NCERT Exemplar Class 12 Physics Solutions Chapter 5 Magnetism and Matter

NCERT Exemplar Class 12 Physics Solutions Chapter 5 MCQ I

Question:1

A toroid of n turns, mean radius R and cross-sectional radius a carries current I. It is placed on a horizontal table taken as x-y plane. Its magnetic moment m
A. is non-zero and points in the z-direction by symmetry.
B. points along the axis of the toroid $m =m\phi$.
C. is zero, otherwise, there would be a field falling as $\frac{1}{r^{3}}$ at large distances outside the toroid.
D. is pointing radially outwards.

Answer:

The answer is option (c)
The toroid is a ring-shaped solenoid, and the magnetic field is confined inside the body of the toroid. The magnetic moment inside and outside the toroid is zero as there is no current enclosed in those spaces.

Question:2

The magnetic field of Earth can be modelled by that of a point dipole placed at the center of the Earth. The dipole axis makes an angle of $11.3^{\circ}$ with the axis of Earth. At Mumbai, declination is nearly zero.

Then,
A. the declination varies between $11.3^{\circ}$ W to $11.3^{\circ}$ E.
B. the least declination is$0^{\circ}$°.
C. the plane defined by dipole axis and Earth axis passes through Greenwich.
D. declination averaged over Earth must always be negative.

Answer:

The answer is the option (a) Earth’s magnetic field is similar to a magnetic dipole and we can simulate the nature of Earth’s magnetic field by assuming a dipole at the center of Earth. The axis of the dipole is at an angle of $11.3^{\circ}$with respect to the axis of rotation.

Question:3

In a permanent magnet at room temperature
A. magnetic moment of each molecule is zero.
B. the individual molecules have a non-zero magnetic moment which is all perfectly aligned.
C. domains are partially aligned.
D. domains are all perfectly aligned

Answer:

The answer is the option (d)
Permanent magnet behaves like a ferromagnetic substance at room temperature and in the absence of an external magnetic field, the domains are spread randomly. However, when put in a magnetic field, the domains are arranged to align with the external field and even when the magnet is taken out of the field, the domains stay intact.

Question:4

Consider the two idealized systems: (i) a parallel plate capacitor with large plates and small separation and (ii) a long solenoid of length L >> R, the radius of the cross-section.

In (i) E is ideally treated as a constant between plates and zero outside.
In (ii) the magnetic field is constant inside the solenoid and zero outside.
These idealized assumptions, however, contradict fundamental laws as below:

A. case (i) contradicts Gauss’s Law for electrostatic fields.
B. case (ii) contradicts Gauss’s Law for magnetic fields.
C. case (i) agrees with $\int E.dl=0$
D. case (ii) contradicts $\int H.dl=l_{en}$

Answer:

The answer is the option (b)
Gauss’ Law for electrostatic field isn’t violated as electric fields don’t require to be in continuous closed paths.
$\oint_{}S E.ds=\frac{q}{\varepsilon_ 0}$
However, Gauss’ Law for magnetic field is violated as magnetic fields need to be in continuous closed paths.
$\oint_{}S B.ds=0$

Question:5

A paramagnetic sample shows a net magnetization of 8 Am–1 when placed in an external magnetic field of 0.6T at a temperature of 4K. When the same sample is placed in an external magnetic field of 0.2 T at a temperature of 16K, the magnetization will be
A. $\frac{32}{3}Am^{-1}$

B. $\frac{2}{3}Am^{-1}$
C. $6Am^{-1}$
D. $2.4Am^{-1}$

Answer:

The answer is the option (b)
$X \propto \frac{B}{T}$
$X_1=8 Am^{-1} and B_1=0.6T and T_1=4K$
At $T_2=16K and B_2=0.2T$
$X_2=\frac{X_1T_1B_2}{T_2B_1}=\frac{2}{}3 Am^{-1}$

NCERT Exemplar Class 12 Physics Solutions Chapter 5 MCQ II

Question:6

S is the surface of a lump of magnetic material.
A. Lines of B are necessarily continuous across S.
B. Some lines of B must be discontinuous across S.
C. Lines of H are necessarily continuous across S.
D. Lines of H cannot all be continuous across S.

Answer:

The correct answer are the options (a, d)
Magnetic field lines form continuous closed loops. Magnetic intensity varies for inside and outside the lump.

