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Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 1 Electric Charges and Fields. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 1 Important Extra Questions Electric Charges and Fields

Electric Charges and Fields Important Extra Questions Very Short Answer Type

Question 1.
What is the value of the angle between the vectors \(\vec{p}\) and \(\vec{E}\) for which the potential energy of an electric dipole of dipole moment \(\vec{p}\), kept in an external electric field \(\vec{E}\), has maximum value.
Answer:
P.E. = –pEcos θ
P.E. is maximum when cos θ = – 1, i.e.
θ = 180°

Question 2.
Define electric field intensity at a point.
Answer:
Electric field intensity at a point is defined as the force experienced by a unit test charge placed at that point. Mathematically
we have

Question 3.
Two equal point charges separated by 1 m distance experience force of 8 N. What will be the force experienced by them, if they are held in water, at the same distance? (Given: K_water = 80) (CBSE Al 2011C)
Answer:
The force in water is given by
F_w = \(\frac{F_{\text {air }}}{K}=\frac{8}{80}\) = 0.1 N

Question 4.
A charge ‘q’ is placed at the center of a cube of side l. What is the electric flux passing through each face of the cube? (CBSE AI 2012) (CBSE Sample Paper 2019)
Answer:
Φ = q/6ε₀

Question 5.
Why do the electric field lines not form closed loops? (CBSE Al 2012C)
Answer:
It is due to the conservative nature of the electric field.

Question 6.
Two equal balls having equal positive charge ‘q’ coulomb are suspended by two insulating strings of equal length. What would be the effect on the force when a plastic sheet is inserted between the two? (CBSE AI 2014)
Answer:
It decreases because force ∝= \(\frac{1}{k}\) and k > 1.

Question 7.
What is the electric flux through a cube of side l cm which encloses an electric dipole? (CBSE Delhi 2015)
Answer:
Zero

Question 8.
Why are electric field lines perpendicular at a point on an equipotential surface of a conductor? (CBSE Al 2015C)
Answer:
So that no net force acts on the charge at the equipotential surface and it remains stationary.

Question 9.
What is the amount of work done in moving a point charge Q. around a circular arc of radius ‘r’ at the centre of which another point charge ‘q’ is located? (CBSE Al 2016)
Answer:
Zero.

Question 10.
How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased? (CBSE Delhi 2016)
Answer:
No change, as flux does not depend upon the size of the Gaussian surface.

Question 11.
Draw the pattern of electric field lines, when a point charge –Q is kept near an uncharged conducting plate. (CBSE Delhi 2019)
Answer:
The pattern is as shown.

Question 12.
Draw a plot showing the variation of the electric field with distance from the center of a solid conducting sphere of radius R, having a charge + Q on its surface. (CBSE Delhi 2017C)
Answer:
The plot is as shown.

Question 13.
Draw a graph to show the variation of E with perpendicular distance r from the line of charge (CBSE Delhi 2018)
A
Answer:
E = \(\frac{\lambda}{2 \pi \varepsilon_{0} r}\)
E ∝ \(\frac{1}{r}\)

The graph between E and r is as shown.

Question 14.
Define the term ‘electric flux’. Write its S.I. unit. (CBSE Delhi 2018)
Answer:
Electric flux: It is the measure of the number of electric field lines crossing a given area normally.

Mathematically the electric flux passing through an area \(\vec{dS}\) is given by
dΦ = \(\vec{E}\) . \(\vec{dS}\)

SI unit of electric flux is Nm²C^-1 or Vm.

Question 15.
Why can the interior of a conductor have no excess charge in the static situation? (CBSE Ai 2019)
Answer:
Since the electric field inside the conductor is zero, electric flux through the closed surface is also zero. Hence by Gauss’s law, there is no net charge enclosed by the closed surface.

Question 16.
Two field lines never cross each other. Why?
Answer:
It is because at the point of intersection two perpendiculars can be drawn. Thus there will be two directions of the electric field at that point which is not possible.

Question 17.
In an electric field, an electron is kept freely. If the electron is replaced by a proton, what will be the force experienced by the proton?
Answer:
The magnitude of force will be the same but the direction will be reversed.

Question 18.
Consider the situation shown in the figure given below. What are the signs of q₁ and q₂?
Answer:

q₁ is negative and q₂ is positive.

Question 19.
In the figure given below, at which point electric field is maximum?
Answer:

The electric field is maximum at point C.

Question 20.
An uncharged insulated conductor A is brought near a charged insulated conductor B. What happens to the charge and potential of B?
Answer:
On bringing uncharged conductor A near a charged conductor B, charges are induced on A as shown in the figure below. As a result, the potential of conductor B is slightly lowered but the charge on it remains unchanged.

Question 21.
In a medium the force of attraction between two point electric charges, distance ‘d’ apart, is F. What distance apart should these be kept in the same medium so that the force between them becomes 3F?
Answer:
Let the new distance be ‘d’, since F ∝\(\frac{1}{r^{2}}\) ,
therefore \(\frac{F}{3 F}=\frac{r^{2}}{d^{2}}=\frac{1}{3} \Rightarrow r=\frac{d}{\sqrt{3}}\)

Question 22.
Find the value of an electric field that would completely balance the weight of an electron.
Answer:
mg = eE ⇒ E = \(\frac{m g}{e}\)
= \(\frac{9.1 \times 10^{-31} \times 9.8}{1.6 \times 10^{-19}}=5.57 \times 10^{-11} \mathrm{Vm}^{-1}\)

Question 23.
Two charges, one +5 µC, and the other -5 µC are placed 1 mm apart. Calculate the electric dipole moment of the system.
Answer:
p = q × 2a = 5 × 10 ^-6 × 10^-3 = 5 × 10^-9 Cm

Question 24.
Two-point charges of+3 µC each are 100 cm apart. At what point on the line joining the charges will the electric intensity be zero?
Answer:
At the mid-point of the line joining the two point charges.

Question 25.
Four charges of came magnitude and same sign are placed at the corners of a square, of each side 0.1 m. What is electric field intensity at the center of the square?
Answer:
Zero.

Question 26.
Why should the electrostatic field be zero ‘ inside a conductor? (CBSE Delhi 2012)
Answer:
Because it does not contain any charge.

Question 27.
A metallic spherical shell has an inner radius R₁ and outer radius R₂. A charge Q is placed at the center of the shell. What will be the surface charge density on the (i) inner surface, and (ii) the outer surface of the shell? (CBSE Delhi 2018)
Answer:
On inner surface

On the outer surface,

Question 28.
An arbitrary surface encloses a dipole. What is the electric flux through this surface? (NCERT Exemplar)
Answer:
Zero.

Electric Charges and Fields Important Extra Questions  Short Answer Type

Question 1.
(a) Electric field inside a conductor is zero. Explain.
(b) The electric field due to a point charge at any point near it is given as
E  =

what is the physical significance of this limit?
Answer:
(a) By Gauss theorem \(\phi \vec{E} \cdot \overrightarrow{d S}=\frac{q}{\varepsilon_{0}}\). Since there is no charge inside a conductor therefore in accordance with the above equation the electric field inside the conductor is zero.
(b) It indicates that the test charge should be infinitesimally small so that it may not disturb the electric field of the source charge.

Question 2.
Define the electric line of force and give its two important properties.
Answer:
It is a line straight or curved, a tangent to which at any point gives the direction of the electric field at that point.
(a) No two field lines can cross, because at the point of intersection two tangents can be drawn giving two directions of the electric field which is not possible.
(b) The field lines are always perpendicular to the surface of a charged conductor.

Question 3.
Draw electric field lines due to (i) two similar charges, (ii) two opposite charges, separated by a small distance.
Answer:
(a) The diagram is as shown.

(b) The diagram is as shown.

Question 4.
An electric dipole is free to move in a uniform electric field. Explain what is the force and torque acting on it when it is placed
(i) parallel to the field
Answer:
When an electric dipole is placed parallel to a uniform electric field, net force, as well as net torque acting on the dipole, is zero and, thus, the dipole remains in equilibrium.

(ii) perpendicular to the field
Answer:
When the dipole is placed perpendicular to the field, two forces acting on the dipole form a couple, and hence a torque acts on it which aligns its dipole along the direction of the electric field.

Question 5.
A small metal sphere carrying charge +Q. is located at the center of a spherical cavity in a large uncharged metallic spherical shell. Write the charges on the inner and outer surfaces of the shell. Write the expression for the electric field at the point P₁(CBSE Delhi 2014C)
Answer:

Charge on inner surface – Q.
Charge on outer surface + Q,
Electric field at point P = E = k\(\frac{Q}{r_{1}^{2}}\)

Question 6.
Two-point charges q and –2q are kept ‘d’ distance apart. Find the location of the point relative to charge ‘q’ at which potential due to this system of charges is zero. (CBSE Al 2014C)
Answer:
Let the potential be zero at point P at a distance x from charge q as shown

Now potential at point P is
V = \(\frac{k q}{x}+\frac{k(-2 q)}{d+x}\) = 0

Solving for x we have
x = d

Question 7.
Two small identical electrical dipoles AB and CD, each of dipole moment ‘p’ are kept at an angle of 120° as shown in the figure. What is the resultant dipole moment of this combination? If this system is subjected to the electric field (\(\vec{E}\)) directed along +X direction, what will be the magnitude and direction of the torque acting on this? (CBSE Delhi 2011)

Answer:
The resultant dipole moment of the combi-nation is
P_R = \(\sqrt{p^{2}+p^{2}+2 p^{2} \cos 120^{\circ}}\) = p

since cos 120° = -1/2
This will make an angle of 30° with the X-axis, therefore torque acting on it is
τ=PE sin 30° = pE/2 (Along Z-direction)

Question 8.
A metallic spherical shell has an inner radius R₁ and outer radius R₂. A charge Q is placed at the center of the spherical cavity. What will be surface charge density on (i) the inner surface, and (ii) the outer surface? (NCERT Exemplar)
Answer:
The induction of charges is as shown.

Therefore surface charge density on the inner and the outer shell is on the outer surface is

Question 9.
If the total charge enclosed by a surface is zero, does it imply that the electric field everywhere on the surface is zero? Conversely, if the electric field everywhere on a surface is zero, does it imply that the net charge inside is zero. (NCERT Exemplar)
Answer:
No, the field may be normal to the surface. However, the converse is true i.e. when the electric field everywhere on the surface be zero then the net charge inside it must be zero.

Electric Charges and Fields Important Extra Questions Long Answer Type

Question 1.
(a) State Gauss theorem in electrostatics. Using it, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.
Answer:
It states, “The net electric flux through any Gaussian surface is equal to \(\frac{1}{\varepsilon_{0}}\) times the net electric charge enclosed by the surface.

Mathematically, Φ =\( \phi \vec{E} \cdot d \vec{A}=\frac{q_{i n}}{\varepsilon_{0}}\)

Consider an infinite plane sheet of charge. Let a be the uniform surface charge density, i.e. the charge per unit surface area. From symmetry, we find that the electric field must be perpendicular to the plane of the sheet and that the direction of E on one side of the plane must be opposite to its direction on the other side as shown in the figure below. In such a case let us choose a Gaussian surface in the form of a cylinder with its axis perpendicular to the sheet of charge, with ends of area A.

The charged sheet passes through the middle of the cylinder’s length so that the cylinder’s ends are equidistant from the sheet. The electric field has a normal component at each end of the cylinder and no normal component along the curved surface of the cylinder. As a result, the electric flux is linked with only the ends and not the curved surface.

Therefore by the definition of eLectric flux, the flux Linked with the Gaussian surface is given by
Φ = \(\oint_{A} \vec{E} \cdot \vec{\Delta} A\)
Φ = E_A + E_A = 2E_A … (1)

But by Gauss’s Law
Φ = \(\frac{q}{\varepsilon_{0}}=\frac{\sigma A}{\varepsilon_{0}}\) [∵ q = σA] … (2)

From equations (1) and (2), we have
2E_A = \(\frac{\sigma A}{\varepsilon_{0}}\) … (3)
E = \(\frac{\sigma}{2 \varepsilon_{0}}\) …. (4)

This gives the electric field due to an infinite plane sheet of charge which is independent of the distance from the sheet.

(b) How is the field directed if (i) the sheet is positively charged, (ii) negatively charged? (C8SE Delhi 2012)
Answer:
(a) directed outwards
(b) directed inwards.

Question 2.
Use Gauss’s law to derive the expression for the electric field (\(\vec{E}\) ) due to a straight uniformly charged infinite line of charge λ Cm^-1. (CBSE Delhi 2018)
Answer:

Consider an infinitely Long, thin wire charged positively and having uniform Linear charge density λ. The symmetry of the charge distribution shows that must be perpendicular to the tine charge and directed outwards. As a result of this symmetry, we consider a Gaussian surface in the form of a cylinder with arbitrary radius r and arbitrary Length L. with its ends perpendicular to the wire as shown in the figure. Applying Gauss’s theorem to curved surface ΔA₁ and circular surface ΔA₂.

Φ EΔA1 cos 0°+ EΔA2 cos 90° = \(\frac{q}{\varepsilon_{0}}\) = \(\frac{\lambda l}{\varepsilon_{0}}\) [∵ λ = \(\frac{q}{e}\)]
Or
E . 2πrl = \(\frac{\lambda l}{\varepsilon_{0}}\) ⇒ E = \(\frac{1}{2 \pi \varepsilon_{0}} \frac{\lambda}{r}\)

This is the expression for the electric field due to an infinitely long thin wire.
The graph is as shown.

Question 3.
Obtain the expression for the potential energy of an electric dipole placed with its axis at an angle (θ) to an external electric field (E). What is the minimum value of the potential energy? (CBSE 2019C)
Answer:
The torque x acting on an electric dipole of dipole moment p placed in a uniform electric field E is given by:
τ = pEsinθ …(i)
where θ is the angle made by the dipole with the electric field E. The torque tends to align the dipole along the direction of the field. If the dipole is rotated through a small angle dθ against the torque, work has to be done, which is stored in the form of the potential energy of the dipole.

Work done in rotating the dipole through the angle dθ against the torque τ is given by dw = τ dθ = pE sin θ dθ

If the dipole is rotated from θ₁ to θ₂, then
Total work is done,

W = – pE(cos θ₂ – cos θ₁)

This work done is stored as potential energy U.

U = – pE(cos θ₂ – cos θ₁)
If the dipole is rotated from θ₁ = \(\frac{\pi}{2}\) to θ₂ =θ, then

U = – pE(cos θ – cos \(\frac{\pi}{2}\) )
= – pE(cos θ – 0)

U = – pE cos θ
minimum value of potential energy
U = – pE When θ = 0°

Question 4.
Why does the electric field inside a dielectric decrease when it is placed in an external electric field?
Answer:
When a dielectric is placed in an electric field (E₀), it gets polarised, i.e. within the dielectric, an electric field (E) is induced in a direction opposite to that of the external field. Therefore, the net field within the dielectric decreases to \(\vec{E}_{0}\) – \(\vec{E}\)

Question 5.
Two-point charges +q and -2q are placed at the vertices ‘B’ and ‘C’ of an equilateral triangle ABC of side ‘a’ as given in the figure. Obtain the expression for (i) the magnitude and (ii) the direction of the resultant electric field at vertex A due to these two charges. (CBSE Al 2014 C)

Answer:
The electric fields due to the two charges placed at B and C are inclined at an angle θ = 120° as shown
Now in magnitude, we have

E_B = k\(\frac{q}{a^{2}}\) and
E_C = k\(\frac{2q}{a^{2}}\) = 2 E_B

Hence E = \(\sqrt{E_{B}^{2}+E_{c}^{2}+2 E_{B} E_{c} \cos \theta}\)
Or E = \(\sqrt{E_{B}^{2}+\left(2 E_{B}\right)^{2}+2 E_{B}\left(2 E_{B}\right) \cos 120^{\circ}}\)

On Solving we have
E = \(\sqrt{3} E_{B}\) = \(\sqrt{3} \frac{k q}{a^{2}}\)

Direction

Therefore B = 90°, the resultant is inclined at an angle of 90° with EB.

Question 6.
Four-point charges Q, q, Q, and q are placed at the corners of a square of side ‘a’ as shown in the figure.
Find the
(a) the resultant electric force on a charge Q and
(b) the potential energy of this system. (CBSEAI, Delhi 2018)
Answer:
(a) Let us find the force on charge Q at point C.

Force due to charge Q placed at point A is
F_A = k\(\frac{Q^{2}}{(a \sqrt{2})^{2}}\) =k\(\frac{Q^{2}}{2 a^{2}}\) along AC

Force due to the charge q placed at D
F_D = k\(\frac{q Q}{a^{2}}\) alongDA

Force due to the charge q placed at B
F_B = k \(\frac{q Q}{a^{2}}\) along BC

The resuLtant of F_D and F_B is
F_BD = K\(\frac{q Q \sqrt{2}}{a^{2}}\) along AC

∴ net force of charge Q placed at point C is

(b) PotentiaL energy of the system

Question 7.
A charge +Q is uniformly distributed within a sphere of radius R. Find the electric field, distant r from the center of the sphere where: (1) r < R and (2) r > R. (CBSEAI 2016C)
Answer:
For a solid sphere p = \( \frac{q}{\frac{4}{3} \pi R^{3}}\) = \(\frac{q}{\text { volume }}\)

Case 1. 0 < r < R The point Lies within the sphere.

Using Gauss’s theorem
Let Q’ be the charge encLosed by Gaussian’s surface of radius r < R.

E(4πr²)=\(\frac{Q^{\prime}}{\varepsilon_{0}}=\frac{Q^{\prime}}{4 \pi \varepsilon_{0} r^{2}}\)

From (i) and (ii)
E = \(\frac{Q r^{3}}{4 \pi \varepsilon_{0} r^{2} R^{3}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q r}{R^{3}}\)

Case 2. For r > R
\(\oint \vec{E} \cdot \overrightarrow{d s}\) = \(\frac{q_{\text {enclosed }}}{\varepsilon_{0}}\)

E(4πr²) = \(\frac{Q}{\varepsilon_{0}}\)
E = \(\frac{Q}{4 \pi \varepsilon_{0} r^{2}}\)

Question 8.
(a) An electric dipole is kept first to the left and then to the right of a negatively charged infinite plane sheet having a uniform surface charge density. The arrows p₁ and p₂ show the directions of its electric dipole moment in the two cases. Identify for each case, whether the dipole is in stable or unstable equilibrium. Justify each answer.
Answer:
p₁: stable equilibrium, p₂: unstable equilibrium. The electric field, on either side, is directed towards the negatively charged sheet and its magnitude is independent of the distance of the field point from the sheet. For position P₁ dipole moment and electric field are parallel. For position p₂, they are antiparallel.

(b) Next, the dipole is kept in a similar way (as shown), near an infinitely long straight wire having uniform negative linear charge density. Will the dipole be in equilibrium at these two positions? Justify your answer. (CBSE Sample Paper 2018-2019)

Answer:
The dipole will not be in equilibrium in any of the two positions.
The electric field due to an infinite straight charged wire is non-uniform (E ∝ 1/r).
Hence there will be a net non-zero force on the dipole in each case.

Question 9.
Two large parallel plane sheets have uniform charge densities +σ and -σ. Determine the electric field (i) between the sheets, and (ii) outside the sheets. (CBSE Delhi 2018)
Answer:
Let us consider two parallel planes charged conductors A and B carrying +ve and -ve charge density σ (charge per unit area). According to Gauss’ theorem, the electric intensity at P due to the charge on sheet A is

EA = \(\frac{\sigma}{2 \varepsilon_{0}}\) (From A and B)

The electric field at P due to charge on sheet B is
E= \(\frac{\sigma}{2 \varepsilon_{0}}\) (From A to B)

The electric field at P is
E = EA + EB
= \(\frac{\sigma}{\varepsilon_{0}}\)

Question 10.
Define electric flux and write its SI unit. The electric field components in the figure shown are : E_x = αx, E_y = 0, E_z = 0 where α = 100N/cm. Calculate the charge within the cube, assuming a = 0.1 m. (CBSE Sample Paper 2019)
Or
An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 10⁴ N/C (figure a)

(a) Calculate the time it takes to fall through this distance starting from rest.

(b) If the direction of the field is reversed (figure b) keeping its magnitude unchanged, calculate the time taken by a proton to fall through this distance starting from rest. (CBSE Delhi 2018C)
Answer:
Electric Flux is the dot product of the electric field and area vector.
Φ = \(\oint \vec{E} \cdot \overrightarrow{d s}\)

SI Unit: Nm²/C or Vm

For a given case
Φ = Φ₁ – Φ₂ = [Ex(atx= 2a) – Ex (atx = a)]a²
= [α(2a)-α(a)]a² = αa³
= 10⁴ × (0.1 )³ = 10 Nm²/C
But
Φ = \(\frac{q}{\varepsilon_{0}}\)

∴ q = ε₀Φ = 8.854 × 10^-12 × 10 C = 8.54 pC
Or
We have
F = qE

Acceleration, a = \(\frac{q E}{m}\)

Also
s = \(\frac { 1 }{ 2 }\) at² [u = 0]
∴ t = \(\sqrt{\frac{2 s}{a}}\)

(i) For the electron
a = \(\frac { eE}{ m }\)

∴ t = \(\sqrt{\frac{2 s m}{e E}}\)

∴ t = \(\sqrt{\frac{3 \times 10^{-2} \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19} \times 2.0 \times 10^{4}}}\) = 2.92 ns

(ii) for proton
t = \(\sqrt{\frac{2 \times 1.5 \times 10^{-2} \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 2 \times 10^{4}}}\)
= -.125 μs

Question 11.
What will be the total flux through the faces of the cube (figure) with the side of length ‘a’ if a charge q is placed at
(a) A: a corner of the cube.
(b) B: mid-point of an edge of the cube.
(c) C: center of the face of the cube.
(d) D: mid-point of B and C. {NCERJ Exemplar)

Answer:
(a) The charge wilt is shared by eight cubes if it has to be enclosed. Therefore the flux through the cube will be one-eighth of the total flux. Φ = q/8ε₀
(b) The charge will be shared by four cubes if it has to be enclosed. Therefore the flux through the cube will be one-fourth of the total flux. Φ = q/ 4ε₀
(c) The charge will be shared by two cubes if it has to be enclosed. Therefore the flux through the cube will be one-half of the total flux. Φ = q/ 2ε₀
(d) The charge will be shared by two cubes if it has to be enclosed. Therefore the flux through the cube will be one-half of the total flux. Φ = q/ 2ε₀

Question 12.
Two charges q and -3q are placed fixed on the x-axis separated by distance ‘d’. Where should a third charge 2q be placed such that it will not experience any force? (NCERT Exemplar)
Answer:
The situation is as shown.

Let the charge 2q be placed at a distance ‘x’ from charge q. For the charge 2q to experience zero force we have
\(\frac{2 q^{2}}{4 \pi \varepsilon_{0} x^{2}}=\frac{6 q^{2}}{4 \pi \varepsilon_{0}(d+x)^{2}}\)
(d + x)² = 3x²

Solving for x we have
x = \(\frac{d}{2} \pm \frac{\sqrt{3} d}{2}\)
(-ve sign would be between q and -3q and hence is unacceptable.)

Therefore, we have
x = \(\frac{d}{2}+\frac{\sqrt{3} d}{2}=\frac{d}{2}(1+\sqrt{3})\) to the left of q.

Question 13.
(a) State Gauss’s law. Use it to deduce the expression for the electric field due to a uniformly charged thin spherical shell at points (i) inside and (ii) outside the shell. Plot a graph showing the variation of the electric field as a function of r > R and r < R. (r being the distance from the center of the shell) (CBSE Delhi 2011, Al 2013)
Answer:
Gauss’s law states that the net outward flux through any closed surface is equal to 1 /ε₀ times the charge enclosed by the closed surface.

Consider a thin spherical shell of radius R and center at O. Let σ be the uniform surface charge density (charge per unit surface area) and q be the total charge on it. The charge distribution is spherically symmetric. Three cases arise

Case 1: at a point outside the spherical shell
In order to find the electric field at a point P outside the shell let us consider a Gaussian surface in the form of a sphere of radius r (r >>R).

By symmetry, we find that the electric field acts radially outwards and has a normal component at alt points on the Gaussian sphere. Therefore by definition of electric flux we have

Φ = E × A, where A is the surface area of the Gaussian sphere therefore
Φ = E × 4πr² …(1)

But by Gauss’s law
Φ = \(\frac{Q}{\varepsilon_{0}}=\frac{\sigma A}{\varepsilon_{0}}=\frac{\sigma \times 4 \pi R^{2}}{\varepsilon_{0}}\) … (2)

from equations (1) and (2) it follows that
E × 4πr² = \(\frac{Q}{\varepsilon_{0}}\) Or E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r^{2}}\)

E × 4πr² = \(\frac{\sigma \times 4 \pi R^{2}}{\varepsilon_{0}}\)

Ē = \(\frac{\sigma R^{2}}{\varepsilon_{0} r^{2}}\) … (3)

It follows that the electric field due to a spherical shell outside it is same as that due to a point charge. Therefore for points Lying outside the spherical shell the shell behaves as if the entire charge is concentrated at the centre.

Case 2: at a point inside the spherical shell
In this case, the Gaussian surface Lies inside the shell. Since no charge is enclosed In this surface therefore we have

E × 4πr²=q₀/ε₀ [∵ q=0]
Therefore E = 0

(b) Two identical metallic spheres A and B having charges +4Q. and -10 Q are kept a certain distance apart. A third identical uncharged sphere C is first placed in contact with sphere A and then with sphere B. Spheres A and B are then brought in contact and then separated. Find the charges on the spheres A and B. (CBSEAI 2011C)
Answer:
The initial charge on the sphere A = + 4 Q.
The initial charge on the sphere B = -10 Q.

Since all the three spheres are identical, i.e. they have the same capacity, therefore when uncharged sphere C is placed in contact with A, the total charge is equally shared between them.

Charge on C after contact with A
= \(\frac{0+4 Q}{2}\) = 2Q

Charge on A after contact with C is 2Q.
When sphere C carrying a charge 2Q is placed in contact with B, again charges are equally shared between C and B equally.

Charge on C after it is in contact with B
= \(\frac{2 Q-10 Q}{2}\) = -4Q

Now when sphere A with a charge of 2Q. is placed in contact with B, with charge -4Q.

Charge are again shared
∴ charge on A or B = \(\frac{2 Q-4 Q}{2}\) = -Q.

Question 14.
(a) 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:
It is defined as the product of the magnitude of either of the two charges and the distance between them.
For derivation see sol. 9(a) of LA-II.

(b) Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential due to the dipole is zero. (CBSE Al 2013)
Answer:
The diagram is as shown

The zero potential points lie on the equatorial line.

