A) Decrease by 30°
B) Decrease by 15°
C) Increase by 15°
D) Increase by 30°
Correct Answer: B
Solution :
In tangent galvanometer, \[I\propto \tan \theta \] \[\therefore \ \frac{{{I}_{1}}}{{{I}_{2}}}=\frac{\tan {{\theta }_{1}}}{\tan {{\theta }_{2}}}\Rightarrow \frac{{{I}_{1}}}{{{I}_{1}}/\sqrt{3}}=\frac{\tan {{45}^{0}}}{\tan {{\theta }_{2}}}\] \[\Rightarrow \sqrt{3}\tan {{\theta }_{2}}=1\Rightarrow \tan {{\theta }_{2}}=\frac{1}{\sqrt{3}}\Rightarrow {{\theta }_{2}}={{30}^{o}}\] So deflection will decrease by 45o- 30o = 15o.
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