Important Questions

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?

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.

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.

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,
F_med = \(\frac {1}{4πε}\) (\(\frac {q_1q_2}{r^2}\)) …………. (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,
F_vac = \(\frac {1}{4πε_0}\) (\(\frac {q_1q_2}{r^2}\)) …………. (1)
Dividing equation (2) by equation (1),

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 F_water and F_vacuum. 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 ε_r, is given by
F_med = \(\frac{1}{4 \pi \varepsilon_{0} \varepsilon_{r}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}\) …………. (1)
If the medium is vacuum,
F_vac = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}\) …………. (2)

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

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.

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

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
m_p = 1.67 × 10^-27 kg.
q_p = 1.67 × 10^-19 C, r = 10^-15
To find:
i. Electrostatic Force (F_E)
ii. Gravitational Force (F_G)
Formula: i. F_E = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
ii. F_E = \(\frac {Gm_1m_2}{r^2}\)
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
∴ F_E ≈ 10³⁶ × F_G

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.

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 ……., q_n 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 = r_n \(\hat{r}\)_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)

Answer:

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,

In ∆ABC 4
cos θ = \(\frac {4}{5}\)
θ = cos^-1 (\(\frac {4}{5}\)) = 36.87°
Forces acting points A are

= 59.36 N
Direction of resultant force is 36.87° (north of west)

[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:

From the formula,
Force on q₂ because of q₁

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.

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.

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.

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)

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.

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

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 = V_q,
but W = Fd
∴ Vq = Fd
∴ \(\frac {F}{q}\) = \(\frac {V}{d}\) = E
∴ Electric field can be defined as E = V/d.

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.

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.

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

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?

Answer:
Electric field at P due to the charges at A, B and C are shown in the figure.

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, E_A = – E_C .
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,

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 × q_e ….[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,
V_A = -10 volt,
V_B = V volt,
W_AB = 100 J
To Find: Potential (V)
Formula: V_B – V_A = \(\frac {W_{AB}}{q_0}\)
Calculation: From formula,
V – (-10) = \(\frac {100}{4}\) = 25
∴ V + 10 = 25
∴ V = 15 volt

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:
V_A = -10V,
V_B = 0,
q = 2 × 10^-6 C
To Find: Work done (W)
Formula: V_BA = \(\frac {W}{q}\)
Calculation: From formula,
W = V_BA × q
= (V_B – V_A) × 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 \(\vec{E}\).

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.

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)
∴ ø_E = \(\int_{\mathrm{s}} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dS}}=\frac{+\mathrm{q}}{\varepsilon_{0}}\)

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 q_n, 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
\(\phi_{\mathrm{E}}=\frac{\mathrm{q}_{1}}{\varepsilon_{0}}+\frac{\mathrm{q}_{2}}{\varepsilon_{0}}+\frac{\mathrm{q}_{3}}{\varepsilon_{0}}+\ldots+\frac{\mathrm{q}_{\mathrm{n}}}{\varepsilon_{0}}=\sum_{\mathrm{i}=1}^{\mathrm{i}=\mathrm{n}} \frac{\mathrm{q}_{\mathrm{i}}}{\varepsilon_{0}}=\frac{\mathrm{Q}}{\varepsilon_{0}}\)

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:

Positive sign indicates that the flux is directed outwards, away from the charge.

If the charge is negative, the flux will be is directed inwards.

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.

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.

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.

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.

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

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 \(\vec{PA}\)

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 E_A and E_B
∴ \(\vec{E}\)_eq = \(\vec{E}\)_A + \(\vec{E}\)_B

iv. Consider ∆ACP
(AP)² = (PC)² + (AC)² = r² + l² = (BP)²

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.

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 E_A and E_B are E_Asin θ and E_B 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 E_A and E_B are E_Acos θ and E_Bcos θ respectively. They are of equal magnitude and are in the same direction.
∴ |\(\vec{E}\)_eq| = E_A cos θ + E_B cos θ From equation (3),
|\(\vec{E}\)_eq| = 2E_A cos θ

x. The direction of this field is along –\(\vec{P}\) (anti-parallel to \(\vec{P}\)).

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.

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.

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.

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.

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., F_E = 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.

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.

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.

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 q_n
(C) Σ q_n
(D) q₁ × q₂ × q₃ × ……… q_n
Answer:
(C) Σ q_n

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

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 15 Hydrocarbons Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 15 Hydrocarbons

Question 1.
What are unsaturated and saturated hydrocarbons?
Answer:
Hydrocarbons which contain carbon-carbon multiple bond (C=C or C≡C) are called unsaturated hydrocarbons, whereas those which contain carbon-carbon single bond (C-C) are called saturated hydrocarbons.

Question 2.
How are hydrocarbons classified?
Answer:

Question 3.
Define alkanes. Write general formula of alkanes.
Answer:

1. Alkanes are aliphatic saturated hydrocarbons containing carbon-carbon and carbon-hydrogen single
   covalent bonds.
2. They have a general formula C_nH_2n+2 where, ‘n’ stands for number of carbon atoms in the alkane molecule.

Question 4.
Give information about isomerism in alkanes. Write the all possible structural isomers of a saturated hydrocarbon containing four carbon atoms along with their IUPAC names.
Answer:
i. Alkanes with more than three carbon atoms generally exhibit, structural isomerism and in particular, the chain isomerism.
ii. The number of possible structural isomers increase rapidly with the number of carbon atoms.
iii. Structural isomers of a saturated hydrocarbon containing four carbon atoms along with their IUPAC names.

Question 5.
Write all the possible structural isomers of a saturated hydrocarbon having molecular formula C₅H₁₂.
Answer:

Question 6.
Define sigma bond.
Answer:
A single covalent bond formed by the coaxial overlap of orbitals is called sigma (σ) bond.

Question 7.
i. Why do C – C bonds in alkanes undergo rotation?
ii. What are conformations?
Answer:
i. a. Alkanes have single covalent bonds (sigma bonds) formed by the coaxial overlap of orbitals.
b. As a direct consequence of coaxial overlap of orbitals, a sigma bond is cylindrically symmetrical and the extent of orbital overlap is unaffected by rotation about the single bond and therefore, C – C bonds undergo rotation.

ii. a. In alkanes, the atoms bonded to one carbon of a C – C single bond change their relative position with reference to the atoms on the other carbon of that bond on rotation of that C – C single bond.
b. The resulting arrangements of the atoms in space about the C – C single bond are called conformations or conformational isomers. Innumerable conformations result on complete rotation of a C – C single bond through 360°.

Question 8.
i. What is conformational isomerism?
ii. Name the two extreme conformations shown by ethane molecule.
Answer:
i. The phenomenon of existence of conformation is a type of stereoisomerism and is known as conformational isomerism.
ii. Ethane molecule shows the following two extreme conformations:

• Staggered conformation
• Eclipsed conformation

Question 9.
Draw structures representing staggered and eclipsed conformations of ethane using:
i. Sawhorse projection
ii. Newman projection
Answer:
i. Sawhorse projection of ethane:

ii. Newman projection of ethane:

Question 10.
How are alkanes obtained from crude oil?
Answer:
Alkanes are obtained by fractional distillation of crude oil in oil refineries.

Question 11.
How are alkanes obtained from alkenes and alkynes?
OR
How are alkanes obtained from catalytic hydrogenation of alkenes and alkynes?
Answer:
i. Catalytic hydrogenation of alkenes or alkynes with dihydrogen gas gives corresponding alkanes.
ii. Finely divided powder of platinum (Pt) or palladium (Pd) catalyse the hydrogenation of alkenes and alkynes at room temperature.
iii. Relatively high temperature and pressure are required with finely divided nickel as the catalyst.
e.g. a. Propene to propane:

b. Ethyne to ethane:

Question 12.
Write the general reactions for the catalytic hydrogenation of alkenes and alkynes.
Answer:
General reaction for catalytic hydrogenation of alkenes:

General reaction for catalytic hydrogenation of alkynes:

Question 13.
Write the structures of alkenes that on catalytic hydrogenation give n-butane.
Answer:

Question 14.
Explain the preparation of alkanes by reduction of alkyl halides with the help of an example.
Answer:
i. Alkanes can be prepared by reduction of alkyl halides using zinc and dilute hydrochloric acid.
ii. The reduction of alkyl halides is due to nascent hydrogen obtained from the reaction between reducing agent Zn and dilute HCl.
e.g. Reduction of methyl iodide to methane.

Question 15.
How are alkanes obtained by Wurtz reaction?
Answer:
Alkyl halides on treatment with reactive sodium metal in dry ether, gives higher alkanes having double the number of carbon atoms. This is called as Wurtz coupling reaction.

Question 16.
How will you convert ethyl chloride into n-butane?
Answer:
Ethyl chloride on heating with sodium metal in presence of dry ether gives n-butane.

Question 17.
Write chemical equations for reactions that take place on treating ethereal solutions of:
i. Methyl iodide with sodium metal
ii. Ethyl iodide with sodium metal
Answer:

Question 18.
Explain the preparation of Grignard reagents.
OR
What is Grignard reagent? Explain its preparation.
Answer:
Grignard reagent are alkyl magnesium halides obtained by treating alkyl halides with dry magnesium metal in the presence of dry ether.

General Reaction:

Question 19.
State the action of water on methyl magnesium bromide in dry ether with the help of a chemical reaction.
Answer:

Question 20.
Write the reagents involved in the following conversions.

Answer:

Question 21.
Straight chain alkanes have higher melting and boiling points as compared to branched isomeric alkanes. Give reason.
Answer:
i. The electronegativity of carbon and hydrogen is nearly the same. Therefore, C-H and C-C bonds are nonpolar covalent bonds and hence, alkanes are nonpolar.
ii. Alkane molecules are held together by weak intermolecular van der Waals forces.
iii. Larger the surface area of molecules, stronger are such intermolecular van der Waals forces.
iv. In straight chain alkane molecules, surface area is relatively larger as compared to branched chain alkanes and as a result, the intermolecular forces are relatively stronger in straight chain alkanes than in branched chain alkanes.

Hence, straight chain alkanes have higher melting and boiling points as compared to branched alkanes.

Question 22.
State physical properties of alkanes.
Answer:

• Alkanes are colourless and odourless.
• At room temperature, the first four alkanes are gases, alkanes having 5 to 17 carbon atoms are liquids while the rest all are solids.
• Alkanes are readily soluble in organic solvents such as chloroform, ether or ethanol while they are insoluble in water.
• Alkanes have low melting and boiling points which increases with an increase in the number of carbon atoms for straight chain molecules. But for branched chain molecules, more the number of branches, lower is the boiling/melting point.

Note: [Melting and boiling points of alkanes]

Question 23.
Define substitution reactions.
Answer:
The reactions in which an atom or group of atoms in a molecule is replaced by another atom or group of atoms is called as substitution reactions.
e.g. Halogenation of alkanes.

Question 24.
i. What is halogenation of alkanes?
ii. Write the order of reactivity of halogens towards alkanes.
Answer:
i. Substitution of H atoms of alkanes by X (halogen, X = Cl, Br, I and F) atom is called halogenation of alkanes.
ii. The reactivity of halogens toward alkanes follows the order: F₂ > Cl₂ > Br₂ > I₂
[Note: The ease of replacement of hydrogen atoms from the carbon in alkanes is in the order: 3 > 2 > 1.]

Question 25.
Explain reactions involved in chlorination of methane.
Answer:
Alkanes react with chlorine gas in presence of UV light or diffused sunlight or at a high temperature (573-773 K) to give a mixture of alkyl halides.

Chlorination of methane:


Tetrachloromethane is a major product when excess of chlorine is used. Chloromethane is obtained as major product when excess of methane is employed.

Question 26.
Predict the products in the following set of reactions.

Answer:

Question 27.
What is the action of Cl₂ and Br₂ on 2-methylpropane?
Answer:

[Note: In bromination, there is high degree of selectivity as to which hydrogen atoms are replaced.]

Question 28.
Explain mechanism of halogenation of alkanes.
Answer:
i. Halogenation of alkanes follows the free radical mechanism.
ii. Homolysis of halogen molecule (X₂) generates halogen atoms, i.e., halogen free radicals.
iii. The mechanism of the first step of chlorination of methane is shown below:

Question 29.
Why are alkanes used as fuels?
Answer:
On heating in the presence of air or dioxygen, alkanes are completely oxidized to carbon dioxide and water with the evolution of a large amount of heat. Hence, alkanes are used as fuels.

Question 30.
What is combustion of alkanes? Write a general equation for alkane combustion.
Answer:
On heating in the presence of air or dioxygen, alkanes are completely oxidized to carbon dioxide and water with the evolution of a large amount of heat. This is known as combustion reaction of alkanes.

General representative equation for combustion is:

Question 31.
Write chemical equations for combustion of butane and methane.
Answer:
i. Combustion of butane:

ii. Combustion of methane:

Question 32.
Write a short note on pyrolysis of alkanes.
Answer:
Alkanes on heating at higher temperature in absence of air decompose to lower alkanes, alkenes and hydrogen, etc. This is known as pyrolysis or cracking.
e.g. Pyrolysis of hexane

Question 33.
Explain aromatization reaction of alkanes. Give its one application.
Answer:
i. Straight chain alkanes containing 6 to 10 carbon atoms are converted to benzene and its homologues, on heating under 10 to 20 atm pressure at about 773 K in the presence of V₂O₅, Cr₂O₃, MO₂O₃, etc. supported over alumina.

ii. The reaction involves simultaneous dehydrogenation and cyclization. This reaction is known as aromatization or reforming.

