Question

A hydraulic automobile lift is designed to lift cars with a maximum mass of $300\text{kg}$. The area of cross-section of the piston carrying the load is $500{\text{cm}}^2$. What maximum pressure would smaller piston have to bear?

• A. ${2.94\times 10}^4\text{Pa}$
• B. ${11.66\times 10}^4\text{Pa}$
• C. ${5.88\times 10}^4\text{Pa}$
• D. ${8.55\times 10}^4\text{Pa}$

Answer 1

The correct option is C ${5.88\times 10}^4\text{Pa}$

The maximum force, which the bigger piston can bear, $F=300\text{kgf}=300\times 9.8\text{N}$
$\therefore$ Maximum pressure on the bigger piston,
$P=\frac{F}{A}=\frac{300\times 9.8}{{500\times 10}^{-4}}={5.88\times 10}^4\text{Pa}$
Since, the liquid transmits pressure equally, therefore the maximum pressure the smaller piston can bear is ${5.88\times 10}^4\text{Pa}$