The ratio of diameters of two wires of the sa | vedclass

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(A) The formula for the extension $(l)$ of a wire is given by $l = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$.Since the material is the same,$Y$ is constan

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$n^2$ times

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(A) The formula for the extension $(l)$ of a wire is given by $l = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$.Since the material is the same,$Y$ is constant. Given that the load $(F)$ and the length $(L)$ are also constant,we have $l \propto \frac{1}{r^2}$.Let $d_1$ and $d_2$ be the diameters of the two wires. We are given $\frac{d_1}{d_2} = n : 1$,which implies $\frac{r_1}{r_2} = n$.For the thin wire (wire $2$),the extension $l_2$ relative to the thick wire (wire $1$) is given by $\frac{l_2}{l_1} = \left( \frac{r_1}{r_2} \right)^2 = n^2$.Therefore,$l_2 = n^2 l_1$.

The ratio of diameters of two wires of the same material is $n : 1$. The length of each wire is $4 \ m$. On applying the same load,the increase in length of the thin wire will be:

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