Important Questions

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 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 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 2 Electrostatic Potential and Capacitance Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance

Very short answer type questions

Question 1.
A charge ‘q’ is moved from a point A above a dipole of dipole moment ‘p’ to a point B below the dipole in equatorial plane without acceleration. Find the work done in the process.

Answer:
Work done in the process is zero. Because, equatorial plane of a dipole is equipotential surface and work done in moving charge oh equipotential surface is zero.
W = qV_AB = q × 0 = 0

Question 2.
A point charge Q is placed at point O as shown in the figure. Is the potential difference (V_A – V_B) positive, negative or zero if Q is
(i) positive
(ii) negative

Answer:
Let the distance of points A and B from charge Q be r_A and r_B, respectively.
∴ Potential difference between points A and B
V_A – V_B = \(\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{r_{A}}-\frac{1}{r_{B}}\right]\)
As r_A = OA, r_B = r_B = OB
and r_A < r_B, \(\frac{1}{r_{A}}>\frac{1}{r_{B}}\)
There,[latex]\frac{1}{r_{A}}-\frac{1}{r_{B}}[/latex] has positive value.
(V_A – V_B) depends on the nature of charge q.
(V_A – V_B) is positive when Q > 0
(V_A – V_B ) is negative when Q < 0

Question 3.
A hollow metal sphere of radius 5 cm is charged such that potential on its surface is 10 V. What is the potential at the centre of the sphere?
Answer:
The electric potential at every point inside the charged spherical shell is same and equal to the electric potential on its surface.
The electric potential at the centre of sphere is 10 V.

Question 4.
Can two equipotential surfaces intersect each other? Justify your answer.
No, two equipotential surfaces cannot intersect each other because

1. Two normals can be drawn at intersecting point on two surfaces which gives two directions of E at the same point which is impossible.
2. Also two values of potential at the same point is not possible.

Question 5.
Why electrostatic potential is constant throughout the volume of the conductor and has the same value as on its surface?
Since, electric field intensity inside the conductor is zero. So, electrostatic potential is a constant.
But, E = –\(\frac{\Delta V}{\Delta r}\)
∴ E = 0, ΔV = 0
or V₂ – V₁ =0, V₂ – V₁
The potential at every point inside the conductor remains same.

Question 6.
In a certain 0.5 cm³ of space, electric potential is found to be 7 V throughout. What is the electric field in this region?
Answer:
Zero, because electric potential is same throughout as
E = \(\frac{d V}{d r}\)

Question 7.
Distinguish between a dielectric and a conductor.
Answer:
Dielectrics are non-conductors and do not have free electrons at all. While conductor has free electrons in its any volume which makes it able to pass the electricity through it.

Question 8.
The given graph shows the variation of charge q versus potential difference V for two capacitors C₁ and C₂. Both the capacitors have same plate separation but plate area of C₂ is greater than that C₁. Which line (A or B) corresponds to C₁ and why?

Answer:
Line B corresponds to Q because slope (q/V) of B is less than slope of A.

Question 9.
Do free electrons travel to region of higher potential or lower potential? (NCERT Exemplar)
Answer:
Free electrons would travel to regions of higher potentials as they are negatively charged.

Question 10.
Can there be a potential difference between two adjacent conductors carrying the same charge? (NCERT Exemplar)
Answer:
Yes, if the sizes are different.

Question 11.
Prove that, if an insulated, uncharged, conductor is placed near a charged conductor and no other conductors are present, the uncharged body must intermediate in potential between that of the charged body and that of infinity. (NCERT Exemplar)
Answer:
Let us take any path from the charged conductor to the uncharged conductor along the direction of electric field. Therefore, the electric potential decrease along this path.
Now, another path from the uncharged conductor to infinity will again continually lower the potential further. This ensures that the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

Question 12.
A test charge q is made to move in the electric field of a point charge Q along two different closed paths [figure]. First path has sections along and perpendicular to lines of electric field.
Second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases? (NCERT Exemplar)

Answer:
As electric field is conservative, work done will be zero in both the cases.

Short answer type questions

Question 1.
What is electrostatic shielding? How is this property used in actual practice? Is the potential in the cavity of a charged conductor zero?
Answer:
Whatever be the charge and field configuration outside, any cavity in a conductor remains shielded from outside electric influence. The field inside a conductor is zero. This is known as electrostatic shielding.

