JEE Rotational Motion Questions

The straight rod is bent until its two ends meet, so the whole length of the rod becomes the circumference of a circle.

Hence, $$\text{circumference}=2\pi R=L$$ $$-(1)$$ which gives $$R=\frac{L}{2\pi}$$ $$-(2)$$

Mass of the ring: linear mass density $$\lambda$$ means $$m=\lambda\times(\text{length})=\lambda L$$ $$-(3)$$

For a thin ring of mass $$m$$ and radius $$R$$:

• Moment of inertia about the axis perpendicular to its plane and passing through the centre is $$I_{z}=mR^{2}$$. • For any diameter (say $$x$$-axis) lying in the plane of the ring, use the perpendicular-axis theorem:   $$I_{x}+I_{y}=I_{z}$$. Because of symmetry, $$I_{x}=I_{y}=I_{d}$$ (moment of inertia about any diameter), so

$$2I_{d}=mR^{2} \;\; \Longrightarrow \;\; I_{d}=\frac{mR^{2}}{2}$$ $$-(4)$$

Substitute $$m$$ from $$(3)$$ and $$R$$ from $$(2)$$ into $$(4)$$:

$$I_{d}=\frac{\lambda L}{2}\left(\frac{L}{2\pi}\right)^{2} =\frac{\lambda L}{2}\cdot\frac{L^{2}}{4\pi^{2}} =\frac{\lambda L^{3}}{8\pi^{2}}$$

Thus, the moment of inertia of the ring about any diameter is $$\boxed{\displaystyle \frac{\lambda L^{3}}{8\pi^{2}}}$$, which corresponds to Option D.

Since the wheel has negligible mass spokes, its mass is concentrated at the rim (treating it as a hoop/thin ring):

$$I = MR^2$$

$$I = 10 \text{ kg} \times (0.1 \text{ m})^2 = 0.1 \text{ kg m}^2$$

The work done by the pull (F) over the unwound distance (s) equals the change in rotational kinetic energy:

$$\text{Work} = \Delta KE_{rot}$$

$$F \cdot s = \frac{1}{2} I \omega^2$$

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

$$\omega = 20 \text{ rad/s}$$

Mass of Removed Part ($$m$$):

$$The\ mass\ of\ a\ uniform\ disc\ is\ proportional\ to\ its\ area\ (\text{Area}=\pi R^2):$$

$$m = \frac{\pi (R/2)^2}{\pi R^2} \times M = \frac{M}{4}$$

Moment of Inertia of the Removed Part ($$I_{rem}$$):

The center of the removed part is at a distance $$d = R/2$$ from the center of the original disc. Using the Parallel Axis Theorem:

$$I_{rem} = I_{cm} + md^2$$

$$I_{rem} = \frac{1}{2} m \left(\frac{R}{2}\right)^2 + m \left(\frac{R}{2}\right)^2 = \frac{3}{2} m \left(\frac{R}{2}\right)^2$$

$$I_{rem} = \frac{3}{2} \left(\frac{M}{4}\right) \frac{R^2}{4} = \frac{3}{32} MR^2$$

Moment of Inertia of the Remaining Part ($$I_{final}$$):

$$I_{final} = I_{original} - I_{rem}$$

$$I_{final} = \frac{1}{2} MR^2 - \frac{3}{32} MR^2$$

$$I_{final} = \frac{16}{32} MR^2 - \frac{3}{32} MR^2 = \frac{13}{32} MR^2$$

The diameter of the ring is given as $$r$$, hence its radius is
$$R = \frac{r}{2}\,.$$

Step 1: Moment of inertia about a diameter (axis in the plane, through centre)
Every mass element of a thin ring is at the same distance $$R$$ from the centre. For an axis along a diameter (say the $$x$$-axis) lying in the plane of the ring, the perpendicular distance of a point $$\bigl(R\cos\theta,\;R\sin\theta\bigr)$$ from that axis is $$R\lvert\sin\theta\rvert$$. Using $$I = \int r_\perp^{\,2}\,dm$$,

$$I_{\text{diameter}} = \int_0^{2\pi} \! \left(R\sin\theta\right)^2 \left(\frac{M}{2\pi}\,d\theta\right) = \frac{MR^2}{2}\,. $$

Step 2: Moment of inertia about a tangential axis in the plane
The required axis is parallel to the diameter axis but touches the ring at one point, so it is shifted by a distance $$R$$ from the centre. Apply the Parallel-Axis Theorem:

$$I_{\text{tangent}} \;=\; I_{\text{diameter}} + M R^2 = \frac{MR^2}{2} + M R^2 = \frac{3}{2}\,M R^2 \,.$$

Step 3: Express in terms of the given diameter
Substitute $$R = \dfrac{r}{2}$$:

$$I_{\text{tangent}} = \frac{3}{2}\,M\left(\frac{r}{2}\right)^2 = \frac{3}{2}\,M\,\frac{r^{2}}{4} = \frac{3}{8}\,M r^{2}\,.$$

Therefore, the moment of inertia of the circular ring about a tangential axis lying in its own plane is $$\frac{3}{8}Mr^{2}$$.

The correct choice is Option B.

The torque $$\vec{\tau}$$ is given by $$\vec{\tau} = \vec{r} \times \vec{F}$$, where $$\vec{r} = \hat{i} + \hat{j} + \hat{k}$$ and $$\vec{F} = 2\hat{i} + \hat{j} + 2\hat{k}$$.

Computing the cross product using the determinant formula:

$$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 1 & 2 \end{vmatrix}$$

$$\hat{i}$$ component: $$(1)(2) - (1)(1) = 2 - 1 = 1$$

$$\hat{j}$$ component: $$-[(1)(2) - (1)(2)] = -(2 - 2) = 0$$

$$\hat{k}$$ component: $$(1)(1) - (1)(2) = 1 - 2 = -1$$

$$\vec{\tau} = \hat{i} - \hat{k}$$

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

Torque (moment of the force) about the chosen origin is defined as
$$\vec{\tau} = \vec{r} \times \vec{F} \quad -(1)$$
where $$\vec{r}$$ is the position vector of the point of application of the force $$\vec{F}$$.

Therefore expression E ( $$\vec{\tau} = \vec{r} \times \vec{F}$$ ) is directly correct.

Using Newton’s second law $$\vec{F} = \dfrac{d\vec{p}}{dt}$$ and substituting in $$(1)$$ gives
$$\vec{\tau} = \vec{r} \times \dfrac{d\vec{p}}{dt} \quad -(2)$$
so expression C is also correct.

The orbital angular momentum of the particle is defined as
$$\vec{L} = \vec{r} \times \vec{p} \quad -(3)$$
Differentiating $$(3)$$ with respect to time (for a fixed origin) yields
$$\dfrac{d\vec{L}}{dt} = \dfrac{d}{dt}(\vec{r} \times \vec{p})$$
By the rotational analogue of Newton’s second law, $$\dfrac{d\vec{L}}{dt} = \vec{\tau}$$, hence
$$\vec{\tau} = \dfrac{d}{dt}(\vec{r} \times \vec{p}) \quad -(4)$$
Thus expression B is correct.

For a rigid body of constant moment of inertia $$I$$ about a fixed axis, the relation
$$\vec{\tau}_{\text{net}} = I \vec{\alpha} \quad -(5)$$
holds, where $$\vec{\alpha}$$ is the angular acceleration. Hence expression D is correct under these usual JEE assumptions.

Check expression A: $$\vec{r} \times \vec{L}$$ has dimensions $$\bigl( \text{m}\bigr)\times \bigl(\text{kg m}^2\!/\text{s}\bigr)=\text{kg m}^3\!/\text{s}$$, whereas torque has dimensions $$\text{kg m}^2\!/\text{s}^2$$. The two are not equal, and there is no general mechanical principle equating them. Hence expression A is incorrect.

Therefore the correct expressions are B, C, D and E.

Option C is the correct choice.

A circular disc of radius 20 cm has a circular hole of radius 5 cm cut such that the edge of the hole touches the edge of the disc. We need the distance of the center of mass of the remaining disc from the origin.

Place the center of the original disc at the origin. Since the edge of the hole touches the edge of the disc, the center of the hole is at distance $$20 - 5 = 15$$ cm from the origin along the x-axis, so the hole center is at $$(15, 0)$$.

Let $$\sigma$$ be the surface mass density. For a uniform disc, the mass of the original disc is $$M = \sigma \pi (20)^2 = 400\pi\sigma$$, the mass of the removed portion is $$m = \sigma \pi (5)^2 = 25\pi\sigma$$, and the mass of the remaining disc is $$M - m = 375\pi\sigma$$.

Using the principle of superposition, we write $$M \cdot 0 = (M-m) \cdot x_{cm} + m \cdot 15$$, which gives $$0 = 375\pi\sigma \cdot x_{cm} + 25\pi\sigma \cdot 15$$ and hence $$x_{cm} = -\frac{25 \times 15}{375} = -\frac{375}{375} = -1$$ cm.

The distance from the origin is $$|x_{cm}| = 1$$ cm, so the correct answer is Option 3: 1.0 cm.

Consider an inclined plane of length $$s$$ and angle of inclination $$\theta$$. Both spheres are released from rest and roll down without slipping.

For any rigid body rolling without slipping, the linear acceleration of the centre of mass is derived from Newton’s 2nd law plus the rotational equation:

$$m a = m g \sin\theta - f$$ $$-(1)$$
$$f R = I \alpha$$ and $$\alpha = \frac{a}{R}$$ $$-(2)$$

Eliminating the static friction $$f$$ using $$(2)$$ and substituting into $$(1)$$ gives the standard result:

$$a = \frac{g \sin\theta}{1 + \dfrac{I}{m R^{2}}}$$ $$-(3)$$

Write $$I = m k^{2}$$, where $$k$$ is the radius of gyration. Then

$$a = \frac{g \sin\theta}{1 + \dfrac{k^{2}}{R^{2}}}$$ $$-(4)$$

Case 1: Solid sphere (uniform density)

Moment of inertia: $$I = \frac{2}{5} m R^{2} \Rightarrow k^{2} = \frac{2}{5} R^{2}$$.

