In a hydraulic lift, the surface area of the input piston is 6cm2 and that of the output piston is 1500cm2. 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
A1F1=A2F2⟹F2=F1⋅A1A2=100×61500=25000 N
The work done in raising the output piston through s=20 cm=0.2 m is
W=F2⋅s=25000×0.2=5000 J=5 kJ
JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Integer answer