NCERT Solutions

Candidates who are searching for NCERT solutions for Class 12 Physics Chapter 12 – Atoms can refer to this page completely. Here, we as a team covered all topics and subtopics solved questions of atoms and presented in these 12th class physics NCERT Solutions Chapter 12 Atoms PDF for a better understanding of the concept.

In NCERT Solutions, each and every question comes with a detailed, step-by-step solution in easy and simple language for all concepts and subjects that helps you to improve your skills. So, go ahead and download the NCERT Solutions for Class 12 Physics Chapter 12 Atoms in pdf and prepare well for all examinations.

Class 12 Physics NCERT Solutions Chapter 12 Atoms

In Chapter 12, Students will acquire the basic knowledge about Atoms in detail. An Atom is the smallest component of an element, portrayed by a sharing of the chemical properties of the element and a nucleus with neutrons, protons, and electrons. You will have a complete grip on atom chapter concepts like Rutherford’s atomic model, atomic spectra, Bohr’s model of the hydrogen atom, and the line spectra in the hydrogen atom, etc with NCERT Solutions.

Ace up your exam preparation with this Class 12 Physics Exercisewise NCERT Solutions for Chapter 12 Atoms and score max. marks in both CBSE Board exams and Competitive exams.

Class	12
Subject	Physics
Book	Physics
Chapter Number	12
Chapter Name	Atoms

NCERT Solutions for Class 12 Physics Chapter 12 Atoms – Solved Exercises

Studying from the NCERT Solutions for Class 12 Physics Ch 12 is the best way to cover all the topics included in the Atoms concept. Having a thorough practice helps you to score well in the Physics CBSE Board Exam paper. Also, you can understand & get familiar with all the topics and subtopics important questions of chapter Atoms preparing from NCERT Solutions Pdf.

Question 1.
Choose the correct alternative from clues given at end of the each statement:
(a) The size of the atom in Thomson’s model is the atomic size in Rutherford’s model.(much greater than/no different from/much less than.)
(b) In the ground state of…………… electrons are in stable equilibrium, while in ……………..  electrons always experience a net force. (Thomson’s model/Rutherford’s model.)
(c) A classical atom based on…………. is doomed to collapse.
(Thomson’s model/Rutherford’s model.)
(d) An atom has a nearly continuous mass distribution in a………….. but has a highly non­uniform mass distribution in  (Thomson’s model/Rutherford’s model.)
(e) The positively charged part of the atom possesses most of the mass in………….
(Rutherford’s model/both the models.)
Answer:
(a) no different from
(b) Thomson’s model; Rutherford’s model.
(c) Rutherford’s model.
(d) Thomson’s model, Rutherford’s model.
(e) both the models.

Question 2.
Suppose you are given a chance to repeat the alpha- particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperature 14 K). What results do you expect ?
Answer:
α-particle is much heavier than hydrogen nuclei. Therefore, α-particles pass through solid hydrogen without deflection. In other words, α-particles are not scattered by solid hydrogen.

Question 3.
What is the shortest wavelength present in the Paschen series of spectral lines ?
Answer:
The wavelength of the spectral lines forming Paschen series is given by

Question 4.
A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom transits from the upper level to the lower level ?
Answer:

Question 5.
The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the electron in this state ?
Answer:
K.E. = -E (Total energy)
= -(-13.6) = 13.6 eV
P.E. = 2 X E = 2 X (-13.6)
= -27.2 eV

Question 6.
A hydrogen atom initially in the ground level absorbs a photon which excites it to the
n = 4 level. Determine the wavelength and frequency of photon.
Answer:
We know, energy of an electron in the nth orbit of hydrogen atom is given by

Question 7.
Using the Bohr’s model, calculate the speed of the electron in a hydrogen atom in the
n = 1,2 and 3 levels. (b) Calculate the orbital period in each of these levels.
Answer:
(a) Speed of an electron in nth orbit of hydrogen atom is given by

Question 8.
The radius of the innermost electron orbit of a hydrogen atom is 5.3 x I0^-11 m. What are the radii of the n = 2 and n = 3 orbits ?
Answer:
We know, radius of nth orbit of hydrogen atom is given by
r_n = r₀n², where r₀ = 5.3 x 10^-u m is the radius of
the innermost orbit of hydrogen atom.
When n = 2, r₂ = 5.3 x 10^-u X 4
= 2.12 X 10-¹⁰m
When n = 3,
= 5.3 x 10-¹¹ x 9
= 4.77 x 10-¹⁰m.

Question 9.
A 12.75 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted ?
Answer:


Question 10.
In accordance with the Bohr’s model, And the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 x 10¹¹ m with orbital speed
3 x 10¹⁴ m s^-1. (Mass of earth = 6.0 x 10²⁴ kg.)
Answer:
According to Bohr’s postulate of quantization of angular momentum

Question 11.
Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.
(a) Is the average angle of deflection of a-particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model ?
(b) Is the probability of backward scattering (i.e., scattering of a-particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model ?
(c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of a-particles scattered at moderate angles is proportional to What clue does this linear independence on t provide ?
(d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of a-particles by a thin foil ?
Ans:
(a) It is about the same because we are considering average deflection angle.
(b) It is much less than the predicted value because the massive core is absent in Thomson’s model.
(c) It suggests that the scattering of a-particles is primarily because of the single collision.
(d) In Thomson’s model of atom. Multiple collisions are required to be considered in this model because positive charge is spread throughout in this model.

Question 12.
The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10^-40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.
Answer:
If electron and proton were bound by gravitational attraction, then

It is astonishing this value of r is much greater than the size of universe.

Question 13.
Obtain an expression for the frequency of radiation emitted when a hydrogen atom
de-excites from level it to level (n – 1). For large it, show that this frequency equals the classical frequency of revolution of the electron in the orbit.
Answer:
The energy of an electron in the nth orbit of hydrogen atom is given by

i.e. frequencies are equal. This is called the Bohr’s correspondence principle.

Question 14.
Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size ? Why is an atom not, say, thousand times bigger than its typical size ? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To stimulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10-¹⁰m).
(a) Construct a quantity with the dimensions of length from the fundamental constants e, m_e, and c. Determine its numerical value.
(b) You will find that the length obtained in («) many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for ‘something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that h, m_e, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, m_g, and e and confirm that its numerical value has indeed the correct order of magnitude.
Answer:
(a) Here, the dimensional formula of e is A¹T¹, the dimensional formula of m_e is M¹, dimensional formula

Question 15.
The total energy of an electron in the first excited state of the hydrogen atom is about
– 3.4 eV.
(a) What is the kinetic energy of the electron in this state ?
(b) What is the potential energy of the electron in this state ?
(c) Which of the answers above would change if the choice of the zero of potential energy is changed ?
Answer:
(a) K.E. = -E = – (-3.4 eV) = 3.4 eV
E. = 2E = 2 x (-3.4 eV)
= -6.8 eV
(b) Kinetic energy does not depend upon the choice of zero of potential energy. Therefore, its value remains unchanged. However, the potential energy gets changed with the change in the zero level of potential energy.

