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NEET PHYSICSRotational motionMedium

Question

The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is:

A

1:21:\sqrt{2}

B

2:12:1

C

2:1\sqrt{2}:1

D

4:14:1

Step-by-Step Solution

Let the mass of the thin uniform disc be MM and its radius be RR. The moment of inertia of the disc about an axis passing through its centre and perpendicular to its plane is I1=12MR2I_1 = \frac{1}{2}MR^2. If K1K_1 is the corresponding radius of gyration, then MK12=12MR2    K1=R2MK_1^2 = \frac{1}{2}MR^2 \implies K_1 = \frac{R}{\sqrt{2}}. The moment of inertia of the disc about its diameter is I2=14MR2I_2 = \frac{1}{4}MR^2. If K2K_2 is the corresponding radius of gyration, then MK22=14MR2    K2=R2MK_2^2 = \frac{1}{4}MR^2 \implies K_2 = \frac{R}{2}. The ratio of their radii of gyration is: K1K2=R2R2=22=21=2:1\frac{K_1}{K_2} = \frac{\frac{R}{\sqrt{2}}}{\frac{R}{2}} = \frac{2}{\sqrt{2}} = \frac{\sqrt{2}}{1} = \sqrt{2}:1.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Rotational motion. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRotational motionradiusgyrationuniformpassingthrough

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