Question

Two identical conducting rods are first connected independently to two vessels, one containing water at ${100}^{\circ }\text{C}$ and the other containing ice at $0^{\circ }\text{C}$. In the second case, the rods are joined end to end and connected to the same vessels. Let $q_1$ and $q_2$ gram per second be the rate of melting of ice in the two cases respectively. The ratio $\frac{q_1}{q_2}$ is

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

Answer 1

The correct option is C $4$

Let $Q$ amount of heat is required to melt ice of mass $m$ having latent heat $L$.
We know that,
$Q=mL$
$\therefore \frac{dQ}{dt}=L\frac{dm}{dt}$
Since, $\frac{dQ}{dt}\propto \frac{1}{R}$ where $R$ is thermal resistance of the rod.
$\Rightarrow \frac{dm}{dt}\propto \frac{1}{R}$
$\because$ In case $1$ both rod is in parallel $\Rightarrow R_1=\frac{R}{2}$
In case $2$ both rod is in series $\Rightarrow R_2=2R$
Therefore, $\frac{q_1}{q_2}=\frac{\frac{d{Q}_1}{dt}}{\frac{d{Q}_2}{dt}}=\frac{R_2}{R_1}=\frac{2R}{\left(\frac{R}{2}\right)}=4$