Put this in $$(4)$$:

$$a_{1} = \frac{g \sin\theta}{1 + \dfrac{2}{5}} = \frac{g \sin\theta}{\dfrac{7}{5}} = \frac{5}{7} g \sin\theta$$ $$-(5)$$

Case 2: Hollow sphere (thin spherical shell)

Moment of inertia: $$I = \frac{2}{3} m R^{2} \Rightarrow k^{2} = \frac{2}{3} R^{2}$$.

Put this in $$(4)$$:

$$a_{2} = \frac{g \sin\theta}{1 + \dfrac{2}{3}} = \frac{g \sin\theta}{\dfrac{5}{3}} = \frac{3}{5} g \sin\theta$$ $$-(6)$$

Numerical comparison:

$$a_{1} = \frac{5}{7} g \sin\theta \approx 0.714 \, g \sin\theta$$
$$a_{2} = \frac{3}{5} g \sin\theta = 0.6 \, g \sin\theta$$.

Thus $$a_{1} \gt a_{2}$$. The solid sphere gains more acceleration.

The time taken to travel the same distance $$s$$ with constant acceleration is

$$t = \sqrt{\frac{2 s}{a}}$$ $$-(7)$$

Therefore

$$t_{1} = \sqrt{\frac{2 s}{a_{1}}}, \qquad t_{2} = \sqrt{\frac{2 s}{a_{2}}}$$.

Since $$a_{1} \gt a_{2}$$, the denominator of $$t_{1}$$ is larger, making $$t_{1}$$ smaller:

$$t_{1} \lt t_{2}$$.

Hence the correct relation is $$t_{1} \lt t_{2}$$, corresponding to Option C.

The force F acts at a distance of 2R from point P, and for a solid sphere, the moment of inertia about the contact point (using Parallel Axis Theorem) is:

$$I_P = I_{cm} + MR^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$$

For rolling without slipping, we can take torque about the point of contact (P) on the ground:

$$\tau_P = I_P \alpha$$

$$F(2R) = \left( \frac{7}{5}MR^2 \right) \frac{a}{R}$$

$$a = \frac{10F}{7M}$$

$$a = \frac{10 \times 49}{7 \times 20}$$

$$a = \frac{490}{140} = 3.5 \text{ m/s}^2$$

We apply the condition of rotational equilibrium (balance of torques) about the pivot at the 40 cm mark.
Torque $$\tau$$ due to a force is given by $$\tau = F \times d$$, where $$F$$ is the weight and $$d$$ is the perpendicular distance from the pivot.

Identify the torques:

Rod’s own weight: mass = 250 g acts at the centre (50 cm).
Lever arm = $$(50 - 40) = 10\text{ cm}$$.
This produces a clockwise torque of magnitude $$250 \times 10\text{ (g·cm)}$$.

Suspended mass 400 g at 10 cm: lever arm = $$(40 - 10) = 30\text{ cm}$$.
This produces an anticlockwise torque of magnitude $$400 \times 30\text{ (g·cm)}$$.

Unknown mass $$m$$ at 90 cm: lever arm = $$(90 - 40) = 50\text{ cm}$$.
This produces a clockwise torque of magnitude $$m \times 50\text{ (g·cm)}$$.

For equilibrium, sum of anticlockwise torques = sum of clockwise torques:

$$400 \times 30 \;=\; 250 \times 10 \;+\; m \times 50 \quad\,-(1)$$

Compute the numerical values in $$(1)$$:

$$12000 \;=\; 2500 \;+\; 50m \quad\,-(2)$$

Rearrange $$(2)$$ to solve for $$m$$:

$$50m \;=\; 12000 - 2500 = 9500$$
$$m = \frac{9500}{50} = 190$$

Therefore, the required mass to be suspended at the 90 cm mark is 190 g. This corresponds to Option A.

For a wheel that rolls without slipping, the centre of mass (CM) moves with linear speed $$v_{CM}$$ and the wheel simultaneously rotates with angular speed $$\omega$$ such that $$v_{CM}=R\omega$$, where $$R$$ is the radius.

The linear velocity of any point on the rim is the vector sum of the translational velocity $$v_{CM}$$ (same for every point, directed horizontally) and the tangential velocity due to rotation $$\omega R$$ (perpendicular to the radius through that point).

The highest point on the rim has its rotational velocity in the same horizontal direction as $$v_{CM}$$, so the speeds add: $$v_{\text{top}} = v_{CM}+R\omega = v_{CM}+v_{CM}=2v_{CM}$$.

Given $$v_{\text{top}} = 8 \text{ m/s}$$, we get
$$2v_{CM}=8 \implies v_{CM}=4 \text{ m/s}$$.

Consider a point on the rim that lies at the same vertical level as the centre (rightmost point if the wheel rolls to the right).
 • Translational velocity: $$4 \text{ m/s}$$ to the right.
 • Rotational (tangential) velocity: $$\omega R = v_{CM}=4 \text{ m/s}$$ vertically downward.

These two perpendicular components give the resultant speed
$$v_{\text{side}}=\sqrt{(4)^2+(4)^2}=4\sqrt{2} \text{ m/s}$$.

Therefore, the speed of the particle on the rim at the same level as the centre is $$4\sqrt{2} \text{ m/s}$$. Hence, Option A is correct.

For the lamina to remain stationary, the net torque about the pivot O must be zero:

$$\sum \tau_O = 0$$

$$\tau=\text{Force}\times\text{Perpendicular distance}$$

$$At\ Point\ A:The\ vertical\ 10\text{ N}\ (down)\ creates\ a\ clockwise\ torque:\ \tau_1=-10\cdot L$$

$$At\ Point\ B:The\ vertical\ 10\text{ N }(up)\ creates\ a\ counter\ clockwise\ torque:\ \tau_2=+10\cdot L$$

$$The\ horizontal\ 10\text{ N}\ (right)\ creates\ a\ clockwise\ torque:\ \tau_3=-10\cdot L$$

$$At\ Point\ C:The\ horizontal\ force\ F\ (left)\ creates\ a\ counter\ clockwise\ torque:\ \tau_4=+F\cdot L$$

$$The\ vertical\ force\ passes\ through\ O,\ so\ its\ torque\ is\ 0.$$

$$\sum \tau_O = (-10 \cdot L) + (10 \cdot L) + (-10 \cdot L) + (F \cdot L) = 0$$

$$-10L + FL = 0$$

$$F = 10 \text{ N}$$

$$For\ an\ object\ rolling\ without\ slipping\ down\ an\ inclined\ plane\ with\ inclination\ \theta,\ $$

$$the\ linear\ acceleration\ (a)\ of\ the\ center\ of\ mass\ is:\ a=\frac{g\sin\theta}{1+\frac{I}{mr^2}}$$

$$For\ a\ uniform\ solid\ cylinder,\ the\ moment\ of\ inertia\ is\ I=\frac{1}{2}mr^2$$

$$Substituting\ this\ and\ \theta=45^{\circ}\ into\ the\ formula:$$

$$a = \frac{g \sin 45^\circ}{1 + \frac{1}{2}mr^2 / mr^2}$$

$$a = \frac{g \cdot \frac{1}{\sqrt{2}}}{1 + \frac{1}{2}}$$

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

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

The given rod has mass $$M$$ and length $$L$$. For a slender rod, the moment of inertia about an axis passing through its centre and perpendicular to its length is

$$I = \frac{1}{12}\,M\,L^{2}$$

The question denotes this value by $$\alpha$$, so write 

$$\alpha = \frac{1}{12}\,M\,L^{2} \; \; -(1)$$

Next, cut the rod into two equal parts. Each part therefore has

length $$=\frac{L}{2}$$
mass $$=\frac{M}{2}$$

These two half-rods are joined at right angles through their centres to form a symmetric cross (a ‘+’ shape). The required axis passes through the common centre and is perpendicular to the plane of the cross.

For one half-rod of length $$\tfrac{L}{2}$$, the moment of inertia about this axis is again given by the same formula $$I=\tfrac{1}{12}ML^{2}$$ with $$M \rightarrow \frac{M}{2}$$ and $$L \rightarrow \frac{L}{2}$$:

$$I_{\text{one half}} = \frac{1}{12}\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^{2} = \frac{1}{12}\,\frac{M}{2}\,\frac{L^{2}}{4} = \frac{M\,L^{2}}{96}$$

The cross consists of two such identical half-rods, so total moment of inertia is

$$I_{\text{cross}} = 2 \times \frac{M\,L^{2}}{96} = \frac{M\,L^{2}}{48}$$

From $$(1)$$ we have $$M\,L^{2} = 12\,\alpha$$. Substituting this into the above result:

$$I_{\text{cross}} = \frac{12\,\alpha}{48} = \frac{\alpha}{4}$$

Hence, the moment of inertia of the cross about the stated axis is $$\alpha/4$$.

Answer: Option B $$(\alpha/4)$$

Given the angular position:

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

Differentiating with respect to time to find the angular velocity:

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

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

$$\alpha = \frac{d\omega}{dt} = 10 \text{ rad/s}^2$$

$$\tau = I\alpha$$

$$I=\frac{1}{2}MR^2$$ for a circular disk rotating about an axis perpendicular to its plane through the center

$$\tau = \left( \frac{1}{2}MR^2 \right) \times 10 = 5MR^2$$ 

Power Delivered ($$P$$):

$$P = \tau \cdot \omega$$

$$P = (5MR^2) \times 12$$
$$P = 60MR^2$$

Solid sphere rolling: KE_linear/KE_rotational.

KE_linear = ½mv². KE_rot = ½Iω² = ½(2/5)mr²(v/r)² = (1/5)mv².

Ratio = (½mv²)/((1/5)mv²) = 5/2.

