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9-1.Fluid Mechanics
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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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