac{(2 / 5) R^2}{R^2}}\)
\(\left[\therefore\right.\) for solid sphere, \(\left.k^2=\frac{2}{5} R^2\right]\)
\(=\frac{g \sin \theta}{7 / 5} \Rightarrow a_1=\frac{5}{7} g \sin \theta\)
For a sphere slipping down an inclined plane
\(
\Rightarrow a_2=g \sin \theta \Rightarrow \frac{a_1}{a_2}=\frac{5 / 7 g \sin \theta}{g \sin \theta}
\)
\(
\Rightarrow \quad \frac{a_1}{a_2}=\frac{5}{7}
\)

Hint

Weight of the rod will produce the torque

Torque on the rod = Moment of weight of the rod about \(P\)
\(
\tau=\operatorname{Mg} \frac{L}{2} \dots(i)
\)
\(\because\) Moment of inertia of rod
\(
\text { about } P=\frac{M L^2}{3} \dots(ii)
\)
As
\(
\tau=\mid \alpha
\)
From Eqs. (i) and (ii), we get
\(
M g \frac{L}{2}=\frac{M L^2}{3} \alpha \Rightarrow \alpha=\frac{3 g}{2 L}
\)

Hint

The correct option is \(\mathrm{c}  100 \mathrm{rad} / \mathrm{s}\)
\(\mathrm{MOI}\) of disc \(D\), about an axis passing through its centre and normal to its plane. \(\mathrm{l}_1=\frac{M R^2}{2}=\frac{2 \times(0.2)^2}{2}=0.04 \mathrm{~kg}-\mathrm{m}^2\)
Initial angular velocity of disc \(D_1\), \(\omega_1=50 \mathrm{rad} / \mathrm{s}\)
\(\mathrm{MOI}\) of disc \(\mathrm{D}_2\) about an axis passing through its centre and normal to the plane.
\(
\begin{aligned}
&I_2=\frac{M R^2}{2}=\frac{4 \times(0.1)^2}{2} \\
&=0.02 \mathrm{~kg}-\mathrm{m}^2
\end{aligned}
\)
Initial angular velocity of disc \(D_2\),
\(
\omega_2=200 \mathrm{rad} / \mathrm{s}
\)
Total angular momentum of the two discs, initially is
\(
L_i=l_1 \omega_1+I_2 \omega_2
\)
When both discs are brought in contact, axes of rotation coincide.
Consider both discs as a system. Hence, final angular momentum of the system is,
\(
L_f=\left(I_1+I_2\right) \omega
\)
Here, \(\omega\) is the final angular speed of the system.
According to the law of conservation of angular momentum, we get,
\(
\mathrm{L}_{\mathrm{i}}=\mathrm{L}_{\mathrm{f}}
\)
\(I_1 \omega_1+I_2 \omega_2=\left(I_1+I_2\right) \omega\)
\(
\begin{aligned}
&\Rightarrow \omega=\frac{(0.04) \times(50)+(0.02) \times(200)}{(0.04+0.02)} \\
&\Rightarrow \omega=100 \mathrm{rad} / \mathrm{s}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&I=M K^2 \therefore K=\sqrt{\frac{I}{M}} \\
&I_{\text {ring }}=M R^2 \text { and } I_{\mathrm{disc}}=\frac{1}{2} M R^2 \\
&\frac{K_1}{K_2}=\sqrt{\frac{I_1}{I_2}}=\sqrt{\frac{M R^2}{\left(\frac{M R^2}{2}\right)}}=\sqrt{2}: 1
\end{aligned}
\)

Hint

From conservation of energy \(\frac{1}{2} m v^2+\frac{1}{2} I \omega^2=m g\left(\frac{3 v^2}{4 g}\right)\)
For pure rolling condition, \(\omega=\frac{V}{R}\)
\(
\begin{aligned}
&\frac{1 \mathrm{I}}{2 \mathrm{R}^2}=\frac{3}{4} \mathrm{~m}-\frac{1}{2} \mathrm{~m} \\
&\Rightarrow \mathrm{I}=\frac{\mathrm{mR}^2}{2}
\end{aligned}
\)
which is the equation of a disc.

Alternate solution

\(
\begin{aligned}
&\mathrm{v}=\sqrt{\frac{2 g h}{1+\frac{k^2}{\mathrm{r}^2}}} \\
&\mathrm{v}^2=\frac{2 \mathrm{~g} 3 \mathrm{v}^2}{4 \mathrm{~g}\left(1+\frac{\mathrm{k}^2}{\mathrm{r}^2}\right)} \\
&\Rightarrow 1+\frac{\mathrm{k}^2}{\mathrm{r}^2}=\frac{3}{2} \\
&\mathrm{k}^2=\frac{1}{2} \mathrm{r}^2 \rightarrow \operatorname{disc}
\end{aligned}
\)

Hint

There is no external force so centre of mass of the system will not shift.

Hint

\(
\begin{aligned}
&X_{\mathrm{cm}}=\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3} \\
&X_{\mathrm{cm}}=\frac{300 \times(0)+500(40)+400 \times 70}{300+500+400} \\
&X_{\mathrm{cm}}=\frac{500 \times 40+400 \times 70}{1200} \\
&X_{\mathrm{cm}}=\frac{50+70}{3}=\frac{120}{3}=40 \mathrm{~cm}
\end{aligned}
\)

Hint

We know \(
\vec{L}=\vec{r} \times \vec{p}
\)
By right hand screw, rule, the direction of \(\vec{L}\) is \(\perp\) to the plane containing \(\vec{r}\) and \(\vec{p}\).

Hint

Let \(r\) be the perpendicular distance of \(F_1, F_2\) and \(F_3\) from O as shown in figure.

The torque of force \(F_3\) about O is clockwise, while torque due to \(F_1\) and \(F_2\) are anticlockwise.
For total torque to be zero about O, we must have
\(
\begin{aligned}
F_1 r+F_2 r-F_3 r &=0 \\
F_3 &=F_1+F_2
\end{aligned}
\)

Hint

(c) 

As the system is initially at rest, therefore,
initial angular momentum \(L_i=0\).
According to the principle of conservation of angular momentum, final angular momentum, \(L_f=0\)

Angular momentum = Angular momentum of man is in opposite direction of platform.
i.e., \(\quad m v R=I \omega\)
or \(\omega=\frac{m v R}{I}=\frac{50 \times 1 \times 2}{200}=\frac{1}{2} \mathrm{rad} \mathrm{s}^{-1}\)
Angular velocity of man relative to platform is
\(
\omega_r=\omega+\frac{v}{R}=\frac{1}{2}+\frac{1}{2}=1 \mathrm{rad} \mathrm{s}^{-1}
\)
Time taken by the man to complete one revolution is
\(
T=\frac{2 \pi}{\omega_r}=\frac{2 \pi}{1}=2 \pi \mathrm{s}
\)

Hint

(a) Moment of inertia of circular disc \(=\frac{1}{2} m R^2\). Thus, as the distance between the centre and the point increases, the moment of inertia increases ( B has the highest distance from the origin).

