A hydraulic lift is shown in the figure. The radii of the movable pistons $P_1$ and $P_2$ are of $2…
A hydraulic lift is shown in the figure. The radii of the movable pistons $P_1$ and $P_2$ are of $2 \mathrm{~m}$ and $8 \mathrm{~m}$ respectively. If a body of mass $2 \mathrm{~kg}$ is placed on piston $P_1$ then the force on piston $P_2$ is
(Ignore atmospheric pressure, acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
$320 \mathrm{~N}$
$80 \mathrm{~N}$
$1280 \mathrm{~N}$
$20 \mathrm{~N}$
Solution
In a hydraulic machine pressure at both pistons is same.
Pressure at piston $P_1=$ Pressure at piston $P_2$
$\Rightarrow \frac{\text { Force on } P_1}{\text { Area of } P_1}=\frac{\text { Force on } P_2}{\text { Area of } P_2}$
$\Rightarrow$ Force on $P_2=\frac{\text { Area } P_2}{\text { Area } P_1} \times$ Force $P_1$
$\Rightarrow$ Force on $P_2=\frac{\pi R_2^2}{\pi R_1^2} \times m g$
$=\frac{R_2^2 \times m g}{R_1^2}=\frac{8^2 \times 2 \times 10}{2^2}=320 \mathrm{~N}$.
