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The lengths of four wires A, B, C and D made of same material are $1 \mathrm{~m}, 2 \mathrm{~m}, 3 \mathrm{~m}$ and $4 \mathrm{~m}$ respectively. The radii of the wires A, B, C and D are $0.2 \mathrm{~mm}, 0.4 \mathrm{~mm}, 0.6 \mathrm{~mm}$ and $0.8 \mathrm{~mm}$ respectively. For the same applied tension, the elongation is more in the wire.
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The correct answer is:
None of these
$Y=\frac{\frac{F}{A}}{\frac{\Delta 1}{1}}$
Elongation, $\Delta \mathrm{l} \propto \frac{1}{\mathrm{r}^2}$
For wire A,
$\Delta \mathrm{l}=\frac{1}{\left(0.2 \times 10^{-3}\right)^2}=5 \times 10^7$
For wire B,
$\Delta 1=\frac{2}{\left(0.4 \times 10^{-3}\right)^2}=5 \times 10^7$
For wire C,
$\Delta \mathrm{l}=\frac{3}{\left(0.6 \times 10^{-3}\right)^2}=5 \times 10^7$
For wire D,
$\Delta \mathrm{l}=\frac{4}{\left(0.8 \times 10^{-3}\right)^2}=5 \times 10^7$
Elongation, $\Delta \mathrm{l} \propto \frac{1}{\mathrm{r}^2}$
For wire A,
$\Delta \mathrm{l}=\frac{1}{\left(0.2 \times 10^{-3}\right)^2}=5 \times 10^7$
For wire B,
$\Delta 1=\frac{2}{\left(0.4 \times 10^{-3}\right)^2}=5 \times 10^7$
For wire C,
$\Delta \mathrm{l}=\frac{3}{\left(0.6 \times 10^{-3}\right)^2}=5 \times 10^7$
For wire D,
$\Delta \mathrm{l}=\frac{4}{\left(0.8 \times 10^{-3}\right)^2}=5 \times 10^7$
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