AP EAMCET · PHYSICS · Laws of Motion
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
- 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 \(R\left[\frac{1}{\sqrt{1-\mu^2}}\right]\)
Answer & Solution
Correct Answer
(A) \(R\left[1-\frac{1}{\sqrt{1+\mu^2}}\right]\)
Step-by-step Solution
Detailed explanation
At maximum height, \(\begin{aligned} & \tan \theta=\mu \\ & \therefore \quad H=R(1-\cos \theta) \\ & =R\left(1-\frac{1}{\sqrt{1+\tan ^2 \theta}}\right)=R\left(1-\frac{1}{\sqrt{1+\mu^2}}\right) \end{aligned}\)
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