Question 24 of 192

Two cylindrical vessels of equal cross sectional area of $2 \mathrm{~m}^2$ contain water up to height 10 m and 6 m , respectively. If the vessels are connected at their bottom then the work done by the force of gravity is : (Density of water is $10^3 \mathrm{~kg} / \mathrm{m}^3$ and $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$)

A$1 \times 10^5 \mathrm{~J}$B$4 \times 10^4 \mathrm{~J}$C$6 \times 10^4 \mathrm{~J}$D$8 \times 10^4 \mathrm{~J}$

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Answer: (D) $8 \times 10^4 \mathrm{~J}$

$\begin{aligned} & \mathrm{U}_1=(\rho \mathrm{A} \times 10) \mathrm{g} \times 5+(\rho \mathrm{A} 6) \mathrm{g} \times 3 \\ & \mathrm{U}_{\mathrm{i}}=\rho \mathrm{Ag}(50+18) \\ & \mathrm{U}_{\mathrm{i}}=68 \rho \mathrm{Ag}\end{aligned}$
$\begin{aligned} & U_f=(\rho A \times 16) g \times 4 \\ & =(\rho A g) \times 64\end{aligned}$
$\begin{aligned} & \omega=\Delta U=4 \times \rho A g \\ & =4 \times 1000 \times 2 \times 10=8 \times 10^4 \mathrm{~J}\end{aligned}$