Answer:
Here,\[F=mg+mr{{\omega }^{2}}=mg+mr4{{\pi
}^{2}}{{v}^{2}}=\] liter = \[=14\cdot 5\times 9\cdot 8+14\cdot 5\times 1\times
4\times {{\left( 22/7 \right)}^{2}}\times {{2}^{2}}\]
\[=142\cdot
1+2291\cdot 6=2433\cdot 7N\]
\[Y=\frac{F}{A}\times
\frac{l}{\vartriangle l}\] or \[\vartriangle l=\frac{Fl}{AY}\]
\[=\frac{2433\cdot 7\times 1}{\left( 0\cdot 065\times
{{10}^{-4}} \right)\times \left( 2\times {{10}^{11}} \right)}\]
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