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(C) At the lowest point of the vertical circle,the tension $T$ in the wire provides the necessary centripetal force and balances the weight of the sph

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$0.7$

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(C) At the lowest point of the vertical circle,the tension $T$ in the wire provides the necessary centripetal force and balances the weight of the sphere.The equation of motion is: $T - mg = m \omega^2 l$,where $m = 4 \,kg$,$l = 1 \,m$,$\omega = 10 \,rad \,s^{-1}$,and $g = 10 \,ms^{-2}$.$T = m(g + \omega^2 l) = 4(10 + 10^2 	imes 1) = 4(10 + 100) = 4(110) = 440 \,N$.The elongation $\Delta l$ is given by Hooke's Law: $\Delta l = \frac{Tl}{AY}$,where $A = \pi r^2$ and $r = 1 \,mm = 10^{-3} \,m$.$A = \pi (10^{-3})^2 = \pi 	imes 10^{-6} \,m^2$.$\Delta l = \frac{440 	imes 1}{\pi 	imes 10^{-6} 	imes 20 	imes 10^{10}} = \frac{440}{20 \pi 	imes 10^4} = \frac{22}{\pi} 	imes 10^{-4} \,m$.Using $\pi \approx 3.14$,$\Delta l \approx \frac{22}{3.14} 	imes 10^{-4} \approx 7.006 	imes 10^{-4} \,m = 0.7 \,mm$.

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

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$B$. Shearing strain $2$. Temporary loss of elastic property
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