Hint

(d) \(\theta(t)=2 t^3-6 t^2\)
\(
\Rightarrow \frac{d \theta}{d t}=6 t^2-12 t \Rightarrow \alpha=\frac{d^2 \theta}{d t^2}=12 t-12=0
\)
\(
\therefore \quad \mathrm{t}=1 \mathrm{sec} .
\)
When angular acceleration \((\alpha)\) is zero then torque on the wheel becomes zero.

Hint

(b) By theorem of parallel axes, \(\mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Md}^2\)
\(
\mathrm{I}=\mathrm{I}_0+\mathrm{M}(\mathrm{L} / 2)^2=\mathrm{I}_0+\mathrm{ML}^2 / 4
\)

Hint

(c) K.E. \(=\frac{\mathrm{L}^2}{2 \mathrm{I}}\)
The angular momentum \(\mathrm{L}\) remains conserved about the centre.
That is, L is constant.
\(
\mathrm{I}=\mathrm{mr}^2
\)
\(
\therefore \mathrm{K} . \mathrm{E} .=\frac{\mathrm{L}^2}{2 \mathrm{mr}^2}
\)
In 2nd case, K.E. \(=\frac{\mathrm{L}^2}{2\left(\mathrm{mr}^{\prime 2}\right)}\)
But \(\mathrm{r}^{\prime}=\frac{\mathrm{r}}{2}\)
\(
\therefore K . E^{\prime}=\frac{L^2}{2 m \cdot \frac{r^2}{4}}=\frac{4 \mathrm{~L}^2}{2 \mathrm{mr}^2} \Rightarrow \mathrm{K} \cdot \mathrm{E}^{\prime}=4 \mathrm{~K} . \mathrm{E}.
\)
\(\therefore\) K.E. is increased by a factor of 4.

Hint

(b) If no external force action a system of particles, the centre of mass remains at rest. So, the speed of centre of mass is zero. Force acting on the rigid body,
\(
\vec{F}=M \vec{a}_{\mathrm{cm}}=\frac{M d^2 \vec{r}}{d t^2}
\)
When external force acting on the rigid body is zero.
\(\vec{F}=M \frac{d}{d t}\left(\vec{V}_{\mathrm{cm}}\right)=0 \Rightarrow \vec{V}_{\mathrm{cm}}=\) Constant
So, when no external force acts on a body its centre of mass will remain at rest or move with constant velocity.

Hint

(d) By conservation of angular momentum, \(\mathrm{I}_{\mathrm{t}} \omega_i=\left(\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right) \omega_f\)
where \(\omega_f\) is the final angular velocity of disks
\(
\therefore \quad \omega_f= \frac{I_t}{\left(I_t+I_b\right)} \omega_i
\)
\(
\begin{aligned}
&\text { Loss in K.E., } \Delta K=\text { Initial K.E. }-\text { Final K.E. }\\
&=\frac{1}{2} I_t \omega_i^2-\frac{1}{2}\left(I_t+I_b\right) \omega_f^2\\
&=\frac{1}{2} I_t \omega_i^2-\frac{1}{2}\left(I_t+I_b\right) \frac{I_t^2}{\left(I_t+I_b\right)^2} \omega_i^2\\
&=\frac{1}{2} \omega_i^2 \frac{I_t}{I_t+I_b}\left(I_t+I_b-I_t\right)=\frac{1}{2} \omega_i^2 \frac{I_t I_b}{I_t+I_b}
\end{aligned}
\)

Hint

(a) The position vector of the centre of mass of two particle system is given by
\(
\begin{aligned}
&\overrightarrow{\mathrm{r}}_{\mathrm{cm}}=\frac{\mathrm{m}_1 \overrightarrow{\mathrm{r}}_1+\mathrm{m}_2 \overrightarrow{\mathrm{r}}_2}{\left(\mathrm{~m}_1+\mathrm{m}_2\right)} \\
&=\frac{1}{4}[-8 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}]=-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}
\end{aligned}
\)

Hint

(c)
\(\tau=\mathbf{r} \times \mathbf{F}\), where \(\mathbf{r}=\) position vector
\(\mathbf{F}=\) force \(\Rightarrow \tau=|\mathbf{r}| \cdot|\mathbf{F}| \sin \theta\)
Torque is perpendicular to both \(\mathbf{r}\) and \(\mathbf{F}\).
So, dot product of two vectors will be zero.
\(
\tau \cdot \mathbf{r}=0 \Rightarrow \mathbf{F} \cdot \tau=0
\)

Hint

(b)

Concept Apply parallel axes theorem of moment of inertia.
According to question by applying conservation of angular momentum
\(
l_1 \omega_1=I_2 \omega_2
\)
In the given case
\(
\begin{aligned}
& I_1=M R^2 \\
I_2 &=M R^2+2 m R^2 \\
\omega_1 &=\omega \\
\text { Then, } \quad \omega_2 &=\frac{I_1}{I_2} \omega=\frac{M}{M+2 m} \omega
\end{aligned}
\)

Hint

(d) Moment of inertia of a thin rod of length \(L\) about an axis passing through centre and perpendicular to the rod \(=\frac{1}{12} \mathrm{ML}^2\).
Thus the moment of inertia of the frame. \(\frac{\mathrm{mL}^2}{12}+\frac{\mathrm{mL}^2}{4}=\frac{4 \mathrm{mL}^2}{12}=\frac{\mathrm{mL}^2}{3}\)
Total M.I. \(=4 \times \frac{\mathrm{mL}^2}{3}\)

Hint

As the rod is bent into two equal halves, the mass and length of each half is \(\frac{M}{2}\) and \(\frac{L}{2}\) respectively.