Question 16.
If Bohr’s quantisation postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun ?
Answer:
Applying the Bohr’s quantization postulate,

i.e., n is very large. Since n is very large, the difference between the two successive energy or angular momentum levels is very small and the levels may be considered continuous.

Question 17.
Obtain the first Bohr’s radius and the ground state energy of a ‘muonic hydrogen atom’ (i.e. an atom in which a negatively charged muon (μ^-1) of mass about 207 m_e orbits around a proton).
Answer:
Here mass of the particle revolving around the proton is

These solutions are prepared by expert teachers as per the NCERT Syllabus based on CBSE Guidelines. So, referring to these NCERT Solved Questions of Ch 12 Atoms can aid you during the exam preparation. Choose Hindi medium or English medium NCERT Solutions PDF for ch 12 Atoms from here and practice daily to score high in the exams.

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NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter Free PDF is provided on this page. All these solutions help students of class 12 in attempting numerical problems and theory questions on the board and entrance exams. You can find all Dual Nature of Radiation and Matter important questions solutions in NCERT Solutions of 12th Physics Ch 11 material.

Subject experts have given all these solutions with a straightforward and detailed description. So that you can rely on these Class 12 Physics NCERT Solutions for Chapter 11 PDF at your exam preparation. Dual Nature of Radiation and Matter NCERT Solutions PDF covers the complete syllabus in a smart manner and solutions are prepared as per the latest CBSE guidelines.

Class 12 Physics NCERT Solutions Chapter 11 Dual Nature of Radiation and Matter

Class 12 Physics Chapter 11 NCERT Solutions covers 9 topics and two main arguments of the Dual Nature of Radiation and Matter. This concept discusses some of the main topics and subtopics like Electron Emission, Photoelectric Effect, Hertz’s observations, Hallwachs’ and Lenard’s observations, the wave theory of light, wave nature of matter, etc.

Brush up your skills & knowledge on chapter 11 Dual Nature of Radiation and Matter by practicing from NCERT class 12 physics Important Questions and Solutions PDF prevailing here.

Class	12
Subject	Physics
Book	Physics
Chapter Number	11
Chapter Name	Dual Nature of Radiation and Matter

NCERT Questions with Solutions for Class 12 Physics Chapter 11

Class 12 Physics NCERT Solutions Ch 11 Dual Nature of Radiation and Matter plays a vital role in scoring max marks in the physics paper. So, candidates of different boards are advised to follow & prepare from the NCERT Solutions for Class 12 physics Ch 11. Because NCERT Solutions are prepared by the subject experts based on the CBSE board prescribed NCERT Textbooks.

Question 1.
Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.
Answer:

Question 2.
The work function of caesium metal is 2.14 eV. When light of frequency 6 x 10¹⁴ Hz is incident on the metal surface, photo emission of electrons occurs. What is the
(а)   maximum kinetic energy of the emitted electrons,
(b)   stopping potential, and
(c) maximum speed of the emitted photo electrons ?
Answer:

Question 3.
The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted ?
Answer:
Here, V₀ = 1.5 V
Maximum kinetic energy. (K.E.)_max
= eV₀ = 1.6 X 10-¹⁹ X 1.5
= 2.4 x 10-¹⁹ J

Question 4.
Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam.
(b) How many photons per second, on the average, arrive at a target irradiated by this beam ? (Assume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon ?
Answer:
Given, λ = 632.8 nm = 632.8 x 10^-9m
Power, P = 9.42 mW = 9.42 x 10^-3 W


Question 5.
The energy flux of sunlight reaching the surface of the earth is 1.388 x 10³ W/m². How many photons (nearly) per square meter are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Answer:

Question 6.
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 x 10^-15 V s. Calculate the value of Planck’s constant.
Answer:

Question 7.
A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the center of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm.
(a) What is the energy per photon associated with the sodium light ?
(b) At what rate are the photons delivered to the sphere ?
Answer:
Power of sodium lamp,
P = 100 W

Question 8.
The threshold frequency for a certain metal is 3.3 x 10¹⁴ Hz. If light of frequency
8.2 x 10¹⁴ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Answer:

Question 9.
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm ?
Answer:

Since energy of incident radiation less than the work function (4.2 eV) of the metal, therefore, the photoelectric emission can not take place from the given metal. 11.10.

Question 10.
Light of frequency 7.21 x 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 x 10⁵ m/s are ejected from the surface. What is the threshold frequency for photo emission of electrons ?
Answer:
Using the relation

Question 11.
Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the cathode, the stopping (cut-off) potential of photo electrons is 0.38 V. Find the work function of the material from which the cathode is made.
Answer:

Question 12.
Calculate the
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Answer:
Given V = 56 V.
(a) momentum of the electron

Question 13.
What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with kinetic energy of 120 eV ?
Answer:
Here, E = 120 eV = 120 x 1.6 x 10^-19J
= 1.92 x 10-¹⁷ J

Question 14.
The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which
(a) an electron, and
(b) a neutron, would have the same de Broglie wavelength.
Answer:

Question 15.
What is the de Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass 1.0 x 10^-9 kg drifting with a speed of 2.2 m/s ?
Answer:
(a) Here, m = 0.040 kg and υ = 1.0 km/s .
= 1000 m/s

Question 16.
An electron and a photon each have a wavelength of 100 nm. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electron.
Answer:

Question 17.
(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be
1.40 x 10-¹⁰ m?  (C.B.S.E. 2008)
(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of \(\frac { 3 }{ 2 } \) kT at 300 K.
Answer:
(a) Here λ = 1.40 x 10^-10m
Also, h = 6.63 X 10^-34 Js
and m = 1.67 x 10^-27kg



Question 18.
Show that the wavelength of electromagnetic radiaion is equal to the de Broglie wavelength of its quantum (photon).
Answer:

Question 19.
What is the de Broglie wavelength of a nitrogen molecule in air at 300 K ? Assume that the molecule is moving with the root mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
Answer:
For nitrogen,


Question 20.
(a) Estimate the speed with which electrons emitted from a heated cathode of an evacuated tube impinge on the anode maintained at a potential difference of 500 V with respect to the cathode. Ignore the small initial speeds of the electrons. The ‘specific charge’ of the electron i.e., its elm is given to be 1.76 x 10¹¹ C kg^-1.
(b) Use the same formula you employ in
(a) to obtain electron speed for an anode potential of 10 MV. Do you see what is
wrong ? In what way is the formula to be modified ?
Answer:




Question 21.
(a) A monoenergetic electron beam with electron speed of 5.20 x 10⁶ ms^-1 is subject to a magnetic field of l.30 x 10^-4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given elm for electron equals 1.76 X 10¹¹ C kg^-1.
(b) Is the formula you employ in
(a) valid for calculating radius of the path of a 20 MeV electron beam ? If not, in what way is it modified ?
Answer:
(a) υ  = 5.20 x 10⁶ms^_1
B = 1.30 x 10 ⁴T

(b) The formula employed in part (a) is not valid because with the increase in velocity, mass varies and in the above formula we have taken m as constant. Instead, m

Question 22.
An electron gun with its anode at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (~ 10^-2 mm of Hg). A magnetic field of
2.83 X 10 ⁴ T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture ; this method is known as the ‘fine beam tube’ method.) Determine e/m from the data.
Answer:

Question 23.
(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at (45 Å. What is the maximum energy of a photon in the radiation ?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube.
Answer:

Question 24.
In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron- positron pair of total energy 10.2 BeV into two y-rays of equal energy. What is the wavelength associated with each γ-ray ?   (1 BeV = 10⁹ eV)
Answer:
Energy carried by the pair of γ-rays = 10 .2 BeV
Energy of each γ ray is

Question 25.
Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons ! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.
(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (~ 10^_1° W m^-2). Take the area of the pupil to be about 0.4 cm², and the average frequency of white light to be about 6 x 10¹⁴
Answer:


This is quite a small number, but still large enough to be counted.
Comparison of cases (a) and (b) tells us that our eye can not count the number of photons individually.

Question 26.
Ultraviolet light of wavelength 2271 A from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is -1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity
(~ 10^s W m^-2) red light of wavelength 6328 A produced by a He-Ne laser ?
[C.B.S.E. 2005, UC, 13]
Answer:

Since the energy of red photon is less than the work function for the metal, photo cell does not respond to red light.

Question 27.
Monochromatic radiation of wavelength 640.2 nm (1 nm = 10^-9 m) from a neon lamp irradiates photosensitive material made of caesium or tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
Answer:

Question 28.
A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used :
λ_x = 3650 Å, λ₂ = 4047 Å, λ₃ = 4358 Å, λ₄ = 5461 Å, λ₅ = 6907 Å.
The stopping voltages, respectively, were measured to be :
V₀₁ = 1.28 V, V₀₂ = 0.95 V, f₀₃ = 0.74 V, = 0.16 V,V₀₅ = 0 V
(a) Determine the value of Planck’s constant
(b) Estimate the threshold frequency and work function for the material.
Answer:
(a) From the Einstein photoelectric equation,




Question 29.
The work function for the following metals is given : Na : 2.75 eV ; K : 2.30 eV ; Mo : 417 eV ; Ni: 515 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell ? What happens if the laser is brought nearer and placed 50 cm away ? (C.B.S.E. 2009)
Answer:
Here, λ = 3300 A = 3300 x 10^-10 m
Distance, r’ = 1 m and r’ = 50 cm = 0.5 m
Using the relation

Since the energy E of the incident photon of light is less than the work functions of Mo and Ni metals, so photoelectric emission will not occur in Mo and Ni. The distance of the source does not increase or decrease the energy of the photon of light incident, therefore, the energy of electrons ejected will not change but the intensity of ejected electrons will increase (1 α 1/r² ) and become four times.

Question 30.
Light of intensity 10^-5 W m^-2 falls on a sodium photo­cell of surface area 2 cm². Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave- picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer ?
Answer:
A = 2 cm² = 2 x 10^-4 m²
φ = 2eV = 2 x 1.6 x 10^-19 J
= 3.2 X 10^-19  J.
Taking the approximate radius of an atom as 10^-10 m. the effective area of sodium atom is ≈ r² = 10^-20 m.
.’. If there is one free electron per atom, then number of electrons in five layers


The answer obtained implies that the time of emission of electron is very large and is not in agreement with the observed time of emission, which is approximately 10 ⁹ s. Thus wave-picture of radiation is not applicable for photo-electric emission.

Question 31.
Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy ? (For quantitative comparison, take the wavelength of the probe equal to 1 Å which is of the order of inter-atomic spacing in the lattice) (m_e = 9.11 x 10^­-31 kg).
Answer:

Clearly, the energy of photon is much greater than the energy of electron.

Question 32.
(a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 44, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable ?
Explain. (m_n = 1.675 x 10^-27 kg).
(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27°C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.
Answer:

This wavelength is about hundred times smaller than the interatomic separation of crystals. Thus, neutrons are not suitable for diffraction experiments in case of crystals.

This wavelength is comparable to the interatomic spacing of crystals. Therefore, thermal electrons are able to interact with the crystal. Since

increasing the temperature, decreases their de Broglie wavelength and they become unsuitable for crystal diffraction. Thus, the fast beam of neutrons needs to be thermalised with the environment for neutron diffraction experiment.

Question 33.
An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow
light ?
Answer:



Question 34.
The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10^-15 m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV).
Answer:

Thus, the energy of the proton ejected out of the linear accelerator is of the order of BeV.

Question 35.
Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure ; and compare it with the mean separation between two atoms under these conditions.
Answer:
Here T = 27 °C = 27 + 273 = 300 K
P = 1 atm = 1.01 x 10⁵ Nm^-2
Also, mass of helium atom,


Question 36.
Compute the typical de Broglie wavelength of an electron in a metal at 27°C and compare it with the mean separation between two electrons in a metal which is given to be about 2 x 10^_10 m.
Answer:
Here, T = 27°C = 27 + 273 = 300 K


Question 37.
Answer the following questions :
(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3) e; (- l/3)e]. Why do they not show up in *Millikan’s oil-drop experiment?
(b) What is so special about the combination elm ? Why do we not simply talk of e and m separately ?
(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures ?
(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic ? Why is there an energy distribution of photoelectrons ?
(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations :

Answer:
(a) In case of the Millikan oil drop experiment, the charge on the electron is measured. The electron revolve outside the nucleus and each has a charge e. Thus, we do not observe the fractional charges

(c) At ordinary pressure, molecules of gas keep on colliding with each other and the ions formed do not have a chance to reach the respective electrodes to constitute a current because of their recombination. At low pressure, however, ions do not collide frequently and are able to reach the respective electrodes to constitute a current.
(d) Work function in fact is the energy required to knock out the electron from highest filled level of conduction band of an emitter. In the conduction band, there are different energy levels which collectively form a continuous band of levels. Therefore, different amounts of energy are required to bring the electrons out of the different levels. Electrons emitted have different kinetic energies according to the energy supplied to the emitter.
(e) Since frequency for a given matter wave remains constant for different layers of the matter but wavelength changes so X is more significant than v.
Similarly energy E = hv = \(\frac { 1 }{ 2 } \) m(λv)² is also constant for a given matter wave so phase λv is also not physically significant.

