On applying a stress of 20 times 10^8 N/m^2, | vedclass

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Question

(B) Young's modulus $(Y)$ is defined as the ratio of stress to strain.$Y = \frac{	ext{stress}}{	ext{strain}}$Let the original length of the wire be

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$20 	imes 10^8 \ N/m^2$

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(B) Young's modulus $(Y)$ is defined as the ratio of stress to strain.$Y = \frac{	ext{stress}}{	ext{strain}}$Let the original length of the wire be $L$. Since the length is doubled,the new length becomes $2L$.Change in length $\Delta L = 2L - L = L$.Strain is defined as the ratio of change in length to the original length:$	ext{strain} = \frac{\Delta L}{L} = \frac{L}{L} = 1$.Given stress = $20 	imes 10^8 \ N/m^2$.Substituting these values into the formula for Young's modulus:$Y = \frac{20 	imes 10^8 \ N/m^2}{1} = 20 	imes 10^8 \ N/m^2$.Therefore,the correct option is $B$.

On applying a stress of $20 \times 10^8 \ N/m^2$,the length of a perfectly elastic wire is doubled. Its Young's modulus will be:

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