- Home
- Standard 11
- Physics
A steel wire $1.5\,m$ long and of radius $1\,mm$ is attached with a load $3\,kg$ at one end the other end of the wire is fixed it is whirled in a vertical circle with a frequency $2\,Hz$ . Find the elongation of the wire when the weight is at the lowest position $(Y = 2 \times 10^{11}\,N/m^2$ and $g = 10\,m/s^2)$
$1.77 \times 10^{-3}\,m$
$7.17 \times 10^{-3}\,m$
$3.17 \times 10^{-7}\,m$
$1.37 \times 10^{-7}\,m$
Solution
At lower most position
Tension in wire $\mathrm{T}=\mathrm{Mg}+\mathrm{m} \omega^{2} \ell$
$\Delta \ell=\frac{\mathrm{FL}}{\mathrm{AY}}=\frac{\left(\mathrm{Mg}+\mathrm{m} \omega^{2} \ell\right) \ell}{\mathrm{AY}}$
$=\frac{\left[3 \times 10+3 \times(2 \times 3.14 \times 2)^{2} \times 1.5\right] \times 1.5}{3.14 \times\left(10^{-3}\right)^{2} \times 2 \times 10^{11}}$
$=1.77 \times 10^{-3} \mathrm{m}$
