$A$ metal sphere of radius $r$ and specific heat $S$ is rotated about an axis passing through its centre at a speed of $n$ rotations per second. It is suddenly stopped and $50 \%$ of its energy is used in increasing its temperature. Then, the rise in temperature of the sphere is

$A$ body kept on a smooth horizontal surface is pulled by a constant horizontal force applied at the top point of the body. If the body rolls purely on the surface,its shape can be:

The position vectors of two $1 \ kg$ particles,$(A)$ and $(B),$ are given by $\overrightarrow{r}_{A} = (\alpha_1 t^2 \hat{i} + \alpha_2 t \hat{j} + \alpha_3 \hat{k}) \ m$ and $\vec{r}_B = (\beta_1 t \hat{i} + \beta_2 t^2 \hat{j} + \beta_3 t \hat{k}) \ m$,respectively. Given $\alpha_1 = 1 \ m/s^2, \alpha_2 = 3n \ m/s, \alpha_3 = 2 \ m, \beta_1 = 2 \ m/s, \beta_2 = -1 \ m/s^2, \beta_3 = 4p \ m/s$,where $t$ is time,$n$ and $p$ are constants. At $t = 1 \ s$,$|\overrightarrow{V}_{A}| = |\overrightarrow{V}_{B}|$ and the velocities $\overrightarrow{V}_{A}$ and $\overrightarrow{V}_{B}$ are orthogonal. At $t = 1 \ s$,the magnitude of angular momentum of particle $(A)$ with respect to particle $(B)$ is $\sqrt{L} \ kg \ m^2/s$. The value of $L$ is:

$A$ uniform rod $AB$ of mass $m$ and length $l$ is at rest on a smooth horizontal surface. An impulse $J$ is applied to the end $B$,perpendicular to the rod in the horizontal direction. Find the speed of particle $P$ at a distance $\frac{l}{6}$ from the centre towards $A$ of the rod after time $t = \frac{\pi m l}{12 J}$.

$A$ particle is moving in a uniform circular motion with angular momentum $L$. If its angular frequency is doubled and its kinetic energy is halved,then the new angular momentum of the particle will be .......

$A$ yo-yo is resting on a rough horizontal table. Forces $F_1, F_2$ and $F_3$ are applied separately as shown. The correct statement is