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, we need to find the inner diameter of a hemispherical bowl that just floats in a liquid of given density. Here’s a step-by-step solution:

### Step 1: Understand the Given Data
- **Outer diameter of the bowl** = 1 m
- **Outer radius (R)** = 1 m / 2 = 0.5 m
- **Density of the bowl (ρ_b)** = 2 × 10^4 kg/m³
- **Density of the liquid (ρ_w)** = 1.2 × 10^3 kg/m³

### Step 2: Write the Equilibrium Condition
For the bowl to float without sinking, the weight of the bowl must equal the buoyant force acting on it. This can be expressed mathematically as:
\[ F_{\text{upthrust}} = mg \]

### Step 3: Calculate the Weight of the Bowl (mg)
The mass (m) of the bowl can be calculated using its volume and density:
\[ m = V \cdot \rho_b \]
The volume (V) of the hemispherical bowl is given by:
\[ V = \frac{2}{3} \pi (R^3 - r^3) \]
Where:
- \( R \) = outer radius = 0.5 m
- \( r \) = inner radius (to be determined)

Thus, the weight of the bowl is:
\[ mg = \frac{2}{3} \pi (R^3 - r^3) \cdot \rho_b \cdot g \]

### Step 4: Calculate the Buoyant Force (F_upthrust)
The buoyant force is equal to the weight of the liquid displaced by the submerged part of the bowl:
\[ F_{\text{upthrust}} = V_{\text{displaced}} \cdot \rho_w \cdot g \]
Where the volume of the displaced liquid (which is the volume of the outer hemisphere) is:
\[ V_{\text{displaced}} = \frac{2}{3} \pi R^3 \]
Thus, the buoyant force becomes:
\[ F_{\text{upthrust}} = \frac{2}{3} \pi R^3 \cdot \rho_w \cdot g \]

### Step 5: Set Up the Equation
Equating the weight of the bowl to the buoyant force:
\[ \frac{2}{3} \pi (R^3 - r^3) \cdot \rho_b \cdot g = \frac{2}{3} \pi R^3 \cdot \rho_w \cdot g \]

### Step 6: Simplify the Equation
Cancel out common terms (including \( \frac{2}{3} \pi g \)):
\[ (R^3 - r^3) \cdot \rho_b = R^3 \cdot \rho_w \]

### Step 7: Substitute Known Values
Substituting \( R = 0.5 \) m, \( \rho_b = 2 \times 10^4 \) kg/m³, and \( \rho_w = 1.2 \times 10^3 \) kg/m³:
\[ (0.5^3 - r^3) \cdot (2 \times 10^4) = (0.5^3) \cdot (1.2 \times 10^3) \]

### Step 8: Calculate \( 0.5^3 \)
\[ 0.5^3 = 0.125 \]
So the equation becomes:
\[ (0.125 - r^3) \cdot (2 \times 10^4) = 0.125 \cdot (1.2 \times 10^3) \]

### Step 9: Solve for \( r^3 \)
Calculating the right side:
\[ 0.125 \cdot (1.2 \times 10^3) = 150 \]
Thus:
\[ (0.125 - r^3) \cdot (2 \times 10^4) = 150 \]
Dividing both sides by \( 2 \times 10^4 \):
\[ 0.125 - r^3 = \frac{150}{2 \times 10^4} = 0.0075 \]
Rearranging gives:
\[ r^3 = 0.125 - 0.0075 = 0.1175 \]

### Step 10: Calculate \( r \)
Taking the cube root:
\[ r = \sqrt[3]{0.1175} \approx 0.49 \text{ m} \]

### Step 11: Calculate Inner Diameter
The inner diameter \( d \) is:
\[ d = 2r \approx 2 \times 0.49 \approx 0.98 \text{ m} \]

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

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