JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Work, Power & Energy

Two cylindrical vessels of equal cross-sectional area

16 cm^{2}

contain water upto heights 100 cm and 150 cm respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process, is [Take density of water

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

and

g = 10 ms^{- 2}

]

(A)0.25 J
(B)1 J✓ Correct
(C)8 J
(D)12 J

Answer: (B)

Solution

The final height in both vessels is

h = \frac{h _{1} + h _{2}}{2} = \frac{100 + 150}{2} = 125

cm. Work done by gravity equals the decrease in potential energy:

W = U_{i} - U_{f} = \frac{ρ A g}{2} [h_{1}^{2} + h_{2}^{2} - \frac{( h _{1} + h _{2} ) ^{2}}{2}]

. Substituting values:

W = \frac{1000 \times 16 \times 1 0 ^{- 4} \times 10}{2} [1 + 2.25 - 3.125] = 0.8 \times 0.125 \times 10 = 1

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Single correct option

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JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Work, Power & Energy

Solution

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