A hemispherical bowl just floats without sinking in a liquid of density `1.2xx10^(3)kg//m^(3)`. If outer diameter and the density of the bowl are `1m` and `2 xx 10^(4) kg//m^(3)` respectively, then the inner diameter of bowl will be

Options

A. `0.94m`
B. `0.97m`
C. `0.98m`
D. `0.99m`

Explanation

To solve the problem of finding the inner diameter of a hemispherical bowl that just floats in a liquid, we will follow these steps:

### Step 1: Understand the Problem
We have a hemispherical bowl with:
- Outer diameter (D) = 1 m, hence the outer radius (R) = 0.5 m
- Density of the bowl (ρ_bowl) = 2 x 10^4 kg/m³
- Density of the liquid (ρ_liquid) = 1.2 x 10^3 kg/m³

We need to find the inner diameter (d) of the bowl.

### Step 2: Apply the Principle of Floatation
According to the principle of floatation, the weight of the bowl must equal the weight of the liquid displaced. This can be expressed as:
\[ \text{Weight of the bowl} = \text{Weight of the liquid displaced} \]

### Step 3: Calculate the Volume of the Bowl
The volume of the hemispherical bowl is given by the formula:
\[ V = \frac{2}{3} \pi R^3 \]
For the outer volume (V_outer):
\[ V_{\text{outer}} = \frac{2}{3} \pi (0.5)^3 = \frac{2}{3} \pi \cdot \frac{1}{8} = \frac{\pi}{12} \, \text{m}^3 \]

### Step 4: Calculate the Volume of the Inner Part
Let the inner radius be \( r \). The volume of the inner hemisphere (V_inner) is:
\[ V_{\text{inner}} = \frac{2}{3} \pi r^3 \]

### Step 5: Calculate the Volume of the Material of the Bowl
The volume of the material of the bowl is the difference between the outer volume and the inner volume:
\[ V_{\text{material}} = V_{\text{outer}} - V_{\text{inner}} = \frac{\pi}{12} - \frac{2}{3} \pi r^3 \]

### Step 6: Calculate the Weight of the Bowl
The weight of the bowl can be calculated using its density and volume:
\[ \text{Weight of the bowl} = \text{Density} \times \text{Volume} \times g \]
\[ = 2 \times 10^4 \times V_{\text{material}} \times g \]

### Step 7: Calculate the Weight of the Liquid Displaced
The weight of the liquid displaced is given by:
\[ \text{Weight of the liquid displaced} = \text{Density of liquid} \times \text{Volume of liquid displaced} \times g \]
The volume of liquid displaced is equal to the outer volume of the bowl:
\[ = 1.2 \times 10^3 \times V_{\text{outer}} \times g \]

### Step 8: Set Up the Equation
Setting the weight of the bowl equal to the weight of the liquid displaced:
\[ 2 \times 10^4 \left( \frac{\pi}{12} - \frac{2}{3} \pi r^3 \right) = 1.2 \times 10^3 \left( \frac{\pi}{12} \right) \]

### Step 9: Simplify and Solve for r
Cancel out \( g \) and \( \pi \) from both sides:
\[ 2 \times 10^4 \left( \frac{1}{12} - \frac{2}{3} r^3 \right) = 1.2 \times 10^3 \left( \frac{1}{12} \right) \]

Now, simplify and solve for \( r^3 \):
\[ 2 \times 10^4 \left( \frac{1}{12} \right) - 2 \times 10^4 \left( \frac{2}{3} r^3 \right) = 1.2 \times 10^3 \left( \frac{1}{12} \right) \]

### Step 10: Calculate r
After solving the equation, we find:
\[ r \approx 0.49 \, \text{m} \]
Thus, the inner diameter \( d = 2r \approx 0.98 \, \text{m} \).

### Final Answer
The inner diameter of the bowl is approximately **0.98 m**.