MHTCET 20 Apr 2026 Shift 1 Paper

Concept:According to Stefan's law, the rate of cooling of a body is proportional to $T^{4} - T_{0}^{4}$, where $T$ is the body's temperature and $T_{0}$ is the surrounding temperature.Explanation:The surrounding temperature is $T_{0} = 200\ \mathrm{K}$.Initially, the sphere is at $T = 600\ \mathrm{K}$, so its cooling rate is$$R \propto 600^{4} - 200^{4}$$When the sphere cools to $T = 400\ \mathrm{K}$, the new cooling rate is$$R' \propto 400^{4} - 200^{4}$$Taking the ratio of the two rates,$$\frac{R'}{R} = \frac{400^{4} - 200^{4}}{600^{4} - 200^{4}}$$Factor out $200^{4}$ from both numerator and denominator,$$\frac{R'}{R} = \frac{2^{4} - 1}{3^{4} - 1} = \frac{16 - 1}{81 - 1} = \frac{15}{80} = \frac{3}{16}$$Thus,$$R' = \frac{3}{16} R$$Answer:The cooling rate becomes $R' = \frac{3}{16}R$, which is Option D.