A man is moving on a circular path of radius R on horizontal ground with constant angular velocity $ω .$ Rain is falling with constant velocity $4 {ms}^{- 1}$ making an angle 45⁰ with vertical, in the plane containing north-south direction. The magnitude of average velocity of rain with respect to man in one full rotation is

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$4 {ms}^{- 1}$

$\sqrt{4^{2} + ω^{2} R^{2} + \sqrt{2} ω R}$

$\sqrt{4^{2} + ω^{2} R^{2} - \sqrt{2} ω R}$

$(4 + ω R)$

answer is A.

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Detailed Solution

${\vec{v}}_{r m} = {\vec{v}}_{r} - {\vec{v}}_{m}$

$\frac{\int v_{r m} d t = \int {\vec{v}}_{r} d t}{\int d t} - \frac{\int {\vec{v}}_{m} d t}{\int d t}$

$< {\vec{v}}_{r m} >= {\vec{v}}_{r} - 0$

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