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JEE Main 20253 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2025 · 3 April · Evening Shift · Physics · Mechanical Properties of Fluids

Two cylindrical vessels of equal cross sectional area of 2 ~m^2...

Two cylindrical vessels of equal cross sectional area of 2 m22 \mathrm{~m}^2 contain water up to 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 103 kg/m310^3 \mathrm{~kg} / \mathrm{m}^3 and g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2)Two cylindrical vessels of equal cross sectional area of 2 ~m^2 contain water up to height 10 m and 6 m , respectively.

Options

  1. A1×105 J1 \times 10^5 \mathrm{~J}
  2. B4×104 J4 \times 10^4 \mathrm{~J}
  3. C6×104 J6 \times 10^4 \mathrm{~J}
  4. D8×104 J8 \times 10^4 \mathrm{~J}

Correct answer

D. 8×104 J8 \times 10^4 \mathrm{~J}

Step-by-step solution

Two cylindrical vessels of equal cross sectional area of 2 ~m^2 contain water up to height 10 m and 6 m , respectively.
U1=(ρA×10)g×5+(ρA6)g×3Ui=ρAg(50+18)Ui=68ρAg\begin{aligned} & \mathrm{U}_1=(\rho \mathrm{A} \times 10) \mathrm{g} \times 5+(\rho \mathrm{A} 6) \mathrm{g} \times 3 \\ & \mathrm{U}_{\mathrm{i}}=\rho \mathrm{Ag}(50+18) \\ & \mathrm{U}_{\mathrm{i}}=68 \rho \mathrm{Ag}\end{aligned}
Uf=(ρA×16)g×4=(ρAg)×64\begin{aligned} & U_f=(\rho A \times 16) g \times 4 \\ & =(\rho A g) \times 64\end{aligned}
ω=ΔU=4×ρAg=4×1000×2×10=8×104 J\begin{aligned} & \omega=\Delta U=4 \times \rho A g \\ & =4 \times 1000 \times 2 \times 10=8 \times 10^4 \mathrm{~J}\end{aligned}
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