A car lift uses a small piston of radius 4cm to lift a 1200kg car on a piston of radius 12cm. What force is applied on t
F1 = A1A2F2, F2 = mg = 1200×10 = 12000N. A1 = π(0.04)2, A2 = π(0.12)2. A1A2 = (0.04)2(0.12)2 = 0.00160.0144 = 19. F1 = 120009 = 1333.33N≈1333N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1333 N. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.