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

(A) The principle of a tangent galvanometer is based on the tangent law, which states that $I = K \tan \theta$, where $I$ is the current, $\theta$ is the deflection, and $K$ is the reduction factor.
From this equation, $K = I / \tan \theta$.
Since $\tan \theta$ is a dimensionless quantity, the unit of the reduction factor $K$ is the same as the unit of current $I$.
Therefore, the unit of the reduction factor of a tangent galvanometer is Ampere $(A)$.

(B) tangent galvanometer measures current based on the principle of the tangent law in magnetism. When a galvanometer is connected in series with a high standard resistance,the combination acts as a voltmeter. This is because the high resistance limits the current flowing through the circuit,allowing the device to measure the potential difference across the combination,which is proportional to the current flowing through it according to Ohm's law $(V = IR)$. Therefore,the correct option is $B$.

(A) The thermoelectric electromotive force $(E)$ in a thermocouple is given by the relation $E = \alpha \theta + \frac{1}{2} \beta \theta^2$,where $\theta$ is the temperature difference between the junctions.
Thermoelectric power is defined as the rate of change of thermoelectric $EMF$ with respect to temperature,given by $P = \frac{dE}{dT}$.
At the neutral temperature $(T_n)$,the thermoelectric $EMF$ reaches its maximum value.
Since the $EMF$ is maximum at the neutral temperature,its derivative with respect to temperature must be zero at that point.
Therefore,$\frac{dE}{dT} = 0$ at the neutral temperature.

(D) The magnetic meridian is a vertical $N-S$ plane,and the Earth's horizontal magnetic field component $(B_H)$ lies within it.
To obtain a neutral point at the centre of the coil,the magnetic field produced by the current in the coil $(B)$ must be equal in magnitude and opposite in direction to the Earth's horizontal magnetic field $(B_H)$.
The magnetic field $(B)$ at the centre of a circular coil is perpendicular to the plane of the coil.
For $(B)$ to be opposite to $(B_H)$,the magnetic field $(B)$ must lie in the magnetic meridian plane.
Since $(B)$ is perpendicular to the plane of the coil,the plane of the coil must be perpendicular to the magnetic meridian plane. Therefore,the angle between the plane of the coil and the magnetic meridian is $90^o$.

(B) When a magnet is placed with its south pole pointing towards the geographic north,the null points are located on the axial line of the magnet.
For a short magnet,the magnetic field $B$ on its axial line at a distance $d$ from its center is given by $B = \frac{2M}{d^3}$,where $M$ is the magnetic moment.
At the null point,the magnetic field of the magnet is equal and opposite to the horizontal component of the Earth's magnetic field $(B_H)$.
Given: $d = 20 \, cm$,$B_H = 0.3 \, gauss$.
Equating the fields: $\frac{2M}{d^3} = B_H$.
Substituting the values: $\frac{2M}{(20)^3} = 0.3$.
$\frac{2M}{8000} = 0.3$.
$2M = 0.3 \times 8000 = 2400$.
$M = 1200 \, e.m.u. = 1.2 \times 10^3 \, e.m.u.$

(B) When a bar magnet is placed with its north pole pointing towards the geographic north,the magnetic field lines due to the magnet are directed from north to south outside the magnet. The horizontal component of the Earth's magnetic field $(B_H)$ is directed from geographic south to geographic north.
At the equatorial line (broadside-on position) of the magnet,the magnetic field due to the magnet is directed from north to south,which is opposite to the direction of the Earth's horizontal magnetic field.
At certain points on the equatorial line,the magnitude of the magnetic field due to the magnet exactly cancels the horizontal component of the Earth's magnetic field. These points are called neutral points.
Therefore,the neutral points lie on the equatorial line,which corresponds to the east and west directions from the center of the magnet.

(D) The magnetic field $B$ at a distance $d$ on the equatorial line (broadside-on position) of a short bar magnet is given by $B = \frac{\mu_0}{4\pi} \frac{M}{d^3}$.
At the neutral point,the magnetic field of the magnet is equal and opposite to the horizontal component of the Earth's magnetic field $(B_H)$.
Thus,$\frac{\mu_0}{4\pi} \frac{M_1}{r^3} = B_H$ and $\frac{\mu_0}{4\pi} \frac{M_2}{(2r)^3} = B_H$.
Equating the two expressions: $\frac{M_1}{r^3} = \frac{M_2}{8r^3}$.
Therefore,the ratio of the magnetic moments is $\frac{M_1}{M_2} = \frac{r^3}{8r^3} = \frac{1}{8}$.

