Magnetic Equipment's (Tangent galvanometer, Vibration magnetometer and Neutral point) Questions in English

(C) The time period of oscillation of a magnet in a uniform magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B$ is the magnetic field.
Since $M = m \times 2l$ (where $m$ is pole strength and $2l$ is magnetic length),we have $T \propto \frac{1}{\sqrt{M}} \propto \frac{1}{\sqrt{m}}$.
Given that the pole strength $m$ becomes $4$ times,the new pole strength $m' = 4m$.
Therefore,the new time period $T'$ is related to the original time period $T$ as $T' = \frac{T}{\sqrt{4}} = \frac{T}{2}$.
Given $T = 2 \ s$,we get $T' = \frac{2}{2} = 1 \ s$.

(B) For a tangent galvanometer,the current $I$ is given by $I = K \tan \theta$,where $K$ is the reduction factor.
Since the coils are connected in series,the current $I$ is the same for both.
Therefore,$K_1 \tan \theta_1 = K_2 \tan \theta_2$.
Given $\theta_1 = 60^o$ and $\theta_2 = 45^o$,we have $K_1 \tan 60^o = K_2 \tan 45^o$.
$K_1 (\sqrt{3}) = K_2 (1) \Rightarrow \frac{K_1}{K_2} = \frac{1}{\sqrt{3}}$.
The reduction factor $K$ is given by $K = \frac{2 R B_H}{\mu_0 N}$,where $R$ is the radius and $N$ is the number of turns.
Since $R$ is the same for both,$K \propto \frac{1}{N}$,which implies $\frac{K_1}{K_2} = \frac{N_2}{N_1}$.
Thus,$\frac{N_2}{N_1} = \frac{1}{\sqrt{3}}$,which means $\frac{N_1}{N_2} = \frac{\sqrt{3}}{1}$.

(B) At the neutral point,the magnetic field produced by the current-carrying wire is equal and opposite to the horizontal component of the Earth's magnetic field $(B_H)$.
Given: Distance $R = 5 \ cm = 0.05 \ m$,$B_H = 0.18 \ G = 0.18 \times 10^{-4} \ T$.
The magnetic field due to a long straight wire is $B = \frac{\mu_0 I}{2\pi R}$.
Equating $B = B_H$,we get $\frac{\mu_0 I}{2\pi R} = B_H$.
Substituting the values: $\frac{2 \times 10^{-7} \times I}{0.05} = 0.18 \times 10^{-4}$.
$I = \frac{0.18 \times 10^{-4} \times 0.05}{2 \times 10^{-7}} = \frac{0.009 \times 10^{-4}}{2 \times 10^{-7}} = 0.0045 \times 10^3 = 4.5 \ A$.
Thus,the current in the wire is $4.5 \ A$.

(B) The time period of a vibration magnetometer is given by the formula $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B_H$ is the horizontal component of the Earth's magnetic field.
At the magnetic poles of the Earth,the magnetic field lines are vertical,meaning the horizontal component of the Earth's magnetic field $B_H$ is zero.
Substituting $B_H = 0$ into the formula,we get $T = 2\pi \sqrt{\frac{I}{M \times 0}} = 2\pi \sqrt{\infty} = \infty$.
Therefore,the time period becomes infinite.

(B) In a tangent galvanometer,the magnetic field $B_{m}$ produced by the coil (or magnet) is perpendicular to the horizontal component of the earth's magnetic field $B_{H}$.
The deflection $\theta$ is given by the tangent law: $B_{m} = B_{H} \tan \theta$.
Given:
$B_{H} = 0.34 \times 10^{-4} \text{ T}$
$\theta = 30^{\circ}$
Substituting the values:
$B_{m} = (0.34 \times 10^{-4}) \times \tan 30^{\circ}$
$B_{m} = (0.34 \times 10^{-4}) \times \frac{1}{\sqrt{3}}$
$B_{m} = \frac{0.34 \times 10^{-4}}{1.732}$
$B_{m} \approx 0.1963 \times 10^{-4} \text{ T}$
$B_{m} \approx 1.96 \times 10^{-5} \text{ T}$

(C) For a short bar magnet placed with its north pole pointing towards the geographic north, the neutral points lie on the equatorial line (East-West line).
At the neutral point, the magnetic field due to the magnet equals the horizontal component of the Earth's magnetic field $(B_H)$.
The formula for the magnetic field on the equatorial line of a short bar magnet is $B = \frac{\mu_0}{4\pi} \cdot \frac{M}{r^3}$.
Given: $r = 30 \, cm = 0.3 \, m$, $B_H = 3.6 \times 10^{-5} \, T$, and $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$.
Equating the fields: $10^{-7} \cdot \frac{M}{(0.3)^3} = 3.6 \times 10^{-5}$.
Solving for $M$: $M = \frac{3.6 \times 10^{-5} \times (0.3)^3}{10^{-7}}$.
$M = 3.6 \times 10^2 \times 0.027$.
$M = 360 \times 0.027 = 9.72 \, A \cdot m^2$.
Thus, the magnetic moment is approximately $9.7 \, A \cdot m^2$.

(A) The formula for the current $i$ in a tangent galvanometer is given by:
$i = \frac{2 r B_{H}}{\mu_{0} N} \tan \theta$
Where:
$r = 15 \times 10^{-2}\, m$ (radius of the coil)
$B_{H} = 3 \times 10^{-5}\, T$ (horizontal component of Earth's magnetic field)
$N = 25$ (number of turns)
$\theta = 45^{\circ}$ (deflection angle)
$\mu_{0} = 4 \pi \times 10^{-7}\, T\cdot m/A$ (permeability of free space)
Substituting the values:
$i = \frac{2 \times 15 \times 10^{-2} \times 3 \times 10^{-5}}{4 \pi \times 10^{-7} \times 25} \times \tan 45^{\circ}$
$i = \frac{90 \times 10^{-7}}{100 \pi \times 10^{-7}} \times 1$
$i = \frac{0.9}{\pi} \approx 0.2865\, A \approx 0.29\, A$

(C) The time period $T$ of a vibration magnetometer is given by $T = 2 \pi \sqrt{\frac{I}{M B_{H}}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B_{H}$ is the horizontal component of the Earth's magnetic field.
From the formula,we see that $T \propto \frac{1}{\sqrt{M}}$.
Therefore,the ratio of the time periods is $\frac{T_{1}}{T_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$.
Given that the magnetic moment is reduced by $19\%$,if $M_{1} = 100$,then $M_{2} = 100 - 19 = 81$.
Substituting these values,we get $\frac{T_{1}}{T_{2}} = \sqrt{\frac{81}{100}} = \frac{9}{10}$.
This implies $T_{2} = \frac{10}{9} T_{1} \approx 1.11 T_{1}$.
The percentage increase in the time period is $\frac{T_{2} - T_{1}}{T_{1}} \times 100 = (1.11 - 1) \times 100 = 11\%$.
Thus,the periodic time increases by $11\%$.

