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QJEEM25Ph0d3b: JEE Main 2025 Physics - Class 11 JEEM25Ph0d3b Solution: 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 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.

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JEE Main 2025 (Online) 29th January Morning Shift Physics - Work Power & Energy Question 8 English Explanation

amp;F1 A1=F2A2,1006=F1500, F=503×1500amp; F=50×500=25×103 Namp;ω=F→⋅ S→=25×103×20100amp;=5×103=5 kJ\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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