JEE Mains 2020 Model Paper 1 New Pattern

A small spherical ball is released from a point at a height h on rough track and the sphere does not slip. Therefore, potential energy it has gained with respect to the surface will be converted to angular kinetic energy about the centre and linear kinetic energy.

Therefore mgh =

1

2

Iω2+

1

2

m2
⇒ mgh =

1

2

×

2

5

mω2R2+

1

2

mv2
⇒ gh =

1

5

v2+

1

2

v2
⇒ v2 = (

10

7

)gh ⇒ v = √(

10gh

7

)