A bob is attached to one end of a string other end of which is fixed at ped `A`. The bob is taken to a position where string makes an angle of `30^(@)` with the horizontal. On the circular path of the bob in vertical plane there is a peg` 'B'` at a symmetrical position with respect to the position of release as shown in the figure. If `V_(c)` and `V_(a)` be the minimum speeds in clockwise and anticlokwise direction respectively, given to the bob in order to hit the peg `'B'` then ratio `V_(c) : V_(a)` is equal to `:`

Options

A. `1: 1`
B. `1 : sqrt(2)`
C. `1 : 2`
D. `1 : 4`

Answer

C

Explanation

For anti-clockwise motion, speed at the highest point should be `sqrt(gR)` Conserving energy at (1) & (2) :

`(1)/(2)mv_(a)^(2)="mg"(R)/(2)+(1)/(2)(gR)`
`rArr V_(a)^(2)=gR+gR=2gR`
`rArr V_(a)=sqrt(2gR)`
For clock-wise motion, the bob must have atleast that much speed initially, so that the string must not become loose any where until it reaches the peg B
At the initial position
`T = mg cos60^(@)=(mv_(0)^(2))/(R)`
`V_(c)`being the initial speed in clockwise direction

For `V_(min)` Put T =0 :
`rArr V = sqrt((gR)/(2))`
`rArr V_(C)//C_(A)=(sqrt((gR)/(2)))/(sqrt(2gR))=(1)/(2)`
`rArr V : V 1 =1:2`