Two communicating vessels contain mercury. Th | vedclass

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(B) Let the diameter of the left vessel be $d$ and the right vessel be $nd$. The cross-sectional area of the left vessel is $A_1 = \pi (d/2)^2$ and th

Vedclass · Accepted Answer

$\frac{h}{(n^2 + 1)s}$

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(B) Let the diameter of the left vessel be $d$ and the right vessel be $nd$. The cross-sectional area of the left vessel is $A_1 = \pi (d/2)^2$ and the right vessel is $A_2 = \pi (nd/2)^2 = n^2 A_1$.When water of height $h$ is poured into the left vessel,the mercury level in the left vessel drops by $h_1$ and rises in the right vessel by $h_2$.Due to the conservation of volume of mercury,$A_1 h_1 = A_2 h_2$,which gives $A_1 h_1 = (n^2 A_1) h_2$,so $h_1 = n^2 h_2$.The pressure at the interface level $A-B$ must be equal in both limbs.Pressure at $A$ (left side) = $h \rho g$ (due to water column).Pressure at $B$ (right side) = $(h_1 + h_2) \rho_{Hg} g$,where $\rho_{Hg} = s \rho$.Equating the pressures: $h \rho g = (h_1 + h_2) s \rho g$.$h = (h_1 + h_2) s$.Substituting $h_1 = n^2 h_2$: $h = (n^2 h_2 + h_2) s = h_2 (n^2 + 1) s$.Therefore,the rise in the right-hand vessel is $h_2 = \frac{h}{(n^2 + 1)s}$.

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