TS EAMCET 6-Aug-2021 Shift 1 Question Paper

Let initial moment of inertia of rod about perpendicular bisector,
I=‌

1

12

ML2. . . (i)
where, M= mass of rod
and L= initial length of rod.
After increasing temperature by ∆T.Now, moment of inertia,
I+∆I‌‌=‌

1

12

M(L+∆L)2
I(1+‌

∆I

I

)‌‌=‌

1

12

ML2(1+‌

∆L

L

)2
[‌ By thermal linear expansion, ‌∆L=Lα∆T
∴‌‌(1+‌

∆I

I

)‌‌=I(1+2‌

∆L

L

)
1+‌

∆I

I

‌‌=1+2α∆T
‌

∆I

I

‌‌=2α∆T