Question

Consider the ratio $r = \dfrac{{\left( {1 - a} \right)}}{{\left( {1 + a} \right)}}$to be determined by measuring a dimensionless quantity a. If the error in the measurement of a is$\Delta a\left( {\dfrac{{\Delta a}}{a} \ll 1} \right)$, then what is the error $\Delta r$in determining r?
A. $\dfrac{{\Delta a}}{{{{\left( {1 + a} \right)}^2}}}$
B. $\dfrac{{2\Delta a}}{{{{\left( {1 + a} \right)}^2}}}$
C. $\dfrac{{2\Delta a}}{{{{\left( {1 - a} \right)}^2}}}$
D. $\dfrac{{2a\Delta a}}{{{{\left( {1 - a} \right)}^2}}}$

Answer 1

Hint: In this question for the ratio $r = \dfrac{{\left( {1 - a} \right)}}{{\left( {1 + a} \right)}}$, we fill find the $\dfrac{{\Delta r}}{r}$ by using ratio error technique and then we will reduce the equation after which we will substitute the value of the $r = \dfrac{{\left( {1 - a} \right)}}{{\left( {1 + a} \right)}}$ to find the error $\Delta r$ in determining r.

Complete step by step answer:
$r = \dfrac{{\left( {1 - a} \right)}}{{\left( {1 + a} \right)}}$
Now we know that the ratio error for$x = \dfrac{A}{B}$ is given by the formula $\dfrac{{\Delta x}}{x} = \dfrac{{\Delta A}}{A} + \dfrac{{\Delta B}}{B}$, hence we can write the ratio (i) as
$\dfrac{{\Delta r}}{r} = \dfrac{{\Delta \left( {1 - a} \right)}}{{\left( {1 - a} \right)}} + \dfrac{{\Delta \left( {1 + a} \right)}}{{\left( {1 + a} \right)}}$
Hence this can be written as
$\dfrac{{\Delta r}}{r} = \dfrac{{\Delta a}}{{\left( {1 - a} \right)}} + \dfrac{{\Delta a}}{{\left( {1 + a} \right)}}$
By further solving we get
$
\dfrac{{\Delta r}}{r} = \dfrac{{\Delta a\left( {1 + a} \right) + \Delta a\left( {1 - a} \right)}}{{\left( {1 - a} \right)\left( {1 + a} \right)}} \\
\implies \dfrac{{\Delta r}}{r} = \dfrac{{\Delta a\left( {1 + a + 1 - a} \right)}}{{\left( {1 - a} \right)\left( {1 + a} \right)}} \\
\implies \dfrac{{\Delta r}}{r} = \dfrac{{2\Delta a}}{{\left( {1 - a} \right)\left( {1 + a} \right)}} \\
 $
Now substitute the value of r from the equation (i), we get
$
\Delta r = \dfrac{{2\Delta a}}{{\left( {1 - a} \right)\left( {1 + a} \right)}}r \\
\implies \Delta r = \dfrac{{2\Delta a}}{{\left( {1 - a} \right)\left( {1 + a} \right)}}\dfrac{{\left( {1 - a} \right)}}{{\left( {1 + a} \right)}} \\
 $
Hence we get
$
  \Delta r = \dfrac{{2\Delta a}}{{\left( {1 + a} \right)\left( {1 + a} \right)}} \\
   = \dfrac{{2\Delta a}}{{{{\left( {1 + a} \right)}^2}}} \\
 $
Therefore the error $\Delta r = \dfrac{{2\Delta a}}{{{{\left( {1 + a} \right)}^2}}}$.

So, the correct answer is “Option B”.

Note:
Students must be careful while expanding the given expression following the error definition. The terms with the raised power puts more error in the calculation in comparison to the addition or subtraction terms.