For the system shown in figure, the cylinder on the left, at L, has a mass of 600 kg and a cross-sectional area of `800 cm^(2)`. The piston on the right, at S, has cross-sectional area `25 cm^(2)` and negligible weight. If the apparatus is filled with oil `(rho=0.78 g//cm^(3))`, what is the force F required to hold the system is equilibrium?
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Answer

A, C

Explanation

Let `A_(1)`=area of cross-section on LHS and
`A_(2)`=area of cross-section on RHS
Equating the pressure on two sides of the dotted line.
Then, `(F)/(A_(2))+ rho gh =(Mg)/(A_(1)) (M=600kg)`
`:. F=A_(2) [(Mg)/(A_(1))-rhogh]`
`=(25xx10^(-4)) [(600xx9.8)/(800xx10^(-4))-780xx9.8xx8]`
`=31N`.