A load of 1 ~kg weight is attached to one end | vedclass

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(A) The Young's modulus $Y$ is defined as the ratio of stress to strain: $Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta l/l}$.Rearrangin

Vedclass · Accepted Answer

$0.3 	imes 10^{-4}$

Vedclass · Answer

(A) The Young's modulus $Y$ is defined as the ratio of stress to strain: $Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta l/l}$.Rearranging for the fractional change in length $\frac{\Delta l}{l}$,we get: $\frac{\Delta l}{l} = \frac{F}{AY}$.At the lowest position,the tension $T$ in the wire is equal to the weight of the load,$T = mg$ (assuming the velocity is negligible or the question implies static equilibrium at the lowest point for the fractional change calculation).Given: $m = 1 ~kg$,$g = 10 ~m/s^2$,$A = 3 ~mm^2 = 3 	imes 10^{-6} ~m^2$,and $Y = 10^{11} ~N/m^2$.Substituting the values: $\frac{\Delta l}{l} = \frac{1 	imes 10}{3 	imes 10^{-6} 	imes 10^{11}} = \frac{10}{3 	imes 10^5} = \frac{1}{3} 	imes 10^{-4} \approx 0.33 	imes 10^{-4}$.Rounding to the provided options,the correct value is $0.3 	imes 10^{-4}$.

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