A tiny sphere of mass m and density x is dropped in a jar of glycerine of density y. When the sphere aquires terminal velocity, the magnitude of the viscous force acting on it is :

Options

A. `(mgx)/(y)`
B. `(mgy)/(x)`
C. `mg[1-(y)/(x)]`
D. `mg[1+(y)/(x)]`

Answer

To solve the problem, we need to analyze the forces acting on the tiny sphere when it reaches terminal velocity in the glycerine.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Sphere**:
- When the sphere is dropped into the glycerine, three forces act on it:
- The gravitational force (weight) acting downwards: \( F_g = mg \)
- The buoyant force acting upwards: \( F_b \)
- The viscous force acting upwards: \( F_v \)

2. **Understand Terminal Velocity**:
- At terminal velocity, the net force acting on the sphere is zero. This means that the downward force (weight) is balanced by the sum of the upward forces (buoyant force and viscous force):
\[ F_g = F_b + F_v \]

3. **Calculate the Buoyant Force**:
- The buoyant force can be calculated using Archimedes' principle:
\[ F_b = \text{density of liquid} \times \text{volume of sphere} \times g \]
- The volume of the sphere can be expressed in terms of its mass and density:
\[ \text{Volume} = \frac{m}{x} \]
- Therefore, the buoyant force becomes:
\[ F_b = y \times \frac{m}{x} \times g = \frac{myg}{x} \]

4. **Set Up the Equation for Terminal Velocity**:
- Substituting the expressions for the forces into the equilibrium equation:
\[ mg = F_b + F_v \]
- Rearranging gives:
\[ F_v = mg - F_b \]
- Substituting for \( F_b \):
\[ F_v = mg - \frac{myg}{x} \]

5. **Factor Out Common Terms**:
- Taking \( mg \) as a common factor:
\[ F_v = mg \left(1 - \frac{y}{x}\right) \]

6. **Final Expression for the Viscous Force**:
- Thus, the magnitude of the viscous force acting on the sphere when it reaches terminal velocity is:
\[ F_v = mg \left(1 - \frac{y}{x}\right) \]

### Conclusion:
The magnitude of the viscous force acting on the sphere at terminal velocity is \( mg \left(1 - \frac{y}{x}\right) \).