Physics Q30 – JEE-MAIN 2025

Home Physics Mechanical Properties of FluidsQuestion #272

Two cylindrical vessels of equal cross sectional area of

2 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} k g / m^{3}

and

g = 10 m / s^{2}

)

Clue 1 of 3: Gravitational Potential Energy Change

The work done by gravity is equal to the negative of the change in gravitational potential energy of the system. Water will flow until the heights in both vessels are equal.

🧠 Full Solution Path

Step 1: Determine Initial and Final States✦ Active

The initial heights of water in the two vessels are $h_{1} = 10 m$ and $h_{2} = 6 m$ . The cross-sectional area of each vessel is $A = 2 m^{2}$ . When connected, water will flow until the height in both vessels is equal. The total volume of water is $V_{t o t a l} = A h_{1} + A h_{2} = A (h_{1} + h_{2})$ . In the final state, this total volume is distributed over two vessels of area $A$ each, so $V_{t o t a l} = 2 A h_{f}$ .

V_{t o t a l} = 2 (10 + 6) = 32 m^{3}

32 = 2 \times 2 \times h_{f} ⟹ 4 h_{f} = 32 ⟹ h_{f} = 8 m

Step 2: Calculate Initial and Final Gravitational Potential Energy○ Expand

The gravitational potential energy of a fluid column of height $h$ and cross-sectional area $A$ is $U = \frac{1}{2} ρ A g h^{2}$ . We use the given values: $ρ = 10^{3} k g / m^{3}$ , $A = 2 m^{2}$ , $g = 10 m / s^{2}$ .

U_{i n i t i a l} = \frac{1}{2} ρ A g h_{1}^{2} + \frac{1}{2} ρ A g h_{2}^{2} = \frac{1}{2} ρ A g (h_{1}^{2} + h_{2}^{2})

U_{i n i t i a l} = \frac{1}{2} \times 10^{3} \times 2 \times 10 \times (10^{2} + 6^{2}) = 10^{4} \times (100 + 36) = 136 \times 10^{4} J

U_{f i n a l} = \frac{1}{2} ρ A g h_{f}^{2} + \frac{1}{2} ρ A g h_{f}^{2} = ρ A g h_{f}^{2}

U_{f i n a l} = 10^{3} \times 2 \times 10 \times 8^{2} = 2 \times 10^{4} \times 64 = 128 \times 10^{4} J

💡 Teacher's Secret Hint

Remember to use the correct formula for the potential energy of a fluid column, which considers the center of mass at $h / 2$ .

Step 3: Calculate Work Done by Gravity○ Expand

The work done by the force of gravity is the negative of the change in potential energy, $W_{g} = - Δ U = U_{i n i t i a l} - U_{f i n a l}$ .

W_{g} = 136 \times 10^{4} J - 128 \times 10^{4} J = (136 - 128) \times 10^{4} J = 8 \times 10^{4} J

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 30 – JEE-MAIN 2025

Clue 1 of 3: Gravitational Potential Energy Change

🧠 Full Solution Path