Important Questions

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter

Very short answer type questions

Question 1.
In the photoelectric effect, why should the photoelectric current increase as the intensity of monochromatic radiation incident on a photosensitive surface is increased? Explain.
Answer:
The photoelectric current increases proportionally with the increase in intensity of incident radiation. Larger the intensity of incident radiation, larger is the number of incident photons and hence larger is the number of electrons ejected from the photosensitive surface.

Question 2.
Define the term ‘threshold frequency’ in relation to photoelectric effect.
Answer:
Threshold frequency is defined as the minimum frequency of incident radiation which can cause photoelectric emission. It is different for different metal.

Question 3.
WrIte the basic features of photon picture of electromagnetic radiation on which Einstein’s photoelectric equation is based.
Answer:
Features of the photons are as:

• Photons are particles of light having energy E = hv and momentum p = \(\frac{h}{\lambda}\), where h is Planck’s constant.
• Photons travel with the speed of light in vacuum, independent of the frame of reference.
• Intensity of light depends on the number of photons crossing unit area in a unit time.

Question 4.
Define Intensity of radiation on the basis of photon picture of light. Write its SI unit.
Answer:
The amount of light energy or photon energy incident per meter square per second is called intensity of radiation1 Its SI unit is \(\frac{\mathrm{W}}{\mathrm{m}^{2}}\) or J/s m

Question 5.
State de Broglie hypothesis.
Answer:
According to the hypothesis of de Brogue “The atomic particles of matter moving with a given velocity, can display the wave-like properties.” i.e., λ = \(\frac{h}{m v}\) (mathematically)

Question 6.
Write the relationship of de Brogue wavelength λ associated with a particle of mass m terms of its kinetic energy E.
Answer:
Kinetic energy E_K = \(\frac{p^{2}}{2 m}\)
where, p = momentum
m = mass and E_K = kinetic energy
⇒ p = \(\sqrt{2 m E_{K}} \)
de Brogue wavelength,
λ = \(\frac{h}{p}\)
Where, p = \(\sqrt{2 m E_{K}} \)
⇒ λ = \(\frac{h}{\sqrt{2 m E_{K}}}\)

Question 7.
Name the phenomenon which shows the quantum nature of electromagnetic radiation.
Answer:
Photoelectric effect.

Question 8.
Do all the electrons that absorb a photon comes out as photoelectrons? (NCERT Exemplar)
Answer:
No, most electrons get scattered into the metal. Only a few come out of the surface of the metal.

Question 9.
There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelengths and emit light of shorter wavelengths? (NCERT Exemplar)
Answer:
In the first case, energy given out is less than the energy supplied. In the second case, the material has to supply the energy as the emitted photon has more energy. This cannot happen for stable substances.

Question 10.
There are two sources of light, each emitting with a power 100W.
One emits X-rays of wavelength 1 nm and the other visible light at 500 nm.
Find the ratio of number of photons of X-rays the photons of visible light of the given wavelength. (NCERT Exemplar) Ans. Total E is constant.
Let n₁ and n₂ be the number of photons of X-rays and visible region.
n₁E₂ = n₂E₂

Short answer type questions

Question 1.
What is meant by work function of a metal? How does the value of work function influence the kinetic energy of electrons liberated during photoelectron emission?
Answer:
Work Function: The minimum energy required to free an electron from metallic surface is called the work function.
Smaller the work function, larger the kinetic energy of emitted electron.

Question 2.
Show mathematically how Bohr’s postulate of quantization of orbital angular momentum in hydrogen atom is explained by de Broglie’s hypothesis.
Answer:
According to de Broglie’s hypothesis,
λ = \(\frac{h}{m v}\) …………………………… (1)
According to de Broglie’s condition of stationary orbits, the stationary orbits are those which contain complete de Broglie wavelength.
2πr = nλ …………………………………… (2)

Substituting value of λ from (1) in (2), we get
2πr = n\(\frac{h}{m v}\)
⇒ mvr = n\(\frac{h}{2 \pi}\) ……………………………………… (3)
This is Bohr’s postulate of quantization of energy levels.

Question 3.
Write briefly the underlying principle used in the Davison-Germer experiment to verify wave nature of electrons experimentally. What is the de Broglie wavelength of an electron with kinetic energy (E_K) 120 eV?
Answer:
Principle: Diffraction effects are observed for beams of electrons scattered by the crystals.
λ = \(\frac{h}{p}=\frac{h}{\sqrt{2 m E_{K}}}=\frac{h}{\sqrt{2 m e V}}\)
= \(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 120}}\)
λ = 0.112 nm.

Question 4.
(i) Describe briefly three experimentally observed features in the phenomenon of photoelectric effect.
(ii) Discuss briefly how wave theory of light cannot explain these features.
Answer:
(i) Three experimentally observed features in the phenomenon of photoelectric effect are as follows :

• Intensity: When intensity of incident light increases as one photon ejects one electron, the increase in intensity will increase the number of ejected electrons. Frequency has no effect on photoelectron.
• Frequency: When the frequency of incident photon increases, the kinetic energy of the emitted electrons increases. Intensity has no effect on kinetic energy of photoelectrons.
• No Time Lag: When energy incident photon is greater than the work function, the photoelectron is immediately ejected. Thus, there is no time lag between the incidence of light and emission of photoelectrons.

(ii) These features cannot be explained in the wave theory of light because wave nature of radiation cannot explain the following :

• The instantaneous ejection of the photoelectrons.
• The existence of threshold frequency for a metal surface.
• The fact that kinetic energy of the emitted electrons is independent of the intensity of light and depends upon its frequency.

Question 5.
A beam of monochromatic radiation is incident on a photosensitive surface. Answer the following questions :
(i) Do the emitted photoelectrons have the same kinetic energy?
(ii) Does the kinetic energy of the emitted electrons depend on the intensity of incident radiation?
(iii) On what factors does the number of emitted photoelectrons depend?
Answer:
In photoelectric effect, an electron absorbs a quantum of energy hv of radiation, which exceeds the work function, an electron is emitted with maximum kinetic energy.
E_{K max} = hv – W
(i) No, all electrons are bound with different forces in different layers of the metal. So, more tightly bound electron will emerge with less kinetic energy. Hence, all electrons do not have same kinetic energy.
(ii) No, because an electron cannot emit out if quantum energy hv is less than the work function of the metal. The K_E depends on the energy of each photon.
(iii) Number of emitted photoelectrons depends on the intensity of the radiations provided the quantum energy hv is greater than the work function of the metal.

Question 6.
Define the term “cut-off frequency” in photoelectric emission. The threshold frequency of a metal is f. When the light of frequency 2f is incident on the metal plate, the maximum velocity of photoelectrons is v₁. When the frequency of the incident radiation is increased to 5f, the maximum velocity of photoelectrons is v₂. Find the ratio v₁: v₂.
Answer:
Cut-off Frequency: It is that maximum frequency of incident radiation below which no photoemission takes place from a photoelectric material. According to Einstein’s photoelectric equation
E_{K max} = \(\frac{h c}{\lambda}-\phi \) = \(h v-\phi\)
Given that threshold frequency of the metal is f. If light of frequency, 2f is incident on metal plate, maximum velocity of photoelectron is v₁ then,
\(\frac{1}{2} m v_{1}^{2}\) = h (2f-f)
⇒ \(\frac{1}{2} m v_{1}^{2}\) = hf …………………….. (1)

If light of frequency, 5f is incident and maximum velocity of photoelectron is v₂.
\(\frac{1}{2} m v_{1}^{2}\) = h(5f-f)
⇒ \(\frac{1}{2} m v_{1}^{2}\) = 4hf ………………………………… (2)
Dividing (1) by (2), we get
\(\left(\frac{v_{1}}{v_{2}}\right)^{2}=\frac{1}{4}\)
⇒ \(\frac{v_{1}}{v_{2}}=\frac{1}{2}\)
∴ v₁ = v₂ = 1:2

Question 7.
Two monochromatic beams A and B of equal intensity I hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then what inference can you make about their frequencies? (NCERTExemplarl
Answer:
Let no. of photons falling per second of beam A = n_A
No. of photons falling per second of beam B = n_B
Energy of beam A = hv_A
Energy of beam B = h v_B

According to question, I = n_Av_A = n_Bv_B
\(\frac{n_{A}}{n_{B}}=\frac{v_{B}}{v_{A}}\) or \(\frac{2 n_{B}}{n_{B}}=\frac{v_{B}}{v_{A}}\)
⇒ v_B = 2 v_A
The frequency of beam B is twice that of A.

Question 8.
Consider Fig. for photoemission.

How would you reconcile with momentum- conservation? No light (Photons) have momentum in a different direction than the emitted electrons. (NCERT Exemplar)
Answer:
The momentum is transferred to the metal. At the microscopic level, atoms absorb the photon and its momentum is transferred mainly to the nucleus and electrons. The excited electron is emitted. Conservation of momentum needs to be accounted for the momentum transferred to the nucleus and electrons.

Long answer type questions

Question 1.
Describe Davisson and Germer’s experiment to demonstrate the wave nature of electrons. Draw a labeled diagram of apparatus used.
Answer:
Davisson and Germer Experiment: In 1927 Davisson and Germer performed a diffraction experiment with electron beam in analogy with X-ray diffraction to observe the wave nature of matter.
Apparatus: It consists of three parts
(i) Electron gun : It gives a fine beam of electrons, de Brogue used electron beam of energy 54 eV. de Brogue wavelength associated with this beam

λ = \(\frac{h}{\sqrt{2 m E_{k}}}\)
Here, m = mass of electron = 9.1 x 10^-31 kg
E_K = Kinetic energy of electron = 54eV
= 54 x 1.6 x 10^-19 J = 86.4 x 10^-19 J
∴ λ = \(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 86.4 \times 10^{-19}}}\)
= 1.66 x 10^-10 = 1.66 Å
(ii) Nickel crystal: The electron beam was directed on nickel crystal against the electron detector. The smallest separation between nickel atoms is O.914Å. Nickel crystal behaves as diffraction grating.

(iii) Electron detector: It measures the intensity of electron beam diffracted from nickel crystal. It may be an ionization chamber fitted with a sensitive galvanometer. The energy of electron beam, the angle of incidence of beam on nickel crystal and the position of detector can all be varied.

Method: The crystal is rotated in small steps to change the angle (α say) between incidence and scattered directions and the corresponding intensity (I) of scattered beam is measured. The variation of the intensity (I) of the scattered electrons with the angle of scattering a is obtained for different accelerating voltages.

The experiment was performed by varying the accelerating voltage from 44 V to 68 V. k was noticed that a strong peak appeared in the intensity (I) of the scattered electron for an accelerating voltage of 54 V at a scattering angle α = 50°.

From Bragg’s law, 2d sinθ = nλ
Here, n =1,d =0.914 Å,θ =65°
∴ λ = \(\frac{2 d \sin \theta}{n}=\frac{2 \times(0.914 \AA) \sin 65^{\circ}}{1}\)
=2 x 0.914 x 0.9063Å =1.65Å
The measured wavelength is in close agreement with the estimated de Broglie wavelength. Thus the wave nature of electrons is verified. Later on G.P. Thomson demonstrated the wave nature of fast electrons. Due to their work Davisson and G.P. Thomson were awarded Nobel Prize in 1937.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 12 Atoms Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 12 Atoms

Very short answer type questions

Question 1.
Why is the classical (Rutherford) model for an atom of electron orbiting around the nucleus not able to explain the atomic structure?
Answer:
The classical method could not explain the atomic structure as the electron revolving around the nucleus are accelerated and emit energy as the result, the radius of the circular paths goes on decreasing. Ultimately electrons fall into the nucleus, which is not in practice.

Question 2.
Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model? (NCERT Exemplar)
Answer:
According to Bohr model electrons having different energies belong to different levels having different values of n. So, their angular momenta will be different, as.
L = \(\frac{n h}{2 \pi} \text { or } L \propto n\)

Question 3.
State Bohr’s quantization condition for defining stationary orbits.
Answer:
According to Bohr’s quantization condition, electrons are permitted to revolve in only those orbits in which the angular momentum of electron is an integral multiple of \(\frac{h}{2 \pi}\) i.e.,
mvr = \(\frac{n h}{2 \pi}\) ,Where n = 1,2,3, ………………
m, y, rare mass, speed, and radius of electron respectively and h being Planck’s constant.

Question 4.
Define ionization energy. What Is Its value for a hydrogen atom?
Answer:
Ionisation Energy: The minimum amount of energy required to remove an electron from the ground state of the atom is known as ionization energy.
Ionisation energy for hydrogen atom =E_∞ – E₁ = – (-13.6 eV) = 13.6 eV

Question 5.
When an electron falls from a higher energy to a lower energy level, the difference in the energies appears in the form of electromagnetic radiation. Why cannot it be emitted as other forms of energy? (NCERT Exemplar)
Answer:
The transition of an electron from a higher energy to a lower energy level can appear in the form of electromagnetic radiation because electrons interact only electromagnetically.

Question 6.
Where is H a -line of the Balmer series in the emission spectrum of hydrogen atom obtained?
Answer:
Hα -line of the Balmer series in the emission spectrum of hydrogen atom is obtained in visible region.

Question 7.
Imagine removing one electron from He⁴ and He³. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why. (NCERT Exemplar)
Answer:
This is because both the nuclei are very heavy as compared to electron mass.

Question 8.
The mass of H-atom is less than the sum of the masses of a proton and electron. Why is this so? (NCERT Exemplar)
Answer:
Einstein’s mass-energy equivalence gives E – mc².
Thus the mass of an H-atom is m_p + m_e – \(\frac{B}{C^{2}}\)
where B ≈ 13.6 eV is the binding energy. It is less than the sum of masses of a proton and an electron.

Question 9.
Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb’s law as usual. Under such circumstances, calculate the ground state energy of a He-atom. (NCERT Exemplar)
Answer:
For a He-nucleus with charge 2 e and electrons of charge -e, the energy level in ground state is -E_n = Z²\(\frac{-13.6 \mathrm{eV}}{n^{2}}=2^{2} \frac{-13.6 \mathrm{eV}}{1^{2}}\)= -54.4eV
Thus, the ground state will have two electrons each of energy E and the total ground state energy would be -(4 x 13.6) eV = -54.4 eV.

Question 10.
Would the Bohr formula for the H-atom remain unchanged if proton had a charge (+ 4/3)e and electron a charge (-3/4)e, where e = 1.6 x 10^-19 C ? Give reasons for your answer. (NCERT Exemplar)
Answer:
Yes, since the Bohr formula involves only the product of the charges.

Short answer type questions

Question 1.
In an experiment of α-particle scattering by a thin foil of gold, draw a plot showing the number of particles scattered versus the scattering angle θ. Why is it that a very small fraction of the particles are scattered at θ > 90°?

