Question

Two communicating vessels contains mercury. The diameter of one vessel is $n$ times larger than the diameter of other. A column of water of height $h$ is poured into the left vessel. The mercury level that will rise in the right hand vessel is ($s$ = relative density of mercury and $\rho$ = density of water)

• A. $\frac{n^{2}h}{(n+1{)}^{2}s}$
• B. $\frac{h}{(n^2+1)s}$
• C. $\frac{h}{(n+1{)}^{2}s}$
• D. $\frac{h}{n^{2}s}$

Answer 1

The correct option is B $\frac{h}{(n^2+1)s}$

If level of mercury in narrow tube goes down by $h_1$ then, the level of mercury in wide tube rises by $h_2$, now $\text{/}\pi r^2{h}_1=\text{/}\pi (nr{)}^2{h}_2$
$h_1=n^2{h}_2$

$P_A=P_B$
$h\rho g=(h_1+h_2)s\rho g$
$(n^2{h}_2+h_2)sg=hg$
$h_2=\frac{h}{(n^2+1)s}$