A sphere of radius R and made of material of relative density `sigma` has a concentric cavity of radius r. It just floats when placed in a tank full of water. The value of the ratio `R//r` will be

Options

A. (a)` ((sigma )/(sigma+ 1 )) ^(1//3) `
B. (b)` ((sigma + 1 ))/(sigma ) `
C. (c)` ((sigma )/(sigma - 1 )) ^(1//3) `
D. (d)` ((sigma + 1 )/(sigma )) ^(1//3) `

Explanation

To solve the problem, we need to apply the principle of flotation, which states that a floating body displaces its own weight of fluid. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a sphere of radius \( R \) made of a material with relative density \( \sigma \).
- There is a concentric cavity of radius \( r \) within the sphere.
- The sphere is floating in water, meaning the weight of the sphere equals the weight of the water displaced.

2. **Weight of the Sphere**:
- The volume of the sphere can be calculated as:
\[ V_{\text{sphere}} = \frac{4}{3} \pi R^3 \]
- The volume of the cavity is:
\[ V_{\text{cavity}} = \frac{4}{3} \pi r^3 \]
- Therefore, the effective volume of the sphere (considering the cavity) is:
\[ V_{\text{effective}} = V_{\text{sphere}} - V_{\text{cavity}} = \frac{4}{3} \pi (R^3 - r^3) \]

3. **Weight of the Sphere**:
- The weight of the sphere can be expressed as:
\[ W_{\text{sphere}} = V_{\text{effective}} \cdot \text{density of sphere} \cdot g = \frac{4}{3} \pi (R^3 - r^3) \cdot (\sigma \cdot \rho_{\text{water}}) \cdot g \]
- Here, \( \rho_{\text{water}} \) is the density of water.

4. **Weight of the Displaced Water**:
- The weight of the water displaced by the sphere is equal to the volume of water displaced times the density of water:
\[ W_{\text{displaced}} = V_{\text{displaced}} \cdot \rho_{\text{water}} \cdot g = \frac{4}{3} \pi R^3 \cdot \rho_{\text{water}} \cdot g \]

5. **Setting the Weights Equal**:
- According to the principle of flotation:
\[ W_{\text{sphere}} = W_{\text{displaced}} \]
- Thus, we have:
\[ \frac{4}{3} \pi (R^3 - r^3) \cdot (\sigma \cdot \rho_{\text{water}}) \cdot g = \frac{4}{3} \pi R^3 \cdot \rho_{\text{water}} \cdot g \]

6. **Canceling Common Terms**:
- Cancel out the common terms \( \frac{4}{3} \pi g \) and \( \rho_{\text{water}} \):
\[ (R^3 - r^3) \cdot \sigma = R^3 \]

7. **Rearranging the Equation**:
- Rearranging gives:
\[ R^3 - r^3 = \frac{R^3}{\sigma} \]
- This simplifies to:
\[ R^3 \left(1 - \frac{1}{\sigma}\right) = r^3 \]

8. **Finding the Ratio**:
- Taking the cube root on both sides:
\[ r = R \left(1 - \frac{1}{\sigma}\right)^{1/3} \]
- Thus, the ratio \( \frac{R}{r} \) can be expressed as:
\[ \frac{R}{r} = \frac{R}{R \left(1 - \frac{1}{\sigma}\right)^{1/3}} = \frac{1}{\left(1 - \frac{1}{\sigma}\right)^{1/3}} \]

9. **Final Expression**:
- Therefore, the final expression for the ratio \( \frac{R}{r} \) is:
\[ \frac{R}{r} = \left(\frac{\sigma}{\sigma - 1}\right)^{1/3} \]

### Conclusion:
The value of the ratio \( \frac{R}{r} \) is \( \left(\frac{\sigma}{\sigma - 1}\right)^{1/3} \).