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 \( 1000 \, \text{N} \).
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