MHTCET 2016 Physics Solved Paper

(b) Moment of inertia of a thin rod whose axis is passing through its middle point and perpendicular to its length is given by
I=

ML2

12

.....(i)
In terms of radius of gyration, it can be written as
I=MK12 .....(ii)
Comparingequation (i) and (ii), we get
MK12=

ML2

12

⇒ K1=

L

2√3

.....(iii)
Moment of inertia of rod whose axis is passing through one of its end is given by
I=

ML2

3

.....(iv)
In terms of radius of gyration, it can be writtenas
I=MK22 .....(v)
Comparing equation (iv) and (v), we get

ML2

3

=MK22
⇒ K2=

L

√3

.....(vi)
Again taking the ratio of K1 and K2 from Eqs. (iii) and (vi),
We get

K1

K2

=

L×√3

2√3×L

=

1

2