A stone tied to the end of a string 0.4 m long is whirled in a horizontal circle with a constant speed. If the stone makes 10 revolutions in 25 s, the magnitude of acceleration is :

Answer

B

Explanation

Time of flight `4=(2u sin theta)/(g cos 60^(@))` …(i)
(angle of projection `=theta`)
Distance travelled by Q on incline in 4 secs is
`=0+1/2xx(sqrt(3)g)/2xx4^(2)=40sqrt(3)`
& the range of particle 'P' is `40 sqrt(3)`
`=u cos thetaxx4+1/2(sqrt(3)g)/2xx4^(2)=40sqrt(3)`
`=u cos theta=0,` so `theta=90^(@)`
from equation (i) `u=10 m//s`