A block of mass $m_{2}$ is placed on a horizontal table and another block of mass $m_{1}$ is placed on top of it. An increasing horizontal force \(F=\alpha t\) is exerted on the upper block but the lower block never moves as a result. If the coefficient of friction between the blocks is $\mu_{i}$ and that between the lower block and the table is $\mu_{2}$, then what is the maximum possible value of $\mu_{1} / \mu_{2} ?$

• A

  $\frac{m_{2}}{m}$
• Correct answer
  B

  $1+\frac{m_{2}}{m_{1}}$
• C

  $\frac{m_{1}}{m_{2}}$
• D

  $1+\frac{m_{1}}{m_{2}}$

Solution

According to the question. $\therefore \quad N_{2}=m_{2} g+N_{1}=\left(m_{1}+m_{2}\right) g$ : Lower block never moves. $\therefore \quad f_{r_{2}} \geq f_{n}$ $\quad \mu_{2} N_{2} \geq \mu_{1} N_{1} \Rightarrow \mu_{2}\left(m_{1}+m_{2}\right) g \geq \mu_{1} m_{1} g$ $\Rightarrow \quad \frac{m_{1}+m_{2}}{m_{1}} \geq \frac{\mu_{1}}{\mu_{2}}$ $\therefore \quad \frac{\mu_{1}}{\mu_{2}} \leq 1+\frac{m_{2}}{m_{1}}$ $\left.\therefore \quad \frac{\mu_{1}}{\mu_{2}}\right|_{\operatorname{man}}=1+\frac{m_{2}}{m_{1}}$

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