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 1500Kg?

Given:

As both pistons are in synchrony, the thrust applied will be equal.

P1 = P2

Pressure is the amount of force acting per unit area. That is, P = F/A.

where:

P = pressure

F = normal force

A is the area of the surface on contact. Let us consider A = \(\pi r^{2} \) \(\frac{F_{1}}{\pi r_{1}^{2}} = \frac{F_{2}}{\pi r_{2}^{2}} \) \(\frac{1500}{20^{2}} = \frac{W}{1^{2}} \)

W = 3.75 kg

Hence, the weight to be placed on the small piston sufficient to lift a car of mass 1500 kg is 3.75 kg.

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