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You will avail the answers in a detailed explanation of each concept included in Ch 4 Moving charge and Magnetism class 12 physics NCERT Solutions. These solutions are prepared by subject experts based on the latest ncert books for physics and CBSE guidelines. Going through with these NCERT Solved Questions of 12th Physics Ch 4 helps you get strong basics on the concept of Moving charge & Magnetism and it leads to score well in the exams.

Class 12 Physics NCERT Solutions Chapter 4 Moving Charge and Magnetism

Chapter 4 is so important because many of the coming learnings will have their base. So, pay some more attention while practicing this ch 4 Moving charge and magnetism concept. This Class 12 Physics NCERT chapter 4 illustrates what a magnetic force is and it’s motion in a magnetic field in a detailed manner. A total of 11 topics are addressed in this Physics Class 12 Moving charge and Magnetism Chapter.

Mostly it talks about ampere’s circuital law and Biot-savarat’s law. Also, it proposes students to the toroid and the solenoid, motion in combined magnetic and electric fields, etc. So, Class 12 Physics NCERT Solutions for Chapter 4 Moving Charges and Magnetism presented here can aid students to clarify all their queries instantly.

Class	12
Subject	Physics
Book	Physics
Chapter Number	4
Chapter Name	Moving Charge and Magnetism

NCERT Exercise Questions & Answers of 12th Class Physics Ch 4

12th Class NCERT Solutions for Physics chapter 4 by Ncertbooks.guru is a reliable study material for your learning process. Class 12 Physics NCERT Solutions of Chapter 4 Moving Charges and Magnetism benefits CBSE class 12 Science students to an extent & make them feel confident while attempting the final exam.

Question 1.
A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the center of the coil ?
Answer:

Question 2.
A long straight wire carries a current of 35 A. What is the magnitude of the Held B at a point 20 cm from the wire ?
Answer:

Question 3.
A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of R at a point 2.5 m east at the wire.
Answer:

Question 4.
A horizontal overhead power line carries a current of 90A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line ?
Answer:

Question 5.
What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T ?
Answer:

Question 6.
A 3.0 cm wire carrying a current of 10A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire ?
Answer:
Using F = BIl, we get
F = 0.27 x 10 x 3 x 10 x 10^-2
= 8.1 x 10-² N
The direction of force is perpendicular to the direction of magnetic field as well as to the direction of current.

Question 7.
Two long and parallel straight wires A and B carrying currents of 8*0 A and 5*0 A in the same direction are separated by a distance of 4-0 cm. Estimate the force on a 10 cm section of wire A.
Answer:

Question 8.
A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its center.
Answer:

Question 9.
A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil ?
Answer:
Using τ= NIAB sin θ, we get
τ = 20 x 12 x 10 x 10 x 10^-4 x 0.8 x sin 30
= 0.96 Nm

Question 10.
Two moving .coil meters, M₁and M₂ have the following particulars:
R₁ = 10 Ω, N₁ = 30, A₁= 3.6 x 10^-3m², B₁= 0.25 T
R₂ = 14 Ω, N₂ = 42, A₂ = 1.8 x 10^-3 m², B₂ = 0.50 T (The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M₁.
Answer:
Using Current sensitivity = NBA/k
For M₁ Current Sensitivity


Question 11.
In a chamber, a uniform magnetic field of 6.5G (1G = RHT) is maintained. An electron is shot into the field with a speed of 4.8 x 10⁶ ms^-1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit,
(e =1.6 x 10^-19 C, m = 9.1 x 10^-31 kg).
Answer:

Question 12.
In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron ? Explain.
Answer:

Question 13.
(a) A circular coil of 30 turns and radius 8-0 cm carrying a current of 6-0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
(b) Would your answer change, if the circular coil in (a) were replaced by a planar toil of some irregular shape that encloses the same area ? (All other particulars are also unaltered)  (C.B.S.E. 1998 C )
Answer:
(a) Using τ = NBIA sin 9, we get
τ= 30 X 1 x 6 x n (8 X 10^-2)² sin 60
= 180 x it (8 x 10-²)² 0.866
= 3.13 N m
The magnitude of the counter torque is 3 .13 N m
(b) Answer will not change because torque does not depend upon shape of the coil provided it encloses same area.

Question 14.
Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A : coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their center.
Answer:
For Coil X

Question 15.
A magnetic field of 100 G (1 G = 10^-4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10^-3 m². The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m^-1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.
Answer:

We may take I = 10 A and n = 800. The given solenoid may have length of 50 cm having 400 turns and area of cross-section = 5 x 10^-3m² (five times the given value.)

Question 16.
For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its center is given by,

(a) Show that this reduces to the familiar result for field at the center of the coil.
(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by,

(Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.)
Answer:

(b) Let there be two coils as mentioned in the statement. Magnetic field in a small region of length 2d about the mid-point of the space between the two coils is given by,


Question 17.
A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field (i) outside the toroid, (ii) inside the core of the toroid, and (iii) in the empty space surrounded by the toroid.
Answer:
(i) Zero

Question 18.
Answer the following questions :
(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle ?

(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment ?

(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflection from its straight line path.
Ans.
(a) The particle goes undeflected so it does not experience magnetic force.
Using F = qυ B sin θ. It is clear that the initial velocity u is either parallel or anti-parallel to magnetic field (B) i.e. θ is either 0° or 180°.

(b) Yes. the final speed is equal to the initial speed because the magnetic force can change the direction of o but the magnitude of o can not be changed by it.

(c) Force acting on electron due to electrostatic field ( E→ ),
F→_e =-eE→

The direction of this force is towards North. Therefore, electron is deflected towards North. Force acting on electron due to magnetic field, F→_m= -e v→ xB→. The electron will not be deflected and move in a straight line path, if the force due to magnetic field on the electron is towards South. According to Fleming’s left hand rule, the direction of B→ is along the vertical in downward direction.

Question 19.
An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30° with the initial velocity.
Answer:
K.E, acquired by electron while passing through V

(b) When electron moves with velocity r making an angle of 30° with the direction of magnetic field, then r cos θ is


Question 20.
A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 x 10⁵ V m^-1, make a simple guess as to what the beam contains. Why is the answer not unique.
Answer:

The given particle may be deutron.
The result is not unique because this e/m ratio can be true for He^{+ +} , Li ^{+ + +} etc.

