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Question: Consider the ratio \(r = \dfrac{{\left( {1 - a} \right)}}{{\left( {1 + a} \right)}}\)to be determine...

Consider the ratio r=(1−a)(1+a)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 isΔa(Δaa≪1)\Delta a\left( {\dfrac{{\Delta a}}{a} \ll 1} \right), then what is the error Δr\Delta rin determining r?
A. Δa(1+a)2\dfrac{{\Delta a}}{{{{\left( {1 + a} \right)}^2}}}
B. 2Δa(1+a)2\dfrac{{2\Delta a}}{{{{\left( {1 + a} \right)}^2}}}
C. 2Δa(1−a)2\dfrac{{2\Delta a}}{{{{\left( {1 - a} \right)}^2}}}
D. 2aΔa(1−a)2\dfrac{{2a\Delta a}}{{{{\left( {1 - a} \right)}^2}}}

Explanation

Solution

In this question for the ratio r=(1−a)(1+a)r = \dfrac{{\left( {1 - a} \right)}}{{\left( {1 + a} \right)}}, we fill find the Δrr\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=(1−a)(1+a)r = \dfrac{{\left( {1 - a} \right)}}{{\left( {1 + a} \right)}} to find the error Δr\Delta r in determining r.

Complete step by step answer:
r=(1−a)(1+a)r = \dfrac{{\left( {1 - a} \right)}}{{\left( {1 + a} \right)}}
Now we know that the ratio error forx=ABx = \dfrac{A}{B} is given by the formula Δxx=ΔAA+ΔBB\dfrac{{\Delta x}}{x} = \dfrac{{\Delta A}}{A} + \dfrac{{\Delta B}}{B}, hence we can write the ratio (i) as
Δrr=Δ(1−a)(1−a)+Δ(1+a)(1+a)\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
Δrr=Δa(1−a)+Δa(1+a)\dfrac{{\Delta r}}{r} = \dfrac{{\Delta a}}{{\left( {1 - a} \right)}} + \dfrac{{\Delta a}}{{\left( {1 + a} \right)}}
By further solving we get
Δrr=Δa(1+a)+Δa(1−a)(1−a)(1+a)   ⟹  Δrr=Δa(1+a+1−a)(1−a)(1+a)   ⟹  Δrr=2Δa(1−a)(1+a)  \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
Δr=2Δa(1−a)(1+a)r   ⟹  Δr=2Δa(1−a)(1+a)(1−a)(1+a)  \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
Δr=2Δa(1+a)(1+a) =2Δa(1+a)2  \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 Δr=2Δa(1+a)2\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.