Question 15.
(a) Using Gauss’s law obtain expressions for the electric field (i) inside, and (ii) outside a positively charged spherical shell.
(c) A Square plane sheet of side 10 cm is inclined at an angle of 30° with the direction of a uniform electric field of 200 NC“1. Calculate the electric flux passing through the sheet. (CBSE 2019C)
Answer:
(a) Spherical Shell
Consider a spherical shell of radius R. Let q be a charge on the shell. Let us find the electric field at a point P at a distance r from the center 0 of the spherical shell.

Case (i): When point P lies inside the spherical shell
From the point, P draws a Gaussian surface which will be a sphere of radius r.

From the Gauss’s Theorem,
\(\oint_{s} \vec{E} \cdot \vec{d} S=\frac{0}{\varepsilon_{0}}\) [∵ No charge exists inside the spherical shell]

Or
E = 0
i. e. electric field inside the charged spherical shell is zero.

Case (ii): When point P is lying outside the shell (i.e. r > R)
From point P, draw a Gaussian surface which will be a spherical shell of radius r. Let dS be a small area element on the Gaussian surface P.


Or

i. e. the electric field outside the spherical shell behaves as if the whole charge is concentrated at the center of the spherical shell.

(b) Show graphically variation of the electric field as a function of the distance r from the center of the sphere.
Answer:
Variation of electric field E with distance
The given figure shows the variation of an electric field with distance from the center of the charged spherical shell.

(c) Here E = 200 N/C, S = 0.1 × 0.1 = 0.01 m²
And θ = 90° – 30° = 60°

The electric flux linked with the square sheet
Φ = E S cos 60°
= 200 × 0.01 × \(\frac { 1 }{ 2 }\) = 1.0 Nm² C^-1

Question 16.
A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at a large distance from the ring, it behaves like a point charge. (CBSE Delhi 2016)
Answer:
Consider a uniformly charged ring of radius ‘a’. Let the total charge on the ring be Q, Let us find the electric field on the axis of the ring at point P distance x from the center of the ring. Consider a segment of charge dQ as shown in the figure.

The magnitude of etectñc field at P due to the segment is
dE=k\(\frac{d Q}{r^{2}}\) …(1)

This field can be resolved into its components: x component dE_x = dE cos α an along the axis of the ring and y component dE perpendicular to the axis. Since these perpendicular components, due to alt the charge segments, are equal and opposite, therefore they cancel out each other. From the diagram we have r = \(\sqrt{x^{2}+a^{2}}\) and cos α = x/r, therefore we have
…(2)

In this case, all the segments of the ring give the same contribution to the field at P since they are all equidistant from this point. Thus we can easily sum over all segments to get the total electric field at point P

If the point of observation is far away, i.e. x >> a, then E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{x^{2}}\). This is the same as that for a point charge. Thus at far-off axial points, a charged ring behaves as if a point charge is situated at the center of the ring.

Question 17.
Two thin concentric and coplanar spherical shells, of radii ‘a’ and ‘b’ (b > a), carry charges, q, and Q respectively. Find the magnitude of the electric field, at a point distant x, from their common center for
(i) O < x < a
(ii) a ≤ x < b
(iii) b ≤ x < ∞ (CBSE Delhi 2016C)
Answer:
The diagram is as shown.

(i) For 0 < x < a
Point Lies inside both the spherical shells.

∴ charge enclosed = 0
Hence, E(x) = 0

(ii) For a < x < b
Point is outside the spherical shell of radius ‘a’ but inside the spherical Shell of radius ‘b’.
Therefore

(iii) For b ≤ x< ∞
The point is outside of both the spherical shells. The total effect we charge at the center equals (Q + q).

Question 18.
(a) Define electric flux. Is It a scalar or a vector quantity?

A point charge q is at a distance of d/2 directly above the center of a square of side d, as shown In the figure. Us. Gauss’s law to obtain the expression for the electric flux through the square.
Answer:
The electric flux through a given surface is defined as the dot product of the electric field and area vector over that surface. It is a scalar quantity.

Constructing a cube of side ‘d’ so that charge ‘q’ gets placed within this cube (Gaussian surface)
According to Gauss’s law the electric flux
Φ = \(\frac{q}{\varepsilon_{0}}\)

This is the total flux through all the six faces of the cube.

Hence electric flux through the square (one face of the cube)
Φ = \(\frac{1}{6} \frac{q}{\varepsilon_{0}}\)

(b) If the point charge is now moved to a distance ‘d’ from the center of the square and the side of the square is doubled, explain how the electric flux will be affected. (CBSE Al, Delhi 2018)
Answer:
If the charge is moved to a distance ‘d’ and the side of the square is doubled the cube will be constructed to have a side 2d but the total charge enclosed in it will remain the same. Hence the total flux through the cube and therefore the flux through the square will remain the same as before.

Question 19.
(a) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.
Answer:
The required graph is

(b) Find the work done in bringing a charge q from perpendicular distance r₁ to r₂ (r₂ > r₁). (CBSEAI, Delhi 2018)
Answer:
Work done in moving the charge “q”. through a smaLL displacement ‘dr’

dW = \(\vec{F}\) . \(\vec{dr}\) =q\(\vec{E}\) .\(\vec{dr}\)
dW = qE dr cos 0 = qEdr
dW= q x \(\frac{\lambda}{2 \pi \varepsilon_{0} r}\) dr

Hence work done in moving the charge from r₁ to r₂ (r₂ > r₁)

Question 20.
Derive an expression for the torque acting on an electric dipole of dipole moment \(\vec{p}\) placed in a uniform electric field \(\vec{E}\). Write the direction along which the torque acts.
OR
Derive an expression for the electric field at a point on the axis of an electric dipole of dipole moment \(\vec{p}\). Also, write its expression when the distance r >> a the length ‘a’ of the dipole. (CBSEAI 2019)
Answer:
Consider an electric dipole consisting of charges -q and +q and dipole length d is placed in a uniform electric field E as shown in the figure.

Let the dipole moment makes an angle θ with the direction of the electric field. The two charges experience force qE each.

These forces are equal, parallel, and opposite. Therefore, the net force acting on the dipole is
F_n = qE – qE = 0

Thus the net force acting on the dipole is zero.

These two forces constitute a couple. This applies a torque on the dipole given by
τ = either force × arm of the couple
τ = qE × d sin θ, where d sin θ is the arm of the couple.
τ = q d Esin θ, where p = qd, dipole moment
τ = pE sin θ.

Consider an electric dipole consisting of -q and +q charges separated by a distance 2a as shown in the figure. Let P be the point of observation on the axial Line where the electric field has to be found. Let it be at a distance r from the center O of the dipole. Let us suppose that the dipole is placed in a vacuum.

Let E and E_B be the electric fields at point P due to the charges at A and B respectively. Therefore,

The two fields at P are in opposite directions. Thus the resultant electric field at P is given by
E = \(\sqrt{E_{A}^{2}+E_{B}^{2}-2 E_{A} E_{B} \cos \theta}=\sqrt{\left(E_{B}-E_{A}\right)^{2}}\) = E_B – E_A

since θ = 180⁰ Therefore, the resultant electric field is

Solving we have

where p = q 2a.
If the dipole is short then r >> a, therefore, ‘a’ is neglected as compared to r, hence

This field is in the direction of the dipole moment.

Question 21.
(a) Derive the expression for the electric field at a point on the equatorial line of an electric dipole. (CBSE 2019C)
Answer:
Consider an electric dipole consisting of charges -q and +q separated by a distance 2a as shown in the figure. Let the point of observation P lie on the. right bisector of the dipole AB at a distance r from its midpoint 0. Let E_A and E_B be the electric field intensities at point P due to charges at A and B, respectively.

The two electric fields have magnitudes

in the direction of AP

in the direction of PB.

The two fields are equal in magnitude but have different directions. Resolving the two fields E_A and E_B into their rectangular components, i.e. perpendicular to and parallel to AB. The components perpendicular to AB, i.e E_A sinG and E_B sinG being equal and opposite cancel out each other while the components parallel to AB, i.e. E_A cos θ and E_B cos θ being in the same direction add up as shown in the figure. Hence the resultant electric field at point P is given by

E = E_Acosθ + E_Bcosθ

∵ cos θ = \(\frac{a}{\left(r^{2}+a^{2}\right)^{1 / 2}}\) and q 2a = p

For a short dipole r²,>>a² therefore
E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{3}}\)

(b) Discuss the orientation of the dipole in (a) stable, (b) unstable equilibrium in a uniform electric field. (CBSE Delhi 2017)
Answer:
For stable equilibrium \(\vec{P}\) is along \(\vec{E}\)
θ = 0°, τ = PE sin 0°, Torque alligns the dipole in the direction of field

For unstable equilibrium \(\vec{P}\) is antiparallel to \(\vec{E}\)
∵ θ = 180°, τ = PE sin 180° = 0, Torque alligns the dipole in a direction opposite to \(\vec{E}\).

Question 22.
(a) An electric dipole of dipole moment \(\vec{p}\) consists of point charges +q and -q separated by a distance 2a apart. Deduce the expression for the electric field due to the dipole at a distance x from the center of the dipole on its axial line in terms of the dipole moment. Hence show that in

the limit r >> a, \(\vec{E}\) → \(\frac{2 \vec{p}}{4 \pi \varepsilon_{0} r^{3}}\)
Answer:

Consider an electric dipole consisting of charges -q and +q and dipole length d is placed in a uniform electric field E as shown in the figure.

Let the dipole moment makes an angle θ with the direction of the electric field. The two charges experience force qE each.

These forces are equal, parallel, and opposite. Therefore, the net force acting on the dipole is
F_n = qE – qE = 0

Thus the net force acting on the dipole is zero.

These two forces constitute a couple. This applies a torque on the dipole given by
τ = either force × arm of the couple
τ = qE × d sin θ, where d sin θ is the arm of the couple.
τ = q d Esin θ, where p = qd, dipole moment
τ = pE sin θ.

Consider an electric dipole consisting of -q and +q charges separated by a distance 2a as shown in the figure. Let P be the point of observation on the axial Line where the electric field has to be found. Let it be at a distance r from the center O of the dipole. Let us suppose that the dipole is placed in a vacuum.

Let E and EB be the electric fields at point P due to the charges at A and B respectively. Therefore,

The two fields at P are in opposite directions. Thus the resultant electric field at P is given by
E = \(\sqrt{E_{A}^{2}+E_{B}^{2}-2 E_{A} E_{B} \cos \theta}=\sqrt{\left(E_{B}-E_{A}\right)^{2}}\) = E_B – E_A

since θ = 1800 Therefore, the resultant electric field is

Solving we have

where p = q 2a.
If the dipole is short then r >> a, therefore, ‘a’ is neglected as compared to r, hence

This field is in the direction of the dipole moment.

(b) Given the electric field in the region E = 2xî, find the net electric flux through the cube and the charge enclosed by it. (CBSE Delhi 2015)

Answer:
Since the electric field has only an x component, for faces perpendicular to the x-direction, the angle between E and ΔS is ± π/2.

Therefore, the flux Φ = E.ΔS is separately zero for each face of the cube except the two faces along the X-axis.

Now the magnitude of the electric field at the left face is
E_L= 0 (x = 0 at the left face)

The magnitude of the electric field at the right face is
E_R = 2x = 2a (x = a at the right face)

The corresponding fluxes are
Φ_L = E_L.ΔS = ΔS (\(\vec{E}_{L} \cdot \hat{n}_{\mathrm{L}}\)) = E_LΔS COS θ
= – E_L ΔS = 0, since θ =180°

Φ_R = E_R.ΔS = E_RΔS cos θ = E ΔS = (2a)a², since θ = 0°

Net flux through the cube
Φ = -Φ_R + Φ_L = 2a³ -0 = 2a³

We can use Gauss’s law to find the total charge q inside the cube.
We have Φ = q/ε₀ or q = Φε₀. Therefore, q = 2a³ × 8.854 × 10^-12C

Question 23.
(a) Derive an expression for the electric field at any point on the equatorial line of an electric dipole.
Answer:
Consider an electric dipole consisting of charges -q and +q separated by a distance 2a as shown in the figure. Let the point of observation P lie on the. right bisector of the dipole AB at a distance r from its midpoint 0. Let EA and EB be the electric field intensities at point P due to charges at A and B, respectively.

The two electric fields have magnitudes

in the direction of AP

in the direction of PB.

The two fields are equal in magnitude but have different directions. Resolving the two fields E_A and E_B into their rectangular components, i.e. perpendicular to and parallel to AB. The components perpendicular to AB, i.e E_A sinG and E_B sinG being equal and opposite cancel out each other while the components parallel to AB, i.e. E_A COS θ and E_B cos θ being in the same direction add up as shown in the figure. Hence the resultant electric field at point P is given by

E = E_A cos θ + E_B cos θ

∵ cos θ = \(\frac{a}{\left(r^{2}+a^{2}\right)^{1 / 2}}\) and q 2a = p

For a short dipole r²,>>a² therefore
E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{3}}\)

(b) Two identical point charges, q each, are kept 2 m apart in the air. A third point charge Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q. (CBSE Delhi 2019)
Answer:
The third charge Q. wilt is in equilibrium if it experiences zero net force. Let it be placed at a distance x meter from the charge q.

Solving for x, we have x= 1 m

For the equilibrium of charges “q”, the nature of charge Q must be opposite to the nature of charge q and should be placed at the center of two charges.

Question 24.
(a) Define electric flux. Write its S.I. unit.
Answer:
It is defined as the total number of electric field lines crossing a given area. The electric flux can be found by multiplying the component of the electric field in the direction of the area vector (or perpendicular to the area) with the area of the closed surface. It is measured in Nm²C-^-1.

(b) A small metal sphere carrying charge +Q is located at the center 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.

Answer:
For point P, using Gauss law we have

Since E and dA are in the same direction therefore we have E = \(\frac{Q}{\varepsilon_{0} A}\)

Point P₂ lies inside the metal, therefore the Gaussian surface drawn at P₂ does not include a charge, hence the electric field at P₂ is zero.

(c) Draw the pattern of electric field lines in this arrangement. (CBSEAI 2012C)
Answer:
The electric field lines are as shown.

Question 25.
(a) Deduce the expression for the torque acting on a dipole of dipole moment \(\vec{P}\) in the presence of a uniform electric field \(\vec{E}\)
Answer:

Consider an electric dipole consisting of charges -q and +q and dipole length d is placed in a uniform electric field E as shown in the figure.

Let the dipole moment makes an angle θ with the direction of the electric field. The two charges experience force qE each.

These forces are equal, parallel, and opposite. Therefore, the net force acting on the dipole is
F_n = qE – qE = 0

Thus the net force acting on the dipole is zero.

These two forces constitute a couple. This applies a torque on the dipole given by
τ = either force × arm of the couple
τ = qE × d sin θ, where d sin θ is the arm of the couple.
τ = q d Esin θ, where p = qd, dipole moment
τ = pE sin θ.

Consider an electric dipole consisting of -q and +q charges separated by a distance 2a as shown in the figure. Let P be the point of observation on the axial Line where the electric field has to be found. Let it be at a distance r from the center O of the dipole. Let us suppose that the dipole is placed in a vacuum.

Let E and EB be the electric fields at point P due to the charges at A and B respectively. Therefore,

The two fields at P are in opposite directions. Thus the resultant electric field at P is given by
E = \(\sqrt{E_{A}^{2}+E_{B}^{2}-2 E_{A} E_{B} \cos \theta}=\sqrt{\left(E_{B}-E_{A}\right)^{2}}\) = E_B – E_A

since θ = 180° Therefore, the resultant electric field is

Solving we have

where p = q 2a.
If the dipole is short then r >> a, therefore, ‘a’ is neglected as compared to r, hence

This field is in the direction of the dipole moment.

(b) Consider two hollow concentric spheres S₁ and S₂ and enclosing charges 2Q and 4Q. respectively as shown in the figure, (i) Find out the ratio of the electric flux through them, (ii) How will the electric flux through the sphere S₁ change if a medium of dielectric constant ‘er’ is introduced in the space inside S1 in place of air? Deduce the necessary expression. (CBSE Al 2014)

Answer:
Φ₁ = \(\frac{2 Q}{\varepsilon_{0}}\) and Φ2 = \(\frac{2 Q+4 Q}{\varepsilon_{0}}=\frac{6 Q}{\varepsilon_{0}}\)

Hence ratio

When a dielectric of dielectric constant εr is introduced then we have
Φ₁ = \(\frac{2 Q}{\varepsilon}=\frac{2 Q}{\varepsilon_{r} \varepsilon_{0}}\)

Question 26.
(a) Use Gauss’s theorem to find the electric field due to a uniformly charged infinitely large plane thin sheet with surface charge density σ.
Answer:
It states, “The net electric flux through any Gaussian surface is equal to \(\frac{1}{\varepsilon_{0}}\) times the net electric charge enclosed by the surface.

Mathematically, Φ =\( \phi \vec{E} \cdot d \vec{A}=\frac{q_{i n}}{\varepsilon_{0}}\)

Consider an infinite plane sheet of charge. Let a be the uniform surface charge density, i.e. the charge per unit surface area. From symmetry, we find that the electric field must be perpendicular to the plane of the sheet and that the direction of E on one side of the plane must be opposite to its direction on the other side as shown in the figure below. In such a case let us choose a Gaussian surface in the form of a cylinder with its axis perpendicular to the sheet of charge, with ends of area A.

The charged sheet passes through the middle of the cylinder’s length so that the cylinder’s ends are equidistant from the sheet. The electric field has a normal component at each end of the cylinder and no normal component along the curved surface of the cylinder. As a result, the electric flux is linked with only the ends and not the curved surface.

Therefore by the definition of eLectric flux, the flux Linked with the Gaussian surface is given by
Φ = \(\oint_{A} \vec{E} \cdot \vec{\Delta} A\)

Φ = E_A + E_A = 2E_A … (1)

But by Gauss’s Law

Φ = \(\frac{q}{\varepsilon_{0}}=\frac{\sigma A}{\varepsilon_{0}}\) [∵ q = σA] … (2)

From equations (1) and (2), we have

2E_A = \(\frac{\sigma A}{\varepsilon_{0}}\) … (3)

E = \(\frac{\sigma}{2 \varepsilon_{0}}\) …. (4)

This gives the electric field due to an infinite plane sheet of charge which is independent of the distance from the sheet.

(b) An infinitely large thin plane sheet has a uniform surface charge density +σ. Obtain the expression for the amount of work done in bringing a point charge q from infinity to a point, distant r, in front of the charged plane sheet. (CBSE Al 2017)
Answer:
Work done

Numerical Problems:

Formulae for solving numerical problems

• q = ±ne
• F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}\)
• Fmed = \(\frac{1}{4 \pi \varepsilon_{0} K} \frac{q_{1} q_{2}}{r^{2}}\) where K is dielectric constant.
• 
• E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{2 p}{r^{3}}\) for an electric dipole on its axial line.
• E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{3}}\) for an electric dipole on its equitorial line.
• Torque on an electric dipole in a uniform electric field, τ = PE sin θ.
• U = -pE cos θ, the potential energy of an electric dipole.
• Φ = \(\oint \vec{E} \cdot d \vec{A}=\frac{q_{\text {in }}}{\varepsilon_{0}}\)
• E = \(\frac{\sigma}{2 \varepsilon_{0}}\) , electric field due to an infinite plane sheet of charge.

Question 1.
An electric dipole of length 2 cm is placed with its axis making an angle of 60° to a uniform electric field of 105 NC^-1. If it experiences a torque of 8\(\sqrt{3}\) Nm, calculate the (i) magnitude of the charge on the dipole and (ii) potential energy of the dipole.
Answer:
(i) Using
q = τ / L E sin Φ

we have
q = 8N\(\sqrt{3}\) × 2 / 2 × 10^-2 × 10⁵ × \(\sqrt{3}\) = 8 × 10^-3 C

(ii) Using U = -pE cos θ, we have
U = – 8 × 10^-3 × 0.02 × 10⁵ × 0.5 = – 8 J

Question 2.
A charge of 17.7 × 10^-4 C is distributed uniformly over a large sheet of area 200 m². Calculate the electric field intensity at a distance of 20 cm from it in the air.
Answer:
Given q = 17.7 × 10^-4C, A = 200 m², r = 20cm = 0.2 m

Using the relation

Question 3.
Two fixed point charges +4e and +e units are separated by a distance ‘a’. Where should the third point charge be placed for it to be in equilibrium?
Answer:
The third charge will be in equilibrium if it experiences zero net force. Let it be placed at a distance x from the charge 4e.

4(a – x)² = x² solving for x we have
2 (a – x) = x or x = a/3

Question 4.
An electric field along the x-axis is given by \(\vec{E}\) = 100 îN/C for x > 0 and \(\vec{E}\) = -100 îN/C for x < 0. A right circular cylinder of length 20 cm and radius 5 cm lies parallel to the x-axis with its centre at the origin and one face at x = +10 cm, the other face at x = -10 cm. Calculate the net outward flux through the cylinder. (CBSE 2019C)

r = 5 cm = 0.05 m
Φ = ?

Net outward flux through the cylinder
Φ = Φ₁ + Φ₂ + Φ₃
= \(\vec{E} \cdot \overrightarrow{d s}_{1}+\vec{E} \cdot \overrightarrow{d s}_{2}+\vec{E} \cdot \overrightarrow{d s}_{3}\)

= Eds₁ cos 180° + Eds₂ cos 90° + Eds3 cos 0°
= – Eds₁ + Eds₂ (0) + Eds cos 0°
= – (-100) ds + 100 ds
= (100 +100) ds
= 200 × πr² = 200 × 3.14 × (0.05)²
= 1.57 Nm² C^-1

Question 5.
A hollow cylindrical box of length 1m and area of cross-section 25 cm² is placed in a three-dimensional coordinate system as shown in the figure. The electric field
in the region is given by \(\vec{E}\) = 50x î, where E is in NC^-1 and x is in meters. Find

(i) Net flux through the cylinder.
Answer:
We can see from the figure that on the left face E and dS are antiparallel. Therefore, the flux is
ΦL = \(\vec{E}\).\(\vec{dS}\) = E dS cos 180°
= -50 × 1 × 25 × 10^-4 = -0.125 N m² C^-1

On the right face, E and dS are parallel and therefore
ΦR = \(\vec{E}\).\(\vec{ds}\) = EdS cos 0°
= 50 × 2 × 25 × 10^-4 = 0.250 N m² C^-1

Therefore net flux is – 0.125 + 0.250 = + 0.125 N m² C^-1

(ii) Charge enclosed by the cylinder. (CBSE Delhi 2013)

Answer:
q = Φε₀ = 0.125 × 8.854 × 10^-12 = 1.1 × 10^-12C

Question 6.
(i) Define electric flux. Write its SI units,
Answer:
(i) It is defined as the number of electric field lines crossing a unit area perpendicular to the given area. It is measured in N m² C^-1

(ii) Consider a uniform electric field \(\vec{E}\) = 5 × 10³î NC^-1. Calculate the flux of this field through a square surface of area 12 cm² when
(a) its plane is parallel to the Y – Z plane, and
(b) the normal to its plane makes a 60° angle with the x-axis. (CBSE Delhi 2013C)
Answer:
Given \(\vec{E}\) = 5 × 10³î N C^-1, A = 12 cm² = 12 × 10^-4 m²

(a) Here \(\vec{E}\) = 5 × 103î N C^-1
Area of square = 12 × 10^-4 m²

The plane of surface area being parallel to YZ plane, hence
A = 12 × 10^-4 î m²

Electric flux of the field
Φ = \(\vec{E}\) \(\vec{A}\) = (5 × 10³ î). (12 × 10^-4î) = 6 N C^-1 m²
(b) When normal to the plane of surface area makes an angle of 60° with the X-axis, the flux is given by
Φ = EA cos 0 = 5 × 10³ × 12 × 10^-4 × 0.5
= 3 NC^-1 m²

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 2 Electrostatic Potential and Capacitance. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 2 Important Extra Questions Electrostatic Potential and Capacitance

Electrostatic Potential and Capacitance Important Extra Questions Very Short Answer Type

Question 1.
Express dielectric constant in terms of the capacitance of a capacitor.
Answer:
It is given by the expression K = \(\frac{c}{c_{0}}\) where C is the capacitance of the capacitor with dielectric and C₀ is the capacitance without the dielectric.

Question 2.
On what factors does the capacitance of a parallel plate capacitor depend?
Answer:

1. Area of plates,
2. The separation between the plates and
3. Nature of dielectric medium between the plates.

Question 3.
What is the ratio of electric field intensities at any two points between the plates of a capacitor?
Answer:
The ratio is one, as the electric field is the same at all points between the plates of a capacitor.

Question 4.
Write a relation between electric displacement vector D and electric field E.
Answer:
\(\vec{D}\) = ε₀ \(\vec{E}\) + \(\vec{P}\)

Question 5.
Write the relation between dielectric constant (K) and electric susceptibility χ_e
Answer:
K = 1 + χ_e

Question 6.
A hollow metal sphere c radius 5 cm is charged such that the potential on its surface is 10 V. What is the potential at the center of the sphere? (CDSE AI 2011)
Answer:
10 V

Question 7.
What is the geometrical shape of equipotential surfaces due to a single isolated charge? (CBSE Delhi 2013)
Answer:
Concentric circles.

Question 8.
Draw the equipotential surfaces due to an isolated point charge. (CBSE Delhi 2019)
Answer:
These areas are shown.

Question 9.
‘For any charge configuration, equipotential surface through a point is normal to the electric field’. Justify. (CBSE Delhi 2014)
Answer:
This is because work done in moving a charge on an equipotential surface is zero. This is possible only if the equipotential surface is perpendicular to the electric field.

Question 10.
The given graph shows the variation of charge ‘q’ versus potential difference ‘V for two capacitors C₁ and C₂. Both the capacitors have the same plate separation but the plate area of C₂ is greater than that of C_y Which line (A or B) corresponds to C₁ and why? (CBSEAI 2014C)

Answer:
Since C = ε₀ A/d, since the area for C₂ is more, therefore capacitance of C₂ is more. From the graph greater the slope greater is than the capacitance, therefore, graph A belongs to capacitor C₂. While graph B belongs to capacitance C_v

Question 11.
Write a relation for polarisation P of dielectric material in the presence of an external electric field E . (CBSE AI 2015)
Answer:
P = χ_e ε₀ \(\vec{E}\).

Question 12.
Why are electric field lines perpendicular at a point on an equipotential surface of a conductor? (CBSE AI 2015C)
Answer:
So that no net force acts on the charge on the equipotential surface and it remains stationary.

Question 13.
Why does a given capacitor store more charge at a given potential difference when a dielectric is filled in between the plates?
Answer:
Because the capacitance of the capacitor increases on filling the dielectric medium in between the plates.