Question 21.
A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.”
(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero ?
(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before ? (Ignore the mass of the wires.) g = 9.8 ms^-2.
Answer:
(a) The tension in the wire is zero if force on the current  carrying wire due to current is equal and opposite to the weight of the wire. This is, BIl= mg

(b) In case current is reversed, the tension is equal to the force acting on the wire due to magnetic field plus weight of the wire. This is,
T = BlL + mg
= 0.26 x 5 x 0.45 + 60 x 10^-3 x 9.8
= 1.18 N.

Question 22.
The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart ? Is the force attractive or repulsive (H.S.E.B.2001)
Answer:

Since the currents in two wires are in opposite direction so the force is repulsive.

Question 23.
A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,
(a) the wire intersects the axis,
(b) the wire is turned from N-S to northeast-northwest direction.
(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm ?
Answer:

Question 24.
A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Figure? What is the force on each case ? Which case corresponds to stable equilibrium ?

Answer:

Question 25.
A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the
(a) total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic field ?
(The coil is made of copper wire of cross-sectional area 10^-5 m², and the free electron density in copper is given to be about 10²⁹ m³)
Answer:

Question 26.
A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its center) normal to its
axis ; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire ? g = 9.8 ms^-2.
Answer:

Question 27.
A galvanometer coil has a resistance of 12 Q and the meter shows full scale deflection for a current of 3 mA. How will you convert the meter into a voltmeter of range 0 to 18 V ?
Answer:
Here 1=3 mA = 3 x 10^-3 A
Galvanometer resistance, G = 12 Ω The galvanometer can be converted into the voltmeter of range 0 to V (here V = 18 V) by connecting a high series resistance R given

Question 28.
A galvanometer coil has a resistance of 15Ω and the meter shows full scale deflection for a current of 4 mA. How will you convert the meter into an ammeter of range 0 to 6 ?Answer:

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Conclusion

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NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity is the best study material that helps you score more marks in the board exams of class 12 physics and also in any entrance exams. Students can get a detailed explanation of answers for all questions included in Ch 3 Current Electricity concept by the NCERT Solutions of 12th physics.

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Class 12 Physics NCERT Solutions Chapter 3 Current Electricity

In Chapter 3, students will deal with the concept called Current Electricity. It is known as the essential chapter in class 12 physics. Therefore, students should pay special attention to this topic called Current Electricity. The chapter covers a total of 16 topics like current density, electric current. Moreover, you will learn how the electric charge varies from liquid conductors to solid metallic conductors.

In NCERT Class 12 Physics Chapter 3 Solutions, you may also find the explanation for the topics such as Ohm’s law, drift velocity, the relationship between relaxation time and resistivity. So, ace up your preparation by referring to these NCERT Exercise Questions of 12th Physics Ch 3 Current Electricity.

Class	12
Subject	Physics
Book	Physics
Chapter Number	3
Chapter Name	Current Electricity

NCERT Questions & Answers of Class 12 Physics Ch 3 Current Electricity – Solved Exercises

Detailed knowledge of the topics covered under NCERT Class 12 Physics Chapter 3 Current Electricity will assist you in learning the information easily. Chapter 3 Current Electricity 12th Physics NCERT Solved Exercises Questions & Answers are given here in PDF free download. Also, you can attain exemplary problems, numerical on current electricity Class 12 PDF along with the answers for each question.

Question 1.
The storage battery of a car has an e.m.f. of 12 V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery ?
Answer:

Question 2.
A battery of e.m.f. 10 V and internal resistance 3Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor ? What is the terminal voltage of the battery when the circuit is closed ?
Answer:

Question 3.
(a) Three resistors IΩ, 2 Ωand 3 Ω are combined in series. What is the total resistance of the combination ?
(b) If the combination is connected to a battery of e.m.f. 12 V and negligible internal resistance, obtain the potential drop across each resistor.
Answer:

Voltage drop across R₃ is given by, V₃ = IR₃ = 2 x 3 = 6V.

Question 4.
(a) Three resistors 2 Ω, 4 Ω, and 5 Ω, are combined in parallel. What is the total resistance of the combination ?
(b) If the combination is connected to a battery of e.m.f. 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.
Answer:

Question 5.
At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 x 10^-4 °C^-1.
Answer:

Question 6.
A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 x 10^-7 m², and its resistance is measured to be 50 Ω. What is the resistivity of the material at the temperature of the experiment ?
Answer:

Question 7.
A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature co-efficient of resistivity of silver.
Answer:

Question 8.
A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C ? Temperature co­efficient of resistance of nichrome averages over the temperature range involved is 1.70 X 10^-4 °C^-1.
Answer:


Question 9.
Determine the current in each branch of the network shown in Figure

Answer:

Question 10.
(a) In a meter bridge (Figure), the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips ?

(b) Determine the balance point of the bridge above if X and Y are interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge ? Would the galvanometer show any current ? (C.B.S.E. 2005)
Answer:


Question 11.
A storage battery of e.m.f. 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging ? What is the purpose of having a series resistor in the charging circuit ?
Answer:
During charging,
V = E + I(r + R)

Question 12.
In a potentiometer arrangement, a cell of e.m.f. 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the e.m.f. of the second cell ?
Answer:

Question 13.
The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 x 10²⁸ m³. How long does an electron take to drift from one end of a wire 3.0 m long to its other end ? The area of cross­section of the wire is 2.0 x 10^-6 m² and it is carrying a current of 3.0 A.
Answer:

Question 14.
The earth’s surface has a negative surface charge density of 10^-9 cm^-2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric Held, how much time (roughly) would be required to neutralise the earth’s surface ? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe.)
(Radius of earth = 6.37 x 10⁶ m.)
Answer:

Question 15.
(a) Six lead-acid type of secondary cells each of e.m.f. 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage ?
(b) A secondary cell after long use has an e.m.f. of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell ? Could the cell drive the starting motor of a car ?
Answer:

It cannot be used for starting motor of a car because large current is needed to start the car.

Question 16.
Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter ? Hence explain why aluminium wires are preferred for overhead power cables. (P_AL = 2.63 x 10⁸ Ω m, P_cu= 1.72 X 1(H Ω m. Relative density of A1 = 2.7, of Cu = 8.9.)
Answer:


Aluminium is lighter, so it is used for overhead power cables.

Question 17.
What conclusion can you draw from the following observations on a resistor made of alloy manganin ?

