Question

A force of $50kgf$ 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. $5cm$ and $25cm$ respectively.

• A. $250kgf$
• B. $4500kgf$
• C. $1250kgf$
• D. $1000kgf$

Answer 1

Answer: (C)

$1250kgf$

Let force and area of smaller piston be $F$ and $A$.

Force and area of larger piston be $F$ and $A$.

Radius of smaller piston $=\frac{5}{2}=2.5cm$

Radius of large piston $=\frac{25}{2}$

Since,

$Pressure=\frac{Force}{Area}$

so,

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

$A_1=\pi (r{)}^2=\pi (2.5{)}^{2}cm$

$A_2=\pi (r{)}^2=\pi (12.5{)}^{2}cm$

$\frac{50}{\pi \times 2.5\times 2.5}=\frac{F_2}{\pi \times 12.5\times 12.5}$

$F_2=\frac{50\times 12.5\times 12.5}{6.25}=1250kgf$

Hence force on large piston $=1250kgf$