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

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(B) The formula for Young's modulus $Y$ is given by $Y = \frac{F/A}{\Delta l/L} = \frac{F \cdot L}{A \cdot \Delta l}$.Rearranging this,we get the rati

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$2$

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(B) The formula for Young's modulus $Y$ is given by $Y = \frac{F/A}{\Delta l/L} = \frac{F \cdot L}{A \cdot \Delta l}$.Rearranging this,we get the ratio of extension to load as $\frac{\Delta l}{F} = \frac{L}{YA}$.Here,the graph plots $\frac{\Delta l}{F}$ on the $y$-axis and $L$ on the $x$-axis.The slope of this graph is $m = \frac{\Delta l/F}{L} = \frac{1}{YA}$.From the graph,we can calculate the slope $m = \frac{0.25 	imes 10^{-5}}{1} = 0.25 	imes 10^{-5}\,m/N$.Given cross-sectional area $A = 2\,mm^2 = 2 	imes 10^{-6}\,m^2$.Substituting these values into the slope equation: $0.25 	imes 10^{-5} = \frac{1}{Y 	imes 2 	imes 10^{-6}}$.$Y = \frac{1}{0.25 	imes 10^{-5} 	imes 2 	imes 10^{-6}} = \frac{1}{0.5 	imes 10^{-11}} = 2 	imes 10^{11}\,N/m^2$.Comparing this with $x 	imes 10^{11}\,N/m^2$,we get $x = 2$.

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