A) 0.6 A
B) 0.5 A
C) 0.4 A
D) 0.3 A
Correct Answer: D
Solution :
In a tangent galvanometer, \[\tan \,\theta \propto I\] \[\therefore \] \[\frac{\tan \,{{\theta }_{1}}}{\tan {{\theta }_{2}}}=\frac{{{I}_{1}}}{{{I}_{2}}}\] \[\frac{\tan \,{{30}^{o}}}{\tan {{60}^{o}}}=\frac{0.1}{{{I}_{2}}}\] \[\frac{(1/\sqrt{3})}{\sqrt{3}}=\frac{0.1}{{{I}_{2}}}\] \[{{I}_{2}}=0.3A\]
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