JEE Physics 2025 Question with Solution

Given: Two cylindrical vessels each have cross-sectional area $A = 2 \, \text{m}^2$. The initial water heights are $h_1 = 10 \, \text{m}$ and $h_2 = 6 \, \text{m}$. The density of water is $\rho = 10^3 \, \text{kg/m}^3$ and acceleration due to gravity is $g = 10 \, \text{m/s}^2$.

Find: The work done by the force of gravity when the vessels are connected at the bottom.

When the vessels are connected, the water levels become equal. Using conservation of volume,

$$V_i = Ah_1 + Ah_2 = 2(10) + 2(6) = 32 \, \text{m}^3$$

Since both vessels have the same area, the final common height $h_f$ satisfies

$$2A h_f = 32$$

With $A = 2 \, \text{m}^2$,

$$4h_f = 32$$

so

$$h_f = 8 \, \text{m}$$

The work done by gravity equals the loss in gravitational potential energy.

For the water in the first vessel:

$$m_1 = \rho A h_1 = 10^3 \times 2 \times 10 = 20000 \, \text{kg}$$

Its center of mass falls by

$$\frac{h_1 - h_f}{2} = \frac{10 - 8}{2} = 1 \, \text{m}$$

Hence,

$$\Delta U_1 = m_1 g \times 1 = 20000 \times 10 = 200000 \, \text{J}$$

For the water in the second vessel:

$$m_2 = \rho A h_2 = 10^3 \times 2 \times 6 = 12000 \, \text{kg}$$

Its center of mass rises by

$$\frac{h_f - h_2}{2} = \frac{8 - 6}{2} = 1 \, \text{m}$$

Hence,

$$\Delta U_2 = m_2 g \times 1 = 12000 \times 10 = 120000 \, \text{J}$$

Therefore, the net work done by gravity is

$$W = \Delta U_1 - \Delta U_2 = 200000 - 120000 = 80000 \, \text{J}$$

Thus, the work done by gravity is $8 \times 10^4 \, \text{J}$. The correct option is D.

Given: The two vessels have equal area $A = 2 \, \text{m}^2$ with initial heights $10 \, \text{m}$ and $6 \, \text{m}$.

Find: The work done by gravity.

Compute the initial potential energy directly using center of mass of each water column:

$$U_i = (\rho A \times 10)g \times 5 + (\rho A \times 6)g \times 3$$ $$U_i = \rho A g (50 + 18) = 68\rho A g$$

The total height after equalization is

$$10 + 6 = 16$$

so the final height in each vessel is

$$h_f = 8 \, \text{m}$$

The final potential energy is

$$U_f = (\rho A \times 16)g \times 4 = 64\rho A g$$

Hence the work done by gravity is the decrease in potential energy:

$$W = U_i - U_f = (68 - 64)\rho A g = 4\rho A g$$

Substituting values,

$$W = 4 \times 1000 \times 2 \times 10 = 8 \times 10^4 \, \text{J}$$

Therefore, the work done by gravity is $8 \times 10^4 \, \text{J}$.