Question 14.
Why is the electrostatic potential inside a charged conducting shell constant throughout the volume of the conductor? (CBSE AI 2019)
Answer:
We know that E = – \(\frac{d V}{d r}\)
Inside a conductor, the electric field is zero and no work is done in moving a charge inside a conductor. Therefore, V is constant.

Question 15.
Does the charge given to a metallic sphere depend on whether it is hollow or solid? (CBSE Delhi 2017)
Answer:
If the capacitance of the two spheres, solid and hollow, is the same, then they will hold the same quantity of charge.

Question 16.
Is the electric potential necessarily zero at a place where the electric field is zero?
Answer:
No, it is not necessary. The electric field inside a hollow metallic conductor is zero but the electric potential is not zero.

Question 17.
In a parallel plate capacitor, the potential difference of 10² V is maintained between the plates. What will be the electric field at points A and B as shown in the figure below?
Answer:
The electric field between the plates of a capacitor is uniform; therefore the electric field at points A and B will be the same.

Question 18.
If the plates of a charged capacitor are suddenly connected to each other by a wire, what will happen?
Answer:
The capacitor is discharged immediately.

Question 19.
Can you place a parallel plate capacitor of one farad capacity in your house?
Answer:
Ordinarily, it is not possible because the surface area of such a capacitor will be extra-large.

Question 20.
Is there any conductor which can be given almost unlimited charge?
Answer:
Yes, the earth.

Question 21.
What would be the work done if a point charge + q is taken from a point A to a point B on the circumference of a circle drawn with another point charge + q at the center?

Answer:
Zero.

Question 22.
Sketch a graph to show how a charge Q, acquired by a capacitor of capacitance, C, varies with the increase in the potential difference between the plates.
Answer:

Question 23.
In a parallel plate capacitor, the capacitance increases from 4 μF to 80 μF, introducing a dielectric medium between the plates. What is the dielectric constant of the medium?
Answer:
The dielectric constant is given by
K = \(\frac{80}{4}\) = 20

Electrostatic Potential and Capacitance Important Extra Questions Short Answer Type

Question 1.
Draw a plot showing the variation of (i) electric field (E) and (ii) electric potential (V) with distance r due to a point charge Q. (CBSE Delhi 2012)
Answer:
The plot is as shown.

Question 2.
Two identical capacitors of 10 pF each are connected in turn (i) in series and (ii) in parallel across a 20 V battery. Calculate the potential difference across each capacitor in the first case and the charge acquired by each capacitor in the second case. (CBSE AI 2019)
Answer:
(i) Since the two capacitors have the same capacitance, therefore, the potential will be divided amongst them. Hence V = 10 V each
(ii) Since the capacitors are connected in parallel, therefore, potential difference = 20 V
Hence charge Q = CV = 10 × 20 = 200 pC

Question 3.
A point charge ‘q’ is placed at O as shown in the figure. Is V_A – V_B positive, negative, or zero, if ‘q’ is an (i) positive, (ii) negative charge? (CBSE Delhi 2011, 2016)

Answer:
If V_A – V_B = \(\frac{q}{4 \pi \varepsilon_{0}}\left(\frac{1}{\mathrm{OA}}-\frac{1}{\mathrm{OB}}\right)\)

As OA < OB
∴ If q is positive then V_A– V_B is positive and
if q is negative V_A – V_B is also negative.

Question 4.
The graph shows the variation of voltage V across the plates of two capacitors A and B versus charge Q stored on them. Which of the two capacitors has higher capacitance? Give a reason for your answer.

Answer:
Capacitor A has higher capacitance. We know that capacitance C = Q/V.

For capacitor A
\(c_{A}=\frac{Q}{V_{A}}\)

For capacitor B
\(c_{B}=\frac{Q}{V_{B}}\)

As V_B > V_A
∴ C_B < C_A
Thus capacitance of A is higher.

Question 5.
A test charge ‘q’ is moved without acceleration from A to C along the path from A to B and then from B to C in electric field E as shown in the figure,

(i) Calculate the potential difference between A and C
Answer:
(i) dV = – E dr = – E (6 – 2) = – 4E

(ii) At which point (of the two) is the electric potential more and why? (CBSE AI 2012)
Answer:
Electric potential is more at point C as dV = – Edr, i.e. the electric potential decreases in the direction of the electric field.

Question 6.
A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor. (CBSE AI 2013)
Answer:
Given t = d/2, C = ?
We know that when a dielectric of thickness ‘t’ is inserted between the plates of a capacitor, its capacitance is given by
C = \(\frac{\varepsilon_{0} A}{d-t+\frac{t}{K}}\)

Hence we have
C = \(\frac{\varepsilon_{0} A}{d-\frac{d}{2}+\frac{d}{2 K}}=\frac{2 K \varepsilon_{0} A}{d(1+K)}\)

Question 7.
Two-point charges q and -2q are kept ‘d’ distance apart. Find the location of the point relative to charge ‘q’ at which potential due to this system of charges is zero. (CBSE Al 2014C)
Answer:
Let the potential be zero at point P at a distance x from charge q as shown

Now potential at point P is
V = \(\frac{k q}{x}+\frac{k(-2 q)}{d+x}\) = 0

Solving for x we have
x = d

Question 8.
Four-point charges Q, q, Q., and q are placed at the corners of a square of side ‘a’ as shown in the figure.

Find the potential energy of this system. (CBSEAI, Delhi 2018)
Answer:
The potential energy of the system
U = \(\frac{1}{4 \pi \varepsilon_{0}}\left(4 \frac{q Q}{a}+\frac{q^{2}}{a \sqrt{2}}+\frac{Q^{2}}{a \sqrt{2}}\right)\)

U = \(\frac{1}{4 \pi \varepsilon_{0} a}\left(4 q Q+\frac{q^{2}}{\sqrt{2}}+\frac{Q^{2}}{\sqrt{2}}\right)\)

Question 9.
Three-point charges q, -4q, and 2q are placed at the vertices of an equilateral triangle ABC of side T as shown in the figure. (CBSE Delhi 2018)

Find out the amount of the work done to separate the charges at infinite distance. (CBSE AI, Delhi 2018)
Answer:
Net potential energy of the system
= \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{l}[-4+2-8]=\frac{5 q^{2}}{2 \pi \varepsilon_{0} l}\)

Electrostatic Potential and Capacitance Important Extra Questions Long Answer Type

Question 1.
Two-point charges 2 μC and —2 μC are placed at points A and B 6 cm apart.
(a) Draw the equipotential surfaces of the system.
Answer:
The diagram is as shown.

(b) Why do the equipotential surfaces get closer to each other near the point charges? (CBSEAI2O11C)
Answer:
We know that E = – dV/dr
Therefore, dr =- dV/E
Since near the charge, electric field E is large, dr will be less.

Question 2.
(a) Obtain the expressions for the resultant capacitance when the three capacitors C₁, C_2, and C₃ are connected (i) in parallel and then (ii) in series.
Answer:
(i) Parallel combination of three capacitors.
Let three capacitors of capacitances C₁, C₂, and C₃ be connected in parallel, and potential difference V be applied across A and B. If q be total charge flowing in the circuit and q₁ q₂ and q₃ be charged flowing across C₁, C_2, and C₃ respectively, then
q = q₁+ q₂ + q₃
or q = C₁v + C₂V + C₃V …(i)

If C_P is the capacitance of the arrangement in parallel, then
q = C_PV

So equation (i) becomes
C_PV = C₁V + C₂V + C₃V
Or
C_P = C₁ + C₂ + C₃

(ii) Series combination of three capacitors Let three capacitors C₁, C₂, and C₃ be connected in series. Let q charge be flowing through the circuit.
If V₁, V₂, and V₃ be potential differences across the plates of the capacitor and V be the potential difference across the series combination, then

V = V₁ + V₂ + V₃
Or
V = \(\frac{q}{C_{1}}+\frac{q}{C_{2}}+\frac{q}{C_{3}}\) … (i)

If Cs is the capacitance of series combination, then V = \(\frac{q}{\mathrm{C}_{\mathrm{s}}}\).

So the equation (i) becomes
\(\frac{q}{\mathrm{C}_{\mathrm{s}}}=\frac{q}{\mathrm{C}_{1}}+\frac{q}{\mathrm{C}_{2}}+\frac{q}{\mathrm{C}_{3}}\)
Or
\(\frac{1}{\mathrm{C}_{\mathrm{s}}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}\)

(b) In the circuit shown in the figure, the charge on the capacitor of 4 μF is 16 μC. Calculate the energy stored in the capacitor of 12 μF capacitance. (CBSE 2019C)

Answer:
Charge q across 4 μF Capacitor is 10 μc Potential difference across the capacitor of capacitance 4 μF will be
V= \(\frac{q}{C}=\frac{16 \mu C}{4 \mu F}=\frac{16 \times 10^{-6} \mathrm{C}}{4 \times 10^{-6} \mathrm{~F}}\)=4V

∴ Potential across 12 μF Capacitors
= 12V – 4V = 8V

Energy stored in the capacitors of capacitance C = 12 μF

U = \(\frac{1}{2}\) CV² = \(\frac{1}{2}\) × 12 × 10^-6 × 8² joule
= 384 × 10^-6 J = 384 μJ

Question 3.
Two identical plane metallic surfaces A and B are kept parallel to each other in the air, separated by a distance of 1 cm as shown in the figure.

A is given a positive potential of 10 V and the outer surface of B is earthed.
(i) What is the magnitude and direction of the uniform electric field between Y and Z?
Answer:
The electric field between the plates is
E = \(\frac{V}{d}\) = 10³ V m^-1
directed from plate A at the higher potential to plate B at a lower potential, i.e. from Y to Z

(ii) What is the work done in moving a charge of 20 µC from X to Y?
Answer:
Since X and Y are on the same plate A, which is an equipotential surface, work done in moving a charge of 20 µC from X to Y on the equipotential surface is zero.

Question 4.
(i) Draw the equipotential surfaces corresponding to a uniform electric field in the z-direction.
Answer:
The equipotential surfaces are as shown.

Equipotential surfaces

(ii) Derive an expression for the electric potential at any point along the axial line of an electric dipole. (CBSE Delhi 2019)
Answer:
Consider an electric dipole of length 2a and having charges +q and -q. Let us find the potential on the axial line at point P at a distance OP = x from the center of the dipole.

Now potential at point P is

Question 5.
A network of four capacitors, each of capacitance 15 µF, is connected across a battery of 100 V, as shown in the figure. Find the net capacitance and the charge on the capacitor C₄. (CBSE Al 2012C)

Answer:
Capacitors C1 C2 and C3 are in series, therefore their equivalent capacitance is
\(\frac{1}{C_{\mathrm{s}}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}=\frac{1}{15}+\frac{1}{15}+\frac{1}{15}=\frac{3}{15}\)

Hence C_S = 5 µF
Now C_S and C₄ are in parallel, hence
C_P = C_S + C₄ = 5 + 15 = 20 µF

Now C4 is connected to 100 V, therefore charge on it is
Q = CV=15 × 10^-6 × 100=15 × 10^-4 C

Question 6.
A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor. (CBSE AI 2014)
Answer:
U_i = \(\frac{1}{2}\) CV²

When the capacitors are connected then the energy stored is
U_F = \(\frac{1}{2}\left(C_{1}+C_{2}\right)\left(\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}\right)^{2}\)

Since C₁ = C₂ = C, and V₂ = 0, we have
Uf = \(\frac{1}{2}(C+C)\left(\frac{C V}{C+C}\right)^{2}=\frac{1}{2}(2 C) \times \frac{V^{2}}{4}=\frac{1}{4} C V^{2}\)

Hence we have
\(\frac{U_{f}}{U_{i}}=\frac{1}{2}\)

Question 7.
An isolated air capacitor of capacitance C₀ is charged to a potential V₀. Now if a dielectric slab of dielectric constant K is inserted between its plates, completely filling the space between the plates, then how to do the following change, when the battery remains connected
(i) capacitance,
Answer:
When the battery remains connected, the potential on the capacitor does not change.
The capacitance of the capacitor becomes K times the original value, i.e. C = K C₀.

(ii) charge
Answer:
Now new charge is Q = CV = K C₀ V = K Q₀.

(iii) the field between the plates
Answer:
The field between the plates becomes
E = \(\frac{V}{d}=\frac{V_{0}}{d}\) = E₀, i.e. no change.

(iv) energy stored by the capacitor?
Answer:
The energy stored becomes
U = \(\frac{1}{2} C V^{2}=\frac{1}{2} K C_{0} V^{2}\) = KU₀.

Question 8.
An isolated air capacitor of capacitance C0 is charged to a potential V0. Now if a dielectric slab of dielectric constant K is inserted between its plates, completely filling the space between the plates, then how to do the following change, when the battery is disconnected
(i) charge
Answer:
If battery is disconnected then charge remains same, Q = Q₀

(ii) electric field between the plates,
Answer:
V = \(\frac{V_{0}}{K}\) ,
∴E = \(\frac{E_{0}}{K}\) ,

(iii) capacitance
C = KCO [∵ C = \(\frac{Q_{0}}{V}=\frac{Q_{0}}{V_{0} / K}=\frac{K Q_{0}}{V_{0}}\)]

(iv) energy stored by the capacitor
U = \(\frac{U_{0}}{K}\left[U=\frac{1}{2} C V^{2}=\frac{1}{2}\left(K C_{0}\right)\left(\frac{V_{0}}{K}\right)^{2}\right]\)

Question 9.
The figure shows two identical capacitors, C₁ and C₂, each of 1 µF capacitance connected to a battery of 6 V. Initially switch ‘S’ is closed. After some time ‘S’ is left open and dielectric slabs of dielectric constant K = 3 are inserted to fill completely the space between the plates of the two capacitors. How will (i) the charge and (II) potential difference between the plates of the capacitors be affected after the slabs are inserted? (CBSE Delhi 2011)

Answer:
When switch S is opened then capacitor C₁ remains connected to the battery white capacitor C₂ is disconnected. Thus for the two capacitors, we have

Physical quantity	Capacitor C1	Capacitor C2
Charge	Becomes K times, i.e. Q = CV = K C0V = 3 × 1 × 10-6 × 6 = 1.8 × 10-5 C	Remains same Q = Q0  = CV = 1 × 10-6 × 6 = 6 × 10-6 C
Potential difference	Remains same V = V0 = 6 V	Becomes 1/K times V =V0/K = 6/3 = 2V

Question 10.
A particle, having a charge +5 µC, is initially at rest at the point x = 30 cm on the x axis. The particle begins to move due to the presence of a charge Q that is kept fixed at the origin. Find the kinetic energy of the particle at the instant it has moved 15 cm from its initial position if (i) Q = +15 µC and (ii) Q = -15 µC (CBSE Sample Paper 2018-19)
Answer:
From energy conservation, U_i + K_i = U_f + K_f
kQq/r_i + 0 = kQq/r_f + K_f
K_f = kQq (1/r_i — 1/r_f)

When Q is +15 µC, q will move 15 cm away from it. Hence r_f = 45 cm
K_f = 9 × 10⁹ × 15 × 10^-6 × 5 × 10^-6 [1/(30 × 10^-6) – 1/(45 × 10^-2)] = 0.75 J

When Q is -15 µC, q will move 15 cm towards it. Hence r_f = 15 cm K_f = 9 × 10⁹ × (-15 × 10^-6) × 5 × 10⁶ [1/(30 × 10^-2) – 1/(15 × 10^-2)] = 2.25 J

Question 11.
Two metal spheres, one of radius R and the other of radius 2R, both have same surface charge density σ. They are brought in contact and separated. What will be new surface charge densities on them? (NCERT Exemplar)
Answer:
Let the two spheres have charges Q₁ and Q₂ respectively. Since σ₁ = σ₂, before contact, we have
\(\frac{Q_{1}}{4 \pi R^{2}}=\frac{Q_{2}}{4 \pi(2 R)^{2}}\)
Or
Q₂ = 4 Q₁

After contact
Let q1 and q2 be the charges on them, then
q₁ + q₂ = Q₁ + Q₂= Q₁ + 4Q₁ = 5Q₁ = 5(σ × 4πr²)

The two will exchange charge till their potentials are equal, therefore we have
\(\frac{q_{1}}{R}=\frac{q_{2}}{2 R}\)
Or
q₂ = 2 q₁

Therefore 3q₁ = 5(σ × 4πr²)
Or
q₁ = \(\frac{5}{3}\)(σ x 4πr²) and q₂ = 2q₁ = \(\frac{10}{3}\)5(σ x 4πr²)

Therefore
σ₁ = \(\frac{q_{1}}{4 \pi R^{2}}=\frac{5}{3}\left(\frac{\sigma \times 4 \pi R^{2}}{4 \pi R^{2}}\right)=\frac{5 \sigma}{3}\)
And
σ₂ = \(\frac{q_{2}}{4 \pi(2 R)^{2}}=\frac{10}{3}\left(\frac{\sigma \times 4 \pi R^{2}}{4 \pi(2 R)^{2}}\right)=\frac{5 \sigma}{6}\)

Question 12.
(a) Derive an expression for the capacitance of a parallel plate capacitor when the space between the plates is partially filled with a dielectric medium of dielectric constant ‘K’.
Answer:
Consider a parallel plate capacitor having each plate of area A and separated by a distance d. When there is a vacuum between the two plates, the capacitance of the parallel plate capacitor is given by
C₀ = ε₀ A/d.

Suppose that when the capacitor is connected to a battery, the electric field of strength E₀ is produced between the two plates of the capacitor. Further, suppose that when a dielectric slab of thickness t (t < d) is introduced between the two plates of the capacitor as shown in the figure, the electric field reduces to E due to the polarisation of the dielectric. Therefore between the two plates of the capacitor, over a distance of t, the strength of the electric field is E, and over the remaining distance (d – f)

the strength is E₀. If V is the potential between the plates of the capacitor, then

V = E_t + E₀(d – t)
Since E = E₀/K where K is the dielectric constant, the above equation becomes

V = \(\frac{E_{0}}{K} t+E_{0}(d-t)=E_{0}\left(d-t+\frac{t}{K}\right)\)

The electric field between the plates of the capacitor is given by
E₀ = σ / ε₀ = Q/ A ε₀

Hence the potential between the two plates becomes.
V = \(E_{0}\left(d-t+\frac{t}{K}\right)=\frac{Q}{\varepsilon_{0} A}\left(d-t+\frac{t}{K}\right)\)

Hence the capacitance of the parallel plate capacitor is given

∴ clearly, C > C₀. Therefore, capacitance increases in the presence of a dielectric medium.

(b) Explain why the capacitance decreases when the dielectric medium is removed from between the plates.
Answer:
On removing the dielectric, the capacitance will decrease.

Question 13.
(a) Obtain the expression for the energy stored per unit volume in a charged parallel plate capacitor.
(b) The electric field inside a parallel plate capacitor is E. Find the amount of work done in moving a charge q over a closed rectangular loop a b c d a. (CBSE Delhi 2014)

Answer:
(a) Suppose the capacitor is charged fully, its final charge is Q. and a final potential difference is V. These are related as Q = CV

Let q and V be the charge and potential difference, respectively.

At an intermediate stage during charging process q = CV. At this stage the small work done dW to transfer an additional charge dq is

dW = Vdq = \(\frac{q d q}{C}\)

The total work W needed to increase the capacitor’s charge q from zero to its final value Q is given by
W = \(\int_{0}^{w} d W=\int_{0}^{Q} \frac{q d q}{C}=\frac{1}{C} \int_{0}^{Q} q d q\)
Or
W = \(\frac{1}{C}\left|\frac{q^{2}}{2}\right|_{0}^{Q}=\frac{Q^{2}}{2 C}\)

This work is stored in the capacitor in the form of its electric potential energy. Hence,
U = \(\frac{Q^{2}}{2 C}\) … (1)

substituting Q = CV in equation (1) we have
U = \(\frac{1}{2}\) CV²

Energy density = Energy stored per unit volume

(b) Since the surface is an equipotential surface, work done is zero.

Question 14.
(a) Derive the expression for the capacitance of a parallel plate capacitor having plate area A and plate separation d.
Answer:
Suppose Q. is the charge on the capacitor, and c is the uniform surface charge density on each plate as shown in the figure. Therefore by Gauss’s theorem, the electric field between the plates of the capacitor (neglecting fringing of electric field at the edges) is given by

E = \(\frac{\sigma}{\varepsilon_{0}}=\frac{Q}{\varepsilon_{0} A}\) …. (1)

The field is uniform, and the distance between the plates is d, so the potential difference between the two plates is
V = Ed = \(\frac{1}{\varepsilon_{0}} \frac{Q d}{A}\) ….. (2)

Therefore by the definition of capacitance we have
C = \(\frac{Q}{V}=\frac{\varepsilon_{0} A}{d}\) …. (3)

This gives the capacitance of a parallel plate capacitor with a vacuum between plates.

(b) Two charged spherical conductors of radii R₁ and R₂ when connected by a conducting wire acquire charges q₁ and q₂ respectively. Find the ratio of their surface charge densities in terms of their radii. (CBSE Delhi 2014)
Answer:

The ratio of the surface charge densities is given by
\(\frac{\sigma_{1}}{\sigma_{2}}=\frac{q_{1}}{4 \pi R_{1}^{2}} \times \frac{4 \pi R_{2}^{2}}{q_{2}}\) ….(4)

Now when they are connected with a wire, their potentiaLs wilt be same: therefore, from the expression.
V = \(\frac{Q}{c}=\frac{Q}{R}\)

∵ C = 4πε₀R for a spherical body
∴\(\frac{q_{1}}{q_{2}}=\frac{R_{1}}{R_{2}}\)

Substituting in equation (4) we have
\(\frac{\sigma_{1}}{\sigma_{2}}=\frac{R_{1}}{R_{1}^{2}} \times \frac{R_{2}^{2}}{R_{2}}=\frac{R_{2}}{R_{1}}\)

Question 15.
(a) Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.
Answer:
The electrons are transferred to the positive terminal of the battery from the metallic plate connected to the positive terminal, leaving behind a positive charge on it. Similarly, the electrons move on to the second plate from the negative terminal, hence it gets negatively charged. This process continues till the potential difference between the two plates becomes equal to the potential of the battery.

Suppose the capacitor is charged fully; its final charge is Q and the final potential difference is V.
These are related as Q = CV

Let q and V be the charge and potential difference respectively, after some time during the charging of the capacitor, then q = CV. At this time, the small work done dW required to transfer an additional charge dq is given by

dW = Vdq = \(\frac{q d q}{C}\)

The total work W needed to increase the capacitor’s charge q from zero to its final value Q is given by

This work is stored in the capacitor in the form of its electric potential energy. Hence,
U = \(\frac{Q^{2}}{2 C}\) …(1)

Substituting Q= CVin equation (1) we have
U = \(\frac{1}{2}\) CV²

(b) A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from the battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor. (CBSE Delhi 2019)
Answer:
The charge stored on the capacitor q = CV, when it is connected to the uncharged capacitor of same capacitance, sharing of charge, takes place between the two capacitors till the potential of both the capacitors becomes V/2.

Energy stored on the combination (U₂)
= \(\frac{1}{2} C\left(\frac{V}{2}\right)^{2}+\frac{1}{2} C\left(\frac{V}{2}\right)^{2}=C\left(\frac{V}{2}\right)^{2}=\frac{C V^{2}}{4}\)

Energy stored on single capacitor before connecting
U₁ = –\(\frac{1}{2}\) CV²

Ratio of energy stored in the combination to that in the single capacitor.
\(\frac{U_{2}}{U_{1}}=\frac{C V^{2} / 4}{C V^{2} / 2}=\frac{1}{2}\)

Question 16.
(a) Obtain the expression for the potential due to an electric dipole of dipole moment p at a point ‘x’ on the axial line.
(b) Two identical capacitors of plate dimensions l × b and plate separation d have dielectric slabs filled In between the space of the plates as shown In the figures.

Obtain the relation between the dielectric constants K, K_1, and K₂. (CBSE Al 2013C)
Answer:
(a) Consider an electric dipole of length 2a and having charges + q and — q. Let us find the potential on the axial Une at point P at a distance OP = x from the center of the dipole.

Now potential at point P is

(b) When there is no dieLectnc then
C = \(\frac{\varepsilon_{0} l b}{d}\)

For the first capacitor
C’ = \(\frac{K \varepsilon_{0} l b}{d}\) = KC

The second case is a case of two capacitors connected in paralleL, therefore
C₁ = \(\frac{K_{1} \varepsilon_{0} l b}{2 d}\) and
C₂ = \(\frac{K_{2} \varepsilon_{0} l b}{2 d}\)

These two are connected in parallel, therefore we have
C” = C₁ + C₂ = \(\frac{K_{1} \varepsilon_{0} l b}{2 d}\) + \(\frac{K_{2} \varepsilon_{0} l b}{2 d}\)
= C\(\left(\frac{K_{1}+K_{2}}{2}\right)\)

If the capacitance in each case be same, then C’ = C”

Hence K= \(\left(\frac{K_{1}+K_{2}}{2}\right)\)

Question 17.
(a) Explain using suitable diagrams the difference in the behavior of a
(i) conductor:
Answer:
The diagram is as shown.

When a conductor is placed in an external electric field, the free charge carriers move and charge distribution in the conductor adjusts itself in such a way that the electric field due to induced charges opposes the external field within the conductor. This happens until in the static situation, the two fields cancel each other and the net electrostatic field in the conductor is zero.

(ii) dielectric In the presence of the external electric field. Define the term polarisation of a dielectric and write its relation with susceptibility.
Answer:

In a dielectric, this free movement of charges is not possible. It turns out that the external field induces dipole moment by stretching or re-orienting molecules of the dielectric. The collective effect of all the molecular dipole moments is that the net charges on the surface of the dielectric produce a field that opposes the external field. However, the opposing field so induced does not exactly cancel the external field. It only reduces it. The extent of the effect depends on the nature of the dielectric.

Dielectric polarization is defined as the dipole moment per unit volume of a dielectric. It is related to susceptibility as P = χ_eε₀\(\vec{E}\)

(b) A thin metallic spherical shell of radius R carries a charge Q on its surface. A point charge is placed at its center C and another charge +2Q. outside the shell, at a distance r from the center as shown in the figure. Find (i) the force on the charge at the center of the shell and at the point A and

Answer:
(i) F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q^{2}}{2 R^{2}}\) and
F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{(Q+Q / 2) \times 2 Q}{r^{2}}\)

(ii) the electric flux through the shell. (CBSE Delhi 2015)
Answer:
Flux = \(\frac{Q}{2 \varepsilon_{0}}\)

Question 18.
(a) Define the SI unit of capacitance.
Answer:
The capacity of a capacitor is said to be one farad when a charge of 1 coulomb is required to raise the potential difference by 1 volt.

