MCQ
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
    $2.63 \times 10^{5} Pa$
  • $6.92 \times 10^{5} Pa$

Answer

Correct option: D.
$6.92 \times 10^{5} Pa$
d
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,

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

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

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

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