A) \[\frac{16}{9}\]
B) 49 : 1
C) \[\frac{9}{16}\]
D) None of these
Correct Answer: C
Solution :
\[I\propto {{a}^{2}}\] \[\Rightarrow \frac{{{I}_{1}}}{{{I}_{2}}}={{\left( \frac{{{a}_{1}}}{{{a}_{2}}} \right)}^{2}}\]\[={{\left( \frac{3}{4} \right)}^{2}}=\frac{9}{16}\]
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