In an experiment determine |ELASTICITYVISCOSITYJEEMain2022

JEE MAIN 2022

26-07-2022 S1

Question

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 $\pm 0.02$ mm when a load of 1kg is applied. The diameter of the wire is measured as 0.4mm with an uncertainty of $\pm 0.01$ mm. The error in the measurement of Young’s modulus $(\Delta \mathrm{Y})$ is found to be $x \times 10^{10} \mathrm{Nm}^{-2}$. The value of $x$ is $\_\_\_\_$ . $\left[\right.$ Take $\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right]$

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Solution

$\begin{aligned} & \mathrm{L}=1 \mathrm{~m} \\ & \Delta \mathrm{~L}=0.4 \times 10^{-3} \mathrm{~m} \\ & \mathrm{~m}=1 \mathrm{~kg} \\ & \mathrm{~d}=0.4 \times 10^{-3} \mathrm{~m} \\ & \frac{\mathrm{~F}}{\mathrm{~A}}=\mathrm{Y}=\frac{\Delta \mathrm{L}}{\mathrm{L}} \\ & \mathrm{Y}=\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}}=\frac{(\mathrm{mg}) .(1)}{\left(\frac{\pi \mathrm{d}^2}{4}\right) 0.4 \times 10^{-3}} \\ & \Rightarrow \frac{10 \times 4}{\pi\left(0.4 \times 10^{-3}\right)^2 \times 0.4 \times 10^{-3}} \\ & \mathrm{Y}=\frac{40}{\pi\left(0.4 \times 10^{-3}\right)^3} \\ & \mathrm{Y}=\frac{40 \times 7}{22 \times 64 \times 10^{-3} \times 10^{-9}} \\ & \mathrm{Y}=0.199 \times 10^{-12} \mathrm{~N} / \mathrm{m}^2 \\ & \frac{\Delta \mathrm{Y}}{\mathrm{Y}}=\frac{\Delta \mathrm{F}}{\mathrm{F}}+\frac{\Delta \mathrm{L}}{\mathrm{L}}+\frac{\Delta \mathrm{A}}{\mathrm{A}}+\frac{\Delta(\Delta \mathrm{L})}{(\Delta \mathrm{L})} \\ & =\frac{0.02}{0.4}+2 \frac{\Delta \mathrm{~d}}{\mathrm{~d}}=\frac{0.2}{4}+2 \times \frac{0.01}{0.4}=\frac{0.1}{2}+\frac{0.1}{2}=0.1 \\ & \Rightarrow \Delta \mathrm{Y}=0.1 \times \mathrm{Y} \\ & =0.199 \times 10^{11}=1.99 \times 10^{10}\end{aligned}$

Question Tags

JEE Main

Physics

Hard

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