In a hydraulic machine, a force of $2 N$ is applied on the piston of area of cross section $10 c m^{2}$ . What force is obtained on its piston of area of cross section $100 c m^{2}$ ?

20 N

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10 N

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5 N

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40 N

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Solution

The correct option is B 20 N
We know that, $P r e s s u r e = \frac{F o r c e}{A r e a}$
$\Rightarrow P r e s s u r e = \frac{2}{10} = 0.2 N c m^{- 2}$
Now, $P r e s s u r e = 0.2$
Again, $F o r c e = P r e s s u r e \times A r e a$
$F o r c e = 0.2 \times 100 = 20 N$

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In a hydraulic machine, a force of 2N is applied on the piston of area of cross section 10cm2. What force is obtained on its piston of area of cross section 100cm2?

The correct option is B 20 NWe know that, Pressure=ForceArea⇒Pressure=210=0.2Ncm−2Now, Pressure=0.2Again, Force=Pressure×AreaForce=0.2×100=20N

In a hydraulic machine, a force of $2 N$ is applied on the piston of area of cross section $10 c m^{2}$ . What force is obtained on its piston of area of cross section $100 c m^{2}$ ?

The correct option is B 20 N
We know that, $P r e s s u r e = \frac{F o r c e}{A r e a}$
$\Rightarrow P r e s s u r e = \frac{2}{10} = 0.2 N c m^{- 2}$
Now, $P r e s s u r e = 0.2$
Again, $F o r c e = P r e s s u r e \times A r e a$
$F o r c e = 0.2 \times 100 = 20 N$