1.

A bead of mass m is atached to a spring of natural length (3R)/(4) with other end of it fixed at O.The bead moves in a track of which part ABC is semicircular of radius R and part CDA varies from R to (R)/(2) and then again to R. The bead is in equilibrium at position D and starts moving downward. Find the ratio of normal reaction on the bead to the cetrifugal force at the bottom most position.

Answer»

2
`(1)/(2)`
`1,0`
`(1)/(3)`

Solution :In position D spring is in equilibrium. ltbr. `N+mg-F=0`. . . (1)
in bottom most position.
`N+f-mg=(MV^(2))/(2)`. . . .(2)
[since spring is ELONGATED]
where F=kx and `x=(3R)/(4)-(R)/(2)=(R)/(4)`in
position D
and `x=R-(3R)/(4)=(R)/(4)` in bottom position
From 1 and 2
`N+(N+mg)-mg=(mv^(2))/(r)`
`2N=(mv^(2))/(r)`
`(N)/(mv^(2)//r)=(1)/(2)`


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