The correct answer is Option 3: 5/2.

$$The\ mass\ per\ unit\ area\ is\ \sigma=\frac{\sigma_0x}{ab}.\ We\ integrate\ over\ the\ area\ ofthe\ plate\ (x\in[0,a],\ y\in[0,b]):$$

$$M = \int_{0}^{a} \int_{0}^{b} \sigma \, dy \, dx = \int_{0}^{a} \int_{0}^{b} \frac{\sigma_0 x}{ab} \, dy \, dx$$

$$M = \frac{\sigma_0}{ab} \left[ \frac{x^2}{2} \right]_0^a \left[ y \right]_0^b = \frac{\sigma_0}{ab} \left( \frac{a^2}{2} \right) (b) = \frac{\sigma_0 a}{2}$$

X-coordinate of Center of Mass:

$$x_{cm} = \frac{1}{M} \int \int x \sigma \, dA$$

$$x_{cm} = \frac{1}{M} \int_{0}^{a} \int_{0}^{b} x \left( \frac{\sigma_0 x}{ab} \right) \, dy \, dx = \frac{1}{M} \frac{\sigma_0}{ab} \int_{0}^{a} x^2 \, dx \int_{0}^{b} dy$$

$$x_{cm} = \frac{2}{\sigma_0 a} \cdot \frac{\sigma_0}{ab} \left( \frac{a^3}{3} \right) (b) = \frac{2}{\sigma_0 a} \cdot \frac{\sigma_0 a^2}{3} = \frac{2}{3}a$$

Y-coordinate of Center of Mass:

$$Since\ the\ density\ \sigma\ depends\ only\ on\ x$$

the mass is distributed uniformly in the y-direction for any given x. Therefore: $$y_{cm}=\frac{b}{2}$$

The center of mass is at: $$(x_{cm},y_{cm})=\left(\frac{2}{3}a,\frac{b}{2}\right)$$

An object of mass $$m$$ is projected from the origin in the vertical xy-plane at an angle of $$45°$$ with initial speed $$v_0$$. To find its angular momentum about the origin at its maximum height, we first determine the time and position at that instant.

The time to reach the maximum height occurs when the vertical component of velocity becomes zero, namely at $$t = \frac{v_0 \sin 45°}{g} = \frac{v_0}{\sqrt{2}g}\,. $$ During this interval the horizontal component of motion remains uniform, so the x-coordinate is $$x = v_0 \cos 45° \times t = \frac{v_0}{\sqrt{2}} \times \frac{v_0}{\sqrt{2}g} = \frac{v_0^2}{2g}\,. $$ The maximum height itself is given by $$y = H = \frac{v_0^2 \sin^2 45°}{2g} = \frac{v_0^2}{4g}\,. $$ Hence the position vector at maximum height is $$\vec{r} = \frac{v_0^2}{2g}\,\hat{i} + \frac{v_0^2}{4g}\,\hat{j}\,. $$

At this instant the vertical velocity vanishes and only the horizontal component remains, so $$\vec{v} = v_0 \cos 45°\,\hat{i} = \frac{v_0}{\sqrt{2}}\,\hat{i}\,. $$

The angular momentum about the origin is given by $$\vec{L} = m\,\bigl(\vec{r}\times\vec{v}\bigr)\,. $$ Evaluating the cross product, $$\vec{r}\times\vec{v} = \Bigl(\frac{v_0^2}{2g}\hat{i} + \frac{v_0^2}{4g}\hat{j}\Bigr) \times \Bigl(\frac{v_0}{\sqrt{2}}\hat{i}\Bigr)\,. $$ Since $$\hat{i}\times\hat{i}=0$$ and $$\hat{j}\times\hat{i}=-\hat{k}$$, it follows that $$\vec{r}\times\vec{v} = \frac{v_0^2}{2g}\frac{v_0}{\sqrt{2}}\bigl(\hat{i}\times\hat{i}\bigr) + \frac{v_0^2}{4g}\frac{v_0}{\sqrt{2}}\bigl(\hat{j}\times\hat{i}\bigr) = -\frac{v_0^3}{4\sqrt{2}g}\,\hat{k}\,. $$ Therefore $$\vec{L} = m\bigl(\vec{r}\times\vec{v}\bigr) = -\frac{m\,v_0^3}{4\sqrt{2}g}\,\hat{k}\,. $$

The magnitude of the angular momentum is $$\frac{m\,v_0^3}{4\sqrt{2}g}$$ directed along the negative z-axis. The correct answer is Option 3: $$\frac{m\,v_0^3}{4\sqrt{2}g}$$ along the negative z-axis.

For a solid sphere rolling without slipping down an inclined plane of length $$L$$, energy conservation relates the loss of gravitational potential energy to the sum of translational and rotational kinetic energies. The moment of inertia of a solid sphere is $$I = \frac{2}{5}mr^2$$.

Specifically, if the sphere descends a vertical height $$h$$, then $$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2\left(1 + \frac{I}{mr^2}\right) = \frac{1}{2}mv^2\left(1 + \frac{2}{5}\right) = \frac{7}{10}mv^2,$$ leading to $$v^2 = \frac{10gh}{7} = \frac{10g \cdot L\sin\theta}{7}.$$

For an incline at $$30°$$, this gives $$v_1^2 = \frac{10gL\sin 30°}{7} = \frac{10gL}{7} \cdot \frac{1}{2},$$ and for $$45°$$, $$v_2^2 = \frac{10gL\sin 45°}{7} = \frac{10gL}{7} \cdot \frac{1}{\sqrt{2}}.$$

Hence the ratio of the squared speeds is $$\frac{v_1^2}{v_2^2} = \frac{\sin 30°}{\sin 45°} = \frac{1/2}{1/\sqrt{2}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}},$$ so that $$v_1^2 : v_2^2 = 1 : \sqrt{2}$$. Therefore, the correct answer is Option 1: $$1:\sqrt{2}$$.

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

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

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

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

For a solid sphere about a diameter, $$I_{sphere}=\tfrac{3}{5}mR^{2}$$, hence $$\dfrac{I}{mR^{2}}=\tfrac{3}{5}$$.

$$v_{sphere}=\sqrt{\frac{2gh}{1+\tfrac{3}{5}}} =\sqrt{\frac{2gh}{\tfrac{8}{5}}} =\sqrt{\frac{5gh}{4}}$$

Required ratio:

$$\frac{v_{ring}}{v_{sphere}} =\sqrt{\frac{gh}{\tfrac{5gh}{4}}} =\sqrt{\frac{4}{5}} =\sqrt{\frac{x}{5}}$$

Hence $$x=4$$.

Thus, the ratio of their velocities at the bottom of the plane is $$\sqrt{\dfrac{4}{5}}$$, giving $$x = 4$$.

The torque produced by the force acting tangentially at the rim is:

$$\tau = F \cdot R$$

$$\tau = 10 \text{ N} \times 0.2 \text{ m} = 2 \text{ N m}$$

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

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

$$I = 1 \text{ kg m}^2$$

Let the initial mass of the solid sphere be $$M$$ and its initial radius be $$R$$. Moment of inertia of a solid sphere about any diameter is

$$I=\frac{2}{5}MR^{2}$$ $$-(1)$$

When the sphere loses mass it keeps its shape, so it remains a uniform solid sphere of a smaller radius. If the radius becomes $$\frac{R}{2}$$, the new volume is

$$V'=\frac{4}{3}\pi\left(\frac{R}{2}\right)^{3}=\frac{1}{8}\left(\frac{4}{3}\pi R^{3}\right)$$

With uniform density, mass is proportional to volume, hence

$$M'=\frac{1}{8}M$$ $$-(2)$$

The new moment of inertia about the same diameter is

$$I'=\frac{2}{5}M'\left(\frac{R}{2}\right)^{2} =\frac{2}{5}\left(\frac{M}{8}\right)\left(\frac{R^{2}}{4}\right) =\frac{2}{5}MR^{2}\left(\frac{1}{32}\right) =\frac{I}{32}$$ $$-(3)$$

The mass is lost symmetrically, so no external torque acts on the sphere. Therefore the angular momentum about the diameter remains constant:

$$I\omega_{1}=I'\omega_{2}$$ $$-(4)$$

Substituting $$I'=\frac{I}{32}$$ from $$(3)$$ into $$(4)$$, we get

$$I\omega_{1}=\frac{I}{32}\,\omega_{2} \;\;\Longrightarrow\;\; \omega_{2}=32\,\omega_{1}$$

Comparing with $$\omega_{2}=x\omega_{1}$$, we find

$$x=32$$

Hence, the angular velocity of the sphere when its radius becomes $$R/2$$ is $$32\omega_{1}$$ and the required value of $$x$$ is 32.

We need to find the magnitude of torque in the z-direction when a force $$\vec{F} = \hat{i} - \hat{j} + \hat{k}$$ acts on a particle at position $$\vec{r} = \hat{i} + \hat{j} + \hat{k}$$ (coordinates (1, 1, 1) m).

Recall that torque is given by $$\vec{\tau} = \vec{r} \times \vec{F}$$.

Substituting the given vectors into the determinant,
$$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix}$$
$$= \hat{i}(1 \cdot 1 - 1 \cdot (-1)) - \hat{j}(1 \cdot 1 - 1 \cdot 1) + \hat{k}(1 \cdot (-1) - 1 \cdot 1)$$
$$= \hat{i}(1 + 1) - \hat{j}(1 - 1) + \hat{k}(-1 - 1)$$
$$= 2\hat{i} - 0\hat{j} - 2\hat{k}$$

The z-component of torque is $$\tau_z = -2$$, so the magnitude of torque in the z-direction is $$|\tau_z| = 2$$.

The answer is 2.