Current	Voltage	Current	Voltage
A	V	A	V
0.2	3.94	3.0	59.2
0.4	78.7	4.0	78.8
0.6	11.8	5.0	98.6
0.8	15.7	6.0	118.5
1.0	19.7	7.0	138.2
2.0	39.4	8.0	158.0

Answer:
It indicates that Ohm’s law i.e.V α I is valid for a wide range.
Resistivity of Manganin remains nearly same with change in temperature.

Question 18.
Answer the following questions :
(a) A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor : current, current density, electric field, drift speed ?
(b) Is Ohm’s law universally applicable for all conducting elements ? If not, give examples of elements which do not obey Ohm’s law.
(c) A low voltage supply from which one needs high currents must have very low internal resistance. Why?
(d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why ?
Answer:
(a) Current, as it given to be steady.The other quantities depend inversely upon area of cross-section.
(b) Ohm’s law is not valid for electrolytes and semi-conductor diodes.
(c) Maximum current drawn

Clearly low r will ensure high current
(d) If internal resistance is not large, then the heavy current drawn during accidental short circuit can damage the supply.

Question 19.
Choose the correct alternative :
(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature co-efficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (10²²/10)³).
Answer:
(a) Greater
(b) Lower
(c) Nearly independent
(d)10²²

Question 20.
(a) Given n resistors each of resistance R, how will you combine them to get the
(i) maximum (ii) minimum effective resistance ? What is the ratio of the maximum to minimum resistance ?
(b) Given the resistances of 1 Q, 2 G, 3 Q, how will be combine them to get an equivalent resistance of (i) (11/3)Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω ?
(c) Determine the equivalent resistance of networks shown in figure.     (N.C,E.R,T.)
Answer:

(i) Maximum resistance can be obtained by combining them in series with each other.
The maximum resistance R_max = R + R + R + ……………..n times = nR.
(ii) Minimum effective resistance can be obtained by combining them in parallel with each other. Minimum resistance R_max is found as

Question 21.
Determine the current drawn from a 12 V supply with internal resistance 0.5Ω by the infinite network shown in figure. Each resistor has 1 Ω resistance.(N.C.E.R.T.)

Answer:
Let the total resistance of the circuit be Z and a set of three resistors of value R each be connected to it as shown in the figure.

Question 22.
Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant e.m.f. of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown e.m.f. e and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(a) What is the value e ?
(b) What purpose does the high resistance of 600 kΩ have ?
(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell ?
(e) Would the method work in the above situation if the driver cell of the potentiometer had an e.m.f. of 1.0 V instead of 2.0 V ?
(f) Would the circuit work cell for determining an extremely small e.m.f., say of the order of a few mV (such as the typical e.m.f. of a thermo ­couple not, how will you modify the circuit?
Answer:

(b) Purpose of using high resistor is that it will allow very small current to flow through the galvanometer when the circuit is not balanced.
(c) No, the balance point is not affected by the high resistance.
(d) No, the balance point is not affected by the internal resistance of the driver cell.
(e) No, the method will not work in this situation because the balance point is not obtained if driver cell of e.m.f. 1V instead of 2V is used.
(f) The circuit will not work well for determining extremely small e.m.f. because in such cases the balance point will be almost on end A.

Question 23.
Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε ?
Answer:

(b) To obtain balance point, either a cell of less e.m.f. be used or a suitable resistor be put in series with R and X so that potential drop across the wire AB is reduced.

Question 24.
Figure shows a 2.0 V potentiometer used for the determination of internal resistance

1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of
9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.
Answer:

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NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance is the most important & scoring chapter in class 12 physics. Students can get more subject knowledge about chapter Electrostatic Potential and Capacitance from these NCERT Solutions of class 12 physics. We presented step by step solutions for each question given in NCERT Class 12 Physics Textbook as per CBSE Board guidelines.

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Class 12 Physics NCERT Solutions Chapter 2 Electrostatic Potential and Capacitance

One of the scoring chapters in 12th physics is chapter 2 Electrostatic Potential and Capacitance. It offers the best weightage for numeric problems and derivations. You can find the effective numerical problems on the topic capacitance of a complex combination of capacitors in the final exams from this chapter. Also, you will deal with some of the important topics like electrostatic potential and its various factors, the difference between the two electric fields, etc in ch 2 Physics.

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Class	12
Subject	Physics
Book	Physics
Chapter Number	2
Chapter Name	Electrostatic Potential and Capacitance

NCERT Exercise Questions of Class 12 Physics Chapter Electrostatic Potential and Capacitance

Class 12 NCERT Physics Solutions for Chapter 2 Electrostatic Potential and Capacitance Questions are given here for all students of CBSE board & other state boards. They can refer to these NCERT solutions for better understanding and clarification of the chapter Electrostatic Potential and Capacitance. Here, we have provided online as well as the PDF version of NCERT solutions for Electrostatic Potential and Capacitance.

Question 1.
Two charges 5 x 10^-8 C and -3 x 10^-8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero ? Take the potential at infinity to be zero. Ans. Let the potential be zero at 0, then
Answer:
Let the potential be zero at 0, then
v_A + V_B= 0,

where V_A is electric potential due to charge q_A and V_B is the electric potential due to charge q_B.

Question 2.
A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the center of the hexagon.
Answer:
Total potential at O is given by,

Question 3.
Two charges 2μC and -2μC are placed at points A and B, 6 cm apart.
(a) Identify an equipotential surface of the system. % What is the direction of the electric field at every through the mid-point. On this plane, potential is zero everywhere.
(b) The direction of electric field is from positive to negative charge i.e. A to B, which is infact perpendicular to the equipotential plane.
Answer:
(a) The eguipotential surface is the plane pependicuIarto the line AB joining the two charges and passing through the mid-point. On this plane, potential is zero everywhere.
(b) The direction of electric field is from positive to negative charge i.e. A to B, which is infact perpendicular to the equipotential plane.