This process is used in refineries to produce high quality gasoline which is used in automobiles as fuel.

Question 34.
Collect the information on CNG and LPG with reference to the constituents and the advantages of CNG over LPG.
Answer:
Constituents of CNG (Compressed Natural Gas):
It mainly consists of methane compressed at a pressure of 200-248 bar.
Constituents of LPG (Liquefied Petroleum Gas):
It contains a mixture of propane and butane liquefied at 15°C and a pressure of 1.7 – 7.5 bar.

Advantages of CNG over LPG:

• CNG is cheaper and cleaner than LPG.
• CNG produces less pollutants than LPG.
• It does not evolve gases containing sulphur and nitrogen.
• Octane rating of CNG is high, hence thermal efficiency is more.
• Vehicles powered by CNG produces less carbon monoxide and hydrocarbon emission.

[Note: Students are expected to collect additional information on their own.]

Question 35.
Write the uses of alkane.
Answer:
Uses of alkanes:

• First four alkanes are used as a fuel mainly for heating and cooking purpose. For example, LPG and CNG.
• CNG, petrol and diesel are used as fuel for automobiles.
• Lower liquid alkanes are used as solvent.
• Alkanes with more than 35 C atoms (tar) are used for road surfacing.
• Waxes are high molecular weight alkanes. They are used as lubricants. They are also used for the preparation of candles and carbon black that is used in manufacture of printing ink, shoe polish, etc.

Question 36.
i. Write the general molecular formula of alkenes.
ii. Why are alkenes also known as olefins?
Answer:
i. Alkenes have general formula C_nH_2n, where, n = 2,3,4… etc.
ii. Alkenes are also known as olefins because the first member ethene/ethylene reacts with chlorine to form oily substance.
[Note: Alkenes with one carbon-carbon double bond, contain two hydrogen atoms less than the corresponding alkanes.]

Question 37.
Define alkadienes and alkatrienes. Give one example for each.
Answer:
i. The aliphatic unsaturated hydrocarbons containing two carbon-carbon double bonds are called as alkadienes.

ii. The aliphatic unsaturated hydrocarbons containing three carbon-carbon double bonds are called as alkatrienes.

Question 38.
Explain structural isomerism in alkenes by giving an example.
Answer:
Alkenes with more than three carbon atoms show structural isomerism.
e.g. Alkene with molecular formula C4H8 is butene. The structural formulae for C₄H₈ can be drawn in three different ways:

Question 39.
Draw structures of chain isomers of butene.
Answer:

Question 40.
Draw structures of position isomers of butene.
Answer:

Question 41.
Define geometrical isomerism.
Answer:
The isomerism which arises due to the difference in spatial arrangement of atoms or groups about doubly bonded carbon (C=C) atoms is called geometrical isomerism.

Question 42.
Explain geometrical isomerism using a general example.
Answer:
i. If the two atoms or groups bonded to each end of the C=C double bond are different, then the molecule can be represented by two different special arrangements of the groups as follows:

ii. In structure (A), two identical atoms or groups lie on the same side of the double bond.
The geometrical isomer in which two identical or similar atoms or groups lie on the same side of the double bond is called cis-isomer.
iii. In structure (B), two identical atoms or groups lie on the opposite side of the double bond.
The geometrical isomer in which two identical or similar atoms or groups lie on the opposite side of the double bond is called trans-isomer.
iv. Due to different arrangement of atoms or groups in space, these isomers differ in their physical properties like melting point, boiling point, solubility, etc.

Question 43.
i. Define cis- and trans-isomer.
ii. Draw geometrical isomers of but-2-ene.
Answer:
i. a. Cis isomer: The geometrical isomer in which two identical or similar atoms or groups lie on the same side of the double bond is called a cis-isomer.
b. Trans isomer: The geometrical isomer in which two identical or similar atoms or groups lie on the opposite side of the double bond is called a trans-isomer.

ii. Geometrical or cis-trans isomers of but-2-ene are represented as:

Question 44.
State whether the following alkenes can exhibit geometrical (or cis-trans) isomerism or not. Give reason for the answer.
i. CH₃ – CH₂ – CH₂ – CH = CH₂
ii. CH₃ – CH₂ – CH = C(CH₃)₂
Answer:
Both the alkenes (i) and (ii) cannot exhibit geometrical isomerism, since 1 alkene is a terminal alkene (containing two H-atoms on the same side of the double bond) while the 2nd alkene is a 1,1-disubstituted alkene (containing two identical alkyl groups on the same side of the double bond).

Question 45.
Write the general formulae of alkenes which exhibit cis-trans isomerism.
Answer:
Alkenes having the following general formulae exhibit cis-trans isomerism:
RCH=CHR, R₁R₂C=CR₁R₃, R₁CH=CR₁R₂, R₁CH=CR₂R₃, R₁CH=CHR₂ and R₁R₂C=CR₃R₄

Question 46.
Draw structures of cis-trans isomers for the following:

Answer:

Question 47.
Which of the following compounds will show geometrical isomerism?

Answer:
Compounds (III), (IV) and (V) will show geometrical isomerism as they have each of the doubly bonded carbon atoms in their structures, attached to different atoms/groups of atoms.

Question 48.
Alkenes can be obtained from which industrial sources?
Answer:

1. The most important alkenes for chemical industry are ethene, propene and buta-1,3-diene.
2. Alkenes containing up to four carbon atoms can be obtained in pure form from the petroleum products.
3. Ethene is produced from natural gas and crude oil by cracking.

Question 49.
What is β-elimination reaction? Explain in brief.
Answer:
The reactions in which two atoms or groups are eliminated from adjacent carbon atoms are called 1,2-elimination reactions. Since the atom/group is removed from β-carbon atom (β to the leaving group) it is called as β-elimination reaction.

The hybridization of each C in the reactant is sp³ while that in the product is sp². This means elimination reactions cause change in hybridization state while forming multiple bonds from single bond.

Question 50.
i. What is dehydrohalogenation reaction?
ii. How is it carried out. Explain with an example.
Answer:
i. a. The reactions in which there is removal of hydrogen (H) atom and halogen (X) atom from adjacent carbon atoms are known as dehydrohalogenation reactions.
b. The carbon carrying X is called α-carbon atom. The hydrogen atom from adjacent carbon called β-carbon atom, is removed and hence, the reaction is known as β-elimination.

ii. When an alkyl halide is boiled with a hot concentrated alcoholic solution of a strong base like KOH or NaOH, alkene is formed with removal of water molecule.

[Note: The ease of dehydrohalogenation of alkyl halides is in the order 3 > 2 > 1.]

Question 51.
State Saytzeff rule.
Answer:
In dehydrohalogenation the preferred product is the alkene that has the greater number of alkyl groups attached to doubly bonded carbon atoms.

Question 52.
Write and explain dehydrohalogenation reaction of 2-chlorobutane.
Answer:

In dehydrohalogenation of 2-chlorobutane, but-2-ene (disubstituted alkene) is the preferred product because it is formed faster than but-1-ene (monosubstituted alkene) which is in accordance with Saytzeff rule.

Question 53.
Write the CORRECT order of stability of alkenes with respect to Saytzeff rule.
R CH = CH₂, CH₂ = CH₂, R₂C = CH₂, R₂C = CR₂, RCH = CHR, R₂C = CHR
Answer:
R₂C = CR₂ > R₂C = CHR > R₂C = CH₂, RCH = CHR > RCH = CH₂ > CH₂ = CH₂

Question 54.
Explain dehydration reaction of alcohols.
Answer:
i. Alcohols on heating with sulphuric acid form alkenes with elimination of water molecule. The reaction is known as catalysed dehydration of alcohols.
ii. The exact conditions of dehydration depend upon the alcohol.
iii. Dehydration of alcohol is an example of β-elimination since -OH group from α-carbon along with H-atom from β-carbon is removed.

The ease of dehydration of alcohol is in the order 3° > 2° > 1°.

Question 55.
Write dehydration reaction of 1°, 2° and 3° alcohols giving one example for each.
Answer:
The ease of dehydration of alcohols is in the order 3° > 2° > 1°.

Question 56.
Explain isomerism with structure in the product obtained by acid catalysed dehydration of pentan-2-ol.
Answer:
i. Pentan-2-ol on acid catalysed dehydration, forms the following isomers.

ii. A and B are position isomers.
iii. Pent-2-ene has the following geometrical isomers:

Question 57.
What is dehalogenation? Write the general reaction for dehalogenation of vicinal dihalides.
Answer:
i. Removal of two halogen atoms from adjacent carbon atoms is called dehalogenation.
ii. The dihalides of alkane in which two halogen atoms are attached to adjacent carbon atoms are called vicinal dihalides.
iii. Vicinal dihalides on heating with zinc metal form an alkene.

Question 58.
How is propene obtained by dehalogenation reaction?
Answer:

Question 59.
How are geometrical isomers of alkenes obtained from alkynes?
Answer:
Alkenes are obtained by partial reduction of alkynes wherein C = C triple bond of alkynes is reduced to a C = C double bond by:
i. using calculated quantity of dihydrogen in presence of Lindlar’s catalyst (palladised charcoal deactivated partially with quinoline or sulphur compound) to give the cis-isomer of alkene.

ii. using sodium in liquid ammonia to give trans-isomer of alkene.

Question 60.
Write physical properties of alkenes.
Answer:

• Alkenes are nonpolar or weakly polar compounds that are insoluble in water, and soluble in nonpolar solvents like benzene, ether, chloroform.
• They are less dense than water.
• The boiling point of alkene rises with increasing number of carbons.
• Branched alkenes have lower boiling points than straight chain alkenes.
• The boiling point of alkene is very nearly the same as that of alkane with the same carbon skeleton.

Question 61.
Arrange the following alkenes in increasing order of their boiling points.
But-1-ene, 2,3-dimethylbut-2-ene, 2-methylpropene, propene, 2-methylbut-2-ene.
Answer:
Propene < 2-methylpropene < but-1-ene < 2-methylbut-2-ene < 2,3-dimethylbut-2-ene.
Note: Melting points and boiling points of alkenes.

Question 62.
What kind of reactions do alkenes undergo? Give reason.
Answer:
Alkenes undergo electrophilic addition reactions since they are unsaturated and contain pi (π) electrons.

Question 63.
Write a note on halogenation of alkenes.
OR
Explain the formation of vicinal dihalides from alkenes with the help of examples.
Answer:
Alkenes are converted into the corresponding vicinal dihalides by addition of halogens (X₂ = Cl₂ or Br₂).

Question 64.
How is carbon-carbon double bond in a compound detected by bromination?
Answer:
When an alkene like ethene is treated with bromine in presence of CCl₄, the red-brown colour of bromine disappears due to following reaction.

Hence, decolourisation of bromine is used to detect the presence of C = C bond in unknown compounds.

Question 65.
Explain the formation of alkyl halides from alkenes.
Answer:
i. Alkenes react with hydrogen halides (HX) like hydrogen chloride, hydrogen bromide and hydrogen iodide to give corresponding alkyl halides (haloalkanes). This reaction is known as hydrohalogenation of alkenes.


ii. The order of reactivity of halogen acids is HI > HBr > HCl.

Question 66.
State Markovnikov’s rule and explain it with the help of an example.
Answer:
i. Markovnikov’s rule: When an unsymmetrical reagent is added to an unsymmetrical alkene, the negative part (X-) of the reagent gets attached to the carbon atom which carries less number of hydrogen atoms.
ii. For example, addition of HBr to unsymmetrical alkenes yield two isomeric products.

iii. Experimentally it has been found that 2-Bromopropane is the major product.
[Note: Addition of HBr to symmetrical alkenes yields only one product.]

Question 67.
Explain Anti-Markovnikov’s addition or peroxide effect or Kharasch-Mayo effect.
Answer:
In 1933, M. S. Kharasch and F. R. Mayo discovered that the addition of HBr to unsymmetrical alkene in the presence of organic peroxide (R-O-O-R) takes place in the opposite orientation to that suggested by Markovnikov’s rule.

Question 68.
Write the structure of major alkyl halide obtained by the action of HCl on pent-1-ene
i. in presence of peroxide
ii. in absence of peroxide.
Answer:
The structures of alkyl halides obtained by the action of hydrogen bromide on pent-1-ene are as follows:
i. In presence of peroxide:

ii. In absence of peroxide:

[Note: Presence/absence of peroxide has no effect on addition of HCl or HI.]

Question 69.
Explain the formation of alcohols from alkenes using conc. sulphuric acid with the help of an example.
Answer:
i. Alkenes react with cold concentrated sulphuric acid to form alkyl hydrogen sulphate (ROSO₃H). The addition takes place according to Markovnikov’s rule as shown in the following steps.


ii. If alkyl hydrogen sulphate is diluted with water and heated, then an alcohol having the same alkyl group as the original alkyl hydrogen sulphate is obtained.

iii. This is an excellent method for the large-scale manufacture of alcohols.

Question 70.
What is hydration of alkenes?
Answer:
i. Reactive alkenes on adding water molecules in the presence of concentrated sulphuric acid, form alcohol.
ii. The addition of water takes place according to Markovnikov’s rule. This reaction is known as hydration of alkenes.

