The area of cross-section of a steel wire (Y | vedclass

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(D) Given: Young's modulus $Y = 2.0 	imes 10^{11} \ N/m^2$,Area $A = 0.1 \ cm^2 = 0.1 	imes 10^{-4} \ m^2 = 10^{-5} \ m^2$.To double the length of t

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$2 	imes 10^6 \ N$

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(D) Given: Young's modulus $Y = 2.0 	imes 10^{11} \ N/m^2$,Area $A = 0.1 \ cm^2 = 0.1 	imes 10^{-4} \ m^2 = 10^{-5} \ m^2$.To double the length of the wire,the change in length $\Delta L$ must be equal to the original length $L$,so $\Delta L = L$.Strain is defined as $\frac{\Delta L}{L} = \frac{L}{L} = 1$.Using the formula for Young's modulus: $Y = \frac{F/A}{\Delta L/L}$.Substituting the values: $2.0 	imes 10^{11} = \frac{F}{10^{-5} 	imes 1}$.$F = 2.0 	imes 10^{11} 	imes 10^{-5} \ N = 2 	imes 10^6 \ N$.

The area of cross-section of a steel wire $(Y = 2.0 \times 10^{11} \ N/m^2)$ is $0.1 \ cm^2$. The force required to double its length will be

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