The diameter of a wire as measured by screw gauge was found to be 2.620, 2.625, 2.630, 2.628 and 2.626 cm. Calculate
(a) mean value of diameter (b) absolute error in each measurement
(c) mean absolute error
(d) fractional error
(e) percentage error
(f) Express the result in terms of percentage error

Explanation

To solve the problem step by step, we will follow the instructions given in the question:

### Step 1: Calculate the Mean Value of Diameter
The mean value (μ) of the diameter can be calculated using the formula:
\[ \mu = \frac{1}{n} \sum_{i=1}^{n} x_i \]
Where \( n \) is the number of measurements and \( x_i \) are the individual measurements.

Given measurements:
- \( x_1 = 2.620 \, \text{cm} \)
- \( x_2 = 2.625 \, \text{cm} \)
- \( x_3 = 2.630 \, \text{cm} \)
- \( x_4 = 2.628 \, \text{cm} \)
- \( x_5 = 2.626 \, \text{cm} \)

Calculating the sum:
\[ \sum_{i=1}^{5} x_i = 2.620 + 2.625 + 2.630 + 2.628 + 2.626 = 13.129 \, \text{cm} \]

Now, calculating the mean:
\[ \mu = \frac{13.129}{5} = 2.6258 \, \text{cm} \]
Rounding to three decimal places:
\[ \mu \approx 2.626 \, \text{cm} \]

### Step 2: Calculate Absolute Error in Each Measurement
The absolute error for each measurement is calculated as:
\[ \text{Absolute Error} = |x_i - \mu| \]

Calculating for each measurement:
- For \( x_1 = 2.620 \):
\[ |2.620 - 2.626| = 0.006 \, \text{cm} \]
- For \( x_2 = 2.625 \):
\[ |2.625 - 2.626| = 0.001 \, \text{cm} \]
- For \( x_3 = 2.630 \):
\[ |2.630 - 2.626| = 0.004 \, \text{cm} \]
- For \( x_4 = 2.628 \):
\[ |2.628 - 2.626| = 0.002 \, \text{cm} \]
- For \( x_5 = 2.626 \):
\[ |2.626 - 2.626| = 0.000 \, \text{cm} \]

### Step 3: Calculate Mean Absolute Error
The mean absolute error (MAE) is calculated as:
\[ \text{MAE} = \frac{1}{n} \sum_{i=1}^{n} \text{Absolute Error}_i \]

Calculating the sum of absolute errors:
\[ 0.006 + 0.001 + 0.004 + 0.002 + 0.000 = 0.013 \, \text{cm} \]

Now, calculating MAE:
\[ \text{MAE} = \frac{0.013}{5} = 0.0026 \, \text{cm} \]

### Step 4: Calculate Fractional Error
The fractional error (FE) is given by:
\[ \text{FE} = \frac{\text{MAE}}{\mu} \]
Substituting the values:
\[ \text{FE} = \frac{0.0026}{2.626} \approx 0.000989 \]

### Step 5: Calculate Percentage Error
The percentage error (PE) is calculated as:
\[ \text{PE} = \text{FE} \times 100\% \]
Substituting the values:
\[ \text{PE} = 0.000989 \times 100 \approx 0.0989\% \]

### Step 6: Express the Result in Terms of Percentage Error
The final result can be expressed as:
\[ \text{Diameter} = 2.626 \pm 0.0989\% \]

### Summary of Results:
(a) Mean value of diameter: \( 2.626 \, \text{cm} \)
(b) Absolute errors: \( 0.006, 0.001, 0.004, 0.002, 0.000 \, \text{cm} \)
(c) Mean absolute error: \( 0.0026 \, \text{cm} \)
(d) Fractional error: \( 0.000989 \)
(e) Percentage error: \( 0.0989\% \)
(f) Result expressed in terms of percentage error: \( 2.626 \pm 0.0989\% \)