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 the viscous drag acting on two metal balls of different radii falling through a fluid at the same velocity, we can follow these steps:

### Step 1: Understand the Formula for Viscous Drag
The viscous drag force \( F \) acting on a sphere moving through a fluid is given by Stokes' law:
\[ F = 6 \pi r \eta v \]
where:
- \( F \) is the viscous drag force,
- \( r \) is the radius of the sphere,
- \( \eta \) is the viscosity of the fluid,
- \( v \) is the velocity of the sphere.

### Step 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 \( r_2 = 2R \).

### Step 3: Write the Drag Force for Each Ball
Using the formula for viscous drag, we can write the drag forces for both balls:
- For the first ball:
\[ F_1 = 6 \pi R \eta v \]
- For the second ball:
\[ F_2 = 6 \pi (2R) \eta v = 12 \pi R \eta v \]

### Step 4: Find the Ratio of the Drag Forces
Now, we can find the ratio of the drag forces \( F_1 \) and \( F_2 \):
\[ \frac{F_1}{F_2} = \frac{6 \pi R \eta v}{12 \pi R \eta v} \]
The terms \( 6 \pi R \eta v \) cancel out:
\[ \frac{F_1}{F_2} = \frac{1}{2} \]

### Step 5: Conclusion
Thus, the ratio of the viscous drag acting on the two balls is:
\[ F_1 : F_2 = 1 : 2 \]

### Final Answer
The correct option is \( 1 : 2 \).

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