Two tangent galvanometers have radii 7.5 cm and 10 cm, number of turns are 15 and 10 and resistances are $8\mathrm{\Omega } \mathrm{and} 12\mathrm{\Omega }$. They are joined in parallel in circuit. If deflection in one is ${60}^{\mathrm{o}}$, the deflection in second galvanometer is

$$\begin{gathered} {\mathrm{r}}_1=7.5 \mathrm{cm}, {\mathrm{r}}_2 = 10 \mathrm{cm} \\ {\mathrm{n}}_1=15, {\mathrm{n}}_2=10 \end{gathered}$$

$$\begin{gathered} {\mathrm{R}}_1=8, {\mathrm{R}}_2=12 \\ {\mathrm{\theta }}_1={60}^{\mathrm{o}}, {\mathrm{\theta }}_2=? \end{gathered}$$

$\frac{{\mathrm{\mu }}_0}{2}\frac{{\mathrm{n}}_1}{{\mathrm{r}}_1} \mathrm{x} {\mathrm{I}}_1=B_H\tan {\mathrm{\theta }}_1 .....\left(\mathrm{i}\right)$

$\frac{{\mathrm{\mu }}_0}{2}\frac{{\mathrm{n}}_2}{{\mathrm{r}}_2} \mathrm{x} {\mathrm{I}}_2=B_H\tan {\mathrm{\theta }}_2 .....\left(\mathrm{ii}\right)$

$\therefore \frac{\tan {\mathrm{\theta }}_1}{\tan {\mathrm{\theta }}_2}= \frac{{\mathrm{n}}_1}{{\mathrm{n}}_2} \mathrm{x} \frac{{\mathrm{r}}_2}{{\mathrm{r}}_1} \mathrm{x} \frac{{\mathrm{I}}_1}{{\mathrm{I}}_2} .....\left(\mathrm{iii}\right)$

But, ${\mathrm{I}}_1=\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2} \mathrm{x} \frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{p}}}$

${\mathrm{I}}_2=\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2} \mathrm{x} \frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{p}}}$

$\frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}=\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2} ......\left(\mathrm{iv}\right)$

From equation (iii), 

$\frac{\tan {\mathrm{\theta }}_2}{\tan {\mathrm{\theta }}_1}=\frac{10}{15} \mathrm{x} \frac{7.5}{10} \mathrm{x} \frac{8}{12}=\frac{1}{3}$

$\therefore \tan {\mathrm{\theta }}_2=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$

$\therefore {\mathrm{\theta }}_2={30}^{\mathrm{o}}$