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Punjab State Board PSEB 12th Class Physics Book Solutions Chapter 1 Electric Charges and Fields Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

PSEB 12th Class Physics Guide Electric Charges and Fields Textbook Questions and Answers

Question 1.
What is the force between two small charged spheres having charges of 2 × 10^-7 C and 3 × 10^-7 C placed 30 cm apart in air?
Answer:
Charge on the first sphere, q₁ = 2 × 10^-7 C
Charge on the second sphere, q₂ = 3 × 10^-7 C
Distance between the spheres, r = 30 cm = 0.3 m
Electrostatic force between the spheres is given by the relation,
F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}\)
where, ε₀ = Permittivity of free space
F = \(\frac{9 \times 10^{9} \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{(0.3)^{2}}\)
= 6 × 10^-3 N
Hence, force between the two small charged spheres is 6 × 10^-3 N. The charges are of same nature. Hence, force between them will be repulsive.

Question 2.
The electrostatic force on a small sphere of charge
0.4 μC due to another small sphere of charge -0.8 μC in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?
Answer:
Electrostatic force on the first sphere, F = 0.2 N
Charge on this sphere, q₁ = 0.4 μC = 0.4 × 10^-6C
Charge on the second sphere, q₂ = -0.8 μC = -0.8 × 10^-6C

(a) Electrostatic force between the spheres is given by the relation,
F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}\)
where, ε₀ = Permittivity of free space
r² = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{F}\)
= \(=\frac{0.4 \times 10^{-6} \times 0.8 \times 10^{-6} \times 9 \times 10^{9}}{0.2}\)
= 16 × 9 × 10^-4 = 144 × 10^-4
= \(\sqrt{144 \times 10^{-4}}\) = 0.12 m
The distance between the two spheres is 0.12 m.

(b) Force on q₂ due to q₁= ?
We know that electrostatic forces always, appear in priors and follow Newton’s 3rd law of motion.
∴ \(\left|\vec{F}_{21}\right|\) = Force on q₂ due to q₁
= 0.2 N and it is attractive in nature
We now that,
F₂₁ = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r_{21}^{2}}\)

On substituting the values, we get
F₂₁ = 9 × 10⁹ × \(\frac{\left(0.4 \times 10^{-6}\right) \times\left(0.8 \times 10^{-6}\right)}{(0.12)^{2}}\)
= 0.2 N

Question 3.
Check that the ratio ke² / Gm_em_p is dimensionless. Look up a
table of physical constants and determine the value of this ratio. What does the ratio signify?
Answer:
The given ratio is \(\frac{k e^{2}}{G m_{e} m_{p}}\)
where, G = Gravitational constant
Its unit is Nm²kg^-2.
m_p and m_p = Masses of electron and proton
Their unit is kg.
e = Electric charge Its unit is C.
k = A constant = \(=\frac{1}{4 \pi \varepsilon_{0}}\)
Its unit is N m²C^-2.
Therefore, unit of the given ratio
\(\frac{k e^{2}}{G m_{e} m_{p}}\) = \(\frac{\left[\mathrm{Nm}^{2} \mathrm{C}^{-2}\right]\left[\mathrm{C}^{-2}\right]}{\left[\mathrm{Nm}^{2} \mathrm{~kg}^{-2}\right][\mathrm{kg}][\mathrm{kg}]}\) = M⁰L⁰T⁰
Hence, the given ratio is dimensionless.
e = 1.6 × 10^-19 C
G = 6.67 × 10^-11 N m²kg^-2
m_e = 9.1 × 10^-31 kg
m_p = 1.67 × 10^-27 kg
Hence, the numerical value of the given ratio is
\(\frac{k e^{2}}{G m_{e} m_{p}}\) = \(\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{6.67 \times 10^{-11} \times 9.1 \times 10^{-31} \times 1.67 \times 10^{-27}}\)
≈ 2.3 × 10³⁹
This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.

Question 4.
(a) Explain the meaning of the statement ‘electric charge of a body is quantised’.
(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i. e., large scale charges?
Answer:
(a) Electric charge of a body is quantised. This means that only integral (1, 2 …………,n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge.

(b) In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantisation of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous.

Question 5.
When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.
Answer:
Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed bodies is zero. This is because equal amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk cloth, opposite natured charges appear on both the bodies. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.

Question 6.
Four point charges q_A = 2 μC, q_B = -5 μC, q_C = 2 μC and q_D) = -5 μC are located at the comers of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?
Answer:
Consider the square ABCD of each side 10 cm and centre O. The charge of 1 μC is placed at O.
Now clearly, OA = OB =OC = OD
AB = BC – 10 cm = 0.1 m
AO = \(\frac{1}{2}\) AC = \(\frac{1}{2} \sqrt{A B^{2}+B C^{2}}\) = \(\frac{1}{2}\) × √2 AB
= \(\frac{1}{\sqrt{2}}\) × 0.1 m = OB = OC = OD

Here,q_A = 2μC, q_B = -5μC,q_C = 2 μC,q_D = -5μC
Clearly q_A = q_C = 2μC = 2 × 10^-6C
and q_B = q_D = -5 μC = -5 × 10^-6C
Since q_A = q_C, the charge of 1 μC will experience equal and opposite forces due to the charges q_A and q_C i. e., along OC and OA respectively. Their magnitudes are
F_A = F_C = \(\frac{1}{4 \pi \varepsilon_{0}} \times \frac{q_{A} \times 1 \mu \mathrm{C}}{A O^{2}}\)
\(\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times 10^{-6}}{\left(\frac{1}{\sqrt{2}} \times 0.1\right)^{2}}\) = 3.6N
∴ \(\vec{F}_{A}=-\vec{F}_{C}\)
Similarly F_B = F_D, the charge of 1 μC will experience equal and opposite forces due to the charge q_B and q_D i. e., along OB and OD respectively, thus
\(\overrightarrow{F_{B}}=-\overrightarrow{F_{D}}\).
Thus the net force on the charge ofl μC due to the given arrangement of charges is zero i.e.,
\(\vec{F}=\vec{F}_{A}+\vec{F}_{B}+\vec{F}_{C}+\vec{F}_{D}\) = 0

Question 7.
(a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
Explain why two field lines never cross each other at any point?
Answer:
(a) An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.

(b) If two field lines cross each other at a point, then electric field intensity will show two directions at that point. This is not possible. Hence, two field lines never cross each other.

Question 8.
Two point charges q_A = 3 μC and q_B = -3 μC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 × 10^-9C is placed at this point, what is the force experienced by the test charge?
Answer:
(a) The situation is represented in the given figure. O is the mid-point of line AB.

Distance between the two charges, AB = 20 cm
∴ AO = OB =10 cm
Net electric field at point O = E
Magnitude of electric field at point 0 caused by + 3 μC charge,

[Since the values of E₁ and E₂ are same, the value is multiplied with 2]
= 5.4 × 10⁶ N/C along OB
Therefore, the electric field at mid-point O is 5.4 × 10⁶ NCT^-1 along OB.

(b) A test charge of amount 1.5 × 10^-9 C is placed at mid-point O.
According to question, q = -1.5 × 10^-9 C
Force experienced by the test charge = F
F = qE
= -1.5 × 10^-9 × 5.4 × 10⁶
= -8.1 × 10^-3 N
The force is directed along line QA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.

Question 9.
A system has two chargesg q_A = 2.5 × 10^-7C andq_A = -2.5 × 10^-7 located at points A : (0, 0, -15 cm) and B: (0, 0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?
Answer:
Both the charges can be located in a coordinate frame of reference as shown in the given figure At A, amount of charge,
q_A =2.5 × 10^-7C
At B, amount of charge,
q_B = -2.5 × 10^-7C

Total charge of the system,
q = q_A + q_B
= 2.5 × 10^-7 C-2.5 × 10^-7C
= 0
Distance between two charges at points A and B,
d. =15 + 15 = 30 cm = 0.3 m
Electric dipole moment of the system is given by,
p = q_A × d = q_B × d.
= 2.5 × 10^-7 × 0.3
= 7.5 × 10^-8 Cm along positive z-axis
Therefore, the electric dipole moment of the system is 7.5 × 10^-8C m along positive z-axis.

Question 10.
An electric dipole with dipole moment 4 × 10^-9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 10⁴NC^-1. Calculate the magnitude of the torque acting on the dipole.
Answer:
Electric dipole moment, p = 4 × 10^-9 C m
Angle made by p with a uniform electric field, 0 = 30°
Electric field, E = 5 × 10⁴NC^-1
Torque acting on the dipole is given by the relation,
τ = pE sinθ
= 4 × 10^-9 × 5 × 10⁴ × sin30
= 20 × 10^-5 \(\frac{1}{2}\)
= 10^-4 Nm
Therefore, the magnitude of the torque acting on the dipole is 10^-4 N m.

