Question

In a typical hydraulic press, a force of 20 N is exerted on a small piston of area 0.050 $m^2$. What is the force exerted by a large piston on load if it has an area of 0.50 $m^2$ ?
A) 200 N
B) 100 N
C) 50 N
D) 10 N

Answer 1

Hint
As, here are two pistons one is small and other is large we have to calculate the force exerted by the large piston. For this we know that the hydraulic pressure is the ratio of force per unit area. We write the hydraulic pressure for both small and large piston. After equating both hydraulic pressure and substitute the given values then we will get the required value of the force by the large piston.

Complete step by step solution
Here, it is given that; Force exerted on small piston is ${F_1} = 20N$
And the area on which force is exerted by small piston is ${A_1} = 0.050{m^2}$
As, hydraulic pressure is the ratio of applied force to the area on which force is applied
Therefore, for small piston hydraulic pressure is ${P_1} = \dfrac{{20}}{{0.050}}$ …………………. (1)
Let the force exerted by the large piston on load is $F_2$.
And area on which force is exerted by the large piston is ${A_2} = 0.50{m^2}$
Therefore, hydraulic pressure for a large piston is ${P_2} = \dfrac{{{F_2}}}{{0.50}}$ ……………………. (2)
As the, both hydraulic pressure is equal so equating the equation (1) and (2), we get
$ \Rightarrow \dfrac{{20}}{{0.05}} = \dfrac{{{F_2}}}{{0.5}}$
$ \Rightarrow {F_2} = 200N$
Thus, the force exerted by the large piston is 200 N.
Hence, (A) option is correct.

Note
A hydraulic press is a machine press using a hydraulic cylinder to generate a compressive force. It uses the hydraulic equivalent of a mechanical lever, and was also known as a Bramah press. The hydraulic press depends on pascal’s principle -the pressure throughout a closed system is constant.