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Mathematics in Physics

JEE Main level, numerical answer, medium

79 of 121

Question 79 of 121

In an experiment of determine the Young's modulus of wire of a length exactly $1m$ , the extension in the length of the wire is measured as $0.4mm$ with an uncertainty of $\pm 0.02mm$ when a load of $1kg$ is applied. The diameter of the wire is measured as $0.4mm$ with an uncertainty of $\pm 0.01mm$ . The error in the measurement of Young's modulus $\left(\Delta Y\right)$ is found to be $x\times {10}^{10}N{m}^{-2}$ . The value of $x$ is _____ .

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Answer: 2

Given here,
Length,

L=1m

and extension in length,

\Delta L=0.4\times {10}^{-3}m

Mass of load,

m=1kg

and diameter of wire,

d=0.4\times {10}^{-3}m

Young's modulus is given by

Y=\dfrac{FL}{A\Delta L}=\dfrac{\left(mg\right)\times \left(1\right)}{\left(\dfrac{\pi {d}^{2}}{4}\right)0.4\times {10}^{-3}}

\Rightarrow Y=\dfrac{10\times 4}{\pi {\left(0.4\times {10}^{-3}\right)}^{2}\times 0.4\times {10}^{-3}}

\Rightarrow Y=\dfrac{40\times 7}{22\times 64\times {10}^{-3}\times {10}^{-9}}

\Rightarrow Y=0.199\times {10}^{-12}N{m}^{-2}

Error in measurement of Young's modulus,

\dfrac{\Delta Y}{Y}=\dfrac{\Delta F}{F}+\dfrac{\Delta L}{L}+\dfrac{\Delta A}{A}+\dfrac{\Delta \left(\Delta L\right)}{\left(\Delta L\right)}

=\dfrac{0.02}{0.4}+2\dfrac{\Delta d}{d}=\dfrac{0.2}{4}+2\times \dfrac{0.01}{0.4}

=\dfrac{0.1}{2}+\dfrac{0.1}{2}=0.1

=0.199\times {10}^{11}\approx 2\times {10}^{10}

Hence, the value of

x=2

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