In a hydraulic machine, a force of 2 N is applied on the piston of area of cross section 10 cm ². What force is obtained on it's piston of area of cross section 100 cm ²?

As we know,

Pressure (P) = $\dfrac{\text {Thrust (F)}}{\text {Area (A)}}$

Given,

A = 10 cm ²

Converting cm ² into m ²

100 cm = 1 m

So, 100 cm x 100 cm = 1 m ²

Hence, 10 cm ² = $\dfrac{1 \times 10}{10000}$

Therefore, A = 10^-3 m ²

F = 2 N at area of cross section 10 cm ²

Substituting the values in the formula we get,

P = $\dfrac{2}{10^{-3}}$     [Equation 1]

Now when, A = 100 cm ²

Converting cm ² into m ²

100 cm = 1 m

So, 100 cm x 100 cm = 1 m ²

Hence, 100 cm ² = $\dfrac{1 \times 100}{10000}$

Therefore, A = 10^-2 m ²

Substituting the value in the formula above we get,

Therefore, P = $\dfrac{F}{10^{-2}}$     [Equation 2]

Equating 1 and 2 we get,

$\dfrac{2}{10^{-3}} = \dfrac{F}{10^{-2}} \\[0.5em] \Rightarrow F = \dfrac{2 \times 10^{-2} }{10^{-3}} \\[0.5em] \Rightarrow F = 20 N$

Hence, force at 100 cm ² = 20 N