(B) The time period of a bar magnet oscillating in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B$ is the magnetic field.
For the original magnet of mass $m$ and length $L$,$I = \frac{mL^2}{12}$ and $M = m_s L$ (where $m_s$ is pole strength).
When cut into $3$ equal parts perpendicular to its length,each part has length $L' = L/3$ and mass $m' = m/3$.
The new moment of inertia is $I' = \frac{m'(L')^2}{12} = \frac{(m/3)(L/3)^2}{12} = \frac{I}{27}$.
The new magnetic moment is $M' = m_s L' = m_s (L/3) = M/3$.
The new time period $T'$ is $T' = 2\pi \sqrt{\frac{I'}{M'B}} = 2\pi \sqrt{\frac{I/27}{(M/3)B}} = 2\pi \sqrt{\frac{I}{9MB}} = \frac{1}{3} \times 2\pi \sqrt{\frac{I}{MB}} = \frac{T}{3}$.

(C) The time period of oscillation of a bar magnet in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B_H$ is the horizontal component of the Earth's magnetic field.
For a bar magnet of mass $m$ and length $l$,the moment of inertia is $I = \frac{ml^2}{12}$.
Since $M$ and $B_H$ remain constant,we have $T \propto \sqrt{I}$.
Since $I \propto m$,we get $T \propto \sqrt{m}$.
Let $T_0$ be the initial time period with mass $m$,and $T'$ be the new time period with mass $2m$.
$\frac{T'}{T_0} = \sqrt{\frac{2m}{m}} = \sqrt{2}$.
Therefore,$T' = \sqrt{2} T_0$.

(D) The time period $T$ of oscillation of a magnet is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B$ is the horizontal magnetic field.
Since $I$ and $M$ are constant,we have $T \propto \frac{1}{\sqrt{B}}$,which implies $\frac{T_2}{T_1} = \sqrt{\frac{B_1}{B_2}}$.
Given $B_1 = 24 \, \mu T$ and $T_1 = 2 \, s$.
The new magnetic field is $B_2 = B_1 - 18 \, \mu T = 24 \, \mu T - 18 \, \mu T = 6 \, \mu T$.
Substituting the values: $\frac{T_2}{2} = \sqrt{\frac{24}{6}} = \sqrt{4} = 2$.
Therefore,$T_2 = 2 \times 2 = 4 \, s$.

(D) The frequency of oscillation in a vibration magnetometer is given by $f = \frac{1}{2\pi} \sqrt{\frac{mB_H}{I}}$.
When like poles are together,the effective magnetic moment is $M_1 = m_1 + m_2$ and the frequency is $f_1 = 12 \text{ oscillations/min}$.
When unlike poles are together,the effective magnetic moment is $M_2 = m_1 - m_2$ and the frequency is $f_2 = 4 \text{ oscillations/min}$.
Since $f \propto \sqrt{M}$,we have $\frac{f_1}{f_2} = \sqrt{\frac{m_1 + m_2}{m_1 - m_2}}$.
Substituting the values: $\frac{12}{4} = \sqrt{\frac{m_1 + m_2}{m_1 - m_2}} \implies 3 = \sqrt{\frac{m_1 + m_2}{m_1 - m_2}}$.
Squaring both sides: $9 = \frac{m_1 + m_2}{m_1 - m_2}$.
$9m_1 - 9m_2 = m_1 + m_2 \implies 8m_1 = 10m_2$.
Therefore,$\frac{m_1}{m_2} = \frac{10}{8} = \frac{5}{4}$.

(D) For a tangent galvanometer,the current $I$ is given by $I = K \tan \theta$,where $K = \frac{2R B_H}{\mu_0 N}$.
Since the coils are connected in series,the current $I$ is the same for both.
Thus,$K_1 \tan \theta_1 = K_2 \tan \theta_2$.
Given $\theta_1 = 60^o$ and $\theta_2 = 45^o$,we have $K_1 \tan 60^o = K_2 \tan 45^o$.
$K_1 (\sqrt{3}) = K_2 (1) \Rightarrow \frac{K_1}{K_2} = \frac{1}{\sqrt{3}}$.
Since $K \propto \frac{1}{N}$ (as radius $R$ is the same),we have $\frac{K_1}{K_2} = \frac{N_2}{N_1}$.
Therefore,$\frac{N_2}{N_1} = \frac{1}{\sqrt{3}}$,which implies $\frac{N_1}{N_2} = \frac{\sqrt{3}}{1}$.

(D) The formula for the current $I$ in a tangent galvanometer is given by $I = K \tan \theta$,where $K = \frac{2r B_H}{\mu_0 n}$.
Here,$n = 100$,$r = 0.20\, m$,$B_H = 3 \times 10^{-5}\, T$,and $\theta = 45^{\circ}$.
Substituting the values:
$I = \frac{2 \times 0.20 \times 3 \times 10^{-5}}{4 \pi \times 10^{-7} \times 100} \times \tan 45^{\circ}$.
$I = \frac{1.2 \times 10^{-5}}{4 \pi \times 10^{-5}} \times 1 = \frac{1.2}{4 \pi} = \frac{0.3}{\pi} \approx 0.09549\, A$.
Rounding to the given options,the current is $0.095\, A$.

