In the previous problem if `t` is the time period of rotation
(i) `t=2pisqrt((L)/(g))`
(ii) `t=2pisqrt((Lcostheta)/(g))`
(iii) `T=(4pi^(2)mL)/(t^(2))`
(iv) The ball is in equilibrium

Options

A. `(i),(ii)`
B. `(ii),(iii)`
C. `(i),(iv)`
D. `(ii),(iv)`

Answer

B

Explanation

`T cos theta=mgimpliesT=mg//costheta`
`T sin theta=momega^(2)r=momega^(2)(Lsintheta)`
`T=momega^(2)L`
`(mg)/(costheta)=momega^(2)Limpliesomega=sqrt((g)/(Lcostheta))`
Time period `t=(2pi)/(omega)=2pisqrt((Lcostheta)/(g))`
`T=momega^(2)L=m((4pi^(2))/(t^(2)))L=(4pi^(2)mL)/(t^(2))`