Question

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 5cm and 25cm respectively.
A. 250 kgf
B. 750 kgf
C. 1250 kgf
D. 1000 kgf

Answer 1

Hint: Use the principle of hydraulic machine, which is pressure on the smaller piston is equal to pressure on a large piston. Pressure is the ratio of force and area. Using this, do find the unknown force of the large piston.

Complete step by step answer:
We are given that a force of 50 kgf is applied to the smaller piston of a hydraulic machine neglecting friction and the diameters of the piston are 5 cm and 25 cm respectively.
We have to find the force exerted on the large piston.
We are given that the diameters of the smaller piston and large piston are 5cm, 25cm respectively.
The ratio of the diameters of smaller and large pistons is $5:25$
Then the ratio of their area will be ${5^2}:{25^2} = 25:625$ as the area of the pistons is $\pi {r^2}$, where r is the radius of the corresponding piston, as pie is constant the squares of the radii will be retained.
By the principle of hydraulic machine (Pascal’s law), pressure applied anywhere to a body of fluid causes a force to be transmitted equally in all directions. Therefore the pressure applied on the smaller piston is equal to pressure applied on the large piston.
${P_s} = {P_l}$
Pressure is defined as the ratio of force and area
$
  {P_s} = {P_l} \\
  \dfrac{{{F_s}}}{{{A_s}}} = \dfrac{{{F_l}}}{{{A_l}}} \\
  \dfrac{{{F_l}}}{{{F_s}}} = \dfrac{{{A_l}}}{{{A_s}}} \\
  {A_l}:{A_s} = 25:625 \\
  {F_s} = 50kgf \\
  \Rightarrow \dfrac{{{F_l}}}{{50}} = \dfrac{{625}}{{25}} \\
  \Rightarrow {F_l} = \dfrac{{625 \times 50}}{{25}} \\
  \Rightarrow {F_l} = 25 \times 50 = 1250kgf \\
 $
Therefore, the force exerted on the large piston is 1250 kgf.
The correct option is Option C, which is 1250 kgf.

Note:Hydraulic machines use liquid fluid power to perform work. In this type of machines, hydraulic fluid is pumped to various hydraulic motors and hydraulic cylinders throughout the machine and becomes pressurized according to the resistance present in them.