A wheel of radius ‘r’ rolls without slipping with a speed ‘v’ on a horizontal road. When it is at point ‘A’ on the road, a small lump of mud separates from the wheel at its highest point B and drops at point ‘C’ on the road. The distance AC will be

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4 $v \sqrt{\frac{r}{g}}$

$2 v \sqrt{\frac{r}{g}}$

$v \sqrt{\frac{3 r}{g}}$

$v \sqrt{\frac{r}{g}}$

answer is C.

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Detailed Solution

Horizontal velocity of the lump of mud will be

$V_{B} = V + Rω = 2 V$

R = (horizontal velocity)x(time of flight)

$= (2 V) \sqrt{\frac{2 h}{g}} = 2 V \sqrt{\frac{2 (2 r)}{g}} \Rightarrow R = 4 V \sqrt{\frac{r}{g}}$

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