OnlineCBT Logo OnlineCBT

Physics · PYQ NEET UG 2021

हिन्दी

From a circular ring of mass 'M' and radius 'R' an arc corresponding to a \( 90^\circ \) sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is 'K' times 'MR^2'. Then the value of 'K' is:

  1. A
    \( \frac{7}{8} \)
  2. B
    \( \frac{1}{4} \)
  3. C
    \( \frac{1}{8} \)
  4. D
    \( \frac{3}{4} \)
View Answer & Solution

Correct answer · Option D

\( \frac{3}{4} \)

Explanation

Mass of remaining \( 270^\circ \) arc is \( M' = \frac{270}{360} M = \frac{3}{4} M \). Moment of inertia \( I = M' R^2 = \frac{3}{4} MR^2 \), so \( K = \frac{3}{4} \).
Address Contact

+91 7877547686

E-mail

onlinecbtportal@gmail.com

Helpline Number

+91 7877547686


Click To Download
Get it on Google Play

ऐप डाउनलोड करने
के लिए Google Play पर
उपलब्ध है

WhatsApp Chat