How much force is required to produce an incr | vedclass

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(C) Given:Diameter $d = 0.6 \, mm = 0.6 	imes 10^{-3} \, m$,so radius $r = 0.3 	imes 10^{-3} \, m$.Young's modulus $Y = 0.9 	imes 10^{11} \, N/m^2$

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Nearly $51 \, N$

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(C) Given:Diameter $d = 0.6 \, mm = 0.6 	imes 10^{-3} \, m$,so radius $r = 0.3 	imes 10^{-3} \, m$.Young's modulus $Y = 0.9 	imes 10^{11} \, N/m^2$.Strain $\frac{\Delta L}{L} = 0.2\% = \frac{0.2}{100} = 0.002$.Area of cross-section $A = \pi r^2 = \pi 	imes (0.3 	imes 10^{-3})^2 = \pi 	imes 0.09 	imes 10^{-6} \, m^2$.Using the formula for Young's modulus: $Y = \frac{F/A}{\Delta L/L}$,we get $F = Y 	imes A 	imes \frac{\Delta L}{L}$.Substituting the values: $F = (0.9 	imes 10^{11}) 	imes (\pi 	imes 0.09 	imes 10^{-6}) 	imes (0.002)$.$F = 0.9 	imes 10^{11} 	imes 3.1416 	imes 0.09 	imes 10^{-6} 	imes 0.002$.$F \approx 0.9 	imes 0.09 	imes 3.1416 	imes 200 \approx 50.89 \, N$.Thus,the force required is nearly $51 \, N$.

How much force is required to produce an increase of $0.2\%$ in the length of a brass wire of diameter $0.6\, mm$ (Young's modulus for brass = $0.9 \times 10^{11} \, N/m^2$)?

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