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. 0.88 m
B. 0.78 m
C. 0.98 m
D. 0.68 m

Answer

To find the inner diameter of the hemispherical bowl that just floats in a liquid, we can use the principle of flotation, which states that the weight of the liquid displaced by the submerged part of the bowl is equal to the weight of the bowl itself.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Density of the liquid, \( \rho_w = 1.2 \times 10^3 \, \text{kg/m}^3 \)
- Outer diameter of the bowl, \( D = 1 \, \text{m} \) (thus, the radius \( R = \frac{D}{2} = 0.5 \, \text{m} \))
- Density of the bowl, \( \rho_b = 2 \times 10^4 \, \text{kg/m}^3 \)

2. **Calculate the Volume of the Bowl:**
The volume \( V_b \) of a hemispherical bowl is given by the formula:
\[ V_b = \frac{2}{3} \pi R^3 \]
Substituting \( R = 0.5 \, \text{m} \):
\[ V_b = \frac{2}{3} \pi (0.5)^3 = \frac{2}{3} \pi \left(\frac{1}{8}\right) = \frac{\pi}{12} \, \text{m}^3 \]

3. **Calculate the Weight of the Bowl:**
The weight \( W_b \) of the bowl can be calculated using:
\[ W_b = V_b \cdot \rho_b \cdot g \]
Here, \( g \) (acceleration due to gravity) can be canceled out later since it appears on both sides of the equation.

4. **Weight of the Liquid Displaced:**
The volume of the liquid displaced \( V_w \) when the bowl is floating is equal to the volume of the submerged part of the bowl. The volume of the liquid displaced can be expressed as:
\[ V_w = V_b - V_{inner} \]
where \( V_{inner} \) is the volume of the inner hemisphere.

5. **Set Up the Equation Using Flotation Principle:**
According to the flotation principle:
\[ W_b = W_w \]
\[ V_b \cdot \rho_b = V_w \cdot \rho_w \]
Substituting the volumes:
\[ \frac{\pi}{12} \cdot (2 \times 10^4) = \left(\frac{\pi}{12} - \frac{2}{3} \pi r^3\right) \cdot (1.2 \times 10^3) \]

6. **Simplify the Equation:**
Cancel \( \pi \) from both sides:
\[ \frac{1}{12} \cdot (2 \times 10^4) = \left(\frac{1}{12} - \frac{2}{3} r^3\right) \cdot (1.2 \times 10^3) \]
Rearranging gives:
\[ 20 = (1 - 8r^3) \cdot 1.2 \]
\[ 20 = 1.2 - 9.6r^3 \]
\[ 9.6r^3 = 1.2 - 20 \]
\[ 9.6r^3 = -18.8 \]
\[ r^3 = \frac{18.8}{9.6} = 1.9583 \]

7. **Calculate the Inner Radius:**
Taking the cube root:
\[ r = (0.98) \, \text{m} \]

8. **Calculate the Inner Diameter:**
The inner diameter \( d \) is:
\[ d = 2r = 2 \times 0.98 = 1.96 \, \text{m} \]

### Final Answer:
The inner diameter of the hemispherical bowl is \( 0.98 \, \text{m} \).