Question 71.
Complete the following conversion.

Answer:

Question 72.
But-1-ene and 2-methylpropene are separately treated with following reagents. Predict the product/products. Indicate major/minor product,
i. HBr
ii. H₂SO₄ / H₂O
Answer:
i. HBr:

ii. H₂SO₄ / H₂O:

Question 73.
Explain: Ozonolysis
Answer:
i. The C = C double bond in alkenes, gets cleaved on reaction with ozone followed by reduction.
ii. The overall process of formation of ozonide by reaction of ozone with alkene in the first step and then decomposing it to the carbonyl compounds by reduction in the second step is called ozonolysis.
iii. When ozone gas is passed into solution of the alkene in an inert solvent like carbon tetrachloride, unstable alkene ozonide is obtained.
iv. This is subsequently treated with water in the presence of a reducing agent zinc dust to form carbonyl compounds, namely, aldehydes and/or ketones.

Question 74.
Write reactions for the ozonolysis of the following alkenes:
i. Ethene
ii. Propene
Answer:

Question 75.
What is the role of zinc dust, in ozonolysis reaction?
Answer:
In ozonolysis, the role of zinc dust is to prevent the formation of hydrogen peroxide which oxidizes aldehydes to corresponding acids.

Question 76.
State TRUE or FALSE. If false, correct the statement.
i. In the cleavage products of ozonide, a carbonyl group (C=O) is formed at each of the original doubly bonded carbon atoms.
ii. In ozonolysis, the structure of original alkene reactant cannot be identified by knowing the number and arrangement of carbon atoms in aldehydes and ketones produced.
iii. Ozonolysis reaction is used to locate the position and determine the number of double bonds in alkenes.
Answer:
i. True
ii. False
In ozonolysis, knowing the number and arrangement of carbon atoms in aldehydes and ketones produced, we can identify the structure of original alkene.
iii. True

Question 77.
Identify the alkene which produces a mixture of methanal and propanone on ozonolysis. Write the reactions involved.
Answer:
i. The structure of alkene which produces a mixture of methanol and propanone on ozonolysis is

ii. Reactions:

Question 78.
Explain the process of hydroboration-oxidation of alkenes.
Answer:
i. Alkenes with diborane in tetrahydrofuran (THF) solvent undergo hydroboration to form trialkylborane, which on oxidation with alkaline peroxide forms primary alcohol.
ii. The overall reaction gives anti-Markovnikov’s product from unsymmetrical alkenes.

Question 79.
Write reactions for the following conversion by hydroboration-oxidation reaction.
Ethene to ethanol
Answer:
Ethene to ethanol:

Question 80.
Define: Polymerization
Answer:
The process in which large number of small molecules join together and form very large molecules with repeating units is called polymerization.

Question 81.
What is the difference between monomer and polymer?
Answer:
The compound having very large molecules made of large number of repeating small units is called polymer while the simple compound forming the repeating units in the polymer is called monomer.

Question 82.
How is ethene converted to polyethene?
Answer:
Ethene at high temperature and under high pressure interacts with oxygen, and undergoes polymerization giving high molecular weight polymer called polyethene.

Here, n represents the number of repeating units and is a large number.

Question 83.
Explain the process of hydroxylation of alkenes.
OR
What is the action of alkaline KMnO₄ on alkenes?
Answer:
Alkenes react with cold and dilute alkaline potassium permanganate to form glycols.

Question 84.
Explain Baeyer’s test giving one example.
Answer:
i. During hydroxylation of alkenes the purple colour of KMnO₄ disappears.
ii. Hence, such reaction serves as a qualitative test for detecting the presence of double bond in the sample compound. This is known as Baeyer’s test.
e.g. As propene contains a double bond, it reacts with alkaline KMnO₄ to give colourless propane-1,2-diol as product. Therefore, the purple colour of alkaline KMnO₄ disappears.

Question 85.
What is the action of following reagents on but-1-ene and but-2-ene?
i. Bromine
ii. Cold and dilute alkaline KMnO₄.
Answer:
i. Action of Br₂:

ii. Action of cold and dilute alkaline KMnO₄:

Question 86.
Describe the action of acidic potassium permanganate on alkenes.
Answer:
Acidic potassium permanganate or acidic potassium dichromate oxidizes alkenes to ketones or acids depending upon the nature of the alkene and the experimental conditions. This is called oxidative cleavage of alkenes.
e.g.

Question 87.
Complete the following conversions.

Answer:

Question 88.
State some important uses of alkenes.
Answer:

• Alkenes are used as starting materials for preparation of alkyl halides, alcohols, aldehydes, ketones, acids, etc.
• Ethene and propene are used to manufacture polythene, polypropylene which are used in polyethene bags, toys, bottles, etc.
• Ethene is used for artificial ripening of fruits, such as mangoes.

Question 89.
What are alkynes? Write their general formula.
Answer:

• Alkynes are aliphatic unsaturated hydrocarbons containing at least one C = C.
• Their general formula is C_nH_2n-2.

Question 90.
Explain position isomerism in alkyne.
Answer:
Alkynes show position isomerism which is a type of structural isomerism.
e.g. But-1-yne and but-2-yne, both are represented by C₄H₆, however, both of them differ in position of triple bond in them.

[Note: 1-Alkynes are also called terminal alkynes.]

Question 91.
Draw the structural isomers of isomers of C₅H₈. Identify position isomers amongst them.
Answer:
i. Structural isomers of C₅H₁₀ (fourth member of homologous series of alkynes):

ii. The compounds pent-1-yne and pent-2-yne are position isomers of each other.

Question 92.
What are alkadiynes and alkatriynes? Give one example of each.
Answer:
The aliphatic unsaturated hydrocarbons containing two and three carbon-carbon triple bonds in their structure are called alkadiynes and alkatriynes, respectively.
e.g. CH ≡ C – CH₂ – C ≡ CH
Alkadiyne (Penta-1,4-diyne)

HC ≡ C- C ≡ C- C ≡ CH
Alkatriyne (Hexa-1,3,5-triyne)

Question 93.
Complete the following table:

Answer:

Question 94.
How is acetylene prepared from the following compounds?
i. Methane
ii. Calcium carbide
Answer:
i. From methane: Ethyne is industrially prepared by controlled, high temperature, partial oxidation of methane.

ii. From calcium carbide: Industrially, ethyne is prepared by reaction of calcium carbide with water.

Question 95.
How are alkynes prepared by dehydrohalogenation of vicinal dihalides? Write general reaction and explain it using an example.
Answer:
Vicinal dihalides react with alcoholic solution of potassium hydroxide to form alkenyl halide which on further treatment with sodamide forms alkyne.

Question 96.
Convert 1,2-dichloropropane to propyne.
Answer:

Question 97.
i. What are terminal alkynes?
ii. How are they converted to higher nonterminal alkynes? Give one example.
Answer:
i. Terminal alkynes are the compounds in which hydrogen atom is directly attached to triply bonded carbon atom.
ii. a. A smaller terminal alkyne first reacts with a very strong base like lithium amide to form metal acetylide (lithium amide is easier to handle than sodamide).
b. Higher alkynes are obtained by reacting metal acetylides (alkyn-1-yl lithium) with primary alkyl halides.

Question 98.
How is pent-2-yne prepared from propyne?
Answer:

Question 99.
Enlist physical properties of alkenes.
Answer:
The physical properties of alkynes are similar to those of alkanes and alkenes.

• They are less dense than water.
• They are insoluble in water and quite soluble in less polar organic solvents like ether, benzene, carbon tetrachloride.
• The melting points and boiling points of alkynes increase with an increase in molecular mass.

Question 100.
Lithium amide (LiNH₂) is very strong base and it reacts with terminal alkynes to form lithium acetylides with the liberation of hydrogen indicating acidic nature of terminal alkynes. Why is it so?
Answer:

• The hydrogen bonded to C ≡ C triple bond has acidic character.
• In terminal alkynes, hydrogen atom is directly attached to sp hybridized carbon atom.
• In sp hybrid orbital, the percentage of s-character is 50%. An electron in s-orbital is very close to the nucleus and is held tightly.
• The sp hybrid carbon atom in terminal alkynes is more electronegative than the sp² carbon in ethene or the sp³ carbon in ethane.
• Due to high electronegative character of carbon in terminal alkynes, hydrogen atom can be given away as proton (H⁺) to very strong base.

Question 101.
Give reason: Acidic nature of alkynes is used to distinguish between terminal and non-terminal alkynes.
Answer:

• Acidic alkynes react with certain heavy metal ions like Ag⁺ and Cu⁺ and form insoluble acetylides.
• On addition of acidic alkyne to the solution of AgNO₃ in alcohol, it forms a precipitate, which indicates that the hydrogen atom is attached to triply bonded carbon.

Hence, this reaction is used to differentiate terminal alkynes and non-terminal alkynes.

Question 102.
Predict the product in the following reactions.
\(\mathbf{H C} \equiv \mathbf{C H}+\mathbf{2 B r}_{2} \stackrel{\mathrm{CCl}_{4}}{\longrightarrow} ?\)
Answer:
Ethyne reacts with bromine in inert solvent such as carbon tetrachloride to give tetrabromoethane.

Question 103.
Write the general reaction for addition of halogens to alkynes.
Answer:

Question 104.
Explain the addition of hydrogen halides to alkynes using a general reaction.
Answer:
i. Hydrogen halides (HCl, HBr and HI) add to alkynes across carbon-carbon triple bond in two steps to form geminal dihalides (in which two halogen atoms are attached to the same carbon atom).
ii. The addition of HX in both the steps takes place according to Markovnikov’s rule as shown in below.
General reaction:

iii. The order of reactivity of hydrogen halides is HI > HBr > HCl.

Question 105.
State the action of HBr on acetylene and methyl acetylene.
Answer:

Question 106.
Explain reactions of alkynes with water using general reaction.
Answer:
Alkynes react with water in presence of 40% sulphuric acid and 1% mercuric sulphate to form aldehydes or ketones, i.e., carbonyl compounds.
General reaction:

Question 107.
Predict the products when ethyne and propyne are treated with 1% mercuric sulphate in H₂SO₄.
Answer:
i. Ethyne:

ii. Propyne:

Question 108.
Convert:
i. But-1-yne to butan-2-one
ii. Hex-3-yne to hexan-3-one
Answer:
i. But-l-yne to butan-2-one:

ii. Hex-3-yne to hexan-3-one:

Question 109.
What are products obtained on hydration of but-1-yne and but-2-yne? Are they same or different? Explain.
Answer:
i. Hydration of but-1-yne:

Hydration of but-2-yne:

The products obtained on hydration of but-1-yne and but-2-yne are same i.e., butan-2-onc since the addition of water to alkyncs takes place according to Markovnikovs rule.

Question 110.
How is ethylene converted into ethylidene dichloride?
Answer:

Question 111.
Write some important uses of acetylene.
Answer:

• Ethyne (acetylene) is used in preparation of ethanal (acetaldehyde), propanone (acetone), ethanoic acid (acetic acid).
• It is used in the manufacture of polymers, synthetic rubber, synthetic fibre, plastic, etc.
• For artificial ripening of fruits.
• In oxy-acetylene (mixture of oxygen and acetylene) flame for welding and cutting of metals.

Question 112.
Many organic compounds obtained from natural sources such as resins, balsams, oil of wintergreen, etc. have pleasant fragrance or aroma. Such compounds are named as aromatic compounds.
i. Name the simplest aromatic compound.
ii. Write the names of any two aromatic compounds.
Answer:
i. Benzene is the simplest aromatic hydrocarbon.
ii. Toluene and naphthalene

Question 113.
Draw structures of any four aromatic hydrocarbons.
Answer:

Question 114.
Write the molecular formula of benzene. Give its boiling point.
Answer:
The molecular formula for benzene is C₆H₆. Its boiling point is 353 K.

Question 115.
State TRUE or FALSE. Correct the false statement.
i. Aromatic hydrocarbons are also called as arenes.
ii. Toluene is a non-aromatic hydrocarbon.
iii. Benzene is colourless liquid having characteristic odour.
Answer:
i. True
ii. False
Toluene is an aromatic hydrocarbon.
iii. True

Question 116.
Name any two large-scale sources of benzene.
Answer:
Coal-tar and petroleum are the two large-scale sources of benzene.
[Note: Other aromatic compounds like toluene, phenol, naphthalene, etc. are also obtained from coal-tar and petroleum.]

Question 117.
Draw the structure of an aromatic compound that resembles benzene but does not have pleasant odour.
Answer:

Question 118.
Name and draw the structures of any three compounds that have pleasant odour but do not resemble benzene.
Answer:

Question 119.
Differentiate between aromatic and aliphatic compounds.
Answer:
Aromatic compounds:

• Aromatic compounds contain higher percentage of carbon.
• They bum with sooty flame.
• They are cyclic compounds with alternate single and double bonds.
• They are not attacked by normal oxidizing and reducing agents.
• They do not undergo addition reactions easily. They do not decolourise dilute alkaline aqueous KMnO₄ and Br₂ in CCl₄, though double bonds appear in their structure.
• They prefer substitution reactions.

