In a typical hydraulic press, a force of 20 N is exerted on small piston of area 0.050 m2. What is force exerted by large piston on load if it has an area of 0.50 m2?

200N

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100N

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50N

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10N

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Solution

The correct option is A 200N
Using $\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$
$∴$ $\frac{20}{0.05} = \frac{F_{2}}{0.5}$ $⟹ F_{2} = 200$ N

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In a typical hydraulic press, a force of 20 N is exerted on small piston of area 0.050 m2. What is force exerted by large piston on load if it has an area of 0.50 m2?

The correct option is A 200NUsing F1A1=F2A2∴ 200.05=F20.5 ⟹F2=200 N

The correct option is A 200N
Using $\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$
$∴$ $\frac{20}{0.05} = \frac{F_{2}}{0.5}$ $⟹ F_{2} = 200$ N