A heavy particle is tied to the end A of a string of length 1.6 m. its other end O is fixed. It revolves as a conical pendulum with the string making `60^(C)` with the vertical then `(g=10(m)/(s^(2)))`

Options

A. its period of revolution is `(2sqrt(2)pi)/(5)` sec.
B. The tension in the string is double the weight of the particle.
C. the velocity of the particle `=sqrt(24)(m)/(s)`
D. the centripetal acceleration of the particle is `10sqrt(3)(m)/(s^(2))`.

Answer

A, B, C, D

Explanation

`(Tsqrt(3))/(2)=(mv^(2))/((lsqrt((3)/(2))))` …(1)
`(T)/(2)=mg` … (2)
Hence `T=2mg`, so (B) holds
From (1) & (2) `V^(2)=(3g^(l))/(2)`
`becauseC=sqrt((3xx10xx1.6)/(2))`
`becauseV=sqrt(24)(m)/(s^(2))` So (C) hold
`a_(C)=V^(2)lr=(((3gl)/(2)))/(((lsqrt(3))/(2)))=10sqrt(3)`
`because` (D) holds
`t=(2pir)/(v)`
`t=(2sqrt(2)pi)/(5)because` (A) holds