As $I$ and $H$ both have same units and dimensions, hence, $\chi$ has no dimensions. Here, in this question, $\chi$ is to be related with $e, m, v, R$ and $\alpha_0$. We know that dimensions of $\propto_0=\left[\mathrm{ML} \theta^{-2}\right]$

From Biot-Savart's law,

$$\begin{aligned} &\begin{aligned} & d B=\frac{\propto_0}{4 \pi} \frac{I d l \sin \theta}{r^2} \\ \Rightarrow \quad \propto_0 & =\frac{4 \pi r^2 d B}{I d l \sin \theta}=\frac{4 \pi r^2}{I d l \sin \theta} \times \frac{f}{q v \sin \theta} \quad \quad\left[\because d B=\frac{F}{q v \sin \theta}\right] \\ \therefore \quad \text { Dimensions of } \propto_0 & =\frac{\mathrm{L}^2 \times\left(\mathrm{ML}^{-2}\right)}{\left(\mathrm{QT}^{-1}\right)(\mathrm{L}) \times 1 \times(\mathrm{Q})\left(\mathrm{LT}^{-1}\right) \times(1)}=\left[\mathrm{MLQ}^{-2}\right] \end{aligned}\\ &\text { where } Q \text { is the dimension of charge. } \end{aligned}$$

As $\chi$ is dimensionless, it should have no involvement of charge $Q$ in its dimensional formula. It will be so if $\alpha_0$ and $e$ together should have the value $\alpha_0 e^2$, as e has the dimensions of charge.

Let $$\chi=\alpha_0 e^2 m^a v^b R^c\quad\text{.... (i)}$$

where $a, b, c$ are the power of $m, v$ and $R$ respectively, such that relation (i) is satisfied.

Dimensional equation of (i) is

$$\begin{aligned} {\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \mathrm{Q}^0\right] } & =\left[\mathrm{MLQ}^{-2}\right] \times\left[\mathrm{Q}^2\right]\left[\mathrm{M}^2\right] \times\left(\mathrm{LT}^{-1}\right)^b \times[\mathrm{L}]^c \\ & =\left[\mathrm{M}^{1+a}+\mathrm{L}^{1+b+c} \mathrm{~T}^{-b} \mathrm{Q}^0\right] \end{aligned}$$

Equating the powers of $M, L$ and $T$, we get

$$\begin{aligned} & 0=1+a \Rightarrow a=-1,0=1+b+c \quad\text{.... (ii)}\\ & 0=-b \Rightarrow b=0,0=1+0+c \text { or } c=-1 \end{aligned}$$

Putting values in Eq. (i), we get

$$\begin{aligned} \chi & =\propto_0 e^2 m^{-1} V^2 R^{-1}=\frac{\alpha_0 e^2}{m R} \quad\text{.... (iii)}\\ \text{Here,}\quad\propto_0 & =4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{~A} A^{-1} \\ e & =1.6 \times 10^{-19} \mathrm{C} \\ m & =9.1 \times 10^{-31} \mathrm{~kg}, R=10^{-10} \mathrm{~m} \\ \chi & =\frac{\left(4 \pi \times 10^{-7}\right) \times\left(1.6 \times 10^{-19}\right)^2}{\left(9.1 \times 10^{-31}\right) \times 10^{-10}} \approx 10^{-4} \\ \therefore\quad\frac{\chi \quad}{\chi_{\text {(given solid) }}} & =\frac{10^{-4}}{10^{-5}}=10 \end{aligned}$$

(a) $$\begin{aligned} B_V & =\frac{\alpha_0}{4 \pi} \frac{2 m \cos \theta}{r^3} \quad\text{.... (i)}\\ B_H & =\frac{\alpha_0}{4 \pi} \frac{\sin \theta m}{r^3}\quad\text{.... (ii)} \end{aligned}$$

Squaring both the equations and adding, we get

$$\begin{aligned} B_V^2+B_H^2 & =\left(\frac{\alpha_0}{4 \pi}\right) \frac{m^2}{r^6}\left[4 \cos ^2 \theta+\sin ^2 \theta\right] \\ B & =\sqrt{B_V^2+B_H^2}=\frac{\alpha_0}{4 \pi} \frac{m}{r^3}\left[3 \cos ^2 \theta+1\right]^{1 / 2}\quad\text{... (iii)} \end{aligned}$$

From Eq. (iii), the value of $B$ is minimum, if $\cos \theta=\frac{\pi}{2}$

$\theta=\frac{\pi}{2}$. Thus, the magnetic equator is the locus.

