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A circular loop of rope of length L rotates with uniform angular velocity $ω$ about an axis through its centre on a horizontal smooth platform. Velocity of pulse (with respect to rope) produced due to slight radial displacement is given by

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$ω L$

$\frac{ω L}{4 π^{2}}$

$\frac{ω L}{π}$

$\frac{ω L}{2 π}$

answer is B.

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Detailed Solution

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$2 T \sin \frac{d θ}{2} = d m . ω^{2} R$
$2 T \frac{d θ}{2} = \frac{m}{l} R d q . ω^{2} R$ ; $T = \frac{m ω^{2} R^{2}}{l}$
$∴ V = \sqrt{\frac{T}{μ}} = \sqrt{\frac{T}{m / l}} = ω R^{2} ∴ V = \frac{ω L}{2 π}$

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