A mass attached to one end of a string crosses top - most point on a vertical circle with critical speed. Its centripetal acceleration when string becomes horizontal will be
(where, g=gravitational acceleration)

Options

A. glancing angle
B. 3g
C. 4g
D. 6g

Answer

B

Explanation

We know that,
velocity of particle at top - most point on vertical circle,
`(v_(top))=sqrt(3rg)` . . . (i)
But centripetal acceleration `(a_(c))=(v^(2))/(r)`
`:." "a_(c)(sqrt(3rg)^(2))/(r)" "` [from Eq. (i)]
`a_(c)=(3rg)/(r)`
`a_(c)=3g`