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 $\_\_\_\_$ kJ.

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Fluid Mechanics

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

Solution

By Pascal's law, pressure is transmitted equally, so

\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}} ⟹ F_{2} = F_{1} \cdot \frac{A _{2}}{A _{1}} = 100 \times \frac{1500}{6} = 25000 N

The work done in raising the output piston through

s = 20 cm = 0.2

m is

W = F_{2} \cdot s = 25000 \times 0.2 = 5000 J = 5 kJ

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Integer answer

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