A) 1.5 A
B) 1 A
C) 500 mA
D) 866 mA
E) 2A
Correct Answer: A
Solution :
If an electric current \[I\] is flowing through the coil of the tangent galvanometer and the magnetic needdle gets deflected and comes to rest making an angle \[\theta \] with the direction of horizontal component of earths magnetic field H, then for the tangent galvanometer \[I=K\tan \theta \] where K is the reduction factor of tangent galvanometer. \[\therefore \] \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{\tan {{\theta }_{1}}}{\tan {{\theta }_{2}}}\] or \[\frac{500}{{{I}_{2}}}=\frac{\tan 30{}^\circ }{\tan 60{}^\circ }\] or \[\frac{500}{{{I}_{2}}}=\frac{\left( \frac{1}{\sqrt{3}} \right)}{\sqrt{3}}\] or \[\frac{500}{{{I}_{2}}}=\frac{1}{3}\] or \[{{I}_{2}}=1500\,mA=1.5A\]
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