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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. $[Y=2 \times 10^{11} N/m^2 \quad g=10m/s^2]$

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

1 Answer

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

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