Answer:
A small fraction of the alpha particles scattered at angle θ > 90° is due to the reason. That if impact parameter ‘b’ reduces to zero, coulomb force increases, hence alpha particles are scattered at angle θ>9O°, and only one alpha particle is scattered at angle 180°.

Question 2.
(i) State Bohr postulate of hydrogen atom that gives the relationship for the frequency of emitted photon in a transition,
(ii) An electron jumps from fourth to first orbit in an atom. How many maximum number of spectral lines can be emitted by the atom? To which series these lines correspond?
Answer:
(i) Bohr’s Third Postulate: It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is given by
hv = E_i – E_f
Where E_i and E_f are the energies of the initial and final states and E_i > E_f.
(ii) Electron jumps from fourth to first orbit in an atom

∴ Maximum number of spectral lines can be
4c₂ = \(\frac{4 !}{2 ! 2 !}=\frac{4 \times 3}{2}\) = 6
The line responds to Lyman series (e^– jumps to 1st orbit), Balmer series (e^– jumps to 2nd orbit), Paschen series (e^– jumps to 3rd orbit).

Question 3.
Using de Broglie’s hypothesis, explain with the help of a suitable diagram, Bohr’s second postulate of quantization of energy levels in a hydrogen atom.
Answer:
According to de Broglie’s hypothesis.
λ = \(\frac{h}{m v}\) ……………………….. (i)
According to de Broglie’s condition of stationary orbits, the stationary orbits are those which contain complete de Broglie wavelength.
2πr = nλ ………………………….. (2)

Substituting value of λ from eq. (2) in eq. (1), we get
2πr = n \(\frac{h}{m v}\)
⇒ mvr = n \(\frac{h}{2 \pi}\) ………………………… (3)
This is Bohr’s postulate of quantisation of energy levels.

Question 4.
In the study of Geiger-Marsden experiment on scattering of α-particles by a thin foil of gold, draw the trajectory of a-particles in the coulomb field of target nucleus. Explain briefly how one gets the information on the size of the nucleus from this study. From the relation R = R₀ A^1/3, where, R₀ is constant and A is the mass number of the nucleus, show that nuclear matter density is independent of A.
Answer:

From this experiment, the following is observed :
1. Most of the α-particles pass straight through the gold foil. It means that they do not suffer any collision with gold atoms.
2. About one α-particle in every 8000 α-particles deflects by more than 90°. As most of the a-particles gounder flected and only a- few get deflected, this shows that most of the space in an atom is empty and at the center of the atom, there exists a nucleus.

By the number of a-particles get deflected, the information regarding size of the nucleus can be known.
If m is the average mass of the nucleus and R is the nuclear radius, then mass of nucleus = mA, where A is the mass number of the element. Volume of the nucleus,

This shows that the nuclear density is independent of A.

Question 5.
Show that the first few frequencies of light that is emitted when electrons fall to nth level from levels higher than n, are approximate harmonics (L e., in the ratio 1: 2: 3,…) when n>> 1. (NCERTExempiar)
Answer:
The frequency of any line in a series in the spectrum of hydrogen-like atoms corresponding to the transition of electrons from (n + p) level to nth level can be expressed as a difference of two terms:
V_mn = \(c R Z^{2}\left[\frac{1}{(n+p)^{2}}-\frac{1}{n^{2}}\right] \)
where, m=n+p,(p=1,2,3,…………………………..)
and R is Rydberg constant.
For p << n

V_mn = \(c R Z^{2}\left[\frac{1}{n^{2}}\left(1+\frac{p}{n}\right)^{-2}-\frac{1}{n^{2}}\right]\)

Thus, the first few frequencies of light that is emitted when electrons fall to the nth level from levels higher than n, are approximate harmonic (i. e., in the ratio 1:2:3,…) when n>>1.

Long answer type questions

Question 1.
Using the postulates of Bohr’s model of hydrogen atom, obtain an expression for the frequency of radiation emitted when atom make a transition from the higher energy state with quantum number n₁ to the lower energy state with quantum number n_f (n_f < n_i).
Or
Using Bohr’s postulates, obtain the expression for the total energy of the electron in the stationary states of the hydrogen atom. Hence draw the energy level diagram showing how the line spectra corresponding to Balmer series occur due to transition between energy levels.
Or
Using Rutherford model of the atom, derive the expression for the total energy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron?
Answer:
Suppose m be the mass of an electron and v be its speed in nth orbit of radius r. The centripetal force for revolution is produced by electrostatic attraction between electron and nucleus.
\(\frac{m v^{2}}{r}=\frac{1}{4 \pi \varepsilon_{0}} \frac{(Z e)(e)}{r^{2}}\) …………………… (1) [Form rutherford Model]
or mv² = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{r}\)
So, Kinetic energy E_k = \(\frac{1}{2} m v^{2}\)
E_k = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{2 r}\)
Potential energy (PE) = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{(Z e)(-e)}{r}=-\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{r}\)
Total energy E = \(E_{K}+P E=\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{2 r}+\left(-\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{r}\right)\)
= \(-\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{2 r}\)

For nth orbit, E can be written as E_n
so,E_n = \(-\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{2 r_{n}}\) …………………. (2)
Negative sign indicates that the electron remains bound with the nucleus (or electron-nucleus form an attractive system) From Bohr’s postulate for quantisation of angular momentum.
mvr = \(\frac{n h}{2 \pi}\)
⇒ v = \(\frac{n h}{2 \pi m r} \)
Substituting this value of v in equation (1), we get

For Bohr’s radius, n = 1
Substituting value of r_n in equation (2), we get
E_n = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{2\left(\frac{\varepsilon_{0} h^{2} n^{2}}{\pi m Z e^{2}}\right)}=-\frac{m Z^{2} e^{4}}{8 \varepsilon_{0} h^{2} n^{2}}\)
R is called Rydberg constant.

For hydrogen atom Z =1, E_n = \(\frac{-R c h}{n^{2}}\)
If n_i and n_f are the quantum numbers of initial and final states and E_i and
E_f are energies of electrons in H-atoms in initial and final state, we have

For Balmer series, n_f=2, while n_i =3, 4, 5, …… ∞.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 10 Wave Optics Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 10 Wave Optics

Very short answer type questions

Question 1.
Define the term ‘wavefront’.
Answer:
It is defined as the locus of all points in a medium vibrating in the same phase.

Question 2.
State Huygen’s principle of diffraction of light.
Answer:
When a wavefront strikes to the corner of an obstacle, lightwave bends around the corner because every point on the wavefront again behaves like a . light source and emit secondary wavelets in all directions (Huygen’s wave theory) including the region of geometrical shadow. This explains diffraction.

Question 3.
Define the term ‘coherent sources’ which are required to produce interference patterns in Young’s double-slit experiment.
Answer:
Two monochromatic sources, which produce light waves, having a constant phase difference are known as coherent sources.

Question 4.
Define Doppler’s effect in light.
Answer:
It states that whenever there is a relative motion between the observer and the source of light, the apparent frequency of light received by the observer is different from the actual frequency of the light emitted by the source of light.

Question 5.
Define Doppler shift.
Answer:
It is defined as the apparent change in the frequency or wavelength of light due to the relative motion between the source and the observer.

Question 6.
Define redshift.
Answer:
It is defined as the shifting of radiations from the source of light towards the red end of the spectrum when the source moves away from the stationary observer. The wavelength increases due to redshift.

Question 7.
Define limit of resolution of an optical instrument.
Answer:
It is defined as the minimum distance by which the timepoint objects are separated so that their images can be seen as just separated by the optical instrument.

Question 8.
Define resolving power of the optical instruments.
Answer:
It is defined as the reciprocal of the limit of resolution of the optical instrument.

Question 9.
How are resolving power of a telescope change by increasing or decreasing the aperture of the objective?
Answer:
We know that the resolving power of telescope is given by
R.P. = \(\frac{D}{1.22 \lambda}\)
As R.p. ∝ D, so by increasing or decreasing D (aperture) of the objective, the resolving power is increased or decreased.

Question 10.
Which of the following waves can be polarised (i) Heat waves (ii) Sound waves? Give reason to support your answer.
Answer:
Heatwaves are transverse or electromagnetic in nature whereas sound waves are not. Polarisation is possible only for transverse waves.

Question 11.
How is linearly polarised light obtain by the process of scattering of light? find the Brewster angle for air-glass interface, when the refractive index of glass = 1.5
Answer:
According to Brewster law
tan i_B = μ
i_B = tan^-1 (μ)
i_B = tan^-1(l. 5)
i_B = 56.30

Question 12.
A polaroid (I) is placed in front of a monochromatic source. Another polaroid (II) is placed in front of this polaroid (I) and rotated till no light passes. A third polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II). Explain. (NCERT Exemplar)
Answer:
Only in the special cases when the pass axis of (III) is parallel to (I) or (II), there shall be no light emerging. In all other cases, there shall be light emerging because the pass axis of (I) is no longer perpendicular to the pass axis of (III).

Question 13.
What is the shape of the wavefront of earth for sunlight? (NCERT Exemplar)
Answer:
Spherical with huge radius as compared to the earth’s radius so that it is almost a plane.

Question 14.
Consider a point at the focal point of a convergent lens. Another convergent lens of short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image? (NCERT Exemplar)
Answer:
The focal point of a convergent lens is the position of real image formed by this lens when object is at infinity. When another convergent lens of short focal length is placed on the other side, the combination will form a real point image at the combined focus of the two lenses. The wavefronts emerging from the final image will be spherical.

Short answer type questions

Question 1.
State two conditions required for obtaining coherent sources. In Young’s arrangement to produce interference pattern, show that dark and bright fringes appearing on the screen are equally spaced.
Answer:
Conditions for obtaining coherent sources:
(i) Coherent sources of light should be obtained from a single source by same device.
(ii) The two sources should give monochromatic light.
The separation between the centres or two consecutive bright fringes is the width of a dark fringe.

Hence, all bright and dark fringes are equally spaced on screen.

Question 2.
How will the interference pattern in Young’s double-slit experiment get affected, when
(i) distance between the slits S₁ and S₂ reduced and
(ii) the entire set-up is immersed in water? Justify your answer in each case.
Answer:
(i) The fringe width of interference pattern increases with the decrease in separation between S₁S₂ as
β ∝ \(\frac{1}{d}\)
(ii) The fringe width decrease as wavelength gets reduced when interference set up is taken from air to water.

Question 3.
What is the minimum angular separation between two stars, if a telescope is used to observe them with an objective of aperture 0.2 m? The wavelength of light used is 5900 A.
Answer:
Here, D = diameter of the objective of telescope = 0.2 m
λ = Wavelength of light used = 5900 Å = 5900 x 10^-10 m
Let dθ = Minimum angular separation between two stars =?
Using the relation,
dθ = \(\frac{1.22 \lambda}{D}\) , we get
dθ = \(\frac{1.22 \times 5900 \times 10^{-10}}{0.2}= \) = 3.6 x 10^-6 rad.

Question 4.
Distinguish between polarised and unpolarised light. Does the intensity of polarised light emitted by a polaroid depend on its orientation? Explain briefly. The vibrations in a beam of polarised light make an angle of 60° with the axis of the polaroid sheet. What percentage of light is transmitted through the sheet?
Answer:
A light which has vibrations in all directions in a plane perpendicular to the direction of propagation is said to be unpolarised light. The light from the sun, an incandescent bulb or a candle is unpolarised. If the electric field vector of a light wave vibrates just in one direction perpendicular to the direction of wave propagation, then it is said to be polarised or linearly polarised light.

Yes, the intensity of polarised light emitted by a polaroid depends on orientation of polaroid. When polarised light is incident on a polaroid, the resultant intensity of transmitted light varies directly as the square of the cosine of the angle between polarisation direction of light and the axis of the polaroid.

I ∝ cos² θ or I = I₀ cos² θ
where I₀ = maximum intensity of transmitted light;
θ = angle between vibrations in light and axis of polaroid sheet.
or I =I₀ cos² 60° = \(\frac{I_{0}}{4}\)
Percentage of light transmitted = \(\frac{I}{I_{0}} \) x 100 = \(\frac{1}{4}\) x 100 = 25%

Question 5.
Find an expression for intensity of transmitted light, when a polaroid sheet is rotated between two crossed polaroids. In which position of the polaroid sheet will the transmitted intensity be maximum?
Answer:
Let us consider two crossed polarizers, P₁ and P₂ with a polaroid sheet P₃ placed between them.

Let I₀ be the intensity of polarised light after passing through the first polarizer P₁.
If θ is the angle between the axes of P₁ and P₃, then the intensity of the polarised light after passing through P₃ will be I =I₀ cos²θ.
As P₁ and P₂ are crossed, the angle between the axes of P₁ and P₂ is 90°.
∴ The angle between the axes of P₂ and P₃ is (90° – 0).
The intensity of light emerging from P₂ will be given by

The intensity of polarised light transmitted from P₂ will be maximum, when ,
sin 2θ = maximum = 1
⇒ sin2θ = sin9O°
⇒ 2θ = 90°
⇒ θ = 45°
Also, the maximum transmitted intensity will be given by I = \(\frac{I_{0}}{4}\)

Question 6.
State Brewster’s law. The value of Brewster angle for a transparent medium is different for light of different colours. Give reason.
Answer:
Brewster’s Law: When unpolarized light is incident on the surface separating two media at polarising angle, the reflected light gets completely polarised only when the reflected light and the refracted light are perpendicular to each other. Now, refractive index of denser (second) medium with respect to rarer (first) medium is given by μ = tan i_B, where i_B = polarising angle.
Since refractive index is different for different colours (wavelengths), Brewster’s angle is different for different colours.

Question 7.
Can reflection result in plane polarised light if the light is incident on the interface from the side with higher refractive index? (NCERT Exemplar)
Answer:
When angle of incidence is equal to Brewster’s angle, the transmitted light is unpolarised and reflected light is plane polarised.

Consider the diagram in which unpolarised light is represented by dot and plane polarised light is represented by arrows.
Polarisation by reflection occurs when the angle of incidence is the Brewster’s angle
i.e., tani_B = ¹μ₂ = \(\frac{\mu_{2}}{\mu_{1}}\) where μ² < μ¹
when the light rays travels in such a medium, the critical angle is
sin i_c = \(\frac{\mu_{2}}{\mu_{1}}\)
where, μ₂ < μ₁
As | tani_B| > | sin i_C| for large angles i_B <i_C.
Thus, the polarisation by reflection occurs definitely.

