A wheel of radius R is rolling on the ground without slipping with a speed u. When it reaches point A on the road a small lump of mud gets detached from its highest point B and drops on the ground at point D. What is the horizontal range AD of the mud?

Options

A. `usqrt(R//g)`
B. `4usqrt(R//g)`
C. `3usqrt(R//g)`
D. `sqrt((3R)/(g))`

Answer

B

Explanation

Here velocity of C.M. = u
`therefore` velocity of `B=u+u`
`=2u`
Also `h=2R` = height of B
Now `h=(1)/(2)g t^(2)ort=sqrt((2h)/(g))`
The horizontal range of the mud is
`X=2u.t`
`=2usqrt((2xx2R)/(g))`
`=4usqrt(R//g)`