if a ball of steel (density `rho=7.8 g//cm^(3)`) attains a terminal velocity of `10 cm//s` when falling in a tank of water (coefficient of viscosity, `eta_(water) =8.5xx10^(-4)`Ps s), then its terminal velocity in glycerine `(rho =1.2 g//cm^(2), eta =13.2 Pa s)` would be nearly

Options

A. `1.6xx10^(-5)cms^(-1)`
B. `6.25xx10^(-4)cms^(-1)`
C. `6.45xx10^(-4)cms^(-1)`
D. `1.5xx10^(-5)cms^(-1)`

Answer

To solve the problem, we need to find the terminal velocity of a steel ball falling in glycerin, given its terminal velocity in water. We will use the relationship between terminal velocity, density, and viscosity.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Density of steel ball, \( \rho_{ball} = 7.8 \, \text{g/cm}^3 \)
- Terminal velocity in water, \( V_1 = 10 \, \text{cm/s} \)
- Coefficient of viscosity of water, \( \eta_{water} = 8.5 \times 10^{-4} \, \text{Pa s} \)
- Density of glycerin, \( \rho_{glycerin} = 1.2 \, \text{g/cm}^3 \)
- Coefficient of viscosity of glycerin, \( \eta_{glycerin} = 13.2 \, \text{Pa s} \)
- Density of water, \( \rho_{water} = 1 \, \text{g/cm}^3 \)

2. **Use the Formula for Terminal Velocity:**
The terminal velocity \( V \) of a sphere falling through a fluid is given by:
\[ V = \frac{2 r^2 (\rho_{ball} - \rho_{fluid}) g}{9 \eta} \]
where:
- \( r \) is the radius of the ball,
- \( \rho_{ball} \) is the density of the ball,
- \( \rho_{fluid} \) is the density of the fluid,
- \( g \) is the acceleration due to gravity,
- \( \eta \) is the viscosity of the fluid.

3. **Establish the Relationship Between Terminal Velocities:**
Since the radius \( r \) and \( g \) are constants for both cases, we can establish a ratio of the terminal velocities:
\[ \frac{V_2}{V_1} = \frac{(\rho_{ball} - \rho_{glycerin}) \eta_{water}}{(\rho_{ball} - \rho_{water}) \eta_{glycerin}} \]

4. **Substitute the Known Values:**
- For water:
\[ \rho_{ball} - \rho_{water} = 7.8 - 1 = 6.8 \, \text{g/cm}^3 \]
- For glycerin:
\[ \rho_{ball} - \rho_{glycerin} = 7.8 - 1.2 = 6.6 \, \text{g/cm}^3 \]
- Substitute these into the ratio:
\[ \frac{V_2}{10} = \frac{6.6 \times (8.5 \times 10^{-4})}{6.8 \times 13.2} \]

5. **Calculate the Right Side:**
\[ \frac{V_2}{10} = \frac{6.6 \times 8.5 \times 10^{-4}}{6.8 \times 13.2} \]
- Calculate \( 6.6 \times 8.5 = 56.1 \)
- Calculate \( 6.8 \times 13.2 = 89.76 \)
- Now, substituting back:
\[ \frac{V_2}{10} = \frac{56.1 \times 10^{-4}}{89.76} \]
- Calculate the fraction:
\[ \frac{56.1}{89.76} \approx 0.625 \]
- Therefore:
\[ \frac{V_2}{10} \approx 0.625 \times 10^{-4} \]

6. **Find \( V_2 \):**
\[ V_2 \approx 0.625 \times 10^{-4} \times 10 = 6.25 \times 10^{-4} \, \text{cm/s} \]

### Final Answer:
The terminal velocity of the ball in glycerin is approximately \( 6.25 \times 10^{-4} \, \text{cm/s} \).

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