Question 8.
Consider a two-slit interference arrangements (figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of λ such that the first minima on the screen falls at a distance D from the centre O. (NCERT Exemplar)

Answer:
From the given figure of two-slit interference arrangements, we can write

The minima will occur when S₂P – S₁P = (2 n -1)\(\frac{\lambda}{2}\)
i.e., [D² +(D + X)²]^1/2 -[D² + (D -x)²]^1/2
= \(\frac{\lambda}{2}\)
[for first minima n = 1]
If x = D
We can write [D² +4D²]^1/2 -[D² +0]^1/2 = \(\frac{\lambda}{2}\)
⇒ [5D²]^1/2 – [D²]^1/2 = \(\frac{\lambda}{2}\)
⇒ \(\sqrt{5}\)D – D = \(\frac{\lambda}{2}\)

Long answer type questions

Question 1.
In Young’s double-slit experiment, deduce the conditions for (i) constructive, and (ii) destructive interference at a point on the screen. Draw a graph showing variation of the resultant intensity in the interference pattern against position ‘X’ on the screen.
Answer:
Conditions for Constructive and Destructive Interference :
When two waves of same frequency and constant initial phase difference travel in the same direction along a straight line simultaneously, they superpose in such a way that the intensity of the resultant wave is maximum at certain points and minimum at certain other points. The phenomenon of redistribution of intensity due to superposition of two waves of same frequency and constant initial phase difference is called the interference.

The waves of same frequency and constant initial phase difference are called coherent waves. At points of medium where the waves arrive in the same phase, the resultant intensity is maximum and the interference at these points is said to be constructive. On the other hand, at points of medium where the waves arrive in opposite phase, the resultant intensity is minimum and the interference at these points is said to be destructive. The positions of maximum intensity are called maxima while those of minimum intensity are called minima. The interference takes place in sound and light both.

Mathematical Analysis: Suppose two coherent waves travel in the same direction along a straight line, the frequency of each wave is \(\frac{\omega}{2 \pi}\) and amplitudes of electric field are a₁ and a₂ respectively. If at any time t, the electric fields of waves at a point are y₁ and y₂ respectively and phase difference is, Φ then equation of waves may be expressed as
y₁ = a₁ sin ωt ………………………. (1)
y₂ = a₂ sin ωt +Φ) ……………………………………….. (2)
According to Young’s principle of superposition, the resultant displacement at that point will be
y = y₁+y₂ ……………………………….. (3)
Substituting values of y₁ and y₂ from (1) and (2) in (3), we get
y = a₁ sin ωt + a₂ sin(ωt + Φ)

Using trigonometric relation,
sin(ωt +Φ) = sinωtcosΦ +cosωtsinΦ
y = a₁ sin ωt + a₂(sinωtcosΦ) + cosωt sin Φ)
= (a₁ +a₂cos Φ) sinωt + (a₂ sinΦ)cosωt …………………………….. (4)
Let a₁ + a₂ cosΦ = A cos θ ……………………………………… (5)
and a₂ sinΦ = A sinθ ………………………………………… (6)

Where A and θ are new constants
Then equation (4) gives
y = A cosθ sinωt + A sinθ cosωt
= A sin (ωt +θ) ……………………………………………. (7)
This is the equation of the resultant disturbance. Clearly the amplitude of resultants disturbance is A and phase difference from first wave is 0. The values of A and 0 are determined by (5) and (6). Squaring (5) and (6) and then adding, we get

∴ Amplitude,
A = \(\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \phi} \) …………………………… (8)
As the intensity of a wave is proportional to its amplitude in arbitrary units I = A²
∴ Intensity of resultant wave,
I = A² = a₁² + a₂² + 2a₁a₂ cosΦ ……………………….. (9)
Clearly, the intensity of the resultant wave at any point depends on the amplitudes of individual waves and the phase difference between the waves at the point.

Constructive Interference: For maximum intensity at any point
cos Φ = +1
or phase difference Φ = 0,2π,4π,6π,……………………….
= 2nπ (n=0,1,2,3,……………………) …………………………………… (10)
The maximum intensity
I_max = a¹₂+a_2²
= (a₁+a₂)² …………………………..(11)
Path difference
Δ = \(\frac{\lambda}{2 \pi}\) x phase difference
= \(\frac{\lambda}{2 \pi} \) x 2nπ …………………………………………. (12)
Clearly, the maximum intensity is obtained in the region of superposition at those points where waves meet in the same phase or the phase difference between the waves is even multiple of π or path difference between them is the integral multiple of λ and maximum intensity is (a₁ +a₂)²

which is greater than the sum intensities of individual waves by an amount 2a₁a₂.
Destructive Interference : For minimum intensity at any point CosΦ = -1
or phase difference,
Φ = π,3π,5π,7π, …………………………..
– (2n-l)π, n = 1,2,3,… …………………………………. (13)
In this case the minimum intensity,
I_min =a₁² +a₂² – 2a₁a₂
= (a₁-a₂)² ………………………… (14)

Path difference, Δ = \(\frac{\lambda}{2 \pi}\) x Phase difference
= \(\frac{\lambda}{2 \pi}\) x (2n – 1)π
= (2n-l) \(\frac{\lambda}{2}\)

Clearly, the minimum intensity is obtained in the region of superposition at those points where waves meet in opposite phase or the phase difference between the waves is odd multiple of π or path difference between the waves is odd multiple of \(\frac{\lambda}{2}\) and minimum intensity = (a₁ -a₂)² which is less than the sum of intensities of the individual waves by an amount 2a₁a₂.

From equations (12) and (14) it is clear that the intensity 2a₁a₂ is transferred from positions of minima to maxima, this implies that the interference is based on conservation of energy i.e., there is no wastage of energy.
Variation of Intensity of light with position x is shown in fig.

Question 2.
Describe diffraction of light due to a single slit. Explain formation of a pattern of fringes obtained on the screen and plot showing variation of intensity with angle θ in single slit diffraction.
Answer:
Diffraction of Light at a Single Slit: When monochromatic light is made incident on a single slit, we get diffraction pattern on a screen placed behind the slit. The diffraction pattern contains bright and dark bands, the intensity of central band is maximum and goes on decreasing on both sides.

Explanation: Let AB be a slit of width ‘a’ and a parallel beam of monochromatic light is incident on it. According to Fresnel, the diffraction pattern is the result of superposition of a large number of waves, starting from different points of illuminated slit.

Let θ be the angle of diffraction for waves reaching at point P of screen and AN the perpendicular dropped from A on wave diffracted from B. The path difference between rays diffracted at points A and B,
Δ = BP – AP = BN
In ΔANB, ∠ANB = 90°
and ∠BAN = θ
∴ sinθ = \(\frac{B N}{A B}\) or BN = AB sinθ
As AB = width of slit = a
Path difference Δ = asinθ ……………………………… (1)

To find the effect of all coherent waves at P, we have to sum up their contribution, each with a different phase. This was done by Fresnel by rigorous calculations, but the main features may be explained by simple arguments given below :
At the central point C of the screen, the angle 0 is zero. Hence the waves starting from all points of slit arrive in the same phase. This gives maximum intensity at the central point C. If point P on screen is such that the path difference between rays starting from edges A and B is λ, then path difference,

If angle θ is small,
sinθ = θ = \(\frac{\lambda}{a}\) ……………………………. (2)
Minima: Now we divide the slit into two equal halves AO and OB, each of width \(\frac{a}{2}\).
Now for every point, M₁ in AO, there is a corresponding point M₂ in OB, such that M₁M₂ = \(\frac{a}{2}\) .
Then path difference between waves arriving at P and starting from M₁ and M₂ will be \(\frac{a}{2}\) sin θ = \(\frac{\lambda}{2}\).

This means that the contributions from the two halves of slit AO and OB are opposite in phase and so cancel each other. Thus equation (2) gives the angle of diffraction at which intensity falls to zero. Similarly it may be shown that the intensity is zero for sin θ = \(\frac{n \lambda}{a}\) , with n as integer. Thus, the general condition of minima is asinθ = nλ ……………………………………… (3)

Secondary Maxima: Let us now consider angle θ such that
sin θ = θ = \(\frac{3 \lambda}{2 a}\)

Which is midway between two dark bands given by
sin θ = θ = \(\frac{\lambda}{a} \) and sin θ = θ = \(\frac{2 \lambda}{a}\)
Let us now divide the slit into three parts. If we take the first two parts of slit, the path difference between rays diffracted from the extreme ends of the first two parts.
\(\frac{2}{3}\) a sin θ = \(\frac{2}{3} a \times \frac{3 \lambda}{2 a}\) = λ

Then the first two parts will have a path difference of \(\frac{\lambda}{2}\) and cancel the effect of each other. The remaining third part will contribute to the intensity at a point between two minima. Clearly, there will be maxima between first two minima, but this maximum will be of much weaker intensity than central maximum.

This is called first secondary maxima. In a similar manner, we can show that there are secondary maxima between any two consecutive minima; and the intensity of maxima will go on decreasing with increase of order of maxima.
In general, the position of nth maxima will be given by
a sin θ = \(\left(n+\frac{1}{2}\right)\) λ (n =1, 2, 3, 4,…) ………………………………… (4)
The intensity of secondary maxima decreases with increase of order n because with increasing n, the contribution of slit decreases.
For n = 2, it is one-fifth, for n = 3, it is one-seventh and so on.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 3 Current Electricity Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 3 Current Electricity

Very short answer type questions

Question 1.
Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with each other across an external resistance R. What is the current through this resistance?
Answer:

⇒ I = \(\frac{E_{\text {eq. }}}{R+r_{\text {eq. }}}\)
Given,internal resistance, r = 0
∴ I = \(\frac{E_{\mathrm{eq} .}}{R}\)

Question 2.
Define mobility of a charge carrier. What is its relation with relaxation time?
Answer:
It is defined as how fast electron moves from one place to another.
It is also defined as drift velocity per unit electric field. The SI unit of mobility is m²/V-sec and it is denoted as μ.
μ = \(\frac{\left|v_{d}\right|}{E}=\frac{e E \tau}{m E}=\frac{e \tau}{m}\)
⇒ μ ∝ τ

Question 3.
Define the term ‘relaxation time’ in a conductor.
Answer:
The average time between successive collisions of electrons conductor is known as relaxation time.

Question 4.
Write the expression for the drift velocity of charge carriers in a conductor of length ‘L’ across which a potential difference ‘V’ is applied.
Answer:
v_d = \(\frac{e V}{m L}\)τ

Question 5.
For household electrical wiring, one uses Cu wires or Al wires. What considerations are kept in mind? (NCERT Exemplar)
Answer:
Two considerations are required: (i) cost of metal, and (ii) good conductivity of metal. Cost factor inhibits silver. Cu and Al are the next best conductors.

Question 6.
Define the current sensitivity of a galvanometer. Write its SI unit.
Answer:
Ratio of deflection produced in the galvanometer and the current flowing through it is called current sensitivity.
Current sensitivity S_i = \(\frac{\theta}{I}\)
SI unit of current sensitivity S_i is division/ampere or radian/ampere.

Question 7.
Nichrome and Copper wires of same length and same radius are connected in series. Current I is passed through them. Which wire gets heated up more? Justify your answer.
Answer:
Nichrome, since its resistance is high.

Question 8.
Why are alloys used for making standard resistance coils?
(NCERT Exemplar)
Answer:
Alloys have:

• low value of temperature coefficient and the resistance of the alloy does not vary much with rise in temperature.
• high resistivity, so even a smaller length of the material is sufficient to design high standard resistance.

Question 9.
What is the advantage of using thick metallic strips to join wires in a potentiometer? (NCERT Exemplar)
Answer:
The metal strips have low resistance and need not be counted in the potentiometer length l of the null point. One measures only their lengths along the straight segments (of length 1 metre each). This is easily done with the help of centimeter rulings or meter ruler and leads to accurate measurements.

Question 10.
Is the motion of a charge across junction momentum conserving? Why or why not? (NCERT Exemplar)
Answer:
When an electron approaches a junction, in addition to the uniform electric field E facing it normally, it keep the drift velocity fixed as drift velocity depend on E by the relation
V_d = \(\frac{e E \tau}{m}\)
This result into accumulation of charges on the surface of wires at the junction. These produce additional field. These fields change the direction of momentum.
Thus, the motion of a charge across junction is not momentum conserving.

Short answer type questions

Question 1.
Sketch a graph showing the variation of resistivity of carbon with temperature.
Or Plot a graph showing temperature dependence of resistivity for a typical semiconductor. How is this behaviour explained?
Answer:
The resistivity of a typical semiconductor (carbon) decreases with increase of temperature. The graph is shown in figure.

Explanation: In semiconductor the number density of free electrons (n) increases with increase in temperature (T) and consequently the relaxation period decreases. But the effect of increase in n has higher impact than decrease of τ. So, resistivity decreases with increase in temperature.

Question 2.
Two cells of emf ε₁ and ε₂ having internal resistances r₁ and r₂ respectively are connected in parallel as shown. Deduce the expressions for the equivalent emf and equivalent internal resistance of a cell which can replace the combination between the points B₁ and B₂.

Answer:
Consider a parallel combination of the cells. I₁ and I₂ are the currents leaving the positive electrodes of the cells. At point B₁, I₁ and I₂ flow in whereas current I flows out. Therefore, we have
I = I₁ + I₂ …………….. (1)
Let V(B₁) and V(B₂) be the potentials at B₁ and B₂ respectively.
Then, considering the first cell, the potential difference across its terminals is V(B₁) – V(B₂). Hence, from equation V = E – Ir we have
V = V(B₁) – V(B₂) = E₁ – I₁r₁ …………… (2)
Points B₁ and B₂ are connected exactly Similarly to the second cell. Hence, considering the second cell, we also have
V = V(B₁) – V(B₂)
= E₂ – I₂r₂ …………… (3)
Combining equations (1), (2) and (3), we have

If we want to replace the combination by a single cell, between Bl and B₂, of emf E_eq and internal resistance r_eq, we would have
V = E_eq – Ir_eq

Question 3.
State Kirchhoff s rules of current distribution in an electrical network.
Or State KirchhofPs rules. Explain briefly how these rules are justified.
Answer:
Junction Rule: In an electric circuit, the algebraic sum of currents at any junction is zero.
At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction.
ΣI = 0
Justification: This rule is based on the law of conservation of charge.
Loop Rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop must be zero.
ΣΔV = 0
or The algebraic sum of emf s in any loop of a circuit is equal to the
sum of products of currents and resistances in it.
ΣΔE = ΣIR
Justification: This rule is based on the law of conservation of energy,

Question 4.
Define the term current density of a metallic conductor. Deduce the relation connecting current density (J) and the conductivity σ of the conductor, when an electric field E, is applied to it.
Answer:
Current density at a point in a conductor is defined as the amount of current flowing per unit area of the conductor around that point provided the area is held in a direction normal to the current,
J = \(\frac{I}{A}\)
Current density is a vector quantity. Its direction is the direction of motion of positive charge. The unit of current density is ampere/metre² or [Am^-2].
Relation between J, σ and E

Question 5.
What is Wheatstone bridge? Deduce the condition for which Wheatstone bridge is balanced.
Or The given figure shows a network of resistances R₁, R₂, R₃ and R₄.

