To solve this problem, we need to understand how the wavelength of a transverse pulse changes as it travels along a rope with varying tension.$\\$The speed of a wave on a string is given by $v = \sqrt{\frac{T}{\mu}},$ where $T$ is the tension and $\mu$ is the linear mass density of the rope.$\\$The linear mass density $\mu$ is given by $\mu = \frac{m_1}{L}. \\$The tension at a point on the rope is due to the weight of the rope below that point and the block.$\\$At the bottom of the rope, the tension $T_{bottom} = m_2 g. \\$At the top of the rope, the tension $T_{top} = (m_1 + m_2) g. \\$The speed of the wave at the bottom of the rope $v_1$ is:$\\ v_1 = \sqrt{\frac{m_2 g}{\mu}} = \sqrt{\frac{m_2 g L}{m_1}}. \\$The speed of the wave at the top of the rope $v_2$ is:$\\ v_2 = \sqrt{\frac{(m_1 + m_2) g}{\mu}} = \sqrt{\frac{(m_1 + m_2) g L}{m_1}}. \\$The wavelength of the wave is related to its speed by $\lambda = \frac{v}{f},$ where $f$ is the frequency.$\\$Since the frequency of the wave remains constant as it travels up the rope, we have:$\\ \frac{\lambda_2}{\lambda_1} = \frac{v_2}{v_1}. \\$Substituting the expressions for $v_1$ and $v_2: \\ \frac{\lambda_2}{\lambda_1} = \frac{\sqrt{\frac{(m_1 + m_2) g L}{m_1}}}{\sqrt{\frac{m_2 g L}{m_1}}}. \\$Simplifying the expression:$\\ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{(m_1 + m_2) g L}{m_1}} \cdot \sqrt{\frac{m_1}{m_2 g L}}. \\ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{m_1 + m_2}{m_2}}. \\$Therefore, the ratio $\frac{\lambda_2}{\lambda_1}$ is $\sqrt{\frac{m_1 + m_2}{m_2}},$ which corresponds to Option 4.