The mass and length of a wire are M and L res | vedclass

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(D) The formula for Young's modulus $Y$ is given by $Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{l/L} = \frac{FL}{Al}$.We know that the mass

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$\frac{Fd{L^2}}{Ml}$

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(D) The formula for Young's modulus $Y$ is given by $Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{l/L} = \frac{FL}{Al}$.We know that the mass $M$ of the wire is given by $M = 	ext{Volume} 	imes 	ext{Density} = (A 	imes L) 	imes d$.From this,we can express the cross-sectional area $A$ as $A = \frac{M}{Ld}$.Substituting the value of $A$ into the formula for $Y$:$Y = \frac{FL}{(\frac{M}{Ld})l} = \frac{FL^2d}{Ml} = \frac{Fd{L^2}}{Ml}$.Thus,the correct option is $D$.

The mass and length of a wire are $M$ and $L$ respectively. The density of the material of the wire is $d$. On applying a force $F$ on the wire,the increase in length is $l$. Then,the Young's modulus of the material of the wire will be:

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