A block $\text{A}$ of mass ${\mathrm{m}}_1$ rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block $\text{B}$ of mass ${\mathrm{m}}_2$ is suspended. The coefficient of kinetic friction between the block and the table is ${\mathrm{\mu }}_{\mathrm{k}}$. When the block $\text{B}$ is sliding on the table, the tension in string is:

• A

  $\frac{\left({\mathrm{m}}_2+{\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_1\right)g}{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}$
• B

  $\frac{\left({\mathrm{m}}_2-{\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_1\right)g}{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}$
• Correct answer
  C

  $\frac{{\mathrm{m}}_1{\mathrm{m}}_2\left(1+{\mathrm{\mu }}_{\mathrm{k}}\right)g}{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}\mathrm{ }$
• D

  $\frac{{\mathrm{m}}_1{\mathrm{m}}_2\left(1-{\mathrm{\mu }}_{\mathrm{k}}\right)g}{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}\mathrm{ }$

Solution

From the figure, \(\begin{aligned} & m_2 g-T=m_2 a \ldots \text { (i) } \\ & T-\mu_k m_1 g=m_1 a \ldots \text { (ii) } \\ & \text {multiply eqn (i) with } \mathrm{m}_1 \text { and eqn (ii) with } \mathrm{m}_2 \\ & m_1 m_2 g-T m_1=m_1 m_2 a \ldots \text { (iii) } \\ & T m_2-\mu_k m_1 m_2 g=m_1 m_2 a \ldots \text { (iv) } \\ & \text{From eqn (iii) and (iv), we get,} \\ & m_1 m_2 g-T m_1=T m_2-\mu_k m_1 m_2 g \\ & m_1 m_2 g\left(1+\mu_k\right)=T\left(m_1+m_2\right) \\ & \therefore T=\frac{m_1 m_2 g\left(1+\mu_k\right)}{m_1+m_2} \end{aligned}\)

Hence, the correct answer is option: 'C'.