A uniform steel rod of mass 1.8 ,kg and lengt | vedclass

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(D) Let the length of each wire be $L = 0.5 \,m$,area $A = 0.01 \,mm^2 = 0.01 	imes 10^{-6} \,m^2 = 10^{-8} \,m^2$,and mass of the rod $M = 1.8 \,kg$

Vedclass · Accepted Answer

$6.25$

Vedclass · Answer

(D) Let the length of each wire be $L = 0.5 \,m$,area $A = 0.01 \,mm^2 = 0.01 	imes 10^{-6} \,m^2 = 10^{-8} \,m^2$,and mass of the rod $M = 1.8 \,kg$.The rod is suspended by two wires forming an isosceles triangle with the nail. The length of the rod is $0.8 \,m$,so the horizontal distance from the center to each end is $0.4 \,m$.The vertical height $h$ of the nail from the rod is $h = \sqrt{L^2 - (0.4)^2} = \sqrt{0.5^2 - 0.4^2} = \sqrt{0.25 - 0.16} = \sqrt{0.09} = 0.3 \,m$.The tension $T$ in each wire is determined by the vertical equilibrium: $2T \cos 	heta = Mg$,where $\cos 	heta = h/L = 0.3/0.5 = 0.6$.$2T(0.6) = 1.8 	imes 10 \implies 1.2T = 18 \implies T = 15 \,N$.The extension in each wire is $\Delta L = \frac{TL}{AY} = \frac{15 	imes 0.5}{10^{-8} 	imes 2 	imes 10^{11}} = \frac{7.5}{2000} = 3.75 	imes 10^{-3} \,m = 3.75 \,mm$.The new length of the wire is $L' = L + \Delta L$. The new vertical height $h'$ is $\sqrt{(L')^2 - (0.4)^2}$.Using binomial approximation,$h' \approx h + \frac{L}{h} \Delta L = 0.3 + \frac{0.5}{0.3} 	imes 3.75 \,mm = 0.3 + 1.666 	imes 3.75 \,mm = 0.3 + 6.25 \,mm$.The increase in vertical distance is $h' - h = 6.25 \,mm$.

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