A uniform rope of length  $12m$ and mass 6 kg hangs vertically from a rigid support. A block of mass 2 kg is attached to free end of the rope. A transverse pulse of wavelength $0.06m$  is produced at the lower end of the rope what is wavelength of pulse when it reaches the top of the rope

The velocity of wave on the string is given by

                                     $V=\sqrt{\frac{T}{\mu }}$

${\displaystyle T}$ is the tension and $\mu$ is the mass per unit length. Since the tension in the string will increase as we move up the string (as the string has mass), therefore the velocity of wave will also increase. 

                        $$\begin{gathered} V\propto \sqrt{T} \\ \Rightarrow \frac{V_1}{V_2}=\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{2\times 9.8}{8\times 9.8}}=\frac{1}{2} \\ \Rightarrow V_2=2V_1 \end{gathered}$$

Since frequency remains the same $$\begin{gathered} V=n\lambda \\ \therefore V\propto \lambda \end{gathered}$$

                          $$\begin{gathered} {\lambda }_2=2{\lambda }_1 \\ \Rightarrow {\lambda }_2=2\times 0.06 \\ =0.12m \end{gathered}$$