A) \[\frac{\sqrt{19}}{2}R\]
B) \[\frac{\sqrt{17}}{2}R\]
C) \[\frac{\sqrt{15}}{2}R\]
D) \[\frac{\sqrt{13}}{2}R\]
Correct Answer: C
Solution :
\[{{I}_{1}}=\frac{M{{R}^{2}}}{4}+M{{(2R)}^{2}}\] \[=\frac{M{{R}^{2}}}{4}+4M{{R}^{2}}\] \[=\frac{17\,M{{R}^{2}}}{4}\] \[{{I}_{2}}=\frac{M{{R}^{2}}}{2}+M{{d}^{2}}\] Given, \[{{I}_{1}}={{I}_{2}}\] \[\therefore \] \[\frac{17\,M{{R}^{2}}}{4}=\frac{M{{R}^{2}}}{2}+M{{d}^{2}}\] \[\Rightarrow \] \[M{{d}^{2}}=\frac{17}{4}M{{R}^{2}}-\frac{M{{R}^{2}}}{2}\] \[\Rightarrow \] \[{{d}^{2}}=\frac{15}{4}{{R}^{2}}\] \[\therefore \] \[d=\sqrt{\frac{15}{2}}R\]You need to login to perform this action.
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