Question

In a hydraulic machine a force of 20 N is applied on the piston of area 10 cm² What force is obtained on its piston of area 100 cm²?​

Answer 1

Soln,

Given,

In a hydraulic machine,

Force applied(F1)= 20 N

Area of piston where force applied(A1)= $10 cm^2$= $10^{-3} m^2$

Area of piston where force is exerted(A2)= $100 cm^2$=$10^{-2} m^2$

Force exerted(F2) =?

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now,

According to pascals law upon hydraulic machine;

$\frac{F1}{A1} = \frac{F2}{A2}$

or, $F2=\frac{F1*A2}{A1}$

          = $\frac{20N *10^{-2} m^2}{10^{-3}m^2}$

          =$20 N* 10^{-2+3}$

         = $20N*10$

      :. F2  =200 N

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Additional Information used while solving question:

$conversion of cm^2 into m^2\\(100 cm)^2=1m^2\\or, 10000 cm^2= 1m^2\\or, 10^{-4} cm^2 = 1 m^2\\:. 1 cm^2= 10^{-4} m^2$

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now, A1= 10$cm^2$= 10$* 1cm^2$= 10$* 10^{-4} m^2$ =$10^{-3} m^2$

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Again,

A2= 100 $cm^2$ =100$*$$1 cm^2$= 100$*10^{-4} m^2$= $10^{-2} m^2$

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