We need to find the magnitude of angular momentum of particle A with respect to the position of particle B at t = 1s.

$$\vec{r_A} = (\alpha_1 t^2 \hat{i} + \alpha_2 t \hat{j} + \alpha_3 t \hat{k})$$ m

$$\vec{r_B} = (\beta_1 t \hat{i} + \beta_2 t^2 \hat{j} + \beta_3 t \hat{k})$$ m

$$\alpha_1 = 1, \alpha_2 = 3n, \alpha_3 = 2, \beta_1 = 2, \beta_2 = -1, \beta_3 = 4p$$

$$\vec{V_A} = \frac{d\vec{r_A}}{dt} = (2\alpha_1 t \hat{i} + \alpha_2 \hat{j} + \alpha_3 \hat{k})$$

At t = 1: $$\vec{V_A} = (2 \hat{i} + 3n \hat{j} + 2 \hat{k})$$ m/s

$$\vec{V_B} = \frac{d\vec{r_B}}{dt} = (\beta_1 \hat{i} + 2\beta_2 t \hat{j} + \beta_3 \hat{k})$$

At t = 1: $$\vec{V_B} = (2 \hat{i} - 2 \hat{j} + 4p \hat{k})$$ m/s

$$4 + 9n^2 + 4 = 4 + 4 + 16p^2$$

$$9n^2 + 4 = 4 + 16p^2$$

$$9n^2 = 16p^2$$ ... (i)

$$\vec{V_A} \cdot \vec{V_B} = 0$$

$$2(2) + 3n(-2) + 2(4p) = 0$$

$$4 - 6n + 8p = 0$$

$$6n - 8p = 4$$

$$3n - 4p = 2$$ ... (ii)

From (i): $$3n = 4p$$ (taking positive root, but let's check both)

If $$3n = 4p$$, substituting in (ii): $$4p - 4p = 2 \Rightarrow 0 = 2$$ (contradiction)

If $$3n = -4p$$, substituting in (ii): $$-4p - 4p = 2 \Rightarrow -8p = 2 \Rightarrow p = -1/4$$

Then $$3n = -4(-1/4) = 1 \Rightarrow n = 1/3$$

$$\vec{r_A}(1) = (1 \hat{i} + 3(1/3) \hat{j} + 2 \hat{k}) = (1 \hat{i} + 1 \hat{j} + 2 \hat{k})$$ m

$$\vec{r_B}(1) = (2 \hat{i} + (-1) \hat{j} + 4(-1/4) \hat{k}) = (2 \hat{i} - 1 \hat{j} - 1 \hat{k})$$ m

$$\vec{V_A}(1) = (2 \hat{i} + 1 \hat{j} + 2 \hat{k})$$ m/s

$$\vec{r} = \vec{r_A} - \vec{r_B} = (1-2)\hat{i} + (1+1)\hat{j} + (2+1)\hat{k} = (-1\hat{i} + 2\hat{j} + 3\hat{k})$$ m

$$\vec{L} = m(\vec{r} \times \vec{V_A}) = 1 \times \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 3 \\ 2 & 1 & 2 \end{vmatrix}$$

$$= \hat{i}(2 \times 2 - 3 \times 1) - \hat{j}((-1)(2) - 3(2)) + \hat{k}((-1)(1) - 2(2))$$

$$= \hat{i}(4-3) - \hat{j}(-2-6) + \hat{k}(-1-4)$$

$$= \hat{i}(1) - \hat{j}(-8) + \hat{k}(-5)$$

$$= (1\hat{i} + 8\hat{j} - 5\hat{k})$$

$$|\vec{L}| = \sqrt{1 + 64 + 25} = \sqrt{90}$$

Since $$|\vec{L}| = \sqrt{L}$$, we get $$L = 90$$.

The answer is 90.

The angular velocity of a rotating body is usually expressed in $$\text{rad s}^{-1}$$.
Given speeds in revolutions per minute (rpm) must first be converted.

Initial speed: $$1800\;\text{rpm} = 1800 \times \frac{2\pi\;\text{rad}}{60\;\text{s}} = 60\pi\;\text{rad s}^{-1}$$.

Final speed: $$2100\;\text{rpm} = 2100 \times \frac{2\pi\;\text{rad}}{60\;\text{s}} = 70\pi\;\text{rad s}^{-1}$$.

The change in angular velocity is therefore
$$\Delta\omega = \omega_f - \omega_i = 70\pi - 60\pi = 10\pi\;\text{rad s}^{-1}$$.

Torque and angular momentum are related through the angular impulse equation:
$$\tau\,t = \Delta L = I\,\Delta\omega$$ $$-(1)$$,
where $$I$$ is the moment of inertia about the rotation axis.

Given values: $$\tau = 25\pi\;\text{N m}$$ and $$t = 40\;\text{s}$$.
Hence the angular impulse is
$$\tau t = 25\pi \times 40 = 1000\pi\;\text{kg m}^2\text{s}^{-1}$$.

Substituting into $$(1)$$ gives
$$I \times 10\pi = 1000\pi$$.

Cancel $$\pi$$ and solve for $$I$$:
$$I = \frac{1000}{10} = 100\;\text{kg m}^2$$.

The axis of rotation is a diameter lying in the plane of the thin solid disk.
For a uniform disk about a diameter, the moment of inertia is
$$I = \frac{1}{4} M R^2$$ $$-(2)$$,
where $$M$$ is the mass and $$R$$ the radius.

With $$M = 1\;\text{kg}$$, substitute $$I = 100\;\text{kg m}^2$$ into $$(2)$$:
$$100 = \frac{1}{4} \times 1 \times R^2 \quad\Longrightarrow\quad R^2 = 400$$.

Thus $$R = 20\;\text{m}$$.
The diameter is twice the radius:

Diameter $$= 2R = 2 \times 20 = 40\;\text{m}$$.

Final Answer: 40 m

Two iron solid discs have radii $$R_1$$ and $$R_2$$ with $$R_2 = 2R_1$$. We first determine the ratio $$I_1/I_2$$.

Next, we recall that the moment of inertia of a solid disc is given by
$$I = \frac{1}{2}MR^2$$

Since both discs are made of iron (same density $$\rho$$) and have negligible thickness $$t$$, their mass can be expressed as
$$M = \rho \cdot \pi R^2 \cdot t$$ so $$M \propto R^2$$.

Substituting into the inertia formula gives
$$I = \frac{1}{2}MR^2 = \frac{1}{2}(\rho \pi R^2 t)R^2 = \frac{\rho \pi t}{2} R^4$$ and hence $$I \propto R^4$$.

It follows that
$$\frac{I_1}{I_2} = \frac{R_1^4}{R_2^4} = \frac{R_1^4}{(2R_1)^4} = \frac{R_1^4}{16R_1^4} = \frac{1}{16}\,. $$

Therefore, $$I_1/I_2 = 1/x$$ where $$x = \boxed{16}$$.

To compare the moments of inertia of a disc, ring, and sphere (all of mass $$M$$ and radius $$R$$) rotating about a diameter, we first recall the standard formulas for each.

For a ring about its diameter we have $$I_{\text{ring}} = \frac{1}{2}MR^2$$, for a disc about its diameter $$I_{\text{disc}} = \frac{1}{4}MR^2$$, and for a sphere about its diameter $$I_{\text{sphere}} = \frac{2}{5}MR^2$$.

Next, we check the condition given in the problem by computing the ratio

$$ \frac{I_{\text{disc}}}{I_{\text{ring}}} = \frac{MR^2/4}{MR^2/2} = \frac{1}{2} $$.

Since the problem states that $$I_{\text{disc}} = 2.5 \times I_{\text{ring}}$$, this standard result does not hold unless the disc and ring have different radii. We are told, however, that the sphere has the same radius as the disc.

Let us denote the ring’s radius by $$r$$ and the disc’s (and sphere’s) radius by $$R$$. Imposing $$I_{\text{disc}} = 2.5 \times I_{\text{ring}}$$ gives

$$ \frac{1}{4}MR^2 = 2.5 \times \frac{1}{2}Mr^2 = \frac{5}{4}Mr^2 $$.

From this we find

$$ R^2 = 5r^2 $$.

Since the sphere has the same radius as the disc, its moment of inertia is

$$ I_{\text{sphere}} = \frac{2}{5}MR^2 = \frac{2}{5}M(5r^2) = 2Mr^2 $$.

Consequently, the ratio of the sphere’s moment of inertia to the ring’s is

$$ n = \frac{I_{\text{sphere}}}{I_{\text{ring}}} = \frac{2Mr^2}{\frac{1}{2}Mr^2} = 4 $$.

Hence the answer is $$n = 4$$.

If two circular loops are concentric and lie in the same plane:

Magnetic field at centre due to loop A:
$$B=\frac{(μ₀I)}{2b}$$

Flux through loop B:
$$Φ=B\times areaofB$$
$$=\frac{μ₀I}{2b}\times(\pi a^2)$$

So mutual inductance:
$$M=\frac{Φ}{I}$$
$$=\frac{(μ₀\pi a^2)}{2b}$$

Two spheres: each mass 2kg, radius 0.5m, separation 1.5m (center to center), so each is 0.75m from middle.

$$I=2\left[\frac{2}{5}mr^2+md^2\right]=2\left[\frac{2}{5}(2)(0.25)+(2)(0.5625)\right]=2[0.2+1.125]=2(1.325)=2.65=\frac{53}{20}$$ kg m².

So $$x=53$$.

The answer is $$\boxed{53}$$.

We need to find the angular momentum of a 5 kg body moving with speed $$3\sqrt{2}$$ m/s along the line $$y = x + 4$$ about the origin.

The formula for angular momentum is $$L = mvd$$, where $$d$$ is the perpendicular distance from the point (origin) to the line of motion.

Rewriting the line $$y = x + 4$$ in standard form gives $$x - y + 4 = 0$$. Hence the perpendicular distance from $$(0, 0)$$ to this line is $$d = \frac{|0 - 0 + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$$ m.

Substituting into the angular momentum formula yields $$L = mvd = 5 \times 3\sqrt{2} \times 2\sqrt{2} = 5 \times 3 \times 2 \times (\sqrt{2} \times \sqrt{2}) = 5 \times 6 \times 2 = 60$$ kg m$$^2$$/s.

The answer is 60 kg m$$^2$$ s$$^{-1}$$.

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):

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

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

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

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

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

For a body rolling down an inclined plane without slipping, the linear acceleration ($$a$$) of the COM is given by:

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

For a solid cylinder, the moment of inertia about its central axis is $$I = \frac{1}{2}MR^2$$.

$$a = \frac{g \sin 60^\circ}{1 + \frac{1}{2}} = \frac{g \sin 60^\circ}{3/2} = \frac{2}{3} g \sin 60^\circ$$

$$a = \frac{2}{3} \times 10 \times \frac{\sqrt{3}}{2} = \frac{10\sqrt{3}}{3} = \frac{10}{\sqrt{3}} \text{ m s}^{-2}$$

$$x = 10$$

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.

For a body rolling down an incline from rest, the total kinetic energy ($$KE$$) at the bottom is equal to the initial gravitational potential energy.

