Two bodies are in equilibrium when suspended in water from the arms of balance. The mass of one body is 36 g and its density is `9 g// cm^3` If the mass of the other is 48 g, its density in `g//cm^3` is

Options

A. (a)`( 4)/( 3)`
B. (b)`(3)/(2)`
C. (c)3
D. (d)5

Answer

To solve the problem step by step, we will use the concept of buoyancy and equilibrium of forces acting on the two bodies submerged in water.

### Step 1: Understand the equilibrium condition
When two bodies are in equilibrium while submerged in water, the apparent weight of both bodies must be equal. The apparent weight of a body is given by the actual weight minus the buoyant force acting on it.

### Step 2: Write down the known values
- Mass of body A (mA) = 36 g
- Density of body A (ρA) = 9 g/cm³
- Mass of body B (mB) = 48 g
- Density of water (ρwater) = 1 g/cm³
- Density of body B (ρB) = ? (to be calculated)

### Step 3: Calculate the volume of body A
Using the formula for density:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
We can rearrange this to find the volume (VA) of body A:
\[ V_A = \frac{m_A}{\rho_A} = \frac{36 \text{ g}}{9 \text{ g/cm}^3} = 4 \text{ cm}^3 \]

### Step 4: Write the equation for apparent weights
The apparent weight of body A (WA) and body B (WB) can be expressed as:
\[ W_A = m_A - \text{Buoyant Force on A} = m_A - \rho_{water} \cdot V_A \]
\[ W_B = m_B - \text{Buoyant Force on B} = m_B - \rho_{water} \cdot V_B \]

### Step 5: Substitute the values into the equation
Since the bodies are in equilibrium:
\[ W_A = W_B \]
Substituting the expressions for apparent weights:
\[ m_A - \rho_{water} \cdot V_A = m_B - \rho_{water} \cdot V_B \]
Substituting known values:
\[ 36 - 1 \cdot 4 = 48 - 1 \cdot V_B \]

### Step 6: Solve for the volume of body B
This simplifies to:
\[ 36 - 4 = 48 - V_B \]
\[ 32 = 48 - V_B \]
Rearranging gives:
\[ V_B = 48 - 32 = 16 \text{ cm}^3 \]

### Step 7: Calculate the density of body B
Now we can find the density of body B using the volume we just calculated:
\[ \rho_B = \frac{m_B}{V_B} = \frac{48 \text{ g}}{16 \text{ cm}^3} = 3 \text{ g/cm}^3 \]

### Final Answer
The density of body B is **3 g/cm³**.

---