A uniform rope of mass 0.1 ~kg and length 2.4 | vedclass

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(D) The speed of a transverse wave in a rope at a distance $x$ from the free end is given by $v = \sqrt{gx}$.Since $v = \frac{dx}{dt}$,we have $\frac{

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(D) The speed of a transverse wave in a rope at a distance $x$ from the free end is given by $v = \sqrt{gx}$.Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \sqrt{gx}$.Rearranging the terms,we get $dt = \frac{dx}{\sqrt{gx}}$.Integrating from $x=0$ to $x=l$,the total time $t$ is:$t = \int_{0}^{l} \frac{dx}{\sqrt{gx}} = \frac{1}{\sqrt{g}} [2\sqrt{x}]_{0}^{l} = 2\sqrt{\frac{l}{g}}$.Substituting the given values $l = 2.45 ~m$ and $g = 9.8 ~m/s^2$:$t = 2 \sqrt{\frac{2.45}{9.8}} = 2 \sqrt{\frac{1}{4}} = 2 	imes 0.5 = 1 ~s$.

$A$ uniform rope of mass $0.1 ~kg$ and length $2.45 ~m$ hangs from a rigid support. The time taken by the transverse wave formed in the rope to travel through the full length of the rope is (Assume $g=9.8 ~m/s^2$) (in $~s$)

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