JEE Main 2025 Apr 3 Shift 2, Physics Q3: Work Energy Theorem And Its Applications

JEE Main2025Apr 3, Shift 2Physics

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

Solution

After connection, the water levels equalise. Conservation of volume gives the common height $h$ :

$A \cdot 10 + A \cdot 6 = 2 A \cdot h \Rightarrow h = 8$ m.

Compute the gravitational potential energy of each cylindrical column using the centre of mass at $h /2$ . With $A = 2$ m $^{2}$ and $ρ g = 1 0^{4}$ N/m $^{3}$ :

Initial: $U_{i} = (ρ A \cdot 10) g \cdot 5 + (ρ A \cdot 6) g \cdot 3 = ρ A g (50 + 18) = 68 ρ A g$ .

Final: $U_{f} = (ρ A \cdot 8) g \cdot 4 + (ρ A \cdot 8) g \cdot 4 = 64 ρ A g$ (each vessel now has height 8, so total mass distributed equally).

$Δ U = U_{f} - U_{i} = - 4 ρ A g$ .

Work done by gravity $= - Δ U = 4 ρ A g = 4 \cdot 1000 \cdot 2 \cdot 10 = 8 \times 1 0^{4}$ J.

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JEE Main 2025 Apr 3 Shift 2, Physics Q3: Work Energy Theorem And Its Applications

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