The given data is
m=5000kgg=10ms−2A=250×10−4m2
Using the formula of pressure,
P=AF=Amg
=250×10−45000×10
=2×106Nm−2=2×106Pa
A hydraulic automobile lift is designed to lift vehicles of mass 5000kg. The area of cross section of the cylinder carrying load is 250cm2. The maximum pressure the smaller piston would have to bear is [Assume g=10ms−2]
Held on 8 Apr 2023 · Verified 6 Jul 2026.
20×106Pa
2×105Pa
200×106Pa
2×106Pa
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