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 ressistance. The deflections produced in the tangent galvanometers A and B are\[30{}^\circ \] and\[60{}^\circ \]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 glavanometer \[I=\frac{2rH}{{{\mu }_{0}}N}\tan \theta \]        ...(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 \[\frac{r\tan \theta }{N}=\frac{{{\mu }_{0}}I}{2H}\] \[\therefore \] \[\frac{{{r}_{A}}\tan {{\theta }_{A}}}{{{N}_{A}}}=\frac{{{r}_{B}}\tan {{\theta }_{B}}}{{{N}_{B}}}\] \[\Rightarrow \] \[\frac{8\times 1}{\sqrt{3}\times 2}=\frac{16\times \sqrt{3}}{{{N}_{B}}}\] \[\therefore \] \[{{N}_{B}}=12\,turns\]