A black rectangular surface of area A emits energy E per unit time at $27^\circ$ C. If length and breadth is reduced to $\left(\dfrac{1}{4}\right)^{th}$ of initial value and temperature is raised to $327^\circ$ C, then energy emitted per unit time becomes

According to Stefan-Boltzmann law, the energy emitted per unit time by a black body is given by $E = \sigma A T^4$.

Initial temperature $T_1 = 27^{\circ}\text{C} = 300\text{ K}$ and initial area $A_1 = A$.

Initial energy emitted per unit time is $E_1 = E = \sigma A (300)^4$.

Final temperature $T_2 = 327^{\circ}\text{C} = 600\text{ K}$.

Since length and breadth are reduced to $\dfrac{1}{4}$ of their initial values, the new area is $A_2 = \dfrac{L}{4} \times \dfrac{B}{4} = \dfrac{A}{16}$.

Final energy emitted per unit time is $E_2 = \sigma A_2 T_2^4$.

Taking the ratio of $E_2$ to $E_1$:

$\dfrac{E_2}{E_1} = \dfrac{A_2}{A_1} \left(\dfrac{T_2}{T_1}\right)^4$

$\dfrac{E_2}{E} = \dfrac{1}{16} \left(\dfrac{600}{300}\right)^4$

$\dfrac{E_2}{E} = \dfrac{1}{16} \times 2^4 = 1$

$E_2 = E$

Answer: $E$