Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 10 Electrostatics Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 10 Electrostatics

Question 1.
Explain: Atoms are electrically neutral.
Answer:

  1. Matter is made up of atoms which in turn consists of elementary particles proton, neutron and electron.
  2. A proton is considered to be positively charged and electron to be negatively charged.
  3. Neutron is electrically neutral i.e., it has no charge.
  4. An atomic nucleus is made up of protons and neutrons and hence is positively charged.
  5. Negatively charged electrons surround the nucleus so as to make an atom electrically neutral.

Question 2.
What does the below diagrams show?
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 1
Answer:

  1. Figure (a) shows insulated conductor.
  2. Figure (b) shows that positive charge is neutralized by electron from Earth.
  3. Figure (c) shows that earthing is removed, negative charge still stays in conductor due to positive charged rod.
  4. Figure (d) shows that when rod is removed, negative charge is distributed over the surface of the conductor.

Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics

Question 3.
Explain concept of charging by conduction.
Answer:

  1. When certain dissimilar substances, like fur and amber or comb and dry hair, are rubbed against each other, electrons get transferred to the other substance making them charged.
  2. The substance receiving electrons develops a negative charge while the other is left with an equal amount of positive charge.
  3. This can be called charging by conduction as charges are transferred from one body to another.

Question 4.
Explain concept of charging by induction.
Answer:

  1. If an uncharged conductor is brought near a charged body, (not in physical contact) the nearer side of the conductor develops opposite charge to that on the charged body and the far side of the conductor develops charge similar to that on the charged body. This is called induction.
  2. This happens because the electrons in a conductor are free and can move easily in presence of charged body.
  3. A charged body attracts or repels electrons in a conductor depending on whether the charge on the body is positive or negative respectively.
  4. Positive and negative charges are redistributed and are accumulated at the ends of the conductor near and away from the charged body.
  5. In induction, there is no transfer of charges between the charged body and the conductor. So when the charged body is moved away from the conductor, the charges in the conductor are free again.

Question 5.
Explain the concept of additive nature of charge.
Answer:

  1. Electric charge is additive, similar to mass. The total electric charge on an object is equal to the algebraic sum of all the electric charges distributed on different parts of the object.
  2. It may be pointed out that while taking the algebraic sum, the sign (positive or negative) of the electric charges must be taken into account.
  3. Thus, if two bodies have equal and opposite charges, the net charge on the system of the two bodies is zero.
  4. This is similar to that in case of atoms where the nucleus is positively charged and this charge is equal to the negative charge of the electrons making the atoms electrically neutral.

Question 6.
State the analogy between the additive property of charge with that of mass.
Answer:

  1. The masses of the particles constituting an object are always positive, whereas the charges distributed on different parts of the object may be positive or negative.
  2. The total mass of an object is always positive whereas, the total charge on the object may be positive, zero or negative.

Question 7.
What is quantization of charge?
Answer:

  1. Protons (+ve) and electrons (-ve) are the charged particles constituting matter, hence the charge on an object must be an integral multiple of ± e i.e., q = ± ne, where n is an integer.
  2. Charge on an object can be increased or decreased in multiples of e.
  3. It is because, during the charging process an integral number of electrons can be transferred from one body to the other body. This is known as quantization of charge or discrete nature of charge.

Question 8.
Explain with an example why quantization of charge is not observed practically.
Answer:
i. The magnitude of the elementary electric charge (e), is extremely small. Due to this, the number of elementary charges involved in charging an object becomes extremely large.

ii. For example, when a glass rod is rubbed with silk, a charge of the order of one µC (10-6 C) appears on the glass rod or silk. Since elementary charge e = 1.6 × 10-19 C. the number of elementary charges on the glass rod (or silk) is given by
n = \(\frac {10^{-6}C}{1.6×10^{-19}C}\) = 6.25 × 1012
Since, it is tremendously large number, the quantization of charge is not observed and one usually observes a continuous variation of charge.

Question 9.
The total charge of an isolated system is always conserved. Explain with an example.
Answer:

  1. When a glass rod is rubbed with silk, it becomes positively charged and silk becomes negatively charged.
  2. The amount of positive charge on glass rod is found to be exactly the same as negative charge on silk.
  3. Thus, the systems of glass rod and silk together possesses zero net charge after rubbing.
    Hence, the total charge of an isolated system is always conserved.

Question 10.
Explain the conclusion when charges are brought close to each other.
Answer:

  1. Unlike charges attract each other.
  2. Like charges repel each other.

Question 11.
How much positive and negative charge is present in 1 g of water? How many electrons are present in it?
(Given: molecular mass of water is 18.0 g)
Answer:
Molecular mass of water is 18 gram, that means the number of molecules in 18 gram of water is 6.02 × 1023
∴ Number of molecules in lgm of water = \(\frac {6.02×10^{23}}{18}\)
One molecule of water (H2O) contains two hydrogen atoms and one oxygen atom. Thus, the number of electrons in ILO is sum of the number of electrons in H2 and oxygen. There are 2 electrons in H2 and 8 electrons in oxygen.
∴ Number of electrons in H2O = 2 + 8 = 10
Total number of protons / electrons in one gram of water
= \(\frac {6.02×10^{23}}{18}\) × 10 = 3.344 × 1023
Total positive charge
= 3.344 × 1023 × charge on a proton
= 3.344 × 1023 × 1.6 × 10-19C
= 5.35 × 104 C
This positive charge is balanced by equal amount of negative charge so that the water molecule is electrically neutral.
∴ Total negative charge = 5.35 × 104C

Question 12.
Define point charge. Which law explains the interaction between charges at rest?
Answer:

  1. A point charge is a charge whose dimensions are negligibly small compared to its distance from another bodies.
  2. Coulomb’s law explains the interaction between charges at rest.

Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics

Question 13.
State and explain Coulomb’s law of electric charge in scalar form.
Answer:
Coulomb’s law:
The force of attraction or repulsion between two point charges at rest is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them. This force acts along the line joining the two charges.
Explanation:
i. Let q1 and q2 be the two point charges at rest with each other and separated by a distance r. F is the magnitude of electrostatic force of attraction or repulsion between them.

ii. According to Coulomb’s law.
F ∝ \(\frac {q_1q_2}{r^2}\)
∴ F = K\(\frac {q_1q_2}{r^2}\)
where, K is the constant of proportionality which depends upon the units of F, q1, q2, r and medium in which charges are placed.

Question 14.
State conditions for electrostatic force to be attractive or repulsive.
Answer:

  1. The force between the two charges will be attractive, if the charges are unlike (one positive and one negative).
  2. The force between the two charges will be repulsive, if the charges are similar (both positive or both negative).

Question 15.
Prove that relative permittivity is the ratio of the force between two point charges placed a certain distance apart in free space or vacuum to the force between the same two point charges when placed at the same distance in the given medium.
Answer:
i. The force between the two charges placed in a medium is given by,
Fmed = \(\frac {1}{4πε}\) (\(\frac {q_1q_2}{r^2}\)) …………. (1)
where, ε is called the absolute permittivity of the medium.

ii. The force between the same two charges placed in free space or vacuum at distance r is given by,
Fvac = \(\frac {1}{4πε_0}\) (\(\frac {q_1q_2}{r^2}\)) …………. (1)
Dividing equation (2) by equation (1),
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 2
Hence, relative permittivity is the ratio of the force between two point charges placed a certain distance apart in free space or vacuum to the force between the same two point charges when placed at the same distance in the given medium.

Question 16.
If relative permittivity of water is 80 then derive the relation between Fwater and Fvacuum. What can be concluded from it?
Answer:
i. The force between two point charges q1 and q2 placed at a distance r in a medium of relative permittivity εr, is given by
Fmed = \(\frac{1}{4 \pi \varepsilon_{0} \varepsilon_{r}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}\) …………. (1)
If the medium is vacuum,
Fvac = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}\) …………. (2)

ii. Dividing equation (2) by equation (1),
\(\frac{\mathrm{F}_{\mathrm{vac}}}{\mathrm{F}_{\text {med }}}\) = εr
For water, εr = 80 ……….. (given)
∴ Fwater = \(\frac {F_{vac}}{80}\)

iii. This means that when two point charges are placed some distance apart in water, the force between them is reduced to (\(\frac {1}{80}\))th of the force between the same two charges placed at the same distance in vacuum.

iv. Thus, it is concluded that a material medium reduces the force between charges by a factor of er, its relative permittivity.

Question 17.
Give conversions of micro-coulomb, nano-coulomb and pico-coulomb to coulomb.
Answer:
1 microcoulomb (µC) = 10-6 C
1 nanocoulomb (nC) = 10-9 C
1 picocoulomb (pC) = 10-12 C

Question 18.
Explain Coulomb’s law in vector form.
Answer:
i. Let q1 and q2 be the two similar point charges situated at points A and B and let \(\vec{r}\)12 be the distance of separation between them.

ii. The force \(\vec{F}\)21 exerted on q2 by q1 is given by,
\(\overrightarrow{\mathrm{F}}_{21}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\left|\mathrm{r}_{12}\right|^{2}} \times \hat{\mathrm{r}}_{12}\)
where, \(\hat{r}\)12 is the unit vector from A to B.
\(\vec{F}\)21 acts on q2 at B and is directed along BA, away from B.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 3

iii. Similarly, the force \(\vec{F}\)12 exerted on q1 by q2 is given by, \(\vec{F}\)12 = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\left|\mathrm{r}_{12}\right|^{2}} \times \hat{\mathrm{r}_{21}}\)
where, \(\hat{r}\)21 is the unit vector from B to A. \(\vec{F}\)12 acts on q1 at A and is directed along BA, away from A.

iv. The unit vectors \(\hat{r}\)12 and \(\hat{r}\)21 are oppositely directed i.e., \(\hat{r}\)12 = –\(\hat{r}\)21
Hence, \(\vec{F}\)21 = –\(\vec{F}\)12
Thus, the two charges experience force of equal magnitude and opposite in direction.

v. These two forces form an action-reaction pair.

vi. As \(\vec{F}\)21 and \(\vec{F}\)12 act along the line joining the two charges, the electrostatic force is a central force.

Question 19.
State similarities and differences of gravitational and electrostatic forces.
Answer:
i. Similarities:
a. Both forces obey inverse square law:
F ∝ \(\frac{1}{r^2}\)
b. Both are central forces and they act along the line joining the two objects.

ii. Differences:
a. Gravitational force between two objects is always attractive while electrostatic force between two charges can be either attractive or repulsive depending on the nature of charges.
b. Gravitational force is about 36 orders of magnitude weaker than the electrostatic force.

Question 20.
Charge on an electron is 1.6 × 10-19 C. How many electrons are required to accumulate a charge of one coulomb?
Answer:
1 electron = 1.6 × 10-19 C
∴ 1 C = \(\frac{1}{1.6 \times 10^{-19}}\) electrons
= 0.625 × 1019 electrons
……. (Taking reciprocal from log table)
= 6.25 × 1018 electrons
Hence, 6.25 × 1018 electrons are required to accumulate a charge of one coulomb.

Question 21.
What is the force between two small charge spheres having charges of 2 × 10-7 C and 3 × 10-7 C placed 30 cm apart in air?
Answer:
Given: q1 = 2 × 10-7 C, q2 = 3 × 10-7 C
r = 30 cm = \(\frac {30}{100}\) m = 0.3 m
To find: Force (F)
Formula: F = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
Calculation: From formula,
F = \(\frac{9 \times 10^{9} \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{(0.3)^{2}}\)
∴ F = 6 × 10-3 N

Question 22.
The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge -0.8 µC in air is 0.2 N. (i) What is the distance between the two spheres? (ii) What is the force on the second sphere due to the first?
Answer:
i. Given: q1 = 0.4 µC = 0.4 × 10-6 C,
q2 = -0.8 µC = -0.8 × 10-6 C, F = 0.2 N
To find: i. Distance (r)
ii. Force on second sphere (F)
Formula: F = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
Calculation:
i. From formula,
r² = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{F}\)
r² = \(\frac{9 \times 10^{9} \times 0.4 \times 10^{-6} \times 0.8 \times 10^{-6}}{0.2}\)
= 0.0144
∴ r = \(\sqrt{0.0144}\) = 0.12 m
∴ r = 12 cm

ii. The force on the second sphere due to the first is also 0.2 N and is attractive in nature.

Question 23.
i. Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion, if the charge on each is 6.5 × 10-7 C? The radii of A and B are negligible compared to the distance of separation,
ii. What is the force of repulsion if each sphere is charged double the above amount and the distance between them is halved?
Answer:
Given: q1 = 6.5 × 10-7 C q2 = 6.5 × 10-7 C
r = 50 cm = 0.50 m
To find: Force of repulsion (F)
Formula: F = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
Calculation:
From formula,
F = \(\frac{9 \times 10^{9} \times 6.5 \times 10^{-7} \times 6.5 \times 10^{-7}}{(0.50)^{2}}\)
F = 1.52 × 10-2 N

ii. When each charge is doubled and the distance between them is reduced to half, then
F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(2 q_{1}\right)\left(2 q_{2}\right)}{(r / 2)^{2}}\)
= 16 × \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\) = 16 × 1.52 × 10-2
∴ F = 0.24 N

Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics

Question 24.
Calculate and compare the electrostatic and gravitational forces between two protons which are 10-15 m apart. Value of G = 6.674 × 10-11 m³ kg-1 s-2 and mass of the porton is 1.67 × 10-27 kg.
Answer:
Given: G = 6.674 × 10-11 m³ kg-1 s-2
mp = 1.67 × 10-27 kg.
qp = 1.67 × 10-19 C, r = 10-15
To find:
i. Electrostatic Force (FE)
ii. Gravitational Force (FG)
Formula: i. FE = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
ii. FE = \(\frac {Gm_1m_2}{r^2}\)
Calculation:
From formula (i),
\(\mathrm{F}_{\mathrm{E}}=9 \times 10^{9} \times \frac{1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{\left(10^{-15}\right)^{2}}\)
= 9 × 1.6 × 1.6 × 10
= 90 × 1.6 × 1.6
= antilog [log 90 + log 1.6 + log 1.6]
= antilog [1.9542 + 0.2041 + 0.2041]
= antilog [2.3624]
= 2.303 × 10² N
From formula (ii),
\(\mathrm{F}_{\mathrm{G}}=6.674 \times 10^{-11} \times \frac{1.67 \times 10^{-27} \times 1.67 \times 10^{-27}}{\left(10^{-15}\right)^{2}}\)
= 6.674 × 1.67 × 1.67 × 10-35
= {antilog [log 6.674 + log 1.67 + log 1.67]} × 10-35
= {antilog [0.8244 + 0.2227 + 0.2227]} × 10-35
= {antilog [1.2698]} × 10-35
= 1.861 × 101 × 10-35
= 1.861 × 10-34 N
Now,
\(\frac{\mathrm{F}_{\mathrm{E}}}{\mathrm{F}_{\mathrm{G}}}=\frac{2.303 \times 10^{2}}{1.861 \times 10^{-34}}\)
= {antilog [log 2.303 – log 1.861]} × 1036
= {antilog [0.3623 – 0.2697]} × 1036
= {antilog [0.0926]}
= 1.238 × 1036
∴ FE ≈ 1036 × FG

Question 25.
State and explain principle of superposition.
Answer:
Statement: When a number of charges are interacting, the resultant force on a particular charge is given by the vector sum of the forces exerted by individual charges.
Explanation:
i. Consider a number of point charges q1, q2, q3 ……………… kept at points A1, A2, A3 ………….. as shown in figure.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 4

ii. The force exerted on the charge q1 by q2 is \(\vec{F}\)12 The value of \(\vec{F}\)12 is calculated by ignoring the presence of other charges. Similarly, force \(\vec{F}\)13, \(\vec{F}\)14 can be found, using the Coulomb’s law.

iii. Total force \(\vec{F}\)1 on charge qi is the vector sum of all such forces.
\(\vec{F}\)1 = \(\vec{F}\)12 + \(\vec{F}\)13 + \(\vec{F}\)14 + …………..
\( =\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\left|\mathrm{r}_{21}\right|^{2}} \times \hat{\mathrm{r}}_{21}+\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{\left|\mathrm{r}_{31}\right|^{2}} \times \hat{\mathrm{r}}_{31}+\ldots .\right]\)

where \(\hat{r}\)21, \(\hat{r}\)31 are unit vectors directed to q1 from q2, q3 respectively and r21, r31, r41 are the distances from q1 to q2, q3 respectively.

iv. If q1, q2, q3 ……., qn are the point charges then the force \(\vec{F}\) exerted by these charges on a test charge q0 is given by,
\(\vec{F}\)test = \(\vec{F}\)1 = \(\vec{F}\)2 + \(\vec{F}\)3 + …. + \(\vec{F}\)n
= \(\sum_{\mathrm{n}=1}^{\mathrm{n}} \mathrm{F}_{\mathrm{n}}=\frac{1}{4 \pi \varepsilon_{0}} \sum_{\mathrm{n}=1}^{\mathrm{n}} \frac{\mathrm{q}_{0} \mathrm{q}_{\mathrm{n}}}{\mathrm{r}_{\mathrm{n}}^{2}} \hat{\mathrm{r}}_{\mathrm{n}}\)
Where, \(\hat{r}\)n, is a unit vector directed from the nth charge to the test charge q0 and r2 is the
separation between them, \(\vec{r}\)n = rn \(\hat{r}\)n

Question 26.
Three charges of 2 µC, 3 µC and 4 µC are placed at points A, B and C respectively, as shown in the figure. Determine the force on A due to other charges.
(Given: AB = 4 cm, BC = 3 cm)
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 5
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 6
Using pythagoras theorem
AC = \(\sqrt {AB^2+BC^2}\)
= \(\sqrt {4^2+3^2}\)
AC = 5 cm
Magnitude of force \(\vec{F}\)AB on A due to B is,
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 7
In ∆ABC 4
cos θ = \(\frac {4}{5}\)
θ = cos-1 (\(\frac {4}{5}\)) = 36.87°
Forces acting points A are
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 8
= 59.36 N
Direction of resultant force is 36.87° (north of west)
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 9
[Note: The question given above is modified considering minimum requirement of data needed to solve the problem.]

Question 27.
There are three charges of magnitude 3 pC, 2 pC and 3 pC located at three corners A, B and C of a square ABCD of each side measuring 2 m. Determine the net force on 2 pC charge.
Answer:
Given: q1 = 3 µC, q2 = 2 µC, q3 = 3 µC, r = 2 m
To find: Net force on q2 (R)
Formula: F = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
Calculation:
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 10
From the formula,
Force on q2 because of q1
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 11
Net force on q2 is the resultant force of \(\vec{F}\)21 and \(\vec{F}\)23 which is given by,
R = \(\sqrt{\mathrm{F}_{21}^{2}+\mathrm{F}_{23}^{2}}\)
= \(\sqrt{\left(1.35 \times 10^{-2}\right)^{2}+\left(1.35 \times 10^{-2}\right)^{2}}\)
∴ R = 1.91 × 10-2 N

Question 28.
Explain the concept of electric field.
Answer:

  1. The space around a charge gets modified when a test charge is brought in that region, it experiences a coulomb force. The region around a charged object in which coulomb force is experienced by another charge is called electric field.
  2. Mathematically, electric field is defined as the force experienced per unit charge.
  3. The coulomb force acts across an empty space (vacuum) and does not need any intervening medium for its transmission.
  4. The electric field exists around a charge irrespective of the presence of other charges.
  5. Since the coulomb force is a vector, the electric field of a charge is also a vector and is directed along the direction of the coulomb force, experienced by a test charge.

Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics

Question 29.
Define electric field. State its SI unit and dimensions.
Answer:

  1. Electric field is the force experienced by a test charge in presence of the given charge at the given distance from it.
    \(\vec{E}\) = \(\lim _{q \rightarrow 0} \frac{\vec{F}}{q}\)
  2. SI unit: newton per coulomb (N/C) or volt per metre (V/m).
  3. Dimensions: [L M T-3 A-1]

Question 30.
Establish relation between electric field intensity and electrostatic force.
Answer:
i. Let Q and q be two charges separated by a distance r.
The coulomb force between them is given by \(\vec{F}\) = \(\frac {1}{4πε_0}\) \(\frac {Qq}{r^2}\) \(\hat{r}\)
where, \(\hat{r}\) is the unit vector along the line joining Q to q.

ii. Therefore, electric field due to charge Q is given \(\vec{F}\) = \(\frac{\vec{F}}{\mathrm{q}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}}{\mathrm{r}^{2}} \hat{\mathrm{r}}\)

iii. Electric field at a point is useful to estimate the force experienced by a charge at that point.

Question 31.
State an expression for electric field on the surface of the sphere due to a positive point charge placed at its centre.
Answer:
The magnitude of electric field at a distance r from a point charge Q is same at all points on the surface of a sphere of radius r as shown in figure.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 12
ii. Magnitude of electric field is given by,
E = \(\frac {1}{4πε_0}\) \(\frac {Q}{r^2}\)
iii. Its direction is along the radius of the sphere, pointing away from its centre if the charge is positive.

Question 32.
Derive expression for electric field intensity due to a point charge in a material medium.
Answer:
i. Consider a point charge q placed at point O in a medium of dielectric constant K as shown in figure.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 13

ii. Consider the point P in the electric field of point charge at distance r from q. A test charge q0 placed at the point P will experience a force which is given by the Coulomb’s law,
\(\vec{F}\) = \(\frac{1}{4 \pi \varepsilon_{0} \mathrm{~K}} \frac{\mathrm{qq}_{0}}{\mathrm{r}^{2}} \hat{\mathrm{r}}\)
where \(\hat{r}\) is the unit vector in the direction of force i.e., along OP.

iii. By the definition of electric field intensity,
\(\vec{F}\) = \(\frac{\vec{F}}{\mathrm{q}_{0}}=\frac{1}{4 \pi \varepsilon_{0} \mathrm{~K}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}}\)
The direction of \(\vec{E}\) will be along OP when q is positive and along PO when q is negative.

iv. The magnitude of electric field intensity in a medium is given by, E = \(\frac{1}{4 \pi \varepsilon_{0} \mathrm{~K}} \frac{\mathrm{q}}{\mathrm{r}^{2}}\)

v. For air or vacuum, K = 1 then
E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}}\)

Question 33.
Show graphical representation of variation of coulomb force and electric field due to point charge with distance.
Answer:
Electrostatic force: F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}\)
Electric field: E : \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}}\)
The coulomb force (F) between two charges and electric field (E) due to a charge both follow the inverse square law.
(F ∝ 1/r², E ∝ 1/r²)
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 14

Question 34.
What is non-uniform electric field?
Answer:
A field whose magnitude and direction is not the same at all points.
For example, field due to a point charge. In this case, the magnitude of field is same at distance r from the point charge in any direction but the direction of the field is not same.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 15

Question 35.
Derive relation between electric field (E) and electric potential (V).
Answer:
i. A pair of parallel plates is connected as shown in the figure. The electric field between them is uniform
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 16

ii. A potential difference V is applied between two parallel plates separated by a distance ‘d’.

iii. The electric field between them is directed from plate A to plate B.

iv. A charge +q placed between the plates experiences a force F due to the electric field.

v. If the charge is moved against the direction of field, i.e., towards the positive plate, some amount of work is done on it.

vi. If the charge is moved +q from the negative plate B to the positive plate A, then the work done against the field is W = Fd; where ‘d’ is the separation between the plates.

vii. The potential difference V between the two plates is given by W = Vq,
but W = Fd
∴ Vq = Fd
∴ \(\frac {F}{q}\) = \(\frac {V}{d}\) = E
∴ Electric field can be defined as E = V/d.

Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics

Question 36.
What are electric lines of force?
Answer:
i. An electric line of force is an imaginary curve drawn in such a way that the tangent at any given point on this curve gives the direction of the electric field at that point.

ii. If a test charge is placed in an electric field it would be acted upon by a force at every point in the field and will move along a path.

iii. The path along which the unit positive charge moves is called a line of force.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 17

iv. A line of force is defined as a curve such that the tangent at any point to this curve gives the direction of the electric field at that point.

v. The density of field lines indicates the strength of electric fields at the given point in space.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 18

Question 37.
State the characteristics of electric lines of force.
Answer:

  1. The lines of force originate from a positively charged object and end on a negatively charged object.
  2. The lines of force neither intersect nor meet each other, as it will mean that electric field has two directions at a single point.
  3. The lines of force leave or terminate on a conductor normally.
  4. The lines of force do not pass through conductor i.e., electric field inside a conductor is always zero, but they pass through insulators.
  5. Magnitude of the electric field intensity is proportional to the number of lines of force per unit area of the surface held perpendicular to the field.
  6. Electric lines of force are crowded in a region where electric intensity is large.
  7. Electric lines of force are widely separated from each other in a region where electric intensity is small
  8. The lines of force of an uniform electric field are parallel to each other and are equally spaced.

Question 38.
Find the distance from a charge of 4 µC placed in air which produces electric field of intensity 9 × 10³ N/C.
Answer:
Given: K = 1, E = 9 × 10³ N/C
q = 4 µC = 4 × 10-6
To Find: Distance (r)
Formula: E = \(\frac{1}{4 \pi \varepsilon_{0} K} \frac{q}{r^{2}}\)
Calculation from formula
9 × 10³ = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{4 \times 10^{-6}}{r^{2}}\)
∴ 9 × 10³ = 9 × 109 \(\frac{4 \times 10^{-6}}{\mathrm{r}^{2}}\)
∴ r² = 4
∴ r = 2 m

Question 39.
What is the magnitude of a point charge chosen so that the electric field 50 cm away has magnitude 2.0 N/C?
Answer:
Given: r = 50 cm – 0.5 m, E = 2 N/C,
To find: Magnitude of charge (q)
Formula: E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}\)
Calculation from formula
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 19

Question 40.
Three point charges are placed at the vertices of a right angled isosceles triangle as shown in the given figure. What is the magnitude and direction of the resultant electric field at point P which is the mid point of its hypotenuse?
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 20
Answer:
Electric field at P due to the charges at A, B and C are shown in the figure.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 21
Let \(\vec{E}\)A be the field at P due to charge at A and \(\vec{E}\)c be the field at P due to charge at C.
Since P is the midpoint of AC and the fields at A and C are equal in magnitudes and are opposite in direction, EA = – EC .
i.e., \(\vec{E}\)A + \(\vec{E}\)C = 0.
Thus, the field at P is only to the charge at B and is given by,
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 22

Question 41.
A simplified model of hydrogen atom consists of an electron revolving about a proton at a distance of 5.3 × 10-11 m. The charge on a proton is +1.6 × 10-19 C. Calculate the intensity of the electric field due to proton at this distance. Also find the force between electron and proton.
Answer:
Given: r = 5.3 × 10-11 m
q = 1.6 × 10-19 C
To Find: i. Intensity of electric field (E)
ii. Force (F)
Formula: i. E = \(\frac{1}{4 \pi \varepsilon_{0}} \times \frac{\mathrm{q}}{\mathrm{r}^{2}}\)
ii. E = \(\frac {F}{q}\)
Calculation from formula (i)
E = 9 × 109 × \(\frac{1.6 \times 10^{-19}}{\left(5.3 \times 10^{-11}\right)^{2}}\)
= 5.126 × 1011 N/C
Force between electron and proton,
Force between electron and proton,
F = E × qe ….[From formula (ii)]
= 5.126 × 10-11 × -1.6 × 10-19
= -8.201 × 108 N

Question 42.
The force exerted by an electric field on a charge of +10 µC at a point is 16 × 10-4 N. What is the intensity of the electric field at the point?
Answer:
Given: q = 10 µC= 10 × 10-6 C, F = 16 × 10-4 N
To find: Electric field intensity (E)
Formula: E = \(\frac {F}{q}\)
Calculation: From formula,
E = \(\frac {16×10^{-4}}{10×10^{-6}}\) = 160 N/C

Question 43.
What is the force experienced by a test charge of 0.20 µC placed in an electric field of 3.2 × 106 N/C?
Answer:
Given: q0 = 0.20 µC = 0.2 × 106 C,
E = 3.2 × 106 N/C
To find: Force (F)
Formula: E = \(\frac {F}{q_0}\)
Calculation: From formula,
F = Eq0
∴ F = 3.2 × 106 × 0.2 × 10-6 = 0.64 N

Question 44.
Gap between two electrodes of the spark-plug used in an automobile engine is 1.25 mm. If the potential of 20 V is applied across the gap, what will be the magnitude of electric field between the electrodes?
Answer:
Given: V = 20 V
d = 1.25 mm = 1.25 × 10-3 m
To Find: Magnitude of electric field (E)
Formula: E = \(\frac {V}{d}\)
Calculation: From formula,
E = \(\frac {20}{1.25×10^{-3}}\)
= 1.6 × 104 V/m

Question 45.
If 100 joules of work must be done to move electric charge equal to 4 C from a place, where potential is -10 volt to another place where potential is V volt, find the value of V.
Answer:
Given: q0 = 4 C,
VA = -10 volt,
VB = V volt,
WAB = 100 J
To Find: Potential (V)
Formula: VB – VA = \(\frac {W_{AB}}{q_0}\)
Calculation: From formula,
V – (-10) = \(\frac {100}{4}\) = 25
∴ V + 10 = 25
∴ V = 15 volt

Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics

Question 46.
Find the work done when a point charge of 2.0 pC is moved from a point at a potential of -10 V to a point at which the potential is zero.
Answer:
VA = -10V,
VB = 0,
q = 2 × 10-6 C
To Find: Work done (W)
Formula: VBA = \(\frac {W}{q}\)
Calculation: From formula,
W = VBA × q
= (VB – VA) × q
= (0 + 10) × 2 × 10-6
= 20 × 10-6 J
∴ W = 2 × 10-5 J

Question 47.
Explain the term: Electric flux
Answer:
i. The number of lines of force per unit area is the intensity of the electric field \(\vec{E}\).
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 23

ii. When the area is inclined at an angle θ with the direction of electric field, the electric flux can be calculated as follows.
Let the angle between electric field \(\vec{E}\), and area vector \(\vec{dS}\) be θ, then the electric flux passing through are dS is given by
dø = (component of dS along \(\vec{E}\)) × (area of \(\vec{dS}\))
dø = EdS cos θ
dø = \(\vec{E}\) .\(\vec{dS}\)
Total flux through the entire surface .
ø = ∫dø = \( \int_{S} \vec{E} \cdot d \vec{S}=\vec{E} \cdot \vec{S}\)

iii. The SI unit of electric flux can be calculated using,
ø = \(\vec{E}\). \(\vec{S}\) = (V/m) m² = V m
[Note: Area vector is a vector whose magnitude is equal to area and is directed normal to its surface]

Question 48.
The electric flux through a plane surface of area 200 cm² in a region of uniform electric field 20 N/C is 0.2 N m²/C. Find the angle between electric field and normal to the surface.
Answer:
Given: ds = 200 cm² = 2 × 10-2 m², E = 20 N/C,
ø = 0.2 N m²/C
To find: Angle between electric field and normal (θ)
Formula: ø = Eds cos θ
Calculation:
From formula,
cos θ = \(\frac {ø}{Eds}\) = \(\frac {0.2}{20×2×10^{-2}}\) = \(\frac {1}{2}\)
∴ θ = cos-1 (\(\frac {1}{2}\))
∴ θ = 60°

Question 49.
A charge of 5.0 C is kept at the centre of a sphere of radius 1 m. What is the flux passing through the sphere? How will this value change if the radius of the sphere is doubled?
Answer:
Given: q = 5C, r = 1 m
To find: Flux (ø)
Formulae: i. E = \(\frac{1}{4 \pi \varepsilon_{0}} \times \frac{\mathrm{q}}{\mathrm{r}^{2}}\)
ii. ø = E × A = E (4πr²)
Calculation: From formula (i),
E = 9 × 109 × \(\frac {5}{1^2}\)
= 4.5 × 1010 N/C
From formula (ii),
ø = E × 4 π r²
= 4.5 × 1010 × 4 × 3.14 × 1²
ø = 5.65 × 1011 Vm
This value of flux will not change if radius of sphere is doubled. Though radius of sphere will increase, increased distance will reduce the electric field intensity. As E ∝ \(\frac {1}{r^2}\) and A × r² net variation in total flux will not be observed.

Question 50.
State and prove Gauss’ law of electrostatics.
Answer:
Statement:
The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by Eo.
\(\int \vec{E} \cdot \overrightarrow{\mathrm{dS}}=\frac{\mathrm{Q}}{\varepsilon_{0}}\)
where Q is the total charge within the surface.

Proof:
i. Consider a closed surface of any shape which encloses number of positive electric charges.

ii. Imagine a small charge +q present at a point O inside closed surface. Imagine an infinitesimal area dS of the given irregular closed surface.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 24

iii. The magnitude of electric field intensity at point P on dS due to charge +q at point O is, E = \(\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{\mathrm{q}}{\mathrm{r}^{2}}\right)\) ………… (1)

iv. The direction of E is away from point O. Let θ be the angle subtended by normal drawn to area dS and the direction of E

v. Electric flux passing through area (dø)
= Ecosθ dS
= \(\frac{\mathrm{q}}{4 \pi \varepsilon_{0} \mathrm{r}^{2}}\) cosθ dS ………….. (from 1)
= \(\left(\frac{\mathrm{q}}{4 \pi \varepsilon_{0}}\right)\left(\frac{\mathrm{d} \mathrm{S} \cos \theta}{\mathrm{r}^{2}}\right)\)
But, dω = \(\frac {dS cos θ}{r^2}\)
where, dco is the solid angle subtended by area dS at a point O.
∴ dø = \(\left(\frac{\mathrm{q}}{4 \pi \varepsilon_{0}}\right)\) dω …………. (2)

vi. Total electric flux crossing the given closed surface can be obtained by integrating equation (2) over the total area.
\(\phi_{\mathrm{E}}=\int_{\mathrm{s}} \mathrm{d} \phi=\int_{\mathrm{s}} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dS}}=\int \frac{\mathrm{q}}{4 \pi \varepsilon_{0}} \mathrm{~d} \omega=\frac{\mathrm{q}}{4 \pi \varepsilon_{0}} \int \mathrm{d} \omega\)

vii. But ∫dω = 4π = solid angle subtended by entire closed surface at point O.
Total Flux = \(\frac {q}{4πε_0}\) (4π)
∴ øE = \(\int_{\mathrm{s}} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dS}}=\frac{+\mathrm{q}}{\varepsilon_{0}}\)

viii. This is true for every electric charge enclosed by a given closed surface.
Total flux due to charge q1, over the given closed surface = + \(\frac {q_1}{ε_0}\)
Total flux due to charge q2, over the given closed surface = + \(\frac {q_2}{ε_0}\)
Total flux due to charge qn, over the given closed surface = +\(\frac {q_n}{ε_0}\)

ix. According to the superposition principle, the total flux c|> due to all charges enclosed within the given closed surface is
\(\phi_{\mathrm{E}}=\frac{\mathrm{q}_{1}}{\varepsilon_{0}}+\frac{\mathrm{q}_{2}}{\varepsilon_{0}}+\frac{\mathrm{q}_{3}}{\varepsilon_{0}}+\ldots+\frac{\mathrm{q}_{\mathrm{n}}}{\varepsilon_{0}}=\sum_{\mathrm{i}=1}^{\mathrm{i}=\mathrm{n}} \frac{\mathrm{q}_{\mathrm{i}}}{\varepsilon_{0}}=\frac{\mathrm{Q}}{\varepsilon_{0}}\)

Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics

Question 51.
With a help of diagram, state the direction of flux due to positive charge, negative charge and charge outside a closed surface.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 25
Positive sign indicates that the flux is directed outwards, away from the charge.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 26
If the charge is negative, the flux will be is directed inwards.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 27
If a charge is outside the closed surface, the net flux through it will be zero.

Question 52.
Explain: Electric flux is independent of shape and size of closed surface.
Answer:
i. The net flux crossing an enclosed surface is equal to \(\frac {q}{ε_0}\) where q is the net charge inside the closed surface.

ii. Consider a charge +q at the centre of concentric circles as shown in figure below.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 28
As the charge inside the sphere is unchanged, the flux passing through a sphere of any radius is the same.

iii. Thus, if the radius of the sphere is increased by a factor of 2, the flux passing through is surface remains unchanged.

iv. As shown in figure same number of lines of force cross both the surfaces.
Hence, total flux is independent of shape of the closed surface radius of the sphere and size of closed surface.

Question 53.
Define the following terms with the help of a diagram.
i. Electric dipole
ii. Dipole axis
iii. Axial line
iv. Equatorial line
Answer:
i. Electric dipole: A pair of equal and opposite charges separated by a finite distance is called an electric dipole.

ii. Dipole axis: Line joining the two charges is called the dipole axis.

iii. Axial line: A line passing through the dipole axis is called axial line.

iv. Equatorial line: A line passing through the centre of the dipole and perpendicular to the axial line is called the equatorial line.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 29
AB : Electric dipole Line joining
AB: Dipole axis
X-Y : Axial line
P-Q : Equatorial line

Question 54.
What are polar molecules? Explain with examples.
Answer:

  1. Polar molecules are the molecules in which the centre of positive charge and the negative charge is naturally separated.
  2. Molecules of water, ammonia, sulphur dioxide, sodium chloride etc. have an inherent separation of centres of positive and negative charges. Such molecules are called polar molecules.

Question 55.
What are non-polar molecules? Explain with examples.
Answer:
i. Non-polar molecules are the molecules in which the centre of positive charge and the negative charge is one and the same. They do not have a permanent electric dipole. When an external electric field is applied to such molecules, the centre of positive and negative charge are displaced and a dipole is induced.

ii. Molecules such as H2, CI2, CO2, CH4, etc., have their positive and negative charges effectively centred at the same point and are called non-polar molecules.

Question 56.
Derive expression for couple acting on an electric dipole in a uniform electric field.
Answer:
i. Consider an electric dipole placed in a uniform electric field E. The axis of electric dipole makes an angle θ with the direction of electric field.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 30

ii. The force acting on charge – q at A is \(\vec{F}\)A = -q\(\vec{E}\) in the direction of\(\vec{E}\) and the force acting on charge +q at B is \(\vec{F}\)B = + q \(\vec{E}\) in the direction opposite to \(\vec{E}\).

iii. Since \(\vec{F}\)A = –\(\vec{F}\)B, the two equal and opposite forces separated by a distance form a couple.

iv. Moment of the couple is called torque and is defined by \(\vec{τ}\) = \(\vec{d}\) × \(\vec{F}\) where, d is the perpendicular distance between the two equal and opposite forces.

v. Magnitude of Torque = Magnitude of force × Perpendicular distance
∴ Torque on the dipole (\(\vec{τ}\)) = \(\vec{BA}\) × q\(\vec{E}\)
= 2lqE sin θ
but p = q2l
∴ τ = pEsin θ
∴ In vector form \(\vec{τ}\) = \(\vec{d}\) × \(\vec{E}\)

vi. If θ = 90° sin θ = 1, then τ = pE
When the axis of electric dipole is perpendicular to uniform electric field, torque of the couple acting on the electric dipole is maximum, i.e., τ = pE.

vii. If θ = 0 then τ = 0, this is the minimum torque on the dipole. Torque tends to align its axis along the direction of electric field.

Question 57.
Derive expression for electric intensity at a point on the axis of an electric dipole.
Answer:
i. Consider an electric dipole consisting of two charges -q and +q separated by a distance 2l.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 31

ii. Let P be a point at a distance r from the centre C of the dipole.

iii. The electric intensity \(\vec{E}\)a at P due to the dipole is the vector sum of the field due to the charge -q at A and +q at B.

iv. Electric field intensity at P due to the charge -q at A = \(\vec{E}\)A = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{(-q)}{(r+l)^{2}} \hat{\mathrm{u}}_{\mathrm{pD}}\),
where, \(\hat{u}\)PD is unit vector directed along \(\vec{PD}\)

v. Electric intensity at P due to charge +q at B
\(\vec{E}\)B = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{(\mathrm{r}-l)^{2}} \hat{\mathrm{u}}_{\mathrm{PQ}}\)
where, \(\hat{u}\)PQ is a unit vector directed along \(\vec{PQ}\)
The magnitude of \(\vec{E}\)B is greater than that of \(\vec{E}\)A since BP < AP

vi. Resultant field \(\vec{E}\)a at P on the axis, due to the dipole is
\(\vec{E}\)a = \(\vec{E}\)B + E\(\vec{E}\)A

vii. The magnitude of \(\vec{E}\)a is given by
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 32

ix. |\(\vec{E}\)a| is directed along PQ, which is the direction of the dipole moment \(\vec{p}\) i.e., from the negative to the positive charge, parallel to the axis.

x. If r >> l, l² can be neglected compared to r²,
|\(\vec{E}\)a| = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{2 p}{r^{3}}\)

The field will be along the direction of the dipole moment \(\vec{p}\).

Question 58.
Drive expression for electric intensity at a point on the equator of an electric dipole.
Answer:
i. Electric field at point P due to charge -q at A is \(\vec{E}\)A = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{(-\mathrm{q})}{(\mathrm{AP})^{2}} \hat{\mathrm{u}}_{\mathrm{PA}}\)
where, \(\hat{u}\)PA is a unit vector directed along \(\vec{PA}\)
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 33

ii. Similarly, electric field at P due to charge +q at B is
\(\vec{E}\)A = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{(\mathrm{BP})^{2}} \hat{\mathrm{u}}_{\mathrm{BP}}\)
where \(\hat{u}\)BP is a unit vector directed along \(\vec{BP}\)

iii. Electric field at P is the sum of EA and EB
∴ \(\vec{E}\)eq = \(\vec{E}\)A + \(\vec{E}\)B

iv. Consider ∆ACP
(AP)² = (PC)² + (AC)² = r² + l² = (BP)²
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 34

v. The resultant of fields \(\vec{E}\)A and \(\vec{E}\)B acting at point P can be calculated by resolving these vectors E\(\vec{E}\)A and E\(\vec{E}\)B along the equatorial line and along a direction perpendicular to it.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 35

vi. Let the Y-axis coincide with the equator of the dipole X-axis will be parallel to dipole axis and the origin is at point P as shown.

vii. The Y-components of EA and EB are EAsin θ and EB sin θ respectively. They are equal in magnitude but opposite in direction and cancel each other. There is no contribution from them towards the resultant.

viii. The X-components of EA and EB are EAcos θ and EBcos θ respectively. They are of equal magnitude and are in the same direction.
∴ |\(\vec{E}\)eq| = EA cos θ + EB cos θ From equation (3),
|\(\vec{E}\)eq| = 2EA cos θ
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 36

x. The direction of this field is along –\(\vec{P}\) (anti-parallel to \(\vec{P}\)).
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 37

Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics

Question 59.
An electric dipole of length 2.0 cm is placed with its axis making an angle of 30° with a uniform electric field of 105 N/C as shown in figure. If it experiences a torque of 10√3 N m, calculate the magnitude of charge on dipole.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 38
Answer:
Given: 2l = 2 cm = 2 × 102 m
E = 105 N/C, τ = 10√3 Nm, θ = 30°
To find: Charge (q)
Formula: τ = q E 2 l sin θ
Calculation: From Formula.
q = \(\frac{τ}{\mathrm{E} \times 2 l \times \sin \theta}\)
= \(\frac{10 \sqrt{3}}{10^{5} \times 2 \times 10^{-2} \times \sin 30^{\circ}}\)
= 1.732 × 10-2 C

Question 60.
Explain the concept of continuous charge distribution.
Answer:
i. A system of charges can be considered as a continuous charge distribution, if the charges are located very close together, compared to their distances from the point where the intensity of electric field is to be found out.

ii. Thus, the charge distribution is said to be continuous for a system of closely spaced charges. It is treated equivalent to a total charge which is continuously distributed along a line or a surface or a volume.

Question 61.
Explain linear charge density.
Answer:
Consider charge q uniformly distributed along a linear conductor of length l, then the linear charge density (λ) is given as,
λ = \(\frac {q}{l}\)
For example, charge distributed uniformly on a straight thin rod or a thin nylon thread. If the charge is not distributed uniformly over the length of thin conductor then charge dq on small element of length dl can be written as dq = λ dl.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 39

Question 62.
Explain surface charge density.
Answer:
i. Consider a charge q uniformly distributed over a surface of area A then the surface charge density c is given as
σ = \(\frac {q}{A}\)
For example, charge distributed uniformly on a thin disc or a synthetic cloth. If the charge is not distributed uniformly over the surface of a conductor, then charge dq on small area element dA can be written as dq = σ dA.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 40

ii. SI unit of σ is (C / m²)

Question 63.
Explain volume charge density.
Answer:
i. Consider a charge q uniformly distributed throughout a volume V, then the volume charge density ρ is given as
ρ = \(\frac {q}{V}\)
For example, charge on a plastic sphere or a plastic cube. If the charge is not distributed uniformly over the volume of a material, then charge dq over small volume element dV can be written as dq = ρ dV.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 41

ii. S.I. unit of p is (C/m³)
[Note: Electric field due to a continuous charge distribution can be calculated by adding electric fields due to all these small charges.]

Question 64.
Explain the concept of static charge.
Answer:

  1. Static charges can be created whenever there is a friction between an insulator and other object.
  2. For example, when an insulator like rubber or ebonite is rubbed against a cloth, the friction between them causes electrons to be transferred from one to the other.
  3. This property of insulators is used in many applications such as photocopier, inkjet printer, painting metal panels, electrostatic precipitation/separators etc.

Question 65.
Explain the disadvantage of static charge.
Answer:

  1. When charge transferred from one body to other is very large, sparking can take place. For example, lightning in sky.
  2. Sparking can be dangerous while refuelling your vehicle.
  3. One can get static shock, if charge transferred is large.
  4. Dust or dirt particles gathered on computer or TV screens can catch static charges and can be troublesome.

Question 66.
State the precautions against static charge.
Answer:

  1. Home appliances should be grounded.
  2. Avoid using rubber soled footwear.
  3. Keep your surroundings humid (dry air can retain static charges).

Question 67.
Two charged particles having charge 3 × 10-8 C each are joined by an insulating string of length 2 m. Find the tension in the string when the system is kept on a smooth horizontal table.
Answer:
Tension (T) in the string is the force of repulsion (F) between the two charges.
According to Coulomb’s law,
F = \(\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{4 \pi \varepsilon_{0} \mathrm{r}^{2}}\)
= \(\frac{9 \times 10^{9} \times 3 \times 10^{-8} \times 3 \times 10^{-8}}{2^{2}}\)
F = 2.025 × 10-6 N
Hence, tension in the string is 2.025 × 10-6 N.

Question 68.
A free pith ball of mass 5 gram carries a positive charge of 0.6 × 10-7 C. What is the nature and magnitude of charge that should be given to second ball fixed 6 cm vertically below the former pith ball so that the upper pith bath is stationary?
Answer:
Let +q2 be the charge on lower pith ball.
Now, the upper pith ball become stationary only when its weight acting downward is balanced by the upward force of repulsion between two pith balls,
i.e., FE = mg
∴ \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}\) = mg
∴ \(\frac{9 \times 10^{9} \times 0.6 \times 10^{-7} \times \mathrm{q}_{2}}{\left(6 \times 10^{-2}\right)^{2}}\) = 5 × 10-3 × 9.8
∴ q2 = 3.27 × 10-7C
Hence, the second pith ball carries a positive charge of 3.27 × 10-7C.

Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics

Question 69.
A water drop of mass 11.0 mg and having a charge of 1.6 × 10-6 C stays suspended in a room. What will be the magnitude and direction of electric Held in the room?
Answer:
As the drop is suspended,
Force (F) due to electric field balances the weight of the drop.
∴ F = mg ………….. (1)
Here, m = 11.0 mg
= 11 × 10-6 kg,
q = 1.6 × 10-6 C
Electric field is given by,
E = \(\frac {F}{q}\)
= \(\frac {mg}{q}\)
= \(\frac {11×10^{-6}×9.8}{1.6×10^{-6}}\)
E = 67.4 N/C
As upward force balances the weight, hence direction of electric field must be vertically upwards.

Question 70.
A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating handle is brought close to A such that the distance between their centres is 10 cm, as shown in figure (a). The resulting repulsion of A is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen.) Spheres A and B are touched by uncharged spheres C and D respectively, as shown in figure (b). C and D are then removed and B is brought closer to A to a distance of 5.0 cm between their centres, as shown in figure (c). What is the expected repulsion of A on the basis of Coulomb’s law? Spheres A and C and spheres B and D have identical sizes. Ignore the sizes of A and B comparison to the separation between their centres.
Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics 42
Answer:
Let the original charge on sphere A be q and that on B be q’. At a distance r between their centres, the magnitude of the electrostatic force on each is given by
F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{qq}^{\prime}}{\mathrm{r}^{2}}\)

Neglecting the sizes of spheres, A and B in comparison to r. When an identical but uncharged sphere C touches A, the charges redistribute on A and C and, by symmetry, each sphere carries a charge q/2. Similarly, after D touches B, the redistributed charge on each is q’/2. Now, if the separation between A and B is halved, the magnitude of the electrostatic force on each is
F’ = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{(\mathrm{q} / 2)\left(\mathrm{q}^{\prime} / 2\right)}{(\mathrm{r} / 2)^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(\mathrm{qq}^{\prime}\right)}{\mathrm{r}^{2}}\) = F

Thus, the electrostatic force on A, due to B, remains unaltered.

Multiple Choice Questions

Question 1.
Force between two charges separated by a certain distance in air is F. If each charge is doubled and the distance between them is also doubled, force would be
(A) F
(B) 2 F
(O’ 4 F
(D) F/4
Answer:
(A) F

Question 2.
For what order of distance is Coulomb7 s law true?
(A) For all distances.
(B) Distances greater than 10-13 m.
(C) Distances less than 10-13 m.
(D) Distance equal to 10-13 m.
Answer:
(B) Distances greater than 10-13 m.

Question 3.
The permittivity of medium is 26.55 × 10-12 C²/Nm². The dielectric constant of the medium will be
(A) 2
(B) 3
(C) 4
(D) 5
Answer:
(B) 3

Question 4.
A glass rod when rubbed with a piece of fur acquires a charge of magnitude 3.2 µC. The number of electrons transferred is
(A) 2 × 10-13 from fur to glass
(B) 5 × 1012 from glass to fur
(C) 2 × 1013 from glass to fur
(D) 5 × 1012 from fur to glass
Answer:
(A) 2 × 10-13 from fur to glass

Question 5.
Choose the correct answer.
(A) Total charge present in the universe is constant.
(B) Total positive charge present in the universe is constant.
(C) Total negative charge present in the universe is constant.
(D) Total number of charged particles present in the universe is constant.
Answer:
(A) Total charge present in the universe is constant.

Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics

Question 6.
If a charge is moved against the Coulomb force of an electric field,
(A) work is done by the electric field
(B) energy is used from some outside source
(C) the strength of the field is decreased
(D) the energy of the system is decreased
Answer:
(B) energy is used from some outside source

Question 7.
Two point charges +4 µC and +2 µC repel each other with a force of 8 N. If a charge of -4 µC is added to each of these charges, the force would be
(A) zero
(B) 8 N
(C) 4 N
(D) 12 N
Answer:
(A) zero

Question 8.
The electric field intensity at a point 2 m from an isolated point charge is 500 N/C. The electric potential at the point is
(A) 0 V
(B) 2.5 V
(C) 250 V
(D) 1000 V
Answer:
(D) 1000 V

Question 9.
The dimensional formula of electric field intensity is
(A) [M1E1T-2A-1]
(B) [M1L1T-3A-1]
(C) [M-1L2T-3A-1]
(D) [M1L2T-3A-2]
Answer:
(B) [M1L1T-3A-1]

Question 10.
A force of 2.25 N acts on a charge of 15 × 10-4C. Calculate the intensity of electric field at that point.
(A) 1500 NC-1
(B) 150 NC-1
(C) 15000NC-1
(D) 2500 NC-1
Answer:
(A) 1500 NC-1

Question 11.
A point charge q produces an electric field of magnitude 2 N C-1 at a point distant 0.25 m from it. What is the value of charge?
(A) 1.39 × 10-11 C
(B) 1.39 × 1011 C
(C) 13.9 × 10-11 C
(D) 13.9 × 1011 C
Answer:
(A) 1.39 × 10-11 C

Question 12.
The electric intensity in air at a point 20 cm from a point charge Q coulombs is 4.5 × 105 N/ C. The magnitude of Q is
(A) 20 µC
(B) 200 µC
(C) 10 µC
(D) 2 µC
Answer:
(D) 2 µC

Question 13.
The charge on the electron is 1.6 × 10-19 C. The number of electrons need to be removed from a metal sphere of 0.05 m radius so as to acquire a charge of 4 × 10-15 C is
(A) 1.25 × 104
(B) 1.25 × 10³
(C) 2.5 × 10³
(D) 2.5 × 104
Answer:
(D) 2.5 × 104

Question 14.
Electric lines of force about a positive point charge and negative point charge are respectively .
(A) circular, clockwise
(B) radially outward, radially inward
(C) radially inward, radially outward
(D) circular, anticlockwise
Answer:
(B) radially outward, radially inward

Question 15.
Which of the following is NOT the property of equipotential surfaces?
(A) They do not intersect each other.
(B) They are concentric spheres for uniform electric field.
(C) Potential at all points on the surface has constant value.
(D) Separation of equipotential surfaces increases with decrease in electric field.
Answer:
(B) They are concentric spheres for uniform electric field.

Question 16.
In a uniform electric field, a charge of 3 C experiences a force of 3000 N. The potential difference between two points 1 cm apart along the electric lines of force will be
(A) 10 V
(B) 3 V
(C) 0.1 V
(D) 20 V
Answer:
(A) 10 V

Question 17.
Gauss’ law helps in
(A) determination of electric field due to symmetric charge distribution.
(B) determination of electric potential due to symmetric charge distribution.
(C) determination of electric flux.
(D) situations where Coulomb’s law fails.
Answer:
(A) determination of electric field due to symmetric charge distribution.

Question 18.
The electric flux over a sphere of radius 1.0 m is ø. If the radius of the sphere is doubled without changing the charge, the flux will be
(A) 4ø
(B) 2ø
(C) ø
(D) 8ø
Answer:
(C) ø

Question 19.
Gauss’ theorem states that total normal electric induction over a closed surface in an electric field is equal to
(A) \( \frac{1}{\varepsilon} \sum \mathrm{q}_{\mathrm{n}}\)
(B) εΣ qn
(C) Σ qn
(D) q1 × q2 × q3 × ……… qn
Answer:
(C) Σ qn

Maharashtra Board Class 11 Physics Important Questions Chapter 10 Electrostatics

Question 20.
Number of lines of induction starting from a conductor holding + q charge surrounded by a medium of permittivity ε is
(A) q and they leave the surface in normal direction.
(B) q and they leave the surface in any direction.
(C) q/ε and they leave the surface normally at every point.
(D) q/ε and they leave the surface in any direction.
Answer:
(C) q/ε and they leave the surface normally at every point.

Question 21.
An electric dipole of moment p is placed in the position of stable equilibrium in a uniform electric field of intensity E. The torque required to rotate, when the dipole makes an angle 0 with the initial position is
(A) pE cosθ
(B) pE sinθ
(C) pE tanθ
(D) pE cotθ
Answer:
(B) pE sinθ

Question 22.
Four coulomb charge is uniformly distributed on 2 km long wire. Its linear charge density is
(A) 2 C/m
(B) 4 C/m
(C) 4 × 10³ C/m
(D) 2 × 10-3 C/m
Answer:
(D) 2 × 10-3 C/m

Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 11 Electric Current Through Conductors Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 11 Electric Current Through Conductors

Question 1.
Define current. State its formula and SI unit.
Answer:

  1. Current is defined as the rate of flow of electric charge.
  2. Formula: I = \(\frac {q}{t}\)
  3. SI unit: ampere (A)

Question 2.
Derive an expression for a current generated due to flow of charged particles
Answer:
i. Consider an imaginary gas of both negatively and positively charged particles moving randomly in various directions across a plane P.
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 1

ii. In a time interval t, let the amount of positive charge flowing in the forward direction be q+ and the amount of negative charge flowing in the forward direction be q. Thus, the net charge flowing in the forward direction is q = q+ – q

Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors

iii. Let I be the current varying with time. Let ∆q be the amount of net charge flowing across the plane P from time t to t + At, i.e. during the time interval ∆t.

iv. Then the current is given by
I(t) = \(\lim _{\Delta t \rightarrow 0} \frac{\Delta \mathrm{q}}{\Delta \mathrm{t}}\)
Flere, the current is expressed as the limit of the ratio (∆q/∆t) as ∆t tends to zero.

Question 3.
Match the amount of current generated A given in column – II with the sources given in column -I.

Column I Column II
1. Lightening a. Few amperes
2. House hold circuits b. 10000 A
c. Order of µA

Answer:

Column I Column II
1. Lightening b. 10000 A
2. House hold circuits a. Few amperes

Question 4.
Which are the most common units of current used in semiconductor devices?
Answer:

  1. milliampere (mA)
  2. microampere (µA)
  3. nanoampere (nA)

Question 5.
Six ampere current flows through a bulb. Find the number of electrons that should flow through the bulb in a time of 4 hrs.
Answer:
Given: I = 6 A, t = 4 hrs = 4 × 60 × 60 s
To find: Number of electrons (N)
Formula: I = \(\frac {q}{t}\) = \(\frac {Ne}{t}\)
Calculation: As we know, e = 1.6 × 10-19 C
From formula,
N = \(\frac {It}{e}\) = \(\frac {6×4×60×60}{1.6×10^{-19}}\) 6x4x60x60 = 5.4 × 1023

Question 6.
Explain flow of current in different conductor.
Answer:

  1. A current can be generated by positively or negatively charged particles.
  2. In an electrolyte, both positively and negatively charged particles take part in the conduction.
  3.  In a metal, the free electrons are responsible for conduction. These electrons flow and generate a net current under the action of an applied electric field.
  4. As long as a steady field exists, the electrons continue to flow in the form of a steady current.
  5. Such steady electric fields are generated by cells and batteries.

Question 7.
State the sign convention used to show the flow of electric current in a circuit.
Answer:
The direction of the current in a circuit is drawn in the direction in which positively charged particles would move, even if the current is constituted by the negatively charged particles, (electrons), which move in the direction opposite to that the electric field.

Question 8.
Explain the concept of drift velocity with neat diagrams.
Answer:
i. When no current flows through a copper rod, the free electrons move in random motion. Therefore, there is no net motion of these electrons in any direction.
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 2

ii. If an electric field is applied along the length of the copper rod, a current is set up in the rod. The electrons inside rod still move randomly, but tend to ‘drift’ in a particular direction.

iii. Their direction is opposite to that of the applied electric field.

iv. The electrons under the action of the applied electric field drift with a drift speed vd.
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 3

Question 9.
What is current density? State its SI unit.
Answer:
i. Current density at a point in a conductor is the amount of current flowing per unit area of the conductor.
Current density, J = \(\frac {I}{A}\)
where, I = Current
A = Area of cross-section

ii. SI unit: A/m²

Question 10.
A metallic wire of diameter 0.02 m contains 10 free electrons per cubic metre. Find the drift velocity for free electrons, having an electric current of 100 amperes flowing through the wire.
(Given: charge on electron = 1.6 × 10-19C)
Answer:
Given: e = 1.6 × 10-19 C, n = 1028 electrons/m³,
D = 0.02 m, r = D/2 = 0.01 m,
I = 100 A
To find: Drift velocity (vd)
Formula: vd = \(\frac {I}{nAe}\)
Calculation: From formula,
vd = \(\frac {I}{nπr^2e}\)
∴ vd = \(\frac {100}{10^{28}×3.142×10^{-4}×1.6×10^{-19}}\)
= \(\frac {10^{-3}}{3.142×1.6}\)
= 1.989 × 10-4 m/s

Question 11.
A copper wire of radius 0.6 mm carries a current of 1 A. Assuming the current to be uniformly distributed over a cross sectional area, find the magnitude of current density. Answer:
Given: r = 0.6 mm = 0.6 × 10-3 m, I = 1 A
To find: Current density (J)
Formula: J = \(\frac {I}{A}\)
Calculation: From formula,
J = \(\frac {1}{3.142×(0.6)^2×10^{-6}}\)
= 0.884 × 106 A/m²

Question 12.
A metal wire of radius 0.4 mm carries a current of 2 A. Find the magnitude of current density if the current is assumed to be uniformly distributed over a cross sectional area.
Answer:
Given: r = 0.4 mm = 0.4 × 10-3 m, I = 2 A
To find: Current density (J)
Formula: J = \(\frac {I}{A}\)
Calculation: From formula,
J = \(\frac {2}{3.142×(0.4)^2×10^{-6}}\)
= 3.978 × 106 A/m²

Question 13.
State and explain ohm’s law.
Answer:
Statement: The current I through a conductor is directly proportional to the potential difference V applied across its two ends provided the physical state of the conductor is unchanged.
Explanation:
According to ohm’s law,
I ∝ V
∴ V = IR or R = \(\frac {V}{I}\)
where, R is proportionality constant and is called the resistance of the conductor.

Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors

Question 14.
Draw a graph showing the I-V curve for a good conductor and ideal conductor.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 4

Question 15.
Define one ohm.
Answer:
If potential difference of 1 volt across a conductor produces a current of 1 ampere through it, then the resistance of the conductor is one ohm.

Question 16.
Define conductance. State its SI unit.
Answer:

  1. Reciprocal of resistance is called conductance.
    C = \(\frac {I}{R}\)
  2. S.I unit statement or Ω-1

Question 17.
Explain the concept of electrical conduction in a conductor.
Answer:

  1. Electrical conduction in a conductor is due to mobile charge carriers (electrons).
  2. These conduction electrons are free to move inside the volume of the conductor.
  3. During their random motion, electrons collide with the ion cores within the conductor. Assuming that electrons do not collide with each other these random motions average out to zero.
  4. On application of an electric field E, the motion of the electron is a combination of the random motion of electrons due to collisions and that due to the electric field \(\vec{E}\).
  5. The electrons drift under the action of the field \(\vec{E}\) and move in a direction opposite to the direction of the field \(\vec{E}\). In this way electrons in a conductor conduct electricity.

Question 18.
Derive expression for electric field when an electron of mass m is subjected to an electric field (E).
Answer:
i. Consider an electron of mass m subjected to an electric field E. The force experienced by the electron will be \(\vec{F}\) = e\(\vec{E}\).

ii. The acceleration experienced by the electron will then be
\(\vec{a}\) = \(\frac {e\vec{E}}{m}\) …………. (1)

iii. The drift velocities attained by electrons before and after collisions are not related to each other.

iv. After the collision, the electron will move in random direction, but will still drift in the direction opposite to \(\vec{E}\).

v. Let τ be the average time between two successive collisions.

vi. Thus, at any given instant of time, the average drift speed of the electron will be,
vd = a τ = \(\frac {eEτ}{m}\) ………………(From 1)
vd = \(\frac {eEτ}{m}\) = \(\frac {J}{ne}\) ……………(2) [∵ vd = \(\frac {J}{ne}\)]

vii. Electric field is given by,
E = (\(\frac {m}{e^2nτ}\))J ………… (from 2)
= ρJ = [∵ ρ = \(\frac {m}{ne^2τ}\)]
where, ρ is resistivity of the material.

Question 19.
A Flashlight uses two 1.5 V batteries to provide a steady current of 0.5 A in the filament. Determine the resistance of the glowing filament.
Answer:
Given: For each battery, V1 = V2 = 1.5 volt,
I = 0.5 A
To find: Resistance (R)
Formula: V = IR
Calculation: Total voltage, V = V1 + V2 = 3 volt
From formula,
R = \(\frac {V}{I}\) = \(\frac {3}{0.5}\) = 6.0 Ω

Question 20.
State an expression for resistance of non-ohmic devices and draw I-V curve for such devices.
Answer:
i. Resistance (R) of a non-ohmic device at a particular value of the potential difference V is given by,
R = \(\lim _{\Delta I \rightarrow 0} \frac{\Delta V}{\Delta I}=\frac{d V}{d I}\)
where, ∆V = potential difference between the
two values of potential V – \(\frac {∆V}{2}\) to V + \(\frac {∆V}{2}\),
and ∆I = corresponding change in the current.
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 5

Question 21.
Derive an expression for decrease in potential energy when a charge flows through an external resistance in a circuit.
Answer:
i. Consider a resistor AB connected to a cell in a circuit with current flowing from A to B.
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 6

ii. The cell maintains a potential difference V between the two terminals of the resistor, higher potential at A and lower at B.

iii. Let Q be the charge flowing in time ∆t through the resistor from A to B.

iv. The potential difference V between the two points A and B, is equal to the amount of work (W) done to carry a unit positive charge from A to B.
∴ V = \(\frac {W}{Q}\)

v. The cell provides this energy through the charge Q, to the resistor AB where the work is performed.

vi. When the charge Q flows from the higher potential point A to the lower potential point B, there is decrease in its potential energy by an amount
∆U = QV = I∆tV
where I is current due to the charge Q flowing in time ∆t.

Question 22.
Prove that power dissipated across a resistor is responsible for heating up the resistor. Give an example for it.
OR
Derive an expression for the power dissipated across a resistor in terms of its resistance R.
Answer:
i. When a charge Q flows from the higher potential point to the lower potential point, its potential energy decreases by an amount,
∆U = QV = I∆tV
where I is current due to the charge Q flowing in time ∆t.

ii. By the principle of conservation of energy, this energy is converted into some other form of energy.

iii. In the limit as ∆t → 0, \(\frac {dU}{dt}\) = IV
Here, \(\frac {dU}{dt}\) is power, the rate of transfer of energy ans is given by p = \(\frac {dU}{dt}\) = IV
Hence, power is transferred by the cell to the resistor or any other device in place of the resistor, such as a motor, a rechargeable battery etc.

iv. Due to the presence of an electric field, the free electrons move across a resistor and their kinetic energy increases as they move.

v. When these electrons collide with the ion cores, the energy gained by them is shared among the ion cores. Consequently, vibrations of the ions increase, resulting in heating up of the resistor.

vi. Thus, some amount of energy is dissipated in the form of heat in a resistor.

vii. The energy dissipated per unit time is actually the power dissipated which is given by,
P = \(\frac {V^2}{R}\) = I²R
Hence, it is the power dissipation across a resistor which is responsible for heating it up.

viii. For example, the filament of an electric bulb heats upto incandescence, radiating out heat and light.

Question 23.
Calculate the current flowing through a heater rated at 2 kW when connected to a 300 V d. c. supply.
Answer:
Given: P = 2 kW = 2000 W, V = 300 V
To find: Current (I)
Formula: P = IV
Calculation: From formula,
I = \(\frac {P}{V}\) = \(\frac {2000}{300}\) = 6.67 A

Question 24.
An electric heater takes 6 A current from a 230 V supply line, calculate the power of the heater and electric energy consumed by it in 5 hours.
Answer:
Given: I = 6 A, V = 230 V, t = 5 hours
To find: Power (P), Energy consumed
Formulae: i. P = IV
ii. Energy consumed = power × time
Calculation: From formula (i),
P = 6 × 230
= 1380 W = 1.38 kW
From formula (ii),
Energy consumed = 1.38 × 5 = 6.9 kWh
= 6.9 units

Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors

Question 25.
When supplied a voltage of 220 V, an electric heater takes 6 A current. Calculate the power of heater and electric energy consumed by its in 2 hours?
Answer:
Given: I = 6 A, V = 220 volt, t = 2 hour
To find: i. Power of heater (P)
ii. Electric energy consumed (E)
Formulae: i. P = IV
ii. Electric energy consumed
= Power × time
Calculation: From formula (i),
P = 6 × 220 = 1320 W = 1.32 kW
From formula (ii),
Electric energy consumed
= 1.32 × 2 = 2.64 kWh = 2.64 units

Question 26.
Explain the colour code system for resistors with an example.
Answer:
i. In colour code system, resistors has 4 bands on it.
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 7

ii. In the four band resistor, the colour code of the first two bands indicate two numbers and third band often called decimal multiplier.

iii. The fourth band separated by a space from the three value bands, indicates tolerance of the resistor.

iv. Following table represents the colour code of carbon resistor.
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 8

v. Example:
Let the colours of the rings of a resistor starting from one end be brown, red and orange and gold at the other end. To determine resistance of resistor we have,
x = 1, y = 2, z = 3 (From colour code table)
∴ Resistance = xy × 10z Ω ± tolerance
= 12 × 10³ Ω ± 5%
= 12 kΩ ± 5%
[Note: To remember the colours in order learn the Mnemonics: B.B. ROY of Great Britain had Very Good Wife]

Question 27.
Explain the concept of rheostat.
Answer:

  1. A rheostat is an adjustable resistor used in applications that require adjustment of current or resistance in an electric circuit.
  2. The rheostat can be used to adjust potential difference between two points in a circuit, change the intensity of lights and control the speed of motors, etc.
  3. Its resistive element can be a metal wire or a ribbon, carbon films or a conducting liquid, depending upon the application.
  4. In hi-fi equipment, rheostats are used for volume control.

