Two cylindrical vessels of equal cross sectional area of $2 \mathrm{~m}^2$ contain water upto heights 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 $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

<p>First, when the two vessels are connected, the water levels equalize at </p> <p>$H = \frac{H_1 + H_2}{2} = \frac{10 + 6}{2} = 8\text{ m}.$ </p> <p>We can get the work done by gravity from the loss of gravitational potential energy: </p> <p><p>Initial potential energy (taking the bottom as zero) </p> <p>For a column of height $H$ and cross-section $A$, </p> <p>$U = \rho g A\int_0^H z\,dz = \frac{\rho g A H^2}{2}.$ </p> <p>So </p> <p>$$U_i = \frac{\rho g A}{2}\bigl(H_1^2 + H_2^2\bigr) = \frac{10^3\cdot 10\cdot 2}{2}(10^2 + 6^2) = 10^4\,(100 + 36) = 1.36\times10^6\text{ J}.$$</p></p> <p><p>Final potential energy (both at height 8 m) </p> <p>$$U_f = 2\;\frac{\rho g A H^2}{2}\;\Big|_{H=8} = \rho g A\,(8^2) = 10^3\cdot10\cdot2\cdot64 = 1.28\times10^6\text{ J}.$$</p></p> <p><p>Work done by gravity = drop in potential energy </p> <p>$$W = U_i - U_f = 1.36\times10^6 - 1.28\times10^6 = 0.08\times10^6 = 8\times10^4\text{ J}.$$</p></p> <p>Answer: 8×10^4 J (Option C).</p>