A) 51
B) 31
C) 41
D) 61
Correct Answer: D
Solution :
From theorem of parallel axis \[I=M{{R}^{2}}+{{I}_{0}}\] where\[{{I}_{0}}\]is moment of inertia through axis. Given, \[{{I}_{0}}=\frac{M{{R}^{2}}}{2}\] \[\therefore \] \[I=M{{R}^{2}}+\frac{M{{R}^{2}}}{2}=\frac{3}{2}M{{R}^{2}}\] ...(i) Also, moment of inertia of disc about a diameter is \[I=\frac{M{{R}^{2}}}{4}\] ?. (ii) From Eqs. (i) and (ii), we get \[\frac{I}{I}=\frac{3}{2}M{{R}^{2}}\times \frac{4}{M{{R}^{2}}}=6\] \[\Rightarrow \] \[I=6I\]You need to login to perform this action.
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