A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle `60^(@)` with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is `(2 - sqrt(3)) //sqrt(10s)` 2– 3 / 10s, then the height of the top of the inclined plane, in metres, is _____________ . Take `g = 10ms^(-2)`

Explanation

`t=sqrt((2(h "cosec"theta))/a), a=(g sin theta)/(1+k^(2)/R^(2))` [K is radius of gyration)
`t_("ring") = sqrt((2h(1+1))/(g sin ^(2)theta)) = sqrt(h) 4/sqrt(30), t_("disc") = sqrt((2h)(1+1/2))/(g sin^(2)theta) = sqrt(h) (2sqrt(3))/(sqrt(30))`

`rArr Delta = t_("ring")-t_("disc") =(4-2sqrt(3))/sqrt(30) sqrt(h) = (2-sqrt(3))/sqrt(10) rArr 2/sqrt(3) sqrt(h)=1 rArr h=3/4 = 0.75`