91Ó°ÊÓ

(i) Diameter of small piston,$\mathrm{d}=3.8\mathrm{cm}\text{or}0.038\mathrm{m}$

(ii) Diameter of large piston,$\mathrm{D}=53\mathrm{cm}\text{or}0.53\mathrm{m}$

(iii) Force on larger piston,$\mathrm{F}=20×{10}^3\mathrm{N}$

We can use Pascal’s law, which states that the change in the pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and the wall of the container. In other words, we can say that the ratio of force to the area across the container would remain constant.

Formula:

Relation of pressure of larger and smaller piston using Pascal’s law,

$\frac{\mathrm{F}}{\mathrm{A}}=\frac{\mathrm{f}}{\mathrm{a}}$ (i)

Where,

$\mathrm{F}$and $\mathrm{A}$are the force and area of larger piston respectively; $\mathrm{f}$and$\mathrm{a}$are the force and area of the smaller piston respectively.

Using equation (i) and the given values, we get

$\mathrm{F}=\left(\begin{array}{c}\mathrm{A}\\ \mathrm{a}\end{array}\right)\mathrm{f}$

Hence, the equation of force that will define the force of larger piston is$\mathrm{F}=\left(\begin{array}{c}\mathrm{A}\\ \mathrm{a}\end{array}\right)\mathrm{f}$

Using equation (i) and the given values, we get the force on smaller piston as:

$\mathrm{F}=\left(\begin{array}{c}\mathrm{A}\\ \mathrm{a}\end{array}\right)\mathrm{f}$

$=\left(\frac{0{.038}^2}{0{.53}^2}\right)20×{10}^3$

$=103\mathrm{N}$

Hence, the force required by smaller piston to balance the large piston is $103\mathrm{N}$