Question:7

The primary origin(s) of magnetism lies in
A. atomic currents.
B. Pauli exclusion principle.
C. polar nature of molecules.
D. intrinsic spin of the electron.

Answer:

The correct answer are the options (a, d)
Magnetism is produced due to the movement of charged particles. Electrons revolving around the nucleus produce a current, which induces the magnetic nature of materials.

Question:8

A long solenoid has 1000 turns per meter and carries a current of 1 A. It has a soft iron core of $\mu _{r}=1000$. The core is heated beyond the Curie temperature, $T_c$.
A. The H field in the solenoid is (nearly) unchanged, but the B field decreases drastically.
B. The H and B fields in the solenoid are nearly unchanged.
C. The magnetization in the core reverses direction.
D. The magnetization in the core diminishes by a factor of about 10⁸

Answer:

The correct answer are the options (a, d)
n=1000turns m
$\mu _{r}=1000$
$H=nI=1000\times 1=1000$
$B=\mu _0 \mu _rnI=\left (\mu _0nI \right )\mu _r=K\mu _r$
$B\propto \mu_r$
Beyond Curie Temperature, ferromagnetic substance behaves like paramagnetic substance.
$(X_m)_{ferro}=10^3$
$(X_m)_{para}=10^{-5}$
Magnetisation diminishes by $10^{-8}$ times.

Question:9

The essential difference between electrostatic shielding by a conducting shell and magnetostatic shielding is due to
A. electrostatic field lines can end on charges and conductors have free charges.
B. lines of B can also end, but conductors cannot end them.
C. lines of B cannot end on any material, and perfect shielding is not possible.
D. shells of high permeability materials can be used to divert lines of B from the interior region.

Answer:

The correct answer is the options (a, c, d)
Magnetic field lines always exist as continuous closed loops. High permeability magnetic materials can repel magnetic field lines to get a region devoid of the magnetic field.

Question:10

Let the magnetic field on Earth be modelled by that of a point magnetic dipole at the center of Earth. The angle of dip at a point on the geographical equator
A. is always zero.
B. can be zero at specific points.
C. can be positive or negative.
D. is bounded.

Answer:

The correct answer are the options (b,c,d)
Earth’s magnetic dipole is at an angle of $11.3^{\circ}$ with respect to the axis of rotation. However, the South of dipole is near the Geographical North Pole and North of dipole is near the Geographical South Pole. Though the magnetic field created by this dipole will be zero at its equatorial plane (not the equator), it will pass the equator at two points, where the magnetic field will be zero. Angle of dip flips in sign at the equatorial plane (+ve on one side and -ve on the other)

NCERT Exemplar Class 12 Physics Solutions Chapter 5 Very Short Answer

Question:11

A proton has spin and magnetic moment just like an electron. Why then its effect is neglected in magnetism of materials.

Answer:

Magnetic moment of electron and proton are:
$M_e=\frac{eh}{4\pi m_e}$ &. $M_p=\frac{eh}{4\pi m_p}$
As $m_p >> m_e$, the Magnetic moment of proton is negligible compared to electron.

Question:12

A permanent magnet in the shape of a thin cylinder of length 10 cm has $M = 10^{6}\frac{A}{m}$ Calculate the magnetization current I_M.

Answer:

$M = 10^{6}\frac{A}{m}$, $l=0.1m$
I_M=Magnetisation current
$M=\frac{I_M}{}l$
$I_M=M\times I=10^5A$

Question:13

Explain quantitatively the order of magnitude difference between the diamagnetic susceptibility of $N_{2}(5 \times 10^{-9})$ (at STP) and $Cu ( 10^{-5})$

Answer:

$\rho_ N=\frac{28g}{22.4L}=\frac{28}{22400}gcm^{-3}$
$\rho _{Cu}=8gcm^{-3}$
$\rho _{N}\rho_{Cu}=\frac{28}{22400\times 8}=1.6\times 10^{-4}$
$\frac{x_N}{x_{Cu}}=\frac{5\times 10^{-9}}{10^{-5}}=5\times 10^{-4}$
$x=\frac{\text{Intensity of Magnetisation(M)}}{\text{Magnetising field Intensity (H)}}=\frac{\frac{\text{Magnetic moment( m)}}{\text{Volume(V)}}}{}H=\frac{m}{HV}$
$V=m'\rho$
$x=m\rho Hm'$
$x\propto \rho$
$\frac{x_{N}}{x_{Cu}}=\frac{\rho _N}{ \rho _ {Cu}}=1.6\times 10^{-4}$

Question:14

From molecular viewpoint, discuss the temperature dependence of susceptibility for diamagnetism, paramagnetism and ferromagnetism.