Aliphatic compounds:

• Aliphatic compounds contain lower percentage of carbon.
• They bum with non-sooty flame.
• They are open chain compounds.
• They are easily attacked by oxidizing and reducing agents.
• Unsaturated aliphatic compounds undergo addition reactions easily. They decolourise dilute aqueous alkaline KMnO₄ and Br₂ in CCl₄.
• The saturated aliphatic compounds give substitution reactions.

Question 120.
Benzene cannot have open chain structure. Explain this statement.
Answer:

• The molecular formula of benzene is C₆H₆. This indicates high degree of unsaturation.
• Open chain or cyclic structure having double and triple bonds can be written for C₆H₆.
• However, benzene does not behave like alkenes or alkynes. This indicates that benzene cannot have the open chain structure.

Question 121.
Compare the reactivity of benzene and alkenes with the following reagents:
i. Dilute alkaline KMnO₄
ii. Br₆ in CCl₄
iii. H₆O in acidic medium
Answer:

Reagent	Alkenes	Benzene
Dilute alkaline aqueous KMnO4	Decolourisation of KMnO4	No decolourisation
Br2 in CCl4	Decolourisation of red brown colour of bromine	No decolourisation
H2O in acidic medium	Addition of H2O molecule	No reaction

Question 122.
Give the evidence for the cyclic structure of benzene.
Answer:
Evidence for the cyclic structure of benzene:
i. Benzene yields only one and no isomeric monosubstituted bromobenzene (C₆H₅Br) when treated with equimolar bromine in FeBr₃. This indicates that all six hydrogen atoms in benzene are identical.

ii. This is possible only if benzene has cyclic structure of six carbons bound to one hydrogen atom each.
iii. Benzene on catalytic hydrogenation gives cyclohexane.

This confirms the cyclic structure of benzene and three C = C in it.

Question 123.
Write a short note on the Kekule structure of benzene.
Answer:
Kekule structure of benzene:
i. August Kekule in 1865 suggested the structure for benzene having a cyclic planar ring of six carbon atoms with alternate single and double bonds and hydrogen atom attached to each carbon atom.

ii. The Kekule structure indicates the possibility of two isomeric 1,2-dibromobenzenes. In one of the isomers, the bromine atoms would be attached to the doubly bonded carbon atoms whereas in the other, they would be attached to single bonded carbons.

iii. However, benzene was found to form only one ortho-disubstituted benzene. This problem was overcome by Kekule by suggesting the concept of oscillating nature of double bonds in benzene as given below.

iv. Even with this modification, Kekule structure of benzene failed to explain unusual stability and preference to substitution reactions rather than addition reactions, which was later explained by resonance.

Question 124.
Explain the resonance phenomenon with respect to benzene.
OR
Explain the resonance hybrid structure of benzene.
Answer:

• Benzene is a hybrid of various resonance structures. The two structures, (A) and (B) given by Kekule are the main contributing structures.
• The resonance hybrid is represented by inserting a circle or a dotted circle inscribed in the hexagon as shown in (C).
• The circle represents six electrons delocalized over the six carbon atoms of benzene ring.
• A double headed arrow between the resonance structures is used to represent the resonance phenomenon.

Question 125.
Why does benzene not prefer to undergo addition reactions?
Answer:

• Benzene is highly unsaturated molecule but despite of this feature, it does not give addition reaction.
• The actual structure of benzene is represented by the resonance hybrid which is the most stable form of benzene than any of its resonance structures.
• This stability due to resonance (delocalization of π electrons) is so high that π-bonds of the molecule becomes strong and thus, resist breaking.

Thus, benzene does not prefer to undergo addition reactions.

Question 126.
Explain the resonance structures of benzene using the orbital overlap concept.
Answer:
The structure of benzene can be better explained by the orbital overlap concept,
i. All six carbon atoms in benzene are sp² hybridized. Two sp² hybrid orbitals of each carbon atom overlap and form carbon-carbon sigma (σ) bond and the remaining third sp² hybrid orbital of each carbon overlaps with s orbital of a hydrogen atom to form six C – H sigma bonds.

ii. The unhybridized p orbitals of carbon atoms overlap laterally forming π bonds. There are two possibilities of forming three π bonds by overlap of p orbitals of C₁ – C₂, C₃ – C₄, C₅ – C₆ or C₂ – C₃, C₄ – C₅, C₆ – C₁, respectively, as shown in the following figure. Both the structures are equally probable.

According to resonance theory, these are two resonance structures of benzene.

Question 127.
Explain the structure of benzene with respect to molecular orbital theory.
Answer:
i. According to molecular orbital (MO) theory, the six p orbitals of six carbons give rise to six molecular orbitals of benzene.
ii. Shape of the most stable MO is as show in the figure below. Three of these π molecular orbitals lie above and the other below those of free carbon atom energies.

iii. The six electrons of the p orbitals cover all the six carbon atoms and are said to be delocalized. Delocalization of π electrons results in stability of benzene molecule.

Question 128.
Give the carbon-carbon bond length in benzene. Explain why benzene shows unusual behaviour.
Answer:
i. X-ray diffraction data indicate that all C – C bond lengths in benzene are equal (139 pm) which is an intermediate between C – C (154 pm) and C = C bond (133 pm).

ii. Thus, absence of pure double bond in benzene accounts for its reluctance to addition reactions under normal conditions, which explains unusual behaviour of benzene.

Question 129.
Write a short note on aromaticity.
Answer:
i. All aromatic compounds undergoes substitution reactions rather than addition reactions and this property is referred to as aromaticity or aromatic character.
ii. The aromatic character of benzene is correlated to its structure.
iii. Aromaticity is due to extensive cyclic delocalization of p electrons in the planar ring structure.
iv. Three rules of aromaticity that is used for predicting whether a particular compound is aromatic or non-aromatic are as follows:

• Aromatic compounds are cyclic and planar (all atoms in ring are sp² hybridized).
• Each atom in aromatic ring has a p orbital. The p orbitals must be parallel so that continuous overlap is possible around the ring.
• Huckel rule: The cyclic π molecular orbital formed by overlap of p orbitals must contain (4n + 2) p electrons, where n = integer 0, 1, 2, 3, … etc.

Question 130.
State and explain the Huckel rule of aromaticity.
Answer:
Huckel rule: The cyclic π molecular orbital formed by overlap of p orbitals must contain (4n + 2) p electrons, where n = integer 0, 1,2,3, … etc.

Explanation:
According to Huckel rule, a cyclic and planar compound is aromatic if it the number of π electrons is equal to (4n + 2), where n = integer 0, 1, 2, 3, … etc.

n	Number of π electrons
n = 0	(4 × 0) – 2 = 2
n = 1	(4 × 1) + 2 = 6
n = 2	(4 × 2) + 2 = 10

e.g. Consider benzene molecule:

Benzene has 6π electrons. According to Huckel rule, if n = 1, then (4n + 2)π = 6π electrons. Hence, benzene is aromatic.

Question 131.
By using the rules of aromaticity, explain whether the following compounds are aromatic or non-aromatic.
i. Benzene
ii. Naphthalene
iii. Cycloheptatriene
Answer:
i. Benzene:
a. It is cyclic and planar.
b. It has three double bonds and six π electrons.
c. It has a p orbital on each carbon of the hexagonal ring. Hence, a continuous overlap above and below the ring is possible.

d. According to Huckcl rule, this compound is aromatic if, 4n + 2 = Number of π electrons.
4n + 2 = 6,
∴ 4n = 6 – 2 = 4
n = 4/4 = 1, here, ‘n’ comes out to be an integer.
Hence, benzene is aromatic.

ii. Naphthalene:
a. It is cyclic and planar.
b. It has 5 double bonds and 10 n electrons.

c. It has a p orbital on each carbon atom of the ring. Hence, a continuous overlap around the ring is possible.
d. According to Huckel rule, this compound is aromatic if, 4n + 2 = Number of π electrons.
4n + 2 = 10,
∴ 4n = 10 – 2 = 8
n = 8/4 = 2, Here ‘n’ comes out to be an integer.
Hence, naphthalene is aromatic.

iii. Cycloheptatriene:
a. It is cyclic and planar.
b. It has three double bonds and 6 π electrons.

c. But one of the carbon atoms is saturated (sp³ hybridized) and it does not have a p orbital.
d. Hence, a continuous overlap around the ring is not possible in cycloheptatriene. Hence, it is non-aromatic.

Question 132.
How does Huckel rule help in determining the aromaticity of pyridine?
Answer:
i. Pyridine has three double bonds and 6 π electrons.
ii. The six p orbitals containing six electrons form delocalized π molecular orbital.
iii. The unused sp² hybrid orbital of nitrogen containing two non-bonding electrons is as it is.

iv. According to Huckel rule, this compound is aromatic if, 4n + 2 = Number of π electrons
4n + 2 = 6,
∴ 4n = 6 – 2 = 4
n = 4/4 = 1, here ‘n’ comes out to be an integer. Hence, pyridine is aromatic.

Question 133.
How is benzene prepared from ethyne/acetylene?
Answer:
From ethyne (By trimerization): Alkynes when passed through a red hot iron tube at 873 K, polymerize to form aromatic hydrocarbons. Ethyne when passed through a red hot iron tube at 873 K undergoes trimerization to form benzene.

Question 134.
How is benzene prepared from sodium benzoate?
OR
Explain preparation of benzene by decarboxylation.
Answer:
From sodium benzoate (by decarboxylation): When anhydrous sodium benzoate is heated with soda lime, it undergoes decarboxylation and gives benzene.

[Note: This reaction is useful for decreasing the length of a carbon chain by one C-atom]

Question 135.
How will you convert phenol to benzene?
Answer:
From phenol (By reduction): When vapours of phenol are passed over heated zinc dust, it undergoes reduction and gives benzene.

Question 136.
Enlist physical properties of benzene.
Answer:
Physical properties of benzene:

• Benzene is a colourless liquid.
• Its boiling point is 353 K and melting point is 278.5 K.
• It is insoluble in water. It forms upper layer when mixed with water.
• It is soluble in alcohol, ether and chloroform.
• Its vapours are highly toxic which on inhalation lead to unconsciousness.

Question 137.
What is the action of chlorine on benzene in the presence of UV light?
Answer:
Addition of chlorine: When benzene is treated with chlorine in the presence of bright sunlight or UV light, three molecules of chlorine gets added to benzene to give benzene hexachloride.

Question 138.
Name the γ-isomer of benzene hexachloride which is used as insecticide.
Answer:
The γ-isomer of benzene hexachloride which is used as insecticide is called as gammaxene or lindane.

Question 139.
How will you convert benzene to cyclohexane?
Answer:
Addition of hydrogen: When a mixture of benzene and hydrogen gas is passed over heated catalyst nickel at 453 K to 473 K, cyclohexane is formed.

Question 140.
What is the action of ozone on benzene?
Answer:
Addition of ozone: When benzene is treated with ozone in the presence of an inert solvent carbon tetrachloride, benzene triozonide is formed, which is then decomposed by zinc dust and water to give glyoxal.

Question 141.
What are the different types of electrophilic substitution reactions of benzene?
Answer:
i. Benzene shows electrophilic substitution reactions, in which one or more hydrogen atoms of benzene ring are replaced by groups like – Cl, – Br, – NO₂, – SO₃H, -R, -COR, etc.
ii. Different types of electrophilic substitution reactions of benzene are as follows:

• Halogenation (chlorination and bromination)
• Nitration
• Sulphonation
• Friedel-Craft’s alkylation and
• Friedel-Craft’s acylation

Question 142.
Write a short note on chlorination reaction of benzene.
Answer:
Chlorination of benzene:
i. In chlorination reaction, hydrogen atom of benzene is replaced by chlorine atom.
ii. Chlorine reacts with benzene in dark in the presence of iron or ferric chloride or anhydrous aluminium chloride or red phosphorus as catalyst to give chlorobenzene (C₆H₅Cl).

iii. Electrophile involved in the reaction: Cl⁺, chloronium ion,
Formation of the electrophile: Cl – Cl + FeCl₃ → Cl⁺ + [FeCl₄]^–

Question 143.
Write a short note on bromination reaction of benzene.
Answer:
Bromination of benzene:
i. In bromination reaction, hydrogen atom of benzene is replaced by bromine atom.
ii. Bromine reacts with benzene in dark in presence of iron or ferric bromide or anhydrous aluminium bromide or red phosphorus as catalyst to give bromobenzene (C₆H₅Br).

iii. Electrophile involved in the reaction: Br⁺
Formation of the electrophile: Br – Br + FeBr₃ → Br⁺ + [FeBr₄]^–

Question 144.
Why direct iodination of benzene is not possible?
Answer:
Direct iodination of benzene is not possible as the reaction is reversible.

[Note: Iodination of benzene can be carried out in the presence of oxidising agents like HIO₃ or HNO₃.]

Question 145.
How will you convert benzene to hexachlorobenzene?
Answer:
When benzene is treated with excess of chlorine in presence of anhydrous aluminium chloride, it gives hexachlorobenzene.

Question 146.
State true or false. Correct the false statement.
i. In halogenation reaction, hydrogen atom of benzene ring is replaced by halogen atom.
ii. The molecular formula of hexachlorobenzene is C₆H₆Cl₆.
iii. Benzene forms the lower layer when mixed with water.
Answer:
i. True
ii. False
The molecular formula of hexachlorobenzene is C₆Cl₆
iii. False
Benzene forms the upper layer when mixed with water.

