At a pressure $p$ and temperature $127^{\circ} \mathrm{C}$, a vessel contains 21 g of a gas. A small hole is made into the vessel, so that the gas in it leaks out. At a pressure of $\frac{2 p}{3}$ and a temperature of $t^{\circ} \mathrm{C}$, the mass of the gas leaked out is 5 g . Then, $t=$

Steam of mass 60 g at a temperature $100^{\circ} \mathrm{C}$ is mixed with water of mass 360 g at a temperature $40^{\circ} \mathrm{C}$. The ratio of the masses of steam and water in equilibrium is (Latent heat of steam is $540 \mathrm{cal} \mathrm{g}^{-1}$ and specific heat capacity of water is $1 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ )

The temperature difference between the ends of two cylindrical rods $A$ and $B$ of the same material is $2: 3$. In steady state the ratio of the rates of flow of heat through the rods $A$ and $B$ is $5: 9$. If the radii of the rods $A$ and $B$ are in the ratio $1: 2$, then the ratio of lengths of the rods $A$ and $B$ is

When $Q_{1}$ amount of heat supplied to a monoatomic gas, the work done by the gas is $W$. When $Q_{2}$ amount of heat is supplied to a diatomic gas, the work done by the gas is $2 W$. Then, $Q_{1}: Q_{2}$.