JEE MAIN - Physics (2025 - 29th January Morning Shift - No. 21)
In a hydraulic lift, the surface area of the input piston is 6 cm2 and that of the output piston is 1500 cm2. 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
Explanation
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$$\begin{aligned} & \frac{\mathrm{F}_1}{\mathrm{~A}_1}=\frac{\mathrm{F}_2}{A_2}, \frac{100}{6}=\frac{\mathrm{F}}{1500}, \mathrm{~F}=\frac{50}{3} \times 1500 \\ & \mathrm{~F}=50 \times 500=25 \times 10^3 \mathrm{~N} \\ & \omega=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{~S}}=25 \times 10^3 \times \frac{20}{100} \\ & =5 \times 10^3=5 \mathrm{~kJ} \end{aligned}$$
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