The temperature of Bhakhra dam water at the ground level with respect to the temperature at a high level should be:

The surface of the lake is at $2^{\circ} C$. The temperature of the bottom of the lake is ....... $^{\circ} C$.

Consider a rectangular sheet of solid material of length $l=9 \ cm$ and width $d=4 \ cm$. The coefficient of linear expansion is $\alpha=3.1 \times 10^{-5} \ K^{-1}$ at room temperature and one atmospheric pressure. The mass of the sheet is $m=0.1 \ kg$ and the specific heat capacity is $C_v=900 \ J \ kg^{-1} K^{-1}$. If the amount of heat supplied to the material is $8.1 \times 10^2 \ J$,then the change in area of the rectangular sheet is:

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

If $60 \%$ of the kinetic energy of water falling from a $210 \ m$ high waterfall is converted into heat,the rise in temperature of the water at the bottom of the fall is nearly (specific heat of water $= 4.2 \times 10^3 \ J \ kg^{-1} \ K^{-1}$ and $g = 10 \ m/s^2$):

$A$ block of steel of mass $2 \,kg$ slides down a rough inclined plane of inclination $\sin ^{-1}\left(\frac{3}{5}\right)$ at a constant speed. Assuming that the mechanical energy lost due to friction is used to increase the temperature of the block, calculate the rise in temperature of the block as it slides through $80 \,cm$. (Specific heat capacity of steel $= 420 \,J kg^{-1} K^{-1}$ and acceleration due to gravity $= 10 \,ms^{-2}$) (in $^{\circ} C$)