A) \[\frac{64}{48}m{{a}^{2}}\]
B) \[\frac{91}{48}m{{a}^{2}}\]
C) \[\frac{27}{48}m{{a}^{2}}\]
D) \[\frac{51}{48}m{{a}^{2}}\]
Correct Answer: B
Solution :
[b] Moment of inertia \[=m{{\left( \frac{3a}{4} \right)}^{2}}+{{m}_{1}}\frac{{{a}^{2}}}{3}\] For the centre of rod \[\left( \frac{{{m}_{1}}{{a}^{2}}}{12}+\frac{{{m}_{1}}{{a}^{2}}}{4} \right)=\frac{{{m}_{1}}{{a}^{2}}}{3}\] \[\therefore \,\,\,{{m}_{1}}=4m\] Total \[I=m{{\left( \frac{3a}{4} \right)}^{2}}+\frac{4m{{a}^{2}}}{3}\] \[=\frac{9m{{a}^{2}}}{16}+\frac{4m{{a}^{2}}}{3}\] \[=\frac{(27+64)}{48}m{{a}^{2}}=\frac{91}{48}m{{a}^{2}}\]You need to login to perform this action.
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