Question 147.
Explain the nitration reaction of benzene.
Answer:
Nitration of benzene:
i. When benzene is heated with a mixture of concentrated nitric acid and concentrated sulphuric acid (nitrating mixture) at about 313 K to 333 K, it gives nitrobenzene.

ii. Electrophile involved in the reaction: \(\mathrm{NO}_{2}^{+}\), nitronium ion
Formation of the electrophile: HO – NO₂ + 2H₂SO₄ ⇌ \(2 \mathrm{HSO}_{4}^{-}\) + H₃O⁺ + \(\mathrm{NO}_{2}^{-}\)

Question 148.
Write a short note on sulphonation of benzene.
Answer:
Sulphonation of benzene:
i. When benzene is heated with fuming sulfuric acid (oleum) at 373 K, it gives benzene sulfonic acid.

ii. Electrophile involved in the reaction: SO₃, free sulphur trioxide
Formation of the electrophile: 2H₂SO₄ → H₃O⁺ + \(\mathrm{HSO}_{4}^{-}\) + SO₃

Question 149.
Write a short note on Friedel-Craft’s alkylation reaction of benzene.
Answer:
Friedel-Craft’s alkylation reaction of benzene:
i. When benzene is treated with an alkyl halide like methyl chloride in the presence of anhydrous aluminium chloride, it gives toluene.

ii. Electrophile involved in the reaction: R⁺
Formation of the electrophile: R – Cl + AlCl₃ → R⁺ + \(\mathrm{AlCl}_{4}^{-}\)
iii. Friedel-Craft’s alkylation reaction is used to extend the chain outside the benzene ring.

Question 150.
Explain Friedel-Craft’s acylation reaction of benzene. Give example reactions.
Answer:
Friedel-craft’s acylation reaction of benzene:
i. When benzene is heated with an acyl halide or acid anhydride in the presence of anhydrous aluminium chloride, it gives corresponding acyl benzene.

ii. Electrophile involved in the reaction: R – C- = O, acylium ion
Formation of the electrophile: R – COCl + AlCl₃ → R – C⁺ = O + \(\mathrm{AlCl}_{4}^{-}\)

Question 151.
Write the general combustion reaction for hydrocarbons.
Answer:
General combustion reaction for any hydrocarbon (C_xH_y) can be represented as follows:

Question 152.
Write the combustion reaction of benzene.
Answer:
When benzene is heated in air, it bums with sooty flame forming carbon dioxide and water.
C₆H₆ + \(\frac {15}{2}\)O₂ → 6CO₂ + 3H₂O

Question 153.
Write a note on the directive influence of substituents (functional groups) in monosubstituted benzene.
Answer:
i. In benzene, all hydrogen atoms are equivalent and so, when it undergoes electrophilic substitution reactions, only one monosubstituted product is possible.
Monosubstituted benzene:

ii. When monosubstituted benzene undergoes further electrophilic substitution, the second substituent (electrophile, E) can occupy any of the five positions available and give three disubstituted products.
But these disubstituted products are not formed in equal amounts.

iii. The position of second substituent (E) is determined by the nature of substituent (S) already present in the benzene ring and not on the nature of second substituent (E).
iv. The groups which direct the incoming group to ortho and para positions are called ortho and para directing groups. The groups which direct the incoming group to meta positions are called meta directing groups. Thus, depending on the nature of the substituent (S) either ortho and para products or meta products are formed as major products.

Question 154.
What are ortho and para directing groups? Enlist few ortho and para directing groups.
Answer:
The groups which direct the incoming group to ortho and para positions are called ortho and para directing groups.
Ortho and para directing groups:

Question 155.
Explain the directive influence of ortho, para directing groups in monosubstituted benzene using suitable example.
OR
Explain the directive influence of -OH group in benzene.
Answer:
i. The directive influence of ortho, para directing groups can be explained with the help of inductive and resonance effects.
ii. phenol has the following resonating structures:

iii. It can be seen from the above resonating structures, that the ortho (o-) and para (p-) positions have a greater electron density than the meta positions.
iv. Therefore, -OH group activates the benzene ring for the attack of second substituent (E) at these electron rich centres. Thus, phenolic -OH group is activating and ortho, para-directing group.
v. In phenol, -OH group has electron withdrawing inductive (-I) effect which slightly decreases the electron density at ortho positions in benzene ring. Thus, resonance effect and inductive effect of -OH group act opposite to each other. However, the strong resonance effect dominates over inductive effect.

Question 156.
Explain the o, p-directive effect of methyl group.
Answer:

• All ortho and para directing groups possess nonbonding electron pair on the atom which is directly attached to the aromatic ring; however, methyl group is an exception.
  Methyl (or alkyl groups) is ortho and para directing, although it has no nonbonding electron pair on the key atom. This is explained on the basis of special type of resonance called hyperconjugation or no bond resonance.

Question 157.
Explain why halide group is an ortho and para directing group.
Answer:
i. In aryl halides, halogens are moderately deactivating. Because of their strong -I effect, overall electron density on the benzene ring decreases, which makes the electrophilic substitution difficult.
ii. However, halogens are ortho and para directing. This can be explained by considering resonance structures.
iii. e.g. Chlorobenzene has the following resonating structures:

iv. Due to resonance, the electron density on ortho and para positions is greater than meta positions and hence, -Cl is ortho and para directing.

Question 158.
What are meta directing groups? Enlist few of them.
Answer:
The groups which direct the incoming group to meta positions are called meta directing groups.

[Note: All meta directing groups have positive (or partial positive) charge on the atom which is directly attached to an aromatic ring.]

Question 159.
Explain the directive influence of nitro group in nitrobenzene.
OR
Explain why nitro group is a meta-directing group.
Answer:
i. Meta directing group withdraws electrons from the aromatic ring by resonance, making the ring electron-deficient. Therefore, meta groups are ring deactivating groups.
ii. Due to -I effect, -NO₂ group reduces electron density in benzene ring on ortho and para positions. So, the attack of incoming group becomes difficult at ortho and para positions. The incoming group can attack on meta positions more easily.
iii. The various resonance structures of nitrobenzene are as shown below:

iv. It is clear from the above resonance structures that the ortho and para positions have comparatively less electron density than at meta positions. Hence, the incoming group/electrophile attacks on meta positions.

Question 160.
What are polycyclic aromatic compounds? How are they produced?
Answer:

• Polycyclic aromatic compounds are the hydrocarbons containing more than two benzene rings fused together.
• They are produced by incomplete combustion of tobacco, coal and petroleum.

Question 161.
Write the harmful effects of benzene.
Answer:

• Benzene is both, toxic and carcinogenic (cancer causing).
• In fact, it might be considered “the mother of all carcinogens” as a large number of carcinogens have structures that include benzene rings.
• In liver, benzene is oxidized to an epoxide and benzopyrene is converted into an epoxy diol. These substances are carcinogenic and can react with DNA and thus, can induce mutation leading to uncontrolled growth of cancer cells.

Multiple Choice Questions

1. Alkanes are represented by the general formula ………….
(A) C_nH_2n-2
(B) C_nH_2n+2
(C) C_nH_2n
(D) C_nH_n
Answer:
(B) C_nH_2n+2

2. Which of the following compound is alkanes?
(A) C₅H₁₀
(B) C₁₀H₂₂
(C) C₁₅H₂₈
(D) C₉H₁₆
Answer:
(B) C₁₀H₂₂

3. Alkanes are commonly called …………
(A) arenes
(B) paraffins
(C) olefins
(D) acetylenes
Answer:
(B) paraffins

4. Every carbon atom in alkanes is …………..
(A) sp hybridized
(B) sp² hybridized
(C) sp³ hybridized
(D) sp³d hybridized
Answer:
(C) sp³ hybridized

5. Isomerism is the phenomenon in which two or more organic compounds have ………….
(A) same molecular formula but different structural formula
(B) same structural formula but different molecular formula
(C) same general formula, but different structural formula
(D) same empirical formula, same structural formula
Answer:
(A) same molecular formula but different structural formula

6. Pentane exhibits …………. chain isomers.
(A) two
(B) three
(C) four
(D) five
Answer:
(B) three

7. Which of the following is NOT an isomer of hexane?
(A) 2-Methylpentane
(B) 2,2-Dimethylbutane
(C) 2,2-Dimethylpentane
(D) 3-Methylpentane
Answer:
(C) 2,2-Dimethylpentane

8. Alkanes can be prepared by ………… of unsaturated hydrocarbons.
(A) hydrogenation
(B) oxidation
(C) hydrolysis
(D) cracking
Answer:
(A) hydrogenation

9. Catalytic hydrogenation of ethene or acetylene gives …………..
(A) ethane
(B) propylene
(C) methane
(D) propane
Answer:
(A) ethane

10. Ethyl iodide when reduced by zinc and dilute HCl, leads to the formation of …………..
(A) Methane
(B) Ethane
(C) Ethylene
(D) Butane
Answer:
(B) Ethane

11. The reaction of alkyl halides with sodium in dry ether to give higher alkanes is called ………..
(A) Wurtz reaction
(B) Kolbe’s reaction
(C) Frankland’s reaction
(D) Williamson’s reaction
Answer:
(A) Wurtz reaction

12. Methane is ………… molecule.
(A) polar
(B) nonpolar
(C) highly polar
(D) none of these
Answer:
(B) nonpolar

13. Alkanes are ………… in water.
(A) soluble
(B) sparingly soluble
(C) insoluble
(D) none of these
Answer:
(C) insoluble

14. As branching increases, boiling point of alkanes ………….
(A) increases
(B) decreases
(C) remains same
(D) None of these
Answer:
(B) decreases

15. Halogenation of alkane is an example of …………. reaction.
(A) dehydration
(B) substitution
(C) addition
(D) elimination
Answer:
(B) substitution

16. Order of reactivity of halogens in halogenation of alkanes is ………….
(A) F₂ > Cl₂ > Br₂ > I₂
(B) I₂ > Br₂ > Cl₂ > F₂
(C) Br₂ < I₂ < F₂ < Cl₂
(D) Cl₂ < I₂ < Br₂ < F₂
Answer:
(A) F₂ > Cl₂ > Br₂ > I₂

17. The thermal decomposition of alkanes in absence of air to give lower alkanes, alkenes and hydrogen is called ………….
(A) vapour phase nitration
(B) pyrolysis
(C) polymerisation
(D) combustion
Answer:
(B) pyrolysis

18. But-1-ene and But-2-ene are …………
(A) chain isomers
(B) position isomers
(C) geometrical isomers
(D) metamers
Answer:
(B) position isomers

19. Hex-2-ene and 2-Methylpent-2-ene exhibit …………
(A) chain isomerism
(B) position isomerism
(C) geometrical isomerism
(D) optical isomerism
Answer:
(A) chain isomerism

20. Which of the following shows position isomerism?
(A) Propene
(B) Ethene
(C) 2-Methylpropene
(D) Pent-2-ene
Answer:
(D) Pent-2-ene

21. When identical atoms or group of atoms are attached to the two carbon atoms on the same side of the double bond, the isomer is called ………… isomer.
(A) cis
(B) trans
(C) position
(D) chain
Answer:
(A) cis

22. Which of the following does NOT exhibit geometrical isomers?
(A) But-2-ene
(B) Pent-2-ene
(C) But-1-ene
(D) Hex-2-ene
Answer:
(C) But-1-ene

23. When ethyl bromide is heated with alcoholic KOH, ………… is formed.
(A) ethane
(B) ethanol
(C) ethene
(D) acetylene
Answer:
(C) ethene

24. Alkenes are insoluble in …………
(A) benzene
(B) water
(C) ether
(D) chloroform
Answer:
(B) water

25. Markownikov’s rule is applicable to …………
(A) symmetrical alkenes
(B) alkanes
(C) unsymmetrical alkenes
(D) alkynes
Answer:
(C) unsymmetrical alkenes

26. When propene is treated with HBr in the dark and in absence of peroxide, then the main product formed is …………
(A) 1-bromopropane
(B) 2-bromopropane
(C) 1,2-dibromopropane
(D) 1,3-dibromopropane
Answer:
(B) 2-bromopropane

27. The product formed by the addition of HCl to propene in presence of peroxide is …………..

Answer:

28. Propene reacts with HBr in presence of peroxide, to form …………..
(A) 2-bromopropane
(B) 1-bromopropane
(C) 3-bromopropane
(D) 1,2-dibromopropane
Answer:
(B) 1-bromopropane

29. Markovnikov’s rule is applicable for …………..
(A) CH₂ = CH₂
(B) CH₃CH = CHCH₃
(C) CH₃CH₂CH = CHCH₂CH₃
(D) (CH₃)₂C = CH₂
Answer:
(D) (CH₃)₂C = CH₂

30. The addition of HCl in presence of peroxide does not follow anti-Markownikov’s rule because …………..
(A) HCl bond is too strong to be broken homolytically
(B) Cl atom is not reactive enough to add on to a double bond
(C) Cl atom combines with H atom to form HCl
(D) HCl is a reducing agent
Answer:
(A) HCl bond is too strong to be broken homolytically

31. An alkene on ozonolysis produces a mixture of acetaldehyde and acetone. Identify the alkene.
(A) But-1-ene
(B) But-2-ene
(C) 2-Methylbut-1-ene
(D) 2-Methylbut-2-ene
Answer:
(D) 2-Methylbut-2-ene

