A small body of mass 'm' is placed at the top of a smooth sphere of radius 'R' placed on a horizontal surface as shown, Now the sphere is given a constant horizontal acceleration `a_(0) = g`. The velocity of the body relative to the sphere at the moment when it loses contact with the sphere is

Options

A. `sqrt((gR)/(3))`
B. `sqrt((2gR)/(3))`
C. `sqrt((5gR)/(3))`
D. `sqrt((7gR)/(3))`

Answer

B

Explanation

Let v is the velocity of the body relative to the sphere when it loses contact with the sphere.
`mg cos theta - N -ma_(0) sin theta (mv^(2))/(R )`
To lose contact with the sphere, `N = 0`
`mg cos theta - ma_(0) sin theta = (mv^(2))/(R )`
Also, `mgR(1 - cos theta) + ma_(0)R sin heta = (1)/(2)mv^(2)`
`mg = (3mv^(2))/(2R)`
`therefore v = sqrt((2gR)/(3))`