JEE Rigid Body Dynamics Questions

Conservation of mechanical energy governs the pure rotation of the disk about its pivot point $$C$$ after the collision, where kinetic energy is entirely converted into gravitational potential energy at the highest point of swing.

$$I_C = \frac{3}{2}MR^2 = 0.06\text{ kg}\cdot\text{m}^2$$

rom conservation of mechanical energy post-collision:

$$E_i = E_f$$

$$\frac{1}{2}I_C\omega^2 = Mg\Delta h_{\text{cm}}$$

Substituting values to find the maximum height change $$\Delta h$$ of center $$O$$:

$$\frac{1}{2}(0.06)(7.138)^2 = (1)(10)\Delta h$$

$$0.03 \times 50.95 = 10\Delta h$$

$$1.5285 = 10\Delta h \implies \Delta h = 0.15285\text{ m} \approx 0.15\text{ m}$$

Angular momentum about the pivot point $$C$$ is conserved during the collision, and the mechanical energy loss is the difference between initial and final kinetic energies of the system.

Using coordinates relative to center $$O(0,0)$$:

$$C = (0, R) = (0, 0.2)$$

$$P = (R\sin 45^\circ, -R\cos 45^\circ) = \left(0.2\frac{1}{\sqrt{2}}, -0.2\frac{1}{\sqrt{2}}\right)$$

Finding the position vector from $$C$$ to $$P$$:

$$\vec{r}_{P/C} = \vec{r}_P - \vec{r}_C = 0.2\frac{1}{\sqrt{2}}\hat{i} - \left(0.2 + 0.2\frac{1}{\sqrt{2}}\right)\hat{j}$$

Finding the initial and final angular momentum of the particle about $$C$$:

$$\vec{L}_{i,p} = \vec{r}_{P/C} \times (m\vec{u}) = \left[0.2\frac{1}{\sqrt{2}}\hat{i} - \left(0.2 + 0.2\frac{1}{\sqrt{2}}\right)\hat{j}\right] \times (-0.02 \times 100\hat{i})$$

$$\vec{L}_{i,p} = -\left(0.2 + 0.2\frac{1}{\sqrt{2}}\right)(2)\hat{k} = -0.4\left(1 + \frac{1}{\sqrt{2}}\right)\hat{k}$$

$$\vec{L}_{f,p} = \vec{r}_{P/C} \times (m\vec{v}) = \left[0.2\frac{1}{\sqrt{2}}\hat{i} - \left(0.2 + 0.2\frac{1}{\sqrt{2}}\right)\hat{j}\right] \times (-0.02 \times 90\hat{j})$$

$$\vec{L}_{f,p} = \left(0.2\frac{1}{\sqrt{2}}\right)(-1.8)\hat{k} = -\frac{0.36}{\sqrt{2}}\hat{k}$$

Using parallel axis theorem for the disk about pivot $$C$$:

$$I_C = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}(1)(0.2)^2 = 0.06\text{ kg}\cdot\text{m}^2$$

From conservation of angular momentum about $$C$$:

$$L_{i,p} = L_{f,p} + I_C\omega$$

$$-0.4 - \frac{0.4}{\sqrt{2}} = -\frac{0.36}{\sqrt{2}} + 0.06\omega$$

$$0.06\omega = -0.4 - \frac{0.04}{\sqrt{2}} = -0.4 - 0.02828 = -0.42828 \implies \omega \approx -7.138\text{ rad}\cdot\text{s}^{-1}$$

Finding kinetic energies:

$$K_i = \frac{1}{2}mu^2 = \frac{1}{2}(0.02)(100)^2 = 100\text{ J}$$

$$K_f = \frac{1}{2}mv^2 + \frac{1}{2}I_C\omega^2 = \frac{1}{2}(0.02)(90)^2 + \frac{1}{2}(0.06)(-7.138)^2 = 81 + 1.53 = 82.53\text{ J}$$

Finding energy loss: $$\Delta K = K_i - K_f = 100 - 82.53 = 17.47\text{ J}$$

For any rigid body rolling without slipping down an inclined plane, conservation of mechanical energy gives

$$mgh \;=\;\frac12\,m v^{2}\;+\;\frac12\,I\omega^{2}$$
Since the body rolls without slipping, $$\omega=\dfrac{v}{R}$$. Thus

$$mgh \;=\;\frac12\,m v^{2}\;+\;\frac12\,I\dfrac{v^{2}}{R^{2}} =\frac12\,m v^{2}\Bigl(1+\frac{I}{mR^{2}}\Bigr)$$

Therefore the speed attained at the bottom is

$$v=\sqrt{\frac{2gh}{\,1+\dfrac{I}{mR^{2}}\,}}$$

Case 1 : Circular ring

For a thin ring about its symmetry axis, $$I_{ring}=mR^{2}$$, so $$\dfrac{I}{mR^{2}}=1$$.

$$v_{ring}=\sqrt{\frac{2gh}{1+1}} =\sqrt{\frac{2gh}{2}} =\sqrt{gh}$$

Case 2 : Solid sphere

For a solid sphere about a diameter, $$I_{sphere}=\tfrac{2}{5}mR^{2}$$, hence $$\dfrac{I}{mR^{2}}=\tfrac{2}{5}$$.

$$v_{sphere}=\sqrt{\frac{2gh}{1+\tfrac{2}{5}}} =\sqrt{\frac{2gh}{\tfrac{7}{5}}} =\sqrt{\frac{10gh}{7}}$$

Required ratio:

$$\frac{v_{ring}}{v_{sphere}} =\sqrt{\frac{gh}{\tfrac{10gh}{7}}} =\sqrt{\frac{7}{10}} =\sqrt{\frac{3.5}{5}}$$

Hence $$x=3.5$$.

Thus, the ratio of their velocities at the bottom of the plane is $$\sqrt{\dfrac{3.5}{5}}$$, giving $$x = 3.5$$.

The outward centrifugal force on the ball is $$F = m\omega^2r$$

$$a_r = \omega^2 r$$

$$a = v \frac{dv}{dr}$$

$$v \frac{dv}{dr} = \omega^2 r$$

Integrate both sides from the point where the ball is placed (r = 1 m) to the edge of the table (r = 3 m):

$$\int_{0}^{v} v \, dv = \omega^2 \int_{1}^{3} r \, dr$$

$$\frac{v^2}{2} = \omega^2 \left[ \frac{r^2}{2} \right]_1^3$$

$$v^2 = \omega^2 (3^2 - 1^2)$$

$$v^2 = 8\omega^2$$

$$v = \sqrt{8}\omega = 2\sqrt{2}\omega$$
$$x = 2$$

We need to find the ratio of rotational kinetic energy to total kinetic energy for a hollow sphere rolling on a plane surface about its axis of symmetry.

Since the sphere is thin-walled, its moment of inertia about an axis through its centre is $$I = \frac{2}{3}mr^2$$, where $$m$$ is the mass and $$r$$ is the radius.

For rolling without slipping with velocity $$v$$ and angular velocity $$\omega = v/r$$, the translational kinetic energy is $$K_{trans} = \frac{1}{2}mv^2$$ and the rotational kinetic energy is $$K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \cdot \frac{2}{3}mr^2 \cdot \frac{v^2}{r^2} = \frac{1}{3}mv^2$$.

Next, the total kinetic energy is $$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{3}mv^2 = mv^2\bigl(\tfrac{1}{2} + \tfrac{1}{3}\bigr) = mv^2 \cdot \frac{5}{6} = \frac{5}{6}mv^2$$.

From this, the ratio of rotational to total kinetic energy is $$\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{3}mv^2}{\frac{5}{6}mv^2} = \frac{\frac{1}{3}}{\frac{5}{6}} = \frac{1}{3} \times \frac{6}{5} = \frac{2}{5}$$. Since this ratio equals $$\frac{x}{5}$$, we obtain $$x = 2$$.

The answer is 2.

A solid circular disc of mass $$m = 50$$ kg rolls along a horizontal floor with center-of-mass speed $$v = 0.4$$ m/s. We need to find the work done to stop it.

We start by recalling that for a body rolling without slipping, the total kinetic energy is the sum of translational and rotational kinetic energies:

$$ KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 $$

Here, $$I$$ is the moment of inertia about the center of mass and $$\omega$$ is the angular velocity.

Next, the rolling condition gives $$v = R\omega$$, so $$\omega = \frac{v}{R}$$, and for a solid disc, $$I = \frac{1}{2}mR^2$$.

Substituting into the rotational kinetic energy term yields:

$$ \frac{1}{2}I\omega^2 = \frac{1}{2} \cdot \frac{1}{2}mR^2 \cdot \frac{v^2}{R^2} = \frac{1}{4}mv^2 $$

Therefore, the total kinetic energy becomes:

$$ KE_{total} = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2 $$

Substituting the numerical values gives:

$$ KE_{total} = \frac{3}{4} \times 50 \times (0.4)^2 = \frac{3}{4} \times 50 \times 0.16 = \frac{3}{4} \times 8 = 6 \text{ J} $$

By the work-energy theorem, the work done to stop the disc equals the total kinetic energy (in magnitude).

The answer is $$\boxed{6}$$ J.

Moment of inertia $$I = 0.40$$ kg m$$^2$$, radius $$r = 10$$ cm = 0.1 m, force $$F = 40$$ N, time $$t = 10$$ s, initial angular velocity = 0.

The torque is calculated as $$\tau = Fr = 40 \times 0.1 = 4$$ N·m.

The angular acceleration is $$\alpha = \frac{\tau}{I} = \frac{4}{0.40} = 10$$ rad/s$$^2$$.

After time $$t$$, the angular velocity is given by:

$$\omega = \omega_0 + \alpha t = 0 + 10 \times 10 = 100 \text{ rad/s}$$

The answer is $$x = \boxed{100}$$.

When the second disc is placed on the rotating one, the internal friction between their surfaces brings them to a common angular velocity. Since no external torque acts on the system about the axis of rotation, the total angular momentum is conserved.

The moment of inertia for a solid disc about its central axis is:

$$I = \frac{1}{2}MR^2 = \frac{1}{2} \times 5 \times (2)^2 = 10 \text{ kg m}^2$$

Initial angular momentum of the single disc equals the final angular momentum of the two discs rotating together:

$$I\omega_1 = (I + I)\omega_f$$

$$10 \times 10 = 20 \times \omega_f \implies \omega_f = 5 \text{ rad s}^{-1}$$

Before the second disc is added, the energy in the system is:

$$E_i = \frac{1}{2}I\omega_1^2 = \frac{1}{2} \times 10 \times (10)^2 = 500 \text{ J}$$

After both discs begin rotating together at the common velocity $$\omega_f$$:

$$E_f = \frac{1}{2}(2I)\omega_f^2 = \frac{1}{2} \times 20 \times (5)^2 = 250 \text{ J}$$

The energy lost due to friction as the discs slip against each other before reaching a common speed is the difference between the initial and final kinetic energies:

$$\Delta E = E_i - E_f = 500 - 250 = 250 \text{ J}$$

Earth's angular momentum is conserved: $$I\omega = const$$. $$I = \frac{2}{5}MR^2$$.

$$I_1\omega_1 = I_2\omega_2 \Rightarrow R^2 \omega_1 = (3R/4)^2 \omega_2 \Rightarrow \omega_2 = \frac{16}{9}\omega_1$$.

$$T_2 = \frac{2\pi}{\omega_2} = \frac{9}{16} T_1 = \frac{9}{16} \times 24 = 13.5$$ hours = 13 hours 30 minutes.

The answer is 13.

We need to find the moment of inertia of three balls placed at the midpoints of the edges of an equilateral triangle, about an axis through the centroid and perpendicular to the plane.

First, recall that the moment of inertia of a system of point masses about an axis is given by $$I = \sum m_i r_i^2$$, where $$r_i$$ is the perpendicular distance of each mass from the axis of rotation.

To determine that distance, note that in an equilateral triangle of side $$a$$, the centroid divides each median in a 2:1 ratio measured from the vertex. Since the median has length $$\frac{\sqrt{3}}{2}a$$, the distance from the centroid to a vertex is $$\frac{2}{3}\times\frac{\sqrt{3}}{2}a = \frac{a}{\sqrt{3}}$$, while the distance from the centroid to the midpoint of a side is $$\frac{1}{3}\times\frac{\sqrt{3}}{2}a = \frac{a}{2\sqrt{3}} = \frac{a\sqrt{3}}{6}$$.

Substituting $$a = 2\text{ m}$$ into the latter expression gives $$r = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}\text{ m}$$.

Because the centroid is equidistant from all three edges, each mass lies at the same distance $$r = \frac{1}{\sqrt{3}}$$ from the axis. Squaring this distance yields $$r^2 = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}\,\text{m}^2.$$

Therefore, the total moment of inertia can be written as $$I = m_1r^2 + m_2r^2 + m_3r^2 = (m_1 + m_2 + m_3)\,r^2$$. Substituting the given masses and $$r^2$$ results in $$I = (2 + 4 + 6)\times\frac{1}{3} = 12\times\frac{1}{3} = 4\,\text{kg m}^2.$$

The correct answer is 4 kg m$$^2$$.

Using conservation of angular momentum:

$$I_1\omega_1 + I_2\omega_2 = (I_1 + I_2)\omega$$

$$4(10) + 2(4) = 6\omega \Rightarrow \omega = \frac{48}{6} = 8$$ rad/s.

Loss in KE = Initial KE - Final KE:

$$= \frac{1}{2}I_1\omega_1^2 + \frac{1}{2}I_2\omega_2^2 - \frac{1}{2}(I_1+I_2)\omega^2$$

$$= \frac{1}{2}(4)(100) + \frac{1}{2}(2)(16) - \frac{1}{2}(6)(64)$$

$$= 200 + 16 - 192 = 24$$ J.

Therefore, the answer is $$\boxed{24}$$.

A circular disc rolls and slides down an inclined plane of length $$l$$. When it slides (no rolling), it takes $$t$$ seconds. When it rolls without slipping, it takes $$\left(\frac{\alpha}{2}\right)^{1/2} t$$ seconds. Find $$\alpha$$.

When the disc slides without friction, the acceleration along the incline is:

$$a_{\text{slide}} = g\sin\theta$$

Using $$l = \frac{1}{2}a_{\text{slide}} t^2$$:

$$t^2 = \frac{2l}{g\sin\theta} \quad \cdots (i)$$

When a disc rolls without slipping, the moment of inertia about the centre is $$I = \frac{1}{2}mR^2$$, so $$\frac{I}{mR^2} = \frac{1}{2}$$.

The translational acceleration along the incline is:

$$a_{\text{roll}} = \frac{g\sin\theta}{1 + I/(mR^2)} = \frac{g\sin\theta}{1 + 1/2} = \frac{2g\sin\theta}{3}$$

Using $$l = \frac{1}{2}a_{\text{roll}} t_{\text{roll}}^2$$:

$$t_{\text{roll}}^2 = \frac{2l}{a_{\text{roll}}} = \frac{2l \cdot 3}{2g\sin\theta} = \frac{3l}{g\sin\theta} \quad \cdots (ii)$$

$$\frac{t_{\text{roll}}^2}{t^2} = \frac{3l/(g\sin\theta)}{2l/(g\sin\theta)} = \frac{3}{2}$$

$$t_{\text{roll}} = t\sqrt{\frac{3}{2}} = \left(\frac{3}{2}\right)^{1/2} t$$

We are given $$t_{\text{roll}} = \left(\frac{\alpha}{2}\right)^{1/2} t$$.

Comparing: $$\frac{\alpha}{2} = \frac{3}{2} \implies \alpha = 3$$.

The answer is 3.

Let the mass of the uniform rod be $$m$$ and let its mid-point be $$C$$. The string of length $$L$$ is attached to the rod at point $$A$$ (one end of the rod) and the other end of the string is fixed to the pivot $$O$$. Hence the points are collinear in the order $$O \;-\; A \;-\; C$$, with $$OA = L$$ and $$AC = L/2$$, so

$$OC = OA + AC = L + \frac{L}{2} = \frac{3L}{2}$$

An impulsive force delivers a (linear) impulse $$P$$ perpendicular to the rod at a point that is a distance $$x = \frac{L}{n}$$ away from the mid-point $$C$$ along the length of the rod.

Linear motion of the centre of mass
The entire impulse $$P$$ is received by the rod, so the centre of mass acquires a linear momentum $$P$$ and hence a speed

$$v_C = \frac{P}{m}$$

Angular motion about the centre of mass
The same impulse produces an angular impulse (about the centre) equal to $$P x$$. For a thin uniform rod, the moment of inertia about its mid-point is

$$I_C = \frac{1}{12}\,mL^{2}$$

Therefore the angular velocity of the rod about its own centre is

$$\omega = \frac{P x}{I_C} = \frac{P x}{\tfrac{1}{12}mL^{2}} = \frac{12\,P\,x}{mL^{2}}$$

Condition for the rod to remain aligned with the string
For the rod and string to revolve rigidly about the fixed point $$O$$, the motion after the impulse must be a pure rotation about $$O$$. In pure rotation, the linear speed of the centre of mass must satisfy

$$v_C = \omega \, OC$$

Substituting $$v_C = \dfrac{P}{m}$$, $$\omega = \dfrac{12Px}{mL^{2}}$$ and $$OC = \dfrac{3L}{2}$$ gives

$$\frac{P}{m} = \frac{12Px}{mL^{2}} \times \frac{3L}{2} = \frac{18P x}{mL}$$

Cancel $$P/m$$ from both sides:

$$1 = \frac{18x}{L} \quad\Longrightarrow\quad x = \frac{L}{18}$$

Since the point of application was specified as $$x = \dfrac{L}{n}$$ from the mid-point, we equate

$$\frac{L}{n} = \frac{L}{18} \quad\Longrightarrow\quad n = 18$$

Hence the required value is

18

A solid sphere rolls without slipping on a horizontal plane. The ratio of angular momentum about its axis of rotation to total energy is $$\pi : 22$$.

Angular momentum about the axis of rotation.

For a solid sphere: $$I = \frac{2}{5}mR^2$$

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

Total kinetic energy.

$$E = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}m(R\omega)^2 + \frac{1}{2} \cdot \frac{2}{5}mR^2\omega^2$$

$$= \frac{1}{2}mR^2\omega^2\left(1 + \frac{2}{5}\right) = \frac{7}{10}mR^2\omega^2$$

Form the ratio.

$$\frac{L}{E} = \frac{\frac{2}{5}mR^2\omega}{\frac{7}{10}mR^2\omega^2} = \frac{2/5}{7\omega/10} = \frac{4}{7\omega}$$

Apply the given ratio.

$$\frac{4}{7\omega} = \frac{\pi}{22}$$

$$\omega = \frac{4 \times 22}{7\pi} = \frac{88}{7\pi}$$

Using $$\pi = \frac{22}{7}$$:

$$\omega = \frac{88}{7 \times \frac{22}{7}} = \frac{88}{22} = 4 \text{ rad s}^{-1}$$

Therefore, the answer is $$\mathbf{4}$$.

Force $$\vec{F} = -P\hat{k}$$ acts at origin. Torque about point (2,-3,0):

$$\vec{r} = (0,0,0) - (2,-3,0) = (-2, 3, 0)$$

$$\vec{\tau} = \vec{r} \times \vec{F} = (-2,3,0) \times (0,0,-P)$$

$$= (3(-P)-0, 0-(-2)(-P), 0) = (-3P, -2P, 0) = P(-3\hat{i} - 2\hat{j})$$

Given $$\vec{\tau} = P(a\hat{i} + b\hat{j})$$: $$a = -3, b = -2$$. $$a/b = 3/2 = x/2$$, so $$x = 3$$.

The answer is 3.

We have a hollow spherical ball rolling up a curved surface with initial velocity $$u = 3$$ m/s. For a hollow sphere, the moment of inertia is $$I = \frac{2}{3}mR^2$$.

Applying conservation of energy, the total kinetic energy at the bottom equals the potential energy at the maximum height:

$$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh$$

Since the ball rolls without slipping, $$v = R\omega$$. Substituting:

$$\frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{3}mR^2\right)\frac{v^2}{R^2} = mgh$$

$$\frac{1}{2}mv^2 + \frac{1}{3}mv^2 = mgh$$

$$\frac{5}{6}mv^2 = mgh$$

Now, solving for $$h$$:

$$h = \frac{5v^2}{6g} = \frac{5 \times 9}{6 \times 10} = \frac{45}{60} = 0.75 \text{ m} = 75 \text{ cm}$$

So, the answer is $$75$$ cm.

A light rope is wound around a hollow cylinder and pulled with a force. We need to find the angular acceleration.

We are given that Mass $$m = 5$$ kg, radius $$R = 70$$ cm $$= 0.7$$ m, force $$F = 52.5$$ N.

To begin,

A hollow cylinder (thin-walled cylindrical shell) has all its mass concentrated at the radius $$R$$. Its moment of inertia about its central axis is:

$$ I = mR^2 $$

This differs from a solid cylinder, which has $$I = \frac{1}{2}mR^2$$. For a hollow cylinder, since all mass is at distance $$R$$ from the axis, we simply have $$I = mR^2$$.

$$ I = 5 \times (0.7)^2 = 5 \times 0.49 = 2.45\;\text{kg m}^2 $$

Next,

The rope is wound around the cylinder at radius $$R$$, so the force $$F$$ acts tangentially at a perpendicular distance $$R$$ from the axis. The torque is:

$$ \tau = F \times R = 52.5 \times 0.7 = 36.75\;\text{N m} $$

From this,

The rotational analogue of Newton's second law states:

$$ \tau = I\alpha $$

where $$\alpha$$ is the angular acceleration. Solving for $$\alpha$$:

$$ \alpha = \frac{\tau}{I} = \frac{36.75}{2.45} = 15\;\text{rad/s}^2 $$

The angular acceleration of the cylinder is 15 rad s$$^{-2}$$.

The torque ($$\tau$$) about the origin $$O$$ is produced by the force of gravity ($$mg$$) acting vertically downward.

$$\tau = \text{Force} \times \text{Perpendicular distance} = (mg) \cdot x$$

In projectile motion, the horizontal distance at any time $$t$$ is given by: $$x = (u \cos \theta)t$$

$$\tau = mg(u \cos \theta)t$$

$$L = \int_{0}^{t} \tau \, dt = \int_{0}^{t} mgu \cos \theta \cdot t \, dt$$

$$L = mgu \cos \theta \left[ \frac{t^2}{2} \right]_0^t = \frac{1}{2} mgu \cos \theta \cdot t^2$$

$$L = \frac{1}{2} \times 0.1 \times 10 \times 20 \times \frac{1}{\sqrt{2}} \times (2)^2$$

$$L = 2 \times \frac{20}{\sqrt{2}} = \frac{40}{\sqrt{2}} = 20\sqrt{2} \text{ kg m}^2 \text{ s}^{-1}$$

$$\sqrt{K} = 20\sqrt{2}$$

$$K = (20\sqrt{2})^2$$

$$K = 400 \times 2 = 800$$

A ring and a solid sphere rotate about an axis passing through their centres with the same radii of gyration. The axis is perpendicular to the plane of the ring.

For a ring about an axis perpendicular to its plane through the center: $$I_{\text{ring}} = MR_1^2$$

Radius of gyration: $$k_1 = R_1$$

For a solid sphere about an axis through its center: $$I_{\text{sphere}} = \dfrac{2}{5}MR_2^2$$

Radius of gyration: $$k_2 = R_2\sqrt{\dfrac{2}{5}}$$

$$k_1 = k_2 \Rightarrow R_1 = R_2\sqrt{\dfrac{2}{5}}$$

$$\dfrac{R_1}{R_2} = \sqrt{\dfrac{2}{5}}$$

$$\dfrac{R_1}{R_2} = \sqrt{\dfrac{2}{x}}$$

$$x = 5$$

The value of $$x$$ is $$\boxed{5}$$.

We have a solid cylinder rolling without slipping down an inclined plane of inclination $$\theta = 30°$$ and length $$l = 60$$ cm $$= 0.6$$ m, with $$g = 10$$ m/s$$^2$$.

The height of the incline is:

$$h = l \sin\theta = 0.6 \times \sin 30° = 0.6 \times 0.5 = 0.3 \text{ m}$$

For a solid cylinder rolling without slipping, using energy conservation:

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

Now, for a solid cylinder $$I = \frac{1}{2}mr^2$$ and $$\omega = \frac{v}{r}$$ (rolling condition), so:

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{1}{2}mr^2 \cdot \frac{v^2}{r^2} = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$$

Solving for $$v$$:

$$v = \sqrt{\frac{4gh}{3}} = \sqrt{\frac{4 \times 10 \times 0.3}{3}} = \sqrt{\frac{12}{3}} = \sqrt{4} = 2 \text{ m/s}$$

Hence, the answer is $$2$$ m/s.

For a solid sphere: $$I = \frac{2}{5}mR^2$$, so $$k_{sph} = R\sqrt{\frac{2}{5}}$$

For a solid cylinder: $$I = \frac{1}{2}mR^2$$, so $$k_{cyl} = R\sqrt{\frac{1}{2}} = \frac{R}{\sqrt{2}}$$

Ratio: $$\frac{k_{sph}}{k_{cyl}} = \frac{R\sqrt{2/5}}{R/\sqrt{2}} = \sqrt{\frac{2}{5}} \times \sqrt{2} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$$

So $$k_{sph} : k_{cyl} = 2 : \sqrt{5}$$

Comparing with $$2 : \sqrt{x}$$, we get $$x = 5$$.

This matches the answer key value of $$\mathbf{5}$$.

For a solid sphere of mass $$m$$ making pure rolling on a horizontal surface, the total kinetic energy is the sum of translational and rotational kinetic energy.

For a solid sphere, the moment of inertia is $$I = \frac{2}{5}mr^2$$, and for pure rolling, $$\omega = \frac{v}{r}$$.

