The velocity of small ball of mass `M` and density (`d_(1)=` when dropped a container filled with glycerine becomes constant after some time. If the density glycerine is `d_(2)`, the viscous force acting on ball is

Options

A. `(Md_(1)g)/(d_(2))`
B. `Mg(1-(d_(2))/(d_(1)))`
C. `(M(d_(1)+d_(2)))/g`
D. `Md_(1)d_(2)`

Explanation

To find the viscous force acting on a small ball of mass \( M \) and density \( d_1 \) when it is dropped in a container filled with glycerine of density \( d_2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Terminal Velocity**:
- When the ball is dropped in glycerine, it eventually reaches a constant velocity known as terminal velocity. At this point, the net force acting on the ball is zero.

2. **Forces Acting on the Ball**:
- The forces acting on the ball are:
- The weight of the ball, \( W = Mg \), where \( g \) is the acceleration due to gravity.
- The buoyant force (upthrust) acting on the ball, which is equal to the weight of the liquid displaced by the ball.

3. **Calculating the Buoyant Force**:
- The volume of the ball can be expressed as \( V = \frac{4}{3} \pi R^3 \) (where \( R \) is the radius of the ball).
- The mass of the liquid displaced by the ball is given by \( \text{mass of liquid} = d_2 \times V \).
- Therefore, the buoyant force \( F_b \) is:
\[ F_b = \text{mass of liquid} \times g = d_2 \times V \times g = d_2 \times \left(\frac{4}{3} \pi R^3\right) g \]

4. **Applying the Equilibrium Condition**:
- At terminal velocity, the viscous force \( F_v \) plus the buoyant force \( F_b \) balances the weight of the ball:
\[ F_v + F_b = W \]
- Rearranging gives us the expression for the viscous force:
\[ F_v = W - F_b \]

5. **Substituting the Values**:
- Substitute \( W = Mg \) and \( F_b = d_2 \times V \times g \):
\[ F_v = Mg - d_2 \left(\frac{4}{3} \pi R^3\right) g \]

6. **Expressing the Volume in Terms of Density**:
- The volume \( V \) of the ball can also be expressed in terms of its mass and density:
\[ V = \frac{M}{d_1} \]
- Substitute this into the expression for the buoyant force:
\[ F_b = d_2 \left(\frac{M}{d_1}\right) g \]

7. **Final Expression for Viscous Force**:
- Now substituting \( F_b \) back into the equation for \( F_v \):
\[ F_v = Mg - d_2 \left(\frac{M}{d_1}\right) g \]
- Factor out \( g \):
\[ F_v = g \left(M - \frac{d_2 M}{d_1}\right) \]
- Simplifying gives:
\[ F_v = Mg \left(1 - \frac{d_2}{d_1}\right) \]

8. **Final Result**:
- Thus, the viscous force acting on the ball is:
\[ F_v = \frac{Mg(d_1 - d_2)}{d_1} \]