Rain drops fall from a certain height with a terminal velocity v on the ground. The viscous force is `F=5pi nerv` Hence `ne` is coefficient of viscosity r the radius of rain drop and v is speed Then work done by all the forces acting on the ball till it reaches the ground is proportional to :

Options

A. `r^(7)`
B. `r^(5)`
C. `r^(3)`
D. `r^(2)`

Answer

To solve the problem step by step, we will analyze the forces acting on the raindrop and calculate the work done by these forces until the raindrop reaches the ground.

### Step 1: Understand Terminal Velocity
The terminal velocity \( v \) of a raindrop is the constant speed that the raindrop eventually reaches when the gravitational force is balanced by the viscous drag force. The viscous force acting on the raindrop is given by:
\[ F = 5 \pi \eta r v \]
where:
- \( \eta \) is the coefficient of viscosity,
- \( r \) is the radius of the raindrop,
- \( v \) is the terminal velocity.

### Step 2: Calculate the Gravitational Force
The gravitational force \( F_g \) acting on the raindrop can be calculated using the formula:
\[ F_g = m g \]
where \( m \) is the mass of the raindrop and \( g \) is the acceleration due to gravity. The mass \( m \) can be expressed in terms of the volume and density of the raindrop:
\[ m = \frac{4}{3} \pi r^3 \rho \]
where \( \rho \) is the density of the raindrop.

### Step 3: Set Up the Equation for Terminal Velocity
At terminal velocity, the gravitational force is equal to the viscous force:
\[ m g = 5 \pi \eta r v \]
Substituting the expression for mass:
\[ \frac{4}{3} \pi r^3 \rho g = 5 \pi \eta r v \]

### Step 4: Solve for Terminal Velocity \( v \)
Rearranging the equation to solve for \( v \):
\[ v = \frac{4 r^2 \rho g}{15 \eta} \]
This shows that terminal velocity \( v \) is proportional to \( r^2 \).

### Step 5: Calculate Kinetic Energy
The kinetic energy \( KE \) of the raindrop when it reaches terminal velocity can be expressed as:
\[ KE = \frac{1}{2} m v^2 \]
Substituting the expression for mass:
\[ KE = \frac{1}{2} \left(\frac{4}{3} \pi r^3 \rho\right) v^2 \]

### Step 6: Substitute Terminal Velocity into Kinetic Energy
Substituting \( v \) from Step 4 into the kinetic energy formula:
\[ KE = \frac{1}{2} \left(\frac{4}{3} \pi r^3 \rho\right) \left(\frac{4 r^2 \rho g}{15 \eta}\right)^2 \]
Calculating \( v^2 \):
\[ v^2 = \left(\frac{4 r^2 \rho g}{15 \eta}\right)^2 = \frac{16 r^4 \rho^2 g^2}{225 \eta^2} \]
Now substituting this back into the kinetic energy expression:
\[ KE = \frac{1}{2} \left(\frac{4}{3} \pi r^3 \rho\right) \left(\frac{16 r^4 \rho^2 g^2}{225 \eta^2}\right) \]

### Step 7: Simplify the Kinetic Energy Expression
Combining the terms:
\[ KE = \frac{32 \pi \rho^3 g^2}{675 \eta^2} r^7 \]
This shows that the kinetic energy (and thus the work done by all forces) is proportional to \( r^7 \).

### Final Conclusion
The work done by all the forces acting on the raindrop until it reaches the ground is proportional to \( r^7 \).