Total KE = Translational KE + Rotational KE:

$$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2$$

$$KE = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$$

Given: $$m = 2$$ kg, $$KE = 2240$$ J.

$$2240 = \frac{7}{10} \times 2 \times v^2$$

$$2240 = \frac{14}{10} v^2 = 1.4\,v^2$$

$$v^2 = \frac{2240}{1.4} = 1600$$

$$v = 40 \text{ m s}^{-1}$$

The velocity of the centre of mass of the sphere is $$\mathbf{40}$$ m s$$^{-1}$$.

A thin uniform rod of length $$L = 2$$ m, cross-sectional area $$A$$, and density $$d$$ rotates about its center. To determine the constant $$\alpha$$ in the relation $$\omega = \sqrt{\frac{\alpha E}{Ad}}$$, we first compute the mass of the rod and its moment of inertia about the center.

The mass of the rod is given by $$m = A \times d \times L = 2Ad$$. Substituting this into the standard formula for the moment of inertia of a rod about its center, we have $$I = \frac{mL^2}{12} = \frac{2Ad \times 4}{12} = \frac{2Ad}{3}$$.

Next, we express the rotational kinetic energy in terms of the angular speed $$\omega$$: $$E = \frac{1}{2}I\omega^2 = \frac{1}{2} \times \frac{2Ad}{3} \times \omega^2 = \frac{Ad\omega^2}{3}$$. Solving for $$\omega^2$$ gives $$\omega^2 = \frac{3E}{Ad}$$, so that $$\omega = \sqrt{\frac{3E}{Ad}}$$.

By comparing this result with the given form $$\omega = \sqrt{\frac{\alpha E}{Ad}}$$, it follows that $$\alpha = \boxed{3}$$.

$$\text{Friction force } (f_k) = \mu mg$$

$$\text{Deceleration } (a) = \frac{f_k}{m} = \mu g$$

$$v = u - at$$

$$v = 18 - (0.3 \times 10) \times 2$$

$$v = 18 - 6 = 12\text{ m/s}$$

$$KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

For a uniform disc, the moment of inertia is $$I = \frac{1}{2}mr^2$$. In pure rolling, $$v = \omega r \implies \omega = \frac{v}{r}$$.

$$KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v}{r}\right)^2$$

$$KE_{total} = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$$

$$KE_{total} = \frac{3}{4} \times 0.5 \times (12)^2$$

$$KE_{total} = 0.375 \times 144 = 54\text{ J}$$

$$I = \frac{2}{3}MR^2$$

$$K_{\text{rot}} = \frac{1}{2}I\omega^2$$

$$K_{\text{rot}} = \frac{1}{2} \left(\frac{2}{3}MR^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{3}Mv^2$$

$$K_{\text{trans}} = \frac{1}{2}Mv^2$$

$$K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2}Mv^2 + \frac{1}{3}Mv^2 = \frac{5}{6}Mv^2$$

$$\frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{\frac{1}{3}Mv^2}{\frac{5}{6}Mv^2} = \frac{1}{3} \times \frac{6}{5} = \frac{2}{5}$$

$$\frac{x}{5} = \frac{2}{5} \implies x = 2$$

If the volume shrinks to $$\frac{1}{64}$$ of the original, and volume $$\propto R^3$$:

$$\frac{R'^3}{R^3} = \frac{1}{64} \Rightarrow R' = \frac{R}{4}$$

By conservation of angular momentum (no external torque):

$$I\omega = I'\omega'$$

$$\frac{2}{5}MR^2 \cdot \omega = \frac{2}{5}MR'^2 \cdot \omega'$$

$$R^2 \omega = \frac{R^2}{16} \omega'$$

$$\omega' = 16\omega$$

Since $$T = \frac{2\pi}{\omega}$$:

$$T' = \frac{T}{16} = \frac{24}{16} = \frac{24}{16} = \frac{3}{2} \text{ h}$$

So $$\frac{24}{x} = \frac{24}{16}$$, giving $$x = 16$$.

The value of x is 16.

A semicircular ring has all its mass distributed at a constant distance $$R$$ from the center.

The moment of inertia about an axis passing through the center and perpendicular to the plane of the ring is:

$$ I = \int r^2\,dm $$

Since every element of mass is at the same distance $$R$$ from the center:

$$ I = R^2 \int dm = MR^2 $$

Comparing with $$\frac{1}{x}MR^2$$: $$\frac{1}{x} = 1 \implies x = 1$$.

The value of $$x$$ is 1.

When the person stretches his hands, the moment of inertia becomes $$3I$$ (triple).

By conservation of angular momentum (no external torque):

$$L = I\omega = 3I\omega'$$

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

Initial kinetic energy: $$E = \frac{1}{2}I\omega^2$$

Final kinetic energy:

$$E_f = \frac{1}{2}(3I)\omega'^2 = \frac{1}{2}(3I)\frac{\omega^2}{9} = \frac{1}{6}I\omega^2 = \frac{E}{3}$$

Therefore $$x = 3$$.

Total kinetic energy of a rigid body that rolls without slipping is the sum of
(i) translational kinetic energy of the centre of mass and
(ii) rotational kinetic energy about the centre of mass.

Let the linear speed of the centre of mass be $$v$$. For a solid sphere of mass $$m$$ and radius $$R$$:

Moment of inertia about its centre is
$$I_{\text{CM}} = \frac{2}{5}\,mR^{2}$$.

Because the sphere rolls without slipping, the rolling condition is
$$v = \omega R$$, where $$\omega$$ is the angular speed.

Write each part of the kinetic energy:

Translational part: $$K_{\text{trans}} = \frac{1}{2}\,m v^{2}$$.

Rotational part: $$K_{\text{rot}} = \frac{1}{2}\,I_{\text{CM}}\omega^{2} = \frac{1}{2}\left(\frac{2}{5}mR^{2}\right)\left(\frac{v}{R}\right)^{2} = \frac{1}{5}\,m v^{2}$$.

Total kinetic energy therefore is
$$K_{\text{total}} = \frac{1}{2}m v^{2} + \frac{1}{5}m v^{2} = \frac{7}{10}\,m v^{2} \quad -(1)$$.

The problem gives $$K_{\text{total}} = 7 \times 10^{-3}\,\text{J}$$ and $$m = 1\,\text{kg}$$. Substitute these into $$(1)$$:

$$7 \times 10^{-3} = \frac{7}{10}\,(1)\,v^{2}$$

Divide both sides by $$\tfrac{7}{10}$$:

$$v^{2} = 10^{-2}$$

Taking the positive square root (speed is positive):

$$v = 0.1\,\text{m s}^{-1}$$.

Convert metres per second to centimetres per second:

$$0.1\,\text{m s}^{-1} = 0.1 \times 100 = 10\,\text{cm s}^{-1}$$.

Hence, the speed of the centre of mass of the sphere is 10 cm s$$^{-1}$$.

We have a solid sphere of mass $$M = 500$$ g $$= 0.5$$ kg, radius $$R = 5$$ cm $$= 0.05$$ m, spinning with angular speed $$\omega = 10$$ rad/s.

The moment of inertia about a diameter is:

$$I_{diameter} = \frac{2}{5}MR^2 = \frac{2}{5}(0.5)(0.05)^2 = 5 \times 10^{-4} \text{ kg m}^2$$

So the angular momentum about the diameter is:

$$L = I_{diameter} \times \omega = 5 \times 10^{-4} \times 10 = 5 \times 10^{-3} \text{ kg m}^2\text{/s}$$

Now, using the parallel axis theorem, the moment of inertia about a tangent parallel to the diameter is:

$$I_{tangent} = I_{diameter} + MR^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$$

$$I_{tangent} = \frac{7}{5}(0.5)(0.0025) = 1.75 \times 10^{-3} \text{ kg m}^2$$

We are given that $$I_{tangent} = x \times 10^{-2} \times L$$. Substituting:

$$1.75 \times 10^{-3} = x \times 10^{-2} \times 5 \times 10^{-3}$$

$$1.75 \times 10^{-3} = x \times 5 \times 10^{-5}$$

$$x = \frac{1.75 \times 10^{-3}}{5 \times 10^{-5}} = 35$$

So, the answer is $$35$$.

For a circular disc of mass $$M$$ and radius $$R$$, the moment of inertia about the central axis perpendicular to its plane ($$I_{CM}$$) is $$I_{CM} = \frac{1}{2}MR^2$$

According to the Parallel Axis Theorem, the moment of inertia about any parallel axis $$AB$$ at a distance $$d$$ is $$I_{AB} = I_{CM} + Md^2$$

Substituting the given distance $$d = \frac{2}{3}R$$:

$$I_{AB} = \frac{1}{2}MR^2 + M\left(\frac{2}{3}R\right)^2$$

$$I_{AB} = \left(\frac{9 + 8}{18}\right)MR^2 = \frac{17}{18}MR^2$$

$$\frac{I_{AB}}{I_{CM}} = \frac{\frac{17}{18}MR^2}{\frac{1}{2}MR^2} = \frac{17}{18} \times \frac{2}{1} = \frac{17}{9}$$

$$x = 17$$

Since the solid sphere has mass $$M_s = 5\text{ kg}$$ and the disc has mass $$M_d = 4\text{ kg}$$, both with radius $$R$$, we first determine their moments of inertia about tangents.

For the disc, the moment of inertia about a diameter is $$\frac{1}{4}M_dR^2$$. Now using the parallel axis theorem to shift the axis to a tangent in its plane (a distance $$R$$ from the center) gives

$$I_{\text{disc}} = \frac{1}{4}M_dR^2 + M_dR^2 = \frac{5}{4}M_dR^2 = \frac{5}{4}\,(4)\,R^2 = 5R^2.$$

Similarly, the moment of inertia of the sphere about a diameter is $$\frac{2}{5}M_sR^2$$. Substituting into the parallel axis theorem at distance $$R$$ yields

$$I_{\text{sphere}} = \frac{2}{5}M_sR^2 + M_sR^2 = \frac{7}{5}M_sR^2 = \frac{7}{5}\,(5)\,R^2 = 7R^2.$$

Therefore, the ratio of the disc’s moment to the sphere’s moment is

$$\frac{I_{\text{disc}}}{I_{\text{sphere}}} = \frac{5R^2}{7R^2} = \frac{5}{7} = \frac{x}{7},$$

which gives $$x = 5\,. $$

For a uniform solid sphere of mass $$M$$ and radius $$R$$: $$I_{cm} = \frac{2}{5}MR^2$$

$$I_{PQ} = I_{cm} + Md^2 = \frac{2}{5}MR^2 + Md^2$$ (Parallel axis theorem)

The radius of gyration ($$k$$) is the distance from the axis at which the entire mass of the body could be concentrated to result in the same moment of inertia. It is defined by the relation $$I = Mk^2$$.

$$Mk^2 = M \left( \frac{2}{5}R^2 + d^2 \right)$$

$$k^2 = \frac{2}{5}R^2 + d^2$$

$$k^2 = \frac{2}{5}(5)^2 + (10)^2 = 10 + 100 = 110$$

We have $$k = \sqrt{x}$$, which implies $$x = k^2$$

The value of $$x$$ is 110.

Two discs of same mass $$M$$, thicknesses $$t_1 = 1$$ cm and $$t_2 = 0.5$$ cm, density ratio $$\rho_1 : \rho_2 = 3 : 5$$.

The mass of a disc is given by $$M = \pi R^2 t \rho$$.

Since $$M_1 = M_2$$:

For a thin disc about its diameter, the moment of inertia is $$I = \frac{1}{4}MR^2$$.

Since both discs have the same mass:

Given that the ratio is $$\frac{x}{6}$$:

The answer is $$\boxed{5}$$.

We have two identical solid spheres, each of mass $$m = 2\;\text{kg}$$ and radius $$r = 10\;\text{cm} = 0.1\;\text{m}$$, fixed at the ends of a light rod with separation between centres = 40 cm. The axis passes through the middle point of the rod, perpendicular to it, so the distance of each sphere's centre from the axis is $$d = 20\;\text{cm} = 0.2\;\text{m}$$.

Using the parallel axis theorem, the moment of inertia of each sphere about the given axis is:

$$I_{each} = \frac{2}{5}mr^2 + md^2$$

Substituting values:

$$I_{each} = \frac{2}{5}(2)(0.1)^2 + 2(0.2)^2 = \frac{2}{5}(2)(0.01) + 2(0.04)$$

$$I_{each} = 0.008 + 0.08 = 0.088\;\text{kg m}^2$$

Now the total moment of inertia is:

$$I = 2 \times I_{each} = 2 \times 0.088 = 0.176\;\text{kg m}^2 = 176 \times 10^{-3}\;\text{kg m}^2$$

So, the answer is $$176$$.

The moment of inertia of a disc of mass $$M$$ and radius $$R$$ about any of its diameters is $$\frac{MR^2}{4}$$. We seek the moment of inertia about an axis perpendicular to the disc and passing through a point on its edge.

By the perpendicular axis theorem for a planar body, $$I_z = I_x + I_y$$. Since each diameter contributes $$I_x = I_y = \frac{MR^2}{4}$$, the moment of inertia about the central axis normal to the plane is $$I_{\text{center}} = \frac{MR^2}{4} + \frac{MR^2}{4} = \frac{MR^2}{2}$$.

Applying the parallel axis theorem to shift the axis to the edge, which is at a distance $$R$$ from the center, gives $$I_{\text{edge}} = I_{\text{center}} + MR^2 = \frac{MR^2}{2} + MR^2 = \frac{3MR^2}{2}$$.

Since the general form is $$I_{\text{edge}} = \frac{x}{2}MR^2$$, equating to $$\frac{3MR^2}{2}$$ yields $$\therefore x = \boxed{3}$$.

The coin is a thin uniform disc of mass $$m = 5\text{ g}=0.005\text{ kg}$$ and radius $$R=\frac{4}{3}\text{ cm}=0.013333\text{ m}$$ that lies in the horizontal $$xy$$-plane.

An impulse of magnitude $$J=\sqrt{\frac{\pi}{2}}\times10^{-2}\ \text{N·s}$$ is applied vertically upward (along +$$z$$) at a point whose distance from the centre is $$r=\frac{2}{3}\text{ cm}=0.006667\text{ m}$$. Because the line of action does not pass through the centre, the impulse produces:

• a translational momentum $$J$$ along +$$z$$,
• an angular momentum (about the centre) whose magnitude is $$L = rJ$$, perpendicular to both $$\vec r$$ and $$\vec J$$, i.e. about a diameter of the disc.

Step 1: Translational upward velocity
Impulse-momentum theorem: $$J = m v_{\text{cm}} \ \Rightarrow\ v_{\text{cm}} = \frac{J}{m}$$ Compute $$J$$ first: $$J = \sqrt{\frac{\pi}{2}}\times10^{-2} = 1.2523\times10^{-2}\ \text{N·s}$$ Hence $$v_{\text{cm}} = \frac{1.2523\times10^{-2}}{0.005}=2.5046\ \text{m s}^{-1}$$

Step 2: Time of flight
The coin is projected upward with speed $$v_{\text{cm}}$$ and returns to its initial level under gravity. Time of flight: $$T = \frac{2v_{\text{cm}}}{g} = \frac{2(2.5046)}{10}=0.5009\ \text{s}$$

Step 3: Angular speed imparted about a diameter
Magnitude of angular momentum: $$L = rJ = (0.006667)(1.2523\times10^{-2}) = 8.3487\times10^{-5}\ \text{kg m}^2\text{ s}^{-1}$$ Moment of inertia of a thin disc about a diameter: $$I = \frac{1}{4}mR^{2}$$ $$R^{2} = (0.013333)^{2}=1.7778\times10^{-4}\ \text{m}^{2}$$ $$I = \frac{1}{4}(0.005)(1.7778\times10^{-4}) = 2.2222\times10^{-7}\ \text{kg m}^{2}$$ Angular speed: $$\omega = \frac{L}{I} = \frac{8.3487\times10^{-5}}{2.2222\times10^{-7}} = 3.7569\times10^{2}\ \text{rad s}^{-1}$$

Step 4: Number of complete rotations during the flight
$$n=\frac{\omega T}{2\pi}= \frac{(3.7569\times10^{2})(0.5009)}{2\pi} =\frac{1.8803\times10^{2}}{6.2832}=29.9\approx30$$

Hence, by the time the coin returns to its initial position, it completes about 30 rotations.

Answer: 30

Let the masses be $$m_1 = 20\ \text{g}=0.02\ \text{kg}$$ and $$m_2 = 30\ \text{g}=0.03\ \text{kg}$$. The distance between them is the length of the rod, $$L = 10\ \text{cm}=0.10\ \text{m}$$.

Since the suspension wire is fixed at the centre of mass (COM), we first locate that COM along the rod. Let the distances of $$m_1$$ and $$m_2$$ from the COM be $$r_1$$ and $$r_2$$ respectively.

By definition of the centre of mass, $$m_1 r_1 = m_2 r_2$$ $$-(1)$$ and $$r_1 + r_2 = L = 0.10\ \text{m}$$ $$-(2)$$.

From $$(1)$$, $$r_1 = \dfrac{m_2}{m_1}\,r_2 = \dfrac{0.03}{0.02}\,r_2 = 1.5\,r_2$$.

Substituting in $$(2)$$ gives $$1.5\,r_2 + r_2 = 2.5\,r_2 = 0.10$$, hence $$r_2 = 0.10/2.5 = 0.04\ \text{m}$$ and $$r_1 = 1.5 \times 0.04 = 0.06\ \text{m}$$.

The moment of inertia of the two-mass system about the COM is
$$I = m_1 r_1^2 + m_2 r_2^2$$

Substituting values:
$$I = 0.02\,(0.06)^2 + 0.03\,(0.04)^2$$
$$I = 0.02 \times 0.0036 + 0.03 \times 0.0016$$
$$I = 7.2 \times 10^{-5} + 4.8 \times 10^{-5}$$
$$I = 1.2 \times 10^{-4}\ \text{kg m}^2$$.

For a torsional pendulum, the angular frequency of small oscillations is
$$\omega = \sqrt{\dfrac{\kappa}{I}}$$
where $$\kappa$$ is the torsional constant of the wire.

Given $$\kappa = 1.2 \times 10^{-8}\ \text{N m rad}^{-1}$$ and $$I = 1.2 \times 10^{-4}\ \text{kg m}^2$$,

$$\dfrac{\kappa}{I} = \dfrac{1.2 \times 10^{-8}}{1.2 \times 10^{-4}} = 10^{-4}\ \text{s}^{-2}$$.

Therefore,
$$\omega = \sqrt{10^{-4}} = 10^{-2}\ \text{rad s}^{-1} = 0.01\ \text{rad s}^{-1}$$.

This can be written as $$\omega = 10 \times 10^{-3}\ \text{rad s}^{-1}$$, hence $$n = 10$$.

Final Answer: 10

For the system to be in equilibrium, the net torque about the pivot point C must be zero ($$\sum \tau_C = 0$$)

$$W_{rod} = m \cdot g = 2\text{ kg} \times 10\text{ m/s}^2 = 20\text{ N}$$

$$\tau_{rod} = 20\text{ N} \times 0.5\text{ m} = \mathbf{10\text{ N m}}$$ (Clockwise)

$$W_{obj} = 8\text{ kg} \times 10\text{ m/s}^2 = 80\text{ N}$$

$$\tau_{obj} = 80\text{ N} \times 1\text{ m} = \mathbf{80\text{ N m}}$$ (Clockwise)

$$\tau_{clockwise} = 10 + 80 = \mathbf{90\text{ N m}}$$

The tension ($$T$$) in cable $$AB$$ provides the counter-clockwise torque to balance the weights. It acts at point $$B$$, which is $$60\text{ cm} = 0.6\text{ m}$$ from the pivot, at an angle of $$30^\circ$$.

$$\tau_{tension} = T \times r \times \sin \theta$$

$$\tau_{tension} = T \times 0.6\text{ m} \times \sin(30^\circ)$$

$$\tau_{tension} = T \times 0.6 \times 0.5 = \mathbf{0.3T}$$

$$\tau_{anticlockwise} = \tau_{clockwise}$$

$$T = \frac{90}{0.3} = \mathbf{300\text{ N}}$$

We have a tube of length $$L = 50$$ cm $$= 0.5$$ m filled with an incompressible liquid of mass $$m = 250$$ g $$= 0.25$$ kg, rotating in a horizontal plane about one end with angular velocity $$\omega$$. We need to find the force exerted by the liquid at the far end.

Consider a small element of the liquid at distance $$r$$ from the axis of rotation, of length $$dr$$. The linear mass density is $$\lambda = \frac{m}{L} = \frac{0.25}{0.5} = 0.5$$ kg/m. The mass of this element is $$dm = \lambda \, dr$$. This element needs a centripetal force $$dF = dm \cdot \omega^2 r = \lambda \omega^2 r \, dr$$.

The force at the far end (at $$r = L$$) is the net force required to keep all the liquid beyond... but actually, the force at the closed end (the far end at distance $$L$$) must provide the centripetal force for the entire liquid column. We integrate the centripetal force contribution from $$r = 0$$ to $$r = L$$: $$F = \int_0^L \lambda \omega^2 r \, dr = \lambda \omega^2 \frac{L^2}{2} = \frac{m}{L} \cdot \omega^2 \cdot \frac{L^2}{2} = \frac{m \omega^2 L}{2}$$

Now we substitute $$\omega = x\sqrt{F}$$: $$F = \frac{m (x\sqrt{F})^2 L}{2} = \frac{m x^2 F L}{2}$$

Dividing both sides by $$F$$ (assuming $$F \neq 0$$): $$1 = \frac{m x^2 L}{2}$$

Solving for $$x^2$$: $$x^2 = \frac{2}{mL} = \frac{2}{0.25 \times 0.5} = \frac{2}{0.125} = 16$$

Hence $$x = 4$$.

Hence, the correct answer is 4.

We need to find the ratio $$I_1/I_2$$ where $$I_1$$ is the moment of inertia of a rod about one end and $$I_2$$ is the moment of inertia of the same rod bent into a ring about its diameter.

For a uniform rod of mass M and length L about a perpendicular axis through one end: $$I_1 = \frac{ML^2}{3}$$

When the rod is bent into a ring, the circumference equals the length of the rod: $$2\pi r = L \implies r = \frac{L}{2\pi}$$

The moment of inertia of a ring of mass M and radius r about its diameter is $$I_2 = \frac{Mr^2}{2} = \frac{M}{2}\left(\frac{L}{2\pi}\right)^2 = \frac{ML^2}{8\pi^2}$$

Therefore, $$\frac{I_1}{I_2} = \frac{ML^2/3}{ML^2/(8\pi^2)} = \frac{8\pi^2}{3}$$ and comparing with $$\frac{x\pi^2}{3}$$ gives $$\frac{x\pi^2}{3} = \frac{8\pi^2}{3}$$ so $$x = 8$$.

The answer is 8.

We are given a pulley of radius $$r = 1.5 \text{ m}$$, a tangential force $$F = (12t - 3t^2) \text{ N}$$, and moment of inertia $$I = 4.5 \text{ kg m}^2$$. We need to find $$K$$ where the number of rotations before the direction reverses is $$\dfrac{K}{\pi}$$.

$$\tau = F \times r = (12t - 3t^2) \times 1.5 = 18t - 4.5t^2$$

$$\alpha = \frac{\tau}{I} = \frac{18t - 4.5t^2}{4.5} = 4t - t^2$$

Since the pulley starts from rest ($$\omega = 0$$ at $$t = 0$$):

$$\omega = \int_0^t \alpha \, dt = \int_0^t (4t - t^2) \, dt = 2t^2 - \frac{t^3}{3}$$

The direction reverses when $$\omega = 0$$ (after the pulley has started rotating):

$$2t^2 - \frac{t^3}{3} = 0$$

$$t^2\left(2 - \frac{t}{3}\right) = 0$$

$$t = 0$$ or $$t = 6 \text{ s}$$

So the direction reverses at $$t = 6 \text{ s}$$.

$$\theta = \int_0^6 \omega \, dt = \int_0^6 \left(2t^2 - \frac{t^3}{3}\right) dt$$

$$\theta = \left[\frac{2t^3}{3} - \frac{t^4}{12}\right]_0^6$$

$$\theta = \frac{2(216)}{3} - \frac{1296}{12} = 144 - 108 = 36 \text{ rad}$$

$$\text{Number of rotations} = \frac{\theta}{2\pi} = \frac{36}{2\pi} = \frac{18}{\pi}$$

Comparing with $$\dfrac{K}{\pi}$$, we get $$K = 18$$.

Concept:
Loss in gravitational PE = Gain in total KE (translation + rotation)

Step 1: Change in Potential Energy

• Wheel (12 kg) moves down by h: 12gh
• Block (3 kg) moves up by h: −3gh

Net loss in PE=12gh − 3gh = 9gh

Step 2: Total Kinetic Energy at bottom

• Block:

$$\frac{1}{2}(3)v^2$$

• Wheel (translation):

$$\frac{1}{2}(12)v^2$$

• Wheel (rotation):

$$\frac{1}{2}I\omega^2 = \frac{1}{2}(12r^2)\left(\frac{v}{r}\right)^2 = \frac{1}{2}(12)v^2$$

Step 3: Apply energy conservation

9gh = $$\frac{1}{2}(3 + 12 + 12)v^2$$

9gh=$$\ \frac{\ 1}{2}(27)v^2$$

$$\Rightarrow v^2 = \frac{2gh}{3}$$ 

v = $$\sqrt{\frac{2gh}{3}}$$ = $$\frac{1}{2}\sqrt{\frac{8gh}{3}}$$

Step 4: Compare with given form

v = $$\frac{1}{2}\sqrt{xgh} \Rightarrow x = \frac{8}{3} \approx 3$$

Final Answer:

3​

For a cylindrical rod (uniform rod) of length $$L$$, the moment of inertia about an axis perpendicular to its length and passing through its center is:

$$I = \frac{ML^2}{12}$$

The radius of gyration $$k$$ is defined by $$I = Mk^2$$, so:

$$Mk^2 = \frac{ML^2}{12}$$

$$k = \frac{L}{\sqrt{12}} = \frac{L}{2\sqrt{3}}$$

Given $$L = 10\sqrt{3} \text{ m}$$:

$$k = \frac{10\sqrt{3}}{2\sqrt{3}} = \frac{10}{2} = 5 \text{ m}$$

The radius of gyration is $$\textbf{5}$$ m.

$$Massofdisc=1kg$$

$$Massofparticle=1kg$$

$$Radius=R$$

Initially the particle is at the top, finally at the bottom

Loss in potential energy of particle:
$$ΔPE=mg(2R)=1\times g\times2R=2gR$$

Disc’s potential energy does not change (its centre stays at same height)

So total energy converted to rotation:
$$2gR=\frac{1}{2}I_{total}ω^2$$

Now moment of inertia:

Disc about centre:
$$I_{disc}=\frac{1}{2}MR^2=\frac{1}{2}R^2$$

Particle at distance R:
$$I_{particle}=MR^2=R^2$$

So:
$$I_{total}=\frac{1}{2}R^2+R^2=\frac{3}{2}R^2$$

Now substitute:

$$2gR=\frac{1}{2}\left(\frac{3}{2}R^2ω^2\right)$$

$$2gR=(3/4)R^2ω^2$$

Solve:

$$ω^2=(8g)/(3R)$$

$$ω=\sqrt{\ \frac{8g}{3r}}$$

Given answer form:
$$ω=4\sqrt{\ \frac{x}{3r}}$$

Compare:
$$x=\frac{g}{2}$$

Given g = 10:

x = 5

Final answer:

5

A metre scale is initially balanced at its centre (the 50 cm mark). Two coins, each of mass 10 g (total mass = 20 g), are placed at the 10.0 cm mark, and the new balance point shifts to the 40.0 cm mark.

Concept: For the scale to be balanced at the new pivot (40 cm mark), the net torque about this point must be zero. The weight of the metre scale acts at its centre of gravity, which is at the 50 cm mark (since it is uniform).

Calculate the torques about the 40 cm mark: the coins (20 g) are at the 10 cm mark, which is $$40 - 10 = 30$$ cm to the left of the pivot.

The centre of gravity of the scale is at the 50 cm mark, which is $$50 - 40 = 10$$ cm to the right of the pivot.

Apply the condition for rotational equilibrium (net torque = 0):: $$\text{Anticlockwise torque} = \text{Clockwise torque}$$

$$20 \times 30 = M \times 10$$

$$600 = 10M$$

$$M = 60 \text{ g} = 60 \times 10^{-3} \text{ kg} = 6 \times 10^{-2} \text{ kg}$$

Since the mass is given as $$x \times 10^{-2}$$ kg, we get $$x = \mathbf{6}$$.

The disk begins from rest and rolls without slipping. Hence the angle turned by the disk in time $$t$$ is obtained from the constant-angular-acceleration relation

$$\theta = \frac12 \alpha t^{2}$$

With $$\alpha = \frac23 \text{ rad s}^{-2}$$ and $$t = \sqrt{\pi}\ \text{s}$$,

$$\theta = \frac12 \left(\frac23\right)(\sqrt{\pi})^{2}= \frac{\pi}{3} \text{ rad}$$

For pure rolling, a point that was at the initial contact position follows the cycloid

$$x = R(\theta - \sin\theta), \qquad y = R(1-\cos\theta)$$

with $$R = 1\ \text{m}$$. At $$\theta = \pi/3$$,

$$y = 1-\cos\left(\frac{\pi}{3}\right)=1-\frac12 = 0.5\ \text{m}$$

Thus the stone is $$0.5\ \text{m}$$ above the plane when it detaches.

The instantaneous angular speed is

$$\omega = \alpha t = \frac23\sqrt{\pi}\ \text{rad s}^{-1}$$

Differentiate the cycloid to find the velocity components (ground frame):

$$v_x = (1-\cos\theta)\,\omega,\qquad v_y = (\sin\theta)\,\omega$$

At $$\theta = \pi/3$$, $$1-\cos\theta = 0.5$$ and $$\sin\theta = \frac{\sqrt3}{2}$$, so

$$v_y = \frac{\sqrt3}{2}\,\omega = \frac{\sqrt3}{2}\left(\frac23\sqrt{\pi}\right) = \frac{\sqrt{3\pi}}{3}\ \text{m s}^{-1}$$

The stone now behaves like a projectile. The extra height it can rise is

$$\Delta h = \frac{v_y^{2}}{2g} = \frac{\left(\frac{\sqrt{3\pi}}{3}\right)^{2}}{2\times10} = \frac{\pi/3}{20}= \frac{\pi}{60}\ \text{m}$$

Total maximum height measured from the plane:

$$H = 0.5 + \frac{\pi}{60}\ \text{m}$$

The question writes this as $$H = \frac12 + \frac{x}{10}$$, hence

$$\frac{x}{10} = \frac{\pi}{60}\;\Longrightarrow\; x = \frac{\pi}{6} \approx 0.5236 \; \text{(m)}$$

Rounded to two decimal places, $$x = 0.52$$.

Answer: 0.52

All four bodies have mass $$M$$ and radius $$2R$$. The standard moment of inertia formulas (with radius $$r$$) are:

Solid sphere about diameter: $$\frac{2}{5}Mr^2$$. So $$I_1 = \frac{2}{5}M(2R)^2 = \frac{8}{5}MR^2$$.

Solid cylinder about its axis: $$\frac{1}{2}Mr^2$$. So $$I_2 = \frac{1}{2}M(2R)^2 = 2MR^2$$.

Solid circular disc about its diameter: $$\frac{1}{4}Mr^2$$. So $$I_3 = \frac{1}{4}M(2R)^2 = MR^2$$.

Thin circular ring about its diameter: $$\frac{1}{2}Mr^2$$. So $$I_4 = \frac{1}{2}M(2R)^2 = 2MR^2$$.

Now we compute $$2(I_2 + I_3) + I_4 = 2(2MR^2 + MR^2) + 2MR^2 = 2(3MR^2) + 2MR^2 = 6MR^2 + 2MR^2 = 8MR^2$$.

Setting this equal to $$xI_1$$: $$8MR^2 = x \cdot \frac{8}{5}MR^2$$.

Dividing both sides by $$\frac{8}{5}MR^2$$: $$x = \frac{8}{\frac{8}{5}} = 5$$.

The value of $$x$$ is $$\mathbf{5}$$.

We need to find the magnitude of the angular momentum of a 1 kg object. The mass is $$m = 1$$ kg, the position vector is $$\vec{r} = 3\hat{i} - \hat{j}$$ m, and the velocity is $$\vec{v} = 3\hat{j} + \hat{k}$$ m/s.

Since linear momentum is $$\vec{p} = m\vec{v} = 1 \times (3\hat{j} + \hat{k}) = 3\hat{j} + \hat{k}$$, substituting into the definition of angular momentum $$\vec{L} = \vec{r} \times \vec{p}$$ gives

$$\vec{L} = \vec{r} \times \vec{p} = (3\hat{i} - \hat{j}) \times (3\hat{j} + \hat{k})$$

Using the determinant method,

$$\vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 0 \\ 0 & 3 & 1 \end{vmatrix}$$

Expanding the determinant yields

$$\vec{L} = \hat{i}[(-1)(1) - (0)(3)] - \hat{j}[(3)(1) - (0)(0)] + \hat{k}[(3)(3) - (-1)(0)]$$

$$\vec{L} = \hat{i}[-1 - 0] - \hat{j}[3 - 0] + \hat{k}[9 - 0]$$

$$\vec{L} = -\hat{i} - 3\hat{j} + 9\hat{k}$$

The magnitude is

$$|\vec{L}| = \sqrt{(-1)^2 + (-3)^2 + (9)^2}$$

$$|\vec{L}| = \sqrt{1 + 9 + 81}$$

$$|\vec{L}| = \sqrt{91} \text{ N m s}$$

Comparing with $$\sqrt{x}$$, we get $$x = 91$$, so the answer is 91.

1. Translational (Orbital) Angular Momentum

• Disk A is fixed at the origin. Since both disks have a radius of $$R$$, the center of mass (CM) of disk B moves in a circular path of radius $$r = R + R = 2R$$.
• The center of mass of disk B moves with a fixed angular speed $\omega$ around the center of A. Therefore, the linear velocity of its center of mass ($$v_{\text{cm}}$$) is: $$v_{\text{cm}} = \omega(2R) = 2\omega R$$
• The orbital angular momentum ($$L_{\text{orbital}}$$) of disk B relative to the center of A is given by: $$L_{\text{orbital}} = M \cdot v_{\text{cm}} \cdot (2R) = M(2\omega R)(2R) = 4M\omega R^2$$

2. Rotational (Spin) Angular Momentum

• Because disk B rolls without slipping on the stationary circumference of disk A, its contact point with disk A must have an instantaneous velocity of zero.
• Let $$\omega_B$$ be the spinning angular velocity of disk B about its own center of mass. The non-slip condition dictates: $$v_{\text{cm}} = \omega_B R \implies 2\omega R = \omega_B R \implies \omega_B = 2\omega$$
• The moment of inertia of a uniform disk about its central axis is $$I_{\text{cm}} = \frac{1}{2}MR^2$$.
• The intrinsic spin angular momentum ($$L_{\text{spin}}$$) about its own center is: $$L_{\text{spin}} = I_{\text{cm}}\omega_B = \left(\frac{1}{2}MR^2\right)(2\omega) = M\omega R^2$$

3. Total Angular Momentum

Since both the orbital revolution and the intrinsic spin occur in the same direction (counter-clockwise), their magnitudes add up directly:

$$L_{\text{total}} = L_{\text{orbital}} + L_{\text{spin}}$$

$$L_{\text{total}} = 4M\omega R^2 + M\omega R^2 = 5M\omega R^2$$

Conclusion

Comparing our result with the expression given in the problem, $$n M \omega R^2$$:

$$n M \omega R^2 = 5 M \omega R^2 \implies n = 5$$

For a solid sphere rolling without slipping, we need to find the ratio of rotational kinetic energy to total kinetic energy.

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

$$K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \times \frac{2}{5}mR^2 \times \omega^2 = \frac{1}{5}mR^2\omega^2$$

For rolling without slipping, $$v = R\omega$$

$$K_{trans} = \frac{1}{2}mv^2 = \frac{1}{2}mR^2\omega^2$$

$$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mR^2\omega^2 + \frac{1}{5}mR^2\omega^2 = \frac{7}{10}mR^2\omega^2$$

$$\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{5}mR^2\omega^2}{\frac{7}{10}mR^2\omega^2} = \frac{\frac{1}{5}}{\frac{7}{10}} = \frac{1}{5} \times \frac{10}{7} = \frac{2}{7}$$

Hence, the correct answer is Option C.

We need to find the torque about the origin given $$\vec{F} = 3\hat{i} + 4\hat{j} - 2\hat{k}$$ and position vector $$\vec{r} = 2\hat{i} + \hat{j} + 2\hat{k}$$.

The torque is given by the cross product of $$\vec{r}$$ and $$\vec{F}$$: $$\vec{\tau} = \vec{r} \times \vec{F}$$.

Using the determinant method, we compute $$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 3 & 4 & -2 \end{vmatrix}$$.

Expanding this determinant gives the components: the $$\hat{i}$$ component is $$(1)(-2) - (2)(4) = -2 - 8 = -10$$; the $$\hat{j}$$ component is $$-[(2)(-2) - (2)(3)] = -[-4 - 6] = 10$$; and the $$\hat{k}$$ component is $$(2)(4) - (1)(3) = 8 - 3 = 5$$.

Therefore, $$\vec{\tau} = -10\hat{i} + 10\hat{j} + 5\hat{k}$$.

The correct answer is Option A: $$-10\hat{i} + 10\hat{j} + 5\hat{k}$$.

We need to match the moment of inertia expressions using standard formulas and the parallel axis theorem.

(A) M.I. of solid sphere about a tangent:

M.I. about the center = $$\frac{2}{5}MR^2$$

By parallel axis theorem (shifting by distance $$R$$):

$$I = \frac{2}{5}MR^2 + MR^2 = \frac{2 + 5}{5}MR^2 = \frac{7}{5}MR^2$$

This matches (II).

(B) M.I. of hollow sphere about a tangent:

M.I. about the center = $$\frac{2}{3}MR^2$$

By parallel axis theorem:

$$I = \frac{2}{3}MR^2 + MR^2 = \frac{2 + 3}{3}MR^2 = \frac{5}{3}MR^2$$

This matches (I).

(C) M.I. of circular ring about its diameter:

By perpendicular axis theorem: $$I_z = I_x + I_y$$

For a ring, $$I_z = MR^2$$ (about axis through center, perpendicular to plane)

By symmetry, $$I_x = I_y$$, so: $$I_x = \frac{MR^2}{2} = \frac{1}{2}MR^2$$

This matches (IV).

(D) M.I. of circular disc about any diameter:

M.I. about axis perpendicular to disc through center = $$\frac{1}{2}MR^2$$

By perpendicular axis theorem and symmetry:

$$I_{\text{diameter}} = \frac{1}{2} \times \frac{1}{2}MR^2 = \frac{1}{4}MR^2$$

This matches (III).

The matching is: A-II, B-I, C-IV, D-III

The correct answer is Option A.

The flywheel starts from rest and accelerates uniformly with angular acceleration $$\alpha$$. Using the equation for angular displacement with uniform angular acceleration $$\theta = \omega_0 t + \frac{1}{2}\alpha t^2$$ and noting that $$\omega_0 = 0$$, in the first second ($$t = 1$$ s) we have $$5 = \frac{1}{2}\alpha(1)^2$$ so $$\alpha = 10 \text{ rad/s}^2$$. In the first two seconds the total angle is $$\theta_2 = \frac{1}{2}(10)(2)^2 = \frac{1}{2}(10)(4) = 20 \text{ rad}$$, and the angle rotated in the second second is $$\theta_{\text{2nd sec}} = 20 - 5 = 15 \text{ rad}$$.

Using the formula for the angle in the $$n$$th second, $$\theta_n = \omega_0 + \frac{\alpha}{2}(2n - 1)$$, for $$n = 2$$ we get $$\theta_2 = 0 + \frac{10}{2}(2 \times 2 - 1) = 5 \times 3 = 15$$ rad, confirming the result. The correct answer is Option B: 15 rad.

We need to find the ratio of the final linear velocity of a rolling solid cylinder to that of a rolling solid sphere when they start from rest and roll down the same inclined plane without slipping.

1. Derive the Velocity Formula for Pure Rolling

When an object of mass $$M$$ and radius $$R$$ rolls down an inclined plane of height $$h$$ without slipping, its initial potential energy is converted into both translational and rotational kinetic energy:

$$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$$

Since the object rolls without slipping, its angular velocity is $$\omega = \frac{v}{R}$$. Substituting this and writing the moment of inertia in terms of its radius of gyration ($$I = MK^2$$):

$$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}(MK^2)\left(\frac{v}{R}\right)^2$$

$$Mgh = \frac{1}{2}Mv^2 \left(1 + \frac{K^2}{R^2}\right)$$

Canceling out the mass ($$M$$) and isolating the velocity ($$v$$) gives:

$$v = \sqrt{\frac{2gh}{1 + \frac{K^2}{R^2}}}$$

Since both objects fall from the same height ($$h$$), their velocities are inversely proportional to their rolling factor:

$$v \propto \frac{1}{\sqrt{1 + \frac{K^2}{R^2}}}$$

2. Calculate the Rolling Factors

• For a Solid Cylinder:

  The moment of inertia is $$I_{\text{cylinder}} = \frac{1}{2}MR^2 \implies \frac{K^2}{R^2} = \frac{1}{2}$$

  $$1 + \frac{K^2}{R^2} = 1 + \frac{1}{2} = \frac{3}{2}$$

• For a Solid Sphere:

  The moment of inertia is $$I_{\text{sphere}} = \frac{2}{5}MR^2 \implies \frac{K^2}{R^2} = \frac{2}{5}$$

  $$1 + \frac{K^2}{R^2} = 1 + \frac{2}{5} = \frac{7}{5}$$

3. Determine the Ratio ($$\frac{v_{\text{cylinder}}}{v_{\text{sphere}}}$$)

Taking the ratio of their final linear velocities:

$$\frac{v_{\text{cylinder}}}{v_{\text{sphere}}} = \sqrt{\frac{1 + \left(\frac{K^2}{R^2}\right)_{\text{sphere}}}{1 + \left(\frac{K^2}{R^2}\right)_{\text{cylinder}}}}$$

Substitute the calculated values into the ratio expression:

$$\frac{v_{\text{cylinder}}}{v_{\text{sphere}}} = \sqrt{\frac{\frac{7}{5}}{\frac{3}{2}}} = \sqrt{\frac{7}{5} \times \frac{2}{3}} = \sqrt{\frac{14}{15}}$$

Conclusion

The ratio of the velocity of the solid cylinder to that of the solid sphere is $$\sqrt{\frac{14}{15}}$$

$$Length=\sqrt{\ 34},base=3m⇒height=5m$$

$$Weight=10g=100Nactingatmidpoint$$

midpoint coordinates:
horizontal = 3/2
vertical = 5/2

Forces:

• at floor: $$F_f$$ (resultant, has horizontal + vertical)
• at wall: $$F_w$$ (horizontal only, since wall is frictionless)

Take moments about the bottom point:

Torque by weight (clockwise):
$$=100\times(horizontaldis\tan ceofCM)$$

$$=100\times(3/2)=150$$

Torque by wall reaction (anticlockwise):
$$=F_w\times height$$
$$=F_w\times5$$

Equilibrium:
$$F_w\times5=150$$
$$F_w=30N$$

Now vertical equilibrium:

Ff(vertical) = 100

Horizontal equilibrium:

$$F_f(horizontal)=F_w=30$$

So resultant floor reaction:

$$Ff=\sqrt{(100^2+30^2)\ }$$
$$=\sqrt{\left(10000+900)\right)}$$
= 10$$\sqrt{109}$$

Now ratio:

$$Fw/Ff=30/(10\sqrt{109}$$
$$=3/\sqrt{109}$$

Final answer:

$$3/\sqrt{109}$$

A thin circular ring of mass $$M$$ and radius $$R$$ rotates with angular velocity $$\omega_0 = 2$$ rad/s. Two objects each of mass $$m$$ are attached gently to opposite ends of a diameter.

Since the masses are attached gently (no external torque), angular momentum is conserved:
$$L_i = L_f$$
$$I_i \omega_0 = I_f \omega_f$$.

Initial moment of inertia (ring only) is
$$I_i = MR^2$$
and final moment of inertia (ring + two point masses at distance $$R$$) is
$$I_f = MR^2 + 2mR^2 = (M + 2m)R^2$$.

Substituting into the conservation equation,
$$MR^2 \times 2 = (M + 2m)R^2 \times \omega_f$$,
this gives
$$\omega_f = \frac{2M}{M + 2m}$$ rad/s.

The correct answer is Option C.

Let us analyze both the Assertion and Reason carefully.

Assertion A: "If we move from poles to equator, the direction of acceleration due to gravity of earth always points towards the center of earth without any variation in its magnitude."

Analysis of Assertion A:

The effective gravitational acceleration at a latitude $$\lambda$$ on the rotating Earth is:

$$g_{eff} = g - R\omega^2\cos^2\lambda$$

where $$g$$ is the gravitational acceleration without rotation, $$R$$ is the radius of the Earth, and $$\omega$$ is the angular velocity of Earth's rotation.

Variation in magnitude: At the poles ($$\lambda = 90°$$), $$g_{eff} = g$$. At the equator ($$\lambda = 0°$$), $$g_{eff} = g - R\omega^2$$, which is smaller. So the magnitude clearly varies as we move from poles to equator. This part of the assertion is false.

Variation in direction: The centrifugal acceleration at latitude $$\lambda$$ is directed perpendicular to Earth's rotation axis (i.e., radially outward from the axis, not from the center of Earth). Only at the equator and at the poles does the centrifugal acceleration align radially with respect to Earth's center. At intermediate latitudes, the centrifugal force has a component tangential to the Earth's surface, causing the net effective gravity to deviate slightly from pointing towards the center. So the direction also varies, making the assertion false.

Reason R: "At the equator, the direction of acceleration due to gravity is towards the center of earth."

Analysis of Reason R:

At the equator, the centrifugal acceleration $$R\omega^2$$ is directed radially outward from Earth's center (since at the equator, the direction perpendicular to the rotation axis coincides with the radial direction from Earth's center). The gravitational pull is directed radially inward towards the center. Since both are along the same line (radial direction), the resultant effective gravity still points towards the center of Earth. Only the magnitude is reduced. This makes Reason R true.

Conclusion:

The Assertion is false (both magnitude and direction vary) but the Reason is true (at the equator specifically, the direction is towards the center).

Hence, the correct answer is Option D.

We have the position vector $$\vec{r} = 2\hat{i} + 2\hat{j} + \hat{k}$$ and the force $$\vec{F} = 5\hat{i} + 3\hat{j} - 7\hat{k}$$. The torque about the origin is $$\vec{\tau} = \vec{r} \times \vec{F}$$.

We compute the cross product using the determinant formula: $$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ 5 & 3 & -7 \end{vmatrix}$$

The $$\hat{i}$$ component is $$(2)(-7) - (1)(3) = -14 - 3 = -17$$. The $$\hat{j}$$ component is $$-\left[(2)(-7) - (1)(5)\right] = -(-14 - 5) = 19$$. The $$\hat{k}$$ component is $$(2)(3) - (2)(5) = 6 - 10 = -4$$.

Therefore $$\vec{\tau} = -17\hat{i} + 19\hat{j} - 4\hat{k}$$.

Hence, the correct answer is Option 3.

Given that it is an spherical shell

We know that moment of inertia of a spherical shell with respect to its centre is $$I=\frac{2}{3}MR^2$$

But with respect to origin it is $$I+MR^2$$ (Here R is the perpendicular distance between origin and centre of the sphere

So moment of inertia with respect to origin is $$I=\frac{5}{3}MR^2$$

We also know that Angular momentum $$L=I\omega\ $$

substitute I

$$L=\frac{5}{3}MR^2\omega\ $$

Given that M=1

$$L=\frac{5}{3}R^2\omega\ $$

So a will be 5

Particle P moves along $$\vec{A} = \hat{i} + \hat{j}$$ and then along a direction normal to the plane containing $$\vec{A}$$ and $$\vec{B} = \hat{j} + \hat{k}$$. So the direction of P after the collision is $$\vec{A} \times \vec{B}$$.

$$\vec{A} \times \vec{B} = (\hat{i} + \hat{j}) \times (\hat{j} + \hat{k}) = \hat{i}\times\hat{j} + \hat{i}\times\hat{k} + \hat{j}\times\hat{j} + \hat{j}\times\hat{k}$$ $$= \hat{k} - \hat{j} + 0 + \hat{i} = \hat{i} - \hat{j} + \hat{k}$$

Particle Q moves along $$\vec{A} = \hat{i} + \hat{j}$$ and then along a direction normal to the plane containing $$\vec{A}$$ and $$\vec{C} = -\hat{i} + \hat{j}$$. So the direction of Q after the collision is $$\vec{A} \times \vec{C}$$.

$$\vec{A} \times \vec{C} = (\hat{i} + \hat{j}) \times (-\hat{i} + \hat{j}) = \hat{i}\times(-\hat{i}) + \hat{i}\times\hat{j} + \hat{j}\times(-\hat{i}) + \hat{j}\times\hat{j}$$ $$= 0 + \hat{k} + \hat{k} + 0 = 2\hat{k}$$

The angle $$\theta$$ between these two directions is found using the dot product formula:

$$\cos\theta = \dfrac{(\hat{i} - \hat{j} + \hat{k}) \cdot (2\hat{k})}{|\hat{i} - \hat{j} + \hat{k}| \times |2\hat{k}|}$$

The numerator is $$2$$, and the denominator is $$\sqrt{1^2 + (-1)^2 + 1^2} \times 2 = \sqrt{3} \times 2$$.

$$\cos\theta = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$$

Comparing with $$\cos^{-1}\!\left(\dfrac{1}{\sqrt{x}}\right)$$, we get $$x = 3$$.

We need to determine the time $$t > 0$$ at which the angular momentum of the particle equals its initial angular momentum at $$t = 0$$.

