KCET 2021 Physics Solved Paper

As we know, kinetic energy of an object in pure rolling motion is given as
E=

1

2

mv2cm(1+

k2

R2

)
where, k is radius of gyration and m is the mass of the object.
⇒

E

v2cm

=

1

2

[1+

k2

R2

] ...(i) [∵ Given, m=1‌kg]
From the given graph, substituting the value of

E

v2

in Eq. (i), we get

3

4

=

1

2

[1+

k2

R2

]
⇒

k2

R2

=

1

2

Since, for a
(a) Sphere,

k2

R2

=

2

5

(b) Ring,

k2

R2

=1
(c) Hollow cylinder,

k2

R2

=1
(d) Disc,

k2

R2

=

1

2

So, the given object is disc.