Calculate the Moment of inertia about the axis shown in figure Column I Column II (A) Uniform Rod (P) 11 8 M R 2 (B) Uniform Semicircular Ring. Axis is perpendicular to plane of ring. [π = 22/7] (Q)

Question

Calculate the Moment of inertia about the axis shown in figure

Column I	Column II
(A) Uniform Rod	(P) $\frac{8 M R ^{2}}{11}$
(B) Uniform Semicircular Ring. Axis is perpendicular to plane of ring. [π = 22/7]	(Q) $\frac{M R ^{2}}{12}$
(C) Uniform Triangular Plate of mass M	(R) $\frac{13 M R ^{2}}{8}$
(D) Uniform disk of initial mass M from which circular portion of radius R is then removed. M.I. of remaining mass about axis which is perpendicular to plane of plate	(S) $\frac{M R ^{2}}{8}$

TestHub · Accepted Answer

(A) I = 3 1 ​ M R 2 sin 2 ( 3 0 ∘ ) I = 3 1 ​ M R 2 ( 2 1 ​ ) 2 = 12 M R 2 ​ → (Q) (B) I = I c m ​ + M ( π 2 R ​ ) 2 = M R 2 Using I = 2 M R 2 ( 1 − π 2 2 ​ ) I = 2 M R 2 ( 1 − 484 2 ⋅ 49 ​ ) ≈ 11 8 M R 2 ​ → (P) (C) Height h = R 2 3 ​ ​ . I = 6 M h 2 ​ = 6 M ​ ( R 2 ⋅ 4 3 ​ ) = 8 M R 2 ​ → (S) (D) I = I t o t a l ​ − I c a v i t y ​ I = [ 2 1 ​ ( 3 4 ​ M ) ( 2 R ) 2 ] − [ 2 1 ​ ( 3 1 ​ M ) R 2 + ( 3 1 ​ M ) R 2 ] I = 3 8 ​ M R 2 − 2 1 ​ M R 2 = 6 13 M R 2 ​ → (R)

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JEE Advanced 2027 (Class 12+)