Pinnaka, I dared to solve d 1st part of ur problem but I am not too sure about the answer. I think it should be (5/7)h.

Explanation:

Let the marble be released from a vertical height h (height of the bowl).

During the downward motion, it is rolling purely. so

    v=rw

 Let the mass be m.

Pot. E at h = K.E. at the bottom

So,  mgh = 1/2 mv² + 1/2 Iw²    (w is the angular speed at the bottom)

       mgh = 1/2 mw²r² + 1/2 X 2/5 mr²w²

         gh = 7/10 r²w²      

    i.e.,   r²w² = 10gh/7    .............................(i)

 

 Since no external torque acts on the upward journey so w remains same.

 

 After reaching max height H

  v becomes 0.

  So, K.E. possessed by the body = 1/2.2/5mr²w²

 So, K.E. lost = 7/10 mr²w² - 1/5 mr²w²  = 1/2 mr²w²

This lost energy is gained in d form of P.E.

So,

   mgH= 1/2 mr²w²

So, H= 1/2 . 10/7gh    (from (i))

      H = 5/7 h

 

Hey everyone out there, please check and tell me if I am correct or wrong.

Now I am sure that the solution of the 1st part is correct and I have solved the 2nd part also. In the 2nd case,  as the ascending surface is also frictionful, so the marble will be pure rolling and both its translational and rotational velocities will become zero when it reaches its max. height. So, all the K.E. is lost thus

 

              loss in K.E. = 7/10 mr²w² 

This is gained as P.E.

So,     mgH = 7/10 mr²w²

                  H = 7/10 . 10h/7

       So,   H= h