A particle slides from rest from the topmost point of a vertical circle of radius `r` along a smooth chord making an angle `theta` with the vertical. The time of descent is .

Options

A. Least for `theta =0`
B. Maximum for `theta =0`
C. Least for `theta =45`
D. Independent of `theta`

Answer

D

Explanation

`a=g sin a= g sin (90^@ -theta)`
`l=2 Rcos theta`
Now using `s ut +(1)/(2) at^(2)`, we get
`l=0t +(1)/(2) g cos theta t^(2)`
`rArr 2R cos theta =(1)/(2) g cos theta t^(2) rArr t=2 sqrt (R)/(g)`
.
this is independent of `theta` .