A concrete sphere of radius `R` has cavity of radius `r` which is packed with sawdust. The specific gravities of concrete and sawdust are respectively `2.4 and 0.3` for this sphere to float with its entire volume submerged under water. Ratio of mass of concrete to mass of swadust will be

Options

A. 8
B. 4
C. 3
D. Zero

Answer

To solve the problem, we need to find the ratio of the mass of concrete to the mass of sawdust in a concrete sphere with a cavity filled with sawdust.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Specific gravity of concrete, \( SG_C = 2.4 \)
- Specific gravity of sawdust, \( SG_S = 0.3 \)
- Density of water, \( \rho_W = 1 \, \text{g/cm}^3 \)

2. **Calculate Densities:**
- Density of concrete, \( \rho_C = SG_C \times \rho_W = 2.4 \times 1 = 2.4 \, \text{g/cm}^3 \)
- Density of sawdust, \( \rho_S = SG_S \times \rho_W = 0.3 \times 1 = 0.3 \, \text{g/cm}^3 \)

3. **Volume of the Sphere:**
- Volume of the concrete sphere (without cavity) = \( V = \frac{4}{3} \pi R^3 \)
- Volume of the cavity = \( v = \frac{4}{3} \pi r^3 \)

4. **Weight of the Sphere:**
- Weight of the concrete part = Volume of concrete × Density of concrete
- Volume of concrete = \( V - v = \frac{4}{3} \pi (R^3 - r^3) \)
- Weight of concrete = \( \frac{4}{3} \pi (R^3 - r^3) \cdot \rho_C \)

5. **Weight of the Sawdust:**
- Weight of sawdust = Volume of sawdust × Density of sawdust
- Volume of sawdust = \( v = \frac{4}{3} \pi r^3 \)
- Weight of sawdust = \( \frac{4}{3} \pi r^3 \cdot \rho_S \)

6. **Buoyant Force:**
- The entire sphere is submerged in water, so the buoyant force equals the weight of the displaced water.
- Buoyant force = Volume of the sphere × Density of water = \( \frac{4}{3} \pi R^3 \cdot \rho_W \)

7. **Setting Up the Equation:**
- According to Archimedes' principle, the weight of the sphere (concrete + sawdust) equals the buoyant force:
\[ \frac{4}{3} \pi (R^3 - r^3) \cdot \rho_C + \frac{4}{3} \pi r^3 \cdot \rho_S = \frac{4}{3} \pi R^3 \cdot \rho_W \]
- Cancel \( \frac{4}{3} \pi \) from both sides:
\[ (R^3 - r^3) \cdot \rho_C + r^3 \cdot \rho_S = R^3 \cdot \rho_W \]

8. **Substituting Densities:**
- Substitute \( \rho_C = 2.4 \), \( \rho_S = 0.3 \), and \( \rho_W = 1 \):
\[ (R^3 - r^3) \cdot 2.4 + r^3 \cdot 0.3 = R^3 \]

9. **Rearranging the Equation:**
\[ 2.4R^3 - 2.4r^3 + 0.3r^3 = R^3 \]
\[ 2.4R^3 - R^3 = 2.4r^3 - 0.3r^3 \]
\[ 1.4R^3 = 2.1r^3 \]
\[ \frac{R^3}{r^3} = \frac{2.1}{1.4} = \frac{21}{14} = \frac{3}{2} \]

10. **Finding the Ratio of Masses:**
- The mass of concrete \( M_C = \frac{4}{3} \pi (R^3 - r^3) \cdot \rho_C \)
- The mass of sawdust \( M_S = \frac{4}{3} \pi r^3 \cdot \rho_S \)
- The ratio of masses:
\[ \frac{M_C}{M_S} = \frac{(R^3 - r^3) \cdot \rho_C}{r^3 \cdot \rho_S} \]
- Substitute \( R^3 = \frac{3}{2} r^3 \):
\[ R^3 - r^3 = \frac{3}{2} r^3 - r^3 = \frac{1}{2} r^3 \]
- Thus,
\[ \frac{M_C}{M_S} = \frac{\left(\frac{1}{2} r^3\right) \cdot 2.4}{r^3 \cdot 0.3} = \frac{1.2}{0.3} = 4 \]

### Final Answer:
The ratio of the mass of concrete to the mass of sawdust is \( 4 \).