Question

In a hydraulic press, a piston of a cross-sectional area $100c{m^2}$ is used to exert a force of $150N$ on water. What cross sectional area of the other piston which will support a car of mass $1000Kg$?

Answer 1

Hint: In this question two pistons are compared. One of the piston’s mass and force exerted on it is given. Another person's mass is given. We can find the force exerted on this piston as force is the product of mass and acceleration. Use the acceleration due to gravity. Now we have to find the cross section area of another piston. It can be found using Pascal’s law.

Complete step by step answer
Pascal’s law
Pascal’s law states that the external static pressure applied on a confined liquid is distributed or transmitted evenly throughout the liquid in all directions.
The law gives an equation
$ \Rightarrow F = PA$
$ \Rightarrow P = \dfrac{F}{A}$
Where,
F is the force applied
P is the pressure transmitted
A is the area of cross section
By Pascal's law, the pressure increased using one piston is communicated everywhere in the fluid. Thus, the same pressure acts on the other piston.
Cross-sectional area of the piston 1, ${A_1} = 100c{m^2} = 100 \times {10^{ - 4}}{m^2}$
Force applied on the piston 1, ${F_1} = 150N$
There is another piston which will support a car of mass, ${M_2} = 1000Kg$
The pressure transmitted on the piston 1, P=?
$ \Rightarrow P = \dfrac{F}{A}$
$ \Rightarrow {P_1} = \dfrac{{150}}{{100 \times {{10}^{ - 4}}}}$
$ \Rightarrow {P_1} = 15 \times {10^3}{\text{ }} \to {\text{1}}$
Force applied on the piston 2:
We know that,
$ \Rightarrow F = ma$
Mass, ${M_2} = 1000Kg$
Acceleration due to gravity $a = 9.8{\text{ m/s}}$
 $ \Rightarrow {F_2} = 1000 \times 9.8$
$ \Rightarrow {F_2} = 9800$
Cross sectional area of the other piston which will support a car, ${A_2} = ?$
The pressure transmitted on the piston 2, P=?
$ \Rightarrow P = \dfrac{F}{A}$
$ \Rightarrow {P_2} = \dfrac{{9800}}{{{A_2}}}{\text{ }} \to {\text{2}}$
Equating the equation 1 and 2
$ \Rightarrow {P_1} = {P_2}$
$ \Rightarrow 15 \times {10^3} = \dfrac{{9800}}{{{A_2}}}$
$ \Rightarrow {A_2} = \dfrac{{9800}}{{15 \times {{10}^3}}}$
$ \Rightarrow {A_2} = 653.33 \times {10^{ - 3}}{m^2}$
$ \Rightarrow {A_2} = 0.6533{m^2}$

Cross sectional area of the other piston which will support a car, ${A_2} = 0.6533{m^2}$

Note According to Pascal’s law, the pressure on a piston creates an equal increase in pressure on another piston in the system. If the second piston has an area five times that of the first, the force on the second piston is five times greater than that of first, though the pressure is the same as that on the first piston.