Answer:

Diamagnetism occurs due to electron’s motion in its orbit and the external magnetic field being opposite to each other, making the net magnetism zero. Temperature does not impact the susceptibility of diamagnetism.
In paramagnetic and ferromagnetic substances, magnetism generated by electron syncs up with the external magnetic field, so there is a net increase in the magnetism. Increase in temperature disturbs the atomic alignment decreasing the susceptibility

Question:15

A ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet. (i) In which direction will it move? (ii) What will be the direction of its magnetic moment?

Answer:

Both liquid nitrogen and the superconducting material are diamagnetic in nature. This does not change even when the superconducting material is dipped into liquid nitrogen. In presence of an external magnetic field, superconducting material will be repelled opposite to the direction of the magnetic field.

NCERT Exemplar Class 12 Physics Solutions Chapter 5 Short Answer

Question:16

Verify the Gauss’s law for magnetic field of a point dipole of dipole moment m at the origin for the surface which is a sphere of radius R.

Answer:

To prove Gauss'Law $B.ds=0.$
Dipole's Magnetic moment at origin O is along the z-axis.
Assume P to be a point at a distance r from Origin and OP is at an angle θ with the z-axis.
Component of M along $OP=M\cos\theta$
Now, the magnetic field induction at P due to the dipole of moment $M \cos\theta is B$
$=\frac{\mu _0}{4\pi} 2M\frac{\cos\theta}{r^3 }\widehat{r}$
From the diagram, r is the radius of a sphere with centre at O lying in yz-plane.
An elementary area dS is taken at P $dS=r\left (r\sin \theta d\theta \right ) \widehat{r}=\left (r^2\sin\theta d\theta \right ) \widehat{r }$
$\oint B.ds=\oint \frac{\mu _{0}}{4 \pi }2M\frac{\cos \theta }{r^{3}}\widehat{r}(r^{2}\sin \theta d \theta )\widehat{r}$
$=\frac{ \frac{\mu _0}{4\pi}M}{r} \int_{0}^{2\pi }2\sin \theta \cos\theta d\theta$
$=\frac{\mu _{0}}{4\pi}\frac{ M}{r} \int_{0}^{2\pi }\sin 2\theta d\theta$
$=-\frac{\mu _{0}}{4\pi}\frac{ M}{2r} \int_{0}^{2\pi }\left ( -\frac{\cos 2 \theta }{2} \right )$
$=-\frac{\mu _{0}}{4\pi}\frac{ M}{2r}[\cos 4 \pi - \cos 0]=-\frac{\mu _{0}}{4\pi}\frac{ M}{2r} [1-1]=0$

Question:17

Three identical bar magnets are riveted together at center in the same plane as shown in Figure. This system is placed at rest in a slowly varying magnetic field. It is found that the system of magnets does not show any motion. The north-south poles of one magnet is shown in the Figure. Determine the poles of the remaining two.

Answer:

The system is considered to be in state of stable equilibrium if both net force and net torque on system is 0. The only way to make such a system feasible is to have the following configuration:

Question:18

Suppose we want to verify the analogy between electrostatic and magnetostatic by an explicit experiment. Consider the motion of (i) electric dipole p in an electrostatic field E and (ii) magnetic dipole m in a magnetic field B. Write down a set of conditions on E, B, p, m so that the two motions are verified to be identical. (Assume identical initial conditions)

Answer:

$p_E \sin \theta =\mu B \sin\theta$
$pE=\mu B$
$E=cB$
$pcB=\mu B$
$p=\mu/ c$

Question:19

A bar magnet of magnetic moment m and moment of inertia I (about center, perpendicular to length) is cut into two equal pieces, perpendicular to length. Let T be the period of oscillations of the original magnet about an axis through the mid-point, perpendicular to length, in a magnetic field B. What would be the similar period T’ for each piece?

