$A$ steel ball of mass $200 \,g$ falls freely from a height of $20 \,m$ and bounces to a height of $10.8 \,m$ from the ground. If the energy lost in this process is absorbed by the ball, the rise in its temperature is ($g = 10 \,ms^{-2}$, specific heat capacity of steel is $460 \,Jkg^{-1} K^{-1}$). (in $^{\circ} C$)

$A$ solid cube of mass $m$ at a temperature $\theta_0$ is heated at a constant rate. It becomes liquid at temperature $\theta_1$ and vapour at temperature $\theta_2$. Let $s_1$ and $s_2$ be specific heats in its solid and liquid states respectively. If $L_f$ and $L_v$ are latent heats of fusion and vaporisation respectively,then the minimum heat energy supplied to the cube until it vaporises is

$A$ lead bullet of $20 \ g$ travelling at $300 \ m/s$ strikes against a block of wood and comes to rest. Assuming $50 \%$ of heat is absorbed by the bullet,the increase in its temperature is $.....^{\circ} C$ (take specific heat of lead $= 150 \ J/kg \cdot K$).

$A$ piece of ice (heat capacity = $2100 \text{ J kg}^{-1} \text{ }^\circ\text{C}^{-1}$ and latent heat = $3.36 \times 10^5 \ J kg^{-1}$) of mass $m$ grams is at $-5^oC$ at atmospheric pressure. It is given $420 \ J$ of heat so that the ice starts melting. Finally, when the ice-water mixture is in equilibrium, it is found that $1 \ gm$ of ice has melted. Assuming there is no other heat exchange in the process, the value of $m$ is ...... $gm$.

'Stem correction' in platinum resistance thermometers is eliminated by the use of

Two objects made of the same material have masses $m$ and $2m$ and are at temperatures $2T$ and $T$ respectively. When heat $Q$ is supplied to the object of mass $2m$, its temperature rises to $2T$. If the same heat $Q$ is supplied to the object of mass $m$, its temperature rises to: