The mass and density of a solid sphere are measured to be `(12.4 +-0.1)kg and (4.6 +- 0.2) kg//m^3`. Calculate the volume of the sphere with error limits .

Explanation

To calculate the volume of a solid sphere given its mass and density along with their respective uncertainties, we can follow these steps:

### Step 1: Write down the given values
The mass \( m \) and density \( \rho \) of the sphere are given as:
- Mass, \( m = 12.4 \pm 0.1 \, \text{kg} \)
- Density, \( \rho = 4.6 \pm 0.2 \, \text{kg/m}^3 \)

### Step 2: Use the formula for volume
The volume \( V \) of the sphere can be calculated using the formula:
\[ V = \frac{m}{\rho} \]
Substituting the values of mass and density, we have:
\[ V = \frac{12.4}{4.6} \]

### Step 3: Calculate the volume
Now, performing the division:
\[ V = 2.69565217391 \, \text{m}^3 \quad (\text{approximately } 2.6956 \, \text{m}^3) \]

### Step 4: Calculate the uncertainty in volume
To find the uncertainty in volume \( \Delta V \), we can use the formula for the propagation of uncertainty for division:
\[ \frac{\Delta V}{V} = \frac{\Delta m}{m} + \frac{\Delta \rho}{\rho} \]
Where:
- \( \Delta m = 0.1 \, \text{kg} \) (uncertainty in mass)
- \( \Delta \rho = 0.2 \, \text{kg/m}^3 \) (uncertainty in density)

Calculating the relative uncertainties:
\[ \frac{\Delta m}{m} = \frac{0.1}{12.4} \approx 0.0080645 \]
\[ \frac{\Delta \rho}{\rho} = \frac{0.2}{4.6} \approx 0.0434783 \]

Now, adding these relative uncertainties:
\[ \frac{\Delta V}{V} \approx 0.0080645 + 0.0434783 \approx 0.0515428 \]

### Step 5: Calculate the absolute uncertainty in volume
Now, we can find \( \Delta V \):
\[ \Delta V = V \times \frac{\Delta V}{V} \approx 2.6956 \times 0.0515428 \approx 0.1390 \, \text{m}^3 \]

### Step 6: Write the final result
Thus, the volume of the sphere with its uncertainty is:
\[ V = 2.6956 \pm 0.1390 \, \text{m}^3 \]