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PhysicsFluidPressureMedium2 minPYQ_2025

PhysicsMediumnumerical

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.

Answer:

5.00

Solution:

$\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}}, \frac{100}{6} = \frac{F}{1500}, F = \frac{50}{3} \times 1500 F = 50 \times 500 = 25 \times 1 0^{3} N ω = F . S = 25 \times 1 0^{3} \times \frac{20}{100} = 5 \times 1 0^{3} = 5 kJ$

Stream:JEESubject:PhysicsTopic:FluidSubtopic:Pressure

⏱ 2mℹ️ Source: PYQ_2025

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