Using Kirchhoffs laws, establish the balance condition for the network.
Or Use Kirchhoffs law to obtain the balance Wheatstone’s bridge.
Answer:

The Wheatstone bridge is an arrangement of four resistances. In this bridge, four resistances are connected on four arms of quadrilateral. In one diagonal, a battery and a key are connected. In second diagonal a galvanometer is connected as shown in fig. Consider P,Q,R and S four resistances are connected on the sidesAB,BC, AD and DC of the quadrilateral respectively.

Galvanometer G is connected between points B and D and a battery E is connected between A and C. Now in balance condition, when the deflection in a galvanometer is zero in closed mesh ABDA, then by applying Kirchhoffs law,
I₁p – IR = 0
or I₁P = I₂R ………….. (1)
In closed mesh CBDC,
I₁Q = I₂S ……………… (2)
Dividing (1) by (2) \(\frac{P}{Q}=\frac{R}{S}[latex]
This is the balanced condition of the Wheatstone bridge.

Question 6.
First a set of n equal resistors of R each are connected in series to a battery of emf E and internal resistance R. A current I is observed to flow. Then the n resistors are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is n ? (NCERT Exemplar)
Answer:
When n resistors are in series, I = [latex]\frac{E}{R+n R}\) ;
When n resistors are in parallel, \(\frac{E}{R+\frac{R}{n}}\) 10I
\(\frac{1+n}{1+\frac{1}{n}}\) = 10 ⇒ \(\frac{1+n}{n+1}\) n = 10
∴ n = 10

Question 7.
Two cells of same emf E but internal resistance r and r₁ and r₂ are connected in series to an external resistor R (Fig.). What should be the value of R so that the potential difference across the terminals of the first cell becomes zero. NCERT Exemplar)
Answer:
I = \(\frac{E+E}{R+r_{1}+r_{2}}\)
V₁ = E – Ir₁ = E – \(\frac{2 E}{r_{1}+r_{2}+R}\) r₁ = 0

Long answer type questions

Question 1.
(i) Find the magnitude and direction of current in 1Ω resistor in given circuit.

(ii)Two students X and Y perform an experiment on potentiometer separately using the circuit diagram shown below.
Keeping other things unchanged (a) X increases the value of resistance R, (b) Y decreases the value of resistance S in the set up. How will these changes affect the position of null point in each case and why?

Answer:
(i) For the mesh APQBA
-6 -1 (I₂ – I₁) + 3I₁ = 0
or -I₂ + 4I₁ = 6 …………… (1)

For the mesh PCDQP
2I₂ – 9 + 3I₂ + 1(I₂ – I₁) = 0
or 6I₂ – I₁ = 9 …………… (2)
Solving eqs. (1) and (2), we get
I = \(\frac{45}{23}\) A
I= \(\frac{42}{23}\) A
∴ Current through the 1Ω resistor = (I₂ – I₁) = \(\) A

(ii) (a) By increasing resistance R, the current in main circuit decreases, so potential gradient decreases. Hence, a greater length of wire would be needed for balancing the same potential difference. So, the null point would shift towards right (i.e., towards B).

(b) By decreasing resistance S, the terminal potential difference V = \(\frac{E}{1+\frac{r}{S}}\) across cell decreases, so balance is obtained at small length i.e., point will be obtained at smaller length. So, the null point would shift towards left (i.e., towards A).

Question 2.
A room has AC run for 5 hours a day at a voltage of 220 V. The wiring of the room consists of Cu of 1 mm radius and a length of 10m. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions? [ρCu = 1.7 × 10^-8Ωm, ρAL = 2.7 × 10^-8Ωm] (NCERT Exemplar)
Answer:
Power consumption in a day i.e., in 5 hours = 10 units
Or power consumption per hour = 2 units
Or power consumption = 2 units = 2 kW = 2000 W
Also, we know that power consumption in resistor,
P = V × I
⇒ 2000 W = 220 V × I
or I = 9 A
Now, the resistance of wire is given by R = ρ \(\frac{l}{A}\)
where, A is cross-sectional area of conductor. Power consumption in first current carrying wire is given by
P = I² . R
ρ \(\frac{l}{A}\) I² = 1.7 × 10^-8 × \(\frac{10}{\pi \times 10^{-6}}\) × 81 = 4J/s A
The fractional loss due to the joule heating in first wire
= \(\frac{4}{2000}\) × 100 = 0.2%
Power loss in Al wire = 4\(\frac{\rho_{A l}}{\rho_{C u}}\) = 1.6 × 4 = 6.4 J/s
The fractional loss due to the joule heating in second wire
= \(\frac{6.4}{2000}\) × 100 =0.32%

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 13 Nuclei Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 13 Nuclei

Very short answer type questions

Question 1.
Write the relationship between the size of a nucleus and Its mass number (A).
Answer:
The relationship is R = RoA^1/3
where, R = radius of nucleus, A = mass number.

Question 2.
Define the activity of a given radioactive substance. Write its SI unit.
Answer:
The activity of a sample is defined as the rate of disintegration taking place in the sample of radioactive substances.
SI unit of activity is Becquerel (Bq).
1 Bq = 1 disintegration/second

Question 3.
Why is it found experimentally difficult to detect neutrinos in nuclear p-decay?
Answer:
Neutrinos are difficult to detect because they are massless, have no charge, and do not interact with nucleons.

Question 4.
A nucleus undergoes p-decay. How does its :
(i) mass number
(ii) atomic number change?
Answer:
During p-decay,
(i) no change in mass number.
(ii) atomic number increases by 1.

Question 5.
In pair annihilation, an electron and a positron destroy each other to produce gamma radiations. How is the momentum conserved? (NCERT Exemplar)
Answer:
In pair annihilation, an electron and a positron destroy each other to produce 2γ photons which move in opposite directions to conserve linear momentum.
The annihilation is shown below ₀e⁺¹ + ₀ e⁺¹ →2γ ray photons.

Question 6.
Which one of the following cannot emit radiation and why? Excited nucleus, excited electron. (NCERT Exemplar)
Answer:
Excited electrons cannot emit radiation because energy of electronic energy levels is in the range of eV and not MeV (mega electron volt). y-radiations have energy of the order of MeV.

Question 7.
He_2³ and He_1³ nuclei have the same mass number.
Do they have the same binding energy? (NCERT Exemplar)
Answer:
Nuclei He_2³ and He_1³ have the same mass number.
He_2³ has two protons and one neutron. He_1³ has one proton arid two neutrons.
The repulsive force between protons is missing in ₁He³, so the binding energy of ₁He³ is greater than that of ₂He³.

Question 8.
Which sample, A or B as shown in figure has shorter mean-life? (NCERT Exemplar)

Answer:
B has shorter mean life as λ is greater for B.

Short answer type questions

Question 1.
(i) Define the terms (a) half-life (b) average life. Find out the relationship with the decay constant (λ).
(ii) A radioactive nucleus has a decay constant ) λ = 0.346 (day)^-1
How long would it take the nucleus of decay to 75% of Its Initial amount?
Answer:
(i)
(a) Half-life of a radioactive element is defined as the time during which half number of atoms present initially in the sample of the element decay or it is the time during which number of atoms left undecayed in the sample is half the total number of atoms present in the sample.
it is represented by T_1/2.
From the equation N = N₀e^-λt ,
At half-life, t = T_1/2,N = \(\frac{N_{0}}{2}\)

On taking log both sides, we get
λT_1/2 = log _e2
T _1/2 = \(\frac{\log _{e} 2}{\lambda}=\frac{\log _{10} 2 \times 2.303}{\lambda}\)
= \(\frac{0.3010 \times 2.303}{\lambda}\)
After n half-life, the number of atoms left undecayed is given by
N = N₀\(\left(\frac{1}{2}\right)^{n}\)
T_1/2 = \(\frac{0.6932}{\lambda}\)

(b) Average life of a radioactive element can be obtained by calculating the total life time of all atoms of the element and dividing it by the total number of atoms present initially in the sample of the element.
Average life or mean life of radioactive element is
τ = \(\frac{\text { Total life of all atoms }}{\text { Total number of atoms }}\)
τ = \(=\int_{0}^{N_{0}} \frac{t d N}{N_{0}}=\int_{\infty}^{0-\lambda N_{0} e^{-\lambda t} d t \times t}{N_{0}}\)
[when N =N₀,t = 0 and when N = 0, t = ∞] [∵dN =-λ(N₀e^–λt)dt]
= λ₀∫^∞ te ^-λtdt

(ii) Given, λ = 0.3465 (day)
According to the radioactive decay law, we have
R = R₀e^-λt

⇒ t = 0.830 s

Question 2.
(i) Write three characteristic properties of nuclear force.
(ii) Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from the graph.
Answer:
(i) Characteristic Properties of Nuclear Force
(a) Nuclear force act between a pair of neutrons, a pair of protons and also between a neutron-proton pair, with the same strength. This shows that nuclear forces are independent of charge.
(b) The nuclear forces are dependent on spin or angular momentum of nuclei.
(c) Nuclear forces are non-central forces. This shows that the distribution of nucleons in a nucleus is not spherically symmetric. From the plot, it is concluded that

(ii)
(a) The potential energy is minimum at a distance r₀(=0.8fm) which means that the force is attractive for distance larger than 0.8 fm and repulsive for the distance less than 0.8 fm between the nucleons.
(b) Nuclear forces are negligible when the distance between the nucleons is more than 10 fm.

Question 3.
Explain giving necessary reaction, how energy is released during:
(i) fission
(ii) fusion
Answer:
(i) Nuclear Fission: The phenomenon of splitting of heavy nuclei (mass number > 120) into smaller nuclei of nearly equal masses is known as nuclear fission. In nuclear fission, the sum of the masses of the product is less than the sum of masses of the reactants. This difference of mass gets converted into energy E = me and hence sample amount of energy is released in a nuclear fission.
e.g., _92²³⁵ U + ₀¹n → ₅₆¹⁴¹Ba + _36⁹²Kr + 3 _0¹ + Q
Masses of reactant = 235.0439 amu + 1.0087 amu
= 236.0526 amu
Masses of product = 140.9139 + 91.8973 + 3.0261
= 235.8373 amu
Mass defect = 236.0526 -235.8373
= 0.2153 amu
∵ 1 amu = 931MeV
⇒ Energy released = 0.2153 x 931
⇒ 200 MeV nearly

(ii) Nuclear Fusion: The phenomenon of conversion of two lighter nuclei into a single heavy nucleus is called.nuclear fusion. Since the mass of the heavier product nucleus is less than the sum of masses of reactant nuclei and therefore certain mass defect occurs which converts into energy as per Einstein’s mass-energy relation. Thus, energy is released during nuclear fusion.
e.g., ₁H¹ + ₁H¹ → ₁H² + e⁺ + v + 0.42 MeV
Also, ₁H² + ₁H² → ₁H³ + ₁H¹ + 4.03 MeV

Question 4.
Give reasons for :
(a) Lighter elements are better moderators for a nuclear reactor than heavier elements.
(b) In a natural uranium reactor, heavy water is preferred moderator as compared to ordinary water.
(c) Cadmium rods are provided in a reactor.
(d) Very high temperatures as those obtained in the interior of the sun are required for fusion reaction.
Answer:
(a) A moderator slows down fast neutrons released in a nuclear reactor. The basic principle of mechanics is that the energy transfer in a collision is the maximum when the colliding particles have equal masses. As lighter elements have mass close to that of neutrons, lighter elements are better moderators than heavier elements.

(b) Ordinary water has hydrogen nuclei (₁¹H) which have greater absorption capture for neutrons; so ordinary water will absorb neutrons rather than slowing them; on the other hand, the heavy hydrogen nuclei (₂¹H) have negligible absorption capture for neutrons, so they share energy with neutrons and neutrons are slowed down.

(c) Cadmium has high absorption capture for neutrons; so cadmium rods are used to absorb extra neutrons; so nuclear fission in a nuclear reactor is controlled; therefore cadmium rods are called control rods.

(d) In nuclear fusion, two positively charge nuclei fuse together. When two positively charged nuclei come near each other to fuse together, strong electrostatic repulsive force acts between them.
To overcome this repulsive force, extremely high temperatures of the order of 10⁸ K are required.
This may be calculated as follows: For fusion of H-nuclei,

The temperature in the interior of sun is about 2 x 10⁷ K.
Therefore, very high temperatures of the order 10⁷ K are required for fusion reaction to take place.

Question 5.
Deuteron is a bound state of a neutron and a proton with a binding energy B 2.2 Mev. A γ-ray of energy E is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the n and p move in the direction of the incident y-ray. If E = B, show that this cannot happen. Hence, calculate how much bigger than B must be E be for such a process to happen. (NCERT Exemplar)
Answer:
Given binding energy B = 2.2 MeV
From the energy conservation law,
E-B=K_n+K_p = \(\frac{p_{n}^{2}}{2 m}+\frac{p_{p}^{2}}{2 m}\) ………………….. (1)
From conservation of momentum,
P_n +P_p = \(\frac{E}{c}\)
As E = B,Eq. (1) p_n²+ P_p² =0 ……………………………. (2)
It only happen if p_n = p_p = 0

So, the Eq. (2) cannot satisfied and the process cannot take place.
Let E = B +X, when X<< B for the process to take place.
Put value of p1 from Eq. (2) in Eq. (1), we get
X = \(\frac{\left(\frac{E}{c}-p_{p}\right)}{2 m}+\frac{p_{p}^{2}}{2 m}\)
or 2p_p² – \(\frac{2 E p_{p}}{c}+\frac{E^{2}}{c^{2}}\) – 2mx = 0

Using the formula of quadratic equation, we get
P_p = \(\frac{\frac{2 E}{c} \pm \sqrt{\frac{4 E^{2}}{c^{2}}-8\left(\frac{E^{2}}{c^{2}}-2 m X\right)}}{4}\)
For the real value p_p, the discriminant is positive

Long answer type questions

Question 1.
Define the term: Half-life period and decay constant of a radioactive sample. Derive the relation between these terms.
Answer:
Half-life Period: The half-life period of an element is defined as the time in which the number of radioactive nuclei decay to half of its initial value.
Decay Constant: The decay constant of a radioactive element is defined as the reciprocal of time in which the number of undecayed nuclei of that radioactive element falls to times of its initial value. Relation between Half-life and Decay Constant:
The radioactive decay equation is N = N₀e^-λt …………………………. (i)
When t = T,N= \(\frac{N_{0}}{2}\)
∴ \(\frac{N_{0}}{2}\) = N₀e^-λT or e^-λT = \(\frac{1}{2}\)
……………………. (2)
Taking log on both sides, we get

Question 2.
Draw the graph showing the variation of binding energy per nucleon with the mass number for a large number of nuclei 2 < A < 240. What are the main inferences from the graph? How do you explain the constancy of binding energy in the range 30<A<170 using the property that the nuclear force is short-ranged? Explain with the help of this plot the release of energy in the processes of nuclear fission and fusion.
Answer:
The variation of binding energy per nucleon versus mass number is shown in figure.

