Question

In the first experiment, two identical conducting rods are joined one after the other and this combination is connected to two vessels, one containing water at ${100}^{\circ }\text{C}$ and the other containing ice at $0^{\circ }\text{C}$ (see Fig). In the second experiment, the two rods are placed one on top of the other and ends are connected to the same vessels. If $q_1$ and $q_2$ (in gram per second) are the respective rates of melting of ice in the two cases, then the ratio $\frac{q_1}{q_2}$ is

• A. $\frac{1}{2}$
• B. $\frac{2}{1}$
• C. $\frac{1}{4}$
• D. $\frac{1}{8}$

Answer 1

The correct option is C $\frac{1}{4}$

Rate of melting of ice = Rate of thermal heat transfer
$L\frac{dm}{dt}=\frac{\Delta T}{R}......\left(1\right)$
Where $L$ = Latent heat of ice
$m$ = mass of ice
$\Delta T$ = temperature difference
$R=$ = Thermal resistance
Lets assume that thermal resistance of each rod is $R$
From equation $\left(1\right)$ we can say that
$\frac{dm}{dt}\propto \frac{1}{R}$
$\Rightarrow q\propto \frac{1}{R}......\left(2\right)$
Experiment 1: Two rods are joined in series
$R_s=R_1+R_2$
$\Rightarrow R_s=R+R=2R........\left(3\right)$

Experiment 2: When two rods are connected in parallel.
$R_p=\frac{R_1{R}_2}{R_1+R_2}$
$\Rightarrow R_p=\frac{R^2}{2R}=\frac{R}{2}.......\left(4\right)$
From $\left(2\right)$ we can write that,
$\frac{q_1}{q_2}=\frac{R_p}{R_s}$
By using ,$\left(3\right)$ and $\left(4\right)$ we get,
$\frac{q_1}{q_2}=\frac{1}{4}$
Thus, option (c) is the correct answer.