32. The ozonolysis of (CH₃)₂C = C(CH₃)₂ followed by treatment with zinc and water will give ……………
(A) acetone
(B) acetone and acetaldehyde
(C) formaldehyde and acetone
(D) acetaldehyde
Answer:
(A) acetone

33. The compound which forms only acetaldehyde on ozonolysis is …………..
(A) ethene
(B) propyne
(C) but-1-ene
(D) but-2-ene
Answer:
(D) but-2-ene

34. Treatment of ethylene with ozone followed by decomposition of the product with Zn/H₂O gives two moles of ………….
(A) formaldehyde
(B) acetaldehyde
(C) formic acid
(D) acetic acid
Answer:
(A) formaldehyde

35. Ozonolysis of 2,3-Dimethylbut-1-ene followed by reduction with zinc and water gives ………….
(A) methanoic acid and 3-methylbutan-2-one
(B) methanal and 2-methylbutan-2-one
(C) methanal and 3-methylbutan-2-one
(D) methanoic acid and 2-methylbutan-2-one
Answer:
(C) methanal and 3-methylbutan-2-one

36. The reaction, CH₂ = CH₂ + H₂O + [O]

is called ……………
(A) hydroxylation
(B) decarboxylation
(C) hydration
(D) dehydration
Answer:
(A) hydroxylation

37. An alkene on vigorous oxidation with KMnO₄ gives only acetic acid. The alkene is …………..
(A) CH₃CH₂CH = CH₂
(B) CH₃CH = CHCH₃
(C) (CH₃)₂C = CH₂
(D) CH₃CH = CH₂
Answer:
(B) CH₃CH = CHCH₃

38. Ethylene reacts with Baeyer’s reagent to give a/an ………….
(A) glycol
(B) aldehyde
(C) acid
(D) alcohol
Answer:
(A) glycol

39. Baeyer’s reagent is ………….
(A) aqueous KMnO₄
(B) neutral KMnO₄
(C) alkaline KMnO₄
(D) aqueous bromine water
Answer:
(C) alkaline KMnO₄

40. Alkynes have general formula ………….
(A) C_nH_2n-2
(B) C_nH_2n
(C) C_nH_2n+2
(D) C_nH_2n+1
Answer:
(A) C_nH_2n-2

41. Aliphatic unsaturated hydrocarbons containing two carbon-carbon triple bonds in their structure are called as ………….
(A) alkadiynes
(B) alkatriynes
(C) alkynes
(D) alkanes
Answer:
(A) alkadiynes

42. Acetylene is prepared in the industry by the action of water on ………….
(A) calcium carbonate
(B) calcium carbide
(C) mercuric chloride
(D) calcium oxide
Answer:
(B) calcium carbide

43. The dihalogen derivatives of alkanes when heated with …………. form corresponding alkynes.
(A) alcoholic water
(B) sodamide
(C) zinc
(D) acids
Answer:
(B) sodamide

44. Alkynes readily undergo …………. reaction.
(A) addition
(B) substitution
(C) elimination
(D) rearrangement
Answer:
(A) addition

45. Liquid bromine reacts with acetylene to form ………….
(A) 1,2-dibromoethene
(B) 1,1,2,2-tetrabromoethane
(C) 1,1-dibromoethene
(D) methyl chloride
Answer:
(B) 1,1,2,2-tetrabromoethane

46. When acetylene is passed through dil H₂SO₄ in the presence of 1% mercuric sulphate, the compound formed is ………….
(A) ethanol
(B) acetone
(C) acetaldehyde
(D) acetic acid
Answer:
(C) acetaldehyde

47. The compounds which contain at least one benzene ring are ………….
(A) aliphatic compounds
(B) aromatic compounds
(C) cycloalkanes
(D) both (A) and (B)
Answer:
(B) aromatic compounds

48. Which of the following compounds does NOT contain any benzene rings in their structure?
(A) Benzaldehyde
(B) Benzoic acid
(C) Naphthalene
(D) Furan
Answer:
(D) Furan

49. Benzene undergoes ………….
(A) only addition reaction
(B) only substitution reaction
(C) both addition and substitution reactions
(D) nucleophilic substitution reactions
Answer:
(C) both addition and substitution reactions

50. If the substituents are on the adjacent carbon atoms in the benzene ring, it is called ………….
(A) meta
(B) para
(C) ortho
(D) beta
Answer:
(C) ortho

51. How many molecules of acetylene are required to form benzene?
(A) 2
(B) 3
(C) 4
(D) 5
Answer:
(B) 3

52. Which of the following compound on reduction gives benzene?
(A) Sodium benzoate
(B) Acetylene
(C) Cyclohexane
(D) Phenol
Answer:
(D) Phenol

53. X-Ray diffraction reveals that benzene is a …………. structure.
(A) triangular
(B) planar
(C) co-planar
(D) 3D
Answer:
(B) planar

54. γ-isomer of BHC is known as ………….
(A) gammene
(B) gammaxane
(C) chlorobenzene
(D) hexachlorobenzene
Answer:
(B) gammaxane

55. Benzene when treated with ozone forms ………….
(A) glyoxal
(B) acetic acid
(C) formaldehyde
(D) benzaldehyde
Answer:
(A) glyoxal

56. …………. is formed as intermediate product in ozonolysis of benzene.
(A) Benzaldehyde
(B) Phenol
(C) Benzene triozonide
(D) Cyclohexane
Answer:
(C) Benzene triozonide

57. Electrophile in chlorination of benzene is ………….
(A) Cl
(B) Cl⁺
(C) Cl^–
(D) Cl₂
Answer:
(B) Cl⁺

58. Benzene when treated with fuming. H₂SO₄ at 373 K forms ………….
(A) ethylbenzene
(B) toluene
(C) benzene sulphonic acid
(D) acetophenone sulphonic acid
Answer:
(C) benzene sulphonic acid

59. Ethyl chloride reacts with benzene in presence of anhydrous aluminium chloride to form ………….
(A) ethyl benzene
(B) chlorobenzene
(C) toluene
(D) acetophenone
Answer:
(A) ethyl benzene

60. The electrophile in Friedel-Craft’s alkylation reaction is ………….
(A) R⁺
(B) R^–
(C) Cl⁺
(D) RCO⁺
Answer:
(A) R⁺

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.

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 +q_m and -q_m 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.
∴ \(\vec{m}\) = q_m (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 (q_m) is Am.
2. SI unit of magnetic dipole moment (m) is Am².

Question 8.
Draw neat labelled diagram for a bar magnet.
Answer:

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.

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
∴ B_axis = 2B_eq

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.

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, B_a = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{\mathrm{r}^{3}}\)
Here
B_a = \(\frac{\mu_{0}}{4 \pi} \frac{2 m \cos \theta}{r^{3}}\) ………….. (1)
directed along m cosθ.

vi. For point on equator,
B_a = \(\frac{\mu_{o}}{4 \pi} \frac{m \sin \theta}{r^{3}}\) …………. (2)
directed opposite to m sin θ

vii. Thus, the magnitude of the resultant magnetic field B, at point P is given by

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 (q_m)
Formula: q_m = \(\frac {m}{2l}\)
Calculation:
From formula.
q_m = \(\frac {5.0}{0.20}\) = 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, q_m = 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

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 (B_a)
ii. Magnetic field on the equatorial point (B_eq)
Formulae:
i. B_a = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{r^{3}}\)
ii. B_a = 2B_eq
Calculation: From formula (i),
B_a = 10^-7 × \(\frac{2 \times 0.5}{(0.2)^{3}}\)
= \(\frac{10^{-7}}{8 \times 10^{-3}}\)
= 0.125 × 10^-4
∴ B_a = 1.25 × 10^-5 Wb/m²
From formula (ii),
B_eq = \(\frac {B_a}{2}\) = \(\frac {1.25×10^{-5}}{2}\)
= 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 (B_a)
ii. Magnetic induction along equator (B_eq)
Formulae:
i. B_a = \(\frac {µ_0}{4π}\) \(\frac {2m}{r^3}\)
ii. B_a = 2 B_eq
Calculation: From formula (i),
B_a = 10^-7 × \(\frac {2×0.48}{10^{-3}}\)
∴ B_a = 0.96 × 10^-4 T along S-N direction
From formula (ii),
B_eq = \(\frac {0.96×106{-4}}{2}\)
∴ B_eq = 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:

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:

Question 22.
Draw a neat labelled diagram for angle of dip.
Answer:

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, B_H = B cos ø, B_v = 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 = B_H along South to North, \(\vec{B}\)_v = 0 and ø = 0

Question 23.
What are magnetic maps?
Answer:
Magnetic elements of the Earth (B_H, 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 (B_H) 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.
∴ B_eq = \(\frac {µ_0}{4π}\) \(\frac {m}{r^3}\) ……….. (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 B_H = 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 = B_H = 3.5 × 10^-5 T ……. (Given)
we have,

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

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°

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°

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 11 Adsorption and Colloids Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 11 Adsorption and Colloids

Question 1.
Explain the phenomenon of adsorption with the help of examples.
Answer:
Consider the following two examples:

• Example 1: When a metal spoon is dipped in milk and taken out, it is observed that a film of milk particles covers the spoon surface.
• Example 2: If a cold water bottle is taken out from the refrigerator and kept on a table for a while, water vapour is seen to condense on the outer surface of the bottle, forming droplets or a film.
• In the above examples, the milk particles or the water molecules from the air get adsorbed on the surface of the spoon and the bottle, respectively.
• Similarly, surfaces of many objects around us are exposed to the atmosphere. Water molecules as well as other gas molecules such as N₂, O₂, from the air form an invisible multimolecular film on these objects.
  This is known as the phenomenon of adsorption.

Question 2.
Why does adsorption occur?
Answer:

• The adsorption phenomenon is caused by dispersion forces (also known as London dispersion forces or van der Waals forces) which are short range and additive. Adsorption force is the sum of all interactions between all the atoms.
• The pulling interactions cause the surface of a liquid to tighten like an elastic film.
• A measure of the elastic force at the surface of a liquid is called surface tension.
• There is a tendency to have minimum surface tension, i.e., decrease of free energy, which leads to adsorption.

Question 3.
Define surface tension.
Answer:
A measure of the elastic force at the surface of a liquid is called surface tension.
OR
Surface tension is the amount of energy required to stretch or increase the surface of a liquid by a unit area.

Question 4.
Define the following terms.
i. Adsorbent
ii. Adsorbate
Answer:
i. Adsorbent: The material or substance present in the bulk, on the surface of which adsorption takes place is called adsorbent.
ii. Adsorbate: The substance getting adsorbed on the adsorbent is called as adsorbate.

Question 5.
Give some examples of adsorption.
Answer:
Following are some examples of adsorption:

• Adsorption of gases like hydrogen and oxygen by finely divided metals, namely, platinum, palladium, copper, nickel, etc.
• Adsorption of gases like nitrogen and carbon dioxide by activated charcoal.
• Removal of colouring matter like an organic dye, for example, methylene blue. When charcoal is added to methylene blue solution and shaken, it becomes colourless after some time as dye molecules accumulate on the surface of charcoal.

Question 6.
What is desorption?
Answer:
The process of removal of an adsorbed substance from a surface on which it was adsorbed is called desorption.

Question 7.
Define sorption.
Answer:
When both adsorption and absorption occur simultaneously, it is known as sorption.
e.g. When a chalk is dipped in ink, the ink molecules are adsorbed at the surface of the chalk while the solvent of the ink goes deeper into the chalk due to absorption.

Question 8.
What is physisorption? State its characteristics.
Answer:
When the adsorbent such as gas molecules are accumulated at the surface of a solid on account of weak van der Waals forces, the adsorption is termed as physical adsorption or physisorption.

Characteristics:

• The van der Waals forces involved in physical adsorption are similar to forces causing condensation of gas into liquid. Thus, heat is released in physisorption.
• The heat released during physisorption is of the same order of magnitude as heat of condensation.
• Due to weak nature of van der Waals forces, physisorption is weak in nature.
• The adsorbed gas forms several layers of molecules at high pressures.
• The extent of adsorption is large at low temperatures.
• The equilibrium is attained rapidly.
• Physisorption is readily reversed by lowering of pressure of gas or by raising temperature.

Question 9.
Define chemisorption, Write its main features.
Answer:
When the gas molecules accumulate at the surface of a solid or adsorbate by means of chemical bonds (covalent or ionic), the adsorption is termed as chemical adsorption or chemisorption.
Features of chemical adsorption:

• Chemisorption is specific in nature.
• Chemisorption involving the gas-solid as the adsorbate and adsorbent is usually exothermic i.e., heat is released during this process (Exception: The adsorption of hydrogen on glass is endothermic).
• The heat evolved in chemisorption per mole of adsorbate is nearly the same order of magnitude as that accompanying chemical bonding.
• Chemisorption involves a large energy of activation and hence, it is also referred as activated adsorption.
• Chemisorption increases with increase in temperature in the beginning, as a greater number of molecules can have activation energy. But after certain temperature chemisorption decreases with increase in temperature as the chemical bonds break.
• Sometimes at low’ temperature, physisorption occurs which passes into chemisorption as the temperature is raised.
• Chemisorption is dependent on surface area of the adsorbent.

