In a hydraulic pressure a piston of cross sectional area 100 cm is used to exert a force of 150 N on water for cross sectional area of the position which will support the mass 100 kg.

Dear Student,

Please find below the solution to the asked query:

Consider the Hydraulic lift as shown in the below figure,



So, if a force f is applied at piston C having area of cross section area is a. Therefore, The pressure given on water at the piston is,

$P = \frac{f}{a}$.

According to Pascal's law, this pressure is equally transmitted to the other piston. So,

So, if the other piston is used to lift a weight W, then $P = \frac{W}{A}$.
Where A = area of cross section of the piston D.

Therefore,

$\frac{f}{a} = \frac{W}{A}$.

Given that,
a = 100 cm²​; f  = 150 N and W = 100 kg.

Therefore, 

$A = \frac{Wa}{f} \Rightarrow A = \frac{100\times 10\times 100}{150} \Rightarrow A = \frac{{10}^4}{15} c{m}^2 \Rightarrow A = 666.67 c{m}^2 = 0.067 m^2$.


 

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