(b) Angle of dip,

$$\begin{aligned} & \tan \delta=\frac{B_V}{B_H}=\frac{\frac{\alpha_0}{4 \pi} \cdot \frac{2 m \cos \theta}{r^3}}{\frac{\alpha_0}{4 \pi} \cdot \frac{\sin \theta \cdot m}{r^3}}=2 \cot \theta \quad\text{.... (iv)}\\ & \tan \delta=2 \cot \theta \end{aligned}$$

For dip angle is zero i.e., $\delta=0$

$$\begin{aligned} \cot \theta & =0 \\ \theta & =\frac{\pi}{2} \end{aligned}$$

It means that locus is again magnetic equator.

(c) $\tan \delta=\frac{B_V}{B_H}$

Angle of dip i.e., $\delta= \pm 45$

$$\begin{aligned} \frac{B_V}{B_H} & =\tan ( \pm 45\Upsilon) \\ \frac{B_V}{B_H} & =1 \\ 2 \cot \theta & =1 \quad\text{[From Eq. (iv)]}\\ \cot \theta & =\frac{1}{2} \\ \tan \theta & =2 \\ \Rightarrow\quad\theta & =\tan ^{-1}(2) \end{aligned}$$

Thus, $\theta=\tan ^{-1}(2)$ is the locus.

$P$ is in the plane $S$, needle is in north, so the declination is zero.

$P$ is also on the magnetic equator, so the angle of dip $=0$, because the value of angle of dip at equator is zero. $Q$ is also on the magnetic equator, thus the angle of dip is zero. As earth tilted on its axis by $11.3^{\circ}$, thus the declination at $Q$ is $11.3^{\circ}$.

$C_1=$ circular coil of radius $R$, length $L$, number of turns per unit length

$$n_1=\frac{L}{2 \pi R}$$

$C_2=$ square of side $a$ and perimeter $L$, number of turns per unit length $n_2=\frac{L}{4 a}$

Magnetic moment of $C_1$

$$\Rightarrow \quad m_1=n_1 I A_1$$

Magnetic moment of $C_2$

$$\Rightarrow \quad m_2=n_2 I A_2$$

$$ \begin{aligned} m_1 & =\frac{L \cdot I \cdot \pi R^2}{2 \pi R} \\ m_2 & =\frac{L}{4 a} \cdot I \cdot a^2 \\ m_1 & =\frac{L I R}{2} \quad\text{.... (i)}\\ m_2 & =\frac{L I a}{4}\quad\text{.... (ii)} \end{aligned}$$

Moment of inertia of $C_1 \Rightarrow I_1=\frac{M R^2}{2}\quad\text{.... (iii)}$

Moment of inertia of $\mathrm{C}_2 \Rightarrow I_2=\frac{\mathrm{Ma}^2}{12}\quad\text{.... (iv)}$

Frequency of $C_1 \Rightarrow f_1=2 \pi \sqrt{\frac{I_1}{m_1 B}}$

Frequency of $C_2 \Rightarrow f_2=2 \pi \sqrt{\frac{I_2}{m_2 B}}$

According to question, $f_1=f_2$

$$\begin{aligned} 2 \pi \sqrt{\frac{I_1}{m_1 B}} & =2 \pi \sqrt{\frac{I_2}{m_2 B}} \\ \frac{I_1}{m_1} & =\frac{I_2}{m_2} \text { or } \frac{m_2}{m_1}=\frac{I_2}{I_1} \end{aligned}$$

Plugging the values by Eqs. (i), (ii), (iii) and (iv)

$$\begin{aligned} \frac{L I a \cdot 2}{4 \times L I R} & =\frac{M a^2 \cdot 2}{12 \cdot M R^2} \\ \frac{a}{2 R} & =\frac{a^2}{6 R^2} \\ 3 R & =a \end{aligned}$$

Thus, the value of $a$ is $3 R$.