Two balls of same material of density `rho` but radius `r_(1)` and `r_(2)` are joined by a light inextensible vertical thread and released from a large height in a medium of coefficient of viscosity `= eta`. Find the terminal velocity acquired by the balls. Also find the tension in the string connecting both the balls when both of them are moving with terminal velocity. Neglect buoyancy and change in acceleration due to gravity.

Answer

To solve the problem step by step, we will first find the terminal velocity of the two balls and then determine the tension in the string connecting them when they are moving with terminal velocity.

### Step 1: Understanding the Forces Acting on the Balls
When the two balls are released, they experience two main forces:
1. The weight (downward force) due to gravity.
2. The viscous drag (upward force) due to the medium.

The weight of each ball can be expressed as:
- For ball 1 (radius \( r_1 \)):
\[ w_1 = \rho \cdot \frac{4}{3} \pi r_1^3 g \]
- For ball 2 (radius \( r_2 \)):
\[ w_2 = \rho \cdot \frac{4}{3} \pi r_2^3 g \]

### Step 2: Viscous Force
The viscous force acting on each ball can be expressed using Stokes' law:
- For ball 1:
\[ F_{v1} = 6 \pi \eta r_1 V_t \]
- For ball 2:
\[ F_{v2} = 6 \pi \eta r_2 V_t \]

### Step 3: Total Forces
When both balls reach terminal velocity, the total downward force (weight) equals the total upward force (viscous drag). Therefore, we can write:
\[ w_1 + w_2 = F_{v1} + F_{v2} \]
Substituting the expressions for weight and viscous force:
\[ \rho \cdot \frac{4}{3} \pi r_1^3 g + \rho \cdot \frac{4}{3} \pi r_2^3 g = 6 \pi \eta r_1 V_t + 6 \pi \eta r_2 V_t \]

### Step 4: Simplifying the Equation
Factoring out common terms:
\[ \rho \cdot \frac{4}{3} \pi g (r_1^3 + r_2^3) = 6 \pi \eta V_t (r_1 + r_2) \]

### Step 5: Solving for Terminal Velocity \( V_t \)
Rearranging the equation to solve for \( V_t \):
\[ V_t = \frac{\rho \cdot \frac{4}{3} \pi g (r_1^3 + r_2^3)}{6 \pi \eta (r_1 + r_2)} \]
Simplifying further:
\[ V_t = \frac{2 \rho g (r_1^3 + r_2^3)}{9 \eta (r_1 + r_2)} \]

### Step 6: Finding the Tension in the String
To find the tension \( T \) in the string connecting the two balls, we consider the forces acting on ball 1:
\[ w_1 = T + F_{v1} \]
Substituting the expressions:
\[ \rho \cdot \frac{4}{3} \pi r_1^3 g = T + 6 \pi \eta r_1 V_t \]
Rearranging to solve for \( T \):
\[ T = \rho \cdot \frac{4}{3} \pi r_1^3 g - 6 \pi \eta r_1 V_t \]

### Step 7: Substitute \( V_t \) into the Tension Equation
Substituting the expression for \( V_t \):
\[ T = \rho \cdot \frac{4}{3} \pi r_1^3 g - 6 \pi \eta r_1 \left(\frac{2 \rho g (r_1^3 + r_2^3)}{9 \eta (r_1 + r_2)}\right) \]
Simplifying this gives:
\[ T = \rho \cdot \frac{4}{3} \pi g r_1^3 - \frac{12 \pi \eta r_1 \rho g (r_1^3 + r_2^3)}{9 \eta (r_1 + r_2)} \]
This can be further simplified to find the final expression for tension.

### Final Results
1. Terminal Velocity:
\[ V_t = \frac{2 \rho g (r_1^3 + r_2^3)}{9 \eta (r_1 + r_2)} \]
2. Tension in the string:
\[ T = \rho \cdot \frac{4}{3} \pi g r_1^3 - \frac{12 \pi \rho g r_1 (r_1^3 + r_2^3)}{9 (r_1 + r_2)} \]