• Sensitive instruments are shielded from outside electrical influences by enclosing them in a hollow conductor.
• During lightning it is safest to sit inside a car, rather than near a tree. The metallic body of a car becomes an electrostatic shielding from lightening.

Potential inside the cavity is not zero. Potential is constant.

Question 2.
Two capacitors of unknown capacitances C₁ and C₂ are connected first in series and then in parallel across a battery of 100 V. If the energy stored in the two combinations is 0.045 J and 0.25 J respectively, determine the value of C₁ and C₂. Also calculate the charge on each capacitor in parallel combination.
Answer:
Energy stored in a capacitor,

Q₁ = C₁V = 38.2 × 10^-6 × 100 = 38.2 × 10^-4 C
Q₂ = C₂V = 11.8 × 10^-6 × 100 = 11.8 × 10^-4C

Question 3.
Calculate the equivalent capacitance between points A and B in the circuit below. If a battery of 10 V is connected across A and B, calculate the charge drawn from the battery by the circuit.

Answer:

Question 4.
A parallel plate capacitor is charged by a battery to a potential. The battery is disconnected and a dielectric slab is inserted to completely fill the space between the plates.
How will
(a) its capacitance
(b) electric field between the plates and
(c) energy stored in the capacitor be affected? Justify your answer giving necessary mathematical expression for each case.
Answer:
On introduction of dielectric slab in a isolated charged capacitor.
(a) The capacitance (C’) becomes K times of original capacitor as
C = \(\frac{\varepsilon_{0} A}{d}\)
and C ‘ = \(\frac{K \varepsilon_{0} A}{d}\)

(b) The total charge on the capacitor remains conserved on introduction of dielectric slab. Also, the capacitance of capacitor increases to K times of original values.
∴ CV = C’V’
CV = (KC)V’
⇒ V’ = \(\frac{V}{K}\)
∴ New electrical field,
E’ = \(\frac{V^{\prime}}{d}=\left(\frac{V / K}{d}\right)=\left(\frac{V}{d}\right) \frac{1}{K}=\frac{E}{K}\)
∴ On introduction of dielectric medium new electric field E’ becomes \(\) times of its original value.
(c) Energy stored initially,
U = \(\frac{q^{2}}{2 C}\)
Energy stored later,
U’ = \(\frac{q^{2}}{2(K C)}\) [< C’ = KG]
where, K = dielectric constant of medium
⇒ U’ = \(\frac{1}{K}\left(\frac{q^{2}}{2 C}\right)\)
⇒ U’ = \(\frac{1}{K}\) × U
The energy stored in the capacitor decreases and becomes \(\frac{1}{K}\) times of original energy.

Question 5.
Two identical parallel plate capacitors A and B are connected to a battery of V volts with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant K. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

Answer:
Let C be the capacitance of each capacitor.
With switch S closed, the two capacitor are in parallel.
∴ Equivalent capacitance is 2 C.
∴ Energy stored = \(\frac{1}{2}\)(2C)V²
U₁ = CV² …………… (1)

Now, when switch is opened and then free space of capacitors are filled with dielectric, the capacitance of each capacitor will be KC. For capacitor B, the charge will remain as before and for A, the potential difference will remain same.

Charge on each capacitor in the previous case will be CV.
∴ Energy stored in capacitor A in circuit case is

Question 6.
Consider two conducting spheres of radii R₁ and R₂ with R₁ > R₂. If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one. (NCERT Exemplar)
Answer:

Question 7.
Calculate potential on the axis of a ring due to charge Q uniformly distributed along the ring of radius R.
(NCERT Exemplar)
Answer:
Let us take point P to be at a distance x from the centre of the ring, as shown in figure. The charge element dq is at a distance x from point P. Therefore, V can be written as

Long answer type questions

Question 1.
(i) Explain using suitable diagrams, the difference in the behaviour of a
(a) conductor and
(b) dielectric in the presence of external electric field. Define the terms polarisation of a dielectric and write its relation with susceptibility.

(ii) A thin metallic spherical shell of radius R carries a charge Q on its surface. A point charge \(\frac{Q}{2}\) is placed at its centre C and an other charge +2Q is placed outside the shell at a distance x from the centre as shown in the figure. Find (a) the force on the charge at the centre of shell and at the point A, (b) the electric flux through the shell.