[Note: Chemisorption was first investigated in 1916 by American Chemist, Irving Langmuir (1881-1957).]

Question 10.
Why is chemisorption also known as activated adsorption?
Answer:
Chemisorption involves a large energy of activation and hence, it is also referred as activated adsorption.

Question 11.
Give reason: Adsorption of hydrogen on glass is an endothermic process.
Answer:
Adsorption of hydrogen on glass is an endothermic process because heat is absorbed during the process due to dissociation of hydrogen.

Question 12.
Explain graphically the effect of the following factors on the adsorption of gases by solids.
i. Temperature of the adsorbent surface
ii. Pressure of the gas (adsorbate)
Answer:
i. Temperature of the adsorbent surface:

• Adsorption is an exothermic process.
• According to Te Chatelier’s principle, it is favoured at low temperature.
• Therefore, the amount of gas adsorbed is inversely proportional to the temperature.
• The graph given below shows plots of volume of N? adsorbed per unit mass of adsorbent against the pressure of a gas at different temperatures.
• As temperature increases from 193 K to 273 K at a constant pressure ‘P’, the amount of gas adsorbed decreases.

ii. Pressure of the gas:

• At any temperature, the extent of gas adsorbed increases with an increase in pressure.
• The extent of adsorption is directly proportional to pressure of the gas.
• At high pressures extent of adsorption becomes independent of the pressure. The surface of adsorbent is then almost fully covered by adsorbed gaseous molecules.

Question 13.
What are the applications of adsorption?
Answer:
Following are the various applications of adsorption:
i. Catalysis (Heterogeneous catalysis):

• The solid catalysts are used in many industrial manufacturing processes.
• For example, iron is used as a catalyst in manufacturing of ammonia, platinum in manufacturing of sulphuric acid, H₂SO₄ (by contact process) while finely divided nickel is employed as a catalyst in hydrogenation of oils.

ii. Gas masks:

• It is a device which consists of activated charcoal or mixture of adsorbents.
• It is used for breathing in coal mines to avoid inhaling of the poisonous gases.

iii. Control of humidity: Silica and alumina gels are good adsorbents of moisture.

iv. Production of high vacuum:

• Lowering of temperature at a given pressure, increases the rate of adsorption of gases on charcoal powder. By using this principle, high vacuum can be attained by adsorption.
• A vessel evacuated by vacuum pump is connected to another vessel containing coconut charcoal cooled by liquid air. The charcoal adsorbs the remaining traces of air or moisture to create a high vacuum.

v. Adsorption indicators: The adsorption is used to detect the end point of precipitation titrations. Dyes such as eosin, fluorescein are used as indicators.
e.g.
a. A solution of sodium chloride containing a small amount of fluorescein is titrated against silver nitrate solution.

b. When chloride ions are over, fluorescein is adsorbed on white silver chloride precipitate and hence, red colour is developed.
c. Thus, colour changes from pale yellow to reddish pink at the end point.

vi. Separation of inert gases:

• In a mixture of noble gases, different gases adsorb to different extent.
• Due to selective adsorption principle, gases can be separated on coconut charcoal.

vii. Froth floatation process:

• A low-grade sulphide ore is concentrated by separating it from silica and other earthy matter using pine oil as frothing agent.
• Hydrophobic pine oil preferentially adsorbs sulphide ore which is taken up in the froth.

viii. Chromatographic analysis:

• It is based on selective adsorption of ions from solution using powdered adsorbents such as silica or alumina gel.
• It has several industrial and analytical applications. Other applications include surface area determination, purification of water, etc.

Question 14.
Explain how high vacuum can be obtained by adsorption.
Answer:

• Lowering of temperature at a given pressure, increases the rate of adsorption of gases on charcoal powder. By using this principle, high vacuum can be attained by adsorption.
• A vessel evacuated by vacuum pump is connected to another vessel containing coconut charcoal cooled by liquid air. The charcoal adsorbs the remaining traces of air or moisture to create a high vacuum.

Question 15.
State whether TRUE or FALSE. Correct if false.
i. The rate of adsorption of gases on charcoal powder decreases on lowering of temperature at a given pressure.
ii. Noble gases can be separated from their mixture using the principle of selective adsorption as they adsorb to different extent.
iii. Pine oil is used as frothing agent in froth floatation process.
Answer:
i. False
The rate of adsorption of gases on charcoal powder increases on lowering of temperature at a given pressure.
ii. True
iii. True

Question 16.
Match the following.

	Column A		Column B
i.	Iron	a.	Hydrogenation of oils
ii.	Nickel	b.	Production of sulphuric acid
iii.	Platinum	c.	Synthesis of ammonia

Answer:
i – c,
ii – a,
iii – b

Question 17.
What is a catalyst?
Answer:
A catalyst is a substance which when added to a reacting system, increases the rate of a reaction without itself undergoing any permanent chemical change.

Question 18.
Explain the importance of catalysts in chemical industries.
Answer:

• A large number of the chemicals manufactured in industries make use of catalysts to obtain specific products.
• The use of catalyst lowers the reaction temperature as well as energy costs significantly.
  Due to these advantages, catalysts are of great importance in chemical industry.

Question 19.
Name two types of catalysis.
Answer:

1. Homogeneous catalysis
2. Heterogeneous catalysis

Question 20.
Define homogeneous catalysis and give any two examples.
Answer:
When the reactants and the catalyst are in the same phase, it is said to be homogeneous catalysis.
e.g.
i. Iodide ion (I^–) is used as homogeneous catalyst in decomposition of aqueous hydrogen peroxide because both I^– and H₂O₂ are present in the same aqueous phase.
ii. Hydrolysis of sugar is catalysed by H⁺ ions furnished by sulphuric acid.

All reactants and catalyst are in same solution phase.
[Note: Enzyme catalysis is also an important type of homogeneous catalysis.]

Question 21.
Justify: Lead chamber process is an example of homogeneous catalysis.
Answer:
i. In the lead chamber process, sulphur dioxide is oxidized to sulphur trioxide with dioxygen (O₂) in the presence of nitric oxide as catalyst.

ii. Since all the reactants as well as the catalyst is present in gaseous state. i.e., in same phase, it is a homogeneous catalysis reaction.
Hence, lead chamber process is an example of homogeneous catalysis.

Question 22.
Describe heterogeneous catalysis with the help of one example.
Answer:
i. When the reactants and catalyst are in different phase, it is said to be heterogeneous catalysis.
ii. The heterogeneous catalyst is generally a solid and the reactants may either be gases or liquids.
iii. When the solid catalyst is added to the reaction mixture, it does not dissolve in the reacting system and the reaction occurs on the surface of the solid catalyst.
e.g. Dinitrogen (N₂) and dihydrogen (H₂) combine to form ammonia in Haber process in the presence of finely divided iron along with K₂O and Al₂O₃.

b. In the above reaction, Al₂O₃ and K₂O are promoters of the Fe catalyst. Al₂O₃ is added to prevent the fusion of Fe particles. K₂O causes chemisorption of nitrogen atoms. Molybdenum is also used as promoter.
c. Since the reactants are present in gaseous phase while the catalyst used is in solid phase, it represents heterogeneous catalysis.

Question 23.

i. State whether the given reaction is an example of heterogeneous or homogeneous catalysis.
ii. What is the role of Fe, K₂O and Al₂O₃ in this reaction?
Answer:
i. This reaction is an example of heterogeneous catalysis.
ii. Fe is used as a catalyst while K₂O and Al₂O₃ are promoters of the Fe catalyst. Al₂O₃ is used to prevent the fusion of Fe particles while K₂O causes chemisorption of nitrogen atoms.

Question 24.
Describe hydrogenation reaction of vegetable oils.
Answer:
i. Hydrogenation reaction of vegetable oils used in food industry to produce solid fats. The reaction is as follows:

ii. The reaction is catalysed by finely divided metals like Ni, Pd or Pt.
iii. Vegetable oil contains one or more carbon-carbon double bonds (C = C) in its structure.
iv. On hydrogenation, a solid product (which contains only carbon-carbon single bonds) is formed. It is called Vanaspati ghee.
v. The hydrogenation reaction of vegetable oils is an example of heterogeneous catalysis as the reactant and the catalyst are not present in the same phase.

Question 25.
i. Explain the role of catalytic converters in automobile exhaust.
ii. Why do automobiles with catalytic converter require unleaded petrol?
Answer:
i. a. An important application of heterogeneous catalysts is in automobile catalytic converters.
b. In automobile exhaust, large number of air pollutants such as carbon monoxide, nitric oxide, etc. are present.
c. The catalytic converter transforms these air pollutants into carbon dioxide, water, nitrogen and oxygen.
ii. The catalyst used in the catalytic converter gets poisoned by the adsorption of lead (Pb) present in the petrol. Hence, the automobiles with catalytic converter requires unleaded petrol.

Question 26.
What are inhibitors? Explain with an example.
Answer:
Inhibitors are substances that decreases the rate of chemical reactions.
e.g. Chloroform forms poisonous substance, carbonyl chloride, by air oxidation.

When 2% ethanol is added to chloroform, the formation of COCl₂ is suppressed because ethanol acts as an inhibitor and retards the above reaction.
[Note: Chloroform was earlier used as an anaesthetic.]

Question 27.
Write decomposition reaction of hydrogen peroxide. Suggest how this decomposition can be prevented.
Answer:
i. Hydrogen peroxide decomposes as,

ii. The reaction can be inhibited by addition of dilute acid or glycerol as they act as inhibitors.

Question 28.
Explain why 2% ethanol is added to chloroform?
Answer:
Inhibitors are substances that decreases the rate of chemical reactions.
e.g. Chloroform forms poisonous substance, carbonyl chloride, by air oxidation.

When 2% ethanol is added to chloroform, the formation of COCl₂ is suppressed because ethanol acts as an inhibitor and retards the above reaction.
[Note: Chloroform was earlier used as an anaesthetic.]

Question 29.
Describe the steps involved in heterogeneous catalysis by solid catalyst.
OR
Explain the mechanism involved in catalytic action of a heterogeneous catalyst.
Answer:
The catalytic action of a heterogeneous catalyst occurs on the surface of a catalyst.
The mechanism involves the following five steps.
i. Diffusion of reactants towards the surface of the catalyst.
ii. Adsorption of reactant molecules on the surface of the catalyst.
iii. Occurrence of chemical reaction on the catalyst surface and formation of an intermediate.
iv. Formation of the products.
v. Desorption of reaction products from the catalyst surface. Products leave the catalyst surface in the following steps.
Steps involved in desorption of reaction products:
Diffusion → Adsorption → Intermediate formation → Product formation → Desorption
vi. Fresh reactant molecules can replace the products to start the cycle again as in first step.
vii. This is why catalyst remains unchanged in mass and chemical composition at the end of the reaction.

Question 30.
Write a short note on catalytic activity.
Answer:

• The catalytic activity of a catalyst depends on the strength of chemisorption.
• If large number of reactant molecules (gas or liquid) are strongly adsorbed on the surface of solid catalyst, the catalyst is said to be active.
• However, the adsorption of reactant molecules on the surface, that is, the bond formed between adsorbate and adsorbent surface should not be very strong so that they are not immobilized.
• d-block metals such as Fe, V and Cr tend to be strongly active towards O₂, C₂H₂, C₂H₄, CO, H₂, CO₂, N₂, etc.
• Mn and Cu are unable to adsorb N₂ and CO₂.
• The metals Mg and Li adsorb O₂ selectively.

Question 31.
Explain catalytic selectivity with suitable examples.
Answer:
i. Some solid catalysts are selective in their action.
ii. The same gaseous reactants produce different products when different catalysts are used.
e.g.
a. The gaseous ethylene and O₂ react to produce different products with different catalysts.

b. The gaseous carbon monoxide and H₂ produce different products by using different catalysts.

Question 32.
i. What are zeolites?
ii. Zeolites are shape selective catalysts. Explain.
iii. What is the use of a zeolite catalyst ZSM-5 in petroleum industry?
Answer:
i. a. Zeolites are aluminosilicates with three-dimensional network of silicates.
b. Some silicon atoms in this network are replaced by aluminium atoms giving Al – O – Si framework which results in microporous structure.

ii. a. The reactions in zeolites are dependent on the size and shape of reactant or products, b. It also depends on the pores and cavities of zeolites.
b. Therefore, zeolites are shape selective catalysts.

iii. In petroleum industry, zeolite catalyst ZSM-5 converts alcohols directly to gasoline (petrol) by dehydration which gives a mixture of hydrocarbons.

Question 33.
State the importance of colloids in day-to-day life.
Answer:

• Colloid chemistry is the chemistry of everyday life.
• A number of substances we use in our day-to-day life are colloids. For example, milk, butter, jelly, whipped cream, mayonnaise.
• Knowledge of colloid chemistry is essential for understanding about many useful materials like cement, bricks, pottery, porcelain, glass, enamels, oils, lacquers, rubber, celluloid and other plastics, leather, paper, textiles, filaments, crayons, inks, road construction material, etc.
• In many daily processes like cooking, washing, dyeing, painting, ore floatation, water purification, sewage disposal, smoke prevention, photography, pharmacy, use of colloids is important.

Question 34.
What are colloids? Explain.
Answer:
i. Colloids are heterogeneous mixtures.
ii. The component of colloid present in the largest proportion is called dispersion medium and the other components are called dispersed phase.
iii. The particles of the dispersed phase are larger than the size of a molecule and smaller than the particles which we can see with naked eye.
e.g.

