AP EAMCET2006PhysicsRotational Motion
The temperature of a thin uniform circular disc, of one metre diameter is increased by 10^ C . The percentage increase in moment of inertia of the disc about an axis passing through its centre and perpendicular to the circular face : (linear coefficient of expansion .=11 10⁻⁶ / ^ C )
Options
- A0.0055
- B0.011
- C0.022
- D0.044
Correct answer
C. 0.022
Step-by-step solution
Increase in area of disc A=A(2 ) t= (0.5)^2 (2 11 10⁻⁶ ) 10=0.000055 New area of the disc, A^ =A+ A A^ = (0.5)^2+0.000055 or r^ 2 =0.250055 m ^2 or r^ =0.500055 ~m Increase in moment of inertia. I^ -I I = (0.500055)^2-(0.5)^2 (0.5)^2 =0.00022=0.022 %