Answer:
(i) When a capacitor is placed in an external electric field, the free charges present inside the conductor redistribute themselves in such a manner that the electric field due to induced charges opposes the external field within the conductor. This happens until a static- situation is achieved, i.e., when the two fields cancels each other and the net electrostatic field in the conductor becomes zero.

In contrast to conductors, dielectrics are non- conducting substances, i.e., they have no charge carriers. Thus, in a dielectric, free movement of charges is not possible. It turns out that the external field induces dipole moment by stretching molecules of the dielectric. The collective effect of all the molecular dipole moments is the net charge on the surface of the dielectric which produces a field that opposes the external field. However, the opposing field is so induced does not exactly cancel the external field, ft only reduces it. The extent of the effect depends on the nature of dielectric.

Both polar and non-polar dielectric develop net dipole moment in the presence of an external field. The dipole moment per unit volume is called polarisation and is denoted by P for linear isotropic dielectrics.
P = %E

(a) At point C, inside the shell.
The electric field inside a spherical shell is zero. Thus, the force experienced by charge at the centre C will also be zero.

Question 2.
(a) Find the ratio of the potential differences that must he applied across the parallel and series combination of two capacitors C ₁and C₂ with their capacitances in the ratio 1:2 so that the energy stored in the two cases becomes the same.

(b) Show that the potential energy of a dipole making angle θ with the direction of the field is given by u(θ) = – p \(\). Hence find out the amount of work done in rotating it from the position of unstable equilibrium to the stable equilibrium.

(b) As charges +q and -q traverse equal distance under equal and opposite forces; therefore, net work done in bringing the dipole in the region of electric field perpendicular to field-direction will be zero, i.e.,W₁ = 0.

Now, the dipole is rotated and brought to orientation making an angle θ with the field direction (i.e., θ₀ = 90° and θ₁ = θ°), therefore, work done
W₂ = p_E (cosθ₀ – cosθ ₁)
= p_E (cos 90° – cos θ) = -pE cos θ

∴Total work done in bringing the electric dipole from infinity, i.e., electric potential energy of electric dipole. Thus, work done by external torque in rotating a dipole in uniform electric field is stored as the potential energy of the system.

U = W₁ + W₂ = 0 – pEcosθ
= -pE cosθ
In vector form
U = – \(\vec{p} \cdot \vec{E}\)
For rotating dipole from position of unstable equilibrium (θ₀ =180°) to the stable equilibrium (θ = 0°)
∴W = pE (cos 180° – cos0°)
= pE (-1 -1) = -2 pE

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 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,

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 15 Communication Systems Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 15 Communication Systems

Very short answer type questions

Question 1.
Name the essential components of a communication system.
Answer:
Transmitter, medium or channel, and receiver.

Question 2.
What is the function of a transducer used in a communication system?
Answer:
Transducer used as a sensor or detector in communication system. It converts the physical signal into electrical signal.

Question 3.
What is the function of a repeater in a communication system?
Answer:
Repeater pick up the signals from the transmitter, amplifies it and transmits it to the receiver. Thus, repeater comprises up of receiver, transmitted and amplifier. Its function is to extend the range of communication.

Question 4.
Define bandwidth and describe briefly its importance in communicating signals.
Answer:
It is defied as the frequency range over which given equipment operates.
Importance: To design the equipment used in communication system for distinguishing different message signals.

Question 5.
Which basic mode of communication is used for telephonic communication?
Answer:
Point to point is a basic mode of communication, which is used for telephonic conversation. In this mode of communication, communication takes place over a link between a single transmitter and a receiver.

Question 6.
How are microwaves produced?
Answer:
A type of electromagnetic wave is microwave whose wavelength ranging from as long as metre to as short as millimeter and having the frequency range 3000 MHz to 300 GHz. This also includes UHF, EHF and various sources with different boundaries.

Question 7.
What is sky wave propagation?
Answer:
Skywave propagation is a mode of propagation in which communication of radiowaves in frequency range 2 MHz-20 MHz takes place due to reflection from the ionosphere.

