Question

The area of cross-section of the two arms of a hydraulic press is $1{\text{cm}}^2$ and $20{\text{cm}}^2$, respectively. A force of $10\text{N}$ is applied on the water, on the liquid in the thinner arm. What force should be applied on the water in the thicker arm, so that the water levels may remain in equilibrium, in both arms?

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

Answer 1

The correct option is D $200\text{N}$

In equilibrium, the pressures at the two surfaces should be equal as the water level in both arms is static. If the atmospheric pressure is $P_a$ and a force $F$ is applied to maintain the equilibrium, the pressure just below the pistons are,

$P_M=P_a+\frac{10}{1\times {10}^{-4}}$

$P_N=P_a+\frac{F}{20\times {10}^{-4}}$

These pressures should be the same i.e., $P_M=P_N$

$\Rightarrow P_a+\frac{10}{1\times {10}^{-4}}=P_a+\frac{F}{20\times {10}^{-4}}$

After solving, we get, $F=200\text{N}$.

Hence, $\left(D\right)$ is the correct answer.

$\begin{array}{c}\text{Why this Question ?}\\ \text{PASCAL'S PRINCIPLE}\\ \text{The pressure applied at one point, in an enclosed liquid, is equally distributed in all directions.}\\ \text{Practical example - Hydraulic lift}\end{array}$