A sphere of solid material of specific gravity `8` has a concentric spherical cavity and just sinks in water. The ratio of radius of cavity to that of outer radius of the sphere must be

Options

A. `7^(1//3)/2`
B. `5^(1//3)/2`
C. `9^(1//3)/2`
D. `3^(1//3)/2`

Explanation

To solve the problem, we need to find the ratio of the radius of the cavity (r) to the outer radius of the sphere (R) given that the sphere just sinks in water.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a sphere with a specific gravity of 8, which means its density is 8 times that of water.
- The sphere has a concentric cavity and just sinks in water, indicating that the weight of the sphere equals the buoyant force acting on it.

2. **Defining Variables**:
- Let the radius of the cavity be \( r \).
- Let the outer radius of the sphere be \( R \).

3. **Buoyant Force Calculation**:
- The buoyant force \( F_b \) acting on the sphere is given by:
\[ F_b = \text{Density of water} \times \text{Volume of the outer sphere} \times g \]
- The volume of the outer sphere is:
\[ V_{\text{outer}} = \frac{4}{3} \pi R^3 \]
- Therefore, the buoyant force is:
\[ F_b = \rho_{\text{water}} \times \frac{4}{3} \pi R^3 \times g \]
- Since the specific gravity of water is 1, we can denote \( \rho_{\text{water}} = 1 \).

4. **Weight of the Sphere Calculation**:
- The weight \( W \) of the sphere can be expressed as:
\[ W = \text{Density of the sphere} \times \text{Volume of the solid part} \times g \]
- The density of the sphere is \( \rho_{\text{sphere}} = 8 \times \rho_{\text{water}} = 8 \).
- The volume of the solid part of the sphere is:
\[ V_{\text{solid}} = V_{\text{outer}} - V_{\text{cavity}} = \frac{4}{3} \pi R^3 - \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (R^3 - r^3) \]
- Therefore, the weight of the sphere is:
\[ W = 8 \times \frac{4}{3} \pi (R^3 - r^3) \times g \]

5. **Setting Up the Equation**:
- Since the sphere just sinks, we have:
\[ W = F_b \]
- Substituting the expressions for \( W \) and \( F_b \):
\[ 8 \times \frac{4}{3} \pi (R^3 - r^3) \times g = \frac{4}{3} \pi R^3 \times g \]

6. **Simplifying the Equation**:
- Cancel out common terms \( \frac{4}{3} \pi g \):
\[ 8(R^3 - r^3) = R^3 \]
- Rearranging gives:
\[ 8R^3 - 8r^3 = R^3 \]
- This simplifies to:
\[ 7R^3 = 8r^3 \]

7. **Finding the Ratio**:
- Dividing both sides by \( R^3 \):
\[ \frac{r^3}{R^3} = \frac{7}{8} \]
- Taking the cube root of both sides:
\[ \frac{r}{R} = \left(\frac{7}{8}\right)^{1/3} \]

### Final Answer:
The ratio of the radius of the cavity to that of the outer radius of the sphere is:
\[ \frac{r}{R} = \frac{7^{1/3}}{2} \]