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Units and Measurements question

2018 · Shift 1 · Q52
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Units and Measurements question

2018 · Shift 1 · Q52

JEE AdvancedPhysicsUnits and MeasurementsMCQ+3 / −1
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.z=x/y. If the errors in x,yx,yx,y and zzz are Δx,Δy\Delta x,\Delta yΔx,Δy and Δz,\Delta z,Δz, respectively, then z±Δz=x±Δxy±Δy=xy(1±Δxx)(1±Δyy)−1.z \pm \Delta z = {{x \pm \Delta x} \over {y \pm \Delta y}} = {x \over y}\left( {1 \pm {{\Delta x} \over x}} \right){\left( {1 \pm {{\Delta y} \over y}} \right)^{ - 1}}.z±Δz=y±Δyx±Δx​=yx​(1±xΔx​)(1±yΔy​)−1. The series expansion for (1±Δyy)−1,{\left( {1 \pm {{\Delta y} \over y}} \right)^{ - 1}},(1±yΔy​)−1, to first power in Δy/y.\Delta y/y.Δy/y. is 1±(Δy/y).1 \pm \left( {\Delta y/y} \right).1±(Δy/y). The relative errors in independent variables are always added. So the error in zzz will be Δz=z(Δxx+Δyy).\Delta z = z\left( {{{\Delta x} \over x} + {{\Delta y} \over y}} \right).Δz=z(xΔx​+yΔy​). The above derivation makes the assumption that Δx/x<<1,Δy/y<<1.\Delta x/x \lt \lt 1,\Delta y/y \lt \lt 1.Δx/x<<1,Δy/y<<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}}r=1+a(1−a)​ to be determined by measuring a dimensionless quantity a.a.a. If the error in the measurement of aaa is Δa(Δa/a<<1.\Delta a\left( {\Delta a/a \lt \lt 1.} \right.Δa(Δa/a<<1. then what is the error Δr\Delta rΔr in determining rrr?
  1. A
    Δa(1+a)2{{\Delta a} \over {{{\left( {1 + a} \right)}^2}}}(1+a)2Δa​
  2. B
    2Δa(1+a)2{{2\Delta a} \over {{{\left( {1 + a} \right)}^2}}}(1+a)22Δa​
  3. C
    2Δa(1−a2){{2\Delta a} \over {\left( {1 - {a^2}} \right)}}(1−a2)2Δa​
  4. D
    2aΔa(1−a2){{2a\Delta a} \over {\left( {1 - {a^2}} \right)}}(1−a2)2aΔa​
View written solutionFree

Correct answer: B

  1. Given relation

    r=1−a1+ar=\frac{1-a}{1+a}r=1+a1−a​

    We need the error in rrr when the measured quantity aaa has an error Δa\Delta aΔa.

  2. Use first-order error propagation

    For a quantity depending on a single variable,

    Δr≈∣drda∣Δa\Delta r \approx \left|\frac{dr}{da}\right|\Delta aΔr≈​dadr​​Δa

    since higher powers of Δa\Delta aΔa are neglected.

  3. Differentiate rrr with respect to aaa

    Using the quotient rule:

    r=1−a1+ar=\frac{1-a}{1+a}r=1+a1−a​

    drda=(1+a)(−1)−(1−a)(1)(1+a)2\frac{dr}{da}=\frac{(1+a)(-1)-(1-a)(1)}{(1+a)^2}dadr​=(1+a)2(1+a)(−1)−(1−a)(1)​

    Simplify the numerator:

    −(1+a)−(1−a)=−1−a−1+a=−2-(1+a)-(1-a)=-1-a-1+a=-2−(1+a)−(1−a)=−1−a−1+a=−2

    Hence,

    drda=−2(1+a)2\frac{dr}{da}=\frac{-2}{(1+a)^2}dadr​=(1+a)2−2​

  4. Magnitude of error

    Therefore,

    Δr=∣−2(1+a)2∣Δa\Delta r=\left|\frac{-2}{(1+a)^2}\right|\Delta aΔr=​(1+a)2−2​​Δa

    Δr=2Δa(1+a)2\Delta r=\frac{2\Delta a}{(1+a)^2}Δr=(1+a)22Δa​

  5. Match with options

    This corresponds to:

    Option B

    2Δa(1+a)2\boxed{\frac{2\Delta a}{(1+a)^2}}(1+a)22Δa​​

  6. Comparison with stored answer

    Stored correct answer: B

    Our derived answer: B

    So they agree.

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