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NCERT Exemplar Class 12 Physics Solutions Chapter 5 Magnetism and Matter

NCERT Exemplar Class 12 Physics Solutions Chapter 5 Magnetism and Matter

Edited By Safeer PP | Updated on Sep 14, 2022 12:39 PM IST | #CBSE Class 12th

NCERT Exemplar Class 12 Physics solutions chapter 5 describes how the phenomena of magnetism is universal. Be it vast, distant galaxies, tiny seemingly insignificant atoms, or men and animals, all are permeated with magnetic fields from various sources. NCERT Exemplar solutions For Class 10 Physics chapter 5 answers questions like - Why does the earth's magnetic field vary from point to point in space? What is the magnetic moment associated with a solenoid? Class 12 NCERT Exemplar Physics chapter 5 solutions also explains how magnetism helps to understand the characteristics and properties of matter, the role of the forces existing between the particles within the substance in determining how the substance behaves when placed in a magnetic field, and classifying the substance based on the kind of response recorded under the influence of the magnetic field.
Also check - NCERT Solutions for Class 12 Physics

This Story also Contains
  1. NCERT Exemplar Class 12 Physics Solutions Chapter 5 MCQI
  2. NCERT Exemplar Class 12 Physics Solutions Chapter 5 MCQII
  3. NCERT Exemplar Class 12 Physics Solutions Chapter 5 Very Short Answer
  4. NCERT Exemplar Class 12 Physics Solutions Chapter 5 Short Answer
  5. NCERT Exemplar Class 12 Physics Solutions Chapter 5 Long Answer
  6. Main Subtopics - NCERT Exemplar Class 12 Physics Solutions Chapter 5 Magnetism and Matter
  7. NCERT Exemplar Class 12 Physics Chapter Wise Links
  8. What will the students learn in NCERT Exemplar Class 12 Physics Solutions Chapter 5?

NCERT Exemplar Class 12 Physics Solutions Chapter 5 MCQI

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 the 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 is0^{\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 MCQII

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 IM.

Answer:

M = 10^{6}\frac{A}{m}, l=0.1m
IM=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 the 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 ofC_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 ofC_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.

In a nutshell, NCERT Exemplar Class 12 Physics solutions chapter 5 explores the laws and derivations concerning magnetism and matter and earths magnetic field. NCERT Exemplar Class 12 Physics solutions chapter 5 pdf download is available for students. The NCERT solutions provided on this page will help understand all the relevant and vital concepts comprehensively and also aid in preparing for the 12 boards and competitive exams.

Main Subtopics - NCERT Exemplar Class 12 Physics Solutions Chapter 5 Magnetism and Matter

  • Introduction
  • The Bar Magnet
  • The magnetic field lines
  • Bar magnet as an equivalent solenoid
  • The dipole in a uniform magnetic field
  • The electrostatic analogue
  • Magnetism And Gauss's Law
  • The Earth's Magnetism
  • Magnetic declination and dip
  • Magnetisation And Magnetic Intensity
  • Magnetic Properties Of Materials
  • Diamagnetism
  • Paramagnetism
  • Ferromagnetism
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.Understanding magnetism is required to answer questions like- Why are some materials magnetic and others not? And why do several substances become magnetised by a field, whereas others remain unaffected? Magnetic materials act as the protagonist in industrial success and technological growth. They have a wide range of uses in the generation of power, transmission and communication, analogue and electronic devices such as clocks or digital wristwatches, scientific and medical equipment, sensors, therapy and delivery. NCERT Exemplar Class 12 Physics chapter 5 solutions discusses the usage of magnets and magnetism

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1. How many questions are solved in the solutions?

The NCERT exemplar Class 12 Physics solutions chapter 5 provided by us will have all the 25 questions of the main exercise.

2. How will these questions and solutions help?

These questions are solved in the most simple and explanative way so that one can grasp the topic. This will help in understanding the basics and how to solve questions in exams.

3. What all crucial topics are covered in magnetism and matter?

This chapter covers various important topics like properties of bar magnet, diamagnetism, ferromagnetism and paramagnetism and earths magnetism etc. These topics are highly crucial from both boards and entrance exam POV.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer
Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Read Complete Answer
Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer
Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read More

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
Read More

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read More

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read More

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read More

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
Read More

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
Read More

With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
Read More

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
Read More

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read More

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