A certain body weighs `22.42 g` and has a measured volume of `4.7 c c`. The possible error in the measurement of mass and volume are `0.01g` and `0.1 c c`. Then, maximum error in the density will be

Options

A. `22%`
B. `2%`
C. `0.2%`
D. `0.02%`

Explanation

To find the maximum error in the density of a body given its mass and volume, we can use the formula for relative error in density. The density \( D \) is defined as:

\[ D = \frac{m}{V} \]

where \( m \) is mass and \( V \) is volume. The relative error in density can be expressed as:

\[ \frac{\Delta D}{D} = \frac{\Delta m}{m} + \frac{\Delta V}{V} \]

where \( \Delta D \) is the error in density, \( \Delta m \) is the error in mass, and \( \Delta V \) is the error in volume.

### Step 1: Identify the given values
- Mass \( m = 22.42 \, g \)
- Volume \( V = 4.7 \, cc \)
- Error in mass \( \Delta m = 0.01 \, g \)
- Error in volume \( \Delta V = 0.1 \, cc \)

### Step 2: Calculate the relative error in mass
\[ \frac{\Delta m}{m} = \frac{0.01 \, g}{22.42 \, g} \]
Calculating this gives:
\[ \frac{\Delta m}{m} \approx 0.000445 \]

### Step 3: Calculate the relative error in volume
\[ \frac{\Delta V}{V} = \frac{0.1 \, cc}{4.7 \, cc} \]
Calculating this gives:
\[ \frac{\Delta V}{V} \approx 0.021277 \]

### Step 4: Combine the relative errors to find the relative error in density
\[ \frac{\Delta D}{D} = \frac{\Delta m}{m} + \frac{\Delta V}{V} \]
Substituting the values we calculated:
\[ \frac{\Delta D}{D} \approx 0.000445 + 0.021277 \approx 0.021722 \]

### Step 5: Convert the relative error to percentage
To find the percentage error, multiply by 100:
\[ \Delta D \text{ (in percentage)} \approx 0.021722 \times 100 \approx 2.1722\% \]

### Conclusion
The maximum error in the density is approximately \( 2.17\% \).