Answer
$1.36\times 10^{-7}m$
Work Step by Step
We can find the required distance follows:
As forces in x-axis are given as
$Tsin\theta=F_g$
and $Tcos\theta=mg$
Dividing above eq(1) by eq (2), we obtain:
$\frac{Tsin\theta}{Tcos\theta}=\frac{F_g}{mg}$
This simplifies to:
$tan\theta=\frac{Gm}{l^2g}$
We know that the decrease in length is $2Lsin\theta$ and for small angle $tan\theta\approx \space sin\theta$
Thus, decrease in length $=2L tan\theta$
We plug in the known values to obtain:
Decrease in length=$2\times 100\times \frac{6.67\times 10^{-11}\times 100}{(1)^2\times 9.8}=1.36\times 10^{-7}m$
