A small metal ball of diameter `4 mm` and density `10.5 g//cm^(3)` in dropped in glycerine of density `1.5 g//cm^(3)`. The ball attains a terminal velocity of `8cm s^(-1)`. The coefficient of viscosity of glycerine is

Options

A. 4.9 poise
B. 9.8 poise
C. 98 poise
D. 980 poise

Explanation

To find the coefficient of viscosity of glycerine, we can use the formula for terminal velocity of a sphere falling through a viscous fluid:

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

Where:
- \( V_t \) = terminal velocity
- \( r \) = radius of the sphere
- \( \sigma \) = density of the sphere
- \( \rho \) = density of the fluid
- \( g \) = acceleration due to gravity
- \( \eta \) = coefficient of viscosity

### Step 1: Convert given values to SI units
1. **Diameter of the ball**: \( 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m} \)
- **Radius**: \( r = \frac{4 \, \text{mm}}{2} = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)

2. **Density of the ball**: \( 10.5 \, \text{g/cm}^3 = 10.5 \times 10^3 \, \text{kg/m}^3 = 10500 \, \text{kg/m}^3 \)

3. **Density of glycerine**: \( 1.5 \, \text{g/cm}^3 = 1.5 \times 10^3 \, \text{kg/m}^3 = 1500 \, \text{kg/m}^3 \)

4. **Terminal velocity**: \( 8 \, \text{cm/s} = 8 \times 10^{-2} \, \text{m/s} \)

5. **Acceleration due to gravity**: \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Substitute values into the formula
We need to rearrange the formula to solve for \( \eta \):

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

Substituting the known values:

\[ \eta = \frac{2(2 \times 10^{-3})^2(10500 - 1500)(9.8)}{9(8 \times 10^{-2})} \]

### Step 3: Calculate each component
1. Calculate \( r^2 \):
\[ r^2 = (2 \times 10^{-3})^2 = 4 \times 10^{-6} \, \text{m}^2 \]

2. Calculate \( \sigma - \rho \):
\[ \sigma - \rho = 10500 - 1500 = 9000 \, \text{kg/m}^3 \]

3. Substitute back into the equation:
\[ \eta = \frac{2 \times 4 \times 10^{-6} \times 9000 \times 9.8}{9 \times 8 \times 10^{-2}} \]

### Step 4: Perform the calculations
1. Calculate the numerator:
\[ 2 \times 4 \times 10^{-6} \times 9000 \times 9.8 = 2 \times 4 \times 9000 \times 9.8 \times 10^{-6} = 705600 \times 10^{-6} = 0.7056 \]

2. Calculate the denominator:
\[ 9 \times 8 \times 10^{-2} = 72 \times 10^{-2} = 0.72 \]

3. Now calculate \( \eta \):
\[ \eta = \frac{0.7056}{0.72} \approx 0.98 \, \text{Pa.s} \]

### Step 5: Convert to deca poise
1. Since \( 1 \, \text{Pa.s} = 10 \, \text{poise} \):
\[ \eta \approx 0.98 \, \text{Pa.s} = 9.8 \, \text{poise} = 0.98 \, \text{deca poise} \]

### Final Answer
The coefficient of viscosity of glycerine is approximately \( 0.98 \, \text{deca poise} \).

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