Question 28.
Explain series combination of resistors.
Answer:
i. In series combination, resistors are connected in single electrical path. Hence, the same electric current flows through each resistor in a series combination.

ii. Whereas, in series combination, the supply voltage between two resistors R1 and R2 is divided into V1 and V2 respectively.
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 9

iii. According to Ohm’s law,
R1 = \(\frac {V_1}{I}\), R2 = \(\frac {v_2}{I}\)
Total Voltage, V = V1 + V2
= I(R1 + R2)
∴ V = I Rs
Thus, the equivalent resistance of the series circuit is, Rs = R1 + R2

iv. When a number of resistors are connected in series, the equivalent resistance is equal to the sum of individual resistances.
For ‘n’ number of resistors,
Rs = R1 + R2 + R2 + ………….. + Rn = \(\sum_{i=1}^{i=n} R_{i}\)

Question 29.
Explain parallel combination of resistors.
Answer:
i. In parallel combination, the resistors are connected in such a way that the same voltage is applied across each resistor.

ii. A number of resistors are said to be connected in parallel if all of them are connected between the same two electrical points each having individual path.

iii. In parallel combination, the total current I is divided into I, and I2 as shown in the circuit diagram.
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 10

iv. Since voltage V across them remains the same,
I = I1 + I2
where I1 is current flowing through R1 and I2 is current flowing through R2.

v. When Ohm’s law is applied to R1,
V = I1R1
i.e. I1 = \(\frac {V}{R_1}\) ………(1)
When Ohm’s law applied to R2,
V = I2R2
i.e., I2 = \(\frac {V}{R_2}\) …………(2)

vi. Total current is given by,
I = I1 + I2
∴ I = \(\frac {V}{R_1}\) + \(\frac {V}{R_2}\) ………[From (1) and (2)]
Since, I = \(\frac {V}{R_p}\)
∴ \(\frac {V}{R_p}\) = \(\frac {V}{R_1}\) + \(\frac {V}{R_2}\)
∴ \(\frac {1}{R_p}\) = \(\frac {1}{R_1}\) + \(\frac {1}{R_2}\)
Where, Rp is the equivalent resistance in parallel combination.

vii. If ‘n’ number of resistors R1, R2, R3, ………….. Rn are connected in parallel, the equivalent resistance of the combination is given by
\(\frac {1}{R_p}\) = \(\frac {1}{R_1}\) + \(\frac {1}{R_2}\) + \(\frac {1}{R_3}\) ……….. + \(\frac {1}{R_n}\) = \(\sum_{i=1}^{\mathrm{i}=\mathrm{n}} \frac{1}{\mathrm{R}}\)
Thus, when a number of resistors are connected in parallel, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of individual resistances.

Question 30.
Colour code of resistor is Yellow-Violet- Orange-Gold. Find its value.
Answer:

Yellow (x) Violet (y) Orange (z) Gold (T%)
Value 4 7 3 ± 5

Value of resistance: xy × 10z Ω ± tolerance
∴ Value of resistance = 47 × 10³ Ω ± 5%
= 47 kΩ ± 5%

Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors

Question 31.
From the given value of resistor, find the colour bands of this resistor.
Value of resistor: 330 Ω
Answer:
Value = 330 Ω = 33 × 101 Ω = xy × 10z Ω

Value 3 3 1
Colour Orange (x) Orange (y) Broen(z)

ii. Given: Green – Blue – Red – Gold

Question 32.
Evaluate resistance for the following colour-coded resistors:
i. Yellow – Violet – Black – Silver
ii. Green – Blue – Red – Gold
ill. Brown – Black – Orange – Gold
Answer:
i. Given: Yellow – Violet – Black – Silver
To find: Value of resistance
Formula: Value of resistance
= (xy × 10z ± T%)Ω

Colour Yellow (x) Violet (y) Black (z) Sliver (T%)
Code 4 7 0 ±10

Hence x = 4, y = 7, z = 0, T = 10%
Value of resistance = (xy ×10z ± T%) Ω
= (47 × 10° ± 10%) Ω
Value of resistance = 47 Ω ± 10%

To find: Value of resistance
Formula: Value of resistance
= (xy × 10z ± T%) Ω
Calculation:

Colour Green (x) Blue (y) Red (z) Gold (T%)
Code 5 6 2 ±5

Hence x = 5, y = 6, z = 2, T = 5%
Value of resistance = (xy × 10z ± T%) Q
= 56 × 102 Ω ± 5%
= 5.6 k Ω ± 5%

iii. Given: Brown – Black – Orange – Gold
To find: Value of the resistance
Formula: Value of the resistance
= (xy × 10z ± T%) Ω
Calculation:

Colour Brown (x) Black (y) Orange (z) Gold (T%)
Code 1 0 3 ±5

Hence x = 1, y = 0, z = 3, T = 5%
Value of resistance = (xy × 10z ± T%) Ω
= 10 × 10³ Ω ± 5%
= 10 kΩ ± 5%

Question 33.
Calculate
i. total resistance and
ii. total current in the following circuit.
R1 = 3 Ω, R2 = 6 Ω, R3 = 5 Ω, V = 14 V
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 11
Answer:
i. R1 and R2 are connected in parallel. This combination (Rp) is connected in series with R3.
∴ Total resistance, RT = Rp + R3
Rp = \(\frac {R_1R_2}{R_1+R_2}\) = \(\frac {3×6}{3+6}\) = 2 Ω
∴ RT = 2+ 5 = 7 Ω

ii. Total current: I = \(\frac {V}{R_T}\) = \(\frac {14}{7}\) = 2 A

Question 34.
State the factors affecting resistance of a conductor.
Answer:
Factors affecting resistance of a conductor:

  1. Length of conductor
  2. Area of cross-section
  3. Nature of material

Question 35.
Derive expression for specific resistance of a material.
Answer:
At a particular temperature, the resistance (R) of a conductor of uniform cross section is
i. directly proportional to its length (l),
i.e., R ∝ l ……….. (1)

ii. inversely proportional to its area of cross section (A),
R ∝ \(\frac {1}{A}\) ……….. (1)
From equations (1) and (2),
R = ρ\(\frac {l}{A}\)
where ρ is a constant of proportionality and it is called specific resistance or resistivity of the material of the conductor at a given temperature.

iii. Thus, resistivity is given by,
ρ = \(\frac {RA}{l}\)

Question 36.
State SI unit of resistivity.
Answer:
SI unit of resistivity is ohm-metre (Ω m).

Question 37.
What is conductivity? State its SI unit.
Answer:
i. Reciprocal of resistivity is called as conductivity of a material.
Formula: σ = \(\frac {1}{ρ}\)
ii. SI unit: (\(\frac {1}{ohm m}\)) or siemens/metre

Question 38.
Explain the similarities between R = \(\frac {V}{I}\) and ρ = \(\frac {E}{J}\)
Answer:

  1. Resistivity (ρ) is a property of a material, while the resistance (R) refers to a particular object.
  2. The electric field \(\vec{E}\) at a point is specified in a material with the potential difference across the resistance and the current density \(\vec{J}\) in a material is specified instead of current I in the resistor.
  3. For an isotropic material, resistivity is given by ρ = \(\frac {E}{J}\)
    For a particular resistor, the resistance R given by, R = \(\frac {V}{I}\)

Question 39.
State expression for current density in terms of conductivity.
Answer:
Current density, \(\vec{J}\) = \(\frac {1}{ρ}\) \(\vec{E}\) = σ \(\vec{E}\)
where, ρ = resistivity of the material
E = electric field intensity
σ = conductivity of the material

Question 40.
Calculate the resistance per metre, at room temperature, of a constantan (alloy) wire of diameter 1.25 mm. The resistivity of constantan at room temperature is 5.0 × 10-7 Ωm.
Answer:
Given: ρ = 5.0 × 10-7 Ω m, d = 1.25 × 10-3 m,
∴ r = 0.625 × 10-3 m
To find: Resistance per metre (\(\frac {R}{l}\))
Formula: ρ = \(\frac {RA}{l}\)
Calculation:
From formula,
\(\frac{\mathrm{R}}{l}=\frac{\rho}{\mathrm{A}}=\frac{\rho}{\pi \mathrm{r}^{2}}\)
= \(\frac{5 \times 10^{-7}}{3.142 \times\left(0.625 \times 10^{-3}\right)^{2}}\)
= \(\frac{5}{3.142 \times 0.625^{2}} \times 10^{-1}\)
= { antilog [log 5 – log 3.142 -2 log 0.625]} × 10-1
= {antilog [ 0.6990 – 0.4972 -2(1.7959)]} × 10-1
= {antilog [0.2018- 1.5918]} × 10-1
= {antilog [0.6100]} × 10-1
= 4.074 × 10-1
∴ \(\frac {R}{l}\) ≈ 0.41 Ω m-1

Question 41.
A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6 × 10-7 m², and its resistance is measured to be 5 Ω. What is the resistivity of the material at the temperature of the experiment?
Answer:
Given: l = 15 m, A = 6.0 × 10-7 m², R = 5 Ω
To find: Resistivity (ρ)
Formula: ρ = \(\frac {RA}{l}\)
Calculation: From formula,
ρ = \(\frac {5×6×10^{-7}}{15}\)
∴ ρ = 2 × 10-7 Ω m

Question 42.
A constantan wire of length 50 cm and 0.4 mm diameter is used in making a resistor. If the resistivity of constantan is 5 × 10-7m, calculate the value of the resistor.
Answer:
Given: l = 50 cm = 0.5 m,
d = 0.4 mm = 0.4 × 10-3 m,
r = 0.2 × 10-3 m, p = 5 × 10-7 Ωm
To Find: Value of resistor (R)
Formula: ρ = \(\frac {RA}{l}\)
Calculation: from formula,
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 12

Question 43.
The resistivity of nichrome is 10-6 Ωm. What length of a uniform wire of this material and of 0.2 mm diameter will have a resistance of 200 ohm?
Answer:
Given: ρ = 10-6 Ω m, d = 0.2 mm,
∴ r = 0.1 mm = 0.1 × 10-3 m, R = 200 Ω
To find: Length (l)
Formula: R = \(\frac {ρl}{A}\) = \(\frac {ρl}{πr^2}\)
Calculation: From formula,
l = \(\frac {πr^2}{ρ}\)
∴ l = \(\frac{200 \times 3.142 \times\left(0.1 \times 10^{-3}\right)^{2}}{10^{-6}}\) = 6 284 m

Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors

Question 44.
A wire of circular cross-section and 30 ohm resistance is uniformly stretched until its new length is three times its original length. Find its resistance.
Answer:
Given: R1 = 30 ohm,
l1 = original length, A1 = original area,
l2 = new length, A2 = new area
l2= 3l1
To find: Resistance (R2)
Formula: R= ρ\(\frac {l}{A}\)
Calculation: From formula,
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 13
The volume of wire remains the same in two cases, we have
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 14

Question 45.
Define temperature coefficient of resistivity. State its SI unit.
Answer:
i. The temperature coefficient of resistivity is defined as the increase in resistance per unit original resistance at the chosen reference temperature, per degree rise in temperature.
α = \(\frac{\rho-\rho_{0}}{\rho_{0}\left(T-T_{0}\right)}\)
= \(\frac{\mathrm{R}-\mathrm{R}_{0}}{\mathrm{R}_{0}\left(\mathrm{~T}-\mathrm{T}_{0}\right)}\)
For small difference in temperatures,
α = \(\frac {1}{R_0}\) \(\frac {dR}{dT}\)

ii. SI unit: °C-1 (per degree Celsius) or K-1 (per kelvin).

Question 46.
Give expressions for variation of resistivity and resistance with temperature. Represent graphically the temperature dependence of resistivity of copper.
Answer:
i. Resistivity is given by,
ρ = ρ0 [1 + α (T – T0)] where,
T0 = chosen reference temperature
ρ0 = resistivity at the chosen temperature
α = temperature coefficient of resistivity
T = final temperature

ii. Resistance is given by,
R = R0 [1+ α (T – T0)]
Where,
T0 = chosen reference temperature
R0 = resistance at the chosen temperature
α = temperature coefficient of resistance
T = final temperature

iii. For example, for copper, the temperature dependence of resistivity can be plotted as shown:
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 15

Question 47.
What is super conductivity?
Answer:

  1. The resistivity of a metal decreases as the temperature decreases.
  2. In case of some metals and metal alloys, the resistivity suddenly drops to zero at a particular temperature (Tc), this temperature is called critical temperature.
  3. Super conductivity is the phenomenon where resistivity of a material becomes zero at particular temperature.
  4. For example, mercury loses its resistance completely to zero at 4.2 K.

Question 48.
A piece of platinum wire has resistance of 2.5 Ω at 0 °C. If its temperature coefficient of resistance is 4 × 10-3/°C. Find the resistance of the wire at 80 °C.
Answer:
Given: R0 = 2.5 Ω
α = 4 × 10-3/°C = 0.004/°C
T = 80 °C
To find: Resistance at 80 °C (RT)
Formula: RT = R0(l + α T)
Calculation: From formula,
RT = 2.5 [1+ (0.004 × 80)]
= 2.5(1 + 0.32)
RT = 2.5 × 1.32
RT = 3.3 Ω

Question 49.
The resistance of a tungsten filament at 150 °C is 133 ohm. What will be its resistance at 500 °C? The temperature coefficient of resistance of tungsten is 0.0045 per °C.
Answer:
Given: Let resistance at 150 °C be R1 and resistance at 500 °C be R2
Thus,
R1= 133 Ω, α = 0.0045 °C-1
To find: Resistance (R2)
Formula: RT = R0 (1 + α∆T)
Calculation:
From formula,
R1 = R0 (1 + α × 150)
∴ 133 = R0(1 + 0.0045 × 150) ……….(i)
R2 = R0 (1 + α × 500)
∴ R2 = R0(1 + 0.0045 × 500) ………(ii)
Dividing equation (ii) by (i), we get
\(\frac{\mathrm{R}_{2}}{133}=\frac{1+(0.0045 \times 500)}{1+(0.0045 \times 150)}=\frac{3.25}{1.675}\)
∴ R2 = \(\frac {3.25}{1.675}\) × 133 = 258 Ω

Question 50.
A silver wire has resistance of 2.1 Ω at 27.5 °C. If temperature coefficient of silver is 3.94 × 10-3/°C, find the silver wire resistance at 100 °C.
Answer:
Given: R1 = 2.1 Ω, T1 = 27.5 °C,
α = 3.94 × 10-3/°C, T2 = 100 °C
To find: Resistance (R2)
Formula: RT = Ro (1 + αT)
Calculation:
From the formula,
R1 = R0(1 + α × 27.5) ……….. (i)
R2 = R0(l + α × 100) ………….. (ii)
Dividing equation (i) by (ii), we get,
\(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1+\left(3.94 \times 10^{-3} \times 27.5\right)}{1+\left(3.94 \times 10^{-3} \times 100\right)}\)
\(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1.10835}{1.394}\) = 0.795
∴ R2 = \(\frac{\mathrm{R}_{1}}{0.795}=\frac{2.1}{0.795}\) = 2.641 Ω

Question 51.
At what temperature would the resistance of a copper conductor be double its resistance at 0 °C?
(a for copper = 3.9 × 10-3/°C)
Answer:
Given: Let the resistance of the conductor at 0°C be R0
R1 = R0 at T1 = 0°C
R2 = 2R0 at T2 = T
To find: Final temperature (T)
Formula: α = \(\frac {R_2-R_1}{R_1(T_2-T_1)}\)
Calculation: From formula,
α = \(\frac {2R_0-R_0}{R_1(T_2-T_1)}\) = \(\frac {1}{T}\)
∴ T = \(\frac {1}{α}\) = \(\frac {1}{3.9×10^{-3}}\) ≈ 256 °C

Question 52.
A conductor has resistance of 15 Ω at 10 °C and 18 Ω at 400 °C. Find the temperature coefficient of resistance of the material.
Answer:
Given: R1 = 15 Ω, T1 = 10 °C, R2 = 18 Ω,
T2 = 400 °C
To find: Temperature coefficient of resistance (α)
Formula: RT = R0 (1 + αT)
Calculation:
From formula,
R1 = R0 (1 + α × 10) ……..(i)
R2 = R0 (1 + α × 400) …….(ii)
Dividing equation (i) by (ii), we get,
\(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1+(\alpha \times 10)}{1+(\alpha \times 400)}\)
∴ \(\frac{15}{18}=\frac{1+10 \alpha}{1+400 \alpha}\)
∴ 18 + 180 α = 15 + 6000 α
∴ 5820 α = 3
∴ α = \(\frac {3}{5820}\) = 5.155 × 10-4/°C

Question 53.
Write short note on e.m.f. devices.
Answer:

  1. When charges flow through a conductor, a potential difference get established between the two ends of the conductor.
  2. For a steady flow of charges, this potential difference is required to be maintained across the two ends of the conductor.
  3. There is a device that does so by doing work on the charges, thereby maintaining the potential difference. Such a device is called an emf device and it provides the emf E.
  4. The charges move in the conductor due to the energy provided by the emf device. This energy is supplied by the e.m.f. device on account of its work done.
  5. Power cells, batteries, Solar cells, fuel cells, and even generators, are some examples of emf devices.

Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors

Question 54.
Explain working of a circuit when connected to emf device.
Answer:
i. A circuit is formed with connecting an emf device and a resistor R. Flere, the emf device keeps the positive terminal (+) at a higher electric potential than the negative terminal (-)
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 16
ii. The emf is represented by an arrow from the negative terminal to the positive terminal.

iii. When the circuit is open, there is no net flow of charge carriers within the device.

iv. When connected in a circuit, the positive charge carriers move towards the positive terminal which acts as cathode inside the emf device.

v. Thus, the positive charge carriers move from the region of lower potential energy, to the region of higher potential energy.

vi. Consider a charge dq flowing through the cross section of the circuit in time dt.

vii. Since, same amount of charge dq flows throughout the circuit, including the emf device. Hence, the device must do work dW on the charge dq, so that the charge enters the negative terminal (low potential terminal) and leaves the positive terminal (higher potential terminal).

viii. Therefore, e.m.f. of the emf device is,
E = \(\frac {dW}{dq}\)
The SI unit of emf is joule/coulomb (J/C).

Question 55.
What is an ideal e.m.f. device?
Answer:

  1. In an ideal e.m.f. device, there is no internal resistance to the motion of charge carriers.
  2. The emf of the device is then equal to the potential difference across the two terminals of the device.

Question 56.
What is a real e.m.f. device?
Answer:

  1. In a real emf device, there is an internal resistance to the motion of charge carriers.
  2. If such a device is not connected in a circuit, there is no current through it.

Question 57.
Derive an expression for current flowing through a circuit when an external resistance is connected to a real e.m.f. device.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 17
i. If a current (I) flows through an emf device, there is an internal resistance (r) and the emf (E) differs from the potential difference across its two terminals (V).
V = E – Ir ……… (1)

ii. The negative sign is due to the fact that the current I flows through the emf device from the negative terminal to the positive terminal.

iii. By the application of Ohm’s law,
V = IR …….(2)
From equations (1) and (2),
IR = E – Ir
∴ \(\frac {E}{R+r}\)

Question 58.
Explain the conditions for maximum current.
Answer:

  1. Current in a circuit is given by, I = \(\frac {E}{R+r}\)
  2. Maximum current can be drawn from the emf device, only when R = 0, i.e.
    Imax = \(\frac {E}{R}\)
  3. Imax is the maximum allowed current from an emf device (or a cell) which decides the maximum current rating of a cell or a battery.

Question 59.
A network of resistors is connected to a 14 V battery with internal resistance 1 Q as shown in the circuit diagram.
i. Calculate the equivalent resistance,
ii. Current in each resistor,
iii. Voltage drops VAB, VBC and VDC.
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 18
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 19
For equivalent resistance (Req):
RAB is given as,
\(\frac{1}{\mathrm{R}_{\mathrm{AB}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}\)
∴ RAB = 2 Ω
RBC = R3 = 1 Ω
Also, RCD is given as,
\(\frac{1}{\mathrm{R}_{\mathrm{CD}}}=\frac{1}{\mathrm{R}_{4}}+\frac{1}{\mathrm{R}_{5}}=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}\)
∴ RCD = 3 Ω
∴ Req = RAB + RBC + RCD
= 2 + 1 + 3 = 6Ω

ii. Current through each resistor:
Total current, I = \(\frac{\mathrm{E}}{\mathrm{R}_{\mathrm{eq}}+\mathrm{r}}\) = \(\frac {14}{6+1}\) = 2 A
Across AB, as, R1 = R2
V1 = V2
∴I1 × 4 = I2 × 4
∴ I1 = I2
But, I1 + I2 = I
∴ 2I1 = I
∴ I1 = I2 =1 A ….(∵I = 2 A)
Similarly, as R4 = R5
I3 = I4 = 1 A
Current through resistor BC is same as I.
∴ IBC = 2 A

iii. Voltage drops across AB, BC and CD:
VAB = IRAB = 2 × 2 = 4 V
VBC = IRBC = 2 × 1 = 2 V
VCD = IRCD = 2 × 3 = 6 V

Question 60.
i. Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?
ii. If the combination is connected to a battery of e.m.f. 20 V and negligible internal resistance, determine the current through each resistor and the total current drawn from the battery.
Answer:
Given: R1 = 2Ω, R2 = 4 Ω, R3 = 5 Ω,
V = 20 V
To Find: i. Total resistance (R)
ii. Current through each resistor (I1, I2, I3 respectively)
iii. Total current (I)
Formulae:
i. \(\frac{1}{\mathrm{R}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}\)
ii. V = IR
iii. Total current, I = I1 +I2 + I3
Calculation
From formula (i):
\(\frac{1}{R}=\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\frac{19}{20}\)
∴ R = \(\frac {20}{19}\) Ω
From formula (ii):
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 20
From formula (iii):
I = 10 + 5 + 4
∴ I = 19 A

Question 61.
i. Three resistors 1 Ω, 2 Ω and 3 Ω are combined in series. What is the total resistance of the combination?
ii. If the combination is connected to a battery of e.m.f. 12 V and negligible internal resistance, obtain the potential drop across each resistor.
Answer:
Given: R1 = 1Ω, R2 = 2 Ω, R3 = 3 Ω,
V = 12 V
To Find: i. Total resistance (R)
ii. P.D Across R1, R2, R3 (V1, V2, V3 respectively)
Formulae:
i. Rs = R1 + R2 + R3
ii. V = IR
Calculation
From formula (i):
Rs = l + 2 + 3 = 6 Ω
From formula (ii),
1 = \(\frac {V}{R}\) = \(\frac {12}{6}\) = 2A
∴ V1 = IR1 = 2 × 1 = 2 V
∴ V2 = IR2 = 2 × 2 = 4 V
∴ V3 = IR3 = 2 × 3 = 6 V

Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors

Question 62.
A voltmeter is connected across a battery of emf 12 V and internal resistance of 10 Ω. If the voltmeter resistance is 230 Ω, what reading will be shown by the voltmeter? Answer:
Given: E = 12 volt, r = 10 Ω, R = 230 Ω
To find: Reading shown by voltmeter (V)
Formula: i. I = \(\frac {E}{R+r}\)
ii. V = E – Ir
Calculation
From formula (i),
I = \(\frac{12}{230+10}=\frac{12}{240}=\frac{1}{20} \mathrm{~A}\)
From formula (ii),
V= 12 – \(\frac {1}{20}\) × 10 = 12 – 0.5
= 11.5 volt

Question 63.
A battery of e.m.f. 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Answer:
Given: E = 10 V, r = 3 Ω, I = 0.5 A
To find: i. Resistance of resistor (R)
ii. Terminal voltage of battery (V)
Formula: I = \(\frac {E}{R+r}\)
Calculation: From formula, R = \(\frac {E}{I}\) – r
∴ R = \(\frac {10}{0.5}\)– 3 = 17 Ω
∴ V = IR = 0.5 × 17 = 8.5 volt

Question 64.
How many cells each of 1.5 V/500 mA rating would be required in series-parallel combination to provide 1500 mA at 3 V?
Answer:
21 = ………… = 1.5 V (given)
I1 = I2 = …………… = 500 mA (given)
1500 mA at 3 V is required.
To determine required number of cells:
For series V = V1 + V2 + ………….., and current remains same.
For parallel I = I1 + I2 + ………, and voltage remains same.
To achieve battery output of 3V, the cells should be connected in series.
If n are the number of cells connected in series, then
V = V1 + V2 + …………. + Vn
∴ V = nV1
∴ 3 = n × 1.5
∴ n = 2 cells in series
The series combination of two cells in series will give a current 500 mA.
To achieve output of 1500 mA, the number of batteries (n) connected in parallel, each one having output 3V is,
I = I1 + I2 + ………. + In
∴ I = nI1
∴ 1500 = n × 500
∴ n = 3 batteries each of two cells
∴ No of cells required are 2 × 3 = 6 .
∴ Number of cells = 6
The six cells must be connected as shown
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 21

Question 65.
Explain the concept of series combination of cells.
Answer:
i. In a series combination, cells are connected in single electrical path, such that the positive terminal of one cell is connected to the negative terminal of the next cell, and so on.

ii. The terminal voltage of batteiy/cell is equal to the sum of voltages of individual cells in series. Example: Given figure shows two 1.5 V cells connected in series. This combination provides total voltage,
V = 1.5 V + 1.5 V = 3 V.
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 22

iii. The equivalent emf of n number of cells in series combination is the algebraic sum of their individual emf.
\(\sum_{i} \mathrm{E}_{\mathrm{i}}\) = E1 + E2 + E2+ …….. + En

iv. The equivalent internal resistance of n cells in a series combination is the sum of their individual internal resistance.
\(\sum_{i} \mathrm{r}_{\mathrm{i}}\) = r1 + r2 + r3 + ……… + rn

Question 66.
State advantages of cells in series.
Answer:

  1. The cells connected in series produce a larger resultant voltage.
  2. Cells which are damaged can be easily identified, hence can be easily replaced.

Question 67.
Explain combination of cells in parallel. Ans:
Answer:
i. Consider two cells which are connected in parallel. Here, positive terminals of all the cells are connected together and the negative terminals of all the cells are connected together.

ii. In parallel connection, the current is divided among the branches i.e. I1 and I2 as shown in figure.
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 23

iii. Consider points A and B having potentials VA and VB, respectively.

iv. For the first cell the potential difference across its terminals is, V = VA – VB = E1 – I1 r1
∴ I1 = \(\frac {E_1V}{r_1}\) ………. (1)

v. Point A and B are connected exactly similarly to the second cell.
Hence, considering the second cell,
V = VA – VB = E2 – I2r2
∴ I2 = \(\frac {E_2V}{r_2}\) ………. (2)

vi. Since, I = I1 + I2 ………….. (3)
Combining equations (1), (2) and (3),
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 24

viii. If we replace the cells by a single cell connected between points A and B with the emf Eeq and the internal resistance req then,
V = Eeq– Ireq
From equations (4) and (5),
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 25

ix. For n number of cells connected in parallel with emf E1, E2, E3, ………….., En and internal resistance r1, r2, r3, …………, rn
\(\frac{1}{\mathrm{r}_{\mathrm{rq}}}=\frac{1}{\mathrm{r}_{1}}+\frac{1}{\mathrm{r}_{2}}+\frac{1}{\mathrm{r}_{3}}+\ldots \ldots \ldots+\frac{1}{\mathrm{r}_{\mathrm{n}}}\)
and \(\frac{\mathrm{E}_{\mathrm{eq}}}{\mathrm{r}_{\mathrm{rq}}}=\frac{\mathrm{E}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{E}_{2}}{\mathrm{r}_{2}}+\ldots \ldots \ldots+\frac{\mathrm{E}_{\mathrm{n}}}{\mathrm{r}_{\mathrm{n}}}\)

Question 68.
State advantages and disadvantages of cells in parallel.
Answer:
Advantages:
For cells connected in parallel in a circuit, the circuit will not break open even if a cell gets damaged or open.

Disadvantages:
The voltage developed by the cells in parallel connection cannot be increased by increasing number of cells present in circuit.

Question 69.
State the basic categories of electrical cells.
Answer:
Electrical cells can be divided into several categories like primary cell, secondary cell, fuel cell, etc.

Question 70.
Write short note on primary cell.
Answer:

  1. A primary cell cannot be charged again. It can be used only once.
  2. Dry cells, alkaline cells are different examples of primary cells.
  3. Primary cells are low cost and can be used easily. But these are not suitable for heavy loads.

Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors

Question 71.
Write short note on secondary cell.
Answer:

  1. The secondary cells are rechargeable and can be reused.
  2. The chemical reaction in a secondary cell is reversible.
  3. Lead acid cell and fuel cell are some examples of secondary cells.
  4. Lead acid battery is used widely in vehicles and other applications which require high load currents.
  5. Solar cells are secondary cells that convert solar energy into electrical energy.

Question 72.
Write short note on fuel cells vehicles.
Answer:

  1. Fuel cells vehicles (FCVs) are electric vehicles that use fuel cells instead of lead acid batteries to power the vehicles.
  2. Hydrogen is used as a fuel in fuel cells. The by product after its burning is water.
  3. This is important in terms of reducing emission of greenhouse gases produced by traditional gasoline fuelled vehicles.
  4. The hydrogen fuel cell vehicles are thus more environment friendly.

Question 73.
What can be concluded from the following observations on a resistor made up of certain material? Calculate the power drawn in each case.

Case Current (A) Voltage (V)
A 0.2 1.6
B 0.4 3.2
C 0.6 4.8
d 0.8 6.4

Answer:
i. As the ratio of voltage and current different readings are same, hence ohm’s is valid i.e., V = IR.

ii. Electric power is given by, P = IV
∴ (a) P1 = 0.2 × 1.6 = 0.32 watt
(b) P2 = 0.4 × 3.2 = 1.28 watt
(c) P3 = 0.6 × 4.8 = 2.88 watt
(d) P4 = 0.8 × 6.4 = 5.12 watt

Question 74.
Answer the following questions from the circuit given below. [S1, S2, S3, S4, S5 ⇒ Switches]. Calculate the current (I) flowing in the following cases:
i. S1, S4 → open; S2, S3, S5 → closed.
ii. S2, S5 → open; S1, S3, S4 → closed.
iii. S3 → open; S1, S2, S4, S5 → closed.
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 26
Answer:
i. Here, the circuit can be represented as,
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 27

ii. Here, the circuit can be represented as,
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 28

iii. Here, the circuit can be represented as,
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 29
∴ As switch S3 is open, no current will flow in the circuit.

Question 75.
An electric circuit with a carton resistor and an electric bulb (60 watt, 300 Ω) are connected in series with a 230 V source.
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 30
i. Calculate the current flowing through the circuit.
ii. If the electric bulb of 60 watt is replaced by an electric bulb (80 watt, 300 Ω), will it glow? Justify your answer.
Answer:
Resistance of carbon resistor (R1)
= 16 × 10 Ω = 160 Ω ….(using colour code)
Resistance of bulb (R2) = 300 Ω
∴ Current through the circuit = \(\frac{V}{R_{1}+R_{2}}\)
∴ I = \(\frac{230}{(160+300)}=\frac{230}{460}\) = 0.5 A

ii. Power drawn through electric bulb
= I²R2 = (0.5)² × 300 = 75 watt
Hence, if the bulb is replaced by 80 watt bulb, it will not glow.

Question 76.
From the graph given below, which of the two temperatures is higher for a metallic wire? Justify your answer.
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 31
Answer:
As R = \(\frac {V}{I}\)
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 32
For constant V,
I2 > I1
∴ R1 > R2
Now, for metallic wire,
R ∝ T
∴ T1 > T2
T1 is greater than T2.

Question 77.
If n identical cells, each of emf E and internal resistance r, are connected in series, write an expression for the terminal p.d. of the combination and hence show that this is nearly n times that of a single cell.
Answer:
i. Let n identical cells, each of emf E and internal resistance r, be connected in series. Let the current supplied by this combination to an external resistance R be I.

ii. The equivalent emf of the combination,
Eeq = E + E + …….. (n times) = nE

iii. The equivalent internal resistance of the combination,
req= r + r + … (n times)
= nr

iv. The terminal p.d. of the combination is
V = Eeq – Ireq = nE – Inr = n (E – Ir)
∴ V = n × terminal p.d. of a single cell
Thus, the terminal p.d. of the series combination is n times that of a single cell.

Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors

Question 78.
If n identical cells, each of emf E and internal resistance r, are connected in parallel, derive an expression for the current supplied by this combination to external resistance R. Prove that the combination supplies current almost n times the current supplied by a single cell, when the external resistance R is much smaller than the internal resistance of the parallel combination of the cells.
Answer:
i. Consider n identical cells, each of emf E and internal resistance r, connected in parallel.

ii. Let the current supplied by the combination to the external resistance R be I.
In this case, the equivalent emf of the combination is E.

iii. The equivalent internal resistance r’ of the combination is,
\(\frac{1}{\mathrm{r}^{\prime}}=\frac{1}{\mathrm{r}}+\frac{1}{\mathrm{r}}\) + …………. (n terms)
∴ \(\frac{1}{\mathrm{r}^{\prime}}=\frac{\mathrm{n}}{\mathrm{r}} \Rightarrow \mathrm{r}^{\prime} \frac{\mathrm{n}}{\mathrm{r}}\)

iv. But V = IR is the terminal p.d. across each cell.

v. Hence, the current supplied by each cell,
I = \(\frac {E-V}{r}\)

vi. This gives the current supplied by the combination to the external resistance as
I = \(\frac {E-V}{r}\) + \(\frac {E-V}{r}\) + …….. (n terms) = n(\(\frac {E-V}{r}\))
Thus, current I = n × current supplied by a single cell
This proves that, the current supplied by the combination is n times the current supplied by a single cell.

Multiple Choice Questions

Question 1.
The drift velocity of the free electrons in a conductor is independent of
(A) length of the conductor.
(B) cross-sectional area of conductor.
(C) current.
(D) electric charge.
Answer:
(A) length of the conductor.

Question 2.
The direction of drift velocity in a conductor is
(A) opposite to that of applied electric field.
(B) opposite to the flow of positive charge.
(C) in the direction of the flow of electrons,
(D) all of these.
Answer:
(D) all of these.

Question 3.
The drift velocity of free electrons in a conductor is vd, when the current is flowing in it. If both the radius and current are doubled, the drift velocity will be
(A) \(\frac {v_d}{8}\)
(B) \(\frac {v_d}{4}\)
(C) \(\frac {v_d}{2}\)
(D) vd
Answer:
(C) \(\frac {v_d}{2}\)

Question 4.
The drift velocity vd of electrons varies with electric field strength E as
(A) vd ∝ E
(B) vd ∝ \(\frac {1}{E}\)
(C) vd ∝ E1/2
(D) vd × E\(\frac {1}{1/2}\)
Answer:
(A) vd ∝ E

Question 5.
When a current I is set up in a wire of radius r, the drift speed is vd. If the same current is set up through a wire of radius 2r, then the drift speed will be
(A) vd/4
(B) vd/2
(C) 2vd
(D) 4vd
Answer:
(A) vd/4

Question 6.
When potential difference is applied across an electrolyte, then Ohm’s law is obeyed at
(A) zero potential
(B) very low potential
(C) negative potential
(D) high potential.
Answer:
(D) high potential.

