Search any question & find its solution
Question:
Answered & Verified by Expert
A load of $1 \mathrm{~kg}$ weight is a attached to one end of a steel wire of area of cross-section $3 \mathrm{~mm}^2$ and Young's modulus $10^{11} \mathrm{~N} / \mathrm{m}^2$. The other end is suspended vertically from a hook on a wall, then the load is pulled horizontally and released. When the load passes through its lowest position the fractional change in length is $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$
Options:
Solution:
1992 Upvotes
Verified Answer
The correct answer is:
$0.3 \times 10^{-4}$
$\begin{aligned} Y & =\frac{m g l}{A \Delta l} \\ \Rightarrow \quad \frac{\Delta l}{l} & =\frac{m g}{A Y}\end{aligned}$
$\begin{aligned} \therefore \quad \frac{\Delta l}{l} & =\frac{1 \times 10}{3 \times 10^{-6} \times 10^{11}} \\ & =0.3 \times 10^{-4}\end{aligned}$
$\begin{aligned} \therefore \quad \frac{\Delta l}{l} & =\frac{1 \times 10}{3 \times 10^{-6} \times 10^{11}} \\ & =0.3 \times 10^{-4}\end{aligned}$
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.