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

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(A) The formula for the extension of a wire is $\Delta l = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$.Since the material $(Y)$,load $(F)$,and length $(L)$

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$n^2$ times

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(A) The formula for the extension of a wire is $\Delta l = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y}$.Since the material $(Y)$,load $(F)$,and length $(L)$ are the same for both wires,the extension $\Delta l$ is inversely proportional to the square of the radius (or diameter),i.e.,$\Delta l \propto \frac{1}{d^2}$.Let $d_1$ and $d_2$ be the diameters of the thick and thin wires respectively. Given $\frac{d_1}{d_2} = n$,so $d_1 = n d_2$.Let $\Delta l_1$ be the extension of the thick wire and $\Delta l_2$ be the extension of the thin wire.Then,$\frac{\Delta l_2}{\Delta l_1} = \left( \frac{d_1}{d_2} \right)^2 = n^2$.Therefore,$\Delta l_2 = n^2 \Delta l_1$.The increase in length of the thin wire will be $n^2$ times the increase in length of the thick wire.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$k$	$2k$	$3k$	$2k$	$k$

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

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