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Exam
JEE Main
Subject
Physics
Chapter
Mathematics in Physics

JEE Main · Physics · Mathematics in Physics

A student calculates the density of a cylindrical wire by measuring...

A student calculates the density of a cylindrical wire by measuring its mass, radius, and length. If the student commits a +1.5%+1.5 \% error in the measurement of mass, a −1.0%-1.0 \% error in the measurement of radius, and a +0.5%+0.5 \% error in the measurement of length, the actual percentage error in the calculated value of density is :

Options

  1. A4.0%4.0 \%
  2. B2.0%2.0 \%
  3. C3.0%3.0 \%
  4. D1.0%1.0 \%

Correct answer

C. 3.0%3.0 \%

Step-by-step solution

The density of a cylinder is given by the formula ρ=Mπr2L\rho = \dfrac{M}{\pi r^2 L}

Taking the natural logarithm and differentiating, we get the algebraic error relation:
Δρρ=ΔMM−2Δrr−ΔLL\dfrac{\Delta \rho}{\rho} = \dfrac{\Delta M}{M} - 2\dfrac{\Delta r}{r} - \dfrac{\Delta L}{L}

The given percentage errors are:
ΔMM%=+1.5%\dfrac{\Delta M}{M}\% = +1.5 \%
Δrr%=−1.0%\dfrac{\Delta r}{r}\% = -1.0 \%
ΔLL%=+0.5%\dfrac{\Delta L}{L}\% = +0.5 \%

Substituting these values into the error equation:
Δρρ%=(+1.5%)−2(−1.0%)−(+0.5%)\dfrac{\Delta \rho}{\rho}\% = (+1.5 \%) - 2(-1.0 \%) - (+0.5 \%)
Δρρ%=1.5%+2.0%−0.5%=3.0%\dfrac{\Delta \rho}{\rho}\% = 1.5 \% + 2.0 \% - 0.5 \% = 3.0 \%

Answer: 3.0%3.0 \%
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