1. Determine the Position and Velocity Vectors

From the problem statement , the position vector $$\vec{r}(t)$$ of the particle of mass $$m$$ is given by:

$$\vec{r}(t) = 10\alpha t^2 \hat{i} + 5\beta(t - 5)\hat{j}$$

To find the velocity vector $$\vec{v}(t)$$, we take the first derivative of the position vector with respect to time ($$\vec{v} = \frac{d\vec{r}}{dt}$$):

$$\vec{v}(t) = \frac{d}{dt}\left(10\alpha t^2\right)\hat{i} + \frac{d}{dt}\left(5\beta t - 25\beta\right)\hat{j}$$

$$\vec{v}(t) = 20\alpha t \hat{i} + 5\beta \hat{j}$$

2. Formulate the Angular Momentum Equation

The angular momentum $$\vec{L}$$ of a particle about the origin is defined by the cross product of its position and linear momentum vectors:

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

Let's calculate the cross product $$\vec{r} \times \vec{v}$$ explicitly using their unit vector components:

$$\vec{r} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 10\alpha t^2 & 5\beta(t - 5) & 0 \\ 20\alpha t & 5\beta & 0 \end{vmatrix}$$

$$\vec{r} \times \vec{v} = \left[(10\alpha t^2)(5\beta) - (5\beta(t - 5))(20\alpha t)\right] \hat{k}$$

$$\vec{r} \times \vec{v} = \left[50\alpha\beta t^2 - 100\alpha\beta t(t - 5)\right] \hat{k}$$

$$\vec{r} \times \vec{v} = \left[50\alpha\beta t^2 - 100\alpha\beta t^2 + 500\alpha\beta t\right] \hat{k} = \left[500\alpha\beta t - 50\alpha\beta t^2\right] \hat{k}$$

Therefore, the angular momentum vector at any time $$t$$ is:

$$\vec{L}(t) = m\alpha\beta\left(500t - 50t^2\right)\hat{k}$$

3. Equate Angular Momentum States

First, evaluate the initial angular momentum at $$t = 0$$:

$$\vec{L}(0) = m\alpha\beta\left(500(0) - 50(0)^2\right)\hat{k} = \vec{0}$$

We are looking for a non-zero time $$t$$ where the angular momentum returns to this initial value ($$\vec{L}(t) = \vec{L}(0)$$):

$$m\alpha\beta\left(500t - 50t^2\right)\hat{k} = \vec{0}$$

Since the mass $$m$$ and dimensional constants $$\alpha, \beta$$ are non-zero, we can divide them out:

$$500t - 50t^2 = 0$$

$$50t(10 - t) = 0$$

This quadratic equation gives two solutions:

• $$t = 0\text{ seconds}$$ (Initial state)
• $$t = 10\text{ seconds}$$

Conclusion

The angular momentum of the particle becomes the same as it was for $$t = 0$$ at time $$t =$$ 10 seconds.

We have a solid disc of mass $$M = 10\;\text{kg}$$ and radius $$R = 20\;\text{cm} = 0.20\;\text{m}$$ rotating with an initial angular velocity of $$600\;\text{rpm}$$ about its central axis. First, we convert this angular speed to SI units.

The relation between revolutions per minute and radians per second is

$$\omega = \text{(rev/min)} \times \frac{2\pi\;\text{rad}}{1\;\text{rev}} \times \frac{1\;\text{min}}{60\;\text{s}}.$$

Substituting the given value,

$$\omega_i = 600 \times \frac{2\pi}{60} = 600 \times \frac{2\pi}{60} = 10 \times 2\pi = 20\pi\; \text{rad s}^{-1}.$$

The disc has to be brought to rest in $$t = 10\;\text{s}$$, so the final angular velocity is $$\omega_f = 0$$. For constant angular deceleration $$\alpha$$ we use

$$\omega_f = \omega_i + \alpha t.$$

Rearranging,

$$\alpha = \frac{\omega_f - \omega_i}{t} = \frac{0 - 20\pi}{10} = -2\pi\;\text{rad s}^{-2}.$$

The minus sign merely shows that the acceleration is retarding; its magnitude is $$|\alpha| = 2\pi\;\text{rad s}^{-2}.$$

Now we need the moment of inertia of a solid disk about its central axis. The formula is stated as

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

Substituting $$M = 10\;\text{kg}$$ and $$R = 0.20\;\text{m},$$

$$I = \frac{1}{2}(10)(0.20)^2 = \frac{1}{2}(10)(0.04) = 0.2\;\text{kg m}^{2}.$$

The magnitude of the required retarding torque $$\tau$$ is connected to angular acceleration by

$$\tau = I\,|\alpha|.$$

Substituting $$I = 0.2\;\text{kg m}^{2}$$ and $$|\alpha| = 2\pi\;\text{rad s}^{-2},$$

$$\tau = 0.2 \times 2\pi = 0.4\pi\;\text{N m}.$$

We express $$0.4\pi$$ in the form $$k \times \pi \times 10^{-1}$$. Noting that $$0.4 = 4 \times 10^{-1},$$ we write

$$\tau = 4\,\pi \times 10^{-1}\;\text{N m}.$$

Thus the numeral to be filled in the blank is $$4$$.

Hence, the correct answer is Option 4.

A uniform thin bar of mass $$M = 6$$ kg and length $$L = 2.4$$ m is bent into a regular hexagon. Since a regular hexagon has 6 equal sides, the length of each side is $$a = \frac{L}{6} = \frac{2.4}{6} = 0.4$$ m, and the mass of each side is $$m = \frac{M}{6} = \frac{6}{6} = 1$$ kg.

For a regular hexagon with side $$a$$, the perpendicular distance from the centre to the midpoint of each side is $$d = \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{2} \times 0.4 = 0.2\sqrt{3}$$ m.

The moment of inertia of each side (a thin rod of mass $$m$$ and length $$a$$) about the axis through the centre of the hexagon and perpendicular to its plane is found using the parallel axis theorem. Each rod has its own moment of inertia about its centre given by $$\frac{ma^2}{12}$$, and using the parallel axis theorem with distance $$d$$, the moment of inertia of each rod about the central axis is $$I_{\text{rod}} = \frac{ma^2}{12} + md^2$$.

Substituting the values: $$I_{\text{rod}} = \frac{1 \times (0.4)^2}{12} + 1 \times (0.2\sqrt{3})^2 = \frac{0.16}{12} + 0.12 = 0.01333 + 0.12 = 0.13333$$ kg m$$^2$$.

Since the hexagon has 6 identical sides, the total moment of inertia is $$I = 6 \times I_{\text{rod}} = 6 \times 0.13333 = 0.8$$ kg m$$^2$$.

Expressing this as $$x \times 10^{-1}$$ kg m$$^2$$: $$0.8 = 8 \times 10^{-1}$$.

Therefore, the answer is $$8$$.

We are given that the angular speed of a truck wheel increases uniformly from $$\omega_1 = 900$$ rpm to $$\omega_2 = 2460$$ rpm in $$t = 26$$ seconds, and we need to find the total number of revolutions during this time.

For uniform angular acceleration, the total angle swept (in revolutions) equals the average angular speed multiplied by the time. The average angular speed is $$\omega_{avg} = \frac{\omega_1 + \omega_2}{2} = \frac{900 + 2460}{2} = \frac{3360}{2} = 1680$$ rpm.

Now we convert the time to minutes: $$t = 26 \text{ seconds} = \frac{26}{60}$$ minutes.

The number of revolutions is $$N = \omega_{avg} \times t = 1680 \times \frac{26}{60} = 1680 \times \frac{13}{30} = \frac{1680 \times 13}{30} = \frac{21840}{30} = 728$$.

Therefore, the number of revolutions by the truck engine during this time is $$\boxed{728}$$.

For a body rolling without slipping down an inclined plane, energy conservation gives:

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

Since rolling without slipping: $$\omega = \frac{v}{R}$$, so:

$$mgh = \frac{1}{2}mv^2\left(1 + \frac{I}{mR^2}\right)$$

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

For a ring: $$I_{\text{ring}} = mR^2$$, so $$\frac{I}{mR^2} = 1$$, giving:

$$v_{\text{ring}} = \sqrt{\frac{2gh}{2}} = \sqrt{gh}$$

For a solid cylinder: $$I_{\text{cylinder}} = \frac{1}{2}mR^2$$, so $$\frac{I}{mR^2} = \frac{1}{2}$$, giving:

$$v_{\text{cylinder}} = \sqrt{\frac{2gh}{\frac{3}{2}}} = \sqrt{\frac{4gh}{3}}$$

The ratio of velocities:

$$\frac{v_{\text{ring}}}{v_{\text{cylinder}}} = \frac{\sqrt{gh}}{\sqrt{\frac{4gh}{3}}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$

Given that this ratio equals $$\frac{\sqrt{x}}{2}$$, we have $$\sqrt{x} = \sqrt{3}$$, so $$x = 3$$.

The initial angular speed is $$\omega_1 = 600 \text{ rpm}$$ and the final angular speed is $$\omega_2 = 1800 \text{ rpm}$$. The time taken is $$t = 10$$ s.

The average angular speed during uniform acceleration is:

$$\omega_{\text{avg}} = \frac{\omega_1 + \omega_2}{2} = \frac{600 + 1800}{2} = 1200 \text{ rpm}$$

The total number of rotations is:

$$N = \omega_{\text{avg}} \times t = 1200 \text{ rpm} \times \frac{10}{60} \text{ min} = 1200 \times \frac{1}{6} = 200 \text{ rotations}$$

The intersection of the $$x = 2$$ plane and the $$x$$-axis is the point $$(2, 0, 0)$$. The force is $$\vec{F} = 4\hat{i} + 3\hat{j} + 4\hat{k}$$ applied at point $$A = (2, 0, 0)$$.

We need the torque about point $$B = (2, 3, 4)$$. The position vector from $$B$$ to $$A$$ is $$\vec{r} = A - B = (2-2)\hat{i} + (0-3)\hat{j} + (0-4)\hat{k} = -3\hat{j} - 4\hat{k}$$.

The torque is $$\vec{\tau} = \vec{r} \times \vec{F}$$:

$$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -3 & -4 \\ 4 & 3 & 4 \end{vmatrix}$$

$$= \hat{i}[(-3)(4) - (-4)(3)] - \hat{j}[(0)(4) - (-4)(4)] + \hat{k}[(0)(3) - (-3)(4)]$$

$$= \hat{i}[-12 + 12] - \hat{j}[0 + 16] + \hat{k}[0 + 12]$$

$$= 0\hat{i} - 16\hat{j} + 12\hat{k}$$

The magnitude is $$|\vec{\tau}| = \sqrt{0^2 + 16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20$$.

The magnitude of torque is $$\boxed{20}$$.

A particle of mass $$m$$ moving with velocity $$u$$ hits one end of a rod of mass $$M$$ and length $$L$$ perpendicularly, with the collision being completely elastic, and the particle comes to rest after the collision.

Let $$V_{cm}$$ be the velocity of the centre of mass of the rod and $$\omega$$ be its angular velocity after the collision. Conservation of linear momentum gives $$mu = MV_{cm}$$, so $$V_{cm} = \frac{mu}{M}$$.

Conservation of angular momentum about the centre of the rod gives $$mu\cdot\frac{L}{2} = I_{rod}\,\omega = \frac{ML^2}{12}\,\omega$$, so $$\omega = \frac{6mu}{ML}$$.

Conservation of kinetic energy (elastic collision): $$\frac{1}{2}mu^2 = \frac{1}{2}MV_{cm}^2 + \frac{1}{2}I_{rod}\,\omega^2$$.

Substituting: $$mu^2 = M\cdot\frac{m^2u^2}{M^2} + \frac{ML^2}{12}\cdot\frac{36m^2u^2}{M^2L^2} = \frac{m^2u^2}{M} + \frac{3m^2u^2}{M} = \frac{4m^2u^2}{M}$$.

Therefore $$m = \frac{M}{4}$$, giving $$\frac{m}{M} = \frac{1}{4} = \frac{1}{x}$$, so $$x = 4$$.

We need to find the number of revolutions $$n$$ made by a uniform circular disk to reach a given angular speed under the action of a constant force.

The moment of inertia $$I$$ of a uniform circular disk of mass $$M$$ and radius $$R$$ about an axis passing through its center is given by: $$I = \frac{1}{2} M R^2$$.

We are given the following values:

Mass of the disk, $$M = 20\text{ kg}$$

Radius of the disk, $$R = 0.2\text{ m}$$

Substituting these values gives: $$I = \frac{1}{2} \times 20 \times (0.2)^2 = 10 \times 0.04 = 0.4\text{ kg m}^2$$.

The torque $$\tau$$ acting on the disk due to the constant force $$F = 20\text{ N}$$ applied at its periphery is: $$\tau = F \times R = 20 \times 0.2 = 4\text{ N m}$$.

According to Newton's second law for rotation, the angular acceleration $$\alpha$$ is given by: $$\alpha = \frac{\tau}{I} = \frac{4}{0.4} = 10\text{ rad s}^{-2}$$.

The disk starts from rest, so its initial angular speed is $$\omega_0 = 0$$. To find the total angular displacement $$\theta$$ required to attain a final angular speed of $$\omega = 50\text{ rad s}^{-1}$$, we use the third equation of rotational motion: $$\omega^2 = \omega_0^2 + 2\alpha\theta$$.

Substituting the known parameters into the equation yields: $$50^2 = 0^2 + 2 \times 10 \times \theta$$, which simplifies to: $$2500 = 20\theta$$.

Solving for the angular displacement gives: $$\theta = \frac{2500}{20} = 125\text{ rad}$$.

Since one complete revolution corresponds to a rotation of $$6.28\text{ rad}$$, the total number of revolutions $$n$$ is: $$n = \frac{\theta}{2\pi} = \frac{125}{6.28} \approx 19.90$$.

Rounding to the nearest integer, the value of $$n$$ is 20.

We need to determine the total moment of inertia of the badminton racket about an axis perpendicular to the handle, lying in the plane of the ring, and located at a distance of

$$\frac{r}{2}$$

from the free end A of the handle.

1. Identify the Given Parameters

From the problem statement , the badminton racket is split into two geometric components:

• The Handle (Linear Rod): Mass = $$M$$, Length = $$6r$$
• The Head (Circular Ring): Mass = $$M$$, Radius = $$r$$ (Diameter = $$2r$$)

The target axis of rotation is positioned at a distance of $$\frac{r}{2}$$ from end A of the handle.

2. Calculate the Moment of Inertia for the Handle

The handle is treated as a uniform thin rod. We determine its moment of inertia using the parallel axis theorem:

• Center of Mass Location:
  The center of mass of the rod lies at its midpoint, which is at a distance of $$3r$$ from end A.

  $$I_{\text{cm, handle}} = \frac{1}{12} M L^2 = \frac{1}{12} M (6r)^2 = 3Mr^2$$

• Parallel Axis Distance Shift:
  The distance from the rod's center of mass to the target axis (at $$\frac{r}{2}$$ from A) is:

  $$d_1 = 3r - \frac{r}{2} = \frac{5r}{2}$$

Applying the Parallel Axis Theorem ($$I = I_{\text{cm}} + Md^2$$):

$$I_{\text{handle}} = 3Mr^2 + M\left(\frac{5r}{2}\right)^2 = 3Mr^2 + \frac{25}{4}Mr^2 = \frac{37}{4}Mr^2$$

3. Calculate the Moment of Inertia for the Ring Head

The head of the racket is treated as a uniform circular ring. The rotation axis lies along the coplanar diameter plane of the ring:

• Center of Mass Axis:
  The moment of inertia of a thin ring about its coplanar diameter axis passing through its center is:

  $$I_{\text{cm, ring}} = \frac{1}{2}Mr^2$$

• Parallel Axis Distance Shift:
  The center of the ring is located at the far end of the handle, a distance of $$6r + r = 7r$$ from end A. The shift distance to the target axis is:

  $$d_2 = 7r - \frac{r}{2} = \frac{13r}{2}$$

Applying the Parallel Axis Theorem ($$I = I_{\text{cm}} + Md^2$$):

$$I_{\text{ring}} = \frac{1}{2}Mr^2 + M\left(\frac{13r}{2}\right)^2 = \frac{2}{4}Mr^2 + \frac{169}{4}Mr^2 = \frac{171}{4}Mr^2$$

4. Determine Total Moment of Inertia

The total moment of inertia of the entire composite racket system is the scalar sum of both individual components:

$$I_{\text{total}} = I_{\text{handle}} + I_{\text{ring}}$$

$$I_{\text{total}} = \frac{37}{4}Mr^2 + \frac{171}{4}Mr^2 = \frac{208}{4}Mr^2 = 52Mr^2$$

Conclusion

The multiplier value for the final total moment of inertia observed for the badminton racket is 52.

We need to determine which of the four given bodies will reach the bottom of the inclined plane first when allowed to roll down simultaneously without slipping.

1. Relate Rolling Acceleration to the Moment of Inertia

When a body rolls down an inclined plane of angle $$\theta$$ without slipping, its linear acceleration ($$a$$) is given by the formula:

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

Where $$I$$ is the moment of inertia, $$m$$ is the mass, and $$R$$ is the radius of the body.

The body with the highest linear acceleration ($$a$$) will take the least amount of time to travel down the incline, meaning it will reach the bottom first. Looking at the formula, acceleration is inversely proportional to the factor $$\frac{I}{mR^2}$$. Therefore, the body with the smallest value of $$\frac{I}{mR^2}$$ will have the highest acceleration.

2. Analyze the Given Bodies

Let's calculate the value of $$\frac{I}{mR^2}$$ for each body mentioned on the  page:

3. Compare the Values

• The solid sphere has the lowest factor of $$\frac{I}{mR^2} = 0.40$$.
• Substituting this value into the acceleration formula gives it the maximum acceleration:

  $$a_{\text{sphere}} = \frac{g \sin\theta}{1 + 0.4} = \frac{5}{7}g \sin\theta \approx 0.714\, g \sin\theta$$

Conclusion

Since the solid sphere has the maximum linear acceleration, it will reach the bottom of the inclined plane first. This matches numbering 4.

We have a uniform steel rod of mass $$m = 2\;\text{kg}$$ and length $$L = 0.6\;\text{m}$$. It is clamped at its lower end, so that end acts as a fixed horizontal axis. The upper (free) end is given a slight push and the rod is then allowed to fall only under the action of gravity. Because the clamp is friction-less, mechanical energy is conserved.

At the instant of release the rod is perfectly vertical, pointing upwards from the pivot. After falling through half a rotation (an angle of $$180^{\circ}$$), the rod becomes vertical again but now points downwards. That latter orientation is the position in which the free (upper) end attains its lowest point in the circular path. We wish to find the linear speed of that free end at this lowest position.

First we calculate the change in gravitational potential energy of the rod’s centre of mass (C.M.). For a uniform rod the C.M. lies at its mid-point, i.e. at a distance $$\dfrac{L}{2}$$ from the pivot.

Initial height of C.M. above the pivot:$$h_i = +\frac{L}{2}$$ Final height of C.M. (after a $$180^{\circ}$$ rotation):$$h_f = -\frac{L}{2}$$

Thus the vertical drop of the centre of mass is

$$\Delta h = h_i - h_f = \frac{L}{2} - \left(-\frac{L}{2}\right)=L.$$

The loss in gravitational potential energy is therefore

$$\Delta U = m g \,\Delta h = m g L.$$

Substituting the numerical values,

$$\Delta U = 2\;\text{kg} \times 10\;\text{m s}^{-2} \times 0.6\;\text{m} = 12\;\text{J}.$$

This entire loss of potential energy appears as rotational kinetic energy about the pivot. The formula for rotational kinetic energy is

$$K = \frac12 I \omega^2,$$

where $$I$$ is the moment of inertia of the rod about the pivot and $$\omega$$ is its angular speed at the considered instant.

For a uniform rod hinged about one end, the moment of inertia is given by

$$I = \frac{1}{3} m L^2.$$

Putting the numbers,

$$I = \frac13 \times 2\;\text{kg} \times (0.6\;\text{m})^2 = \frac13 \times 2 \times 0.36\;\text{kg m}^2 = 0.24\;\text{kg m}^2.$$

Conserving mechanical energy, we equate the gravitational energy lost to the rotational kinetic energy gained:

$$m g L = \frac12 I \omega^2.$$

Substituting $$m g L = 12\;\text{J}$$ and $$I = 0.24\;\text{kg m}^2,$$

$$12 = \frac12 \times 0.24 \times \omega^2,$$ $$12 = 0.12\,\omega^2,$$ $$\omega^2 = \frac{12}{0.12} = 100,$$ $$\omega = 10\;\text{rad s}^{-1}.$$

The linear speed of the free end is related to the angular speed by

$$v = \omega L.$$

Hence,

$$v = 10\;\text{rad s}^{-1} \times 0.6\;\text{m} = 6\;\text{m s}^{-1}.$$

So, the answer is $$6$$.

For a solid disc rolling without slipping on an inclined plane, the acceleration is given by $$a = \frac{g\sin\theta}{1 + \frac{I}{mR^2}}$$.

The moment of inertia of a solid disc about its center is $$I = \frac{1}{2}ma^2$$, where $$a$$ is the radius. So $$\frac{I}{ma^2} = \frac{1}{2}$$.

Therefore, $$a = \frac{g\sin\theta}{1 + \frac{1}{2}} = \frac{g\sin\theta}{\frac{3}{2}} = \frac{2}{3}g\sin\theta$$.

Comparing with the given expression $$\frac{2}{b}g\sin\theta$$, we get $$b = 3$$.

The value of $$b$$ is $$\boxed{3}$$.

We need to find the value of $$x$$ representing the ratio of the rotational inertias ($$\frac{I_1}{I_2}$$) for two connected wheels having the same rotational kinetic energy.

1. Analyze the Belt Connection (Linear Velocity)

From the problem, the two wheels $$P$$ and $$Q$$ are linked by a non-slipping belt $$B$$. Because the belt moves at a uniform speed, the linear tangential velocities at the rims of both wheels must be equal:

$$v_P = v_Q$$

Using the rotational relationship $$v = r\omega$$, we can rewrite this equality as:

$$r_1\omega_1 = r_2\omega_2$$

Where:

• $$r_1 = 3R$$ (radius of wheel $$P$$)
• $$r_2 = R$$ (radius of wheel $$Q$$)
• $$\omega_1$$ and $$\omega_2$$ are their respective angular velocities.

Substituting the radius values gives:

$$(3R)\omega_1 = R\omega_2 \implies \frac{\omega_1}{\omega_2} = \frac{1}{3}$$

2. Apply the Rotational Kinetic Energy Condition

The problem states that both wheels possess the same rotational kinetic energy ($$K_{\text{rot}}$$). The formula for rotational kinetic energy is:

$$K_{\text{rot}} = \frac{1}{2}I\omega^2$$

Equating the kinetic energies of both wheels:

$$\frac{1}{2}I_1\omega_1^2 = \frac{1}{2}I_2\omega_2^2$$

3. Calculate the Ratio of Rotational Inertias ($$x$$)

Rearranging the kinetic energy equation to isolate the ratio of moments of inertia ($$\frac{I_1}{I_2}$$):

$$\frac{I_1}{I_2} = \left(\frac{\omega_2}{\omega_1}\right)^2$$

Since we found earlier that $$\frac{\omega_1}{\omega_2} = \frac{1}{3}$$, its reciprocal is:

$$\frac{\omega_2}{\omega_1} = 3$$

Now, substitute this value back into our ratio equation:

$$\frac{I_1}{I_2} = (3)^2 = 9$$

The problem states that the ratio $$\frac{I_1}{I_2}$$ is expressed as $$x : 1$$. Comparing the terms:

$$x = 9$$

Conclusion

The value of $$x$$ is 9.

For a wheel of radius $$R$$ rolling without slipping on a flat surface, the centre of the wheel moves with speed $$v_0$$. The rolling condition gives the angular velocity as $$\omega = v_0/R$$, directed (for rightward motion) clockwise, i.e. $$\vec{\omega} = -\dfrac{v_0}{R}\hat{k}$$.

A particle on the rim at the same height as the centre is positioned either to the left or right of the centre. Consider the particle at position $$\vec{r} = R\hat{i}$$ relative to the centre (i.e., at the rightmost point of the wheel, at the same height as the centre).

The velocity of this particle is the sum of the centre-of-mass velocity and the rotational velocity:

$$\vec{v} = v_0\hat{i} + \vec{\omega} \times \vec{r} = v_0\hat{i} + \left(-\frac{v_0}{R}\hat{k}\right) \times (R\hat{i}) = v_0\hat{i} - v_0(\hat{k}\times\hat{i}) = v_0\hat{i} - v_0\hat{j}$$

The speed is:

$$|\vec{v}| = \sqrt{v_0^2 + v_0^2} = v_0\sqrt{2} = \sqrt{2}\,v_0$$

Comparing with the given form $$\sqrt{x}\,v_0$$, the value of $$x$$ is $$2$$.

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

$$K_i = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega_0^2$$

$$K_i = \frac{1}{2}mv_0^2 + \frac{1}{2}\left(\frac{2}{5}ma^2\right)\left(\frac{v_0}{a}\right)^2 = \frac{1}{2}mv_0^2 + \frac{1}{5}mv_0^2 = \frac{7}{10}mv_0^2$$

$$K_i = mgh \implies \frac{7}{10}mv_0^2 = mgh \implies h = \frac{7v_0^2}{10g}$$

Distance traveled along the inclined plane ($$s$$):

$$h = s\sin\theta \implies s = \frac{h}{\sin\theta} = \frac{7v_0^2}{10g\sin\theta}$$

We need to find the moments of inertia of four bodies, all having the same mass $$M$$ and radius $$R$$.

$$I_1$$ is the M.I. of a thin circular ring about its diameter. For a thin ring, the mass is distributed uniformly at distance $$R$$ from the centre. By the perpendicular axis theorem, the M.I. about an axis perpendicular to the ring through its centre is $$MR^2$$, and this equals the sum of M.I. about two perpendicular diameters. By symmetry, both diameters give the same M.I. So $$MR^2 = I_d + I_d = 2I_d$$, giving $$I_1 = I_d = \frac{1}{2}MR^2$$.

$$I_2$$ is the M.I. of a circular disc about an axis perpendicular to the disc and passing through its centre. Using the standard integration result for a uniform disc, $$I_2 = \frac{1}{2}MR^2$$.

$$I_3$$ is the M.I. of a solid cylinder about its own axis. Each thin disc element of the cylinder has M.I. $$\frac{1}{2}(dm)R^2$$ about the axis. Summing over all elements, $$I_3 = \frac{1}{2}MR^2$$.

$$I_4$$ is the M.I. of a solid sphere about its diameter. Using the standard result, $$I_4 = \frac{2}{5}MR^2$$.

Now we compare: $$I_1 = I_2 = I_3 = \frac{1}{2}MR^2 = 0.5\,MR^2$$ and $$I_4 = \frac{2}{5}MR^2 = 0.4\,MR^2$$.

Since $$0.5\,MR^2 > 0.4\,MR^2$$, we have $$I_1 = I_2 = I_3 > I_4$$.

