Question

Find out the fractional error for the following measurements of diameter (in $\text{cm}$) of a wire : $2.620,2.625,2.628,2.630$ and $2.626$.

• A. $1$
• B. $0.001$
• C. $0.1$
• D. $0.01$

Answer 1

The correct option is B $0.001$

Arithmetic mean,
$a_{mean}=\frac{a_1+a_2+a_3+.....+a_n}{n}$
$\Rightarrow a_{mean}=\frac{2.620+2.625+2.628+2.630+2.626}{5}$
$\Rightarrow a_{mean}=2.6258\text{cm}$
On rounding off to correct decimal places, we get, $a_{mean}=2.626\text{cm}$
Now, absolute error,
$|\Delta a_1|=|a_1-a_{mean}|=|2.620-2.626|=0.006\text{cm}$
Similarly,
$|\Delta a_2|=0.001\text{cm}$
$|\Delta a_3|=0.002\text{cm}$
$|\Delta a_4|=0.004\text{cm}$
$|\Delta a_5|=0.000\text{cm}$
So, mean absolute error,
$\Delta a_{mean}=\frac{|\Delta a_1|+|\Delta a_2|+|\Delta a_3|+.....+|\Delta a_n|}{n}$
$\Rightarrow \Delta a_{mean}=\frac{0.006+0.001+0.002+0.004+0.000}{5}$
$\Rightarrow \Delta a_{mean}=0.0026\text{cm}$
On rounding off to correct decimal places, we get, $\Delta a_{mean}=0.003\text{cm}$
Hence, fractional error $=\frac{\Delta a_{mean}}{a_{mean}}=\frac{0.003}{2.626}=0.00114$
On rounding off to correct decimal places, we get, $0.001$.