JEE Main 2022 — Mechanical Properties of Fluids Question with Solution

Initially, the centre of mass of the right column will be at height of $75 \mathrm{cm}$ and the centre of mass of the left column will be at $50 \mathrm{cm}$. After opening of the valve both columns will have height of $125 \mathrm{cm}$ and their centre of mass will be at height $\frac{125}{2} \mathrm{cm}$. Therefore,

$W=-∆U=-\left[\left(m_1+m_2\right)g\frac{h_1+h_2}{4}-m_{1}g\frac{h_1}{2}-m_{2}g\frac{h_2}{2}\right]$

Now, $m_1=\rho A{h}_1$ and $m_2=\rho A{h}_2$

Putting $\rho =1000 \mathrm{kg} {\mathrm{m}}^{-3}, g\mathit{=}10 \mathrm{m} {\mathrm{s}}^{-2}, h_1=100 \mathrm{cm}, h_2=150 \mathrm{cm}$ and $A=16\times {10}^{-4} {\mathrm{m}}^2$, we get,

$W=1 \mathrm{J}$