Energy Conservation: Since both bodies descend the same vertical height $$h$$:

$$KE_{\text{ring}} = M_{\text{ring}}gh$$

$$KE_{\text{sphere}} = M_{\text{sphere}}gh$$

Assuming identical masses for comparison:

$$KE_{\text{ring}} = KE_{\text{sphere}}$$

$$\frac{KE_{\text{ring}}}{KE_{\text{sphere}}} = 1$$

Given the ratio is $$\frac{7}{x}$$:

$$\frac{7}{x} = 1 \implies x = 7$$

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 $$x = \boxed{100}$$.

A uniform rod AB of mass 2 kg and length 30 cm (0.3 m) is initially at rest on a smooth horizontal surface. When an impulse of 0.2 N s is applied horizontally at end B, the rod acquires both linear and angular motion, and we seek the time required for it to rotate by 90° (π/2 radians), expressed in the form π/x seconds.

The impulse-momentum theorem gives the change in linear momentum of the rod’s center of mass according to $$J = M \cdot v_{\text{cm}}$$, where $$J = 0.2$$ N s and $$M = 2$$ kg. Substituting the values, $$0.2 = 2 \cdot v_{\text{cm}}$$, so $$v_{\text{cm}} = \frac{0.2}{2} = 0.1\ \text{m/s}$$. Thus the center of mass moves at 0.1 m/s in the direction of the impulse.

The same impulse also imparts angular momentum about the center of mass. The perpendicular distance from the center of mass to B is $$L/2 = 0.15$$ m, so the angular impulse is $$J \cdot \frac{L}{2} = 0.2 \cdot 0.15 = 0.03\ \text{N s m}$$. The moment of inertia of a uniform rod about its center of mass is $$I = \frac{M L^2}{12} = \frac{2 \cdot (0.3)^2}{12} = \frac{2 \cdot 0.09}{12} = \frac{0.18}{12} = 0.015\ \text{kg m}^2\,. $$ Equating angular impulse to the change in angular momentum via $$J \cdot \frac{L}{2} = I \cdot \omega$$ gives $$0.03 = 0.015 \cdot \omega$$ and hence $$\omega = \frac{0.03}{0.015} = 2\ \text{rad/s}\,.$$

Since no external torques act after the impulse, the rotation proceeds at constant angular velocity $$\omega = 2\ \text{rad/s}$$. The angle rotated satisfies $$\theta = \omega \cdot t$$, so $$\frac{\pi}{2} = 2 \cdot t$$ and thus $$t = \frac{\pi}{4}\ \text{seconds}\,, $$ giving $$x = 4$$.

The translation of the center of mass does not affect the rotation angle, as the angular motion is independent; hence the time taken for the rod to rotate by 90° is $$\pi/4$$ seconds.

Since the plate is uniform, we can find the COM by treating the object as a large rectangle with a smaller section removed from its top.

We consider the full $$3 \times 2$$ rectangle and subtract the $$1 \times 1$$ empty square region in the upper middle.

Total Rectangle ($$A_1$$): The area is $$3 \times 2 = 6$$, with its geometric center at $$(1.5, 1)$$.

Removed Square ($$A_2$$): The area is $$1 \times 1 = 1$$, with its geometric center at $$(1.5, 1.5)$$.

Coordinates of the Center of Mass

Because the figure is symmetric about the line $$x = 1.5$$, the $$x$$-coordinate is:

$$x_{cm} = \frac{A_1x_1 - A_2x_2}{A_1 - A_2} = \frac{6(1.5) - 1(1.5)}{5} = 1.5$$

The $$y$$-coordinate is determined by the distribution of the remaining area along the vertical axis:

$$y_{cm} = \frac{A_1y_1 - A_2y_2}{A_1 - A_2} = \frac{6(1) - 1(1.5)}{6 - 1} = \frac{4.5}{5} = 0.9$$

Taking the ratio of the $$x$$ and $$y$$ coordinates:

$$\text{Ratio} = \frac{x_{cm}}{y_{cm}} = \frac{1.5}{0.9}$$

$$\frac{15}{9} = \frac{n}{9}$$

The value of $$n$$ is 15.

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:

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

Four particles each of mass 1 kg at corners of a square with side 2 m. The axis passes through one vertex, perpendicular to the plane.

Taking the vertex through which the axis passes as origin, the distances of the four particles from the axis are:

- Particle at the vertex: $$r_1 = 0$$

- Two adjacent particles: $$r_2 = r_3 = 2$$ m (side length)

- Diagonally opposite particle: $$r_4 = 2\sqrt{2}$$ m (diagonal)

$$I = m(r_1^2 + r_2^2 + r_3^2 + r_4^2) = 1(0 + 4 + 4 + 8) = 16 \text{ kg m}^2$$

The answer is $$\boxed{16}$$ kg m$$^2$$.

We need to find how far $$m_2$$ must move to keep the centre of mass at its original position. The position of the centre of mass is $$x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$, and for $$x_{\text{cm}}$$ to remain unchanged the net displacement must satisfy $$m_1 \Delta x_1 + m_2 \Delta x_2 = 0$$.

Here $$m_1 = 3$$ kg moves towards the centre of mass by $$\Delta x_1 = 2$$ cm, so we have $$3 \times 2 = 2 \times d_2$$, giving $$d_2 = \frac{6}{2} = 3 \text{ cm}$$. Both particles move towards the CM, so both displacements reduce the distance to the CM.

The correct answer is 3 cm.

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.

The electric field ($$E$$) due to an infinitely long thin charged sheet with surface charge density $$\sigma$$ is  $$E = \frac{\sigma}{2\epsilon_0}$$.

Field points away from a positive charge ($$+\sigma$$)

Field points towards a negative charge ($$-\sigma$$)

Sheet 1 (at $$x = -a$$, charge $$-\sigma$$):

Point $$P$$ is to the right of this sheet. Since it is negatively charged, the field points left. $$\vec{E}_1 = -\frac{\sigma}{2\epsilon_0} \hat{i}$$

Sheet 2 (at $$x = a$$, charge $$-2\sigma$$):

Point $$P$$ is to the right of this sheet. Since it is negatively charged, the field points left. $$\vec{E}_2 = -\frac{2\sigma}{2\epsilon_0} \hat{i} = -\frac{\sigma}{\epsilon_0} \hat{i}$$

Sheet 3 (at $$x = 3a$$, charge $$\sigma$$):

Point $$P$$ is to the left of this sheet. Since it is positively charged, the field points left (away from the sheet). $$\vec{E}_3 = -\frac{\sigma}{2\epsilon_0} \hat{i}$$

$$\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3$$

$$\vec{E}_{net} = \left( -\frac{\sigma}{2\epsilon_0} - \frac{\sigma}{\epsilon_0} - \frac{\sigma}{2\epsilon_0} \right) \hat{i}$$

$$\vec{E}_{net} = \left( -\frac{2\sigma}{2\epsilon_0} - \frac{\sigma}{\epsilon_0} \right) \hat{i} = -\frac{2\sigma}{\epsilon_0} \hat{i}$$

$$|\vec{E}_{net}| = \frac{2\sigma}{\epsilon_0}$$

$$x = 2$$

Take the vertical axis through the centre of the larger disc as the reference (positive sense anticlockwise).
Initially both discs are at rest ⇒ total angular momentum about this axis is zero and, because no external torque acts, it must remain zero.

Moments of inertia about the vertical axis
Large disc (mass $$M$$, radius $$R$$): $$I_L = \frac{1}{2}MR^{2}$$
Small disc (mass $$M$$, radius $$\dfrac{R}{2}$$) about its own centre: $$I_{s,\;{\rm cm}} = \frac{1}{2}M\!\left(\frac{R}{2}\right)^{2}= \frac{1}{8}MR^{2}$$

The small disc is fixed to the circumference of the big disc, so its centre is at a distance $$R$$ from the main axis. Hence, using the parallel-axis theorem, the small disc’s moment of inertia about the main axis is

$$I_s = I_{s,\;{\rm cm}} + MR^{2}= \frac{1}{8}MR^{2}+MR^{2}= \frac{9}{8}MR^{2}$$

Angular velocities
Motor keeps the small disc spinning at a constant angular speed $$\omega$$ (anticlockwise) about its own axis. Reaction torque makes the large disc rotate clockwise with angular speed $$\dfrac{\omega}{n}$$ (-ve sign). Because the small disc is rigidly fixed to the large disc, its centre also moves in a circle of radius $$R$$ with the same angular speed as the large disc.

Angular momenta about the main axis

1. Large disc: $$L_L = I_L\!\left(-\frac{\omega}{n}\right)= -\frac{1}{2}MR^{2}\frac{\omega}{n}$$
2. Small disc (two parts)    a) Spin about its own axis: $$L_{s,\;{\rm spin}} = I_{s,\;{\rm cm}}\;\omega = \frac{1}{8}MR^{2}\,\omega$$ (anticlockwise)
   b) Orbital motion of its centre: $$L_{s,\;{\rm orb}} = MR^{2}\!\left(-\frac{\omega}{n}\right)= -MR^{2}\frac{\omega}{n}$$

Total angular momentum must stay zero:

$$L_L + L_{s,\;{\rm spin}} + L_{s,\;{\rm orb}} = 0$$ $$-\frac{1}{2}MR^{2}\frac{\omega}{n} + \frac{1}{8}MR^{2}\,\omega - MR^{2}\frac{\omega}{n}=0$$

Divide by $$MR^{2}\omega$$:

$$-\frac{1}{2n} + \frac{1}{8} - \frac{1}{n}=0$$ $$-\frac{3}{2n} + \frac{1}{8}=0$$

$$\frac{1}{n}= \frac{1}{8}\times\frac{2}{3}= \frac{1}{12} \;\Longrightarrow\; n = 12$$

Hence, the large disc rotates with angular speed $$\dfrac{\omega}{12}$$ opposite to that of the smaller disc.

