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Two wires $A$ and $B$ are of the same material. Their lengths are in the ratio of $1: 2$ and the diameter are in the ratio $2: 1$. If they are pulled by the same force, then increase in length will be in the ratio of
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The correct answer is:
$1: 8$
We know that Young's modulus
$$
\mathrm{Y}=\frac{\mathrm{F}}{\pi \mathrm{r}^2} \times \frac{\mathrm{L}}{\ell}
$$
Since Y, F are same for both the wires, we have,
$$
\begin{aligned}
& \frac{1}{\mathrm{r}_1^2} \frac{\mathrm{L}_1}{\ell_1}=\frac{1}{\mathrm{r}_2^2} \frac{\mathrm{L}_2}{\ell_2} \quad \text { or, } \frac{\ell_1}{\ell_2}=\frac{\mathrm{r}_2^2 \times \mathrm{L}_1}{\mathrm{r}_1^2 \times \mathrm{L}_2}= \\
& \frac{\left(\mathrm{D}_2 / 2\right)^2 \times \mathrm{L}_1}{\left(\mathrm{D}_1 / 2\right)^2 \times \mathrm{L}_2} \\
& \text { or, } \frac{\ell_1}{\ell_2}=\frac{\mathrm{D}_2^2 \times \mathrm{L}_1}{\mathrm{D}_1^2 \times \mathrm{L}_2}=\frac{\mathrm{D}_2^2}{\left(2 \mathrm{D}_2\right)^2} \times \frac{\mathrm{L}_2}{2 \mathrm{~L}_2}=\frac{1}{8} \\
& \text { So, } \ell_1: \ell_2=1: 8
\end{aligned}
$$
$$
\mathrm{Y}=\frac{\mathrm{F}}{\pi \mathrm{r}^2} \times \frac{\mathrm{L}}{\ell}
$$
Since Y, F are same for both the wires, we have,
$$
\begin{aligned}
& \frac{1}{\mathrm{r}_1^2} \frac{\mathrm{L}_1}{\ell_1}=\frac{1}{\mathrm{r}_2^2} \frac{\mathrm{L}_2}{\ell_2} \quad \text { or, } \frac{\ell_1}{\ell_2}=\frac{\mathrm{r}_2^2 \times \mathrm{L}_1}{\mathrm{r}_1^2 \times \mathrm{L}_2}= \\
& \frac{\left(\mathrm{D}_2 / 2\right)^2 \times \mathrm{L}_1}{\left(\mathrm{D}_1 / 2\right)^2 \times \mathrm{L}_2} \\
& \text { or, } \frac{\ell_1}{\ell_2}=\frac{\mathrm{D}_2^2 \times \mathrm{L}_1}{\mathrm{D}_1^2 \times \mathrm{L}_2}=\frac{\mathrm{D}_2^2}{\left(2 \mathrm{D}_2\right)^2} \times \frac{\mathrm{L}_2}{2 \mathrm{~L}_2}=\frac{1}{8} \\
& \text { So, } \ell_1: \ell_2=1: 8
\end{aligned}
$$
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