When a weight of $10\ kg$ is suspended from a copper wire of length $3\ metres$ and diameter $0.4\ mm$, its length increases by $2.4\ cm$. If the diameter of the wire is doubled, then the extension in its length will be

Question

Dr. Abubacker · Accepted Answer

(d) $l \propto \frac{FL}{\pi r^{2}Y} \Rightarrow l \propto \frac{L}{r^{2}}\ \ \lbrack F$ and $Y$ are constant]
$\frac{l_{1}}{l_{2}} = \frac{L_{1}}{L_{2}}\left( \frac{r_{2}}{r_{1}} ight)^{2} = \frac{1}{2}\left( \sqrt{2} ight)^{2}\ 	herefore\frac{l_{1}}{l_{2}} = 1:1$

Q. When a weight of $10\ kg$ is suspended from a copper wire of length $3\ metres$ and diameter $0.4\ mm$, its length increases by $2.4\ cm$. If the diameter of the wire is doubled, then the extension in its length will be

Explanation

Reviewed by Dr. Abubacker

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