NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

NCERT Solutions for Class 12 Physics Chapter 8 – Free PDF Download

The NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves provide detailed answers to textbook theory questions, numerical problems, worksheets and exercises. In Class 12 Physics, there are many complicated formulas and equations. In order to score good marks in the Class 12 term – II examination, it is important to solve the exercise questions provided at the end of each chapter using the NCERT Solutions for Class 12 Physics.

Most frequently, questions that are asked in CBSE Class 12 Physics term-wise exams directly appear from the NCERT textbook. Electromagnetism is one of the most frequently asked topics in the second term exam. Hence, students are suggested to refer to the NCERT Solutions for Class 12 Physics to attained a firm grip over the chapter as well as the subject. Now, download the NCERT Solutions for Class 12 Physics Chapter 8 from the link mentioned below.

Download NCERT Solutions Class 12 Physics Chapter 8 PDF:-Download Here

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Class 12 Physics NCERT Solutions Electromagnetic Waves Important Questions


Q 8.1) The Figure shows a capacitor made of two circular plates each of radius 12 cm and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A.

(a) Calculate the capacitance and the rate of change of the potential difference between the plates.

(b) Obtain the displacement current across the plates.

(c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves Question 1

Answer 8.1:

Given Values:

The radius of each circular plate (r) is 12 cm or 0.12 m

The distance between the plates (d) is 5 cm or 0.05 m

The charging current (I) is 0.15 A

The permittivity of free space is ε0=8.85×1012  C2N1m2\varepsilon_{0} = 8.85\times 10^{-12}\; C^{2}N^{-1}m^{-2}

(a) The capacitance between the two plates can be calculated as follows:

C=ε0AdC = \frac{\varepsilon _{0} A}{d}

where,

A = Area of each plate = πr2\pi r^{2} C=ε0πr2dC = \frac{\varepsilon _{0} \pi r^{2}}{d}

= 8.85×1012×π(0.12)20.05\frac{8.85\times 10^{-12}\times \pi (0.12)^{2}}{0.05}

= 8.0032×1012  F8.0032\times 10^{-12}\; F

= 80.032 pF

The charge on each plate is given by,

q = CV

where,

V is the potential difference across the plates

Differentiation on both sides with respect to time (t) gives:

dqdt=CdVdt\frac{\mathrm{d} q}{\mathrm{d} t} = C \frac{\mathrm{d} V}{\mathrm{d} t}

But, dqdt\frac{\mathrm{d} q}{\mathrm{d} t} = Current (I)

dVdt=IC∴\frac{\mathrm{d} V}{\mathrm{d} t} = \frac{I}{C}

=>0.1580.032×1012=1.87×109  V/s\frac{0.15}{80.032\times 10^{-12}} = 1.87\times 10^{9}\; V/s

Therefore, the change in the potential difference between the plates is 1.87×109  V/s1.87\times 10^{9}\; V/s.

(b) The displacement current across the plates is the same as the conduction current. Hence, the displacement current, id is 0.15 A.

(c) Yes

Kirchhoff’s first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.

 

Q 8.2) A parallel plate capacitor made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with an (angular) frequency of 300 rad s–1.

(a) What is the rms value of the conduction current?

(b) Is the conduction current equal to the displacement current?

(c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

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Answer 8.2:

Radius of each circular plate, R = 6.0 cm = 0.06 m

Capacitance of a parallel plate capacitor, C = 100 pF = 100×1012  F100\times 10^{-12}\; F

Supply voltage, V = 230 V

Angular frequency, ω=300  rad  s1\omega = 300\; rad\;s^{-1}

 

(a) Rms value of conduction current, I = VXc\frac{V}{X_{c}}

Where,

XcX_{c} = Capacitive reactance = 1ωC\frac{1}{\omega C}

 

I=V×ωC∴ I = V\times \omega C

= 230×300×100×1012230\times 300\times 100\times 10^{-12}

= 6.9×106A6.9\times 10^{-6} A

= 6.9  μA6.9\; \mu A

Hence, the rms value of conduction current is 6.9  μA6.9\; \mu A.

(b) Yes, conduction current is equivalent to displacement current.

(c) Magnetic field is given as:

B=μ0r2πR2I0B = \frac{\mu_{0}r}{2\pi R^{2}}I_{0}

Where,

μ0\mu_{0} = Permeability of free space = 4π×107  N  A24\pi \times 10^{-7}\; N\;A^{-2}

 

I0I_{0} = Maximum value of current = 2  I\sqrt{2}\; I

r = Distance between the plates from the axis = 3.0 cm = 0.03 m

B=4π×107×0.03×2×6.9×1062π×(0.06)2∴B = \frac{4\pi\times 10^{-7}\times 0.03\times \sqrt{2}\times 6.9\times 10^{-6}}{2\pi \times (0.06)^{2}}

= 1.63×1011  T1.63\times 10^{-11}\; T

Hence, the magnetic field at that point is 1.63×1011  T1.63\times 10^{-11}\; T.

