Question

In a hydraulic lift, force $F_1$ isexerted on small cylinder $A$ of radius $7cm$. This pressure is transmitted to a second cylinder of radius $14cm$ through a pipe $C$. If the car is to be lifted is $1500kg.$ What is the pressure applied to cylinder $A?$

• A. $1kPa$
• B. $3\times 10^{6}Pa$
• C. $354\times 10^{4}Pa$
• D. $382\times 10^{4}Pa$

Answer 1

Answer: (D)

$F_1{A}_1=F_2{A}_2$ (Pascal's law)
$F_2=1500kgf=1500\times 9.8$
$=14700N$
$r_2=14cm=14\times 10^{-2}m$
$F_1=\frac{F_2\times A_2}{A_1}=F_2\times \frac{r_2^2}{r_1^2}$
$={\left(\frac{14}{7}\right)}^2\times 14700$
$F_1=14700\times 4=58800N$
Pressure $=\frac{F_1}{A_1}=\frac{58800}{\frac{22}{7}\times 7\times 7\times 10^{-4}}$
$=\frac{58800}{154}\times 10^4$
$P=381.8\times 10^{4}Pa$