Hence, the correct answer is Option C.

At the north pole, the body weighs 49 N. Using $$W = mg$$, the mass of the body is $$m = \frac{49}{9.8} = 5$$ kg.

At the equator, the effective acceleration due to gravity is reduced because of Earth's rotation. The effective gravity at the equator is $$g' = g - R\omega^2$$, where $$R = 6400 \times 10^3$$ m is the radius of Earth and $$\omega = \frac{2\pi}{T}$$ is the angular velocity of Earth with $$T = 24 \times 3600 = 86400$$ s.

Calculating $$\omega = \frac{2\pi}{86400} = 7.27 \times 10^{-5}$$ rad/s. Then $$R\omega^2 = 6.4 \times 10^6 \times (7.27 \times 10^{-5})^2 = 6.4 \times 10^6 \times 5.285 \times 10^{-9} = 0.0338$$ m/s$$^2$$.

So the effective gravity at the equator is $$g' = 9.8 - 0.0338 = 9.766$$ m/s$$^2$$.

The weight at the equator is $$W' = mg' = 5 \times 9.766 = 48.83$$ N.

The correct answer is 48.83 N.

We need to determine the total time it takes for a solid rolling sphere to travel up an inclined plane and return to its starting point $$A$$.

1. Identify the Physical Properties and System Parameters

From the problem statement, we have:

• Mass of the sphere ($$m$$) = $$2\text{ kg}$$
• Radius of the sphere ($$R$$) = $$0.5\text{ m}$$
• Initial linear speed ($$u$$) = $$1\text{ m s}^{-1}$$
• Angle of inclination ($$\theta$$) = $$30^\circ$$
• Acceleration due to gravity ($$g$$) = $$9.8\text{ m s}^{-2}$$ (standard value used for precise decimal options)

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

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

2. Calculate Retardation During Pure Rolling Up the Incline

When an object rolls up an inclined plane without slipping, both gravity and static friction act on it. The linear acceleration ($$a$$) of a rolling body down an incline is given by the formula:

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

Substitute the ratio $$\frac{I}{mR^2} = \frac{2}{5}$$ into the acceleration formula:

$$a = \frac{g \sin 30^\circ}{1 + \frac{2}{5}} = \frac{g \left(\frac{1}{2}\right)}{\frac{7}{5}} = \frac{5}{14}g$$

Plugging in $$g = 9.8\text{ m s}^{-2}$$:

$$a = \frac{5}{14} \times 9.8 = 5 \times 0.7 = 3.5\text{ m s}^{-2}$$

3. Determine Time to Return to Point A

Let $$t_1$$ be the time taken by the sphere to come to rest at the highest point of its climb. Using the first equation of motion ($$v = u - at_1$$) where final velocity $$v = 0$$:

$$0 = 1 - 3.5 \cdot t_1 \implies t_1 = \frac{1}{3.5} = \frac{2}{7}\text{ s}$$

Since the system is conservative and rolls without slipping in both directions, the time of ascent ($$t_1$$) equals the time of descent ($$t_2$$). Therefore, the total round-trip time ($$T$$) to return to point $$A$$ is:

$$T = 2 \cdot t_1 = 2 \times \frac{2}{7} = \frac{4}{7}\text{ s} \approx 0.571\text{ s}$$

Rounding to two decimal places closest to the calculated metric options yields approximately $$0.56\text{ s}$$.

Conclusion

The total time the sphere takes to return to the starting point is 0.56 s, which corresponds to Option C.

We need to determine the torque acting at point $$P$$ with respect to point $$O$$ and point $$Q$$ on an equilateral triangular plate.

1. Identify the Geometry and Vector Positions

From the diagram on the, the plate forms an equilateral triangle with side lengths of $$10\text{ cm} = 0.1\text{ m}$$ and interior angles of $$60^\circ$$. Let's establish the coordinate positions by placing the origin at point $$O(0,0)$$:

• Point O: $$(0, 0)$$
• Point Q: $$(10, 0)\text{ cm} = (0.1, 0)\text{ m}$$
• Point P: Located using trigonometric components along the side $$OP$$:

  $$x_P = 10 \cos 60^\circ = 5\text{ cm} = 0.05\text{ m}$$

  $$y_P = 10 \sin 60^\circ = 5\sqrt{3}\text{ cm} = 0.05\sqrt{3}\text{ m}$$

  Therefore, the position vector of $$P$$ relative to $$O$$ is: $$\vec{r}_{P/O} = 0.05\hat{i} + 0.05\sqrt{3}\hat{j}$$

The applied force vector given in the problem is:

$$\vec{F} = 4\hat{i} - 3\hat{j}$$

2. Calculate Torque with respect to Point O ($$\vec{\tau}_O$$)

Torque is defined by the cross product $$\vec{\tau} = \vec{r} \times \vec{F}$$:

$$\vec{\tau}_O = \vec{r}_{P/O} \times \vec{F} = \det \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0.05 & 0.05\sqrt{3} & 0 \\ 4 & -3 & 0 \end{bmatrix}$$

$$\vec{\tau}_O = \left[ 0.05 \times (-3) - 0.05\sqrt{3} \times 4 \right] \hat{k}$$

$$\vec{\tau}_O = \left( -0.15 - 0.20\sqrt{3} \right) \hat{k}\text{ N}\cdot\text{m}$$

Converting the units into $$\text{N}\cdot\text{cm}$$ to match the option formats (by multiplying the values by $$100$$):

$$\tau_O = -15 - 20\sqrt{3}$$

3. Calculate Torque with respect to Point Q ($$\vec{\tau}_Q$$)

First, find the position vector of $$P$$ relative to $$Q$$:

$$\vec{r}_{P/Q} = \vec{r}_P - \vec{r}_Q = (0.05 - 0.1)\hat{i} + 0.05\sqrt{3}\hat{j} = -0.05\hat{i} + 0.05\sqrt{3}\hat{j}$$

Now, compute the cross product for torque about $$Q$$:

$$\vec{\tau}_Q = \vec{r}_{P/Q} \times \vec{F} = \det \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ -0.05 & 0.05\sqrt{3} & 0 \\ 4 & -3 & 0 \end{bmatrix}$$

$$\vec{\tau}_Q = \left[ (-0.05) \times (-3) - 0.05\sqrt{3} \times 4 \right] \hat{k}$$

$$\vec{\tau}_Q = \left( 0.15 - 0.20\sqrt{3} \right) \hat{k}\text{ N}\cdot\text{m}$$

Converting the units into $$\text{N}\cdot\text{cm}$$ gives:

$$\tau_Q = 15 - 20\sqrt{3}$$

Conclusion

The torque values at point $$P$$ relative to $$O$$ and $$Q$$ are $$-15 - 20\sqrt{3}$$ and $$15 - 20\sqrt{3}$$ respectively. This corresponds perfectly to Option A.

We begin by recalling the definition of moment of inertia. For any rigid body about a given axis we write the relation

$$I = M\,k^{2}$$

where $$I$$ is the moment of inertia, $$M$$ is the total mass of the body, and $$k$$ is the radius of gyration about that axis. The radius of gyration therefore depends directly on the numerical value of the moment of inertia about the chosen axis.

Now we consider a uniform thin circular disc of mass $$M$$ and radius $$R$$. Its centre is the origin, its plane is the X-Y plane, so the Z-axis passes through its centre perpendicular to the plane. The X- and Y-axes are two mutually perpendicular diameters lying in the plane of the disc.

First, we write down the standard results for the moments of inertia of such a disc (these results can be obtained by integration, but are quoted here as standard):

$$$I_{Z} = \frac12\,M R^{2} \qquad\text{(about the Z-axis)}$$$

$$$I_{X} = I_{Y} = \frac14\,M R^{2} \qquad\text{(about any diameter in the plane, e.g. X or Y axis)}$$$

We next find the radii of gyration corresponding to each axis by substituting these values into the defining equation $$I = M\,k^{2}$$.

For the Z-axis we have

$$$k_{Z}^{2} = \frac{I_{Z}}{M} = \frac{\dfrac12\,M R^{2}}{M} = \frac12\,R^{2}$$$

so

$$k_{Z} = \sqrt{\frac12}\,R = \frac{R}{\sqrt2}\,.$$

For the X-axis we write

$$$k_{X}^{2} = \frac{I_{X}}{M} = \frac{\dfrac14\,M R^{2}}{M} = \frac14\,R^{2}$$$

and therefore

$$k_{X} = \sqrt{\frac14}\,R = \frac{R}{2}\,.$$

The Y-axis expression is identical because $$I_{Y} = I_{X}$$, giving

$$k_{Y} = \frac{R}{2}\,.$$

So we finally have

$$k_{X} = k_{Y} = \frac{R}{2}, \qquad k_{Z} = \frac{R}{\sqrt2}\,. $$

These three values are clearly not equal because

$$\frac{R}{2} \neq \frac{R}{\sqrt2}\,. $$

Therefore the assertion “the respective radii of gyration about all the three axes will be the same” is false.

We now examine the reason. The statement “A rigid body making rotational motion has fixed mass and shape” is always true; the mass and the geometrical size of a rigid body do not change during rotation. Hence the reason is a correct factual statement, although it has no logical connection with the equality (or inequality) of the radii of gyration.

We thus have: Assertion A is not correct, while Reason R is correct but does not explain A.

Hence, the correct answer is Option B.

We recall the standard formula for the moment of inertia of a uniform slender rod of length $$l$$ and mass $$m$$ about an axis perpendicular to the rod:

1. When the axis passes through the mid-point of the rod, the formula is

$$I_{\text{mid}}=\frac{1}{12}\,m\,l^{2}.$$

2. When the axis passes through one end of the rod, the formula is

$$I_{\text{end}}=\frac{1}{3}\,m\,l^{2}.$$

With these two relations ready, we now evaluate each item in List-I one by one and match with List-II.

Term (a): The rod has length $$L$$ and mass $$M$$. The axis is perpendicular to the rod and passes through its mid-point, so we use the mid-point formula.

$$I_{(a)} \;=\;\frac{1}{12}\,M\,L^{2}.$$

Checking List-II, we see that $$\frac{ML^{2}}{12}$$ is entry (iii). So, (a) corresponds to (iii).

Term (b): The rod has the same length $$L$$ but double the mass, i.e. $$2M$$. The axis is at one end, so we use the end-point formula.

First write the formula: $$I_{\text{end}}=\frac{1}{3}\,m\,l^{2}.$$ Now substitute $$m=2M$$ and $$l=L$$:

$$I_{(b)} \;=\;\frac{1}{3}\,(2M)\,L^{2} =\;\frac{2\,M\,L^{2}}{3}.$$

Looking at List-II, $$\dfrac{2ML^{2}}{3}$$ is entry (iv). Therefore, (b) matches (iv).

Term (c): The rod now has length $$2L$$ while the mass is $$M$$. The axis again passes through the mid-point, so we use the mid-point formula.

Formula: $$I_{\text{mid}}=\frac{1}{12}\,m\,l^{2}.$$ Substituting $$m=M$$ and $$l=2L$$ gives

$$I_{(c)} \;=\;\frac{1}{12}\,M\,(2L)^{2} =\;\frac{1}{12}\,M\,(4L^{2}) =\;\frac{4}{12}\,M\,L^{2} =\;\frac{1}{3}\,M\,L^{2}.$$

In List-II, $$\dfrac{ML^{2}}{3}$$ is entry (ii). Thus, (c) corresponds to (ii).

Term (d): The rod now has length $$2L$$ and mass $$2M$$, with the axis at one end. Hence we again use the end-point formula.

Formula: $$I_{\text{end}}=\frac{1}{3}\,m\,l^{2}.$$ Substituting $$m=2M$$ and $$l=2L$$ yields

$$I_{(d)} \;=\;\frac{1}{3}\,(2M)\,(2L)^{2} =\;\frac{2}{3}\,M\,(4L^{2}) =\;\frac{8}{3}\,M\,L^{2}.$$

The expression $$\dfrac{8ML^{2}}{3}$$ appears in List-II as entry (i). Consequently, (d) matches with (i).

Gathering all the matches we have obtained:

(a) → (iii),  (b) → (iv),  (c) → (ii),  (d) → (i).

When we inspect the four options, we see that this exact sequence is given in Option C.

Hence, the correct answer is Option C.

We need to analyze the characteristics of the angular momentum of a mass $$M$$ about two different reference points, $$A$$ and $$B$$, during steady conical pendulum-like circular motion.

1. Analyze Angular Momentum about Point A ($$\vec{L}_A$$)

Point $$A$$ lies at the center of the circular path of the mass $$M$$ along the vertical axis of rotation ($$z$$-axis).

• The position vector $$\vec{r}_A$$ of the mass relative to point $$A$$ lies entirely within the horizontal plane of rotation.
• The linear velocity vector $$\vec{v}$$ is tangential to the circle at every point.
• By definition, the angular momentum vector is:

  $$\vec{L}_A = \vec{r}_A \times \vec{p} = m(\vec{r}_A \times \vec{v})$$

• Using the right-hand rule, the direction of $$\vec{L}_A$$ points straight up along the positive $$z$$-axis at all times. Since the speed and radius are constant, its magnitude ($$m v r_A$$) and its direction remain completely fixed throughout the motion.

2. Analyze Angular Momentum about Point B ($$\vec{L}_B$$)

Point $$B$$ is located on the vertical rotation axis but at a height above the plane of the circle (the suspension point of the rod).

• The position vector $$\vec{r}_B$$ points from $$B$$ down to the mass $$M$$, tracing out the surface of a cone during motion.
• The angular momentum about $$B$$ is:

  $$\vec{L}_B = m(\vec{r}_B \times \vec{v})$$

• Since $$\vec{r}_B$$ and $$\vec{v}$$ are always perpendicular, the magnitude of $$\vec{L}_B$$ is constant ($$m v l$$).
• However, by the right-hand rule, the vector $$\vec{L}_B$$ is always perpendicular to the rod. As the mass rotates, the vector $$\vec{L}_B$$ sweeps out a cone around the vertical $$z$$-axis. Because its orientation is continuously changing in space, $$\vec{L}_B$$ is not constant in direction.

Conclusion

Therefore, $$\vec{L}_A$$ remains constant both in magnitude and direction, while $$\vec{L}_B$$ changes its direction continuously. This corresponds to the correct choice on the  page:

$$\vec{L}_A$$ is constant, both in magnitude and direction (Option D).

We are asked to find the moment of inertia of the whole system about an axis that is perpendicular to the light rod and passes through the midpoint of the rod. The system consists of two identical solid spheres attached to the two ends of the rod, so we have to calculate the contribution of each sphere and then add the two contributions.

First, recall the two formulas we will need:

1. For a solid sphere about its own centre, the moment of inertia is given by $$I_{\text{cm}}=\dfrac{2}{5}\,mR^{2}.$$

2. The parallel-axis theorem (Steiner’s theorem) states that if an axis is displaced from the centre of mass by a distance $$d,$$ then the new moment of inertia is $$I = I_{\text{cm}} + m d^{2}.$$

Now we insert the numerical data. For each sphere:

Mass: $$m = 1.5\ \text{kg}.$$

Radius: $$R = 50\ \text{cm} = 0.5\ \text{m}.$$

The moment of inertia about its own centre is therefore

$$I_{\text{cm}} = \dfrac{2}{5}\,mR^{2} = \dfrac{2}{5}\,(1.5)\,(0.5)^{2}.$$

We calculate step by step:

$$(0.5)^{2} = 0.25,$$

$$\dfrac{2}{5} = 0.4,$$

$$0.4 \times 0.25 = 0.10,$$

$$0.10 \times 1.5 = 0.15.$$

So $$I_{\text{cm}} = 0.15\ \text{kg m}^{2}.$$

The distance of the centre of each sphere from the given axis: the two centres are 5 m apart, and the axis is at the midpoint of the rod, so each centre is

$$d = \dfrac{5\ \text{m}}{2} = 2.5\ \text{m}$$

away from the axis.

Using the parallel-axis theorem, the moment of inertia of one sphere about the required axis is

$$I_{\text{one}} = I_{\text{cm}} + m d^{2} = 0.15 + (1.5)(2.5)^{2}.$$

Calculating the term $$m d^{2}:$$

$$d^{2} = (2.5)^{2} = 6.25,$$

$$m d^{2} = 1.5 \times 6.25 = 9.375.$$

Adding $$I_{\text{cm}}$$ to this:

$$I_{\text{one}} = 0.15 + 9.375 = 9.525\ \text{kg m}^{2}.$$

Because there are two identical spheres, the total moment of inertia of the system is

$$I_{\text{total}} = 2\,I_{\text{one}} = 2 \times 9.525 = 19.05\ \text{kg m}^{2}.$$

Hence, the correct answer is Option C.

We start with the definition of angular momentum of a single particle about the centre of the circular path.

By definition, $$\vec L = \vec r \times \vec p,$$ where $$\vec r$$ is the position vector of the particle with respect to the centre, and $$\vec p = m\vec v$$ is its linear momentum.

For uniform circular motion the radius is constant, so $$|\vec r| = R,$$ and the speed is constant, so $$|\vec v| = v.$$ The position vector $$\vec r$$ is always radial, while the velocity $$\vec v$$ is always tangential, making the angle between them $$90^\circ.$$

Using the formula for the magnitude of a cross-product, $$|\vec A \times \vec B| = |\vec A|\,|\vec B|\,\sin\theta,$$ we substitute $$\vec A = \vec r,\ \vec B = \vec p,\ \theta = 90^\circ.$$ This gives

$$|\vec L| = |\vec r|\,|\vec p|\,\sin 90^\circ = R\,(mv)\,(1) = m v R.$$

The quantities $$m,\,v,$$ and $$R$$ are each constant during the motion, so the magnitude $$|\vec L|$$ remains fixed.

The direction of $$\vec L$$ is given by the right-hand rule applied to $$\vec r \times \vec p.$$ Because $$\vec r$$ is always in the plane of the circle and $$\vec p$$ is always in that same plane but perpendicular to $$\vec r,$$ the cross-product is always perpendicular to the plane of the circle. If the motion is anticlockwise, $$\vec L$$ points steadily “out of the page”; if the motion is clockwise, it points steadily “into the page.” In either case the direction is fixed and does not change as the particle moves around.

We can confirm this with torque. Torque about the centre is $$\vec\tau = \vec r \times \vec F.$$ The only force needed for uniform circular motion is the centripetal force $$\vec F_c = -\dfrac{mv^2}{R}\,\hat r,$$ directed along $$-\hat r.$$ Since $$\vec r$$ is parallel to $$\hat r,$$ their cross-product is zero:

$$\vec\tau = \vec r \times \vec F_c = 0.$$

With zero external torque, the rotational analogue of Newton’s second law, $$\vec\tau = \dfrac{d\vec L}{dt},$$ immediately gives $$\dfrac{d\vec L}{dt} = 0.$$ Therefore the vector $$\vec L$$ is completely constant—both its magnitude and its direction stay unchanged.

Hence, the correct answer is Option B.

When a rigid body rolls down an inclined plane without slipping, we use energy conservation. Starting from rest at height $$h$$, all potential energy converts to translational and rotational kinetic energy:

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

Using the rolling condition $$\omega = v/R$$ and writing $$I = kmR^2$$ where $$k$$ is a constant depending on the body's geometry:

$$mgh = \frac{1}{2}mv^2(1 + k)$$ $$v = \sqrt{\frac{2gh}{1+k}}$$

The moment of inertia factors are: for a ring, $$I = mR^2$$ so $$k = 1$$; for a solid cylinder, $$I = \frac{1}{2}mR^2$$ so $$k = \frac{1}{2}$$; for a solid sphere, $$I = \frac{2}{5}mR^2$$ so $$k = \frac{2}{5}$$.

The velocity at the bottom is: $$v_{\text{ring}} = \sqrt{\frac{2gh}{1+1}} = \sqrt{gh}$$ $$v_{\text{cylinder}} = \sqrt{\frac{2gh}{1+\frac{1}{2}}} = \sqrt{\frac{4gh}{3}}$$ $$v_{\text{sphere}} = \sqrt{\frac{2gh}{1+\frac{2}{5}}} = \sqrt{\frac{10gh}{7}}$$

Comparing: $$\frac{10}{7} > \frac{4}{3} > 1$$, so $$v_{\text{sphere}} > v_{\text{cylinder}} > v_{\text{ring}}$$. The sphere has the greatest and the ring has the least velocity at the bottom.

A wire of mass $$M$$ and length $$L$$ is bent into a semicircle. The relationship between the length and the radius is $$L = \pi R$$, giving $$R = \frac{L}{\pi}$$.

For a semicircular wire, every element of mass $$dm$$ lies at the same distance $$R$$ from the centre. Therefore, the moment of inertia about an axis perpendicular to the plane passing through the centre is $$I = MR^2$$.

Substituting $$R = \frac{L}{\pi}$$: $$I = M\left(\frac{L}{\pi}\right)^2 = \frac{ML^2}{\pi^2}$$.

Let us place the square with vertices $$A$$, $$B$$, $$C$$, $$D$$ at coordinates $$A = (0, 0)$$, $$B = (l, 0)$$, $$C = (l, l)$$, and $$D = (0, l)$$. The diagonal $$DB$$ goes from $$(0, l)$$ to $$(l, 0)$$, so its direction vector is $$(1, -1)$$ (or unit vector $$\frac{1}{\sqrt{2}}(1, -1)$$).

The axis passes through $$A = (0, 0)$$ and is parallel to $$DB$$. The line through the origin with direction $$(1, -1)$$ has equation $$x + y = 0$$, so the perpendicular distance of a point $$(x, y)$$ from this axis is $$d = \frac{|x + y|}{\sqrt{2}}$$.

For mass at $$A = (0,0)$$: $$d_A = 0$$, so $$d_A^2 = 0$$.

For mass at $$B = (l, 0)$$: $$d_B = \frac{l}{\sqrt{2}}$$, so $$d_B^2 = \frac{l^2}{2}$$.

For mass at $$C = (l, l)$$: $$d_C = \frac{2l}{\sqrt{2}} = l\sqrt{2}$$, so $$d_C^2 = 2l^2$$.

For mass at $$D = (0, l)$$: $$d_D = \frac{l}{\sqrt{2}}$$, so $$d_D^2 = \frac{l^2}{2}$$.

The total moment of inertia is $$I = m(d_A^2 + d_B^2 + d_C^2 + d_D^2) = m\left(0 + \frac{l^2}{2} + 2l^2 + \frac{l^2}{2}\right) = m \times 3l^2 = 3ml^2$$.

We are asked to find the moment of inertia of a uniform square plate of side $$l$$ about an axis which (i) passes through one of its corners and (ii) is perpendicular to the plane of the plate. To proceed, we shall first recall the standard result for a simpler, symmetrical axis and then use the Parallel-Axis Theorem to shift that result to the required corner axis.

For a thin rectangular plate of mass $$M$$, sides $$a$$ and $$b$$, the moment of inertia about an axis through its centre and perpendicular to its plane is given by the well-known formula

$$I_{\text{CM}}=\frac{M}{12}\left(a^{2}+b^{2}\right).$$

In the present problem the plate is a square, so $$a=b=l$$. Substituting $$a=l$$ and $$b=l$$ into the above formula we obtain

$$I_{\text{CM}}=\frac{M}{12}\left(l^{2}+l^{2}\right)=\frac{M}{12}\,2l^{2}=\frac{M l^{2}}{6}.$$

Thus the moment of inertia about the central (centroidal) axis, which is perpendicular to the plane, equals $$\displaystyle\frac{M l^{2}}{6}$$.

However, the axis demanded in the question is not through the centre but through a corner. To relate the two axes we invoke the Parallel-Axis Theorem. The theorem states:

$$I_{\text{new}} \;=\; I_{\text{CM}} + M d^{2},$$

where $$d$$ is the distance between the centre of mass and the new axis. In our square, the centre lies at the midpoint of both diagonals, i.e.\ at coordinates $$\left(\tfrac{l}{2}, \tfrac{l}{2}\right)$$ when one corner is taken as the origin. The distance from this centre to that same corner is therefore

$$d=\sqrt{\left(\frac{l}{2}\right)^{2}+\left(\frac{l}{2}\right)^{2}}=\sqrt{\frac{l^{2}}{4}+\frac{l^{2}}{4}}=\sqrt{\frac{l^{2}}{2}}=\frac{l}{\sqrt{2}}.$$

We next square this distance, because the parallel-axis term uses $$d^{2}$$:

$$d^{2}=\left(\frac{l}{\sqrt{2}}\right)^{2}=\frac{l^{2}}{2}.$$

Now we substitute $$I_{\text{CM}}=\dfrac{M l^{2}}{6}$$ and $$d^{2}=\dfrac{l^{2}}{2}$$ into the Parallel-Axis Theorem:

$$I_{\text{corner}} \;=\; I_{\text{CM}} + M d^{2} =\frac{M l^{2}}{6} + M\left(\frac{l^{2}}{2}\right).$$

We combine the two fractions by expressing both over a common denominator of 6:

$$I_{\text{corner}} =\frac{M l^{2}}{6} + \frac{3 M l^{2}}{6} =\frac{4 M l^{2}}{6}.$$

Finally, we reduce $$\dfrac{4}{6}$$ to its simplest form $$\dfrac{2}{3}$$, giving

$$I_{\text{corner}}=\frac{2}{3}M l^{2}.$$

This value exactly matches Option B in the list provided. Hence, the correct answer is Option B.

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We need to find the ratio of the moment of inertia (MI) of a larger solid disc to that of a smaller solid disc based on the conditions given.