 

 

Q 8.3) What physical quantity is the same for X-rays of wavelength 10–10m, the red light of wavelength 6800 Å and radiowaves of wavelength 500m?

Answer 8.3:

The speed of light (3×1083\times 10^{8} m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum.

 

Q 8.4) A plane electromagnetic wave travels in vacuum along the z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its
wavelength?

Answer 8.4:

The electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) and the magnetic field (H) are in the x-y plane. They are mutually perpendicular.

Frequency of the wave, v = 30 MHz = 30×106  s130\times 10^{6}\;s^{-1}

Speed of light in vacuum, C = 3×1083\times 10^{8} m/s

Wavelength of a wave is given as:

λ=cv\lambda = \frac{c}{v}

= 3×10830×106\frac{3\times 10^{8}}{30\times 10^{6}} = 10 m

Q 8.5) A radio can tune in to any station in the 7.5 MHz to 12 MHz bands. What is the corresponding wavelength band?

Answer 8.5:

A radio can tune to minimum frequency, v1=7.5  MHz=7.5×106  Hzv_{1} = 7.5\; MHz = 7.5\times 10^{6}\; Hz

Maximum frequency, v2=12  MHz=12×106  Hzv_{2} = 12\; MHz = 12\times 10^{6}\; Hz

Speed of light, c = 3×108  m/s3\times 10^{8}\; m/s

Corresponding wavelength for v1v_{1} can be calculated as:

λ1=cv1\lambda_{1} = \frac{c}{v_{1}}

 

=3×1037.5×106=40  m=\frac{3\times 10^{3}}{7.5\times 10^{6}} = 40\;m

Corresponding wavelength for v2v_{2} can be calculated as:

λ2=cv2\lambda_{2} = \frac{c}{v_{2}}

 

=3×10312×106=25  m=\frac{3\times 10^{3}}{12\times 10^{6}} = 25\;m

Thus, the wavelength band of the radio is 40 m to 25 m.

 

Q 8.6) A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator?

Answer 8.6:

The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., 109 Hz.

 

Q 8.7) The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0=510 nT. What is the amplitude of the electric field part of the wave?

Answer 8.7:

Amplitude of magnetic field of an electromagnetic wave in a vacuum,

B0=510  nT=510×109  TB_{0} = 510\; nT = 510\times 10^{-9}\; T

Speed of light in vacuum, c = 3×108  m/s3\times 10^{8}\; m/s

Amplitude of electric field of an electromagnetic wave is given by the relation,

E=cB0=3×108×510×109=153  N/CE = cB_{0} = 3\times 10^{8}\times 510\times 10^{-9} = 153\; N/C

Therefore, the electric field part of the wave is 153 N/C.

 

Q 8.8) Suppose that the electric field amplitude of an electromagnetic wave is E0=120  N/CE_{0} = 120\; N/C and that its frequency is v = 50 MHz.(a) Determine B0,  ω,  k  and  λB_{0},\; \omega,\; k\;and\; \lambda (b) Find expressions for E and B.

Answer 8.8:

Electric field amplitude, E0=120  N/CE_{0} = 120\; N/C

Frequency of source, v = 50 MHz = 50×10650\times 10^{6} Hz

Speed of light, c = 3×1083\times 10^{8} m/s

(a) Magnitude of magnetic field strength is given as:

B0=E0cB_{0} = \frac{E_{0}}{c}

= 1203×108\frac{120}{3\times 10^{8}}

= 40×108  =400×109T=400  nT40\times 10^{-8}\;=400\times 10^{-9} T = 400\; nT

Angular frequency of source is given by:

ω=2πv=2π×50×106=3.14×108rads1\omega =2\pi v=2\pi \times 50\times 10^{6}=3.14\times 10^{8}\,rads^{-1}

= 3.14×1083.14\times 10^{8} rad/s

Propagation constant is given as:

k=ωck = \frac{\omega }{c}

= 3.14×1083×108=1.05  rad/m\frac{3.14\times 10^{8}}{3\times 10^{8}} = 1.05\; rad/m

Wavelength of wave is given by:

λ=cv\lambda = \frac{c}{v}

= 3×10850×106\frac{3\times 10^{8}}{50\times 10^{6}} = 6.0 m

 

(b) Suppose the wave is propagating in the positive x-direction. Then, the electric field vector will be in the positive y-direction and the magnetic field vector will be in the positive z-direction. This is because all three vectors are mutually perpendicular.

