A particle of mass `m` starts to slide down from the top of the fixed smooth sphere. What is the tangential acceleration when it break off the sphere ?

Options

A. `(2g)/3`
B. `(sqrt(5)g)/3`
C. g
D. g/3

Answer

B

Explanation

At loose contact N=0
`mg cos theta=(mv^(2))/R.....(1)`
from energy conservation
`mgR(1- cos theta)=1/2 mv^(2)......(2)`
from (1) and (2)
`cos theta=2/3rArr sin theta=(sqrt(5))/3 `
tangential acceleration `=g sin theta=(sqrt(5)g)/3`