An object of mass 26 kg floats in the air and it is in the equilibrium state.Air density is `1.3 kg m^(-3)`. The volume of the object is

Options

A. `10 m^3`
B. `20 m^3`
C. `13 m^3`
D. `26 m^3`

Answer

To find the volume of the object that is floating in the air and is in equilibrium, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Equilibrium Condition**:
Since the object is floating in the air, it is in equilibrium. This means that the weight of the object is equal to the weight of the air displaced by it.

2. **Weight of the Object**:
The weight (W) of the object can be calculated using the formula:
\[ W = m \cdot g \]
where \( m \) is the mass of the object (26 kg) and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
\[ W = 26 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 254.06 \, \text{N} \]

3. **Weight of the Displaced Air**:
The weight of the air displaced by the object can also be expressed in terms of the volume of the object (V) and the density of the air (\( \rho \)):
\[ \text{Weight of displaced air} = \rho \cdot V \cdot g \]
where \( \rho \) (density of air) is given as \( 1.3 \, \text{kg/m}^3 \).

4. **Setting up the Equation**:
Since the object is in equilibrium, we can set the weight of the object equal to the weight of the displaced air:
\[ m \cdot g = \rho \cdot V \cdot g \]
We can cancel \( g \) from both sides (assuming \( g \neq 0 \)):
\[ m = \rho \cdot V \]

5. **Solving for Volume (V)**:
Rearranging the equation to solve for volume:
\[ V = \frac{m}{\rho} \]
Substituting the known values:
\[ V = \frac{26 \, \text{kg}}{1.3 \, \text{kg/m}^3} \]
\[ V = 20 \, \text{m}^3 \]

### Final Answer:
The volume of the object is \( 20 \, \text{m}^3 \).

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