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, we need to understand the relationship between the terminal velocity of a sphere and its volume. Here’s a step-by-step solution:

### Step 1: Understand Terminal Velocity
The terminal velocity \( v_t \) of a sphere falling through a viscous fluid is reached when the downward gravitational force is balanced by the upward buoyant force and the viscous drag force.

### Step 2: Write the Equation for Terminal Velocity
The terminal velocity \( v_t \) can be expressed as:
\[ v_t = \frac{2}{9} \frac{( \rho - \sigma ) g r^2}{\eta} \]
where:
- \( \rho \) = density of the sphere
- \( \sigma \) = density of the fluid
- \( g \) = acceleration due to gravity
- \( r \) = radius of the sphere
- \( \eta \) = coefficient of viscosity of the fluid

### Step 3: Relate Volume to Radius
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
If the volume of the sphere increases from \( V \) to \( 8V \), we can relate the initial radius \( r \) to the new radius \( r_1 \):
\[ 8V = \frac{4}{3} \pi r_1^3 \]
From the initial volume \( V = \frac{4}{3} \pi r^3 \), we can write:
\[ \frac{8V}{V} = \frac{r_1^3}{r^3} \implies 8 = \left( \frac{r_1}{r} \right)^3 \]

### Step 4: Solve for the New Radius
Taking the cube root of both sides gives:
\[ \frac{r_1}{r} = 2 \implies r_1 = 2r \]

### Step 5: Substitute the New Radius into the Terminal Velocity Formula
Now, substituting \( r_1 = 2r \) into the terminal velocity formula:
\[ v_{t1} = \frac{2}{9} \frac{( \rho - \sigma ) g (2r)^2}{\eta} \]
This simplifies to:
\[ v_{t1} = \frac{2}{9} \frac{( \rho - \sigma ) g (4r^2)}{\eta} = 4 \left( \frac{2}{9} \frac{( \rho - \sigma ) g r^2}{\eta} \right) = 4v_t \]

### Conclusion
Thus, the terminal velocity \( v_{t1} \) of the sphere with volume \( 8V \) is:
\[ \boxed{4v_t} \]