A hydraulic automobile lift is designed to lift cars with a maximum mass of $3000\; kg$. The area of cross-section of the piston carrying the load is $425 \;cm ^{2} .$ What maximum pressure would the smaller piston have to bear?
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The maximum mass of a car that can be lifted, $m=3000 kg$ Area of cross-section of the load-carrying piston, $A=425 cm ^{2}=425 \times 10^{-4} m ^{2}$ The maximum force exerted by the load,
$F=m g =3000 \times 9.8=29400 N$
The maximum pressure exerted on the load-carrying piston, $ P=\frac{F}{A}$
$=\frac{29400}{425 \times 10^{-4}}=6.917 \times 10^{5} Pa$
Pressure is transmitted equally in all directions in a liquid. Therefore, the maximum pressure that the smaller piston would have to bear is $6.917 \times 10^{5} \;Pa$
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