Figure shows a smooth vertical circular track AB of radius R. A block slides along the surface AB when it is given a velocity equal to `sqrt(6gR)` at point A. The ratio of the force exerted by the track on the block at point A to that at point B is

Options

A. (a) `0.25`
B. (b) `0.35`
C. (c) `0.45`
D. (d) `0.55`

Answer

D

Explanation

`N_A+mgcos60^@=(mv_A^2)/(R)`
`N_A=(mv_A^2)/(R)-mgcos60^@`
`=6mg-0.5mg=5.5mg`
`1/2mv_A^2+mgR(1+cos 60^@)=1/2mv_B^2`
`v_A^2+2gR(1+1/2)=v_B^2`
`6gR+3gR=v_B^2`
`v_B^2=9gRimpliesv_B=sqrt(9gR)`
`N_B=mg+(mv_B^2)/(R)=10mg`