Answer:
The cross-sectional area of the larger piston is 800 cm².
Explanation:
Given;
mass of the automobile, m = 4000 kg
force applied on the small piston, F₁ = 500 N
area of the smaller piston, A₁ = 10 cm²
load lifted by the larger piston, F₂ = 4000 x 10 = 40,000 N
Pressure experienced by each piston is given as;
[tex]\frac{F_1}{A_1} = \frac{F_2}{A_2}[/tex]
Where;
A₂ is the cross-sectional area of the larger piston
[tex]\frac{F_1}{A_1} = \frac{F_2}{A_2} \\\\A_2 = \frac{F_2A_1}{F_1} \\\\A_2 = \frac{40,000 \ \times\ 10}{500} \\\\A_2=800 \ cm^2[/tex]
Therefore, the cross-sectional area of the larger piston is 800 cm².
