A hydraulic automobile lift is designed to lift vehicles of mass 5000 kg. The area of cross section of the cylinder carrying the load is 250 cm². The maximum pressure the smaller piston would have to bear is [Assume g = 10m/s²]

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$20 \times 10^{+ 6} Pa$

$2 \times 10^{+ 6} Pa$

$2 \times 10^{+ 5} Pa$

$200 \times 10^{+ 6} Pa$

answer is A.

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Detailed Solution

Wt. to be lifted = 5000 kg
$5000 g = (ΔP) (A)$
$\begin{array}{l} \Rightarrow ΔP = \frac{5000 g}{A} = \frac{5000 \times 10}{250 \times 10^{- 4}} \\ = 200 \times 10^{4} = 2 \times 10^{6} pascal \end{array}$

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