Final answer: 12

An electron with KE = 5 eV enters a magnetic field of 3 $$\mu$$T. Find the electric field $$E$$ for the electron to move undeflected.

$$KE = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2 \times KE}{m}}$$

Convert KE to Joules: $$5 \text{ eV} = 5 \times 1.6 \times 10^{-19} = 8 \times 10^{-19}$$ J.

$$v = \sqrt{\frac{2 \times 8 \times 10^{-19}}{9 \times 10^{-31}}} = \sqrt{\frac{16 \times 10^{-19}}{9 \times 10^{-31}}} = \sqrt{1.778 \times 10^{12}}$$

$$v \approx 1.33 \times 10^6 \text{ m/s}$$

$$qE = qvB \implies E = vB$$

$$E = 1.33 \times 10^6 \times 3 \times 10^{-6} = 3.99 \approx 4 \text{ N/C}$$

The correct answer is 4 NC$$^{-1}$$.

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 disc of mass $$M$$ and radius $$R$$ rotates with angular velocity $$\omega$$. When another disc of mass $$M/2$$ and the same radius is placed gently on it coaxially, no external torque acts on the system and therefore angular momentum is conserved: $$L_{\text{before}} = L_{\text{after}}$$.

The moment of inertia of a disc about an axis through its centre perpendicular to its plane is given by $$I = \frac{1}{2}MR^2$$. Hence the initial moment of inertia is $$I_1 = \frac{1}{2}MR^2$$ while the moment of inertia of the second disc is $$I_2 = \frac{1}{2} \cdot \frac{M}{2} \cdot R^2 = \frac{MR^2}{4}$$. The combined moment of inertia therefore becomes $$I_1 + I_2 = \frac{MR^2}{2} + \frac{MR^2}{4} = \frac{3MR^2}{4}$$.

Since angular momentum is conserved, we have $$I_1 \omega = (I_1 + I_2)\omega'$$. Substituting the expressions for the moments of inertia gives $$\frac{MR^2}{2} \cdot \omega = \frac{3MR^2}{4} \cdot \omega'$$, which simplifies to $$\omega' = \frac{MR^2/2}{3MR^2/4} \cdot \omega = \frac{2}{3}\omega$$.

Therefore, the new angular velocity is $$\frac{2}{3}\omega$$.

Condition at the Topmost Point ($$B$$)

For the bob to just complete the circle, the tension in the string at point B must be zero. The centripetal force at this point is provided entirely by the gravitational force:

$$mg = \frac{mv_B^2}{L}$$

$$v_B^2 = gL$$

Kinetic Energy at Point $$B$$

The kinetic energy at the highest point is: 

$$(K.E.)_B=\frac{1}{2}mv_B^2=\frac{1}{2}mgL$$

Conservation of Mechanical Energy

$$(K.E.)_A = (K.E.)_B + (P.E.)_B$$

$$(K.E.)_A = \frac{1}{2}mgL + mg(2L)$$

$$(K.E.)_A = \frac{5}{2}mgL$$

The ratio is $$5 : 1$$.

Initial Total Energy:

On the horizontal track, the disc has both translational and rotational kinetic energy:

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

Final Total Energy:

At the maximum height $$h$$, the translational velocity is zero, but the rotational kinetic energy remains unchanged due to the lack of friction:

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

Conservation of Mechanical Energy (since there is no friction):

Equating the initial and final energy states ($$E_i = E_f$$):

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

$$\frac{1}{2}Mv^2 = Mgh$$

$$h = \frac{v^2}{2g}$$

A heavy iron bar (12 kg) has one end on the ground and the other on a man’s shoulder, making 60° with the horizontal. We wish to find the normal force applied by the man.

Let the bar have length $$L$$. The weight $$W = 12$$ kg-wt acts at the center of the bar, which is a distance $$L/2$$ from the ground end.

The component of the weight perpendicular to the bar is $$W\cos 60° = 12 \times \tfrac{1}{2} = 6$$ kg-wt, acting at distance $$L/2$$ from the pivot at the ground end. The normal force $$N$$ applied by the man is also perpendicular to the bar and acts at distance $$L$$ from the pivot.

Balancing torques about the ground end gives $$N \times L = W\cos 60° \times \frac{L}{2}$$ and therefore $$N = \frac{6}{2} = 3$$ kg-wt.

The correct answer is Option C: 3 kg-wt.

For a hollow sphere of mass $$M$$ and radius $$R$$, the moment of inertia about its diameter is:

$$I_1 = \frac{2}{3}MR^2$$

The radius of gyration is $$k_1 = \sqrt{\frac{I_1}{M}} = \sqrt{\frac{2}{3}R^2}$$

For the solid cylinder of mass $$M$$, radius $$R$$, and length $$L = 4R$$, the axis AB passes through one end. Using the Parallel Axis Theorem:

$$I_2 = I_{cm} + M\left(\frac{L}{2}\right)^2$$

$$I_2 = M\left(\frac{R^2}{4} + \frac{L^2}{12}\right) + M\frac{L^2}{4} = M\left(\frac{R^2}{4} + \frac{L^2}{3}\right)$$

$$I_2 = M\left(\frac{R^2}{4} + \frac{(4R)^2}{3}\right) = M\left(\frac{R^2}{4} + \frac{16R^2}{3}\right)$$

$$I_2 = MR^2 \left(\frac{3 + 64}{12}\right) = \frac{67}{12}MR^2$$

The radius of gyration is $$k_2 = \sqrt{\frac{I_2}{M}} = \sqrt{\frac{67}{12}R^2}$$

$$\text{Ratio} = \frac{k_1}{k_2} = \frac{\sqrt{\frac{2}{3}R^2}}{\sqrt{\frac{67}{12}R^2}}$$ = $$\sqrt{\frac{8}{67}}$$

$$x\ =\ 67$$

For a planet of mass $$m$$ moving in a circular orbit of radius $$r$$ around the Sun (mass $$M_\odot$$), the necessary centripetal force is provided by gravitation:

$$\frac{G M_\odot m}{r^{2}} = \frac{m v^{2}}{r}$$

Solving for the orbital speed,

$$v = \sqrt{\frac{G M_\odot}{r}} \quad -(1)$$

Angular momentum about the Sun is

$$L = m v r$$

Substituting $$v$$ from $$(1)$$,

$$L = m r \sqrt{\frac{G M_\odot}{r}} = m \sqrt{G M_\odot\, r} \quad -(2)$$

Applying $$(2)$$ to planets $$A$$ and $$B$$:

$$L_A = m_1 \sqrt{G M_\odot\, r_1}$$
$$L_B = m_2 \sqrt{G M_\odot\, r_2}$$

Given $$L_B = 3L_A$$,

$$m_2 \sqrt{G M_\odot\, r_2} = 3 m_1 \sqrt{G M_\odot\, r_1}$$

Canceling the common factor $$\sqrt{G M_\odot}$$,

$$m_2 \sqrt{r_2} = 3 m_1 \sqrt{r_1} \quad -(3)$$

From $$(3)$$,

$$\frac{\sqrt{r_2}}{\sqrt{r_1}} = \frac{3 m_1}{m_2}$$
Squaring both sides: $$\frac{r_2}{r_1} = \frac{9 m_1^{2}}{m_2^{2}} \quad -(4)$$

Kepler’s third law for circular orbits gives the time period:

$$T = 2\pi \sqrt{\frac{r^{3}}{G M_\odot}} \propto r^{3/2} \quad -(5)$$

Therefore,

$$\frac{T_A}{T_B} = \left(\frac{r_1}{r_2}\right)^{3/2} \quad -(6)$$

Insert $$\frac{r_1}{r_2}$$ from $$(4)$$ into $$(6)$$:

$$\frac{T_A}{T_B} = \left(\frac{m_2^{2}}{9 m_1^{2}}\right)^{3/2}$$

Compute the exponent:

$$\left(\frac{m_2^{2}}{9 m_1^{2}}\right)^{3/2} = \left(\frac{m_2}{m_1}\right)^{3} \left(\frac{1}{9}\right)^{3/2}$$

Since $$9 = 3^{2}$$, we have $$9^{3/2} = 3^{3} = 27$$. Thus

$$\frac{T_A}{T_B} = \frac{1}{27}\left(\frac{m_2}{m_1}\right)^{3}$$

Hence, the correct ratio is $$\frac{1}{27}\left(\frac{m_2}{m_1}\right)^{3}$$, which corresponds to Option B.

The rod is hinged at one end, therefore the pivot supplies no external impulse about itself during the very short collision. Hence both angular momentum and kinetic energy of the two-body system are conserved.

Data: $$M = 1.00\;\text{kg},\; L = 0.20\;\text{m},\; m = 0.10\;\text{kg},\; u = 5.00\;\text{m s}^{-1},\; r = \frac{L}{2}=0.10\;\text{m}$$

Moment of inertia of a uniform rod about one end $$I = \frac{1}{3}ML^{2} = \frac{1}{3}\times 1.00\times (0.20)^{2} = 0.01333\;\text{kg m}^{2}$$ $$-(1)$$

Conservation of angular momentum about the pivot:
Initial angular momentum $$L_i = m u r$$ (mass moves perpendicular to the rod).
After collision the mass rebounds, so its angular momentum is $$-m v r$$ and the rod carries $$I\omega$$. $$m u r = I\omega - m v r$$ $$-(2)$$

Conservation of kinetic energy (elastic impact):
$$\frac{1}{2}m u^{2} = \frac{1}{2}I\omega^{2} + \frac{1}{2}m v^{2}$$ Multiplying by 2, $$m u^{2} = I\omega^{2} + m v^{2}$$ $$-(3)$$

Substituting known numbers in $$-(2)$$:
$$0.10 \times 5.00 \times 0.10 = 0.01333\,\omega - 0.10\,v \times 0.10$$
$$0.05 = 0.01333\,\omega - 0.01\,v$$
$$\omega = 75\,(0.05 + 0.01\,v) = 3.75 + 0.75\,v$$ $$-(4)$$

Put $$-(4)$$ in the energy equation $$-(3)$$:
$$0.10\,(5.00)^{2} = 0.01333\,(3.75 + 0.75\,v)^{2} + 0.10\,v^{2}$$
$$2.50 = \frac{1}{75}(3.75 + 0.75\,v)^{2} + 0.10\,v^{2}$$

Multiplying by 75 to clear the denominator:
$$187.5 = (3.75 + 0.75\,v)^{2} + 7.5\,v^{2}$$

Expanding and simplifying gives the quadratic
$$8.0625\,v^{2} + 5.625\,v - 173.4375 = 0$$

Discriminant $$D = 5.625^{2} + 4\times 8.0625 \times 173.4375 = 75^{2}$$ $$\sqrt{D} = 75$$

Therefore $$v = \frac{-5.625 + 75}{2\times 8.0625} = 4.30\;\text{m s}^{-1}$$ (negative root would give a negative speed and is rejected).

