question_answer

Two tangent galvanometers A and B have coils of radii 8 cm and 16 cm respectively and resistance \[8\Omega \] each. They are connected in parallel with a cell of emf 4 V and negligible internal resistance. The deflections produced in the tangent galvanometers A and B are \[{{30}^{o}}\]and \[{{60}^{o}}\] respectively. If A has 2 turns, then B must have

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                 \[\frac{r\,\tan \theta }{N}=\frac{{{\mu }_{0}}I}{2\,H}\] \[\therefore \]  \[\frac{{{r}_{A}}\,\tan {{\theta }_{A}}}{{{N}_{A}}}=\frac{{{r}_{B}}\tan {{\theta }_{B}}}{{{N}_{B}}}\] \[\Rightarrow \]               \[\frac{8\times 1}{\sqrt{3}\times 2}=\frac{16\times \sqrt{3}}{{{N}_{B}}}\] \[\therefore \]  \[{{N}_{B}}=12\] turns