Two bodies having volumes V and 2V are suspended from the two arms of a common balance and they are found to balance each other. If larger body is immersed in oil (density `d_(1)=0.9gm//cm^(3)`) and the smaller body is immersed in an unknown liquid, then the balance remain in equilibrium. The density of unknown liquid is given by :

Options

A. `2.4gm//cm^(3)`
B. `1.8gm//cm^(3)`
C. `0.45gm//cm^(3)`
D. `2.7gm//cm^(3)`

Explanation

To solve the problem, we need to analyze the forces acting on the two bodies when they are immersed in their respective liquids.

### Step-by-Step Solution:

1. **Understanding the Balance Condition**:
Since the two bodies balance each other when suspended, the weight of the larger body (when immersed in oil) must equal the buoyant force acting on it, and the weight of the smaller body (when immersed in the unknown liquid) must equal the buoyant force acting on it.

2. **Weight of the Bodies**:
- Let the density of the smaller body be \( \rho_s \) and the larger body be \( \rho_l \).
- The weight of the larger body (volume \( 2V \)) is \( W_l = \rho_l \cdot 2V \cdot g \).
- The weight of the smaller body (volume \( V \)) is \( W_s = \rho_s \cdot V \cdot g \).

3. **Buoyant Force on the Larger Body**:
- The buoyant force \( F_b \) on the larger body when immersed in oil (density \( d_1 = 0.9 \, \text{g/cm}^3 \)) is given by:
\[ F_{b,l} = V_l \cdot d_1 \cdot g = 2V \cdot 0.9 \cdot g = 1.8Vg \]

4. **Buoyant Force on the Smaller Body**:
- The buoyant force on the smaller body when immersed in the unknown liquid (density \( d_2 \)) is:
\[ F_{b,s} = V_s \cdot d_2 \cdot g = V \cdot d_2 \cdot g \]

5. **Setting Up the Equilibrium Condition**:
- For the balance to remain in equilibrium, the weight of the larger body must equal the buoyant force acting on it, and the weight of the smaller body must equal the buoyant force acting on it:
\[ W_l = F_{b,l} \quad \text{and} \quad W_s = F_{b,s} \]
- This gives us two equations:
\[ \rho_l \cdot 2V \cdot g = 1.8Vg \]
\[ \rho_s \cdot V \cdot g = V \cdot d_2 \cdot g \]

6. **Simplifying the Equations**:
- From the first equation, we can simplify:
\[ \rho_l \cdot 2 = 1.8 \implies \rho_l = 0.9 \, \text{g/cm}^3 \]
- From the second equation, we can simplify:
\[ \rho_s = d_2 \]

7. **Finding the Density of the Unknown Liquid**:
- Since the larger body is made of the same material as the smaller body, we can equate the densities:
\[ d_2 = \frac{\rho_l \cdot 2}{1} = 0.9 \cdot 2 = 1.8 \, \text{g/cm}^3 \]

### Final Answer:
The density of the unknown liquid is \( 1.8 \, \text{g/cm}^3 \).

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