(b) Obtain the expression for the capacitance of a parallel plate capacitor.
Answer:
Let Q. be the charge on the capacitor, and o be the uniform surface charge density on each plate as shown in the figure.

d = distance between plates of the capacitor.

Therefore by Gauss’ theorem, the electric field between the plates of the capacitor is given by
E = \(\frac{\sigma}{\varepsilon_{0}}=\frac{Q}{\varepsilon_{0} A}\)

The field is uniform, so the potential difference between the two plates is
V = Ed = \(\frac{1}{\varepsilon_{0}} \frac{Q d}{A}\)

Therefore, by the definition of capacitance we have
C = \(\frac{Q}{V}=\frac{\varepsilon_{0} A}{d}\)

This gives the capacitance of a parallel plate capacitor.

(c) Derive the expression of the effective capacitance of a series combination of n capacitance (CBSE Delhi 2016C)
Answer:
Capacitors in series. Capacitors are said to be connected in series if the second plate of one capacitor is connected to the first plate of the next and so on as shown in the figure.

Resultant capacitance will be
\(\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\ldots \ldots \ldots \ldots=+\frac{1}{C_{n}}\)

Question 19.
A charge Q is distributed over the surfaces of two concentric hollow spheres of radii r and R (R >> r), such that their surface charge densities are equal. Derive the expression for the potential at the common center.
Or
Three concentric metallic shells A, B, and C of radii a, b, and c (a <b < C) have surface charge densities +a, -a, and + o respectively as shown. Obtain the expressions for the potential of three shells A, B, and C. If shells A and C are at the same potential, obtain the relation between a, b and c. (CBSE Al 2019)

Answer:
Let the charges on the spheres be q, and q2 such that
Q=q₁ + q₂
= 4πσ(r² + R²)
Or
σ = \(\frac{Q}{4 \pi\left(r^{2}+R^{2}\right)}\)

Now potential at the common centre V=V₁ + V₂
V = \(\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{1}}{r}+\frac{q_{2}}{R}\right)\)
= \(\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{4 \pi r^{2} \sigma}{r}+\frac{4 \pi R^{2} \sigma}{R}\right)\)

V = \(\frac{(r+R) \sigma}{\varepsilon_{0}}\)

Substituting for o, we have

Numerical Problems :

Formulae for solving numerical problems

• V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\) electric potential at a point.
• Q = CV for a capacitor.
• C = 4πε₀R for a spherical conductor.
• C =\(\frac{Q}{V}=\frac{\varepsilon_{0} A}{d}\) capacitance of a parallel plate capacitor with air as dielectric.
• C = \(\frac{K \varepsilon_{0} A}{d}\) capacitance of a parallel plate capacitor with a dielectric.
• For capacitors in senes \(\frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}\) and for capacitors in parallel C_p = C₁ + C₂
  
• U = \(\frac{1}{2} \frac{Q^{2}}{c}\) = \(\frac{1}{2}\)CV² = \(\frac{1}{2}\)QV Energy stored in a capacitor.
• Energy density u = \(\frac{1}{2}\) ε₀E₂
  
• Capacitance of a parallel plate capacitor with a conducting slab of thickness t between plates is C = \(\frac{\varepsilon_{0} A}{d-t}\)
• Capacitance of a capacitor with dielectric slab of thickness t << d , C = \(\frac{\varepsilon_{0} A}{d+t\left(\frac{1}{K}-1\right)}\)
• Common potential V = \(\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}\)
• Loss of energy when two conductors are combined, U₁ – U₂ = \(\frac{C_{1} C_{2}}{2\left(C_{1}+C_{2}\right)}\left(V_{1}-V_{2}\right)^{2}\)
• If n small drops each having a charge Q, capacitance C, and potential V coalesce to form a big drop, then

1. The charge on the big drop = nQ
2. The capacitance of the big drop = n^1/3 C
3. Potential of the big drop = n^2/3 V
4. The potential energy of the big drop = n^5/3 U

Question 1.
Electric field intensity at point B due to a point charge Q kept at point A is 24 N C^-1 and the electric potential at point B due – to the same charge is 12 J C^-1. Calculate the distance AB and also the magnitude of the charge Q.
Answer:
Given E = 24 N C^-1 , V = 12 J C^-1 , r = ? and Q = ?

Using the relation
E = \(\frac { V }{ r }\)
Or
\(\frac { V }{ E }\) = \(\frac { 12 }{ 24 }\) = 0.5 m

Also V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}\)

Therefore 12 = 9 × 10⁹ × \(\frac { Q }{ 0.5 }\) , solving for Q

we have Q = 6.67 × 10^-10C

Question 2.
A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 360 μC. When the potential across the capacitor is reduced by 120 V, the charge stored in it becomes 120 μC. Calculate:
(i) The potential V and the unknown capacitance C.
(ii) What will be the charge stored in the capacitor, if the voltage applied had increased by 120 V? (CBSE Delhi 2013)
Answer:
Given Q₁ = 360 μC, Q₂ = 120 μC,
V₂ = (V- 120) volt, V₁ = V

(i) We know that C = \(\frac { Q }{ V }\)

Since capacitance is same, we have
\(\frac{Q_{1}}{V_{1}}=\frac{Q_{2}}{V_{2}}\)
Or
\(\frac{360 \times 10^{-6}}{V}=\frac{120 \times 10^{-6}}{V-120}\)

Solving for V we have V= 180 volt

Also C = Q₁/ V₁ = 360 × 10^-6 / 180 = 2 × 10^-6 C = 2 pF

(ii) V= 180 + 120 = 300 V
Therefore Q = CV = 2 x 10^-5 × 300 = 600 x 10^-6 = 600 pC

Question 3.
Two identical capacitors of 12 pF each are connected in series across a 50 V battery. Calculate the electrostatic energy stored in the combination. If these were connected in parallel across the same battery, find out the value of the energy stored in this combination. (CBSE Al 2019)
Answer:
Given C = 12 pF = 12 × 10^-12 F, V= 50 V,
In series
C_s = \(\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\frac{12 \times 12}{12+12}\) = 6 pF

Hence energy stored
U_s = \(\frac{1}{2}\) C_sV₂ = \(\frac{1}{2}\) × 6 × 10^-12× (50)²
U_s = 7.5 × 10^-9 J

In parallel
C_P = C₁ + C₂ = 12 + 12 = 24 pF
U_P = \(\frac{1}{2}\) C_PV² = \(\frac{1}{2}\) × 24 × 10^-12 × (50)²
U_P = 3 × 10^-8 J

Question 4.
The figure shows a network of three capacitors C₁ = 2 μF; C₂ = 6 μF and C₃ = 3 μF connected across a battery of 10 V. If a charge of 6 μC is acquired by the capacitor C₃, calculate the charge acquired by C₁ (CBSE Al 2019)

Answer:
Capacitors C₂ and C₃ are connected in parallel, therefore, the net capacitance of the combination.
C₂₃ = (6 + 3) = 9 μF

Let V1, be the potential across C1 and V2 be the potential across C₂₃
Now C₁ = Q/V, or V₁ = Q/C₁ = Q/2

Also V₂ = Q/9
But V = V₁ + V₂

Solving for Q we get
Q= 16.4 μC

Question 5.
(a) Find equivalent capacitance between A and B in the combination given below. Each capacitor is of 2 μF capacitance.

Answer:
This can be redrawn as

Like C₂, C₃ and C₄ are in parallel,
C₂₃₄ = C₂ + C₃ + C₄ = 6 μF

Further, C₁, C₂₃₄ and C₅ are in series
\(\frac{1}{C_{\text {net }}}=\frac{1}{2}+\frac{1}{6}+\frac{1}{2}=\frac{3+1+3}{6}=\frac{7}{6}\)
\(C_{\text {net }}=\frac{6}{7}\)μF

(b) If a DC source of 7 V is connected across AB, how much charge is drawn from the source and what is the energy stored in the network? (CBSE Delhi 2017)
Answer:
∴ C_net = \(\frac { 6 }{ 7 }\) μF ,

q = C_net V = \(\frac { 6 }{ 7 }\) × 10^-6 × 7 = 10^-6C

∴ Energy stored = \(\frac{1}{2}\)qV

= \(\frac{1}{2}\) × 6 × 10^-6 × 7 = 21 × 10^-6J

Question 6.
Two identical capacitors of 12 pF each are connected in series across a battery of 50 V. How much electrostatic energy is stored in the combination? If these were connected in parallel across the same battery, how much energy will be stored in the combination now? Also, find the charge drawn from the battery in each case. (CBSE Delhi 2017)
Answer:
(i) Net capacitance C_net = \(\frac{C_{1} C_{2}}{C_{1}+C_{2}}\)
= \(\frac{12 \times 12}{12+12}\)pF = 6 pF

∴ Energy stored = \(\frac{1}{2}\)C_net V²
= \(\frac{1}{2}\) × 6 × 10^-12 × (50)² = 75 × 10^-10J
q = Chargedrawn = C_net V=6 × 10^-12 × 50 = 3 × 10^-10 C

(ii) C_net=12 + 12 = 24pF
∴ Energy stored = \(\frac{1}{2}\) C_net V²
= \(\frac{1}{2}\) × 24 × 10^-12 × (50)²
= 3 × 10^-8J

Charge drawn, q = C_netV
= 24 × 10^-12 × 50
= 1200 × 10^-12
= 12 × 10^-10 C

Question 7.
In the following arrangement of capacitors, the energy stored in the 6 μF capacitor is E. Find the value of the following. (Foreign 2016)
(a) Energy stored in 12 pF capacitor.
(b) Energy stored in 3 pF capacitor.
(c) Total energy drawn from the battery.

(a) Energy stored in 6 μ capacitor is E. Capacitors 6 μF and 12 μF are connected in parallel.
So, the voltage across 6 μF capacitor
= voltage across 12 μF capacitor
= voltage across 12 μF capacitor
= V (say)

As E = \(\frac{1}{2}\) × 6 × V²
∴ V = \(\sqrt{\frac{E}{3}}\)

Similarly energy U’ stored in 12 pF capacitor
= \(\frac{1}{2}\) × 12 × V²
= \(\frac{1}{2}\) ×12 × \(\frac{E}{3}\) = 2E

U’ = 2 E

(b) Equivalent capacitance of 6 µF and 12 µF is 6 + 12 = 18 µF

Charge on 18 µF and 3 µF is same as they are in series as they are in series
∴ Q = CV= 18 × V

∴ Energy in 3 µF capacitor
U” = \(\frac{1}{2} \frac{Q^{2}}{C}=\frac{1}{2} \frac{(18 V)^{2}}{3}\)

U” = \(\frac{1}{2} \times \frac{18 \times 18}{3} \frac{E}{3}\)
U” = 18E

(c) Total energy drawn from battery U = E + 2E + 18E = 21E

Question 8.
Two parallel plate capacitors X and Y have the same area of plates and the same separation between them, X has air between the plates, while Y contains a dielectric medium of ε_r = 4.
(a) Calculate the capacitance of each capacitor if the equivalent capacitance of the combination is 4 μF.

C_X = \(\frac{\varepsilon_{0} A}{d}\) = C(say)
C_Y= 4\(\frac{\varepsilon_{0} A}{d}\) = 4C

EquivaLent capacitance = 4 μF
C_eq = \(\frac{C_{X} C_{Y}}{C_{X}+C_{Y}}\) = 4
= \(\frac{C \times 4 C}{C+4 C}\)
Or
= \(\frac{4C}{5}\) = 4

C = 5 μF
∴ C_X = C = 5 μF and
C_Y = 4C = 20 μF

(b) Total charge, q = C_eq V = 4 × 12 = 48μC
C_X and C_Y are in series. Hence, charge on both is 48 μC each.

∴ The potential difference across C_X,
V_X = \(\frac{q}{C_{X}}=\frac{48}{5}\) = 9.6 V0lt

The potential difference across C_Y,
V_Y = \(\frac{q}{C_{Y}}=\frac{48}{20}\) = 2.4 Volt

(c) U_X=\(\frac{1}{2}\)CXVX² ; UY = \(\frac{1}{2}\)CYVy²

∴ \(\frac{U_{x}}{U_{Y}}=\frac{5 \times(9.6)^{2}}{20 \times(2.4)^{2}}\)
= \(\frac{1}{4} \times \frac{9.6 \times 9.6}{2.4 \times 2.4}\) = 4

\(\frac{U_{x}}{U_{r}}\) = 4

Question 9.
Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C₁ and C₂ with their capacitances in the ratio 1:2 so that the energy stored in these two cases becomes the same. (CBSE Al 2016)
Answer:
(i) Let C₁ = C, ∴ C₂ = 2C
For series combination equivalent capacitance is
\(C_{s}=\frac{C_{1} \times C_{2}}{C_{1}+C_{2}}=\frac{C \times 2 C}{3 C}=\frac{2}{3} C\) …. (1)
and energy stored,
\(U_{\mathrm{s}}=\frac{1}{2} C_{s} V_{\mathrm{s}}^{2}=\frac{1}{2} \times \frac{2}{3} C V_{\mathrm{s}}^{2}\) …. (2)

For parallel combination equivalent capacitance
C_p = C + 2C = 3C ….(3)
and energy stored
∴ \(U_{D}=\frac{1}{2} 3 C \times V_{p}^{2}\) …(4)

But U_s = U_p [Given]

Question 10.
Calculate the potential difference and the energy stored in the capacitor C₂ in the circuit shown in the figure. Given potential at A is 90 V, C₁ = 20 μF, C₂ = 30 μF, and C₃= 15 μF. (CBSEAI 2015)

Answer:

C = \(\frac{60}{9}\) μF = \(\frac{20}{3}\) μF

The potential at A = 90V

∴ Charge on each capacitor
Q = C x V= \(\frac{20}{3}\) x 90 =600 μC.

∴ Charge on C₂ = 600 μC

∴ V₂ = \(\frac{Q}{C_{2}}=\frac{600}{30}\) ; V₂ = 20 V

Energy stored in C₂ = \(\frac{1}{2}\)C₂V₂²

U₂ = \(\frac{1}{2}\) × 3o × 10^-6 × 20 × 20
U₂ = 6000 × 10^-6J = 6 × 10^-3J

Here we are providing Class 12 Biology Important Extra Questions and Answers Chapter 13 Organisms and Populations. Important Questions for Class 12 Biology are the best resource for students which helps in Class 12 board exams.

Class 12 Biology Chapter 13 Important Extra Questions Organisms and Populations

Organisms and Populations Important Extra Questions Very Short Answer Type

Question 1.
Define microclimate.
Answer:
Microclimate represents the climatic

Question 2.
Define habitat and niche.
Answer:
Habitat is the native environment of conditions that prevail at a local scale or in animal or place. area of limited size. Niche is the position or function of an organism in a community of plants and animals.

Question 3.
How is thermoregulation achieved in polar bears?
Answer:
Thermoregulation in the polar bear. By developing blubber in a subcutaneous zone prevents loss of body heat.

Question 4.
What are osmoconformers? Give one example.
Answer:
Osmoconformers. Animals that can change the osmolarity of their body fluids according to that of the surrounding medium are termed osmoconformers, e.g. Myxine.

Question 5.
What factors cause annual variations in the intensity and duration of temperature?
Answer:

1. Rotation of the earth around the Sun.
2. The tilt of the earth on its axis.

Question 6.
Species that can tolerate a narrow range of temperature are called ………………..
Answer:
Stenothermic.

Question 7.
What are eurythermic species?
Answer:
Species that can tolerate a wide range of temperature variations are called eurythermic.

Question 8.
Species that can tolerate a wide range of salinity are called ……………………
Answer:
Euryhaline.

Question 9.
Define stenohaline species.
Answer:
It is a species that lives within a narrow range of salinity.

Question 10.
Name two factors that cause the formation of major biomes.
Answer:

1. Variation in the intensity and duration of temperature.
2. Variation in precipitation.

Question 11.
What is the main cause of salinity?
Answer:
Salinity is due to the accumulation of soluble minerals on the surface or beneath the surface of the earth.

Question 12.
What does the stratification of community depict?
Answer:
Stratification of a community depicts vertical layering of vegetation.

Question 13.
From where are individual organisms derived?
Answer:
Individual organisms are always derived from pre-existing organisms through the mechanism of reproduction-may be vegetative, asexual or sexual.

Question 14.
List two negative interactions between two species.
Answer:
Parasitism, Predation.

Question 15.
What is the other term for facultative mutualism?
Answer:
Protocooperation.

Question 16.
Name the association in which one species produces a poisonous substance or a change in environmental conditions that is harmful to another species. (CBSE Delhi 2019)
Answer:
Amensalism.

Question 17.
What is mycorrhiza?
Answer:
It is a symbiotic or mutually beneficial association between a fungus and roots of higher plants.

Question 18.
Emergent land plants that can tolerate the salinity of the sea are called ………………..
Answer:
Mangrove plants.

Question 19.
Give an example of a commensal relationship.
Answer:

1. Epiphytes and tree,
2. Remora fish and shark.

Question 20.
Define community periodicity.
Answer:
The recurrence of daily or seasonal or lunar changes in a community is called community periodicity.

Question 21.
Mention any two significant roles predation plays in nature. (CBSE 2008)
Answer:
Role of predators:

1. Transferring energy to a higher trophic level.
2. They keep the prey population under control.

Question 22.
Why is the polar region not a suitable habitat for a tiny hummingbird? (CBSE 2008)
Answer:
Tiny animals like hummingbird have more surface area compared to their volume. So heat loss may occur which will not be good for the tiny bird in polar regions.

In polar regions, only animals with a thick layer of fat below the skin can survive.

Question 23.
Between birds and amphibians which will be able to cope with global warming? Give reason. (CBSE2008)
Answer:
Birds will be able to cope with global warming because they are homeotherms as body temperature remains constant irrespective to change in surroundings.

Question 24.
Which one of the two stenothermal or eurythermal shows a wide range of distribution on the earth? (CBSE 2008)
Answer:
Eurythermal shows a wide range of distribution on the earth because they can tolerate a wide range of temperature.

Question 25.
In a pond, there were 30 Hydrilla plants. Through reproduction 10 new Hydrilla plants were added in a year. Calculate the birth rate of the population. (CBSE 2010)
Answer:
A number of individuals added = 10 per 20 Hydrilla plants.

\frac{10}{30}= 0.33 offspring per hydrilla plant per year.

Question 26.
How is ‘stratification’ represented in a forest ecosystem? (CBSE Outside Delhi 2014)
Answer:
Stratification is a grouping of plants in a forest into two or more well-defined layers depending upon height like tall trees, medium-sized trees, small trees, bushes and herbs. In the rainforest, it is multi-storeyed.

Question 27.
Give an example of an organism that enters ‘diapause’ and why? (CBSE Delhi 2014)
Answer:
Under unfavourable conditions, many zooplanktonic species in lakes and ponds enter a stage of suspended development called diapause.

Organisms and Populations Important Extra Questions Short Answer Type

Question 1.
Certain species of wasps are seen to
frequently visit flowering fig trees. What type of interaction is seen between them and why? (CBSE 2008, 2014)
Answer:

1. Mutualism.
2. Flowers of a fig tree can only be pollinated by wasp species. In return, female wasp visits the fruit of fig as not only an egg-laying site but also uses the developing seed in fruit for nourishing its larvae.

Question 2.
Write a short note on microclimate.
Answer:
Microclimate: It represents the climatic conditions that prevail at a local scale, or in areas of limited size such as the immediate surroundings of plants and animals. Microclimate generally differs from the prevailing regional climatic conditions. For example, in a forest, dense foliage reduces the amount of light reaching the ground. This also results in a changed air temperature profile. The day-time air temperature inside the forests is lower than the outside temperature. Also, the interior of a forest may be more humid than a nearby non-forested area.

Question 3.
What are the effects of organisms on habitat?
Answer:
Effects of organisms on habitat:

1. Excessive growth may lead to the death of the habitat as in the case of water hyacinth- an increase in population had led to the death of Hussain Sagar lake of Hyderabad.
2. The predominance of a predator will reduce the population of prey.
3. Man manipulates the habitat and has affected the forest causing serious problems due to deforestation. More so man has created pollution of air, water and land.

Question 4.
How do you differentiate habitat from the environment?
Answer:
Differences between habitat and environment:

Habitat	Environment
(i) It is the living place of organisms.	(i) It Is a specific region surrounding the population of organisms.
(ii) It is a part of the total environment.	(ii) It is larger than a habitat.
(iii) It must offer food, shelter and climate condition suitable for the organism to live and flourish.	(iii) It has a special condition suitable for specific organisms. Example. Desert rats of Rajasthan are not found in plains of U.P and M.P

Question 5.
Differentiate microhabitat. between habitat and
Answer:
Differences between habitat and microhabitat:

Habitat	Microhabitat
(i) It is a living place of an organism.	(i) It is a more localised part of the habitat.
(ii) It is a part of the total environment of the region.	(ii) It is a part of the habitat.
(iii) It has a common climate for all organisms. Example. Sundarban forests are the habitat of Bengal tigers.	(iii) It is mostly suited for specific animals, e.g. sediments of pond or togs.

Question 6.
List the means by which organisms survive at freezing temperatures.
Answer:

1. The animals are usually white or light coloured. The light colouration of animals helps in camouflage with snow and thermal regulation.
2. Animals are thickly coated with fur.
3. Presence of a thick layer of fat below the skin.

Question 7.
Differentiate eurythermal, stenothermal and euryhaline animals.
Answer:
Differences between eurythermal, stenothermal and euryhaline animals:

Eurythermal Animals	Stenothermal Animals	Euryhaline Animals
These are the animals/ organisms which can tolerate a wide range of temperatures.	These are the animals/ organisms which can tolerate only: a narrow range of temperatures.	These are the animals that can tolerate a wide range of salinity of the medium.

Question 8.
Explain how is an orchid plant adapted to changes in temperature and humidity.
Answer:
An orchid plant, e.g.Venda, an epiphyte, is a native of tropical forests of India and South Asia.

Its adaptations are:

• Stem stores water and leaves are adapted to keep water in and dryness out.
• Roots of the plant secure their attachment to the bark of the tree and help in the absorption of moisture from humid air or rain.

Question 9.
Why do submerged plants receive weaker illuminations than exposed floating plants in a lake?
Answer:
Because light intensities of stronger illuminations are absorbed by the exposed floating plants of the lake/water-bodies, so only the weaker illuminations reach the submerged, Some light is reflected at the water surface and a part is absorbed by upper layers of water.

Question 10.
Categorise the following plants into hydrophytes, halophytes, mesophytes and xerophytes. Give reasons for your answers.
(i) Salvinia
Answer:
Hydrophyte. Because it is not able to tolerate deficiency of water. It is partially or completely submerged.

(ii) Opuntia
Answer:
Xerophyte. Because it prefers a dry and hot climate with low rainfall, thus has succulent leaves.

(iii) Rhizophora
Answer:
Halophyte. Because it grows in a saline habitat.

(iv) Marigifera
Answer:
Mesophyte. Because it prefers areas with high moisture content.

Question 11.
If a freshwater fish is placed in an aquarium containing seawater, will the fish be able to survive? Explain giving reasons. (CBSE Delhi 2015)
Answer:
No, a freshwater fish placed in the aquarium containing sea-water will not be able to survive. Because its body system is adapted to function normally in a narrow range of salinity and it cannot survive in the high salinity of sea-water.

Question 12.
Distinguish between population and community.
Answer:
Differences between population and community:

Population	Community
(i) Group of individuals of same species inhabiting the same area.	(i) Group of species lying in the same area.
(ii) Organism in a population undergoes the same life cycle.	(ii) Different species have a different life cycle.

Question 13.
How does population size increase or decrease?
Answer:

• Population (size/density) increases by birth, hatching, germination and immigration that add individuals.
• Population (size/density) decreases by death and emigration. The population is regulated by food, space, disease, natural calamities and environmental factors.

Question 14.
Differentiate between natality rate and death/mortality rate.
Answer:
Differences between natality rate and mortality rate:

Character	Natality rate	Morality rate
(i) Definition	A number of births per 1,000 individuals of a population per year.	A number of death per 1,000 individuals of a population per year.
(ii) Population size and population density.	Increases	Decreases

Question 15.
Discuss the role of predators in an ecosystem.
Answer:
Predators are of great importance as they play the following important roles in an ecosystem:

1. They act as ‘conduits’ for energy transfer to higher trophic levels.
2. They keep the prey population under control, which otherwise can reach very high population density and cause an imbalance in the ecosystem,
3. They help in maintaining species diversity in a community by reducing the intensity of competition among the competing prey species.

Question 16.
What is brood parasitism? Give an example.
Answer:
Brood parasitism. It refers to the phenomenon in which one (parasitic) bird species lays its eggs in the nest of another bird species.

Evolution has occurred in such a way that the eggs of the parasitic birds resemble those of the host bird in size, colour, etc. to avoid the host bird detecting the foreign eggs and ejecting them from the nest. Example. Cuckoo birds lay eggs in the nests of crows.

Question 17.
List a few adaptations that parasites have developed. (CBSE Delhi 2019)
Answer:
Parasites have evolved one or more of the following adaptations:

1. Loss of unnecessary sense organs.
2. Presence of hooks/adhesive organs and suckers.
3. Loss of digestive system.
4. High reproductive capacity.
5. Produces antitoxins to counter toxins to the host.

Question 18.
How do parasites harm the host?
Answer:
The parasites harm the host in the following ways:

1. Reduces the survival of the host.
2. Growth and reproductive rates of hosts are reduced.
3. Render the most vulnerable to its predators by making them physically weak.

Question 19.
Justify the statement “Predators and scavengers are markedly different.”
Answer:
Predators feed on another organism, i.e. prey, whereas scavenger feeds on dead animals or an animal killed by another animal. A predator can be a prey also, e.g. a frog eats insects and the frog may be eaten by a snake. But a scavenger such as jackals, hyenas and vultures cannot kill zebra or giraffe or deer but a lion kills them and leaves a part of it to be eaten by such scavengers.

Question 20.
Mention any two significant roles predators play in nature. (CBSE 2008, 2016)
Answer:
Role of predators:

1. Transferring energy to a higher trophic level.
2. Keeping the prey population under control.

Question 21.
How do organisms manage stressful conditions existing in their habitat for a short duration? Explain with the help of one example each. (CBSE (Delhi) 2008)
Answer:
Physiological adaptations allow some organisms to respond quickly to stressful conditions. At high altitude, mountain (altitude) sickness is experienced resulting in nausea, fatigue and heart palpitations. It is due to low atmospheric pressure at high altitude and the body does not receive the proper amount of oxygen. It is accommodated by increasing RBC production, decreasing the binding capacity of haemoglobin and increasing breathing rate.

