To find the density of the second body, we will use the principle of buoyancy and the condition of equilibrium. Here’s a step-by-step solution:
### Step 1: Understand the Equilibrium Condition
When two bodies are suspended in water and are in equilibrium, the effective weight of both bodies must be equal. The effective weight is the actual weight of the body minus the buoyant force acting on it.
### Step 2: Define the Variables
Let:
- \( m_1 = 60 \, \text{g} \) (mass of the first body)
- \( \rho_1 = 12 \, \text{g/cm}^3 \) (density of the first body)
- \( m_2 = 72 \, \text{g} \) (mass of the second body)
- \( \rho_2 \) = ? (density of the second body)
- \( \rho_L = 1 \, \text{g/cm}^3 \) (density of water)
### Step 3: Calculate the Volume of the First Body
The volume \( V_1 \) of the first body can be calculated using the formula:
\[
V_1 = \frac{m_1}{\rho_1}
\]
Substituting the values:
\[
V_1 = \frac{60 \, \text{g}}{12 \, \text{g/cm}^3} = 5 \, \text{cm}^3
\]
### Step 4: Calculate the Buoyant Force on the First Body
The buoyant force \( F_b \) on the first body is given by:
\[
F_b = V_1 \cdot \rho_L \cdot g
\]
Since \( g \) cancels out in the equilibrium equation, we can ignore it for now. Thus:
\[
F_b = V_1 \cdot \rho_L = 5 \, \text{cm}^3 \cdot 1 \, \text{g/cm}^3 = 5 \, \text{g}
\]
### Step 5: Calculate the Effective Weight of the First Body
The effective weight \( W_{eff1} \) of the first body is:
\[
W_{eff1} = m_1 - F_b = 60 \, \text{g} - 5 \, \text{g} = 55 \, \text{g}
\]
### Step 6: Calculate the Volume of the Second Body
The volume \( V_2 \) of the second body can be expressed as:
\[
V_2 = \frac{m_2}{\rho_2}
\]
### Step 7: Calculate the Buoyant Force on the Second Body
The buoyant force \( F_b \) on the second body is:
\[
F_b = V_2 \cdot \rho_L = \frac{m_2}{\rho_2} \cdot 1 \, \text{g/cm}^3 = \frac{72 \, \text{g}}{\rho_2}
\]
### Step 8: Calculate the Effective Weight of the Second Body
The effective weight \( W_{eff2} \) of the second body is:
\[
W_{eff2} = m_2 - F_b = 72 \, \text{g} - \frac{72 \, \text{g}}{\rho_2}
\]
### Step 9: Set the Effective Weights Equal
Since the two bodies are in equilibrium:
\[
W_{eff1} = W_{eff2}
\]
Substituting the values:
\[
55 \, \text{g} = 72 \, \text{g} - \frac{72 \, \text{g}}{\rho_2}
\]
### Step 10: Solve for \( \rho_2 \)
Rearranging the equation:
\[
\frac{72 \, \text{g}}{\rho_2} = 72 \, \text{g} - 55 \, \text{g}
\]
\[
\frac{72 \, \text{g}}{\rho_2} = 17 \, \text{g}
\]
Cross-multiplying gives:
\[
72 = 17 \cdot \rho_2
\]
\[
\rho_2 = \frac{72}{17} \approx 4.24 \, \text{g/cm}^3
\]
### Final Answer
The density of the second body is approximately \( 4.24 \, \text{g/cm}^3 \).
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