Question

In a hydraulic lift, force $F_{1}$ isexerted on small cylinder $A$ of radius $7 cm$. This pressure is transmitted to a second cylinder of radius $14 cm$ through a pipe $C$. If the car is to be lifted is $1500 kg .$ What is the pressure applied to cylinder $A ?$

• A. $1\,kPa$
• 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)
$F _{2}=1500 kgf =1500 \times 9.8$
$=14700 N$
$r _{2}=14 cm =14 \times 10^{-2} m$
$F _{1}=\frac{ F _{2} \times A _{2}}{ A _{1}}= F _{2} \times \frac{ r _{2}^{2}}{ r _{1}^{2}}$
$=\left(\frac{14}{7}\right)^{2} \times 14700$
$F _{1}=14700 \times 4=58800 N$
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 $