A hollow sphere of external radius R and thickness t(`lt lt`R) is made of a metal of density `rho`. The sphere will float in water if (Take density of water = `"1 g cm"^(-3)` )

Options

A. `t le R/rho`
B. `t le R/(3rho)`
C. `t le R/(2rho)`
D. `t ge R/(3rho)`

Answer

To determine the condition for a hollow sphere to float in water, we start by analyzing the forces acting on the sphere.

### Step-by-Step Solution:

1. **Understanding the Problem**: We have a hollow sphere with an external radius \( R \) and thickness \( t \) (where \( t \ll R \)). The density of the metal used to make the sphere is \( \rho \), and the density of water is \( 1 \, \text{g/cm}^3 \).

2. **Floating Condition**: For the sphere to float, the buoyant force acting on it must be equal to or greater than its weight. The buoyant force is equal to the weight of the water displaced by the submerged part of the sphere.

3. **Calculating the Volume of the Sphere**:
- The volume \( V \) of the hollow sphere can be calculated as the difference between the volume of the outer sphere and the inner sphere.
- The volume of the outer sphere is given by:
\[ V_{\text{outer}} = \frac{4}{3} \pi R^3 \]
- The inner radius \( R_{\text{inner}} \) is \( R - t \). Thus, the volume of the inner sphere is:
\[ V_{\text{inner}} = \frac{4}{3} \pi (R - t)^3 \]
- Therefore, the volume of the hollow part of the sphere is:
\[ V = V_{\text{outer}} - V_{\text{inner}} = \frac{4}{3} \pi R^3 - \frac{4}{3} \pi (R - t)^3 \]

4. **Simplifying the Volume**:
- We can simplify the expression for the volume of the hollow sphere:
\[ V = \frac{4}{3} \pi \left( R^3 - (R - t)^3 \right) \]
- Using the binomial expansion for \( (R - t)^3 \):
\[ (R - t)^3 = R^3 - 3R^2t + 3Rt^2 - t^3 \]
- Thus, we have:
\[ V = \frac{4}{3} \pi \left( 3R^2t - 3Rt^2 + t^3 \right) \approx \frac{4}{3} \pi (3R^2t) \quad \text{(since \( t \ll R \))} \]
- Therefore:
\[ V \approx 4 \pi R^2 t \]

5. **Weight of the Sphere**:
- The weight \( W \) of the sphere can be calculated using its volume and density:
\[ W = V \cdot \rho = 4 \pi R^2 t \cdot \rho \]

6. **Buoyant Force**:
- The buoyant force \( F_b \) is equal to the weight of the water displaced:
\[ F_b = \text{Volume of water displaced} \cdot \text{Density of water} = V \cdot 1 \, \text{g/cm}^3 = 4 \pi R^2 t \cdot 1 \]

7. **Setting Up the Equation**:
- For the sphere to float:
\[ F_b \geq W \]
- Therefore:
\[ 4 \pi R^2 t \geq 4 \pi R^2 t \cdot \rho \]

8. **Solving for Thickness \( t \)**:
- Dividing both sides by \( 4 \pi R^2 \) (assuming \( R \neq 0 \)):
\[ t \geq t \cdot \rho \]
- Rearranging gives:
\[ t \cdot (1 - \rho) \geq 0 \]
- Since \( \rho \) (density of the metal) must be less than or equal to the density of water for the sphere to float:
\[ \rho \leq 1 \, \text{g/cm}^3 \]

### Final Condition:
The thickness \( t \) must satisfy:
\[ t \leq \frac{R}{3} \cdot (1 - \rho) \]