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JEE Advanced 2020 Paper 1 · Physics · Units and Measurements · easy

In an experiment to find the Young's modulus of a wire using Y=(4FL)/(π d^2 l), the following readings are taken: load F=10.0 N measured with least…

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In an experiment to find the Young's modulus of a wire using Y=4FLπd2lY=\dfrac{4FL}{\pi d^2 l}, the following readings are taken: load F=10.0 NF=10.0\ \mathrm{N} measured with least count 0.1 N0.1\ \mathrm{N}; original length L=1.000 mL=1.000\ \mathrm{m} with least count 1 mm1\ \mathrm{mm}; wire diameter d=0.50 mmd=0.50\ \mathrm{mm} with a screw gauge of least count 0.01 mm0.01\ \mathrm{mm}; and extension l=0.40 mml=0.40\ \mathrm{mm} measured with least count 0.01 mm0.01\ \mathrm{mm}. Taking the maximum possible error in each reading equal to its least count, the maximum percentage error in YY is (give the answer correct to two decimal places)
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Answer: 7.60

For a product or quotient of powers, the maximum fractional errors add with weights equal to the powers: ΔYY=ΔFF+ΔLL+2Δdd+Δll\frac{\Delta Y}{Y}=\frac{\Delta F}{F}+\frac{\Delta L}{L}+2\frac{\Delta d}{d}+\frac{\Delta l}{l}. Substituting: 0.110.0=1%\frac{0.1}{10.0}=1\%; 1 mm1000 mm=0.1%\frac{1\ \mathrm{mm}}{1000\ \mathrm{mm}}=0.1\%; 2×0.010.50=2×2%=4%2\times\frac{0.01}{0.50}=2\times2\%=4\%; 0.010.40=2.5%\frac{0.01}{0.40}=2.5\%. The total is 1+0.1+4+2.5=7.6%1+0.1+4+2.5=7.6\%. The diameter enters squared, so its error is doubled, and the small extension ll together with the thin diameter dominate the uncertainty. Hence the maximum percentage error in YY is 7.607.60. Since the question asks for the maximum error, the individual errors are added and never combined in quadrature, and the least count of each instrument is the error assumed in its reading. Therefore the answer is 7.60%7.60\%.
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