A stone of mass 1 kg tied at the end of a string of length 1 m and is whirled in a verticle circle at a constant speed of `3 ms^(-1)`. The tension in the string will be 19 N when the stone is `(g=10 ms^(-1))`

Options

A. at the top most point on the vertical circle
B. at the bottom most point on the vertical circle
C. half way down
D. making an angle `30^(@)` with the vertical

Answer

B

Explanation

`T=(mv^(2))/(r)+mg cos theta`
`mg cos theta=T-(mv^(2))/(r)=19-(1xx9)/(1)=19-9`
`mg cos theta =10`
`therefore cos theta =(10)/(mg)=(10)/(1xx10)`
`cos theta =1`
`theta= cos ^(-1)(1)=0`
At the bottom