On a disc fixed in a vertical plane, we have a string with its one end tied to the circumference of the disc. It is given that length of sting `L=2 piR`, where R is the raduis of the disc. Keeping the string horizontal, the particle of mass M is projected down with a velocity `v=sqrt((2gR))`. the tension in the string when it again becomes horizontal is

Options

A. `4 Mg//pi`
B. `6 Mg//pi`
C. `2 Mg//pi`
D. `Mg//pi`

Answer

B

Explanation

`mgxx2R=1/2 m(v^(2)-2gR)`
`4gR+2gR=v^(2)`
`T=(mv^(2))/(piR)=(6mgR)/(piR)`