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(D) The formula for Young's modulus $Y$ is given by $Y = \frac{F/A}{\Delta \ell / \ell}$,where $F$ is the force,$A$ is the cross-sectional area,$\ell$

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$4: 3$

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(D) The formula for Young's modulus $Y$ is given by $Y = \frac{F/A}{\Delta \ell / \ell}$,where $F$ is the force,$A$ is the cross-sectional area,$\ell$ is the original length,and $\Delta \ell$ is the change in length.Rearranging for $\Delta \ell$,we get $\Delta \ell = \frac{F \ell}{AY}$.Since the area $A = \pi r^2 = \pi (d/2)^2$,where $d$ is the diameter,we have $\Delta \ell = \frac{F \ell}{\pi (d/2)^2 Y} = \frac{4F \ell}{\pi d^2 Y}$.Given that the wires are made of the same material ($Y$ is constant) and are stretched by the same force ($F$ is constant),the change in length is proportional to $\frac{\ell}{d^2}$.Therefore,$\frac{\Delta \ell_A}{\Delta \ell_B} = \left( \frac{\ell_A}{\ell_B} ight) 	imes \left( \frac{d_B}{d_A} ight)^2$.Given $\frac{\ell_A}{\ell_B} = \frac{1}{3}$ and $\frac{d_A}{d_B} = \frac{1}{2}$ (which implies $\frac{d_B}{d_A} = 2$),we substitute these values:$\frac{\Delta \ell_A}{\Delta \ell_B} = \left( \frac{1}{3} ight) 	imes (2)^2 = \frac{1}{3} 	imes 4 = \frac{4}{3}$.Thus,the ratio of the increase in their lengths is $4: 3$.

Two wires $A$ and $B$ are made of the same material. Their diameters are in the ratio of $1: 2$ and their lengths are in the ratio of $1: 3$. If they are stretched by the same force,then the increase in their lengths will be in the ratio of:

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