Question 7.
A current of 1.6 A is passed through an electric lamp for half a minute. If the charge on the electron is 1.6 × 10-19 C, the number of electrons passing through it is
(A) 1 × 1019
(B) 1.5 × 1020
(C) 3 × 1019
(D) 3 × 1020
Answer:
(D) 3 × 1020

Question 8.
The SI unit of the emf of a cell is
(A) V/m
(B) V/C
(C) J/C
(D) C/J
Answer:
(C) J/C

Question 9.
The unit of specific resistance is
(A) Ω m-1
(B) Ω-1 m-1
(C) Ω m
(D) Ω m-2
Answer:
(C) Ω m

Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors

Question 10.
If the length of a conductor is halved, then its conductivity will be
(A) doubled
(B) halved
(C) quadrupled
(D) unchanged
Answer:
(D) unchanged

Question 11.
The resistance of a metal conductor increases with temperature due to
(A) change in current carriers.
(B) change in the dimensions of the conductor.
(C) increase in the number of collisions among the current carriers.
(D) increase in the rate of collisions between the current carriers and the vibrating atoms of the conductor.
Answer:
(D) increase in the rate of collisions between the current carriers and the vibrating atoms of the conductor.

Question 12.
The resistivity of Nichrome is 10-6 Ω-m. The wire of this material has radius of 0.1 mm with resistance 100 Ω, then the length will be
(A) 3.142 m
(B) 0.3142 m
(C) 3.142 cm
(D) 31.42 m
Answer:
(A) 3.142 m

Question 13.
Given a current carrying wire of non-uniform cross-section. Which of the following is constant throughout the length of the wire?
(A) Current, electric field and drift speed
(B) Drift speed only
(C) Current and drift speed
(D) Current only
Answer:
(D) Current only

Question 14.
A cell of emf E and internal resistance r is connected across an external resistance R (R >> r). The p.d. across R is A 1
(A) \(\frac {E}{R+r}\)
(B) E(I – \(\frac {r}{R}\))
(C) E(I + \(\frac {r}{R}\))
(D) E (R + r)
Answer:
(B) E(I – \(\frac {r}{R}\))

Question 15.
The e.m.f. of a cell of negligible internal resistance is 2 V. It is connected to the series combination of 2 Ω, 3 Ω and 5 Ω resistances. The potential difference across 3 Ω resistance will be
(A) 0.6 V
(B) 10 V
(C) 3 V
(D) 6 V
Answer:
(A) 0.6 V

Question 16.
A P.D. of 20 V is applied across a conductance of 8 mho. The current in the conductor is
(A) 2.5 A
(B) 28 A
(C) 160 A
(D) 45 A
Answer:
(C) 160 A

Question 17.
If an increase in length of copper wire is 0.5% due to stretching, the percentage increase in its resistance will be
(A) 0.1%
(B) 0.2%
(C) 1 %
(D) 2 %
Answer:
(C) 1 %

Question 18.
If a certain piece of copper is to be shaped into a conductor of minimum resistance, its length (L) and cross-sectional area A shall be respectively
(A) L/3 and 4 A
(B) L/2 and 2 A
(C) 2L and A2
(D) L and A
Answer:
(A) L/3 and 4 A

Question 19.
A given resistor has the following colour scheme of the various strips on it: Brown, black, green and silver. Its value in ohm is
(A) 1.0 × 104 ± 10%
(B) 1.0 × 105 ± 10%
(C) 1.0 × 106 + 10%
(D) 1.0 × 107 ± 10%
Answer:
(C) 1.0 × 106 + 10%

Question 20.
A given carbon resistor has the following colour code of the various strips: Orange, red, yellow and gold. The value of resistance in ohm is
(A) 32 × 104 ± 5%
(B) 32 × 104 ± 10%
(C) 23 × 105 ± 5%
(D) 23 × 105 ± 10%
Answer:
(A) 32 × 104 ± 5%

Question 21.
A typical thermistor can easily measure a change in temperature of the order of
(A) 10-3 °C
(B) 10-2 °C
(C) 10² °C
(D) 10³ °C
Answer:
(A) 10-3 °C

Question 22.
Thermistors are usually prepared from
(A) non-metals
(B) metals
(C) oxides of non-metals
(D) oxides of metals
Answer:
(D) oxides of metals

Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors

Question 23.
On increasing the temperature of a conductor, its resistance increases because
(A) relaxation time decreases.
(B) mass of the electron increases.
(C) electron density decreases.
(D) all of the above.
Answer:
(A) relaxation time decreases.

Question 24.
Which of the following is used for the formation of thermistor?
(A) copper oxide
(B) nickel oxide
(C) iron oxide
(D) all of the above
Answer:
(D) all of the above

Question 25.
Emf of a cell is 2.2 volt. When resistance R = 5 Ω is connected in series, potential drop across the cell becomes 1.8 volt. Value of internal resistance of the cell is
(A) 10/9 Ω
(B) 7/12 Ω
(C) 9/10 Ω
(D) 12/7 Ω
Answer:
(A) 10/9 Ω

Question 26.
A strip of copper, another of germanium are cooled from room temperature to 80 K. The resistance of
(A) copper strip decreases germanium decreases. and that of
(B) copper strip decreases germanium increases. and that of
(C) Both the strip increases.
(D) copper strip increases germanium decreases. and that of
Answer:
(B) copper strip decreases germanium increases. and that of

Question 27.
The terminal voltage of a cell of emf E on short circuiting will be
(A) E
(B) \(\frac {E}{2}\)
(C) 2E
(D) zero
Answer:
(D) zero

Question 28.
If a battery of emf 2 V with internal resistance one ohm is connected to an external circuit of resistance R across it, then the terminal p.d. becomes 1.5 V. The value of R is
(A) 1 Ω
(B) 1.5 Ω
(C) 2 Ω
(D) 3 Ω
Answer:
(D) 3 Ω

Question 29.
A hall is used 5 hours a day for 25 days in a month. It has 6 lamps of 100 W each and 4 fans of 150 W. The total energy consumed for the month is
(A) 1500 kWh
(B) 150 kWh
(C) 15 kWh
(D) 1.5 kWh
Answer:
(B) 150 kWh

Question 30.
The internal resistance of a cell of emf 2 V is 0.1 Ω. It is connected to a resistance of 3.9 Ω. The voltage across the cell will be
(A) 0.5 V
(B) 1.5 V
(C) 1.95 V
(D) 2 V
Answer:
(C) 1.95 V

Question 31.
The emf of a cell is 12 V. When it sends a current of 1 A through an external resistance, the p.d. across the terminals reduces to 10 V. The internal resistance of the cell is
(A) 0.1 Ω
(B) 0.5 Ω
(C) 1 Ω
(D) 2 Ω
Answer:
(D) 2 Ω

Question 32.
Three resistors, 8 Ω, 4 Ω and 10 Ω connected in parallel as shown in figure, the equivalent resistance is
Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors 33
(A) \(\frac {19}{40}\) Ω
(B) \(\frac {40}{19}\) Ω
(C) \(\frac {80}{19}\) Ω
(D) \(\frac {34}{23}\) Ω
Answer:
(B) \(\frac {40}{19}\) Ω

Question 33.
A potential difference of 20 V is applied across the ends of a coil. The amount of heat generated in it is 800 cal/s. The value of resistance of the coil will be
(A) 12 Ω
(B) 1.2 Ω
(C) 0.12 Ω
(D) 0.012 Ω
Answer:
(C) 0.12 Ω

Question 34.
In a series combination of cells, the effective internal resistance will
(A) remain the same.
(B) decrease.
(C) increase.
(D) be half that of the 1st cell.
Answer:
(C) increase.

Maharashtra Board Class 11 Physics Important Questions Chapter 11 Electric Current Through Conductors

Question 35.
The terminal voltage across a cell is more than its e.m.f., if another cell of
(A) higher e.m.f. is connected parallel to it.
(B) less e.m.f. is connected parallel to it.
(C) less e.m.f. is connected in series with it.
(D) higher e.m.f. is connected in series with it.
Answer:
(A) higher e.m.f. is connected parallel to it.

Question 36.
A 100 W, 200 V bulb is connected to a 160 volt supply. The actual’ power consumption would be
(A) 64 W
(B) 125 W
(C) 100 W
(D) 80 W
Answer:
(A) 64 W

Maharashtra Board Class 11 Physics Solutions Chapter 11 Electric Current Through Conductors

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 11 Electric Current Through Conductors Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 11 Electric Current Through Conductors

1. Choose the correct Alternative.

Question 1.
You are given four bulbs of 25 W, 40 W, 60 W and 100 W of power, all operating at 230 V. Which of them has the lowest resistance?
(A) 25 W
(B) 40 W
(C) 60 W
(D) 100 W
Answer:
(D) 100 W

Question 2.
Which of the following is an ohmic conductor?
(A) transistor
(B) vacuum tube
(C) electrolyte
(D) nichrome wire
Answer:
(D) nichrome wire

Maharashtra Board Class 11 Physics Solutions Chapter 11 Electric Current Through Conductors

Question 3.
A rheostat is used
(A) to bring on a known change of resistance in the circuit to alter the current.
(B) to continuously change the resistance in any arbitrary manner and there by alter the current.
(C) to make and break the circuit at any instant.
(D) neither to alter the resistance nor the current.
Answer:
(B) to continuously change the resistance in any arbitrary manner and there by alter the current.

Question 4.
The wire of length L and resistance R is stretched so that its radius of cross-section is halved. What is its new resistance?
(A) 5R
(B) 8R
(C) 4R
(D) 16R
Answer:
(D) 16R

Question 5.
Masses of three pieces of wires made of the same metal are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5 : 3 : 1. The ratios of their resistances are
(A) 1 : 3 : 5
(B) 5 : 3 : 1
(C) 1 : 15 : 125
(D) 125 : 15 : 1
Answer:
(D) 125 : 15 : 1

Question 6.
The internal resistance of a cell of emf 2 V is 0.1 Ω, it is connected to a resistance of 0.9 Ω. The voltage across the cell will be
(A) 0.5 V
(B) 1.8 V
(C) 1.95 V
(D) 3V
Answer:
(B) 1.8 V

Question 7.
100 cells each of emf 5 V and internal resistance 1 Ω are to be arranged so as to produce maximum current in a 25 Ω resistance. Each row contains equal number of cells. The number of rows should be
(A) 2
(B) 4
(C) 5
(D) 100
Answer:
(A) 2

Question 8.
Five dry cells each of voltage 1.5 V are connected as shown in diagram
Maharashtra Board Class 11 Physics Solutions Chapter 11 Electric Current Through Conductors 1
What is the overall voltage with this arrangement?
(A) 0 V
(B) 4.5 V
(C) 6.0 V
(D) 7.5 V
Answer:
(B) 4.5 V

2. Give reasons / short answers

Question 1.
In given circuit diagram two resistors are connected to a 5V supply.
Maharashtra Board Class 11 Physics Solutions Chapter 11 Electric Current Through Conductors 2
i. Calculate potential difference across the 8Q resistor.
ii. A third resistor is now connected in parallel with 6 Ω resistor. Will the potential difference across the 8 Ω resistor be larger, smaller or same as before? Explain the reason for your answer.
Answer:
Total current flowing through the circuit,
I = \(\frac {V}{R_s}\)
= \(\frac {5}{8+6}\)
= \(\frac {5}{14}\) = 0.36 A
∴ Potential difference across 8 f2 (Vi) = 0.36 × 8
= 2.88 V

Maharashtra Board Class 11 Physics Solutions Chapter 11 Electric Current Through Conductors

ii. Potential difference across 8 Ω resistor will be larger.
Reason: As per question, the new circuit diagram will be
Maharashtra Board Class 11 Physics Solutions Chapter 11 Electric Current Through Conductors 3
When any resistor is connected parallel to 6 Ω resistance. Then the resistance across that branch (6 Ω and R Ω) will become less than 6 Ω. i.e., equivalent resistance of the entire circuit will decrease and hence current will increase. Since, V = IR, the potential difference across 8 Ω resistor will be larger.

Question 2.
Prove that the current density of a metallic conductor is directly proportional to the drift speed of electrons.
Answer:
i. Consider a part of conducting wire with its free electrons having the drift speed vd in the direction opposite to the electric field \(\vec{E}\).

ii. All the electrons move with the same drift speed vd and the current I is the same throughout the cross section (A) of the wire.

iii. Let L be the length of the wire and n be the number of free electrons per unit volume of the wire. Then the total number of free electrons in the length L of the conducting wire is nAL.

iv. The total charge in the length L is,
q = nALe ………….. (1)
where, e is the charge of electron.

v. Equation (1) is total charge that moves through any cross section of the wire in a certain time interval t.
∴ t = \(\frac {L}{v_d}\) ………….. (2)

vi. Current is given by,
I = \(\frac {q}{t}\) = \(\frac {nALe}{L/v_d}\) ……………. [From Equations (1) and (2)]
= n Avde
Hence
vd = \(\frac {1}{nAe}\)
= \(\frac {J}{ne}\) …………. (∵ J = \(\frac {1}{A}\))
Hence for constant ‘ne’, current density of a metallic conductor is directly proportional to the drift speed of electrons, J ∝ vd.

3. Answer the following questions.

Question 1.
Distinguish between ohmic and non ohmic substances; explain with the help of example.
Answer:

Ohmic substances Non-ohmic substances
1. Substances which obey ohm’s law are called ohmic substances. Substances which do not obey ohm’s law are called non-ohmic substances.
2. Potential difference (V) versus current (I) curve is a straight line. Potential difference (V) versus current (I) curve is not a straight line.
3. Resistance of these substances is constant i.e. they follow linear I-V characteristic. Resistance of these substances
Expression for resistance is, R = \(\frac {V}{I}\) Expression for resistance is,
R = \(\lim _{\Delta I \rightarrow 0} \frac{\Delta V}{\Delta I}=\frac{d V}{d I}\)
Examples: Gold, silver, copper etc. Examples:  Liquid electrolytes, vacuum tubes, junction diodes, thermistors etc.

Maharashtra Board Class 11 Physics Solutions Chapter 11 Electric Current Through Conductors

Question 2.
DC current flows in a metal piece of non uniform cross-section. Which of these quantities remains constant along the conductor: current, current density or drift speed?
Answer:
Drift velocity and current density will change as it depends upon area of cross-section whereas current will remain constant.

4. Solve the following problems.

Question 1.
What is the resistance of one of the rails of a railway track 20 km long at 20°C? The cross-section area of rail is 25 cm² and the rail is made of steel having resistivity at 20°C as 6 × 10-8 Ω m.
Answer:
Given: l = 20 km = 20 × 10³ m,
A = 25 cm² = 25 × 10-4 m²,
ρ = 6 × 10-8 Ω m
To find: Resistance of rail (R)
Formula: ρ = \(\frac {RA}{l}\)
Calculation: From formula.
R = ρ\(\frac {l}{A}\)
∴ R = \(\frac {6×10^{-8}×20×10^3}{A}\) = \(\frac {6×4}{5}\) × 10-1
= 0.48 Ω

Question 2.
A battery after a long use has an emf 24 V and an internal resistance 380 Ω. Calculate the maximum current drawn from the battery. Can this battery drive starting motor of car?
Answer:
E = 24 V, r = 380 Ω
i. Maximum current (Imax)
ii. Can battery start the motor?
Formula: Imax = \(\frac {E}{r}\)
Calculation:
From formula,
Imax = \(\frac {24}{380}\) = 0,063 A
As, the value of current is very small compared to required current to run a starting motor of a car, this battery cannot be used to drive the motor.

Question 3.
A battery of emf 12 V and internal resistance 3 O is connected to a resistor. If the current in the circuit is 0.5 A,
i. Calculate resistance of resistor.
ii. Calculate terminal voltage of the battery when the circuit is closed.
Answer:
Given: E = 12 V, r = 3 Ω, I = 0.5 A
To find:
i. Resistance (R)
ii. Terminal voltage (V)
Formulae:
i. E = I (r + R)
ii. V = IR
Calculation: From formula (i),
E = Ir + IR
∴ R = \(\frac {E-Ir}{l}\)
= \(\frac {12-0.5×3}{0.5}\)
= 21 Ω
From formula (ii),
V = 0.5 × 21
= 10.5 V

Maharashtra Board Class 11 Physics Solutions Chapter 11 Electric Current Through Conductors

Question 4.
The magnitude of current density in a copper wire is 500 A/cm². If the number of free electrons per cm³ of copper is 8.47 × 1022, calculate the drift velocity of the electrons through the copper wire (charge on an e = 1.6 × 10-19 C)
Answer:
Given: J = 500 A/cm² = 500 × 104 A/m²,
n = 8.47 × 1022 electrons/cm³
= 8.47 × 1028 electrons/m³
e = 1.6 × 10-19 C
To Find: Drift velocity (vd)
Formula: vd = \(\frac {J}{ne}\)
Calculation:
From formula,
vd = \(\frac {500×10^4}{8.47×10^{28}×1.6×10^{-19}}\)
= \(\frac {500}{8.47×1.6}\) × 10-5
= {antilog [log 500 – log 8.47 – log 1.6]} × 10-5
= {antilog [2.6990 – 0.9279 -0.2041]} × 10-5
= {antilog [1.5670]} × 10-5
= 3.690 × 101 × 10-5
= 3.69 × 10-4 m/s

Question 5.
Three resistors 10 Ω, 20 Ω and 30 Ω are connected in series combination.
i. Find equivalent resistance of series combination.
ii. When this series combination is connected to 12 V supply, by neglecting the value of internal resistance, obtain potential difference across each resistor.
Answer:
Given: R1 = 10 Ω, R2 = 20 Ω,
R3 = 30 Ω, V = 12 V
To Find: i. Series equivalent resistance(Rs)
ii. Potential difference across each resistor (V1, V2, V3)
Formula: i. Rs = R1 + R2 + R3
ii. V = IR
Calculation:
From formula (i),
Rs = 10 + 20 + 30 = 60 Ω
From formula (ii),
I = \(\frac {V}{R}\) = \(\frac {12}{60}\) = 0.2 A
∴ Potential difference across R1,
V1 = I × R1 = 0.2 × 10 = 2 V
∴ Potential difference across R2,
V2 = 0.2 × 20 = 4 V
∴ Potential difference across R3,
V3 = 0.2 × 30 = 6 V

Question 6.
Two resistors 1 Ω and 2 Ω are connected in parallel combination.
i. Find equivalent resistance of parallel combination.
ii. When this parallel combination is connected to 9 V supply, by neglecting internal resistance, calculate current through each resistor.
Answer:
R1 = 1 kΩ = 10³ Ω,
R2 = 2 kΩ = 2 × 10³ Ω, V = 9 V
To find:
i. Parallel equivalent resistance (Rp)
ii. Current through 1 kΩ and 2 kΩ (I1 and I2)
Formula:
i. \(\frac {1}{R_p}\) = \(\frac {1}{R_1}\) + \(\frac {1}{R_2}\)
ii. V = IR
Calculation: From formula (i),
\(\frac {1}{R_p}\) = \(\frac {1}{10^3}\) + \(\frac {1}{2×10^3}\)
= \(\frac {3}{2×10^3}\)
∴ Rp = \(\frac {2×10^3}{3}\) = 0.66 kΩ
From formula (ii),
I1 = \(\frac {V}{R_1}\) + \(\frac {9}{10^3}\)
= 9 × 10-3 A
= 3 mA
I2 = \(\frac {V}{R_2}\) + \(\frac {9}{2×10^3}\)
= 4.5 × 10-3 A
= 4.5 mA

Question 7.
A silver wire has a resistance of 4.2 Ω at 27°C and resistance 5.4 Ω at 100°C. Determine the temperature coefficient of resistance.
Answer:
Given: R1 =4.2 Ω, R2 = 5.4 Ω,
T, = 27° C, T2= 100 °C
To find: Temperature coefficient of resistance (α)
Formula: α = \(\frac {R_2-R_1}{R_1(T_2-T_1)}\)
Calculation:
From Formula
α = \(\frac {5.4-4.2}{4.2(100-27)}\) = 3.91 × 10-3/°C

Maharashtra Board Class 11 Physics Solutions Chapter 11 Electric Current Through Conductors

Question 8.
A 6 m long wire has diameter 0.5 mm. Its resistance is 50 Ω. Find the resistivity and conductivity.
Answer:
Given: l = 6 m, D = 0.5 mm,
r = 0.25 mm = 0.25 × 10-3 m, R = 50 Ω
To find:
i. Resistivity (ρ)
ii. Conductivity (σ)
Formulae:
i. ρ = \(\frac {RA}{l}\) = \(\frac {Rπr^2}{l}\)
ii. σ = \(\frac {1}{ρ}\)
Calculation:
From formula (i),
ρ = \(\frac {50×3.142×(0.25×10^{-3})^2}{6}\)
= {antilog [log 50 + log 3.142 + 21og 0.25 -log 6]} × 10-6
= {antilog [ 1.6990 + 0.4972 + 2(1.3979) -0.7782]} × 10-6
= {antilog [2.1962 + 2 .7958 – 0.7782]} × 10-6
= {antilog [0.9920 – 0.7782]} × 10-6
= {antilog [0.2138]} × 10-6
= 1.636 × 10-6 Ω/m
From formula (ii),
σ = \(\frac {1}{1.636×10^{-6}}\)
= 0.6157 × 106
….(Using reciprocal from log table)
= 6.157 × 105 m/Ω

Question 9.
Find the value of resistances for the following colour code.
i. Blue Green Red Gold
ii. Brown Black Red Silver
iii. Red Red Orange Gold
iv. Orange White Red Gold
v. Yellow Violet Brown Silver
Answer:
i. Given: Blue – Green – Red – Gold
To find: Value of resistance
Formula: Value of resistance
= (xy × 10z ± T%)Ω
Calculation:

Colour Blue (x) Green (y) Red (z) Gold (T%)
Code 6 5 2 ± 5

From formula,
Value of resistance = (65 × 10² ± 5%) Ω
Value of resistance = 6.5 kΩ ± 5%

ii. Given: Brown – Black – Red – Silver
To find: Value of resistance
Formula: Value of resistance
= (xy × 10z + T%) Ω
Calculation:

Colour Brown (x) Black (y) Red (z) sliver (T%)
Code 1 0 2 ± 10

From formula,
Value of resistance = (10 × 10² ± 10%) Ω
Value of resistance = 1.0 kΩ ± 10%

iii. Given: Red – Red – Orange – Gold
To find: Value of the resistance
Formula: Value of the resistance
= (xy × 10z ± T%)
Calculation:

Colour Red (x) Red (y) Orange (z) Gold (T%)
Code 2 2 3 ± 5

From formula,
Value of resistance = (22 × 10³ ± 5%)Ω
Value of resistance = 22 kΩ ± 5%
[Note: The answer given above is presented considering correct order of magnitude.]

iv. Given: Orange – White – Red – Gold
To find: Value of the resistance
Formula: Value of the resistance
= (xy × 10z ± T%)
Calculation:

Colour Ornage (x) White (y) Red (z) Gold (T%)
Code 3 9 2 ± 5

From formula,
Value of resistance = (39 × 10² ± 5%) Ω
Value of resistance = 3.9 kΩ ± 5%

v. Given: Yellow-Violet-Brown-Silver
To find: Value of the resistance
Formula: Value of the resistance
= (xy × 10z ± T%)
Calculation:

Colour Yellow (x) violet (y) Brown (z) Sliver (T%)
Code 4 7 1 ± 10

From formula,
Value of resistance = (47 × 10 ± 10%) Ω
Value of resistance = 470 Ω ± 10%
[Note: The answer given above is presented considering correct order of magnitude.]

Question 10.
Find the colour code for the following value of resistor having tolerance ± 10%.
i. 330 Ω
ii. 100 Ω
iii. 47 kΩ
iv. 160 Ω
v. 1 kΩ
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 11 Electric Current Through Conductors 4

Maharashtra Board Class 11 Physics Solutions Chapter 11 Electric Current Through Conductors

Question 11.
A current of 4 A flows through an automobile headlight. How many electrons flow through the headlight in a time of 2 hrs?
Answer:
Given: I = 4 A, t = 2 hrs = 2 × 60 × 60 s
To find: Number of electrons (N)
Formula: I = \(\frac {q}{t}\) = \(\frac {Ne}{t}\)
Calculation: As we know, e = 1.6 × 10-19 C
From formula,
N = \(\frac {It}{e}\) = \(\frac {4×2×60×60}{1.6×10^{-19}}\) = 18 × 10-23

Question 12.
The heating element connected to 230 V draws a current of 5 A. Determine the amount of heat dissipated in 1 hour (J = 4.2 J/cal).
Answer:
Given: V = 230 V, I = 5 A,
At = 1 hour = 60 × 60 sec
To find: Heat dissipated (H)
Formula: H = ∆U = I∆tV
Calculation: From formula,
H = 5 × 60 × 60 × 230
= 4.14 × 106 J
Heat dissipated in calorie,
H = \(\frac {4.14×10^6}{4.2}\) = 985.7 × 10³ cal
= 985.7 kcal

11th Physics Digest Chapter 11 Electric Current Through Conductors Intext Questions and Answers

Can you recall? (Textbookpage no. 207)

An electric current in a metallic conductor such as a wire is due to flow of electrons, the negatively charged particles in the wire. What is the role of the valence electrons which are the outermost electrons of an atom?
Answer:
i. The valence electrons become de-localized when large number of atoms come together in a metal.
ii. These electrons become conduction electrons or free electrons constituting an electric current when a potential difference is applied across the conductor.

Maharashtra Board Class 11 Physics Solutions Chapter 11 Electric Current Through Conductors

Internet my friend (Textbook page no. 218)

https://www.britannica.com/science/supercond uctivityphysics

[Students are expected to visit the above mentioned website and Collect more information about superconductivity.]

Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 14 Semiconductors Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 14 Semiconductors

Question 1.
What are the factors on which electrical conductivity of any solid depends?
Answer:
Electrical conduction in a solid depends on its temperature, the number of charge carriers, how easily these carries can move inside a solid (mobility), its crystal structure, types and the nature of defects present in a solid.

Question 2.
Why are metals good conductor of electricity?
Answer:
Metals are good conductors of electricity due to the large number of free electrons (≈ 1028 per m³) present in them.

Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors

Question 3.
Give the formula for electrical conductivity of a solid and give significance of the terms involved.
Answer:
Electrical conductivity (σ) of a solid is given by a = nqµ,
where, n = charge carrier density (number of charge carriers per unit volume)
q = charge on the carriers
µ = mobility of carriers

Question 4.
Explain in brief temperature dependence of electrical conductivity of metals and semiconductors with the help of graph.
Answer:
i. The electrical conductivity of a metal decreases with increase in its temperature.
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 1

ii. When the temperature of a semiconductor is increased, its electrical conductivity also increases
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 2

Question 5.
Mention the broad classification of semiconductors along with examples.
Answer:
A broad classification of semiconductors can be:

  1. Elemental semiconductors: Silicon, germanium
  2. Compound Semiconductors: Cadmium sulphide, zinc sulphide, etc.
  3. Organic Semiconductors: Anthracene, doped pthalocyanines, polyaniline etc.

Question 6.
What are some electrical properties of semiconductors?
Answer:

  1. Electrical properties of semiconductors are different from metals and insulators due to their unique conduction mechanism.
  2. The electronic configuration of the elemental semiconductors plays a very important role in their electrical properties.
  3. They are from the fourth group of elements in the periodic table.
  4. They have a valence of four.
  5. Their atoms are bonded by covalent bonds. At absolute zero temperature, all the covalent bonds are completely satisfied in a single crystal of pure semiconductor like silicon or germanium.

Question 7.
Explain in detail the distribution of electron energy levels in an isolated atom with the help of an example.
Answer:

  1. An isolated atom has its nucleus at the centre which is surrounded by a number of revolving electrons. These electrons are arranged in different and discrete energy levels.
  2. Consider the electronic configuration of sodium (atomic number 11) i.e, 1s², 2s², 2p6, 3s1. The outermost level 3s can take one more electron but it is half filled in sodium,
  3. The energy levels in each atom are filled according to Pauli’s exclusion principle which states that no two similar spin electrons can occupy the same energy level.
  4. That means any energy level can accommodate only two electrons (one with spin up state and the other with spin down state)
  5. Thus, there can be two states per energy level.
  6. Figure given below shows the allowed energy levels of a sodium atom by horizontal lines. The curved lines represent the potential energy of an electron near the nucleus due to Coulomb interaction.
    Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 3

Question 8.
Explain formation of energy bands in solid sodium with neat labelled energy band diagrams.
Answer:
i. For an isolated sodium atom (atomic number 11) the electronic configuration is given as 1s², 2s², 2p6, 3s1. The outermost level 3s is half filled in sodium.

ii. The energy levels are filled according to Pauli’s exclusion principle.

iii. Consider two sodium atoms close enough so that outer 3s electrons can be considered equally to be part of any atom.

iv. The 3s electrons from both the sodium atoms need to be accommodated in the same level.

v. This is made possible by splitting the 3 s level into two sub-levels so that the Pauli’s exclusion principle is not violated. Figure given below shows the splitting of the 3 s level into two sub levels.
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 4

vi. When solid sodium is formed, the atoms come close to each other such that distance between them remains of the order of 2 – 3 Å. Therefore, the electrons from different atoms interact with each other and also with the neighbouring atomic cores.

vii. The interaction between the outer most electrons is more due to overlap while the inner most electrons remain mostly unaffected. Each of these energy levels is split into a large number of sub levels, of the order of Avogadro’s number due to number of atoms in solid sodium is of the order of this number.

viii. The separation between the sublevels is so small that the energy levels appear almost continuous. This continuum of energy levels is called an energy band. The bands are called 1 s band, 2s band, 2p band and so on. Figure shows these bands in sodium metal.
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 5

Question 9.
Explain concept of valence band and conduction band in solid crystal.
Answer:
A. Valence band (V.B):

  1. The topmost occupied energy level in an atom is the valence level. The energy band formed by valence energy levels of atoms in a solid is called the valence band.
  2. In metallic conductors, the valence electrons are loosely attached to the nucleus. At ordinary room temperature, some valence electrons become free. They do not leave the metal surface but can move from atom to atom randomly.
  3. Such free electrons are responsible for electric current through conductors.

B. Conduction band (C.B):

  1. The immediately next energy level that electrons from valence band can occupy is called conduction level. The band formed by conduction levels is called conduction band.
  2. It is the next permitted energy band beyond valence band.
  3. In conduction band, electrons move freely and conduct electric current through the solids.
  4. An insulator has empty conduction band.

Question 10.
Draw neat labelled diagram showing energy bands in sodium. Why broadening of higher bands is different than that of the lower energy bands?
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 6
Broadening of valence and higher bands is more since interaction of these electrons is stronger than the inner most electrons.

Question 11.
State the conditions when electrons of a semiconductor can take part in conduction.
Answer:

  1. All the energy levels in a band, including the topmost band, in a semiconductor are completely occupied at absolute zero.
  2. At some finite temperature T, few electrons gain thermal energy of the order of kT, where k is the Boltzmann constant.
  3. Electrons in the bands between the valence band cannot move to higher band since these are already occupied.
  4. Only electrons from the valence band can be excited to the empty conduction band, if the thermal energy gained by these electrons is greater than the band gap.
  5. Electrons can also gain energy when an external electric field is applied to a solid. Energy gained due to electric field is smaller, hence only electrons at the topmost energy level gain such energy and participate in electrical conduction.

Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors

Question 12.
Define 1 eV.
Answer:
1 eV is the energy gained by an electron while it overcomes a potential difference of one volt. 1 eV= 1.6 × 10-19 J.

Question 13.
C, Si and Ge have same lattice structure. Why is C insulator while Si and Ge intrinsic semiconductors?
Answer:

  1. The 4 valence electrons of C, Si or Ge lie respectively in the second, third and fourth orbit.
  2. Energy required to take out an electron from these atoms (i.e., ionisation energy Eg) will be least for Ge, followed by Si and highest for C.
  3. Hence, number of free electrons for conduction in Ge and Si are significant but negligibly small for

Question 14.
What is intrinsic semiconductor?
Answer:
A pure semiconductor is blown as intrinsic semiconductor.

Question 15.
Explain characteristics and structure of silicon using a neat labelled diagram.
Answer:

  1. Silicon (Si) has atomic number 14 and its electronic configuration is 1s² 2s² 2p6 3s² 3p².
  2. Its valence is 4.
  3. Each atom of Si forms four covalent bonds with its neighbouring atoms. One Si atom is surrounded by four Si atoms at the comers of a regular tetrahedron as shown in the figure.
    Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 7

Question 16.
Describe in detail formation of holes in ii. intrinsic semiconductor.
Answer:
i. In intrinsic semiconductor at absolute zero temperature, all valence electrons are tightly bound to respective atoms and the covalent bonds are complete.

ii. Electrons are not available to conduct electricity through the crystal because they cannot gain enough energy to get into higher energy levels.
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 8

iii. At room temperature, however, a few covalent bonds are broken due to heat energy produced by random motion of atoms. Some of the valence electrons can be moved to the conduction band. This creates a vacancy in the valence band as shown in figure.
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 9

iv. These vacancies of electrons in the valence band are called holes. The holes are thus absence of electrons in the valence band and they carry an effective positive charge.

Question 17.
How does electric conduction take place inside a pure silicon?
Answer:

  1. There are two different types of charge carriers in a pure semiconductor. One is the electron and the other is the hole or absence of electron.
  2. Electrical conduction takes place by transportation of both carriers or any one of the two carriers in a semiconductor.
  3. When a semiconductor is connected in a circuit, electrons, being negatively charged, move towards positive terminal of the battery.
  4. Holes have an effective positive charge, and move towards negative terminal of the battery. Thus, the current through a semiconductor is carried by two types of charge carriers moving in opposite directions.
  5. Figure given below represents the current through a pure silicon.
    Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 10

Question 18.
Why do holes not exist in conductor?
Answer:

  1. In case of semiconductors, there is one missing electron from one of the covalent bonds.
  2. The absence of electron leaves an empty space called as hole; each hole carries an effective positive charge.
  3. In case of an conductor, number of free electrons are always available for conduction. There is no absence of electron in it. Hence holes do not exist in conductor.

Question 19.
What is the need for doping an intrinsic semiconductor?
Answer:
The electric conductivity of an intrinsic semiconductor is very low at room temperature; hence no electronic devices can be fabricated using them. Addition of a small amount of a suitable impurity to an intrinsic semiconductor increases its conductivity appreciably. Hence, intrinsic semiconductors are doped with impurities.

Question 20.
Explain what is doping.
Answer:

  1. The process of adding impurities to an intrinsic semiconductor is called doping.
  2. The impurity atoms are called dopants which may be either trivalent or pentavalent. The parent atoms are called hosts.
  3. The dopant material is so selected that it does not disturb the crystal structure of the host.
  4. The size and the electronic configuration of the dopant should be compatible with that of the host.
  5. Doping is expressed in ppm (parts per million), i.e., one impurity atom per one million atoms of the host.
  6. Doping significantly increases the concentration of charge carriers.