The moment of inertia about an axis passing through its edge and perpendicular to the rod
\(
\begin{aligned}
&=\frac{\text { Mass } \times(\text { length })^2}{3} \\
I &=2 \times \frac{1}{3} \times \frac{M}{2}\left(\frac{L}{2}\right)^2=\frac{M L^2}{12}
\end{aligned}
\)

Alternate Solution:

Mass of each part \(=M / 2\)
Length of each part \(=\mathrm{L} / 2\)
Total M.I. \(=\) Sum of M.I.s of both parts
\(
\begin{aligned}
&=\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^2 \times \frac{1}{3}+\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^2 \times \frac{1}{3} \\
&=2 \times \frac{M}{2} \times \frac{L^2}{4} \times \frac{1}{3}=\frac{M L^2}{I 2}
\end{aligned}
\)

Hint

(b) is the right answer.
As we know the radius of gyration
\(
k=\sqrt{\frac{1}{m}}
\)
So, for two different cases
\(\frac{k_{\text {ring }}}{k_{\text {disc }}}=\sqrt{\frac{l_{\text {ting }}}{l_{\text {disc }}}}=\sqrt{\frac{M R^2}{\frac{1}{2} M R^2}}\)
\(\therefore \quad \frac{k_{\text {ring }}}{k_{\text {disc }}}=\sqrt{2} \Rightarrow \frac{k_{\text {disc }}}{k_{\text {ring }}}=\frac{1}{\sqrt{2}}\)

Hint

(a) Since, \(\theta=\omega_0 t+\frac{1}{2} \alpha t^2\)
Where \(\alpha\) is angular aceleration, \(\omega_0\) is the initial angular speed.
\(
t=2 \mathrm{~s}
\)
\(
\theta=2 \times 2+\frac{1}{2} \times 3(2)^2=4+6=10 \mathrm{rad}
\)

Hint

Weight of the rod will produce torque,
\(
\tau=m g \times \frac{\ell}{2}
\)
Also, \(\tau=I \alpha\)
where, \(\mathrm{I}\) is the moment of inertia \(=\frac{m \ell^2}{3}\) and \(\alpha\) is the angular acceleration
\(
\therefore \frac{m \ell^2}{3} \alpha=m g \times \frac{\ell}{2} \Rightarrow \alpha=\frac{3 g}{2 \ell}
\)

Hint

From the definition of angular momentum,
\(
\begin{array}{r}
\mathbf{L}=\mathbf{r} \times \mathbf{p}=r \mathrm{mv} \sin \phi(-\mathrm{k}) \\
{\left[\begin{array}{l}
\mathbf{r}=\text { position vector } \\
\mathbf{p}=\text { momentum }
\end{array}\right]}
\end{array}
\)
Therefore, the magnitude of \(L\) is \(L=m v r \sin \phi=m v d\)
where, \(d=r \sin \phi\) is the distance of closest approach of the particle to the origin. As \(d\) is same for both the particles, hence \(L_A=L_B\).

Hint

(a) is the correct answer.

Let the length of a small element of the tube be \(d x\).
Mass of this element
where, \(M\) is mass of filled liquid and \(L\) is length of tube. Force on this element
\(d F=d m \times \times \omega^2\)
\(\int_0^F d F=\frac{M}{L} \omega^2 \int_0^L x d x\)
or \(\quad F=\frac{M}{L} \omega^2\left[\frac{L^2}{2}\right]=\frac{M L \omega^2}{2}\)
or \(\quad F=\frac{1}{2} M L \omega^2\)

Hint

(b) M.I. of a uniform circular disc of radius ‘ \(\mathrm{R}\) ‘ and mass ‘ \(M\) ‘ about an axis passing though C.M. and normal to the disc is
\(
\mathrm{I}_{\text {C.M. }}=\frac{1}{2} \mathrm{MR}^2
\)
From the parallel axis theorem,
\(
I_T=I_{C . M .}+M R^2=\frac{1}{2} M R^2+M R^2=\frac{3}{2} M R^2
\)

Hint

(d) The relation between K.E. of rotation and angular momentum,
\(
\begin{aligned}
&K=\frac{L^2}{2 I} \Rightarrow L^2=2 K I \Rightarrow L=\sqrt{2 K I} \\
&\frac{L_1}{L_2}=\sqrt{\left(\frac{K_1}{K_2}\right)\left(\frac{I_1}{I_2}\right)}=\sqrt{\left(\frac{K}{K}\right)\left(\frac{I}{2 I}\right)}=\frac{1}{\sqrt{2}} \\
&L_1: L_2=1: \sqrt{2}
\end{aligned}
\)

Hint

(d) Net work done by frictional force when the drum rolls down without slipping is zero.
\(
W_{\text {net }}=0
\)
\(
\begin{aligned}
W_{\text {trans. }}+W_{\text {rot. }} &=0 ; \Delta K_{\text {trans. }}+\Delta K_{\text {rot. }}=0 \\
\Delta K_{\text {trans }} &=-\Delta K_{\text {rot }}
\end{aligned}
\)
i.e., converts translation energy to rotational energy.

Hint

(d)

\(\mathrm{I}=2 \mathrm{kgm}^2, v_0=60 \mathrm{rpm}=1 \mathrm{rps}\)

\(\omega_0=2 \pi v_0=2 \pi \mathrm{rad} / \mathrm{sec}\)
\(\omega_f=0\) and \(t=1 \mathrm{~min}=60 \mathrm{sec}\)
So, \(\propto=\frac{\omega_f-\omega_0}{t}=\frac{0-2 \pi}{60}\)
\(\alpha=\frac{-\pi}{30} \mathrm{rad} / \sec ^2\)
and Torque, \(\tau=\) I. \(\alpha\)
\(
\tau=2 \cdot\left(\frac{-\pi}{30}\right)=\frac{-\pi}{15}
\)
\(
\tau=\frac{\pi}{15} \mathrm{~N}-\mathrm{m}
\)

Hint

The system of two given particles of masses \(m_1\) and \(m_2\) are shown in the figure.

Initially the centre of mass
\(
r_{C M}=\frac{m_1 r_1+m_2 r_2}{m_1+m_2} \dots(i)
\)
When mass \(m_1\) moves towards centre of mass by a distance, then let mass \(m_2\) moves a distance \(d^{\prime}\) away from CM to keep the CM in its initial position.
So, \(\quad r_{\mathrm{CM}}=\frac{m_1\left(r_1-d\right)+m_2\left(r_2+d^{\prime}\right)}{m_1+m_2} \dots(ii)\)
Equating Eqs. (i) and (ii), we get
\(
\frac{m_1 r_1+m_2 r_2}{m_1+m_2}=\frac{m_1\left(r_1-d\right)+m_2\left(r_2+d^{\prime}\right)}{m_1+m_2}
\)
\(
\begin{array}{lrl}
\Rightarrow & -m_1 d+m_2 d^{\prime} & =0 \\
\Rightarrow & & d^{\prime}=\frac{m_1}{m_2} d .
\end{array}
\)
If both the masses are equal i.e., \(m_1=m_2\), then second mass will move a distance equal to the distance at which first mass is being displaced.

