The bob of a `0.2` m pendulum describes an arc of circle in a vertical plane. If the tension in the cord is `sqrt3` times the weight of the bob when the cord makes an angle `30^(@)` with the vertical, the acceleration of the bob in that position is

Options

A. g
B. `(g)/(2)`
C. `(sqrt3g)/(2)`
D. `(g)/(4)`

Answer

A

Explanation

`a=sqrt(a_(n)^(2)+a_(t)^(2))=sqrt(((T-"mg cos 30"^(@))/(m))^(2)+("g sin 30")^(2))`
`=gsqrt((sqrt3-(sqrt3)/(2))^(2)+((1)/(4)))=g`