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A person with his hands in his pocket is skating on ice at the rate of $10 {ms}^{- 1}$ and describes a circle of radius 50 m. His inclination to the vertical is $(g = 10 {ms}^{- 2})$

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$\tan^{- 1} (\frac{1}{2})$

$\tan^{- 1} (\frac{1}{5})$

$\tan^{- 1} (\frac{3}{5})$

$\tan^{- 1} (\frac{1}{10})$

answer is B.

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Detailed Solution

$v = 10 m/s, r = 50 m, g = 10 {m/s}^{2}, θ = ?$ $θ = \tan^{- 1} (\frac{v^{2}}{rg})$

$= \tan^{- 1} (\frac{10 \times 10}{50 \times 10})$

$θ = \tan^{- 1} (\frac{1}{5})$

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