0 votes
in Kinematics by (12.2k points)
From a circular ring of mass \(^{\prime} M^{\prime}\) and radius \(^{\prime} R^{\prime}\) 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 ' \(K^{\prime}\) times ' \(M R^{2}\). Then the value of \(^{\prime} K^{\prime}\) is :
(A) \(\frac{3}{4}\)
B \(\frac{7}{8}\)
c \(\frac{1}{4}\)
D \(\frac{1}{8}\)

3 Answers

0 votes
by (12.2k points)
(A) \(\frac{3}{4}\)
0 votes
by
order tadalafil 10mg online cheap <a href="https://ordergnonline.com/">buy generic cialis 5mg</a> ed remedies
0 votes
by
cialis tadalafil 40mg <a href="https://ordergnonline.com/">tadalafil without prescription</a> where can i buy ed pills

Related questions

...