Equation of electric field vector is given as:

E=E0  sin(kxωt)  j^\overline{E} = E_{0}\;sin(kx – \omega t)\;\widehat{j}

= 120  sin[1.05x3.14×108t]  j^120\;sin[1.05x – 3.14\times 10^{8}t]\;\widehat{j}

And, magnetic field vector is given as:

B=B0  sin(kxωt)  k^\overline{B} = B_{0}\;sin(kx – \omega t)\;\widehat{k}

 

B=(400×109)sin[1.05x3.14×108t]  k^\overline{B} = (400 \times 10^{-9}) sin[1.05x – 3.14\times 10^{8}t]\;\widehat{k}

 

Q 8.9) The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hν (for the energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for
different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?

Answer 8.9:

The energy of a photon is given as:

E = hv = hcλ\frac{hc}{\lambda}

Where,

h = Planck’s constant = 6.6×1034  Js6.6\times 10^{-34}\;Js

c = Speed of light = 3×108  m/s3\times 10^{8}\;m/s

If the wavelength λ is in metre and the energy is in Joule, then by dividing E by 1.6 × 10-19 will convert the energy into eV.

E=hcλ×1.6×1019eVE=\frac{hc}{\lambda \times 1.6\times 10^{-19}}\,eV

a) For Gamma rays, the wavelength ranges from 10-10 to 10-14 m, therefore the photon energy can be calculated as follows:

E=6.62×1034×3×1081010×1.6×1019=12.4×103104eVE=\frac{6.62\times 10^{-34}\times 3\times 10^8}{10^{-10}\times 1.6\times 10^{-19}}=12.4\times 10^3\approx 10^4\,eV
.
Therefore,

λ=1010m,energy=104eV\lambda =10^{-10}\,m, energy = 10^{4}\,eV

 

λ=1014m,energy=108eV\lambda =10^{-14}\,m, energy = 10^{8}\,eV

The energy for Gamma rays ranges from 104 to 108 eV.

b) The wavelength for X-rays ranges between 10-8 m to 10-13 m

For λ = 10-8,

Energy=6.62×1034×3×108108×1.6×1019=124102eVEnergy = \frac{6.62\times 10^{-34}\times 3\times 10^8}{10^{-8}\times 1.6\times 10^{-19}}=124\approx 10^2\,eV

For λ = 10-13 m, energy = 107 eV

c) For ultraviolet radiation, the wavelength ranges from 4 × 10-7 m to 6 × 10-7 m.

For 4 × 10-7 m,

Energy=6.62×1034×3×1084×107×1.6×1019=3.11010eVEnergy = \frac{6.62\times 10^{-34}\times 3\times 10^8}{4\times 10^{-7}\times 1.6\times 10^{-19}}=3.1\approx 10^{10}\,eV

For 6 × 10-7 m, the energy is equal to 103 eV.

The energy of the ultraviolet radiation varies between 1010 to 103 eV.

d) For visible light, the wavelength ranges from 4 × 10-7 m to 7 × 10-7 m.

For 4 × 10-7, the energy is the same as above, that is 1010 eV

For 7 × 10-7 m, the energy is 100 eV

e) For infrared radiation, the wavelength ranges between 7 × 10-7 m to 7 × 10-14 m.

The energy for 7 × 10-7 m is 100 eV

The energy for 7 × 10-14 m is 10-3 eV

f) For microwaves, the wavelength ranges from 1 mm to 0.3 m.

For 1 mm, the energy is 10-3 eV.
For 0.3 m, the energy is 10-6 eV.

g) For radio waves, the wavelength ranges from 1 m to few km.

For 1 m, the energy is 10-6 eV.

