Uniform speed of 2 cm diameter ball is `20cm//s` in a viscous liquid. Then, the speed of 1 cm diameter ball in the same liquid is

Options

A. `5cms^(-1)`
B. `10cms^(-1)`
C. `40cms^(-1)`
D. `80cms^(-1)`

Explanation

To solve the problem, we need to find the speed of a 1 cm diameter ball in a viscous liquid, given that a 2 cm diameter ball moves at a uniform speed of 20 cm/s. This speed is the terminal velocity, which is influenced by the size of the ball.

### Step-by-Step Solution:

1. **Understand the relationship between terminal velocity and radius:**
The terminal velocity (V) of a sphere in a viscous fluid is directly proportional to the square of its radius (R). This can be expressed as:
\[ V \propto R^2 \]
This means that if we have two balls with different diameters, we can set up a ratio of their terminal velocities based on the square of their radii.

2. **Define the variables:**
- Let \( V_1 \) be the terminal velocity of the first ball (2 cm diameter).
- Let \( V_2 \) be the terminal velocity of the second ball (1 cm diameter).
- The radius of the first ball \( R_1 = \frac{2 \text{ cm}}{2} = 1 \text{ cm} \).
- The radius of the second ball \( R_2 = \frac{1 \text{ cm}}{2} = 0.5 \text{ cm} \).
- Given \( V_1 = 20 \text{ cm/s} \).

3. **Set up the ratio of terminal velocities:**
Using the relationship \( V \propto R^2 \), we can write:
\[ \frac{V_1}{V_2} = \frac{R_1^2}{R_2^2} \]

4. **Substituting the values:**
Substitute \( R_1 = 1 \text{ cm} \) and \( R_2 = 0.5 \text{ cm} \):
\[ \frac{20}{V_2} = \frac{(1)^2}{(0.5)^2} = \frac{1}{0.25} = 4 \]

5. **Solve for \( V_2 \):**
Rearranging the equation gives:
\[ 20 = 4 \times V_2 \]
Therefore,
\[ V_2 = \frac{20}{4} = 5 \text{ cm/s} \]

6. **Conclusion:**
The speed of the 1 cm diameter ball in the same viscous liquid is \( 5 \text{ cm/s} \).

### Final Answer:
The speed of the 1 cm diameter ball in the same liquid is **5 cm/s**.

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