$A$ heated body emits radiation which has maximum intensity at frequency $f_m$. If the temperature of the body is doubled,then:

Fill in the blanks:
$(a)$ $0.49 \frac{\text{cal}}{\text{cm} \cdot \text{K} \cdot \text{s}} = \dots \frac{\text{J}}{\text{m} \cdot \text{K} \cdot \text{s}}$
$(b)$ If the rate of emission of heat of a substance is less than its rate of absorption,then its temperature $\dots$.
$(c)$ The rate of emission of heat of a substance is directly proportional to $\dots$ of temperature of it and surroundings.

Power radiated by a black body at temperature $T_1$ is $P$ and it radiates maximum energy at a wavelength $\lambda_1$. If the temperature of the black body is changed from $T_1$ to $T_2$,it radiates maximum energy at a wavelength $\frac{\lambda_1}{2}$. The power radiated at $T_2$ is (in $P$)

$A$ rod of length $L$ and uniform cross-sectional area has varying thermal conductivity which changes linearly from $2K$ at end $A$ to $K$ at the other end $B$. The ends $A$ and $B$ of the rod are maintained at constant temperatures $100^oC$ and $0^oC$, respectively. At steady state, the graph of temperature $T = T(x)$, where $x$ is the distance from end $A$, will be:

What are the modes of heat transfer due to a temperature difference?

The power radiated by a black body is $P$ and it radiates maximum energy around the wavelength $\lambda_0$. If the temperature of the black body is now changed so that it radiates maximum energy around wavelength $\frac{3}{4}\lambda_0$,the power radiated by it will increase by a factor of