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QJEEAdv18Ph1956: JEE Advanced 2018 Physics - Class 11 JEEAdv18Ph1956 Solution: If the measurement errors in all the independent quantities are …

If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation z=x/y.z = x/y. If the errors in x,yx,y and zz are Δx,Δy\Delta x,\Delta y and Δz,\Delta z, respectively, then
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The series expansion for (1±Δyy)−1,{\left( {1 \pm {{\Delta y} \over y}} \right)^{ - 1}}, to first power in Δy/y.\Delta y/y. is 1±(Δy/y).1 \pm \left( {\Delta y/y} \right). The relative errors in independent variables are always added. So the error in zz will be
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The above derivation makes the assumption that \Delta x/x &lt; &lt; 1, \Delta y/y &lt; &lt; 1. Therefore, the higher powers of these quantities are neglected.

Consider the ratio r=(1−a)1+ar = {{\left( {1 - a} \right)} \over {1 + a}} to be determined by measuring a dimensionless quantity a.a. If the error in the measurement of aa is \Delta a\left( {\Delta a/a &lt; &lt; 1.} \right. then what is the error Δr\Delta r in determining rr ?

Multiple Choice Options

  • A Δa(1+a)2{{\Delta a} \over {{{\left( {1 + a} \right)}^2}}}
  • B 2Δa(1+a)2{{2\Delta a} \over {{{\left( {1 + a} \right)}^2}}}
  • C 2Δa(1−a2){{2\Delta a} \over {\left( {1 - {a^2}} \right)}}
  • D 2aΔa(1−a2){{2a\Delta a} \over {\left( {1 - {a^2}} \right)}}

Step-by-Step Verified Solution

Given, ratio r=1−a1+ar = {{1 - a} \over {1 + a}} ..... (1)

Given, error in measurement of a is Δ\Delta a ( Δ\Delta a/a << 1)

Taking natural log of Eq. (1), we get

ln⁡r=ln⁡(1−a1+a)\ln r = \ln \left( {{{1 - a} \over {1 + a}}} \right)

⇒ln⁡r=ln⁡(1−a)−ln⁡(1+a)\Rightarrow \ln r = \ln (1 - a) - \ln (1 + a)

Δrr=Δa1−a+Δa1+a=(1+a)Δa+(1−a)Δa(1−a)(1+a){{\Delta r} \over r} = {{\Delta a} \over {1 - a}} + {{\Delta a} \over {1 + a}} = {{(1 + a)\Delta a + (1 - a)\Delta a} \over {(1 - a)(1 + a)}}

⇒Δrr=2Δa(1−a)(1+a)\Rightarrow {{\Delta r} \over r} = {{2\Delta a} \over {(1 - a)(1 + a)}}

⇒Δr=r . 2Δa(1−a)(1+a)\Rightarrow \Delta r = r\,.\,{{2\Delta a} \over {(1 - a)(1 + a)}}

Substituting value of r, we get

Δr=1−a1+a2Δa(1−a)(1+a)=2Δa(1+a)2\Delta r = {{1 - a} \over {1 + a}}{{2\Delta a} \over {(1 - a)(1 + a)}} = {{2\Delta a} \over {{{(1 + a)}^2}}}

Therefore, error Δ\Delta r in determining r is 2Δa(1+a)2{{2\Delta a} \over {{{(1 + a)}^2}}}

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