Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

A steel wire $1.5\;m$ long and radius $1\;mm$ is attached with a load of $3 \;Kg$ at one end and the other end of he wire is fixed. It is wirled in a vertical circle with frequency $2\;Hz$. find the elongation of the wire when the weight is at the lowest position. $[Y=2 \times 10^{11} N/m^2 \quad g=10m/s^2]$

\[\begin {array} {1 1} (a)\;1.77 \times 10^{-3}\;m \\ (b)\;7.17 \times 10^{-3}\;m \\ (c)\;3.17 \times 10^{-7}\;m \\ (d)\;1.37 \times 10^{-7}\;m \end {array}\]

Can you answer this question?

1 Answer

0 votes
At the lowest position Let T be the tension in the wire.
$\qquad= mw^2r$----------(1)
Also $Y=\large\frac{T/A}{\Delta l/l}$-----------(2)
$\quad=m(w^2r +g)$
$\quad=3 \bigg( \large\frac{2 \pi}{2}\bigg)^2$$ \times 1.5 +10$
$T=4.5 \pi ^2+30$
From (2)
$2 \times 10^{11}= \large\frac{3 \pi ^2 +30}{\Large\frac{\pi (1 \times 10^{-3})^2}{\Delta l/1.5}}$
Solving we get
$\Delta l= 7.17 \times 10^{-3}\;m$
Hence b is the correct answer. 
answered Oct 4, 2013 by meena.p
edited Mar 14, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App