Question

A force of 500 N is applied to the smaller piston of a hydraulic machine. Find the force exerted on the larger piston if the diameter of pistons are 5 cm and 25 cm respectively​

Answer 1

We know that Pressure on both the Piston should be same,

∴Pressure on smaller Piston=Pressure on larger Piston

Area of smaller

Force on smaller

=

Area of larger

Force on larger

∴ Force on Larger=

Area of smaller

Force on smaller

×Area of larger

⇒

2.5×2.5×3.14

50

×12.5×12.5×3.14=1250kgf

Answer 2

Answer:

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We know that the pressure on both the pistons should be the same.

Pressure on the smaller piston = Pressure on the larger piston

That is,

$\tt \dfrac{Force \: on \: smaller}{Area \: of \: smaller } = \dfrac{Force \: on \: larger}{Area \: of \: larger }$

Therefore,

$\tt Force \: on \: larger = \dfrac{Force \: on \: smaller}{Area \: of \: smaller } \times {Area \: of \: larger }$

$\tt \implies \dfrac{50}{2.5 \times 2.5 \times 3.14 } \times 12.5 \times 12.5 \times 3.14$

$\implies \boxed{\tt{1250 \: N}}$

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