A) decrease by \[30{}^\circ \]
B) decrease by \[15{}^\circ \]
C) decrease by \[15{}^\circ \]
D) increase by \[30{}^\circ \]
Correct Answer: B
Solution :
\[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{\tan {{\theta }_{1}}}{\tan {{\theta }_{2}}}\] \[\frac{{{I}_{1}}}{\frac{{{I}_{1}}}{\sqrt{3}}}=\frac{\tan {{45}^{o}}}{\tan {{\theta }_{2}}}\] \[\sqrt{3}\tan {{\theta }_{2}}=1\] \[\Rightarrow \] \[\tan {{\theta }_{2}}=\frac{1}{\sqrt{3}}\] \[{{\theta }_{2}}={{\tan }^{-1}}\left( \frac{1}{\sqrt{3}} \right)={{30}^{o}}\] So, deflection will decrease by \[=45{}^\circ -30{}^\circ =15{}^\circ \]
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