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(D) The tension $T$ in the wire is given by the formula for an Atwood machine:$T = \frac{2 m_1 m_2}{m_1 + m_2} g = \frac{2 	imes 2 	imes 4}{2 + 4} \

Vedclass · Accepted Answer

$12$

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(D) The tension $T$ in the wire is given by the formula for an Atwood machine:$T = \frac{2 m_1 m_2}{m_1 + m_2} g = \frac{2 	imes 2 	imes 4}{2 + 4} 	imes 10 = \frac{16}{6} 	imes 10 = \frac{80}{3} \ N$The cross-sectional area $A$ of the wire is:$A = \pi r^2 = \pi (4.0 	imes 10^{-5})^2 = 16 \pi 	imes 10^{-10} \ m^2$Longitudinal strain is defined as:$	ext{Strain} = \frac{\Delta \ell}{\ell} = \frac{F}{AY} = \frac{T}{AY}$Substituting the values:$	ext{Strain} = \frac{80/3}{16 \pi 	imes 10^{-10} 	imes 2.0 	imes 10^{11}}$$	ext{Strain} = \frac{80/3}{32 \pi 	imes 10} = \frac{80}{3 	imes 320 \pi} = \frac{80}{960 \pi} = \frac{1}{12 \pi}$Comparing this with $\frac{1}{\alpha \pi}$,we get $\alpha = 12$.

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