A) 18 turns
B) 12 turns
C) 6 turns
D) 2 turns
Correct Answer: B
Solution :
Current in tangent galvanometer \[I=\frac{2\,rH}{{{\mu }_{0}}N}\tan \theta \] ... (i) Here, \[{{R}_{1}}\] and \[{{R}_{2}}\] are in parallel \[\therefore \] \[\frac{1}{{{R}_{net}}}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}\] \[=\frac{{{R}_{2}}+{{R}_{1}}}{{{R}_{1}}{{R}_{2}}}=\frac{8+8}{8\times 8}\] \[{{R}_{net}}=4\,\Omega \] Hence, \[I=\frac{V}{R}=\frac{4}{4}=1\,A\] From Eq. (i), we get \[\frac{r\,\tan \theta }{N}=\frac{{{\mu }_{0}}I}{2\,H}\] \[\therefore \] \[\frac{{{r}_{A}}\,\tan {{\theta }_{A}}}{{{N}_{A}}}=\frac{{{r}_{B}}\tan {{\theta }_{B}}}{{{N}_{B}}}\] \[\Rightarrow \] \[\frac{8\times 1}{\sqrt{3}\times 2}=\frac{16\times \sqrt{3}}{{{N}_{B}}}\] \[\therefore \] \[{{N}_{B}}=12\] turnsYou need to login to perform this action.
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