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

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At the lowest position Let T be the tension in the wire.

$T-mg=\large\frac{mv^2}{r}$

$\qquad= mw^2r$----------(1)

Also $Y=\large\frac{T/A}{\Delta l/l}$-----------(2)

$T=mw^2r+mg$

$\quad=m(w^2r +g)$

$\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.

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