A heavy particle is tied to the end A of a string of the length`1.6` m. Its other end O is fixed. It revolves as a conical pendulum with the string making `60^(@)` with the horizontal. Then,

Options

A. its period of revolution is `(4pi)/(7)s`
B. the tension is the string is `(1)/(sqrt3)` times the weight of the particle
C. the speed of the particle is `2.8sqrt3ms^(-1)`
D. the centripetal acceleration of the particle is `(9.8)/(sqrt3)ms^(-2)`

Answer

D

Explanation

`R=1.6cos60^(@)=0.8 m`

`Tsin60^(@)="mg or T"(2mg)/(sqrt3)rArrTcos 60^(@)=(mv^(2))/(R)`
`therefore" "(v^(2))/(R)=(T)/(2m)=(g)/(sqrt3)=(9.8)/(sqrt3)ms^(-2) rArrv=sqrt((9.8)/(sqrt3)xx0.8)ms^(-1)`
Time period `=2pi(R)/(v)`