A metallic sphere of mass M falls through glycerine with a terminal velocity v. If we drop a ball of mass 8M of same metal into a column of glycerine, the terminal velocity of ball will be

Options

A. 2v
B. 4v
C. 8v
D. 16v

Explanation

To solve the problem, we need to understand the relationship between the mass of the sphere, its volume, and the terminal velocity when it falls through a fluid like glycerine.

### Step-by-Step Solution:

1. **Understanding Terminal Velocity**:
The terminal velocity \( v_t \) of an object falling through a fluid is given by the balance of forces acting on it. The gravitational force \( F_g \) acting downwards is balanced by the drag force \( F_d \) acting upwards when the object reaches terminal velocity.

\[ F_g = F_d \]

2. **For the first sphere**:
For a metallic sphere of mass \( M \) and radius \( r \), the gravitational force is:
\[ F_g = Mg \]
where \( g \) is the acceleration due to gravity.

The drag force \( F_d \) can be expressed using Stokes' law for small Reynolds numbers:
\[ F_d = 6 \pi \eta r v \]
where \( \eta \) is the viscosity of glycerine and \( v \) is the terminal velocity.

Setting these equal gives:
\[ Mg = 6 \pi \eta r v \]

3. **For the second sphere**:
Now consider a ball of mass \( 8M \) and the same material. The mass of the second sphere is \( 8M \), and if we denote its radius as \( R \), we can relate the radii based on the volume since they are made of the same material.

The volume of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Therefore, for the first sphere:
\[ V_1 = \frac{4}{3} \pi r^3 \]
For the second sphere:
\[ V_2 = \frac{4}{3} \pi R^3 \]

Since the mass is proportional to the volume and density, we have:
\[ 8M = \rho V_2 = \rho \left(\frac{4}{3} \pi R^3\right) \]

4. **Relating the radii**:
Since the mass of the second sphere is 8 times that of the first, we can relate the radii:
\[ 8M = \rho \left(\frac{4}{3} \pi R^3\right) \quad \text{and} \quad M = \rho \left(\frac{4}{3} \pi r^3\right) \]
Dividing these equations gives:
\[ \frac{8M}{M} = \frac{R^3}{r^3} \implies 8 = \frac{R^3}{r^3} \implies R = 2r \]

5. **Calculating the terminal velocity for the second sphere**:
Now we can find the terminal velocity \( V' \) for the second sphere:
\[ F_g = 8Mg = 6 \pi \eta R V' \]
Substituting \( R = 2r \):
\[ 8Mg = 6 \pi \eta (2r) V' \]
Simplifying gives:
\[ 8Mg = 12 \pi \eta r V' \implies V' = \frac{8Mg}{12 \pi \eta r} = \frac{2}{3} \cdot \frac{Mg}{6 \pi \eta r} \]
From the first sphere, we know:
\[ \frac{Mg}{6 \pi \eta r} = v \implies V' = \frac{2}{3} \cdot v \]

### Final Answer:
The terminal velocity of the ball of mass \( 8M \) will be \( 4v \).