A wire has a mass `(0.3pm0.003)` g, radius `(0.5pm0.005)` mm and length `(6pm0.06)`cm. The maximum percentage error in the measurement of its density is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the maximum percentage error in the measurement of the density of the wire, we can follow these steps:

### Step 1: Understand the formula for density
The density (ρ) of the wire is given by the formula:
\[ \rho = \frac{m}{V} \]
where \( m \) is the mass and \( V \) is the volume of the wire. The volume \( V \) of a cylinder (which we can assume the wire to be) is given by:
\[ V = \pi r^2 L \]
where \( r \) is the radius and \( L \) is the length.

### Step 2: Write the formula for density in terms of mass, radius, and length
Substituting the volume formula into the density formula, we have:
\[ \rho = \frac{m}{\pi r^2 L} \]

### Step 3: Determine the percentage error in density
The formula for the maximum percentage error in density can be expressed as:
\[ \text{Percentage Error in } \rho = \frac{\Delta \rho}{\rho} \times 100 \]
Where \( \Delta \rho \) is the total error in density.

### Step 4: Calculate the individual errors
The percentage error in density can be calculated using the errors in mass, radius, and length:
\[ \text{Percentage Error in } \rho = \frac{\Delta m}{m} + 2 \frac{\Delta r}{r} + \frac{\Delta L}{L} \]
Here:
- \( \Delta m \) = error in mass
- \( \Delta r \) = error in radius
- \( \Delta L \) = error in length

### Step 5: Substitute the values
Given:
- Mass \( m = 0.3 \) g with an error \( \Delta m = 0.003 \) g
- Radius \( r = 0.5 \) mm with an error \( \Delta r = 0.005 \) mm
- Length \( L = 6 \) cm with an error \( \Delta L = 0.06 \) cm

Now we convert the radius from mm to cm for consistency:
\[ r = 0.5 \text{ mm} = 0.05 \text{ cm} \]

Now, substituting the values into the formula:
\[ \text{Percentage Error in } \rho = \frac{0.003}{0.3} + 2 \frac{0.005}{0.05} + \frac{0.06}{6} \]

### Step 6: Calculate each term
1. For mass:
\[ \frac{0.003}{0.3} = 0.01 \]
2. For radius:
\[ 2 \frac{0.005}{0.05} = 2 \times 0.1 = 0.2 \]
3. For length:
\[ \frac{0.06}{6} = 0.01 \]

### Step 7: Sum the percentage errors
Now, summing these values:
\[ \text{Total Percentage Error} = 0.01 + 0.2 + 0.01 = 0.22 \]

### Step 8: Convert to percentage
To express this as a percentage:
\[ \text{Total Percentage Error} = 0.22 \times 100 = 22\% \]

### Final Answer
The maximum percentage error in the measurement of the density of the wire is **22%**.

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