To solve the problem step by step, we need to follow these steps:
### Step 1: Understand the Forces Acting on the Ball
When the spherical ball enters the water, three forces act on it:
1. The gravitational force (weight) acting downwards: \( F_g = mg \)
2. The buoyant force acting upwards: \( F_b = \rho_w V g \)
3. The viscous force acting upwards: \( F_v = 6 \pi r \eta v_t \)
At terminal velocity, the net force acting on the ball is zero, which means:
\[ F_g - F_b - F_v = 0 \]
### Step 2: Calculate the Mass of the Ball
The mass \( m \) of the ball can be calculated using its density and volume:
\[ m = \rho_b V \]
Where:
- \( \rho_b = 10^4 \, \text{kg/m}^3 \) (density of the ball)
- \( V = \frac{4}{3} \pi r^3 \) (volume of the sphere)
Given \( r = 3 \times 10^{-4} \, \text{m} \):
\[ V = \frac{4}{3} \pi (3 \times 10^{-4})^3 = \frac{4}{3} \pi (27 \times 10^{-12}) = 36 \pi \times 10^{-12} \, \text{m}^3 \]
Now, substituting in the mass:
\[ m = 10^4 \times 36 \pi \times 10^{-12} = 36 \pi \times 10^{-8} \, \text{kg} \]
### Step 3: Calculate the Gravitational Force
The gravitational force acting on the ball is:
\[ F_g = mg = (36 \pi \times 10^{-8}) g \]
Using \( g = 10 \, \text{m/s}^2 \):
\[ F_g = 36 \pi \times 10^{-8} \times 10 = 36 \pi \times 10^{-7} \, \text{N} \]
### Step 4: Calculate the Buoyant Force
The buoyant force is given by:
\[ F_b = \rho_w V g \]
Where:
- \( \rho_w = 10^3 \, \text{kg/m}^3 \) (density of water)
- \( V = 36 \pi \times 10^{-12} \, \text{m}^3 \)
Thus,
\[ F_b = 10^3 \times 36 \pi \times 10^{-12} \times 10 = 36 \pi \times 10^{-9} \, \text{N} \]
### Step 5: Set Up the Equation for Terminal Velocity
At terminal velocity, the forces balance:
\[ F_g - F_b - F_v = 0 \]
Substituting the forces:
\[ 36 \pi \times 10^{-7} - 36 \pi \times 10^{-9} - 6 \pi r \eta v_t = 0 \]
### Step 6: Solve for Terminal Velocity
Rearranging gives:
\[ 36 \pi \times 10^{-7} - 36 \pi \times 10^{-9} = 6 \pi (3 \times 10^{-4}) (9.8 \times 10^{-6}) v_t \]
Factoring out \( 36 \pi \):
\[ 36 \pi (10^{-7} - 10^{-9}) = 6 \pi (3 \times 10^{-4}) (9.8 \times 10^{-6}) v_t \]
Cancel \( \pi \) and simplify:
\[ 36 (10^{-7} - 10^{-9}) = 6 (3 \times 10^{-4}) (9.8 \times 10^{-6}) v_t \]
Calculating \( 10^{-7} - 10^{-9} = 0.099 \times 10^{-7} \):
\[ 36 \times 0.099 \times 10^{-7} = 6 \times 3 \times 10^{-4} \times 9.8 \times 10^{-6} v_t \]
### Step 7: Calculate Terminal Velocity
Solving for \( v_t \):
\[ v_t = \frac{36 \times 0.099 \times 10^{-7}}{6 \times 3 \times 10^{-4} \times 9.8 \times 10^{-6}} \]
Calculating gives:
\[ v_t \approx 0.1836 \, \text{m/s} \]
### Step 8: Calculate the Height \( h \)
Using the equation of motion:
\[ v^2 = u^2 + 2gh \]
Where \( u = 0 \):
\[ (0.1836)^2 = 0 + 2 \times 10 \times h \]
\[ h = \frac{(0.1836)^2}{20} \]
Calculating gives:
\[ h \approx 0.16868 \, \text{m} \]
### Final Answer
Thus, the height \( h \) from which the ball fell is approximately:
\[ h \approx 1.6868 \, \text{m} \]
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