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 $250\,cm ^2$. The maximum pressure the smaller piston would have to bear is [Assume $g=10\,m / s ^2$]
JEE MAIN 2023, Easy
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$\text { Force }= mg =5000\,g$
$\text { Area of cross section }=250\,cm ^2=250 \times 10^{-4}\,m ^2$
$\text { max imum pressure }=\frac{\text { Force }}{\text { area of cross section }}$
$\quad=\frac{5000\,g }{250 \times 10^{-4}}=\frac{20 \times g }{10^{-4}}=2 \times 10^6\,Pa$
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