1. Relate Mass to Area

From the problem details, both solid discs have the same mass per unit area (surface mass density, $$\sigma$$):

$$\sigma = \frac{\text{Mass}}{\text{Area}}$$

• Larger Disc (Radius $$R$$):
  Its mass $$M$$ is given by:

  $$M = \sigma \cdot (\pi R^2)$$

• Smaller Disc (Radius $$r$$):
  Its mass $$m$$ is given by:

  $$m = \sigma \cdot (\pi r^2)$$

Taking the ratio of their masses eliminates the constant density $$\sigma$$:

$$\frac{M}{m} = \frac{R^2}{r^2} \implies m = M\frac{r^2}{R^2}$$

2. Formulate the Moments of Inertia

• Larger Disc ($I_1$):
  The moment of inertia of a solid disc around axis $$AB$$ (perpendicular to its plane and passing through its center) is:

  $$I_1 = \frac{1}{2}MR^2$$

• Smaller Disc ($I_2$):
  The moment of inertia of a solid disc around one of its diameters lying on its plane is:

  $$I_2 = \frac{1}{4}mr^2$$

3. Calculate the Ratio ($I_1 : I_2$)

Substitute the expression for the smaller mass $$m$$ into the equation for $$I_2$$:

$$I_2 = \frac{1}{4} \left( M\frac{r^2}{R^2} \right) r^2 = \frac{1}{4}M\frac{r^4}{R^2}$$

Now, write out the explicit ratio of $$I_1$$ to $$I_2$$:

$$\frac{I_1}{I_2} = \frac{\frac{1}{2}MR^2}{\frac{1}{4}M\frac{r^4}{R^2}}$$

Cancel out the mass variable $$M$$ and simplify the fractions:

$$\frac{I_1}{I_2} = \frac{4}{2} \cdot \frac{R^2}{\left(\frac{r^4}{R^2}\right)} = 2 \cdot \frac{R^4}{r^4}$$

Expressing this back as a clean ratio format:

$$I_1 : I_2 = 2R^4 : r^4$$

Conclusion

The ratio of the moments of inertia between the two discs is $$2R^4 : r^4$$, which corresponds exactly to the D option

.

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For a body rolling without slipping, the ratio of rotational kinetic energy to translational kinetic energy is $$\frac{KE_{rot}}{KE_{trans}} = \frac{\frac{1}{2}I\omega^2}{\frac{1}{2}mv^2}$$. Using $$v = R\omega$$ and $$I = kmR^2$$ (where $$k$$ is the moment of inertia factor), this ratio equals $$k$$.

Given that $$KE_{rot} = 50\%$$ of $$KE_{trans}$$, we have $$\frac{KE_{rot}}{KE_{trans}} = \frac{1}{2}$$, so $$k = \frac{1}{2}$$.

Checking known bodies: a solid sphere has $$I = \frac{2}{5}mR^2$$ ($$k = 2/5$$), a solid cylinder has $$I = \frac{1}{2}mR^2$$ ($$k = 1/2$$), a hollow cylinder has $$I = mR^2$$ ($$k = 1$$), and a ring also has $$I = mR^2$$ ($$k = 1$$).

Since $$k = 1/2$$ corresponds to a solid cylinder, the body is a solid cylinder.

We use conservation of energy. The weight $$mg$$ falls through a distance $$h$$, losing potential energy $$mgh$$. This energy is converted into the rotational kinetic energy of the wheel and the translational kinetic energy of the falling mass.

The velocity of the falling mass is related to the angular velocity of the wheel by $$v = r\omega$$, since the cord is wound around the wheel of radius $$r$$.

By conservation of energy: $$mgh = \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2 = \frac{1}{2}I\omega^2 + \frac{1}{2}m(r\omega)^2 = \frac{1}{2}(I + mr^2)\omega^2$$.

Solving for $$\omega^2$$: $$\omega^2 = \frac{2mgh}{I + mr^2}$$.

The angular momentum of the planet about the sun is $$L = M v_\perp r$$, where $$v_\perp$$ is the component of velocity perpendicular to the radius vector and $$r$$ is the distance from the sun.

The areal velocity is the rate at which the radius vector sweeps out area: $$\frac{dA}{dt} = \frac{1}{2} r \, v_\perp$$.

Since $$L = M r \, v_\perp$$, we get $$r \, v_\perp = \frac{L}{M}$$. Substituting into the areal velocity expression: $$\frac{dA}{dt} = \frac{1}{2} \cdot \frac{L}{M} = \frac{L}{2M}$$.

The two discs are coupled coaxially, so no external torque acts about the common axis. Therefore, by the principle of conservation of angular momentum, the total angular momentum before contact equals the angular momentum after the discs have come to a common angular speed.

Initially, disc 1 has angular momentum $$I_1\omega_1$$ and disc 2 has angular momentum $$I_2\omega_2$$. Hence the total initial angular momentum is $$L_{\text{initial}} \;=\; I_1\omega_1 \;+\; I_2\omega_2.$$ After they come in contact and rotate together, let the common angular speed be $$\omega_f$$. The combined moment of inertia is $$I_1+I_2,$$ so the final angular momentum is $$L_{\text{final}} \;=\; (I_1+I_2)\,\omega_f.$$ Setting $$L_{\text{initial}}=L_{\text{final}}$$ we have $$(I_1+I_2)\,\omega_f \;=\; I_1\omega_1 + I_2\omega_2.$$ Now solving for the common angular speed, $$\omega_f \;=\; \dfrac{I_1\omega_1 + I_2\omega_2}{I_1+I_2}.$$

The kinetic energy of rotation is given by the formula $$K \;=\; \tfrac12 I\omega^2.$$ Using this, the initial kinetic energy of the system is $$K_{\text{initial}} \;=\; \tfrac12 I_1\omega_1^{\,2} \;+\; \tfrac12 I_2\omega_2^{\,2}.$$ The final kinetic energy, when both discs rotate together with angular speed $$\omega_f,$$ is $$K_{\text{final}} \;=\; \tfrac12 (I_1+I_2)\,\omega_f^{\,2}.$$

We substitute the value of $$\omega_f$$ obtained earlier: $$K_{\text{final}} \;=\; \tfrac12 (I_1+I_2) \left( \dfrac{I_1\omega_1 + I_2\omega_2}{I_1+I_2} \right)^{\!2} \;=\; \tfrac12 \dfrac{(I_1\omega_1 + I_2\omega_2)^{2}}{I_1+I_2}.$$

The loss in kinetic energy is the difference between the initial and final values: $$\Delta K \;=\; K_{\text{initial}} - K_{\text{final}} \;=\; \tfrac12 I_1\omega_1^{\,2} + \tfrac12 I_2\omega_2^{\,2} \;-\; \tfrac12 \dfrac{(I_1\omega_1 + I_2\omega_2)^{2}}{I_1+I_2}.$$

For easier algebra, we factor out $$\tfrac12$$ and bring everything over the common denominator $$I_1+I_2$$: $$\Delta K \;=\; \tfrac12\, \dfrac{(I_1+I_2)(I_1\omega_1^{\,2} + I_2\omega_2^{\,2}) - (I_1\omega_1 + I_2\omega_2)^{2}} {\,I_1+I_2\,}.$$ Now we expand the numerator term by term.

First expand the product: $$(I_1+I_2)(I_1\omega_1^{\,2} + I_2\omega_2^{\,2}) \,=\, I_1^{2}\omega_1^{\,2} \;+\; I_1I_2\omega_2^{\,2} \;+\; I_1I_2\omega_1^{\,2} \;+\; I_2^{2}\omega_2^{\,2}.$$ Next expand the square: $$(I_1\omega_1 + I_2\omega_2)^{2} \,=\, I_1^{2}\omega_1^{\,2} \;+\; I_2^{2}\omega_2^{\,2} \;+\; 2I_1I_2\omega_1\omega_2.$$ Substituting these back into the numerator gives $$$ \bigl[I_1^{2}\omega_1^{\,2}+I_1I_2\omega_2^{\,2}+I_1I_2\omega_1^{\,2}+I_2^{2}\omega_2^{\,2}\bigr] \;-\; \bigl[I_1^{2}\omega_1^{\,2}+I_2^{2}\omega_2^{\,2}+2I_1I_2\omega_1\omega_2\bigr]. $$$

Now we cancel identical terms and collect the remaining ones: $$I_1^{2}\omega_1^{\,2} \text{ and } I_2^{2}\omega_2^{\,2}$$ appear in both brackets and therefore cancel out. What is left is $$I_1I_2\omega_2^{\,2} + I_1I_2\omega_1^{\,2} - 2I_1I_2\omega_1\omega_2 \;=\; I_1I_2\bigl(\omega_1^{\,2} - 2\omega_1\omega_2 + \omega_2^{\,2}\bigr) \;=\; I_1I_2(\omega_1 - \omega_2)^{2}.$$ Hence the numerator simplifies neatly to $$I_1I_2(\omega_1 - \omega_2)^{2}.$$

Putting this back, we obtain $$\Delta K \;=\; \tfrac12\, \dfrac{I_1I_2(\omega_1 - \omega_2)^{2}}{I_1 + I_2}.$$ Thus the loss in kinetic energy is $$\Delta K \;=\; \frac{I_1I_2}{2\,(I_1+I_2)}\,(\omega_1 - \omega_2)^{2}.$$

Comparing with the options, this matches Option C.

Hence, the correct answer is Option C.

Moment of inertia of the rod-and-block system about pivot O:

$$I = I_{\text{rod}} + I_{\text{block}} = \frac{1}{3}Ml^2 + ml^2$$

$$I = \frac{1}{3}(2)(1)^2 + (1)(1)^2 = \frac{2}{3} + 1 = \frac{5}{3}\ \text{kg m}^2$$

Conservation of angular momentum about pivot O during collision:

$$L_i = L_f \implies mvl = I\omega$$

$$(1)(6)(1) = \frac{5}{3}\omega \implies \omega = \frac{18}{5} = 3.6\ \text{rad s}^{-1}$$

Conservation of mechanical energy after collision:

$$K_f = \Delta U \implies \frac{1}{2}I\omega^2 = Mgh_{\text{cm, rod}} + mgh_{\text{block}}$$

Height gained by the rod's centre of mass and the block:

$$h_{\text{cm, rod}} = \frac{l}{2}(1 - \cos\theta)$$

$$h_{\text{block}} = l(1 - \cos\theta)$$

$$\frac{1}{2}I\omega^2 = \left[M\frac{l}{2} + ml\right]g(1 - \cos\theta)$$

$$\frac{1}{2}\left(\frac{5}{3}\right)(3.6)^2 = \left[2\left(\frac{1}{2}\right) + 1(1)\right](10)(1 - \cos\theta)$$

$$\frac{5}{6}(12.96) = (1 + 1)(10)(1 - \cos\theta)$$

$$10.8 = 20(1 - \cos\theta)$$

$$1 - \cos\theta = \frac{10.8}{20} = 0.54 \implies \cos\theta = 0.46$$

$$\theta = \cos^{-1}(0.46) \approx 62.6^\circ \approx 63^\circ$$

We have two identical axes, so the external torque about that common axis is zero. Whenever external torque is zero, the law of conservation of angular momentum tells us that the total angular momentum $$L$$ of the system remains constant. Mathematically the law is stated as $$L_{\text{initial}} = L_{\text{final}}$$, where $$L = I\,\omega$$ for each rigid body. Here $$I$$ is the moment of inertia and $$\omega$$ is the angular speed.

For the first disc the data are $$I_1 = 0.1\ \text{kg·m}^2$$ and $$\omega_1 = 10\ \text{rad·s}^{-1}$$. Hence its angular momentum is

$$L_1 = I_1\,\omega_1 = 0.1 \times 10 = 1\ \text{kg·m}^2\text{·s}^{-1}.$$

For the second disc we are given $$I_2 = 0.2\ \text{kg·m}^2$$ and $$\omega_2 = 5\ \text{rad·s}^{-1}$$, so its angular momentum equals

$$L_2 = I_2\,\omega_2 = 0.2 \times 5 = 1\ \text{kg·m}^2\text{·s}^{-1}.$$

Both discs rotate in the same direction, hence their angular momenta add algebraically. The total initial angular momentum of the system therefore is

$$L_{\text{initial}} = L_1 + L_2 = 1 + 1 = 2\ \text{kg·m}^2\text{·s}^{-1}.$$

Now the discs suddenly stick together. Because no external torque acts, the final angular momentum is the same:

$$L_{\text{final}} = 2\ \text{kg·m}^2\text{·s}^{-1}.$$

After sticking, the two discs behave like a single rigid body whose moment of inertia is simply the sum of the individual moments, since both masses are still distributed at the same radii from the common axis. Thus

$$I_{\text{combined}} = I_1 + I_2 = 0.1 + 0.2 = 0.3\ \text{kg·m}^2.$$

If $$\omega_f$$ is the common angular speed after they join, conservation of angular momentum gives

$$I_{\text{combined}}\;\omega_f = L_{\text{final}}.$$

Substituting the known numbers,

$$0.3\;\omega_f = 2$$

and therefore

$$\omega_f = \frac{2}{0.3} = \frac{20}{3}\ \text{rad·s}^{-1}.$$

We now compute the rotational kinetic energy of the combined system. The formula for rotational kinetic energy is

$$K = \frac{1}{2}\,I\,\omega^2.$$

Using $$I = 0.3\ \text{kg·m}^2$$ and $$\omega = \dfrac{20}{3}\ \text{rad·s}^{-1}$$, we get

$$K = \frac{1}{2}\times 0.3 \times \left(\frac{20}{3}\right)^2.$$

First square the angular speed:

$$\left(\frac{20}{3}\right)^2 = \frac{400}{9}.$$

Now multiply step by step:

$$\frac{1}{2}\times 0.3 = 0.15,$$

so

$$K = 0.15 \times \frac{400}{9} = \frac{0.15 \times 400}{9}.$$

Multiplying $$0.15 \times 400$$ gives $$60$$, hence

$$K = \frac{60}{9} = \frac{20}{3}\ \text{joule}.$$

Hence, the correct answer is Option B.

Mass and dimensions of each partitioned triangle: $$m_1 = \frac{M}{4},\ a_1 = \frac{a}{2}$$

Moment of inertia of the central inverted triangle about the common centroid $$G$$: $$I_{\text{central}} = \frac{I_0}{16}$$

By rotational symmetry, the three corner triangles have equal moments of inertia $$I_1$$ about $$G$$:

$$I_0 = I_{\text{central}} + 3I_1$$

$$I_0 = \frac{I_0}{16} + 3I_1 \implies 3I_1 = \frac{15I_0}{16} \implies I_1 = \frac{5I_0}{16}$$

Moment of inertia of the remaining portion after removing one corner piece ($$ADE$$):

$$I_{\text{remaining}} = I_0 - I_1 = I_0 - \frac{5I_0}{16} = \frac{11I_0}{16}$$

For a uniform circular disc, the moment of inertia about the central axis is given by the formula $$I=\frac12\,M R^{2}.$$

We have one disc of mass $$M$$ and radius $$R$$ already rotating with angular speed $$\omega_1.$$ For this disc

$$I_1=\frac12\,M R^{2}.$$

The second disc is identical in mass but has radius $$\dfrac{R}{2}$$ and is initially at rest. Using the same formula

$$I_2=\frac12\,M\left(\frac{R}{2}\right)^{2} =\frac12\,M\left(\frac{R^{2}}{4}\right) =\frac18\,M R^{2}.$$

When the stationary disc is dropped coaxially, the two discs interact frictionally until they spin together with a common angular speed $$\omega_2.$$ Because the external torque about the axis is zero, angular momentum is conserved:

$$I_1\,\omega_1 = \bigl(I_1+I_2\bigr)\,\omega_2.$$

Substituting the values of the moments of inertia,

$$\frac12\,M R^{2}\,\omega_1 =\left(\frac12\,M R^{2}+\frac18\,M R^{2}\right)\omega_2 =\left(\frac58\,M R^{2}\right)\omega_2.$$

Canceling the common factor $$M R^{2}$$ and solving for $$\omega_2$$ gives

$$\omega_2=\frac{\frac12}{\frac58}\,\omega_1 =\frac12\cdot\frac85\,\omega_1 =\frac45\,\omega_1.$$

The initial rotational kinetic energy of the system is only that of the first disc:

$$E_i=\frac12\,I_1\,\omega_1^{2} =\frac12\left(\frac12\,M R^{2}\right)\omega_1^{2} =\frac14\,M R^{2}\,\omega_1^{2}.$$

After both discs rotate together, the total kinetic energy is

$$E_f=\frac12\bigl(I_1+I_2\bigr)\omega_2^{2} =\frac12\left(\frac58\,M R^{2}\right)\left(\frac45\,\omega_1\right)^{2}.$$ Simplifying step by step,

$$E_f=\frac12\cdot\frac58\,M R^{2}\cdot\frac{16}{25}\,\omega_1^{2} =M R^{2}\,\omega_1^{2}\cdot\frac{1}{2}\cdot\frac{5}{8}\cdot\frac{16}{25} =M R^{2}\,\omega_1^{2}\cdot\frac{80}{400} =\frac15\,M R^{2}\,\omega_1^{2}.$$

The mechanical energy lost in the process is therefore

$$\Delta E = E_i - E_f =\left(\frac14-\frac15\right)M R^{2}\,\omega_1^{2} =\left(\frac5{20}-\frac4{20}\right)M R^{2}\,\omega_1^{2} =\frac1{20}\,M R^{2}\,\omega_1^{2}.$$

To find the percentage loss, compare $$\Delta E$$ with the initial energy $$E_i$$:

$$p=\left(\frac{\Delta E}{E_i}\right)\times100 =\left(\frac{\frac1{20}\,M R^{2}\,\omega_1^{2}} {\frac14\,M R^{2}\,\omega_1^{2}}\right)\times100 =\left(\frac1{20}\times\frac4{1}\right)\times100 =\frac4{20}\times100 =20\%.$$

So, the answer is $$20$$.

We have a force vector given by $$\vec F = \hat i + 2\hat j + 3\hat k\;{\rm N}$$ and the point of application of this force is at the position vector $$\vec r_1 = 4\hat i + 3\hat j - \hat k\;{\rm m}.$$

The torque (or moment of force) about some reference point with position vector $$\vec r_0$$ is defined by the vector formula

$$\vec\tau = (\vec r_1 - \vec r_0)\times\vec F.$$

Here the reference point is $$\vec r_0 = \hat i + 2\hat j + \hat k\;{\rm m}.$$ Subtracting, we obtain the position vector of the line of action of the force with respect to the reference point:

$$\vec r = \vec r_1 - \vec r_0 = \bigl(4\hat i + 3\hat j - \hat k\bigr) - \bigl(\hat i + 2\hat j + \hat k\bigr) = (4-1)\hat i + (3-2)\hat j + (-1-1)\hat k = 3\hat i + 1\hat j - 2\hat k\;{\rm m}.$$

Now we calculate the cross product $$\vec\tau = \vec r \times \vec F.$$ Writing the two vectors component-wise,

$$\vec r = 3\hat i + 1\hat j - 2\hat k, \qquad \vec F = 1\hat i + 2\hat j + 3\hat k.$$ Using the determinant form for the cross product,

$$\vec\tau = \begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & 1 & -2\\ 1 & 2 & 3 \end{vmatrix}.$$

Expanding this determinant step by step,

$$\vec\tau = \hat i\,(1\cdot 3 - (-2)\cdot 2) - \hat j\,(3\cdot 3 - (-2)\cdot 1) + \hat k\,(3\cdot 2 - 1\cdot 1).$$

Simplifying each term inside the parentheses,

$$1\cdot 3 - (-2)\cdot 2 = 3 + 4 = 7,$$ $$3\cdot 3 - (-2)\cdot 1 = 9 + 2 = 11,$$ $$3\cdot 2 - 1\cdot 1 = 6 - 1 = 5.$$

Substituting these values back, we get

$$\vec\tau = 7\hat i - 11\hat j + 5\hat k\;{\rm N\!-\!m}.$$

The magnitude of the torque vector is obtained through the relation

$$|\vec\tau| = \sqrt{(7)^2 + (-11)^2 + (5)^2}.$$

Carrying out the squaring and addition,

$$(7)^2 = 49,\quad (-11)^2 = 121,\quad (5)^2 = 25,$$ $$49 + 121 + 25 = 195.$$

Thus,

$$|\vec\tau| = \sqrt{195}\,{\rm N\!-\!m}.$$

Comparing with the given form $$\sqrt{x}\,{\rm N\!-\!m},$$ we see that $$x = 195.$$

Hence, the correct answer is Option C.

We have a circular platform of mass $$M = 200 \text{ kg}$$ that can be treated as a uniform solid disc. For a solid disc of radius $$R$$, the moment of inertia about the central axis is given by the standard formula

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

The person standing on the rim has mass $$m = 80 \text{ kg}$$. A human can be modelled as a point mass for rotational calculations, so when the person is at a distance $$R$$ from the axis, the moment of inertia contributed by the person is

$$I_{\text{person (initial)}} = m R^{2}.$$

Initially, therefore, the total moment of inertia of the system “platform + person” is the sum of the two individual contributions:

$$$I_{1} = I_{\text{disc}} + I_{\text{person (initial)}} = \frac{1}{2} M R^{2} + m R^{2} = R^{2}\!\left(\frac{1}{2} M + m\right).$$$

The platform is rotating at an initial speed of $$$n_{1} = 5 \text{ revolutions per minute}$$$ (rpm). Converting to angular speed in rad s−1 is unnecessary because the same unit cancels out, but we shall denote the initial angular velocity by $$\omega_{1}$$, so

$$\omega_{1} = 5 \text{ rpm}.$$

Now the person walks from the rim straight to the centre. When the person reaches the centre, his distance from the axis becomes practically zero, so his moment of inertia becomes

$$I_{\text{person (final)}} \approx 0.$$

Therefore the final moment of inertia of the system is only that of the disc:

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

Because no external torque acts on the system, angular momentum is conserved. We state the law first:

$$$\text{Angular momentum } L = I \omega \quad\text{remains constant if the external torque is zero.}$$$

So we set initial angular momentum equal to final angular momentum:

$$I_{1}\,\omega_{1} = I_{2}\,\omega_{2}.$$

Substituting the expressions for $$I_{1}$$ and $$I_{2}$$, we get

$$$\biggl[R^{2}\!\left(\dfrac{1}{2} M + m\right)\biggr] \omega_{1} \;=\; \bigl[\tfrac{1}{2} M R^{2}\bigr] \omega_{2}.$$$

We observe that $$R^{2}$$ cancels from both sides, giving

$$\left(\frac{1}{2} M + m\right) \omega_{1} \;=\; \frac{1}{2} M \,\omega_{2}.$$

Solving for $$\omega_{2}$$, we write

$$\omega_{2} \;=\; \frac{\left(\dfrac{1}{2} M + m\right)}{\dfrac{1}{2} M}\;\omega_{1}.$$

Now we substitute the numerical values $$M = 200 \text{ kg}$$ and $$m = 80 \text{ kg}$$. First compute the denominator:

$$\frac{1}{2} M = \frac{1}{2} \times 200 = 100.$$

Then compute the numerator:

$$\frac{1}{2} M + m = 100 + 80 = 180.$$

Hence the required ratio is

$$\frac{180}{100} = 1.8.$$

Substituting this factor and the initial speed $$\omega_{1} = 5 \text{ rpm}$$, we obtain

$$\omega_{2} = 1.8 \times 5 \text{ rpm} = 9 \text{ rpm}.$$

Therefore, when the person arrives at the centre, the platform-person system speeds up to 9 revolutions per minute.

So, the answer is $$9 \text{ rpm}$$.

$$L_i = m v L$$

$$I = I_{\text{rod}} + I_{\text{particle}}$$

$$I = \frac{1}{3}ML^2 + mL^2$$

$$I = \left(\frac{0.9}{3} + 0.1\right) \times 1^2 = 0.3 + 0.1 = 0.4\text{ kg m}^2$$

$$L_f = I\omega$$

$$m v L = I\omega \implies 0.1 \times 80 \times 1 = 0.4 \times \omega \implies 8 = 0.4\omega \implies \omega = 20\text{ rad s}^{-1}$$

Perpendicular distance of mass at E from axis EX: $$r_E = 0$$

Perpendicular distance of mass at G from axis EX: $$r_G = a$$

Perpendicular distance of mass at F from axis EX: $$r_F = a \cos(60^\circ) = \frac{a}{2}$$

Total moment of inertia of the system about axis EX: $$I = m r_E^2 + m r_G^2 + m r_F^2$$

$$I = m(0)^2 + m(a)^2 + m\left(\frac{a}{2}\right)^2 = ma^2 + \frac{ma^2}{4} = \frac{5}{4}ma^2$$

$$I = \frac{5 \times 5}{4 \times 5}ma^2 = \frac{25}{20}ma^2$$

$$\frac{N}{20}ma^2 = \frac{25}{20}ma^2 \implies N = 25$$

Minimum Required Theory

• Sliding Condition: To initiate sliding, the minimum applied force $$F$$ must overcome limiting static friction ($$f = \mu mg$$).
• No Toppling Condition: For the box not to tip over, the overturning torque about the pivoting edge (bottom front corner) must be less than or equal to the restoring torque provided by the box's weight.
• Overturning Torque: The force $$F$$ is applied at a height of $$(\frac{a}{2} + b)$$ from the ground. Torque $$= F\left(\frac{a}{2} + b\right)$$
• Restoring Torque: The weight $$mg$$ acts through the center of mass, a horizontal distance of $$\frac{a}{2}$$ from the edge. Torque $$= mg\left(\frac{a}{2}\right)$$
• The variable $$b$$ represents the distance above the center of mass.
• The center of mass is strictly at height $$0.5a$$.
• The total height of the cubical box is $$a$$.
• Therefore, the physical top edge of the box is at a maximum distance of $$0.5a$$ above the center of mass. You cannot apply a force higher than the box itself.

Step-by-Step Solution

Step 1: Determine the minimum force for sliding

For the box to just start moving:

$$F = \mu mg$$

Step 2: Set up the torque inequality for "No Toppling"

Calculate the torques about the bottom front edge.

