Question

One end of the uniform rod of length $1$ m is placed in boiling water while its other end is placed in melting ice. A point $P$ on the rod is maintained at a constant temperature of ${800}^{o}C$, The mass of steam produced per second is equal to the mass of ice melted per second. If specific latent heat of steam is $7$ times the specific latent heat of ice, the distance of $P$ from the steam chamber must be

• A. $\frac{1}{9}m$
• B. $\frac{1}{8}m$
• C. $\frac{1}{7}m$
• D. $\frac{1}{10}m$

Answer 1

Answer: (A)

$\frac{1}{9}m$

As it is given that the two end of the rod is maintained at a different temperature and a point P will lie at distance $X$ from the left end so it's distance from the right end will be $1-X$.
Now, we know that heat produced is given as $Q=\frac{KA\Delta T}{L}$

Now, for left end $Q_1=\frac{KA(800-100)}{X}=m{l}_{steam}$ -----(1)

And for right end $Q_2=\frac{KA(800-0)}{L-X}=m{l}_{ice}$ --------(2)

It is also given that: $l_{steam}=7\times l_{ice}$

So, dividing (1) by (2) we get:

$\frac{7L-7X}{8X}=7$

$X=\frac{1}{9}$