Two metal ball of radius R and 2 R falling through a fluid have same velocity at some point. The viscous drag acting on them at that instant are in the ratio

Options

A. `1 : 2`
B. `1 : 4`
C. `1 : sqrt(2)`
D. `sqrt(2) : 1`

Answer

To solve the problem of finding the ratio of viscous drag acting on two metal balls of radius R and 2R falling through a fluid with the same velocity, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Viscous Drag**:
The viscous drag (F) acting on a spherical object moving through a fluid is given by Stokes' Law, which states:
\[ F = 6 \pi \eta r V \]
where:
- \( F \) = viscous drag
- \( \eta \) = coefficient of viscosity of the fluid
- \( r \) = radius of the spherical object
- \( V \) = velocity of the object

2. **Identify the Radii of the Balls**:
Let the radius of the first ball be \( r_1 = R \) and the radius of the second ball be \( r_2 = 2R \).

3. **Write the Expressions for Viscous Drag**:
For the first ball (radius \( R \)):
\[ F_1 = 6 \pi \eta R V \]
For the second ball (radius \( 2R \)):
\[ F_2 = 6 \pi \eta (2R) V = 12 \pi \eta R V \]

4. **Calculate the Ratio of Viscous Drag**:
To find the ratio of the viscous drag on the two balls, we can set up the following equation:
\[ \frac{F_1}{F_2} = \frac{6 \pi \eta R V}{12 \pi \eta R V} \]
Simplifying this gives:
\[ \frac{F_1}{F_2} = \frac{6}{12} = \frac{1}{2} \]

5. **Conclusion**:
The ratio of the viscous drag acting on the two balls is:
\[ F_1 : F_2 = 1 : 2 \]

### Final Answer:
The ratio of the viscous drag acting on the two balls is \( 1 : 2 \).