A solid shell loses half, its weight in water. Relative density of shell is 5. what fraction of its volume is hollow ?

Options

A. `(3)/(5)`
B. `(2)/(5)`
C. `(1)/(5)`
D. `(4)/(5)`

Explanation

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

### Step 1: Understand the given data
- The relative density of the shell is given as 5. This means the density of the shell is 5 times the density of water.
- Let the density of water be \( \rho_w = a \). Therefore, the density of the shell \( \rho_s = 5a \).

### Step 2: Define the volume of the shell
- Let the total volume of the shell be \( V \).
- Let the fraction of the volume that is hollow be \( x \). Thus, the volume of the hollow part is \( xV \) and the volume of the solid part is \( (1 - x)V \).

### Step 3: Calculate the weight of the solid shell
- The mass of the solid part of the shell is given by:
\[ \text{Mass of solid part} = \text{Density} \times \text{Volume} = \rho_s \times (1 - x)V = 5a \times (1 - x)V \]
- The weight of the solid shell is:
\[ W_s = \text{Mass} \times g = 5a(1 - x)Vg \]

### Step 4: Calculate the weight of the water displaced
- According to Archimedes' principle, the weight of the water displaced by the shell is equal to the weight of the shell when it is submerged.
- The volume of water displaced is equal to the total volume of the shell \( V \), so the weight of the displaced water is:
\[ W_d = \text{Density of water} \times \text{Volume of water displaced} \times g = aVg \]

### Step 5: Set up the equation based on the problem statement
- The problem states that the solid shell loses half of its weight in water. Therefore, we can write:
\[ W_d = \frac{1}{2} W_s \]
- Substituting the expressions for \( W_d \) and \( W_s \):
\[ aVg = \frac{1}{2} \times 5a(1 - x)Vg \]

### Step 6: Simplify the equation
- We can cancel \( a \), \( V \), and \( g \) from both sides (assuming they are not zero):
\[ 1 = \frac{1}{2} \times 5(1 - x) \]
- Multiplying both sides by 2:
\[ 2 = 5(1 - x) \]

### Step 7: Solve for \( x \)
- Expanding the equation:
\[ 2 = 5 - 5x \]
- Rearranging gives:
\[ 5x = 5 - 2 \]
\[ 5x = 3 \]
\[ x = \frac{3}{5} \]

### Conclusion
The fraction of the volume that is hollow is \( \frac{3}{5} \).

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