(C) When the $N$-pole of a bar magnet points towards the geographic south and the $S$-pole points towards the geographic north,the magnetic field lines of the magnet and the horizontal component of the Earth's magnetic field $(B_H)$ are in the same direction along the axial line.
However,along the equatorial line (the perpendicular bisector of the magnetic axis),the magnetic field of the bar magnet is directed opposite to the Earth's horizontal magnetic field $(B_H)$.
$A$ null point (or neutral point) is defined as a point where the net magnetic field is zero,meaning the magnetic field of the bar magnet is equal in magnitude and opposite in direction to the Earth's horizontal magnetic field.
Therefore,the null points are located on the perpendicular bisector of the magnetic axis.

(C) At the neutral point, the magnetic field produced by the magnet is equal in magnitude and opposite in direction to the horizontal component of the Earth's magnetic field.
For a point on the axial line of a short bar magnet, the magnetic field $B$ is given by:
$B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d^3}$
Given:
Magnetic moment $M = 6.75 \, Am^2$
Horizontal component of Earth's magnetic field $B_H = 5 \times 10^{-5} \, Wb/m^2$
$\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$
Equating the fields:
$10^{-7} \times \frac{2 \times 6.75}{d^3} = 5 \times 10^{-5}$
$\frac{13.5 \times 10^{-7}}{d^3} = 5 \times 10^{-5}$
$d^3 = \frac{13.5 \times 10^{-7}}{5 \times 10^{-5}} = 2.7 \times 10^{-2} = 0.027 \, m^3$
$d = \sqrt[3]{0.027} = 0.3 \, m$
Converting to centimeters: $d = 0.3 \times 100 = 30 \, cm$.

(B) When a bar magnet is placed with its axis along the east-west direction,its North pole points towards the East (or West) and its South pole points towards the West (or East).
In this configuration,the magnetic field of the magnet and the horizontal component of the Earth's magnetic field $(B_H)$ are perpendicular to each other at all points on the equatorial line of the magnet.
Since the magnetic field of the magnet and the Earth's magnetic field are never in opposite directions along the same line,they cannot cancel each other out to produce a zero magnetic field.
Therefore,there are no neutral points in this orientation.

(A) In the given case,the magnetic field $H$ produced by the coil and the horizontal component of the Earth's magnetic field $H_0$ are perpendicular to each other.
When a magnetic needle is placed in two mutually perpendicular magnetic fields $H$ and $H_0$,it aligns itself along the resultant magnetic field.
From the geometry of the vector addition,we can form a right-angled triangle where the base is $H$ and the perpendicular side is $H_0$.
The angle $\theta$ that the magnet makes with $H$ is given by the trigonometric relation:
$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{H_0}{H}$
Therefore,$\theta = \tan^{-1}\left(\frac{H_0}{H}\right)$.

(A) Neutral points are points in space where the compass needle does not point in any specific direction. Such points are obtained when the magnetic field due to the magnet and the Earth's magnetic field cancel each other out.
If a magnet is kept vertical,only one pole of the magnet is near the plane of the paper (magnetic meridian).
Consequently,only one point where the resultant magnetic field is zero lies in the plane of the paper.

(D) The time period $T$ of a freely suspended magnet in a uniform magnetic field is given by the formula:
$T = 2 \pi \sqrt{\frac{I}{MH}}$
Where:
$I$ is the moment of inertia of the magnet about the axis of suspension.
$M$ is the magnetic dipole moment of the magnet.
$H$ is the horizontal component of the Earth's magnetic field.
From the formula,we can see that the time period $T$ depends on the moment of inertia $(I)$,the magnetic moment $(M)$,and the horizontal component of the Earth's magnetic field $(H)$.
It does not depend on the length of the suspension thread. Therefore,option $(d)$ is correct.