(A) The time period of a vibration magnetometer is given by the formula:
$T = 2 \pi \sqrt{\frac{I}{MB}}$
where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B$ is the magnetic field.
Given the initial state: $T_0 = 2 \pi \sqrt{\frac{I_0}{M_0 B}}$.
For the new magnet,the moment of inertia $I_1 = 3I_0$ and the magnetic moment $M_1 = \frac{M_0}{3}$.
The new time period $T_1$ is:
$T_1 = 2 \pi \sqrt{\frac{I_1}{M_1 B}} = 2 \pi \sqrt{\frac{3I_0}{(M_0/3) B}}$
$T_1 = 2 \pi \sqrt{\frac{9I_0}{M_0 B}} = 3 \times (2 \pi \sqrt{\frac{I_0}{M_0 B}})$
$T_1 = 3T_0$.

(D) The time period of oscillation of a magnet in a magnetic field $H$ is given by $T = 2\pi \sqrt{\frac{I}{MH}}$.
Since the same magnet is used, its moment of inertia $I$ and magnetic moment $M$ remain constant.
Therefore, $T \propto \frac{1}{\sqrt{H}}$, which implies $T^2 \propto \frac{1}{H}$ or $H \propto \frac{1}{T^2}$.
In the first case:
Frequency $f_1 = 40 \, \text{oscillations/min} = \frac{40}{60} \, \text{Hz}$.
Time period $T_1 = \frac{1}{f_1} = \frac{60}{40} = 1.5 \, s$.
Magnetic field $H_1 = 0.1 \times 10^{-5} \, T = 10^{-6} \, T$.
In the second case:
Time period $T_2 = 2.5 \, s$.
Magnetic field $H_2 = ?$.
Using the relation $\frac{H_2}{H_1} = \left( \frac{T_1}{T_2} \right)^2$:
$\frac{H_2}{10^{-6}} = \left( \frac{1.5}{2.5} \right)^2 = \left( \frac{3}{5} \right)^2 = \frac{9}{25} = 0.36$.
$H_2 = 0.36 \times 10^{-6} \, T$.

(A) The time period of oscillation $T$ is given by the total time divided by the number of oscillations: $T = \frac{6.70 \; s}{10} = 0.67 \; s$.
The formula for the time period of a magnetic needle oscillating in a magnetic field is $T = 2\pi \sqrt{\frac{I}{mB}}$.
Rearranging this formula to solve for the magnetic field $B$,we get: $B = \frac{4\pi^2 I}{m T^2}$.
Substituting the given values: $B = \frac{4 \times (3.14)^2 \times 7.5 \times 10^{-6}}{(6.7 \times 10^{-2}) \times (0.67)^2}$.
Calculating the denominator: $(6.7 \times 10^{-2}) \times 0.4489 \approx 0.0300763$.
Calculating the numerator: $4 \times 9.8596 \times 7.5 \times 10^{-6} \approx 0.000295788$.
Thus,$B = \frac{0.000295788}{0.0300763} \approx 0.00983 \; T$,which is approximately $0.01 \; T$.

(D) The horizontal component of the Earth's magnetic field is $H = 0.36 \; G$.
At the null point on the axial line,the magnetic field of the magnet $B_{axial}$ is equal to the horizontal component of the Earth's magnetic field $H$:
$B_{axial} = \frac{\mu_0}{4\pi} \frac{2M}{d^3} = H = 0.36 \; G \dots (i)$
where $d = 14 \; cm$ is the distance from the center.
The magnetic field on the equatorial line (normal bisector) at the same distance $d$ is given by:
$B_{equatorial} = \frac{\mu_0}{4\pi} \frac{M}{d^3} = \frac{H}{2} = \frac{0.36}{2} = 0.18 \; G$.
Since the magnetic field of the magnet on the equatorial line is in the same direction as the Earth's horizontal magnetic field,the total magnetic field $B_{total}$ at that point is:
$B_{total} = B_{equatorial} + H = 0.18 \; G + 0.36 \; G = 0.54 \; G$.

(A) The magnetic field on the axis of the magnet at a distance $d_{1} = 14 \; cm$ is given by:
$B_{1} = \frac{\mu_{0}}{4 \pi} \frac{2M}{d_{1}^{3}} = H \dots (1)$
Where $M$ is the magnetic moment and $H$ is the horizontal component of the Earth's magnetic field.
When the magnet is turned by $180^o$,the neutral points shift to the equatorial line of the magnet.
The magnetic field on the equatorial line at a distance $d_{2}$ is given by:
$B_{2} = \frac{\mu_{0}}{4 \pi} \frac{M}{d_{2}^{3}} = H \dots (2)$
Equating $(1)$ and $(2)$:
$\frac{2}{d_{1}^{3}} = \frac{1}{d_{2}^{3}}$
$d_{2}^{3} = \frac{d_{1}^{3}}{2}$
$d_{2} = \frac{d_{1}}{\sqrt[3]{2}} = \frac{14}{1.26} \approx 11.1 \; cm$
The new null points will be located at $11.1 \; cm$ on the equatorial line.

(N/A) The periodic time $T$ of a bar magnet oscillating in a uniform magnetic field $B$ is given by the formula:
$T = 2 \pi \sqrt{\frac{I}{mB}}$
Where:
$T$ is the time period of oscillation.
$I$ is the moment of inertia of the bar magnet.
$m$ is the magnetic dipole moment of the magnet.
$B$ is the strength of the uniform magnetic field.

(A) The time period of a bar magnet oscillating in a magnetic field $B$ is given by $T = 2 \pi \sqrt{\frac{I}{MB}}$.
For the original magnet,$I = \frac{m l^2}{12}$,where $m$ is the mass and $l$ is the length.
When the magnet is cut into two equal pieces perpendicular to its length,the length of each piece becomes $l' = l/2$ and the mass becomes $m' = m/2$.
The new moment of inertia $I'$ for each piece about its center is $I' = \frac{m' (l')^2}{12} = \frac{(m/2) (l/2)^2}{12} = \frac{m l^2}{12 \times 8} = \frac{I}{8}$.
The new magnetic moment $M'$ for each piece is $M' = M/2$.
The new time period $T'$ is $T' = 2 \pi \sqrt{\frac{I'}{M' B}} = 2 \pi \sqrt{\frac{I/8}{(M/2) B}} = 2 \pi \sqrt{\frac{1}{4} \frac{I}{MB}} = \frac{1}{2} \left( 2 \pi \sqrt{\frac{I}{MB}} \right) = \frac{T}{2}$.

(C) The length of the magnet $2l = 14 \, cm$,so $l = 7 \, cm = 0.07 \, m$.
The distance of the neutral point from the center is $d = 18 \, cm = 0.18 \, m$.
The distance $r$ from each pole to the neutral point is $r = \sqrt{d^2 + l^2} = \sqrt{18^2 + 7^2} = \sqrt{324 + 49} = \sqrt{373} \, cm$.
The magnetic field due to one pole at the neutral point is $B_0 = \frac{\mu_0}{4\pi} \frac{m}{r^2}$.
At the neutral point,the horizontal component of the Earth's magnetic field $B_H$ is balanced by the resultant magnetic field of the two poles: $B_H = 2 B_0 \sin \theta$,where $\sin \theta = \frac{l}{r}$.
Substituting the values: $B_H = 2 \left( \frac{\mu_0}{4\pi} \frac{m}{r^2} \right) \left( \frac{l}{r} \right) = \frac{\mu_0}{4\pi} \frac{2ml}{r^3} = \frac{\mu_0}{4\pi} \frac{M}{r^3}$,where $M = m(2l)$ is the magnetic moment.
$0.4 \times 10^{-4} = 10^{-7} \times \frac{M}{(r \times 10^{-2})^3} = 10^{-7} \times \frac{M}{(373 \times 10^{-4})^{3/2}}$.
$M = \frac{0.4 \times 10^{-4} \times (373 \times 10^{-4})^{3/2}}{10^{-7}} = 0.4 \times 10^3 \times (373)^{3/2} \times 10^{-6} = 0.4 \times 10^{-3} \times 7203.82 \approx 2.88 \, J \, T^{-1}$.

(A) The time period $T$ of a magnetic needle oscillating in a uniform magnetic field $B$ is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB}}$.
Given: Magnetic moment $M = 9.85 \times 10^{-2} \, A \cdot m^{2}$,Moment of inertia $I = 5 \times 10^{-6} \, kg \cdot m^{2}$.
The needle performs $10$ oscillations in $5 \, s$,so the time period $T = \frac{5}{10} = 0.5 \, s$.
Substituting the values into the formula: $0.5 = 2\pi \sqrt{\frac{5 \times 10^{-6}}{9.85 \times 10^{-2} \times B}}$.
Squaring both sides: $0.25 = 4\pi^{2} \left( \frac{5 \times 10^{-6}}{9.85 \times 10^{-2} \times B} \right)$.
Using $\pi^{2} = 9.85$: $0.25 = 4 \times 9.85 \times \frac{5 \times 10^{-6}}{9.85 \times 10^{-2} \times B}$.
$0.25 = 4 \times \frac{5 \times 10^{-6}}{10^{-2} \times B} = \frac{20 \times 10^{-6}}{10^{-2} \times B} = \frac{20 \times 10^{-4}}{B}$.
$B = \frac{20 \times 10^{-4}}{0.25} = 80 \times 10^{-4} = 8 \times 10^{-3} \, T$.
Since $1 \, T = 1000 \, mT$,$B = 8 \, mT$.

(A) The time period of an oscillation magnetometer is given by $T = 2\pi \sqrt{\frac{I}{M B_H}}$,where $B_H = B \cos \delta$ is the horizontal component of the Earth's magnetic field and $\delta$ is the angle of dip.
Frequency $f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{M B \cos \delta}{I}}$. Thus,$f \propto \sqrt{B \cos \delta}$.
At place $P$: $f_P = 20 \text{ oscillations/min}$,$\delta_P = 30^{\circ}$.
At place $Q$: $f_Q = 10 \text{ oscillations/min}$,$\delta_Q = 60^{\circ}$.
Taking the ratio:
$\frac{f_P}{f_Q} = \sqrt{\frac{B_P \cos 30^{\circ}}{B_Q \cos 60^{\circ}}}$
$\frac{20}{10} = \sqrt{\frac{B_P (\sqrt{3}/2)}{B_Q (1/2)}}$
$2 = \sqrt{\frac{B_P \sqrt{3}}{B_Q}}$
Squaring both sides: $4 = \frac{B_P \sqrt{3}}{B_Q}$
Therefore,$\frac{B_Q}{B_P} = \frac{\sqrt{3}}{4}$.

(B) The time period of a bar magnet oscillating in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$.
Given $T_1 = 3\,s$,$T_2 = 4\,s$,and the ratio of moments of inertia $\frac{I_1}{I_2} = \frac{3}{2}$.
Taking the ratio of the time periods:
$\frac{T_1}{T_2} = \sqrt{\frac{I_1}{I_2} \cdot \frac{M_2}{M_1}} = \frac{3}{4}$.
Squaring both sides:
$\frac{I_1}{I_2} \cdot \frac{M_2}{M_1} = \left(\frac{3}{4}\right)^2 = \frac{9}{16}$.
Substituting $\frac{I_1}{I_2} = \frac{3}{2}$:
$\frac{3}{2} \cdot \frac{M_2}{M_1} = \frac{9}{16}$.
$\frac{M_2}{M_1} = \frac{9}{16} \cdot \frac{2}{3} = \frac{3}{8}$.
Therefore,the ratio of magnetic moments $\frac{M_1}{M_2} = \frac{8}{3}$.

(A) neutral point is a point where the net magnetic field due to the magnets is zero. This occurs where the magnetic field vectors from the two magnets are equal in magnitude and opposite in direction.
By analyzing the magnetic field lines of the two bar magnets,we can determine the direction of the magnetic field in each zone.
In zone $I$,the magnetic field lines from the horizontal magnet point towards the left,and the magnetic field lines from the vertical magnet also point in a direction that opposes the field from the horizontal magnet. Thus,the fields can cancel each other out in this region.
Therefore,the neutral point is located in zone $I$.

(A) The time period of a magnetic needle oscillating in a magnetic field $B$ is given by $T = 2 \pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia and $M$ is the magnetic dipole moment.
When the magnet is broken into $n$ equal parts perpendicular to its length,the length of each part becomes $l' = l/n$.
The new mass of each part is $m' = m/n$.
The new moment of inertia of each part is $I' = \frac{m' (l')^2}{12} = \frac{(m/n) (l/n)^2}{12} = \frac{I}{n^3}$.
The new magnetic moment of each part is $M' = M/n$.
The new time period $T'$ is given by $T' = 2 \pi \sqrt{\frac{I'}{M' B}} = 2 \pi \sqrt{\frac{I/n^3}{(M/n) B}} = 2 \pi \sqrt{\frac{I}{n^2 MB}} = \frac{1}{n} \left( 2 \pi \sqrt{\frac{I}{MB}} \right) = \frac{T}{n}$.

(B) The period of oscillation of a magnet in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia and $M$ is the magnetic moment.
Since the magnets are of the same size,their moments of inertia $I$ are the same. Let the moments of inertia be $I_1 = I_2 = I$. The total moment of inertia is $I_{total} = I + I = 2I$.
Case $1$: Similar poles together. The net magnetic moment is $M_{net} = M_1 + M_2$. Given $M_1 : M_2 = 1 : 2$,let $M_1 = M$ and $M_2 = 2M$. So,$M_{net} = M + 2M = 3M$.
The period $T_1 = 3 \, s = 2\pi \sqrt{\frac{2I}{3MB}}$.
Case $2$: One magnet is reversed. The net magnetic moment is $M'_{net} = |M_2 - M_1| = |2M - M| = M$.
The new period $T_2 = 2\pi \sqrt{\frac{2I}{MB}}$.
Dividing the two equations: $\frac{T_2}{T_1} = \sqrt{\frac{2I/MB}{2I/3MB}} = \sqrt{3}$.
Therefore,$T_2 = 3 \times \sqrt{3} = 3\sqrt{3} \, s$.

(D) In the magnetic meridian,the apparent dip $\delta$ is related to the vertical component $B_V$ and horizontal component $B_H$ of the Earth's magnetic field by $\tan \delta = \frac{B_V}{B_H}$.
When the dip circle is rotated by an angle $\alpha$ in the horizontal plane,the effective horizontal component becomes $B_H^{\prime} = B_H \cos \alpha$,while the vertical component $B_V$ remains unchanged.
The new apparent angle of dip $\delta^{\prime}$ is given by $\tan \delta^{\prime} = \frac{B_V}{B_H^{\prime}} = \frac{B_V}{B_H \cos \alpha}$.
Substituting $\tan \delta = \frac{B_V}{B_H}$,we get $\tan \delta^{\prime} = \frac{\tan \delta}{\cos \alpha}$.
Therefore,the ratio $\frac{\tan \delta^{\prime}}{\tan \delta} = \frac{1}{\cos \alpha}$.

(D) When a bar magnet is placed such that its magnetic moment $\vec{M}$ is parallel to the Earth's horizontal magnetic field $\vec{B}_H$,the magnetic field produced by the magnet opposes the Earth's magnetic field in the equatorial region.
The locus of points where the magnetic field of the magnet is equal and opposite to the Earth's magnetic field forms a circle (or ring) around the magnet in the equatorial plane.
Since there are infinite points on this circle where the net magnetic field is zero,there are infinite neutral points.
In a $2D$ plane (like a sheet of paper),we only observe two of these points,which are the intersection of the ring with the plane of the paper.

(A) The time period of a magnetic needle oscillating in a horizontal plane is given by $T = 2\pi \sqrt{\frac{I}{M B_H}}$,where $B_H = B \cos \delta$ is the horizontal component of the Earth's magnetic field and $\delta$ is the angle of dip.
Given $\delta = 60^{\circ}$,we have $T = 2\pi \sqrt{\frac{I}{M B \cos 60^{\circ}}}$.
When the needle oscillates in the vertical plane coinciding with the magnetic meridian,it experiences the total magnetic field $B$. The time period $T'$ is given by $T' = 2\pi \sqrt{\frac{I}{M B}}$.
Dividing the two equations: $\frac{T'}{T} = \sqrt{\frac{M B \cos 60^{\circ}}{M B}} = \sqrt{\cos 60^{\circ}}$.
Since $\cos 60^{\circ} = \frac{1}{2}$,we get $\frac{T'}{T} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Therefore,$T' = \frac{T}{\sqrt{2}}$.

(A) The time period of a vibration magnetometer is given by $T = 2 \pi \sqrt{\frac{I}{MH}}$,where $I$ is the moment of inertia and $M$ is the magnetic moment.
When like poles are joined,the effective magnetic moment is $M_1 + M_2$. Given $T_1 = 5 \, s$,we have $5 = 2 \pi \sqrt{\frac{I}{M_1 + M_2}}$.
When unlike poles are joined,the effective magnetic moment is $M_1 - M_2$. Given $T_2 = 15 \, s$,we have $15 = 2 \pi \sqrt{\frac{I}{M_1 - M_2}}$.
Dividing the two equations: $\frac{5}{15} = \sqrt{\frac{M_1 - M_2}{M_1 + M_2}} \Rightarrow \frac{1}{3} = \sqrt{\frac{M_1 - M_2}{M_1 + M_2}}$.
Squaring both sides: $\frac{1}{9} = \frac{M_1 - M_2}{M_1 + M_2}$.
Cross-multiplying: $M_1 + M_2 = 9M_1 - 9M_2 \Rightarrow 10M_2 = 8M_1$.
Therefore,the ratio $\frac{M_1}{M_2} = \frac{10}{8} = \frac{5}{4}$.

(C) The reduction factor $K$ of a tangent galvanometer is given by the formula $K = \frac{2 R H}{N}$.
Given:
Number of turns $N = 80$.
Internal diameter $d_1 = 19 \, cm$, External diameter $d_2 = 21 \, cm$.
Mean radius $R = \frac{d_1 + d_2}{4} = \frac{19 + 21}{4} = 10 \, cm = 0.1 \, m$.
Horizontal component of Earth's magnetic field $H = 0.32 \, \text{oersted} = 0.32 \times 80 \, A/m = 25.6 \, A/m$.
Using $K = \frac{2 R H}{N}$ where $H$ is in $A/m$:
$K = \frac{2 \times 0.1 \times 25.6}{80} = \frac{5.12}{80} = 0.064 \, A$.

