A cylindrical wire of mass `(0.4+-0.01)g` has lenght `(8+-0.04)cm` and radius `(6+-0.03)mm`. The maximum error in its density will be :

Options

A. `1%`
B. `4%`
C. `5%`
D. `3.5%`

Answer

To find the maximum error in the density of a cylindrical wire, we can use the formula for density and the concept of maximum relative error. The density \( \rho \) of a cylindrical wire is given by the formula:

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

where \( m \) is the mass and \( V \) is the volume. The volume \( V \) of a cylinder is given by:

\[ V = \pi r^2 h \]

where \( r \) is the radius and \( h \) is the height (length) of the cylinder.

### Step 1: Write the formula for density
The density of the cylindrical wire can be expressed as:

\[ \rho = \frac{m}{\pi r^2 h} \]

### Step 2: Identify the variables and their uncertainties
Given:
- Mass \( m = 0.4 \pm 0.01 \, \text{g} \)
- Length \( h = 8 \pm 0.04 \, \text{cm} = 0.08 \pm 0.0004 \, \text{m} \) (converting cm to m)
- Radius \( r = 6 \pm 0.03 \, \text{mm} = 0.006 \pm 0.00003 \, \text{m} \) (converting mm to m)

### Step 3: Calculate the volume and its uncertainty
The volume \( V \) can be calculated as:

\[ V = \pi r^2 h \]

To find the maximum error in \( V \), we use the formula for the propagation of uncertainty:

\[ \frac{\Delta V}{V} = 2 \frac{\Delta r}{r} + \frac{\Delta h}{h} \]

### Step 4: Calculate the relative errors
1. **Calculate \( \frac{\Delta r}{r} \)**:
\[ \frac{\Delta r}{r} = \frac{0.00003}{0.006} = 0.005 \]

2. **Calculate \( \frac{\Delta h}{h} \)**:
\[ \frac{\Delta h}{h} = \frac{0.0004}{0.08} = 0.005 \]

### Step 5: Substitute into the volume error formula
Now substituting into the volume error formula:

\[ \frac{\Delta V}{V} = 2(0.005) + 0.005 = 0.01 + 0.005 = 0.015 \]

### Step 6: Calculate the relative error in density
The relative error in density \( \frac{\Delta \rho}{\rho} \) can be expressed as:

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

1. **Calculate \( \frac{\Delta m}{m} \)**:
\[ \frac{\Delta m}{m} = \frac{0.01}{0.4} = 0.025 \]

2. **Substituting into the density error formula**:
\[ \frac{\Delta \rho}{\rho} = 0.025 + 0.015 = 0.04 \]

### Step 7: Convert to percentage
To find the maximum error in density as a percentage:

\[ \text{Maximum Error in Density} = 0.04 \times 100\% = 4\% \]

### Final Answer
The maximum error in the density of the cylindrical wire is \( \pm 4\% \).

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