A hollow sphere of external radius R and thickness t (< < R) is made of a metal of density `rho`. The sphere will float in water if

Options

A. `t lt= R/rho`
B. `t lt= (R )/(3 rho)`
C. `t lt= (R )/(2rho)`
D. `t gt= (R )/(3rho)`

Explanation

To solve the problem of whether a hollow sphere of external radius \( R \) and thickness \( t \) will float in water, we need to analyze the conditions for flotation based on the principles of buoyancy and density.

### Step-by-Step Solution:

1. **Understand the Condition for Floating**:
A hollow sphere will float if the weight of the sphere is less than or equal to the weight of the water displaced by the volume of the sphere. This can be expressed mathematically using the principle of buoyancy.

2. **Calculate the Volume of the Hollow Sphere**:
The volume \( V \) of the hollow sphere can be calculated using the formula for the volume of a sphere. The volume of the outer sphere (with radius \( R \)) minus the volume of the inner sphere (with radius \( R - t \)) gives us the volume of the hollow part:
\[ V = \frac{4}{3} \pi R^3 - \frac{4}{3} \pi (R - t)^3 \]
Simplifying this, we get:
\[ V = \frac{4}{3} \pi \left( R^3 - (R - t)^3 \right) \]

3. **Calculate the Mass of the Hollow Sphere**:
The mass \( M \) of the hollow sphere can be calculated using the density \( \rho \) of the metal and the volume \( V \):
\[ M = \rho \cdot V \]

4. **Weight of the Hollow Sphere**:
The weight of the hollow sphere \( W_s \) is given by:
\[ W_s = M \cdot g = \rho \cdot V \cdot g \]
where \( g \) is the acceleration due to gravity.

5. **Weight of the Water Displaced**:
The weight of the water displaced \( W_w \) when the sphere is submerged is given by the weight of the volume of water equal to the volume of the sphere:
\[ W_w = \text{Volume of water displaced} \cdot \text{Density of water} \cdot g = V \cdot \rho_w \cdot g \]
where \( \rho_w \) is the density of water (approximately \( 1 \, \text{g/cm}^3 \) or \( 1000 \, \text{kg/m}^3 \)).

6. **Set Up the Inequality for Floating**:
For the sphere to float, we need:
\[ W_s \leq W_w \]
This implies:
\[ \rho \cdot V \cdot g \leq V \cdot \rho_w \cdot g \]
Dividing both sides by \( V \cdot g \) (assuming \( V \neq 0 \)):
\[ \rho \leq \rho_w \]

7. **Relate Density, Thickness, and Radius**:
Since the thickness \( t \) is much smaller than the radius \( R \), we can approximate the volume of the hollow sphere. The mass of the hollow sphere can be approximated as:
\[ M \approx 4 \pi R^2 t \cdot \rho \]
The volume of the outer sphere is:
\[ V \approx \frac{4}{3} \pi R^3 \]
Setting the mass of the sphere less than or equal to the mass of the displaced water gives:
\[ 4 \pi R^2 t \cdot \rho \leq \frac{4}{3} \pi R^3 \cdot \rho_w \]
Simplifying this leads to:
\[ t \cdot \rho \leq \frac{R \cdot \rho_w}{3} \]

8. **Final Condition**:
Rearranging gives us the condition for the thickness \( t \):
\[ t \leq \frac{R \cdot \rho_w}{3 \cdot \rho} \]

### Conclusion:
The hollow sphere will float in water if the thickness \( t \) is less than or equal to \( \frac{R \cdot \rho_w}{3 \cdot \rho} \).