A sphere of radius R is gently dropped into liquid of viscosity `eta` in a vertical uniform tube. It attains a terminal velocity v. Another sphere of radius 2R when dropped into the same liquid, will attain its teriminal velocity.

Options

A. v
B. 2v
C. 4v
D. 9v

Answer

To solve the problem, we need to determine the terminal velocity of a sphere with radius \(2R\) when it is dropped into a viscous liquid, given that a sphere of radius \(R\) has a terminal velocity \(v\).

### Step-by-Step Solution:

1. **Understand the relationship of terminal velocity with radius**:
According to Stokes' law, the terminal velocity \(v_t\) of a sphere falling through a viscous fluid is given by the formula:
\[ v_t = \frac{2R^2 (\rho - \sigma) g}{9\eta} \]
where:
- \(R\) = radius of the sphere
- \(\rho\) = density of the sphere
- \(\sigma\) = density of the fluid
- \(g\) = acceleration due to gravity
- \(\eta\) = viscosity of the fluid

2. **Identify the terminal velocity for the first sphere**:
For the first sphere of radius \(R\), the terminal velocity is:
\[ v = \frac{2R^2 (\rho - \sigma) g}{9\eta} \]

3. **Determine the terminal velocity for the second sphere**:
For the second sphere of radius \(2R\), we can substitute \(2R\) into the terminal velocity formula:
\[ v_{t2} = \frac{2(2R)^2 (\rho - \sigma) g}{9\eta} \]
Simplifying this gives:
\[ v_{t2} = \frac{2 \cdot 4R^2 (\rho - \sigma) g}{9\eta} = \frac{8R^2 (\rho - \sigma) g}{9\eta} \]

4. **Relate the terminal velocities**:
Now, we can relate \(v_{t2}\) to \(v\):
\[ v_{t2} = 4 \cdot \frac{2R^2 (\rho - \sigma) g}{9\eta} = 4v \]

5. **Conclusion**:
Thus, the terminal velocity of the second sphere (radius \(2R\)) is:
\[ v_{t2} = 4v \]

### Final Answer:
The terminal velocity of the sphere with radius \(2R\) is \(4v\).

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