Hint

(d) Apply conservation of angular momentum
The angular momentum of a disc of moment of inertial \(l_1\) and rotating about its axis with angular velocity \(\omega\) is
\(
L_1=l_1 \omega
\)
When a round disc of moment of inertia \(I_2\) is placed on first disc, then angular momentum of the combination is
\(
L_2=\left(l_1+I_2\right) \omega^{\prime}
\)
In the absence of any external torque, angular momentum remains conserved i.e.,
\(
\begin{aligned}
L_1 &=L_2 \\
l_1 \omega &=\left(l_1+I_2\right) \omega^{\prime} \\
\Rightarrow \quad \omega^{\prime} &=\frac{l_1 \omega}{I_1+I_2}
\end{aligned}
\)

Hint

(d) The moment of inertia of a disc and circular ring about a tangential axis in their planes are respectively. Momentum inertia of disc about the tangential axis
\(
I_d=\frac{5}{4} M_d R^2
\)
Moment of inertia of ring about a tangential axis
\(
\begin{aligned}
&I_r=\frac{3}{2} M_r R^2 \\
&\text { but } \quad I=M k^2 \Rightarrow k=\sqrt{\frac{I}{M}} \\
&\therefore \quad \frac{k_d}{k_r}=\sqrt{\frac{I_d}{I_r} \times \frac{M_r}{M_d}} \\
&\text { or } \frac{k_d}{k_r}=\sqrt{\frac{(5 / 4) M_d R^2}{(3 / 2) M_r R^2} \times \frac{M_r}{M_d}}=\sqrt{\frac{5}{6}} \\
&\therefore k_d: k_r=\sqrt{5}: \sqrt{6}
\end{aligned}
\)

Hint

Moment of inertia of the system about \(A X\) is given by

Moment of inertia \(=m_A r_A^2+m_B r_B^2+m_C r_C^2\)
\(
\left[\begin{array}{l}
r_A=0 \\
r_B=l \\
r_c=l \sin 30^{\circ}
\end{array}\right]
\)
Moment of inertia
\(
\begin{aligned}
&=m(0)^2+m(l)^2+m\left(l \sin 30^{\circ}\right)^2. \\
&=m l^2+\frac{m l^2}{4}=\frac{5}{4} m l^2
\end{aligned}
\)

Hint

Initial angular momentum of ring \(=I \omega=M R^2 \omega\)
If four objects each of mass \(m\), and kept gently to the opposite ends of two perpendicular
diameters of the ring then final angular momentum \(=\left(M R^2+4 m R^2\right) \omega^{\prime}\)
By the conservation of angular momentum
Initial angular momentum = Final angular momentum
\(
M R^2 \omega=\left(M R^2+4 m R^2\right) \omega^{\prime} \Rightarrow \omega^{\prime}=\left(\frac{M}{M+4 m}\right) \omega .
\)

Hint

(a)
The kinetic energy of rotation is
\(
K_{\text {rot }}=\frac{1}{2} I \omega^2=\frac{1}{2} M K^2 \frac{v^2}{R^2} \quad\left[K=\sqrt{\frac{I}{M}}\right]
\)
where, \(k\) is radius of gyration.
Kinetic energy of translation is
\(
K_{\text {trans }}=\frac{1}{2} M v^2
\)
Thus, total energy
\(
\begin{aligned}
E &=K_{\text {rot }}+K_{\text {trans }} \\
&=\frac{1}{2} M k^2 \frac{v^2}{R^2}+\frac{1}{2} M v^2 \\
&=\frac{1}{2} M v^2\left(\frac{k^2}{R^2}+1\right) \\
&=\frac{1}{2} \frac{M v^2}{R^2}\left(k^2+R^2\right)
\end{aligned}
\)
Hence, \(\frac{K_{\text {rot }}}{E}=\frac{\frac{1}{2} M k^2 \frac{v^2}{R^2}}{\frac{1}{2} \frac{M v^2}{R^2}\left(k^2+R^2\right)}\)
\(
=\frac{k^2}{k^2+R^2}
\)

Hint

(c) \(\quad K . E .=\frac{1}{2} \mathrm{I} \omega^2+\frac{1}{2} m v^2\)
\(K . E .=\frac{1}{2}\left(\frac{1}{2} m r^2\right) \omega^2+\frac{1}{2} m v^2\)
\(
=\frac{1}{4} m v^2+\frac{1}{2} m v^2=\frac{3}{4} m v^2
\)
Now, gain in K.E. = Loss in P.E.
\(
\frac{3}{4} m v^2=m g h \Rightarrow v=\sqrt{\left(\frac{4}{3}\right) g h}
\)

Hint

(d) Consider an element of length \(d x\) at a distance \(x\) from end \(A\).
Here, mass per unit length \(\lambda\) of rod
\(\lambda \propto x \Rightarrow \lambda=k x\)
\(\therefore d m=\lambda d x=k x d x\)
Position of centre of gravity of rod from end \(A\).
\(x_{C G}=\frac{\int_0^L x d m}{\int_0^L d m}\)
\(
x_{C G}=\frac{\int_0^3 x(k x d x)}{\int_0^3 k x d x}=\frac{\left[\frac{x^3}{3}\right]_0^3}{\left[\frac{x^2}{2}\right]_0^3}=\frac{\frac{(3)^3}{3}}{\frac{(3)^3}{2}}=2 m
\)

Hint

(d) As friction is absent at the point of contact, Acceleration \(=\frac{\text { Force }}{\text { Mass }}\) It is independent of \(h\)

Hint

(b) If external torque is zero, angular momentum remains conserved. \(L=I \omega=\) constant. When a child sits on a rotating disc, then no torque is applied (the weight of the child acts downward), so angular momentum will remain conserved.

Hint

(b) As net torque applied is zero.
Hence, \(\tau=\frac{d L}{d t}\)
\(\frac{d L}{d t}=0, L=\) constant.
\(L\) (angular momentum) remains conserved.

Hint

(b) Moment of inertia \(=\int r^2 d m\).
\(\therefore\) Since, \(\rho_{\text {iron }}>\rho_{\text {aluminium }}\)
So, the whole of aluminium is kept in the core, and the iron is at the outer rim of the disc.

The moment of inertia depends on the distribution of mass and about the axis of rotation. The density of iron is more than that of aluminium, therefore for the moment of inertia to be maximum, the iron should be far away from the axis. Thus, aluminium should be at the interior and iron surrounds it.