Question 11.
A polythene piece rubbed with wool is found to have a negative charge of 3 × 10^-7C.
(a) Estimate the number of electrons transferred (from which to which?)
(b) Is there a transfer of mass from wool to polythene?
Answer:
(a) When polythene is rubbed with wool, a number of electrons get transferred from wool to polythene. Hence, wool becomes positively charged and polythene becomes negatively charged.
Amount of charge on the polythene piece, q = -3 × 10^-7 C
Amount of charge on an electron, e = -1.6 × 10^-19C
Number of electrons transferred from wool to polythene = n
n can be calculated using the relation,
> q = ne
n = \(\frac{q}{e}\)
= \(\frac{-3 \times 10^{-7}}{-1.6 \times 10^{-19}}\)
= 1.87 × 10¹²
Therefore, the number of electrons transferred from wool to polythene is 1.87 × 10¹².

(b) Yes.
There is a transfer of mass taking place. This is because an electron has mass,
m_e = 9.1 × 10^-31 kg
Total mass transferred to polythene from wool,
m = m_e × n
= 9.1 × 10^-31 × 1.87 × 10¹²
= 1.706 × 10^-18 kg
Hence, a negligible amount of mass is transferred from wool to polythene.

Question 12.
(a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10^-7 C ? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
Answer:
(a) Charge on sphere A, q_A = Charge on sphere B, q_B = 6.5 × 10^-7 C
Distance between the spheres, r = 50 cm = 0.5 m
Force of repulsion between the two spheres,
F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{A} q_{B}}{r^{2}}\)
∴ F = \(=\frac{9 \times 10^{9} \times\left(6.5 \times 10^{-7}\right)^{2}}{(0.5)^{2}}[latex]
= 1.52 × 10^-2 N
Therefore, the force between the two spheres is 1.52 × 10^{-2 N.}

(b) After doubling the charge, charge on sphere A, q_A = charge on sphere
B, q_B = 2 × 6.5 × 10^-7 C = 1.3 × 10<>-6 C
The distance between the spheres is halved
∴ r = [latex]\frac{0.5}{2}\) = 0.25 m
Force of repulsion between the two spheres,
F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{A} q_{B}}{r^{2}}\)
∴ F = \(\frac{9 \times 10^{9} \times 1.3 \times 10^{-6} \times 1.3 \times 10^{-6}}{(0.25)^{2}}\)
= 16 × 1.52 × 10^-2 = 0.243 N Therefore, the force between the two spheres is 0.243 N.

Question 13.
Suppose the spheres A and B in exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B ?
Answer:
Distance between the spheres, A and B,r = 0.5 m
Initially, the charge on each sphere, q = 6.5 × 10 ^-7 C
When sphere A is touched with an uncharged sphere C, \(\frac{q}{2}[latex] amount of
charge from A will transfer to sphere C. Hence, charge on each of the spheres, A and C is [latex]\frac{q}{2}\).
When sphere C with charge \(\frac{q}{2}\) is brought in contact with sphere B with
charge q, total charges on the system will divide into two equal halves given as,
\(\frac{\frac{q}{2}+q}{2}=\frac{3 q}{4}\)
Each sphere will each half. Hence, charge on each of the spheres, C and B, is \(\frac{3 q}{4}\).
Force of repulsion between sphere A having charge \(\frac{q}{2}\) and sphere B having
charge \(\frac{3 q}{4}\)

= 5.703 × 10^-3 N
Therefore, the force of attraction between the two spheres is 5.703 × 10^-3 N.

Question 14.
Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Answer:
Opposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged. The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.

Question 15.
Consider a uniform electric field E = 3 × 10³ î N/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?
Answer:
Here, \(\overrightarrow{\vec{E}}\) = 3 × 10³ î NC^-1 i.e., the electric field acts along positive direction of x-axis.
Side of square = 10 cm
∴ Its surface area, ΔS = (10 cm)² = 10^-2 m²
or Δ\(\overrightarrow{\vec{S}}\) = 10^-2 î m²
as normal to the square is along x-axis.

(a) If Φ be the electric flux through the square, then

Φ = \(\vec{E} \cdot \Delta \vec{S}\)
= (3 × 10³î) .(10^-2)
= 3 × 10³ × 10^-2 î .î
= 3 × 10 = 30 Nm²C^-1

(b) Here, angle between normal to the square i.e., area vector and the electric field is 60°. i.e.,
θ = 60°
∴ Φ = \(\overrightarrow{\vec{E}}\). Δ \(\overrightarrow{\vec{S}}\) = E. Δ S cos60°
= 3 × 10³ × 10^-2 × \(\frac{1}{2}\)
= 15Nm²C^-1

Question 16.
What is the net flux of the uniform electric field of exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?
Answer:
All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero.

Question 17.
Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 10³ N m² / C. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?
Answer:
(a) Net outward flux through the surface of the box Φ = 8.0 × 10³ N m² / C
For a body containing net charge q, flux is given by the relation,
Φ = \(\)
q = ε₀Φ)
= 8.854 × 10^-12 × 8.0 × 10³
(∵ ε₀ = 8.854 × 10^-12N^-1C²m^-2) = 7.08 × 10^-8
= 0.07 µC
Therefore, the net charge inside the box is 0.07 pC.

(b) No.
Net flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that net charge inside the body is zero. The body may have equal amount of positive and negative charges.

Question 18.
A point charge +10 µC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint : Think of the square as one face of a cube with edge 10 cm.)

Answer:
The square can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed. According to Gauss’s theorem for a cube, total electric flux is through all its six faces.
Φ total = \(\frac{q}{\varepsilon_{0}}\)

Hence, electric flux through one face of the cube i. e., through the square,
Φ = \(\frac{\phi_{\text {Total }}}{6}=\frac{1}{6} \frac{q}{\varepsilon_{0}}\)
ε₀ = 8.854 × 10^-12 N^-1C²m^-2
q = 10 µC =10 × 10^-6C
Φ = \(\frac{1}{6} \times \frac{10 \times 10^{-6}}{8.854 \times 10^{-12}}\)
= 1.88 × 10⁵ N m² C^-1
Therefore, electric flux through the square is 1.88 × 10⁵ N m² C^-1.

Question 19.
A point charge of 2.0 µC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
Answer:
Net electric flux (Φ _Net) through the cubic surface is given by,
Φ _Net = \(\frac{q}{\varepsilon_{0}}\)
ε₀ = 8.854 × 10^-12 N^-1C²m^-22 q q = Net charge contained inside the cube
= 20 µC = 2 × 10^-6C
∴ Φ _Net = \(\frac{2 \times 10^{-6}}{8.854 \times 10^{-12}}\)
∴ Φ _Net = \(\frac{2 \times 10^{-6}}{8.854 \times 10^{-12}}\)
= 2.26 × 10⁵ Nm²C^-1
The net electric flux through the surface is = 2.26 × 10⁵ Nm²C^-1

Question 20.
A point charge causes an electric flux of -1.0 × 10³ Nm² /C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge, (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?
Answer:
Electric flux, Φ = -1.0 × 10³ Nm²/C
Radius of the Gaussian surface,
r = 10.0 cm

(a) Electric flux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i. e., -1.0 × 10³ 

(b) Electric flux is given by the relation,
Φ = \(\frac{q}{\varepsilon_{0}}\)
q = Φε₀
= -1.0 × 10³ × 8.854 × 10^-12
= -8.854 × 10^-9 C
= -8.854 nC
Therefore, the value of the point charge is -8.854 nC.

Question 21.
A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 10³ N/C and points radially inward, what is the net charge on the sphere?
Answer:
Here R = radius of the conducting sphere = 10 cm = 0.10 m
r = distance of the point from centre of sphere = 20 cm = 0.20 m Clearly
r > R
E = electric field at a point at a distance of 20 cm from the sphere = 1.5 × 10³ NC ^-1acting inward.
q = net charge on the sphere = ?
Using the formula, E = \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r^{2}}\) We get
q = 4πε₀Er² = \(\frac{1}{9 \times 10^{9}}\) × 1.5 × 10<>3 × (0.20)²
= 6.67 × 10^-9C =6.67 nC
Also as E acts in the inward direction, so charge on the sphere is negative.
q = -6.67 × 10^-9C -6.67nC

Question 22.
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC / m². (a) Find the charge on the sphere, (b) What is the total electric flux leaving the surface of the sphere?
Answer:
Diameter of the sphere, d = 2.4 m
Radius of the sphere, r = 1.2 m
Surface charge density, σ = 80.0 μC / m² = 80 × 10 ^-6C / m²

(a) Total charge on the surface of the sphere,
q = Charge density × Surface area = σ × 4πr²
= 80 × 10^-6 × 4 × 3.14 × (1.2)² = 1.447 × 10^-3 C
Therefore, the charge on the sphere is 1.447 × 10^-3 C.

(b) Total electric flux (Φ _Total) leaving out the surface of a sphere containing net charge q is given by the relation,
Φ _Total = \(\frac{q}{\varepsilon_{0}}\)
ε₀ = 8.854 × 10^-12 N^-1C²m^-2
q = 1.447 × 10^-3 C
Φ _Total = \(\frac{1.44 \times 10^{-3}}{8.854 \times 10^{-12}}\)
= 1.63 × 10⁸ NC^-1 m²
Therefore, the total electric flux leaving the surface of the sphere is 1.63 × 10⁸ NC^-1 m².

