A force of $50 k g f$ is applied to the smaller piston of a hydraulic machine. Neglecting friction, The force exerted on the large piston, if the diameters of the piston are $5 c m$ and $25 c m$ respectively is $^{'} X^{'} N$ . Find $\frac{X}{250}$

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Solution

We know that Pressure on both the Piston should be same,
$∴ P r e s s u r e o n s m a l l e r P i s t o n = P r e s s u r e o n l a r g e r P i s t o n$
$\frac{F o r c e o n s m a l l e r}{A r e a o f s m a l l e r} = \frac{F o r c e o n l a r g e r}{A r e a o f l a r g e r}$
$∴ F o r c e o n L a r g e r = \frac{F o r c e o n s m a l l e r}{A r e a o f s m a l l e r} \times A r e a o f l a r g e r$
$\Rightarrow \frac{50}{2.5 \times 2.5 \times 3.14} \times 12.5 \times 12.5 \times 3.14 = 1250 k g f$

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A force of 50kgf is applied to the smaller piston of a hydraulic machine. Neglecting friction, The force exerted on the large piston, if the diameters of the piston are 5cm and 25cm respectively is ′X′ N. Find X250

We know that Pressure on both the Piston should be same, ∴Pressure on smaller Piston=Pressure on larger PistonForce on smallerArea of smaller=Force on largerArea of larger∴ Force on Larger=Force on smallerArea of smaller×Area of larger⇒502.5×2.5×3.14×12.5×12.5×3.14=1250kgf

A force of $50 k g f$ is applied to the smaller piston of a hydraulic machine. Neglecting friction, The force exerted on the large piston, if the diameters of the piston are $5 c m$ and $25 c m$ respectively is $^{'} X^{'} N$ . Find $\frac{X}{250}$