A force of 50 kgf is applied to the smaller piston of a hydraulic machine Neglecting friction, then find the force exerted on the large piston, the diameters of the piston being 5 cm and 25 cm respectively.

250 kgf

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750 kgf

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1250 kgf

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1000 kgf

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Solution

The correct option is B 1250 kgf
We know that the pressure on both the piston should be same.

$P r e s s u r e o n s m a l l e r p i s t o n = P r e s s u r e o n l a r g e r p i s t o n$

That is,

$\frac{F o r c e o n s m a l l e r}{A r e a o f s m a l l e r} = \frac{F o r c e o n l a r g e r}{A r e a o f l a r g e r}$

Therefore,

$F o r c e o n l a r g e r = \frac{F o r c e o n s m a l l e r}{A r e a o f s m a l l e r} \times A r e a o f l a r g e r$

$⟹ \frac{50}{2.5 \times 2.5 \times 3.14} \times 12.5 \times 12.5 \times 3.14 = 1250 k g f$

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