A spherical ball of radius `3xx10^(-4)`m and density `10^(4)kg//m^(3)` falls freely under gravity through a distance h before entering a tank of water. If after entering the water the velocity of the ball does not change, find h the viscosity of water is `9.8xx10^(-6)N-s//m^(2)`

Options

A. `1.65xx10^(3)m`
B. `2.65xx10^(2)m`
C. `3.65xx10^(4)m`
D. `1.45xx10^(2) m`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have a spherical ball falling through a distance \( h \) before entering a tank of water. The ball has a radius of \( 3 \times 10^{-4} \) m and a density of \( 10^4 \) kg/m³. The viscosity of water is given as \( 9.8 \times 10^{-6} \) N·s/m². We need to find the distance \( h \) that the ball falls before it enters the water, where its velocity remains constant.

### Step 2: Apply the concept of terminal velocity
When the ball enters the water, it reaches a terminal velocity \( v_t \). According to Stokes' law, the terminal velocity for a sphere falling in a viscous fluid is given by:

\[ v_t = \frac{2}{9} \cdot \frac{(\rho - \sigma) \cdot g \cdot r^2}{\eta} \]

Where:
- \( \rho \) = density of the ball = \( 10^4 \) kg/m³
- \( \sigma \) = density of water = \( 10^3 \) kg/m³
- \( g \) = acceleration due to gravity = \( 9.8 \) m/s²
- \( r \) = radius of the ball = \( 3 \times 10^{-4} \) m
- \( \eta \) = viscosity of water = \( 9.8 \times 10^{-6} \) N·s/m²

### Step 3: Calculate the terminal velocity \( v_t \)
Substituting the values into the terminal velocity formula:

\[ v_t = \frac{2}{9} \cdot \frac{(10^4 - 10^3) \cdot 9.8 \cdot (3 \times 10^{-4})^2}{9.8 \times 10^{-6}} \]

Calculating \( (10^4 - 10^3) = 9 \times 10^3 \):

\[ v_t = \frac{2}{9} \cdot \frac{9 \times 10^3 \cdot 9.8 \cdot (9 \times 10^{-8})}{9.8 \times 10^{-6}} \]

Simplifying:

\[ v_t = \frac{2}{9} \cdot \frac{9 \times 10^3 \cdot 9 \times 10^{-8}}{10^{-6}} = \frac{2}{9} \cdot 81 \times 10^{-3} = \frac{162 \times 10^{-3}}{9} = 18 \times 10^{-3} = 0.018 \text{ m/s} \]

### Step 4: Relate the distance fallen \( h \) to the velocity
Before entering the water, the ball falls a distance \( h \) under gravity. The velocity \( v \) just before entering the water can be calculated using the equation of motion:

\[ v^2 = u^2 + 2gh \]

Where \( u = 0 \) (initial velocity). Thus:

\[ v^2 = 2gh \implies v = \sqrt{2gh} \]

### Step 5: Set the velocities equal
Since the velocity of the ball just before entering the water is equal to the terminal velocity in water:

\[ \sqrt{2gh} = v_t \]

Squaring both sides:

\[ 2gh = v_t^2 \]

### Step 6: Solve for \( h \)
Rearranging gives:

\[ h = \frac{v_t^2}{2g} \]

Substituting \( v_t = 0.018 \) m/s and \( g = 9.8 \) m/s²:

\[ h = \frac{(0.018)^2}{2 \cdot 9.8} \]

Calculating:

\[ h = \frac{0.000324}{19.6} \approx 0.0000165 \text{ m} = 1.65 \times 10^{-5} \text{ m} \]

### Step 7: Final answer
Thus, the distance \( h \) that the ball falls before entering the water is approximately:

\[ h \approx 1.65 \times 10^{-5} \text{ m} \]