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(B) The speed of a transverse pulse at any point on the rope is given by $v = \sqrt{\frac{T(x)}{\mu}}$,where $T(x)$ is the tension at that point and $

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$A, B, D$

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(B) The speed of a transverse pulse at any point on the rope is given by $v = \sqrt{\frac{T(x)}{\mu}}$,where $T(x)$ is the tension at that point and $\mu$ is the linear mass density. $1$. For Pulse $1$ moving from $O$ to $A$,the tension at a distance $x$ from $O$ is $T(x) = (M + \mu x)g$. $2$. For Pulse $2$ moving from $A$ to $O$,the tension at a distance $x$ from $O$ is also $T(x) = (M + \mu x)g$. Since the tension profile along the rope is identical for both paths,the speed at any given point $x$ is the same for both pulses. Thus,the time taken $T_{OA} = T_{AO}$. Option $A$ is correct. $3$. Since the speed $v(x)$ is the same at any point $x$ for both pulses,option $B$ is correct. $4$. The speed of a transverse wave on a string is $v = \sqrt{T/\mu}$,which is independent of frequency and wavelength. Thus,option $D$ is correct. $5$. As the pulse moves from $O$ to $A$,the tension increases,so the speed $v$ increases. Since $v = f\lambda$ and the frequency $f$ remains constant,the wavelength $\lambda$ must increase. Thus,option $C$ is correct. Therefore,options $A, B, C,$ and $D$ are all correct.

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