Question 22.
Explain with the help of an example each of any three ways the ecologists use to measure the population density of different organisms rather than by calculating their absolute number. (CBSE Delhi 2019 C)
Answer:
Method to measure population density: Population density is the number of individuals of a species per unit area or volume.

P_D= N/S

Where N = The no. of the individual in the region
S = No. of the unit area in a region

1. Number of animals per square kilometre
2. Number of trees per hectare
3. A number of phytoplanktons per cubic litre of water. Sometimes the relative densities also serve the purpose, that is

• The number of fish caught per trap is good enough to measure the population of fish.
• Tiger census is based on pug marks and faecal pellets.

Question 23.
Name the interaction in each of the following:
(i) The cuckoo lays her eggs in the crow’s nest.
Answer:
Brood parasitism.

(ii) Orchid grows on a mango tree.
Answer:
Commensalism.

(ii) Ticks live on the skin of dogs.
Answer:
Parasitism.

(iv) Sea anemone is often found on the shell of a hermit crab. (CBSE Delhi 2008)
Answer:
Mutualism.

Question 24.
Name the interaction in each of the following:
(i) Cuscuta growing on a shoe flower plant.
Answer:
Parasitism.

(ii) Mycorrhizae living on the roots of higher plants.
Answer:
Mutualism.

(iii) Clownfish living among the tentacles of a sea anemone.
Answer:
CommensaLism.

(iv) Koel laying its eggs in the crow’s nest. (CBSE Delhi 2011)
Answer:
Brood parasitism.

Question 25.
Name the interaction In each of the following:
(i) Ascaris worms living in the intestine of human.
Answer:
ParasItism

(ii) Suckerfish attached to the shark.
Answer:
Commensalism

(iii) Smaller barnacles disappeared when Balanus dominated the coast of Scotland.
Answer:
Competition

(iv) Wasp pollinating fig inflorescence. (CBSE 2008)
Answer:
Mutualism.

Question 26.
Why do clownfish and sea anemone pair up? What is this relationship called? (CBSE 2012)
Answer:

1. Commensalism,
2. Interaction between sea-anemone that has stinging tentacles and the clownfish that lives among them in which fish gets protection from predators. The anemone does not appear to derive any benefit from the clownfish.

Question 27.
Some organisms suspend their metabolic activities to survive in unfavourable conditions. Explain with the help of any four examples. (CBSE 2012)
Answer:

1. Thick-walled spores of bacteria and fungi help them to overcome unfavourable conditions.
2. Frogs and lizards undergo hibernation during the winter season.
3. Snails and fish undergo aestivation during summer to avoid summer-related problems of heat and desiccation.
4. Species of zooplankton are known to enter a suspended state of development termed-diapause.

Question 28.
A moss plant is unable to complete its life cycle in a dry environment. State reason. (CBSE Delhi 2015)
Answer:

1. Due to the absence of vascular tissue, water and minerals cannot be transported to various parts.
2. Transfer of flagellated antherozooids to oospore of archegonium depends upon water.

Question 29.
Plants that inhabit a rain-forest are not found in wetlands. Explain. (CBSE Delhi 2016)
Answer:
A habitat is a specific place or area or locality having a combination of factors, physical features and barrier where a community resides. In rainforest plants adapt themselves to the specific conditions of their habitat and the characteristics adaptations related to a particular habitat are not found in those plants living in wetland and vice versa. Thus they can not inhabit a new habitat.

Question 30.
Answer:
In certain seasons we sweat profusely while in some other season we shiver. Explain. (CBSE Delhi 2016)
Answer:
Humans are a warm-blooded or homeothermic or endothermal organism. They maintain a fixed body temperature despite changes in the surrounding. The optimum temperature for maximum efficiency of enzymes is 37°C. Thus the human body maintains a body temperature near it.

Homoeostasis is the phenomenon of maintaining a constant internal environment. Thus during heavy exercise or summer, sweating occurs. It helps in cooling the body. During winter, shivering helps in retaining heat within the body.

Question 31.
How did David Tillman show that the “Stability of a community depends on its species richness”? Explain. (Outside Delhi 2019)
Answer:
The stability of a community depends on its species richness. David Tilman’s long-term ecosystem experiments using outdoor plots provided the answer for species richness. Tilman found that plots with more species showed less year-to-year variations in total biomass. He also showed that increased diversity contributed to higher productivity.

Thus he established that the stability of a community depends on its species richness.

Organisms and Populations Important Extra Questions Long Answer Type

Question 1.
(i) “Organisms may be conformers or regulators.” Explain this statement and give one example of each.
(ii) Why are there more conformers than regulators in the animal world? (CBSE 2017)
Or
Human is categorised as ‘regulators’. Explain how they maintain a constant body temperature. (CBSE Outside Delhi 2019)
Answer:
(i) Regulators are organisms that are able to maintain homoeostasis by physiological and behavioural means. All birds and mammals and few lower vertebrate and invertebrate species maintain homoeostasis by thermoregulation and osmoregulation, e.g. birds and mammals.

However, a majority (99%) of animals and nearly all plants cannot maintain a constant internal environment, i. e. their body temperature is not constant. They are conformers, e.g. fishes, frogs, etc.

(ii) There are more conformers than regulators in the animal world because conformers lack the capability to maintain a constant internal environment or homoeostasis. It is limited to birds, mammals and few lower vertebrate and invertebrate species only because it is a very energy-expensive process.

Question 2.
Explain the ecological hierarchy.
Answer:
Ecological hierarchy: It is a series of graded ecological categories.
Characteristics of ecological hierarchy:

1. A biological unit at each level has a specific structure and function.
2. In this hierarchy, smaller biological units coordinate to form the next higher level of organisation.
3. Only the organisms show free existence.
4. Organisms cannot live in isolation.
5. At each level, different units show interdependence.
6. At each level, the unit shows interaction with the physical environment (energy and matter).
7. The biological units are successfully adapted to their environment.

Question 3.
In a seashore, the benthic animals live in sandy, muddy and rocky substrata and accordingly developed the following adaptations.
Find suitable substratum against each adaptation.
(i) Burrowing
Answer:
In a seashore, the benthic animals adapt to their mode of life according to the nature of the seafloor, e.g. Benthic animals become fossorial (burrowing) in the sandy substratum, e.g. tube worm.

(ii) Building cubes
Answer:
Benthic animals build cubes in the muddy substratum.

(iii) Holdfasts/peduncle
Answer:
Benthic animals develop holdfast/ peduncle if the substratum is rocky.

Question 4.
In a pond, we see plants that are free. floating, rooted-submerged, rooted-emergent, rooted with floating leaves. Write the type of plant against the following examples:
(i) Hydrilla
Answer:
Hydrilla is Rooted submerged.

(ii) Typha
Answer:
Typha is Rooted emergent.

(iii) Nymphaea
Answer:
Nymphaea is Rooted with floating leaves.

(iv) Lemna
Answer:
Lemna is Free-floating.

(v) Vallisneria.
Answer:
Vallisneria is Rooted submerged.

Question 5.
Why do all the freshwater organisms have contractile vacuoles whereas the majority of marine organisms lack them?
Answer:
Contractile vacuole helps in maintaining salt and water level called osmoregulation. Because of the cellular environment of a freshwater organism such as Amoeba, Paramecium etc. being hypertonic, the water diffuses inside the cell constantly and gets collected in the contractile vacuole, which squeezes the extra water out of the cell periodically.

While in the case of marine protozoans organisms, this does not occur due to high salt concentration. These organisms live in isotonic conditions in seawater. Thus there is no need for contractile vacuole.

Question 6.
Explain the following terms:
(i) Mimicry
Answer:
Mimicry: It is a phenomenon in which a living organism modifies its form, appearance, structure or behaviour and looks like another living organism or some inanimate (non-living) object so as to defend from its predators, or to increase the chances of capturing the prey.

The individual which shows mimicry is called a mimic, while the animate or inanimate object with which a mimic resembles is called a model. The concept of mimicry was first observed by an English naturalist, Henry Bates (1862 A.D.), so the phenomenon is also called Batesian mimicry.

Types: Mimicry is of three types:
(a) Protective mimicry,
(b) Aggressive mimicry,
(c) Feigning death or Conscious mimicry.

(ii) Acclimatisation
Answer:
Acclimatisation: The gradual physiological adjustment to slowly changing new environmental conditions is known as acclimatisation. If some factors shift beyond the tolerance range, the organism can come to a tolerance range or migrate to acclimatise.

Question 7.
DepIct the temperature-based thermal stratification in lakes.
Answer:
Thermal stratification in Lakes:

Thermal stratification occurs in lakes, the seasonal mixing patterns of a lake are determined by its temperature profile.

Question 8.
1. How does age distribution help in the study of the population?
2. How does an age pyramid, for the human population at a given point of time helps the policymakers in planning the future? (CBSE Delhi 2016)
Or
Draw a stable human age pyramid. (CBSE Outside Delhi 2019)
Answer:
Age distribution (Age composition). The relative abundance of the organisms of various age groups in the population is called the age distribution of the population. With regard to age distribution, there are three kinds of populations:

1. Rapidly growing population. It has a high birth rate and low death rate, so there is more number of young individuals in the population. According to a recent survey, more than 42% of the Indian population consists of children below the age of 14 years, so the Indian population is called the young population.

2. Stationary population: It has equal birth and death rates, so the population shows zero population growth.

3. Declining population. It has a higher death rate than the birth rate, so the population of young members is lower than that of old members.
(a) For the human population, the age pyramids generally show the age distribution of males and females in a combined diagram.
(b) The shape of the age pyramid reflects the growth status of the population. Thus age pyramid for the human population at a given time helps the policymakers in planning for the future.

Question 9.
Discuss life-history traits of an organism that have evolved in relation to the constraints imposed by biotic and abiotic factors in their habitat.
Answer:
According to ecologists, life-history traits of an organism have evolved in relation to the constraints imposed by the biotic and abiotic factors in their habitats.

1. It can be illustrated with vast variations and life history.
2. The evolution of populations aims at improving reproductive fitness or Darwinian fitness to the maximum in their habitats.
3. They evolve towards the most efficient reproductive strategy.
4. Organisms like Pacific Salmon fish and bamboo breed only once in their lifetime.
5. Most birds and mammals breed many times during their lifetime.
6. Oysters and pelagic fishes produce a large number of small-sized offspring.
7. Birds and mammals produce a small number of large-sized offspring.

Question 10.
What is the predator-prey relationship? Give example. (CBSE 2016)
Answer:
Predation. It is an interspecific interaction, where one animal (called predator) kills Organisms and Populations and consumes the other weaker animal (called prey). Herbivores are predators of plants.

Predator-prey relationship. It is a kind of direct food relationship between two species of animals in which larger species, called predator, attacks, kills and feeds on the smaller species, called prey. It was proved by G.F. Gause (1934).

Although predation and competition appear to be very harmful processes, they are essential to keep a check on the size of the population of other species.

Predation is nature’s way of transferring the energy fixed by plants, to higher trophic levels.
Examples:

1. A tiger killing and eating a deer.
2. A snake eating a frog.
3. A sparrow eating a fruit/seed.

Question 11.
Many prey organisms have developed different defence mechanism. Give a few examples.
Answer:
Prey species have developed various defence mechanisms to reduce the impact of predation; some of them are listed below:

1. Certain insect species and frogs have camouflage (cryptic colouration) to avoid detection by their predators.
2. Some animals (e.g. monarch butterfly) are highly distasteful to their predators. This butterfly species accumulates a chemical by feeding on a poisonous weed during its caterpillar stage.
3. Some prey species are poisonous and hence are avoided by predators, e.g. Dart frogs like Phyllobates bicolour and Dendrobats.

Question 12.
“Herbivores are the predators of plants”. Discuss a few defence mechanisms of plants against herbivory.
Or
Write what do phytophagous insects feed on? (CBSE2012)
Answer:
Herbivores are the predators of plants. The problem of predation is more severe for plants than animals as the plants cannot move away from the predators. About 25% of the known insects are phytophagous and feed on the sap and other parts of plants.

Plants have developed certain morphological and chemical defence mechanisms against herbivores; a few of them are listed below:

1. Morphological: Thorns (Bougainvillaea) and spines (Acacia, Cactus, etc.) are the most common morphological means of defence.
2. Chemical: Plants produce and store certain chemicals which function in one or more of the following ways:
   (a) They make the animal feel sick.
   (b) They inhibit them from feeding.
   (c) They interfere with digestion.
   (d) They even directly kill them, e.g. Calotropis produces a highly poisonous glycoside, that is a cardiac poison.
   (e) Nicotine, strychnine, opium, quinine, etc. are the chemicals produced by plants for their defence against herbivores.

Question 13.
What is parasitism? Define parasite and host. What are the kinds of parasite?
Answer:
Parasitism: This is a relationship between two organisms in which one obtains its nourishment from the other and harms it at the same time.

Parasite: The organism which obtains its food from the other without directly killing it is known as a parasite. Parasites are host specific and parasite and host tend to co-evolve.

Host: The second organism which provides food to the parasite is named the host. Endoparasite. The parasite that lives inside the body of the host is known as endoparasite.

Ectoparasite. The parasite lives on the outside of the body of the host, e.g. Leech, Louse, Bedbug.

Question 14.
Define commensalism. Give examples.
Answer:
Commensalism: It is defined as the interspecific interaction where one species is benefitted while the other species is neither benefitted nor harmed.
Examples:

1. Orchids grow as epiphytes on mango or other fruit trees. Orchids are benefitted by getting shelter, while the tree is neither benefitted nor harmed.
2. The clownfish living among sea anemones get protection from their predators, which stay away from the stinging tentacles of the sea anemone.
3. Barnacles growing on the whale are benefitted to move to where food is available.
4. The cattle egrets always forage near where the cattle are grazing. As the cattle animals stir up, the insects are flushed out from the vegetation. The egrets are benefitted from this as otherwise, it might be difficult for the birds to detect and catch the insects.

Question 15.
Differentiate between the following:
(i) Mutualism commensalism.
Answer:
Differences between mutualism and commensalism:

Mutualism	commensalism
(i) It is the inter-specific interaction, in which both the interacting species are benefitted.	(i) It is the interspecific interaction, in which one species is benefitted white the other is neither benefitted nor harmed.
(ii) It may or may not involve physical association.	(ii) The two individuals come in close physical contact.

(ii) Commensalism and amensalism.
Answer:
Differences between commensalism and amensalism:

Commensalism	Amensalism
It is the interspecific interaction in which one species is benefitted while the other one is neither harmed nor benefitted.	It is the interspecific interaction in which one species is harmed I affected, while the other is neither benefitted nor harmed.

(iii) Predators and parasites.
Answer:
Differences between predators and parasites:

Predators	Parasites
(1) Predators are Larger and stronger animals that kill and consume the prey.	(i) Parasites are small or microscopic organisms that depend on the host.
(ii) They do not take shelter on the prey	(ii) They take shelter on the host.
(iii) Their biotic potential is Low.	(iii) They have higher biotic potential.
(iv) They are mobile to capture the prey.	(iv) They have poor means of dispersal.
(v) They are not specific for the Prey	(v) They are host-specific.

Question 16.
Study the graph given below and answer the questions that follow:

(i) Write the status of food and space in the curves (a) and (b).
Answer:
Food and space are unlimited in the curve (a), while they are limited in the curve (b).

(ii) In the absence of predators, which one of the two curves would appropriately depict the prey population?
Answer:
In the absence of a predator, any species will grow exponentially and show curve (b).

(iii) Time has been shown on X-axis and there is a parallel dotted line above it. Give the significance of this dotted line. (CBSE 2014)
Answer:
The dotted line represents (k). It is carrying capacity.

Question 17.
Explain parasitism and co-evolution with the help of one example of each. (CBSE Outside Delhi 2016)
Answer:

1. Many parasites have evolved to be host-specific. They can live as parasite only in a single species of host.
2. These organisms parasitize in such a way that both the host and the parasite tend to co-evolve; that is if the host evolves a special mechanism for rejecting or resisting the parasite, in such conditions parasite has to evolve a mechanism to counteract and neutralise them, in order to be successful with the same host species.
3. In accordance with their mode of living, parasites evolve special adaptations such as loss of unnecessary sense organs, presence of adhesive organs or sucker so as to cling to the host. It also loses the digestive system. Parasites have a high reproductive capacity.
4. The human liver fluke depends on the intermediate host (a snail and a fish ) to complete the life cycle.
5. The malarial parasite needs a female anopheles mosquito as a vector to spread to other hosts.

Question 18.
(i) In a pond, there were 200 frogs. 40 more were born in the year. Calculate the birth rate of the population.
Answer:
Birth rate = No. of individuals born/ Total no. of individuals = 40/200 = 0.2 = 0.2 frog per year.

(ii) Population in terms of number is not always a necessary parameter to measure population density. Justify with two examples. (CBSE Outside Delhi 2019)
Answer:
To measure population density, the number is not always a necessary parameter.

For example:
(a) If there are 200 Parthenium plants but only a single huge banyan tree with a large canopy, the population density of banyan is low relative to that of Parthenium which amounts to underestimating the enormous role of the banyan in that community. In such cases, the per cent cover or biomass is a more meaningful measure of the population size.
(b) In a dense laboratory culture of a microbial population in a Petri dish, the total number of microbes is again not an easily adaptable measure because as the population is huge, counting is impossible and time¬consuming.

Question 19.
How do organisms which cannot migrate tend to overcome adverse environ¬mental conditions? Explain taking one example each from vertebrates and angiosperms respectively. (CBSE Delhi 2009)
Answer:
Organisms that cannot migrate tend to overcome adverse environmental conditions by developing several methods/ features. For example, some vertebrates escape the stress caused by unfavourable environmental conditions by escaping in time like bears go into hibernation during the winter months.

In angiosperms, seeds and some other vegetative reproductive structures serve as means to tide over periods of stress. They reduce their metabolic activity and go into an inactive, i.e. ‘dormant’, state. They germinate to form new plant when the favourable conditions return.

Question 20.
Explain how tolerance to environmental factors determines the distribution of species.
Answer:
Range of tolerance: Biological species can show a range of tolerance to environmental factors. These factors show variation in their effects and anyone who is present in the least amount may become limiting. The response of an organism to a range of gradient of a single environmental factor such as temperature, sunlight or nutrient concentration forms a bell-shaped curve as shown in the figure.

The response of an organism to a range or gradient of an environmental factor, (temperature, Light, nutrient)

Question 21.
Discuss in detail various adaptations found in plants and animals in snowy winter of polar regions.
Answer:
Adaptations in plant and animals of polar regions:

1. Organisms living in polar regions have to face the severe snowy winter and show entirely different types of adaptations. Animals in polar regions are generally white or tight coloured. This light colouration helps in camouflage and in thermal regulation.
2. During extreme winter they undergo hibernation.
3. Animals eat a lot of food during summer and autumn and store a lot of energy in the form of fat.
4. During hibernation, their metabolic activities are reduced considerably,
5. The plants growing in these regions remain dormant during extreme winter months.
6. They bear narrow leaves, spines, etc. which are shed easily and rapidly,
7. Plants remain dwarf. Trees are usually not found in these regions as they cannot withstand the low temperatures.

Question 22.
How do desert plants prevent loss of water?
Answer:
Xeric adaptations of desert plants:

1. Some xerophytes remain dormant either as seeds or as roots and during rains, they sprout up. It is termed ephemerals, e.g. Cassia, Argemone.
2. Some desert plants develop succulent organs (stems, leaves and roots), e.g. Asparagus, Begonia, Bryophyilum.
3. Presence of extensive root system.
4. Either leaf is absent or small¬sized leaves present to minimise the rate of transpiration.
5. The stomata get sunken to reduce the rate of transpiration.
6. Presence of thick cuticle on stem and leaf surfaces. This reduces the rate of transpiration.
7. Shedding of leaves in some desert plants. This helps in reducing the surface area and water loss.

Question 23.
(i) What is the competition?
Answer:
Competition is an interaction between organisms for life requirements (nutrition, shelter, sunlight, etc.). It is of two types: intraspecific and interspecific. The effort of a tiger and a leopard for prey is an interspecific competition. In general, it is believed that competition occurs among closely related species when they compete for the same resources that are limited.

(ii) Why is it not true always?
Answer:
But it is found to be not true always for the following reasons:
(a) Completely unrelated species can also compete for the same sources. For example in certain shallow lakes of South America, the visiting flamingoes and the native fishes compete for the same zooplankton as their food.

(b) Resources need not be limiting for the competition to occur, the feeding efficiency of one species might be reduced due to the interfering and inhibitory presence of the other species, even if resources are plenty.

For example, the Abingdon tortoise in Galapagos island became extinct within a decade after goats were introduced into the island: this was due to the fact that the goats had greater browsing efficiency than the tortoise.

(iii) Explain competitive release and Gause’s competitive exclusion principle.
Answer:
Another evidence for the competition is competitive release, a phenomenon in which a species whose distribution is restricted to a small geographical area due to the presence of a competitively superior species, expands its distributional range when the competing species is experimentally removed.

Gause’s competitive exclusion principle states that two closely related species competing for the same resources cannot exist together as the competitively inferior one will be eliminated, but this is true only when the resources are limiting and not otherwise.

They have also pointed out that species facing competition might evolve mechanisms that promote co¬existence, rather than exclusion.

(iv) Write contribution of Mac Arthur.
Answer:
Mac Arthur had shown that five closely related species of warblers living on the same tree were able to avoid competition and co-exist to behavioural differences in their foraging activities.

Question 24.
Illustrate symbiosis with any four examples.
Answer:
Symbiosis: It is the relationship between two individuals where both partners are benefited.

That following are examples of symbiosis:

1. The Rhizobium bacteria present in the root nodules of leguminous plant roots are an example of symbiosis. Bacteria live in the roots in the form of nodules and avail carbohydrate and other food substances. In exchange, bacteria fix the nitrogen present in the atmosphere and make it useful for plants.
2. Trichonympha, a protozoan parasite, lives inside the intestine of termites. The presence of this protozoan helps the termites to digest the cellulose food and the parasite gets food and shelter.
3. Lichen plants. Lichen is the result of a symbiotic relationship between algae and fungi. In this, algae depend on fungi for water, minerals, salts and safety, whereas fungi get food material prepared by algae.
4. The human intestine contains a large number of symbiotic bacteria that help in the synthesis of vitamin B-complex. Bacteria get necessary nutritional substances from the human intestine and a safe place for living.

Question 25.
Define phenotypic adaptation. Give one example.
Answer:
Phenotypic adaptation involves non-genetic changes in individuals such as physiological modifications, acclimatisation or behavioural changes. Some organisms possess adaptations that are phenotypic, which allow them to respond quickly to a stressful situation.

Example. If a person had ever been to any high altitude place (Rohtang pass near Manali or Mansrover in Tibet), he or she must have altitude sickness because the body does not get enough oxygen at such a height and protect itself from atmospheric hypoxia. But after some time, the body acclimatises itself to the situation by increasing the number of RBC and decreasing the binding affinity of Hb and oxygen.

Question 26.
Name important defence mechanisms in plants against herbivores.
Answer:
Defence mechanisms of plants against herbivorous animals:

1. Formation of thick cuticle on their leaf surface.
2. Formation of leaf spines, e.g. leaf spines in opuntia
3. Modification of leaves into thorns, e.g. Bougainvillea and Duranta
4. Development of spiny margins on leaves.
5. Development of sharp silicated edges in leaves.
6. Many plants produce and store toxic chemicals which cause discomforts to herbivores, e.g. cardiac glucoside by Calotropis, nicotine by tobacco.

Question 27.
Distinguish between the following:
(i) Hibernation and aestivation
Answer:
Differences between hibernation and aestivation:

Hibernation	Aestivation
(i) Spending the winter in an inactive? dormant state, Is called hibernation (winter sleep).	(i) Spending dr hot period (summer) in an inactive period is called aestivation (summer sleep).
(ii) Example. Northern ground squirrels.	(ii) Example. Ground squirrels in the South West deserts.

(ii) Ectotherms and endotherms.
Answer:
Differences between ectotherms and endotherms:

Ectotherms	Endotherms
Ectotherms are those animals whose body temperature changes to match with that of the surroundings, in which they are living. They cannot maintain their internal environment constant.	Endotherms are those animals whose body temperature is maintained relatively constant by physiological regulation.

Question 28.
Write short notes on:
(i) adaptations of desert plants and animals. (CBSE2011)
Answer:
Adaptations of desert plants and animals:
(a) Adaptations of desert animals.

• Animals faced with water scarcity as found in arid or desert areas, show two types of adaptations, reducing water loss and the ability to tolerate arid conditions. Kangaroo/Desert rat seldom drinks water. It has a thick coat to minimise evaporative desiccation. The animal seldom comes out of its comparatively humid and cool burrow during the daytime. 90% of its water requirement is met from metabolic water (water produced by respiratory breakdown) while 10% is obtained from food.
• Loss of water is minimised by producing nearly solid urine and faeces.
• Spiny skin and highly cornified in Phrynosoma (horned toad) and Moloch horridus.
• Camels have long legs to stay away from the hot desert surface.

(b) Adaptations of desert plants.

• Plants have thick cuticle, succulent organs where water and mucilage are stored.
• Stomata are sunken to prevent water loss.
• They have a well-developed branched root system.
• They possess a waxy coating on the surface.
• Crassulacean pathway of photosynthesis.

(ii) behavioural adaptations in animals. (CBSE 2011)
Answer:
Adaptations of plants to water scarcities. They are called xerophytes. The above-mentioned adaptation of plants are applicable (see part (a)).

(iii) Behavioural adaptations in animals,
(a) Hibernation
(b) Aestivation
(c) Periodic activity
(d) Camouflage
(e) Migration.

(iv) importance of light to plants.
Answer:
Importance of light to plants.
(a) Source of energy for photosynthesis
(b) Photoperiodism
(c) Pigmentation
(d) Daily rhythm
(e) Plant movements and (f) Growth.

(v) effect of temperature or water scarcity and the adaptation of animals.
Answer:
Effect of temperature or water scarcity and adaptation of animals. Optimum temperature is necessary for animals, survival as all the metabolic activities are driven by enzymes. And enzymes work actively only in a certain optimum range of temperature. Regulators can regulate their body temperature in case of temperature fluctuations in the external environment. Conformers also try to maintain temperature by certain methods but not internally.

Similarly, water is also necessary for the metabolic activities of animals. Various adaptations seen in animals to deal with temperature fluctuations are

• thick fur,
• sweating, short limbs and ears.

Various adaptation to counter the scarcity of water is the ability to use metabolic water, store water, reduced water loss, etc.

