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 ?

Open in App

Solution

Given, the maximum mass of the automobile is 3000 kg and the area of cross section of the piston is 425 cm 2 .

The pressure due to the piston on the car is given by the equation,

P= W A = mg A

Substituting the values in the given equation, we get:

P= 3000 kg×9.8 m/ s 2 425× 10 −4 m 2 =6.92× 10 5 Pa

Since, the pressure is same for both the pistons, the pressure exerted on the small piston is 6.92× 10 5 Pa.

Suggest Corrections

Similar questions

Q. A hydraulic automobile lift is designed to lift cars with a maximum mass of

300 kg

. The area of cross-section of the piston carrying the load is

500 {cm}^{2}

. What maximum pressure would smaller piston have to bear?

Q. A hydraulic automobile lift is designed to lift cars with a maximum mass of

300 kg

. The area of cross-section of the piston carrying the load is

500 {cm}^{2}

. What maximum pressure would smaller piston have to bear?

Q. A hydraulic lift is used to lift a car of mass

3000 k g

. The cross-sectional area of the lift on which the car is supported is

5 \times 10^{- 2} m

. What is the pressure on the smaller piston, if both the pistons are at the same horizontal level?
Take

g = 10 m s^{- 2}

In a motor garage, a hydraulic lift is used to lift heavy automobiles such as trucks. The cross-sectional areas of the piston carrying the load and the piston where force is applied are in the ratio 5 : 3. The mass of a truck is 15000 kg.

What is the magnitude of the force applied on the piston to just lift the truck?

Q. To lift an automobile of

1600 k g

, a hydraulic pump with a larger piston

800 c m^{2}

in area is employed. If the area of the smaller piston is

10 c m^{2}

. The force that must be applied on it is

Join BYJU'S Learning Program