Question 21.
What is extrinsic semiconductors?
Answer:
The semiconductor with impurity is called a doped semiconductor or an extrinsic semiconductor.

Question 22.
Draw neat diagrams showing schematic electronic structure of:
i. A pentavalent atom [Antimony (Sb)]
ii. A trivalent atom [Boron (B)]
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 11
[Note: Electronic structure of antimony is drawn as per its electronic configuration in accordance with Modern Periodic Table.]

Question 23.
With the help of neat diagram, explain the structure of n-type semiconductor in detail.
Answer:
i. When silicon or germanium crystal is doped with a pentavalent impurity such as phosphorus, arsenic, or antimony we get n-type semiconductor.

ii. When a dopant atom of 5 valence electrons occupies the position of a Si atom in the crystal lattice, 4 electrons from the dopant form bonds with 4 neighbouring Si atoms and the fifth electron from the dopant remains very weakly bound to its parent atom
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 12

iii. To make this electron free even at room temperature, very small energy is required. It is 0.01 eV for Ge and 0.05 eV for Si.

iv. As this semiconductor has large number of electrons in conduction band and its conductivity is due to negatively charged carriers, it is called n-type semiconductor.

v. The n-type semiconductor also has a few electrons and holes produced due to the thermally broken bonds.

vi. The density of conduction electrons (ne) in a doped semiconductor is the sum total of the electrons contributed by donors and the thermally generated electrons from the host.

vii. The density of holes (nh) is only due to the thermally generated holes of the host Si atoms.

viii. Thus, the number of free electrons exceeds the number of holes (ne >> nh). Thus, in n-type semiconductor electrons are the majority carriers and holes are the minority carriers.

Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors

Question 24.
What are some features of n-type semiconductor?
Answer:

  1. These are materials doped with pentavalent impurity (donors) atoms.
  2. Electrical conduction in these materials is due to majority charge carriers i.e., electrons.
  3. The donor atom loses electrons and becomes positively charged ions.
  4. Number of free electrons is very large compared to the number of holes, ne >> nh. Electrons are majority charge carriers.
  5. When energy is supplied externally, negatively charged free electrons (majority charges carries) and positively charged holes (minority charges carries) are available for conduction.

Question 25.
With the help of neat diagram, explain the structure of p-type semiconductor in detail.
Answer:
i. When silicon or germanium crystal is doped with a trivalent impurity such as boron, aluminium or indium, we get a p-type semiconductor.

ii. The dopant trivalent atom has one valence electron less than that of a silicon atom. Every trivalent dopant atom shares its three electrons with three neighbouring Si atoms to form covalent bonds. But the fourth bond between silicon atom and its neighbour is not complete.

iii. The incomplete bond can be completed by another electron in the neighbourhood from Si atom.

iv. The shared electron creates a vacancy in its place. This vacancy or the absence of electron is a hole.
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 13

v. Thus, a hole is available for conduction from each acceptor impurity atom.

vi. Holes are majority carriers and electrons are minority carriers in such materials. Acceptor atoms are negatively charged ions and majority carriers are holes. Therefore, extrinsic semiconductor doped with trivalent impurity is called a p-type semiconductor.

vii. For a p-type semiconductor, nh >> ne.

Question 26.
What are some features of p-type semiconductors?
Answer:

  1. These are materials doped with trivalent impurity atoms (acceptors).
  2. Electrical conduction in these materials is due to majority charge carriers i.e., holes.
  3. The acceptor atoms acquire electron and become negatively charged-ions.
  4. Number of holes is very large compared to the number of free electrons. nh >> ne. Holes are majority charge carriers.
  5. When energy is supplied externally, positively charged holes (majority charge carriers) and negatively charged free electrons (minority charge carriers) are available for conduction.

Question 27.
What are donor and acceptor impurities?
Answer:

  1. Every pentavalent dopant atom which donates one electron for conduction is called a donor impurity.
  2. Each trivalent atom which can accept an electron is called an acceptor impurity.

Question 28.
Explain the energy levels of both donor and acceptor impurities with a schematic band structure.
Answer:
i. The free electrons donated by the donor impurity atoms occupy energy levels which are in the band gap and are close to the conduction band.
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 14

ii. The vacancies of electrons or the extra holes are created in the valence band due to addition of acceptor impurities. The impurity levels are created just above the valence band in the band gap.
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 15

Question 29.
Distinguish between p-type and n-type semiconductor.
Answer:

p-type semiconductor n-type semiconductor
1. The impurity of some trivalent element like B, Al, In, etc. is mixed with semiconductor. The impurity of some pentavalent element like P, As, Sb, etc. is mixed
2. The impurity atom accepts one electron hence the impurities The impurity atom donates – one electron, hence the impurities added are known as donor impurities.
3. The holes are majority charge carriers and electrons are minority charge carriers. The electrons are j majority charge carriers and holes are minority charge carriers.
4. The acceptor energy level is close to the valence band and far away from conduction band. Donor energy level is close to the conduction band and far away from valence band.

Question 30.
What is the charge on a p-type and n-type semiconductor?
Answer:
n-type as well as p-type semiconductors are electrically neutral.

Question 31.
Explain the transportation of holes inside a p-type semiconductor.
Answer:
i. Consider a p-type semiconductor connected to terminals of a battery as shown.

ii. When the circuit is switched on, electrons at 1 and 2 are attracted to the positive terminal of the battery and occupy nearby holes at x and y. This creates holes at the positions 1 and 2 previously occupied by electrons.
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 16

iii. Next, electrons at 3 and 4 move towards the positive terminal and create holes in their previous positions.

iv. But, the holes are captured at the negative terminal by the electrons supplied by the battery.

v. In this way, holes are transported from one place to other and density of holes is kept constant so long as the battery is working.

Question 32.
A pure Si crystal has 4 × 1028 atoms m-3. It is doped by 1 ppm concentration of antimony. Calculate the number of electrons and holes. Given n1 = 1.2 × 1016/m³.
Answer:
As, the atom is doped with 1 ppm concentration of antimony (Sb).
1 ppm = 1 parts per one million atoms. = 1/106
∴ no. of Si atoms = \(\frac {Total no. of Si atoms}{10^6}\)
= \(\frac {4×10^{28}}{10^6}\) = 4 × 1022 m-3
i.e., total no. of extra free electrons (ne)
= 4 × 1022 m-3
ni2 = ne nh
∴ nh = \(\frac {n_i^2}{n_e}\) = \(\frac {(1.2×10^{16})^2}{4×10^{22}}\)
= \(\frac {144×10^{30}{4×10^{22}}\)
= 36 × 10-8
= 3.6 × 109 m-3.

Question 33.
A pure silicon crystal at temperature of 300 K has electron and hole concentration 1.5 × 1016 m-3 each. (ne = nh). Doping bv indium increases nh to 4.5 × 1022 m-3. Calculate ne for the doped silicon crystal.
Answer:
Given: At 300 K, ni = ne = nh = 1.5 × 1016 m-3
After doping nh = 4.5 × 1022 m-3
To find: Number density of electrons (ne)
Formula: ni² = ne nnh
Calculation From formula:
ne = \(\frac {n_i^2}{n_h}\) = \(\frac {(1.5×10^{16})^2}{4×10^{22}}\)
= \(\frac {255×10^{30}{45×10^{21}}\)
= 5 × 10-9 m-3.

Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors

Question 34.
A Ge specimen is doped with A/. The concentration of acceptor atoms is ~1021 atoms/m³. Given that the intrinsic concentration of electron-hole pairs is ~10 19/m³, calculate the concentration of electrons in the specimen.
Answer:
Given: At room temperature,
ni = ne = nh = 1019 m-3
After doping nh = 1021 m-3
To find: Number density of electrons (nc)
Formulae: ni2 = nenh
Calculation: From formula,
ne = \(\frac {n_i^2}{n_h}\) = \(\frac {(10^{19})^2}{10^{21}}\)
= \(\frac {255×10^{30}{45×10^{21}}\)
= 1017 m-3.

Question 35.
A semiconductor has equal electron and hole concentration of 2 × 108 m-3. On doping with a certain impurity, the electron concentration increases to 4 × 1010 m-3, then calculate the new hole concentration of the semiconductor.
Answer:
Given: ni = 2 × 108 m-3, n = 4 × 1010 m-3
After doping nh = 1021 m-3
To find: Number density of holes (nh)
Formulae: ni 2= nenh
Calculation: From formula.
nh = \(\frac {n_i^2}{n_e}\) = \(\frac {(2×10^{8})^2}{4×10^{10}}\) = 106 m-3

Question 36.
What is a p-n junction?
Answer:
When n-type and p-type semiconductor materials are fused together, the junction formed is called as p-n junction.

Question 37.
Explain the process of diffusion in p-n junction.
Answer:
i. The transfer of electrons and holes across the p-n junction is called diffusion.

ii. When an n-type and a p-type semiconductor materials are fused together, initially, the number of electrons in the n-side of a junction is very large compared to the number of electrons on the p-side. The same is true for the number of holes on the p-side and on the n-side.

iii. Thus, a large difference in density of carriers exists on both sides of the p-n junction. This difference causes migration of electrons from the n-side to the p-side of the
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 17

iv. They fill up the holes in the p-type material and produce negative ions.

v. When the electrons from the n-side of a junction migrate to the p-side, they leave behind positively charged donor ions on the n- side. Effectively, holes from the p-side migrate into the n-region.

vi. As a result, in the p-type region near the junction there are negatively charged acceptor ions, and in the n-type region near the junction there are positively charged donor ions.

vii. The extent up to which the electrons and the holes can diffuse across the junction depends on the density of the donor and the acceptor ions on the n-side and the p-side respectively, of the junction.

Question 38.
Define potential barrier.
Answer:
The diffusion of carriers across the junction and resultant accumulation of positive and negative charges across the junction builds a potential difference across the junction. This potential difference is called the potential barrier.

Question 39.
Draw neat labelled diagrams for potentials barrier and depletion layer in a p-n junction.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 18

Question 40.
Explain in brief electric field across a p-n junction with a neat labelled diagram.
Answer:

  1. When p-type semiconductor is fused with n-type semiconductor, a depletion region is developed across the junction.
  2. The n-side near the boundary of a p-n junction becomes positive with respect to the p-side because it has lost electrons and the p-side has lost holes.
  3. Thus, the presence of impurity ions on both sides of the junction establishes an electric field across this region such that the n-side is at a positive voltage relative to the p-side.
    Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 19

Question 41.
What is the need of biasing a p-n junction?
Answer:

  1. Due to potential barrier across depletion region, charge carriers require extra energy to overcome the barrier.
  2. A suitable voltage needs to be applied to the junction externally, so that these charge carriers can overcome the potential barrier and move across the junction.

Question 41.
Explain the mechanism of forward biased p-n junction.
Answer:

  1. In forward bias, a p-n junction is connected in an electric circuit such that the p-region is connected to the positive terminal and the n-region is connected to the negative terminal of an external voltage source.
  2. The external voltage effectively opposes the built-in potential of the junction. The width of potential barrier is thus reduced.
  3. Also, negative charge carriers (electrons) from the n-region are pushed towards the junction.
  4. A similar effect is experienced by positive charge carriers (holes) in the p-region and they are pushed towards the junction.
  5. Both the charge carriers thus find it easy to cross over the barrier and contribute towards the electric current.
    Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 20

Question 42.
Explain the mechanism of reverse biased p-n junction.
Answer:
i. In reverse biased, the p-region is connected to the negative terminal and the n-region is connected to the positive terminal of the external voltage source. This external voltage effectively adds to the built-in potential of the junction. The width of potential barrier is thus increased.

ii. Also, the negative charge carriers (electrons) from the n-region are pulled away from the junction.

iii. Similar effect is experienced by the positive charge carriers (holes) in the p-region and they are pulled away from the junction.

iv. Both the charge carriers thus find it very difficult to cross over the barrier and thus do not contribute towards the electric current.
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 21

Question 43.
State some important features of the depletion region.
Answer:

  1. It is formed by diffusion of electrons from n-region to the p-region. This leaves positively charged ions in the n-region.
  2. The p-region accumulates electrons (negative charges) and the n-region accumulates the holes (positive charges).
  3. The accumulation of charges on either sides of the junction results in forming a potential barrier and prevents flow of charges.
  4. There are no charges in this region.
  5. The depletion region has higher potential on the n-side and lower potential on the p-side of the junction.

Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors

Question 44.
What is p-n junction diode? Draw its circuit symbol.
Answer:
A p-n junction, when provided with metallic connectors on each side is called a junction diode
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 22

Question 45.
Explain asymmetrical flow of current in p-n junction diode in detail.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 23
i. The barrier potential is reduced in forward biased mode and it is increased in reverse biased mode.

ii. Carriers find it easy to cross the junction in forward bias and contribute towards current because the barrier width is reduced and they are pushed towards the junction and gain extra energy to cross the junction.

iii. The current through the diode in forward bias is large and of the order of a few milliamperes (10-3 A) for a typical diode.

iv. When connected in reverse bias, width of the potential barrier is increased and the carriers are pushed away from the junction so that very few carriers can cross the junction and contribute towards current.

v. This results in a very small current through a reverse biased diode. The current in reverse biased diode is of the order of a few microamperes (10-6 A).

vi. When the polarity of bias voltage is reversed, the width of the depletion layer changes. This results in asymmetrical current flow through a diode as shown in figure.
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 24

Question 46.
What is knee voltage?
Answer:
In forward bias mode, the voltage for which the current in a p-n junction diode rises sharply is called knee voltage.

Question 47.
What is a forward current in case of zero biased p-n junction diode?
Answer:
When the diode terminals are shorted together, some holes (majority carriers) in the p-side have enough thermal energy to overcome the potential barrier. Such carriers cross the barrier potential and contribute to current. This current is known as the forward current.

Question 48.
Define reverse current in zero biased p-n junction diode.
Answer:
When the diode terminals are shorted together some holes generated in the n-side (minority carriers), move across the junction and contribute to current. This current is known as the reverse current.

Question 49.
Explain the I-V characteristics of a reverse biased junction diode.
Answer:
i. The positive terminal of the external voltage is connected to the cathode (n-side) and negative terminal to the anode (p-side) across the diode.

ii. In case of reverse bias the width of the depletion region increases and the p-n junction behaves like a high resistance.

iii. Practically no current flows through it with an increase in the reverse bias voltage. However, a very small leakage current does flow through the junction which is of the order of a few micro amperes, (µA).

iv. When the reverse bias voltage applied to a diode is increased to sufficiently large value, it causes the p-n junction to overheat. The overheating of the junction results in a sudden rise in the current through the junction. This is because covalent bonds break and a large number of carries are available for conduction. The diode thus no longer behaves like a diode. This effect is called the avalanche breakdown.

v. The reverse biased characteristic of a diode is shown in figure.
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 25

Question 50.
Explain zero biased junction diode.
Answer:
i. When a diode is connected in a zero bias condition, no external potential energy is applied to the p-n junction.

li. The potential barrier that exists in a junction prevents the diffusion of any more majority carriers across it. However, some minority carriers (few free electrons in the p-region and few holes in the n-region) drift across the junction.

iii. An equilibrium is established when the majority carriers are equal in number (ne = nh) and both moving in opposite directions. The net current flowing across the junction is zero. This is a state of‘dynamic equilibrium’.

iv. The minority carriers are continuously generated due to thermal energy.

v. When the temperature of the p-n junction is raised, this state of equilibrium is changed.

vi. This results in generating more minority carriers and an increase in the leakage current. An electric current, however, cannot flow through the diode because it is not connected in any electric circuit
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 26

Question 51.
What is dynamic equilibrium?
Answer:
An equilibrium is established when the majority carriers are equal in number (ne = nh) and both moving in opposite directions. The net current flowing across the junction is zero. This is a state of‘dynamic equilibrium’.

Question 52.
Draw a neat diagram and state I-V characteristics of an ideal diode.
Answer:
An ideal diode offers zero resistance in forward biased mode and infinite resistance in reverse biased mode.
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 27

Question 53.
What do you mean by static resistance of a diode?
Answer:
Static (DC) resistance:

  1. When a p-n junction diode is forward biased, it offers a definite resistance in the circuit. This resistance is called the static or DC resistance (Rg) of a diode.
  2. The DC resistance of a diode is the ratio of the DC voltage across the diode to the DC current flowing through it at a particular voltage.
  3. It is given by, Rg = \(\frac {V}{I}\)

Question 54.
Explain dynamic resistance of a diode.
Answer:

  1. The dynamic (AC) resistance of a diode, rg, at a particular applied voltage, is defined as
    rg = \(\frac {∆V}{∆I}\)
  2. The dynamic resistance of a diode depends on the operating voltage.
  3. It is the reciprocal of the slope of the characteristics at that point.

Question 55.
Draw a graph representing static and dynamic resistances of a diode.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 28

Question 56.
Refer to the figure as shown below and find the resistance between point A and B when an ideal diode is (i) forward biased and (ii) reverse biased.
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 29
Answer:
We know that for an ideal diode, the resistance is zero when forward biased and infinite when reverse biased.
i. Figure (a) shows the circuit when the diode is forward biased. An ideal diode behaves as a conductor and the circuit is similar to two resistances in parallel.
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 30
∴ RAB = (30 × 30)/(30 +30) = 900/60 = 15 Ω

ii. Figure (b) shows the circuit when the diode is reverse biased.
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 31
It does not conduct and behaves as an open switch along path ACB. Therefore, RAB = 30 Ω. the only resistance in the circuit along the path ADB.

Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors

Question 57.
State advantages of semiconductor devices.
Answer:

  1. Electronic properties of semiconductors can be controlled to suit our requirement.
  2. They are smaller in size and light weight.
  3. They can operate at smaller voltages (of the order of few mV) and require less current (of the order of pA or mA), therefore, consume lesser power.
  4. Almost no heating effects occur, therefore these devices are thermally stable.
  5. Faster speed of operation due to smaller size.
  6. Fabrication of ICs is possible.

Question 58.
State disadvantages of semiconductor devices.
Answer:

  1. They are sensitive to electrostatic charges.
  2. Not very useful for controlling high power.
  3. They are sensitive to radiation.
  4. They are sensitive to fluctuations in temperature.
  5. They need controlled conditions for their manufacturing.
  6. Very few materials are semiconductors.

Question 59.
Explain applications of semiconductors.
Answer:
i. Solar cell:

  1. It converts light energy into electric energy.
  2. t is useful to produce electricity in remote areas and also for providing electricity for satellites, space probes and space stations.

ii. Photo resistor: It changes its resistance when light is incident on it.

iii. Bi-polar junction transistor:

  1. These are devices with two junctions and three terminals.
  2. A transistor can be a p-n-p or n-p-n transistor.
  3. Conduction takes place with holes and electrons.
  4. Many other types of transistors are designed and fabricated to suit specific requirements.
  5. They are used in almost all semiconductor devices.

iv. Photodiode: It conducts when illuminated with light.

v. LED (Light Emitting Diode):

  1. It emits light when current passes through it.
  2. House hold LED lamps use similar technology.
  3. They consume less power, are smaller in size and have a longer life and are cost effective.

vi. Solid State Laser: It is a special type of LED. It emits light of specific frequency. It is smaller in size and consumes less power.

vii. Integrated Circuits (ICs): A small device having hundreds of diodes and transistors performs the work of a large number of electronic circuits.

Question 60.
Explain any four application of p-n junction diode.
Answer:
1. Solar cell:

  1. It converts light energy into electric energy.
  2. It is useful to produce electricity in remote areas and also for providing electricity for satellites, space probes and space stations.

ii. Photodiode: It conducts when illuminated with light.

iii. LED (Light Emitting Diode):

  1. It emits light when current passes through it.
  2. House hold LED lamps use similar technology.
  3. They consume less power, are smaller in size and have a longer life and are cost effective.

iv. Solid State Laser: It is a special type of LED. It emits light of specific frequency. It is smaller in size and consumes less power.

Question 61.
What is thermistor?
Answer:
Thermistor is a temperature sensitive resistor. Its resistance changes with change in its temperature.

Question 62.
What are different ty pes of thermistor and what are their applications?
Answer:
There are two types of thermistor:
i. NTC (Negative Thermal Coefficient) thermistor: Resistance of a NTC thermistor decreases with increase in its temperature. Its temperature coefficient is negative. They are commonly used as temperature sensors and also in temperature control circuits.

ii. PTC (Positive Thermal Coefficient) thermistor: Resistance of a PTC thermistor increases with increase in its temperature. They are commonly used in series with a circuit. They are generally used as a reusable fuse to limit current passing through a circuit to protect against over current conditions, as resettable fuses.

Question 63.
How are thermistors fabricated?
Answer:
Thermistors are made from thermally sensitive metal oxide semiconductors. Thermistors are very sensitive to changes in temperature.

Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors

Question 64.
Enlist any two features of thermistor.
Answer:

  1. A small change in surrounding temperature causes a large change in their resistance.
  2. They can measure temperature variations of a small area due to their small size.

Question 65.
Write a note on:
i. Electric devices
ii. Electronic devices
Answer:
i. Electric devices:

  1. These devices convert electrical energy into some other form.
  2. Examples: Fan, refrigerator, geyser etc. Fan converts electrical energy into mechanical energy. A geyser converts it into heat energy.
  3. They use good conductors (mostly metals) for conduction of electricity.
  4. Common working range of currents for electric circuits is milliampere (mA) to ampere.
  5. Their energy consumption is also moderate to high. A typical geyser consumes about 2.0 to 2.50 kW of power.
  6. They are moderate to large in size and are costly.

ii. Electronic devices:

  1. Electronic circuits work with control or sequential changes in current through a cell.
  2. A calculator, a cell phone, a smart watch or the remote control of a TV set are some of the electronic devices.
  3. Semiconductors are used to fabricate such devices.
  4. Common working range of currents for electronic circuits it is nano-ampere to µA.
  5. They consume very low energy. They are very compact, and cost effective.

Question 66.
Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductor to get p-n junction?
Answer:

  1. No. Any slab, howsoever flat, will have roughness much larger than the inter-atomic crystal spacing (~2 to 3 Å).
  2. Hence, continuous contact at the atomic level will not be possible. The junction will behave as a discontinuity for the flowing charge carriers.

Question 67.
What is Avalanche breakdown and zener breakdown?
Answer:
i. Avalanche breakdown: In high reverse bias, minority carriers acquire sufficient kinetic energy and collide with a valence electron. Due to collisions the covalent bond breaks. The valence electron enters conduction band. A breakdown occurring in such a manner is avalanche breakdown. It occurs with lightly doped p-n junctions.

ii. Zener breakdown: It occurs in specially designed and highly doped p-n junctions, viz., zener diodes. In this case, covalent bonds break directly due to application of high electric field. Avalanche breakdown voltage is higher than zener voltage.

Question 68.
Indicators on platform, digital clocks, calculators make use of seven LEDs to indicate a number. How do you think these LEDs might be arranged?
Answer:
i. The indicators on platforms, digital clocks, calculators are made using seven LEDs arranged in such a way that when provided proper signal they light up displaying desired alphabet or number.

ii. This arrangement of LEDs is called Seven Segment Display.
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 32

Multiple Choice Questions

Question 1.
The number of electrons in the valence shell of semiconductor is ……………
(A) less than 4
(B) equal to 4
(C) more than 4
(D) zero
Answer:
(B) equal to 4

Question 2.
If the temperature of semiconductor is increased, the number of electrons in the valence band will ……………….
(A) decrease
(B) remains same
(C) increase
(D) either increase or decrease
Answer:
(A) decrease

Question 3.
When N-type semiconductor is heated, the ……………..
(A) number of electrons and holes remains same.
(B) number of electrons increases while that of holes decreases.
(C) number of electrons decreases while that of holes increases.
(D) number of electrons and holes increases equally.
Answer:
(D) number of electrons and holes increases equally.

Question 4.
In conduction band of solid, there is no electron at room temperature. The solid is ……………
(A) semiconductors
(B) insulator
(C) conductor
(D) metal
Answer:
(B) insulator

Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors

Question 5.
In the crystal of pure Ge or Si, each covalent bond consists of …………..
(A) 1 electron
(B) 2 electrons
(C) 3 electrons
(D) 4 electrons
Answer:
(B) 2 electrons

Question 6.
A pure semiconductor is ……………..
(A) an extrinsic semiconductor
(B) an intrinsic semiconductor
(C) p-type semiconductor
(D) n-type semiconductor
Answer:
(B) an intrinsic semiconductor

Question 7.
For an extrinsic semiconductor, the valency of the donor impurity is …………..
(A) 2
(B) 1
(C) 4
(D) 5
Answer:
(D) 5

Question 8.
In a semiconductor, acceptor impurity is
(A) antimony
(B) indium
(C) phosphorous
(D) arsenic
Answer:
(B) indium

Question 9.
What are majority carriers in a semiconductor?
(A) Holes in n-type and electrons in p-type.
(B) Holes in n-type and p-type both.
(C) Electrons in n-type and p-type both.
(D) Holes in p-type and electrons in n-type.
Answer:
(D) Holes in p-type and electrons in n-type.

Question 10.
When a hole is produced in P-type semiconductor, there is ……………….
(A) extra electron in valence band.
(B) extra electron in conduction band.
(C) missing electron in valence band.
(D) missing electron in conduction band.
Answer:
(C) missing electron in valence band.

Question 11.
The number of bonds formed in p-type and n-type semiconductors are respectively
(A) 4,5
(B) 3,4
(C) 4,3
(D) 5,4
Answer:
(B) 3,4

Question 12.
The movement of a hole is brought about by the valency being filled by a ………………..
(A) free electrons
(B) valence electrons
(C) positive ions
(D) negative ions
Answer:
(B) valence electrons

Question 13.
The drift current in a p-n junction is
(A) from the p region to n region.
(B) from the n region to p region.
(C) from n to p region if the junction is forward biased and from p to n region if the junction is reverse biased.
(D) from p to n region if the junction is forward biased and from n to p region if the junction is reverse biased.
Answer:
(B) from the n region to p region.

Question 14.
If a p-n junction diode is not connected to any circuit, then
(A) the potential is same everywhere.
(B) potential is not same and n-type side has lower potential than p-type side.
(C) there is an electric field at junction direction from p-type side to n-type side.
(D) there is an electric field at the junction directed from n-type side to p-type side.
Answer:
(D) there is an electric field at the junction directed from n-type side to p-type side.

Question 15.
In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(A) free electrons in the n-region attract them.
(B) they move across the junction by the potential difference.
(C) hole concentration in p-region is more as compared to n-region.
(D) all the above.
Answer:
(C) hole concentration in p-region is more as compared to n-region.

Question 16.
The width of depletion region ……………
(A) becomes small in forward bias of diode
(B) becomes large in forward bias of diode
(C) is not affected upon by the bias
(D) becomes small in reverse bias of diode
Answer:
(A) becomes small in forward bias of diode

Question 17.
For p-n junction in reverse bias, which of the following is true?
(A) There is no current through P-N junction due to majority carriers from both regions.
(B) Width of potential barriers is small and it offers low resistance.
(C) Current is due to majority carriers.
(D) Both (B) and (C)
Answer:
(A) There is no current through P-N junction due to majority carriers from both regions.

Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors

Question 18.
In the circuit shown below Di and D2 are two silicon diodes. The current in the circuit is …………….
Maharashtra Board Class 11 Physics Important Questions Chapter 14 Semiconductors 33
(A) 2 A
(B) 2 mA
(C) 0.8 mA
(D) very small (approx 0)
Answer:
(D) very small (approx 0)

Question 19.
For an ideal junction diode,
(A) forward bias resistance is infinity.
(B) forward bias resistance is zero.
(C) reverse bias resistance is infinity.
(D) both (B) and (C).
Answer:
(D) both (B) and (C).

Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 14 Semiconductors Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 14 Semiconductors

1. Choose the correct option.

Question 1.
Electric conduction through a semiconductor is due to:
(A) Electrons
(B) holes
(C) none of these
(D) both electrons and holes
Answer:
(D) both electrons and holes

Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors

Question 2.
The energy levels of holes are:
(A) in the valence band
(B) in the conduction band
(C) in the band gap but close to valence band
(D) in the band gap but close to conduction band
Answer:
(C) in the band gap but close to valence band

Question 3.
Current through a reverse biased p-n junction, increases abruptly at:
(A) Breakdown voltage
(B) 0.0 V
(C) 0.3V
(D) 0.7V
Answer:
(A) Breakdown voltage

Question 4.
A reverse biased diode, is equivalent to:
(A) an off switch
(B) an on switch
(C) a low resistance
(D) none of the above
Answer:
(A) an off switch

Question 5.
The potential barrier in p-n diode is due to:
(A) depletion of positive charges near the junction
(B) accumulation of positive charges near the junction
(C) depletion of negative charges near the junction,
(D) accumulation of positive and negative charges near the junction
Answer:
(D) accumulation of positive and negative charges near the junction

2. Answer the following questions.

Question 1.
What is the importance of energy gap in a semiconductor?
Answer:

  1. The gap between the bottom of the conduction band and the top of the valence band is called the energy gap or the band gap.
  2. This band gap is present only in semiconductors and insulators.
  3. Magnitude of the band gap plays a very important role in the electronic properties of a solid.
  4. Band gap in semiconductors is of the order of 1 eV.
  5. If electrons in valence band of a semiconductor are provided with energy more than band gap energy (in the form of thermal energy or electrical energy), then the electrons get excited and occupy energy levels in conduction band. These electrons can easily take part in conduction.

Question 2.
Which element would you use as an impurity to make germanium an n-type semiconductor?
Answer:
Germanium can be made an n-type semiconductor by doping it with pentavalent impurity, like phosphorus (P), arsenic (As) or antimony (Sb).

Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors

Question 3.
What causes a larger current through a p-n junction diode when forward biased?
Answer:
In case of forward bias the width of the depletion region decreases and the p-n junction offers a low resistance path allowing a high current to flow across the junction.

Question 4.
On which factors does the electrical conductivity of a pure semiconductor depend at a given temperature?
Answer:
For pure semiconductor, the number density of free electrons and number density of holes is equal. Thus, at a given temperature, the conductivity of pure semiconductor depends on the number density of charge carriers in the semiconductor.

Question 5.
Why is the conductivity of a n-type semiconductor greater than that of p-type semiconductor even when both of these have same level of doping?
Answer:

  1. In a p-type semiconductor, holes are majority charge carriers.
  2. When a p-type semiconductor is connected to terminals of a battery, holes, which are not actual charges, behave like a positive charge and get attracted towards the negative terminal of the battery.
  3. During transportation of hole, there is an indirect movement of electrons.
  4. The drift speed of these electrons is less than that in the n-type semiconductors. Mobility of the holes is also less than that of the electrons.
  5. As, electrical conductivity depends on the mobility of charge carriers, the conductivity of a n-type semiconductor is greater than that of p-type semiconductor even when both of these have same level of doping.

3. Answer in detail.

Question 1.
Explain how solids are classified on the basis of band theory of solids.
Answer:
i. The solids can be classified into conductors, insulators and semiconductors depending on the distribution of electron energies in each atom.

ii. As an outcome of the small distances between atoms, the resulting interaction amongst electrons and the Pauli’s exclusion principle, energy bands are formed in the solids.

iii. In metals, conduction band and valence band overlap. However, in a semiconductor or an insulator, there is gap between the bottom of the conduction band and the top of the valence band. This is called the energy gap or the band gap.
Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors 1

iv. For metals, the valence band and the conduction band overlap and there is no band gap as shown in figure (b). Therefore, electrons can easily gain electrical energy when an external electric field is applied and are easily available for conduction.
Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors 2

v. In case of semiconductors, the band gap is fairly small, of the order of 1 eV or less as shown in figure (c). Hence, with application of external electric field, electrons get excited and occupy energy levels in conduction band. These can take part in conduction easily.
Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors 3

vi. Insulators, on the contrary, have a wide gap between valence band and conduction band of the order of 5 eV (for diamond) as shown in figure (d). Therefore, electrons find it very difficult to gain sufficient energy to occupy energy levels in conduction band.
Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors 4

vii. Thus, an energy band gap plays an important role in classifying solids into conductors, insulators and semiconductors based on band theory of solids.

Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors

Question 2.
Distinguish between intrinsic semiconductors and extrinsic semiconductors
Answer:

Intrinsic semiconductors Extrinsic semiconductors
1. A pure semiconductor is known as intrinsic semiconductors. The semiconductor, resulting
2. Their conductivity is low Their conductivity is high even at room temperature.
3. Its electrical conductivity is a function of temperature alone. Its electrical conductivity depends upon the temperature as well as on the quantity of impurity atoms doped in the structure.
4. The number density of holes (nh) is same as the number density of free electron (ne) (nh = ne). The number density of free electrons and number density of holes are unequal.

Question 3.
Explain the importance of the depletion region in a p-n junction diode.
Answer:
i. The region across the p-n junction where there are no charges is called the depletion layer or the depletion region.

ii. During diffusion of charge carriers across the junction, electrons migrate from the n-side to the p-side of the junction. At the same time, holes are transported from p-side to n-side of the junction.

iii. As a result, in the p-type region near the junction there are negatively charged acceptor ions, and in the n-type region near the junction there are positively charged donor ions.

iv. The potential barrier thus developed, prevents continuous flow of charges across the junction. A state of electrostatic equilibrium is thus reached across the junction.

v. Free charge carriers cannot be present in a region where there is a potential barrier. This creates the depletion region.

vi. In absence of depletion region, all the majority charge carriers from n-region (i.e., electron) will get transferred to the p-region and will get combined with the holes present in that region. This will result in the decreased efficiency of p-n junction.

vii. Hence, formation of depletion layer across the junction is important to limit the number of majority carriers crossing the junction.

Question 4.
Explain the I-V characteristic of a forward biased junction diode.
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors 5

  1. Figure given below shows the I-V characteristic of a forward biased diode.
  2. When connected in forward bias mode, initially, the current through diode is very low and then there is a sudden rise in the current.
  3. The point at which current rises sharply is shown as the ‘knee’ point on the I-V characteristic curve.
  4. The corresponding voltage is called the knee voltage. It is about 0.7 V for silicon and 0.3 V for germanium.
  5. A diode effectively becomes a short circuit above this knee point and can conduct a very large current.
  6. To limit current flowing through the diode, resistors are used in series with the diode.
  7. If the current through a diode exceeds the specified value, the diode can heat up due to the Joule’s heating and this may result in its physical damage.

