A solid ball rolls down a parabolic path `ABC` from a height `h` as shown in figure. Portion `AB` of the path is rough while `BC` is smooth. How high will the ball climb in `BC`?

Options

A. `(3h)/5`
B. `(4h)/9`
C. `(5h)/7`
D. None of these

Answer

C

Explanation

At B, total kinetic energy `=mgh`
Here, `m=` mass of ball
The ratio of ratational to translational kinetic energy would be.
`(K_(R))/(K_(T))=(2)/(5)impliesthereforeK_(R)=(2)/(7)mgh` and `K_(T)=(5)/(7)mgh`
In portiona BC, friction is absent. therefore, rotational kinetic energy will remain constant and translational kinetic energy will convert into potential energy. Hence, if `H` be the height to which ball climbs in BC, then
`mgH=K_(T)` or `mgH=(5)/(7)mgh` or `H=(5)/(7)h`