(C) The magnetic moments of two bar magnets can be compared using a Deflection magnetometer by measuring the deflection of the magnetic needle in different positions (Tan-$A$ or Tan-$B$).
Alternatively,they can be compared using a Vibration magnetometer by measuring the time period of oscillations of the magnets in a uniform magnetic field,where the ratio of magnetic moments is inversely proportional to the square of the time periods $(M_1/M_2 = T_2^2/T_1^2)$.

(A) freely suspended bar magnet in a uniform magnetic field $B_H$ experiences a restoring torque $\tau = -M B_H \sin \theta$.
For small oscillations,$\sin \theta \approx \theta$,so $\tau = -M B_H \theta$.
The equation of motion is $I \alpha = \tau$,where $I$ is the moment of inertia and $\alpha$ is the angular acceleration.
$I \frac{d^2 \theta}{dt^2} = -M B_H \theta$.
This is the equation of simple harmonic motion $\frac{d^2 \theta}{dt^2} + \omega^2 \theta = 0$,where $\omega^2 = \frac{M B_H}{I}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{I}{M B_H}}$.

(B) In the sum position,the time period is given by $T_S = 2\pi \sqrt{\frac{I_1 + I_2}{(M_1 + M_2)B_H}}$.
In the difference position,the time period is given by $T_d = 2\pi \sqrt{\frac{I_1 + I_2}{(M_1 - M_2)B_H}}$.
Since the denominator in the difference position $(M_1 - M_2)$ is smaller than the denominator in the sum position $(M_1 + M_2)$,the time period $T_d$ is greater than $T_S$.
The difference position occurs when opposite poles of the two magnets are placed on the same side,resulting in a net magnetic moment of $(M_1 - M_2)$.

(D) The time period of a magnetic dipole oscillating in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$.
This implies $T \propto \frac{1}{\sqrt{B_H}}$.
Given at the first place,$n_1 = 30 \, \text{oscillations/min} = 0.5 \, \text{Hz}$.
Thus,the time period $T_1 = \frac{1}{n_1} = \frac{1}{0.5} = 2 \, \text{sec}$.
Let the magnetic field at the first place be $B_H$. Then at the second place,the magnetic field is $2B_H$.
Using the relation $\frac{T_2}{T_1} = \sqrt{\frac{B_{H1}}{B_{H2}}}$,we get:
$T_2 = T_1 \sqrt{\frac{B_H}{2B_H}} = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2} \, \text{sec}$.

(D) vibration magnetometer is a device used to compare the magnetic moments $(M)$ of two magnets and the horizontal components of magnetic fields $(B_H)$.
It operates on the principle that when a freely suspended magnet in a uniform magnetic field is slightly disturbed from its equilibrium position, it performs simple harmonic oscillations about its mean position.
The time period of oscillation is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$, where $I$ is the moment of inertia of the magnet.
By measuring the time period, one can compare the magnetic moments of different magnets or the strengths of different magnetic fields.

(A) The time period of an oscillating 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.
Since the magnets have the same size and mass,their moment of inertia $I$ is the same.
Thus,$T \propto \frac{1}{\sqrt{M}}$,which implies $M \propto \frac{1}{T^2}$.
Given the frequency $f_1 = 10 \text{ oscillations/min}$ and $f_2 = 15 \text{ oscillations/min}$,the time periods are $T_1 = \frac{60}{10} = 6 \text{ s}$ and $T_2 = \frac{60}{15} = 4 \text{ s}$.
The ratio of the magnetic moments is $\frac{M_1}{M_2} = \frac{T_2^2}{T_1^2} = \left(\frac{4}{6}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$.

(C) The time period of an oscillating magnet 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 magnetic field.
When the magnet is cut into four equal parts (two along the axis and two perpendicular to the axis),for each part:
The new pole strength $m' = \frac{m}{2}$ and the new length $l' = \frac{l}{2}$.
The new magnetic moment $M' = m' \times l' = \frac{m}{2} \times \frac{l}{2} = \frac{M}{4}$.
The new mass $w' = \frac{w}{4}$.
The new moment of inertia $I' = \frac{w' (l')^2}{12} = \frac{(w/4) (l/2)^2}{12} = \frac{1}{16} \times \frac{wl^2}{12} = \frac{I}{16}$.
Substituting these into the formula for the new time period $T'$:
$T' = 2\pi \sqrt{\frac{I'}{M'B_H}} = 2\pi \sqrt{\frac{I/16}{(M/4)B_H}} = 2\pi \sqrt{\frac{I}{4MB_H}} = \frac{1}{2} \times 2\pi \sqrt{\frac{I}{MB_H}} = \frac{T}{2}$.