The photon energies for the different parts of the spectrum of a source indicate the spacing of the relevant energy levels of the source

 

Q 8.10) In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m–1.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the B field. [ c = 3×108  m  s13\times 10^{8}\;m\;s^{-1} ]

Answer 8.10:

Frequency of the electromagnetic wave, v = 2×1010  Hz2\times 10^{10}\;Hz

Electric field amplitude, E0=48  V  m1E_{0} = 48\;V\;m^{-1}

Speed of light, c = 3×108  m/s3\times 10^{8}\;m/s

 

(a) Wavelength of a wave is given as:

λ=cv\lambda = \frac{c}{v}

= 3×1082×1010=0.015  m\frac{3\times 10^{8}}{2\times 10^{10}} = 0.015\; m

 

(b) Magnetic field strength is given as:

B0=E0cB_{0} = \frac{E_{0}}{c}

= 483×108=1.6×107  T\frac{48}{3\times 10^{8}} = 1.6\times 10^{-7}\; T

 

(c) Energy density of the electric field is given as:

UE=12  ϵ0  E2U_{E} = \frac{1}{2}\; \epsilon _{0} \;E^{2}

And, energy density of the magnetic field is given as:

UB=12μ0B2U_{B} = \frac{1}{2\mu_{0}}B^{2}

Where,

ϵ0\epsilon _{0} = Permittivity of free space

 

μ0\mu_{0} = Permeability of free space

E = cB  …(1)

Where,

c=1ϵ0  μ0c = \frac{1}{\sqrt{\epsilon_{0}\; \mu_{0}}}  …(2)

Putting equation (2) in equation (1), we get

E=1ϵ0  μ0  BE = \frac{1}{\sqrt{\epsilon_{0}\; \mu_{0}}}\; B

Squaring on both sides, we get

E2=1ϵ0  μ0  B2E^{2} = \frac{1}{\epsilon_{0}\; \mu_{0}}\; B^{2}

 

ϵ0  E2=B2μ0\epsilon_{0}\; E^{2} = \frac{B^{2}}{\mu_{0}}

 

12  ϵ0  E2=12  B2μ0\frac{1}{2}\; \epsilon_{0}\; E^{2} = \frac{1}{2}\; \frac{B^{2}}{\mu_{0}}

=> UE=UBU_{E} = U_{B}

Q 8.11) Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 106 rad/s)t]} ˆi .
(a) What is the direction of propagation?
(b) What is the wavelength λ?
(c) What is the frequency ν?
(d) What is the amplitude of the magnetic field part of the wave?
(e) Write an expression for the magnetic field part of the wave.

Answer:

(a) The direction of motion is along the negative y-direction. i.e., along -j.

(b) The given equation is compared with the equation,

E = E0 cos (ky + ωt)

⇒ k = 1.8 rad/s

ω = 5.4 x 106 rad/s

λ = 2π/k = (2 x 3.14)/1.8 = 3.492 m

(c) Frequency, ν = ω/2π =  5.4 x 106/(2 x 3.14) = 0.859  x 10Hz

(d) Amplitude of the magnetic field, B0 = E0/c

= 3.1/(3 x 108) = 1.03 x 10-8 T= 10.3 x 10-9 T= 10.3 nT

(e) Bz = B0 cos (ky + ωt)ˆk ={(10.3 nT) cos[(1.8 rad/m)y + (5.4 × 106 rad/s)t]} kˆ

Q 8. 12) About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation
(a) at a distance of 1m from the bulb?
(b) at a distance of 10 m?
Assume that the radiation is emitted isotropically and neglect reflection.

Answer:

(a) Average intensity of the visible radiation, I = P’/4πd2

Here, the power of the visible radiation, P’ = (5/100) x 100 = 5 W

At d = 1 m

I = P’/4πd2 = 5/(4 x 3.14 x 12) = 5/12.56 = 0.39 W/m2

(b) At d = 10 m

I = P’/4πd2 = 5/(4 x 3.14 x 102)  = 5/1256 = 0.39 x 10-2 W/m2

Q 8.13) Use the formula λ m T = 0.29 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?

We have the equation,  λ m T = 0.29 cm K

⇒ T = (0.29/λ m )cm K

Here, T is the temperature

λ m is the maximum wavelength of the wave

For λ m = 10-4 cm

T = (0.29/10-4)cm K = 2900 K

For the visible light, λ m = 5 x 10-5 cm

T = (0.29/ 5 x 10-5 )cm K ≈ 6000 K

Note: a lower temperature will also produce wavelength but not with maximum intensity.

Q 8. 14) Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs.
(a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space).
(b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift).
(c) 2.7 K [temperature associated with the isotropic radiation filling all space-thought to be a relic of the ‘big-bang’ origin of the universe].
(d) 5890 Å – 5896 Å [double lines of sodium]
(e) 14.4 keV [energy of a particular transition in 57Fe nucleus
associated with a famous high-resolution spectroscopic method (Mössbauer spectroscopy)].

