Question

answer ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀​

Answer 1

Question :-

Reading of length of a pole are 2.63m, 2.56m, 2.42m, 2.71m, and 2.80m. Calculate the absolute error relative error and percentage error . What do you think of the actual value of the length and it's limits ?

AnSwer :-

The mean value of length,

$\sf L = \dfrac{(2.63 + 2.56 + 2.42 + 2.71 + 2.80)}{5}$

$\sf = \dfrac{13.12}{5} m = 2.62m$

As the length are measured to a resolution of 0.01m, all lengths are given to the second place of decimal , it is proper to round off this mean length also to the second place of decimal.

In the first measurement

Error = $\sf \Delta x_{1}$= 2.63m - 2.62m = 0.01m

Absolute error = 0.01m

Relative error = 0.01/2.62 = 0.0038

%error = relative error × 100 = 0.38%

In the second measurement

Error =$\sf \Delta x_{2}$= 2.56m - 2.62m = -0.06,m

Absolute error = 0.06m

Relative error = 0.06/2.62 = 0.023

%error = relative error × 100 = 2.3%

In the third measurement

Error =$\sf \Delta x_{3}$= 2.42m - 2.62m = -0.2m

Absolute error = 0.2m

Relative error = 0.2/2.62 = 0.076

%error = relative error × 100 = 7.6%

In the fourth measurement

Error = $\sf \Delta x_{4}$= 2.71m - 2.62m = 0.09m

Absolute error = 0.09m

Relative error = 0.09/2.62 = 0.034

%error = relative error × 100 = 3.4%

In the fifth measurement

Error = $\sf \Delta x_{5}$ = 2.80m - 2.62m = 0.18m

Absolute error = 0.18m

Relative error = 0.18/2.62 = 0.068

%error = relative error × 100 = 6.8%

Now,

$\sf mean \: or \: final \: absolute \: error = \dfrac{ \sum \limits_{i = 1} ^{5} | \Delta</p><p>x_{1}| }{5}$

$\sf = \dfrac{(0.01 + 0.06 + 0.20 + 0.09 + 0.18)}{5}$

$\sf = \dfrac{0.54}{5} = 0.108m = 0.11m$

This means that the length is (2.62 ± 0.11m) ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀ .i.e., it lies between (2.62 - 0.11m) and (2.62 + 0.11m)

thus, the length is between 2.51 and 2.73m.