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

$\mathrm{\left(\text { take } \mathrm{g}=10 \mathrm{~ms}^{-2}\right)}$
Option: 1
2

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{L=1 \mathrm{~m} }$

$\mathrm{Extension=l=0.4mm}$

$\mathrm{=4\times 10^{-4}m}$

$\mathrm{\Delta L =0.02 \mathrm{~mm}}$

$\mathrm{=2\times 10^{-5}m}$

$\mathrm{m =1 \mathrm{~kg} }$

$\mathrm{d =0.4 \mathrm{~mm}=4 \times 10^{-4} \mathrm{~m} }$

$\mathrm{\Delta d =1 \times 10^{-5} \mathrm{~m}}$

$\mathrm{ Y =\frac{m g}{A} \times \frac{L}{l} }$

$\mathrm{\frac{\Delta Y}{Y} =\frac{\Delta m}{m}+\frac{\Delta L}{L}+\frac{2 \Delta d}{d}+\frac{\Delta l}{l} }$

$\mathrm{=0+0+2\left(\frac{0.01}{0.4}\right)+\left(\frac{0.02}{0.4}\right) }$

$\mathrm{=\frac{1}{20}+\frac{1}{20}=\frac{2}{20} }$

$\mathrm{\Delta Y =\frac{Y}{10}= \frac{m g \times L}{(A l) 10} }$

$\mathrm{\Delta Y=\frac{1\times 10\times 1}{3.14\times \frac{16\times 10^{-8}}{4}\times 0.4\times 10^{-3}\times 10}}$

$\mathrm{\Delta Y=2\times 10^{10}}$

The Value of $\mathrm{x}$ is $\mathrm{2}$

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Devendra Khairwa

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