Spherical rain drop evaporates at a rate proportional to its surface area. The differential equation corresponding to the rate of change of the radius of the rain drop if the constant of proportionality is `K >0` is (a) `( b ) (c) (d)(( e ) dy)/( f )(( g ) dt)( h ) (i)+K=0( j )` (k) (b) `( l ) (m) (n)(( o ) d r)/( p )(( q ) dt)( r ) (s)-K=0( t )` (u) (c) `( d ) (e) (f)(( g ) d r)/( h )(( i ) dt)( j ) (k)=K r (l)` (m) (d) None of these

Options

A. `(dr)/(dt)+ K=0`
B. `(dr)/(dt) -K =0`
C. `(dr)/(dt) = Kr`
D. None of these

Explanation

To solve the problem of a spherical raindrop evaporating at a rate proportional to its surface area, we will derive the corresponding differential equation step by step.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We know that the volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
- The surface area \( S \) of a sphere is given by:
\[ S = 4 \pi r^2 \]
- The problem states that the rate of change of volume is proportional to the surface area.

2. **Setting Up the Differential Equation**:
- The rate of change of volume with respect to time \( t \) can be expressed as:
\[ \frac{dV}{dt} = -k S \]
- Substituting the expression for surface area \( S \):
\[ \frac{dV}{dt} = -k (4 \pi r^2) \]

3. **Relating Volume and Radius**:
- We know that:
\[ V = \frac{4}{3} \pi r^3 \]
- Therefore, the derivative of volume with respect to time can be expressed using the chain rule:
\[ \frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} \]
- Calculating \( \frac{dV}{dr} \):
\[ \frac{dV}{dr} = 4 \pi r^2 \]
- Thus, we can write:
\[ \frac{dV}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt} \]

4. **Equating the Two Expressions for \( \frac{dV}{dt} \)**:
- Now we equate the two expressions for \( \frac{dV}{dt} \):
\[ 4 \pi r^2 \cdot \frac{dr}{dt} = -k (4 \pi r^2) \]

5. **Simplifying the Equation**:
- We can cancel \( 4 \pi r^2 \) from both sides (assuming \( r \neq 0 \)):
\[ \frac{dr}{dt} = -k \]

6. **Rearranging the Equation**:
- We can rearrange this equation to:
\[ \frac{dr}{dt} + k = 0 \]

### Final Differential Equation:
The final differential equation corresponding to the rate of change of the radius of the raindrop is:
\[ \frac{dr}{dt} + k = 0 \]