Practice question
Question
A hydraulic lift raises a 2300kg load using a small piston of radius 5.0cm and a large piston of radius 20cm. What force is applied on the small piston? (Take g\=9.8m/s2)
Explanation
F1 = A1A2F2, F2 = 2300×9.8 = 22540N. A1 = π(0.05)2, A2 = π(0.2)2, A1A2 = 0.00250.04 = 116. F1 = 2254016 = 1408.75N≈1409N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1409 N. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.
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