Answer:

Time period in this type of S.H.M is
$T=2\pi \sqrt{\frac{I}{MB} }$
where, T=time period
I=moment of Inertia
m=mass of magnet
B=magnetic field
$I=\frac{ml^2}{12}$
When the magnet is cut into two equal pieces, perpendicular to length the M.O.I of each piece
of magnet about an axis perpendicular to the length passing through its centre is
$I'=\frac{m/2}{12}\left (\frac{l}{2} \right )^2\times =\frac{ml^2}{12}\times \frac{1}{}8=\frac{I}{}8$
Magnetic dipole momen
t $M'=\frac{M}{}2$
Its time period of oscillation is

$T'=\frac{T}{2}$

Question:20

Use (i) the Ampere’s law for H and (ii) continuity of lines of B, to conclude that inside a bar magnet, (a) lines of H run from the N pole to S pole, while (b) lines of B must run from the S pole to N pole.

Answer:

Consider a magnetic field line going through the bar magnet. We know that it must be a closed-loop.

$\int_{Q}^{P}H.dl=\int_{Q}^{P}\frac{\overrightarrow{B}}{\overrightarrow{\mu_{0}}}\overrightarrow{dl}$
B and dl are at an acute angle to each other inside the magnet.
i.e., $\int_{Q}^{P}H.dl=\int_{Q}^{P}\frac{\overrightarrow{B}}{\overrightarrow{\mu_{0}}}\overrightarrow{dl}>0$ i.e. positive
Hence, the lines of B must run from south pole S to north pole N inside the bar magnet.
According to ampere's law
$\\\oint \overrightarrow{H}.\overrightarrow{dl} =0\\\oint \overrightarrow{H}.\overrightarrow{dl} =\int_{P}^{Q}\overrightarrow{H}.\overrightarrow{dl}+\int_{Q}^{P}\overrightarrow{H}.\overrightarrow{dl}$
As $\int_{P}^{Q}\overrightarrow{H}.\overrightarrow{dl}>0 so ,\int_{Q}^{P}\overrightarrow{H}.\overrightarrow{dl}<0$ negative
If angle between H and dl is more than $90^{\circ}$, so that $\cos \theta$ is negative
(meaning that the lines run from N pole to S pole inside the bar magnet)

NCERT Exemplar Class 12 Physics Solutions Chapter 5 Long Answer

Question:21

Verify the Ampere’s law for magnetic field of a point dipole of dipole moment $m = \widehat{m}k$. Take C as the closed curve running clockwise along (i) the z-axis from z = a > 0 to z = R; (ii) along the quarter circle of radius R and center at the origin, in the first quadrant of x-z plane; (iii) along the x-axis from x = R to x = a, and (iv) along the quarter circle of radius an and center at the origin in the first quadrant of x-z plane.

Answer:

Assume the x-z plane (shown below). All points from P to Q lie on the axial line NS placed at the origin.


The magnetic field at a distance r is
$B=\frac{\mu _02\left |M \right |}{4\pi r^3}=\frac{\mu_0M}{2\pi r^3 }$
Along z-axis from P to Q
$\int_{P}^{Q}B.dl=\int_{P}^{Q} B.dl\cos 0^{\circ}=\int_{a}^{R}Bdz=\int_{a}^{R}\frac{\mu_{0}M}{2\pi r^3}dz=\frac{\mu _0M}{2\pi} \left (-\frac{1}{}2 \right )\left (\frac{1}{R^2}-\frac{1}{a^2} \right )=\frac{\mu _0M}{4\pi} \left (\frac{1}{a^2}-\frac{1}{R^2} \right )$
ii Along the quarter circle QS (radius R)


Consider point A to lie on the equatorial line of magnetic dipole of moment M sinθ.
Magnetic field at A is
$B_t=\frac{\frac{\mu _0}{4\pi} M\sin\theta}{R^{3}} ;dl=Rd\theta$
$B_r=\frac{\frac{\mu _0}{2\pi} m\cos \theta}{R^{3}}$
$\int_{0}^{\frac{\pi}{2}}B.dl=\int_{0}^{\frac{\pi}{2}} B_tdl\cos 0^{\circ}+ \int_{0}^{\frac{\pi}{2}} B_ r dl\cos 90^{\circ} =\int_{0}^{\frac{\pi}{2}}\frac{\mu _0m}{4\pi R^3}\sin \theta (Rd\theta )=\frac{\mu _{0}m}{4 \pi R^2} \int_{0}^{\frac{\pi}{2}}\sin\theta d\theta =\frac{\mu_{0}m}{4\pi R^2 }$
iii Along x-axis over the path ST, consider the figure given below