For the box not to topple:

$$F\left(\frac{a}{2} + b\right) \le mg\left(\frac{a}{2}\right)$$

Step 3: Solve the mathematical inequality for $$b$$

Substitute $$F = \mu mg$$ into the torque equation:

$$\mu mg\left(\frac{a}{2} + b\right) \le mg\left(\frac{a}{2}\right)$$

Cancel $$mg$$ from both sides:

$$\mu\left(\frac{a}{2} + b\right) \le \frac{a}{2}$$

Substitute the given coefficient of friction ($$\mu = 0.4$$):

$$0.4\left(0.5a + b\right) \le 0.5a$$

$$0.2a + 0.4b \le 0.5a$$

$$0.4b \le 0.3a$$

$$b \le 0.75a$$

Mathematical Conclusion: To avoid toppling, $$b$$ can be up to $$0.75a$$.

Step 4: Apply the physical geometric constraint

Since $$0.5a$$ is smaller than the limiting mathematical value of $$0.75a$$, the geometry dictates the maximum limit:

$$b_{max} = 0.5a$$

Step 5: Calculate the final requested value

The question asks for the maximum possible value of $$100 \times \frac{b}{a}$$.

$$\frac{b}{a} = 0.5$$

$$100 \times \left(\frac{b}{a}\right) = 100 \times 0.5 = 50$$

Final Answer: 50

Moment of inertia of a uniform bar about an end pivot: $$I = \frac{1}{3}ML^2$$

Conservation of mechanical energy: $$E_i = E_f \implies U_i + K_i = U_f + K_f$$

Taking the horizontal table surface as the reference potential level:

$$Mg\left(\frac{L}{2}\sin 30^\circ\right) + 0 = 0 + \frac{1}{2}I\omega^2$$

$$Mg\frac{L}{4} = \frac{1}{2}\left(\frac{1}{3}ML^2\right)\omega^2 \implies \frac{g}{4} = \frac{1}{6}L\omega^2$$

$$\frac{10}{4} = \frac{1}{6}(1)\omega^2 \implies \omega^2 = \frac{60}{4} = 15$$

$$\omega = \sqrt{15}$$

We first note that the wire will break when the tension in it produces a stress equal to the breaking stress. By definition, the stress $$\sigma$$ in a wire is given by the formula

$$\sigma = \dfrac{F}{A},$$

where $$F$$ is the tensile force (tension) and $$A$$ is the cross-sectional area of the wire. From this relation the maximum permissible tension before breaking is

$$F_{\text{max}} = \sigma A.$$

Substituting the given numerical values, we have

$$\sigma = 4.8 \times 10^{7}\; \text{N m}^{-2},$$ $$A = 10^{-2}\; \text{cm}^2.$$

Because the standard SI unit for area is square metres, we convert the area from square centimetres to square metres. Remember that $$1\; \text{cm} = 10^{-2}\; \text{m}$$, so

$$1\; \text{cm}^2 = (10^{-2}\; \text{m})^2 = 10^{-4}\; \text{m}^2.$$

Therefore,

$$A = 10^{-2}\; \text{cm}^2 = 10^{-2} \times 10^{-4}\; \text{m}^2 = 10^{-6}\; \text{m}^2.$$

Now we calculate the breaking force:

$$F_{\text{max}} = \sigma A = (4.8 \times 10^{7}\; \text{N m}^{-2})\,(10^{-6}\; \text{m}^2) = 4.8 \times 10^{1}\; \text{N} = 48\; \text{N}.$$

This $$48\; \text{N}$$ is the greatest tension the wire can withstand without snapping.

Next, we relate this tension to the required angular speed. In a space station there is no effective weight, so the only force providing the necessary centripetal acceleration for circular motion is the tension in the wire. For a body of mass $$m$$ moving in a circle of radius $$L$$ with angular speed $$\omega$$, the centripetal force is given by the well-known formula

$$F_{\text{centripetal}} = m\,\omega^{2}\,L.$$

The wire will be on the verge of breaking when this centripetal force equals $$F_{\text{max}}$$ calculated above. Hence we set

$$m\,\omega^{2}\,L = F_{\text{max}}.$$

Substituting the known quantities,

$$m = 10\; \text{kg}, \qquad L = 0.3\; \text{m}, \qquad F_{\text{max}} = 48\; \text{N},$$

we obtain

$$10 \times \omega^{2} \times 0.3 = 48.$$

Simplifying step by step,

$$10 \times 0.3 = 3,$$

so the equation becomes

$$3\,\omega^{2} = 48.$$

Dividing both sides by $$3$$, we find

$$\omega^{2} = \dfrac{48}{3} = 16.$$

Taking the positive square root (because speed is positive),

$$\omega = \sqrt{16} = 4\; \text{rad s}^{-1}.$$

Thus the body can be rotated with a maximum angular speed of $$4\; \text{rad}\,\text{s}^{-1}$$ without the wire breaking.

Hence, the correct answer is Option 4.

Moment of inertia of the flywheel (disc): $$I = \frac{1}{2}mr^2$$

No-slip string unwinding condition: $$v = r\omega$$

Total initial mechanical energy of the system: $$E_i = 0$$

Total final mechanical energy after bob drops by height $$h$$: $$E_f = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 - mgh$$

From $$E_i = E_f$$: $$mgh = \frac{1}{2}m(r\omega)^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\omega^2$$

$$mgh = \frac{1}{2}mr^2\omega^2 + \frac{1}{4}mr^2\omega^2 = \frac{3}{4}mr^2\omega^2$$

$$gh = \frac{3}{4}r^2\omega^2 \implies \omega^2 = \frac{4gh}{3r^2}$$

$$\omega = \frac{1}{r}\sqrt{\frac{4gh}{3}}$$

Torque equilibrium about the step corner:  $$\tau_F = \tau_{Mg}$$

Perpendicular distance for $$F$$:  $$r_{\perp, F} = R$$

Perpendicular distance for $$Mg$$:  $$r_{\perp, Mg} = \sqrt{R^2 - (R-a)^2} = R\sqrt{1 - \left(\frac{R-a}{R}\right)^2}$$

Equating torques:  $$F \cdot R = Mg \cdot R\sqrt{1 - \left(\frac{R-a}{R}\right)^2}$$

$$F = Mg\sqrt{1 - \left(\frac{R-a}{R}\right)^2}$$

We have a wheel of moment of inertia $$I$$ spinning freely with angular speed $$\omega$$ on a shaft whose own moment of inertia is negligible. The rotational kinetic energy initially stored in this wheel is given by the well-known formula for rotational kinetic energy

$$K = \tfrac12 I \omega^2.$$

So, the initial angular momentum and kinetic energy of the system are, respectively,

$$L_{\text{initial}} = I\,\omega,$$

$$K_{\text{initial}} = \tfrac12\,I\,\omega^2.$$

Now another wheel, whose moment of inertia is $$3I$$ and which is initially at rest, is suddenly coupled rigidly to the same shaft. Because the coupling is assumed to occur without any external torque acting on the combined system, we invoke the conservation of angular momentum. We first compute the total moment of inertia after coupling:

$$I_{\text{total}} = I + 3I = 4I.$$

If the common angular speed after coupling is $$\omega_f,$$ conservation of angular momentum gives

$$I\,\omega = (4I)\,\omega_f.$$

Solving for $$\omega_f$$ (by dividing both sides by $$4I$$), we obtain

$$\omega_f = \frac{I\,\omega}{4I} = \frac{\omega}{4}.$$

Next we calculate the rotational kinetic energy of the combined system after coupling. Using the same energy formula, but with the new moment of inertia $$4I$$ and angular speed $$\omega_f = \omega/4,$$ we get

$$\begin{aligned} K_{\text{final}} &= \tfrac12 \,(4I)\,\left(\frac{\omega}{4}\right)^{\!2} \\ &= 2I \,\left(\frac{\omega^2}{16}\right) \\ &= \frac{2I\omega^2}{16} \\ &= \frac{I\omega^2}{8}. \end{aligned}$$

For easier comparison with the initial energy, we rewrite $$K_{\text{final}}$$ in terms of $$K_{\text{initial}} = \tfrac12 I\omega^2$$:

$$\begin{aligned} K_{\text{final}} &= \frac{I\omega^2}{8} \\ &= \frac{1}{4}\left(\tfrac12 I\omega^2\right) \\ &= \frac{1}{4}\,K_{\text{initial}}. \end{aligned}$$

Thus the final kinetic energy is one-quarter of the initial kinetic energy. The fractional loss of kinetic energy is therefore

$$\frac{K_{\text{initial}} - K_{\text{final}}}{K_{\text{initial}}} = 1 - \frac{1}{4} = \frac{3}{4}.$$

Hence, the correct answer is Option D.

The surface mass density of the disc is given as $$\sigma(r)=A+Br,$$ where $$r$$ is the distance of a point from the centre. We have to find the moment of inertia about the axis perpendicular to the plane of the disc and passing through its centre.

We consider a thin concentric ring of radius $$r$$ and radial thickness $$dr$$.

The elementary area of this ring is obtained from the geometry of a circle:

$$dA = 2\pi r\,dr.$$

Mass of this elementary ring is obtained by multiplying the surface density with the area:

$$dm = \sigma(r)\,dA = (A+Br)\, (2\pi r\,dr).$$

For a ring, every particle is at the same distance $$r$$ from the axis, so the elemental moment of inertia is

$$dI = r^{2}\,dm.$$

Substituting the expression for $$dm$$, we get

$$dI = r^{2}\,(A+Br)\,(2\pi r\,dr).$$

Simplifying the powers of $$r$$:

$$dI = 2\pi\,r^{3}(A+Br)\,dr.$$

To obtain the total moment of inertia, we integrate this expression from the centre (where $$r=0$$) to the rim (where $$r=a$$):

$$I = \int_{0}^{a} 2\pi\,r^{3}(A+Br)\,dr.$$

We separate the integral into two standard parts:

$$I = 2\pi\left[A\int_{0}^{a} r^{3}\,dr \;+\; B\int_{0}^{a} r^{4}\,dr\right].$$

Now we evaluate the individual integrals, using the power-rule formula $$\displaystyle\int r^{n}\,dr=\frac{r^{n+1}}{n+1}$$.

For the first integral:

$$\int_{0}^{a} r^{3}\,dr = \left[\frac{r^{4}}{4}\right]_{0}^{a} = \frac{a^{4}}{4}.$$

For the second integral:

$$\int_{0}^{a} r^{4}\,dr = \left[\frac{r^{5}}{5}\right]_{0}^{a} = \frac{a^{5}}{5}.$$

Substituting these results back, we obtain

$$I = 2\pi\left[A\left(\frac{a^{4}}{4}\right) + B\left(\frac{a^{5}}{5}\right)\right].$$

We now factor out $$a^{4}$$ from both terms:

$$I = 2\pi\,a^{4}\left(\frac{A}{4} + \frac{aB}{5}\right).$$

Thus, the moment of inertia of the disc is

$$I = 2\pi a^{4}\left(\frac{A}{4} + \frac{aB}{5}\right).$$

Hence, the correct answer is Option A.

By geometry of the cone:

$$\frac{r}{R} = \frac{H-y}{H} \implies r = \frac{R}{H}(H-y)$$

$$dl = \sqrt{dy^2 + dr^2} = dy\sqrt{1 + \left(\frac{dr}{dy}\right)^2} = dy\sqrt{1 + \frac{R^2}{H^2}} = \frac{L}{H}dy \quad \text{(where } L = \sqrt{R^2+H^2}\text{)}$$

Mass element $$dm$$ of the hollow cone surface area:

$$dm = \frac{M}{\pi R L} \cdot (2\pi r \cdot dl) = \frac{2M}{RL} \cdot r \cdot \left(\frac{L}{H}dy\right) = \frac{2M}{RH} r dy$$

Using $$dy = -\frac{H}{R}dr$$ from differentiating $$y = H - \frac{H}{R}r$$:

$$dm = \frac{2M}{RH} r \left(-\frac{H}{R}dr\right) = -\frac{2M}{R^2} r dr$$

Integrating from the base vertex ($$r=0$$) to the top surface ($$r=R$$):

$$I = \int r^2 dm \implies I = \int_{0}^{R} r^2 \left(\frac{2M}{R^2} r dr\right)$$

$$I = \frac{2M}{R^2} \int_{0}^{R} r^3 dr = \frac{2M}{R^2} \left[ \frac{r^4}{4} \right]_{0}^{R} = \frac{2M}{R^2} \cdot \frac{R^4}{4} = \frac{MR^2}{2}$$

We begin by translating the information given in the statement into mathematical form. The linear mass density (mass per unit length) of the rod at a point that is at a distance $$x$$ from the end A is

$$\lambda(x)=\lambda_0\left(1+\frac{x}{L}\right).$$

To calculate the moment of inertia about an axis passing through A and perpendicular to the rod, two main steps are needed:

1. Express the constant $$\lambda_0$$ in terms of the total mass $$M$$ and the length $$L$$ of the rod.
2. Substitute this expression for $$\lambda_0$$ into the general integral formula for moment of inertia.

We tackle each in detail.

Step 1: Relating $$\lambda_0$$ to the total mass $$M$$.

By definition, the total mass of the rod is obtained by integrating its linear density along its entire length. Hence

$$M=\int_{0}^{L}\lambda(x)\,dx.$$

Substituting the given density function, we have

$$M=\int_{0}^{L}\lambda_0\left(1+\frac{x}{L}\right)\,dx.$$

Because $$\lambda_0$$ is a constant, it can be taken outside the integral:

$$M=\lambda_0\int_{0}^{L}\left(1+\frac{x}{L}\right)\,dx.$$

We now integrate term by term:

$$\int_{0}^{L}1\,dx = \left[x\right]_{0}^{L}=L,$$

$$\int_{0}^{L}\frac{x}{L}\,dx=\frac{1}{L}\int_{0}^{L}x\,dx =\frac{1}{L}\left[\frac{x^{2}}{2}\right]_{0}^{L} =\frac{1}{L}\cdot\frac{L^{2}}{2} =\frac{L}{2}.$$

Adding the two contributions gives

$$\int_{0}^{L}\left(1+\frac{x}{L}\right)\,dx = L+\frac{L}{2}=\frac{3L}{2}.$$

Therefore,

$$M=\lambda_0\left(\frac{3L}{2}\right).$$

Solving for $$\lambda_0$$ gives

$$\lambda_0 = \frac{2M}{3L}.$$

Step 2: Calculating the moment of inertia $$I$$.

The standard continuous-mass formula for moment of inertia is

$$I=\int r^{2}\,dm,$$

where here the perpendicular distance from the axis (through A) to a small element at position $$x$$ is just $$r=x$$. Also, for a thin rod, an infinitesimal mass element $$dm$$ is related to its length element $$dx$$ through $$dm=\lambda(x)\,dx$$.

Thus

$$I=\int_{0}^{L}x^{2}\,dm =\int_{0}^{L}x^{2}\,\lambda(x)\,dx =\int_{0}^{L}x^{2}\,\lambda_0\left(1+\frac{x}{L}\right)\,dx.$$

Pulling out the constant $$\lambda_0$$ gives

$$I=\lambda_0\int_{0}^{L}\left(x^{2}+\frac{x^{3}}{L}\right)\,dx.$$

We now integrate each term individually.

First term: $$\int_{0}^{L}x^{2}\,dx=\left[\frac{x^{3}}{3}\right]_{0}^{L}=\frac{L^{3}}{3}.$$

Second term: $$\int_{0}^{L}\frac{x^{3}}{L}\,dx=\frac{1}{L}\int_{0}^{L}x^{3}\,dx =\frac{1}{L}\left[\frac{x^{4}}{4}\right]_{0}^{L} =\frac{1}{L}\cdot\frac{L^{4}}{4} =\frac{L^{3}}{4}.$$

Adding these two integrals yields

$$\int_{0}^{L}\left(x^{2}+\frac{x^{3}}{L}\right)\,dx =\frac{L^{3}}{3}+\frac{L^{3}}{4} =L^{3}\left(\frac{1}{3}+\frac{1}{4}\right) =L^{3}\left(\frac{4}{12}+\frac{3}{12}\right) =L^{3}\left(\frac{7}{12}\right) =\frac{7L^{3}}{12}.$$

Therefore,

$$I=\lambda_0\left(\frac{7L^{3}}{12}\right).$$

Substituting $$\lambda_0=\dfrac{2M}{3L}$$ into this expression, we get

$$I=\left(\frac{2M}{3L}\right)\left(\frac{7L^{3}}{12}\right) =\frac{2\cdot7\,M\,L^{2}}{3\cdot12} =\frac{14\,M\,L^{2}}{36} =\frac{7\,M\,L^{2}}{18}.$$

Thus, the moment of inertia of the rod about the given axis is

$$I=\frac{7}{18}ML^{2}.$$

Comparing with the options provided, we see that this matches option B.

Hence, the correct answer is Option B.

For a uniform rod of length $$l$$ and mass $$M$$, the standard formula for the moment of inertia about an axis perpendicular to the rod and passing through its centre (the midpoint) is stated first: $$I_{\text{centre}}=\dfrac{1}{12}\,M\,l^{2}.$$

Now the required axis is still perpendicular to the rod but is displaced along the length by a distance $$d=\dfrac{l}{4}$$ from the centre. To shift the axis, we invoke the Parallel-Axis Theorem, which states: $$I = I_{\text{centre}} + M\,d^{2}.$$

Substituting the known values, we have $$I = \dfrac{1}{12}\,M\,l^{2} + M\left(\dfrac{l}{4}\right)^{2}.$$

Evaluating the square, $$\left(\dfrac{l}{4}\right)^{2}=\dfrac{l^{2}}{16},$$ so $$I = \dfrac{1}{12}\,M\,l^{2} + M\left(\dfrac{l^{2}}{16}\right).$$

Collecting the terms with a common denominator of $$48$$, $$I = M\,l^{2}\left(\dfrac{4}{48}+\dfrac{3}{48}\right)=M\,l^{2}\left(\dfrac{7}{48}\right).$$

The radius of gyration $$k$$ is defined through $$I = M\,k^{2}.$$ Setting $$I = M\,k^{2}$$ and cancelling the common factor $$M$$, we obtain $$k^{2}=l^{2}\left(\dfrac{7}{48}\right).$$

Taking the positive square root, $$k = l\sqrt{\dfrac{7}{48}}.$$

Hence, the correct answer is Option C.

For a solid sphere of mass $$m$$ and diameter $$d$$ (radius $$R = \frac{d}{2}$$), the moment of inertia about its center of mass axis is:  $$I_{\text{cm}} = \frac{2}{5}mR^2 = \frac{2}{5}m\left(\frac{d}{2}\right)^2 = \frac{m d^2}{10}$$

For the distance from the centroid $$O$$ of the equilateral triangle of side $$d$$ to the center of any sphere:

$$x = \frac{d}{\sqrt{3}}$$

For the total moment of inertia $$I_0$$ about an axis passing through the centroid $$O$$:

$$I_0 = 3 \times \left( I_{\text{cm}} + m x^2 \right) = 3 \times \left( \frac{m d^2}{10} + m \left(\frac{d}{\sqrt{3}}\right)^2 \right)$$

$$I_0 = 3 \times \left( \frac{m d^2}{10} + \frac{m d^2}{3} \right) = 3 \times \left(\frac{13}{30}\right) m d^2 = \frac{13}{10} m d^2$$

For the total moment of inertia $$I_A$$ about an axis passing through the center of sphere A:

$$I_A = I_{\text{cm}} + \left(I_{\text{cm}} + m d^2\right) + \left(I_{\text{cm}} + m d^2\right) = 3 I_{\text{cm}} + 2 m d^2$$

$$I_A = 3\left(\frac{m d^2}{10}\right) + 2 m d^2 = \frac{23}{10} m d^2$$

$$\frac{I_0}{I_A} = \frac{\frac{13}{10} m d^2}{\frac{23}{10} m d^2} = \frac{13}{23}$$

$$F_v = mg$$

$$F_H = m a_{\text{cm}} = m \omega^2 \left(\frac{l}{2}\sin\theta\right)$$

Torque about the centre of mass due to forces acting at the pivot:

$$\tau_{\text{cm}} = F_H \cdot \left(\frac{l}{2}\cos\theta\right) - F_v \cdot \left(\frac{l}{2}\sin\theta\right)$$

$$\tau_{\text{cm}} = \left(m \omega^2 \frac{l}{2}\sin\theta\right)\left(\frac{l}{2}\cos\theta\right) - (mg)\left(\frac{l}{2}\sin\theta\right)$$

Equating to the given rate of change of angular momentum (taking magnitudes in direction):

$$\frac{m l^2}{12} \omega^2 \sin\theta \cos\theta = (mg)\left(\frac{l}{2}\sin\theta\right) - \left(m \omega^2 \frac{l}{2}\sin\theta\right)\left(\frac{l}{2}\cos\theta\right)$$

$$\frac{ml^2}{12}\omega^2 \sin\theta \cos\theta + \frac{ml^2}{4}\omega^2 \sin\theta \cos\theta = \frac{mgl}{2}\sin\theta$$

$$\frac{ml^2}{3}\omega^2 \sin\theta \cos\theta = \frac{mgl}{2}\sin\theta$$

$$\frac{l}{3}\omega^2 \cos\theta = \frac{g}{2} \implies \cos\theta = \frac{3g}{2l\omega^2}$$

We are given two uniform circular discs having the same thickness and made of the same material. Their radii are $$R_1 = R$$ and $$R_2 = \alpha R$$, respectively. We denote their moments of inertia about the central axes (perpendicular to the plane of the discs) by $$I_1$$ and $$I_2$$.

First we recall the standard formula for the moment of inertia of a uniform disc about its central axis. The formula is

$$I = \dfrac{1}{2} M R^2,$$

where $$M$$ is the mass of the disc and $$R$$ is its radius.

Because the two discs have the same thickness and are made of the same material, their mass per unit volume (density) is the same. Let us call the mass density $$\rho$$ and the common thickness $$t$$. For any disc, the mass equals the product of density, volume, and area. The volume of a disc is the area of its face times the thickness, so the mass of a disc of radius $$R_i$$ is

$$M_i = \rho\,(\text{volume}) = \rho \, (\text{area}) \times (\text{thickness}) = \rho \,(\pi R_i^2)\,t.$$

Now we calculate each moment of inertia separately.

For the first disc, having radius $$R_1 = R$$, its mass is

$$M_1 = \rho\,\pi R^2\,t.$$

The corresponding moment of inertia is

$$I_1 = \dfrac{1}{2} M_1 R^2 = \dfrac{1}{2}\bigl(\rho \pi R^2 t\bigr) R^2 = \dfrac{1}{2}\,\rho\,\pi\,t\,R^4.$$

For the second disc, whose radius is $$R_2 = \alpha R$$, its mass becomes

$$M_2 = \rho\,\pi\,(\alpha R)^2\,t = \rho\,\pi\,\alpha^2 R^2\,t.$$

Using the same moment-of-inertia formula, we write

$$I_2 = \dfrac{1}{2} M_2 (\alpha R)^2 = \dfrac{1}{2} \bigl(\rho\,\pi\,\alpha^2 R^2\,t\bigr) (\alpha^2 R^2) = \dfrac{1}{2}\,\rho\,\pi\,t\,\alpha^4 R^4.$$

We are told that the ratio of the moments of inertia satisfies

$$I_1 : I_2 = 1 : 16.$$

Therefore we set up the equation

$$\dfrac{I_1}{I_2} = \dfrac{1}{16}.$$

Substituting the explicit expressions for $$I_1$$ and $$I_2$$, we have

$$\dfrac{\dfrac{1}{2}\,\rho\,\pi\,t\,R^4}{\dfrac{1}{2}\,\rho\,\pi\,t\,\alpha^4 R^4} = \dfrac{1}{16}.$$

Notice that the common factors $$\dfrac{1}{2}\,\rho\,\pi\,t\,R^4$$ cancel from numerator and denominator, leaving

$$\dfrac{1}{\alpha^4} = \dfrac{1}{16}.$$

Cross-multiplying, we obtain

$$\alpha^4 = 16.$$

Taking the positive fourth root (since radius is positive), we get

$$\alpha = 16^{1/4} = 2.$$

The numerical value of $$\alpha$$ is therefore $$2$$, which corresponds to Option C.

Hence, the correct answer is Option C.

Let the four corners be labeled such that:

Mass 1 is on the axis: $$r_1 = 0$$

Masses 2 and 3 are at adjacent corners, each at a perpendicular distance equal to half the diagonal: $$r_2 = r_3 = \frac{l}{\sqrt{2}}$$

Mass 4 is at the opposite corner, at a perpendicular distance equal to the full length of the diagonal: $$r_4 = \sqrt{2}l$$

$$I = m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2 + m_4 r_4^2$$

$$I = m(0)^2 + m\left(\frac{l}{\sqrt{2}}\right)^2 + m\left(\frac{l}{\sqrt{2}}\right)^2 + m\left(\sqrt{2}l\right)^2$$

$$I = 0 + \frac{1}{2}ml^2 + \frac{1}{2}ml^2 + 2ml^2 = 3ml^2$$

$$L = I\omega \implies L = 3ml^2\omega$$

We are given a uniform solid sphere having mass $$m = 500 \text{ g} = 0.5\ \text{kg}$$. It rolls without slipping on a horizontal surface, and the linear speed of its centre is $$v = 5.00\ \text{cm s}^{-1} = 0.05\ \text{m s}^{-1}$$.

For a rolling body, the total kinetic energy is the sum of translational and rotational parts. We therefore write the basic relation

$$K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}}.$$

According to the definition, the translational kinetic energy is

$$K_{\text{trans}} = \dfrac12 m v^{2}.$$

For rotational kinetic energy about the centre of mass, the standard formula is

$$K_{\text{rot}} = \dfrac12 I \omega^{2},$$

where $$I$$ is the moment of inertia of the body about its centre and $$\omega$$ is the angular speed.

