Correct Answer is Option (C)

$\begin{array}{rl}& \text{ Displacement of }CM=R\omega \frac{T}{2}=\pi r\\ \\ & \Rightarrow \text{ Displacement of }A=\sqrt{(2R{)}^2+(\pi R{)}^2}\\ \\ & =R\sqrt{{\pi }^2+4}\end{array}$

Correct answer is (4)

Given that the solid sphere is rolling without slipping, we have:

Angular momentum $L=\left(I_{\text{com}}\right)(\omega )$

Kinetic energy $K=\frac{1}{2}(I_{\text{com}})({\omega }^2)+\frac{1}{2}M{V}_{\text{com}}^2$

For a solid sphere, the moment of inertia is $I_{\text{com}}=\frac{2}{5}M{R}^2$, and the relationship between linear and angular velocity is $V_{\text{com}}=R\omega$.

Substituting these values into the expressions for $L$ and $K$:

$L=\frac{2}{5}M{R}^2\frac{V_{\text{com}}}{R}=\frac{2MR{V}_{\text{com}}}{5}$

$K=\frac{1}{2}\left(\frac{2}{5}M{R}^2\right)\frac{V_{\text{com}}^2}{R^2}+\frac{1}{2}M{V}_{\text{com}}^2=\frac{7}{10}M{V}_{\text{com}}^2$

Now, the given ratio of $\frac{L}{K}$ is $\frac{\pi }{22}$:

$\frac{L}{K}=\frac{4}{7}\frac{R}{V_{\text{com}}}=\frac{\pi }{22}$

Since $V_{\text{com}}=R\omega$, we can substitute this relationship into the equation and solve for $\omega$:

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

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

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

Thus, the value of the angular speed is $4\text{rad/s}$.

Correct answer is (2)

For a rolling spherical shell, we must consider the fact that it has both translational and rotational kinetic energy. The total kinetic energy ($K_{total}$) can be expressed as the sum of the translational kinetic energy ($K_{trans}$) and the rotational kinetic energy ($K_{rot}$):

$K_{total}=K_{trans}+K_{rot}$

The translational kinetic energy of an object with mass (m) and linear velocity (v) is given by:

$K_{trans}=\frac{1}{2}m{v}^2$

The rotational kinetic energy of a rolling spherical shell with moment of inertia (I) and angular velocity (ω) is given by:

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

For a rolling object without slipping, the relationship between linear velocity (v) and angular velocity (ω) is:

$v=R\omega$

Where R is the radius of the spherical shell.

The moment of inertia for a spherical shell is given by:

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

Now, we can substitute the moment of inertia and the relationship between linear and angular velocity into the equation for rotational kinetic energy:

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

Simplifying the equation:

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

$K_{rot}=\frac{1}{3}m{v}^2$

Now, we can find the ratio of rotational kinetic energy to total kinetic energy:

$\frac{K_{rot}}{K_{total}}=\frac{\frac{1}{3}m{v}^2}{\frac{1}{2}m{v}^2+\frac{1}{3}m{v}^2}$

Simplifying the equation:

$\frac{K_{rot}}{K_{total}}=\frac{\frac{1}{3}}{\frac{1}{2}+\frac{1}{3}}=\frac{\frac{1}{3}}{\frac{5}{6}}$

Multiplying both the numerator and the denominator by 6:

$\frac{K_{rot}}{K_{total}}=\frac{2}{5}$

Comparing this to the given ratio of $\frac{x}{5}$, we can determine that the value of $x$ is 2.

Correct answer is (75)

Total initial kinetic energy

$=\frac{1}{2}m{\mathrm{v}}^2+\frac{1}{2}\mathrm{I}{\omega }^2$

$\mathrm{v}=\mathrm{R}\omega$ (for pure rolling)

$\mathrm{K}.EC=\frac{1}{2}m{\mathrm{v}}^2+\frac{1}{2}\times \frac{2}{3}m{\mathrm{R}}^2\times \frac{{\mathrm{v}}^2}{{\mathrm{R}}^2}=\frac{5}{6}m{\mathrm{v}}^2$

Energy remains conserve during whole journey.

${\mathrm{K}}_{\cdot }{\mathrm{E}}_i+{\mathrm{P}.\mathrm{E}}_{\cdot i}={\mathrm{K}}_{\cdot }{\mathrm{E}}_f+{\mathrm{P}.\mathrm{E}}_{\cdot f}$

$\begin{array}{rl}& \Rightarrow \frac{5}{2}m{\mathrm{v}}^2=mg\mathrm{H}(\because {K.E.}_f=0)\\ \\ & \Rightarrow \mathrm{H}=\frac{5}{6}\times \frac{{\mathrm{v}}^2}{g}\\ \\ & =\frac{5\times (3{)}^2}{6\times 10}\\ \\ & =\frac{15}{20}\mathrm{m}=0.75\mathrm{m}=75\mathrm{cm}\end{array}$

Correct answer is (2)

Loss in potential energy $=mgh=mg[60\sin {30}^{\circ }\mathrm{cm}]$

$\Rightarrow mg\left[\frac{30}{100}\right]=\frac{1}{2}m{v}^2+\frac{1}{2}\frac{m{v}^2}{2}$

$\Rightarrow 0.3\times 10=\frac{3}{4}{v}^2$

$\Rightarrow v^2=4$

$\Rightarrow v=2\mathrm{m}/\mathrm{s}$