The load versus elongation graph for four wir | vedclass

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(A) The formula for elongation $l$ is given by $l = \frac{FL}{AY}$,where $F$ is the load,$L$ is the original length,$A$ is the cross-sectional area,an

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$OD$

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(A) The formula for elongation $l$ is given by $l = \frac{FL}{AY}$,where $F$ is the load,$L$ is the original length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.Since the wires are of the same material ($Y$ is constant) and have the same length ($L$ is constant),for a given load ($F$ is constant),the elongation $l$ is inversely proportional to the cross-sectional area $A$ $(l \propto \frac{1}{A})$.Since $A = \pi r^2$,we have $l \propto \frac{1}{r^2}$,where $r$ is the radius of the wire.This means that for the same load,the thickest wire (largest $r$) will show the minimum elongation.In the given graph,for a constant load on the $y$-axis,the line $OD$ corresponds to the minimum elongation on the $x$-axis.Therefore,the line $OD$ represents the thickest wire.

Column $I$	Column $II$
$A$. Growth	$I$. Production of offspring.
$B$. Reproduction	$II$. Composed of one or more cells.
$C$. Metabolism	$III$. Increase in mass and increase in number of individuals.
$D$. Cellular organization	$IV$. Sum total of all chemical reactions occurring in the body.

The load versus elongation graph for four wires of the same material is shown in the figure. The thickest wire is represented by the line

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