JEE Main 2025 Jan 29 Shift 1, Physics Q24: Hydrostatics

JEE Main2025Jan 29, Shift 1Physics

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.

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Solution

Pascal's principle gives equal pressure on both pistons: $\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}} .$

Output force: $F_{2} = F_{1} \cdot \frac{A _{2}}{A _{1}} = 100 \cdot \frac{1500}{6} = 25000 N .$

The output piston rises $s = 20 cm = 0.20 m .$

Work done on the load: $W = F_{2} \cdot s = 25000 \cdot 0.20 = 5000 J = 5 kJ .$

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JEE Main 2025 Jan 29 Shift 1, Physics Q24: Hydrostatics

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