(C) The time period $T$ of a vibration magnetometer for two magnets placed in a magnetic field is given by the formula:
$T = 2\pi \sqrt{\frac{I_1 + I_2}{(M_1 + M_2)B_H}}$ for like poles together,and $T = 2\pi \sqrt{\frac{I_1 + I_2}{(M_1 - M_2)B_H}}$ for dissimilar poles together.
Given that the magnets have equal pole strength and length,their magnetic moments are equal,i.e.,$M_1 = M_2 = M$.
Substituting this into the formula for dissimilar poles:
$T = 2\pi \sqrt{\frac{I_1 + I_2}{(M - M)B_H}} = 2\pi \sqrt{\frac{I_1 + I_2}{0 \cdot B_H}} = 2\pi \sqrt{\infty}$.
Therefore,the time period $T = \infty$.

(B) The time period $T$ 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,the horizontal component of the Earth's magnetic field $B_H$ is zero.
Substituting $B_H = 0$ into the formula,we get $T = 2\pi \sqrt{\frac{I}{M \times 0}} = \infty$.
Therefore,the time period becomes infinity at the magnetic poles.

(C) In a vibration magnetometer,the magnet is suspended such that it oscillates under the influence of an external magnetic field. The time period of oscillation is given by $T = 2\pi \sqrt{\frac{I}{MH}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $H$ is the horizontal component of the Earth's magnetic field.
If there are twists in the suspension fibre,an additional restoring torque is introduced due to the elasticity of the fibre. This increases the effective restoring force,which decreases the time period of oscillation.
Therefore,to ensure the magnet oscillates solely due to the magnetic torque (i.e.,to measure the true time period),the twists in the suspension fibre must be removed so that the magnet may vibrate freely without the influence of the fibre's torsional constant.

(A) The time period of oscillation of a magnet in 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.
From this formula,we can 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 $T_1 = 2 \, s$ and $M_2 = 4M_1$.
Substituting these values,we get $\frac{2}{T_2} = \sqrt{\frac{4M_1}{M_1}} = \sqrt{4} = 2$.
Solving for $T_2$,we get $T_2 = \frac{2}{2} = 1 \, s$.

(A) The time period of an oscillating magnetic needle is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$.
Given:
$I = 40 \, g \cdot cm^2 = 40 \times 10^{-3} \, kg \times (10^{-2} \, m)^2 = 40 \times 10^{-7} \, kg \cdot m^2 = 4 \times 10^{-6} \, kg \cdot m^2$.
$T = 3 \, s$.
$B_H = 3.6 \times 10^{-5} \, Wb/m^2$.
Substituting the values into the formula:
$3 = 2\pi \sqrt{\frac{4 \times 10^{-6}}{M \times 3.6 \times 10^{-5}}}$
Squaring both sides:
$9 = 4\pi^2 \left( \frac{4 \times 10^{-6}}{M \times 3.6 \times 10^{-5}} \right)$
Using $\pi^2 \approx 10$:
$9 = 40 \left( \frac{4 \times 10^{-6}}{M \times 3.6 \times 10^{-5}} \right)$
$9 = \frac{160 \times 10^{-6}}{M \times 3.6 \times 10^{-5}}$
$M = \frac{160 \times 10^{-6}}{9 \times 3.6 \times 10^{-5}} = \frac{160 \times 10^{-1}}{32.4} \approx 0.5 \, A \cdot m^2$.

(A) vibration magnetometer is used to determine the magnetic moment of a magnet or the horizontal component of the Earth's magnetic field. To ensure that the magnet oscillates under the influence of the Earth's horizontal magnetic field $(B_H)$ alone,the instrument must be set in the magnetic meridian. If it is not aligned with the magnetic meridian,additional forces or torques may act on the magnet,leading to inaccurate measurements.

