An ice cube of dimensions $60\,cm \times 50\,cm \times 20\,cm$ is placed in an insulation box of wall thickness $1\,cm$. The box keeping the ice cube at $0^{\circ}\,C$ of temperature is brought to a room of temperature $40^{\circ}\,C$. The rate of melting of ice is approximately. (Latent heat of fusion of ice is $3.4 \times 10^{5}\,J\,kg ^{-1}$ and thermal conducting of insulation wall is $0.05\,Wm ^{-10} C ^{-1}$ )

Two vessels of different materials are similar in size in every respect. The same quantity of ice filled in them gets melted in $20$ minutes and $40$ minutes respectively. The ratio of thermal conductivities of the materials is 

The two opposite faces of a cubical piece of iron (thermal conductivity $= 0.2\, CGS$ units) are at ${100^o}C$ and ${0^o}C$ in ice. If the area of a surface is $4c{m^2}$, then the mass of ice melted in $10$ minutes will be …… $gm$

If $K_{1}$ and $K_{2}$ are the thermal conductivities $L_{1}$ and $L _{2}$ are the lengths and $A _{1}$ and $A _{2}$ are the cross sectional areas of steel and copper rods respectively such that $\frac{K_{2}}{K_{1}}=9, \frac{A_{1}}{A_{2}}=2, \frac{L_{1}}{L_{2}}=2$.

Then, for the arrangement as shown in the figure. The value of temperature $T$ of the steel – copper junction in the steady state will be ……….. $^{\circ} C$

The ends $\mathrm{Q}$ and $\mathrm{R}$ of two thin wires, $\mathrm{PQ}$ and $RS$, are soldered (joined) togetker. Initially each of the wires has a length of $1 \mathrm{~m}$ at $10^{\circ} \mathrm{C}$. Now the end $\mathrm{P}$ is maintained at $10^{\circ} \mathrm{C}$, while the end $\mathrm{S}$ is heated and maintained at $400^{\circ} \mathrm{C}$. The system is thermally insulated from its surroundings. If the thermal conductivity of wire $\mathrm{PQ}$ is twice that of the wire $RS$ and the coefficient of linear thermal expansion of $P Q$ is $1.2 \times 10^{-5} \mathrm{~K}^{-1}$, the change in length of the wire $\mathrm{PQ}$ is