Correct answer is (120)

In case of rotational motion of a rigid body like this, the kinetic energy is not just due to the linear motion, but also due to its rotation.

For a solid cylinder, the moment of inertia I is given by $\frac{1}{2}m{r}^2$. The kinetic energy due to rotation is given by $\frac{1}{2}I{\omega }^2$. But we also know that $\omega =\frac{v}{r}$, hence the rotational kinetic energy can be written as $\frac{1}{2}\frac{1}{2}m{v}^2=\frac{1}{4}m{v}^2$, where $\frac{1}{2}$ is from the moment of inertia of the solid cylinder.

Therefore, the total kinetic energy (linear + rotational) when the string snaps is $\frac{1}{2}m{v}^2+\frac{1}{4}m{v}^2=\frac{3}{4}m{v}^2$.

Equating this to the potential energy $mgh$ and solving for h gives the result :

$h=\frac{v^2}{2g}\times \frac{3}{2}=1.2\mathrm{m}=120\mathrm{cm}$

Alternate Method :

Applying COE, we get

$mgh=\frac{1}{2}m{v}^2(1+\frac{{\mathrm{K}}^2}{r^2})$

$\mathrm{K}=$ radius of gyration

For a solid cylinder, $\frac{{\mathrm{K}}^2}{r^2}=\frac{1}{2}$

$\begin{array}{rl}\therefore h& =\frac{v^2}{2g}(1+\frac{1}{2})\\ \\ & =\frac{16}{2\times 10}\times \frac{3}{2}\\ \\ & =1.2\mathrm{m}\\ \\ & =120\mathrm{cm}\end{array}$

Correct answer is (5)

Loss in P.E. = Gain in K.E.

$2mgR=\frac{1}{2}\left[\frac{1}{2}m{R}^2+m{R}^2\right]w^2$

$2mgR=\frac{1}{2}\times \frac{3}{2}m{R}^2{w}^2$

$w^2=\frac{8g}{3R}$

$w=\sqrt{\frac{8g}{3R}}=4\sqrt{\frac{g}{2\times 3R}}$

$\Rightarrow x=\frac{g}{2}=5$

Correct Answer is Option (C)

$\frac{K_T}{K_R}=\frac{M{R}^2}{I_{CM}}$

I_CM is maximum for ring.

$\Rightarrow$ v is least for ring.

Correct Answer is Option (B)

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

$I=\frac{1}{2}m{R}^2$

Body is solid cylinder

Correct Answer is Option (D)

$a=\frac{g\sin \theta }{1+c}$

$a=\frac{9.8\sin {30}^{\circ }}{1+\frac{2}{5}}$

a = 3.5 m/sec²

Time of accent

V = u + at

0 = 1 $-$ 3.5 t

$t=\frac{1}{3.5}$ sec.

Time of decent

$t=\frac{1}{3.5}$ sec.

Total time $T=\frac{2}{3.5}$ = 0.57 sec.