The period of oscillation of a simple pendulum in the experiment is recorded as 2.63 s, 2.56 s, 2.42 s, 2.71 s and 2.80 s respectively. The average absolute error is

The correct answer/option is 0.11 s.

To find the average absolute error of the measurements, we follow these steps:

Step 1: Calculate the mean value of the period of oscillation
The recorded values for the period of oscillation ($T_i$) are:
$T_1=2.63\text{s}$
$T_2=2.56\text{s}$
$T_3=2.42\text{s}$
$T_4=2.71\text{s}$
$T_5=2.80\text{s}$

The mean period ($T_{\text{mean}}$) is the sum of all measurements divided by the total number of measurements (5):

$T_{\text{mean}}=\frac{2.63+2.56+2.42+2.71+2.80}{5}$

$T_{\text{mean}}=\frac{13.12}{5}=2.624\text{s}$

Rounding to two decimal places (since the measurements are to two decimal places), we get:
$T_{\text{mean}}\approx 2.62\text{s}$

Step 2: Calculate the absolute errors for each measurement
The absolute error ($\Delta T_i$) for each measurement is calculated as the absolute difference between the measured value and the mean value:
$\left|\Delta T_i\right|=|T_i-T_{\text{mean}}|$

For each observation:
$\left|\Delta T_1\right|=|2.63-2.62|=0.01\text{s}$
$\left|\Delta T_2\right|=|2.56-2.62|=0.06\text{s}$
$\left|\Delta T_3\right|=|2.42-2.62|=0.20\text{s}$
$\left|\Delta T_4\right|=|2.71-2.62|=0.09\text{s}$
$\left|\Delta T_5\right|=|2.80-2.62|=0.18\text{s}$

Step 3: Calculate the average absolute error
The average (mean) absolute error ($\Delta T_{\text{mean}}$) is the sum of these absolute errors divided by the total number of measurements:

$\Delta T_{\text{mean}}=\frac{0.01+0.06+0.20+0.09+0.18}{5}$

$\Delta T_{\text{mean}}=\frac{0.54}{5}=0.108\text{s}$

Rounding to two decimal places gives:
$\Delta T_{\text{mean}}\approx 0.11\text{s}$

Thus, the average absolute error is 0.11 s.