(B) The time period of a vibrating magnet is given by the formula $T = 2\pi \sqrt{\frac{I}{MH}}$, where $I$ is the moment of inertia, $M$ is the magnetic moment, and $H$ is the horizontal component of the Earth's magnetic field.
When a brass bar is placed on the vibrating magnet, the total moment of inertia $(I)$ of the system increases.
Since the time period $T$ is directly proportional to the square root of the moment of inertia $(T \propto \sqrt{I})$, an increase in $I$ leads to an increase in the time period $T$.
Therefore, the time period of the vibrating magnet increases.

(B) The time period of a magnetic needle vibrating in a uniform magnetic field $B_H$ is given by the formula:
$T = 2\pi \sqrt{\frac{I}{MB_H}}$
where $I$ is the moment of inertia and $M$ is the magnetic moment.
From this relation,we can see that $T \propto \frac{1}{\sqrt{B_H}}$.
Therefore,the ratio of the time periods is:
$\frac{T_1}{T_2} = \sqrt{\frac{B_{H2}}{B_{H1}}}$
Given $B_{H1} = H$ and $B_{H2} = 4H$,we have:
$\frac{T}{T_2} = \sqrt{\frac{4H}{H}} = \sqrt{4} = 2$
$T_2 = \frac{T}{2}$
Thus,the new time period is $T/2$.

(B) 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 and $M$ is the net magnetic moment.
Since both magnets have the same mass,length,and breadth,their moment of inertia $I$ is the same.
In the sum position,the net magnetic moment is $M_{sum} = M + 2M = 3M$. Given $T_1 = 3 \, s$,we have $3 \propto \frac{1}{\sqrt{3M}}$.
In the difference position,the net magnetic moment is $M_{diff} = 2M - M = M$. Let the new time period be $T_2$.
Thus,$T_2 \propto \frac{1}{\sqrt{M}}$.
Taking the ratio: $\frac{T_2}{T_1} = \sqrt{\frac{3M}{M}} = \sqrt{3}$.
Therefore,$T_2 = 3 \times \sqrt{3} = 3\sqrt{3} \, s$.

(D) In the 'sum and difference method' for comparing magnetic moments ($M_1$ and $M_2$) using a vibration magnetometer,the magnets are placed in the same orientation (sum) and opposite orientation (difference).
The ratio is given by $\frac{M_1}{M_2} = \frac{T_2^2 + T_1^2}{T_2^2 - T_1^2}$.
This method is advantageous because:
$1$. The moment of inertia $(I)$ of the magnets does not need to be calculated,which eliminates errors associated with its determination.
$2$. It requires fewer experimental observations compared to other methods.
$3$. The mathematical calculations involved are relatively simpler and fewer in number.

(C) The time period $T$ of a magnet in 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 horizontal component of the Earth's magnetic field.
Since the magnets are similar in size,shape,and mass,their moment of inertia $I$ remains the same.
Thus,$T \propto \frac{1}{\sqrt{M}}$.
Given $T_1 = 1.5 \ s$ and $M_2 = \frac{M_1}{4}$.
Using the ratio formula: $\frac{T_1}{T_2} = \sqrt{\frac{M_2}{M_1}}$.
Substituting the values: $\frac{1.5}{T_2} = \sqrt{\frac{M_1/4}{M_1}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,$T_2 = 1.5 \times 2 = 3 \ s$.

(C) The frequency of oscillation $\nu$ of a bar magnet in a vibration magnetometer is given by the formula: $\nu = \frac{1}{2\pi} \sqrt{\frac{MB_H}{I}}$.
Assuming the moment of inertia $I$ is the same for both magnets,we have $\nu \propto \sqrt{M}$.
Given that the frequency of magnet $A$ is twice that of magnet $B$,we have $\nu_A = 2\nu_B$.
Therefore,$\frac{\nu_A}{\nu_B} = \sqrt{\frac{M_A}{M_B}}$.
Substituting the values: $2 = \sqrt{\frac{M_A}{M_B}}$.
Squaring both sides: $4 = \frac{M_A}{M_B}$,which implies $M_A = 4M_B$.