Question 4.
A spherical conductor of radius 12 cm has a charge of 1.6 x 10^-7 C distributed uniformly on its surface. What is the electric field
(a) inside the sphere
(b) just outside the sphere
(c) at a point 18 cm from the center of the sphere?
Answer:
(a) inside a conductor, the electric field is zero because the charge resides on the surface of a conductor.
(b) Electric field just outside the sphere is given by

Question 5.
A parallel plate capacitor with air between the plates has a capacitance of 8 pF
(1 pF = 10^-12 F.) What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Answer:

Question 6.
Three capacitors each of capacitance 9 pF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply ?
Answer:

Question 7.
Three capacitors of capacitances 2 pF, 3 pF and 4pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Answer:
(a) Total capacitance
= c₁ + c₂ + c₃
= 2 + 3+ 4 = 9pF.
(b) Using C = qv we get q = CV
∴ q_x = C₁V = 2 x 10-¹² x 100
= 2 x 10-¹⁰ C = 200 pC
q₂ = c₂V
= 3 x 10^-12 x 100
= 3 x 10-¹⁰ C = 300 pC
q₃ = c₃v
= 4 x 10-¹² x 100
= 4 x 10-¹⁰ C = 400 pC

Question 8.
In a parallel plate capacitor with air between the plates, each plate has an area of
6 x 10^-3 in² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor ?
Answer:

Question 9.
Explain what would happen if in the capacitor given in Q. 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,
(a) while the voltage supply remained connected.
(b) after the supply was disconnected.
Answer:

Question 10.
A 12 pF capacitor is connected to a 50 V battery.How much electrostatic energy is stored in the capacitor ?
Answer:
E = 12 CV² = 12 x 12 x 10-¹² x 50 x 50 = 1.5 2 2 x 10-⁸J.

Question 11.
A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from die supply and is connected  to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process ?
Ans.
Here, C₁ = 600pF = 6 x 10^-10 F, C₂ =
6 x 10-¹⁰ F, V₁ = 200V, V₂ = 0

Question 12.
A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of -2 x 10^-9 C from a point P(0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point
R(0, 6 cm, 9 cm.)
Answer:
The work done by electrostatic force on a charge is independent of the path followed by the charge. It depends only on the initial and final positions of the charge.

Question 13.
A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the center of the cube.
Answer:
(1) Distance of the center of the cube from vertex is half of the diagonal of the cube

(2) From symmetry, it is clear that electric field at center of the cube is zero.

Question 14.
Two tiny spheres carrying charges 1.5 μc and 2.5 μc are located 30 cm apart. Find the potential and electric field :
(a) at the mid point of the line joining the two charges,
(b) at a point 10 cm from this mid-point in a plane normal to the line and passing through the mid­point.
Answer:

Question 15.
A spherical conducting shell of inner radius r_l and outer radius r₂ has a charge Q.
(a) A charge q is placed at the center of the shell. What is the surface charge density on the inner and outer surfaces of the shell ?
(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape ? Explain.
Answer:
(a) Charge Q appears on the outer surface.

When charge q is placed at the center, it induces – q charge on the inner surface and +q on the outer surface.
.’. charge density of the inner surface,

and charge density of the outer surface,

Consider a cavity of irregular shape with net charge to be zero inside it. Let a closed loop be partially inside and the rest outside the cavity. The field inside the conductor is zero, so some work is done by the field to carry a test charge in the closed loop, but this is against the provisions of an electrostatic field because as per Gauss’ law, the net charge inside a Gaussian surface must be zero. Thus, there cannot be field lines inside the cavity irrespective of its shape.

Question 16.
(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

where n^ is a unit vector normal to the surface at a point and a is the surface charge density at that point. (The direction of h is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σn^ /ε₀
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.[Hint. For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]
Answer:
Consider a sheet of charge having charge density a. E on either side of the sheet, perpendicular to the plane of sheet, has same magnitude at all points equidistant from the sheet.
Electric field intensity on the left side of the sheet,

The electric field tangential to the plate is continuous throughout.

Question 17.
long charged cylinder of linear charged density A. is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders ?
Answer:
A cylinder P has linear charge density, λ, length l and radius r₁
The charge on cylinder P, q = XL A hollow co-axial conducting cylinder of length / and radius r₂ surrounds the cylinder P. Charge on cylinder Q = – q.

Consider a Gaussian surface in the form of a cylinder of radius rand length /. The electric flux through the curved surface of the Gaussian surface,

Question 18.
In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å :
(a) Estimate the potential Energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.
(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a) ?
(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 A separation?
Answer:

Question 19.
If one of the two electrons of a H₂ molecule is removed, we get a hydrogen molecular ion H⁺₂ .In the ground state of an H⁺₂, the two protons are separated by roughly 1.5 Å, and the electron is roughly Å from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.
Answer:

Question 20.
Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric field at the surfaces of the two spheres ? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.
Answer:
Two charged conducting spheres of radii a and b connected by a wire will reach to same potential.

Clearly electric charge density for the pointed surface will be more because a flat surface can be equated to a spherical surface of large radius and a pointed portion to a spherical surface of small radius.

Question 21.
Two charges -q and + q are located at points (0, 0, -a) and (0, 0, a), respectively.
(a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0) ?
(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a > > 1.
(c) How much work is done in moving a small test charge from the point
(5, 0, 0) to (-7, 0, 0) along the x-axis ? Does the answer change if the path of the test charge between the same points is not along the x-axis ?
Answer:

The point (x, y, 0) is perpendicular to Z-axis, there­fore, potential at (x, y, 0) is zero.
(b) Consider P to be the point of observation at a distance r from the center (O) of the electric dipole.
Let OP make an angle 0 with the dipole moment p→
and r_1, r₂ be the distances of point P from – q charge and + q charge respectively. Potential at P due to – q charge,



Answer does not change because in electrostatics, the work done does not depend upon actual path, it simply depends upon the initial and final positions.

Question 22.
Figure shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/a > > l, and contrast your results with that due to an electric dipole, and an electric monopole
(i.e., a single charge.)

Answer:

(2) Due to electric dipole, the potential is of 1/r² type.
(3) Due to an electric monopole, the potential is of 1/r type.

Question 23.
An electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1 kV. A large number of 1 μF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.
Answer:
Let N capacitors be used in m rows when each row has n capacitor i.e. N = mn
In series

Question 24.
What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm ? (You will realise from your answer why ordinary capacitors are in the range of μF or less. However, electrolytic capacitors do have a much larger capacitance ((0.1 F) because of very minute separation between the conductors.)
Answer:

Question 25.
Obtain the equivalent capacitance of the network in figure. For a 300 V supply, determine the charge and voltage across each capacitor. (C.B.S.E. 2008)

Answer:
The equivalent circuit is as shown below :

Potential difference across C₄ is in the ratio 2:1
i.e., 200 V
.’. Charge on C₄ = C₄V₄
= 100 x 200 x 10^-12 = 2 x 10^-8 C
Potential difference across C₁= 100 V
Charge on C₁ = C₁x V₁
=100 x 100 x 10-¹² = 1 x 10^-8 C
Potential difference across C₂ and C₃ is 50 v each
∴ Charge on C₂ or C₃ = C₂V₂
= 200 X 50 x 10^-12 = 10-⁸ C.

