⇒Rotational Dynamics

Correct Option is (D)

Translational kinetic energy of a ring is:

${\mathrm{KE}}_{\text{Trans }}=\frac{1}{2}{\mathrm{mv}}^2$

Rotational kinetic energy of the ring is:

$\begin{array}{rl}{\mathrm{KE}}_{\text{Rolling }}& =\frac{1}{2}{\mathrm{I}}^2\\ & =\frac{1}{2}{\mathrm{mR}}^2{\omega }^2\\ {\mathrm{KE}}_{\text{Rolling }}& =\frac{1}{2}{\mathrm{mv}}^2(\because \mathrm{v}=\omega \mathrm{R})\end{array}$

Correct Option is (D)

At maximum compression, the solid cylinder will stop.

So loss in K.E. of cylinder = Gain in P.E. of spring

$\begin{array}{ll}\therefore & \frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}\mathrm{I}{\omega }^2=\frac{1}{2}{\mathrm{kx}}^2\\ \therefore & \frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}\frac{{\mathrm{mR}}^2}{2}{\left(\frac{\mathrm{v}}{\mathrm{R}}\right)}^2=\frac{1}{2}{\mathrm{kx}}^2\\ \therefore & \frac{3}{4}{\mathrm{mv}}^2=\frac{1}{2}{\mathrm{kx}}^2\\ \therefore & \frac{3}{4}\times 3\times (4{)}^2=\frac{1}{2}\times 200\times {\mathrm{x}}^2\\ \therefore & \frac{36}{100}={\mathrm{x}}^2\Rightarrow \mathrm{x}=0.6\mathrm{m}\end{array}$

Correct Option is (A)

Moment of Inertia of a solid sphere, $I=\frac{2}{5}{\mathrm{MR}}^2$

Since there is no slipping,

$\mathrm{v}=\mathrm{R}\omega$

$\therefore$ Rotational kinetic energy

$\begin{array}{r}{\mathrm{E}}_{\mathrm{rot}}=\frac{1}{2}\mathrm{I}{\omega }^2\\ =\frac{1}{2}\times \frac{2}{5}\times \mathrm{M}\times {\mathrm{R}}^2\times {\omega }^2\\ =\frac{{\mathrm{MR}}^2{\omega }^2}{5}\\ =\frac{{\mathrm{MV}}^2}{5}\text{.... (i)}\end{array}$

Total kinetic energy

$\begin{array}{rl}{E}_{\mathrm{K}}& =\frac{1}{2}I{\omega }^2+\frac{1}{2}{\mathrm{MV}}^2\\ & =\frac{{\mathrm{MV}}^2}{5}+\frac{{\mathrm{MV}}^2}{2}\\ & =\frac{7{\mathrm{MV}}^2}{10}\text{..... (ii)}\end{array}$

Dividing (ii) by (i), we get,

$\frac{{\mathrm{E}}_{\mathrm{K}}}{{\mathrm{E}}_{\mathrm{rot}}}=\frac{\left(\frac{7{\mathrm{MV}}^2}{10}\right)}{\left(\frac{{\mathrm{MV}}^2}{5}\right)}=\frac{7}{2}$

Correct Option is (C)

Kinetic energy $k=\frac{1}{2}I{\omega }^2$

$\begin{array}{rl}\therefore \frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}& =\frac{{\mathrm{I}}_2{\omega }_2^2}{{\mathrm{I}}_1{\omega }_1^2}\therefore \frac{1}{4}=\frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}\cdot 4\therefore \frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}=\frac{1}{16}\\ \frac{{\mathrm{L}}_2}{{\mathrm{L}}_1}& =\frac{{\mathrm{I}}_2{\omega }_2}{{\mathrm{I}}_1{\omega }_1}=\frac{1}{16}\times 2=\frac{1}{8}\\ \therefore {\mathrm{L}}_2& =\frac{{\mathrm{L}}_1}{8}\end{array}$

Correct Option is (A)

$\mathrm{R}=0.4\mathrm{m},\mathrm{M}=1\mathrm{kg},\alpha =10\mathrm{rad}/{\mathrm{s}}^2$

$I=\frac{M^2}{2}=\frac{1\times (0.4{)}^2}{2}=0.08\mathrm{kg}{\mathrm{m}}^2$

Torque, $\tau =\mathrm{I}\alpha =0.08\times 10=0.8\mathrm{N}-\mathrm{m}$

Also, $\tau =FR$

or $F=\frac{\tau }{R}=\frac{0.8}{0.4}=2\mathrm{N}$