The resistivity of a wire is found…

The resistivity of a wire is found from ρ=(π d^2V)/(4IL). The measurements are: potential difference V=5.00 V with least count 0.05 V, current I=0.50…

Numerical answer

The resistivity of a wire is found from

\rho=\dfrac{\pi d^{2}V}{4IL}

. The measurements are: potential difference

V=5.00\ \mathrm{V}

with least count

0.05\ \mathrm{V}

, current

I=0.50\ \mathrm{A}

with least count

0.005\ \mathrm{A}

, diameter

d=1.00\ \mathrm{mm}

by a screw gauge of least count

0.01\ \mathrm{mm}

, and length

L=50.0\ \mathrm{cm}

by a metre scale with an uncertainty of

0.5\ \mathrm{cm}

. Taking the error in each quantity equal to its least count (or stated uncertainty), the maximum percentage error in

\rho

is what integer?

Show answer and solution

Answer: 5

For a product of powers, the maximum fractional error is the sum of each exponent times the fractional error of that quantity. Here

\rho\propto d^{2}V\,I^{-1}L^{-1}

, so

\frac{\Delta\rho}{\rho}=2\frac{\Delta d}{d}+\frac{\Delta V}{V}+\frac{\Delta I}{I}+\frac{\Delta L}{L}

. The individual terms are

\frac{\Delta V}{V}=\frac{0.05}{5.00}=1\%

\frac{\Delta I}{I}=\frac{0.005}{0.50}=1\%

\frac{\Delta d}{d}=\frac{0.01}{1.00}=1\%

2\times1\%=2\%

, and

\frac{\Delta L}{L}=\frac{0.5}{50.0}=1\%

. The errors in the numerator and denominator quantities all add in the worst case. Therefore

\frac{\Delta\rho}{\rho}\times100=1+1+2+1=5\%

. The constants

\pi

and

4

are exact and contribute nothing.

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Original question written in the JEE Advanced 2021 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake