Fluid Mechanics

JEE Main level, single correct, medium

9 of 192

Question 9 of 192

A hydraulic automobile lift is designed to lift vehicles of mass 5000kg5000kg. The area of cross section of the cylinder carrying load is 250cm2250{cm}^{2}. The maximum pressure the smaller piston would have to bear is [Assume g=10ms−2g=10m{s}^{-2}]

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Answer: (D) 2×106Pa2\times {10}^{6}Pa

The given data is
m=5000kgg=10ms−2A=250×10−4m2m=5000kg g=10m{s}^{-2} A=250\times {10}^{-4}{m}^{2}
Using the formula of pressure,
P=FA=mgAP=\dfrac{F}{A}=\dfrac{mg}{A}
=5000×10250×10−4=\dfrac{5000\times 10}{250\times {10}^{-4}}

=2×106Nm−2=2×106Pa=2\times {10}^{6}N{m}^{-2} =2\times {10}^{6}Pa

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