PhysicsMedium148×since 2002Q7918A hydraulic automobile lift is designed to lift vehicles of mass 5000 kg5000 \mathrm{~kg}5000 kg. The area of cross section of the cylinder carrying the load is 250 cm2250 \mathrm{~cm}^{2}250 cm2. The maximum pressure the smaller piston would have to bear is [\left[\right.[ Assume g=10 m/s2]\left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right]g=10 m/s2]A20×10+6 Pa20 \times 10^{+6} \mathrm{~Pa}20×10+6 PaB200×10+6 Pa200 \times 10^{+6} \mathrm{~Pa}200×10+6 PaC2×10+5 Pa2 \times 10^{+5} \mathrm{~Pa}2×10+5 PaD2×10+6 Pa2 \times 10^{+6} \mathrm{~Pa}2×10+6 PaCheck answerSkip