Inferences from Graph
1. The nuclei having mass numbers below 20 and above 180 have relatively small binding energy and hence they are unstable.
2. The nuclei having mass numbers 56 and about 56 have maximum binding energy – 5.8 MeV and so they are most stable.
3. Some nuclei have peaks, e.g., ₂He⁴, ₆C¹², ₈O¹⁶; this indicates that these nuclei are relatively more stable than their neighbors.
(i) Explanation of constancy of binding energy: Nuclear force is short-ranged, so every nucleus interacts with its neighbors only, therefore binding energy per nucleon remains constant.

(ii) Explanation of nuclear fission: When a heavy nucleus (A ≥ 235 say) breaks into two lighter nuclei (nuclear fission), the binding energy per nucleon increases i.e., nucleons get more tightly bound. This implies that energy would be released in nuclear fission,

(iii) Explanation of nuclear fusion: When two very light nuclei (A ≤ 10) join to form a heavy nucleus, the binding energy per nucleon of fused heavier nucleus is more than the binding energy per nucleon of lighter nuclei, so again energy would be released in nuclear fusion.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 5 Magnetism and Matter Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 5 Magnetism and Matter

Very short answer type questions

Question 1.
What is the angle of dip at a place where the horizontal and vertical components of the earth’s magnetic field are equal?
Answer:
The angle of dip is given by
θ = tan^-1 (\(\frac{B_{V}}{B_{H}}\))
B_V = vertical component of the earth’s magnetic field.
B_H = horizontal component of the earth’s magnetic field.
So, as B_V = B_H
Then, θ = tan^-1 (1) = 45°
∴ The angle of dip will be θ = 45°.

Question 2.
The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping?
Answer:
As the copper plates oscillate in the magnetic field between the two plates of the magnet, there is a continuous change of magnetic flux linked with the pendulum. Due to this, eddy currents are set up in the copper plate which try to oppose the motion of the pendulum according to the Lenz’s law and finally bring it to rest.

Question 3.
Relative permeability of a material μ_r = 0.5. Identify the nature of the magnetic material and write its relation of magnetic susceptibility.
Answer:
The nature of magnetic material is a diamagnetic.
μ_r = 1 + χ_m

Question 4.
Which of the following substances are diamagnetic?
Bi, Al, Na, Cu, Ca and Ni
Answer:
Diamagnetic substances are (i) Bi (ii) Cu.

Question 5.
The susceptibility of a magnetic material is -4.2 × 10^-6. Name the type of magnetic material, it represents.
Answer:
Negative susceptibility represents diamagnetic substance.

Question 6.
What are permanent magnets? Give one example.
Answer:
Substances that retain their attractive property for a long period of time at room temperature are called permanent magnets.
Examples: Those pieces which are made up of steel, alnico, cobalt and ticonal.

Question 7.
Why is the core of an electromagnet made of ferromagnetic materials?
Answer:
Ferromagnetic material has a high retentivity. So on passing current through windings it gains sufficient magnetism immediately.

Question 8.
From molecular view point, discuss the temperature dependence of susceptibility for diamagnetism, paramagnetism and ferromagnetism. (NCERT Exemplar)
Answer:
Diamagnetism is due to orbital motion of electrons developing magnetic moments opposite to applied field and hence is not much affected by temperature.

Paramagnetism and ferromagnetism is due to alignments of atomic magnetic moments in the direction of the applied field. As temperature increases, this alignment is disturbed and hence susceptibilities of both decrease as temperature increases.

Question 9.
A ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet.
(i) In which direction will it move?
(ii) What will be the direction of its magnetic moment? (NCERT Exemplar)
Answer:
When a diamagnetic material is dipped in liquid nitrogen, it again behaves as a diamagnetic material. Thus, superconducting material will again behave as a diamagnetic material. When this diamagnetic material is placed near a bar magnet, it will be feebly magnetised opposite to the direction of magnetising field.

(i) Thus, it will be repelled.
(ii) Also its direction of magnetic moment will be opposite to the direction of magnetic field of magnet.

Question 10.
Consider the plane S formed by the dipole axis and the axis of earth. Let P be point on the magnetic equator and in S. Let Q be the point of intersection of the geographical and magnetic equators. Obtain the declination and dip angles at P and Q. (NCERT Exemplar)
Answer:
In adjoining figure:

(i) P is in S (needle will point both north)
Declination = 0
P is also on magnetic equator.
∴ Dip = 0

(ii) Q is on magnetic equator.
∴ Dip = 0
But declination = 11.3.

Short answer type questions

Question 1.
Explain the following:
(i) Why do magnetic field lines form continuous closed loops?
(ii) Why are the field lines repelled (expelled) when a
diamagnetic material is placed in an external uniform magnetic field?
Answer:

(i) Magnetic lines of force form continuous closed loops because a magnet is always a dipole and as a result, the net magnetic flux of a magnet is always zero.
(ii) When a diamagnetic substance is placed in an external magnetic field, a feeble magnetism is induced in opposite direction. So, magnetic lines of force are repelled.

Question 2.
How does a circular loop carrying current behaves as a magnet?
Answer:
The current round in the face of the coil is in anti-clockwise direction, then this behaves like a North pole, whereas when it viewed from other scale, then current round in it is in clockwise direction necessarily forming South pole of magnet.

Hence, current loop have both magnetic poles and therefore, behaves like a magnetic dipole.

Question 3.
Give two points to distinguish between a paramagnetic and diamagnetic substance.
Answer:

Paramagnetic substance	Diamagnetic substance
1. A paramagnetic substance is feebly attracted by magnet.	A diamagnetic substance is feebly repelled by a magnet.
2. For a paramagnetic substance, the intensity of magnetisation has a small positive value.	For a diamagnetic substance, the intensity of magnetism has a small negative value.

Question 4.
(a) How is an electromagnet different from a permanent magnet?
Write two properties of a material which makes it suitable for making (i) a permanent magnet, and (ii) an electromagnet.
Answer:
(a) An electromagnet consists of a core made of a ferromagnetic material placed inside a solenoid. It behaves like a strong magnet when current flows through the solenoid and effectively loses its magnetism when the current is switched off.

A permanent magnet is also made up of a ferromagnetic material but it retains its magnetism at room temperature for a long time after being magnetised one.

(b) Properties of material are as below:
(i) Permanent magnet

• Retentivity and coercivity should be large
• Magnetically hard

(ii) An electromagnet

• Magnetically soft
• Coercivity should be low.

Question 5.
A bar magnet of magnetic moment M and moment of inertia I (about centre, perpendicular to length) is cut into two equal pieces, perpendicular to length. Let T be the period of oscifiations of the original magnet about an axis through the mid-point, perpendicular to length, in a magnetic field B. What would be the similar period T’ for each piece? (NCERT Exemplar)
Answer:
Given, I = moment of inertia of the bar magnet
m = mass of bar magnet
l = length of magnet about an any passing through its centre and perpendicular to its length
M = magnetic moment of the magnet
B = uniform magnetic field in which magnet is oscillating, we get time period of oscillation is
T = 2π\(\sqrt{\frac{I}{M B}}\)
Here I = \(\frac{m l^{2}}{12}\)
When magnet is cut into two equal pieces, perpendicular to length, then moment of inertia of each piece of magnet about an axis perpendicular to length passing through its centre is

Question 6.
A uniform conducting wire of length 12 a and resistance R is wound up as a current carrying coil in the shape of (i) an equilateral triangle of side a; (ii) a square of sides a and, (iii) a regular hexagon of sides a. The coil is connected to a voltage source V₀. Find the magnetic moment of the coils in each case. (NCERT Exemplar)

(i) Area of equilateral triangle, A = \(\frac{\sqrt{3}}{4}\) a²
(ii) Area of square, A = a²

Long answer type questions

Question 1.
Verify the Ampere’s law for magnetic field of a point dipole of dipole moment M = Mk̂. Take C as the closed curve running clockwise along
(i) the z-axis from z a > 0 to z = R,
(ii) along the quarter circle of radius R and centre at the origin in the first quadrant of vz-plane,
(iii) along the x-axis from x = R to x = a, and
(iv) along the quarter circle of radius a and centre at the origin
in the first quadrant of xz-plane (NCERT Exemplar)
Answer:
From P to Q, every point on the z-axis lies at the axial line of magnetic
dipole of moment \(\vec{M}\). Magnetic field induction at a point distance z from the magnetic dipole of moment is

(ii) Along the quarter circle QS of radius R as-.given in the figure below

The point A lies on the equatorial line of the magnetic dipole of moment M sin0. Magnetic field at point A on the circular arc is
B = \(\frac{\mu_{0}}{4 \pi} \frac{M \sin \theta}{R^{3}}\) ; \(\overrightarrow{d l}\) = Rdθ

(iii) Along x-axis over the path ST, consider the figure given below.

From figure, every point lies on the equatorial line of magnetic dipole. Magnetic field induction at a point distance x from the dipole is

(iv) Along the quarter circle TP of radius a. Consider the figure given below

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 4 Moving Charges and Magnetism Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 4 Moving Charges and Magnetism

Very short answer type questions

Question 1.
Using the concept of force between two infinitely long parallel current carrying conductors define one ampere of current.
Answer:
One ampere is that value of current which flows through two straight, parallel infinitely long current carrying conductors placed in air or. vacuum at a distance of 1 m and they experience a force of attractive or repulsive nature of magnitude 2 × 10^-7 N/m on their unit length.

Question 2.
State Ampere’s circuit law.
Answer:
It states that the line integral of the magnetic field \(\vec{B}\) around any closed circuit is equal to p0 times the total current passing through this closed circuit.
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = μ₀ I

Question 3.
A narrow beam of protons and deuterons, each having the same momentum, enters a region of uniform magnetic field directed perpendicular to their direction of momentum. What would be the ratio of the radii of the circular path described by them?
Answer:
For the given momentum of charge particle, radius of circular paths depends on charge and magnetic field as
r = \(\) ⇒ r ∝ \(\)
For given momentum,
∴ r_proton : r_deuteron = 1 : 1
As they have same momentum and charge moving in a small magnetic field.

Question 4.
Write the expression, in a vector form, for the Lorentz magnetic force \(\overrightarrow{\boldsymbol{F}}\) due to a charge moving with velocity \(\vec{v}\) in a magnetic field \(\overrightarrow{\boldsymbol{B}}\). What is the direction of the magnetic force?
Answer:
Force, \(\vec{F}=q(\vec{v} \times \vec{B})\)
Obviously, the force on charged particle is perpendicular to both velocity \(\vec{v}\) and magnetic field \(\vec{B}\).

Question 5.
When a charged particle moving with velocity \(\vec{v}\) is subjected to magnetic field \(\overrightarrow{\boldsymbol{B}}\), the force acting on it is non-zero. Would the particle gain any energy?
Answe:
No. (i) This is because the charge particle moves on a circular path.
(ii) \(\vec{F}=q(\vec{\nu} \times \vec{B})\)
and power dissipated p = \(\vec{F} \times \vec{V}\)
= q \((\vec{v} \times \vec{B}) \times \vec{y}\) = p\((\vec{v} \times \vec{v}) \times \vec{B}\)
The particle does not gain any energy.

Question 6.
A square coil OPQR of side a carrying a current 7, is placed in the Y-Z plane as shown here. Find the magnetic moment associated with this coil.

Answer:
The magnetic moment associated with the coil, is \(\vec{\mu}\)_m = Ia²î

Question 7.
Under what condition is the force acting on a charge moving through a uniform magnetic field is minimum?
Answer:
F_m = qvB sinθ; for minimum force sinθ = 0. i.e., force is minimum when charged particle move parallel or anti-parallel to the field.

Question 8.
What is the nature of magnetic field in a moving coil galvanometer?
Answer:
The nature of magnetic field in a moving coil galvanometer is radial.

Question 9.
Verify that the cyclotron frequency ω = eB/m has the correct dimensions of [T]^-1. (NCERT Exemplar)
Or A charged particle of charge e and mass m is moving in an electric field E and magnetic field B. Construct dimensionless quantities and quantities of dimension [T]^-1. (NCERT Exemplar)
Answer:
For a charge particle moving perpendicular to the magnetic field, the magnetic Lorentz forces provides necessary centripetal force for revolution.
\(\frac{m v^{2}}{R}\) = qvB
On simplifying the terms, we have
∴ \(\frac{q B}{m}=\frac{v}{R}\) = ω
Finding the dimensional formula of angu
∴ [ω] = \(\left[\frac{q B}{m}\right]=\left[\frac{v}{R}\right]\) = [T] ^-1

Question 10.
Show that a force that does no work must be a velocity dependent force. (NCERT Exemplar)
Answer:
Let no work is done by a force, so we have
dW = F.dl = 0
⇒ F. v dt = 0 (Since, dl = v dt and dt ≠ 0)
⇒ F.v = 0
Thus, F must be velocity dependent which implies that angle between F and v is 90°. If v changes (direction), then (directions) F should also change so that above condition is satisfied.

Question 11.
The magnetic force depends on v which depends on the inertial frame of reference. Does then the magnetic force differ from inertial frame to frame? Is it reasonable that the net acceleration has a different value in different frames of reference? (NCERT Exemplar|
Answer:
Yes, the magnetic force differ from inertial frame to frame. The magnetic force is frame dependent.
The net acceleration which comes into existing out of this is however, frame independent (non-relativistic physics) for inertial frames.

Question 12.
An electron enters with a velocity υ = υ₀î into a cubical region (faces parallel to coordinate planes) in which there are uniform electric and magnetic fields. The orbit of the electron is found to spiral down inside the cube in plane parallel to the x-y plane. Suggest a configuration of fields E and B that can lead to it. (NCERT Exemplar)
Answer:
Considering magnetic field B = B₀k̂, and an electron enters with a velocity v = v₀î into a cubical region (faces parallel to coordinate planes). The force on electron, using magnetic Lorentz force, is given by
F = -e(v₀î x B₀k̂) = ev₀B₀î
which revolves the electron in x-y plane.
The electric force F = -eE₀k̂ accelerates e along z-axis which in turn increases the radius of circular path and hence particle traversed on spiral path.

Short answer type questions

Question 1.
Write any two important points of similarities and differences each between Coulomb’s law for the electrostatic field and . Biot-Savart’s law for the magnetic field.
Answer:
Similarities: Both electrostatic field and magnetic field

• follows the principle of superposition.
• depends inversely on the square of distance from source to the point of interest.

