A block of wood is floating in water such that `1//2` of it is submerged in water when the same block is floated in alcohol, `1//3^(rd)` of it's volume is submerged Now a mixture of water and alcohol is made taking equal volume of both and block is floated in it. What is the % of it's volume that is now submerged?

Options

A. 50
B. 40
C. 25
D. 60

Answer

To solve the problem step by step, we will analyze the conditions given for the block of wood floating in water, alcohol, and then in a mixture of both.

### Step 1: Understand the buoyancy principle
When an object floats, the weight of the object is balanced by the buoyant force acting on it. The buoyant force is equal to the weight of the fluid displaced by the submerged part of the object.

### Step 2: Analyze the block in water
Let the volume of the block be \( V \).
- When the block is floating in water, half of it is submerged:
\[ \text{Volume submerged in water} = \frac{V}{2} \]
- The buoyant force due to water is:
\[ F_b = \text{density of water} \times g \times \frac{V}{2} \]
- This buoyant force equals the weight of the block:
\[ \text{Weight of block} = \text{density of block} \times V \times g \]
- Setting the two forces equal gives:
\[ \text{density of water} \times g \times \frac{V}{2} = \text{density of block} \times V \times g \]
- Canceling \( g \) and \( V \) (assuming \( V \neq 0 \)):
\[ \text{density of water} = 2 \times \text{density of block} \quad \text{(1)} \]

### Step 3: Analyze the block in alcohol
- When the block is floating in alcohol, one-third of it is submerged:
\[ \text{Volume submerged in alcohol} = \frac{V}{3} \]
- The buoyant force due to alcohol is:
\[ F_b = \text{density of alcohol} \times g \times \frac{V}{3} \]
- Setting this equal to the weight of the block:
\[ \text{density of alcohol} \times g \times \frac{V}{3} = \text{density of block} \times V \times g \]
- Canceling \( g \) and \( V \):
\[ \text{density of alcohol} = 3 \times \text{density of block} \quad \text{(2)} \]

### Step 4: Analyze the block in the mixture of water and alcohol
- The mixture is made of equal volumes of water and alcohol. Let’s denote the density of the block as \( \rho_b \).
- From equations (1) and (2):
\[ \text{density of water} = 2\rho_b \quad \text{and} \quad \text{density of alcohol} = 3\rho_b \]
- The density of the mixture can be calculated as:
\[ \text{density of mixture} = \frac{\text{density of water} + \text{density of alcohol}}{2} = \frac{2\rho_b + 3\rho_b}{2} = \frac{5\rho_b}{2} \]

### Step 5: Calculate the volume submerged in the mixture
- Let \( V_d \) be the volume submerged in the mixture.
- The buoyant force in the mixture is:
\[ \text{density of mixture} \times g \times V_d = \text{density of block} \times V \times g \]
- Substituting the density of the mixture:
\[ \frac{5\rho_b}{2} \times g \times V_d = \rho_b \times V \times g \]
- Canceling \( g \) and \( \rho_b \):
\[ \frac{5}{2} V_d = V \]
- Rearranging gives:
\[ V_d = \frac{2}{5} V \]

### Step 6: Calculate the percentage of volume submerged
- The percentage of the block submerged is given by:
\[ \text{Percentage submerged} = \left(\frac{V_d}{V}\right) \times 100 = \left(\frac{2}{5}\right) \times 100 = 40\% \]

### Final Answer
Thus, the percentage of the block's volume that is submerged in the mixture of water and alcohol is **40%**.

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