Practice question
Question
A hydraulic lift raises a 2100kg load using a small piston of radius 3.0cm and a large piston of radius 12cm. What force is applied on the small piston? (Take g\=9.8m/s2)
Explanation
F1 = A1A2F2, F2 = 2100×9.8 = 20580N. A1 = π(0.03)2, A2 = π(0.12)2, A1A2 = 0.00090.0144 = 116. F1 = 2058016 = 1286.25N≈1286N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1286 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
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