JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Work, Power & Energy

Two cylindrical vessels of equal cross sectional area of

2 m^{2}

contain water upto height 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

1 0^{3} kg / m^{3}

and

g = 10 m / s^{2}

)

(A) $1 \times 1 0^{5} J$
(B) $4 \times 1 0^{4} J$
(C) $6 \times 1 0^{4} J$
(D) $8 \times 1 0^{4} J$ ✓ Correct

Answer: (D)

Solution

When the vessels are joined the water settles to a common level of

\frac{10 + 6}{2} = 8 m

. Treating each column's weight as acting at its mid-height, the initial potential energy is

ρ A g (\frac{1 0 ^{2}}{2} + \frac{6 ^{2}}{2}) = 68 ρ A g

and the final is

2 ρ A g \frac{8 ^{2}}{2} = 64 ρ A g

. Gravity does work equal to the fall in potential energy,

4 ρ A g = 4 \times 1 0^{3} \times 2 \times 10 = 8 \times 1 0^{4} J

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Work, Power & Energy

Solution

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