Question 29.
With the help of a suitable diagram describe the logistic population growth curve. (CBSE Delhi 2019 C)
Answer:
Logistic popuLation growth curve.

S-shaped Population growth curve (Verhulst-Pearl logistic growth curve)

Unlimited resources result in exponential growth. Many countries have realised this fact and introduced various restraints to limit human population growth. In nature, a given habitat has enough resources to support a maximum possible number, beyond which no further growth is possible. This limit is called carrying capacity (K) for that species in that habitat.

A population growing in a habitat with limited resources shows initially a lag phase, followed by phases of increase and decrease and finally the population density reaches the carrying capacity. A plot of N in relation to time (t) results in a sigmoid curve. This type of population growth is called Verhulst-Pearl Logistic Growth as explained by the following equation:

\(\frac{d \mathrm{~N}}{d t}=r \mathrm{~N}\left(\frac{\mathrm{K}-\mathrm{N}}{\mathrm{K}}\right)\)

where N = Population density at a time t; r = Intrinsic rate of natural increase and K = Carrying capacity,

\(\left[\frac{k-N}{k}\right]\) = Environmental resistance.

Since resources for growth for most animal populations become limiting sooner or later, the logistic growth model is considered a more realistic one.

Question 30.
List any three important characteristics of a population and explain. (CBSE Delhi 2019)
Answer:
Three important characteristics of a population are:

1. Population size and population density
2. Birth or natality rate
3. Death or mortality rate.

• Population size: It is the actual number of individuals in the population. The size of the population keeps changing with time depending on the factors like:
  (a) Food availability
  (b) Weather
  (c) Predation pressure and
  (d) Competition.

Population density is a measure of population size per unit area. The population density in a given habitat during a given period changes due to four basic processes, namely

1. Natality,
2. Mortality,
3. Immigration and
4. Emigration.

While natality and immigration contribute to an increase in the size and density of a population, mortality and emigration contribute to a decrease in them.

So the equation for population growth is: Nt + 1 = Nt + [(B + I) – (D + E)],
where Nt = population density at time t.
B = No of birth I = No of Immigration D = No of death E = No of emigration

• If B + I is more than D – E, the population density increases.
• If B + I is less than D – E, the population density decreases.

Birth or Natality rate. It is generally expressed as the number of births per 1,000 individuals of a population per year. It increases the population size (total number of individuals of a population) and population density. Natal or Birth Rate =

Death or Mortality rate. It is the opposite of the natality rate. It is commonly expressed as the number of deaths per 1,000 individuals of a population per year.

Question 31.
Define mutualism. Give examples.
Answer:
Mutualism: It is a symbiotic relationship between the members of two species in which the two partners are mutually benefited. There is a complete dependence of the partners on each other, and one cannot survive in the absence of the other.

Sometimes the term symbiosis is used as a synonym with mutualism.
Examples:
(a) Mutualism between animal and bacteria. Symbiotic bacteria like Ruminococcus are found in the rumen part of the compound stomach of cud-chewing mammals like cattle, sheep, goat, camel, etc. and secrete cellulase enzyme to digest the cellulose of plant food eaten by the ruminants which provide food and shelter to the bacteria.

(b) Mutualism between crab and sea anemone. In this case, a sea anemone gets attached to the back of the hermit crab.

(c) The Mediterranean orchid, Ophrys, employs sexual deceit to get its flowers pollinated. In this orchid, one petal of the flower resembles the female of a bee species in size, colour, markings, etc. The male bee perceives it as a female and pseudo copulate with it. During the process, the pollen grains get dusted on its body.

Question 32.
It is observed that plant-animal interactions often involve co-evolution. Explain with the help of a suitable example. (CBSE Delhi 2019 C)
Or
Mention four significant services that a healthy forest ecosystem provides. (CBSE Delhi 2019)
Answer:
Plants and animals interact for mutual ben¬efit. Plant-animal interactions often involve co-evolution of mutualism, that is, the evolution of flower and its pollinator species are tightly linked with one another. In many species of fig trees, there is a tight one to one relationship with the pollinator species of wasp. It means that a given fig species can be pollinated only by its partner wasp species and no other species.

The female wasp uses the fruit not only at the oviposition site but also uses the developing seeds within the fruit for nourishing its larvae. The wasp pol¬linates the fig inflorescence while searching for suitable egg-laying sites. In return for the favour of pollination, the fig offers the wasp some of its developing seeds as food for the developing wasp larvae.

Question 33.
(i) Compare, giving reasons, the J-shaped and S-shaped models of population growth of a species.
Answer:
There are two models of population growth-exponential growth and logistic growth.
(a) Exponential growth: This growth occurs where the resources, i.e. food and space, are unlimited. The equation can be represented as follows:
\(\frac{\mathrm{dN}}{\mathrm{dt}}\) = (b-d) x N

Let (b-d) = r
\(\frac{\mathrm{dN}}{\mathrm{dt}}\) = rN or N_t=N_oe^rt

N = population size
N_t = population density after time t.
N_o = population density at time zero
r = growth rate
e = base of natural log (2.71828)
b = birth rate
d = death rate
In this growth, when N in relation to time is plotted on a graph, the curve becomes J-shaped.

(b) Logistics growth model: This is a realistic approach as the resources become limited at a certain point in time.

(a) J-shaped curve exponential growth
(b) S-shaped curve logistics growth Every ecosystem has limited resources to support a particular maximum carrying capacity (K). When N is plotted in relation to time t, a sigmoid S-shaped curve is obtained. It is also called Verhulst-Pearl logistic growth.

The equation is:
\(\frac{\mathrm{dN}}{\mathrm{dt}}\)=rN\(\left(\frac{K-N}{K}\right)\)

N = population density at time t
r = growth rate
K = carrying capacity.

(ii) Explain “fitness of a species” as mentioned by Darwin. (CBSE 2017)
Answer:
According to Darwin “Fitness of a species” means reproductive fitness. All organisms after reaching reproductive age have a varying degree of reproductive potential. Some organisms produce more offspring, whereas some organisms produce few offspring only. This phenomenon is also called differential reproduction. The species which produces more offspring are selected by nature.

Question 34.
(i) The following are the responses of different animals to various abiotic factors. Describe each one with the help of an example.
(a) Regulate
Answer:
Regulate: Some organisms are able to maintain homoeostasis by regulating their body temperatures. The mechanisms used by most mammals to regulate their body temperature are similar to what we humans use.

For example, our body temperature remains constant at 37°C. In summer, when the outside temperature is more than our body temperature, we sweat profusely to cool down and when the temperature is much lower than 37° C, we shiver to generate heat. Thus body temperature remains constant.

(b) Conform
Answer:
Confirm: Most of the animals cannot maintain a constant internal environment. Their body temperature changes with the ambient temperature. These are called conformers. Heat loss or heat gain is a function of surface area. Since small animals have a larger surface area relative to their volume, they tend to lose body heat very fast when it is cold outside, for example, shrews and hummingbirds.

(c) Migrate
Answer:
Migrate: Organism can move away temporarily from the stressful habitat to a more hospitable area and then return when the stressful period is over.

For example, every winter, the famous Keoladeo National Park in Bharatpur hosts thousands of migratory birds coming from Siberia and other northern regions,

(d) Suspend
Answer:
Suspend: Bacteria, fungi and some lower plants survive unfavourable conditions by forming thick-walled spores. In higher plants, seeds and some other vegetative reproductive structures serve as means to tide over periods of stress. They do so by reducing their metabolic activity and going into a state of dormancy.

(ii) If 8 individuals in a population of 80 butterflies die in a week, calculate the death rate of the population of butterflies during that period. (CBSE 2018)
Answer:
Death Rate = \(\frac{8}{80}\) = 0.1 per week

Question 35.
(a) Identify the features of a stable biological community.
Answer:
(a) Features of the stable biological community.

• It is stable.
• A stable biological community is not replaced by any other community.
• The environment becomes moister and shadier.
• Communities should have greater biodiversity for greater stability.
• It should be able to prevent invasion by alien species.
• It should be able to restore itself in a short period of time.

(b) How did David Tilman’s findings link the stability of a biological community to its species richness? (CBSE Delhi 2019 C)
Answer:
David Tilman’s with his experiments established that the stability of a community depends on its species richness.

Here we are providing Class 12 Biology Important Extra Questions and Answers Chapter 10 Microbes in Human Welfare. Important Questions for Class 12 Biology are the best resource for students which helps in Class 12 board exams.

Class 12 Biology Chapter 10 Important Extra Questions Microbes in Human Welfare

Microbes in Human Welfare Important Extra Questions Very Short Answer Type

Question 1.
Which bacterium Is responsible for the formation of curd from milk?
Answer:
Lactobacillus but Agaricus (Lactic acid bacteria).

Question 2.
What is brewing?
Answer:
Brewing is a complex fermentation process, which involves the production of malt beverages such as beer, ale, porter, and stout with the help of strains of Saccharomyces cerevisiae.

Question 3.
Name the type of association that genus Glomus exhibits with the higher plant. (CBSE2014)
Answer:
Mycorrhiza- Symbiotic association.

Question 4.
Which one of the following is the baker’s yeast used in fermentation-Saccharum Barberi, Saccharomyces cerevisiae or Sonalika? (CBSE2009)
Answer:
Saccharomyces cerevisiae.

Question 5.
Milk starts to coagulate when Lactic Acid Bacteria (LAB) is added to milk as a starter. Mention two benefits that LAB provides. (CBSE 2009)
Answer:

1. LAB checks the growth of disease-causing microbes.
2. LAB converts milk into curd and also increases nutritional quality by increasing vitamin B12.

Question 6.
Give the scientific name of the source organism from which the first antibiotic was produced. (CBSE Sample paper 2018-19)
Answer:
Penicillium Notatum

Question 7.
Name the different vitamins which are produced by micro-organisms.
Answer:

1. Riboflavin or Vitamin B2 is produced by yeast and bacteria.
2. Vitamin B12 or cobalamine is produced by bacteria and actinomycetes.

Question 8.
Name the original wild strain of the mold by which vitamin B2 is produced.
Answer:
Ashbya Gossypii.

Question 9.
What is a single-cell protein (SCP)?
Answer:
Single-cell protein (SCP) refers to any microbial biomass produced by uni and multi-cellular organisms and can be used as food or feed additives.

Question 10.
Name a microbe used for statin production. How do statins lower blood cholesterol levels?
Answer:
Microbe:
Monascus Purpureus Mechanism: Statins are competitive inhibitors of enzymes required for cholesterol synthesis. Therefore, play role in decreasing cholesterol level in the body.

Question 11.
‘Swiss cheese’ is characterized by the presence of large holes. Name the bacterium responsible for it. (CBSE Delhi Outside 2019)
Answer:
Propionibacterium sharmanii

Question 12.
What for Nudeopolyhedra viruses (NVP) are being used nowadays? (CBSE, Delhi 2014, 2019C)
Answer:
Nudeopolyhedro viruses are being used to kill insects and other arthropods pests of crops. The viruses have no effect on plants and non-target animals. Thus used in biological control of pests.

Question 13.
How has the discovery of antibiotics helped mankind in the field of medicine?
Answer:
Antibiotics have helped mankind in treating most of the deadly bacterial and fungal diseases of humans.

Question 14.
Why is distillation required for producing certain alcoholic drinks?
Answer:
For increasing the alcohol strength or concentration of the drinks.

Question 15.
What is the primary sludge?
Answer:
All the solids that settle from the sewage on primary treatment constitute primary sludge.

Question 16.
What is the relationship between BOD and organic matter in sewage?
Answer:
The greater the BOD of wastewater more is the amount of organic matter in sewage.

Question 17.
Name two gases produced during secondary treatment by sewage.
Answer:

1. Carbon dioxide and
2. Hydrogen sulfide.

Question 18.
What are bioreactors?
Answer:
In the pilot plant, the glass vessels are replaced by stainless steel vessels. They are called bioreactors.

Question 19.
Name the bacteria which can be used for yogurt formation.
Answer:

1. Lactobacillus bulsaricus.
2. Streptococcus Thermophilus.

Question 20.
What is Bacitracin?
Answer:
It is an antibiotic obtained from Bacillus Licheniformis.

Question 21.
Name the group of organisms and the substrate they act on to produce biogas. (CBSE 2009)
Answer:
Methanogens such as Methanol bacterium act on activated sludge to produce biogas.

Question 22.
WrIte the scientific name of the microbe used for fermenting malted cereals and fruit juices. (CBSE 2011)
Answer:
Saccharomyces cerevisiae

Question 23.
Write an alternate source of protein for animal and human nutrition. (CBSE 2014)
Answer:
Single-cell proteins.

Question 24.
How are the members of the genus Glomus useful to organic farmers? (CBSE Delhi Outside 2019)
Answer:
Many members of the genus Glomus form mycorrhizae- symbiotic associations with roots of higher plants. The fungal component of these associations helps in the absorption of phosphorus from soil. It also makes the plant drought-resistant.

Microbes in Human Welfare Important Extra Questions Short Answer Type

Question 1.
Expand the ‘LAB’. How are LABs beneficial to humans? (Write any two benefits) (CBSE 2019 C)
Answer:
LAB-Lactic Acid Bacteria Benefits:

• Found in curd. They improve the nutritional quality of food.
• Yogurt is prepared from milk by Lactobacillus Bulgaricus.

Question 2.
What is cyclosporin A? What is its importance?
Answer:
Cyclosporin A. It is an eleven-membered cyclic oligopeptide obtained through the fermentative activity of fungus Trichoderma Polysporum.

Importance. It has antifungal, anti-inflammatory, and immunosuppressive properties. It inhibits the activation of T-cells and therefore, prevents rejection reactions in organ transplantation.

Question 3.
How do antibiotics act?
Answer:
Antibiotics do not have identical effects on all harmful microbes. All of them inhibit growth or destroy bacteria, viruses, and fungi. Actually, antibiotic molecules should disrupt a vital link in the microbe’s metabolism and this link is their target or point of impact.

Question 4.
Write the various steps of fermentation.
Answer:
The major steps of fermentation are:

1. Sterilization of the fermenter and medium in steam. It is carried out under pressure and high temperature.
2. Inoculation of a selected strain of the yeast.
3. Recovery of the product.

Question 5.
What are the two ways by which micro-organisms can be grown in bioreactors?
Answer:
Micro-organisms can be grown in the bioreactors in two ways:

1. As a layer or film on the surface of the nutrient medium. It is known as a support growth system.
2. By suspending cells or mycelia in a liquid medium contained in the growth vessel. It is known as a suspended growth system.

Question 6.
What is sewage? In which way can this be harmful?
Answer:
Sewage is used and wastewater consisting of human excreta, wash waters, industrial and agricultural wastes that enter the sewage system. In general, sewage contains 95.5% water and 0.1 to 0.5% organic and inorganic matter. They are very harmful to us due to the presence of a variety of micro¬organisms in them, most of which are highly pathogenic. Sewage has a high BOD value, which develops anaerobic conditions in water resulting in the death of water animals and emitting foul smell due to incomplete oxidation of organic materials in the sewage.

Question 7.
What is the key difference between primary and secondary sewage treatment?
Answer:
Primary treatment of wastes is the screening and removal of insoluble particulate materials, by addition of alum and other coagulants. It is the physical removal of 20-30% of organic materials present in sewage in particulate form. Secondary treatment of waste is the biological removal of dissolved organic matter through trickling filters, activated sludge, lagoons, extended aeration systems, and anaerobic digestors.

Question 8.
Draw a simple diagram to show an anaerobic sludge digester.
Answer:

Anaerobic sludge digester.

Question 9.
GIve the full form of Bt. Name the insects killed by It.
Answer:
The full form of Bt is Bacillus Ttiuringiensis. It kills a wide range of Insects Like moths, beetles, mosquitoes, aphids, and termites.

Question 10.
Why are biofertilizers or biopesticides preferred to chemical fertilizers or pesticides? (CBSE Delhi 2011)
Answer:
Biofertilizers or biopesticides are preferred to chemical fertilizers or pesticides because

• They are safe to use and are biological in origin.
• They do not spoil the quality of the soil and are target-specific.
• They do not pollute the atmosphere and are non-poisonous.
• They are less expensive and are biodegradable.

Question 11.
Name the blank spaces a, b, c, and d from the table given below: (CBSE 2008)

Type of microbe	Scientific  name	Product	Medical application
(i) Fungus	a	Cyclosporin	B
(ii) c	Mascus Purpureus	Statin	d

Answer:
(a) Trichoderma polypore
(b) Organ transplantation (Immunosuppressant)
(c) Yeast
(d) Blood cholesterol-lowering agent

Question 12.
How does the addition of a small amount of curd to fresh milk help the formation of curd? Mention a nutritional quality that gets added to the curd. (CBSE Delhi 2010 and Outside Delhi 2019)
Answer:

1. Curd is prepared from milk.
2. Microorganisms such as Lactobacillus and others commonly called lactic acid bacteria (LAB) grow in milk and convert it to curd.
3. During growth, the LAB produces acids that coagulate and partially digest the milk proteins.
4. A small amount of curd added into the fresh milk as inoculum or starter contains millions of LAB which at suitable temperatures multiply, thus converting milk to curd, which also improves its nutritional quality by increasing vitamin B12.
5. In our stomach too, the LAB plays a very beneficial role in checking disease-causing microbes.

Question 13.
Name a free-living and symbiotic bacterium that serves as a biofertilizer. Why are they called so? (CBSE Outside Delhi 2016)
Answer:
Free-living nitrogen-fixing bacteria Azotobacter and Bacillus Polymyxa Symbiotic nitrogen-fixing bacteria. Rhizobium.

These micro-organisms enrich the soil by fixing nitrogen. They enhance the availability of nutrients to crops, thus called biofertilizers.

Question 14.
(i) Why are fruit juices bought from the market clearer as compared to those made at home?
Answer:
Bottled juices are clarified by the use of pectinases and proteases.

(ii) Name the bioactive molecules produced by Trichoderma Polysporum and Monascus Purpureus. (CBSE Delhi 2013)
Answer:
(a) Bioactive molecules produced by Trichoderma polypore are cyclosporin A. It is used as an immunosuppressive agent in organ- transplant patients.
(b) Bioactive molecules produced by Monascus Purpureus are statins. It is a blood cholesterol-lowering agent.

Question 15.
Your advice is sought to improve the nitrogen content of the soil to be used for the cultivation of a non-leguminous terrestrial crop.
(i) Recommend two microbes that can enrich the soil with nitrogen.
Answer:
Azospirillum, Azotobacter, Anabaena, Oscillatoria (Any two)

(ii) Why do leguminous crops not require such enrichment of the soil? (CBSE 2018)
Answer:
Leguminous crops do not require such enrichment of the soil because they have a symbiotic association with Rhizobium bacteria which traps nitrogen directly from the atmosphere and provides it to the plant and in turn gets food and shelter.

Question 16.
What are ‘floes’, formed during secondary treatment of sewage? (CBSE Delhi 2019)
Answer:
Floes are masses of bacteria, associated with fungal filaments to form mesh-like structures.

Question 17.
Write any two places where methanogens can be found. (CBSE Delhi 2019)
Answer:
Methanogens can be found in the following places:

1. In anaerobic sludge (digester) of a sewage treatment plant
2. In rumen (gut/stomach) of cattle or ruminants
3. Marshy areas
4. Flooded paddy fields
5. Biogas plant Methane, H₂S, and C0₂ are produced during microbial digestion of organic compounds in case of secondary treatment of sewage.
6. The dung of the cattle produces methane gas in the biogas plants.

Microbes in Human Welfare Important Extra Questions Long Answer Type

Question 1.
Give examples to prove that microbes release gases during metabolism.
Answer:

1. Large holes in ‘Swiss Cheese’ are due to the production of a large amount of C0₂ by a bacterium named Propionibacterium shamanic.
2. The puffed-up appearance of dough is due to the production of C0₂ gas by yeast, Saccharomyces cerevisiae.
3. Methane, H₂S, and CO₂ are produced during microbial digestion of organic compounds in the case of secondary treatment of sewage.
4. The dung of the cattle produces methane gas in the biogas pLants.

Question 2.
Make a table showing industrial products obtained from activities of bacteria.
Answer:
industrial products obtained from use activities of Bacteria:

Products	Bacteria
Acetone, butanol	Clostridium Aceto bretylium
Acetic acid, Vinegar	Acetobacter acetic
Curing of tobacco	Bacillus megatherium micrococcus
Curing of tea leaves	Micrococcus conditions
Lactic acid	Lactobacillus Delbreuckii
Lysine	Micrococcus glutamic
Retting of fibers	Clostridium but lyricism
Riboflavin (Vitamin B2)	Clostridium but lyricism
Cobalamin (Vit. B17)	Bacillus megatherium

Question 3.
What are Baculo viruses? Write their significance.
Answer:
Baculoviruses are those viruses, which attack insects and other arthropods, e.g. Nuclepolyhedrovirus.

Significance:

• Baculoviruses are species-specific and narrow-spectrum insecticides.
• They have no negative impacts on plants, birds, mammals, or even other non-target insects.
• The desirable aspect In conservation of beneficial insects in overall integrated pest management (IPM) program as in an ecologically sensitive area.

Question 4.
Which nitrogen fixers are available on a commercial basis In the market? Also, name the beneficial crop.
Answer:

Products	Microbe used	Beneficial crop
1. Nitragin TM	Rhizobium	Soybean
2. Rhizocote	Rhizobium	Legumes
3. Nodosit	Rhizobium	Legumes

Question 5.
Distinguish between the roles of floes and anaerobic sludge digester in sewage treatment. (CBSE Delhi 2016)
Answer:
Floes are masses of bacteria associated with fungal filaments to form mesh-like structures. These microbes digest a lot of organic matter, converting it into microbial biomass and releasing a lot of minerals. Anaerobic sludge digester is a large tank in which anaerobic microbes digest the anaerobic mass as well as aerobic microbes of sludge. Biogas is produced by methanogens. It is inflammable and a source of energy.

Question 6.
Tabulate the list of common antibiotics, organisms producing them, and organisms sensitive to these antibiotics.
Answer:

Name of Antibiotic	Name of Producing Organism	Sensitive Organisms
(i) Penicillin	Penicillium	Most of the Gram+ve bacteria, Clostridium actinomycetes, Spirochaetae, and Corynebacterium.
(ii) Streptomycin	Streptomyces griseus	Gram + ve and Gram – ve bacteria; Mycobacterium tuberculosis
(iii) Bacitracin	Bacillus licheniformis	Treponema, Histoplasma, Clostridium.
(iv) Tetracycline and Chlorotetracycline	Streptomyces aureofaciens	Rickettsiae, Klebsiella pneumonia, Streptococcus.
(v) Synthromycin	Streptomyces erythematous	Gram+ve; Gram-ve bacteria and many viruses.
(vi) Chloromycetin	Streptomyces Venezuelae	Gram+ve; Gram-ve bacteria; Entamoeba, Borrelia.

Question 7.
Give a flow chart of sewage treatment.
Answer:
Flow chart of sewage treatment:

Flow chart of sewage treatment

Question 8.
List the events that lead to the production of biogas from wastewater whose BOD has been reduced significantly. (CBSE Dethi 2016)
Answer:

1. During secondary treatment of wastewater, sewage fungus forms focus.
2. BOD decreases. As it decreases to 10-15% of originaL sewage, the wastewater Is taken to a Large settling tank where the focus of sewage fungus settles down.
3. The supernatant can be passed into water bodies or treated further.
4. The organic sediment is passed into an anaerobic sludge digester where anaerobic microbes methanogens decompose organic matter.
5. It is accompanied by the production of blogs and the formation of manure or compost.

Question 9.
Explain the basis of biological control of weeds.
Answer:
Basis of biological control of weeds:

1. Biological control of weeds involves breeding of insects that would feed selectively a weed or use of certain micro-organisms which will produce diseases in the weeds and eliminate them.
2. Certain crop plants do not allow the growth of weeds nearby. They are called smoother plants such as Barley, Rye, Sorghum, Millet, etc. They eliminate weeds through chemicals.
3. In some cases, specially tailored plants called transgenic plants have been introduced which have tolerance against weeds.
4. In India and Australia, the overgrowth of cacti was checked by the introduction of the cochineal insect (Cactoblastis cactorum).
5. The latest technique is to use fungal spores to control weeds. These are suitable because they can be kept for a long time and also resist adverse conditions.

Question 10.
What are biofertilizers? What are the main sources of biological nitrogen fixation? Name two organisms that fix nitrogen symbiotically and two organisms that fix symbiotically.
Answer:
Biofertilizers are organisms that can bring about soil nutrient enrichment by their biological activity.

• Sources of biofertilizers: Bacteria, cyanobacteria, and fungi.
• Biological nitrogen fixation: The conversion of atmospheric nitrogen into nitrogenous compounds through the agency of living organisms is called biological nitrogen fixation.

Symbiotically nitrogen-fixing organisms:

• Rhizobium leguminosarum, Frankia Bacillus radicicola.
• Free-living/Asymbiotic nitrogen-fixing organisms-Cyanobacteria, Azotobacter.

Question 11.
(a) What is biogas? What are its components? What is the calorific value of biogas? (CBSE Outside Delhi 2013)
Answer:
Biogas is a methane-rich fuel gas produced by anaerobic breakdown or digestion of biomass with the help of methanogenic bacteria.

Components of biogas: Methane, Carbon dioxide, Hydrogen sulfide, hydrogen, and nitrogen.
Calorific value 23-28 MJ/m³.

(b) Why is a slurry of cattle dung (gobar) added to bio-wastes in the tank of a gobar gas plant for the generation of biogas? (CBSE Delhi 2019 C)
Answer:
Slurry consisting of excreta dung of cattle commonly called gobar is rich in methanogen bacteria. It is used for the generation of biogas. These bacteria called methane bacterium grow anaerobically and break down the cellulose of dung to liberate gases such as methane, C0₂, and H₂.

Question 12.
(?) Name the toxin produced by B. Thuringiensis.
Answer:
∝-exotoxin, β-exotoxin, γ-exotoxin, and louse factor

(ii) Nitrogen fixers are available on a commercial basis in the market? Also, name the beneficial crop and microbes used in the following table.

