One end of a steel wire of radius r is fixed | vedclass

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(B) The longitudinal strain $\epsilon$ is given by $\epsilon = \frac{	ext{Stress}}{Y} = \frac{F}{A \cdot Y}$,where $F$ is the tension,$A$ is the cros

Vedclass · Accepted Answer

$1: 6$

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(B) The longitudinal strain $\epsilon$ is given by $\epsilon = \frac{	ext{Stress}}{Y} = \frac{F}{A \cdot Y}$,where $F$ is the tension,$A$ is the cross-sectional area,and $Y$ is the Young's modulus.For the steel wire: Tension $F_s = (3 + 2) \ kg 	imes g = 5g$. Area $A_s = \pi r^2$. Young's modulus $Y_s = 20 	imes 10^{10} \ Nm^{-2}$.Strain in steel $\epsilon_s = \frac{5g}{\pi r^2 \cdot 20 	imes 10^{10}}$.For the copper wire: Tension $F_c = 2 \ kg 	imes g = 2g$. Area $A_c = \pi (2r)^2 = 4\pi r^2$. Young's modulus $Y_c = 12 	imes 10^{10} \ Nm^{-2}$.Strain in copper $\epsilon_c = \frac{2g}{4\pi r^2 \cdot 12 	imes 10^{10}} = \frac{g}{2\pi r^2 \cdot 12 	imes 10^{10}} = \frac{g}{24\pi r^2 	imes 10^{10}}$.Ratio $\frac{\epsilon_c}{\epsilon_s} = \frac{g}{24\pi r^2 	imes 10^{10}} 	imes \frac{20\pi r^2 	imes 10^{10}}{5g} = \frac{20}{24 	imes 5} = \frac{20}{120} = \frac{1}{6}$.Thus,the ratio is $1: 6$.

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Two wires each of radius $0.2\,cm$ and negligible mass, one made of steel and the other made of brass, are loaded as shown in the figure. The elongation of the steel wire is $.........\times 10^{-6}\,m$. [Young's modulus for steel $= 2 \times 10^{11}\,N/m^2$ and $g = 10\,m/s^2$]

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