Question 8.
Would sky waves be suitable for transmission of TV signals of 60 MHz frequency? (NCERT Exemplar)
Answer:
A signal to be transmitted through sky waves must have a frequency range of 1710 kHz to 40 MHz. But, here the frequency of TV signals are 60 MHz which is beyond the required range. So, sky waves will not be suitable for transmission of TV signals of 60 MHz frequency.

Question 9.
How are sidebands produced?
Answer:
Sidebands are produced due to the superposition of carrier wave of frequency ω_c over modulating or audio signal of frequency ω_m. The frequency of lower sideband is ω_c -ω_m and the upper side band is ω_c+ω_m

Question 10.
Why are broadcast frequencies (carrier waves) sufficiently spaced in amplitude modulated wave?
Answer:
To avoid mixing up of signals from different transmitters. This can be done by modulating the signals on high-frequency carrier waves, e.g., frequency band for satellite communication is 5.925 – 6.425 GHz.

Question 11.
Why is the amplitude of modulating signal kept less than the amplitude of carrier wave?
Answer:
The amplitude of modulating signal is kept less than the amplitude of carrier wave to avoid distortion.

Question 12.
What is the function of a bandpass filter used in a modulator for obtaining AM signal?
Answer:
Bandpass filter rejects DC and sinusoid of frequency ω_m, 2ω_m, and 2ω_c and retains frequencies ω_c + ω_m.
Thus, it allows only the desired frequencies to pass through it.

Question 13.
On radiating (sending out) and AM modulated signal, the total radiated power is due to energy carried by ω_c, ω_c -ω_m and ω_c +ω_m. Suggest ways to minimise cost of radiation without compromising on information. (NCERT Exemplar)
Answer:
In amplitude modulated signals, only sideband frequencies contain information.
Thus only (ω_c +ω_m) and (ω_c – ω_m) contain information.

Now, according to question, the total radiated power is due to energy carried by
ω_c, (ω_c – ω_m) and (ω_c +ω_m)
Thus to minimize the cost of radiation without compromising on information, ω_c can be left and transmitting
(ω_c +ω_m), (ω_c – ω_m) or both (ω_c +ω_m) and (ω_c – ω_m).

Question 14.
Two waves A and B of frequencies %MHz and 3 MHz, respectively are beamed in the sartie direction for communication via skywave. Which one of these is likely to travel longer distance in the ionosphere before suffering total internal reflection? (NCERT Exemplar)
Answer:
As the frequency of wave B is more than wave A, it means the refractive index of wave B is more than refractive index of wave A (as refractive index increases with frequency increases). For higher frequency wave (i.e., higher refractive index) the angle of refraction is less i.e., bending is less. So, wave B travels longer distance in the ionosphere before suffering total internal reflection.

Short answer type questions

Question 1.
Draw a block diagram of a generalized communication system. Write the functions of each of the following:
(a) Transmitter
(b) Channel
(c) Receiver
Answer:
The block diagram of a generalized communication system is shown in figure

Functions are as follows :
(a) Transmitter: It comprises of message signal source, modulator and transmitting antenna. Transmitter makes signals compatible for communication channel via modulator and antenna.
(b) Channel: It is a link for propagating the signal from transmitter to receiver.
(c) Receiver: It recovers the desired original message signals from the received signals at the end of channel.

Question 2.
Explain the terms
(i) Attenuation and
(ii) Demodulation used in communication system.
Answer:
Attenuation: The loss in strength of a signal while propagating through a medium is known as attenuation.
Demodulation: The process of retrieval of information, from the carrier wave at the receiver end. This is the reverse process of modulation.

Question 3.
(a) Distinguish between ‘Analog and Digital signals’.
(b) Explain briefly two commonly used applications of the Internet.
Answer:
(a) A signal that varies continuously with time (e. g., sine waveform) is called an analog signal.
A signal that is discrete is called a digital signal. The presence of signal is denoted by digit 1 and absence is denoted by digit 0.

(b) Uses of Internet: email, e-banking, e-shopping, e-ticketing, charting, surfing, file transfer, etc.

Question 4.
What is ground wave communication? Explain why this mode cannot be used for long-distance communication using high frequencies.
Answer:
The mode of wave propagation in which wave guided along the surface of the earth is called ground wave communication.
The maximum range of propagation in this mode depends on
(i) transmitted power and
(ii) frequency (less than a few MHz)
At high frequencies, the rate of energy dissipation of the signal increases and the signal gets attenuated over a short distance.