Product	Microbe used	Beneficial crop
1. NitraginTM	(A)	Soybean
2. Rhizocote	Rhizobium	(B)
3. Nodosit	Rhizobium	(C)

Answer:
A. Rhizobium B. Legume C. Legume

(iii) Expand BOD and COD
Answer:
BOD- Biological Oxygen Demand COD- Chemical Oxygen Demand

Question 13.
By a flow chart showing the stages in anaerobic digestion during the production of biogas.
Answer:

Stages in Anaerobic Digestion during biogas formation

Question 14.
Given below is a list of six microorganisms. State their usefulness to humans.
(i) Nucleopolyhedrovirus
(ii) Saccharomyces cerevisiae
(iii) Monascus Purpureus
(iv) Trichoderma polypore
(v) Penicillium Notatum
(vi) Propionibacterium shamanic. (CBSE Delhi 2016)
Answer:

Name of Micro-organisms	Uses
(i) Nucleopolyhedrovirus	Used in biocontrol of insects
(ii) Saccharomyces cerevisiae	Bread making, Ethanol making
(iii) Monascus Purpureus	Produces Statin used as blood cholesterol-lowering agent
(iv) Trichoderma polypore	Preparation of cyclosporin having antifungal, anti-inflammatory, immuno-suppressive properties
(v) Penicillium Notatum	Production of antibiotic, Penicillin
(vi) Propionibacterium shamanic	Preparation of large-holed swiss cheese.

Question 15.
Explain the different steps involved in the secondary treatment of sewage. (CBSE Sample paper 2018—19)
Or
Secondary treatment of sewage is also called biological treatment. Justify this statement and explain the process. (CBSE 2018)
Answer:

1. Secondary treatment of sewage is a biological process that employs the heterotrophic bacteria naturally present in the sewage.
2. The effluent from the primary treatment is passed into large aeration tanks, where it is constantly agitated and the air is pumped into it.
3. This allows the rapid growth of aerobic microbes into ‘floes’ which consume the organic matter of the sewage and reduce the biological oxygen demand (BOD). The greater is the BOD of wastewater, the more is its polluting potential.
4. When the BOD of sewage is reduced significantly, the effluent is passed into a settling tank, where the ‘floes’ are allowed to sediment forming the activated sludge.
5. A small part of the activated sludge is pumped back into the aeration tanks.
6. The remaining major part of the sludge is pumped into anaerobic sludge digesters, where the anaerobic bacteria digest the bacteria and fungi in the sludge-producing methane, hydrogen sulfide, and carbon dioxide,
   i. e. biogas. This is why secondary treatment of sewage is also called biological treatment.
7. The effluent after secondary treatment is released into water-bodies like streams or rivers.

Question 16.
Microbes can be used to decrease the use of chemical fertilizers. Explain how this can be accomplished. (CBSE Delhi 2019)
Answer:

1. Rhizobium bacteria present in the root nodules of leguminous plants (pea family) forms a symbiotic association and fixes atmospheric nitrogen into organic forms as nitrates/nitrites which are used by the plant as nutrient.
2. Free-living bacteria in the soil Azospirillum and Azotobacter can fix atmospheric nitrogen thus enriching the nitrogen content of the soil.
3. Many members of the genus Glomus (Fungi) form mycorrhizal symbiotic associations with higher plants. In these, the fungal symbiont absorbs phosphorus from soil and passes it to the plant.

Question 17.
(?) Organic farmers prefer biological control of diseases and pests to the use of chemicals for the same purpose. Justify.
Answer:
Chemical methods often kill both useful and harmful living beings indiscriminately. The organic farmer holds the view that the eradication of the creatures that are often described as pests is not only possible but also undesirable, for without them the beneficial predatory and parasitic insects which depend upon them as food or hosts would not be able to survive. Thus, the use of biocontrol measures will greatly reduce our dependence on toxic chemicals and pesticides.

(ii) Give an example of a bacterium, a fungus, and an insect that are used as biocontrol agents. (CBSE 2018)
Answer:
Insects = Ladybird and Dragonflies. Bacteria = Bacillus thuringiensis. Fungus = Trichoderma

Question 18.
The three microbes are listed below. Name the product produced by each one of them and mention their use.
(i) Aspergillus niger
(ii) Trichoderma polypore
(iii) Monascus Purpureus (CBSE Delhi 2018C)
Or
(i) A patient had suffered myocardial infarction and clots were found in his blood vessels. Name a ‘clot buster’ that can be used to dissolve clots and the microorganism from which it is obtained.
(ii) A woman had just undergone a kidney transplant. A bioactive molecular drug is administered to oppose kidney rejection by the body. What is the bioactive molecule? Name the microbe from which this is extracted.
(iii) What do doctors prescribe to lower the blood cholesterol level in patients with high blood cholesterol? Name the source organism from which this drug can be obtained. (CBSE Outside Delhi 2019)
Answer:
(i) Aspergillus niger produces citric acid. Citric acid is used as a flavoring agent and as a food preservative.
(ii) Trichoderma Polysporum produces a bioactive molecule cyclosporin A. It is used as an immunosuppressive agent in organ transplant patients.
(iii) Monascus Purpureus produces statins. Statins are capable of competitive inhibition of enzymes required for cholesterol synthesis. Hence, it is used as blood cholesterol-lowering agents.
Or
(i) Streptokinase-‘Clot buster’ can be used to dissolve clots. It is obtained from the bacteria Streptococcus.
(ii) The bioactive molecule is Cyclosporin A which is used as an immunosuppressive agent in organ transplantation. It is produced by the fungus Trichoderma Polysporum.
(iii) Doctors prescribe Statins to lower blood cholesterol. It is obtained from the fungus Monascus Purpureus.

Question 19.
Baculoviruses are good examples of biocontrol agents. Justify giving reasons. (CBSE Delhi 2018C)
Answer:
Baculoviruses kill insects and other arthropods, hence they are used as biocontrol agents especially Nucleopolyhedrovirus.

Reasons for their use:

• These viruses are species-specific and have narrow spectrum insecticidal applications.
• They do not harm non-target organisms like other harmless insects, birds, animals, etc.
• It is very useful in integrated pest management programs or treatment of ecologically sensitive areas.

Question 20.
Describe the primary and secondary treatment of domestic sewage before it is released for reuse. (CBSE, 2014)
Answer:
Treatment of domestic sewage. The municipal wastewaters are treated in Effluent Treatment Plant (ETP) prior to disposal in water bodies.

It consists of 3 steps: primary, secondary, and tertiary.
1. Primary treatment. It includes physical processes, such as sedimentation, floatation, shredding (fragmenting and filtering). These processes remove most of the large debris.

2. Secondary treatment. It is a biological method. Activated sludge method. Sewage, after primary treatment, is pumped into aeration tanks or oxidation ponds. Here, it is mixed with air and sludge containing algae and bacteria. Bacteria consume organic matter. The process results in the release of C02 and the formation of sludge or biosolid. Algae produce oxygen for the bacteria. The water, which is now almost clear of organic matter, is chlorinated to kill microorganisms.

3. Tertiary treatment. It involves. removal of nitrates and phosphates. The water, after the above treatment, is then released. It can be reused.

Question 21.
Explain biological control of pests and plant pathogens with examples.
Answer:
The very familiar beetle with red and black markings the Ladybird, and Dragonflies are useful to get rid of aphids and mosquitoes, respectively.

Role of Bacillus Thuringinesis:
Bt Coming to microbial biocontrol agents that can be introduced in order to control butterfly caterpillars is the bacteria Bacillus thuringiensis (often written as Bt). These are available in sachets of dried spores which are mixed with water and sprayed onto vulnerable plants such as Brassica and fruit trees, where these are eaten by the insect larvae. In the gut of the larvae, the toxin is released and the larvae get killed.

The bacterial disease will kill the caterpillars, but leave other insects unharmed. Because of the development of the methods of genetic engineering in the last decade or so, scientists have introduced B. thuringiensis toxin genes into plants. Such plants are resistant to attack by insect pests. Bt-cotton is one such example which is being cultivated in some states of our country.

Biological control of plant pathogens: A biological control developed for use in the treatment of plant disease is the fungus Trichoderma. Trichoderma sp. are free-living fungi that are very common in soil and root ecosystems. They are effective biocontrol agents of several plant pathogens.

Baculoviruses are pathogens that attack insects and other arthropods. The majority of baculoviruses used as biological control agents are in the genus Nucleopolyhedrovirus. These viruses are excellent candidates for species-specific, narrow spectrum insecticidal applications.

They have been shown to have no negative impacts on plants, mammals, birds, fish, or even on non-target insects. This is especially desirable when beneficial insects are being conserved to aid in an overall IPM (integrated pest management) program, or when an ecologically sensitive area is being treated.

Question 22.
How do biofertilizers enrich the soil?
Answer:
Biofertilizers play a vital role to solve the problems of soil fertility and soil productivity.

1. Anabaena azollae, a cyanobacterium, lives in symbiotic association with the free-floating water fern, Azolla. The symbiotic system Azolla-Anabaena complex is known to contribute 40-60 mg N ha-1 per rice crop. In addition to this, cyanobacteria add organic matter, secretes growth-promoting substances like auxins and vitamins, mobilizes insoluble phosphate, and thus improves the physical and chemical nature of the soil.
2. Rhizobium Leguminoserum and Azospirillum fix atmospheric nitrogen as nitrates and nitrites.
3. Mycorrhizae formed by an association of bacteria and roots of higher plants increase soil fertility.

Question 23.
Discuss the role of Microbes as Biofertilizers. (CBSE Delhi 2011, 2015, 2019)
Answer:
Role of microbes as biofertilizers:
Bacteria, cyanobacteria, and fungi (mycorrhiza) are the three groups of organisms used as biofertilizers.
1. Bacteria:
(a) Symbiotic bacteria Rhizobium.
(b) Free-living bacteria Azospirillum and Azotobacter.
(c) They fix the atmospheric nitrogen and enrich soil nutrients.

2.  Cyanobacteria, e.g. Anabaena, Nostoc, Aulosira, Oscillatoria, etc.
(a) They function as biofertilizers by fixing atmospheric nitrogen and
(b) Increasing the organic matter of the soil through their photosynthetic activity.

3. Fungi/mycorrhizae:
(a) Fungi form a symbiotic association with roots of higher plants (mycorrhizae), e.g. Glomus.
(b) The fungus absorbs phosphorus and passes it on to the plant.
(c) Other benefits of mycorrhizae are :

• resistance to root-borne pathogens.
• tolerance to salinity.
• tolerance to drought.
• the overall increase in the plant growth and development

Question 24.
You have been deputed by your school principal to train local villagers in the use of biogas plants. With the help of a labeled sketch explain the various parts of the biogas plant. (CBSE Outside Delhi 2013)
Answer:
Biogas plant:

Biogas plant

1. The plant consists of a well-like underground tank made of bricks. This tank is called a digester.
2. The roof of the digester is dome-shaped made of cement and bricks. The dome of the digester tank acts as a gas holder (or storage tank for biogas). Thus, the gas holder in this type of gas-plant is fixed.
3. There is a gas outlet at the top of the dome for the supply of biogas.
4. On the one side of the digester tank, there is a slopping inlet tank and on the other side, there is a rectangular outlet tank or overflow tank. Both these tanks are made of cement and bricks.
5. The mixing tank is connected with a slopping inlet chamber (or tank) while the outflow tank is connected with a rectangular outlet chamber (or tank).
6. The inlet-chamber is for introducing fresh dung slurry into the main digester tank whereas the outlet chamber is for taking out spent dung slurry after extraction of biogas.

Here we are providing Class 12 Biology Important Extra Questions and Answers Chapter 16 Environmental Issues. Important Questions for Class 12 Biology are the best resource for students which helps in Class 12 board exams.

Class 12 Biology Chapter 16 Important Extra Questions Environmental Issues

Environmental Issues Important Extra Questions Very Short Answer Type

Question 1.
What are pollutants?
Answer:
The substances causing pollution are termed pollutants.

Question 2.
What is the source of aerosols?
Answer:
Smoke, ash, soot.

Question 3.
What photochemical oxidants pollute the air?
Answer:
Peroxyacetyl nitrate, ozone, and aldehyde.

Question 4.
The use of lead-free petrol or diesel is recommended to reduce the pollutants emitted by automobiles. What role does leadership play?
Answer:
Lead inactivates the catalytic converter.

Question 5.
In which year was the Air (Prevention and Control of Pollution) Act amended to include noise as an air pollutant?
Answer:
In 1987.

Question 6.
Name the city in our country where the entire public road transport runs on CNG.
Answer:
Delhi.

Question 7.
It is a common practice to undertake desilting of overhead water tanks. What is the possible source of silt that gets deposited in the water tanks?
Answer:
The soil particles carried by water from a source of the water supply.

Question 8.
What is cultural eutrophication?
Answer:
Faster aging of a lake due to the presence of large amounts of industrial, agricultural wastes, and domestic sewage produced by human activities is called cultural or accelerated eutrophication.

Question 9.
List any two adverse effects of particulate matter on human health.
Answer:

1. Breathing and respiratory problems like bronchitis, asthma, inflammation of lungs, etc.
2. Cardiovascular disorders and premature death.

Question 10.
Why are tall chimneys recommended for factories?
Answer:
To reduce air pollution at ground level.

Question 11.
Name the biggest source of air pollution in large cities.
Answer:
Automobiles.

Question 12.
Why is CNG better than diesel? (CBSE Outside Delhi 2016)
Answer:
CNG burns most efficiently in automobiles and very little of it is left unburnt. It is cheaper and cannot be easily adulterated.

Question 13.
What is the frequency of infrasound and ultrasound?
Answer:

1. Infrasound below 50 Hz.
2. Ultrasound-Above 20000 Hz.

Question 14.
Name the layer of atmosphere that is associated with ‘good ozone’.(CBSE 2019)
Answer:
Stratosphere

Question 15.
What would have been the temperature at the surface of the earth without the greenhouse effect? (HOTS)
Answer:
18°C as compared to the present average of 15°C.

Question 16.
How much average global temperature will rise by the year 2100?
Answer:
1.4 – 5.8°C.

Question 17.
What are Dobson units? (CBSE Delhi 2011)
Answer:
Dobson units (DU) measure the thickness of ozone in a column of air from the ground to the top of the atmosphere.

Question 18.
Explain polar vortex.
Answer:
Antarctic air is completely isolated from the rest of the world by the natural circulation of wind called the polar vortex.

Question 19.
Name the world’s most problematic aquatic weed. What is the nature of the water body in which the weeds grow abundantly? (CBSE Delhi 2008)
Answer:

1. Eichhornia
2. Highly polluted freshwater bodies containing organic waste.

Question 20.
Write the name of the organism that is referred to as ‘Terror of Bengal’. (CBSE Delhi 2014)
Answer:
Eichhornia (Water hyacinth).

Question 21.
State the cause of Accelerated Eutrophication. (CBSE Delhi 2014)
Answer:
Nutrient enrichment due to the addition of pollutants from industries and homes into water bodies causes accelerated eutrophication.

Environmental Issues Important Extra Questions Short Answer Type

Question 1.
An electrostatic precipitator in a thermal power plant is not able to generate a high voltage of several thousand. Write the ecological implication because of it. (CBSE 2017)
Answer:
It will not be able to remove particulate matter present in the exhaust of thermal power plants. Hence it cannot control pollution as dust particles will be released into the air.

Question 2.
List four benefits to human life by eliminating the use of CFCs. Suggest two practices giving one example of each that helps protects rare or threatened species. (CBSE 2017)
Answer:
Benefits to human life by eliminating the use of CFCs:

1. Ozone depletion will be prevented.
2. The greenhouse effect will be controlled.
3. Global warming will be controlled.
4. It will prevent old climatic changes that take place due to a rise in temperature.

Two practices that help, protect rare or threatened species are:

1. Ex-situ conservation: In this, organisms are taken out of their natural habitat and kept at a special care location. Example: Botanical garden, zoo, etc.
2. In situ conservation: Here the species are conserved in their natural habitat. Example: Natural park, sanctuaries.

Question 3.
Why are microbes like Spirulina being produced on a commercial scale? Mention its two advantages. (CBSE Delhi 2018C)
Answer:
Spirulina is produced on a commercial scale due to the following reasons:

1. Spirulina is a rich source of protein. It can be used to solve the problem of hunger and malnutrition.
2. It also reduces environmental pollution as it utilizes the waste as raw material which otherwise pollutes the environment.

Question 4.
Classify pollution on the basis of origin.
Answer:
Types of pollution on the basis of origin:

1. Natural, e.g. volcanic eruption.
2. Anthropogenic, e.g. man-made such as industrial pollution.

Question 5.
What are the two main sources of air pollution?
Answer:

1. Fixed sources: Which include large factories, electrical power plants, mineral smelters, small industries.
2. Mobile sources: They include all sorts of transports.

Question 6.
What are the effects of carbon monoxide?
Answer:
Carbon monoxide (CO) is a product of the incomplete combustion of fossil fuels. Nearly 50% of all CO emission originates from automobiles. It is also present in cigarette smoke. CO is short-lived in the atmosphere and gets oxidized to CO₂. Carbon monoxide is highly poisonous to most animals. When inhaled, CO reduces the oxygen-carrying capacity of the blood.

Question 7.
Write the names of four main gases which pollute the air.
Answer:
C0₂, CO, N0_2, and S02 pollute the air.

Question 8.
Differentiate between primary air pollutants and secondary air pollutants.
Answer:
Differences between primary air pollutants and secondary air pollutants:

Primary air pollutant	Secondary air pollutant
1. Pollutant persists in the form in which it is released in the environment.	1. It is formed from another pollutant due to change or reaction.
2. Example. Glass, D.D.T., Carbon monoxide, Nitrogen oxide, Hydrocarbons.	2. Example. Ozone, Peroxyacyl nitrate (PAN)

Question 9.
What are the effects of air pollution? How can it be minimized?
Answer:
Effects of air pollution: Atmospheric pollution causes respiratory and vascular diseases in humans, produces fluorosis in livestock, adversely affects plants and buildings, and poses a threat to the climate.

Control of air pollution: Air pollution can be minimized by separating pollutants from harmless gases, or by converting them to harmless substances before releasing the industrial and motor vehicle exhausts into the atmosphere.

Question 10.
What is particulate matter? How does particulate matter affect the biological world?
Answer:
Suspended Particulate Matter (SPM): The solid particles or liquid droplets (aerosols) remain suspended in the air. For example, smoke, soot, dust, asbestos, etc. SPM above the size of 10 mm is trapped by nasal hair, pharyngeal, tracheal, and bronchial mucus. The same is coughed out or deposited in the nose as nasal scales. Smaller particles of SPM reach the alveoli.

There they may be attacked by phagocytes or pass into living cells. SPM causes irritation in the respiratory tract, bronchitis, and lung diseases. These particles may cause asbestosis, pneumoconiosis, etc. They also result in reduced visibility.

Question 11.
What is photochemical smog? How smog affects the biological world?
Answer:
1. Photochemical Smog: It is a secondary pollutant. It is yellowish-brown smog that is formed under oxidizing conditions and high temperatures over cities and towns that are releasing a lot of nitrogen oxides and unburnt hydrocarbons. In still air, the two interact photochemically to produce photochemical oxidants-ozone, PAN, aldehydes, and phenols. Photochemical smog does not have any appreciable amount of primary pollutants. It is also called Los Angeles smog.

2. Effect on the biological world: It causes bleaching of foliage of certain plants. It also causes silvering, glazing, and necrosis of the leaves. It causes respiratory problems in animals.

Question 12.
What is acid rain? What are its effects on plants?
Answer:
Acid rains: Sulphur dioxide and sulfur trioxide are produced by the oxidation of sulfur in fossil fuels. Similarly, nitrogen dioxide or nitrogen monoxide are released through vehicular emissions. These gases react with water and form sulphuric acid, sulfurous acid, or nitric acid. These acids when precipitated as rain or snow create acid rain or acid precipitation. The pH of acid rain is less than 5-6 and could be as low as 4. Acid rain adversely affects plant vegetation by causing chlorosis and necrosis.

Question 13.
Write the effects of hydrocarbons (HCs).
Answer:
Effects of hydrocarbons:

1. Benzene and its derivatives are carcinogens.
2. Formaldehyde causes indoor pollution. It contributes to photochemical smog. Inhaling formaldehyde can cause respiratory irritation.
3. Some reactive HCs contribute to the formation of secondary pollutants.

Question 14.
What is the relation between epiphytic lichen and air pollution?
Answer:
Epiphytic lichens are reliable indicators of air pollution and can often be ranked on a sensitivity scale to estimate changing level of atmospheric pollution, especially of S02. The presence or absence of certain epiphytic lichens in a given locality and the pattern of their distribution can be related to the extent of pollution in the area.

Question 15.
How is air pollution responsible for changing the ill effects of ultraviolet radiation?
Answer:
Air pollution is enhancing the ill effects of UV radiation. Fluorocarbons are threatening and depleting the ozone layer of the stratosphere, thus allowing the UV radiations to reach this planet. The UV radiations cause serious damages to life by damaging nucleic acids in living organisms.

Question 16.
Mention any two examples of plants used as windbreakers in the agricultural fields.
Answer:
Windbreakers or shelterbelts provide shelter from wind and protect soil from erosion. Jamun and Imli and some other trees like Babul, Lawsonia, Thevetia, and Calotropis act as windbreakers in the agricultural field.

Question 17.
It is true that carpets and curtains/drapes placed on the floor or wall surfaces can reduce the noise levels. Explain briefly.
Answer:
Yes, it is true that placing carpets on the floor and curtains on the wall surface, windows, reduce the noise level. This is because the curtains and carpets on the wall surface act as muffling devices and absorb sounds of moderate level.

Question 18.
What is hybrid vehicle technology? Explain its advantages with a suitable example.
Answer:
The technology used to run vehicles on dual-mode like petrol or compressed natural gas is said to be hybrid vehicle technology. These vehicles run on either petrol or CNG. As CNG is a clean and green fuel so it is helpful to reduce environmental pollution and also to conserve petrol and other fossil fuels.

Question 19.
Is it true that if the dissolved oxygen level drops to zero the water will become septic? Give an example that could lower the dissolved oxygen content of an aquatic body.
Answer:
Yes, the water becomes septic if the dissolved oxygen drops to zero. Organic pollution (biodegradable) is an example.

Question 20.
Name any one of the greenhouse gases and their possible source of production on a large scale. What is the harmful effect of it?
Answer:
C02 and Methane. C02 levels are increasing due to the burning of fossil fuels leading to global warming.

Question 21.
It is a common practice to plant trees and shrubs near the boundary walls of buildings. What purpose do they serve?
Answer:
The plants growing near the boundary wall act as barriers to sound pollution and act as dust catchers.

Question 22.
Why has the National Forest Commission of India recommended a relatively larger forest cover for hills than the plains?
Answer:
National Forest Commission in its National Forest Policy (1988) has recommended 33 percent forest cover for the plains and 67 percent for the hills because plains mainly provide land for human settlements. Forests on the hills prevent soil erosion and landslides.

Question 23.
How can slash and burn agriculture become environment-friendly?
Answer:
Slash and burn agriculture can become environment-friendly if rows of trees and shrubs are left intact while clearing the area for cultivation. This will prevent soil erosion and the invasion of weeds. There will be a quicker recovery of the forest after the area is abandoned.

Question 24.
What is the main idea behind the “Joint Forest Management Concept” introduced by the Government of India?
Answer:
It is government-private entrepreneurship for upkeeping forests. The forest department prepares the plan, procures saplings and equipment for planting and plant protection. The locals take care of the plants till they become mature. For this, they get an honorarium and share in plant products. This sort of management provides livelihood to locals and protection to plants against stealing and illegal felling of trees.

Question 25.
What do you understand by snow blindness?
Answer:
Snow blindness is temporary blindness caused by inflammation of the cornea due to absorption of UV-B radiations. It is accompanied by photo burning and dimming of eyesight after which cataract develops. Regular exposure to UV-B radiations causes permanent damage to the cornea resulting in blindness.

Question 26.
What is water pollution?
Answer:
Water pollution: The addition of harmful materials to water is termed water pollution. The sources of inland water pollution are community wastewater (sewage) and wastes from industries and agricultural practices. Water pollutants include organic matter, pathogens, chemicals and minerals, solid particles, radioactive wastes, and heat.

Question 27.
What are two serious problems with the use of nuclear energy?
Answer:
Nuclear energy has two most serious problems:

1. The accidental leakage of radioactivity
2. The safe disposal of radioactive wastes.

Question 28.
Particulate and gaseous pollutants along with harmless gases are released from the thermal power plants.
(i) Name any two harmless gases released.
Answer:
Nitrogen and oxygen.

(ii) Name the most widely used device for removing particulate pollutants from the air. Explain how the device is used? (CBSE 2009)
Answer:
Electrostatic precipitator: Working of electrostatic precipitator The device has electrode wires kept at very high voltage in order to produce corona that releases electrons. These electrons provide dust particles a negative charge. The device has a collecting plate that is grounded. So when air with dust rushes through the device, the negatively charged dust is trapped at the collecting plate.

Question 29.
Study the graph given below and answer the questions that follow:
(i) What is the relationship between dissolved oxygen and biochemical oxygen demand (BOD)?
(ii) Mention their effect on aquatic life in the river. (CBSE 2008)

Answer:
(i) BOD indicates the quantity of wastewater. BOD refers to the amount of dissolved oxygen required by bacteria in decomposing oxygen.

Thus greater the BOD, the lesser will be dissolved oxygen. A sudden decline in dissolved oxygen causes the death of many aquatic organisms.

(ii) There is a sharp decline in the dissolved organic waste downstream from the point of discharge. More the dissolved oxygen, the healthier will be the aquatic life and vice versa.

Question 30.
A crane had a DDT level of 5 ppm in its body. What would happen to the population of such birds? Explain giving reasons. (CBSE 2009)
Answer:

1. Fish-eating bird crane has a DDT level of 5 ppm due to biomagnification.
2. The high concentration of DDT disturbs calcium metabolism in birds.
3. It causes thinning of eggshells and premature breaking of the egg.
4. There will be a decline in the bird population.

Question 31.
Study the graph given below. Explain how is oxygen concentration affected in the river when the sewage is discharged into it. (CBSE Delhi 2011)
Answer:
The figure shows the changes as a result of sewage discharged into the river. Micro-organisms involved in the biodegradation of organic matter in the receiving water body consume a lot of oxygen. As a result, there is a sharp decline in dissolved oxygen downstream. It causes the death of a large number of fishes and other aquatic creatures.

Question 32.
“Determination of Biochemical Oxygen Demand (BOD) can help in suggesting the quality of a water body” Explain. (CBSE Delhi 2015)
Answer:
The quality of the water body depends upon the organic wastes present in it. More is the amount of organic wastes present, poorer is the quality of water body for human consumption. The number of organic wastes can be determined by the BOD of that water body, e.g. BOD of pure drinking water is less than one, below 1500 mg/liter for weak organic wastewater but is more than 4000 mg/liter for a highly polluted water body.

Question 33.
Discuss the role of women and communities in the protection and conservation of forests.
Answer:
Role of women and communities in the protection and conservation of forests:

1. In 1731, the Bishnoi community led by a woman Amrita Devi obstructed the cutting of trees by hugging the tree and asked the workers of the king to cut her before cutting the tree. Her three daughters and hundreds of other Bishnois lost their lives in saving trees.
2. Chipko movement started by Chandi Prasad Bhatt and Sunder Lai Bahuguna of Silyara in the Tehri region when workers of the contractor were not allowed to cut the trees by village people by hugging around them.

