A small sphere of volume V falling in a viscous fluid acquires a terminal velocity ` v_t`. The terminal velocity of a sphere of volume 8V of the same material and falling in the same fluid will be :

Options

A. ` (v_t )/(2) `
B. ` v _ t `
C. ` 2 v_t `
D. ` 4 v _ t `

Explanation

To solve the problem of determining the terminal velocity of a sphere of volume \(8V\) falling in a viscous fluid, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Terminal Velocity**:
The terminal velocity \(v_t\) of a sphere in a viscous fluid is reached when the gravitational force acting on the sphere is balanced by the viscous drag force. The forces acting on the sphere can be expressed as:
\[ F_{\text{gravity}} = mg - F_{\text{buoyant}} \]
where \(m\) is the mass of the sphere, and \(F_{\text{buoyant}}\) is the buoyant force.

2. **Expressing Forces**:
The gravitational force can be expressed as:
\[ mg = \rho_s V g \]
where \(\rho_s\) is the density of the sphere, \(V\) is its volume, and \(g\) is the acceleration due to gravity. The buoyant force is given by:
\[ F_{\text{buoyant}} = \rho_f V g \]
where \(\rho_f\) is the density of the fluid.

3. **Setting Up the Equation**:
At terminal velocity, the drag force equals the net force acting on the sphere:
\[ 6\pi \eta r v_t = mg - F_{\text{buoyant}} \]
Substituting the expressions for \(mg\) and \(F_{\text{buoyant}}\):
\[ 6\pi \eta r v_t = \rho_s V g - \rho_f V g \]
This simplifies to:
\[ 6\pi \eta r v_t = (\rho_s - \rho_f) V g \]

4. **Relating Volume and Radius**:
The volume \(V\) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
For a sphere of volume \(8V\), we can express the new volume as:
\[ V' = 8V = \frac{4}{3} \pi R^3 \]
From this, we can find the new radius \(R\):
\[ 8 \left(\frac{4}{3} \pi r^3\right) = \frac{4}{3} \pi R^3 \]
Simplifying gives:
\[ 8r^3 = R^3 \implies R = 2r \]

5. **Finding the New Terminal Velocity**:
We can express the terminal velocity in terms of the radius:
\[ v_t = k \cdot r^2 \]
where \(k\) is a constant. For the new sphere:
\[ v_t' = k \cdot R^2 = k \cdot (2r)^2 = k \cdot 4r^2 \]
Thus, we have:
\[ v_t' = 4(k \cdot r^2) = 4v_t \]

6. **Conclusion**:
Therefore, the terminal velocity of the sphere of volume \(8V\) is:
\[ v_t' = 4v_t \]

### Final Answer:
The terminal velocity of a sphere of volume \(8V\) of the same material and falling in the same fluid will be \(4v_t\).