Maharashtra Board Class 11 Physics Solutions Chapter 14 Semiconductors

Question 5.
Discuss the effect of external voltage on the width of depletion region of a p-n junction.
Answer:

  1. A p-n junction can be connected to an external voltage supply in two possible ways.
  2. A p-n junction is said to be connected in a forward bias when the p-region connected to the positive terminal and the n-region is connected to the negative terminal of an external voltage source.
  3. In forward bias connection, the external voltage effectively opposes the built-in potential of the junction. The width of depletion region is thus reduced.
  4. The second possibility of connecting p-n junction is in reverse biased electric circuit.
  5. In reverse bias connection, the p-region is connected to the negative terminal and the n-region is connected to the positive terminal of the external voltage source. This external voltage effectively adds to the built-in potential of the junction. The width of potential barrier is thus increased

11th Physics Digest Chapter 14 Semiconductors Intext Questions and Answers

Internet my friend (Textbookpage no. 256)

i. https://www.electronics-tutorials.ws/diode
ii. https://www.hitachi-hightech.com
iii. https://nptel.ac.in/courses
iv. https://physics.info/semiconductors
v. http://hyperphysics.phy- astr.gsu.edu/hbase/Solids/semcn.html

[Students are expected to visit above mentioned links and collect more information regarding semiconductors.]

Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System

Question 1.
Describe Gauss’ law of electrostatics in brief.
Answer:
i. Gauss’ law of electrostatics states that electric flux through any closed surface S is equal to the total electric charge Qin enclosed by the surface divided by so.
\(\int \vec{E} \cdot \overrightarrow{\mathrm{dS}}=\frac{\mathrm{Q}_{\text {in }}}{\varepsilon_{0}}\)
where, \(\vec{E}\) is the electric field and e0 is the permittivity of vacuum. The integral is over a closed surface S.

ii. Gauss’ law describes the relation between an electric charge and electric field it produces.

Question 2.
Describe Gauss’ law of magnetism in brief.
Answer:
i. Gauss’ law for magnetism states that magnetic monopoles which are thought to be magnetic charges equivalent to the electric charges, do not exist. Magnetic poles always occur in pairs.

ii. This means, magnetic flux through a closed surface is always zero, i.e., the magnetic field lines are continuous closed curves, having neither beginning nor end.
\(\int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{dS}}\) = 0
where, B is the magnetic field. The integral is over a closed surface S.

Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System

Question 3.
Describe Faraday’s law along with Lenz’s law.
Answer:
i. Faraday’s law states that, time varying magnetic field induces an electromotive force (emf) and an electric field.

ii. Whereas, Lenz’s law states that, the direction of the induced emf is such that the change is opposed.

iii. According to Faraday’s law with Lenz’s law,
\(\int \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{d} l}=-\frac{\mathrm{d} \phi_{\mathrm{m}}}{\mathrm{dt}}\)
where, øm is the magnetic flux and the integral is over a closed loop.

Question 4.
What does Ampere’s law describe?
Answer:
Ampere’s law describes the relation between the induced magnetic field associated with a loop and the current flowing through the loop.

Question 5.
Describe Ampere-Maxwell law in brief.
Answer:
According to Ampere-Maxwell law, magnetic field is generated by moving charges and also by varying electric fields.
\(\int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}=\mu_{0} \mathrm{I}+\varepsilon_{0} \mu_{0} \frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}\)
where, p0 and e0 are the permeability and permittivity of vacuum respectively and the integral is over a closed loop, I is the current flowing through the loop, E is the electric flux linked with the circuit.

Question 6.
What are Maxwell’s equations for charges and currents in vacuum?
Answer:
\(\int \vec{E} \cdot \overrightarrow{\mathrm{dS}}=\frac{\mathrm{Q}_{\text {in }}}{\varepsilon_{0}}\)
\(\int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{dS}}\) = 0
\(\int \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{d} l}=-\frac{\mathrm{d} \phi_{\mathrm{m}}}{\mathrm{dt}}\)
\(\int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}=\mu_{0} \mathrm{I}+\varepsilon_{0} \mu_{0} \frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}\)

Question 7.
Explain the origin of displacement current?
Answer:

  1. Maxwell pointed a major flaw in the Ampere’s law for time dependant fields.
  2. He noticed that the magnetic field can be generated not only by electric current but also by changing electric field.
  3. Therefore, he added one more term to the equation describing Ampere’s law. This term is called the displacement current.

Question 8.
In the following table, every entry on the left column can match with any number of entries on the right side. Pick up all those and write respectively against (i), (ii), and (iii).

Name of the Physicist Work
i. H. Hertz a. Existence of EM waves
ii. J. Maxwell b. Properties of EM waves
iii. G. Marconi c. Wireless communication
d. Displacement current

Answer:
(i – a, b), (ii – d), (iii – c)

Question 9.
Varying electric and magnetic fields regenerate each other. Explain.
Answer:

  1. According to Maxwell’s theory, accelerated charges radiate EM waves.
  2. Consider a charge oscillating with some frequency. This produces an oscillating electric field in space, which produces an oscillating magnetic field which in turn is a source of oscillating electric field.
  3. Thus, varying electric and magnetic fields regenerate each other.

Question 10.
Draw a neat diagram representing electromagnetic wave propagating along Z-axis.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System 1

Question 11.
How can energy be transported in the form of EM waves?
Answer:

  1. Maxwell proposed that an oscillating electric charge radiates energy in the form of EM wave.
  2. EM waves are periodic changes in electric and magnetic fields, which propagate through space.
  3. Thus, energy can be transported in the form of EM waves.

Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System

Question 12.
State the main characteristics of EM waves.
Answer:
i. The electric and magnetic fields, \(\vec{E}\) and \(\vec{B}\) are always perpendicular to each other and also to the direction of propagation of the EM wave. Thus, the EM waves are transverse waves.

ii. The cross product (\(\vec{E}\) × \(\vec{B}\)) gives the direction in which the EM wave travels. (\(\vec{E}\) × \(\vec{B}\)) also gives the energy carried by EM wave.

iii. The \(\vec{E}\) and \(\vec{B}\) fields vary sinusoidally and are in phase.

iv. EM waves are produced by accelerated electric charges.

v. EM waves can travel through free space as well as through solids, liquids and gases.

vi. In free space, EM waves travel with velocity c, equal to that of light in free space.
c = \(\frac {1}{\sqrt{µ_0ε_0}}\) = 3 × 108 m/s,
where µ0 is permeability and ε0 is permittivity of free space.

vii. In a given material medium, the velocity (vm) of EM waves is given by vm = \(\frac {1}{\sqrt{µε}}\)
where µ is the permeability and ε is the permittivity of the given medium.

viii. The EM waves obey the principle of superposition.

ix. The ratio of the amplitudes of electric and magnetic fields is constant at any point and is equal to the velocity of the EM wave.
\( \left|\overrightarrow{\mathrm{E}}_{0}\right|=\mathrm{c}\left|\overrightarrow{\mathrm{B}}_{0}\right| \text { or } \frac{\left|\overrightarrow{\mathrm{E}}_{0}\right|}{\left|\overrightarrow{\mathrm{B}_{0}}\right|}=\mathrm{c}=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}\)
where, |\(\vec{E_0}\)| and |\(\vec{B_0}\)| are the amplitudes of \(\vec{E}\) and \(\vec{B}\) respectively.

x. As the electric field vector (\(\vec{E_0}\)) is more prominent than the magnetic field vector (\(\vec{B_0}\)), it is responsible for optical effects due to EM waves. For this reason, electric vector is called light vector.

xi. The intensity of a wave is proportional to the square of its amplitude and is given by the equations
\(\mathrm{I}_{\mathrm{E}}=\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2}, \mathrm{I}_{\mathrm{B}}=\frac{1}{2} \frac{\mathrm{B}_{0}^{2}}{\mu_{0}}\)

xii. The energy of EM waves is distributed equally between the electric and magnetic fields. IE = IB.

Question 13.
Give reason: Electric vector is called light vector.
Answer:
As the electric field vector (\(\vec{E_0}\)) is more prominent than the magnetic field vector (\(\vec{B_0}\)), it is responsible for optical effects due to EM waves. For this reason, electric vector is called light vector.

Question 14.
Explain the equations describing an EM wave.
Answer:
i. In an EM wave, the magnetic field and electric field both vary sinusoidally with x.

ii. For a wave travelling along X-axis having \(\vec{E}\) along Y-axis and \(\vec{B}\) along the Z-axis,
Ey = E0 sin (kx – ωt)
Bz = B0 sin (kx – ωt)
where, E0 is the amplitude of the electric field (Ey) and B0 is the amplitude of the magnetic field (Bz).

iii. The propagation constant is given by k = \(\frac {2π}{λ}\) and λ is the wavelength of the wave. The angular frequency of oscillations is given by ω = 2πv, v being the frequency of the wave.
Hence, Ey = E0 sin (\(\frac {2πx}{λ}\) – 2πvt)
Bz = B0 sin (\(\frac {2πx}{λ}\) – 2πvt)

iv. Both the electric and magnetic fields attain their maximum or minimum values at the same time and at the same point in space, i.e., \(\vec{E}\) and \(\vec{B}\) oscillate in phase with the same frequency.

Question 15.
A radio wave of frequency of 1.0 × 107 Hz propagates with speed 3 × 108 m/s. Calculate its wavelength.
Answer:
Given: v= 1.0 × 107 Hz, c = 3 × 108 m/s
To find: Wavelength (λ)
Formula: vλ
Calculation: From formula,
λ = \(\frac {c}{v}\) = \(\frac {3×10^8}{1.0×10^7}\) = 30 m

Question 16.
A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Answer:
Given: V1 = 7.5 MHz = 7.5 × 106 Hz,
V1 = 12 MHz = 12 × 106 Hz.
To find: Wavelength band
Formula: λ = \(\frac {c}{v}\)
Calculation: From formula,
V1 = \(\frac {3×10^8}{7.5×10^6}\) = 40 m
V1 = \(\frac {3×10^8}{12×10^6}\) = 25 m
∴ Wavelength band = 40 m to 25 m

Question 17.
Calculate the ratio of the intensities of the two waves, if amplitude of first beam of light is 1.5 times the amplitude of second beam of light.
Answer:
a1 = 1.5 a2
To find: \(\frac {I_1}{I-2}\)
Formula: I ∝ a²
Calculation: From formula,
\(\frac{I_{1}}{I_{2}}=\left(\frac{a_{1}}{a_{2}}\right)^{2}=\left(\frac{1.5 a_{2}}{a_{2}}\right)^{2}\) = (1.5)² = 2.25

Question 18.
A beam of red light has an amplitude 2.5 times the amplitude of second beam of the same colour. Calculate the ratio of the intensities of the two waves.
Answer:
a1 = 2.5 a2
To find: \(\frac {I_1}{I-2}\)
Formula: I ∝ a²
Calculation: From formula,
\(\frac{I_{1}}{I_{2}}=\left(\frac{a_{1}}{a_{2}}\right)^{2}=\left(\frac{2.5 a_{2}}{a_{2}}\right)^{2}\) = (2.5)² = 6.25

Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System

Question 19.
Calculate the velocity of EM waves in vacuum.
Answer:
Given: ε0 = 8.85 × 10-12 C²/Nm²
µ0 = 4π × 10-7 Tm/A
To find: Velocity of EM waves (c)
Formula: c = \(\frac {1}{\sqrt{µ_0ε_0}}\)
Calculation: From formula,
Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System 2
………… (Taking square roots using log table)
= 0.2998 × 109 ≈ 3 × 108 m/s

Question 20.
In free space, an EM wave of frequency 28 MHz travels along the X-direction. The amplitude of the electric field is E = 9.6 V/m and its direction is along the Y-axis. What is amplitude and direction of magnetic field B?
Answer:
Given: v = 28 MHz, E = 9.6 V/m,
c = 3 × 108 m/s
To find:
i. Amplitude of magnetic field (B)
ii. Direction of B
Formula:
|B| = \(\frac {|E|}{c}\)
Calculation: From formula,
|B| = \(\frac {9.6}{3×10^8}\) = 3.2 × 10-8 T
Since that E is along Y-direction and the wave propagates along X-axis. The magnetic induction, B should be in a direction perpendicular to both X and Y axes, i.e., along the Z-direction.

Question 21.
An EM wave of frequency 50 MHz travels in vacuum along the positive X-axis and \(\vec{E}\) at a particular point, x and at a particular instant of time t is 9.6 j V/m. Find the magnitude and direction of \(\vec{B}\) at this point x and at instant of time t.
Answer:
Given: \(\vec{E}\) = 9.6 j V/m
i. e., Electric field E is directed along +Y axis Magnitude of \(\vec{B}\).
|B| = \(\frac {|E|}{c}\) = \(\frac {9.6}{3×10^8}\) = 3.2 × 10-8 T
As the wave propagates along +X axis and E is along +Y axis, direction of B will be along +Z-axis i.e. B = 3.2 × 10-8 \(\hat{k}\)T.

Question 22.
A plane electromagnetic wave travels in vacuum along Z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?
Answer:
Since the electromagnetic waves are transverse in nature, the electric and magnetic field vectors are mutually perpendicular to each other as well as perpendicular to the direction of propagation of wave.
As the wave is travelling along Z-direction,

\(\vec{E}\) and \(\vec{B}\) are in XY plane.
For v = 30 MHz = 30 × 106 Hz
Wavelength, λ = \(\frac {c}{v}\) = \(\frac {3×10^8}{30×10^6}\) = 10 m

Question 23.
For an EM wave propagating along X direction, the magnetic field oscillates along the Z-direction at a frequency of 3 × 1010 Hz and has amplitude of 10-9 T.
i. What is the wavelength of the wave?
ii. Write the expression representing the corresponding electric field.
Answer:
Given: v = 3 × 1010 Hz, B = 10-9 T
i. For wavelength of the wave:
λ = \(\frac{\mathrm{c}}{\mathrm{v}}=\frac{3 \times 10^{8}}{3 \times 10^{10}}\) = 10-2 m

ii. Since B acts along Z-axis, E acts along Y-axis. Expression representing the oscillating electric field is
Ey = E0 sin (kx – ωt)
Ey = E0 sin [(\(\frac {2π}{λ}\))x – (2πv)t]
Ey = E0 sin 2π [\(\frac {x}{λ}\) – vt]
Ey = E0 sin 2π [\(\frac {x}{10^{-2}}\) – 3 × 1010 t]
Ey = E0 sin 2π [100x – 3 × 1010 t] V/m

Question 24.
The magnetic field of an EM wave travelling along X-axis is
\(\vec{B}\) = \(\hat{k}\) [4 × 10-4 sin (ωt – kx)]. Here B is in tesla, t is in second and x is in m. Calculate the peak value of electric force acting on a particle of charge 5 µC travelling with a velocity of 5 × 105 m/s along the Y-axis.
Answer:
Expression for EM wave travelling along
X-axis, \(\vec{B}\) = \(\hat{k}\) [4 × 10-4 sin (ωt – kx)]
Here, B0 = 4 × 10-4
Given: q = 5 µC = 5 × 10-6 C
v = 5 × 105 m/s along Y-axis
∴ E0 = cB0 = 3 × 108 × 4 × 10-4
= 12 × 104 N/C
Maximum electric force = qE0
= 5 × 10-6 × 12 × 104
= 0.6 N

Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System

Question 25.
The amplitude of the magnetic field part of harmonic electromagnetic wave in vaccum is B0 = 510 nT. What is the amplitude of the electric field part of the wave?
Answer:
Given: B0 = 510 nT = 510 × 10-9 T
To find: Amplitude of electric field (E0)
Formula: E0 = B0C
Calculation: From formula,
E0 = 510 × 10-9 × 3 × 108
= 153V/m

Question 26.
Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is v = 50.0 MHz. (i) Determine, B0, ω, k, and λ. (ii) Find expressions for \(\vec{E}\) and \(\vec{B}\).
Solution:
For E0 = 120 N/C, v = 50 MHz = 50 × 106 Hz
i. λ = \(\frac {c}{v}\) = \(\frac {3×10^8}{50×10^6}\) = 6 m
B0 = \(\frac {E_0}{v}\) = \(\frac {120}{3×10^8}\) = 4 × 10-7 T = 400 nT
k = \(\frac {2π}{λ}\) = \(\frac {2π}{6}\) = 1.0472 rad/m
ω = 2πv = 2π × 50 × 106
= 3.14 × 108 rad/s.

ii. Assuming motion of em wave along X-axis, expression for electric field vector may lie along Y-axis,
∴ \(\vec{E}\) = E0 sin (kx – ωt)
= 120 sin (1.0472 × – 3.14 × 108 t) \(\hat{j}\) N/C
Also, magnetic field vector will lie along Z-axis, expression for magnetic field vector,
∴ \(\vec{E}\) = B0 sin (kx – ωt)
= 4 × 10-7 sin (1.0472 × – 3.14 × 108 t) \(\hat{k}\) T.

Question 27.
What is electromagnetic spectrum?
Answer:
The orderly distribution (sequential arrangement) of EM waves according to their wavelengths (or frequencies) in the form of distinct groups having different properties is called the EM spectrum.

Question 28.
State various units used for frequency of electromagnetic waves.
Answer:

  1. SI unit of frequency of electromagnetic waves is hertz (Hz).
  2. Higher frequencies are represented by kHz, MHz, GHz etc.
    [Note: 1 kHz = 10³ Hz, 1 MHz =106 Hz. 1 GHz = 109 Hz]

Question 29.
State different units used for wavelength of electromagnetic waves.
Answer:

  1. The SI unit of wavelength of electromagnetic waves is metre (m).
  2. Small wavelengths are represented by micrometre (µm), angstrom (Å), nanometre (nm) etc.
    [Note:l A = 10-10 m = 10-8 cm, 1 µm = 10-6 m, 1 nm = 10-9 m.]

Question 30.
How are radio waves produced? State their properties and uses.
Answer:
Production:

  1. Radio waves are produced by accelerated motion of charges in a conducting wire. The frequency of waves produced by the circuit depends upon the magnitudes of the inductance and the capacitance.
  2. Thus, by choosing suitable values of the inductance and the capacitance, radio waves of desired frequency can be produced.

Properties:

  1. They have very long wavelengths ranging from a few centimetres to a few hundreds of kilometres.
  2. The frequency range of AM band is 530 kHz to 1710 kHz. Frequency of the waves used for TV-transmission range from 54 MHz to 890 MHz, while those for FM radio band range from 88 MHz to 108 MHz.

Uses:

  1. Radio waves are used for wireless communication purpose.
  2. They are used for radio broadcasting and transmission of TV signals.
  3. Cellular phones use radio waves to transmit voice communication in the ultra high frequency (UHF) band.

Question 31.
How are microwaves produced? State their properties and uses.
Answer:
Production:

  1. Microwaves are produced by oscillator electric circuits containing a capacitor and an inductor.
  2. They can be produced by special vacuum tubes.

Properties:

  1. They heat certain substances on which they are incident.
  2. They can be detected by crystal detectors.

Uses:

  1. Used for the transmission of TV signals.
  2. Used for long distance telephone communication.
  3. Microwave ovens are used for cooking.
  4. Used in radar systems for the location of distant objects like ships, aeroplanes etc,
  5. They are used in the study of atomic and molecular structure.

Question 32.
How are infrared waves produced? State their properties and uses.
Answer:
Production:

  1. All hot bodies are sources of infrared rays. About 60% of the solar radiations are infrared in nature.
  2. Thermocouples, thermopile and bolometers are used to detect infrared rays.

Properties:

  1. When infrared rays are incident on any object, the object gets heated.
  2. These rays are strongly absorbed by glass.
  3. They can penetrate through thick columns of fog, mist and cloud cover.

Uses:

  1. Used in remote sensing.
  2. Used in diagnosis of superficial tumours and varicose veins.
  3. Used to cure infantile paralysis and to treat sprains, dislocations and fractures.
  4. They are used in solar water heaters and solar cookers.
  5. Special infrared photographs of the body called thermograms, can reveal diseased organs because these parts radiate less heat than the healthy organs.
  6. Infrared binoculars and thermal imaging cameras are used in military applications for night vision.
  7. Used to keep green house warm.
  8. Used in remote controls of TV, VCR, etc.

Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System

Question 33.
Write short note on visible light.
Answer:

  1. It is the most familiar form of EM waves.
  2. These waves are detected by human eye. Therefore this wavelength range is called the visible light.
  3. The visible light is emitted due to atomic excitations.
  4. Visible light emitted or reflected from objects around us provides us information about those objects and hence about the surroundings.
  5. Different wavelengths give rise to different colours as shown in the table given below.
    Colour Wavelength
    Violet 380-450 nm
    Blue 450-495 nm
    Green 495-570 nm
    Yellow 570-590 nm
    Orange 590-620 nm
    Red 620-750 nm

Question 34.
How are ultraviolet rays produced? State their properties and uses.
Answer:
Production:

  1. Ultraviolet rays can be produced by the mercury vapour lamp, electric spark and carbon arc lamp.
  2. They can also be obtained by striking electrical discharge in hydrogen and xenon gas tubes.
  3. The Sun is the most important natural source of ultraviolet rays, most of which are absorbed by the ozone layer in the Earth’s atmosphere.

Properties:

  1. They produce fluorescence in certain materials, such as ‘phosphors’.
  2. They cause photoelectric effect.
  3. They cannot pass through glass but pass through quartz, fluorite, rock salt etc.
  4. They possess the property of synthesizing vitamin D, when skin is exposed to them.

Uses:

  1. Ultraviolet rays destroy germs and bacteria and hence they are used for sterilizing surgical instruments and for purification of water.
  2. Used in burglar alarms and security systems.
  3. Used to distinguish real and fake gems.
  4. Used in analysis of chemical compounds.
  5. Used to detect forgery.

Question 35.
How are X-rays produced? State their properties and uses.
Answer:
Production:

  1. German physicist W. C. Rontgen discovered X-rays while studying cathode rays. Hence, X-rays are also called Rontgen rays.
  2. Cathode ray is a stream of electrons emitted by the cathode in a vacuum tube.
  3. X-rays are produced when cathode rays are suddenly stopped by an obstacle.

Properties:

  1. They are high energy EM waves.
  2. They are not deflected by electric and magnetic fields.
  3. X-rays ionize the gases through which they pass.
  4. They have high penetrating power.
  5. Their over dose can kill living plant and animal tissues and hence are harmful.

Uses:

  1. Useful in the study of the structure of crystals.
  2. X-ray photographs are useful to detect bone fracture. X-rays have many other medical uses such as CT scan.
  3. X-rays are used to detect flaws or cracks in metals.
  4. These are used for detection of explosives, opium etc.

Question 36.
X-rays are used in medicine and industry. Explain.
Answer:
X-rays have many practical applications in medicine and industry. Because X-ray photons are of such high energy, they can penetrate several centimetres of solid matter and can be used to visualize the interiors of materials that are opaque to ordinary light.

Question 37.
How are Gamma rays produced? State their properties and uses.
Answer:
Production:
Gamma rays are emitted from the nuclei of some radioactive elements such as uranium, radium etc.

Properties:

  1. They are highest energy (energy range keV – GeV) EM waves.
  2. They are highly penetrating.
  3. They have a small ionising power.
  4. They kill living cells.

Uses:

  1. Used as insecticide and disinfectant for wheat and flour.
  2. Used for food preservation.
  3. Used in radiotherapy for the treatment of cancer and tumour.
  4. They are used to produce nuclear reactions.

Question 38.
Identify the name and part of electromagnetic spectrum and arrange these wavelengths in ascending order of magnitude:
Electromagnetic waves with wavelength
i. λ1 are used by a FM radio station for broad casting.
ii. λ2 are used to detect bone fracture.
iii. λ3 are absorbed by the ozone layer of atmosphere.
iv. λ4 are used to treat muscular strain.
Answer:
i. λ1 belongs to radiowaves.
ii. λ2 belongs to X-rays.
iii. λ3 belongs to ultraviolet rays.
iv. λ4 belongs to infrared radiations.
Ascending order of magnitude of wavelengths:
λ3 < λ3 < λ4 < λ1

Question 39.
Explain how different types of waves emitted by stars and galaxies are observed?
Answer:
i. Stars and galaxies emit different types of waves. Radio waves and visible light can pass through the Earth’s atmosphere and reach the ground without getting absorbed significantly. Thus, the radio telescopes and optical telescopes can be placed on the ground.

ii. All other type of waves get absorbed by the atmospheric gases and dust particles. Hence, the y-ray, X-ray, ultraviolet, infrared, and microwave telescopes are kept aboard artificial satellites and are operated remotely from the Earth.

iii. Even though the visible radiation reaches the surface of the Earth, its intensity decreases to some extent due to absorption and scattering by atmospheric gases and dust particles. Optical telescopes are therefore located at higher altitudes.

Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System

Question 40.
In communication using radiowaves, how are EM waves propagated?
Answer:
In communication using radio waves, an antenna in the transmitter radiates the EM waves, which travel through space and reach the receiving antenna at the other end.

Question 41.
Draw a schematic structure of earth’s atmosphere describing different atmospheric layers.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System 3

Question 42.
Draw a diagram showing different types of EM waves.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System 4

Question 43.
Explain ground wave propagation.
Answer:

  1. When a radio wave from a transmitting antenna propagates near surface of the Earth so as to reach the receiving antenna, the wave propagation is called ground wave or surface wave propagation.
  2. In this mode, radio waves travel close to the surface of the Earth and move along its curved surface from transmitter to receiver.
  3. The radio waves induce currents in the ground and lose their energy by absorption. Therefore, the signal cannot be transmitted over large distances.
  4. Radio waves having frequency less than 2 MHz (in the medium frequency band) are transmitted by ground wave propagation.
  5. This is suitable for local broadcasting only. For TV or FM signals (very high frequency), ground wave propagation cannot be used.

Question 44.
Explain space wave propagation.
Answer:
i. When the radio waves from the transmitting antenna reach the receiving antenna either directly along a straight line (line of sight) or after reflection from the ground or satellite or after reflection from troposphere, the wave propagation is called space wave propagation.

ii. The radio waves reflected from troposphere are called tropospheric waves.

iii. Radio waves with frequency greater than 30 MHz can pass through the ionosphere (60 km – 1000 km) after suffering a small deviation. Hence, these waves cannot be transmitted by space wave propagation except by using a satellite.

iv. Also, for TV signals which have high frequency, transmission over long distance is not possible by means of space wave propagation.

Question 45.
Explain the concept of range of the signal.
Answer:
i. The maximum distance over which a signal can reach is called its range.

ii. For larger TV coverage, the height of the transmitting antenna should be as large as possible. This is the reason why the transmitting and receiving antennas are mounted on top of high rise buildings.

iii. Range is the straight line distance from the point of transmission (the top of the antenna) to the point on Earth where the wave will hit while travelling along a straight line.
Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System 5

iv. Let the height of the transmitting antenna (AA’) situated at A be h. B represents the point on the surface of the Earth at which the space wave hits the Earth.

v. The triangle OA’B is a right angled triangle. From ∆OA’ B,
(OA’)² = A’B² + OB²
(R + h)² = d² + R²
or R² + h² + 2Rh = d² + R² As
h << R, neglecting h²
d ≈ \(\sqrt{2Rh}\)

vi. The range can be increased by mounting the receiver at a height h’ say at a point C on the surface of the Earth. The range increases to d + d’ where d’ is 2Rh’. Thus
Total range = d + d’ = \(\sqrt{2Rh}\) + \(\sqrt{2Rh’}\)

Question 46.
Explain sky wave propagation.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System 6

  1. When radio waves from a transmitting antenna reach the receiving antenna after reflection in the ionosphere, the wave propagation is called sky wave propagation.
  2. The sky waves include waves of frequency between 3 MHz and 30 MHz.
  3. These waves can suffer multiple reflections between the ionosphere and the Earth. Therefore, they can be transmitted over large distances.

Question 47.
What is critical frequency?
Answer:
Critical frequency is the maximum value of the frequency of radio wave which can be reflected back to the Earth from the ionosphere when the waves are directed normally to ionosphere.

Question 48.
What is skip distance (zone)?
Answer:
Skip distance is the shortest distance from a transmitter measured along the surface of the Earth at which a sky wave of fixed frequency (if greater than critical frequency) will be returned to the Earth so that no sky waves can be received within the skip distance.

Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System

Question 49.
A radar has a power of 10 kW and is operating at a frequency of 20 GHz. It is located on the top of a hill of height 500 m. Calculate the maximum distance upto which it can detect object located on the surface of the Earth.
(Radius of Earth = 6.4 × 106 m)
Answer:
Given: h = 500 m, R = 6.4 × 106 m
To find: Maximum distance or range (d)
Formula: d = \(\sqrt{2Rh}\)
Calculation: From formula,
d = \(\sqrt{2Rh}\) = \(\sqrt{2×64×10^6×500}\)
= 8 × 104
= 80 km

Question 50.
If the height of a TV transmitting antenna is 128 m, how much square area can be covered by the transmitted signal if the receiving antenna is at the ground level? (Radius of the Earth = 6400 km)
Answer:
Given: h = 128 m, R = 6400 km – 6400 × 10³ m
To find: Area covered (A)
Formulae: i. d = \(\sqrt{2Rh}\) ii. A = πd²
Calculation:
From formula (i),
Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System 7
= 4.048 × 104
= 40.48 km
From formula (ii).
Area covered = 3.142 × (40.48)²
= antilog [log 3.142 + 2log 40.48]
= antilog [0.4972 + 2(1.6073)]
= antilog [3.7118]
= 5.150 × 10³
= 5150 km²

Question 51.
The height of a transmitting antenna is 68 m and the receiving antenna is at the top of a tower of height 34 m. Calculate the maximum distance between them for satisfactory transmission in line of sight mode. (Radius of Earth = 6400 km)
Answer:
Given: ht = 68 m, hr = 34 m,
R = 6400 km = 6.4 × 106 m
To find: Maximum distance or range (d)
Formula: d = \(\sqrt{2Rh}\)
Calculation:
From formula,
Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System 8
= 2.086 × 104
= 20.86 km
d = dt + dr = 29.51 + 20.86 = 50.37 km

Question 52.
Explain block diagram of communication system.
Answer:
i. There are three basic (essential) elements of every communication system:

  1. Transmitter
  2. Communication channel
  3. Receiver

ii. In a communication system, the transmitter is located at one place and the receiver at another place.
Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System 9

iii. The communication channel is a passage through which signals transfer in between a transmitter and a receiver.

iv. This channel may be in the form of wires or cables, or may also be wireless, depending on the types of communication system.

Question 53.
What are the two different modes of communication?
Answer:
i. There are two basic modes of communication:
a. point to point communication
b. broadcast communication

ii. In point to point communication mode, communication takes place over a link between a single transmitter and a receiver e.g. telephony.

iii. In the broadcast mode, there are large number of receivers corresponding to the single transmitter e.g., Radio and Television transmission.

Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System

Question 54.
Explain the following terms:
i. Signal
ii. Analog signal
iii. Digital signal
iv. Transmitter
v. Transducer
vi. Receiver
vii. Attenuation
viii. Amplification
ix. Range
x. Repeater
Answer:
i. Signal: The information converted into electrical form that is suitable for transmission is called a signal. In a radio station, music and speech are converted into electrical form by a microphone for transmission into space. This electrical form of sound is the signal. A signal can be analog or digital.

ii. Analog signal: A continuously varying signal (voltage or current) is called an analog signal. Since a wave is a fundamental analog signal, sound and picture signals in TV are analog in nature.
Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System 10

iii. Digital signal: A signal (voltage or current) that can have only two discrete values is called a digital signal. For example, a square wave is a digital signal. It has two values viz, +5 V and 0 V.
Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System 11

iv. Transmitter: A transmitter converts the signal produced by a source of information into a form suitable for transmission through a channel and subsequent reception.

v. Transducer: A device that converts one form of energy into another form of energy is called a transducer. For example, a microphone converts sound energy into electrical energy. Therefore, a microphone is a transducer. Similarly, a loudspeaker is a transducer which converts electrical energy into sound energy.

vi. Receiver: The receiver receives the message signal at the channel output, reconstructs it in recognizable form of the original message for delivering it to the user of information.

vii. Attenuation: The loss of strength of the signal while propagating through the channel is known as attenuation. It occurs because the channel distorts, reflects and refracts the signals as it passes through it.

viii. Amplification: Amplification is the process of raising the strength of a signal, using an electronic circuit called amplifier.

ix. Range: The maximum (largest) distance between a source and a destination up to which the signal can be received with sufficient strength is termed as range.

x. Repeater: It is a combination of a transmitter and a receiver. The receiver receives the signal from the transmitter, amplifies it and transmits it to the next repeater. Repeaters are used to increase the range of a communication system.

Question 55.
Explain the role of modulation.
Answer:

  1. Low frequency signals cannot be transmitted over large distances. Because of this, a high frequency wave, called a carrier wave, is used.
  2. Some characteristic (e.g. amplitude, frequency or phase) of this wave is changed in accordance with the amplitude of the signal. This process is known as modulation.
  3. Modulation also helps avoid mixing up of signals from different transmitters as different carrier wave frequencies can be allotted to different transmitters.
  4. Without the use of these waves, the audio signals, if transmitted directly by different transmitters, would have got mixed up.

Question 56.
Explain the different types of modulation.
Answer:

  1. Modulation can be done by modifying the amplitude (amplitude modulation), frequency (frequency modulation), and phase (phase modulation) of the carrier wave in proportion to the intensity of the signal wave keeping the other two properties same.
  2. The carrier wave is a high frequency wave while the signal is a low frequency wave.
  3. Waveform (a) in the figure shows a carrier wave and waveform (b) shows the signal.
  4. Amplitude modulation, frequency modulation and phase modulation of carrier waves are shown in waveforms (c), (d) and (e) respectively.
    Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System 12

Question 57.
State advantages and disadvantages of amplitude modulation.
Answer:
Advantages:

  1. It is simple to implement.
  2. It has large range.
  3. It is cheaper.

Disadvantages:

  1. It is not very efficient as far as power usage is concerned.
  2. It is prone to noise.
  3. The reproduced signal may not exactly match the original signal.

In spite of this, these are used for commercial broadcasting in the long, medium and short wave bands.

Question 58.
State uses and limitations of frequency modulation.
Answer:

  1. Frequency modulation (FM) is more complex as compared to amplitude modulation and, therefore is more difficult to implement.
  2. However, its main advantage is that it reproduces the original signal closely and is less susceptible to noise.
  3. This modulation is used for high quality broadcast transmission.

Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System

Question 59.
State benefits of phase modulation.
Answer:

  1. Phase modulation (PM) is easier than frequency modulation.
  2. It is used in determining the velocity of a moving target which cannot be done using frequency modulation.

Question 60.
Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs.
i. 21 cm (wavelength emitted by atomic hydrogen in interstellar space).
ii. 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen known as Lamb Shift).
iii. 5890 A – 5896 A [double lines of sodium]
Answer:
i. Radio waves (short wavelength or high frequency end)
ii. Radio waves (short wavelength or high frequency end)
iii. Visible region (yellow light)

Question 61.
Vidhya and Vijay were studying the effect of certain radiations on flower plants. Vidhya exposed her plants to UV rays and Vijay exposed his plants to infrared rays. After few days, Vidhya’s plants got damaged and Vijay’s plants had beautiful bloom. Why did this happen?
Answer:
Frequency of UV rays is greater than infrared rays, hence UV rays are much more energetic than infrared rays. Plants cannot tolerate the exposure of high energy rays. As a result, Vidhya’s plants got damaged and Vijay’s plants had a beautiful bloom.

