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Byju's Answer

Standard IX

Physics

Pascal's Law

In a hydrauli...

Question

In a hydraulic machine, the two pistons are of the area of cross-section in the ratio 1 : 20. What force is needed on the narrow piston to overcome a force of 200 N on the wider piston?

1 : 1

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2 : √3

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√3 : 2

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2 : 1

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Solution

The correct option is B
2 : √3

Given, $A_{1}$ : $A_{2}$ = 1 : 20
$F_{2}$ = 200 N
By the principle of hydraulic machine,
Pressure on narrow piston = Pressure on wider piston
$\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$
Thus, $F_{1} = F_{2} \times \frac{A_{1}}{A_{2}} = 200 \times \frac{1}{20} = 10 N$

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In a hydraulic machine, the two pistons are of the area of cross-section in the ratio 1 : 20. What force is needed on the narrow piston to overcome a force of 200 N on the wider piston?

The correct option is B 2 : √3 Given, A1 : A2 = 1 : 20 F2 = 200 N By the principle of hydraulic machine, Pressure on narrow piston = Pressure on wider piston F1A1=F2A2 Thus, F1=F2×A1A2=200×120=10N