A string of cross-sectional area 4 times 10^{ | vedclass

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(C) At the bottom of the vertical circular path,the tension $T$ in the string is given by the sum of the gravitational force and the centripetal force

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(C) At the bottom of the vertical circular path,the tension $T$ in the string is given by the sum of the gravitational force and the centripetal force:$T = mg + \frac{mv^{2}}{R}$Given: $m = 2 \, kg$,$v = 5 \, m/s$,$R = 0.5 \, m$,$g = 10 \, m/s^{2}$,$A = 4 	imes 10^{-6} \, m^{2}$,$Y = 10^{11} \, N/m^{2}$.$T = (2 	imes 10) + \frac{2 	imes (5)^{2}}{0.5} = 20 + \frac{50}{0.5} = 20 + 100 = 120 \, N$.From Hooke's Law,Stress $= Y 	imes 	ext{Strain}$,so $	ext{Strain} = \frac{	ext{Stress}}{Y} = \frac{T}{AY}$.$	ext{Strain} = \frac{120}{(4 	imes 10^{-6}) 	imes 10^{11}} = \frac{120}{4 	imes 10^{5}} = 30 	imes 10^{-5}$.Thus,the strain is $30 	imes 10^{-5}$.

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