A meter scale of mass m,Young's modulus Y,and | vedclass

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(B) Let the total length of the scale be $L = 100\ cm$. The tension $T$ at a distance $x$ from the lower end is given by $T = \frac{mgx}{L}$.Using You

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$20\frac{mg}{AY}\ cm + 40\ cm$

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(B) Let the total length of the scale be $L = 100\ cm$. The tension $T$ at a distance $x$ from the lower end is given by $T = \frac{mgx}{L}$.Using Young's modulus $Y = \frac{	ext{stress}}{	ext{strain}} = \frac{T/A}{dy/dx}$,we have $dy = \frac{T}{AY} dx = \frac{mgx}{LAY} dx$.The elongation $\Delta y$ between the $30\ cm$ and $70\ cm$ marks (measured from the lower end,where $x=0$ is the bottom) corresponds to the segment from $x_1 = 100-70 = 30\ cm$ to $x_2 = 100-30 = 70\ cm$.$\Delta y = \int_{30}^{70} \frac{mgx}{LAY} dx = \frac{mg}{LAY} \left[ \frac{x^2}{2} ight]_{30}^{70} = \frac{mg}{LAY} \left( \frac{4900 - 900}{2} ight) = \frac{mg}{LAY} 	imes 2000 = \frac{mg}{AY 	imes 100} 	imes 2000 = 20 \frac{mg}{AY}$.The original separation is $70\ cm - 30\ cm = 40\ cm$.Thus,the new separation is $40\ cm + 20\frac{mg}{AY}\ cm$.

$A$ meter scale of mass $m$,Young's modulus $Y$,and cross-sectional area $A$ is hung vertically from the ceiling at the zero mark. The separation between the $30\ cm$ and $70\ cm$ marks will be: (Assume $\frac{mg}{AY}$ is dimensionless)

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