A circular table of radius `0.5 m` has a smooth diametrical groove. A half of mass `90g` is is placed inside the groove along with a spring constant `10^(2)N cm^(-1)` One end of the spring is tied to the table and the other end to the ball The ball is at a distance of `.0.1 m` from the center when the table is at rest On rotating the table with a constant angular frequency of `10^(2)(rad)/s^(-1)` the ball moves away from the center by a distance nearly equal to

Options

A. `10^(-1) m`
B. `10^(-2) m`
C. `10^(-3) m`
D. `2 xx10^(-1) m`

Answer

b

Explanation

`K = 10^(2)N cm^(-1) = 10^(4)Nm^(-1)`
Let the ball move distance x along from the center as shown in figure

`f le f_(1) rArr m omega^(2) r le mu mg`
`rArr mu ge (omega^(2)r)/(g) = (2^(2) xx 50//100)/(10)`
`rArr mu ge 0.2`