(A) The reduction factor $K$ of a tangent galvanometer is given by the formula:
$K = \frac{2 R B_H}{\mu_0 N}$
where $R$ is the radius of the coil,$B_H$ is the horizontal component of the Earth's magnetic field,$\mu_0$ is the permeability of free space,and $N$ is the number of turns.
From the formula,we can see that $K \propto \frac{R}{N}$.
Given that the new radius $R' = 2R$ and the new number of turns $N' = 2N$,the new reduction factor $K'$ is:
$K' = \frac{2 R' B_H}{\mu_0 N'} = \frac{2 (2R) B_H}{\mu_0 (2N)} = \frac{2 R B_H}{\mu_0 N} = K$
Therefore,the reduction factor remains unchanged.

(B) When a bar magnet is placed with its north pole pointing towards the geographic north,the magnetic field of the magnet and the horizontal component of the Earth's magnetic field $(B_H)$ oppose each other along the equatorial line (the east-west line passing through the centre of the magnet).
At the neutral point,the magnetic field due to the magnet $(B_{mag})$ is equal and opposite to the horizontal component of the Earth's magnetic field $(B_H)$.
Since the field lines of the magnet emerge from the north pole and enter the south pole,on the equatorial line,the field points from north to south (opposite to the Earth's $B_H$).
Thus,the neutral points are obtained on the east-west line passing through the centre of the magnet.

(D) The time period of a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{M B_H}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B_H$ is the horizontal component of the earth's magnetic field.
From the formula,$T \propto \frac{1}{\sqrt{B_H}}$,which implies $B_H \propto \frac{1}{T^2}$.
Given $T_1 = 2.45 \,s$ and $T_2 = 4.9 \,s$.
Therefore,$\frac{B_{H_1}}{B_{H_2}} = \left( \frac{T_2}{T_1} \right)^2$.
Substituting the values: $\frac{B_{H_1}}{B_{H_2}} = \left( \frac{4.9}{2.45} \right)^2 = (2)^2 = 4$.
Thus,the ratio is $4: 1$.

(C) The time period of oscillation for a magnetic needle in a uniform magnetic field is given by $T = 2 \pi \sqrt{\frac{I}{MB}}$, where $I$ is the moment of inertia, $M$ is the magnetic moment, and $B$ is the magnetic field strength.
Given: $B = 0.049 \ T$, $I = 9.8 \times 10^{-5} \ kg \ m^2$, and the needle performs $20$ oscillations in $5 \ s$.
The time period $T = \frac{5 \ s}{20} = 0.25 \ s = \frac{1}{4} \ s$.
Substituting the values into the formula: $\frac{1}{4} = 2 \pi \sqrt{\frac{9.8 \times 10^{-5}}{M \times 0.049}}$.
Squaring both sides: $\frac{1}{16} = 4 \pi^2 \times \frac{9.8 \times 10^{-5}}{M \times 0.049}$.
Rearranging for $M$: $M = \frac{4 \pi^2 \times 9.8 \times 10^{-5} \times 16}{0.049} = \frac{4 \pi^2 \times 9.8 \times 10^{-5} \times 16}{0.049}$.
Since $9.8 / 0.049 = 200$, we get $M = 4 \pi^2 \times 200 \times 16 \times 10^{-5} = 12800 \pi^2 \times 10^{-5} \ A \ m^2$.
Wait, re-calculating: $M = \frac{4 \pi^2 \times 9.8 \times 10^{-5} \times 16}{49 \times 10^{-3}} = 4 \pi^2 \times 0.2 \times 16 \times 10^{-2} = 12.8 \pi^2 \times 10^{-2} = 1280 \pi^2 \times 10^{-5} \ A \ m^2$.
Thus, $x = 1280 \pi^2$.

(B) When a current $I$ is passed through the galvanometer coil,a magnetic field $B$ is produced at right angles to the plane of the coil,i.e.,at right angles to the horizontal component of the Earth's magnetic field $H$. Under the influence of two crossed magnetic fields $B$ and $H$,the magnetic needle of the galvanometer undergoes a deflection $\theta$,which is given by the tangent law. Using the tangent law,we can find the relation $I = K \tan \theta$,where $K$ is the reduction factor. This clearly indicates that a tangent galvanometer is an instrument used for the measurement of electric current in a circuit. Note: $A$ tangent galvanometer is most accurate when its deflection is $45^{\circ}$.

(A) The time period of an oscillating magnet is given by $T = 2 \pi \sqrt{\frac{I}{MB}}$,which implies $T \propto \frac{1}{\sqrt{B}}$.
Given $n_1 = 30 \text{ oscillations/minute}$,the time period $T_1 = \frac{60}{30} = 2 \text{ s}$.
Let the magnetic induction at the first place be $B_1$ and at the second place be $B_2 = 2B_1$.
Using the relation $\frac{T_2}{T_1} = \sqrt{\frac{B_1}{B_2}}$,we get:
$\frac{T_2}{2} = \sqrt{\frac{B_1}{2B_1}} = \frac{1}{\sqrt{2}}$.
Therefore,$T_2 = \frac{2}{\sqrt{2}} = \sqrt{2} \text{ s}$.
Note: If the problem implies the induction is increased *by* two times (i.e.,$B_2 = B_1 + 2B_1 = 3B_1$),then $T_2 = \frac{2}{\sqrt{3}} \text{ s}$. Given the standard phrasing of such problems,we assume $B_2 = 2B_1$ leads to $\sqrt{2} \text{ s}$,but since $\frac{2}{\sqrt{3}}$ is an option,the intended meaning is $B_2 = 3B_1$.

(C) The frequency of oscillation for a bar magnet in a uniform magnetic field is given by $f = \frac{1}{2 \pi} \sqrt{\frac{MB}{I}}$.
For magnet $P$: $f_{P} = \frac{1}{2 \pi} \sqrt{\frac{M_{P} B}{I_{P}}}$.
For magnet $Q$: $f_{Q} = \frac{1}{2 \pi} \sqrt{\frac{M_{Q} B}{I_{Q}}}$.
Given $f_{P} = 2 f_{Q}$ and $I_{P} = 2 I_{Q}$.
Substituting these into the ratio: $\frac{f_{P}}{f_{Q}} = \sqrt{\frac{M_{P} I_{Q}}{M_{Q} I_{P}}} = 2$.
Squaring both sides: $\frac{M_{P} I_{Q}}{M_{Q} I_{P}} = 4$.
Since $I_{P} = 2 I_{Q}$,we have $\frac{I_{Q}}{I_{P}} = \frac{1}{2}$.
Therefore,$\frac{M_{P}}{M_{Q}} \cdot \frac{1}{2} = 4$,which implies $\frac{M_{P}}{M_{Q}} = 8$,or $M_{P} = 8 M_{Q}$.

