Q 11-09-053JEE MainJEE Main 2025 (29 Jan, Shift 1)Easy
In a hydraulic lift, the surface area of the input piston is $6\ \text{cm}^2$ and that of the output piston is $1500\ \text{cm}^2$. If a $100\ \text{N}$ force is applied to the input piston to raise the output piston by $20\ \text{cm}$, then the work done is ______ kJ.
Numerical value type. Enter your answer.
Answer: 5
By Pascal's law the output force is
$$F_2 = F_1\frac{A_2}{A_1} = 100\times\frac{1500}{6} = 25000\ \text{N}$$
Work done on the load (equal to the work done at the input, the lift being ideal):
$$W = F_2h = 25000\times0.20 = 5000\ \text{J} = 5\ \text{kJ}$$
Solution by Sreeraj P, M.Sc Physics