A 4 kg stone attached at the end of a steel wire is being whirled at a constant speed $12 \mathrm{~ms}^{-1}$…
A 4 kg stone attached at the end of a steel wire is being whirled at a constant speed $12 \mathrm{~ms}^{-1}$ in a horizontal circle. The wire is 4 m long with a diameter 2.0 mm and young's modules of the steel is $2 \times 10^{11} \mathrm{Nm}^{-2}$. The strain in the wire is.
$2.3 \times 10^{-4}$
$2.3 \times 10^{-5}$
$4.6 \times 10^{-4}$
$6.9 \times 10^{-4}$
Solution
$\mathrm{M}=4 \mathrm{~kg}, \mathrm{v}=12 \mathrm{~m} / \mathrm{s}, \mathrm{l}=4 \mathrm{~m}, \mathrm{~d}=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$ $\mathrm{Y}=2 \times 10^{11} \mathrm{Nm}^{-2}$
Force developed in the wire,
$F=\frac{m v^2}{1}=\frac{4 \times 12 \times 12}{4}=144 N$
$\therefore$ The strain produced in the wire is
$\begin{aligned}
& \mathrm{E}=\frac{\mathrm{F}}{\mathrm{AY}}=\frac{4 \mathrm{~F}}{\pi \mathrm{~d}^2 \mathrm{Y}}=\frac{4 \times 144}{\pi \times\left(2 \times 10^{-3}\right)^2 \times 2 \times 10^{11}} \\
& =2.3 \times 10^{-4}
\end{aligned}$