MHTCET 11 May 2024 Shift 1 Paper

Concept:For equal emissive power and equal surface area, the Stefan-Boltzmann law relates emissivity and temperature as $e \propto \frac{1}{T^4}$.Explanation:The spheres have the same radii, so their surface areas are equal: $A_1 = A_2$.Since their emissive powers are the same, we equate the radiated power for both spheres.Using the Stefan-Boltzmann law, $E = e \sigma A T^4$, we get:$$e_1 \sigma A T_1^4 = e_2 \sigma A T_2^4$$Cancelling $\sigma$ and $A$, we have:$$e_1 T_1^4 = e_2 T_2^4$$Given the emissivity ratio $e_1 : e_2 = 1 : 4$, we substitute $e_2 = 4e_1$:$$e_1 T_1^4 = 4e_1 T_2^4$$This simplifies to:$$T_1^4 = 4T_2^4$$Taking the fourth root on both sides:$$\frac{T_1}{T_2} = 4^{\frac{1}{4}} = \sqrt{2}$$Therefore, the required ratio is $T_1 : T_2 = \sqrt{2} : 1$.Answer:$T_1 : T_2 = \sqrt{2} : 1$Correct option: B. $\sqrt{2} : 1$