An insect is crawling in a hemi-spherical bowl of radius ' $R$ '. If the coefficient of friction between the insect and bowl is ' $\mu$ ', then the maximum height to which tile insect can crawl the bowl is

• Correct answer
  A

  $R\left[1-\frac{1}{\sqrt{1+\mu^2}}\right]$
• B

  $R\left[1+\frac{1}{\sqrt{1+\mu^2}}\right]$
• C

  $R\left[\frac{1}{\sqrt{1+\mu^2}}\right]$
• D

  $R\left[\frac{1}{\sqrt{1-\mu^2}}\right]$

Solution

At maximum height, $\begin{aligned} & \tan \theta=\mu \\ & \therefore \quad H=R(1-\cos \theta) \\ & =R\left(1-\frac{1}{\sqrt{1+\tan ^2 \theta}}\right)=R\left(1-\frac{1}{\sqrt{1+\mu^2}}\right) \end{aligned}$

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