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A wheel of radius r rolls without slipping on the ground with speed v. When it is at point p, a piece of mud flies off tangentially from its highest point, and lands on the ground at a point Q. The distance PQ is

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

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

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

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

answer is C.

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Detailed Solution

$t = \sqrt{\frac{2 \times 2 r}{g}}$

$PQ = 2 v \times \sqrt{\frac{4 r}{g}} = 4 v \sqrt{\frac{r}{g}}$

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