Question 23.
An infinite line charge produces a field of 9 × 10⁴ N/C^-1 at a distance of 2 cm. Calculate the linear charge density.
Answer:
Here E = electric field produced by infinite line charge = 9 × 10⁴ NC^-1.
r = distance of the point where E is produce = 2 cm = 0.02 m.
λ = linear charge density = ?

Question 24.
Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10^-22 C/m². What is E : (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?
Answer:
The situation is represented in the following figure:

A and B are two parallel plates close to each other. Outer region of plate A is labelled as I, outer region of plate B is labelled as III, and the region between the plates, A and B is labelled as II.

(a) Charge density of plate A,
σ = 17.0 × 10^-22 C/m²
Charge density of plate B,
σ = -17.0 × 10^-22 C/ m²
In the regions, I and III, electric field E is zero. This is because charge is not enclosed by the respective plates.

(b) Electric field E in region II is given by the relation,
E = \(\frac{\sigma}{\varepsilon_{0}}\)
where, ε₀ = Permittivity of free space
= 8.854 × 10^-12N^-1C²m^-2
E = \(\frac{17.0 \times 10^{-22}}{8.854 \times 10^{-12}}\)
= 1.92 × 10^-10 N/C
Therefore, electric field between the plates is 1.92 × 10^-10 N/C.

Question 25.
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10⁴ NC^-1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm^-3. Estimate the radius of the drop. (g = 9.81 ms^-2, e = 1.60 × 10^-9 C).
Answer:
Here, E = constant electric field = 2.55 × 10⁴ NC^-1, e = charge of an electron = 1.6 × 10^-19 C, n = no. of electrons = 12
If q = charge on the drop, then
q = ne = 12 × 1.6 × 10^-19 C = 19.2 × 10^-19 C
If F_e be the electrostatic force on the oil drop due to electric field, then
F_e = qE = 19.2 × 10^-19 × 2.55 × 10⁴ …………..(1)
Also let F_g = Force on the drop due to gravity, then
F_g = mg = \(\frac{4}{3}\) πr³ρg …………… (2)
Here ρ = density of oil = 1.26 g cm^-3
= 1.26 × 10^-3 kg(10^-2 m)^-3
= 1.26 × 10³ kg m^-3 g
g = 9.81 ms^-2
r = radius of the drop = ?
Putting these values in equation (2), we get

Question 26.
Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?


Answer:
(a) The field lines showed in (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor.
(b) The field lines showed in (b) do not represent electrostatic field lines because the field lines cannot emerge from a negative charge and cannot terminate at a positive charge.
(c) The field lines showed in (c) represent electrostatic field lines. This is because the field lines emerge from the positive charges and repel each other.
(d) The field lines showed in (d) do not represent electrostatic field lines because the field Hnes should not intersect each other.
(e) The field lines showed in (e) do not represent electrostatic field lines because closed loops are not formed in the area between the field lines.

Question 27.
In a certain region of space, electric Held is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 10⁵ NC^-1 per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10^-7 Cm in the negative z-direction?
Answer:
Dipole moment of the system, p = q × dl = -10^-7Cm
Rate of increase of electric field per unit length,
\(\frac{d E}{d l}\) = 10⁺⁵ NC^-1
Force (F) experienced by the system is given by the relation,
F = qE
F = q\(\frac{d E}{d l}\) × dl
= p × \(\frac{d E}{d l}\)
= -10^-7 × 10⁵ .
= -10^-2 N
The force is -10^-2 N in the negative z-direction i.e., opposite to the direction of electric field. Hence, the angle between electric field and dipole moment is 180°.
Torque (τ) is given by the relation,
τ = pEsin180°
= 0
Therefore, the torque experienced by the system is zero.

Question 28.
(a) A conductor A with a cavity as shown in Fig. 1.36 (a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.

(b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q q [Fig. 1.36 (b)].

(c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Answer:
(a) Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing the cavity. The electric field intensity E inside the charged conductor is zero.
Let q is the charge inside the conductor and e 0 is the permittivity of free space.
According to Gauss’s law,
Flux, Φ = \(\vec{E} \cdot \overrightarrow{d s}\) = \(\frac{q}{\varepsilon_{0}}\)
Here, E = 0
\(\frac{q}{\varepsilon_{0}}\) = 0
∵ ε₀ ≠ 0
∴ q = 0
Therefore, charge inside the conductor is zero.
The entire charge Q appears on the outer surface of the conductor.

The outer surface of conductor A has a charge of amount Q. Another conductor B having charge +q is kept inside conductor A and it is insulated from A. Hence, a charge of amount -q will be induced in the inner surface of conductor A and + q is induced on the outer surface of conductor A. Therefore, total charge on the outer surface of conductor A is Q + q.

A sensitive instrument can be shielded from the strong electrostatic field in its environment by enclosing it fully inside a metallic surface. A closed metallic body acts as an electrostatic shield.

Question 29.
A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (\(\)) n̂, where n̂ is the unit vector in the outward normal direction and a is the surface charge density near the hole.
Answer:
Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero.

Let E is the electric field just outside the conductor, q is the electric charge, σ is the charge density, and ε₀ is the permittivity of free space.
Charge \(|q|=\vec{\sigma} \times \overrightarrow{d s}\)
According to Gauss’s law,
Flux Φ = \(\vec{E} \cdot \overrightarrow{d s}\) = \(\frac{|q|}{\varepsilon_{0}}\)
Eds = \(\frac{\vec{\sigma} \times \overrightarrow{d s}}{\varepsilon_{0}}\)
∴ E = \(\frac{\sigma}{\varepsilon_{0}} \hat{n}\)

Therefore, the electric field just outside the conductor is \(\frac{\sigma}{\varepsilon_{0}} \hat{n}\) . This field is

a superposition of field due to the cavity (\(\vec{E}\)) and the field due to the rest of the charged conductor (E). These fields are equal and opposite inside the conductor, and equal in magnitude and direction outside the conductor.
∴ E’ + E’ = E
E’ = \(=\frac{E}{2}=\frac{\sigma}{2 \varepsilon_{0}} \hat{n}\)
Therefore, the field due to the rest of the conductor is \(\frac{\sigma}{\varepsilon_{0}} \hat{n}\) .
Hence, proved.

Question 30.
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]
Answer:
Take a long thin wire XY (as shown in the figure) of uniform linear charge density λ.

Consider a point A at a perpendicular distance l from the mid-point O of the wire, as shown in the following figure:

Let E be the electric field at point A due to the wire, XY.
Consider a small length element dx on the wire section with OZ = x
Let q be the charge on this piece.
∴ q = λ dx
Electric field due to the piece,
dE = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\lambda d x}{(A Z)^{2}}\)
However, AZ = \(\sqrt{\left(l^{2}+x^{2}\right)}\)
∴ dE = \(\frac{\lambda d x}{4 \pi \varepsilon_{0}\left(l^{2}+x^{2}\right)}\)

The electric field is resolved into two rectangular components. dE cos θ is the perpendicular component and dE sin0 is the parallel component. When the whole wire is considered, the component dE sin0 is cancelled. Only the perpendicular component dE cosO affects point A.

Hence, effective electric field at point A due to the element dx is dE₁.
∴ dE₁ = \(\frac{\lambda d x \cos \theta}{4 \pi \varepsilon_{0}\left(x^{2}+l^{2}\right)}\) ……………. (1)
In Δ AZO, tanθ = \(\frac{x}{l}\)
x = ltanθ ……………. (2)

On differentiating equation (2), we obtain
\(\frac{d x}{d \theta}\) = l sin²θ
dx = lsin²θ dθ ……………… (3)

Question 31.
It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up* quark (denoted by u) of charge (+2/3)e, and the ‘down’ quark (denoted by d) of charge (-1/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.
Answer:
A proton has three quarks. Let there be n up quarks in a proton, each having a charge of + \(\frac{2}{3}\) e.
Charge due to n up quarks = (\(\frac{2}{3}\)e)n
Number of down quarks in a proton = 3 – n
Each down quark has a charge of – \(\frac{1}{3}\) e.
Charge due to (3 – n) down quarks = (-\(\frac{1}{3}\)e) (3 – n)
Total charge on a proton = + e
e = (\(\frac{2}{3}\)e)n + (-\(\frac{1}{3}\)e) (3 – n)
e = (\(\frac{2 \text { ne }}{3}\)) – e + \(\frac{n e}{3}\)
2e = ne
n = 2
Number of up quarks in a proton, n = 2
Number of down quarks in a proton = 3 – n = 3 – 2 = 1
Therefore, a proton can be represented as ‘uud’.
A neutron also has three quarks. Let there be n up quarks in a neutron,
each having a charge of + \(\frac{2}{3}\) e.
Charge on a neutron due to n up quarks (+\(\frac{2}{3}\)e) n

Number of down quarks is 3 -n, each having a charge of (-\(\frac{1}{3}\))e
Charge on a neutron due to (3 – n) down quarks = (-\(\frac{1}{3}\)e) (3 – n)
Total charge on a neutron = 0
0 = (\(\frac{2}{3}\)e)n + (-\(\frac{1}{3}\)e) (3 – n)
0 = \(\frac{2}{3}\) en – e + \(\frac{n e}{3}\)
e = ne
n = 1
Number of up quarks in a neutron, n = 1
Number of down quarks in a neutron = 3 – n = 3 – 1 = 2
Therefore, a neutron can be represented as ‘udd’.