(C) Let the magnetic moments of magnets $A$ and $B$ be $M_A = 2M$ and $M_B = M$ respectively. Let their moments of inertia be $I_A = I_B = I$ (since they are geometrically similar and have the same mass/dimensions).
When like poles are kept together,the net magnetic moment is $M_{net1} = M_A + M_B = 2M + M = 3M$. The total moment of inertia is $I_{net} = I_A + I_B = 2I$.
The time period is given by $T = 2\pi \sqrt{\frac{I_{net}}{M_{net}B_H}}$.
Thus,$T_1 = 2\pi \sqrt{\frac{2I}{3MB_H}}$.
When unlike poles are kept together,the net magnetic moment is $M_{net2} = M_A - M_B = 2M - M = M$.
The total moment of inertia remains $I_{net} = 2I$.
Thus,$T_2 = 2\pi \sqrt{\frac{2I}{MB_H}}$.
Taking the ratio $T_1 / T_2 = \sqrt{\frac{2I}{3MB_H} / \frac{2I}{MB_H}} = \sqrt{\frac{1}{3}} = 1 : \sqrt{3}$.

(C) The time period $T$ of a bar magnet oscillating in a magnetic field is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic dipole moment,and $B$ is the magnetic field strength.
Since the mass and volume are the same,the moment of inertia $I$ remains constant.
For the new magnet,the magnetic dipole moment $M' = 9M$.
The new time period $T'$ is given by: $T' = 2\pi \sqrt{\frac{I}{M'B}} = 2\pi \sqrt{\frac{I}{9MB}}$.
Simplifying this,we get: $T' = \frac{1}{\sqrt{9}} \times 2\pi \sqrt{\frac{I}{MB}} = \frac{1}{3} T$.
Therefore,the new time period is $\frac{T}{3}$.

(D) For a tangent galvanometer,the current $I$ is given by $I = K \tan \theta$,where $K = \frac{2r B_{H}}{n \mu_{0}}$.
Since the galvanometers are connected in parallel,the voltage $V$ across them is the same. Thus,$I_1 R_1 = I_2 R_2$,which implies $\frac{I_1}{I_2} = \frac{R_2}{R_1} = \frac{3}{1}$.
Using the formula $I = K \tan \theta$,we have $\frac{I_1}{I_2} = \frac{n_2}{n_1} \cdot \frac{\tan \theta_1}{\tan \theta_2}$.
Substituting the given values: $\frac{3}{1} = \frac{n_2}{n_1} \cdot \frac{\tan 30^{\circ}}{\tan 60^{\circ}}$.
$\frac{3}{1} = \frac{n_2}{n_1} \cdot \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{n_2}{n_1} \cdot \frac{1}{3}$.
Therefore,$\frac{n_1}{n_2} = \frac{1}{3 \times 3} = \frac{1}{9}$.
Wait,re-evaluating: $\frac{3}{1} = \frac{n_2}{n_1} \cdot \frac{1}{3} \implies \frac{n_2}{n_1} = 9 \implies \frac{n_1}{n_2} = \frac{1}{9}$.
Given the options,let's re-check the ratio: $\frac{R_2}{R_1} = 3$. $\frac{I_1}{I_2} = 3$. $\frac{n_2}{n_1} = \frac{I_1 \tan \theta_2}{I_2 \tan \theta_1} = 3 \cdot \frac{\sqrt{3}}{1/\sqrt{3}} = 3 \cdot 3 = 9$. The ratio $n_1:n_2$ is $1:9$. Since $1:9$ is not an option,let's re-verify the resistance ratio logic. If $R_1:R_2 = 1:3$,then $I_1:I_2 = 3:1$. The correct ratio is $1:9$.

(B) The time period of oscillation for a magnetic needle in a magnetic field is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB}}$.
Given: $T = 2 \text{ s}$,$I = 8 \times 10^{-6} \text{ kg m}^2$,$M = 5 \times 10^{-2} \text{ A m}^2$.
Substituting the values into the formula: $2 = 2\pi \sqrt{\frac{8 \times 10^{-6}}{5 \times 10^{-2} \times B}}$.
Dividing by $2$: $1 = \pi \sqrt{\frac{8 \times 10^{-6}}{5 \times 10^{-2} \times B}}$.
Squaring both sides: $1 = \pi^2 \left( \frac{8 \times 10^{-6}}{5 \times 10^{-2} \times B} \right)$.
Rearranging for $B$: $B = \frac{\pi^2 \times 8 \times 10^{-6}}{5 \times 10^{-2}}$.
Using $\pi^2 \approx 9.86$ (or approximately $10$): $B = \frac{9.86 \times 8 \times 10^{-6}}{5 \times 10^{-2}} \approx 1.577 \times 10^{-4} \text{ T}$.
Rounding to the nearest option,$B \approx 1.6 \times 10^{-4} \text{ T}$.

(B) The principle of a tangent galvanometer is given by the formula $I = K \tan \theta$,where $I$ is the current,$K$ is the reduction factor,and $\theta$ is the deflection angle.
Given current $I = \frac{2}{\sqrt{3}} \text{ A}$ and deflection $\theta = 60^{\circ}$.
Rearranging the formula to solve for the reduction factor $K$:
$K = \frac{I}{\tan \theta}$
Substituting the given values:
$K = \frac{2 / \sqrt{3}}{\tan 60^{\circ}}$
Since $\tan 60^{\circ} = \sqrt{3}$,we have:
$K = \frac{2 / \sqrt{3}}{\sqrt{3}} = \frac{2}{\sqrt{3} \times \sqrt{3}} = \frac{2}{3} \text{ A}$.
Thus,the reduction factor is $\frac{2}{3} \text{ A}$.

