Exam

JEE Main

Year

2025

Date

29 January

Shift

Morning Shift

Subject

Physics

Chapter

Mechanical Properties of Fluids

JEE Main 2025 · 29 January · Morning Shift · Physics · Mechanical Properties of Fluids

In a hydraulic lift, the surface area of the input piston is (6...

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.

Step-by-step solution

$$\begin{aligned} & \frac{\mathrm{F}_1}{\mathrm{~A}_1}=\frac{\mathrm{F}_2}{\mathrm{~A}_2}, \frac{100}{6}=\frac{\mathrm{F}}{1500}, \mathrm{~F}=\frac{50}{3} \times 1500 \\ & \mathrm{~F}=50 \times 500=25 \times 10^3 \mathrm{~N} \\ & \omega=\overrightarrow{\mathrm{F}} . \overrightarrow{\mathrm{S}}=25 \times 10^3 \times \frac{20}{100} \\ & =5 \times 10^3=5 \mathrm{~kJ}\end{aligned}$$