Question 32.
(a) Consider an arbitrary electrostatic field configuration. A small test charge isplaced at a null point (i. e., where E = 0) of the configuration. Shew that the equilibrium of the test charge is necessarily unstable.
(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.
Answer:
(a) An arbitrary electrostatic configuration consists of two charges of unequal charges of unequal magnitude and of same sign. e. g., consider a system of two fixed charges + ve and + e separated by a distance r placed at point A and B respectively. Let a test charge q0 be placed at a point C at a distance x from + 4 e. C is the point is resultant field or force on 90 is zero.

or 4(r – x)² = x² or 2(r – x) = ±x
∴ x = \(\frac{2 r}{3}\) or 2r
For equilibrium, the charge q0 can be either + ve or – ve.

Case I: If q0 is – ve, then it experiences equal attractive force due to both the charges. When it is displaced on either side along the line joining the two charges from its equilibrium position, the attractive force due to one charge gets increased while due to the other, it is decreased. As a result of this, charge -q0 no longer returns to its equilibrium position i.e., equilibrium of – ve charge is necessarily unstable.

Case II: If q0 is the + ve but displaced perpendicular to line joining the two charges, then resultant force tends to displace it further more i.e., it will not come back to its equilibrium position i. e., equilibrium will be unstable.

(b) Let the simple configuration consists of two equal charges +q at point A and B. As now the two charges are of same nature and have same magnitude hence their resultant \(\vec{E}\) will be zero at the mid-point O of the line joining them being equal and opposite and system will be unstable if the charge is slightly displaced, it executes S.H.M.

Question 33.
A particle of mass m and charge (~q) enters the region between the two charged plates initially moving along x-axis with speed vx (like particle 1 in Fig. 1.33). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is q EL² (2m v²x).
Compare this motion with motion of a projectile in gravitational field discussed in section 4.10 of Class XI Textbook of Physics.
Answer:
Charge on a particle of mass m = -q
Velocity of the particle = v_x
Length of the plates = L
Magnitude of the uniform electric field between the plates = E
Mechanical force,F = Mass (m) x Acceleration (a)
a = \(\frac{F}{m}\)
However, electric force, F = qE
Therefore, acceleration, a = \(\frac{q E}{m}\) …………. (1)
Time taken by the particle to cross the field of length L is given by,
= \(\frac{\text { Length of the plate }}{\text { Velocity of the particle }}=\frac{L}{v_{x}}\) …………… (2)

Velocity of the particle vx In the vertical direction, initial velocity, u = 0
According to the third equation of motion, vertical deflection s of the particle can be obtained as,
s = ut + \(\frac{1}{2}\)at²
s = 0 + \(\frac{1}{2}\) (\(\frac{q E}{m}\))(\(\frac{L}{v_{x}}\))²
s = \(\frac{q E L^{2}}{2 m v_{x}^{2}}\) ……………. (3)

Hence, vertical deflection of the particle at the far edge of the plate is qEL² / (2m v_x²). This is similar to the motion of horizontal projectiles under gravity.

Question 34.
Suppose that the particle in Exercise in 1.33 is an electron projected with velocity v_x = 2.0 x 10⁶ ms^-1. If E between the plates separated by 0.5 cm is 9.1 x 10² N/C, where will the electron strike the upper plate? (|e| = 1.6 x 10^-19 C, me = 9.1 x 10^-31 kg.)
Answer:
Velocity of the particle, v_x = 2.0 x 10⁶ m/ s
Separation of the two plates, d = 0.5 cm = 0.005 m
Electric field between the two plates, E = 9.1 x 10² N / C
Charge on an electron, q = 1.6 x 10^-19 C
Mass of an electron, m_e = 9.1 x 10^-31 kg
Let the electron strike the upper plate at the end of plate L, when deflection is s. Therefore,

Therefore, the electron will strike the upper plate after travelling 1.6 cm.

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

PSEB Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

PSEB 12th Class Physics Guide Semiconductor Electronics: Materials, Devices, and Simple Circuits Textbook Questions and Answers

Question 1.
In an n-type silicon, which of the following statement is true:
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
Answer:
The correct statement is (c).
In an n-type silicon, the electrons are the majority carriers, while the holes are the minority carriers.
An n-type semiconductor is obtained when pentavalent atoms, such as phosphorus, are doped in silicon atoms.

Question 2.
Which of the statements given in Exercise 14.1 is true for p-type semiconductors.
Answer:
The correct statement is (d).
In a p-type semiconductor, the holes are majority carriers, while the electrons are the minority carriers.
A p-type semiconductor is obtained when trivalent atoms, such as aluminum, are doped in silicon atoms.

Question 3.
Carbon, silicon, and germanium have four valence electrons each.
These are characterized by valence and conduction bands separated by energy bandgap respectively equal to (E_g)_C, (E_g) _si, and (E_g)_Ge Which of the following statements is true?
(a) (E_g)_si<(E_g)_Ge<(E_g)_C
(b) (E_g)_C<(E_g)_Ge>(E_g)_Si
(c) (E_g)_C > (E_g)_Si > (E_g)_Ge
(d) (E_g)_C =(E_g)_Si = (E_g)_Ge
Answer:
The correct statement is (c).
Out of the three given elements, the energy band gap of carbon is the maximum and that of germanium is the least.
The energy band gap of these elements are related as (E_g)_C > (E_g)_Si > (E_g)_Ge.

Question 4.
In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) free electrons in the n-region attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in p-region is more as compared to n-region.
(d) All of the above.
Answer:
The correct statement is (c).
The diffusion of charge carriers across a junction takes place from the region of higher concentration to the region of lower concentration. In this case, the p-region has greater concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from the p-region to the n-region.

Question 5.
When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier,
(b) reduces the majority carrier current to zero,
(c) lowers the potential barrier.
(d) None of the above.
Answer:
The correct statement is (c).
When a forward bias is applied to a p-n junction, it lowers the value of potential barrier.
In the case of a forward bias, the potential barrier opposes the applied voltage.
Hence, the potential barrier across the junction gets reduced.

Question 6.
For transistor action which of the following statements are correct:
(a) Base, emitter and collector regions should have similar size and doping concentrations.
(b) The base region must be very thin and lightly doped.
(c) The emitter junction is forward biased and collector junction is reverse biased.
(d) Both the emitter junction as well as the collector junction are forward biased.
Answer:
The correct statements are (b) and (c).
For a transistor action, the junction must be lightly doped so that the base region is very thin.
Also, both the emitter junction must be forward-biased and collector junction should be reverse-biased.

Question 7.
For a transistor amplifier, the voltage gain
(a) remains constant for all frequencies.
(b) is high at high and low frequencies and constant in the middle-frequency range.
(c) is low at high and low frequencies and constant at mid frequencies.
(d) None of the above.
Answer:
The correct statement is (c).
The voltage gain of a transistor amplifier is constant at mid-frequency range only. It is low at high and low frequencies.

Question 8.
In half-wave rectification, what is the output frequency if the input frequency is 50 Hz.
What is the output frequency of a full-wave rectifier for the same input frequency.
Answer:
Input frequency = 50 Hz
For a half-wave rectifier, the output frequency is equal to the input frequency.
∴ Output frequency = 50 Hz
For a full-wave rectifier, the output frequency is twice the input frequency
∴ Output frequency = 2 x 50 = 100 Hz

Question 9.
For a CE-transistor amplifier, the audio signal voltage across the collected resistance of 2 kΩ is 2V.
Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ.
Answer:
Given, R_c = 2 kΩ, R_B = 1 kΩ,
V₀ = 2V, input voltage V_i = ?
β = \(\frac{I_{C}}{I_{B}}\) = 100
V₀=I_CR_C= 2V
i_c = \(\frac{2 V}{R_{C}}=\frac{2 \mathrm{~V}}{2 \times 10^{3} \Omega}\) = 1.0^-3 A.

Base current,
I_B = \(\frac{I_{C}}{\beta}=\frac{10^{-3}}{100}\) = 10 x 10^-6 A = 10 μA
Base current, R_B = \(\frac{V_{B B}}{I_{B}}\)
Input Voltage, V_i = R_B x I_B
= 1 x 10³ x 10 x 10^-6
= 0.01 V

Question 10.
Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20. If the input signal is 0.01 volt, calculate the output ac signal.
Answer:
Voltage gain of the first amplifier, V₁ = 10
Voltage gain of the second amplifier; V₂ = 20
Input signal voltage, V_i = 0.01V
Output AC signal voltage = V₀
The total voltage gain of a two-stage cascaded amplifier is given by the product of voltage gains of both the stages, Le.,
V = V₁ x V₂=10 x 20 =200
We have the relation
V₀ =V x V_i= 200 x 0.01 = 2 V
Therefore, the output AC signal of the given amplifier is 2 V.

