question_answer

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)             \[\frac{h}{({{n}^{2}}+1)\,s}\]

C)             \[\frac{h}{{{(n+1)}^{2}}s}\]     

D)             \[\frac{h}{{{n}^{2}}s}\]

Correct Answer: B

Solution :

                   If the level in narrow tube goes down by h1 then in wider tube goes up to h2, Now, \[\pi {{r}^{2}}{{h}_{1}}=\pi {{(nr)}^{2}}{{h}_{2}}\]Þ \[{{h}_{1}}={{n}^{2}}{{h}_{2}}\] Now, pressure at point A = pressure at point B \[h\rho g=({{h}_{1}}+{{h}_{2}})\rho 'g\] Þ h = \[({{n}^{2}}{{h}_{2}}+{{h}_{2}})sg\] \[\left( \text{As}\ s=\frac{\rho '}{\rho } \right)\] Þ \[{{h}_{2}}=\frac{h}{({{n}^{2}}+1)s}\]