Question

In a hydraulic lift, compressed air exerts a force $F_1$ on a small piston having a radius of $5\text{cm}$. This pressure is transmitted to a second piston of radius $15\text{cm}$. If the mass of the load to be lifted is $1350\text{kg}$, calculate the pressure acting on the smaller piston.
[Take $g=10{\text{m/s}}^2$]

• A. $3\times {10}^5\text{Pa}$
• B. $2\times {10}^5\text{Pa}$
• C. $1.9\times {10}^5\text{Pa}$
• D. $1.9\text{Pa}$

Answer 1

The correct option is C $1.9\times {10}^5\text{Pa}$

Since pressure is transmitted undiminished throughout the fluid (Pascal's law).
$\therefore P_1=P_2$
Thus, pressure acting on smaller piston
$P_1=\frac{F_1}{A_1}=\frac{F_2}{A_2}=\frac{(1350\times 10)}{\pi (15\times 15)\times {10}^{-4}}$
$\approx 1.9\times {10}^5\text{Pa}$