The length of a uniform rod is 100.0 cm and radius is 1.00 cm if length is measured with a meter rod having least count 1mm and radus is measured with vernier callipers having least count 0.1 mm the percentage error in calculated volume of cylinder is

Options

A. `2.1%`
B. 0.03
C. `2.01%`
D. `3.2%`

Answer

To solve the problem of finding the percentage error in the calculated volume of a cylinder given the measurements of length and radius, we can follow these steps:

### Step 1: Identify the given values
- Length of the rod (L) = 100.0 cm
- Radius of the rod (R) = 1.00 cm
- Least count of the meter rod (for length) = 1 mm = 0.1 cm
- Least count of the vernier caliper (for radius) = 0.1 mm = 0.01 cm

### Step 2: Determine the absolute errors
- The absolute error in length (ΔL) = ±0.1 cm (since the least count is 0.1 cm)
- The absolute error in radius (ΔR) = ±0.01 cm (since the least count is 0.01 cm)

### Step 3: Write the formula for the volume of a cylinder
The volume (V) of a cylinder is given by the formula:
\[ V = \pi R^2 L \]

### Step 4: Calculate the percentage error in volume
The formula for percentage error in volume due to errors in radius and length is given by:
\[ \frac{\Delta V}{V} \times 100 = 2 \frac{\Delta R}{R} \times 100 + \frac{\Delta L}{L} \times 100 \]

### Step 5: Substitute the values into the formula
1. Calculate the percentage error due to radius:
\[ 2 \frac{\Delta R}{R} \times 100 = 2 \left(\frac{0.01 \text{ cm}}{1.00 \text{ cm}}\right) \times 100 = 2 \times 0.01 \times 100 = 2 \% \]

2. Calculate the percentage error due to length:
\[ \frac{\Delta L}{L} \times 100 = \left(\frac{0.1 \text{ cm}}{100.0 \text{ cm}}\right) \times 100 = 0.001 \times 100 = 0.1 \% \]

### Step 6: Add the percentage errors
Now, we can add the two percentage errors together:
\[ \text{Total Percentage Error} = 2\% + 0.1\% = 2.1\% \]

### Final Answer
The percentage error in the calculated volume of the cylinder is **2.1%**.

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