question_answer

Two cylindrical vessels having equal cross-sectional areas a containing water upto levels \[{{H}_{1}}\] and \[{{H}_{2}}\]. Vessels are connected through a pipe to make level of contained water same in both. If the density of water is d then what will be work done by the gravity in the process?

A)  \[gd\left( \frac{{{H}_{1}}-{{H}_{2}}}{2} \right)\]

B)  \[{{g}^{2}}{{d}^{2}}{{a}^{2}}{{\left( \frac{{{H}_{1}}-{{H}_{2}}}{2} \right)}^{2}}\]

C)  \[adg{{\left( \frac{{{H}_{1}}-{{H}_{2}}}{2} \right)}^{2}}\]

D)  \[\frac{{{({{H}_{1}}-{{H}_{2}})}^{2}}}{2ga}\]

Correct Answer: C

Solution :

                Consider the diagram As volume of water is constant, so finally, height water level in each vessel will be \[\frac{{{H}_{1}}+{{H}_{2}}}{2}.\] The mass per unit volume of water \[m=d\,a\,\left[ {{H}_{1}}-\frac{{{H}_{1}}+{{H}_{2}}}{2} \right]\] \[=ad\left[ \frac{{{H}_{1}}-{{H}_{2}}}{2} \right]\] Now, work done by gravity \[=mg\cdot x=ad\left( \frac{{{H}_{1}}-{{H}_{2}}}{2} \right)\cdot g\cdot \left( \frac{{{H}_{1}}-{{H}_{2}}}{2} \right)\] \[\left( \because \,x=\frac{{{H}_{1}}-{{H}_{2}}}{2} \right)\] \[\Rightarrow \]               \[{{W}_{mg}}=adg{{\left( \frac{{{H}_{1}}-{{H}_{2}}}{2} \right)}^{2}}\]