Calculate the density of a body of mass `(10+-0.6)g` and volume `(4.4+-0.7) cm^(3)`, to the correct significant figures?

Explanation

To calculate the density of a body given its mass and volume with uncertainties, we will follow these steps:

### Step 1: Write down the formula for density
The formula for density (ρ) is given by:
\[ \rho = \frac{m}{V} \]
where \( m \) is the mass and \( V \) is the volume.

### Step 2: Substitute the values
Given:
- Mass \( m = 10 \, \text{g} \)
- Volume \( V = 4.4 \, \text{cm}^3 \)

Substituting these values into the formula:
\[ \rho = \frac{10 \, \text{g}}{4.4 \, \text{cm}^3} \]

### Step 3: Perform the calculation
Calculating the density:
\[ \rho = \frac{10}{4.4} \approx 2.27 \, \text{g/cm}^3 \]

### Step 4: Calculate the relative uncertainty
To find the uncertainty in density, we use the formula for the propagation of uncertainty:
\[ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta V}{V} \]
where:
- \( \Delta m = 0.6 \, \text{g} \) (uncertainty in mass)
- \( \Delta V = 0.7 \, \text{cm}^3 \) (uncertainty in volume)

Substituting the values:
\[ \frac{\Delta \rho}{\rho} = \frac{0.6}{10} + \frac{0.7}{4.4} \]

### Step 5: Calculate each term
Calculating each term:
1. \( \frac{0.6}{10} = 0.06 \)
2. \( \frac{0.7}{4.4} \approx 0.1591 \)

Adding these:
\[ \frac{\Delta \rho}{\rho} = 0.06 + 0.1591 \approx 0.2191 \]

### Step 6: Calculate the absolute uncertainty in density
Now, we can find the absolute uncertainty \( \Delta \rho \):
\[ \Delta \rho = \rho \times \frac{\Delta \rho}{\rho} = 2.27 \times 0.2191 \approx 0.497 \]

### Step 7: Round the results
Now we round the density and the uncertainty to the correct significant figures:
- The density \( \rho \approx 2.27 \) rounds to \( 2.3 \) (2 significant figures).
- The uncertainty \( \Delta \rho \approx 0.497 \) rounds to \( 0.5 \) (1 decimal place).

### Final Answer
Thus, the density of the body is:
\[ \rho = 2.3 \pm 0.5 \, \text{g/cm}^3 \]

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