Q. Two communicating vessels contain mercury. The diameter of one vessel is $n$ times larger than the diameter of the other. A column of water of height $h$ is poured into the left vessel. The mercury level will rise in the right-hand vessel ($s$ = relative density of mercury and $\rho$ = density of water) by
A) $\frac{n^{2}h}{(n + 1)^{2}s}$
B) $\frac{h}{\left( n^{2} + 1 \right)s}$
C) $\frac{h}{(n + 1)^{2}s}$
D) $\frac{h}{n^{2}s}$
Explanation
(b) If the level in narrow tube goes down by $h_{1}$ then in wider tube goes up to $h_{2}$
Now, $\pi r^{2}h_{1} = \pi(nr)^{2}h_{2} \Rightarrow h_{1} = n^{2}h_{2}$
Now, pressure at point $A =$ pressure at point $B$
$h\rho g = \left( h_{1} + h_{2} \right)\rho'g$
$\Rightarrow h = (n^{2}h_{2} + h_{2})s$ (As $s = \frac{\rho^{'}}{\rho}$ ) $\Rightarrow h_{2} = \frac{h}{(n^{2} + 1)s}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.