The power radiated by a black body is $P$ and it radiates maximum energy around the wavelength $\lambda_0$. Now the temperature of the black body is changed so that it radiates maximum energy around wavelength $\left(\frac{\lambda_0}{2}\right)$. The power radiated by it will now increase by a factor of

  • A
    $2$
  • B
    $8$
  • C
    $16$
  • D
    $32$

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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.

Four rods of identical cross-sectional area and made from the same metal form the sides of a square. The temperatures of two diagonally opposite points $A$ and $B$ are $\sqrt{2}T$ and $T$ respectively in the steady state. Assuming that only heat conduction takes place,what will be the temperature difference between the other two points $C$ and $D$?

$A$ black body radiates maximum energy at wavelength $\lambda$ and its emissive power is $E$. Now, due to a change in the temperature of that body, it radiates maximum energy at wavelength $\frac{2 \lambda}{3}$. At that temperature, the emissive power is:

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