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The moment of inertia of a uniform circular disc of radius R and mass M about an axis touching the disc at its diameter and normal to the disc
Options
(a) 1/2 MR²
(b) MR²
(c) 2/5 MR²
(d) 3/2 MR²
Correct Answer:
3/2 MR²
Explanation:
Moment of inertia of a uniform circular disc about an axis through its centre and perpendicular of its plane is,
Ic = (1/2) MR²
By the theorem of parallel axes,
Therefore, moment of inertia of a uniform circular disc about an axis touching the disc at its diameter and normal to the disc is I.
I = Ic + Mh² = (1/2) MR² + MR² = (3/2) MR².
Related Questions: - A disc of radius 0.1 m is rotating with a frequency 10 rev/sec in a normal magnetic
- A transistor is working in common emitter mode. Its amplification factor is 80
- A thin semicircular conducting ring (PQR) of radius r is falling with its plane
- The acceleration of an electron in an electric field of magnitude 50 V/cm,
- A mass m moves in a circle on a smooth horizontal plane with velocity v₀
Topics: Motion of system of Particles and Rigid Body
(73)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A disc of radius 0.1 m is rotating with a frequency 10 rev/sec in a normal magnetic
- A transistor is working in common emitter mode. Its amplification factor is 80
- A thin semicircular conducting ring (PQR) of radius r is falling with its plane
- The acceleration of an electron in an electric field of magnitude 50 V/cm,
- A mass m moves in a circle on a smooth horizontal plane with velocity v₀
Topics: Motion of system of Particles and Rigid Body (73)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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