A sphere of mass 4 ,kg is attached to one end | vedclass

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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 $\pi \approx 3.14$,$\Delta l \approx \frac{22}{3.14} 	imes 10^{-4} \approx 7.006 	imes 10^{-4} \,m = 0.7 \,mm$.

List-$I$	List-$II$
$A$. Young's Modulus	$I$. $\frac{Ad}{\Delta L}$
$B$. Compressibility	$II$. $\frac{FL}{A\Delta L}$
$C$. Bulk Modulus	$III$. $-\frac{1}{\Delta P}(\frac{\Delta V}{V})$
$D$. Poisson's Ratio	$IV$. $-\frac{\Delta D/D}{\Delta L/L}$

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$A$. Young's Modulus $I$. $\frac{Ad}{\Delta L}$

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