Question 11.
A p-n photodiode is fabricated from a semiconductor with bandgap of 2.8 eV. Can it detect a wavelength of 6000 nm?
Answer:
Energy band gap of the given photodiode, E_g = 2.8 eV
Wavelength, λ = 6000 nm= 6000 x 10^-9m
The energy of a signal is given by the relation
E = \(\frac{h c}{\lambda}\)
where, h = Planck’s constant = 6.63 x 10^-34 Js
c = Speed of light = 3 x 10⁸ m/s
E = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{6000 \times 10^{-9}}\) = 3.313 x 10^-20 J
But 1.6 x 10^-19 J = 1 eV
E = \(\frac{3.313 \times 10^{-20}}{1.6 \times 10^{-19}}\) = 0.207 eV

Additional Exercises

Question 12.
The number of silicon atoms per m³ is 5 x 10²⁸.
This is doped simultaneously with 5 x 10²² atoms per m³ of Arsenic and 5 x 10²⁰ per m³ atoms of Indium.
Calculate the number of electrons and holes. Given that n_i= 1.5 x 10¹⁶m^-3. Is the material n-type or p-type?
Answer:
Arsenic is n-type impurity and indium is p-type impurity.
Number of electrons, n_e = n_D – n_A
= 5 x 10²² – 5 x 10²⁰ = 4.95 x10²² m^-3
Also, n_i = n_en_h
Given, n_i =1.5 x 10¹⁶ m^-3
Number of holes, n_h=\(\frac{n_{i}^{2}}{n_{e}}\) = \(\frac{\left(1.5 \times 10^{16}\right)^{2}}{4.95 \times 10^{22}} \)
⇒ n_h =4.54 x 10⁹ m^-3
As n_e > n_h. so the material is an n-type semiconductor.

Question 13.
In an intrinsic semiconductor, the energy gap E_g is 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600K and that at 300K?  Assume that the temperature dependence of intrinsic carrier concentration n_i is given by n_i = n₀ exp\(\left[-\frac{E_{g}}{2 k_{B} T}\right]\) where n₀ is a constant.
Answer:
Energy gap of the given intrinsic semiconductor, E = 1.2 eV
The temperature dependence of the intrinsic carrier-concentration is written as
n_i = n₀ exp\(\left[-\frac{E_{g}}{2 k_{B} T}\right]\)

where, k_B = Boltzmann constant = 8.62 x 10^-5eV/K
T = Temperature, n₀ = Constant
Initial temperature, T₁ = 300 K
The intrinsic carrier-concentration at this temperature can be written as
n_i1 =n₀exp \(\left[-\frac{E_{g}}{2 k_{B} \times 300}\right]\)
………………………….. (1)
Final temperature, T₂ = 600 K
The intrinsic carrier-concentration at this temperature can be written as
n_i1 = n₀exp \(\left[-\frac{E_{g}}{2 k_{B} \times 600}\right]\)
………………………………… (2)

The ratio between the conductivities at 600 K and at 300 K is equal to the ratio between the respective intrinsic carrier- concentrations at these temperatures.

Therefore, the ratio between the conductivities is 1.09 x 10⁵.

Question 14.
In a p.n junction diode, the current I can be expressed as
I = I₀exp\(\left(\frac{\mathrm{eV}}{2 k_{B} T}-1\right)\) where I₀ is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, k_B is the Boltzmann constant (8.6 x 10^-5 eV/K) and T is the absolute temperature. if for a given diode I₀ = 5 x 10^-12 A and T = 300K, then
(a) What will be the forward current at a forward voltage of 0.6V?
(b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V?
(c) What is the dynamic resistance?
(d) What will be the current if reverse bias voltage changes from 1V to 2 V?
Answer:
Given, I₀ = 5 x 10^-12 A, T = 300 K
k_B =8.6 x 10^-5 eV/K
= 8.6 x 10^-5 x 1.6 x 10^-19J/K
(a) Given, voltage V = 0.6 V
∴ \(\frac{e V}{k_{B} T}=\frac{1.6 \times 10^{-19} \times 0.6}{8.6 \times 10^{-5} \times 1.6 \times 10^{-19} \times 300}\) = 23.26
The current I through a junction diode is given by

(b) Given, voltage V = 0.7 V
∴ \( \frac{e V}{k_{B} T}=\frac{1.6 \times 10^{-19} \times 0.7}{8.6 \times 10^{-5} \times 1.6 \times 10^{-19} \times 300}\) = 27.14
Now,

Change in current ΔI = 3.035 — 0.063 = 2.9 A
(c) ΔI =2.9 A, voltage ΔV=0.7—0.6=0.1 V
Dynamic resistance R_d =\(\frac{\Delta V}{\Delta I}=\frac{0.1}{2.9}\) = 0.0336 Ω
(d) As the voltage changes from 1 V to 2 V, the current I will be almost
equal to I₀ = 5 x 10^-12 A.
It is due to that the diode possesses practically infinite resistance in the reverse bias.

Question 15.
You are given the two circuits as shown in Fig. 14.44. Show that
circuit (a) acts as OR gate while the circuit (b) acts as AND gate.

Answer:
(a) A and B are the inputs and Y is the output of the given circuit. The left half of the given figure acts as the NOR Gate, while the right half acts as the NOT Gate. This is shown in the following figure

Hence, the output of the NOR Gate = \(\overline{A+B}\)
This will be the input for the NOT Gate. Its output will be \(\overline{\overline{A+B}}\) = A+B
∴ Y = A + B
Hence, this circuit functions as an OR Gate.

(b) A and B are the inputs and Y is the output of the given circuit. It can be observed from the following figure that the inputs of the right half NOR Gate are the outputs of the two NOT Gates.

Hence, the output of the given circuit can be written as
Y = \(\overline{\bar{A}+\bar{B}}=\overline{\bar{A}} \cdot \overline{\bar{B}}= \) = A.B
Hence, the circuit functions as an AND Gate.

Question 16.
Write the truth table for a NAND gate connected as given in Fig. 14.45. Hence identify the exact logic operation carried out by this circuit.

Answer:
A acts as the two inputs of the NAND gate and Y is the output, as shown in the following figure

Hence, the output can be written as
Y =\(\overline{A \cdot A}=\bar{A}+\bar{A}=\bar{A}\) ……………………… (1)
The truth table for equation (1) can be drawn as

A	Y = \(\bar{A}\)
0	1
1	0

This circuit functions as a NOT gate. The symbol for this logic circuit is shown as

Question 17.
You are given two circuits as shown in Fig. 14.46, which consist of NAND gates. Identify the logic operation carried out by the two circuits.

Answer:
In both the given circuits, A and B are the inputs and Y is the output
(a) The output of the left NAND gate will be A. B, as shown in the following figure

Hence, the output of the combination of the two NAND gates is given as
Y= \(\overline{(\overline{A \cdot B}) \cdot(\overline{A \cdot B})}=\overline{\overline{A B}}+\overline{\overline{A B}}\) =AB
Hence, this circuit functions as an AND gate.

(b) \(\bar{A}\) is the output of the upper half of the NAND gate and B is the output of the lower half of the NAND gate, as shown in the following figure

Hence, the output of the combination of theNAND gates will be given as
Y = \(\bar{A} \cdot \bar{B}=\overline{\bar{A}}+\overline{\bar{B}}\) =A + B
Hence, this circuit functions as an OR gate.

Question 18.
Write the truth table for circuit given in Fig 14.47 below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.

(Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence Y = 1. Similarly, work out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND, NOT gates and find the correct one.)
Answer:
A and B are the mputs of the given circuit The output of the first NOR gate ’ is \(\overline{A+B}\) .
It can be observed from the following figure that the inputs of the second NOR gate become the out put of the first one

Hence, the output of the combination is given as
Y= \(\overline{\overline{A+B}+\overline{A+B}}\)
= \(\overline{\bar{A}+\bar{B}}+\overline{\bar{A} \cdot \bar{B}}\)
= \(\overline{\bar{A} \cdot \bar{B}}=\overline{\bar{A}}+\overline{\bar{B}}\) = A+B
The truth table for this operation is given as

A	B	Y(= A+B)
0	0	0
0	1	1
1	0	1
1	1	1

This is the truth table of an OR gate. Hence, this circuit functions as an OR gate.

Question 19.
Write the truth table for the circuits given in Fig.14.48 consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.

Answer:
(a) A acts as the two inputs of the NOR gate and Y is the output, as shown in the following figure.
Hence, the output of the circuit is \(\overline{A+A}\)

The truth table for the same is given as

A	Y = \(\bar{A}\)
0	1
1	0

This is the truth table of a NOT gate. Hence, this circuit functions as a NOT gate.

