A hollow metallic sphere, the outer and inner diameters of which are `d_(1) and d_(2)`. floats on the surface of a liquid. The density of metal is `rho_(1)` and the density of liquid is `rho_(2)`. What weight must be added inside the sphere in order for it to float below the level of liquid?

Options

A. `pi[d_(1)^(3)(rho_(2)-rho_(1))+d_(2)^(3)rho_(1)]g`
B. `(pi)/(6)[d_(1)^(3)(rho_(2)-rho_(1))+d_(2)^(3)rho_(1)]g`
C. `(4pi)/(3)[d_(1)^(3)(rho_(2)-rho_(1))+d_(2)^(3)rho_(1)]g`
D. `(4pi)/(3)d_(1)^(3)(rho_(1)-rho_(2))g`

Explanation

To solve the problem of determining the weight that must be added inside a hollow metallic sphere for it to float just below the surface of a liquid, we can follow these steps:

### Step 1: Understand the Conditions for Floating
For the sphere to float just below the surface of the liquid, the buoyant force acting on the sphere must equal the total weight of the sphere plus the weight added inside it.

### Step 2: Calculate the Buoyant Force
The buoyant force \( F_b \) can be calculated using Archimedes' principle, which states that the buoyant force is equal to the weight of the liquid displaced by the submerged part of the sphere. The volume of the outer sphere is given by:

\[ V_{outer} = \frac{4}{3} \pi \left( \frac{d_1}{2} \right)^3 = \frac{\pi d_1^3}{6} \]

The buoyant force can then be expressed as:

\[ F_b = \rho_2 \cdot V_{outer} \cdot g = \rho_2 \cdot \frac{\pi d_1^3}{6} \cdot g \]

### Step 3: Calculate the Weight of the Sphere
The weight of the hollow metallic sphere can be calculated by finding the volume of the metal used in the sphere, which is the difference between the outer and inner volumes:

\[ V_{metal} = V_{outer} - V_{inner} = \frac{4}{3} \pi \left( \frac{d_1}{2} \right)^3 - \frac{4}{3} \pi \left( \frac{d_2}{2} \right)^3 \]

This simplifies to:

\[ V_{metal} = \frac{\pi}{6} (d_1^3 - d_2^3) \]

The weight of the metal is then:

\[ W_{metal} = \rho_1 \cdot V_{metal} \cdot g = \rho_1 \cdot \frac{\pi}{6} (d_1^3 - d_2^3) \cdot g \]

### Step 4: Set Up the Equation for Floating Condition
For the sphere to float just below the surface, we set the buoyant force equal to the total weight (weight of the sphere plus the added weight \( W_{added} \)):

\[ F_b = W_{metal} + W_{added} \]

Substituting the expressions we derived:

\[ \rho_2 \cdot \frac{\pi d_1^3}{6} \cdot g = \rho_1 \cdot \frac{\pi}{6} (d_1^3 - d_2^3) \cdot g + W_{added} \]

### Step 5: Solve for the Added Weight
Rearranging the equation to find \( W_{added} \):

\[ W_{added} = \rho_2 \cdot \frac{\pi d_1^3}{6} \cdot g - \rho_1 \cdot \frac{\pi}{6} (d_1^3 - d_2^3) \cdot g \]

Factoring out \( \frac{\pi g}{6} \):

\[ W_{added} = \frac{\pi g}{6} \left( \rho_2 d_1^3 - \rho_1 (d_1^3 - d_2^3) \right) \]

### Final Expression
Thus, the weight that must be added inside the sphere is:

\[ W_{added} = \frac{\pi g}{6} \left( \rho_2 d_1^3 - \rho_1 (d_1^3 - d_2^3) \right) \]