Differences I

• Electrostatic field is produced by a scalar source (q) and the magnetic
  field is produced by a vector source (I\(\overrightarrow{d l}\)).
• Electrostatic field is along the displacement vector between source and point of interest; while magnetic field is perpendicular to the plane, containing the displacement vector and vector source.
• Electrostatic field is angle independent, while magnetic field is angle
  dependent between source vector and displacement vector.

Question 2.
State the underlying principle of a cyclotron. Write briefly how this machine is used to accelerate charged particles to high energies.
Or State the principle of the working of a cyclotron. Write two uses of this machine.
Answer:
The combination of crossed electric and magnetic fields is used to increase the energy of the charged particle. Cyclotron uses the fact that the frequency of revolution of the charged particle in a magnetic field is independent of its energy.

Inside the dees the particle is shielded from the electric field and magnetic field acts on the particle and makes it to go round in a circular path inside a dee.

Every time, the particle moves from one dee to the other it comes under the influence of electric field which ensures to increase the energy of the particle as the sign of the electric field changed alternately.
The increased energy increases the radius of the circular path so the accelerated particle moves in a spiral path.
Since, radius of trajectory
r = \(\frac{v m}{q B}\)
∴ v = \(\frac{r q B}{m}\)
Hence, the kinetic energy of ions
= \(\frac{1}{2}\)mv² = \frac{1}{2}\(\)m\(\frac{r^{2} q^{2} B^{2}}{m^{2}}\)
⇒ KE = \(\frac{1}{2}\)\(\frac{r^{2} q^{2} B^{2}}{m}\)

Question 3.
(i) State Ampere’s circuital law expressing it in the integral form.
(ii) Two long co-axial insulated solenoids S₁ and S₂ of equal length are wound one over the other as shown in the figure. A steady current I flows through the inner solenoid S₁ to the other end B which is connected to the outer solenoid S₂ through which the some current I flows in the opposite direction so, as to come out at end A. If n₁ and n₂ are the number of turns per unit length, find the magnitude and direction of the net magnetic field at a point
(a) inside on the axis and
(b) outside the combined system.

Answer:
(i) Ampere’s circuital law states that the line integral of magnetic field
(B) around any closed path in vacuum is μ₀ times the net current (I) threading the area enclosed by the curve.
Mathematically, \(\oint \vec{B} \cdot d \vec{l}\) = μ₀I
Ampere’s law is applicable only for an Amperian loop as the Gauss’s law is used for Gaussian surface in electrostatics.

(ii) According to Ampere’s circuital law, the net magnetic field is given by
B = μ₀nî.
(a) The net magnetic field is given by
B_net = B₂ – B₁
μ₀n₂I₂ – ₀n₁I₁
= μ₀I(n₂ – n₁)
The direction is from B to A.

(b) As the magnetic field due to Sx is confined solely inside S₁ as the solenoids are assumed to be very long. So, there is no magnetic field outside S₁ due to current in S₁, similarly there is no field outside S₂.
B_net = 0

Question 4.
(a) State Biot-Savart law and express this law in the vector
form.
(b) Two identical circular coils, P and Q each of radius R, carrying currents 1 A and √3 A, respectively, are placed concentrically and perpendicular to each other lying in the XY and YZ planes. Find the magnitude and direction of the net magnetic field at the centre of the coils. ’
Answer:
(a) According to Biot-Savart’s law the magnitude of the magnetic field \(\overrightarrow{d B}\) due to a small element of length dl of a current carrying wire at a point P, is proportional to the current I, the element length dl and is inversly proportional to the square of the distance r. It is also proportional to sinθ,

where θ is the angle between \(\overrightarrow{d l}\) and \(\vec{r}\).

Its direction is perpendicular to the plane containing \(\overrightarrow{d l}\) and \(\vec{r}\) in vector form
\(\overrightarrow{d B}\) ∝ \(\frac{I \overrightarrow{d l} \times \vec{r}}{r^{3}}\)
⇒ \(\overrightarrow{d B}\) = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{I d \vec{l} \cdot \vec{r}}{r^{3}}\)
(\(\overrightarrow{d l}\) is directed along the length of the wire in the direction of current and
\(\vec{r}\) is the vector joining the centre of current element to the point P) (b) Field due to current in coil P is
\(\vec{B}\)₂ = \(\frac{\mu_{0} I_{1}}{2 R}\) .k̂
(Assuming current to be anticlockwise as seen form + ve Z-axis) and that due to current in coil Q is
\(\vec{B}\)₂ = \(\frac{\mu_{0} I_{2}}{2 R} \hat{i}\)
(Assuming current to be anticlockwise as seen form positive X-axis)

Question 5.
Explain, giving reasons, the basic difference in converting a galvanometer into (i) a voltmeter and (ii) an ammeter.
Answer:
(i) In converting a galvanometer into a voltmeter, a very high suitable resistance is connected in series to its coil. So, the galvanometer gives full scale deflection.
(ii) In converting a galvanometer into an ammeter, a very small suitable resistance is connected in parallel to its coil. The remaining pair of the current i.e., (I – I_g) flows through the resistance. Here I = Circuit current
and I_g = Current through galvanometer.

Question 6.
A current carrying loop consists of 3 identical quarter circles of radius R, lying in the positive quadrants of the x-y, y-z and z-x planes with their centres at the origin, joined together. Find the direction and magnitude of B at the origin. (NCERT Exemplar)
Answer:
For the current carrying loop quarter circles of radius R, lying in the positive quadrants of the x-y plane
B₁ = \(\frac{\mu_{0}}{4 \pi} \frac{I(\pi / 2)}{R}\) = k̂\(\frac{\mu_{0}}{4} \frac{I}{2 R}\) k̂
For the current carrying loop quarter circles of radius R, lying in the positive quadrants of the y-z plane
B₂ = \(\frac{\mu_{0}}{4} \frac{I}{2 R}\)î
For the current carrying loop quarter circles of radius R, lying in the positive quadrants of the z-x plane
B₃ = \(\frac{\mu_{0}}{4} \frac{I}{2 R}\)
Current carrying loop consists of 3 identical quarter circles of radius R, lying in the positive quadrants of the x-y, y-y and z-z planes with their centres at the origin, joined together is equal to the vector sum of magnetic field due to each quarter and given by,
B = \(\frac{1}{4 \pi}\) (î + ĵ + k̂)\(\frac{\mu_{0} I}{2 R}\)

Question 7.
A long straight wire carrying current of 25 A rests on a table as shown in figure. Another wire PQ of length 1 m, mass 2.5 g carries the same current but in the opposite direction.
The wire PQ is free to slide up and down. To what height will PQ rise? (NCERT Exemplar)

Answer:
The magnetic field produced by long straight wire carrying current of 25 A rests on a table on small wire
B = \(\frac{\mu_{0} I}{2 \pi h}\)
The magnetic force on small conductor is , F = BIl sin θ = BIl
Force applied on PQ balance the weight of small current carrying wire.

Long answer type questions

Question 1.
Derive an expression for the force per unit length between two long straight parallel current carrying conductors. Hence define SI unit of current (ampere).
Answer:
Suppose two long thin straight conductors (or wires) PQ and RS are placed parallel to each other in vacuum (or air) carrying currents I₁ and Isub>2 respectively. It has been observed experimentally that when the currents in the wire are in the same direction, they experience an attractive force (fig. a) and when they carry currents in opposite directions, they experience a repulsive force (fig. b).

Let the conductors PQ and RS carry currents I₁ and I₂ in same direction and placed at separation r.

Consider a current-element ‘ab’ of length ΔL of wire RS. The magnetic field produced by current- carrying conductor PQ at the location of other wire RS.
B₁ = \(\frac{\mu_{0} I_{1}}{2 \pi r}\) ………….(1)

According to Maxwell’s right hand rule or right hand palm rule number 1, the direction of B₁ will be perpendicular to the plane of paper and directed downward. Due to this magnetic field, each element of other wire experiences a force. The direction of current element is perpendicular to the magnetic field; therefore the magnetic force on element ab of length ΔL
ΔF = B₁I₂ΔL sin90° = \(\frac{\mu_{0} I_{1}}{2 \pi r}\) I₂ ΔL
∴ The total force on conductor of length L will be
F = \(\frac{\mu_{0} I_{1} I_{2}}{2 \pi r}\) ΔΣL = \(\frac{\mu_{0} I_{1} I_{2}}{2 \pi r}\)L
∴ Force acting per unit length of conductor
f = \(\frac{F}{L}\) = \(\frac{\mu_{0} I_{1} I_{2}}{2 \pi r}\) N/M …………… (2)

According to Fleming’s left hand rule, the direction of magnetic force will be towards PQ, i.e., the force will be attractive.

On the other hand if the currents I₁ and I₂ in wires are in opposite directions, the force will be repulsive. The magnitude of force in each case remains the same.

Definition of SI Unit of Current (Ampere) : In SI system of fundamental unit of current ‘ampere’ is defined assuming the force between the two current carrying wires as standard.

The force between two parallel current carrying conductors of separation r is
\(\frac{F}{L}\) = \(\frac{\mu_{0} I_{1} I_{2}}{2 \pi r}\) N/M

If I₁ = I₂ = 1A, r = lm, then
f = \(\frac{\mu_{0}}{2 \pi}\) = 2 x 10^-7 N/m
Thus, 1 ampere is the current which when flowing in each of parallel conductors placed at separation 1 m in vacuum exert a force of 2 x 10^-7 on 1 m length of either wire.

Question 2.
Draw the labelled diagram of a moving coil galvanometer. Prove that in a radial magnetic field, the deflection of the coil is directly proportional to the current flowing in the coil.
Or (a) Draw a labelled diagram of a moving coil galvanometer.
Describe briefly its principle and working.
(b) Answer the following:
(i) Why is it necessary to introduce a cylindrical soft iron core inside the coil of a galvanometer?
(ii) Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity. Explain, giving reason.
Or Explain, using a labelled diagram, the principle and working of a moving coil galvanometer. What is the function of (i) uniform radial magnetic field, (ii) soft iron core?
Or Define the terms (i) current sensitivity and (ii) voltage sensitivity of a galvanometer. Why does increasing the current sensitivity not necessarily increase voltage sensitivity?
Answer:
(a) Moving Coil Galvanometer: A galvanometer is used to detect current in a circuit.

Construction: It consists of a rectangular coil wound on a non-conducting metallic frame and is suspended by phosphor bronze strip between the pole-pieces (N and S) of a strong permanent magnet. A soft iron core in cylindrical form is placed between the coil.

One end of coil is attached to suspension wire which also serves as one terminal (Tx) of galvanometer. The other end of coil is connected to a loosely coiled strip, which serves as the other terminal (T2). The other end of the suspension is attached to a torsion head which can be rotated to set the coil in zero position. A mirror (M) is fixed on the phosphor bronze strip by means of which the deflection of the coil is measured by the lamp and scale arrangement. The levelling screws are also provided at the base of the instrument.

The pole pieces of the permanent magnet are cylindrical so that the magnetic field is radial at any position of the coil.

Principle and Working : When current (I) is passed in the coil, torque τ acts on the coil, given by
τ = NIABsinθ

where θ is the angle between the normal to plane of coil and the magnetic field of strength B, N is the number of turns in a coil.

When the magnetic field is radial, as in the case of cylindrical pole pieces and soft iron core, then in every position of coil the plane of the coil, is parallel to the magnetic field lines, so that θ = 90° and sin 90 ° = 1. The coil experiences a uniform coupler.
Deflecting torque, τ = NIAB
If C is the torsional rigidity of the wire and θ is the twist of suspension strip, then restoring torque = C0. For equilibrium, deflecting torque = restoring torque
i.e., NIAB = Cθ
θ = \(\frac{N A B}{C}\)I ………… (1)
i.e., θ ∝ I
Deflection of coil is directly proportional to current flowing in the coil and hence we can construct a linear scale.

Importance (or Function) of Uniform Radial Magnetic Field
Torque as current carrying coil in a magnetic field is τ = NIAB sinθ In radial magnetic field sinθ = 1, so torque is τ = NIAB.
This makes the deflection (θ) proportional to current. In other words, the radial magnetic field makes the scale linear.

(b)
(i) The cylindrical, soft iron core makes the (1) field radial and (2) increases the strength of the magnetic field, i.e., the magnitude of the torque.

(ii) Sensitivity of Galvanometer
Current sensitivity: It is defined as the deflection of coil per unit current flowing in it.
Sensitivity,
_I = (\(\frac{\theta}{I}\)) = \(\frac{N A B}{C}\)………… (1)
Voltage sensitivity: It is defined as the deflection of coil per unit potential difference across its ends.
i.e., S_V = \(\frac{\theta}{V}\) = \(\frac{N A B}{R_{g} \cdot C}\) …………. (2)
where R_g is resistance of galvanometer.
Clearly for greater sensitivity number of turns N, area A and magnetic field strength B should be large and torsional rigidity C of suspension should be small.
Dividing eqs. (2) by (1)
\(\frac{S_{V}}{S_{I}}=\frac{1}{G}\) = 1 ⇒ S_V = \(\frac{1}{G}\) SI
Clearly, the voltage sensitivity depends on current sensitivity and the resistance of galvanometer. If we increase current sensitivity and resistance G is larger, then it is not certain that voltage sensitivity will be increased. Thus, the increase of current sensitivity does not imply the increase of voltage sensitivity.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits

Very short answer type questions

Question 1.
The graph shown in the figure represents a plot of current versus voltage for a given semiconductor. Identify the region at which 5 the semiconductor has a negative resistance.
Answer:
Resistance of a material can be found out by the slope of the curve V versus I. Part BC of the curve u shows the negative resistance as with the increase in current and decrease in voltage.

Question 2.
Sn, C, SI and Ge are all group 14 elements, Yet, Sn is a conductor, C is an insulator while Si and Ge are semiconductors. Why? (NCERT Exemplar)
Answer:
If the valence and conduction bands overlap (no energy gap), the substance is referred as a conductor. For insulators the energy gap is large and for semiconductors, the energy gap is moderate. The energy gap of Sn is O eV, for C is 5.4 eV, for Si is 1.1 eV and for Ge is 0.7 eV. Accordingly, their electrical conductivity varies.

Question 3.
Why are elemental dopants for Silicon or Germanium usually chosen from group 13 or group 15? (NCERTExemplar)
Answer:
The size of dopant atoms should be such as not to distort the pure semiconductor lattice structure and yet easily contribute a charge camer on forming covalent bonds with Si or Ge.

Question 4.
What happens to the width of depletion Layer of a p.n junction when it is
(i) forward biased,
(ii) reverse biased
Answer:
(i) When forward biased, the width of depletion layer decreases.
(ii) When reverse biased, the width of depletion layer increases.

Question 5.
In a transistor, doping level in base is increased slightly. How will it affect
(i) collector current and
(ii) base current?
Answer:
When doping level in base is increased slightly;
(i) Collector current decreases slightly and
(ii) Base current increases slightly.

