A hydraulic automobile lift is designed to lift vehicles of mass 5000 kg . The area of cross section of the cylinder carrying load is 250 cm 2 . The maximum pressure the smaller piston would have to bear is [Assume g = 10 m s - 2 ]

Question

A hydraulic automobile lift is designed to lift vehicles of mass 5000 kg . The area of cross section of the cylinder carrying load is 250 cm 2 . The maximum pressure the smaller piston would have to bear is [Assume g = 10 m s - 2 ], 20 × 10 6 Pa, 2 × 10 5 Pa, 200 × 10 6 Pa, 2 × 10 6 Pa

MARKS App · Accepted Answer

The given data is m = 5000 kg g = 10 m s - 2 A = 250 × 10 - 4 m 2 Using the formula of pressure, P = F A = m g A = 5000 × 10 250 × 10 - 4 = 2 × 10 6 N m - 2 = 2 × 10 6 Pa

Search any question & find its solution

Looking for more such questions to practice?