(C) The time period $T$ of a magnet in 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.
Since the magnets are identical in mass,length,and breadth,their moment of inertia $I$ is the same.
Thus,$T \propto \frac{1}{\sqrt{M}}$,which implies $T^2 \propto \frac{1}{M}$ or $M \propto \frac{1}{T^2}$.
Given that the time period of $B$ is twice the time period of $A$,we have $T_B = 2T_A$.
Therefore,the ratio of the magnetic moments is $\frac{M_A}{M_B} = \left(\frac{T_B}{T_A}\right)^2 = \left(\frac{2T_A}{T_A}\right)^2 = 2^2 = 4$.
Hence,the ratio $M_A/M_B$ is $4$.

(C) The frequency of oscillation $n$ for a magnet in a horizontal magnetic field $B_H$ is given by the formula: $n = \frac{1}{2\pi} \sqrt{\frac{MB_H}{I}}$,where $M$ is the magnetic moment and $I$ is the moment of inertia.
From this relation,we can see that $n \propto \sqrt{MB_H}$.
Given that the magnetic moment becomes $M' = 4M$ and the magnetic field becomes $B_H' = 2B_H$.
Substituting these values into the proportionality,we get the new frequency $n'$:
$n' \propto \sqrt{M' B_H'} = \sqrt{(4M)(2B_H)} = \sqrt{8MB_H} = 2\sqrt{2} \sqrt{MB_H}$.
Since $n \propto \sqrt{MB_H}$,it follows that $n' = 2\sqrt{2} n$.

(C) When a magnetic needle is suspended horizontally by an unspun silk fibre,it acts as a magnetic dipole in the Earth's magnetic field.
When the needle is slightly displaced from its equilibrium position,it experiences a restoring torque due to the interaction between its magnetic moment $M$ and the horizontal component of the Earth's magnetic field,denoted as $B_H$.
The torque is given by $\tau = M \times B_H = MB_H \sin \theta$.
For small angles,$\sin \theta \approx \theta$,so $\tau \approx MB_H \theta$.
This torque acts to bring the needle back to its equilibrium position,causing it to oscillate in the horizontal plane. The unspun silk fibre is chosen specifically to minimize any torsion,ensuring that the restoring force is primarily due to the Earth's magnetic field.

(A) The time period of a vibration magnetometer is given by the formula: $T = 2\pi \sqrt{\frac{I}{MH}}$.
Here,$I$ is the moment of inertia and $M$ is the magnetic moment of the magnet,both of which remain constant.
Thus,$T \propto \frac{1}{\sqrt{H}}$,which implies $H \propto \frac{1}{T^2}$.
Given $T_A = 2 \ s$ and $T_B = 3 \ s$,we have the ratio:
$\frac{H_A}{H_B} = \frac{T_B^2}{T_A^2} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$.
Therefore,the ratio $H_A : H_B$ is $9:4$.

(A) The time period $T$ of a bar magnet oscillating in a uniform magnetic field $B_H$ is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB_H}}$.
Here,$I$ is the moment of inertia of the bar magnet,$M$ is the magnetic moment,and $B_H$ is the horizontal component of the earth's magnetic field.
The moment of inertia $I$ for a rectangular bar magnet of mass $m$,length $l$,and breadth $b$ is given by $I = \frac{m(l^2 + b^2)}{12}$.
Substituting $I$ into the time period formula,we get $T = 2\pi \sqrt{\frac{m(l^2 + b^2)}{12MB_H}}$.
Since $T \propto \sqrt{m}$,the time period is directly proportional to the square root of its mass.

(C) The time period of oscillation for 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.
For two magnets $A$ and $B$ with moments $M_1 = 2M$ and $M_2 = M$,and moments of inertia $I_1$ and $I_2$,when placed together:
When like poles are together,the effective magnetic moment is $M_{sum} = M_1 + M_2 = 3M$. The time period is $T_1 = 2\pi \sqrt{\frac{I_1 + I_2}{(M_1 + M_2)B_H}}$.
When unlike poles are together,the effective magnetic moment is $M_{diff} = M_1 - M_2 = 2M - M = M$. The time period is $T_2 = 2\pi \sqrt{\frac{I_1 + I_2}{(M_1 - M_2)B_H}}$.
Taking the ratio: $\frac{T_1}{T_2} = \sqrt{\frac{M_1 - M_2}{M_1 + M_2}} = \sqrt{\frac{2M - M}{2M + M}} = \sqrt{\frac{M}{3M}} = \frac{1}{\sqrt{3}}$.