(b) A and B are the inputs and Y is the output of the given circuit. By using the result obtained in solution
(a), we can infer that the outputs of the
first two NOR gates are \(\bar{A}\) and \(\bar{B}\), as shown in the following figure

\(\bar{A}\) and \(\bar{B}\) are the inputs for the last NOR gate.
Hence, the output for the circuit can be written as
Y = \(\overline{\bar{A}+\bar{B}}=\overline{\bar{A}} \cdot \overline{\bar{B}}\) = A.B
The truth table for the same can be written as

A	B	Y(=A.B)
o	o	o
o	1	o
1	0	0
1	1	1

This is the truth table of an ANL) gate. Hence, this circuit functions as an AND gate.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 11 Biotechnology: Principles and Processes Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

PSEB 12th Class Biology Guide Biotechnology: Principles and Processes Textbook Questions and Answers
Question 1.
Can you list 10 recombinant proteins which are used in medical ; practice? Find out where they are used as therapeutics (use the internet).
Answer:

Recombinant proteins	Therapeutic uses
(a) Insulin	Used for diabetes mellitus
(b) OKT-3	Therapeutic antibody, used for reversal
(c) DNase	Treatment of cystic fibrosis
(d) Reo Pro	Prevention of blood clots
(e) Blood clotting factor VIII	Treatment of haemophilia A
(f) Blood clotting factor IX	Treatment of haemophilia B
(g) Tissue plasminogen activator	For acute myocardial infarction
(h) Interferon alpha (INF alpha)	Used for hepatitis C
(i) Interferon beta (INF beta)	Used for multiple sclerosis
(j) Interferon gamma (INF gamma)	Used for granulomatous disease

Question 2.
Make a chart (with diagrammatic representation) showing a restriction enzyme, the substrate DNA on which it acts, the site at which it cuts DNA and the product it produces.
Answer:

Question 3.
From what you have learnt, can you tell whether enzymes are bigger or DNA is bigger in molecular size? How did you know?
Answer:
The molecular size of DNA molecules is more than that of enzymes. It is because an enzyme (protein) is synthesised from a segment of DNA called the gene.

Question 4.
What would be the molar concentration of human DNA in a human cell? Consult your teacher.
Answer:
The molar concentration of human DNA in a human diploid cell is as follows:
Total number of chromosomes × 6.023 × 10²³
= 46 × 6.023 × 10²³
= 2.77 × 1018moles

Question 5.
Do eukaryotic cells have restriction endonucleases? Justify your answer.
Answer:
No. Eukaryotic cells do not have restriction endonucleases. All the restriction endonucleases have been isolated from the various strains of bacteria and they are also named according to the genus and species of prokaryotes. The first letter of the enzyme comes from the genus and the second two letters come from the species of the prokaryotic cell from which they have been isolated. For example, EcoRI comes from Escherichia coliRY 13. In EcoRI, the letter ‘R’ is derived from the name of the strain. Roman numbers following the names indicate the order in which the enzymes were isolated from that strain of bacteria.

Question 6.
Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors have over shake flasks?
Answer:
Shake flasks are used for growing microbes and mixing the desired materials on a small scale in the laboratory. However, the large-scale production of a desired biotechnological product requires large stirred tank bioreactors.

Besides better aeration and mixing properties, bioreactors have the following advantages:

• It has an oxygen delivery system.
• It has a foam control, temperature and pH control system.
• Small volumes of culture can be withdrawn periodically.

Question 7.
Collect 5 examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following base-pair rules.
Answer:
Some palindromic DNA sequences and the restriction enzymes which act on them are as follows :

• 5′-AGCT-3′ Alul (Arthrobacter luteus)
  3′-TCGA-5′
• 5′-GAATTC-3′ EcoRI (Escherichia coli)
  3′-CTTAAG-5′
• 5′-AAGCTT-3′ Hindlil (Haemophilus influenzae)
  3′-TTCGAA-5′
• 5′-GTCGAC-3′ Sail (Streptomyces albus)
  3′-CAGCTG-5′
• 5′-CTGCAG-3′ PstI (Providencia stuartii)
  3′-GACGTC-5′

Question 8.
Can you recall meiosis and indicate at what stage a recombinant DNA is made?
Answer:
A recombinant DNA is made in the pachytene stage of prophase I by f crossing over.

Question 9.
Can you think and answer how a reporter enzyme can be used to
monitor transformation of host cells by foreign DNA in addition to a selectable marker?
Answer:
Reporter enzyme can differentiate recombinants from non-recombinants on the basis of their ability to produce a specific colour in the presence of a chromogenic substrate. DNA is inserted within the coding sequence of the enzyme p-galactosidase. This results into inactivation of the enzyme which is referred to as insertional inactivation.

The presence of a chromogenic substrate gives blue-coloured colonies if the plasmid in the bacteria does not have an insert. The presence of the insert results in insertional inactivation of α-galactosidase and the colonies do not produce any colour. These are identified as recombinant colonies.

Question 10.
Describe briefly the following:
(a) Origin of replication
(b) Bioreactors
(c) Downstream processing
Answer:
(a) Origin of Replication: This is a sequence from where replication starts and any piece of DNA when linked to this sequence can be made to replicate within the host cells. This sequence is also responsible for controlling the copy number of the linked DNA. So, if one wants to recover many copies of the target DNA it should be cloned in a vector whose origin supports a high copy number.

(b) Bioreactors: Bioreactors are vessels in which raw materials are biologically converted into specific products, individual enzymes etc. using microbial plant, animal or human cells. A bioreactor provides the optimal conditions for achieving the desired product by providing optimum growth conditions (temperature, pH, substrate, salts, vitamins, oxygen). The most commonly used bioreactors are of stirring type. A biogas plant is a good example of a bioreactor.

(c) Downstream Processing: After completion of the biosynthetic stage, the product is subjected through a series of processes before it is ready for marketing as a finished product. The processes include separation and purification, which are collectively referred to as downstream processing. The product has to be formulated with suitable preservatives. Such formulation has to undergo thorough clinical trials as in the case of drugs. Strict quality control testing for each product is also required. Downstream processing and quality control testing vary from product to product.

Question 11.
Explain briefly
(a) PCR
(b) Restriction enzymes and DNA
(c) Chitinase
Answer:
(a) PCR: PCR stands for Polymerase Chain Reaction. In this reaction multiple copies of the gene (or DNA) of interest is synthesised in vitro using two sets of primers and the enzyme DNA polymerase. The enzyme extends the primers using the nucleotides provided in the reaction and the genomic DNA as template. The segment of DNA can be amplified to approximately billion times if the process of replication of DNA is repeated many times.

(b) Restriction Enzymes and DNA: Restriction enzymes are synthesised by microbes as a defence mechanism and are specifically endonucleases which cleave the double-stranded DNA with the desired genes. This activity occurs at a limited number of sites depending on the number of recognition sequences in DNA. Lysing enzymes, synthesising enzymes (DNA polymerase and reverse transcriptase) and ligases are also tools of genetic engineering.

(c) Chitinase: During the isolation of DNA in the processes of recombinant DNA technology, the fungal cell is heated with an enzyme called chitinase. The chitinase enzyme dissolves the chitin membrane to open the cell for release of DNA along with other macromolecules such as RNA, proteins, polysaccharides and lipids.

Question 12.
Discuss with your teacher and find out how to distinguish between
(a) Plasmid DNA and Chromosomal DNA
(b) RNA and DNA
(c) Exonuclease and Endonuclease
Answer:
(a)

Plasmid DNA	Chromosomal DNA
Plasmid DNA is the naked double-stranded DNA which forms a circle with no free ends. It is associated with few proteins but contains RNA polymerase enzyme. They are smaller than the host chromosomes and can be easily separated.	Chromosomal DNA is a double-stranded linear DNA molecule associated with large proteins. This DNA exists in relaxed and supercoiled forms and provides a template for replication and transcription. It has free ends represented as 3′-5′.

(b)

DNA	RNA
(i) It is mainly confined to the nucleus. A small quantity occurs in mitochondria and chloroplasts.	It mainly occurs in the cytoplasm. A small quantity is found in the nucleus.
(ii) Its quantity is constant in each cell of a species.	Its quantity varies in different cells.
(iii) It contains deoxyribose sugar.	It contains ribose sugar.
(iv) Its pyrimidines are adenine and thymine.	Its pyrimidines are adenine and uracil.
(v) The amount of adenine is equal to the amount of thymine. Also, the amount of cytosine is equal to the amount of guanine.	Adenine and uracil are not necessarily in equal amounts, nor are cytosine and guanine necessarily in equal amounts.
(vi) It can replicate itself.	It cannot replicate itself. It is formed by DNA. Some RNA viruses (paramyxo virus) can produce RNA from an RNA template.

(c) Exonucleases are nucleases which cut off the nucleotides from the 5′ or 3′ ends of a DNA molecule, whereas endonucleases are nucleases which cleave the DNA duplex at any point except at the ends.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 10 Microbes in Human Welfare Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 10 Microbes in Human Welfare

PSEB 12th Class Biology Guide Microbes in Human Welfare Textbook Questions and Answers

Question 1.
Bacteria cannot be seen with the naked eyes, but these can be seen with the help of a microscope. If you have to carry a sample from your home to your biology laboratory to demonstrate the presence of microbes under a microscope, which sample would you carry and why?
Answer:
Curd can be used as a sample for the study of microbes. Curd contains numerous lactic acid bacteria (LAB) or Lactobacillus. These bacteria produce acids that coagulate and digest milk proteins. A small drop of curd contains millions of bacteria, which can be easily observed under a microscope.

Question 2.
Give examples to prove that microbes release gases during metabolism.
Answer:
The examples of microbes that release gases during metabolism are as follows :
(a) Bacteria and fungi carry out the process of fermentation and during this process, they release carbon dioxide. Fermentation is the process of converting a complex organic substance into a simpler substance with the action of bacteria or yeast. Fermentation of sugar produces alcohol with the release of carbon dioxide and very little energy.

