Question

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
cm
^2. The maximum pressure the smaller piston
has to bear is:
(g=10m/s^2)


​

Answer 1

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

cm

^2. The maximum pressure the smaller piston

has to bear is:

(g=10m/s^2)

Answer 2

Given :-

Maximum mass that can be lifted = 3000 kg

Area of cross-section of the load-carrying piston = 425 cm²

To Find :-

Maximum pressure would the smaller piston have to bear.

Solution :-

We know that,

• m = Mass
• a = Area
• f = Force
• p = Pressure

Given that,

Maximum mass that can be lifted (m) = 3000 kg

Area of cross-section of the load-carrying piston (a) = $\sf 425 \ cm^{2}= 425 \times 10^{-4} \ m^{2}$

According to the question,

The maximum force exerted by the load,

$\sf F = mg = 3000 \times 9.8 = 30.0 \times 10^{3} \ Pa$

The maximum pressure  on the load carrying piston,

$\sf Pressure=\dfrac{Force}{Area}$

Substituting their values,

$\sf P=\dfrac{3000}{500 \times 10^{-4}}$

$\sf P=4.9 \times 10^{5} \ Pa$

In a liquid, the pressure is transmitted equally in all directions.

Therefore, the maximum pressure on the smaller is 4.9 × 10⁵ Pa