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(A) Let the original linear dimension be $L$. The new linear dimension is $L' = 9L$.Since the density $\rho$ remains constant,the mass $m$ is proporti

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$9$

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(A) Let the original linear dimension be $L$. The new linear dimension is $L' = 9L$.Since the density $ho$ remains constant,the mass $m$ is proportional to the volume $V = L^3$.Therefore,the ratio of the new mass to the original mass is $\frac{m'}{m} = \left(\frac{L'}{L}ight)^3 = 9^3 = 729$.The cross-sectional area $A$ of the leg is proportional to the square of the linear dimension,$A \propto L^2$.Therefore,the ratio of the new area to the original area is $\frac{A'}{A} = \left(\frac{L'}{L}ight)^2 = 9^2 = 81$.Stress $\sigma$ is defined as the force per unit area,where the force is the weight $mg$.$\sigma = \frac{mg}{A}$.The ratio of the new stress $\sigma'$ to the original stress $\sigma$ is:$\frac{\sigma'}{\sigma} = \left(\frac{m'}{m}ight) 	imes \left(\frac{A}{A'}ight) = \frac{9^3}{9^2} = 9$.Thus,the stress in the leg increases by a factor of $9$.

Column $I$	Column $II$
$A$. Hooke's law	$1$. Tangential strain
$B$. Shearing strain	$2$. Temporary loss of elastic property
$C$. Bulk strain	$3$. Elastic limit
$D$. Elastic fatigue	$4$. $3$ times the linear strain

$A$ man grows into a giant such that his linear dimensions increase by a factor of $9$. Assuming that his density remains the same,the stress in the leg will change by a factor of

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