The ratio of masses of a cubical block of wood and a chunk of concrete is 3/5 so that the combination just floats with entire volume submerged under water. If the specific gravity of wood is 0.5 then specific gravity of concrete will be

Options

A. 2.5
B. 25
C. 15
D. 3

Answer

To solve the problem step by step, we will follow the logic presented in the video transcript.

### Step 1: Understand the Given Information
We have a cubical block of wood and a chunk of concrete with a mass ratio of \( \frac{3}{5} \). The specific gravity of wood is given as \( 0.5 \). We need to find the specific gravity of concrete.

### Step 2: Define Specific Gravity
Specific gravity (SG) is defined as the ratio of the density of a substance to the density of a reference substance (usually water). Mathematically:
\[ SG = \frac{\text{Density of substance}}{\text{Density of water}} \]

### Step 3: Calculate the Density of Wood
Given that the specific gravity of wood is \( 0.5 \), we can calculate the density of wood (\( \rho_1 \)):
\[ \rho_1 = SG_{\text{wood}} \times \rho_{\text{water}} = 0.5 \times 1000 \, \text{kg/m}^3 = 500 \, \text{kg/m}^3 \]

### Step 4: Set Up the Masses
Let the mass of the wood be \( m_1 \) and the mass of the concrete be \( m_2 \). According to the problem, the ratio of their masses is:
\[ \frac{m_1}{m_2} = \frac{3}{5} \]
This implies:
\[ m_1 = 3k \quad \text{and} \quad m_2 = 5k \quad \text{for some constant } k. \]

### Step 5: Calculate the Volumes
The volume of the wood (\( V_1 \)) and the volume of the concrete (\( V_2 \)) can be expressed as:
\[ V_1 = \frac{m_1}{\rho_1} = \frac{3k}{500} \quad \text{and} \quad V_2 = \frac{m_2}{\rho_2} = \frac{5k}{\rho_2} \]

### Step 6: Apply the Condition for Floating
For the combination to float with the entire volume submerged, the upward buoyant force must equal the total weight of the wood and concrete:
\[ \rho_{\text{water}} \cdot (V_1 + V_2) \cdot g = (m_1 + m_2) \cdot g \]
Cancelling \( g \) from both sides gives:
\[ \rho_{\text{water}} \cdot (V_1 + V_2) = m_1 + m_2 \]
Substituting the values:
\[ 1000 \left( \frac{3k}{500} + \frac{5k}{\rho_2} \right) = 3k + 5k \]
This simplifies to:
\[ 1000 \left( \frac{3k}{500} + \frac{5k}{\rho_2} \right) = 8k \]

### Step 7: Simplify the Equation
Dividing through by \( k \) (assuming \( k \neq 0 \)):
\[ 1000 \left( \frac{3}{500} + \frac{5}{\rho_2} \right) = 8 \]
This simplifies to:
\[ \frac{3000}{500} + \frac{5000}{\rho_2} = 8 \]
\[ 6 + \frac{5000}{\rho_2} = 8 \]
\[ \frac{5000}{\rho_2} = 2 \]
Thus:
\[ \rho_2 = \frac{5000}{2} = 2500 \, \text{kg/m}^3 \]

### Step 8: Calculate the Specific Gravity of Concrete
Now, we can find the specific gravity of concrete:
\[ SG_{\text{concrete}} = \frac{\rho_2}{\rho_{\text{water}}} = \frac{2500}{1000} = 2.5 \]

### Final Answer
The specific gravity of concrete is \( 2.5 \).

---