Consider an expanding sphere of instantaneous radius ? whose total mass remains constant. The expansion is such that the instantaneous density `rho` remains uniform throughout the volume. The rate of fractional change in density `((dp)/(rhodt))` is constant. The velocity v of any point on the surface of the expanding sphere is proportional to

Options

A. `R^3`
B. `1/R`
C. R
D. `R^(2/3)`

Answer

To solve the problem, we need to analyze the relationship between the radius of the sphere, its mass, and the density as the sphere expands. Let's break this down step by step.

### Step-by-Step Solution:

1. **Understanding the Mass of the Sphere**:
The mass \( m \) of a sphere with radius \( R \) and uniform density \( \rho \) is given by the formula:
\[ m = \rho \cdot V = \rho \cdot \left( \frac{4}{3} \pi R^3 \right) \]
Here, \( V \) is the volume of the sphere.

2. **Differentiating the Mass**:
Since the mass \( m \) remains constant as the sphere expands, we can differentiate the mass with respect to time \( t \):
\[ \frac{dm}{dt} = 0 \]
Applying the product rule, we differentiate \( m \):
\[ \frac{dm}{dt} = \frac{d\rho}{dt} \cdot \left( \frac{4}{3} \pi R^3 \right) + \rho \cdot \frac{dV}{dt} \]
The volume \( V \) changes as the radius changes, so we need to differentiate the volume:
\[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi R^3 \right) = 4 \pi R^2 \frac{dR}{dt} \]

3. **Setting Up the Equation**:
Substituting the expression for \( \frac{dV}{dt} \) back into the differentiated mass equation gives:
\[ 0 = \frac{d\rho}{dt} \cdot \left( \frac{4}{3} \pi R^3 \right) + \rho \cdot \left( 4 \pi R^2 \frac{dR}{dt} \right) \]

4. **Rearranging the Equation**:
Rearranging the equation, we can isolate \( \frac{dR}{dt} \):
\[ -\frac{d\rho}{dt} \cdot \left( \frac{4}{3} \pi R^3 \right) = \rho \cdot \left( 4 \pi R^2 \frac{dR}{dt} \right) \]
Dividing both sides by \( 4 \pi R^2 \) (assuming \( R \neq 0 \)):
\[ -\frac{d\rho}{dt} \cdot \frac{R}{3} = \rho \cdot \frac{dR}{dt} \]

5. **Finding the Relationship**:
From the equation above, we can express \( \frac{dR}{dt} \) in terms of \( R \) and \( \rho \):
\[ \frac{dR}{dt} = -\frac{1}{3} \frac{d\rho}{dt} \cdot \frac{R}{\rho} \]
Since it is given that the rate of fractional change in density \( \left( \frac{d\rho}{\rho dt} \right) \) is constant, we can denote this constant as \( k \):
\[ \frac{d\rho}{dt} = k \rho \]

6. **Substituting Back**:
Substituting \( \frac{d\rho}{dt} = k \rho \) into the equation for \( \frac{dR}{dt} \):
\[ \frac{dR}{dt} = -\frac{1}{3} k R \]
This shows that the velocity \( v \) of any point on the surface of the expanding sphere is proportional to \( R \):
\[ v \propto R \]

### Conclusion:
The velocity \( v \) of any point on the surface of the expanding sphere is proportional to the radius \( R \).