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)
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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)
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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)
Verified answer
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,
Given that,
Maximum mass that can be lifted (m) = 3000 kg
Area of cross-section of the load-carrying piston (a) = [tex]\sf 425 \ cm^{2}= 425 \times 10^{-4} \ m^{2}[/tex]
According to the question,
The maximum force exerted by the load,
[tex]\sf F = mg = 3000 \times 9.8 = 30.0 \times 10^{3} \ Pa[/tex]
The maximum pressure on the load carrying piston,
[tex]\sf Pressure=\dfrac{Force}{Area}[/tex]
Substituting their values,
[tex]\sf P=\dfrac{3000}{500 \times 10^{-4}}[/tex]
[tex]\sf P=4.9 \times 10^{5} \ Pa[/tex]
In a liquid, the pressure is transmitted equally in all directions.
Therefore, the maximum pressure on the smaller is 4.9 × 10⁵ Pa