Hint

(d) Linear distance moved by the wheel in half revolution \(=\pi r\). Point \(P_1[latex] after half revolution reaches at [latex]P_2\) vertically \(2 \mathrm{~m}\) above the ground.
\(\therefore\) Displacement \(P_1 P_2\)
\(
=\sqrt{\pi^2 r^2+2^2}=\sqrt{\pi^2+4} \quad[\because r=1 m]
\)

Hint

(d) \(\tau=1000 \mathrm{~N}-m, I=200 \mathrm{~kg}-\mathrm{m}^2\)
\(
\begin{aligned}
&\tau=\text { I. } \alpha \text { and } \alpha=\left(\frac{\omega_{\mathrm{f}}-\omega_0}{\mathrm{t}}\right) \\
&\Rightarrow \alpha=\frac{1000}{200}=5 \mathrm{rad} / \mathrm{sec}^2 \\
&\omega_1=\omega_0+\alpha t=0+3 \times 5=15 \mathrm{rad} / \mathrm{s}
\end{aligned}
\)

Hint

(a)
Similarly, the velocity of point \(A\) is given by
\(v_A=\) velocity of centre of mass
\(\left(v_{C M}\right)+\) Linear velocity of point \(A(R \omega)\)
\(=v_{\mathrm{CM}}+v_{\mathrm{CM}} \quad\left(\because v_{\mathrm{CM}}=R \omega\right)\)
\(=2 v_{\mathrm{CM}}\)
Velocity of point \(B\) is,
\(
v_B=v_{C M}-R \omega=v_{C M}-v_{C M}=0
\)
Thus, the velocity of point \(A\) is \(2 v_{C M}\) and velocity of point \(B\) is zero.

Hint

(b)
The moment of inertia of a body about an axis depends not only on the mass of the body but also on the distribution of mass from the axis. For a given body, mass is the same, so it will depend only on the distribution of mass from the axis. The mass is farthest from axis \(B C\), so \(I_2\) is maximum. Mass is nearest to axis \(A C\), so \(l_3\) is minimum.

Hence, the correct sequence will be
\(I_2>I_1>I_3\)
In a rotational motion, the moment of inertia is also known as rotational inertia.

Moment of Inertia depend upon mass and distribution of masses as \(I=\Sigma m r^2\).
Further, as the distance of masses is more, more is the moment of Inertia.
\(
\therefore I_2>I_1, I_2>I_3
\)

Hint

\(
\text { Torque, I } \alpha=f . R \text {. }
\)
Using Newton’s IInd law, \(m g \sin \theta-f=m a\)
\(\because\) pure rolling is there, \(a=R \alpha\)
\(\Rightarrow m g \sin \theta-\frac{I \alpha}{R}=m a\)
\(\Rightarrow m g \sin \theta-\frac{I a}{R^2}=m a \quad\left(\because \alpha=\frac{\mathrm{a}}{\mathrm{R}}\right)\) or, acceleration, \(a=\frac{m g \sin \theta}{\left(I / R^2+m\right)}\)
Using, \(s=u t+\frac{1}{2} a t^2\)
or, \(s=\frac{1}{2} a t^2 \Rightarrow t \alpha \frac{1}{\sqrt{\mathrm{a}}}\)
\(\mathrm{t}\) minimum means a should be more. This is possible when \(I\) is minimum which is the case for solid cylinder.
Therefore, solid cylinder will reach the bottom first.

Hint

(a) As the time taken by both cars to complete one revolution is the same.
As \(\omega=\frac{2 \pi}{T} \Rightarrow \omega \propto \frac{1}{T}\), as \(T\) is same in both cases. Hence ‘ \(\omega\) ‘ will also be same.

Hint

The moment of inertia of a disc about its diameter is
\(
I_d=\frac{1}{4} M R^2
\)
Now, according to the perpendicular axis theorem, the moment of inertia of disc about a tangent passing through the rim and in the plane of the disc is
\(
I=I_d+M R^2=\frac{1}{4} M R^2+M R^2=\frac{5}{4} M R^2
\)

Hint

(d)
The whole mass of the ball will be concentrated at the centre of the ball. All the three balls are identical, i.e., the balls have same mass. On each vertex of equilateral \(\triangle P O R\), same mass is kept. Therefore, centre of mass of the triangle is the centre of mass of the system which is point of intersection of the medians of the triangle.

Hint

(d) \(A B\) is the ladder, let \(\mathrm{F}\) be the horizontal force and \(\mathrm{W}\) is the weight of man. Let \(N_1\) and \(N_2\) be normal reactions of ground and wall, respectively. Then for vertical equilibrium \(W=N_1 \dots(1)\)
For horizontal equilibrium, \(N_2=F \quad\)…..(2)
Taking moments about \(A\), \(N_2\left(A B \sin 60^{\circ}\right)-W\left(A C \cos 60^{\circ}\right)=0 \ldots \ldots(3)\)
Using (2) and \(A B=20 \mathrm{ft}, B C=4 \mathrm{ft}\), we get

\(
\begin{aligned}
&F\left(20 \times \frac{\sqrt{3}}{2}\right)-W\left(16 \times \frac{1}{2}\right)=0 \\
&\Rightarrow F=\frac{8 W \times 2}{20 \sqrt{3}}=\frac{4 W}{5 \sqrt{3}}=\frac{150 \times 4}{5 \sqrt{3}} \text { pound } \\
&=40 \sqrt{3}=40 \times 1.73=69.2 \text { pound }
\end{aligned}
\)

Hint

The correct option is \(\mathbf{A}\) forces acting on the particles
From the formula for the coordinates of COM of a system:
\(x_{C O M}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}\),
\(\mathrm{y}_{\mathrm{COM}}=\frac{\mathrm{m}_1 \mathrm{y}_1+\mathrm{m}_2 \mathrm{y}_2}{\mathrm{~m}_1+\mathrm{m}_2}\)
\(z_{C O M}=\frac{m_1 z_1+m_2 z_2}{m_1+m_2}\)
\(\therefore \mathrm{x}_{\mathrm{COM}}\) or \(\mathrm{y}_{\mathrm{COM}}\) or \(\mathrm{z}_{\mathrm{COM}}\) only depends on the mass of the particles and the position of the particles. It will also depend on the relative position of the particles (i.e distance between them).
The position of COM(Centre Of Mass) is independent of the forces acting on particles.

Hint

Let the distance of the centre of mass from the carbon atom be \(x_{\mathrm{cm}}\).
The mass of carbon, \(m_1=12 \mathrm{amu}\)
The mass of oxygen, \(m_2=16\) amu
[atomic mass unit]
From the definition of centre of mass
\(
\begin{aligned}
x_{\mathrm{cm}} &=\frac{m_1 x_1+m_2 x_2}{m_1+m_2} \\
&=\frac{(12 \mathrm{amu}) \times 0+(16 \mathrm{amu}) \times r}{12 \mathrm{amu}+16 \mathrm{amu}} \\
&=\frac{16}{28} r=\frac{16}{28} \times 1.12 \times 10^{-10} \mathrm{~m} \\
&=0.64 \times 10^{-10} \mathrm{~m}
\end{aligned}
\)

Hint

A couple consists of two equal and opposite forces acting at a separation so that the net force becomes zero. When a couple acts on a body it rotates the body but does not produce any translatory motion. Hence, only rotational motion is produced.

