The mass of a body is `20.000 g` and its volume is `10.00 cm^(3)`. If the measured values are expressed to the correct significant figures , the maximum error in the value of density is

Options

A. `0.001 g cm^(-3)`
B. `0.010 g cm^(-3)`
C. `0.100 g cm^(-3)`
D. None of these

Explanation

To find the maximum error in the value of density based on the given mass and volume, we can follow these steps:

### Step 1: Understand the formula for density
Density (ρ) is defined as the mass (m) of an object divided by its volume (V):
\[ \rho = \frac{m}{V} \]

### Step 2: Identify the given values
From the question, we have:
- Mass, \( m = 20.000 \, \text{g} \)
- Volume, \( V = 10.00 \, \text{cm}^3 \)

### Step 3: Calculate the density
Using the formula for density:
\[ \rho = \frac{20.000 \, \text{g}}{10.00 \, \text{cm}^3} = 2.000 \, \text{g/cm}^3 \]

### Step 4: Determine significant figures
- The mass (20.000 g) has 5 significant figures.
- The volume (10.00 cm³) has 4 significant figures.

The density value should be expressed to the least number of significant figures from the measurements used, which is 4 significant figures (from the volume).

### Step 5: Round the density to the correct significant figures
The calculated density is \( 2.000 \, \text{g/cm}^3 \), which can be rounded to 4 significant figures as:
\[ \rho = 2.000 \, \text{g/cm}^3 \]

### Step 6: Calculate the maximum error
The maximum error in the density can be determined from the least count of the measurements used:
- The mass has an uncertainty of \( 0.001 \, \text{g} \) (since it has 5 decimal places).
- The volume has an uncertainty of \( 0.01 \, \text{cm}^3 \) (since it has 2 decimal places).

To find the maximum error in density, we can use the formula for propagation of errors:
\[ \text{Error in density} = \rho \times \left( \frac{\Delta m}{m} + \frac{\Delta V}{V} \right) \]
Where:
- \( \Delta m \) is the uncertainty in mass
- \( \Delta V \) is the uncertainty in volume

Calculating the relative errors:
\[ \frac{\Delta m}{m} = \frac{0.001 \, \text{g}}{20.000 \, \text{g}} = 0.00005 \]
\[ \frac{\Delta V}{V} = \frac{0.01 \, \text{cm}^3}{10.00 \, \text{cm}^3} = 0.001 \]

Now substituting back into the error formula:
\[ \text{Error in density} = 2.000 \, \text{g/cm}^3 \times (0.00005 + 0.001) = 2.000 \, \text{g/cm}^3 \times 0.00105 = 0.0021 \, \text{g/cm}^3 \]

### Step 7: Round the error to the correct significant figures
The maximum error should be rounded to 2 decimal places (since the density is expressed to 4 significant figures):
\[ \text{Maximum error} \approx 0.01 \, \text{g/cm}^3 \]

### Final Answer
The maximum error in the value of density is:
\[ \text{Maximum Error} = 0.01 \, \text{g/cm}^3 \]

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