Question

The small piston of a hydraulic lift has an area of $0.20m^2$. A car weighing $1.20\times {10}^{4}N$ sits on a rack mounted on the large piston. The large piston has an area of $0.90m^2$. How large a force must be applied to the small piston to support the car?

• A. $2.67\times {10}^{3}N$
• B. $2.67\times {10}^{4}N$
• C. $2.67\times {10}^{5}N$
• D. $2.67\times {10}^{6}N$

Answer 1

The correct option is A $2.67\times {10}^{3}N$

Hydraulic lift works on the principle of pascal's law
$P=\frac{F_1}{a_1}=\frac{F_2}{a_2}$
applying the given formula to the question.
$\frac{1.2\times {10}^4}{0.90}=\frac{F}{0.2}F=\frac{0.24\times {10}^4}{0.90}=2.67\times {10}^{3}N$