A solid floats with `2//3` of its volume immersed in a liquid and with `3//4` of its volume immersed in another liquid what fraction of its volume will be immersed of it floats in a homogenous mixture formed of equal volumes of the liquids?

Options

A. `6//7`
B. `8//11`
C. `11//16`
D. `12//17`

Explanation

To solve the problem, we need to find the fraction of the volume of a solid that will be immersed when it floats in a homogeneous mixture formed of equal volumes of two different liquids. Let's break down the solution step by step.

### Step 1: Understanding the Given Information
We know that:
- In the first liquid, the solid is floating with \( \frac{2}{3} \) of its volume immersed.
- In the second liquid, the solid is floating with \( \frac{3}{4} \) of its volume immersed.

Let \( V \) be the total volume of the solid, \( \rho_s \) be the density of the solid, \( \rho_1 \) be the density of the first liquid, and \( \rho_2 \) be the density of the second liquid.

### Step 2: Applying the Principle of Buoyancy
According to Archimedes' principle, the weight of the solid is equal to the weight of the liquid displaced by the immersed volume of the solid.

1. For the first liquid:
\[ \text{Weight of solid} = V \cdot \rho_s \cdot g \]
\[ \text{Weight of liquid displaced} = \left(\frac{2}{3} V\right) \cdot \rho_1 \cdot g \]
Setting these equal gives:
\[ V \cdot \rho_s \cdot g = \left(\frac{2}{3} V\right) \cdot \rho_1 \cdot g \]
Cancelling \( V \) and \( g \) from both sides:
\[ \rho_s = \frac{2}{3} \rho_1 \]

2. For the second liquid:
\[ \text{Weight of liquid displaced} = \left(\frac{3}{4} V\right) \cdot \rho_2 \cdot g \]
Setting the weights equal gives:
\[ V \cdot \rho_s \cdot g = \left(\frac{3}{4} V\right) \cdot \rho_2 \cdot g \]
Cancelling \( V \) and \( g \):
\[ \rho_s = \frac{3}{4} \rho_2 \]

### Step 3: Relating the Densities
From the equations we derived:
1. \( \rho_1 = \frac{3}{2} \rho_s \)
2. \( \rho_2 = \frac{4}{3} \rho_s \)

### Step 4: Finding the Density of the Mixture
When equal volumes of the two liquids are mixed, the density of the mixture \( \rho_m \) can be calculated as:
\[ \rho_m = \frac{\rho_1 + \rho_2}{2} \]
Substituting the values of \( \rho_1 \) and \( \rho_2 \):
\[ \rho_m = \frac{\frac{3}{2} \rho_s + \frac{4}{3} \rho_s}{2} \]
Finding a common denominator (which is 6):
\[ \rho_m = \frac{\frac{9}{6} \rho_s + \frac{8}{6} \rho_s}{2} = \frac{\frac{17}{6} \rho_s}{2} = \frac{17}{12} \rho_s \]

### Step 5: Finding the Immersed Volume in the Mixture
Let \( V' \) be the volume of the solid immersed in the mixture. Using the principle of buoyancy again:
\[ V' \cdot \rho_m \cdot g = V \cdot \rho_s \cdot g \]
Cancelling \( g \) and rearranging gives:
\[ V' \cdot \frac{17}{12} \rho_s = V \cdot \rho_s \]
Cancelling \( \rho_s \) from both sides:
\[ V' \cdot \frac{17}{12} = V \]
Thus,
\[ V' = V \cdot \frac{12}{17} \]

### Step 6: Finding the Fraction of Volume Immersed
The fraction of the volume of the solid that is immersed in the mixture is:
\[ \text{Fraction immersed} = \frac{V'}{V} = \frac{12}{17} \]

### Final Answer
The fraction of the volume of the solid that will be immersed when it floats in the homogeneous mixture is \( \frac{12}{17} \).

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