(A) vibration magnetometer operates based on the restoring torque acting on a suspended bar magnet in a uniform magnetic field.
When a bar magnet is suspended in a horizontal magnetic field (like the Earth's magnetic field) and slightly displaced from its equilibrium position,a restoring torque $\tau = -MB \sin \theta$ acts on it.
This torque causes the magnet to perform simple harmonic oscillations about the magnetic meridian.
Therefore,the principle is the torque acting on the bar magnet.

(A) When a steady current $I$ is passed through the coil of a tangent galvanometer,it produces a magnetic field $B$ at the center of the coil,which is perpendicular to the horizontal component of the Earth's magnetic field $H$.
Under the influence of these two mutually perpendicular magnetic fields,the magnetic needle deflects by an angle $\theta$.
According to the tangent law,$B = H \tan \theta$.
Since $B$ is directly proportional to the current $I$ $(B = kI)$,we get $I = (H/k) \tan \theta$,which means $I \propto \tan \theta$.
Thus,the tangent galvanometer is primarily used to measure steady electric currents.

(B) For a tangent galvanometer,the current $I$ is given by the formula: $I = \frac{2rB_H}{\mu_0 n} \tan \theta$,where $r$ is the radius,$B_H$ is the horizontal component of the earth's magnetic field,$n$ is the number of turns,and $\theta$ is the deflection.
Rearranging for $\tan \theta$: $\tan \theta = \frac{I \mu_0 n}{2r B_H}$.
Given: $I = 0.1$ $A$,$n = 50$,$r = 4$ $cm = 0.04$ $m$,$B_H = 7 \times 10^{-5}$ $T$,and $\mu_0 = 4\pi \times 10^{-7}$ $T \cdot m/A$.
Substituting the values: $\tan \theta = \frac{0.1 \times 4\pi \times 10^{-7} \times 50}{2 \times 0.04 \times 7 \times 10^{-5}}$.
$\tan \theta = \frac{0.1 \times 4 \times 3.14159 \times 10^{-7} \times 50}{0.08 \times 7 \times 10^{-5}} = \frac{6.283 \times 10^{-6}}{0.56 \times 10^{-5}} = \frac{0.6283}{0.56} \approx 1.122$.
Therefore,$\theta = \tan^{-1}(1.122) \approx 48.2^o$.

(D) The time period $T$ of a bar magnet oscillating in a magnetic field is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB}}$.
Here,$M = 5 \times 10^{-5} \, \text{Wb} \cdot \text{m}$,$B = 8\pi \times 10^{-4} \, \text{T}$,and $T = 15 \, \text{s}$.
Rearranging the formula to solve for the moment of inertia $I$:
$I = \frac{T^2 MB}{4\pi^2}$.
Substituting the given values:
$I = \frac{(15)^2 \times (5 \times 10^{-5}) \times (8\pi \times 10^{-4})}{4\pi^2}$.
$I = \frac{225 \times 40\pi \times 10^{-9}}{4\pi^2} = \frac{225 \times 10 \times 10^{-9}}{\pi} = \frac{2250 \times 10^{-9}}{3.14159}$.
$I \approx 7.16 \times 10^{-7} \, \text{kg} \cdot \text{m}^2$.

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

(D) For a tangent galvanometer,the current $I$ is given by $I = K \tan \theta$,where $K = \frac{2rB_H}{\mu_0 n}$.
Since the radius $r$ and horizontal component of Earth's magnetic field $B_H$ are the same for both,$K \propto \frac{1}{n}$.
Thus,$I = \frac{C}{n} \tan \theta$,where $C$ is a constant.
Since the galvanometers are connected in series,the current $I$ is the same for both: $I_1 = I_2$.
Therefore,$\frac{C}{n_1} \tan 60^{\circ} = \frac{C}{n_2} \tan 45^{\circ}$.
$\frac{\sqrt{3}}{n_1} = \frac{1}{n_2}$.
Rearranging the terms,we get $\frac{n_1}{n_2} = \frac{\sqrt{3}}{1}$.