Question 6.
Can the potential barrier across a p-n junction be measured by simply connecting a voltmeter across the junction (NCERT Exemplar)
Answer:
No, because the voltmeter must have a resistance very high compared to the junction resistance, the latter being nearly infinite.

Question 7.
Explain why elemental semiconductors cannot be used to make visible LEDs.(NCERT Exemplar)
Answer:
Elemental semiconductor’s band-gap is such that’ electromagnetic emissions are in infrared region.

Question 8.
Draw the logic circuit of a NAND gate and write its truth table.
Answer:
Logic circuit of a NAND gate

Truth table

A	B	Y = A.B
0	0	1
0	1	1
1	0	1
1	1	0

Short answer type questions

Question 1.
Write any two distinguishing features between conductors, semiconductors and insulators on the basis of energy band diagrams.
or
Draw the necessary energy band diagrams to distinguish between conductors, semiconductors and insulators. How does the change in temperature affect the behaviour of these materials? Explain briefly.
Answer:

Distinguishing Features
(a) In conductors: Valence band and conduction band overlap each other.
In semiconductors: Valence band and conduction band are separated by a small energy gap.
In insulators: Valence band and conduction band are separated by a large energy gap.
(b) In conductors: Large number of free electrons are available in conduction band.
In semiconductors: A very small number of electrons are available for electrical conduction.
In insulators: Conduction band is almost empty i.e., no electron is available for conduction.

Effect of temperature

• In conductors: At high temperature, the collision of electrons become more frequent with the atoms/molecules at lattice site in the metals as a result the conductivity decreases (or resistivity increases).
• In semiconductors: As the temperature of the semiconducting material increases, more electron-hole pairs becomes available in the conduction band and valence band, and hence the conductivity increases or the resistivity decreases.
• In insulators: The energy band between conduction band and valence band is very large, so it is unsurpassable for small temperature rise. So, there is no change in their behaviour.

Question 2.
Distinguish between ‘intrinsic’ and ‘extrinsic’ semiconductors?
Answer:

Intrinsic semiconductor	Extrinsic semiconductor
1. It is a pure semiconductor material with no impurity atoms in it.	It is prepared by doping a small quantity of impurity atoms to the pure semiconductor.
2. The number of free electrons in the conduction band and the number of holes in valence band is exactly equal.	The number of free electrons and holes is never equal. There is an excess of electrons in n-type semicoñductors and excess of holes in p-tpe semiconductors.

Question 3.
The circuit shown in the figure has two oppositely connected ‘ ideal diodes connected in parallel. Find the current flowing through each diode in the circuit.

Answer:
(i) Diode D₁ is reverse biased, so it offers an infinite resistance. So, no current flows in the branch of diode D₁.
(ii) Diode D₂ is forward biased and offers no resistance in the circuit. So, current in the branch,
I = \(\frac{V}{R_{e q}}=\frac{12 \mathrm{~V}}{2 \Omega+4 \Omega}=\frac{12 \mathrm{~V}}{6 \Omega}\) = 2A.

Question 4.
A Zener of power rating 1W is to be used as a voltage regulator. If Zener has a breakdown of 5V and it has to regulate voltage which fluctuated between 3V and 7 V, what should be the value of Rs for safe operation (see figure)? (NCERT Exemplar)

Answer:
Here, P =1W, V_z = 5 V
V_s = 3V to 7 V
I_Zmax = \(\frac{P}{V_{Z}}=\frac{1}{5}=0.2 \mathrm{~A}\) = 200 mA
R_s = \(\frac{V_{s}-V_{z}}{I_{Z_{\max }}}=\frac{7-5}{0.2}=\frac{2}{0.2}\) = 10Ω

Question 5.
Draw V-I characteristics of a p-n junction diode. Answer the following questions, giving reasons:
(i) Why is the current under reverse bias almost independent of the applied potential upto a critical voltage?
(ii) Why does the reverse current show a sudden increase at the critical voltage?
Name any semiconductor device which operates under the reverse bias in the breakdown region.
Answer:

(i) In the reverse biasing, the current of order of μA is due to movement/drifting of minority charge carriers from one region to another through the junction. A small applied voltage is sufficient to sweep the minority charge carriers through the junction. So, reverse current is almost independent of critical voltage.

(ii) At critical voltage (or breakdown voltage), a large number of covalent bonds break, resulting in the increase of large number of charge carriers. Hence, current increases at critical voltage. Semiconductor device that is used in reverse biasing, is Zener diode.

Question 6.
How is a light-emitting diode fabricated? Briefly state its working. Write any two important advantages of LEDs over the conventional incandescent low power lamps.
Answer:
LED is fabricated by:
(i) heavy doping of both the p and n regions.
(ii) providing a transparent cover so that light can come out.

Working: When the diode is forward biased, electrons are sent from n →b p and holes from p → n. At the junction boundary, the excess minority carriers on either side of junction recombine with majority carriers. This releases energy in the form of photon hv = E_g
Advantages:

• Low operational voltage and less power consumption.
• Fast action and no warm-up time required.
• Long life and ruggedness.
• Fast on-off switching capability.

Question 7.
Explain, with the help of a circuit diagram, the working of a photodiode. Write briefly how it is used to detect the optical signals.
Or
(a) How is photodiode fabricated?
(b) Briefly explain its working. Draw its VI characteristics for two different intensities of illumination.
Or
With what considerations in view, a photodiode is fabricated?
State its working with the help of a suitable diagram.
Even though the current in the forward bias is known to be
more than in the reverse bias, yet the photodiode works in reverse bias. What is the reason?
Answer:
A photodiode is fabricated using photosensitive semiconducting material with a transparent window to allow light to fall on the junction of the diode.

Working: In diode (any type of diode), an electric field ‘E’ exists across the junction from n-side to p-side, when light with energy hv greater than energy gap E₂ (hv> Eg ) illuminate the junction, then electron-hole pairs are generated due to absorption of photons, in or near the depletion region of the diode. Due to existing electric fIeld, electrons are collected on n-side and holes are collected on p-side, giving rise to an emf. Due to the generated emf, an electric current of μA order flows through the external resistance.

Detection of Optical Signals: It is easier to observe the change in the current with change in the light intensity if a reverse bias is applied. Thus, photodiode can be used as a photodetector to detect optical signals. The characteristic curves of a photodiode for two different illuminations I₁ and I₂ (I₂ > I₁)are shown below

Question 8.
Describe briefly using the necessary circuit diagram, the three basic processes which take place to generate the emf in a solar cell when light falls on it. Draw the V-I characteristics of a solar cell. Write two important criteria required for the selection of a material for solar cell fabrication.
Answer:
The three basic processes which take place to generate the emf in a solar cell are generation, separation and collection.
(i) Generation of electron-hole pairs due to the light incident (with hv>Eg) close to the junction.
(ii) Separation of electrons and holes due to the electric field of the depletion region.
(iii) Collection of electrons and holes by n-side and p-side, respectively.

Important criteria for the selection of a material for solar cell fabrication are:
(i) bandgap (~ 1.0 to 1.8 eV).
(ii) high optical absorption (-10 4 cm^-1),
(iii) electrical conductivity.
(iv) availability of the raw material, and
(v) cost

Long answer type questions

Question 1.
(a) State briefly the process involved in the formation of p-n junction explaining clearly how the depletion region is formed.
(b) Using the necessary circuit diagrams, show how the V-I characteristics of a p-n junction are obtained in:
(i) Forward biasing
(ii) Reverse biasing
How are these characteristics made use of in rectification?
Or
Draw the circuit arrangement for studying the V-I characteristics of a p-n junction diode
(i) in forward bias and
(ii) in reverse bias. Draw the typical V-I characteristics of a silicon diode.
Describe briefly the following terms:
(i) “minority carrier injection” in forward bias
(ii) “breakdown voltage” in reverse bias.
Answer:
(a) Two processes occur during the formation of a p-n junction are diffusion and drift. Due to the concentration gradient across p and n-sides of the junction, holes diffuse from p-side to n-side (p → n) and electrons diffuse from n-side to p-side (n → p). This movement of charge carriers leaves behind ionised acceptors (negative charge Φ -immobile) on the p-side and donors (positive charge immobile) on the n-side of the junction. This space charge region on either side of the junction together is known as depletion region.

(b) The circuit arrangement for studying the V-I characteristics of a diode are shown in figure (a) and (b). For different values of voltages, the value of current is noted. A graph between V and I is obtained as in figure (c). From the V-I characteristic of a junction diode, it is clear that it allows current to pass only when it is forward biased. So, if an alternative voltage is applied across a diode, the current flows only in that part of the cycle when the diode is forward biased. This property is used to rectify alternating voltages.


(i) Minority Carrier Injection: Due to the applied voltage, electrons from n-side cross the depletion region and reach p-side (where they are minority carriers). Similarly, holes from p-side cross this junction and reach the n-side (where they are minority carriers). This process under forward bias is known as minority carrier injection.

(ii) Breakdown Voltage: It is critical reverse bias voltage at which current is independent of applied voltage.

Question 2.
State the principle of working of p-n diode as a rectifier. Explain with the help of a circuit diagram, the use of p-n diode as a full-wave rectifier. Draw a sketch of the input and output waveforms.
Or
Draw a circuit diagram of a full-wave rectifier. Explain the working principle. Draw the input/output waveforms indicating clearly, the functions of the two diodes used.
Or
With the help of a circuit diagram, explain the working of a junction diode as a full-wave rectifier. Draw its Input and output waveforms. Which characteristic property makes the junction diode suitable for rectification?
Answer:
Rectification: Rectification means conversion of AC into DC. A p-n diode acts as a rectifier because an AC changes polarity periodically and a p-n diode allows the current to pass only when it is forward biased. This makes the diode suitable for rectification.

Working: The AC input voltage across secondary s₁ and s₂ changes polarity after each half cycle. Suppose during the first half cycle of input AC signal, the terminal S₁ is positive relative to centre tap O and s₂ is negative relative to O. Then diode D₁ is forward biased and diode D₂ is reverse biased. Therefore, diode D₁ conducts while diode D₂ does not. The direction of current (i₁) due to diode D₁ in load resistance R_L is direction from A to B. In next half cycle, the terminal s₁ is negative and s₂ is positive relative to centre tap O.

The diode D₁ is reverse biased and diode D₂ is forward biased. Therefore, diode D₂ conducts while D₁ does not.
The direction of current (i₂) due to diode D₂ in load resistance R_L is still from A to B. Thus, the current in load resistance R_L is in the same direction for both half cycles of input AC voltage. Thus for input AC signal the output current is a continuous series of unidirectional pulses. In a full-wave rectifier, if input frequency is f hertz, then output frequency, will be 2f hertz because for each cycle of input, two positive half cycles of output are obtained.

Question 3.
Why is a Zener diode considered as a special purpose semiconductor diode? Draw the I-V characteristics of Zener diode and explain briefly, how reverse current suddenly Increase at the breakdown voltage? Describe briefly with the help of a circuit diagram, how a Zener diode works to obtain a constant DC voltage from the unregulated DC output of a rectifier.
Answer:
Zener diode works only in reverse breakdown region that is why it is, considered as a special purpose semiconductor. I – V characteristics of Zener diode is given below.

Reverse current is due to the flow of electrons from n → p and holes from p → n. As, the reverse-biased voltage increase the electric field across the junction, increases significantly and when reverse bias voltage V = V_Z, then the electric field strength is high enough to pull the electrons from p-side and accelerated it to n-side. These electrons are responsible for the high current at the breakdown.

Voltage regulator converts an unregulated DC output of rectifier into a constant regulated DC voltage, using Zener diode. The unregulated voltage is connected to the Zener diode through a series resistance Rs such that the Zener diode is reverse biased. If the input voltage increase, then-current through R_s and Zener diode increases.

Thus, the voltage drop across R_s increase without any change in the voltage drop across zener diode. This is because of the breakdown region, Zener voltage remains constant even though the current through Zener diode changes.
Similarly, if the input voltage decreases, the current through R_s and Zener diode decreases. The voltage drop across Rs decreases without any change in the voltage across the Zener diode. Now, any change in input voltage results the change in voltage drop across Rs, without any changes; in voltage across the Zener diode. Thus, Zener diode acts as a voltage regulator.

Question 4.
(a) Differentiate between three segments of a transistor on the basis of their size and level of doping.
(b) How is a transistor biased to be in active state?
(c) With the help of necessary circuit diagram, describe briefly how n-p-n transistor in CE configuration amplifies a small sinusoidal input voltage. Write the expression for AC current gain.
Or
Explain with the help of a circuit diagram the working of n-p-n transistor as a common emitter amplifier.
Or
Draw the circuit diagram of a common-emitter amplifier using an n-p-n transistor. What is the phase difference between the input signal and output voltage? Draw the input and output waveforms of the signal. Write the expression for its voltage gain. State two reasons why a common emitter amplifier is preferred to a common base amplifier.
Answer:
(a) Emitter: It is of moderate size and heavily doped.
Base: It is very thin and lightly doped.
Collector: The collector side is moderately doped and larger in size as compared to the emitter.

(b) Transistor is said to be inactive state when its emitter-base junction is suitably forward biased and base-collector junction is suitably reverse biased.
(c) The circuit of common emitter amplifier using n-p-n transistor is shown below :

Working: If a small sinusoidal voltage, with amplitude V_S is superposed on DC basic bias (by connecting the sinusoidal voltage in series with base supply V_BB), the base current will have sinusoidal variations superimposed on the base current I_B.

As a consequence the collector current is also sinusoidal variations superimposed on the value of collector current I_c, this will produce corresponding amplified changes in the value of output voltage V₀.
The AC variations across input and output terminals may be measured by blocking the DC voltage by large capacitors. The phase difference between input signal and output voltage is 180°. The input and output waveforms are shown in figure.

Voltage gain,
A_v = \(\beta \frac{R_{L}}{R_{i}}\)
AC current gain, β_AC = \(\left(\frac{\Delta I_{C}}{\Delta I_{B}}\right)_{V_{C E}}\)
Reasons for Using a Common Emitter Amplifier
(i) Voltage gain is quite high.
(ii) Voltage gain is uniform over a wider frequency range or power gain is high.

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 12 Biotechnology and its Applications Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 12 Biotechnology and its Applications

Very short answer type questions

Question 1.
State the role of transposons in silencing of mRNA in eukaryotic cells.
Answer:
Transposons or mobile genetic elements in viruses are the sources of the complementary dsRNA, which in turn bind to specific mRNA and cause RNA interference of the parasite.

Question 2.
How do interferons protect us?
Answer:
Interferons protect non-infected cells from further viral infections, by creating cytokine barriers.

Question 3.
State the role of C peptide in human insulin.
Answer:
C-peptide (extra stretch of polypeptide) makes the insulin inactive.

