A solid shell loses half of its weight in water. Relative density of shell is 5.0 What fraction of its volume is hollow ?

Explanation

To solve the problem step by step, we will use the given information about the solid shell and apply the principles of buoyancy and relative density.

### Step 1: Understand the Given Data
- The relative density (specific gravity) of the shell is given as 5.0.
- The shell loses half of its weight when submerged in water.

### Step 2: Define Variables
Let:
- \( V \) = total volume of the shell
- \( X \) = fraction of the volume that is hollow
- The volume of the solid part of the shell = \( V(1 - X) \)

### Step 3: Calculate the Weight of the Shell
The weight of the shell can be calculated using the formula:
\[ \text{Weight of shell} = \text{Volume} \times \text{Density} \times g \]
Since the relative density of the shell is 5.0, its density is:
\[ \text{Density of shell} = 5 \times 10^3 \, \text{kg/m}^3 \]
Thus, the weight of the shell is:
\[ W_{\text{shell}} = V(1 - X) \times 5 \times 10^3 \times g \]

### Step 4: Calculate the Weight of Water Displaced
When the shell is submerged, it displaces a volume of water equal to its total volume \( V \). The weight of the water displaced is:
\[ W_{\text{water}} = V \times 10^3 \times g \]

### Step 5: Set Up the Equation
According to the problem, the shell loses half of its weight in water. Therefore, we can write:
\[ W_{\text{shell}} - W_{\text{water}} = \frac{1}{2} W_{\text{shell}} \]
Substituting the expressions for weight:
\[ V(1 - X) \times 5 \times 10^3 \times g - V \times 10^3 \times g = \frac{1}{2} \left( V(1 - X) \times 5 \times 10^3 \times g \right) \]

### Step 6: Simplify the Equation
Dividing through by \( g \) and \( V \) (assuming \( V \neq 0 \)):
\[ (1 - X) \times 5 \times 10^3 - 10^3 = \frac{1}{2} \left( (1 - X) \times 5 \times 10^3 \right) \]

### Step 7: Rearranging the Equation
This simplifies to:
\[ (1 - X) \times 5 - 1 = \frac{5}{2} (1 - X) \]
Multiplying through by 2 to eliminate the fraction:
\[ 2(1 - X) \times 5 - 2 = 5(1 - X) \]
This simplifies to:
\[ 10(1 - X) - 2 = 5(1 - X) \]
Rearranging gives:
\[ 10 - 10X - 2 = 5 - 5X \]
Combining like terms:
\[ 8 - 10X = 5 - 5X \]
Bringing all terms involving \( X \) to one side:
\[ 8 - 5 = 10X - 5X \]
This results in:
\[ 3 = 5X \]
Thus:
\[ X = \frac{3}{5} \]

### Step 8: Conclusion
The fraction of the volume that is hollow is:
\[ \frac{3}{5} \]