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

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

`(1)/(2) ma_(a)^(2)=mg``(R)/(2)+(1)/(2)m(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 cos 60^(@)=(mv_(c)^(2))/(R),V_(C)` being the intial speed in clockwise direction.
For `V_(c mi n):`Put `T=0,`
`V_(C min): T=0`
`rArr V_(C)=ssqrt((gR)/(2))`
`rArr V_(C)//V_(a)=(sqrt((gR)/(2)))/(sqrt(2gR))=(1)/(2)` `Ans`.