Question 4.
How are two short polypeptide chains of insulin linked [ together?
Answer:
Two short polypeptide chains of insulin are limced together by disulphide bridges.

Question 5.
How can bacterial DNA be released from the bacterial cell for biotechnology experiments?
Answer:
The bacterial cell wall is digested by the enzyme lysozyme to release 1 DNA from the cell.

Question 6.
Suggest any two possible treatments that can be given to a patient exhibiting adenosine deaminase deficiency.
Answer:

• Enzyme replacement therapy (in which functional ADA is injected)
• Bone marrow transplantation
• Gene therapy/Culturing the lymphocytes followed by introduction of functional ADA cDNA into it and returning it into the patient’s body. (Any two)

Question 7.
Name a molecular diagnostic technique to detect the presence
of a pathogen in its early stage of infection.
Answer:
ELISA (Enzyme Linked Immunosorbent Assay)

Question 8.
What are transgenic animals. Uive an example.
Answer:
The transgenic animals are those that have their DNA manipulated to possess and express/foreign genes e.g.. transgenic cow-Rosie, rats, pigs fish, rabbits and mice.

Question 9.
What was the speciality of the milk produced by the transgenic cow, Rosie?
Answer:
The first transgenic cow, Rosie, produced milk with human alpha-lactalbumin which was nutritionally, more balanced product for human babies than natural cow milk.

Question 10.
What is biopiracy?
Answer:
Biopiracy refers to the use of bioresources by multinational companies and other organisations without proper authorisation from the country and people without compensatory payment.

Question 11.
Name the following:
(a) The semi-dwarf variety of wheat which is high-yielding and disease-resistant.
(b) Any one inter-specific hybrid mammal.
Answer:
(a) Kalyan Sona/Sonalika
(b) Mule/Hinny/Liger/Tigon

Question 12.
For which variety of Indian rice, patent was filed by a USA Company? [NCERT Exemplar]
Answer:
Indian Basmati was crossed with semi-dwarf variety and was claimed as a new variety for which the patent was filed by a USA company.

Question 13.
Name the enzymes that are used for the isolation of DNA from bacterial and fungal cells for recombinant DNA technology.
Answer:
Bacteria: lysozyme; fungi : chitinase.

Question 14.
Bt cotton is resistant to pest, such PS lepidopteran, dipterans and coleopter&ns. Is Bt cotton resistant to other pests as well? [NCERT Exemplar]
Answer:
Bt cotton is made resistant to only certain specific taxa of pests. It is quite likely that in future, some other pests may infest the Bt cotton plants. It is similar to immunisation against small-pox which does not provide immunity against other pathogens like those that cause cholera, typhoid, etc.

Short answer type questions

Question 1.
Explain the application of biotechnology in producing Bt-cotton.
Or One of the major contributions of biotechnology is to develop pest-resistant varieties of cotton plants. Explain how it has been made possible.
Or One of the main objectives of biotechnology is to minimise the use of insecticides on cultivated crops. Explain with the help of a suitable example how insect resistant crops have been developed using techniques of biotechnology.
Answer:
Production of Bt-cotton, a Pest Resistant Crop : Soil bacterium Bacillus thuringiensis possess gene called Cry-gene which synthesises an endotoxin protein called Cry-protein. Now, by biotechnology technique, the Cry gene from B. thuringiensis have been isolated, cloned, introduced and incorporated into cotton-plant using recombinant DNA technology. In the genetically modified cotton crop plants, the Cry or Bt-toxin gene expresses to produce a toxic insecticidal protein in an inactive form called prototoxin in crystalline state.

As an insect feeds over the plant, the inactive prototoxin crystals pass into the gut where alkaline pH and digestive enzymes solubilise the crystals and convert the prototoxin into an active toxin. The activated toxin creates pores in the midgut epithelial cells by lysing that cause death of the insect. Thus, the GM cotton plants do not require protection of expensive insecticides as they themselves act as bioinsecticides.

Since there are a number of Cry genes and Cry IAc and Cry II Ab controls the cotton bollworms while Cry IAb controls corn borer.

Question 2.
Write the functions of
(a) cry lAc gene
(b) RNA interference (RNAi)
Or Explain the process of RNA interference.
Answer:
(a) Cry IAc gene is present in Bacillus thuringiensisThe gene encodes for a toxic insecticidal protein during a particular phase of their growth. Cry genes are isolated and incorporated in the crop plant. The insect feeding on transgenic crop die because of the presence of toxin protein. Cry IAc produces Bt-toxins specific for cotton bollworm insect group.

(b) RNAi involves the silencing of a specific m-RNA due to the complementary ds-RNA molecule that binds to and prevents translation of m-RNA. As a result, parasite a favourable protein are not produced
and it could not infest, multiply and survive in a transgenic host expressing specific RNA interference. The transgenic plant, therefore, gets itself protected from the parasite such as nematode.

Question 3.
How did Eli Lilly synthesise the human insulin? Mention one difference between this insulin and the one produced by the human pancreas.
Or How did Eli Lilly Company go about preparing the human insulin? How is the insulin thus produced different from that produced by the functional human insulin gene?
Or How did an American Company, Eli Lilly use the knowledge of r-DNA technology to produce human insulin?
Or Explain how Eli Lilly, an American Company, Produced insulin by Recombinant DNA technology?
Answer:
Eli Lilly prepared two DNA sequences corresponding to A and B chains of human insulin and introduced them in plasmids of E. coli to produce insulin chains. Chains A and B were produced separately, extracted and combined by creating disulphide bonds to form human insulin.

Insulin in human pancreas is synthesised as a pro-hormone containing the C peptide, which is removed to form mature hormone. The synthesised insulin did not contain C peptide and was directly prepared
in mature form.

Question 4.
How is a mature functional insulin hormone different from its pro-hormone form? [NCERT Exemplar]
Answer:
Mature functional insulin is obtained by processing of pro-hormone which contains extra peptide called C-peptide. This C-peptide is removed during maturation of pro-insulin to insulin.

Question 5.
How does a transgenic organism differ from the rest of its population? Give any two examples of such organism for human advantage.
Answer:
A transgenic organism contains foreign gene, hence it differs from the rest of the population in having one or more extra genes apart from the gene pool of that population showing an additional phenotype. Example, (i) Transgenic E. coli, with gene for human insulin, (ii) Transgenic mouse with gene for human growth hormone.

Question 6.
What is GEAC and what are its objectives? [NCERT Exemplar]
Or Mention two objectives of setting up GEAC by our Government.
Or State the purpose for which the Indian Government has set up GEAC.
Or Describe the responsibility of GEAC, set up by the Indian Government.
Answer:
GEAG (Genetic Engineering Approval Committee) is an Indian government organisation. Its objective are to:
(a) examine the validity of GM (Genetic modification of organism) research.
(b) inspect the safety of introducing GM for public services.

Long answer type questions

Question 1.
List the disadvantages of insulin obtained from the pancreas of slaughtered cow and pigs. [NCERT Exemplar]
Answer:

• Insulin being a hormone is produced in very little amounts in the body. Hence, a large number of animals need to be sacrificed for obtaining small quantities of insulin. This makes the cost of insulin very high, demand being manyfold higher than supply.
• Slaughtering of animal is also not ethical.
• There is potential of immune response in humans against the administered insulin which is derived from animals.
• There is possibility of slaughtered animals being infested with some infectious micro-organism which may contaminate insulin.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 1 Electric Charges and Fields Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 1 Electric Charges and Fields

Very short answer type questions

Question 1.
Define electrostatics.
Answer:
Electrostatics deals with the study of forces, fields and potentials arising from static charges.

Question 2.
Define charge.
Answer:
It is defined as the basic and characteristic property of elementary particles of matter in the form of which certain force of interaction energies may be explained.

Question 3.
What are the behaviour of charges?
Answer:
Like charges repel and unlike charges attract each other.

Question 4.
Define the polarity of charge.
Answer:
The property which differentiates the two kinds of charges is called the polarity of charge.

Question 5.
Who first assigned the positive and negative signs to charge?
Answer:
Benjamin Franklin

Question 6.
What are conductors?
Answer:
The substances which allow electricity to pass through them easily are called conductors e.g., metals, human, earth etc.

Question 7.
What are insulators?
Answer:
Most of the non-metals like glass, porcelain, plastic, nylon, wood offer high resistance to the passage of electricity through them, they are called insulators.

Question 8.
Define the semiconductors.
Answer:
The substances which offer resistance to the movement of charges are called semiconductors. They are intermediate between conductors and insulators,

Question 9.
What are point charges?
Answer:
If the sizes of charged bodies are very small as compared to the distances between them, they are called point charges.

Question 10.
Write the law of conservation of charges.
Answer:
“The total charge of the isolated system is always conserved.” It is not possible to create or destroy net charge carried by isolated system although the charge carrying particles may be created or destroyed in a process.

Question 11.
The dimensions of an atom are of the order of an Angstrom. Thus there must be large electric fields between the protons and electrons. Why then is the electrostatic field inside a conductor zero? (NCERT Exemplar)
Answer:
The electric fields bind the atoms to neutral entity. Fields are caused by excess charges. There can be no excess charge on the inner surface of an isolated conductor. So, the electrostatic field inside a conductor is zero.

Question 12.
An arbitrary surface encloses a dipole. What is the electric flux through this surface? (NCERT Exemplar)
Answer:
Net charge on a dipole = -q + q = 0. According to Gauss’s theorem, electric flux through the surface,

Question 13.
Sketch the electric field lines for a uniformly charged hollow cylinder shown in figure. (NCERT Exemplar)

Answer:

Short answer type questions

Question 1.
Describe the gold-leaf electroscope.
Question
A simple apparatus to detect charge on a body is the gold-leaf electroscope. It consists of a vertical metal rod housed in a box, with two thin gold leaves attached to its bottom end. When a charged object touches the metal knob at the top of the rod, charge flows on to the leaves and they diverge. The degree of divergence is an indicator of the amount of charge.

Question 2.
Define the grounding or earthing.
Answer:
When we bring a charged body in contact with the earth, all the excess charge on the body disappears by causing a momentary current to pass to the ground through the connecting conductor (such as our body). This process of sharing the charges with the earth is called grounding or earthing.

Question 3.
What is the importance of earthing in buildings?
Answer:
Earthing provides a safety measure for electrical circuits and appliances. A thick metal plate is buried deep into the earth and thick wires are drawn from this plate; these are used in buildings for the purpose of earthing near the mains supply. The electric wiring in our houses has three wires; live, neutral and earth. The first two carry electric current from the power station and the third is earthed by connecting it to the buried metal plate. Metallic bodies of the electric appliances such as electric iron, refrigerator, TV are connected to the earth wire. When any fault occurs or live wire touches the metallic body the charge flows to the earth without damaging the appliance and without causing any injury to the humans; this would have otherwise been unavoidable since the human body is a conductor of electricity.

Question 4.
Describe the additivity of charges in brief.
Answer:
If a system contains two point charges q₁ and q₂, the total charge of the system is obtained simply by adding algebraically qi and q2, i. e., charges add up like real numbers or they are scalars like the mass of a body. If a system contains n charges q₁, q₂, q₃,……, q_n, then the total charge of the system is q₁ + q₂ + q₃ + … + q_n. Charge has magnitude but no direction, similar to the mass. However, there is one difference between mass and charge. Mass of a body is always positive whereas a charge can be either positive or negative. Proper signs have to be used while adding the charges in a system. For example, the total charge of a system containing five charges +1, +2, -3, +4 and -5, in some arbitrary unit, is (+1) + (+2) + (-3) + (+4) + (-5) = -1 in the same unit.

Question 5.
(a) Define electric flux. Write its SI unit.
(b) A small metal sphere carrying charge +Q is located at the centre of a spherical cavity inside a large uncharged metallic spherical shell as shown in the figure. Use Gauss’s law to find the expressions for the electric field at points P₁ and P₂.

(c) Draw the pattern of electric field lines in this arrangement.
Answer:
(a) Electric flux over an area in an electric field represents the total number of electric field lines crossing this area and is given by the product of surface area and the component of electric field intensity normal to the area.
The SI unit of flux is Nm²C^-1.

(b) Let point P₁ is at distance R from the centre 0.
S₁ is the Gaussian surface, then according to Gauss’s theorem

Inside the shell the charge is zero, so the field is also zero

(c) The direction of electric field is shown in figure.

Question 6.
A metallic spherical shell has an inner radius R₁ and outer radius R₂– A charge Q is placed at the centre of the spherical cavity. What will be surface charge density on (i) the inner surface, and (ii) the outer surface? (NCERT Exemplar)
Answer:
When a charge +Q is placed at the centre of spherical cavity,

The charge induced on the inner surface = -Q
The charge induced on the outer surface = +Q
∴ Surface charge density on the inner surface = \(\frac{-Q}{4 \pi R_{1}^{2}}\)
Surface charge density on the outer surface = \(\frac{+Q}{4 \pi R_{2}^{2}}\)

Question 7.
Two charges q and -3q are placed fixed on x-axis separated by distance id\ Where should a third charge 2q be placed such that it will not experience any force? (NCERT Exemplar)
Answer:

Let the charge 2q be placed at point P as shown. The force due to q is to the left and that due to -3q is to the right.

Long answer type questions

Question 1.
How can you charge a metal sphere positively without touching it?
Answer:
Figure (a) shows an uncharged metallic sphere on an insulating metal stand. Bring a negatively charged rod close to the metallic sphere, as shown in Fig. (b). As the rod is brought close to the sphere, the free electrons in the sphere move away due to repulsion and start piling up at the farther end. The near end becomes positively charged due to deficit of electrons. This process of charge distribution stops when the net force on the free electrons inside the metal is zero. Connect the sphere to the ground by a conducting wire. The electrons will flow to the ground while the positiye charges at the near end will remain held there due to the attractive force of the negative charges on the rod, as shown in Fig. (c). Disconnect the sphere from the ground. The positive charge continues to be held at the near end [Fig.(d)]. Remove the electrified rod. The positive charge will spread uniformly over the sphere as shown in Fig. (e).

In this experiment, the metal sphere gets charged by the process of induction and the rod does not lose any of its charge.

Similar steps are involved in charging a metal sphere negatively by induction, by bringing a positively charged rod near it. In this case the electrons will flow from the ground to the sphere when the sphere is connected to the ground with a wire.

Question 2.
Define electric dipole moment. Is it a scalar or a vector? Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole.
Answer:
Electric Dipole Moment : The strength of an electric dipole is measured by the quantity electric dipole moment. Its magnitude is equal to the product of the magnitude of either charge and the distance between the two charges.
Electric dipole moment,
p = q × d
It is a vector quantity.
In vector form it is written as \(\), where the direction of \(\) is from negative charge to positive charge.
Electric field of dipole at points on the equatorial plane:

The magnitudes of the electric field due to the two charges + q and – q are given by,