Question

Two identical conducting rods are first connected independently to two vessels, one containing water at ${100}^o$C and the other containing ice at $0^o$C. In the second case the rods are joined end to end and connected to the same vessels. Let $Q_1$ and $Q_2$ be the rate of melting of ice in the two cases respectively. Then the ratio $\frac{Q_1}{Q_2}$ is :

• A. $\frac{1}{2}$
• B. $\frac{2}{1}$
• C. $\frac{1}{4}$
• D. $\frac{4}{1}$

Answer 1

Answer: (C)

$\frac{1}{4}$

As we know that $Q\propto K$.
So, here we have to find thermal conductivity for two different cases:
1) For series connection equivalent thermal resistance $R_1=\frac{l}{k}+\frac{l}{k}=\frac{2l}{k}$
2) For parallel connection equivalent thermal resistance $R_2=\frac{\frac{l}{k}\times \frac{l}{k}}{\frac{l}{k}+\frac{l}{k}}=\frac{l}{2k}$
Hence, $\frac{Q_1}{Q_2}=\frac{K_1}{K_2}=\frac{k}{2l}\times \frac{l}{2k}=\frac{1}{4}$