Hint

(c)
If the man pulls the cart by the rope, the man and cart will meet at the centre of mass.
\(
\therefore \quad x_{C M}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}
\)
Taking axis at the point where man is present
\(
\begin{aligned}
&=\frac{M \times 0+M \times 10}{M+M}\left[\begin{array}{l}
x_1=0, x_2=10 \\
m_1=m_2=M
\end{array}\right] \\
&=\frac{10 M}{2 M}=5 \mathrm{~m}
\end{aligned}
\)

Hint

(d)
Given, \(\mathrm{r}=7 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathrm{k}}, \mathbf{F}=-3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+5 \hat{\mathbf{k}}\) \(\therefore \quad \tau=\mathbf{r} \times \mathbf{F}=|\mathbf{r}||\mathbf{F}| \sin \theta\) where, \(\theta\) is the angle between \(r\) and \(\mathbf{F}\)
\(
\begin{aligned}
&=(7 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \times(-3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \\
&=\left|\begin{array}{rrr}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
7 & 3 & 1 \\
-3 & 1 & 5
\end{array}\right| \\
&=\hat{\mathbf{i}}(15-1)-\hat{\mathbf{j}}(35+3)+\hat{\mathbf{k}}(7+9) \\
&=14 \hat{\mathbf{i}}-38 \hat{\mathbf{j}}+16 \hat{\mathbf{k}}
\end{aligned}
\)

Alternative

\(\therefore \tau=\mathrm{r} \times \mathrm{F}\)
\(=(7 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \times(-3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+5 \hat{\mathbf{k}})\)
\(=-21(\hat{\mathbf{i}} \times \hat{\mathbf{i}})+7(\hat{\mathbf{i}} \times \hat{\mathbf{j}})+35(\hat{\mathbf{i}} \times \hat{\mathbf{k}})\)
\(-9(\hat{\mathbf{j}} \times \hat{\mathbf{i}})+3(\hat{\mathbf{j}} \times \hat{\mathbf{j}})+15(\hat{\mathbf{j}} \times \hat{\mathbf{k}})\)
\(-3(\hat{\mathbf{k}} \times \hat{\mathbf{i}})+(\hat{\mathbf{k}} \times \hat{\mathbf{j}})+5(\hat{\mathbf{k}} \times \hat{\mathbf{k}})\)
\(=0+7 \hat{\mathbf{k}}-35 \hat{\mathbf{j}}+9 \hat{\mathbf{k}}+0+15 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\hat{\mathbf{i}}+0\)
\(=14 \hat{\mathbf{i}}-38 \hat{\mathbf{j}}+16 \hat{\mathbf{k}}\)

Hint

\(
\text { (c) Angular speed, } \omega=\frac{120 \times 2 \pi}{60}=4 \pi \mathrm{rad} / \mathrm{sec}
\)

Hint

(a) Let body contain \(m_1, m_2, m_3 \ldots \ldots . m_n\) masses at distance \(r_1, r_2, r_3 \ldots \ldots \ldots r_n\) from axis \(O A\).
Angular momentum of the body
\(
\begin{aligned}
&=m_1 v_1 r_1+m_2 v_2 r_2 \ldots . .+m_n v_n r_n \\
&=m_1\left(\omega r_1\right) r_1+m_2\left(\omega r_2\right) r_2 \ldots \ldots+m_n\left(\omega r_n\right) r_n \\
&=\left(m_1 r_1^2\right) \omega+\left(m_2 r_2^2\right) \omega \ldots \ldots+\left(m_n r_n^2\right) \omega \\
&=\left(\sum_{i=1}^n m_i r_i^2\right) \omega=I \omega
\end{aligned}
\)

Hint

(b) Number of revolutions made by the engine wheel \((n)=90\) per minute.
The angular speed of the engine wheel
\((\omega)=\frac{2 \pi n}{60}=\frac{2 \pi \times 90}{60}=3 \pi \mathrm{rad} / \mathrm{s}\)

Hint

Angular velocity is the rate of change of angular displacement with time. Given that both the cars complete one circle in the same time ‘ \(t\) ‘
Angular speed \(=\omega=\frac{\text { angle subtended by the car for one revolution }(2 \pi)}{\text { Time Taken( }(t)}\)
Angular speed of first car, \(=\omega_1=\frac{2 \pi}{t}\)
Angular speed of second car, \(=\omega_2=\frac{2 \pi}{t}\)
The ratio of angular speed of the first car to that of the second \(=\frac{\omega_1}{\omega_2}=1: 1\)

Hint

The moment of inertia of the triangular plate is maximum about the shortest side because the effective distance of mass distribution about this side is maximum.
The intersection of medians is the centre of mass of the triangle. Since the distances of centre of mass from the sides is related as \(\mathrm{x}_{\mathrm{BC}}\lt \mathrm{x}_{\mathrm{AB}} \lt \mathrm{x}_{\mathrm{AC}}\)

Therefore \(I_{B C}>I_{A B}>I_{A C}\)
or \(\mathrm{I}_{\mathrm{BC}}>\mathrm{I}_{\mathrm{AB}}\)

Hint

(a) Angular momentum \(\vec{L}\) is defined as \(\vec{L}=\vec{r} \times m(\vec{v})\)
So, \(\vec{L}\) is an axial vector.

Hint

(c)

\(\frac{K_r}{E}=\frac{\frac{1}{2} M K^2 \omega^2}{\frac{1}{2} M \omega^2\left[K^2+R^2\right]}=\frac{K^2}{K^2+R^2}\) \(=\frac{2 / 5}{1+2 / 5}=\frac{2}{7}\)
Here, \(K^2=\frac{2}{5} R^2\)
When a body is executing only linear motion, its K.E. is given by \(E_t=\frac{1}{2} m v^2\) when a body is rolling without slipping, its centre of mas has linear motion too.
Therefore, its total kinetic energy
\(
\begin{aligned}
&E=E_t+E_r \\
&=\frac{1}{2} m v^2+\frac{1}{2} \mathrm{I} \omega^2 \\
&=\frac{1}{2} m v^2+\frac{1}{2} m\left(\mathrm{~K}^2\right) \frac{v^2}{R^2} \\
&=\frac{1}{2} m v^2\left(1+\frac{K^2}{R^2}\right)
\end{aligned}
\)

Hint

Acceleration of the centre of mass of the rolling body is given by
\(
a=\frac{g \sin \theta}{1+\left(\frac{1}{M R^2}\right)}
\)
Moment of inertia of the ring about an axis perpendicular to the plane of the ring and passing through its centre is given by
\(
I=M R^2
\)
\(
\begin{aligned}
a &=\frac{g \sin \theta}{1+M R^2 / M R^2} \\
&=\frac{g \sin 30^{\circ}}{1+1}=\frac{g}{4}
\end{aligned}
\)

Hint

Concept Magnitude of moment of inertia depends on the distribution of mass taken from the axis.
From the axis EG, the distribution of masses is at minimum distance while from the axis BD the distribution of masses is at maximum distance. Hence, the moment of inertia is minimum along the axis through EG.

