1. Given data

• Cross-sectional area of each vessel: $A = 2\,\text{m}^2$
• Initial water heights: $h_1 = 10\,\text{m}$ and $h_2 = 6\,\text{m}$
• Density of water: $\rho = 10^3\,\text{kg/m}^3$
• Acceleration due to gravity: $g = 10\,\text{m/s}^2$

Since the vessels have equal cross-sectional area and are connected at the bottom, the final water level in both vessels will be the same.

2. Find the final common height

Total initial volume of water: $V = A h_1 + A h_2 = A(h_1+h_2)$

After connection, both vessels have the same final height $h$: $2Ah = A(h_1+h_2)$ $2h = h_1+h_2 = 10+6=16$ $h = 8\,\text{m}$

So the final height in each vessel is: $h_f = 8\,\text{m}$

3. Compute initial gravitational potential energy

For a liquid column of height $h$ and area $A$, mass is $m=\rho Ah$, and its center of mass is at height $h/2$.

So potential energy is: $U = m g \frac{h}{2} = \rho A h g \frac{h}{2} = \frac{1}{2}\rho A g h^2$

Thus initial total potential energy: $U_i = \frac{1}{2}\rho A g (h_1^2 + h_2^2)$

Substitute values: $U_i = \frac{1}{2}(10^3)(2)(10)(10^2+6^2)$ $U_i = 10^4(100+36)$ $U_i = 10^4 \times 136 = 1.36\times 10^6\,\text{J}$

4. Compute final gravitational potential energy

After equilibrium, both vessels have height $8\,\text{m}$: $U_f = 2\left(\frac{1}{2}\rho A g h_f^2\right)$ $U_f = \rho A g h_f^2$

Substitute values: $U_f = (10^3)(2)(10)(8^2)$ $U_f = 2\times 10^4 \times 64$ $U_f = 1.28\times 10^6\,\text{J}$

5. Work done by gravity

Work done by gravity equals decrease in gravitational potential energy: $W_g = U_i - U_f$ $W_g = 1.36\times 10^6 - 1.28\times 10^6$ $W_g = 0.08\times 10^6$ $W_g = 8\times 10^4\,\text{J}$

6. Check options

• A: $1\times 10^5\,\text{J}$ ❌
• B: $4\times 10^4\,\text{J}$ ❌
• C: $8\times 10^4\,\text{J}$ ✅
• D: $6\times 10^4\,\text{J}$ ❌

Therefore, the correct option is C.