From $$-(4)$$, $$\omega = 3.75 + 0.75 \times 4.30 = 6.98\;\text{rad s}^{-1}$$

Thus the bar acquires an angular speed of $$6.98\;\text{rad s}^{-1}$$ and the small mass rebounds with $$4.30\;\text{m s}^{-1}$$, matching

Option A which is: $$\omega = 6.98\;\text{rad s}^{-1}$$ and $$v = 4.30\;\text{m s}^{-1}$$

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.

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 $$\alpha = \boxed{3}$$.

A uniform solid cylinder of radius $$R$$ and length $$L$$ has moment of inertia $$I_1$$ about its axis, and a concentric cylinder of radius $$R' = \frac{R}{2}$$ and length $$L' = \frac{L}{2}$$ is carved out.

Since the moment of inertia of a solid cylinder about its axis is given by $$I = \frac{1}{2}MR^2$$, for the original cylinder the mass is $$M = \rho \pi R^2 L$$, so its moment of inertia becomes $$I_1 = \frac{1}{2}MR^2 = \frac{1}{2}\rho \pi R^4 L$$. Substituting the corresponding dimensions for the carved-out portion gives its mass as $$M' = \rho \pi \left(\frac{R}{2}\right)^2 \cdot \frac{L}{2} = \frac{\rho \pi R^2 L}{8} = \frac{M}{8}$$, and therefore its moment of inertia is $$I_2 = \frac{1}{2}M'\left(\frac{R}{2}\right)^2 = \frac{1}{2} \cdot \frac{M}{8} \cdot \frac{R^2}{4} = \frac{MR^2}{64}$$.

From the above, the ratio of the moments of inertia is $$\frac{I_1}{I_2} = \frac{\frac{MR^2}{2}}{\frac{MR^2}{64}} = \frac{64}{2} = 32$$. Therefore, the correct answer is 32.

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.

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 are given angular acceleration $$\alpha = 6t^2 - 2t$$ rad s$$^{-2}$$, with initial conditions $$\omega(0) = 10$$ rad s$$^{-1}$$ and $$\theta(0) = 4$$ rad.

Integrating the angular acceleration with respect to time gives $$\omega = \int \alpha \, dt = \int (6t^2 - 2t) \, dt = 2t^3 - t^2 + C_1$$. Imposing $$\omega(0) = 10$$ rad s$$^{-1}$$ yields $$C_1 = 10$$, so $$\omega = 2t^3 - t^2 + 10$$. Further integrating this angular velocity yields $$\theta = \int \omega \, dt = \int (2t^3 - t^2 + 10) \, dt = \frac{2t^4}{4} - \frac{t^3}{3} + 10t + C_2$$, hence $$\theta = \frac{t^4}{2} - \frac{t^3}{3} + 10t + C_2$$. Applying $$\theta(0) = 4$$ rad gives $$C_2 = 4$$, and thus $$\theta = \frac{t^4}{2} - \frac{t^3}{3} + 10t + 4$$. The correct answer is Option B.

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.

Three identical spheres, each of mass $$M$$, are placed at the corners of a right-angled triangle with mutually perpendicular sides of $$3 \text{ m}$$ each. Taking the point of intersection of the perpendicular sides as the origin, the three vertices are at:

$$A = (0, 0)$$, $$B = (3, 0)$$, $$C = (0, 3)$$

The coordinates of the centre of mass are:

$$x_{cm} = \dfrac{M(0) + M(3) + M(0)}{3M} = \dfrac{3}{3} = 1 \text{ m}$$

$$y_{cm} = \dfrac{M(0) + M(0) + M(3)}{3M} = \dfrac{3}{3} = 1 \text{ m}$$

The magnitude of the position vector of the centre of mass is:

$$r = \sqrt{x_{cm}^2 + y_{cm}^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \text{ m}$$

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

Therefore, the answer is $$\boxed{2}$$.

$$Massofdisc=1kg$$

$$Massofparticle=1kg$$

$$Radius=R$$

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

We have a one-dimensional rod AB of length $$L$$ with mass density $$\rho = \rho_0\left(1 - \frac{x^2}{L^2}\right)$$ kg/m, where $$x$$ is measured from end A. We need to find the distance of the centre of mass from end A.

The centre of mass is given by $$x_{\text{cm}} = \frac{\int_0^L x \, \rho \, dx}{\int_0^L \rho \, dx}$$.

First, let us compute the denominator (total mass). We have $$\int_0^L \rho_0\left(1 - \frac{x^2}{L^2}\right) dx = \rho_0\left[x - \frac{x^3}{3L^2}\right]_0^L = \rho_0\left(L - \frac{L}{3}\right) = \frac{2\rho_0 L}{3}$$.

Now for the numerator, $$\int_0^L x \cdot \rho_0\left(1 - \frac{x^2}{L^2}\right) dx = \rho_0\int_0^L \left(x - \frac{x^3}{L^2}\right) dx = \rho_0\left[\frac{x^2}{2} - \frac{x^4}{4L^2}\right]_0^L = \rho_0\left(\frac{L^2}{2} - \frac{L^2}{4}\right) = \frac{\rho_0 L^2}{4}$$.

So the centre of mass position is $$x_{\text{cm}} = \frac{\rho_0 L^2 / 4}{2\rho_0 L / 3} = \frac{L^2}{4} \times \frac{3}{2L} = \frac{3L}{8}$$.

Comparing with the given expression $$\frac{3L}{\alpha}$$, we get $$\alpha = 8$$.

Hence, the value of $$\alpha$$ is $$\textbf{8}$$.

The sphere moves down the plane, so take the downward direction along the incline as positive for translation and the clockwise sense as positive for rotation.

Forces along the incline
• Weight component: $$mg\sin\theta = 1 \times 10 \times \sin 30^\circ = 5\,$$N (down the plane).
• Static friction $$f$$ acts up the plane (opposite the motion) to maintain rolling without slipping.
• Two equal forces of $$1\,$$N are applied. One is up the plane and the other is down the plane, so their net translational effect cancels: $$+1\,$$N - $$1\,$$N = 0. They therefore constitute a couple that produces only a torque.

Translational equation
Using Newton’s second law, $$ma = mg\sin\theta - f$$ $$\Rightarrow\; a = 5 - f$$ $$-(1)$$

Torque equation about the centre
Moment of inertia of a solid sphere: $$I = \frac{2}{5}mR^{2} = \frac{2}{5}\times1\times1^{2} = 0.4\,$$kg m$$^{2}$$.
Rolling condition: $$a = \alpha R \Longrightarrow \alpha = a/R = a$$ because $$R = 1\,$$m.

Clockwise torques are taken positive.
• Torque due to friction: $$fR = f \times 1 = f$$ (clockwise, +).
• Torque due to the couple: each $$1\,$$N force acts at a perpendicular distance $$r = 0.5\,$m, giving a total anticlockwise torque
$$\tau_c = 2Fr = 2 $$\times$$ 1 $$\times$$ 0.5 = 1\,$$N m (anticlockwise, −).

Applying $$$$\sum$$\tau = I$$\alpha$$$$: $$f - 1 = I$$\alpha$$ = 0.4\,a$$ $$-(2)$$

Solve simultaneous equations
From $$(1):\; f = 5 - a$$.
Substitute in $$(2):$$ $$\bigl(5 - a\bigr) - 1 = 0.4\,a$$
$$4 - a = 0.4\,a$$
$$4 = 1.4\,a$$
$$a = $$\frac{4}{1.4} = \frac{20}{7}$$ $$\approx$$ 2.857 $$\text{ m\,s}^{-2}$$$$

Rounded to two decimal places, the acceleration of the sphere is $$\boxed{2.85$$\text{ m\,s}^{-2}}$$$$.

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.

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.

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

$$Weight=10g=100Nactingatmidpoint$$

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):
$$=100\times(horizontaldis\tan ceofCM)$$

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:

$$3/\sqrt{109}$$

For the centre of mass to remain unchanged, the total displacement weighted by mass must be zero.

Let the 10 kg block move a distance of 6 cm towards the 30 kg block (i.e., in the positive x-direction), and let the 30 kg block move a distance $$d$$ in some direction.

For the centre of mass position to remain unchanged:

$$m_1 \Delta x_1 + m_2 \Delta x_2 = 0$$

$$10 \times 6 + 30 \times \Delta x_2 = 0$$

$$60 + 30 \Delta x_2 = 0$$

$$\Delta x_2 = -2 \text{ cm}$$

The negative sign indicates that the 30 kg block must move in the opposite direction to the 10 kg block's movement. Since the 10 kg block moved towards the 30 kg block, the 30 kg block must move towards the 10 kg block.

Therefore, the 30 kg block must be moved 2 cm towards the 10 kg block.

The correct answer is Option C.

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.

We have two bodies of masses $$m_1 = 1$$ kg and $$m_2 = 3$$ kg with position vectors $$\vec{r}_1 = \hat{i} + 2\hat{j} + \hat{k}$$ and $$\vec{r}_2 = -3\hat{i} - 2\hat{j} + \hat{k}$$ respectively.

The position vector of the centre of mass is $$\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2} = \frac{1(\hat{i} + 2\hat{j} + \hat{k}) + 3(-3\hat{i} - 2\hat{j} + \hat{k})}{1 + 3}$$.

Computing the numerator: $$(1 - 9)\hat{i} + (2 - 6)\hat{j} + (1 + 3)\hat{k} = -8\hat{i} - 4\hat{j} + 4\hat{k}$$.

Dividing by 4: $$\vec{r}_{cm} = -2\hat{i} - \hat{j} + \hat{k}$$.

The magnitude is $$|\vec{r}_{cm}| = \sqrt{(-2)^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$$.

