$A$ steel pendulum clock manufactured at $32^{\circ} C$ and working at $47^{\circ} C$ is nearly (Coefficient of linear expansion of steel $= 12 \times 10^{-6} /{ }^{\circ} C$)

Heat is being supplied at a constant rate to a sphere of ice which is melting at the rate of $0.1 \,g/s$. It melts completely in $100 \,s$. The rate of rise of temperature of the resulting water thereafter will be ............ $^{\circ}C/s$.

Water falls from a height of $200 \ m$ into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool. ($g = 10 \ m/s^2$,specific heat of water $s = 4200 \ J/(kg \ K)$) (in $K$)

$A$ 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

$A$ beaker of height $H$ is made up of a material whose coefficient of linear thermal expansion is $3\alpha$. It is filled up to the brim by a liquid whose coefficient of thermal expansion is $\alpha$. If now the beaker along with its contents is uniformly heated through a small temperature $T$,the level of liquid will reduce by (given $\alpha T << 1$):

$Assertion :$ In the pressure-temperature $(P-T)$ phase diagram of water,the slope of the melting curve is found to be negative.
$Reason :$ Ice contracts on melting to water.