Q. A hydraulic lift is designed to life cars of maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is$4.25 \times 10^{- 12}\ m^{2}$. What maximum pressure the smaller piston have to bear?
A) $6.92 \times 10^{5}\ Nm^{- 2}$
B) $7.82 \times 10^{7}\ Nm^{- 2}$
C) $9.63 \times 10^{9}\ Nm^{- 2}$
D) $13.76 \times 10^{11}\ Nm^{- 2}$
Explanation
(a) In hydraulic life, the pressure of smaller piston = pressure of bigger piston $= F/A$
$= (3000 \times 9.8)/(4.25 \times 10^{- 2})$
$= 6.92 \times 10^{5}\ Nm^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.