Question 34.
Explain the different steps involved during the primary treatment phase of sewage. (CBSE Delhi 2015)
Answer:
Primary treatment of wastewater in sewage treatment plants involves mechanisms like floatation, Alteration, and sedimentation so as to remove insoluble and stable solid wastes. The final product after primary treatment is called primary sludge.

Question 35.
How have human activities caused desertification? Explain. (CBSE Delhi 2013)
Answer:
Desertification caused by human activities. The development of the fertile topsoil takes centuries. But it can be removed very easily due to human activities.

• Over-cultivation
• Unrestricted grazing
• Deforestation
• Poor irrigation practices.

All these human activities result in arid patches on land. When large barren patches extend and meet over time a desert is created.

Thus desertification is a major problem particularly due to increased urbanization.

Environmental Issues Important Extra Questions Long Answer Type

Question 1.
Define pollution. Compare the biodegradable and non-biodegradable pollutants.
Answer:
Pollution: “Environmental pollution is an unfavorable alteration of our surroundings largely as a by-product of man’s actions through direct or indirect effects of changes in energy patterns, radiation levels, chemical and physical conditions and abundance of organisms.”

Differences between biodegradable and non-biodegradable pollutants:

Biodegradable Pollutants	Non-biodegradable Pollutants
1. These wastes can be broken down into harmless substances naturally.	1. These cannot be broken down into harmless substances naturally.
2. The disposal of biodegradable wastes is easy and therefore, maintains balance in the ecosystem. Example. Cattle dung	2. Their disposal is not easy and is a problem. Example. D.D.T., plastics.

Question 2.
What is air pollution? List various air pollutants.
Answer:
Air pollution: The release of harmful materials into the air is called air or atmospheric pollution. It is the degradation of air quality and atmospheric conditions.

Air pollutants include gaseous materials, suspended particles, and radioactive substances.
1. The gaseous pollutants of the air come from combustion in motor vehicles and industries. These include CO, C0₂, NO, N0₂, S0₂, S0₃, hydrocarbons, fluorides, and photochemical oxidants such as peroxyacetyl nitrate (PAN), ozone (0₃), and aldehydes. PAN is a secondary pollutant formed in the air by the interaction between nitrogen oxides and hydrocarbons in the presence of sunlight. It is more toxic than the primary pollutants. An increase in toxicity by the reaction among pollutants is called synergism.

2. The particulate contaminants of air also come from industries and automobiles. These include fly-ash, soot, metal dust, cotton fibers, asbestos, fibers, lead, aerosols (chlorofluorocarbons or CFCs), polychlorinated biphenyls (PCBs), tobacco smoke, smog, pollen, spores, cysts, and bacteria.
(in) Radioactive substances released by nuclear explosions and war explosives are also very harmful air pollutants.

Question 3.
Define a pollutant. How many types of pollutants are there?
Answer:
Pollutant: It is a substance (e.g. dust, smoke), chemical (e.g. S0₂), or factor (e.g. heat, noise) that on release into the environment has an actual or potentially adverse effect on human interests. A pollutant can also be defined as a constituent in the wrong amount, at the wrong place, or at the wrong time.

Question 4.
What measures do you suggest to control pollution from automobile exhaust?
Answer:
Control of pollution from automobile exhaust:

1. Efficient engines can reduce the number of unburnt hydrocarbons from vehicuLar emissions.
2. Use of cataLytic converters to convert harmful gases to harmless.
3. Use of good quality fuel.
4. Unleaded petroL can reduce the amount of lead in the exhaust.
5. The use of CNG (compressed natural gas) Lowers the toxic contaminants in the exhaust.

Question 5.
Blends of polyblend and bitumen, when used, help to increase road life by a factor of three. What is the reason?
Answer:
Polyblend is a fine powder of recycled modified plastic. The binding property due to increased cohesion and enhanced water-repelling property of plastic makes the road last longer besides giving added strength to withstand more loads.

This is because:

• Plastic increases the melting point of the bitumen which would prevent it from melting in India’s hot and extremely humid climate, where temperature frequently crosses 45°C.
• Rainwater will not seep through because of the plastic in the tar.

Question 6.
“Indiscriminate human activities have strengthened the greenhouse effect resulting in Global Warming.” Give the relative contribution of various Greenhouse Gases in the form of a pie chart and explain the rate of the energy of sunlight reaching the earth’s surface contributing towards Global Warming. (CBSE Sample Paper 2020)
Answer:
a)
(b) Rate of the energy of sunlight:

• Clouds and gases reflect one-fourth of incoming solar radiations.
• Some of the energy is absorbed by clouds and gases.
• Thus only half of the solar energy falls on the surface of the earth.
• The small amount of energy is reflected back.
• Earth surface re-emits heat in the form of infrared radiations.
• The major fraction of solar energy is absorbed by atmospheric gases.

Question 7.
How does a scrubber function?
Answer:
Scrubber: A scrubber can remove gases like sulfur dioxide. In a scrubber, the exhaust is passed through a spray of water or lime. Water dissolves gases and lime reacts with sulfur dioxide to form a precipitate of calcium sulfate or sulfite.

Question 8.
Discuss the causes and effects of global warming.
Answer:
Global warming: The increase in global mean temperature due to the enhanced greenhouse effect is called global warming.

Effects of global warming:
1. Effect on weather and climate.
(а) The average temperature of the earth may increase by 1.4° to 5.8°C by the year 2100.
(б) Winter precipitation may decrease at lower altitudes.
(c) Frequency of droughts, floods may increase.
(d) Climatic change is a threat to human health in tropical and subtropical countries.

2. Sea level change. Sea level had been raised by 1 to 2 mm per year during the 20th century. It is predicted that by the year 2100, the global mean sea level can increase up to 0.88 m over the 1990 level. Global warming may contribute to sea-level rise due to the thermal expansion of the ocean.

3. Effect on a range of species distribution. Vegetation may extend 250-600 km poleward with a global rise in temperature by 2 to 5°C during the 21st century.

4. Effect on food production. Increased temperature will cause an eruption of plant diseases and pests and vast growth of weeds.

Question 9.
Show the relative contribution of greenhouse gases to total global warming.
Answer:
Greenhouse gases:

The relative contribution of various greenhouse gases to total Global warming.

Question 10.
Make a chart illustrating the kinds of water pollutants.
Answer:
Classification of water pollutants:

Question 11.
List four laws that enforce control of pollution.
Answer:
1. The Environment (Protection) Act, 1986: This act clearly brings the protection of air, water, and soil quality, and the control of environmental pollutants including noise under its purview.

2. The Insecticide Act, 1968: This act deals with the regulation of import, manufacture, sale, transport, distribution, and use of insecticides with a view to preventing risk to human health and other organisms.

3. The Water (Prevention and Control of Pollution) Act, 1974: This act deals with the preservation of water quality and the control of water pollution with a concern for the detrimental effects of water pollutants on human health and also on the biological world.

4. The Air (Prevention and Control of Pollution) Act, 1981: The act deals with the preservation of air quality and the control of air pollution with a concern for the detrimental effects of air pollutants on human health and also on the biological world. In 1987, important amendments of the Air Act 1981 were made and the noise was recognized as an air pollutant.

5. Many countries have enacted legislation to control noise. India enacted the Air (Prevention and Control of Pollution) Act, 1981 and as per this, noise pollution has been declared as an offense.

Question 12.
A factory drains its wastewater into the nearby lake causing algal bloom.
(i) How was the algal bloom caused?
Answer:
An algal bloom is caused due to enrichment of water with nutrients such as nitrogen and phosphorus. As fertility increases plant life increases.

(ii) What would be the consequences?
Answer:
Water becomes unfit for aquatic fauna because of scum and unpleasant odors. The fish and other organisms die because of a shortage of oxygen.

(iii) Name the phenomenon that caused it. (CBSE2008)
Answer:
Eutrophication.

Question 13.
Explain accelerated eutrophication. Mention any two consequences of this phenomenon. (CBSE 2009)
Answer:
Accelerated eutrophication: The natural aging of a lake by the biological enrichment of its water is called eutrophication. However, if pollutants from man’s activities like effluents from industries and homes radically accelerate the aging process, it is called accelerated cultural eutrophication.

Consequences:

• There is excess growth of algae causing algal bloom that gives an unpleasant odor.
• There is a shortage of oxygen which causes the death of the fauna of the water body.
• Ultimately death of the lake occurs.

Question 14.
How does algal bloom destroy the quality of a freshwater body? Explain. (CBSE Delhi 2013)
Answer:
Algal bloom destroys the quality of the freshwater body. Phosphorus and nitrates dissolved in polluted water act as nutrients for the green algal plants. These pollutants accelerate the growth of algae that may form a mat on the water surface. It is termed an algal bloom.

Effect:

1. The algae use oxygen at night and deoxygenate the water enough to kill the fish and other aquatic animals.
2. The algal mat on the water surface blocks light, making it unavailable for submerged plants.
3. The algae and dead organisms feed the oxygen-consuming bacteria.
4. Silt and decaying matters accumulate and finally fill the water body (lake or pond). This process is called senescence.
5. It is a natural stage in the change of water body into dry land and it becomes barren. Ultimately death of the water body occurs.

Question 15.
How did a citizen group called Friends of Areata March, Areata, California, USA, help to improve the water quality of the marshland using Integrated Waste Water Treatment? Explain in four steps. (CBSE 2018)
Answer:
Wastewater including sewage can be treated in an integrated manner by utilizing a mix of artificial and natural processes, which are as follows:

1. First of all the conventional sedimentation, filtering, and chlorine treatments are given. However, after this stage, a lot of dangerous pollutants like dissolved heavy metals still remain.
2. To combat this, an innovative approach was taken. The biologists developed a series of six connected marshes over 60 hectares of marshland.
3. Appropriate plants, algae, fungi, and bacteria were seeded into this area, which neutralizes, absorbs, and assimilates the pollutants. Hence, as the water flows through the marshes, it gets purified naturally.
4. The marshes also constitute a sanctuary, with a high level of biodiversity in the form of fishes, animals, and birds.

Question 16.
List the control measures for minimizing air pollution.
Answer:
Control measures for minimizing air pollution:

1. Simple combustible solid wastes should be burnt in incinerators.
2. Automobiles must be either made to eliminate the use of gasoline and diesel oil or complete combustion is obtained in the engine so that harmful products are omitted.
3. There should be a cut in the use of agricultural sprays.
4. Excessive and undesirable burning of vegetation should be stopped.
5. Smoking should be stopped.
6. Use of electrostatic precipitators and filters in the factories to minimize atmospheric pollution.
7. The use of tall chimneys can substantially reduce the concentration of pollutants at the ground level.
8. Weather forecasts can help in controlling air pollution as the meteorological conditions affect the dispersion, dilution, and mixing of various emissions and proper operation of factory schedule, e.g. when atmospheric stagnation or calm is predicted, a power plant should switch over from coal to gas.
9. Separation of pollutants from harmless gases.
10. Dispersion of pollutants to innocuous (harmless) products before releasing into the atmosphere.

Question 17.
Why is the ozone layer in the stratosphere called a protective layer?
Answer:
Ozone layer as a protective layer: The ozone layer in the stratosphere is very useful to human beings because it absorbs the major part of harmful ultraviolet radiation coming from the sun. Therefore, it is called a protective layer. However, it has been observed that the ozone layer is getting depleted. One of the reasons for the depletion of the ozone layer is the action of aerosols spray propellants.

Chemicals such as fluorocarbons and chlorofluorocarbons are used as aerosol propellants. These compounds react with ozone gas in the atmosphere thereby depleting it. Scientists all over the world are worried about the destruction of the ozone layer. If the ozone layer in the atmosphere is significantly decreased, these harmful radiations would reach the earth and would cause many damages such as skin cancer, genetic disorders in man and other living forms. Efforts are being made to find substitutes for these chemicals that do not react with ozone.

Question 18.
Why industrial effluents are more difficult to manage than municipal sewage? Name a disease that is caused by heavy metal contamination.
Answer:
Industrial effluents released into water contain toxic substances, such as arsenic, cadmium, lead, zinc, copper, mercury, and cyanides, besides some salts, acids, and alkalies. All these materials can prove harmful to health. They may reach the human body. Minamata is a disease caused by heavy metal (Mercury) contaminated water.

Question 19.
What is deforestation? List the causes of deforestation.
Answer:
Deforestation: According to an estimate, almost 40% of forests have been lost in the tropics and 1% of forests in the temperate region. In India, at the beginning of the twentieth century, forests covered about 30% of land whereas, by the end of the century, it reduced to 19.4%. The National Forest Policy (1988) has recommended 33% forest cover for plains and 67% for hills.

Causes of Deforestation:

• Forests are converted into agricultural land to feed the growing human population.
• Forests are cleared for making homes and establishing industries.
• Trees are felled for timber, firewood, etc.
• Jhum cultivation in the north¬eastern states has contributed to deforestation.

Question 20.
List all the wastes that you generate, at home, school, or during your trips to other places that you could very easily reduce. Which would be difficult or rather impossible to reduce?
Answer:
List of wastes:

1. Papers, clothes, polythene bags
2. Disposable crockery
3. Aluminum foil, cans
4. Leftover of tiffins
5. Wood
6. Sewage

Wastes that can be reduced:

1. Papers, clothes,
2. Leftover of tiffins.

Wastes that cannot be reduced:

1. Aluminum foil cans
2. Disposable crockery
3. Polythene bags

Question 21.
Why ozone hole form over Antarctica? How will enhanced UV radiations affect us?
Answer:
Ozone hole: During the period 1956-1970 the springtime O₃ layer thickness above Antarctica varied from 280 to 325 Dobson unit. Thickness was sharply reduced to 225 DU in 1979 and 136 DU in 1985. Antarctic air is completely isolated from the rest of the world by the natural circulation of wind called the polar vortex. The decline in ozone layer thickness during that springtime is called ozone hole. It was first noted in 1985 over Antarctica.

Effects of UV radiation on humans:

• In humans, the increase in UV radiation increases the incidence of cancer (including melanoma).
• Reduces the functioning of the immune system.
• The cornea absorbs UV-B radiations, and a high dose of UV-B causes inflammation of the cornea called snow blindness, cataract, etc. Exposure may permanently damage the cornea and cause blindness.

Question 22.
Write critical notes on:
(i) Eutrophication
Answer:
Eutrophication: The process by which a body of water becomes barren either by natural means or by pollution, extensively rich in dissolved nutrients. It results in the increased primary productivity that often leads to seasonal deficiency in dissolved oxygen. Less dissolved oxygen ultimately affects aquatic life.

Algal bloom: Phosphorus and nitrates dissolved in water act as nutrients and accelerate the growth of algae that may form a mat on the water surface. It is termed algal bloom.

Effects: The algae use oxygen at night and may deoxygenate the water enough to kill the fish and other animals. The algal mat at the water surface may block light to the submerged plants. The algae may die and sink, and feed the oxygen-consuming bacteria. They may be pushed onto the shore by wind and decompose, releasing foul gases, such as hydrogen sulfide. Silt and decaying matter may accumulate and finally fill the lake or pond. This process is called senescence. It is a natural stage in the change of a lake into dry land and it becomes barren.

(ii) Biological magnification (CBSE 2015, 2019 C)
Answer:
Biological magnification: The phenomenon in which the harmful pollutants (such as pesticides) enter the food chain and get concentrated more and more at each successive trophic level of organisms is called biological magnification.

This phenomenon is well known for mercury and DDT. The figure given ahead shows the biomagnification of DDT in an aquatic food chain. Zooplankton (tiny floating animals in the food chain), accumulated modest levels of DDT. However, small fish, forming the next level of the food chain, must eat zooplankton several times their own weight, and thus they accumulated more DDT.

In this manner, the concentration of DDT magnified at successive trophic levels, starting with 0.003 ppb (ppb = parts per billion) in water it ultimately reached 25 ppm (ppm = parts per million) in fish-eating birds. High concentrations of DDT disturb calcium metabolism in birds, which causes thinning of eggshells and their premature breaking eventually causing a decline in the bird population.

Biomagnification of DDT in an aquatic food chain.

(iii) Groundwater depletion and ways for its replenishment. (CBSE 2012)
Answer:
Groundwater depletion and ways for its replenishment: Underground water is most pure and safe for drinking. It is getting polluted particularly in industrial towns. The common sources of underground water pollution are sewage and industrial effluents spilled over the ground. The fertilizers and pesticides used in fields also act as pollutants. Pollution is also increased due to seepage from refuse dumps, septic tanks, and seepage pits. Method to control. Sewage and factory wastes should be treated to clean them before their release into water sources.

Question 23.
List various measures for control of noise pollution.
Answer:
Control of noise pollution:

1. Construction of soundproof rooms for noisy machines in industries.
2. Radios and transistors should be kept at low volume.
3. The use of horns with jarring sounds should be banned.
4. Noise-producing industries, aerodromes, railway stations, etc. should be shifted away from the inhabited areas.
5. Proper laws should be enforced to check the misuse of loudspeakers and public announcement systems,
6. Need to enforce a silence zone around institutions like educational institutes, residential areas, hospitals, etc.
7. Sound-absorbing techniques like acoustical furnishing should be extensively employed.
8. Noise should be deflected away from the receiver by using mechanical devices.
9. The green muffler scheme involves growing green plants along roadsides to reduce noise pollution.

Question 24.
What measures, as an individual, you would take to reduce environmental pollution? (CBSE Delhi 2011)
Answer:
Role of individuals in reducing pollution:

1. Use of unleaded petrol or CNG in vehicles as fuel.
2. Use of reformulated gasoline to save ozone in the atmosphere.
3. The use of power generators in residential areas should be avoided.
4. Plantation of trees.
5. Excessive and unplanned use of fertilizers should be avoided.
6. Biodegradable material should be used.
7. Do not blow the horn with a jarring sound.
8. Radio, transistors, T.V., Music systems should be kept at low volume to control noise pollution.

Question 25.
Discuss briefly the following:
(i) Radioactive wastes
Answer:
(i) Radioactive wastes: Radioactive wastes are of three types depending on the amount of radioactivity.
(a) Low-level radioactive wastes.
(b) Intermediate level radioactive wastes.
(c) High-level radioactive wastes.

The use of nuclear energy has two most serious problems. The first is accidental leakage and the second is safe disposal of radioactive wastes. Wastes from atomic power plants come in the form of spent fuels of uranium and plutonium. People working in such power plants, nuclear reactors, fuel processors, etc. are vulnerable to their exposure. These also undergo biological magnification and may reach 75,000 times in birds.

Radioisotopes. Many radioactive isotopes like C¹⁴, I¹²⁵, P^32, and their compounds are used in scientific researches. The wastewater of these research centers contains radioactive elements which may reach human beings through water and food chains.

Disposal of radioactive wastes. Such wastes are first concentrated to reduce the volume and then kept for 50-100 years in small ponds within the premises of nuclear power plants. This interim storage causes considerable decay of radioactivity and lessening of heat problem. It has been recommended that subsequent storage should be done in suitably shielded containers buried within the rocks, about 500 m deep inside Earth. However, this method of disposal is meeting stiff opposition from the public.

(ii) Defunct ships and e-wastes.
Answer:
Defunct ships and e-wastes: Defunct ships are a kind of solid waste requiring proper disposal. Such ships are broken down in developing countries because of cheap labor and scrap metal. They often contain toxic substances such as asbestos, polychlorinated biphenyls, tributyltin, lead, and mercury.

Similarly, irreparable computers and electronic goods are known as electronic wastes of e-wastes. Most of them are shipped to developing countries for metals like copper, nickel, silicon, gold and are recovered by recycling. Recycling is the only solution for the treatment of such wastes, provided it is carried out in an environment-friendly manner. An international treaty called Basel Convention drafted in Basel (Switzerland) as a result of great demand from developing countries.

(iii) Municipal solid wastes.
Answer:
Municipal Solid Wastes: Solid wastes refer to everything that goes out in the trash. Municipal solid wastes are wastes from homes, offices, stores, schools, etc. that are collected and disposed of by the municipality. The municipal solid wastes generally comprise paper, food wastes, glass, metals, rubber, leather, textile, etc. Burning reduces the volume of the wastes, although it is generally not complete and open dumps often serve as the breeding ground for rats and flies.

Management of municipal solid wastes:

1. Sanitary landfills were adopted as the substitute for open-burning dumps. In a sanitary landfill, wastes are dumped in a depression or trench after compaction and covered with dirt every day.
2. Municipal solid wastes, containing biodegradable organic wastes, can be transformed into organic manure for agriculture.
3. Sewage sludge and industrial solid wastes are used as landfills.
4. Hazardous metal-containing wastes are used as bedding material for road construction.
5. Other options are incineration of wastes and using emitted heat in electricity generation, and recycling various components of wastes.

Question 26.
What initiatives were taken for reducing vehicular air pollution in Delhi? Has the air quality improved in Delhi?
Or
Explain any three measures which will control vehicular air pollution in Indian cities. (CBSE 2009)
Answer:
Automobiles are the major sources of air pollution in Delhi because it has a very high number of cars.

Some specific measures taken to reduce vehicular air pollution are as follows.

• Use of CNG (Compressed Natural Gas) for its public transport system.
• Phasing out of old vehicles.
• Use of unleaded petrol.
• Use of low sulfur petrol and diesel.
• Use of catalytic converter in vehicles.
• Application of Euro II norms for vehicles.

Because of the above-mentioned steps taken up by the Government, the air quality of Delhi has improved with a substantial fall in carbon monoxide, oxides of sulfur, and nitrogen levels between 1997 and 2005.

Question 27.
Discuss briefly the following:
(i) Greenhouse effect
Answer:
Greenhouse effect: Earth’s temperature is maintained by reradiated infrared radiations by CO₂, CH₄, O₃, NO, and N0₂ and slightly by water vapors in the atmosphere. These gases prevent heat from escaping to outer space, so are functionally comparable to glass panels of a greenhouse and are called greenhouse gases (GHGs) and phenomena called the greenhouse effect. The CO₂ is added to the atmosphere mainly by burning fossil fuels, volcanic activities, etc.

Greenhouse gases are useful in keeping the earth warm with an average temperature of 15° C. In their absence, the surface temperature of the earth will be as low as 18° C. This temperature will freeze all water and kill most life. However, excess greenhouse gases are equally harmful. Over 7 × 10¹² kg of CO₂ is being added annually to the atmosphere by the burning of fossil fuels. As a result, the CO₂ concentration of the atmosphere has risen from 0.028% in 1800 to .0359% in 1994. Now enhanced greenhouse effect is resulting in Global Warming.

(ii) Catalytic converters
Answer:
Catalytic converters: Catalytic converters, having expensive metals namely platinum-palladium and rhodium as catalysts, are fitted into automobiles for reducing the emission of harmful gases. As the exhaust passes through the catalytic converter, unburnt hydrocarbons are converted into carbon dioxide and water, and carbon monoxide and nitric oxide are changed to carbon dioxide and nitrogen gas, respectively. Motor vehicles equipped with catalytic converters should use unleaded petrol because leaded petrol inactivates the catalyst.

It has been established that the installation of catalytic converters can slash carbon monoxide emissions from 90 grams to 3.4 grams per mile run. So if half the vehicles on Delhi and Mumbai roads are made to install such catalytic converters, then total CO emission in India can be reduced by 70 percent.

(iii) Ultraviolet B
Answer:
Ultraviolet-B radiations: These are high energetic UV-radiations that are mostly blocked by the ozonosphere located in the stratosphere of the atmosphere. But due to the increased production of ozone-depleting substances (ODS) like CFCs, halons, etc., the ozone shield is becoming thinner and thinner. This is increasing the amount of UV-B radiations reaching the earth’s surface. These radiations are carcinogenic.

The cornea absorbs UV-B radiation. High dose of UV-B causes inflammation of the cornea called snow blindness cataract. Exposure may permanently damage the cornea.

Question 28.
Looking at the deteriorating air quality because of air pollution in many cities of the country, the citizens are very much worried and concerned about their health. The doctors have declared a health emergency in the cities where the air quality is very severely poor.
(i) Mention any two major causes of air pollution.
Answer:
Two causes of air pollution are:
(a) Burning of fossil fuels.
(b) Industrial effluents

(ii) Write the two harmful effects of air pollution on plants and humans.
Answer:
Harmful effects of air pollution are:
(a) It affects the respiratory system of humans and animals.
(b) It also reduces the growth and yield of crops and causes premature death of plants.

(iii) As a captain of your school Eco¬club, suggest any two programs you would plan to organize in the school so as to bring awareness among the students on how to check air pollution in and around the school. (CBSE2018)
Answer:
As a captain of your school Eco-club, I shall suggest:
(a) Encouraging public transport, i.e. buses, metro, etc and using CNG/electric vehicles instead of diesel and petrol vehicles.
(b) Planting more trees to curb pollution.

Question 29.
While studying pollution of water, a group of students observed mortality of fish in the river flowing through the city and also in the pond which was away from the city but was adjacent to the crop fields. They further found that drains of the city discharged sewage into the river and the water from farms flowed into the pond. Explain how these could be the cause of fish mortality. (CBSE Delhi 2019 C)
Answer:

• Pollutants from man’s activities cause eutrophication in ponds and rivers.
• The prime contaminants are nitrates and phosphate which act as plant nutrients.
• They overstimulate the growth of algae causing the formation of a thick layer of scum on the surface and an unpleasant odor.
• The dissolved oxygen in the water decreases. It is essential for all forms of aquatic life.
• At the same time, other pollutants flowing from farms into pond poison the whole population of fish.
• As the decomposition of dead fish occurs, the dissolved oxygen content of water further decreases.
• Ultimately pond chokes to death.

Question 30.
A young sperm whale, 33-foot long was found dead off the coast. It had a large amount of human trash like trash bags, polypropylene sacks, ropes, net segments, etc. amounting to 29 kilograms in its digestive system. The whale died because of inflammation of the abdominal lining. Analyze the possible reasons for such mishaps and suggest measures that can be taken to reduce such incidents. (CBSE Sample Paper 2018-19)
Answer:

1. We are increasing the use of non-biodegradable products. In most of the products we buy, there is at least one layer of plastic in it.
2. We have started packaging even our daily use products like milk and water in polybags.
3. In cities, fruits and vegetables are packed in polystyrene and plastic packaging and we contribute heavily to environmental pollution.
4. Most of the wastes are dumped in water bodies without segregation and treatment. One has to reduce the generation of non-biodegradable wastes and stop the irresponsible dumping of wastes into water bodies.