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

Please find this answer

3.75 kg

Hey


◆ 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...

Answer-
m2 = 3.75 kg

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

 Solution-
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