Question 5.
Define the term modulation. Draw a block diagram of a simple modulator for obtaining AM signals.
Or
Draw a block diagram of a simple modulator to explain how the AM wave is produced. Can the modulated signal be transmitted as such? Explain.
Answer:
Modulation is the process in which low-frequency message signal is superimposed on high-frequency carrier wave so that they can be transmitted over long distances. The block diagram for a simple modulator for obtaining AM signal is shown as below :

Question 6.
Define modulation index. Why is it kept low? What is the role of a bandpass filter? Give its physical significance.
Answer:
Modulation index is the ratio of the amplitude of modulating signal to that of carrier wave. Mathematically,
µ = \(\frac{A_{m}}{A_{c}}\)
Reason: It is kept low to avoid distortion.
Role: A bandpass filter rejects low and high frequencies and allows a band of desired frequencies to pass through it.
Physical Significance: It signifies the level of distortion or noise. A lower value of modulation index indicates a lower distortion in the transmitted signal.

Question 7.
State the concept of mobile telephony and explain its working.
Answer:
Concept of mobile telephony is to divide the service area into a suitable number of cells centered on an office MTSO (Mobile Telephone Switching Office). Mobile telephony means that you can talk to any person from anywhere.
Explanation:

• Entire service area is divided into smaller parts called cells or cell zones.
• Each cell has a base station to receive and send signals to all the mobile phones present inside that cell.
• Each base station is linked to MTSO. MTSO coordinates between base station and TCO (Telephone Control Office).

Question 8.
What is space wave propagation? State the factors which limit its range of propagation. Derive an expression for the maximum line of sight distance between two antennas for space wave propagation.
Answer:
Space Wave Propagation
The mode of propagation in which radio waves travel, along a straight line, from the transmitting to the receiving antenna.
Limiting Factors
(i) Curvature of the earth
(ii) Insufficient height of the receiving antenna
(iii) LOS distance (> 40 MHz) travel in straight line

Derivation : In right angled triangle BOD, ∠BDO =90°
∴BO² = (OD)² +(BD)²
le., (R_e+h)² =R_e² +(BD)² ………………. (1)
As height h of the tower is very small as compared to radius (R_e) of earth the point B will be very close to A, so that
BD ≈AD = d(say)
∴ Equation (1) given (R_e + h)² = R_e² + d²

or height of transmitting antenna, h =\(\frac{d^{2}}{2 R_{e}}\)
or covering range of TV transmitting tower, d = \(\sqrt{2 R_{e} h}\)
Thus, covering range of TV signal can be increased by increasing the height of transmission antenna.
For a transmitting antenna of height h_T, and a receiving antenna of height h_R, the maximum line of sight distance
becomes
d_M= \(\sqrt{2 R h_{T}}+\sqrt{2 R h_{R}}\)

Question 9.
(a) Explain any two factors which justify the need of modulating a low-frequency signal.
(b) Write two advantages of frequency modulation over amplitude modulation
Answer:
(a) (i) If λ is the wavelength of the signal then the antenna should have a length at least \(\frac{\lambda}{4}\)
For an electromagnetic wave of frequency 20 kHz, the wavelength λ is 15 km. Such a long antenna is not possible to construct and operate. So, there is need to modulate the wave in order to reduce the height of antenna to a reasonable height.

(ii) The power radiated by a linear antenna (length l) is proportional to \(\left(\frac{l}{\lambda}\right)^{2}\).
This shows that power radiated increases with decreasing λ. So, for effective power radiation by antenna, there is need to modulate the wave.
(b)

• High frequency
• Less noise
• Maximum use of transmitted power

Question 10.
Given reasons for the following :
(i) For ground wave transmission, size of antenna (l) should the comparable to wavelength (λ) of signal, i.e. I = \(\frac{\lambda}{4}\)
(ii) Audio signals converted into an electromagnetic wave are not directly transmitted.
(iii) The amplitude of a modulating signal is kept less than the amplitude of carrier wave.
Answer:
(i) To radiate the signals with high efficiency.
(ii) Because they are of large wavelength and power radiated by antenna is very small as
P ∝ \(1 / \lambda^{4}\)
(iii) It is so to avoid making over modulated carrier wave. In that situation, the negative half cycle of the modulating signal is dipped and distortion occurs in reception.