Multiple Choice Questions

Question 1.
Which of the following type of radiations are radiated by an oscillating electric charge?
(A) Electric
(B) Magnetic
(C) Thermoelectric
(D) Electromagnetic
Answer:
(D) Electromagnetic

Question 2.
If \(\vec{E}\) and \(\vec{B}\) are the electric and magnetic field vectors of e.m. waves, then the direction of propagation of e.m. direction of wave is along the
(A) \(\vec{E}\)
(B) \(\vec{B}\)
(C) \(\vec{E}\) × \(\vec{B}\)
(D) \(\vec{E}\) • \(\vec{B}\)
Answer:
(C) \(\vec{E}\) × \(\vec{B}\)

Question 3.
The unit of expression µ0o ε0 is
(A) m / s
(B) m² / s²
(C) s² / m²
(D) s / m
Answer:
(C) s² / m²

Question 4.
According to Maxwell’s equation the velocity of light in any medium is expressed as
(A) \(\frac {1}{\sqrt{µ_0ε_0}}\)
(B) \(\frac {22}{\sqrt{µε}}\)
(C) \(\sqrt{\frac {µ}{ε}}\)
(D) \(\sqrt{\frac {µ_0}{ε}}\)
Answer:
(B) \(\frac {22}{\sqrt{µε}}\)

Question 5.
The electromagnetic waves do not transport.
(A) energy
(B) charge
(C) momentum
(D) pressure
Answer:
(B) charge

Question 6.
In an electromagnetic wave, the direction of the magnetic induction \(\vec{B}\) is
(A) parallel to the electric field \(\vec{E}\).
(B) perpendicular to the electric field \(\vec{E}\).
(C) antiparallel to the pointing vector \(\vec{S}\).
(D) random.
Answer:
(B) perpendicular to the electric field \(\vec{E}\).

Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System

Question 7.
Which of the following electromagnetic waves have the longest wavelength?
(A) heat waves
(B) light waves
(C) radio waves
(D) microwaves.
Answer:
(C) radio waves

Question 8.
Radio waves do not penetrate in the band of
(A) ionosphere
(B) mesosphere
(C) troposphere
(D) stratosphere
Answer:
(A) ionosphere

Question 9.
Which of the following electromagnetic wave has least wavelength?
(A) Gamma rays
(B) X- rays
(C) Radio waves
(D) microwaves
Answer:
(A) Gamma rays

Question 10.
If E is an electric field and \(\vec{B}\) is the magnetic induction, then the energy flow per unit area per unit time in an electromagnetic field is given by
(A) \(\frac {1}{µ_0}\) \(\vec{E}\) × \(\vec{B}\)
(B) \(\vec{E}\).\(\vec{B}\)
(C) E² + B²
(D) \(\frac {E}{B}\)
Answer:
(A) \(\frac {1}{µ_0}\) \(\vec{E}\) × \(\vec{B}\)

Question 11.
Out of the X-rays, microwaves, ultra-violet rays, the shortest frequency wave is ……………
(A) X-rays
(B) microwaves
(C) ultra-violet rays
(D) γ-rays
Answer:
(B) microwaves

Question 12.
The part of electromagnetic spectrum used in operating radar is ……………
(A) y-rays
(B) visible rays
(C) infra-red rays
(D) microwaves
Answer:
(D) microwaves

Question 13.
The correct sequence of descending order of wavelength values of the given radiation source is …………..
(A) radio waves, microwaves, infra-red, γ- rays
(B) γ-rays, infra-red, radio waves, microwaves
(C) Infra-red, radio waves, microwaves, γ- rays
(D) microwaves, γ-rays, infra-red, radio waves
Answer:
(A) radio waves, microwaves, infra-red, γ- rays

Question 14.
The nuclei of atoms of radioactive elements produce ……………
(A) X-rays
(B) γ-rays
(C) microwaves
(D) ultra-violet rays
Answer:
(B) γ-rays

Question 15.
The electronic transition in atom produces
(A) ultra violet light
(B) visible light
(C) infra-red rays
(D) microwaves
Answer:
(B) visible light

Question 16.
When radio waves from transmitting antenna reach the receiving antenna directly or after reflection in the ionosphere, the wave propagation is called ………………
(A) ground wave propagation
(B) space wave propagation
(C) sky wave propagation
(D) satellite propagation
Answer:
(C) sky wave propagation

Question 17.
Basic components of a transmitter are ……………..
(A) message signal generator and antenna
(B) modulator and antenna
(C) signal generator and modulator
(D) message signal generator, modulator and antenna
Answer:
(D) message signal generator, modulator and antenna

Question 18.
The process of changing some characteristics of a carrier wave in accordance with the incoming signal is called …………..
(A) amplification
(B) modulation
(C) rectification
(D) demodulation
Answer:
(B) modulation

Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System

Question 19.
The process of superimposing a low frequency signal on a high frequency wave is …………….
(A) detection
(B) mixing
(C) modulation
(D) attenuation
Answer:
(C) modulation

Question 20.
A device that converts one form of energy into another form is termed as ……………
(A) transducer
(B) transmitter
(C) amplifier
(D) receiver
Answer:
(A) transducer

Question 21.
A microphone which converts sound into electrical signal is an example of .
(A) a thermistor
(B) a rectifier
(C) a modulator
(D) an electrical transducer
Answer:
(D) an electrical transducer

Question 22.
The process of regaining of information from carries wave at the receiver is called
(A) modulation
(B) transmission
(C) propagation
(D) demodulation
Answer:
(D) demodulation

Question 23.
Range of communication can be increased by
(A) increasing the heights of transmitting and receiving antennas.
(B) decreasing the heights of transmitting and receiving antennas.
(C) increasing height of transmitting antenna and decreasing the height of receiving antenna.
(D) increasing height of receiving antenna only.
Answer:
(A) increasing the heights of transmitting and receiving antennas.

Question 24
Ionosphere mainly consists of
(A) positive ions and electrons
(B) water vapour and smoke
(C) ozone layer
(D) dust particles
Answer:
(A) positive ions and electrons

Question 25.
The reflected waves from the ionosphere are
(A) ground waves.
(B) sky waves.
(C) space waves.
(D) very high frequency waves.
Answer:
(B) sky waves.

Question 26.
Communication is the process of
(A) keeping in touch.
(B) exchanging information.
(C) broadcasting.
(D) entertainment.
Answer:
(B) exchanging information.

Question 27.
The message fed to the transmitter are generally
(A) radio signals
(B) audio signals
(C) both (A) and (B)
(D) optical signals
Answer:
(B) audio signals

Question 28.
Line of sight propagation is also called as ……………. propagation.
(A) sky wave
(B) ground wave
(C) sound wave
(D) space wave
Answer:
(D) space wave

Question 29.
The ozone layer in the atmosphere absorbs
(A) only the radio waves.
(B) only the visible light.
(C) only the y rays.
(D) X-rays and ultraviolet rays.
Answer:
(D) X-rays and ultraviolet rays.

Question 30.
Modem communication systems consist of
(A) electronic systems
(B) electrical system
(C) optical system
(D) all of these
Answer:
(D) all of these

Question 31.
What determines the absorption of radio waves by the atmosphere?
(A) Frequency .
(B) Polarisation
(C) Interference
(D) Distance of receiver
Answer:
(A) Frequency .

Question 32.
The portion of the atmosphere closest to the earth’s surface is ……………
(A) troposphere
(B) stratosphere
(C) mesosphere
(D) ionosphere
Answer:
(A) troposphere

Maharashtra Board Class 11 Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System

Question 33.
An antenna behaves as resonant circuit only when its length is ………………
(A) λ/2
(B) λ/4
(C) λ
(D) n λ/2
Answer:
(D) n λ/2

Question 34.
Space wave travels through …………………
(A) ionosphere
(B) mesosphere
(C) troposphere
(D) stratosphere
Answer:
(C) troposphere

Question 35.
Transmission lines start radiating
(A) at low frequencies
(B) at high frequencies.
(C) at both high and low frequencies.
(D) none of the above.
Answer:
(B) at high frequencies.

Question 36.
If ‘ht‘ and ‘hr’ are height of transmitting and receiving antennae and ‘R’ is radius of the earth, the range of space wave is
(A) \(\sqrt {2R}\) (ht + hr)
(B) 2R \(\sqrt {(ht + hr)}\)
(C) \(\sqrt {2R(ht + hr)}\)
(D) \(\sqrt {2R}\) (√ht + √hr)
Answer:
(D) \(\sqrt {2R}\) (√ht + √hr)

Question 37.
In a communication system, noise is most likely to affect the signal ………..
(A) at the transmitter
(B) in the transmission medium
(C) in the information source
(D) at the destination
Answer:
(B) in the transmission medium

Question 38.
The power radiated by linear antenna of length 7’ is proportional to (A = wavelength)
(A) \(\frac {λ}{l}\)
(B) (\(\frac {λ}{l}\))²
(C) \(\frac {l}{λ}\)
(D) (\(\frac {l}{λ}\))²
Answer:
(D) (\(\frac {l}{λ}\))²

Question 39.
For efficient radiation and reception of signal with wavelength λ, the transmitting antennas would have length comparable to ……………….
(A) λ of frequency used
(B) λ/2 of frequency used
(C) λ/3 of frequency used
(D) λ/4 of frequency used
Answer:
(A) λ of frequency used

Maharashtra Board Class 11 Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 13 Electromagnetic Waves and Communication System Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

1. Choose the correct option.

Question 1.
The EM wave emitted by the Sun and responsible for heating the Earth’s atmosphere due to green house effect is
(A) Infra-red radiation
(B) X ray
(C) Microwave
(D) Visible light
Answer:
(A) Infra-red radiation

Question 2.
Earth’s atmosphere is richest in
(A) UV
(B) IR
(C) X-ray
(D) Microwaves
Answer:
(B) IR

Maharashtra Board Class 11 Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

Question 3.
How does the frequency of a beam of ultraviolet light change when it travels from air into glass?
(A) depends on the values of p and e
(B) increases
(C) decreases
(D) remains same
Answer:
(D) remains same

Question 4.
The direction of EM wave is given by
(A) \(\bar{E}\) × \(\bar{B}\)
(B) \(\bar{E}\).\(\bar{B}\)
(C) along \(\bar{E}\)
(D) along \(\bar{B}\)
Answer:
(A) \(\bar{E}\) × \(\bar{B}\)

Question 5.
The maximum distance upto which TV transmission from a TV tower of height h can be received is proportional to
(A) h½
(B) h
(C) h3/2
(D) h²
Answer:
(A) h½

Question 6.
The waves used by artificial satellites for communication purposes are
(A) Microwave
(B) AM radio waves
(C) FM radio waves
(D) X-rays
Answer:
(A) Microwave

Question 7.
If a TV telecast is to cover a radius of 640 km, what should be the height of transmitting antenna?
(A) 32000 m
(B) 53000 m
(C) 42000 m
(D) 55000 m
Answer:
(A) 32000 m

2. Answer briefly.

Question 1.
State two characteristics of an EM wave.
Answer:
i. The electric and magnetic fields, \(\vec{E}\) and \(\vec{B}\) are always perpendicular to each other and also to the direction of propagation of the EM wave. Thus, the EM waves are transverse waves.

ii. The cross product (\(\vec{E}\) × \(\vec{B}\)) gives the direction in which the EM wave travels. (\(\vec{E}\) × \(\vec{B}\)) also gives the energy carried by EM wave.

Question 2.
Why are microwaves used in radar?
Answer:
Microwaves are used in radar systems for identifying the location of distant objects like ships, aeroplanes etc.

Maharashtra Board Class 11 Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

Question 3.
What are EM waves?
Answer:
Waves that are caused by the acceleration of charged particles and consist of electric and magnetic fields vibrating sinusoidally at right angles to each other and to the direction of propagation are called EM waves or EM radiation.

Question 4.
How are EM waves produced?
Answer:

  1. According to quantum theory, an electron, while orbiting around the nucleus in a stable orbit does not emit EM radiation even though it undergoes acceleration.
  2. It will emit an EM radiation only when it falls from an orbit of higher energy to one of lower energy.
  3. EM waves (such as X-rays) are produced when fast moving electrons hit a target of high atomic number (such as molybdenum, copper, etc.).
  4. An electric charge at rest has an electric field in the region around it but has no magnetic field.
  5. When the charge moves, it produces both electric and magnetic fields.
  6. If the charge moves with a constant velocity, the magnetic field will not change with time and hence, it cannot produce an EM wave.
  7. But if the charge is accelerated, both the magnetic and electric fields change with space and time and an EM wave is produced.
  8. Thus, an oscillating charge emits an EM wave which has the same frequency as that of the oscillation of the charge.

Question 5.
Can we produce a pure electric or magnetic wave in space? Why?
Answer:
No.
In vacuum, an electric field cannot directly induce another electric field so a “pure” electric field wave cannot exist and same can be said for a “pure” magnetic wave.

Question 6.
Does an ordinary electric lamp emit EM waves?
Answer:
Yes, ordinary electric lamp emits EM waves.

Question 7.
Why light waves travel in vacuum whereas sound wave cannot?
Answer:
Light waves are electromagnetic waves which can travel in vacuum whereas sound waves travel due to the vibration of particles of medium. Without any particles present (like in a vacuum) no vibrations can be produced. Hence, the sound wave cannot travel through the vacuum.

Question 8.
What are ultraviolet rays? Give two uses.
Answer:
Production:

  1. Ultraviolet rays can be produced by the mercury vapour lamp, electric spark and carbon arc lamp.
  2. They can also be obtained by striking electrical discharge in hydrogen and xenon gas tubes.
  3. The Sun is the most important natural source of ultraviolet rays, most of which are absorbed by the ozone layer in the Earth’s atmosphere.

Uses:

  1. Ultraviolet rays destroy germs and bacteria and hence they are used for sterilizing surgical instruments and for purification of water.
  2. Used in burglar alarms and security systems.
  3. Used to distinguish real and fake gems.

Maharashtra Board Class 11 Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

Question 9.
What are radio waves? Give its two uses.
Answer:

  1. Radio waves are produced by accelerated motion of charges in a conducting wire. The frequency of waves produced by the circuit depends upon the magnitudes of the inductance and the capacitance.
  2. Thus, by choosing suitable values of the inductance and the capacitance, radio waves of desired frequency can be produced.

Uses:

  1. Radio waves are used for wireless communication purpose.
  2. They are used for radio broadcasting and transmission of TV signals.
  3. Cellular phones use radio waves to transmit voice communication in the ultra high frequency (UHF) band.

Question 10.
Name the most harmful radiation entering the Earth’s atmosphere from the outer space.
Answer:
Ultraviolet radiation.

Question 11.
Give reasons for the following:
i. Long distance radio broadcast uses short wave bands.
ii. Satellites are used for long distance TV transmission.
Answer:
i. Long distance radio broadcast uses short wave bands because electromagnetic waves only in the frequency range of short wave bands only are reflected by the ionosphere.

ii. a. It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere.
b. Hence, satellites are helpful in long distance TV transmission.

Question 12.
Name the three basic units of any communication system.
Answer:
Three basic (essential) elements of every communication system are transmitter, communication channel and receiver.

Question 13.
What is a carrier wave?
Answer:
The high frequency waves on which the signals to be transmitted are superimposed are called carrier waves.

Question 14.
Why high frequency carrier waves are used for transmission of audio signals?
Answer:
An audio signal has low frequency (<20 kHz) and low frequency signals cannot be transmitted over large distances. Because of this, a high frequency carrier waves are used for transmission.

Question 15.
What is modulation?
Answer:
The signals in communication system (e.g. music, speech etc.) are low frequency signals and cannot be transmitted over large distances. In order to transmit the signal to large distances, it is superimposed on a high frequency wave (called carrier wave). This process is called modulation.

Question 16.
What is meant by amplitude modulation?
Answer:
When the amplitude of carrier wave is varied in accordance with the modulating signal, the process is called amplitude modulation.

Question 17.
What is meant by noise?
Answer:

  1. A random unwanted signal is called noise.
  2. The source generating the noise may be located inside or outside the system.
  3. Efforts should be made to minimize the noise level in a communication system.

Question 18.
What is meant by bandwidth?
Answer:
The bandwidth of an electronic circuit is the range of frequencies over which it operates efficiently.

Maharashtra Board Class 11 Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

Question 19.
What is demodulation?
Answer:
The process of regaining signal from a modulated wave is called demodulation. This is the reverse process of modulation.

Question 20.
What type of modulation is required for television broadcast?
Answer:
Amplitude modulation is required for television broadcast.

Question 21.
How does the effective power radiated by an antenna vary with wavelength?
Answer:

  1. To transmit a signal, an antenna or an aerial is needed.
  2. Power radiated from a linear antenna of length l is, P ∝ (\(\frac {l}{λ}\))²
    where, λ is the wavelength of the signal.

Question 22.
Why should broadcasting programs use different frequencies?
Answer:
If broadcasting programs run on same frequency, then the information carried by these waves will get mixed up with each other. Hence, different broadcasting programs should run on different frequencies.

Question 23.
Explain the necessity of a carrier wave in communication.
Answer:

  1. Without a carrier wave, the input signals could be carried by very low frequency electromagnetic waves but it will need quite a bit of amplification in order to transmit those very low frequencies.
  2. The input signals themselves do not have much power and need a fairly large antenna in order to transmit the information.
  3. Hence, it is necessary to impose the input signal on carrier wave as it requires less power in order to transmit the information.

Question 24.
Why does amplitude modulation give noisy reception?
Answer:
i. In amplitude modulation, carrier is varied in accordance with the message signal.

ii. The higher the amplitude, the greater is magnitude of the signal. So even if due to any reason, the magnitude of the signal changes, it will lead to variation in the amplitude of the signal. So its easy for noise to disturb the amplitude modulated signal.

Question 25.
Explain why is modulation needed.
Answer:
Modulation helps in avoiding mixing up of signals from different transmitters as different carrier wave frequencies can be allotted to different transmitters. Without the use of these waves, the audio signals, if transmitted directly by different transmitters, would get mixed up.

3. Solve the numerical problem.

Question 1.
Calculate the frequency in MHz of a radio wave of wavelength 250 m. Remember that the speed of all EM waves in vacuum is 3.0 × 108 m/s.
Answer:
Given: λ = 250 m, c = 3 × 108 m/s
To find: Frequency (v)
Formula: c = v8
Calculation: From formula,
v = \(\frac {c}{λ}\) = \(\frac {3×10^8}{250}\) = 1.2 × 106 Hz
= 1.2 MHz

Maharashtra Board Class 11 Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

Question 2.
Calculate the wavelength in nm of an X-ray wave of frequency 2.0 × 1018 Hz.
Solution:
Given: c = 3 × 108, v = 2 × 1018 Hz
To find: Wavelength (λ)
Formula: c = vλ
Calculation. From formula,
λ = \(\frac {c}{v}\) = \(\frac {3×10^8}{2×10^{18}}\) = 1.5 × 10-10
= 0.15 nm

Question 3.
The speed of light is 3 × 108 m/s. Calculate the frequency of red light of wavelength of 6.5 × 10-7 m.
Answer:
Given: c = 3 × 108 m/s, λ = 6.5 × 10-7 m
To find: Frequency (v)
Formula: c = vλ
Calculation: From formula,
v = \(\frac {c}{λ}\) = \(\frac {3×10^8}{6.5×10^{-7}}\) = 4.6 × 1014 Hz

Question 4.
Calculate the wavelength of a microwave of frequency 8.0 GHz.
Answer:
Given: v = 8 GHz = 8 × 109 Hz,
c = 3 × 108 m/s
To find: Wavelength (λ)
Formula: c = vλ
Calculation: From formula,
λ = \(\frac {c}{λ}\) = \(\frac {3×10^8}{8×10^9}\) = 3.75 × 10-2
= 3.75 cm

Question 5.
In a EM wave the electric field oscillates sinusoidally at a frequency of 2 × 1010 What is the wavelength of the wave?
Answer:
Given: v = 2 × 1010 Hz, c = 3 × 108 m
To find: Wavelength (λ)
Formula: c = vλ
Calculation: From formula,
λ = \(\frac {c}{λ}\) = \(\frac {3×10^8}{2×10^{10}}\) = 1.5 × 10-2

Question 6.
The amplitude of the magnetic field part of a harmonic EM wave in vacuum is B0 = 5 X 10-7 T. What is the amplitude of the electric field part of the wave?
Answer:
Given: B0 = 5 × 10-7 T, c = 3 × 108
To find: Amplitude of electric field (E0)
Formula: c = \(\frac {E_0}{B_0}\)
Calculation /From formula,
E0 = c × B0
= 3 × 108 × 5 × 10-7
= 150 V/m

Question 7.
A TV tower has a height of 200 m. How much population is covered by TV transmission if the average population density around the tower is 1000/km²? (Radius of the Earth = 6.4 × 106 m)
Answer:
Given: h = 200 m,
Population density (n)
= 1000/km² = 1000 × 10-6/m² = 10-3/m²
R = 6.4 ×106 m
To find: Population covered
Formulae: i. A = πd² = π(\(\sqrt{2Rh}\))² = 2πRh
ii. Population covered = nA
Calculation /From formula (i),
A = 2πRh
= 2 × 3.142 × 6.4 × 106 × 200
≈ 8 × 109
From formula (ii),
Population covered = nA
= 10-3 × 8 × 109
= 8 × 106

Maharashtra Board Class 11 Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

Question 8.
Height of a TV tower is 600 m at a given place. Calculate its coverage range if the radius of the Earth is 6400 km. What should be the height to get the double coverage area?
Answer:
Given: h = 600 m, R = 6.4 × 106 m
To find: Range (d)
Height to get the double coverage (h’)
Formula: d = \(\sqrt{2hR}\)
Calculation: From formula,
d = \(\sqrt{2×600×6.4×10^6}\) = 87.6 × 10³ = 87.6 km
Now, for A’ = 2A
π(d’)² = 2 (πd²)
∴ (d’)² = 2d²
From formula,
h’ = \(\frac{(d’)^2}{2R}\)
= \(\frac{2d^2}{2R}\)
= 2 × h ……….. (∵ h = \(\frac{d^2}{2R}\))
= 2 × 600
=1200 m

Question 9.
A transmitting antenna at the top of a tower has a height 32 m and that of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in line of sight mode? Given radius of Earth is 6.4 × 106 m.
Answer:
Given: ht = 32 m, hr = 50 m, R = 6.4 × 106 m
To find: Maximum distance or range (d)
Formula: d = \(\sqrt{2Rh}\)
Calculation: From formula,
dt = \(\sqrt{2Rh_t}\) = \(\sqrt{2×6.4×10^6×32}\)
= 20.238 × 10³ m
= 20.238 km
dr = \(\sqrt{2Rh_t}\)
= \(\sqrt{2×6.4×10^6×50}\)
= 25.298 × 10³ m
= 25.298 km
Now, d = dt + dr
= 20.238 + 25.298
= 45.536 km

11th Physics Digest Chapter 13 Electromagnetic Waves and Communication System Intext Questions and Answers

Can you recall? (Textbookpage no. 229)

Question 1.
i. What is a wave?
Answer:
Wave is an oscillatory disturbance which travels through a medium without change in its form.

ii. What is the difference between longitudinal and transverse waves?
Answer:
a. Transverse wave: A wave in which particles of the medium vibrate in a direction perpendicular to the direction of propagation of wave is called transverse wave.
b. Longitudinal wave: A wave in which particles of the medium vibrate in a direction parallel to the direction of propagation of wave is called longitudinal wave.

iii. What are electric and magnetic fields and what are their sources?
Answer:
a. Electric field is the force experienced by a test charge in presence of the given charge at the given distance from it.
b. A magnetic field is produced around a magnet or around a current carrying conductor.

iv. By which mechanism heat is lost by hot bodies?
Answer:
Hot bodies lose the heat in the form of radiation.

Maharashtra Board Class 11 Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

Question 2.
What are Lenz’s law, Ampere’s law and Faraday’s law?
Answer:
Lenz’s law:
Whereas, Lenz’s law states that, the direction of the induced emf is such that the change is opposed.

Ampere’s law:
Ampere’s law describes the relation between the induced magnetic field associated with a loop and the current flowing through the loop.

Faraday’s law:
Faraday’s law states that, time varying magnetic field induces an electromotive force (emf) and an electric field.

Internet my friend. (Tpxtboakpage no. 240)

https//www.iiap.res.in/centers/iao
[Students are expected to visit the above mentioned website and collect more information about different EM wave propagations used by astronomical observatories.]

Maharashtra Board Class 11 Physics Important Questions Chapter 12 Magnetism

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 12 Magnetism Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 12 Magnetism

Question 1.
What are some commonly known facts about magnetism?
Answer:
Some commonly known facts about magnetism:

  1. Every magnet regardless of its size and shape has two poles called north pole and south pole.
  2. Isolated magnetic monopoles do not exist. If a magnet is broken into two or more pieces then each piece behaves like an independent magnet with some what weaker magnetic field.
  3. Like magnetic poles repel each other, whereas unlike poles attract each other.
  4. When a bar magnet/ magnetic needle is suspended freely or is pivoted, it aligns itself in geographically north-south direction.

Question 2.
What are some properties of magnetic lines of force?
Answer:

  1. Magnetic lines of force originate from the north pole and end at the south pole.
  2. The magnetic lines of force of a magnet or a solenoid form closed loops. This is in contrast to the case of an electric dipole, where the electric lines of force originate from the positive charge and end on the negative charge.
  3. The direction of the net magnetic field \(\vec{B}\) at a point is given by the tangent to the magnetic line of force at that point.
  4. The number of lines of force crossing per unit area decides the magnitude of magnetic field \(\vec{B}\).
  5. The magnetic lines of force do not intersect. This is because had they intersected, the direction of magnetic field would not be unique at that point.

Maharashtra Board Class 11 Physics Important Questions Chapter 12 Magnetism

Question 3.
What is magnetic flux? What is unit of magnetic flux in SI system?
Answer:

  1. The number of lines of force per unit area is called magnetic flux (ø).
  2. SI unit of magnetic flux (ø) is weber (Wb).

Question 4.
How do we determine strength of magnetic field at a given point due to a magnet? Write down units of magnetic field in SI and CGS system and their interconversion.
Answer:
i. Density of lines of force i.e., the number of lines of force per unit area around a particular point determines the strength of the magnetic field at that point.

ii. The magnitude of magnetic field strength B at a point in a magnetic field is given by,
Magnetic Field = \(\frac {magnetic flux}{area}\)
i.e., B = \(\frac {ø}{A}\)

iii. SI unit of magnetic field (B) is expressed as weber/m² or Tesla.

iv. 1 Tesla = 10⁴ Gauss

Question 5.
What is the unit of magnetic intensity?
Answer:
SI unit: weber/m² or Tesla.

Question 6.
Explain the pole strength and magnetic dipole moment of a bar magnet.
Answer:
i. The bar magnet said to have pole strength +qm and -qm near the north and south poles respectively.

ii. As bar magnet has two poles with equal and opposite pole strength, it is called as a magnetic dipole.

iii. The two poles are separated by a distance equal to 2l.

iv. The product of pole strength and the magnetic length is called as magnetic dipole moment.
∴ \(\vec{m}\) = qm (2\(\vec{l}\))
where, 2\(\vec{l}\) is a vector from south pole to north pole.

Question 7.
State the SI units of pole strength and magnetic dipole moment.
Answer:

  1. SI unit of pole strength (qm) is Am.
  2. SI unit of magnetic dipole moment (m) is Am².

Question 8.
Draw neat labelled diagram for a bar magnet.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 12 Magnetism 1

Question 9.
Define and explain the following terms in case of a bar magnet:
i. Axis
ii. Equator
iii. Magnetic length
Answer:
i. Axis: It is the line passing through both the poles of a bar magnet. There is only one axis for a given bar magnet.

ii. Equator:

  • A line passing through the centre of a magnet and perpendicular to its axis is called magnetic equator.
  • The plane containing all equators is called the equatorial plane.
  • The locus of points, on the equatorial plane, which are equidistant from the centre of the magnet is called the equatorial circle.
  • The popularly known ‘equator’ of the planet is actually an ‘equatorial circle’. Such a circle with any diameter is an equator.

iii. Magnetic length (2l)
It is the distance between the two poles of a magnet.
Magnetic length (2l) = \(\frac {5}{6}\) × Geometric length.

Maharashtra Board Class 11 Physics Important Questions Chapter 12 Magnetism

Question 10.
State the expression for magnetic induction at a point due to a very short bar magnet along its axis.
Answer:
For very short bar magnet, the magnetic induction at point on the axis is given as,
\(\overrightarrow{\mathrm{B}}_{\mathrm{axis}}=\frac{\mu_{0}}{4 \pi} \frac{2 \overrightarrow{\mathrm{m}}}{\mathrm{r}^{3}}\)

Question 11.
State the expression for the magnetic induction at any point along the equator of a very short bar magnet.
Answer:
For very short bar magnet, the magnetic induction at point on the equator is given as,
\(\overrightarrow{\mathrm{B}}_{\text {equator }}=-\frac{\mu_{0}}{4 \pi} \frac{\overrightarrow{\mathrm{m}}}{\mathrm{r}^{3}}\)

Question 12.
Show that the magnitude of magnetic induction at a point on the axis of a short bar magnet is twice the magnitude of magnetic induction at a point on the equator at the same distance.
Answer:
i. Magnitude of magnetic induction at a point along the axis of a short magnet is given by,
\(\mathrm{B}_{\mathrm{axis}}=\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{~m}}{\mathrm{r}^{3}}\) ………….. (1)

ii. Magnitude of magnetic induction at a point on equatorial line is given by
\(\mathrm{B}_{\text {equator }}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{m}}{\mathrm{r}^{3}}\) …………… (2)

iii. Dividing equation (1) by (2), we get,
\(\frac{\mathrm{B}_{\mathrm{axis}}}{\mathrm{B}_{\mathrm{eq}}}=\frac{\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{~m}}{\mathrm{r}^{3}}}{\frac{\mu_{0}}{4 \pi} \frac{\mathrm{m}}{\mathrm{r}^{3}}}\)
∴ \(\frac{B_{\text {axis }}}{B_{e q}}\) = 2
∴ Baxis = 2Beq

Question 13.
Derive an expression for the magnetic field due to a bar magnet at an arbitrary point.
Answer:
i. Consider a bar magnet of magnetic moment \(\vec{m}\) with centre at O as shown in figure and let P be any point in its magnetic field.
Maharashtra Board Class 11 Physics Important Questions Chapter 12 Magnetism 2

ii. Magnetic moment \(\vec{m}\) is resolved into components along \(\vec{r}\) and perpendicular to \(\vec{r}\).

iii. For the component m cos θ along \(\vec{r}\), the point P is an axial point.

iv. For the component m sinθ perpendicular to \(\vec{r}\), the point P is an equatorial point at the same distance \(\vec{r}\).

v. For a point on the axis, Ba = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{\mathrm{r}^{3}}\)
Here
Ba = \(\frac{\mu_{0}}{4 \pi} \frac{2 m \cos \theta}{r^{3}}\) ………….. (1)
directed along m cosθ.

vi. For point on equator,
Ba = \(\frac{\mu_{o}}{4 \pi} \frac{m \sin \theta}{r^{3}}\) …………. (2)
directed opposite to m sin θ

vii. Thus, the magnitude of the resultant magnetic field B, at point P is given by
Maharashtra Board Class 11 Physics Important Questions Chapter 12 Magnetism 3

viii. Let a be the angle made by the direction of \(\vec{B}\) with \(\vec{r}\). Then, by using equation (1) and equation (2),
tan α = \(\frac {B_{eq}}{B_a}\) = \(\frac {1}{2}\) (tan θ)
The angle between directions of \(\vec{B}\) and \(\vec{m}\) is then (θ + a).

Question 14.
A bar magnet of magnetic moment 5.0 Am² has the poles 20 cm apart. Calculate the pole strength.
Solution:
Given: m = 5.0 Am², 2l = 20 cm = 0.20 m
To find: Pole strength (qm)
Formula: qm = \(\frac {m}{2l}\)
Calculation:
From formula.
qm = \(\frac {5.0}{0.20}\) = 25 Am

Question 15.
A bar magnet has magnetic moment 3.6 Am² and pole strength 10.8 Am. Determine its magnetic length and geometric length.
Answer:
Given: m = 3.6 Am², qm = 10.8 Am
To find:
i. Magnetic length
ii. Geometric length
Formulae:
i. Magnetic length = \(\frac {m}{q_m}\)
ii. Geometric length = \(\frac {6}{5}\) × magnetic length.
Calculation: From formula (i),
Magnetic length = \(\frac {3.6}{10.8}\) = 0.33 m
From formula (ii),
Geometric length = \(\frac {6}{5}\) × 0.33
= 0.396 m ≈ 0.4 m

Maharashtra Board Class 11 Physics Important Questions Chapter 12 Magnetism

Question 16.
A short magnetic dipole has magnetic moment 0.5 A m². Calculate its magnetic field at a distance of 20 cm from the centre of magnetic dipole on (i) the axis (ii) the equatorial line (Given µ0 = 4π × 10-7 SI units)
Answer:
Given: m = 0.5 Am², r = 20 cm = 20 × 10-2 m
To Find: i. Magnetic field on the axial point (Ba)
ii. Magnetic field on the equatorial point (Beq)
Formulae:
i. Ba = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{r^{3}}\)
ii. Ba = 2Beq
Calculation: From formula (i),
Ba = 10-7 × \(\frac{2 \times 0.5}{(0.2)^{3}}\)
= \(\frac{10^{-7}}{8 \times 10^{-3}}\)
= 0.125 × 10-4
∴ Ba = 1.25 × 10-5 Wb/m²
From formula (ii),
Beq = \(\frac {B_a}{2}\) = \(\frac {1.25×10^{-5}}{2}\)
= 0.625 × 10-5 Wb/m²

Question 17.
A short bar magnet has a magnetic moment of 0.48 JT-1. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (i) the axis (ii) the equatorial lines (normal bisector) of the magnet.
Answer:
Given: m = 0.48 JT-1, r = 10 cm = 0.1 m
To find:
i. Magnetic induction along axis (Ba)
ii. Magnetic induction along equator (Beq)
Formulae:
i. Ba = \(\frac {µ_0}{4π}\) \(\frac {2m}{r^3}\)
ii. Ba = 2 Beq
Calculation: From formula (i),
Ba = 10-7 × \(\frac {2×0.48}{10^{-3}}\)
∴ Ba = 0.96 × 10-4 T along S-N direction
From formula (ii),
Beq = \(\frac {0.96×106{-4}}{2}\)
∴ Beq = 0.48 × 10-4 T along N-S direction

Question 18.
Define the following magnetic parameters.
i. Magnetic axis
ii. Magnetic equator
iii. Magnetic Meridian
Answer:
i. Magnetic axis: The Earth is considered to be a huge magnetic dipole. The straight line joining the two poles is called the magnetic axis.

ii. Magnetic equator: A great circle in the plane perpendicular to magnetic axis is magnetic equatorial circle.

iii. Magnetic Meridian: A plane perpendicular to surface of the Earth (Vertical plane) and passing through the magnetic axis is magnetic meridian. Direction of resultant magnetic field of the Earth is always along or parallel to magnetic meridian.

