A wooden cube is found to float in water with `½cm` of its vertical side above the water. On keeping a weight of 50 gm over its top, it is just submerged in the water. The specific gravity of wood is

Options

A. 0.8
B. 0.9
C. 0.85
D. 0.95

Explanation

To find the specific gravity of the wooden cube, we can follow these steps:

### Step 1: Understand the problem
We have a wooden cube that floats in water with `½ cm` of its vertical side above the water. When a weight of `50 g` is placed on top of the cube, it becomes fully submerged. We need to find the specific gravity of the wood.

### Step 2: Define the dimensions of the cube
Let the side length of the cube be `x cm`. Since `½ cm` of the cube is above the water when it is floating, the submerged height of the cube is:
\[ \text{Submerged height} = x - 0.5 \, \text{cm} \]

### Step 3: Calculate the buoyant force when the cube is floating
The buoyant force (Fb1) acting on the cube when it is floating is equal to the weight of the water displaced by the submerged part of the cube:
\[ F_{b1} = \text{Volume of submerged part} \times \text{Density of water} \times g \]

The volume of the submerged part is:
\[ \text{Volume} = x^2 \times (x - 0.5) \]

Since the density of water is `1 g/cm³`, we have:
\[ F_{b1} = x^2 \times (x - 0.5) \times g \]

### Step 4: Set up the equation for the first case
The weight of the cube (Wc) is equal to the buoyant force when the cube is floating:
\[ W_c = F_{b1} \]
\[ W_c = x^2 \times (x - 0.5) \]

### Step 5: Calculate the buoyant force when the weight is added
When a `50 g` weight is added, the cube is fully submerged. The total weight acting downwards is:
\[ W_{total} = W_c + 50 \, \text{g} \]

The buoyant force (Fb2) when the cube is fully submerged is:
\[ F_{b2} = \text{Volume of cube} \times \text{Density of water} \times g \]
\[ F_{b2} = x^3 \times g \]

### Step 6: Set up the equation for the second case
Now we can write the equation for the second case:
\[ W_c + 50 = F_{b2} \]
\[ W_c + 50 = x^3 \]

### Step 7: Substitute Wc from the first case into the second case
From the first case, we have:
\[ W_c = x^2 \times (x - 0.5) \]

Substituting this into the second equation:
\[ x^2 \times (x - 0.5) + 50 = x^3 \]

### Step 8: Simplify the equation
Rearranging gives:
\[ x^2 \times (x - 0.5) + 50 = x^3 \]
\[ x^3 - x^2 \times (x - 0.5) = 50 \]
\[ x^3 - x^3 + 0.5x^2 = 50 \]
\[ 0.5x^2 = 50 \]
\[ x^2 = 100 \]
\[ x = 10 \, \text{cm} \]

### Step 9: Calculate the mass of the cube
Now we can find the mass of the cube:
\[ W_c = x^2 \times (x - 0.5) \]
\[ W_c = 10^2 \times (10 - 0.5) = 100 \times 9.5 = 950 \, \text{g} \]

### Step 10: Calculate the specific gravity
The specific gravity (SG) of the wood is given by the ratio of the density of the wood to the density of water:
\[ \text{Density of wood} = \frac{\text{Mass}}{\text{Volume}} = \frac{950 \, \text{g}}{10^3 \, \text{cm}^3} = 0.95 \, \text{g/cm}^3 \]

Since the density of water is `1 g/cm³`, the specific gravity is:
\[ \text{Specific Gravity} = \frac{0.95 \, \text{g/cm}^3}{1 \, \text{g/cm}^3} = 0.95 \]

### Final Answer
The specific gravity of the wood is **0.95**.

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