A sphere of mass 2 ,kg and diameter 4.5 ,cm i | vedclass

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Question

(A) Given: Mass $M = 2 \,kg$, Diameter $d = 4.5 \,cm$, Radius $r = 2.25 \,cm = 0.0225 \,m$, Wire length $L = 2 \,m$, Cross-sectional area $A = 0.24

Vedclass · Accepted Answer

$10$

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(A) Given: Mass $M = 2 \,kg$, Diameter $d = 4.5 \,cm$, Radius $r = 2.25 \,cm = 0.0225 \,m$, Wire length $L = 2 \,m$, Cross-sectional area $A = 0.24 	imes 10^{-6} \,m^2$, Ceiling height $H = 205 \,cm = 2.05 \,m$, Young's modulus $Y = 2 	imes 10^{11} \,Nm^{-2}$, $g = 10 \,ms^{-2}$.At the lowest position, the total length of the system (wire + sphere) is $2.05 \,m$. The unstretched length of the wire is $2 \,m$ and the diameter of the sphere is $4.5 \,cm = 0.045 \,m$. The extension $\Delta L$ in the wire is $\Delta L = 2.05 \,m - (2 \,m + 0.045 \,m) = 0.005 \,m$.The tension $T$ in the wire at the lowest point is the sum of the weight of the sphere and the centrifugal force: $T = Mg + \frac{Mv^2}{R}$, where $R$ is the distance from the ceiling to the center of the sphere, $R = L + \Delta L + r = 2 + 0.005 + 0.0225 = 2.0275 \,m$.Using Hooke's Law: $Y = \frac{T L}{A \Delta L} \Rightarrow T = \frac{Y A \Delta L}{L}$.Substituting the values: $T = \frac{(2 	imes 10^{11}) 	imes (0.24 	imes 10^{-6}) 	imes 0.005}{2} = 120 \,N$.Now, $Mg + \frac{Mv^2}{R} = 120 \Rightarrow (2 	imes 10) + \frac{2 v^2}{2.0275} = 120$.$20 + \frac{2 v^2}{2.0275} = 120 \Rightarrow \frac{2 v^2}{2.0275} = 100 \Rightarrow v^2 = 50 	imes 2.0275 = 101.375$.$v = \sqrt{101.375} \approx 10.07 \,ms^{-1}$. The closest option is $10 \,ms^{-1}$.

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