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Two tangent galvanometers $A$ and $B$ have coils of radii $8 \mathrm{~cm}$ and $16 \mathrm{~cm}$ respectively and having resistance of $8 \Omega$ each. They are connected in parallel with a cell of emf $4 \mathrm{~V}$ and negligible internal resistance. 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
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The correct answer is:
12 turns
The given situation is shown below
Current through tangent galvanometer and deflection given by it are related as,
$I=\frac{2 r}{\mu_0 N} \cdot B_H \cdot \tan \theta$
where, $r=$ radius of coil of galvanometer.
If $R=$ resistance of galvanometer, then potential drop across it is
$V=I R=\frac{2 r \cdot B_H \cdot \tan \theta \cdot R}{\mu_0 \cdot N}$
In given case, $V_A=V_B$
$\Rightarrow \frac{2 r_1 \tan \theta_1}{N_1}=\frac{2 r_2 \tan \theta_2}{N_2} \quad\left[\because R_1=R_2=R=10 \Omega\right]$
$N_2=\frac{N_1 \cdot r_2 \cdot \tan \theta_2}{r_1 \cdot \tan \theta_1}=\frac{2 \times 16 \times 10^{-2} \times \tan 60^{\circ}}{8 \times 10^{-2} \times \tan 30^{\circ}}$
$=2 \times 2 \times 3=12$ turns
Current through tangent galvanometer and deflection given by it are related as,
$I=\frac{2 r}{\mu_0 N} \cdot B_H \cdot \tan \theta$
where, $r=$ radius of coil of galvanometer.
If $R=$ resistance of galvanometer, then potential drop across it is
$V=I R=\frac{2 r \cdot B_H \cdot \tan \theta \cdot R}{\mu_0 \cdot N}$
In given case, $V_A=V_B$
$\Rightarrow \frac{2 r_1 \tan \theta_1}{N_1}=\frac{2 r_2 \tan \theta_2}{N_2} \quad\left[\because R_1=R_2=R=10 \Omega\right]$
$N_2=\frac{N_1 \cdot r_2 \cdot \tan \theta_2}{r_1 \cdot \tan \theta_1}=\frac{2 \times 16 \times 10^{-2} \times \tan 60^{\circ}}{8 \times 10^{-2} \times \tan 30^{\circ}}$
$=2 \times 2 \times 3=12$ turns
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