(C) Given: $T_1 = 0.1 \, s$, $B_H = 24 \, \mu T = 24 \times 10^{-6} \, T$, $I = 18 \, A$, $r = 20 \, cm = 0.2 \, m$.
The magnetic field $B$ due to the vertical wire at the position of the magnet is given by $B = \frac{\mu_0 I}{2 \pi r}$.
$B = \frac{4 \pi \times 10^{-7} \times 18}{2 \pi \times 0.2} = 18 \times 10^{-6} \, T = 18 \, \mu T$.
Since the wire is to the east and the current is downward, by the right-hand rule, the magnetic field $B$ produced by the wire at the magnet's location points towards the north (same direction as $B_H$).
The resultant magnetic field is $B_{net} = B_H + B = 24 \, \mu T + 18 \, \mu T = 42 \, \mu T$.
The time period of oscillation is $T = 2 \pi \sqrt{\frac{I}{MB}}$, so $T \propto \frac{1}{\sqrt{B}}$.
Therefore, $\frac{T_2}{T_1} = \sqrt{\frac{B_H}{B_{net}}} = \sqrt{\frac{24}{42}} = \sqrt{\frac{4}{7}} = \frac{2}{\sqrt{7}}$.
$T_2 = T_1 \times \frac{2}{\sqrt{7}} = 0.1 \times \frac{2}{2.645} \approx 0.076 \, s$.

(D) The frequency of oscillation of a magnetic needle in the Earth's magnetic field is given by $f = \frac{1}{2 \pi} \sqrt{\frac{\mu B_H}{I}}$,where $B_H = B \cos \theta$ is the horizontal component of the Earth's magnetic field,$\mu$ is the magnetic moment,and $I$ is the moment of inertia.
Since the same needle is used,$\mu$ and $I$ are constant. Thus,$f \propto \sqrt{B \cos \theta}$.
Given $f_1 = 20$ oscillations/min at $\theta_1 = 45^{\circ}$ and $f_2 = 30$ oscillations/min at $\theta_2 = 30^{\circ}$.
Taking the ratio: $\frac{f_1}{f_2} = \sqrt{\frac{B_1 \cos 45^{\circ}}{B_2 \cos 30^{\circ}}} \Rightarrow \frac{20}{30} = \sqrt{\frac{B_1 (1/\sqrt{2})}{B_2 (\sqrt{3}/2)}}$.
Squaring both sides: $\frac{4}{9} = \frac{B_1}{B_2} \times \frac{1}{\sqrt{2}} \times \frac{2}{\sqrt{3}} = \frac{B_1}{B_2} \times \frac{\sqrt{2}}{\sqrt{3}}$.
Therefore,$\frac{B_1}{B_2} = \frac{4}{9} \times \frac{\sqrt{3}}{\sqrt{2}} = \frac{2 \times 2 \times \sqrt{3}}{3 \times 3 \times \sqrt{2}} = \frac{2 \sqrt{2} \times \sqrt{2} \times \sqrt{3}}{3 \sqrt{3} \times \sqrt{3} \times \sqrt{2}} = \frac{2 \sqrt{2}}{3 \sqrt{3}}$.
Thus,$B_1: B_2 = 2 \sqrt{2}: 3 \sqrt{3}$.

(A) The time period of a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MB_{H}}}$, where $I$ is the moment of inertia, $M$ is the magnetic moment, and $B_{H}$ is the horizontal component of the earth's magnetic field.
From the formula, we see that $T \propto \frac{1}{\sqrt{B_{H}}}$.
Given that $T_{A} = 2T_{B}$, we can write the ratio as $\frac{T_{A}}{T_{B}} = 2$.
Using the proportionality, $\frac{T_{A}}{T_{B}} = \sqrt{\frac{B_{HB}}{B_{HA}}}$.
Substituting the values, $2 = \sqrt{\frac{32 \times 10^{-6}}{B_{HA}}}$.
Squaring both sides, $4 = \frac{32 \times 10^{-6}}{B_{HA}}$.
Therefore, $B_{HA} = \frac{32 \times 10^{-6}}{4} = 8 \times 10^{-6} \,T$.

(B) For a tangent galvanometer,the current $I$ is given by $I = \frac{2r B_H \tan \theta}{\mu_0 N}$.
Given: Resistance $R = 9 \ \Omega$,Potential difference $V = 4.5 \ V$,Number of turns $N = 10$,Deflection $\theta = 30^{\circ}$,Horizontal component of Earth's magnetic field $B_H = 3.14 \times 10^{-5} \ T$.
First,calculate the current $I$ using Ohm's law: $I = \frac{V}{R} = \frac{4.5}{9} = 0.5 \ A$.
Rearranging the tangent galvanometer formula for radius $r$: $r = \frac{\mu_0 N I}{2 B_H \tan \theta}$.
Substituting the values: $r = \frac{4 \pi \times 10^{-7} \times 10 \times 0.5}{2 \times 3.14 \times 10^{-5} \times \tan 30^{\circ}}$.
Since $\mu_0 = 4 \pi \times 10^{-7} \approx 4 \times 3.14 \times 10^{-7}$,we have $r = \frac{4 \times 3.14 \times 10^{-7} \times 5}{2 \times 3.14 \times 10^{-5} \times (1/\sqrt{3})}$.
$r = \frac{2 \times 10^{-7} \times 5}{10^{-5} \times (1/\sqrt{3})} = 10 \times 10^{-2} \times \sqrt{3} \ m = 10 \sqrt{3} \times 10^{-2} \ m$.

(A) The time period $T$ of a bar magnet oscillating in a magnetic field is given by the formula:
$T = 2 \pi \sqrt{\frac{I}{MB}}$
where $I$ is the moment of inertia,$M$ is the magnetic dipole moment,and $B$ is the magnetic field.
Since the moment of inertia $I = mk^2$ (where $m$ is mass and $k$ is the radius of gyration),we have:
$T = 2 \pi \sqrt{\frac{mk^2}{MB}}$
This implies that $T \propto \sqrt{m}$.
Given that the mass is increased by $9$ times $(m_2 = 9m_1)$,the new time period $T_2$ is:
$\frac{T_2}{T_1} = \sqrt{\frac{m_2}{m_1}} = \sqrt{\frac{9m_1}{m_1}} = \sqrt{9} = 3$
Therefore,$T_2 = 3T_1 = 3T$.