If S be the surface area of V be the volume of a sphere of radius r. Then `(dv)/(dt) = (r)/(k) (ds)/(dt)`. Value of k is

Options

A. `(1)/(2)`
B. `(1)/(4)`
C. 2
D. 4

Answer

To solve the problem, we need to find the value of \( k \) in the equation

\[ \frac{dV}{dt} = \frac{r}{k} \frac{dS}{dt} \]

where \( S \) is the surface area and \( V \) is the volume of a sphere with radius \( r \).

### Step 1: Write the formulas for surface area and volume of a sphere.

The surface area \( S \) of a sphere is given by:

\[ S = 4\pi r^2 \]

The volume \( V \) of a sphere is given by:

\[ V = \frac{4}{3}\pi r^3 \]

### Step 2: Differentiate the surface area with respect to time.

To find \( \frac{dS}{dt} \), we first differentiate \( S \) with respect to \( r \):

\[ \frac{dS}{dr} = \frac{d}{dr}(4\pi r^2) = 8\pi r \]

Now, applying the chain rule, we get:

\[ \frac{dS}{dt} = \frac{dS}{dr} \cdot \frac{dr}{dt} = 8\pi r \frac{dr}{dt} \]

### Step 3: Differentiate the volume with respect to time.

Next, we differentiate \( V \) with respect to \( r \):

\[ \frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2 \]

Now, applying the chain rule again, we have:

\[ \frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt} \]

### Step 4: Substitute \( \frac{dS}{dt} \) and \( \frac{dV}{dt} \) into the equation.

Now we substitute \( \frac{dS}{dt} \) and \( \frac{dV}{dt} \) into the original equation:

\[ 4\pi r^2 \frac{dr}{dt} = \frac{r}{k} (8\pi r \frac{dr}{dt}) \]

### Step 5: Simplify the equation.

We can cancel \( \frac{dr}{dt} \) from both sides (assuming \( \frac{dr}{dt} \neq 0 \)):

\[ 4\pi r^2 = \frac{r}{k} (8\pi r) \]

Now, we can simplify further:

\[ 4\pi r^2 = \frac{8\pi r^2}{k} \]

### Step 6: Solve for \( k \).

To isolate \( k \), we can multiply both sides by \( k \):

\[ 4\pi r^2 k = 8\pi r^2 \]

Now, divide both sides by \( 4\pi r^2 \) (assuming \( r \neq 0 \)):

\[ k = \frac{8\pi r^2}{4\pi r^2} = \frac{8}{4} = 2 \]

### Conclusion

Thus, the value of \( k \) is:

\[ \boxed{2} \]