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(A) Let $Q$ be the heat required to melt the ice. The rate of heat flow $P$ is given by $P = \frac{Q}{t}$.For experiment $I$: $P_1 = \frac{Q}{t_1} = \

Vedclass · Accepted Answer

The value of $t_{10}$ is $100$ minutes

Vedclass · Answer

(A) Let $Q$ be the heat required to melt the ice. The rate of heat flow $P$ is given by $P = \frac{Q}{t}$.For experiment $I$: $P_1 = \frac{Q}{t_1} = \frac{Q}{20}$.For experiment $II$: $P_2 = \frac{Q}{t_2} = \frac{Q}{80}$.In experiment $III$ (series): The equivalent thermal resistance $R_{eq} = R_1 + R_2$. Since $P = \frac{\Delta T}{R}$,we have $\frac{1}{P_{eq}} = \frac{1}{P_1} + \frac{1}{P_2}$.$\frac{t_{10}}{Q} = \frac{t_1}{Q} + \frac{t_2}{Q} \Rightarrow t_{10} = t_1 + t_2 = 20 + 80 = 100 	ext{ minutes}$.In experiment $IV$ (parallel): The equivalent thermal resistance is $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$,which implies $P_{eq} = P_1 + P_2$.$\frac{Q}{t_{20}} = \frac{Q}{t_1} + \frac{Q}{t_2} \Rightarrow \frac{1}{t_{20}} = \frac{1}{20} + \frac{1}{80} = \frac{4+1}{80} = \frac{5}{80} = \frac{1}{16}$.Thus,$t_{20} = 16 	ext{ minutes}$.Comparing with the options,the correct statement is that $t_{10} = 100 	ext{ minutes}$.

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