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 use the principles of buoyancy and the relationship between mass, density, and volume.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the object, \( m = 26 \, \text{kg} \)
- Density of air, \( \rho = 1.3 \, \text{kg/m}^3 \)

2. **Understand the equilibrium condition:**
- When the object is floating in the air and is in equilibrium, the buoyant force acting on it is equal to its weight.
- The weight of the object can be expressed as:
\[ \text{Weight} = m \cdot g \]
- The buoyant force can be expressed as:
\[ \text{Buoyant Force} = \rho \cdot V \cdot g \]
- Since the object is in equilibrium:
\[ m \cdot g = \rho \cdot V \cdot g \]

3. **Cancel out \( g \) (acceleration due to gravity):**
- Since \( g \) appears on both sides of the equation, we can cancel it out:
\[ m = \rho \cdot V \]

4. **Rearrange the equation to solve for volume \( V \):**
- We can express the volume \( V \) as:
\[ V = \frac{m}{\rho} \]

5. **Substitute the known values into the equation:**
- Substitute \( m = 26 \, \text{kg} \) and \( \rho = 1.3 \, \text{kg/m}^3 \):
\[ V = \frac{26 \, \text{kg}}{1.3 \, \text{kg/m}^3} \]

6. **Calculate the volume:**
- Perform the division:
\[ V = 20 \, \text{m}^3 \]

7. **Conclusion:**
- The volume of the object is \( V = 20 \, \text{m}^3 \).

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

---