A metallic rod of length $l$ , linear mass density $μ$ rotates about one of its end ‘O’ in a smooth horizontal plane with an angular velocity $ω$ about an end axis which is perpendicular to plane of rotation shown in figure. A transverse pulse generated at the free end P to reach the axis of rotation. (neglect the gravitational effect).

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The speed of the transverse pulse when it reaches the midpoint of sting with respect to ground is
$\frac{\sqrt{5} l ω}{\sqrt{8}}$

The tension in the string at a distance ‘x’ from the free end $\frac{μ [l^{2} - {(l - x)}^{2}]}{2} ω^{2}$

The speed of the transverse pulse just after generated at the free end point P with respect to ground is $l ω$

The time taken by the transverse pulse to reach the axis of rotation from free end is $\frac{π}{\sqrt{2} ω}$

answer is A, B, C, D.

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Detailed Solution

Tension $= \frac{μ (l^{2} - {(l - x)}^{2}) ω^{2}}{2}$
Transverse velocity of pulse w.r.t to the string $= ω \sqrt{\frac{l^{2} - {(l - x)}^{2}}{2}}$
Transverse velocity of pulse at the free and $= l ω$
Transverse velocity of pulse at the mid of string $= \sqrt{\frac{3 l^{2} ω^{2}}{8} + \frac{l^{2} ω^{2}}{4}} = \sqrt{\frac{5}{8}} l ω$

$\begin{array}{l} \frac{d x}{d t} = \frac{ω}{\sqrt{2}} \sqrt{l^{2} - (l - x)^{2}} \\ \int \frac{\sqrt{2} d x}{ω \sqrt{l^{2} - (l - x)^{2}}} = \int_{0}^{t} d t \\ t = \frac{π}{\sqrt{2} ω} \end{array}$