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Q.

A hydraulic automobile lift is designed to lift vehicles of mass 5000kg. The area of cross section of the cylinder carrying the load is 250 cm2. The maximum pressure the smaller piston would have to bear is (g=10 m/sec2)

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a

20×106 Pa

b

2×105 Pa

c

200×106 Pa

d

2×106 Pa

answer is D.

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Detailed Solution

From Passcal law

 P=FA=mgA=5000×10250×10−4=2×106Pa

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