A) 4 units
B) 8 units
C) 16 units
D) 32 units
Correct Answer: B
Solution :
\[M=niA=ni(\pi {{r}^{2}})\] \[\therefore \] \[\frac{{{M}_{1}}}{{{M}_{2}}}=\frac{{{i}_{1}}{{r}_{1}}^{2}}{{{i}_{2}}{{r}_{2}}^{2}}\] \[\Rightarrow \] \[{{M}_{2}}={{M}_{1}}\left( \frac{{{i}_{2}}{{r}_{2}}^{2}}{{{i}_{2}}{{r}_{1}}^{2}} \right)\] Given: \[{{i}_{2}}=\frac{{{i}_{1}}}{2}\]and\[{{r}_{2}}=2{{r}_{1}}\] \[\therefore \] \[{{M}_{2}}=4\frac{({{i}_{1}}/2){{(2{{r}_{1}})}^{2}}}{{{i}_{1}}{{r}_{1}}^{2}}=8\]
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