Question

A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of $0.1$ gm of ice per second. If the rod is replaced by another rod with half the length and double the radius of the first and if the thermal conductivity of the material of the second rod is $0.25$ times that of first, the rate at which ice melts in gm/second will be :

• A. 0.1
• B. 0.2
• C. 1.6
• D. 3.2

Answer 1

The correct option is B 0.2

The thermal conductance is directly proportional to the area and conductivity, inversely proportional to the length. So if the radius is doubled, the area becomes four times. The length is halved and conductivity becomes one-fourth. So the conductance becomes $0.25\times 4/0.5=2$ times.

Now, the rate of heat transfer is directly proportional to the thermal conductance, it is also doubles

The rate of melting of ice $m=rateofheattransfer/LatentHeatoffusion$, so it is also doubled.

Its value is $2\times 0.1=0.2gm$

$\left(\frac{m}{t}\right)L=\frac{k\pi r^2({\theta }_1-{\theta }_2)}{\ell }$

$\left(\frac{m}{t}\right)L\propto \frac{k{r}^2}{\ell }$