Question

A hydraulic press containing water has two arms with diameters as mentioned in the figure. A force of 10 N is applied on the surface of water in the thinner arm. The force required to be applied on the surface of water in the thicker arm to maintain equilibrium of water is _______ N.

Answer 1

Given data:
- Force applied on the thinner arm, $F_2 = 10 \, \text{N}$
- Diameter of the thinner arm, $d_2 = 1.4 \, \text{cm}$
- Diameter of the thicker arm, $d_1 = 14 \, \text{cm}$

Step 1: Calculating the Cross-Sectional Areas
The cross-sectional area of each arm is given by:

$A = \pi \left( \frac{d}{2} \right)^2.$

For the thinner arm:

$A_2 = \pi \left( \frac{1.4}{2} \right)^2 = \pi (0.7)^2 \, \text{cm}^2.$

For the thicker arm:

$A_1 = \pi \left( \frac{14}{2} \right)^2 = \pi (7)^2 \, \text{cm}^2.$

Step 2: Applying Pascal’s Law
According to Pascal’s law, the pressure exerted on both arms must be equal for equilibrium:

$\frac{F_1}{A_1} = \frac{F_2}{A_2}.$

Rearranging to find $F_1$:

$F_1 = F_2 \times \frac{A_1}{A_2}.$

Step 3: Substituting the Values
Substituting the given values:

$F_1 = 10 \times \frac{\pi (7)^2}{\pi (0.7)^2}.$

Simplifying:

$F_1 = 10 \times \frac{49}{0.49}.$

Calculating:

$F_1 = 10 \times 100 = 1000 \, \text{N}.$

Therefore, the force required to be applied on the surface of water in the thicker arm is $1000 \, \text{N}$.