A hydraulic automobile lift is designed to lift cars . . .

Question : A hydraulic automobile lift is designed to lift cars with 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?


Doubt by Ananya

Solution : 
m = 3000 kg
A = 425 cm2
   = 425×10-4 m2

As per the pascal's law
Pressure at larger piston would be equal to the pressure at smaller piston.

Pressure at Smaller Piston 
= Pressure at Larger Piston
= Force/Area
= Weight/Area
= mg/A
=(3000×9.8)/(425×10-4)
= (29400×104)/425
= 69.176×104
= 6.9176×105
= 6.92×105 N/m2