(b) The dough used for making idli and dosa gives a puffed appearance. This is because of the action of bacteria which releases carbon dioxide. This CO₂ released from the dough gets trapped in the dough, thereby giving it a puffed appearance.

Question 3.
In which food would you find lactic acid bacteria? Mention some of their useful applications.
Answer:
Lactic acid bacteria can be found in curd. It is this bacterium that promotes the formation of milk into curd. The bacterium multiplies and increases its number, which converts the milk into curd. They also increase the content of vitamin B₁₂ in curd.
Lactic acid bacteria are also found in our stomach where it keeps a check on the disease-causing micro-organisms.

Question 4.
Name some traditional Indian foods made of wheat, rice and Bengal gram (or their products) which involve use of microbes.
Answer:

1. Wheat Product: Bread, cake etc.
2. Rice Product: Idli, dosa etc.
3. Bengal Gram Product: Dhokla, khandvi etc.

Question 5.
In which way have microbes played a m^jor role in controlling diseases caused by harmful bacteria?
Answer:
Several micro-organisms are used for preparing medicines. Antibiotics are medicines produced by certain micro-organisms to kill other disease-causing micro-organisms. These medicines are commonly obtained from bacteria and fungi. They either kill or stop the growth of disease-causing micro-organisms. Streptomycin, tetracycline, and penicillin are common antibiotics. Penicillium notatum produces chemical penicillin, which checks the growth of Staphylococci bacteria in the body. Antibiotics are designed to destroy bacteria by weakening their cell walls. As a result of this weakening, certain immune cells such as the white blood cells enter the bacterial cell and cause cell lysis. Cell lysis is the process of destroying cells such as blood cells and bacteria.

Question 6.
Name any two species of fungus, which are used in the production of the antibiotics.
Answer:
Antibiotics are medicines that are produced by certain micro-organisms to kill other disease-causing micro-organisms. These medicines are commonly obtained from bacteria and fungi.
The species of fungus used in the production of antibiotics are as follows:

Antibiotic	Fungus source
1. Penicillin	Penicillium notatum
2. Cephalosporin	Cephalosporium acremonium

Question 7.
What is sewage? In which way can sewage be harmful to us?
Answer:
Sewage is the municipal waste matter that is carried away in sewers and drains. It includes both liquid and solid wastes, rich in organic matter and microbes. Many of these microbes are pathogenic and can cause several water borne diseases. Sewage water is a major cause of polluting drinking water. Hence, it is essential that sewage water is properly collected, treated, and disposed. If untreated sewage is disposed into rivers and streams, it will pollute the water bodies.

Question 8.
What is the key difference between primary and secondary sewage treatment?
Answer:

Primary sewage treatment	Secondary sewage treatment
1. It is a mechanical process involving the removal of coarse solid materials.	It is a biological process involving the action of microbes.
2. It is inexpensive and relatively less complicated.	It is a very expensive and complicated process.

Question 9.
Do you think microbes can also be used as source of energy? If yes, how?
Answer:
Yes, microbes can be used as a source of energy. Bacteria such as Methane bacterium is used for the generation of gobar gas or biogas.

The generation of biogas is an anaerobic process in a biogas plant, which consists of a concrete tank (10-15 feet deep) with sufficient outlets and inlets. The dung is mixed with water to form the slurry and thrown into the tank. The digester of the tank is filled with numerous anaerobic methane-producing bacteria, which produce biogas from the slurry. Biogas can be removed through the pipe which is then used as a source of energy, while the spent slurry is removed from the outlet and is used as a fertiliser.

Question 10.
Microbes can be used to decrease the use of chemical fertilisers and pesticides. Explain how this can be accomplished.
Answer:
Microbes play an important role in organic farming, which is done without the use of chemical fertilisers and pesticides. Bio-fertilisers are living organisms which help increase the fertility of soil. It involves the ’ selection of beneficial micro-organisms that help in improving plant growth through the supply of plant nutrients. Bio-fertilisers are introduced in seeds, roots, or soil to mobilise the availability of nutrients. Thus, they are extremely beneficial in enriching the soil with organic nutrients. Many species of bacteria and cyanobacteria have the ability to fix free atmospheric nitrogen. Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. Azpirillum and Azotobacter are free living nitrogen-fixing bacteria whereas Anabaena, Nostoc and Oscillatoria are examples of nitrogen-fixing cyanobacteria. Bio-fertilisers are cost effective and eco-friendly.

Microbes can also act as bio-pesticides to control insect pests in plants.
An example of bio-pesticides is Bacillus thuringiensis, which produces a toxin that kills the insect pests. Dried bacterial spores are mixed in water and sprayed in agricultural fields. When larvae of insects feed on crops, these bacterial spores enter the gut of the larvae and release r toxins, thereby it. Similarly, Trichoderma are free living fungi. They live in the roots of higher plants and protect them from various pathogens.

Baculoviruses is another bio-pesticide that is used as a biological control I agent against insects and other arthropods.

Question 11.
Three water samples namely river water, untreated sewage water and secondary effluent discharged from a sewage treatment plant were subjected to BOD test. The samples were labelled A, B and C; but the laboratory attendant did not note which was which. The BOD values of the three samples A, B and C were recorded as 20mg/L, 8mg/L and 400mg/L, respectively. Which sample of the water is most polluted? Can you assign the correct label to each assuming the river water is relatively clean?
Answer:
Biological oxygen demand (BOD) is the method of determining the amount of oxygen required by micro-organisms to decompose the waste present in the water supply. If the quantity of organic wastes in the water supply is high, then the number of decomposing bacteria present in the water will also be high. As a result, the BOD value will increase. Therefore, it can be concluded that if the water supply is more polluted, then it will have a higher BOD value. Out of the above three samples, sample C is the most polluted since it has the maximum BOD value of 400 mg/L.

After untreated sewage water, secondary effluent discharge from a sewage treatment plant is most polluted. Thus, sample A is secondary effluent discharge from a sewage treatment plant and has the BOD value of 20 mg/L, while sample B is river water and has the BOD value of 8 mg/L.
Hence, the correct label for each sample is:

Label	BOD value	Sample
A.	20 mg/L	Secondary effluent discharge from a sewage treatment plant
B.	8 mg/L	River water
C.	400 mg/L	Untreated sewage water

Question 12.
Find out the name of the microbes from which Cyclosporin A (an immunosuppressive drug) and Statins (blood cholesterol lowering agents) are obtained.
Answer:

Drug	Function	Microbe
1. Cyclosporin-A	Immuno suppressive drug	Trichoderma polysporum
2. Statin	Blood cholesterol lowering agent	Monascus purpureus

Question 13.
Find out the role of microbes in the following and discuss it with your teacher.
(a) Single cell protein (SCP)
(b) Soil
Answer:
(a) Single Cell Protein (SCP): A single cell protein is a protein obtained from certain microbes, which forms an alternate source of proteins in animal feeds. The microbes involved in the preparation of single cell proteins are algae, yeast, or bacteria. These microbes are grown on an industrial scale to obtain the desired protein. For example, Spirulina can be grown on waste materials obtained from molasses, sewage, and animal manures. It serves as a rich supplement of dietary nutrients such as proteins, carbohydrate, fats, minerals, and vitamins. Similarly, micro-organisms such as Methylophilus and methylotrophus have a large rate of biomass production. Their growth can produce a large amount of proteins.

(b) Soil: Microbes play an important role in maintaining soil fertility. They help in the formation of nutrient-rich humus by the process of decomposition. Many species of bacteria and cyanobacteria have the ability to fix atmospheric nitrogen into usable form. Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. Azospirillum and Azotobacter are free living nitrogen-fixing bacteria, whereas Anabaena, Nostoc, and Oscillatoria are examples of nitrogen-fixing cyanobacteria.

Question 14.
Arrange the following in the decreasing order (most important first) of their importance, for the welfare of human society. Give reasons for your answer.
Biogas, Citric acid, Penicillin and Curd
Answer:
The order of arrangement of products according to their decreasing importance is :
Penicillin > Biogas > Citric acid > Curd
Penicillin is the most important product for the welfare of human society. It is an antibiotic, which is used for controlling various bacterial diseases. The second most important product is biogas. It is aneco-friendly source of energy. The next important product is citric acid, which is used as a food preservative. The least important product is curd, a food item obtained by the action of Lactobacillus bacteria on milk.

