In an experiment to determine the Young's mod | vedclass

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(C) Young's modulus is given by $Y = \frac{FL}{A\Delta L} = \frac{mgL}{(\pi d^2/4)\Delta L} = \frac{4mgL}{\pi d^2 \Delta L}$.Given: $L = 1\;m$,$m = 1\

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$2$

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(C) Young's modulus is given by $Y = \frac{FL}{A\Delta L} = \frac{mgL}{(\pi d^2/4)\Delta L} = \frac{4mgL}{\pi d^2 \Delta L}$.Given: $L = 1\;m$,$m = 1\;kg$,$g = 10\;m/s^2$,$\Delta L = 0.4 	imes 10^{-3}\;m$,$d = 0.4 	imes 10^{-3}\;m$.$Y = \frac{4 	imes 1 	imes 10 	imes 1}{\pi 	imes (0.4 	imes 10^{-3})^2 	imes 0.4 	imes 10^{-3}} = \frac{40}{\pi 	imes 0.064 	imes 10^{-9}} \approx \frac{40}{3.14 	imes 0.064 	imes 10^{-9}} \approx 1.99 	imes 10^{11}\;N/m^2$.The relative error in $Y$ is given by $\frac{\Delta Y}{Y} = \frac{\Delta m}{m} + \frac{\Delta L}{L} + 2\frac{\Delta d}{d} + \frac{\Delta(\Delta L)}{\Delta L}$.Assuming $\Delta m = 0$ and $\Delta L = 0$ (as length is exact),$\frac{\Delta Y}{Y} = 2\frac{\Delta d}{d} + \frac{\Delta(\Delta L)}{\Delta L} = 2 	imes \frac{0.01}{0.4} + \frac{0.02}{0.4} = 0.05 + 0.05 = 0.1$.$\Delta Y = 0.1 	imes Y = 0.1 	imes 1.99 	imes 10^{11} = 1.99 	imes 10^{10}\;N/m^2$.Comparing with $x 	imes 10^{10}$,we get $x \approx 2$.

Similar Questions

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