Advertisement Remove all ads

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?

Advertisement Remove all ads

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.

  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 10 Mechanical Properties of Fluids
Q 8 | Page 269
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×