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

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

Very short answer type questions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Short answer type questions

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

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

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

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

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

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

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

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

• Retentivity and coercivity should be large
• Magnetically hard

(ii) An electromagnet

• Magnetically soft
• Coercivity should be low.

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

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

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

Long answer type questions

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

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

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

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

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

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

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

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

Very short answer type questions

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

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

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

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

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

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

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

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

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

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

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

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

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

Short answer type questions

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

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

Differences I

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Long answer type questions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Very short answer type questions

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

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

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

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

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

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

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

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

Truth table

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:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Long answer type questions

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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Very short answer type questions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Short answer type questions

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

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

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

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

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

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

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

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

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

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

Long answer type questions

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

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

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

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

Very short answer type questions

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer:

Short answer type questions

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

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

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

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

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

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

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

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

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

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

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

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

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

Long answer type questions

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

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

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

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

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

Punjab State Board PSEB 12th Class Biology Important Questions Chapter 11 Biotechnology: Principles and Processes Important Questions and Answers.

PSEB 12th Class Biology Important Questions Chapter 11 Biotechnology: Principles and Processes

Very short answer type questions

Question 1.
Write the two components of the first artificial recombinant DNA molecule constructed by Cohen and Boyer.
Answer:

• Antibiotic resistance gene and
• Plasmid vector of Salmonella typhimurium.

Question 2.
Suggest a technique to a researcher who needs to separate fragments of DNA.
Answer:
Gel Electrophoresis.

Question 3.
Mention the uses of cloning vector in biotechnology.
Answer:
Cloning vectors are used for transferring fragments of foreign DNA into a suitable host. They are also used to select recombinants from non-recombinants.

Question 4.
Why is it not possible for an alien DNA to become part of a chromosome anywhere along its length and replicate normally?
Answer:
Alien DNA must be linked to ori or origin of replication site to start replication.

Question 5.
Why is it essential to have a ‘selectable marker’ in a cloning vector?
Answer:
Selectable markers are essential to identify and eliminate non-transformants, by selectively permitting the growth of the transformant.

Question 6.
Biotechnologists refer to Agrobacterium tumifaciens as a natural genetic engineer of plants. Give reasons to support the statement.
Answer:
This is because A. tumifaciens can transfer genes naturally by delivering a piece of T-DNA to plant cells. It has a tumour inducing plasmid.

Question 7.
Name the host cells in which micro-injection technique is used to introduce an alien DNA.
Answer:
Animal cells.

Question 8.
How does an alien DNA gain entry into a plant cell by ‘biolistics’ method?
Answer:
In biolistics method cells are bombarded with high velocity micro-particles of gold or tungsten coated with DNA.

Question 9.
What is the host called that produces a foreign gene product? What is this product called?
Answer:
The host that produces a foreign gene product is called competent host. The product is called recombinant protein.

Question 10.
Give any two microbes that are useful in biotechnology. [NCERT Exemplar]
Answer:
E. coli and Saccharomyces cerevisiae.

Question 11.
What is EcoRI? How does EcoRI differ from an exonuclease?
Answer:
EcoRI is restriction endonuclease enzyme. Exonuclease removes nucleotides from the ends of DNA while EcoRI makes cut at specific position within the DNA.

Question 12.
What is the significance of adding proteases at the time of isolation of genetic material (DNA)? [NCERT Exemplar]
Answer:
Role of proteases is to degrade the proteins present inside a cell (from which DNA is being isolated). If the proteins are not removed from DNA preparation then they could interfere with any downstream treatment of DNA.

Question 13.
While doing a PCR, ‘denaturation’ step is missed. What will be its effect on the process? [NCERT Exemplar]
Answer:
If denaturation of double-stranded DNA does not take place, then primers will not be able to anneal to the template, no extension will take place, hence no amplification will occur.

Question 14.
What would happen when you grow a recombinant in a bioreactor but forget to add antibiotic to the medium in which the recombinant is growing? [NCERT Exemplar]
Answer:
In the absence of antibiotic, there will be no pressure on recombinants to retain the plasmid (containing the gene of your interest). Since, maintaining a high copy number of plasmids is a metabolic burden to the microbial cells, it will thus tend to lose the plasmid.

Short answer type questions

Question 1.
What is meant by gene cloning? [NCERT Exemplar]
Answer:
Gene cloning refers to a process in which a gene of interest is ligated to a vector. The recombinant DNA thus produced is introduced in a host cell by transformation. Each cell gets one DNA molecule and when the transformed cell grows to a bacterial colony, each cell in the colony has a copy of the gene.

