1. Given data

• Cross-sectional area of each vessel: $A=16\,\text{cm}^2=16\times 10^{-4}\,\text{m}^2=1.6\times 10^{-3}\,\text{m}^2$
• Initial heights: $h_1=100\,\text{cm}=1.0\,\text{m},\qquad h_2=150\,\text{cm}=1.5\,\text{m}$
• Density of water: $\rho=10^3\,\text{kg/m}^3$
• Acceleration due to gravity: $g=10\,\text{m/s}^2$

2. Final common height

Since the vessels have equal cross-sectional area and are interconnected, total volume is conserved.

Initial total volume: $V=A h_1 + A h_2 = A(h_1+h_2)$

If final common height is $h$, then $2Ah=A(h_1+h_2)$ $h=\frac{h_1+h_2}{2}=\frac{1.0+1.5}{2}=1.25\,\text{m}$

3. Change in gravitational potential energy

For a liquid column of height $h$, mass is $m=\rho Ah$, and its center of mass is at height $h/2$. So its gravitational potential energy is $U=mg\frac{h}{2}=\rho Ah g\frac{h}{2}=\frac{1}{2}\rho A g h^2$

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

Final potential energy is $U_f=2\times \frac{1}{2}\rho A g h^2=\rho A g h^2$

Now substitute values:

$U_i=\frac{1}{2}(10^3)(1.6\times10^{-3})(10)\left(1^2+1.5^2\right)$ $=\frac{1}{2}(16)(1+2.25)$ $=8\times 3.25=26\,\text{J}$

$U_f=(10^3)(1.6\times10^{-3})(10)(1.25)^2$ $=16\times 1.5625=25\,\text{J}$

4. Work done by gravity

Work done by gravity equals decrease in gravitational potential energy: $W=U_i-U_f=26-25=1\,\text{J}$

5. Option check

• A: $0.25\,\text{J}$ ❌
• B: $1\,\text{J}$ ✅
• C: $8\,\text{J}$ ❌
• D: $12\,\text{J}$ ❌

Therefore, the correct answer is: $\boxed{1\,\text{J}}$