Ncert Exercises & Solutions

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

$P r e s s u r e (P) o n t h e p i s t o n d u e t o c a r = \frac{W e i g h t o f c a r}{A r e a o f p i s t o n}$
$= \frac{3000 \times 9.8}{425 \times 10^{- 4}}$
$= 6.92 \times 10^{5} P a .$

The pressure is transmitted equally in all directions in a liquid. Therefore, the maximum pressure that the smaller piston would have to bear is 6.92 × 10⁵ Pa.

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