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A Hydraulic Automobile Lift is Designed to Lift Cars with a Maximum Mass of 3000 Kg. the Area of Cross-section of the Piston Carrying the Load is 425 Cm2. What Maximum Pressure Would the Smaller Piston Have to Bear? - Physics

A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm2. What maximum pressure would the smaller piston have to bear?

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Solution 1

The maximum mass of a car that can be lifted, m = 3000 kg

Area of cross-section of the load-carrying piston, A = 425 cm= 425 × 10–4 m2

The maximum force exerted by the load, F = mg

= 3000 × 9.8

= 29400 N

The maximum pressure exerted on the load-carrying piston, `P = F/A`

`= 29400/(425xx10^(-4))`

= 6.917 × 105 Pa

Pressure is transmitted equally in all directions in a liquid. Therefore, the maximum pressure that the smaller piston would have to bear is 6.917 × 105 Pa.


Solution 2

Pressure on the piston due to car = `"Weight of car"/"Area of piston"`

`P = (3000xx9.8)/(425xx10^(-4)) Nm^(-2) = 6.92 xx 10^5 "Pa"`

This is also the maximum pressure that the smaller piston would have to bear.

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NCERT Class 11 Physics Textbook
Chapter 10 Mechanical Properties of Fluids
Q 8 | Page 269
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