Question

The container $A$ contains ice at $0^{\circ }\text{C}$. A conducting uniform rod $PQ$ of length $4R$ is used to transfer heat to the ice in the container. The end $P$ of the rod is maintained at ${100}^{\circ }\text{C}$ and the other end $Q$ is kept inside the container $A$. The complete ice melts in $23\text{minutes}$. In another experiments, two conductors in the shape of quarter circles of radii $2R$ and $R$ are welded to the conductor $PQ$ at $M$ and $N$ respectively and their other ends are inserted inside the container $A$. All conductors are made of the same material and have the same cross sectional area. Once again, the end $P$ is maintained at ${100}^{\circ }\text{C}$ and this time, the complete ice melts in $t\text{minutes}$. Find $t$.
[Assume no heat loss from the curved surface of the rods and take $\pi \simeq 3.0$]

• A. $35\text{min}$
• B. $21\text{min}$
• C. $17.5\text{min}$
• D. $12\text{min}$

Answer 1

The correct option is C $17.5\text{min}$

Thermal resistance of length $R$ of the rod is
$r=\frac{R}{kA}$ [$k=$ thermal conductivity]
With only rod $PQ$, the thermal resistance is $4r$.
Heat current $H_1=\frac{\Delta T}{4r}$
Since $H=\frac{dQ}{dt}=m{L}_f$,
Time taken to melt the ice completely is inversely proportional to $H$
Given, $t_1=23\text{min}$ & $t_1\propto 4r$ ... (i)

In second experiment, when circular parts are added, the equivalent thermal resistance will be obtained by the series-parallel combination shown in the figure.


where $r_1=\frac{\pi R}{kA}=\pi r\simeq 3r$
$r_2=\frac{\pi R/2}{kA}=\frac{\pi }{2}r\simeq \frac{3}{2}r$
Since $r$ in parallel to $r_2$, we get equivalent resistance $\frac{r{r}_2}{r+r_2}=\frac{\frac{3}{2}r}{1+\frac{3}{2}}=\frac{3r}{5}$
$\frac{3r}{5}$ and $r$ are in series which gives $\frac{8r}{5}$
$\frac{8r}{5}$ is in parallel to $r_1$, which gives $\frac{\frac{8r}{5}\times 3r}{\frac{8r}{5}+3r}=\frac{24r}{23}$
Finally, $\frac{24r}{23}$ and $2r$ are in series; which gives $\frac{70r}{23}$ as net equivalent resistance.

Therefore, heat current in second case
$H_2=\frac{\Delta T}{70r/23}$
$\Rightarrow t_2\propto \frac{70r}{23}$
From (i) and (ii),
$\frac{t_2}{t_1}=\frac{70}{23\times 4}=\frac{t}{23\text{min}}$
$\Rightarrow t=\frac{70}{4}\text{min}$ is the time taken in second case.