Question 11.
Which of the following would produce analog signals and which would produce digital signals?(NCERT Exemplar)
(a) A vibrating tuning fork
(b) Musical sound due to a vibrating sitar string
(c) Light pulse
(d) Output of NAND gate
Answer:
Analog and digital signals are used to transmit information, usually through electric signals. In both these technologies, the information such as any audio or video is transformed into electric signals. The difference between analog and digital technologies is that in analog technology, information is translated into electric pulses of varying amplitude. In digital technology, translation of information is into binary formal (zero or one) where each bit is representative of two distinct 4 amplitudes. Thus, (a) and (b) would produce analog signals and (c) and (d) would produce digital signals.

Question 12.
Why is AM signal likely to be more noisy than a FM signal upon transmission through a channel? (NCERT Exemplar)
Answer:
In case of AM, the instantaneous voltage of carrier waves is varied by the modulating wave voltage. So, during the transmission, noise signals can also be added and receiver assumes noise a part of the modulating signal.
In case of FM, the frequency of carrier waves is changed as the change in the instantaneous voltage of modulating waves. This can be done by mixing and not while the signal is transmitting in channel. So, noise does not affect FM signal.

Long answer type questions

Question 1.
What does the term LOS communication mean? Name the types of waves that are used for this communication. What is the range of their frequencies? Give typical, examples, with the help of suitable figure of communication systems that use space wave mode propagation.
Answer:
LOS Communication: It means “Line of sight communication”. Space wave are used for LOS communication.
In this communication, the space waves (radio or microwaves) travel directly from transmitting antenna to receiving antenna. Frequency for LOS communication must be more than 40 MHz.

If transmitting antenna and receiving antenna have heights h_T and h_R respectively, then radio horizon of transmitting antenna.
d_T = \( \sqrt{2 R_{e} h_{T}}\)
where R_e is radius of earth and radius horizon of receiving antenna,
d_R = \(\sqrt{2 R_{e} h_{R}}\)
∴ Maximum line of sight distance,
d_M = d_T + d_R = \(\sqrt{2 R_{e} h_{T}}+\sqrt{2 R_{e} h_{R}}\)

Television, broadcast, microwave links and satellite communication.
The satellite communication is shown in figure. The space wave used is microwave.

Question 2.
(a) Distinguish between sinusoidal and pulse-shaped signals,
(b) Explain, showing graphically, how a sinusoidal carrier wave is superimposed on a modulating signal to obtain the resultant amplitude modulated (AM) wave.
Answer:
(a) In the process of modulation, some specific characteristics of the carrier wave is varied in accordance with the information or message signal. The carrier wave maybe
(i) Continuous (sinusoidal) wave, or
(ii) Pulse, which is discontinuous.
A continuous sinusoidal carrier wave can be expressed as,
E = E_o sin (ωt +Φ)
Three distinct characteristics of such a wave are amplitude (E₀), angular frequency (ω) and phase angle fcj)).
Any one of these three characteristics can be varied in accordance with the modulating baseband (AF) signal, giving rise to the respective Amplitude Modulation;
Frequency Modulation and Phase Modulation.

Again, the significant characteristics of a pulse are Pulse Amplitude, Pulse Duration or Pulse Width and Pulse Position (representing the time of rise or fall of the pulse amplitude). Any one of these characteristics can be varied in accordance with the modulating baseband (AF) signal, giving rise to the respective. Pulse Amplitude
Modulation (PAM), Pulse Duration Modulation (PDM), Pulse Width Modulation (PWM) and Pulse Position Modulation (PPM).

(b) Amplitude Modulation: When a modulating AF wave is superimposed on a high-frequency carrier wave in a manner that the frequency of modulated wave is same as that of the carrier wave, but its amplitude is made proportional to the instantaneous amplitude of the audio, frequency modulating voltage, the process is called amplitude modulation (AM).