Question 2.
List the key tools used in recombinant DNA technology.
Answer:
The key tools used in recombinant DNA technology are as follows:

• Restriction enzymes
• Polymerase enzyme
• Ligase enzyme ‘
• Vectors
• Host organism/cell.

Question 3.
(a) What are “molecular scissors”? Give one example.
(b) Explain their role in recombinant DNA technology.
Or Why are molecular scissors so called? Write their use in biotechnology.
Answer:
(a) The restriction endonucleases are called molecular scissors, as they cut the DNA segments at particular locations, e.g., EcoRI. They are so called because they cut DNA at specific points.

(b) The restriction enzymes cut the DNA strands a little away from the centre of the palindromic sites, but between the same two bases on the opposite strands. This leaves single stranded portions with overhanging stretches called sticky ends on each strand as they form hydrogen bonds with their complementary cut counterparts. This stickiness at the ends facilitates the action of the enzyme DNA ligase.

Question 4.
Name the natural source of agarose. Mention one role of agarose in biotechnology.
Answer:
The natural source of agarose is sea weed. Agarose is a natural polymer. It is used to develop the matrix for gel electrophoresis. It helps in the separation of DNA fragments based on their size.

Question 5.
For producing a recombinant protein (for therapeutic purpose) in large scale, which vector would you choose – a low copy number or high-copy number? [NCERT Exemplar]
Answer:
High-copy number, because higher the copy number of vector plasmid, higher the copy number of gene and consequently, protein coded by the gene is produced in high amount.

Question 6.
What modification is done in the Ti plasmid of Agrobacterium tumefaciens to convert it into a cloning vector? [NCERT Exemplar]
Answer:
T-DNA is the only essential part required to make Ti plasmid a cloning vector. The plasmid is disarmed by deleting the tumour inducing genes in the plasmid so that it becomes an effective cloning vector and remove it harmful effect.

Question 7.
Describe the role of CaCl₂ in preparation of competent cells.
[NCERT Exemplar]
Answer:
CaCl₂ is known to increase the efficiency of DNA uptake to produce transformed bacterial cells. The divalent Ca⁺² ions create transient pores in the bacterial cell wall, by which the entry of foreign DNA is facilitated into the bacterial cells.

Question 8.
(a) Why must a cell be made ‘competent’ in biotechnology experiments? How does calcium ion help in doing so?
(b) State the role of ‘biolistic gun’ in biotechnology experiments?
Answer:
(a) A recombinant DNA transfer into the host cell, needs that the recipient cell must be made competent in order to receive and absorb the DNA, present in the surrounding. The calcium ions in the medium increase the efficiency with which DNA enters the bacterium through the pores in the cell wall.

(b) Biolistic or Gene Gun : It is a vectorless method in which DNA is directly introduced in the nucleus of plant cell. Plant cells are bombarded with high velocity micro particles of gold or tungsten coated with DNA.

Long answer type questions

Question 1.
(a) Name the selectable markers in the cloning vector pBR322? Mention the role they play. ‘
(b) Why is the coding sequence of an enzyme β-galactosidase a preferred selectable marker in comparison to the ones nam£d above?
Answer:
(a) In pBR322, amp^R and Tet^R, the two antibiotic resistant genes act as selectable markers. If an alien DNA ligates at the Bam HI site of tetracycline resistant gene in the vector pBR322, the recombinant loses the tetracycline resistance. Non-recombinant will grow on both the media containing tetracycline/ampicillin whereas recombinant will grow on ampicillin medium but not on medium containing tetracycline. In this case, one antibiotic resistance gene helps in selecting the transformants and the other antibiotic gene gets inactivated due to insertion of alien DNA.

Thus, selectable markers (amp^R) help in indentifying and eliminating non-transformants and help in selecting those host cells which contain the recombinant vector i.e., transformants.

(b) The selection of recombinant by the inactivation of one of the antibiotic resistance gene is a cumbersome, complicated, time consuming technique involving plating both the recombinant and non-recombinant on the ampicillin medium and then on tetracycline containing medium. In insertional inactivation, the DNA inserted in the coding sequence of an enzyme p-galactosidase results in inactivation of the enzyme and the bacterial colony, with insert shows no colouration while those without inserted plasmid, form blue colour colonies. This is a simple, less cumbersome technique.