(B) 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.
When bar $Q$ is placed on top of $P$,the new moment of inertia becomes $I' = I + I = 2I$.
For the time period to remain unchanged $(T' = T)$,the new magnetic moment $M'$ must satisfy the relation $T' = 2\pi \sqrt{\frac{2I}{M'B_H}} = 2\pi \sqrt{\frac{I}{MB_H}}$.
Squaring both sides,we get $\frac{2I}{M'} = \frac{I}{M}$,which implies $M' = 2M$.
This indicates that the magnetic moment of the system has doubled,which happens if $Q$ is an identical magnet placed with its poles aligned (north on north,south on south) with $P$.

(A) The frequency of oscillation $\nu$ of a vibration magnetometer is given by the formula $\nu = \frac{1}{2\pi} \sqrt{\frac{MB}{I}}$,where $M$ is the magnetic moment,$B$ is the magnetic field strength,and $I$ is the moment of inertia.
From this relation,we can see that $\nu \propto \sqrt{B}$.
Let the initial frequency be $\nu_1$ and the initial magnetic field be $B_1$. Let the final frequency be $\nu_2$ and the final magnetic field be $B_2 = 4B_1$.
Then,$\frac{\nu_2}{\nu_1} = \sqrt{\frac{B_2}{B_1}} = \sqrt{\frac{4B_1}{B_1}} = \sqrt{4} = 2$.
Therefore,$\nu_2 = 2\nu_1$.
The frequency becomes twice its original value.

(B) In a tangent galvanometer,the current $I$ is proportional to the tangent of the deflection angle $\theta$,given by $I = K \tan \theta$,where $K$ is the reduction factor.
Given $I_1 \propto \tan \theta_1$ and $I_2 \propto \tan \theta_2$.
We have $\frac{I_1}{I_2} = \frac{\tan \theta_1}{\tan \theta_2}$.
Given $\theta_1 = 45^{\circ}$ and $I_2 = \frac{I_1}{\sqrt{3}}$,so $\frac{I_1}{I_1/\sqrt{3}} = \frac{\tan 45^{\circ}}{\tan \theta_2}$.
This simplifies to $\sqrt{3} = \frac{1}{\tan \theta_2}$,which means $\tan \theta_2 = \frac{1}{\sqrt{3}}$.
Therefore,$\theta_2 = 30^{\circ}$.
The change in deflection is $\theta_1 - \theta_2 = 45^{\circ} - 30^{\circ} = 15^{\circ}$.
Thus,the deflection decreases by $15^{\circ}$.

(D) When a magnetic needle is placed in two mutually perpendicular magnetic fields $F$ and $H$,it aligns itself along the direction of the resultant magnetic field.
Let the angle made by the needle with the field $F$ be $\theta = 60^\circ$.
From the geometry of the vector components,the tangent of the angle $\theta$ is given by the ratio of the perpendicular component to the base component:
$\tan \theta = \frac{H}{F}$
Substituting the given value $\theta = 60^\circ$:
$\tan 60^\circ = \frac{H}{F}$
Since $\tan 60^\circ = \sqrt{3}$,we have:
$\sqrt{3} = \frac{H}{F}$
Therefore,the ratio $\frac{F}{H}$ is:
$\frac{F}{H} = \frac{1}{\sqrt{3}}$
Thus,the correct option is $(d)$.

(D) Let the initial magnetic field be $B_1 = 1 \, T$ and the perpendicular magnetic field be $B_2 = \sqrt{3} \, T$.
When a magnetic needle is placed in two mutually perpendicular magnetic fields,it aligns itself along the direction of the resultant magnetic field.
The angle $\theta$ that the needle makes with the direction of $B_1$ is given by the tangent law:
$\tan \theta = \frac{B_2}{B_1}$
Substituting the given values:
$\tan \theta = \frac{\sqrt{3}}{1} = \sqrt{3}$
Since $\tan 60^o = \sqrt{3}$,we have:
$\theta = 60^o$

(C) When a current $I$ is passed through the coil of a tangent galvanometer,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 these two crossed magnetic fields $B$ and $H$,the magnetic needle of the galvanometer undergoes a deflection $\theta$,which is governed by the tangent law.
According to the tangent law,the relationship is given by $B = H \tan \theta$.
Since $B \propto I$,we get $I = k \tan \theta$,where $k$ is the reduction factor of the galvanometer.
This clearly indicates that a tangent galvanometer is an instrument used for the measurement of electric current in a circuit.