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?
The maximum mass of a car that can be lifted, m = 3000 kg
Area of cross-section of the load-carrying piston, A = 425 cm2 = 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 / 425 x 10-4
= 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.
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Welcome to the NCERT Solutions for Class 11 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 8: A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cro....
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