Hint

(a) For solid sphere, \(\frac{K^2}{R^2}=\frac{2}{5}\)
For disc and solid cylinder, \(\frac{K^2}{R^2}=\frac{1}{2}\) As \(\frac{K^2}{R^2}\) for solid sphere is smallest, it takes minimum time to reach the bottom of the incline

When bodies of different shapes are allowed to roll down an incline, then velocity at the bottom of incline, acceleration and time taken in reaching the bottom depends on \(\frac{K^2}{R^2}\). If the body has less moment of inertia then \(\frac{K^2}{R^2}\) will be less, then it will take less time in reaching the bottom.

Hint

a)
When a solid sphere rolls on the inclined plane, then it has both rotational as well as translational kinetic energy Total kinetic energy
\(
K=K_{\mathrm{rot}}+K_{\text {trans }}=\frac{1}{2} I \omega^2+\frac{1}{2} m v^2
\)
For sphere, moment of inertia about its diameter
\(
\begin{aligned}
I &=\frac{2}{5} m r^2 \\
\therefore \quad K &=\frac{1}{2}\left(\frac{2}{5} m r^2\right) \omega^2+\frac{1}{2} m v^2 \\
&=\frac{1}{5} m r^2 \omega^2+\frac{1}{2} m v^2 \\
&=\frac{1}{5} m v^2+\frac{1}{2} m v^2 \quad(\text { as } v=r \omega) \\
&=\frac{7}{10} m v^2
\end{aligned}
\)

(On reaching sphere at bottom , it has only kinetic energy
\(
\begin{aligned}
\therefore & P E=\text { Total KE } \\
m g h=\frac{7}{10} m v^2 \\
\Rightarrow \quad v &=\sqrt{\frac{10 g h}{7}}
\end{aligned}
\)

Hint

\(Y=X+4\) line is shown in the figure.
To find the slope of this line comparing this with the equation of the line
\(
\begin{aligned}
& y =m^{\prime} x+c \\
\therefore & \text { Slope, } m^{\prime} &=\tan \theta=1 \\
\Rightarrow \quad \theta & =45^{\circ}
\end{aligned}
\)
Length of perpendicular \(=O P\)
In \(\triangle P S O, \quad \frac{O P}{O S}=\sin 45^{\circ}\)
\(\therefore \quad O P=O S \sin 45^{\circ}\)
\(=4 \times \frac{1}{\sqrt{2}}=\frac{4}{\sqrt{2}}\)
\(\therefore\) Angular momentum of particle going along this line
\(
\begin{aligned}
&=m v R \\
&=5 \times 3 \sqrt{2} \times \frac{4}{\sqrt{2}} \\
&=60 \text { unit }
\end{aligned}
\)

Hint

When sphere rolls, then it has both translational and rotational kinetic energy
\(
\begin{aligned}
\therefore \quad K &=K_{\text {rot }}+K_{\text {trans }} \\
&=\frac{1}{2} I \omega^2+\frac{1}{2} m v^2
\end{aligned}
\)
\(\because\) Moment of inertia of the sphere about its diameter is
\(
\begin{aligned}
I &=\frac{2}{5} m r^2 \\
\therefore \quad K &=\frac{1}{2}\left(\frac{2}{5} m r^2\right) \omega^2+\frac{1}{2} m v^2
\end{aligned}
\)
(as \(v=r \omega)\)
\(
\begin{aligned}
&K=\frac{1}{5} m v^2+\frac{1}{2} m v^2=\frac{7}{10} m v^2 \\
&\frac{K_t}{K}=\frac{\frac{1}{2} m v^2}{\frac{7}{10} m v^2}=\frac{5}{7}
\end{aligned}
\)

Hint

or this type of problem, always apply parallel and perpendicular axes theorem of moment of inertia.
Moment of inertia of uniform circular disc about its diameter \(=I\)

According to theorem of perpendicular axes,
Moment of inertia of disc about its axis \(=2I\)
Applying theorem of parallel axes,
Moment of inertia of disc about the given axis
\(
\begin{aligned}
&=2I+m r^2 \\
&=2I+4I \\
& \quad\left(\operatorname{as} 2I=\frac{1}{2} m r^2 \therefore m r^2=4I\right) \\
\therefore \quad &=6I
\end{aligned}
\)

Hint

Given,
Moment of inertia, \(I=1.2 \mathrm{~kg}-\mathrm{m}^2\)
Rotational kinetic energy, \(K_r=1500 \mathrm{~J}\)
Angular acceleration
\(
\alpha=25 \mathrm{rad} / \mathrm{s}^2, \omega_0=0, t=\text { ? }
\)
Kinetic energy of rotation is given by
\(K_{r o t}=\frac{1}{2} l \omega^2\)
\(\therefore \quad \omega=\sqrt{\frac{2 K_r}{1}}=\sqrt{\frac{2 \times 1500}{1.2}}\)
\(=50 \mathrm{rad} / \mathrm{s}\)
Now, from equation of rotational motion
\(
\begin{aligned}
\omega &=\omega_0+\alpha t \\
t &=\frac{\omega-\omega_0}{\alpha} \\
&=\frac{50-0}{25}=2 \mathrm{~s}
\end{aligned}
\)

Hint

A flywheel is a large heavy wheel with a Iong cylindrical axle supported on ball bearings. Its centre of mass lies on its axis of rotation, so that it remains at rest in any position. Rotational kinetic energy of flywheel is given by
\(
K_{\text {rot }}=\frac{1}{2} I \omega^2
\)
where,\(I=\) moment of inertia of the wheel about the axis of rotation
\(\omega=\) angular velocity of flywheel
Given, Rotational kinetic energy
\(
K_r=360 \mathrm{~J}
\)
Angular velocity \(\omega=30 \mathrm{rad} / \mathrm{s}\)
\(
\begin{aligned}
\therefore \quad I &=\frac{2 K}{\omega^2}=\frac{2 \times 360}{(30)^2} \\
&=0.8 \mathrm{~kg}-\mathrm{m}^2
\end{aligned}
\)

Alternate

(c) \(\quad E_r=\frac{1}{2} I \omega^2\)
\(
I=\frac{2 E_r}{\omega^2}=\frac{2 \times 360}{30 \times 30}=0.8 \mathrm{~kg} \mathrm{~m}^2
\)

Hint

(a)
Since, in this case, instantaneous axis of rotation is always below the centre of mass. This is possible only when point of contact moves with a velocity equal to centre of mass.