Let the instantaneous carrier voltage (e_c) and modulating voltage (e_m) be represented by,
e_c = E_c sinω_c t …………………… (1)
e_m = E_m sinω_mt …………………….. (2)
Thus, in amplitude modulation, amplitude A of modulated wave is made proportional to the instantaneous modulating voltage e_m i.e.,
A = E_c+ke_m ……………………….. (3)
where k is a constant of proportionality.
In amplitude modulation, the proportionality constant k is made equal to unity. Therefore, maximum positive amplitude of AM wave is given
by.
A = E_c +e_m =E_c +E_m sinω_m t …………………….. (4)
It is called top envelope.
The maximum negative amplitude of AM wave is given by,
-A = – E_c – e_m
= – (E_c +E_m sinω_m t) …………………………. (5)

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 14 Ecosystem Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 14 Ecosystem

Very Short Answer Type Questions

Question 1.
How is ‘stratification’ represented in a forest ecosystem?
Answer:
Stratification is the vertical distribution of species at different levels. Trees occupy top vertical strata or layer, shrubs the second layer and herbs/grasses occupy the bottom layers.

Question 2.
Write a difference between net primary productivity and gross primary productivity.
Answer:
Gross primary productivity (GPP) is the rate of production of organic matter during photosynthesis. Net primary productivity (NPP) is the available biomass for the consumption by heterotrophs.
GPP – R = NPP.

Question 3.
Why is the rate of assimilation of energy at the herbivore level called secondary productivity? [NCERT Exemplar]
Answer:
It is because the biomass available to the consumer for consumption is a resultant of the primary productivity from plants.

Question 4.
Why is an earthworm called a detritivore?
Answer:
This is because earthworm breaks down detritus into smaller particles.

Question 5.
Justify the pitcher plant as a producer. [NCERT Exemplar]
Answer:
Pitcher plant is chlorophyllous and is thus capable of photosynthesis and act as producer.

Question 6.
Name any two organisms which occupy more than one trophic level in an ecosystem? [NCERT Exemplar]
Answer:
Man and sparrow.

Question 7.
What is common to earthworm, mushroom, soil mites, and dung beetle in an ecosystem? [NCERT Exemplar]
Answer:
They are all detritivores, i.e., decomposing organisms which feed on dead remains of plants and animals.

Question 8.
“Man can be a primary as well as a secondary consumer.” Justify this statement.
Answer:
Man has a varied diet. When on a vegetarian diet, they are primary consumers, and when on a non-vegetarian diet, they are secondary consumers.

Question 9.
Name an omnivore which occurs in both grazing food chain and the decomposer food chain. [NCERT Exemplar]
Answer:
Sparrow/crow.

Question 10.
Differentiate between standing state and standing crop in an ecosystem.
Answer:
In an ecosystem, standing crop is the mass of living material in each trophic level at a particular time. Whereas standing state refers to the amount- of nutrients in the soil at any given time.

Question 11.
Under what conditions would a particular stage in the process of succession revert back to an earlier stage? [NCERT Exemplar]
Answer:
Natural or human-induced disturbances like fire, deforestation, etc.

Question 12.
Climax stage is achieved quickly in secondary succession as compared to primary succession. Why? [NCERT Exemplar]
Answer:
The rate of succession is much faster in secondary succession as the substratum (soil) is already present as compared to primary succession where the process starts from a bare area (rock).

Short answer type questions

Question 1.
What is an incomplete ecosystem? Explain with the help of a suitable example. [NCERT Exemplar]
Answer:
An ecosystem is a functional unit with biotic and abiotic factors interacting with one another resulting in a physical structure. Absence of any component will make an ecosystem incomplete as it will hinder the functioning of the ecosystem. Examples of such an ecosystem can be a fish tank or deep aphotic zone of the oceans where producers are absent.

Question 2.
Justify the importance of decomposers in an ecosystem.
Answer:
Decomposers which are heterotrophic organisms, mainly fungi and bacteria break down complex organic matter into inorganic substances like carbon dioxide, water, and nutrients. They meet their energy and nutrient requirements by degrading dead organic matter or detritus. These are also known as saprotrophs. Decomposers secrete digestive enzymes that break down dead and waste into simple, inorganic materials, which are subsequently absorbed by them.

Question 3.
“In a food chain, a trophic level represents a functional level, not a species.” Explain.
Answer:
Trophic level in a food chain is a level at which organisms obtain their food. Each trophic level has a specific mode of obtaining food. Thus, trophic level represents the mode of obtaining food and not the species. The species has no significance in the food chain. In the food chain, there are generally four trophic levels-producers or autotrophs, the first trophic level; primary consumer/herbivore secondary trophic level; secondary consumer/carnivore third trophic level and finally tertiary consumers (top carnivores) representing fourth trophic level. Thus, in a food chain, trophic levels represent a functional level and which species represent the trophic level, does not matter.