From the figure, every point lies on the equatorial line of the magnetic dipole.
Magnetic field induction at a point distance x from the dipole is
$B=\frac{\frac{\mu _0}{4\pi} M}{x^3 }$
$\int_{S}^{T}B.dl=\int_{R}^{a}-\frac{\mu _0M}{4\pi x^3} =0$ angle between-M and dl is $90^{\circ}$
iv Along the quarter circle TP of radius a.
Let's consider the figure given below


From case ii we get line integral of B along the quarter circle TP of radius a
is circular arc TP.
$\int B.dl=\int_{\frac{\pi}{2}}^{0} \frac{\mu _{0}}{4\pi} M\frac{\sin \theta }{a^3}ad\theta =\frac{\mu _0M}{4\pi a^2} \int_{\frac{\pi}{2}}^{0} \sin\theta d\theta =\frac{\frac{\mu _{0}}{4\pi} M}{a^2} \int_{\frac{\pi}{2}}^{0}[-\cos\theta ]$
$=-\frac{\frac{\mu _0}{4\pi} M}{a^2 }$
$\int B.dl= \int_{P}^{Q}B.dl+\int_{Q}^{S}B.dl+\int_{S}^{T}B.dl+\int_{T}^{P}B.dl$
$=\frac{\mu _0M}{4}\left [\frac{1}{a^2}-\frac{1}{R^2} \right ]+\frac{\mu _0M}{4\pi R^2}+0+\left (-\frac{\mu _0M}{4\pi R^2} \right )=0$

Question:22

What are the dimensions of χ, the magnetic susceptibility? Consider an H-atom. Guess an expression for χ, up to a constant by constructing a quantity of dimensions of χ, out of parameters of the atom: e, m, v, R and μ0. Here, m is the electronic mass, v is electronic velocity, R is Bohr radius. Estimate the number so obtained and compare with the value of $\left | x \right |\sim 10^{5}$ for many solid materials.

Answer:

$X_m=\frac{I}{}H=\frac{\text{Intensity of magnetisation}}{\text{Magnetising force }}$
As I and H both have same units and dimensions, hence χ has no dimensions.
Here, χ is related with e, m, v, R and $\mu _{0}$
From Biot-Savart'slaw,
$dB=\frac{\mu _0}{4\pi} Idl\frac{\sin \theta }{r^2 }$
$\mu _0=\frac{4\pi r^2dB}{Idl\sin \theta }=\frac{4\pi r^2}{Idl \sin\theta }\times \frac{F}{qv \sin \theta }$
Dimensions of $\mu _{0}=\frac{L^2\times [MLT^{-2}]}{[QT^{-1}][L]\times 1\times Q[LT^{-1}]\times 1}=[MLQ^{-2} ]$
where Q is the dimension of charge
As χ is dimensionless, it should have no involvement of charge Q in its dimensional formula.
It will be so if μ0 and e together should have the value $\mu _0e^2$, as e has the dimensions of charge.
Let $X= \mu _0e^2 m^av^bR^c .......i$
where a, b ,c are the power of m,v and R respectively
$[M^0L^0T^0A^0T^0]=[MLA^{-2}T^{-2}]\times [A^2T^2][M]^a\times [LT^{-1}]^b\times [L]^c =[M^{1+a}L^{1+b+c}T^{-b}A^0 ]$
Equating the powers we get 0=1+a
a=-1
0=1+b+c
0=-b
b=0
1+0+c=0
c=-1
Putting values in equation i we get $X=\mu _{0}e^2m^{-1}v^2R^{-1}=\frac{\mu _0e^2}{mR }$
Here, $\mu _0=4\pi \times 10^{-7} Tm A^{-1 }$
$e=1.6\times 10^{-19} C$
$m=9.1\times 10^{-31} kg$
$R=10^{-10}m X=\left (4\pi \times 10^{-7} \right )\times \frac{(1.6\times 10^{-19})^{2}}{9.1\times 10^{-31}\times 10^{-10}}\approx 10^{-4}$

Question:23

Assume the dipole model for earth’s magnetic field B which is given by BV = vertical component of magnetic field
$=\frac{\mu _{0}}{ 4 \pi}\frac{2m \cos \theta }{r^{3}}$

