An object floats in water such that `(1)/(4)`th of its volume is above the surface of water. When the same object is made to float in a sample of impure alcohol, the volume that remains outside the liquid is of `(1)/(10)` total volume. The relative density of the sample of alcohol will be

Options

A. 83
B. 8.3
C. 0.83
D. 0.083

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have an object that floats in water with \( \frac{1}{4} \) of its volume above the surface. When the same object is placed in impure alcohol, \( \frac{1}{10} \) of its volume is above the surface. We need to find the relative density of the alcohol.

### Step 2: Define Variables
Let the total volume of the object be \( V \).

### Step 3: Calculate Volume Submerged in Water
Since \( \frac{1}{4} \) of the volume is above water, the volume submerged in water is:
\[ V_{\text{submerged in water}} = V - \frac{1}{4}V = \frac{3}{4}V \]

### Step 4: Write the Buoyant Force in Water
The buoyant force acting on the object when submerged in water can be expressed as:
\[ F_b = V_{\text{submerged}} \times \text{density of water} \times g \]
Substituting the values:
\[ F_b = \frac{3}{4}V \times \rho_W \times g \quad \text{(Equation 1)} \]

### Step 5: Calculate Volume Submerged in Alcohol
When the object is in alcohol, \( \frac{1}{10} \) of its volume is above the surface, so the volume submerged in alcohol is:
\[ V_{\text{submerged in alcohol}} = V - \frac{1}{10}V = \frac{9}{10}V \]

### Step 6: Write the Buoyant Force in Alcohol
The buoyant force acting on the object when submerged in alcohol can be expressed as:
\[ F_b = V_{\text{submerged}} \times \text{density of alcohol} \times g \]
Substituting the values:
\[ F_b = \frac{9}{10}V \times \rho_A \times g \quad \text{(Equation 2)} \]

### Step 7: Equate the Buoyant Forces
Since the buoyant force is the same in both cases, we can set Equation 1 equal to Equation 2:
\[ \frac{3}{4}V \times \rho_W \times g = \frac{9}{10}V \times \rho_A \times g \]

### Step 8: Simplify the Equation
Cancel \( V \) and \( g \) from both sides:
\[ \frac{3}{4} \rho_W = \frac{9}{10} \rho_A \]

### Step 9: Solve for the Density of Alcohol
Rearranging the equation gives:
\[ \rho_A = \frac{3}{4} \times \frac{10}{9} \rho_W = \frac{30}{36} \rho_W = \frac{5}{6} \rho_W \]

### Step 10: Calculate the Relative Density
The relative density (RD) of the alcohol is given by:
\[ \text{Relative Density} = \frac{\rho_A}{\rho_W} = \frac{5}{6} \approx 0.833 \]

### Final Answer
The relative density of the sample of alcohol is approximately \( 0.83 \).

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