Hint

When solid cylinder rolls down on an inclined plane, then it has both rotational and translational kinetic energy.

Total kinetic energy \(K=K_{\text {zot }}+K_{\text {trans }}\) or \(\quad K=\frac{1}{2} I \omega^2+\frac{1}{2} m v^2\)
where, \(I=\) moment of inertia of solid cylinderabout its axis
\(=\frac{1}{2} m r^2\)
\(\therefore \quad K=\frac{1}{2}\left(\frac{1}{2} m r^2\right) \omega^2+\frac{1}{2} m v^2\)
\(=\frac{1}{4} m v^2+\frac{1}{2} m v^2 \quad(\) as \(v=r \omega)\)
\(=\frac{3}{4} m v^2\)
Now, gain in \(\mathrm{KE}=\) loss in \(\mathrm{PE}\)
\(\therefore \quad \frac{3}{4} m v^2=m g h\)
\(\Rightarrow \quad v=\sqrt{\left(\frac{4}{3} g h\right)}\)

Hint

(b) Angular momentum about the point of contact with the surface includes the angular momentum about the centre. Due to friction linear momentum is conserved.

Hint

(a) Kinetic energy of rotation of a body is the energy possessed by the body on account of its rotation about a given axis. If \(I\) is the moment of inertia of the body about the given axis of rotation, \(\omega\) is angular velocity of the body, then kinetic energy of rotation
\(
K_{\text {rot }}=\frac{1}{2} I \omega^2
\)
Moment of inertia of the ring about an axis perpendicular to the plane of the ring and passing through its centre is
\(
\begin{aligned}
I &=m r^2 \\
K_{\text {rot }} &=\frac{1}{2} m r^2 \omega^2
\end{aligned}
\)

Hint

Angular acceleration \(\alpha=\frac{\omega_f-\omega_i}{t}\)
\(\omega_f=3120 \times \frac{2 \pi}{60} \mathrm{rad} / \mathrm{s}\)
\(\omega_i=1200 \times \frac{2 \pi}{60} \mathrm{rad} / \mathrm{s}\)
\(\Rightarrow \alpha=\frac{(3120-1200)}{16} \times \frac{2 \pi}{60}=4 \pi\)

Hint

Momentum of the system would remain conserved.
Initial momentum \(=0\)
Final momentum should also be zero.
Let masses be \(2 m, 2 m\), and \(m\)
Momentum along \(x\)-direction \(=2 m v \hat{i}\)
Momentum along \(y\)-direction \(=2 m v \hat{j}\)
Net momentum \(=\sqrt{(2 m v)^2+(2 m v)^2}=\sqrt{2} \cdot 2 m v\)
Now, \(2 \sqrt{2} m v=m v^{\prime}\)
\(
v^{\prime}=2 \sqrt{2} v
\)

Hint

\(
\begin{aligned}
& X_{\mathrm{cm}}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2} \\
& =\frac{10 \times 0+20 \times 10}{10+20} \\
& =\frac{200}{30} \\
& =\frac{20}{3} \mathrm{~m}
\end{aligned}
\)

Hint

The radius of gyration is defined as the square root of the ratio of moment of inertia and mass and it is written as;
\(
\mathrm{k}=\sqrt{\frac{I}{m}} \dots(1)
\)
Here \(k\) is the radius of gyration, ‘ \(I\) ‘ is the moment of inertia and \(m\) is the mass. The figure of a thin uniform disc and disc is shown below;

The radius of gyration of a thin uniform disc=\(k=\sqrt{\frac{I}{m}}\) and moment of inertia of the disc, \(I=\frac{m R^2}{4}\)
Here we have \(R\) as the radius, \(m\) is the mass
Now, on putting the values in equation (1) we have;
The radius of gyration of the uniform disc;
\(
\begin{aligned}
& \mathrm{k}_1=\sqrt{\frac{\frac{m R^2}{2}}{m}} \\
& \mathrm{k}_1=\sqrt{\frac{R^2}{2}} \dots(2)
\end{aligned}
\)
and the radius of gyration of the uniform disc;
\(
\begin{aligned}
& \mathrm{k}_2=\sqrt{\frac{\frac{m R^2}{4}}{m}} \\
& \mathrm{k}_2=\sqrt{\frac{R^2}{4}} \dots(3)
\end{aligned}
\)

Now, on dividing equation (2) by equation (3) we have;
\(
\begin{aligned}
& \frac{k_1}{k_2}=\frac{\sqrt{\frac{R^2}{2}}}{\sqrt{\frac{R^2}{4}}} \\
& \Rightarrow \mathrm{k}_1: \mathrm{k}_2=\sqrt{2}: 1
\end{aligned}
\)

Hint

\(
\mathrm{L}_1=121 \mathrm{~cm}=\frac{121}{100}=1.21 \mathrm{~m}
\)
\(
\mathrm{L}_2=100 \mathrm{~cm}=\frac{100}{100}=1 \mathrm{~m}
\)
We have to find the vibrations made by the shorter pendulum, such that both will be in same phase from the reaction,
\(\mathrm{T}_1=\) longer pendulum
\(\mathrm{T}_2=\) shorter pendulum
\(
\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}
\)
\(\mathrm{T} \propto \sqrt{\mathrm{L}}\)
\(
\frac{\mathrm{T}_1}{\mathrm{~T}_2} \propto \sqrt{\mathrm{L}_1 / \mathrm{L}_2}
\)
\(
\frac{\mathrm{T}_1}{\mathrm{~T}_2} \propto \sqrt{\frac{1.21}{1}}
\)
\(
\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{1.1}{1}
\)
\(
10 \mathrm{~T}_1=11 \mathrm{~T}_2
\)
10 vibrations of longer pendulum \(=11\) vibrations of shorter pendulum.

Hint

Radius of gyration : \(\mathrm{K}=\sqrt{\frac{I}{m}}\)
\(
\begin{aligned}
& \frac{k_{\text {solid sphere }}}{k_{\text {hollow sphere }}}=\sqrt{\frac{2 m R^2 / 5 m}{2 m R^2 / 3 m}} \\
& =\sqrt{3}: \sqrt{5}
\end{aligned}
\)

Hint

Angular acceleration of a body, moving along the circumference of a circle is along the axis of rotation.