NTA Abhyas JEE Main2020PhysicsRotational MotionPractice
An equilateral triangle ABC is cut from a thin solid sheet of wood. (See figure) D , E and F are the mid-points of its sides as shown and G is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane of the triangle is I 0 . If the smaller triangle DEF is removed from ABC , the moment of inertia of the remaining figure about the same axis is I .
Options
- AI = 3 4 I 0
- BI = 15 16 I 0
- CI = 9 16 I 0
- DI = I 0 4
Correct answer
B. I = 15 16 I 0
Step-by-step solution
Dimension analysis I o = K M a 2 Now for small lamina I ′ = K M 4 a 2 2 = k m a 2 16 I ′ = I o 16 So moment of Inertia of remaining part I L = I o - I ′ = I o - I o 16 I L = 15 I o 16