Question 30 of 192

Two cylindrical vessels of equal cross-sectional area $16{cm}^{2}$ contain water upto heights $100cm$ and $150cm$ respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process, is [Take density of water $={10}^{3}kg{m}^{-3}$ and $g=10{ms}^{-2}$]

A$0.25J$B$1J$C$8J$D$12J$

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Answer: (B) $1J$

Initially, the centre of mass of the right column will be at height of $75cm$ and the centre of mass of the left column will be at $50cm$. After opening of the valve both columns will have height of $125cm$ and their centre of mass will be at height $\dfrac{125}{2}cm$. Therefore,
$W=-∆U=-\left[\left({m}_{1}+{m}_{2}\right)g\dfrac{{h}_{1}+{h}_{2}}{4}-{m}_{1}g\dfrac{{h}_{1}}{2}-{m}_{2}g\dfrac{{h}_{2}}{2}\right]$
Now, ${m}_{1}=\rho A{h}_{1}$ and ${m}_{2}=\rho A{h}_{2}$
Putting $\rho =1000kg{m}^{-3},g=10m{s}^{-2},{h}_{1}=100cm,{h}_{2}=150cm$ and $A=16\times {10}^{-4}{m}^{2}$, we get,
$W=1J$