Question 4.
How does primary succession start in water to the climax community? Explain.
Answer:
In primary succession in water, the pioneers are the small phytoplanktons. They are replaced with time by free-floating angiosperms, then by rooted hydrophytes; sedges, grasses, and finally the trees. The climax again would be a forest. With time the water body is converted into land.

Question 5.
Why are nutrient cycles in nature called biogeochemical cycles? [NCERT Exemplar]
Answer:
Nutrient cycles are called biogeochemical cycles because ions/molecules of a nutrient are transferred from the environment (rocks, air and water) to organisms (life) and then brought back to the environment in a cyclic pathway.
The literal meaning of biogeochemical is bio-living organisms and geo-rocks, air, and water.

Question 6.
(a) State any two differences between phosphorus and carbon cycles in nature.
(b) Write the importance of phosphorus in living organisms.
Answer:
(a)

Phosphorus cycle	Carbon cycle
1. It is a sedimentary cycle.	It is a gaseous cycle.
2. Atmospheric inputs through rainfall are much smaller.	Atmospheric inputs through rainfall are more.
3. Gaseous exchange of phosphorus between organism and environment is nil.	Gaseous exchange of carbon between organism and environment is much more.

(b) Phosphorus is a major constituent of biological membranes, nucleic acids, and cellular energy transfer systems.

Long answer type questions

Question 1.
Explain succession of plants in xerophytic habitat until it reaches climax community.
Answer:
Xerarch Succession:

• It starts in primary bare, dry area such as rocks or sand dunes, etc.
• The pioneer species on the rock are usually lichens and blue-green algae under humid conditions.
• They secrete acids to dissolve rocks, help in weathering and soil formation.
• Little soil formation makes the way to the appearance of small plants like bryophytes (mosses). They accumulate more soil and organic matter.
• With time, they are succeeded by bigger plants; perennial grasses and shrubs.
• After several more stages, ultimately a stable climax forest community is formed.
• Climax community remains stable as long as environment remains unchanged.
• With time, the xerophytic habitat gets converted into mesic habitat.

Question 2.
Describe the advantages for keeping the ecosystems healthy.
Answer:
An healthy ecosystem is stable and have a functional balance amongst different populations found in ecosystem. Ecosystem advantages are benefits provided by ecosystem processes to environment in its cleaning and maintenance, enhancing aesthetic beauty, maintenance of biodiversity, protection of soil, water and land sources besides providing a habitat to wildlife, tribals and grazing areas. The important advantages are given ahead :

(i) Oxygen Release : (Purify air) Suspended particulate matter is intercepted by vegetation and made to settle down. Air is thus removed of its pollutants. It is further purified by removal of carbon dioxide and release of oxygen during photosynthetic activity of vegetation.

(ii) Water: Most of rainwater is held over the soil by plant litter like a sponge. It slowly percolates down and becomes the source of all springs, rivulets and rivers. The water is clean and fresh.

(iii) Prevention of Floods: As there is little runoff, flood water is not formed.

(iv) Protection of Soil: Maintenance of soil fertility depends upon a good soil cover and optimum nutrient cycling. Soil cover also protects the soil from air and water erosion. Soil particles remain bound together by plant roots. Landslides are rare.

(v) Biodiversity: Natural ecosystems are a source of biodiversity with a variety of genes, gene pools, species and habitats.

(vi) Climate: Ecosystems, especially forests, maintain good climatic conditions by increasing humidity, reducing extremes of temperature and increasing periodicity of rainfall.

(vii) Nutrient Cycling: It is one of the most important ecosystem services which maintains the continuity of life on earth. Through cycling, biogenetic nutrients are made available all the time for absorption.

(viii) Pollination: Insects, especially bees, and birds visit areas around the forests for pollination of crop plants, bushes, and trees.

(ix) Pest Control: In natural ecosystem, pests are kept under control by their natural predators. Maintenance of natural ecosystems will allow the predators to free the nearby areas of pests.

(x) Wildlife: Ecosystems provide habitats to wildlife.

(xi) Aesthetic Value: Ecosystems also provide aesthetic, cultural, and spiritual value.

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.