Two walls of thicknesses $d_1$ and $d_2$ and thermal conductivities $k_1$ and $k_2$ are in contact. In the steady state, if the temperatures at the outer surfaces are ${T_1}$ and ${T_2}$, the temperature at the common wall is

A constant potential difference is applied to the ends of a graphite rod, whose resistance decreases with a rise of temperature. The rod can be $(1)$ covered with asbestos or $(2)$ left open to atmosphere. Answer for steady state.

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity $K$ and $2K$ and thickness $x$ and $4x$ , respectively are $T_2$ and $T_1$ ($T_2$ > $T_1$). The rate of heat transfer through the slab, in a steady state is $\left( {\frac{{A({T_2} - {T_1})K}}{x}} \right)f$, with $f $ which equal to

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

Four conducting rods are joined to make a square. All rods are identical and ends $A, B$ and $C$ are maintained at given temperatures. choose $INCORRECT$ statement for given arrangement in steady state. (value of $\frac {KA}{L}$ is $1\frac{J}{{{S^o}C}}$ , symbols , have their usual meaning) 

The heat is flowing through two cylindrical rods of same material. The diameters of the rods are in the ratio $1 : 2$ and their lengths are in the ratio $2 : 1$ . If the temperature difference between their ends is the same, the ratio of rate of flow of heat through them will be