Answer:

(a) Radio waves (short-wavelength end)
(b) Radio waves (short-wavelength end)
(c) Microwave
(d) Visible light (Yellow)
(e) X-rays (or soft γ-rays) region

Q 8.15) Answer the following questions:
(a) Long-distance radio broadcasts use short-wave bands. Why?
(b) It is necessary to use satellites for long-distance TV transmission. Why?
(c) Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why?
(d) The small ozone layer on top of the stratosphere is crucial for human survival. Why?
(e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?
(f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction?

Answer:

(a) Ionosphere reflects waves in the shortwave bands.
(b) Television signals have high frequency and high energy. Therefore, it is not properly reflected by the ionosphere. Satellites are used to reflect the TV signals.
(c) Atmosphere absorbs X-rays, while visible and radio waves can penetrate it.
(d) Ozone layer absorbs the ultraviolet radiations from the sunlight and prevents it from reaching the surface of the earth and causing damage to life.
(e) If the atmosphere is not present, there would be no greenhouse effect. As a result, the temperature of the earth would decrease.
(f) The smoke clouds produced by global nuclear war would perhaps cover substantial parts of the sky preventing solar light from reaching many parts of the globe. This would cause a ‘winter’.

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

Chapter 8 Electromagnetic Waves of Class 12 Physics is categorized under the term – II CBSE Syllabus for 2021-22. Electromagnetism is a physical attraction that occurs in electrically charged particles. When a capacitor is charged using an external source, there can be a potential difference between two capacitive plates, we will show how to calculate that along with the displacement. This question will be solved using Kirchhoff’s rules. Students can make use of the NCERT Solutions for Class 12 to learn the correct methods of solving the exercise problems.

Concepts involved in Class 12 Physics Chapter 8 Electromagnetic Waves

  1. Introduction
  2. Displacement Current
  3. Electromagnetic Waves
    1. Sources of electromagnetic waves
    2. Nature of electromagnetic waves
  4. Electromagnetic Spectrum
    1. Radio waves
    2. Microwaves
    3. Infrared waves
    4. Visible rays
    5. Ultraviolet rays
    6. X-rays
    7. Gamma rays

We will determine the RMS value of the conduction current and we will be analysing the similarities between conduction current and displacement current. We will be analysing the similarities among the wavelengths of X-rays, red lights and radio waves. In this solution, you will be seeing questions on the wavelength of electromagnetic waves traveling in a vacuum.

Do you want to know what the frequency of electromagnetic waves produced by the oscillator is? Want to know about the electric field part of the harmonic electromagnetic wave in a vacuum? Check out the answers in the NCERT Solutions. We will be obtaining photo-energy of different parts of the electromagnetic spectrum and perceiving how to obtain different scales of photon energies of electromagnetic radiation.

We will be gaining knowledge on how to prove that the energy density of one field is equal to the average energy density of another field. We know that there are more fundamental forces such as weak and strong nuclear force and gravitational force. You will be finding questions on them in a different chapter. The questions mentioned in this chapter are very common in the second term exam and if prepared thoroughly, will definitely help you understand electromagnetism with ease.

BYJU’S provides class-wise NCERT Solutions, along with study materials, notes, books, assignments and sample papers prepared by top-notch subject experts of the country who have been involved in teaching the CBSE Syllabus for decades.

Frequently Asked Questions on NCERT Solutions for Class 12 Physics Chapter 8

Do NCERT Solutions for Class 12 Physics Chapter 8 have answers for all the textbook questions?

The NCERT Solutions for Class 12 Physics Chapter 8 is available in PDF format designed by the subject experts. These solutions are completely based on the latest term – II CBSE Syllabus 2021-22 and cover all the important concepts for the exam. The textbook problems are solved in a stepwise manner as per the marks weightage in the second term exam. Both chapter wise and exercise wise PDF links are present in BYJU’S which can be accessed by the students to get their doubts clarified instantly.

How can we score full marks in NCERT Solutions for Class 12 Physics Chapter 8?

The NCERT Solutions for Class 12 Physics Chapter 8 are designed by experts at BYJU’S after conducting vast research on each concept. Every minute detail is explained in a comprehensive manner to help students score well in the class test as well as in term – II exams. It also helps students in doing their assignments given to them on time without any difficulty.

Are NCERT Solutions for Class 12 Physics Chapter 8 PDF enough to score well in the term – II exams?

NCERT Solutions for Class 12 Physics Chapter 8 are available in PDF format which can be downloaded and used by the students without any time constraints. The solutions are created by the highly experienced faculty at BYJU’S based on the latest term – II CBSE Syllabus and its guidelines. The exercise-wise solutions help students to gain an overall idea about the concepts which are important for the second term exams. Practising these questions on a regular basis will improve the time management and problem-solving abilities of students.

 

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