A partical moves from rest at A on the surface of a smooth circule cylinder of radius `r` as shown . At `B` it leavels the cylinder. The equation releaseder . The equation relating `alpha and beta` is

Options

A. `"3 sin"alpha="2 cos"beta`
B. `"2 sin"alpha="3 cos"beta`
C. `"3 sin"beta="2 cos"alpha`
D. `"2 sin"beta="3 cos"alpha`

Answer

C

Explanation

`h_(AB)=(rcos alpha-rsinbeta)`
Velocity of particle at B,
`v=sqrt(2gh_(AB))=sqrt(2g(r cos alpha -r sin beta))`
Particle will leave contact at B, if component of weight is just equal to centripetal force (toward centre).
or `mg sinbeta(mv^(2))/(r)"or sin"beta=2 cosalpha=2sinbeta`
`therefore" "3sinbeta=2cosalpha`