The area of pistons in a hydraulic machine are $5 c m^{2}$ and $625 c m^{2}$ . The force on the smaller piston so that a load of $1250 N$ on the larger piston can be supported, is X N. Find $\frac{X}{2}$ N.

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1250

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Solution

The correct option is A 5
We know that, $P r e s s u r e = \frac{F o r c e}{A r e a}$
For Larger Piston, $P_{l} = \frac{1250}{625} = 2 N c m^{- 2}$
So, Pressure on the smaller piston should be equal to $2 N c m^{- 2}$
So, Force on smaller piston $X = P r e s s u r e \times A r e a = 2 \times 5 = 10 N$
$∴ \frac{X}{2} = \frac{10}{2} = 5 N$

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The area of pistons in a hydraulic machine are 5cm2 and 625cm2. The force on the smaller piston so that a load of 1250N on the larger piston can be supported, is X N. Find X2 N.

The correct option is A 5We know that, Pressure=ForceAreaFor Larger Piston, Pl=1250625=2Ncm−2So, Pressure on the smaller piston should be equal to 2Ncm−2So, Force on smaller piston X=Pressure×Area=2×5=10N∴X2=102=5N

The area of pistons in a hydraulic machine are $5 c m^{2}$ and $625 c m^{2}$ . The force on the smaller piston so that a load of $1250 N$ on the larger piston can be supported, is X N. Find $\frac{X}{2}$ N.