MHTCET 22 Apr 2025 Shift 2 Paper

Concept:Stefan-Boltzmann law for thermal radiation.Explanation:For a body, the heat radiated per unit area per unit time is given by $E = e \sigma T^{4}$, where $e$ is the emissivity and $T$ is the absolute temperature.Since both bodies have the same dimensions, their surface areas are equal.Given the emissivities are in the ratio $16 : 1$, let the emissivity of body A be $e_A = 16k$ and that of body B be $e_B = k$.For body A:$$E_A = 16k \sigma T_1^{4}$$For body B:$$E_B = k \sigma T_2^{4}$$Since they radiate the same heat per unit area per unit time, we equate $E_A = E_B$:$$16k \sigma T_1^{4} = k \sigma T_2^{4}$$Cancelling $k\sigma$ from both sides:$$16 T_1^{4} = T_2^{4}$$$$\left(\frac{T_1}{T_2}\right)^{4} = \frac{1}{16}$$Taking the fourth root on both sides:$$\frac{T_1}{T_2} = \left(\frac{1}{16}\right)^{1/4} = \frac{1}{2}$$Given $T_1 = x T_2$, we get $x = \frac{1}{2} = 0.5$.Answer:The value of $x$ is $0.5$.Therefore, the correct option is D.