An object of mass 15 ,kg is attached to the e | vedclass

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(B) 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 ob

Vedclass · Accepted Answer

$0.39$

Vedclass · Answer

(B) 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 object.The net force $F$ acting on the wire is given by $F = mg + m\omega^2l$.Given values: $m = 15 \,kg$, $l = 1.0 \,m$, $\omega = 4 \,rad/s$, $A = 0.05 \,cm^2 = 0.05 	imes 10^{-4} \,m^2$, $Y = 2 	imes 10^{11} \,N/m^2$, and $g = 10 \,m/s^2$.Calculating the force $F$:$F = (15 	imes 10) + (15 	imes 4^2 	imes 1) = 150 + 240 = 390 \,N$.Using the formula for Young's modulus $Y = \frac{F/A}{\Delta l/l}$, the elongation $\Delta l$ is:$\Delta l = \frac{Fl}{AY} = \frac{390 	imes 1.0}{(0.05 	imes 10^{-4}) 	imes (2 	imes 10^{11})}$$\Delta l = \frac{390}{0.1 	imes 10^7} = \frac{390}{10^6} = 390 	imes 10^{-6} \,m = 0.39 \,mm$.Thus, the correct option is $B$.

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