Two masses $m_1$ and $m_2$ are connected by a light string passing over smooth pulley. When set free $m_1$ moves downwards by 3 m in 3 s . The ratio of $\frac{m_1}{m_2}$ is $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$

• A

  $\frac{9}{7}$
• Correct answer
  B

  $\frac{8}{7}$
• C

  $\frac{10}{7}$
• D

  $\frac{15}{13}$

Solution

$a=\left(\frac{m_1-m_2}{m_1+m_2}\right) g$ Now, $S=u t+\frac{1}{2} a t^2$ $\Rightarrow 3=0+\frac{1}{2}\left(\frac{m_1-m_2}{m_1+m_2}\right) \times 10 \times(3)^2$ $\begin{aligned} & \Rightarrow \frac{\mathrm{m}_1-\mathrm{m}_2}{\mathrm{~m}_1+\mathrm{m}_2}=\frac{1}{15} \\ & \therefore \frac{\mathrm{~m}_1}{\mathrm{~m}_2}=\frac{8}{7}\end{aligned}$

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