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A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of the cross-section of the piston carrying the board is 425 cm². What maximum pressure would a smaller piston have to bear?

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$6.92 \times 10^{4} Pa$

$5.92 \times 10^{4} Pa$

$4.92 \times 10^{4} Pa$

$3.92 \times 10^{4} Pa$

answer is A.

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Detailed Solution

The maximum force, which the bigger piston can bear,

$F = 30000 kgf = 3000 \times 9.8 N$

Area of the piston, $A = 425 {cm}^{2} = 425 \times 10^{- 4} m^{2}$

$∴ Maximum pressure on the bigger piston, p = \frac{F}{A} = \frac{300 \times 9.8}{425 \times 10^{- 4}} = 6.92 \times 10^{4} Pa$

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