In a hydraulic lift, the surface area of the input piston is 6 cm 2 and that of the output piston is 1500 cm 2 . If 100 N force is applied to the input piston to raise the output piston by 20 cm , then the work done is kJ.

Physics · Fluid Mechanics

JEE Main 2025 — 29 January, Morning Shift — Question 43

In a hydraulic lift, the surface area of the input piston is $6 \mathrm{~cm}^{2}$

and that of the output piston is $1500 \mathrm{~cm}^{2}$ .

If 100 N force is applied to the input piston to raise the output piston by 20 cm ,

then the work done is $\qquad$ kJ.

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

$\frac{\mathrm{F}_{1}}{\mathrm{~A}_{1}}=\frac{\mathrm{F}_{2}}{\mathrm{~A}_{2}}, \frac{100}{6}=\frac{\mathrm{F}}{1500}, \mathrm{~F}=\frac{50}{3} \times 1500$

$\mathrm{F}=50 \times 500=25 \times 10^{3} \mathrm{~N}$

$\omega=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{S}}=25 \times 10^{3} \times \frac{20}{100}$ $=5 \times 10^{3}=5 \mathrm{~kJ}$

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Exam: JEE Main 2025
Paper: 29 January, Morning Shift
Subject: Physics
Chapter: Fluid Mechanics
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