Question 19.
Draw neat labelled diagram representing the Earth as a magnet.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 12 Magnetism 4

Question 20.
Define magnetic declination.
Answer:
Angle between the geographic and the magnetic meridian at a place is called magnetic declination (α).

Question 21.
Draw a neat labelled diagram showing the magnetic declination at a place.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 12 Magnetism 5

Question 22.
Draw a neat labelled diagram for angle of dip.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 12 Magnetism 6

Write a short note on Earth’s magnetic field. Mention the extreme values of magnetic field at magnetic poles and magnetic equator.
Ans:
i. Magnetic force experienced per unit pole strength is magnetic field \(\vec{B}\) at that place.

ii. This field can be resolved in components along the horizontal (\(\vec{B}_H\)) and along vertical (\(\vec{B}_v\)).

iii. The two components are related with the angle of dip (ø) as, BH = B cos ø, Bv = B sin ø
\(\frac {B_v}{B_H}\) = tan ø
B² = B\(_v^2\) + B\(_H^2\)
∴ B = \( \sqrt{\mathrm{B}_{\mathrm{V}}^{2}+\mathrm{B}_{\mathrm{H}}^{2}}\)

iv. At the magnetic North pole: \(\vec{B}\) = \(\vec{B}\)v, directed upward, \(\vec{B}\)H = 0 and ø = 90°.

v. At the magnetic south pole: \(\vec{B}\) = \(\vec{B}\)v, directed downward, \(\vec{B}\)H = 0 and ø = 270°.

vi. Anywhere on the magnetic equator (magnetic great circle): B = BH along South to North, \(\vec{B}\)v = 0 and ø = 0

Maharashtra Board Class 11 Physics Important Questions Chapter 12 Magnetism

Question 23.
What are magnetic maps?
Answer:
Magnetic elements of the Earth (BH, α and ø) vary from place to place and also with time. The maps providing these values at different locations are called magnetic maps.

Question 24.
Define following terms in case of magnetic maps:
i. Isomagnetic charts
ii. Isodynamic lines
iii. Isogonic lines
iv. Aclinic lines
Answer:
i. Isomagnetic charts: Magnetic maps drawn by joining places with the same value of a particular element are called isomagnetic charts.
ii. Isodynamic lines: Lines joining the places of equal horizontal components (BH) on magnetic maps are known as isodynamic lines.
iii. Isogonic lines: Lines joining the places of equal declination (α) on magnetic maps are called isogonic lines.
iv. Aclinic lines: Lines joining the places of equal inclination or dip (ø) on magnetic maps are called aclinic lines.

Question 25.
Magnetic equator and geographical equator of the earth are same. Is this true or false?
Answer:
False. Magnetic equator and geographical equator of the earth are not same. By definition, they are different. Magnetic declination is the angle between magnetic equator and geographical equator of the earth.

Question 26.
Earth’s magnetic field at the equator is approximately 4 × 10-5 T. Calculate Earth’s dipole moment. (Radius of Earth = 6.4 × 106 m, µ0 = 4π × 10-7 SI units)
Answer:
Consider earth’s magnetic field as due to a bar magnet at the centre of earth, held along the polar axis of earth.
∴ Beq = \(\frac {µ_0}{4π}\) \(\frac {m}{r^3}\) ……….. (where, R = radius of earth)
∴ m = \(\frac{\mathrm{B}_{\mathrm{eq}} \times \mathrm{R}^{3}}{\mu_{0} / 4 \pi}\) = \(\frac{4 \times 10^{-5} \times\left(6.4 \times 10^{6}\right)^{3}}{10^{-7}}\)
= 4 × (6.4)³ × 1020
= 1048 × 1020
∴ M = 1.048 × 1023 Am²

Question 27.
At a given place on the Earth, a bar magnet of magnetic moment \(\vec{m}\) is kept horizontal in the East-West direction. P and Q are the two neutral points due to magnetic field of this magnet and \(\vec{B}\)H is the horizontal component of the Earth’s magnetic field.
i. Calculate the angles between position vectors of P and Q with the direction of \(\vec{m}\).
ii. Points P and Q are 1 m from the centre of the bar magnet and BH = 3.5 × 10-5 T. Calculate magnetic dipole moment of the bar magnet.
Neutral point is that point where the resultant magnetic field is zero.
Answer:
i. The direction of magnetic field \(\vec{B}\) due to the bar magnet is opposite to \(\vec{B}\)H at the neutral points P and Q such that (θ + α) = 90° at P and (θ + α) = 270° at Question
∴ tan α = \(\frac {1}{2}\) tan θ
∴ tan θ = 2 tan α
= 2 tan (90 – θ) and 2 tan (270 – θ)
∴ tan θ = ± 2 cot θ
∴ tan²θ = 2 …….. (1)
∴ tanθ = ±√2
∴ θ = tan-1 (±√2)
∴ θ = 54°44′ and 180° – 54° 44° = 125°16′

ii. For magnetic dipole moment of the bar magnet:
From equation (2), tan² θ = 2
∴ sec² θ = 1 + tan² θ = 1 + 2 = 3
∴ cos² θ = \(\frac {1}{3}\)
r = 1 m and B = BH = 3.5 × 10-5 T ……. (Given)
we have,
Maharashtra Board Class 11 Physics Important Questions Chapter 12 Magnetism 7

Question 28.
A bar magnet is cut into two equal parts vertically and half part of bar magnet is kept on the other such that opposite poles align each other. Calculate the magnetic moment of the combination, if m is the magnetic moment of the original magnet.
Answer:
When bar magnet is cut into two equal parts, then magnetic moment of each part becomes half of the original directed from S to N pole.
∴ Magnetic moment of the combination = \(\frac {m}{2}\) – \(\frac {m}{2}\) = 0
∴ The net magnetic moment of the combination is zero.

Question 29.
Answer the following questions regarding earth’s magnetism:
i. Which direction would a compass needlepoint to, if located right on the geomagnetic north or south pole?
ii. Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all?
Answer:
i. At the poles, earth’s magnetic field is exactly vertical. As the compass needle is free to rotate in a horizontal plane only, it may point out in any direction.
ii. The earth’s magnetic field is only approximately a dipole field. Hence the local N-S poles may lie oriented in different directions. This is possible due to deposits of magnetised minerals in the earth’s crust.

Choose the correct option.

Question 1.
The ratio of magnetic induction along the axis to magnetic induction along the equator of a magnet is
(A) 1 : 1
(B) 1 : 2
(C) 2 : 1
(D) 4 : 1
Answer:
(C) 2 : 1

Maharashtra Board Class 11 Physics Important Questions Chapter 12 Magnetism

Question 2.
Magnetic field lines
(A) do not intersect each other.
(B) intersect each other at 45°.
(C) intersect each other at 90°.
(D) intersect each other at 60°.
Answer:
(A) do not intersect each other.

Question 3.
The points A and B are situated perpendicular to the axis of 2 cm long bar magnet at large distances x and 3 x from the centre on opposite sides. The ratio of magnetic fields at A and B will be approximately equal to
(A) 27 : 1
(B) 1 : 27
(C) 9 : 1
(D) 1 : 9
Answer:
(A) 27 : 1

Question 4.
A compass needle is placed at the magnetic pole. It
(A) points N – S.
(B) points E – W.
(C) becomes vertical.
(D) may stay in any direction.
Answer:
(D) may stay in any direction.

Question 5.
Magnetic lines of force originate from …………… pole and end at …………….. pole outside the magnet.
(A) north, north
(B) north, south
(C) south, north
(D) south, south
Answer:
(B) north, south

Question 6.
Two isolated point poles of strength 30 A-m and 60 A-m are placed at a distance of 0.3 m. The force of repulsion between them is
(A) 2 × 10-3 N
(B) 2 × 10-4 N
(C) 2 × 105 N
(D) 2 × 10-5 N
Answer:
(A) 2 × 10-3 N

Question 7.
The magnetic dipole moment has dimensions of
(A) current × length.
(B) charge × time × length.
(C) current × area.
(D) \(\frac {current}{area}\)
Answer:
(C) current × area.

Question 8.
A large magnet is broken into two pieces so that their lengths are in the ratio 2:1. The pole strengths of the two pieces will have the ratio
(A) 2 : 1
(B) 1 :2
(C) 4 : 1
(D) 1 : 1
Answer:
(A) 2 : 1

Question 9.
The magnetic induction B and the force F on a pole of strength m are related by
(A) B = m F
(B) F = nIABm
(C) F = m B
(D) F = \(\frac {m}{B}\)
Answer:
(C) F = m B

Question 10.
A magnetic dipole has magnetic length 10 cm and pole strength 100 Am. Its magnetic dipole moment is ………………. Am².
(A) 1000
(B) 500
(C) 10
(D) 5
Answer:
(C) 10

Question 11.
The geometric length of a bar magnet having half magnetic length 5 cm is …………… cm.
(A) 12
(B) 10
(C) 6
(D) 4.2
Answer:
(A) 12

Question 12.
The angle of dip at the equator is
(A) 90°
(B) 45°
(C) 30°
(D) 0°
Answer:
(D) 0°

Maharashtra Board Class 11 Physics Important Questions Chapter 12 Magnetism

Question 13.
The angle of dip at the magnetic poles of the earth is
(A) 90°
(B) 45°
(C) 30°
(D) 0°
Answer:
(A) 90°

Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 12 Magnetism Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 12 Magnetism

1. Choose the correct option.

Question 1.
Let r be the distance of a point on the axis of a bar magnet from its center. The magnetic field at r is always proportional to
(A) \(\frac {1}{r^2}\)
(B) \(\frac {1}{r^3}\)
(C) \(\frac {1}{r}\)
(D) Not necessarily \(\frac {1}{r^3}\) at all points
Answer:
(B) \(\frac {1}{r^3}\)

Question 2.
Magnetic meridian is the plane
(A) perpendicular to the magnetic axis of Earth
(B) perpendicular to geographic axis of Earth
(C) passing through the magnetic axis of Earth
(D) passing through the geographic axis
Answer:
(C) passing through the magnetic axis of Earth

Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism

Question 3.
The horizontal and vertical component of magnetic field of Earth are same at some place on the surface of Earth. The magnetic dip angle at this place will be
(A) 30°
(B) 45°
(C) 0°
(D) 90°
Answer:
(B) 45°

Question 4.
Inside a bar magnet, the magnetic field lines
(A) are not present
(B) are parallel to the cross sectional area of the magnet
(C) are in the direction from N pole to S pole
(D) are in the direction from S pole to N pole
Answer:
(D) are in the direction from S pole to N pole

Question 5.
A place where the vertical components of Earth’s magnetic field is zero has the angle of dip equal to
(A) 0°
(B) 45°
(C) 60°
(D) 90°
Answer:
(A) 0°

Question 6.
A place where the horizontal component of Earth’s magnetic field is zero lies at
(A) geographic equator
(B) geomagnetic equator
(C) one of the geographic poles
(D) one of the geomagnetic poles
Answer:
(D) one of the geomagnetic poles

Question 7.
A magnetic needle kept nonparallel to the magnetic field in a nonuniform magnetic field experiences
(A) a force but not a torque
(B) a torque but not a force
(C) both a force and a torque
(D) neither force nor a torque
Answer:
(C) both a force and a torque

2. Answer the following questions in brief.

Question 1.
What happens if a bar magnet is cut into two pieces transverse to its length/ along its length?
Answer:
i. When a magnet is cut into two pieces, then each piece behaves like an independent magnet.

ii. When a bar magnet is cut transverse to its length, the two pieces generated will behave as independent magnets of reduced magnetic length. However, the pole strength of all the four poles formed will be same as that of the original bar magnet. Thus, the new dipole moment of the smaller magnets will be,
Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism 1

iii. When the bar magnet is cut along its length, the two pieces generated will behave like an independent magnet with reduced pole strength. However, the magnetic length of both the new magnets will be same as that of the original bar magnet. Thus, the new dipole moment of the smaller magnets will be,
Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism 2

Question 2.
What could be the equation for Gauss’ law of magnetism, if a monopole of pole strength p is enclosed by a surface?
Answer:
i. According to Gauss’ law of electrostatics, the net electric flux through any Gaussian surface is proportional to net charge enclosed in it. The equation is given as,
øE = ∫\(\vec{E}\) . \(\vec{dS}\) = \(\frac {q}{ε_0}\)

ii. Similarly, if a monopole of a magnet of pole strength p exists, the Gauss’ law of magnetism in S.I. units will be given as,
øE = ∫\(\vec{B}\) . \(\vec{dS}\) = µ0P

Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism

3. Answer the following questions in detail.

Question 1.
Explain the Gauss’ law for magnetic fields.
Answer:
i. Analogous to the Gauss’ law for electric field, the Gauss’ law for magnetism states that, the net magnetic flux (øB) through a closed Gaussian surface is zero. øB = ∫\(\vec{B}\) . \(\vec{dS}\) = 0

ii. Consider a bar magnet, a current carrying solenoid and an electric dipole. The magnetic field lines of these three are as shown in figures.
Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism 3

iii. The areas (P) and (Q) are the cross – sections of three dimensional closed Gaussian surfaces. The Gaussian surface (P) does not include poles while the Gaussian surface (Q) includes N-pole of bar magnet, solenoid and the positive charge in case of electric dipole.

iv. The number of lines of force entering the surface (P) is equal to the number of lines of force leaving the surface. This can be observed in all the three cases.

v. However, Gaussian surface (Q) of bar magnet, enclose north pole. As, even thin slice of a bar magnet will have both north and south poles associated with it, the number of lines of Force entering surface (Q) are equal to the number of lines of force leaving the surface.

vi. For an electric dipole, the field lines begin from positive charge and end on negative charge. For a closed surface (Q), there is a net outward flux since it does include a net (positive) charge.

vii. Thus, according to the Gauss’ law of electrostatics øE = ∫\(\vec{E}\) . \(\vec{dS}\) = \(\frac {q}{ε_0}\), where q is the positive charge enclosed.

viii. The situation is entirely different from magnetic lines of force. Gauss’ law of magnetism can be written as øB = ∫\(\vec{B}\) . \(\vec{dS}\) = 0
From this, one can conclude that for electrostatics, an isolated electric charge exists but an isolated magnetic pole does not exist.

Question 2.
What is a geographic meridian? How does the declination vary with latitude? Where is it minimum?
Answer:
A plane perpendicular to the surface of the Earth (vertical plane) and passing through geographic axis is geographic meridian.

i. Angle between the geographic and the magnetic meridian at a place is called magnetic declination (a).
ii. Magnetic declination varies with location and over time. As one moves away from the true north the declination changes depending on the latitude as well as longitude of the place. By convention, declination is positive when magnetic north is east of true north, and negative when it is to the west. The declination is small in India. It is 0° 58′ west at Mumbai and 0° 41′ east at Delhi.

Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism

Question 3.
Define the angle of dip. What happens to angle of dip as we move towards magnetic pole from magnetic equator?
Answer:
Angle made by the direction of resultant magnetic field with the horizontal at a place is inclination or angle of dip (ø) at the place.
At the magnetic pole value of ø = 90° and it goes on decreasing when we move towards equator such that at equator value of (ø) = 0°.

4. Solve the following problems.

Question 1.
A magnetic pole of bar magnet with pole strength of 100 Am is 20 cm away from the centre of a bar magnet. Bar magnet has pole strength of 200 Am and has a length 5 cm. If the magnetic pole is on the axis of the bar magnet, find the force on the magnetic pole.
Answer:
Given that, (qm)1 = 200 Am
and (2l) = 5 cm = 5 × 10-2 m
∴ m = 200 × 5 × 10-2 = 10 Am²
For a bar magnet, magnetic dipole moment is,
m = qm (21)
For a point on the axis of a bar magnet at distance, r = 20 cm = 0.2 m,
Ba = \(\frac{\mu_{0}}{4 \pi} \times \frac{2 m}{r^{3}}\)
= 10-7 × \(\frac{2 \times 10}{(0.2)^{3}}\)
= 0.25 × 10-3
= 2.5 × 10-4 Wb/m²
The force acting on the pole will be given by,
F = qm Ba = 100 × 2.5 × 10-4
= 2.5 × 10-2 N

Question 2.
A magnet makes an angle of 45° with the horizontal in a plane making an angle of 30° with the magnetic meridian. Find the true value of the dip angle at the place.
Answer:
Let true value of dip be ø. When the magnet is kept 45° aligned with declination 30°, the horizontal component of Earth’s magnetic field.
B’H = BH cos 30° Whereas, vertical component remains unchanged.
∴ For apparent dip of 45°,
tan 45° = \(\frac{\mathrm{B}_{\mathrm{V}}^{\prime}}{\mathrm{B}_{\mathrm{H}}^{\prime}}=\frac{\mathrm{B}_{\mathrm{V}}}{\mathrm{B}_{\mathrm{H}} \cos 30^{\circ}}=\frac{\mathrm{B}_{\mathrm{v}}}{\mathrm{B}_{\mathrm{H}}} \times \frac{1}{\cos 30^{\circ}}\)
But, real value of dip is,
tan ø = \(\frac {B_V}{B_H}\)
∴ tan 45° = \(\frac {tan ø}{cos 30°}\)
∴ tan ø = tan 45° × cos 30°
= 1 × \(\frac {√3}{2}\)
∴ ø = tan-1 (0.866)

Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism

Question 3.
Two small and similar bar magnets have magnetic dipole moment of 1.0 Am² each. They are kept in a plane in such a way that their axes are perpendicular to each other. A line drawn through the axis of one magnet passes through the centre of other magnet. If the distance between their centres is 2 m, find the magnitude of magnetic field at the midpoint of the line joining their centres.
Answer:
Let P be the midpoint of the line joining the centres of two bar magnets. As shown in figure, P is at the axis of one bar magnet and at the equator of another bar magnet. Thus, the magnetic field on the axis of the first bar magnet at distance of 1 m from the centre will be,
Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism 4
Ba = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{r^{3}}\)
= 10-7 × \(\frac {2×1.0}{(1)^3}\)
= 2 × 10-7 Wb/m²
Magnetic field on the equator of second bar magnet will be,
Beq = \(\frac{\mu_{0}}{4 \pi} \frac{m}{r^{3}}\)
= 10-7 × \(\frac {1.0}{(1)^3}\)
= 1 × 10-7 Wb/m²
The net magnetic field at P,
Bnet = \(\sqrt {B_a^2+B_{eq}^2}\)
= \(\sqrt {(2×10^{-7})^2+(1×10^{-7})^2}\)
= \(\sqrt {(10^{-7})^2×(4+1)}\)
= √5 × 10-7 Wb/m²

Question 4.
A circular magnet is made with its north pole at the centre, separated from the surrounding circular south pole by an air gap. Draw the magnetic field lines in the gap. Draw a diagram to illustrate the magnetic lines of force between the south poles of two such magnets.
Answer:
i. For a circular magnet:
Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism 5

Question 5.
Two bar magnets are placed on a horizontal surface. Draw magnetic lines around them. Mark the position of any neutral points (points where there is no resultant magnetic field) on your diagram.
Answer:
The magnetic lines of force between two magnets will depend on their relative positions. Considering the magnets to be placed one besides the other as shown in figure, the magnetic lines of force will be as shown.
Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism 6

11th Physics Digest Chapter 12 Magnetism Intext Questions and Answers

Can you recall? (Textbook page no. 221)

Question 1.
What are the magnetic lines of force?
Answer:
The magnetic field around a magnet is shown by lines going from one end of the magnet to the other. These lines are named as magnetic lines of force.

Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism

Question 2.
What are the rules concerning the lines of force?
Answer:
i. Magnetic lines of force originate from the north pole and end at the south pole.
ii. The magnetic lines of force of a magnet or a solenoid form closed loops. This is in contrast to the case of an electric dipole, where the electric lines of force originate from the positive charge and end on the negative charge.
iii. The direction of the net magnetic field \(\vec {B}\) at a point is given by the tangent to the magnetic line of force at that point.
iv. The number of lines of force crossing per unit area decides the magnitude of magnetic field \(\vec {B}\)
v. The magnetic lines of force do not intersect. This is because had they intersected, the direction of magnetic field would not be unique at that point.

Question 3.
What is a bar magnet?
Answer:
Bar magnet is a magnet in the shape of bar having two poles of equal and opposite pole strengths separated by certain distance (2l).

Question 4.
If you freely hang a bar magnet horizontally, in which direction will it become stable?
Answer:
A bar magnet suspended freely in air always aligns itself along geographic N-S direction.

Try this (Textbook page no. 221)

You can take a bar magnet and a small compass needle. Place the bar magnet at a fixed position on a paper and place the needle at various positions. Noting the orientation of the needle, the magnetic field direction at various locations can be traced.
Answer:
When a small compass needle is kept at any position near a bar magnet, the needle always aligns itself in the direction parallel to the direction of magnetic lines of force.
Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism 7
Hence, by placing it at different positions, A, B, C, D,… as shown in the figure, the direction of magnetic lines of force can be traced. The direction of magnetic field will be a tangent at that point.

Maharashtra Board Class 11 Physics Solutions Chapter 12 Magnetism

Internet my friend: (Text book page no. 227)

https://www.ngdc.noaa.gov
[Students are expected to visit above mentioned link and collect more information about Geomagnetism.]

Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 10 Electrostatics Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 10 Electrostatics

1. Choose the correct option.

Question 1.
A positively charged glass rod is brought close to a metallic rod isolated from ground. The charge on the side of the metallic rod away from the glass rod will be
(A) same as that on the glass rod and equal in quantity
(B) opposite to that on the glass of and equal in quantity
(C) same as that on the glass rod but lesser in quantity
(D) same as that on the glass rod but more in quantity
Answer:
(A) same as that on the glass rod and equal in quantity

Question 2.
An electron is placed between two parallel plates connected to a battery. If the battery is switched on, the electron will
(A) be attracted to the +ve plate
(B) be attracted to the -ve plate
(C) remain stationary
(D) will move parallel to the plates
Answer:
(A) be attracted to the +ve plate

Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics

Question 3.
A charge of + 7 µC is placed at the centre of two concentric spheres with radius 2.0 cm and 4.0 cm respectively. The ratio of the flux through them will be
(A) 1 : 4
(B) 1 : 2
(C) 1 : 1
(D) 1 : 16
Answer:
(C) 1 : 1

Question 4.
Two charges of 1.0 C each are placed one meter apart in free space. The force between them will be
(A) 1.0 N
(B) 9 × 109 N
(C) 9 × 10-9 N
(D) 10 N
Answer:
(B) 9 × 109 N

Question 5.
Two point charges of +5 µC are so placed that they experience a force of 80 × 10-3 N. They are then moved apart, so that the force is now 2.0 × 10-3 N. The distance between them is now
(A) 1/4 the previous distance
(B) double the previous distance
(C) four times the previous distance
(D) half the previous distance
Answer:
(B) double the previous distance

Question 6.
A metallic sphere A isolated from ground is charged to +50 µC. This sphere is brought in contact with other isolated metallic sphere B of half the radius of sphere A. The charge on the two sphere will be now in the ratio
(A) 1 : 2
(B) 2 : 1
(C) 4 : 1
(D) 1 : 1
Answer:
(D) 1 : 1

Question 7.
Which of the following produces uniform electric field?
(A) point charge
(B) linear charge
(C) two parallel plates
(D) charge distributed an circular any
Answer:
(C) two parallel plates

Question 8.
Two point charges of A = +5.0 µC and B = -5.0 µC are separated by 5.0 cm. A point charge C = 1.0 µC is placed at 3.0 cm away from the centre on the perpendicular bisector of the line joining the two point charges. The charge at C will experience a force directed towards
(A) point A
(B) point B
(C) a direction parallel to line AB
(D) a direction along the perpendicular bisector
Answer:
(C) a direction parallel to line AB

2. Answer the following questions.

Question 1.
What is the magnitude of charge on an electron?
Answer:
The magnitude of charge on an electron is 1.6 × 10-19 C

Question 2.
State the law of conservation of charge.
Answer:
In any given physical process, charge may get transferred from one part of the system to another, but the total charge in the system remains constant”
OR
For an isolated system, total charge cannot be created nor destroyed.

Question 3.
Define a unit charge.
Answer:
Unit charge (one coulomb) is the amount of charge which, when placed at a distance of one metre from another charge of the same magnitude in vacuum, experiences a force of 9.0 × 109 N.

Question 4.
Two parallel plates have a potential difference of 10V between them. If the plates are 0.5 mm apart, what will be the strength of electric charge.
Answer:
V=10V
d = 0.5 mm = 0.5 × 10-3 m
To find: The strength of electric field (E)
Formula: E = \(\frac {V}{d}\)
Calculation: From formula,
E = \(\frac {10}{0.5×10^{-3}}\)
20 × 103 V/m

Question 5.
What is uniform electric field?
Answer:
A uniform electric field is a field whose magnitude and direction are same at all points. For example, field between two parallel plates as shown in the diagram.
Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics 1

Question 6.
If two lines of force intersect of one point. What does it mean?
Answer:
If two lines of force intersect of one point, it would mean that electric field has two directions at a single point.

Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics

Question 7.
State the units of linear charge density.
Answer:
SI unit of λ is (C / m).

Question 8.
What is the unit of dipole moment?
Answer:
i. Strength of a dipole is measured in terms of a quantity called the dipole moment.
Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics 2

ii. Let q be the magnitude of each charge and 2\(\vec{l}\) be the distance from negative charge to positive charge. Then, the product q(2\(\vec{l}\)) is called the dipole moment \(\vec{p}\).

iii. Dipole moment is defined as \(\vec{p}\) = q(2\(\vec{l}\))

iv. A dipole moment is a vector whose magnitude is q (2\(\vec{l}\)) and the direction is from the negative to the positive charge.

v. The unit of dipole moment is coulomb-metre (C m) or debye (D).

Question 9.
What is relative permittivity?
Answer:
i. Relative permittivity or dielectric constant is the ratio of absolute permittivity of a medium to the permittivity of free space.
It is denoted as K or εr.
i.e., K or εr = \(\frac {ε}{ε_0}\)

ii. It is the ratio of the force between two point charges placed a certain distance apart in free space or vacuum to the force between the same two point charges when placed at the same distance in the given medium.
i.e., K or εr = \(\frac {F_{vacuum}}{F_{medium}}\)

iii. It is also called as specific inductive capacity or dielectric constant.

3. Solve numerical examples.

Question 1.
Two small spheres 18 cm apart have equal negative charges and repel each other with the force of 6 × 10-8 N. Find the total charge on both spheres.
Solution:
Given: F = 6 × 10-8 N, r = 18 cm = 18 × 10-2 m
To find: Total charge (q1 + q2)
Formula: F = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
Calculation: From formula,
Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics 3
Taking square roots from log table,
∴ q = -4.648 × 10-10 C
….(∵ the charges are negative)
Total charge = q1 + q2 = 2q
= 2 × (-4.648) × 10-10
= -9.296 × 10-10 C

Question 2.
A charge + q exerts a force of magnitude – 0.2 N on another charge -2q. If they are separated by 25.0 cm, determine the value of q.
Answer:
Given: q1 = + q, q2 = -2q, F = -0.2 N
r = 25 cm = 25 × 10-2 m
To find: Charge (q)
Formula: F = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
Calculation: From formula,
Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics 4
[Note: The answer given above is calculated in accordance with textual method considering the given data]

Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics

Question 3.
Four charges of +6 × 10-8 C each are placed at the corners of a square whose sides are 3 cm each. Calculate the resultant force on each charge and show in direction on a diagram drawn to scale.
Answer:
Given: qA = qB = qC = qD = 6 × 10-8 C, a = 3 cm
Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics 5
∴ Resultant force on ‘A’
= FAD cos 45 + FAB cos 45 + FAC
= (3.6 × 10-2 × \(\frac {1}{√2}\)) + (3.6 × 10-2 × \(\frac {1}{√2}\)) + 1.8 × 10-2
= 6.89 × 10-2 N directed along \(\vec{F_{AC}}\)

Question 4.
The electric field in a region is given by \(\vec{E}\) = 5.0 \(\hat{k}\) N/C Calculate the electric flux through a square of side 10.0 cm in the following cases
i. The square is along the XY plane
ii. The square is along XZ plane
iii. The normal to the square makes an angle of 45° with the Z axis.
Answer:
Given: \(\vec{E}\) = 5.0 \(\hat{k}\) N/C, |E| = 5 N/C
l = 10 cm = 10 × 10-2 m = 10-1 m
A = l² – 10-2
To find: Electric flux in three cases.
1) (ø2) (ø3)
Formula: (ø1) = EA cos θ
Calculation:
Case I: When square is along the XY plane,
∴ θ = 0
ø1 = 5 × 10-2 cos 0
= 5 × 10-2 V m

Case II: When square is along XZ plane,
∴ θ = 90°
ø1 = 5 × 10-2 cos 90° = 0 V m

Case III: When normal to the square makes an angle of 45° with the Z axis.
∴ 0 = 45°
∴ ø3 = 5 × 10-2 × cos 45°
= 3.5 × 10-2 V m

Question 5.
Three equal charges of 10 × 10-8 C respectively, each located at the corners of a right triangle whose sides are 15 cm, 20 cm and 25 cm respectively. Find the force exerted on the charge located at the 90° angle.
Answer:
Given: qA = qB = qC = 10 × 10-8
Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics 6
Force on B due to A,
Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics 7

Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics

Question 6.
A potential difference of 5000 volt is applied between two parallel plates 5 cm apart. A small oil drop having a charge of 9.6 x 10-19 C falls between the plates. Find (i) electric field intensity between the plates and (ii) the force on the oil drop.
Answer:
Given: V = 5000 volt, d = 5 cm = 5 × 10-2 m
q = 9.6 × 10-19 C
To find:
i. Electric field intensity (E)
ii. Force (F)
Formula:
i. E = \(\frac {V}{d}\) \(\frac {q}{r}\)
ii. E = \(\frac {F}{q}\)
Calculation: From formula (i),
E = \(\frac {F}{q}\) = 105 N/C
From formula (ii)
F = E x q
= 105 × 9.6 × 10-19
= 9.6 × 10-14 N

Question 7.
Calculate the electric field due to a charge of -8.0 × 10-8 C at a distance of 5.0 cm from it.
Answer:
Given: q = – 8 × 10-8 C, r = 5 cm = 5 × 10-2 m
To Find: Electric field (E)
Formula: E = \(\frac {1}{4πε_0}\) \(\frac {q}{r^2}\)
Calculation: From formula,
E = 9 × 109 × \(\frac {(-8×10^{-8})}{(5×10^{-2})^2}\)
= -2.88 × 105 N/C

11th Physics Digest Chapter 10 Electrostatics Intext Questions and Answers

Can you recall? (Textbookpage no. 188)

Question 1.
Have you experienced a shock while getting up from a plastic chair and shaking hand with your friend?
Answer:
Yes, sometimes a shock while getting up from a plastic chair and shaking hand with friend is experienced.

Question 2.
Ever heard a crackling sound while taking out your sweater in winter?
Answer:
Yes, sometimes while removing our sweater in winter, some crackling sound is heard and the sweater appears to stick to body.

Question 3.
Have you seen the lightning striking during pre-monsoon weather?
Answer:
Yes, sometimes lightning striking during pre-pre-monsoon weather seen.

Can you tell? (Textbook page no. 189)

i. When a petrol or a diesel tanker is emptied in a tank, it is grounded.
ii. A thick chain hangs from a petrol or a diesel tanker and it is in contact with ground when the tanker is moving.
Answer:
i. When a petrol or a diesel tanker is emptied in a tank, it is grounded so that it has an electrically conductive connection from the petrol or diesel tank to ground (Earth) to allow leakage of static and electrical charges.

ii. Metallic bodies of cars, trucks or any other big vehicles get charged because of friction between them and the air rushing past them. Hence, a thick chain is hanged from a petrol or a diesel tanker to make a contact with ground so that charge produced can leak to the ground through chain.

Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics

Can you tell? (Textbook page no. 194)

Three charges, q each, are placed at the vertices of an equilateral triangle. What will be the resultant force on charge q placed at the centroid of the triangle?
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics 8
Since AD. BE and CF meets at O, as centroid of an equilateral triangle.
∴ OA = OB = OC
∴ Let, r = OA = OB = OC
Force acting on point O due to charge on point A,
\(\overrightarrow{\mathrm{F}}_{\mathrm{OA}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}_{i}^{2}} \hat{\mathrm{r}}_{\mathrm{AO}}\)
Force acting on point O due to charge on point B,
\(\overrightarrow{\mathrm{F}}_{O \mathrm{~B}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}}_{\mathrm{BO}}\)
Force acting on point O due to charge on point C,
\(\overrightarrow{\mathrm{F}}_{\mathrm{OC}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}} \mathrm{co}\)
∴ Resultant force acting on point O,
F = \(\vec{F}\)OA + \(\vec{F}\)OB + \(\vec{F}\)OC
On resolving \(\vec{F}\)OB and \(\vec{F}\)OC, we get –\(\vec{F}\)OA
i.e., \(\vec{F}\)OB + \(\vec{F}\)OC = –\(\vec{F}\)OA
∴ \(\vec{F}\) = \(\vec{F}\)OA – \(\vec{F}\)OA = 0
Hence, the resultant force on the charge placed at the centroid of the equilateral triangle is zero.

Can you tell? (Textbook page no. 197)

Why a small voltage can produce a reasonably large electric field?
Answer:

  1. Electric field produced depends upon voltage as well as separation distance.
  2. Electric field varies linearly with voltage and inversely with distance.
  3. Hence, even if voltage is small, it can produce a reasonable large electric field when the gap between the electrode is reduced significantly.

Can you tell? (Textbook page no. 198)

Lines of force are imaginary; can they have any practical use?
Answer:
Yes, electric lines of force help us to visualise the nature of electric field in a region.

Can you tell? (Textbook page no. 204)

The surface charge density of Earth is σ = -1.33 nC/m². That is about 8.3 × 109 electrons per square metre. If that is the case why don’t we feel it?
Answer:
The Earth along with its atmosphere acts as a neutral system. The atmosphere (ionosphere in particular) has nearly equal and opposite charge.

As a result, there exists a mechanism to replenish electric charges in the form of continual thunder storms and lightning that occurs in different parts of the globe. This makes average charge on surface of the Earth as zero at any given time instant. Hence, we do not feel it.

Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics

Internet my friend (Textbook page no. 205)

i. https://www.physicsclassroom.com/class
ii. https://courses.lumenleaming.com/physics/
iii. https://www,khanacademy.org/science
iv. https://www.toppr.com/guides/physics/
[Students are expected to visit the above mentioned websites and collect more information about electrostatics.]