In a hydraulic lift used at a service station

Asked by admin @ 17/08/2021 in Physics viewed by 326 People

In a hydraulic lift used at a service station the radius of large and small piston are in ratio 20:1. What weight placed on the small piston will be sufficient to lift a car of mass 1500 kg

Answered by admin @ 17/08/2021

Hey dear,

◆ Answer-
m2 = 3.75 kg

◆ Explaination-
# Given-
r1/r2 = 20/1
m1 = 1500 kg

# Solution-
As both pistons are in synchrony, thrust applied will be equal.
P1 = P2
F1 / F2 = A1 / A2

We know, F = mg & A = πr^2 ,
m1g / m2g = πr1^2 / πr2^2
m2 = m1 (r2/r1)^2
m2 = 1500 × (1/20)^2
m2 = 3.75 kg

Mass applied on small piston is 3.75 kg

Hope this helps you...

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