Question

The period of oscillation of a simple pendulum in the experiment is recorded as $2.63s,$ $2.56s,$ $2.42s,$ $2.71s,$ and $2.80s$ respectively. The average absolute error is:
(A) $0.1s$
(B) $0.11s$
(C) $0.01s$
(D) $1.0s$

Answer 1

Hint: To solve this question, we need to find out the mean of the values given in the problem. From the mean, we can find out the individual absolute errors. Then we can determine the mean of the absolute errors.

Formula used: The formula used in solving this question is
 $\mu ={\displaystyle \frac{x_1+x_2+x_3+..........+x_n}{n}}$ , where $\mu$ is the average of the $n$ values $x_1$ , $x_2$ , ………, $x_n$ .

Complete step by step solution:
To find the absolute errors of each of the measurements, we need to find out the mean of all the measurements given in the problem.
The mean $\mu$ of the measurements is given by
 $\mu ={\displaystyle \frac{T_1+T_2+T_3+T_4+T_5}{5}}$
 $\mu ={\displaystyle \frac{2.63+2.56+2.42+2.71+2.80}{5}}$
On solving we get
 $\mu =2.624$
We need to round off this value to the same number of decimal places as the values are given in the problem
 $\therefore \mu =2.62$
Now, to find the absolute errors, we need to subtract the mean from each of the values given.
For the first measurement
 $\mathrm{\Delta }{T}_1=|2.63-2.62|=0.01s$
For the second measurement
 $\mathrm{\Delta }{T}_2=|2.56-2.62|=0.06s$
For the third measurement
 $\mathrm{\Delta }{T}_3=|2.42-2.62|=0.20s$
For the fourth measurement
 $\mathrm{\Delta }{T}_4=|2.71-2.62|=0.09s$
For the fifth measurement
 $\mathrm{\Delta }{T}_5=|2.80-2.62|=0.18s$
Now, the mean of the absolute or the average absolute error can be calculated as
 $\therefore \mathrm{\Delta }{T}_{avg}={\displaystyle \frac{\mathrm{\Delta }{T}_1+\mathrm{\Delta }{T}_2+\mathrm{\Delta }{T}_3+\mathrm{\Delta }{T}_4+\mathrm{\Delta }{T}_5}{5}}$
Putting the above values, we get
 $\mathrm{\Delta }{T}_{avg}={\displaystyle \frac{0.01+0.06+0.20+0.09+0.18}{5}}$
 $\mathrm{\Delta }{T}_{avg}=0.108s$
Rounding off to two decimal places, we get
 $\mathrm{\Delta }{T}_{avg}=0.11s$
So we get the average absolute error to be equal to $0.11s$
Hence, the correct answer is option B.

Note:
While calculating the absolute errors, we should not forget to take the modulus of the differences. As the name suggests, the absolute error is the magnitude of the deflection of a measurement from its mean value. If the modulus is not taken, then we may get the mean of the absolute errors as incorrect.