Two rods are connected as shown. The rods are of the same length and same cross-sectional area. In the steady state,the temperature $(\theta)$ of the interface will be........ $^oC$

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

$A$ copper rod and a steel rod of equal cross-sections and lengths $L$ are joined side by side and connected between two heat baths as shown in the figure. If heat flows through them from $x = 0$ to $x = 2L$ at a steady rate and thermal conductivities of the metals are $K_{Cu}$ and $K_{Steel}$ $(K_{Cu} > K_{Steel})$,then the temperature varies as (convection and radiation are negligible):

$A$ metallic rod of cross-sectional area $9.0 \, cm^2$ and length $0.54 \, m$,with the surface insulated to prevent heat loss,has one end immersed in boiling water and the other in an ice-water mixture. The heat conducted through the rod melts the ice at the rate of $1 \, g$ for every $33 \, s$. The thermal conductivity of the rod is ....... $W m^{-1} K^{-1}$.

$A$ cylindrical rod has temperatures $\theta_1$ and $\theta_2$ at its ends. The rate of heat flow is $Q$. All the linear dimensions of the rod are doubled while keeping the temperatures constant. The new rate of flow of heat is

$A$ composite slab is prepared with two different materials $A$ and $B$. The relation between their coefficients of thermal conductivity and thickness is given as $K_A = \frac{K_B}{2}$ and $X_A = 2 X_B$, respectively. If the temperatures of the outer faces of $A$ and $B$ are $75^{\circ} C$ and $50^{\circ} C$ respectively, what will be the temperature of the common surface (in $^{\circ} C$)?