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. (a)`pi[d_(1)^(3)(rho_(2)-rho_(1))+d_(2)^(3)rho_(1)]g`
B. (b)`(pi)/(6)[d_(1)^(3)(rho_(2)-rho_(1))+d_(2)^(3)rho_(1)]g`
C. (c)`(4pi)/(3)[d_(1)^(3)(rho_(2)-rho_(1))+d_(2)^(2)rho_(1)]g`
D. (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 below the level of a liquid, we can follow these steps:

### Step 1: Understand the Problem
We have a hollow metallic sphere with outer diameter \(d_1\) and inner diameter \(d_2\). The density of the metal is \(\rho_1\) and the density of the liquid is \(\rho_2\). We need to find the additional weight \(W\) that must be added inside the sphere for it to float below the surface of the liquid.

### Step 2: Calculate the Volume of the Sphere
The volume \(V\) of the hollow metallic sphere can be calculated using the formula for the volume of a sphere:

\[ V = \frac{4}{3} \pi R^3 \]

Where \(R\) is the radius. The volume of the hollow part can be found by subtracting the inner volume from the outer volume:

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

### Step 3: Calculate the Weight of the Sphere
The weight of the metallic sphere \(W_{\text{sphere}}\) can be calculated using the density of the metal:

\[ W_{\text{sphere}} = V_{\text{sphere}} \cdot \rho_1 \cdot g \]

Substituting the volume from Step 2:

\[ W_{\text{sphere}} = \left(\frac{4}{3} \pi \left(\frac{d_1^3}{8} - \frac{d_2^3}{8}\right)\right) \cdot \rho_1 \cdot g \]

### Step 4: Calculate the Weight of Liquid Displaced
According to Archimedes' principle, the weight of the liquid displaced \(W_{\text{displaced}}\) is equal to the weight of the liquid that would occupy the volume of the outer sphere:

\[ W_{\text{displaced}} = V_{\text{outer}} \cdot \rho_2 \cdot g = \frac{4}{3} \pi \left(\frac{d_1}{2}\right)^3 \cdot \rho_2 \cdot g \]

### Step 5: Apply the Law of Floatation
For the sphere to float, the total weight (weight of the sphere plus the added weight \(W\)) must equal the weight of the liquid displaced:

\[ W_{\text{sphere}} + W = W_{\text{displaced}} \]

Substituting the expressions from Steps 3 and 4:

\[ \left(\frac{4}{3} \pi \left(\frac{d_1^3}{8} - \frac{d_2^3}{8}\right) \cdot \rho_1 \cdot g\right) + W = \frac{4}{3} \pi \left(\frac{d_1^3}{8}\right) \cdot \rho_2 \cdot g \]

### Step 6: Solve for the Added Weight \(W\)
Rearranging the equation to solve for \(W\):

\[ W = \frac{4}{3} \pi \left(\frac{d_1^3}{8}\right) \cdot \rho_2 \cdot g - \left(\frac{4}{3} \pi \left(\frac{d_1^3}{8} - \frac{d_2^3}{8}\right) \cdot \rho_1 \cdot g\right) \]

This simplifies to:

\[ W = \frac{4}{3} \pi g \left(\frac{d_1^3}{8} \rho_2 - \left(\frac{d_1^3}{8} - \frac{d_2^3}{8}\right) \rho_1\right) \]

### Step 7: Final Expression for \(W\)
Combining terms gives:

\[ W = \frac{4}{3} \pi g \left(\frac{d_1^3}{8} \rho_2 - \frac{d_1^3 \rho_1}{8} + \frac{d_2^3 \rho_1}{8}\right) \]