Search any question & find its solution
Question:
Answered & Verified by Expert
A $5 \mathrm{~m}$ long aluminium wire $\left(Y=7 \times 10^{10} \mathrm{Nm}^{-2}\right.$ ) of diameter $3 \mathrm{~mm}$ supports a $40 \mathrm{~kg}$ mass. In order to have the same elongation in the copper wire $\left(Y=12 \times 10^{10} \mathrm{Nm}^{-2}\right)$ of the same
length under the same weight, the diameter should now be (in $\mathrm{mm}$ )
Options:
length under the same weight, the diameter should now be (in $\mathrm{mm}$ )
Solution:
1142 Upvotes
Verified Answer
The correct answer is:
$2.5$
$l=\frac{F L}{\pi r^{2} Y}$
$\Rightarrow r^{2} \propto \frac{1}{Y}(F, L$ and $l$ are constant) $\frac{r_{2}}{r_{1}}=\left[\frac{Y_{1}}{Y_{2}}\right]^{1 / 2}=\left[\frac{7 \times 10^{10}}{12 \times 10^{10}}\right]^{1 / 2}$
$\Rightarrow \quad r_{2}=1.5 \times\left(\frac{7}{12}\right)^{1 / 2}=1.145 \mathrm{~mm}$
$\therefore$ diameter $=2.29 \mathrm{~mm}$.
$\Rightarrow r^{2} \propto \frac{1}{Y}(F, L$ and $l$ are constant) $\frac{r_{2}}{r_{1}}=\left[\frac{Y_{1}}{Y_{2}}\right]^{1 / 2}=\left[\frac{7 \times 10^{10}}{12 \times 10^{10}}\right]^{1 / 2}$
$\Rightarrow \quad r_{2}=1.5 \times\left(\frac{7}{12}\right)^{1 / 2}=1.145 \mathrm{~mm}$
$\therefore$ diameter $=2.29 \mathrm{~mm}$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.