Now we check the magnitudes of each option. For Option A: $$\hat{i} - 2\hat{j} + \hat{k}$$, the magnitude is $$\sqrt{1 + 4 + 1} = \sqrt{6}$$. This matches our answer.

Hence, the correct answer is Option A.

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 $$x = 4$$.

For a uniform quarter disc of radius $$a$$ lying in the first quadrant (from $$\theta = 0$$ to $$\theta = \frac{\pi}{2}$$), we find the center of mass coordinates by integration.

Using polar coordinates, the $$x$$-coordinate of the center of mass is $$\bar{x} = \frac{\int_0^{\pi/2}\int_0^a r\cos\theta \cdot r \, dr \, d\theta}{\int_0^{\pi/2}\int_0^a r \, dr \, d\theta}$$. The denominator (total mass divided by $$\sigma$$) is $$\int_0^{\pi/2}\int_0^a r \, dr \, d\theta = \frac{a^2}{2} \cdot \frac{\pi}{2} = \frac{\pi a^2}{4}$$.

The numerator is $$\int_0^{\pi/2}\int_0^a r^2 \cos\theta \, dr \, d\theta = \frac{a^3}{3} \cdot [\sin\theta]_0^{\pi/2} = \frac{a^3}{3} \cdot 1 = \frac{a^3}{3}$$.

Therefore $$\bar{x} = \frac{a^3/3}{\pi a^2/4} = \frac{4a}{3\pi}$$. By symmetry, $$\bar{y} = \frac{4a}{3\pi}$$ as well.

The center of mass is at $$\left(\frac{4a}{3\pi}, \frac{4a}{3\pi}\right)$$. Comparing with the given form $$\left(\frac{xa}{3\pi}, \frac{xa}{3\pi}\right)$$, we get $$x = 4$$.

Consider a uniform semicircular wire of radius $$R$$ lying in the $$x$$-$$y$$ plane with its centre at the origin and its diameter along the $$x$$-axis. A small element of the wire at angle $$\theta$$ (measured from the positive $$x$$-axis) has coordinates $$(R\cos\theta,\, R\sin\theta)$$ and arc length $$dl = R\,d\theta$$, where $$\theta$$ ranges from $$0$$ to $$\pi$$ (upper semicircle).

By symmetry, the $$x$$-coordinate of the centre of mass is zero. For the $$y$$-coordinate:

$$y_{\text{cm}} = \frac{\displaystyle\int_0^{\pi} (R\sin\theta)\, R\,d\theta}{\displaystyle\int_0^{\pi} R\,d\theta} = \frac{R^2 \displaystyle\int_0^{\pi} \sin\theta\, d\theta}{R\pi}$$

$$\int_0^{\pi} \sin\theta\, d\theta = [-\cos\theta]_0^{\pi} = -\cos\pi + \cos 0 = 1 + 1 = 2$$

$$y_{\text{cm}} = \frac{R^2 \times 2}{R\pi} = \frac{2R}{\pi}$$

The position of the centre of mass is $$\left(0,\, \dfrac{2R}{\pi}\right)$$. Comparing with the given form $$\left(0,\, \dfrac{xR}{\pi}\right)$$, we get $$x = 2$$, so $$|x| = 2$$.

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 $$\boxed{3}$$.

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 $$2$$.

Let us consider Centroid 1 as

$$C_1=\left(x_1,\ y_1\right)$$

where $$C_1=\left(a,\ 0\right)$$

and Centroid 2 as

$$C_2=\left(x_2,\ y_2\right)$$

where 

$$C_2=\left(\frac{3a}{2},\ 0\right)$$

Area of bigger circle is $$A_1=\ \pi\ R^2$$

Area of the cut part is 

$$A_2=\ \ \frac{\pi R^2}{4}$$

Now formula is $$X\ =\ \frac{\left(A_1x_1+A_2x_2\right)}{A_1+A_2}$$

$$X\ =\ \frac{\left(\pi\ R^2\left(a\right)-\frac{\pi R^2}{4}\left(\frac{3a}{2}\right)\right)}{\pi\ R^2-\frac{\pi R^2}{4}}$$

$$X\ =\ \frac{\frac{5\pi R^2}{8}}{\frac{3\pi R^2}{4}}$$

$$X\ =\frac{5}{6}$$

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.

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 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.

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 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.

We are told that a solid cylinder of mass $$m$$, length $$L$$ and radius $$R$$ has its moment of inertia about an axis through its centre and perpendicular to the cylinder axis given by

$$I \;=\; m\left(\frac{R^{2}}{4} + \frac{L^{2}}{12}\right).$$

The problem says that a fixed mass of material is to be used. For a uniform material the mass is the product of density and volume. If $$\rho$$ is the density, then

$$m \;=\; \rho \,(\text{Volume}) \;=\; \rho\;(\pi R^{2}L).$$

Since $$m$$ and $$\rho$$ are both fixed, the product $$R^{2}L$$ must be a constant. Let us write this constant as

$$k \;=\; \frac{m}{\rho\pi}, \qquad\text{so that}\qquad R^{2}L \;=\; k.$$

Solving for the length we obtain

$$L \;=\; \frac{k}{R^{2}}.$$

We now substitute this expression for $$L$$ into the formula for the moment of inertia.

$$\begin{aligned} I &= m\left(\frac{R^{2}}{4} + \frac{L^{2}}{12}\right) \\ &= m\left(\frac{R^{2}}{4} + \frac{1}{12}\left(\frac{k}{R^{2}}\right)^{2}\right) \\ &= m\left(\frac{R^{2}}{4} + \frac{k^{2}}{12\,R^{4}}\right). \end{aligned}$$

Because the factor $$m$$ is a positive constant, minimising $$I$$ is equivalent to minimising the function

$$f(R) \;=\; \frac{R^{2}}{4} \;+\; \frac{k^{2}}{12\,R^{4}}.$$

To find the minimum we differentiate with respect to $$R$$ and set the derivative to zero.

Using the power-rule derivative $$\frac{d}{dR}(R^{n}) = nR^{n-1},$$ we have

$$\frac{df}{dR} \;=\; \frac{2R}{4} \;-\; \frac{4k^{2}}{12\,R^{5}} \;=\; \frac{R}{2} \;-\; \frac{k^{2}}{3\,R^{5}}.$$

Setting $$\dfrac{df}{dR}=0$$ gives

$$\frac{R}{2} \;=\; \frac{k^{2}}{3\,R^{5}}.$$ Multiplying both sides by $$6R^{5}$$ we get

$$3\,R^{6} \;=\; 2k^{2},$$

hence

$$R^{6} \;=\; \frac{2k^{2}}{3} \quad\Longrightarrow\quad R^{3} \;=\; k\sqrt{\frac{2}{3}}.$$ (The positive root is taken because radius is positive.)

We now compute the desired ratio $$\dfrac{L}{R}.$$ Since $$L = \dfrac{k}{R^{2}},$$ we have

$$\frac{L}{R} \;=\; \frac{k}{R^{3}}.$$ Substituting $$R^{3} = k\sqrt{\dfrac{2}{3}},$$ we get

$$\frac{L}{R} \;=\; \frac{k}{k\sqrt{\dfrac{2}{3}}} \;=\; \frac{1}{\sqrt{\dfrac{2}{3}}} \;=\; \sqrt{\frac{3}{2}}.$$

Hence, the correct answer is Option C.

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.

We take the solid hemisphere of radius $$R=8\,\text{cm}$$. Because of symmetry, its centre of mass will lie on the central vertical axis passing through the centre of the flat circular face. We measure every distance $$z$$ from this flat face, so $$z=0$$ at the plane of the face and $$z=R$$ at the pole of the hemisphere.

To locate the centre of mass we use the continuous form of the centre-of-mass formula for volume distributions:

$$z_{\text{cm}}=\dfrac{\displaystyle\int z\,dV}{\displaystyle\int dV}.$$

We slice the hemisphere into thin circular discs perpendicular to the axis. A slice at distance $$z$$ from the base has thickness $$dz$$ and radius $$r$$. From the equation of a sphere we have

$$r^{2}+z^{2}=R^{2}\;\Longrightarrow\;r^{2}=R^{2}-z^{2}.$$

The volume of this thin disc is

$$dV=\pi r^{2}\,dz=\pi\bigl(R^{2}-z^{2}\bigr)\,dz.$$

Now we write the numerator of the centre-of-mass expression:

$$\int_{0}^{R} z\,dV=\int_{0}^{R} z\,\pi\bigl(R^{2}-z^{2}\bigr)\,dz =\pi\int_{0}^{R}\bigl(R^{2}z - z^{3}\bigr)\,dz.$$

Integrating term by term,

$$\pi\left[\frac{R^{2}z^{2}}{2}-\frac{z^{4}}{4}\right]_{0}^{R} =\pi\left(\frac{R^{2}R^{2}}{2}-\frac{R^{4}}{4}\right) =\pi\left(\frac{R^{4}}{2}-\frac{R^{4}}{4}\right) =\pi\,\frac{R^{4}}{4}.$$

Next we find the denominator (the total volume of the hemisphere):

$$\int_{0}^{R} dV=\int_{0}^{R}\pi\bigl(R^{2}-z^{2}\bigr)\,dz =\pi\int_{0}^{R}\bigl(R^{2}-z^{2}\bigr)\,dz =\pi\left[R^{2}z-\frac{z^{3}}{3}\right]_{0}^{R} =\pi\left(R^{3}-\frac{R^{3}}{3}\right) =\pi\,\frac{2R^{3}}{3}.$$

Substituting these two results into the centre-of-mass formula,

$$z_{\text{cm}}=\frac{\pi\,\dfrac{R^{4}}{4}} {\pi\,\dfrac{2R^{3}}{3}} =\frac{R^{4}}{4}\times\frac{3}{2R^{3}} =R\,\frac{3}{8} =\frac{3R}{8}.$$

Putting $$R=8\,\text{cm}$$, we obtain

$$z_{\text{cm}}=\frac{3\times8}{8}\,\text{cm}=3\,\text{cm}.$$

So, the answer is $$3\,\text{cm}$$.

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 $$9 \text{ rpm}$$.