9-1.Fluid Mechanics
medium

A hydraulic automobile lift is designed to lift cars with a maximum mass of $3000\, kg$. The area of cross section a of piston carrying the load is $425\, cm ^{2}$. What is the maximum pressure () would smaller piston have to bear ?

A

$15.82 \times 10^{5} Pa$

B

$1.12 \times 10^{5} Pa$

C

$2.63 \times 10^{5} Pa$

D

$6.92 \times 10^{5} Pa$

(AIIMS-2019)

Solution

The force acting on bigger piston is,

$F=m g$

$F=3000 kg \times 9.8 m / s ^{2}$

$F=29400 N$

The area of the piston is,

$A=425 cm ^{2}=425 \times 10^{-4} m ^{2}$

The pressure acting on the bigger piston is,

$P=\frac{F}{A}$

$P=\frac{29400 N }{425 \times 10^{-4} m ^{2}}$

$P=6.92 \times 10^{5} Pa$

Standard 11
Physics

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