NTA Abhyas JEE Main2020PhysicsRotational MotionPractice
Line PQ is parallel to y -axis and moment of inertia of a rigid body about P Q line is given by I = 2 x 2 - 12 x + 27 , where x is in meter and I is in k g m 2 . The minimum value of I is:
Options
- A27 k g m 2
- B11 k g m 2
- C17 k g m 2
- D9 k g m 2
Correct answer
D. 9 k g m 2
Step-by-step solution
I = 2 x 2 - 12 x + 27 d I d x = 4 x - 12 For minimum or maximum d I d x = 0 ⇒ x = 3 d 2 I d x 2 = 4. Hence I is minimum at x = 3 I = 9 k g m 2