The areas of pistons in a hydraulic machine are $5 c m^{2} a n d 625 c m^{2}$ What force on the smaller piston will support a load of 1250N on the larger piston? State any assumption which you make in your calculation.
Assumption:There is no friction and no leakage of liquid.

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Solution

Given:
Area of narrow piston = $5 c m^{2} = A_{1}$ , Let force applied be $F_{1}$ ,
Area of wider piston = $625 c m^{2} = A_{2}$ , let force applied be $F_{2}$ =1250N
We know from the hydraulic machine,
$\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$
$\Rightarrow \frac{F_{1}}{5} = \frac{1250}{625}$
$\Rightarrow F_{1} = 10 N$

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The areas of pistons in a hydraulic machine are 5cm2and625cm2 What force on the smaller piston will support a load of 1250N on the larger piston? State any assumption which you make in your calculation.Assumption:There is no friction and no leakage of liquid.

Given:Area of narrow piston = 5cm2=A1 , Let force applied be F1 , Area of wider piston =625cm2=A2 , let force applied be F2 =1250NWe know from the hydraulic machine, F1A1=F2A2⇒F15=1250625⇒F1=10N

The areas of pistons in a hydraulic machine are $5 c m^{2} a n d 625 c m^{2}$ What force on the smaller piston will support a load of 1250N on the larger piston? State any assumption which you make in your calculation.
Assumption:There is no friction and no leakage of liquid.