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Standard IX

Physics

Application of Pascal's Law

In a hydrauli...

Question

In a hydraulic machine a force of 0.6 N acts normally on the smaller piston of area 0.2 m the force that acts on the larger piston whose area 200 m will be

60 gf

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60 kg wt

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600 kgf

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60 N

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Solution

The correct option is B 60 kg wt
$\begin{matrix} e s s u r e = \frac{F o r c e}{A r e a} p r e s s u r e u n d e r s m a l l e r p i s t o n = \frac{0.6}{0.2} = 3 N / m^{2} F o r c e o n 200 m a r e a i s = 3 \times 200 = 600 N = 60 k g w t \end{matrix}$
Hence,
option $(B)$ is correct answer.

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Similar questions

Q. The areas of the pistons in a hydraulic machine are

5 {c m}^{2}

and

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. What force on the smaller piston will support a load of

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Q. The area of pistons in a hydraulic machine are

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Q. The area of the piston in hydraulic machine are

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. The force required on the smaller piston to support a load of

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The areas of pistons in a hydraulic machine are 5 $c m^{2}$ and 625 $c m^{2}$ . What force on the smaller piston will support a load of 1250 N on the larger piston ? State any assumption which you make in your calculation.

Q. A force of 2 x

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In a hydraulic machine a force of 0.6 N acts normally on the smaller piston of area 0.2 m the force that acts on the larger piston whose area 200 m will be

The correct option is B 60 kg wtessure=ForceAreapressureundersmallerpiston=0.60.2=3N/m2Forceon200mareais=3×200=600N=60kgwtHence,option (B) is correct answer.

The correct option is B 60 kg wt
$\begin{matrix} e s s u r e = \frac{F o r c e}{A r e a} p r e s s u r e u n d e r s m a l l e r p i s t o n = \frac{0.6}{0.2} = 3 N / m^{2} F o r c e o n 200 m a r e a i s = 3 \times 200 = 600 N = 60 k g w t \end{matrix}$
Hence,
option $(B)$ is correct answer.