A body of mass \( m_{1} \mathrm{~kg} \), is placed on a rough horizontal table, the coefficient of kinetic friction between them is \( \mu_{\mathrm{k}} \) the body connected by a light string passing over a smooth pulley fixed at the edge of the table and another body of mass \( m_{2} \mathrm{~kg} \). is hanging vertically at the other end of the string, if the system moves with uniform velocity, then the coefficient of kinetic friction between the body and the table \( =\ldots \ldots \ldots \ldots \ldots \). \( \begin{array}{llll}\text { (a) } \frac{m_{1}}{m_{2}} & \text { (b) } m_{2}-m_{1} & \text { (c) } \frac{m_{2}}{m_{1}} & \text { (d) } m_{1} m_{2}\end{array} \)

**Step 1: Analyze the forces on \( m_1 \) (on the table)** For the body on the table, the forces acting horizontally are: - The tension \( T \) in the string pulling the body. - The kinetic friction force \( f_k \) opposing the motion, where \[ f_k = \mu_k m_1 g. \] Since the system moves with uniform velocity, the net force is zero, so: \[ T - \mu_k m_1 g = 0 \quad \Longrightarrow \quad T = \mu_k m_1 g. \] **Step 2: Analyze the forces on \( m_2 \) (hanging vertically)** For the hanging body, the forces are: - The gravitational force \( m_2 g \) downward. - The tension \( T \) in the string upward. Since the velocity is uniform (i.e., no acceleration), we have: \[ m_2 g - T = 0 \quad \Longrightarrow \quad T = m_2 g. \] **Step 3: Equate the expressions for \( T \)** From the equations for \( m_1 \) and \( m_2 \): \[ \mu_k m_1 g = m_2 g. \] Cancel \( g \) (assuming \( g \neq 0 \)): \[ \mu_k m_1 = m_2 \quad \Longrightarrow \quad \mu_k = \frac{m_2}{m_1}. \] **Answer:** The coefficient of kinetic friction is \[ \mu_k = \frac{m_{2}}{m_{1}}, \] which corresponds to option (c).