A) \[2M{{R}^{2}}\]
B) \[M{{R}^{2}}\]
C) \[\frac{M{{R}^{2}}}{2}\]
D) \[\frac{M{{R}^{2}}}{4}\]
Correct Answer: D
Solution :
Moment of inertia of thin circular disc about its centre\[{{I}_{zz}}=\frac{M{{R}^{2}}}{2}\] Now from the theorem of perpendicular axis Moment of inertia of the disc about \[x-\]axis or \[x-y-\]axis \[i.e.\] any diameter \[=\frac{{{I}_{zz}}}{2}=\frac{M{{R}^{2}}/2}{2}=\frac{M{{R}^{2}}}{4}\]
You need to login to perform this action.
You will be redirected in
3 sec
