A thin uniform metallic rod of length 0.5 m and radius 0.1 m with an angular velocity 400 rad s^-1 in a horizontal plane about a vertical axis passing through one of its ends. Elongation in the rod is (in m) [Given density of material of the rod is $10^{4}$ kg $m^{- 3}$ and Y = 2 X10¹¹ Nm^-2]

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$\frac{10^{- 3}}{3}$

$\frac{3}{10^{- 3}}$

$\frac{2}{10^{2}}$

$\frac{10^{2}}{2}$

answer is A.

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Detailed Solution

Consider an element of length dx at a distance x from the axis of rotation,
$\begin{array}{l} dF = {dmxω}^{2} \\ = ρAdx \cdot x \cdot ω^{2} \\ F = {ρAω}^{2} \int_{x}^{L} xdx \\ = \frac{1}{2} {ρAω}^{2} (L^{2} - x^{2}) \end{array}$

If dy is the elongation in the element of length dx, then
$\begin{array}{l} \frac{dy}{dx} = \frac{F}{AY} \\ dy = {\frac{Fdx}{A \cdot Y}|}_{Δl} = \frac{{ρω}^{2}}{2 Y} \int_{0}^{L} (L^{2} - x^{2}) dx \\ Δl = \frac{{ρω}^{2} \cdot L^{3}}{3 Y} \\ Δl = \frac{10^{4} \times (400)^{2} \times {(\frac{1}{2})}^{3}}{2 \times 10^{11}} = \frac{10^{- 3}}{3} m \end{array}$