Question 26.
The plates of a parallel plate capacitor have an area of 90 cm² each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.
(a) How much electrostatic energy is stored by the capacitor ?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume Hence arrive at a relation between u and the magnitude of electric field E between the plates.
Answer:

Question 27.
A 4 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ?
Answer:

Question 28.
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1/2) QE, where Q is the charge on the capacitor and E is the magnitude of electric field between the plates. Explain the origin of the factor (1/2).
Answer:
Let F be the force on each plate of the capacitor. If the distance between the plates of the capacitor is increased by dx, then work done = F dx. This work done is stored as the potential energy of the capacitor. The increase in the volume of capacitor = A dx

Question 29.
A spherical capacitor consists of two concentric spherical conductors held in position by suitable insulating supports (Figure.) Show that the capacitance of a spherical capacitor is given by

where r_x and r₂ are the radii of outer and inner spheres, respectively.
Answer:
It consists of two concentric spherical shells A and B of radii a and b with charge +q and charge -q respectively. (Outer sphere is grounded)

Question 30.
A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 pC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.
(a) Determine the capacitance of the capacitor.
(b) What is the potential of the inner sphere ?
(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.
Answer:

Question 31.
Answer carefully :
(a)Two large conducting spheres carrying charges Q₁ and Q₂ are brought close to each other. Is the magnitude of electrostatic force between them exactly given by Q₁Q₂/4Πε₀r², where r is the distance between their centers ?
(b) If Coulomb’s law involved 1/r³ dependence (instead of 1/r²), would Gauss’s law be still true ?
(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?
(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron ? What if the orbit is elliptical ?
(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there ?
(f) What meaning would you give to the capacitance of a single conductor ?
(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (=6).
Answer:
(a) No, because the distribution of charge on the conducting spheres becomes non-uniform when they are brought closer to each other. Coulomb’s law is valid for point charges only or for spheres on which charge distribution is uniform.
(b) If dependence of Coulomb’s law involves 1/r³ dependence then Gauss’ law in the existing state will not be true.
(c) It is true only if field line is a straight line.
(d) Whatever may be the shape of the closed orbit, it is zero.
(e) Potential is continuous.
(f) It means that a single conductor is a capacitor whose other plate can be considered to be at infinity.
(g) A water molecule has permanent dipole moment
so its dielectric constant is high.
Mica does not have polar molecules.

Question 32.
A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 pC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects, (i.e., bending of field lines at the end)
Answer:
Capacitance of a cylindrical capacitor is given by


Question 33.
A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 10⁷ Vm^_1 . (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF ?
Answer:
10% of the given field i.e. 10⁷ V m¹ gives E = 0.1 X 10⁷ Vnr¹

Question 34.
Describe schematically the equi-potential surfaces corresponding to
(a) a constant electric field in the z-direction,
(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,
(c) a single positive charge at the origin, and
(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.
Answer:
(a) A plane parallel to xy plane.
(b) Plane parallel to xy plane but the planes having different fixed potential will become closer with the increase in field intensity.
(c) Concentric spheres with origin as center.
(d) A time dependent changing shape nearer to grid which slowly becomes planar and parallel to the grid at far off distances from the grid.

Question 35.
In a van de Graaff type generator a spherical metal shell is to be a 15 X 10⁶ V electrode. The dielectric strength of the gas surrounding the electrode is 5 x 10⁷ vm^_1. What is the minimum radius of the spherical shell required ? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.) (C.B.S.E. 2008)
Answer:
Minimum radius of the shell of van de Graaff generator is given by the relation

Question 36.
A small sphere of radius r₁ and charge q₁ is enclosed by a spherical shell of radius r₂ and charge q₂ .Show that if q₁ is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge
q₂ on the shell is.
Answer:
Charge resides on the outer surface of a conductor. So the charge on inner sphere will flow towards the shell through the conducting wire. Moreover, from Gauss’ law, no electric field exists inside a Gaussian surface and also the charges enclosed by a closed surface only contribute towards the field. So q₂ does not matter in this case.If is positive, potential difference is also positive.

Question 37.
Answer the following:
(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decrases with altitude. Near the surface of the earth, the field is about 100 Vm^_1. Why then do we not get an electric shock as we step out of our house into the open ? (Assume the house to be a stell cage so there is no field inside !)
(b) A man fixes outside his house one evening a two meter high insulating slab carrying on its top a large aluminium sheet of area 1 m². Will he get an electric shock if he touches the metal sheet next morning ?
(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral ? In other words, what keeps the atmosphere charged ?
(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning ?
[Hint. The earth has an electric field of about 100 Vm^-1 at its surface in the downward direction, corresponding to a surface charge density = -10^-9 C m^-2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about +1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.]
Answer:
(a) Our body and the earth surface become equipotential (at equal potential). It means there is no potential difference between the earth and our body. Hence no current flows through our body and therefore we do not experience an electric shock.

(b) The aluminium sheet and the earth form a capacitor with the insulating slab as dielectric. The down pour of the atmospheric charge will raise the potential of the sheet of aluminium. When we touch the aluminium sheet, charge will flow to the earth through our body. This flow of charge constitutes on electric current and we will experience a shock.

(c) No doubt the atmosphere continuously gets charged due to lightning, thunder storms but simultaneously it gets discharged through normal weather zones. This keeps the system balanced.

(d) Electrical energy of the atmosphere appears as light, sound and heat energies during thunder storms and lightning.

Chapter 2 Electrostatic Potential and Capacitance Physics Class XII NCERT Exercise Questions & Solutions hold special importance in Engineering entrance exams also. NCERT Solutions of 12th Physics chapter 2 have 16 topics along with few subtopics which deal with the electrostatics concept. Therefore, Students are recommended to refer to all topics & subtopics conceptual explanation from NCERT Physics Books and then start solving the questions with ease.

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Class 11 Maths NCERT Solutions Chapter 1 Sets

Class 11 Maths Chapter 1 of NCERT has basic definitions and Operations involving Sets. In general, Sets are used to define the concepts of functions and relations. It is necessary to have basic knowledge of Sets as they are needed in the further study of sequences, geometry, probability. Sets is the easiest chapter among all the chapters and can be quite scoring. You will learn about Empty, Finite, Infinite, Equal, Subsets in this Chapter. You will also have Operations on Sets like Difference of Sets, Intersection of Sets and Union of Sets, etc. covered.

Avail the CBSE 11th Class NCERT Solutions for Chapter 1 Sets Ex 1.1, Ex 1.2, Ex 1.3, Ex 1.4, Ex 1.5, Ex 1.6, and brush up your fundamentals.

Class	11
Book	Mathematics
Subject	Maths
Chapter Number	1
Name of the Chapter	Sets

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