Question

The area of cross-section of the two arms of a hydraulic press is $1{\text{cm}}^2$ and $10{\text{cm}}^2$, respectively as shown in figure. A force of $10\text{N}$ is applied on the water in the thinner arm. What force should be applied on the water in the thicker arms so that the water may remain in equilibrium?

• A. $10\text{N}$
• B. $20\text{N}$
• C. $100\text{N}$
• D. $200\text{N}$

Answer 1

The correct option is C $100\text{N}$

Given,
$F_1=10\text{N};F_2=F\text{N}{A}_1=1{\text{cm}}^2;A_2=10{\text{cm}}^2$


At the equilibrium condition, the pressure at the two surfaces should be equal as they lie in the same horizontal level if fluid is same throughout the container.

So pressure at section $1$,

$P_{!}=\frac{F_{!}}{A_1}=P_2$

$\therefore F_2=P_2{A}_2=\frac{A_2}{A_1}{F}_1$

$F_2=\frac{10}{1}\times 10$

$\therefore F=100\text{N}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{C}\right)$ is the correct answer.

Why this question ? Key concept : The pressure applied at one point in an enclosed fluid is transmitted uniformly to every part of the fluid and to the walls of the container.