A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane is x5 where x = .
Answer
4 or 3.5
Solution
For a body rolling without slipping down a height h:
mgh = 12mv2 (1 + k2R2 ) v = 2gh1 + k2/R2
Ring: k2R2 = 1 vring = gh
Solid sphere: k2R2 = 25 vsphere = 10gh7
vringvsphere = 710 = 3.55
x = 3.5 (4 when rounded off to the nearest integer; both values are accepted)
Answer: 4 or 3.5