Question

A science student takes 100 observations in an experiment. Second time he took 500 observations in the same experiment. By doing so the possible error becomes:
A) 5 times
B) 1/5 times
C) Unchanged
D) None of these

Answer 1

As we know that the higher number of observations makes less error means if you take less observation than the error may be higher error defined as error= true value – measured value and when we take the average of error is called average or final absolute error. So you can observe which one should be correct here.

Complete step by step answer:
As we know error means the difference between true value and measured value of any physical quantity.
Absolute error =true value – measured value
Let’s assume we take n observation take in first observation quantity measured ${a_1}$ and true value is $a$ then error in first observation defined as
$\vartriangle {a_1} = a - {a_1}$Where $\vartriangle {a_1} \Rightarrow$ absolute error in first observation
In same manner we can find errors in all observation which is $\vartriangle {a_2},\vartriangle {a_3},\vartriangle {a_4}..........\vartriangle {a_n}$
So the average or final absolute error in 100 observation ${\left( {\overline {\vartriangle a} } \right)_n}$
${\left( {\overline {\vartriangle a} } \right)_n} = \dfrac{{\vartriangle {a_2} + \vartriangle {a_3} + \vartriangle {a_4} + .......... + \vartriangle {a_n}}}{n}$
From this we can say that average error is inversely proportional to number of observation
${\left( {\overline {\vartriangle a} } \right)_n}\alpha \dfrac{1}{n}$ ........ (1)
Using this equation error in 100 observation
Put $n = 100$ in equation (1)
${\left( {\overline {\vartriangle a} } \right)_{100}}\alpha \dfrac{1}{{100}}$............ (2)
For second time student take 500 observations
$n = 500$
${\left( {\overline {\vartriangle a} } \right)_{500}}\alpha \dfrac{1}{{500}}$......... (3)
Divide equation (3) by (2)
$\dfrac{{{{\left( {\overline {\vartriangle a} } \right)}_{500}}}}{{{{\left( {\overline {\vartriangle a} } \right)}_{100}}}} = \dfrac{{100}}{{500}}$
Solving it
$\dfrac{{{{\left( {\overline {\vartriangle a} } \right)}_{500}}}}{{{{\left( {\overline {\vartriangle a} } \right)}_{100}}}} = \dfrac{1}{5}$
So the possible error become $\dfrac{1}{5}times$ of error in 100 observation
${\left( {\overline {\vartriangle a} } \right)_{500}} = \dfrac{1}{5}{\left( {\overline {\vartriangle a} } \right)_{100}}$

So option B is correct

Note: By this method we can calculate the absolute error in any physical quantity. Hence it is clear from above if we increase the number of observations in any experiment we get more precious value of measured quantity.