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)`

Answer

0,75

Explanation

Acceleration of object rolling down the inclined plane is given by
`a=(mgsintheta)/(m+(I)/(R^(2)))`
`a_("ring")=(mgsintheta)/(m+(mR^(2))/(R^(2)))=(1)/(2)gsintheta`
`a_("disc")=(mgsintheta)/(m+(mR^(2))/(2R^(2)))=(2)/(3)gsintheta`
If h is the height of inclined plane, then length of the incline will be `l=h//sintheta`.
If t is the time taken to reach the bottom, then
`(h)/(sintheta)=(1)/(2)at^(2)impliest=sqrt((2h)/(asintheta))`
`impliest_("ring")sqrt((2h)/((gsintheta)/(2)sintheta))=sqrt((4h)/(10sin^(2)theta))`
`impliest_("disc")=sqrt((2h)/((2gsintheta)/(3)sintheta))=sqrt((3h)/(10sin^(2)theta))`
It is given that
`impliessqrt((4h)/(10sintheta))-sqrt((3h)/(10sin^(2)theta))=(2-sqrt3)/(sqrt10)`
`impliesh=sin^(2)theta=sin^(2)60=(3)/(4)=0.75m`