Question 15.
How do biofertilisers enrich the fertility of the soil?
Answer:
Bio-fertilisers are living organisms which help in increasing the fertility of soil. It involves the selection of beneficial micro-organisms that help in improving plant growth through the supply of plant nutrients. These are introduced to seeds, roots, or soil to mobilise the availability of nutrients by their biological activity. Thus, they are extremely beneficial jf in enriching the soil with organic nutrients. Many species of bacteria and cyanobacteria have the ability to fix free atmospheric nitrogen. Rhizobium is a symbiotic bacteria found in the root nodules of leguminous plants. Azospirillum and Azotobacter are free living nitrogen-fixing bacteria, whereas Anabaena, Nostoc, and Oscillatoria are examples of nitrogen-fixing cyanobacteria. Bio-fertilisers are cost effective and eco-friendly.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 9 Strategies for Enhancement in Food Production Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 9 Strategies for Enhancement in Food Production

PSEB 12th Class Biology Guide Strategies for Enhancement in Food Production Textbook Questions and Answers

Question 1.
Explain in brief the role of animal husbandry in human welfare.
Answer:
Animal husbandry deals with the scientific management of livestock. It includes various aspects such as feeding, breeding, and control diseases to raise the population of animal livestock. Animal husbandry usually includes animals such as cattle, pig, sheep, poultry, honeybee, silkworm and fish which are useful for humans in various ways. These animals are managed for the production of commercially important products such as milk, meat, wool, egg, honey, silk, etc. The increase in human population has increased the demand of these products. Hence, it is necessary to improve the management of livestock scientifically.

Question 2.
If your family owned a dairy farm, what measures would you undertake to improve the quality and quantity of milk production?
Answer:
Dairy farm management deals with processes which aim at improving the quality and quantity of milk production. Milk production is primarily dependent on choosing improved cattle breeds, provision of proper feed for cattle, maintaining proper shelter facilities, and regular cleaning of cattle.

Choosing improved cattle breeds is an important factor of cattle management. Hybrid cattle breeds are produced for improved productivity. Therefore, it is essential that hybrid cattle breeds should have a combination of various desirable genes such as high milk production and high resistance to diseases. Cattle should also be given healthy and nutritious food consisting of roughage, fibre concentrates, and high levels of proteins and other nutrients.

Cattle should be housed in proper cattle-houses and should be kept in well ventilated roofs to prevent them from harsh weather conditions such as heat, cold, and rain. Regular baths and proper brushing should be ensured to control diseases. Also, time-to-time check ups by a veterinary doctor for symptoms of various diseases should be undertaken.

Question 3.
What is meant by the term ‘breed’? What are the objectives of animal breeding?
Answer:
A breed is a special variety of animals within a species. It is similar in most characters such as general appearance, size, configuration, and features with other members of the same species. Jersey and Brown
Swiss are examples of foreign breeds of cattle. These two varieties of cattle have the ability to produce abundant quantities of milk. This milk is very nutritious with high protein content.
Objectives of animal breeding are as follows:

• To increase thd yield of animals.
• To improve the desirable qualities of the animal produce.
• To produce disease-resistant varieties of animals.

Question 4.
Name the methods employed in animal breeding. According to you which of the methods is best? Why?
Answer:
Animal breeding is the method of mating closely related individuals. There are several methods employed in animal breeding, which can be classified into the following categories:

(A) Natural Methods of Breeding Include Inbreeding and Out-breeding : Breeding between animals of the same breed is known as inbreeding, while breeding between animals of different breeds is known as out-breeding. Out-breeding of animals is of three types:
(a) Out-crossing: In this type of out-breeding, the mating of animals occurs within the same breed. Thus, they have no common ancestors up to the last 4-5 generations.

(b) Cross-breeding: In this type of out-breeding, the mating occurs between different breeds of the same species, thereby producing a hybrid.

(c) Interspecific hybridisation: In this type of out-breeding, the mating occurs between different species.

(B) Artificial Methods of Breeding Include Modern Techniques of Breeding : It involves controlled breeding experiments, which are of two types:
(a) Artificial insemination : It is a process of introducing the semen (collected from the male) into the oviduct or the uterus of the female body by the breeder. This method of breeding helps the breeder overcome certain problems faced in abnormal mating.

(b) Multiple ovulation embryo technology (MOET) : It is a technique for cattle improvement in which super-ovulation is induced by a hormone injection. Then, fertilisation is achieved by artificial insemination and early embryos are collected. Each of these embryos are then transplanted into the surrogate mother for further development of the embryo.

The best method to carry out animal breeding is the artificial method of breeding, which includes artificial insemination and MOET technology. These technologies are scientific in nature. They help overcome problems of normal mating and have a high success rate of crossing between mature males and females. Also, it ensures the production of hybrids with the desired qualities. This method is highly economical as a small amount of semen from the male can be used to inseminate several cattle.

Question 5.
What is apiculture? How is it important in our lives?
Answer:
Apiculture is the practise of bee-keeping for the production of various products such as honey, bee’s wax, etc. Honey is a highly nutritious food source and is used as an indigenous system of medicines. It is
useful in the treatment of many disorders such as cold, flu, and 1 dysentery. Other commercial products obtained from honey bees include bee’s wax and bee pollen. Bee’s wax is used for making cosmetics, polishes, and is even used in several medicinal preparations. Therefore, to meet the increasing demand of honey, people have started practicing bee-keeping on a large scale. It has become an f income generating activity for farmers since it requires a low investment and is labour intensive.

Question 6.
Discuss the role of fishery in enhancement of food production.
Answer:
Fishery is an industry which deals with catching, processing, and marketing of fishes and other aquatic animals that have a high economic value. Some commercially important aquatic animals are prawns, crabs, oysters, lobsters, and octopus. Fisheries play an important role in the Indian economy. This is because a large part of the Indian population is dependent on fishes as a source of food, which is both cheap and high in animal protein. Fishery is an employment generating industry especially for people staying in the coastal areas. Both fresh water fishes (such as Catla, Rohu, etc.) and marine fishes (such as tuna, mackerel, pomfret, etc.) are of high economic value.

Question 7.
Briefly describe various steps involved in plant breeding.
Answer:
Plant breeding is the process in which two genetically dissimilar varieties are purposely crossed to produce a new hybrid variety. As a result, characteristics from both parents can be obtained in the hybrid plant variety. Thus, it involves the production of a new variety with the desired characteristics such as resistance to diseases, climatic adaptability, and better productivity. The various steps involved in plant breeding are as follows:

(a) Collection of Genetic Variability: Genetic variability from various wild relatives of the cultivated species are collected to maintain the genetic diversity of a species. The entire collection of the diverse alleles of a gene in a crop is called the germplasm collection.

(b) Evaluation of Germplasm and Selection of Parents: The germplasm collected is then evaluated for the desirable genes. The selected plants with the desired genes are then used as parents in plant breeding experiments and are multiplied by the process of hybridisation.

(c) Cross-hybridisation between Selected Parents: The next step in plant breeding is to combine the desirable characters present in two different parents to produce hybrids. It is a tedious job as one has to ensure that the pollen grains collected from the male parent reach the stigma of the female parent.

(d) Selection of Superior Hybrids: The progenies of the hybrids having the desired characteristics are selected through scientific evaluation. The selected progenies are then self-pollinated for several generations to ensure homozygosity.

(e) Testing, Release, and Commercialisation of New Cultivars: The selected progenies are evaluated for characters such as yield, resistance to diseases, performance, etc. by growing them in research fields for at least three growing seasons in different parts of the country. After thorough testing and evaluation, the selected varieties are given to the farmers for growing in fields for a large-scale production.

Question 8.
Explain what is meant by biofortification.
Answer:
Biofortification is a process of breeding crops with higher levels of vitamins, minerals, proteins, and fat content. This method is employed to improve public health. Breeding of crops with improved nutritional quality is undertaken to improve the content of proteins, oil, vitamins, minerals, and micro-nutrients in crops. It is also undertaken to upgrade the quality of oil and proteins. An example of this is a wheat variety known as Atlas 66, which has high protein content in comparison to the existing wheat. In addition, there are several other improved varieties of crop plants such as rice, carrots, spinach etc. which have more nutritious value and more nutrients than the existing varieties.

Question 9.
Which part of the plant is best suited for making virus-free plants and why?
Answer:
Apical and axillary meristems of plants is used for making virus-free plants. In a diseased plant, only this region is not infected by the virus as compared to the rest of the plant region. Hence, the scientists remove axillary and apical meristems of the diseased plant and grow it in vitro to obtain a disease-free and healthy plant.
Virus-free plants of banana, sugarcane, and potato have been obtained using this method by scientists.

Question 10.
What is the major advantage of producing plants by micropropagation?
Answer:
Micropropagation is a method of producing new plants in a short duration using plant tissue culture.
Some major advantages of micropropagation are as follows:

• Micropropagation helps in the propagation of a large number of plants in a short span of time.
• The plants produced are identical to the mother plant.
• It leads to the production of healthier plantlets, which exhibit better disease-resisting powers.

Question 11.
Find out what the various components of the medium used for propagation of an explant in vitro are?
Answer:
The major components of medium used for propagation of explants in vitro are carbon sources such as sucrose, inorganic salts, vitamins, amino acids, water, agar-agar, and certain growth hormones such as auxins, gibberellins and cytokinins etc.

Question 12.
Name any five hybrid varieties of crop plants which have been developed in India.
Answer:
The five hybrid varieties of crop plants which have been developed in India are as follow:

Crop plant	Hybrid variety
Wheat	Sonalika and Kalyan Sona
Rice	Jaya and Ratna
Cauliflower	Pusa Shubhra and Pusa Snowball K-1
Cowpea	Pusa Komal
Mustard	Pusa Swarnim