Practice question
Question
A hydraulic lift raises a 1700kg load using a small piston of radius 3.5cm and a large piston of radius 14cm. What force is applied on the small piston? (Take g\=10m/s2)
Explanation
F1 = A1A2F2, F2 = 1700×10 = 17000N. A1 = π(0.035)2, A2 = π(0.14)2, A1A2 = 0.0012250.0196 = 116. F1 = 1700016 = 1062.5N≈1063N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1063 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
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