A) 185
B) 65
C) 285
D) 140
Correct Answer: D
Solution :
\[{{I}_{1}}=\frac{\left( \frac{7M}{8} \right){{(2R)}^{2}}}{2}=\left( \frac{7}{16}\times 4 \right)M{{R}^{2}}=\frac{7}{4}M{{R}^{2}}\] \[{{I}_{2}}=\frac{2}{5}\left( \frac{M}{8} \right)R_{1}^{2}=\frac{2}{5}\left( \frac{M}{8} \right)\frac{{{R}^{2}}}{4}=\frac{M{{R}^{2}}}{80}\] \[\frac{4}{3}\pi {{R}^{3}}=8\left( \frac{4}{3}\pi R_{1}^{3} \right)\] \[{{R}^{3}}=8R_{1}^{3}\] \[R=2{{R}_{1}}\] \[\therefore \]\[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{7/4M{{R}^{2}}}{\frac{M{{R}^{2}}}{80}}=\frac{7}{4}\times 80=140\]You need to login to perform this action.
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