Because the sphere is uniform (solid), its moment of inertia about its centre is well-known:

$$I = \dfrac{2}{5} m r^{2}.$$

Rolling without slipping provides the kinematic link

$$\omega = \dfrac{v}{r},$$

since each point on the rim that touches the ground momentarily has zero velocity relative to the ground.

Now we substitute $$I$$ and $$\omega$$ into the rotational energy expression:

$$K_{\text{rot}} = \dfrac12 \left(\dfrac{2}{5} m r^{2}\right) \left(\dfrac{v}{r}\right)^{2} = \dfrac12 \left(\dfrac{2}{5} m r^{2}\right) \dfrac{v^{2}}{r^{2}} = \dfrac12 \left(\dfrac{2}{5}\right) m v^{2} = \dfrac{1}{5} m v^{2}.$$

Hence the total kinetic energy becomes

$$K_{\text{total}} = \dfrac12 m v^{2} + \dfrac15 m v^{2} = \left(\dfrac12 + \dfrac15\right) m v^{2} = \left(\dfrac{5}{10} + \dfrac{2}{10}\right) m v^{2} = \dfrac{7}{10} m v^{2}.$$

Substituting the numerical values $$m = 0.5\ \text{kg}$$ and $$v = 0.05\ \text{m s}^{-1}$$, we get

$$K_{\text{total}} = \dfrac{7}{10}\,(0.5)\,(0.05)^{2}.$$

First we square the speed:

$$(0.05)^{2} = 0.0025.$$

Next we multiply step by step:

$$\dfrac{7}{10} \times 0.5 = 0.7 \times 0.5 = 0.35,$$

and then

$$0.35 \times 0.0025 = 0.000875.$$

Expressing this result in scientific notation, we write

$$K_{\text{total}} = 8.75 \times 10^{-4}\ \text{J}.$$

Examining the options, this numerical value matches Option A.

Hence, the correct answer is Option A.

We begin by noting that the wheel can rotate without any slipping of the string, so the linear speed $$v$$ of each block is related to the angular speed $$\omega$$ of the wheel by the no-slip condition

$$v = R\,\omega.$$

The system is released from rest, therefore its initial kinetic energy is zero. Block $$m_1$$ descends a vertical distance $$h$$ while block $$m_2$$ rises the same distance. The change in gravitational potential energy of the two blocks is

$$\Delta U \;=\; -m_1gh \;+\; m_2gh \;=\; -(m_1 - m_2)gh.$$

Because energy is lost by the heavier block and gained by the lighter one, the net decrease in potential energy available to produce motion is

$$ (m_1 - m_2)gh. $$

We now invoke the principle of conservation of mechanical energy, which states:

“Loss of gravitational potential energy = Gain in kinetic energy of all moving parts.”

The kinetic energy acquired has two contributions:

• The translational kinetic energies of the two blocks, each moving with speed $$v$$.
• The rotational kinetic energy of the wheel, which is rotating with angular speed $$\omega$$.

Writing each term explicitly, we have

$$\text{Total K.E.} \;=\;\frac12 m_1v^2\;+\;\frac12 m_2v^2\;+\;\frac12 I\omega^2.$$

Combining the first two terms gives

$$\frac12(m_1+m_2)v^2 + \frac12 I \omega^2.$$

Substituting $$\omega = \dfrac{v}{R}$$ into the rotational part, we obtain

$$\frac12(m_1+m_2)v^2 \;+\;\frac12 I\left(\frac{v}{R}\right)^2 \;=\;\frac12\left[(m_1+m_2) + \frac{I}{R^2}\right]v^2.$$

Equating the loss in potential energy to this total kinetic energy, we write

$$ (m_1 - m_2)gh \;=\;\frac12\left[(m_1+m_2) + \frac{I}{R^2}\right]v^2.$$

Solving for $$v^2$$ gives

$$v^2 \;=\;\frac{2(m_1 - m_2)gh}{(m_1+m_2) \;+\;\dfrac{I}{R^2}}.$$

Multiplying numerator and denominator of the right-hand side by $$R^2$$ to prepare for finding $$\omega$$, we write

$$v^2 \;=\;\frac{2(m_1 - m_2)gh\,R^2}{(m_1+m_2)R^2 + I}.$$

Finally, using $$\omega = \dfrac{v}{R}$$, we have

$$\omega^2 \;=\;\frac{v^2}{R^2} \;=\;\frac{2(m_1 - m_2)gh}{(m_1+m_2)R^2 + I},$$

and therefore

$$\boxed{\displaystyle \omega \;=\;\left[\frac{2(m_1 - m_2)gh}{(m_1 + m_2)R^2 + I}\right]^{\tfrac12}}.$$

Comparing this result with the options provided, we see that it matches Option A.

Hence, the correct answer is Option A.

Moment of inertia of a rectangular sheet of mass $$M$$ and dimensions $$a\times b $$ about an axis passing through its center of mass $$O$$ and perpendicular to its plane:

$$I_O = \frac{1}{12}M(a^2 + b^2)$$

$$I_O = \frac{1}{12}M(60^2 + 80^2) = \frac{1}{12}M(3600 + 6400) = \frac{10000}{12}M = \frac{2500}{3}M$$

$$\text{By the parallel axis theorem}, I_{O'} = I_O + Md^2$$

$$\text{where } d \text{ is the distance from the center of mass } O\left(\frac{a}{2}, \frac{b}{2}\right) \text{ to the corner point } O'(a, b)\text{:}$$

$$d^2 = \left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2 = \frac{a^2 + b^2}{4} = \frac{60^2 + 80^2}{4} = \frac{10000}{4} = 2500\text{ cm}^2$$

$$I_{O'} = \frac{2500}{3}M + 2500M = 2500M\left(\frac{1}{3} + 1\right) = \frac{4}{3}(2500)M = \frac{10000}{3}M$$

$$\text{Ratio of the moments of inertia: } \frac{I_O}{I_{O'}} = \frac{\frac{2500}{3}M}{\frac{10000}{3}M} = \frac{2500}{10000} = \frac{1}{4}$$

Length of uniform rod $$AB$$: $$L = 100\text{ cm} = 1\text{ m}$$

Positions of forces measured from end A:

$$\text{Tension } T_A \text{ at } x = 0\text{ m}$$

$$\text{Weight of uniform rod } mg \text{ at } x = 0.5\text{ m}$$

$$\text{Hung mass } 2mg \text{ at } x = 0.75\text{ m}$$

$$\text{Tension } T_B \text{ at } x = 1\text{ m}$$

Taking torque about end B for rotational equilibrium ($$\Sigma \tau_B = 0$$):

$$T_A(1) - mg(1 - 0.5) - 2mg(1 - 0.75) = 0$$

$$T_A - mg(0.5) - 2mg(0.25) = 0$$

$$T_A - 0.5mg - 0.5mg = 0 \implies T_A = 1mg$$

Let the ring have mass $$M$$ and radius $$R$$. The nail goes through a point on the circumference, so the distance from the pivot to the centre of mass is simply $$d = R$$.

For any rigid body oscillating with small amplitude about a fixed horizontal axis, the time period is given by the physical-pendulum formula

$$T = 2\pi \sqrt{\dfrac{I_P}{M g d}},$$

where $$I_P$$ is the moment of inertia about the chosen pivot axis and $$d$$ is the distance between the pivot and the centre of mass.

We now evaluate $$I_P$$ for the two different oscillations.

Oscillation in the plane of the ring : time period $$T_1$$

If the ring swings in its own plane, the axis through the nail is perpendicular to the plane. For a thin ring, the moment of inertia about the axis through the centre and perpendicular to the plane is

$$I_C = M R^{2}.$$

Using the parallel-axis theorem, $$I_{P1} = I_C + M R^{2} = M R^{2} + M R^{2} = 2 M R^{2}.$$

Substituting in the time-period formula, we obtain

$$T_1 = 2\pi \sqrt{\dfrac{2 M R^{2}}{M g R}} = 2\pi \sqrt{\dfrac{2 R}{g}}.$$

Oscillation perpendicular to the plane of the ring : time period $$T_2$$

Here the ring tilts forward and backward, so the axis through the nail lies in the plane of the ring and is tangential to the circle. First, for a thin ring the moment of inertia about any diameter (an axis in the plane through the centre) is, by the perpendicular-axis theorem,

$$I_\text{diameter} = \dfrac{1}{2} M R^{2}.$$

The tangent axis is parallel to this diameter but displaced by a distance $$R$$, so using the parallel-axis theorem again,

$$I_{P2} = I_\text{diameter} + M R^{2} = \dfrac{1}{2} M R^{2} + M R^{2} = \dfrac{3}{2} M R^{2}.$$

Therefore

$$T_2 = 2\pi \sqrt{\dfrac{\dfrac{3}{2} M R^{2}}{M g R}} = 2\pi \sqrt{\dfrac{3 R}{2 g}}.$$

Ratio of the two periods

Dividing the two expressions, the common factors $$2\pi$$, $$M$$ and $$R/g$$ cancel out:

$$\dfrac{T_1}{T_2} \;=\; \sqrt{\dfrac{I_{P1}}{I_{P2}}} \;=\; \sqrt{\dfrac{2 M R^{2}}{\dfrac{3}{2} M R^{2}}} = \sqrt{\dfrac{2}{\dfrac{3}{2}}} = \sqrt{\dfrac{4}{3}} = \dfrac{2}{\sqrt{3}}.$$

Hence, the correct answer is Option A.

We begin by noting that the position vector of the particle with respect to the origin is

$$\vec r = x\hat i + y\hat j = \bigl(x_0 + a\cos\omega_1 t\bigr)\hat i + \bigl(y_0 + b\sin\omega_2 t\bigr)\hat j.$$

The torque about the origin is defined as

$$\vec\tau = \vec r \times \vec F,$$

where $$\vec F = m\vec a$$ is the net force and $$\vec a$$ is the acceleration. Therefore, we first calculate the acceleration components.

Differentiating the given coordinates twice with respect to time, we have

$$x = x_0 + a\cos\omega_1 t \;\;\Longrightarrow\;\; \dot x = -a\omega_1\sin\omega_1 t, \qquad \ddot x = -a\omega_1^2\cos\omega_1 t,$$

$$y = y_0 + b\sin\omega_2 t \;\;\Longrightarrow\;\; \dot y = b\omega_2\cos\omega_2 t, \qquad \ddot y = -b\omega_2^2\sin\omega_2 t.$$

Hence the acceleration vector is

$$\vec a = \ddot x\,\hat i + \ddot y\,\hat j = \bigl(-a\omega_1^2\cos\omega_1 t\bigr)\hat i + \bigl(-b\omega_2^2\sin\omega_2 t\bigr)\hat j.$$

Multiplying by the mass $$m$$ gives the force

$$\vec F = m\vec a = \bigl(-ma\omega_1^2\cos\omega_1 t\bigr)\hat i + \bigl(-mb\omega_2^2\sin\omega_2 t\bigr)\hat j.$$

Now we set up the cross product $$\vec\tau = \vec r \times \vec F$$. Using the determinant form,

$$ \vec\tau = \begin{vmatrix} \hat i & \hat j & \hat k\\ x & y & 0\\ F_x & F_y & 0 \end{vmatrix} = \bigl(xF_y - yF_x\bigr)\,\hat k. $$

Substituting $$x = x_0 + a\cos\omega_1 t,\; y = y_0 + b\sin\omega_2 t,\; F_x = -ma\omega_1^2\cos\omega_1 t,\; F_y = -mb\omega_2^2\sin\omega_2 t,$$ we get

$$\vec\tau = \Bigl[(x_0 + a\cos\omega_1 t)\!\bigl(-mb\omega_2^2\sin\omega_2 t\bigr) - (y_0 + b\sin\omega_2 t)\!\bigl(-ma\omega_1^2\cos\omega_1 t\bigr)\Bigr]\hat k.$$

We have to evaluate this at $$t = 0$$. At that instant, $$\cos 0 = 1$$ and $$\sin 0 = 0$$, so

$$x(0) = x_0 + a,\qquad y(0) = y_0,$$

$$F_x(0) = -ma\omega_1^2,\qquad F_y(0) = 0.$$

Therefore

$$\vec\tau(0) = \bigl[(x_0 + a)\cdot 0 - y_0\!\cdot(-ma\omega_1^2)\bigr]\hat k = \bigl[0 + my_0a\omega_1^2\bigr]\hat k = +my_0a\omega_1^2\,\hat k.$$

Hence, the correct answer is Option A.

For moment of inertia of uniform rod about its center O: $$I = \frac{m l^2}{12}$$

For restoring torque after small angular displacement $$\theta$$:

$$\text{Elongation/compression of each spring at the ends: } x = \frac{l}{2}\theta$$

$$\text{Restoring force from each spring: } F = kx = k\frac{l}{2}\theta$$

$$\text{Net restoring torque about O: } \tau = -2 \times \left(F \times \frac{l}{2}\right) = -2 \times \left(k\frac{l}{2}\theta \times \frac{l}{2}\right) = -\frac{k l^2}{2}\theta$$

For frequency of oscillation: $$\omega = \sqrt{\frac{C}{I}} = \sqrt{\frac{\frac{k l^2}{2}}{\frac{m l^2}{12}}} = \sqrt{\frac{6k}{m}}$$

$$f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{6k}{m}}$$

We have a uniform rod of mass $$M$$ and length $$2L$$ hung at its mid-point by a thin wire, so it performs torsional oscillations about a vertical axis through its centre. The oscillation frequency of a torsional pendulum is determined by the formula

$$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{I}},$$

where $$k$$ is the torsional constant of the wire and $$I$$ is the moment of inertia of the system about the axis of rotation.

For the bare rod the moment of inertia about a perpendicular axis through its centre is obtained from the standard result for a slender rod:

$$I_0=\int x^2\,dm=\dfrac{1}{12}M(2L)^2=\dfrac{1}{3}ML^{2}.$$

Hence the original frequency is

$$f_0=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{I_0}}.$$

Now two small masses, each of mass $$m$$, are attached symmetrically at a distance $$\dfrac{L}{2}$$ from the centre. The contribution of each attached mass to the moment of inertia is $$mr^{2}=m\left(\dfrac{L}{2}\right)^{2}=\dfrac{mL^{2}}{4}$$, so for both masses together we add

$$I_{\text{masses}}=2\times\dfrac{mL^{2}}{4}=\dfrac{mL^{2}}{2}.$$

The new total moment of inertia therefore becomes

$$I'=I_0+\dfrac{mL^{2}}{2}=\dfrac{1}{3}ML^{2}+\dfrac{1}{2}mL^{2}.$$

Because the wire is unchanged, the torsional constant $$k$$ remains the same. The modified frequency is

$$f'=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{I'}}.$$

According to the statement of the problem, attaching the masses reduces the frequency by 20 %, i.e.

$$f' = 0.8\,f_0.$$

Taking the ratio of the expressions for $$f'$$ and $$f_0$$ we get

$$\dfrac{f'}{f_0}=\sqrt{\dfrac{I_0}{I'}}=0.8.$$

Squaring both sides gives

$$\dfrac{I_0}{I'}=0.64 \quad\Longrightarrow\quad I'=\dfrac{I_0}{0.64}=\dfrac{25}{16}I_0.$$

Substituting $$I'$$ and $$I_0$$ explicitly,

$$\dfrac{1}{3}ML^{2}+\dfrac{1}{2}mL^{2}=\dfrac{25}{16}\left(\dfrac{1}{3}ML^{2}\right).$$

Cancelling the common factor $$L^{2}$$ we have

$$\dfrac{1}{3}M+\dfrac{1}{2}m=\dfrac{25}{48}M.$$

Rearranging to isolate $$m$$,

$$\dfrac{1}{2}m=\dfrac{25}{48}M-\dfrac{1}{3}M =\left(\dfrac{25}{48}-\dfrac{16}{48}\right)M =\dfrac{9}{48}M =\dfrac{3}{16}M.$$

Multiplying by 2,

$$m=\dfrac{3}{8}M.$$

Therefore the sought ratio is

$$\dfrac{m}{M}=\dfrac{3}{8}=0.375\approx 0.37.$$

Hence, the correct answer is Option D.

We begin by recalling that for any surface with surface mass density $$\sigma(r)$$, an infinitesimal ring of radius $$r$$ and width $$dr$$ has an area $$dA = 2\pi r\,dr$$ (because the circumference is $$2\pi r$$ and the thickness is $$dr$$). The small mass of this ring is therefore $$dm = \sigma(r)\,dA$$.

The given surface mass density for the disc is $$\sigma(r) = k r^{2}$$, where $$k$$ is a constant that we shall determine from the total mass $$M$$ of the disc.

First, we write the expression for the total mass. By definition,

$$M = \int_{0}^{R} dm = \int_{0}^{R} \sigma(r)\,dA.$$

Substituting $$\sigma(r) = k r^{2}$$ and $$dA = 2\pi r\,dr$$, we get

$$M = \int_{0}^{R} k r^{2}\,(2\pi r\,dr) = 2\pi k \int_{0}^{R} r^{3}\,dr.$$

We now integrate $$r^{3}$$:

$$\int r^{3}\,dr = \frac{r^{4}}{4}.$$ Hence

$$M = 2\pi k\left[\frac{r^{4}}{4}\right]_{0}^{R} = 2\pi k\left(\frac{R^{4}}{4}\right) = \frac{\pi k R^{4}}{2}.$$

Solving this relation for the constant $$k$$ gives

$$k = \frac{2M}{\pi R^{4}}.$$

Now we calculate the moment of inertia about the central axis (perpendicular to the plane). The general formula for a surface is

$$I = \int r^{2}\,dm,$$ where again $$dm = \sigma(r)\,dA.$$ Substituting both $$\sigma(r)$$ and $$dA$$, we obtain

$$I = \int_{0}^{R} r^{2}\,\bigl(k r^{2}\bigr)\,(2\pi r\,dr) = 2\pi k \int_{0}^{R} r^{5}\,dr.$$

We now integrate $$r^{5}$$: $$\int r^{5}\,dr = \frac{r^{6}}{6}.$$ Therefore

$$I = 2\pi k\left[\frac{r^{6}}{6}\right]_{0}^{R} = 2\pi k\left(\frac{R^{6}}{6}\right) = \frac{\pi k R^{6}}{3}.$$

Substituting the value of $$k = \dfrac{2M}{\pi R^{4}}$$ that we found earlier, we get

$$I = \frac{\pi}{3}\left(\frac{2M}{\pi R^{4}}\right)R^{6} = \frac{2M R^{2}}{3}.$$

This simplifies neatly to

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

Hence, the correct answer is Option D.

We recall the definition of torque produced by a force about the origin. In vector form, the torque is given by $$\vec \tau = \vec r \times \vec F$$, where $$\vec r$$ is the position vector (whose magnitude is the distance of the particle from the origin) and $$\vec F$$ is the force vector.

For magnitudes alone, the cross-product gives

$$|\vec \tau| = |\vec r|\,|\vec F|\,\sin\theta,$$

where $$\theta$$ is the angle between $$\vec r$$ and $$\vec F$$. This is the relationship we shall use.

Now we substitute the given numerical values. The magnitude of torque is $$|\vec \tau| = 2.5\ \text{N\,m}$$, the magnitude of the force is $$|\vec F| = 1\ \text{N}$$, and the distance from the origin is $$|\vec r| = 5\ \text{m}$$.

So we write

$$2.5 = 5 \times 1 \times \sin\theta.$$

Dividing both sides by 5, we obtain

$$\sin\theta = \frac{2.5}{5}.$$

The fraction simplifies to

$$\sin\theta = 0.5 = \frac{1}{2}.$$

We now look for an angle whose sine is $$\tfrac12$$. The principal angle in the first quadrant with this sine value is

$$\theta = \frac{\pi}{6}\ \text{radians}.$$

(Another mathematical solution is $$\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$, but that corresponds to an obtuse angle of 150°, whereas the usual convention for the angle between two vectors in such problems is the acute angle. Hence we select $$\theta = \frac{\pi}{6}$$.)

Hence, the correct answer is Option A.

Velocity at Point B:

Using conservation of energy between point A and point B:

$$\frac{1}{2}mv_A^2 + mgh = \frac{1}{2}mv_B^2$$

$$v_B = \sqrt{v_A^2 + 2gh}$$

$$v_B = \sqrt{5^2 + 2(10)(10)} = 15 \text{ m s}^{-1}$$

Angular Momentum about O:

The perpendicular distance from point O to the horizontal velocity vector at B is:

$$r_{\perp} = a + h$$ = 10 + 10 = 20 m

$$L_O = m \cdot v_B \cdot r_{\perp}$$

$$L_O = 0.02 \times 15 \times 20$$

$$L_O = 6 \text{ kg m}^2 \text{ s}^{-1}$$

$$\text{For force } \vec{F}_1:$$ $$\text{Position vector: } \vec{r}_1 = 2\hat{i} + 3\hat{j}$$

$$\text{Force vector (acting along z-axis): } \vec{F}_1 = F\hat{k}$$

$$\vec{\tau}_1 = \vec{r}_1 \times \vec{F}_1 = (2\hat{i} + 3\hat{j}) \times (F\hat{k}) = 2F(-\hat{j}) + 3F(\hat{i}) = (3\hat{i} - 2\hat{j})F$$

$$\text{For force } \vec{F}_2:$$ $$\text{From the geometry of the slab, the point of application is on the y-axis at } y = 6\text{ m}:$$ $$\vec{r}_2 = 6\hat{j}$$

$$\vec{F}_2 = -F\sin(30^\circ)\hat{i} - F\cos(30^\circ)\hat{j} = -\frac{F}{2}\hat{i} - \frac{F\sqrt{3}}{2}\hat{j}$$

$$\vec{\tau}_2 = \vec{r}_2 \times \vec{F}_2 = (6\hat{j}) \times \left(-\frac{F}{2}\hat{i} - \frac{F\sqrt{3}}{2}\hat{j}\right) = -3F(\hat{j} \times \hat{i}) - 3\sqrt{3}F(\hat{j} \times \hat{j})$$

$$\text{Using the unit vector properties } \hat{j} \times \hat{i} = -\hat{k} \text{ and } \hat{j} \times \hat{j} = 0:$$ $$\vec{\tau}_2 = 3F\hat{k}$$

$$\text{Total moment about point O:}$$

$$\vec{\tau}_{\text{net}} = \vec{\tau}_1 + \vec{\tau}_2 = (3\hat{i} - 2\hat{j})F + 3F\hat{k} = (3\hat{i} - 2\hat{j} + 3\hat{k})F$$

Let us denote the density of the original solid sphere by $$\rho$$. Because the sphere is uniform, we have

$$\rho=\dfrac{\text{mass}}{\text{volume}}=\dfrac{M}{\dfrac{4}{3}\pi R^{3}}\;.$$

The sphere is cut into two parts.

First part ⇒ uniform disc

The first part has mass $$\dfrac{7M}{8}$$ and is recast into a uniform disc of radius $$2R$$. For a uniform disc about an axis perpendicular to its plane and passing through its centre, the standard formula is

$$I=\dfrac{1}{2}MR^{2}\;.$$

Here the mass is $$\dfrac{7M}{8}$$ and the radius is $$2R$$, so

$$\begin{aligned} I_{1}&=\dfrac{1}{2}\left(\dfrac{7M}{8}\right)(2R)^{2} \\ &=\dfrac{1}{2}\left(\dfrac{7M}{8}\right)\cdot4R^{2} \\ &=\left(\dfrac{7M}{16}\right)4R^{2} \\ &=\dfrac{7M}{4}\,R^{2}. \end{aligned}$$

Second part ⇒ new solid sphere

The second part has mass $$\dfrac{M}{8}$$ and is melted into another uniform solid sphere of (unknown) radius $$r$$. Using the same density $$\rho$$, we write

$$\dfrac{M}{8}=\rho\left(\dfrac{4}{3}\pi r^{3}\right).$$

Substituting $$\rho=\dfrac{M}{\dfrac{4}{3}\pi R^{3}}$$ from above, we get

$$\dfrac{M}{8}=\left[\dfrac{M}{\dfrac{4}{3}\pi R^{3}}\right]\!\left(\dfrac{4}{3}\pi r^{3}\right) =M\left(\dfrac{r^{3}}{R^{3}}\right).$$

Dividing both sides by $$M$$,

$$\dfrac{1}{8}=\dfrac{r^{3}}{R^{3}} \;\;\Longrightarrow\;\; r^{3}=\dfrac{R^{3}}{8} \;\;\Longrightarrow\;\; r=\dfrac{R}{2}.$$

For a uniform solid sphere about any diameter, the standard formula is

$$I=\dfrac{2}{5}MR^{2}.$$

Replacing $$M$$ by $$\dfrac{M}{8}$$ and $$R$$ by the new radius $$r=\dfrac{R}{2}$$, we have

$$\begin{aligned} I_{2}&=\dfrac{2}{5}\left(\dfrac{M}{8}\right)\left(\dfrac{R}{2}\right)^{2} \\ &=\dfrac{2}{5}\left(\dfrac{M}{8}\right)\dfrac{R^{2}}{4} \\ &=\dfrac{2M R^{2}}{5\cdot8\cdot4} \\ &=\dfrac{2M R^{2}}{160} \\ &=\dfrac{M R^{2}}{80}. \end{aligned}$$

Ratio of the two moments of inertia

$$\begin{aligned} \dfrac{I_{1}}{I_{2}} &=\dfrac{\dfrac{7M}{4}R^{2}}{\dfrac{M R^{2}}{80}} \\ &=\dfrac{7M R^{2}}{4}\times\dfrac{80}{M R^{2}} \\ &=7\times\dfrac{80}{4} \\ &=7\times20 \\ &=140. \end{aligned}$$

Hence, the correct answer is Option A.