BH = Horizontal component of magnetic field
$=\frac{\mu _{0}}{ 4 \pi}\frac{ \sin \theta m }{r^{3}}$
$\theta = 90^{\circ}$– latitude as measured from magnetic equator.
Find loci of points for which (i) B is minimum; (ii) dip angle is zero; and (iii) dip angle is $\pm 45^{\circ}$

Answer:

$(a)B_v=\mu _{0}2m\frac{\cos \theta}{ 4\pi r^3 }$
$B_H=\frac{\mu _{0}}{4 \pi}\frac{m\sin \theta}{ r^{3} }$
These are the components of B to a net magnetic field will be
$B=\sqrt{B_v^2+ B_H^2}=\frac{\mu_{0}m}{4\pi r^3}[3\cos 2 \theta +1]^{\frac{1}{}2}$
From the above equation, the value of B is minimum, if
$\cos \theta =\frac{\pi}{}2$
$\theta =\frac{\pi}{}2$
Thus, B is minimum at the magnetic equator.
(b) Angle of dip
$\tan \delta =\frac{B_v}{B_H}=\frac{\mu _{0}}{4 \pi }\frac{.2m \frac{\cos \theta }{r^3}}{\frac{\mu _{0}}{4\pi }\sin \theta .\frac{m}{r^{3}}} =2\cot \theta .........i$
$\tan \delta =2\cot \theta$
For dip angle is zero $\tan \delta =0$
$\cot \theta =0 , \theta =\frac{\pi}{}2$
For this value of θ angle of dip is vanished. It means that locus is again magnetic equator.

(c)

$tan \delta =\frac{B_v}{B_H}$

Angle of dip

$\delta =\pm 45^{\circ}$
$\frac{ B_v}{B_H}=tan \pm 45^{\circ}$
$\frac{ B_v}{B_H}=1$
$2 \cot \theta =1$
$\tan \theta =\frac{1}{}2$
$\theta =\tan ^{-1}({\frac{1}{}2})$
Thus, $\theta =\tan ^{-1}({\frac{1}{}2})$ is the locus.

Question:24

Consider the plane S formed by the dipole axis and the axis of earth. Let P be the point on the magnetic equator and in S. Let Q be the point of intersection of the geographical and magnetic equators. Obtain the declination and dip angles at P and Q.

Answer:

The angle of declination is zero on the plane formed by the dipole axis and Earth’s axis of rotation. So, the angle of declination is 0 at P.
The angle of dip is zero on the magnetic equator. So, point Q has an angle of dip 0 and an angle of declination 11.3°.

Question:25

There are two current carrying planar coils made each from identical wires of length L. $C_1$ is circular (radius R) and $C_2$ is square (side a). They are so constructed that they have the same frequency of oscillation when they are placed in the same uniform B and carry the same current. Find an in terms of R.

Answer:

For $C_1$,
Radius=R
length=L
number of turns per unit length
$n_1=\frac{L}{2\pi R}$
For $C_2$
side=a
perimeter=L
number of turns per unit length
$n_2=\frac{L}{4a}$

Let Magnetic moment of $C_1$ be $m_1=n_1iA_1$ where i is the current in the coil and
Magnetic moment of $C_{2}$ be $m_2=n_2iA_2$ where i is the current in the coil
$m_1=L.i.\pi .\frac{R^2}{2\pi R } ; m_2=\frac{L}{4a}.i.a^2$
$m_1=\frac{LiR}{2 } ; m_2=\frac{Lia}{4}$
Moment of inertia of $C_1=I_1=\frac{MR^2}{}2$
Moment of inertia of$C_2=I_2=\frac{Ma^2}{12}$….ii where M is the mass of coil
Frequency of $C_1=f_1=2\pi \sqrt{\frac{I_1}{m_1B}}$…iii
Frequency of$C_2=f_2=2\pi \sqrt{\frac{I_2}{m_2B}}$….iv
According to problem, $f_{1}=f_{2}$
$2\pi \sqrt{\frac{I_1}{m_1B}} =2\pi \sqrt{\frac{I_2}{m_2B}}$
$\frac{I_1}{m_1}=\frac{I_2}{m_2} or \frac{m_2}{m_1}=\frac{I_2}{I_1}$
On substituting the values from eqn i, ii, iiiand iv we get
$Lia.\frac{2}{4 \times LiR}=Ma^{2}.\frac{2}{12 MR^2}$
$\frac{ a}{2R}=\frac{a^2}{6R^2 }$
$3R=a$
Thus, the value of a is 3R.