• Observe the formation of solution of salt and water. Salt dissolves completely in water and forms homogeneous system.
• On the other hand, ground coffee or tea leaves with milk form suspension.
• Between the two extremes of solution and suspension exists a large group of systems called colloidal dispersions or simply colloids.

Question 35.
State the differences between colloids and solutions.
Answer:
Colloids:

1. Colloids contain particles of dispersed phase with diameters in the range of 2 to 500 nm.
2. They are translucent to light.
3. e.g. Milk, fog, etc.

Solutions:

1. Solutions contain solute particles with diameters in the range of 0.1 to 2 nm.
2. They are transparent or may be coloured.
3. e.g. NaCl solution

Question 36.
Explain: Natural phenomena of colloids observed in daily life.
Answer:
Following are some examples of colloids observed in daily life.
i. Blue colour of the sky: The sky appears blue to us because minute dust particles along with minute water droplets dispersed in air scatter blue light which reaches our eyes.
ii. Blood: It is a colloidal dispersion of plasma proteins and antibodies in water arid at the same time blood is also a suspension of blood cells and platelets in water.
iii. Soils: Fertile soils are colloidal in nature where humus acts as a protective colloid. Soil adsorbs moisture and nourishing materials due to its colloidal nature.
iv. Fog, mist and rain:

• Mist is caused by small droplets of water dispersed in air.
• Fog is formed whenever there is temperature difference between ground and air.
• A large portion of air containing dust particles gets cooled below its dew point, the moisture from the air condenses on the surface of these particles which form fine droplets, which are colloidal particles and float in the air as fog or mist.

Question 37.
State different ways to classify colloids.
Answer:
Colloids can be classified in three different ways:

• Physical states of dispersed phase and dispersion medium
• Interaction or affinity of phases
• Molecular size

Question 38.
Name the types of colloids based on the physical states of dispersed phase and dispersion medium. Give two examples of each.
Answer:
There are eight types of colloids based on the physical states of dispersed phase and dispersion medium as given below.

Sr. No.	Type of Colloids	Examples
i.	Solid sol (solid dispersed in solid)	Coloured glasses, gemstones
ii.	Sols and gels (solid in liquid)	Gelatin, muddy water
iii.	Aerosol (solid in gas)	Smoke, dust
iv.	Gel (liquid in solid)	Cheese, jellies
v.	Emulsion (liquid in liquid)	Milk, hair cream
vi.	Aerosol (liquid in gas)	Fog, mist
vii.	Solid sol (gas in solid)	Foam rubber, plaster
viii.	Foam (gas in liquid)	Froth, soap lather

Question 39.
Complete the following chart.

Answer:

[Note: Students can write any one example of the given type of colloids.]

Note: Types of colloids based on the physical states of dispersed phase and dispersion medium.

Question 40.
Describe classification of colloids based on the interaction or affinity of phases.
Answer:
On the basis of interaction or affinity of phases, a colloidal solution is classified as lyophilic and lyophobic.
i. Lyophilic colloids:

• A colloidal solution in which the particles of dispersed phase have a great affinity for the dispersion medium are lyophilic colloids.
• If the lyophilic sol is evaporated, the dispersed phase separates. However, if it is remixed with the medium, the sol. can be formed again and hence, such sols are called reversible sols.
• They are stable and difficult to coagulate.

ii. Lyophobic colloids:

• Colloidal solution in which the particles of the dispersed phase have no affinity for the dispersion
  medium are called lyophobic colloids.
• The common examples are Ag, Au, hydroxides like Al(OH)₃, Fe(OH)₃, metal sulphides.
• Once precipitated or coagulated they have little tendency or no tendency to revert back to colloidal state.

[Note: Lyo means liquid and philic means loving whereas phobic means fearing and hence liquid hating. If water is the dispersion medium, the terms hydrophilic and hydrophobic are used.]

Question 41.
Give reason: Lyophilic sols are called reversible sols.
Answer:

• When lyophilic sol is evaporated, the dispersed phase separates.
• However, if the dispersed phase is remixed with the medium, the sol can be formed again.

Hence, lyophilic sols are called reversible sols.

Question 42.
How are colloids classified based on their molecular size?
Answer:
Colloids are classified into three types based on their molecular size as described below.
i. Multimolecular colloids:

• In multimolecular colloids, the individual particles consist of an aggregate of atoms or small molecules with size less than 10³ pm.
  e.g. Gold sol consists of particles of various sizes having several gold atoms.
• Colloidal solution in which particles are held together with van der Waals force of attraction is called multimolecular colloid.
  e.g. S₈ sulphur molecules

ii. Macromolecular colloids: In this type of colloids, the molecules of the dispersed phase are sufficiently large in size (macro) to be of colloidal dimensions.
e.g. Starch, cellulose, proteins, polythene, nylon, plastics.

iii. Associated colloids or micelles:

• The substances behave as normal electrolytes at low concentration and associated in higher concentration forming a colloidal solution.
• The associated particles are called micelles, e.g. Soaps and detergents

Question 43.
How can be colloids prepared by chemical methods?
Answer:
i. Colloidal dispersions can be prepared by chemical reactions leading to formation of molecules by double decomposition, oxidation, reduction or hydrolysis.
ii. Molecules formed in these reactions are water-insoluble and thus, they aggregate leading to the formation of colloids.
e.g.

Question 44.
Describe the process involved in peptization?
Answer:

• During peptization a precipitate is converted into colloidal sol by shaking with dispersion medium in the presence of a small amount of an electrolyte. The electrolyte used is known as peptizing agent.
• During the process, the precipitate adsorbs one of the ions of the electrolyte on its surface and as a result, positive or negative charge is developed on the precipitate which finally breaks up into small particles of colloidal size.

[Note: This method is generally applied to convert a freshly prepared precipitate into a colloidal sol.]

Question 45.
Why is it necessary to purify colloidal solutions?
Answer:

• Colloidal solution generally contains excessive amount of electrolytes and some other soluble impurities.
• A small quantity of an electrolyte is necessary for the stability of colloidal solution, however, a large quantity of electrolyte may result in coagulation.
• It is also necessary to reduce soluble impurities.

Hence, it is necessary to purify colloidal solutions.

Question 46.
i. What is purification of colloidal solution?
ii. How can a colloidal solution be purified using the method of dialysis?
Answer:
i. The process used for reducing the amount of impurities to a requisite minimum is known as purification of colloidal solution.

ii. a. Dialysis is a process of removing a dissolved substance from a colloidal solution by diffusion through a suitable membrane.
b. Purification of colloidal solution can be carried out using dialysis by the following method.

• The apparatus used is dialyser.
• A bag of suitable membrane containing the colloidal solution is suspended in a vessel through which fresh water is continuously flowing.
• The molecules and ions diffuse through membrane into the outer water and pure colloidal solution is left behind.

Question 47.
What are the general properties exhibited by colloidal dispersions?
Answer:
General properties exhibited by colloidal dispersions:

• Colloidal system is heterogeneous and consists of two phases, dispersed phase and dispersion medium.
• The dispersed phase particles pass slowly through parchment paper or animal membrane, but readily pass through ordinary filter paper.
• Colloidal particles are usually not detectable by powerful microscope.

Question 48.
Discuss the factors that influence the colour of colloidal solutions.
Answer:

• Colour of colloidal solution depends on the wavelength of light scattered by dispersed particles.
• The colour of colloidal dispersion also changes with the manner in which the observer receives the light.
  e.g. Mixture of a few drops of milk and large amount of water appears blue when viewed by the scattered light and red when viewed by transmitted light.
• It also depends on size of colloidal particles.
  e.g. Finest gold sol is red in colour whereas with increase in size it appears purple.

Question 49.
Give three examples each:
i. Positively charged sols
ii. Negatively charged sols
Answer:
i. Positively charged sols: Al₂O₃. xH₂O, haemoglobin, TiO₂ sol
ii. Negatively charged sols: Au sols, Congo red sol, clay

Note: Some common sols with the nature of charge on the particles are listed in the table below.

Positively charged sols	Negatively charged sols
Hydrated metallic oxides: Al2O3.xH2O, CrO3.xH2O, Fe2O3.xH2O.	Metals: Cu, Ag. Au sols Metallic sulphides: As2S3, Sb2S3, CdS
Basic dye stuff, methylene blue sols	Acid dye stuff, eosin, Congo red sol
Haemoglobin (blood)	Sols of starch, gum
Oxides: TiO2 sol	Gelatin, clay, gum sols

Question 50.
Explain the term electroosmosis.
Answer:

• Movement of dispersed particles can be prevented by suitable means such as use of membrane.
• On doing so, it is observed that the dispersion medium begins to move in an electric field. This is known as electroosmosis.

Question 51.
What is coagulation?
Answer:
The precipitation of colloids by removal of charge associated with colloidal particles is called coagulation.

Question 52.
How can we bring about precipitation of lyophobic colloids?
Answer:

• The charge on the colloidal particles is due to the preferential adsorption of ions on their surface.
• Hence, lyophobic colloids can be precipitated out by removing the charge on the colloidal particles (dispersed phase).

Question 53.
Discuss various methods that are used to bring about coagulation of lyophobic sols.
Answer:
Coagulation of the lyophobic sols can be carried out in the following ways.

• By electrophoresis: The colloidal particles move towards oppositely charged electrodes, get discharged and precipitate.
• By mixing two oppositely charged sols: Oppositely charged sols when mixed in almost equal proportions neutralize their charges and get precipitated.
  e. g. Mixing of hydrated ferric oxide (positive sol) and arsenious sulphide (negative sol) brings them in the precipitated forms. This type of coagulation is called mutual coagulation.
• By boiling: When a sol is boiled, the adsorbed layer is disturbed as a result of increased collisions with molecules in the dispersion medium. This reduces charge on the particles and subsequently particles settle down as a precipitate.
• By persistent dialysis: On prolonged dialysis, traces of the electrolyte present in the sol are removed almost completely. The colloids then become unstable and finally precipitate.
• By addition of electrolytes: When excess of an electrolyte is added, the colloidal particles are precipitated.

Question 54.
Write Hardy-Schulze rule.
Answer:
Generally, greater the valency of the flocculating ion added, greater is its power to cause precipitation. This is known as Hardy-Schulze rule.

Question 55.
Differentiate between oil in water and water in oil emulsions.
Answer:
Oil in water:

1. Oil is the dispersed phase and water is the dispersion medium.
2. If water is added, it will be miscible with the emulsion.
3. Addition of small amount of an electrolyte makes the emulsion conducting.
4. Continuous phase is water.
5. Basic metal sulphates, water soluble alkali metal soaps are used as emulsifiers.

Water in oil:

1. Water is the dispersed phase and oil is the dispersion medium.
2. If oil is added, it will be miscible with the emulsion.
3. Addition of small amount of an electrolyte has no effect on conducting power.
4. Continuous phase is oil.
5. Water insoluble soaps such as those of Zn, Al, Fe, alkaline earth metals are used as emulsifiers.

Question 56.
What are the properties of emulsion?
Answer:
Properties of emulsion:

• Emulsion can be diluted with any amount of the dispersion medium. On the other hand, the dispersed liquid when mixed forms a separate layer.
• The droplets in emulsions are often negatively charged and can be precipitated by electrolytes.
• Emulsions show Brownian movement and Tyndall effect.
• The two liquids in emulsions can be separated by heating, freezing, centrifuging, etc.

Question 57.
Give applications of colloids.
Answer:
Applications of colloids:
i. Electrical precipitation of smoke:

• Smoke is a colloidal solution of solid particles of carbon, arsenic compound, dust, etc. in the air.
• When smoke is allowed to pass through chamber containing charged plates, smoke particles lose their charge and get precipitated. The particles then settle down on the floor of the chamber.
• The precipitator used is called Cottrell precipitator.

ii. Purification of drinking water:

• Water obtained from natural sources contains colloidal impurities.
• By addition of alum to such water, colloidal impurities get coagulated and settle down. This makes water potable.

iii. Medicines:

• Usually medicines are colloidal in nature.
• Colloidal medicines are more effective owing to large surface area to volume ratio of a colloidal particle and easy assimilation.
  e.g. Argyrol is a silver sol used as an eye lotion. Milk of magnesia, an emulsion is used in stomach disorders.

iv. Rubber industry: Rubber is obtained by coagulation of latex.
v. Cleansing action of soaps and detergents.
vi. Photographic plates, films, and industrial products like paints, inks, synthetic plastics, rubber, graphite lubricants, cement, etc. are colloids.

Question 58.
Match column A with column B.

	Column A		Column B
i.	Tyndall effect	i.	Kinetic property
ii.	Electrophoresis	ii.	Argyrol
iii.	Silver sol	iii.	Optical property
iv.	Brownian motion	iv.	Coagulation

Answer:
i – c,
ii – d,
iii – b,
iv – a

Question 59.
In drinking water treatment, often alum is added for the complete removal of suspended impurities. On complete dissolution, alum produces positive charge which neutralizes the charge on the suspended particles and thus, impurities are easily removed.
i. Name and define the process involved due to which charge on particles get neutralized.
ii. What is the role of alum in the above mentioned process?
Answer:
i. a. Charge on particles get neutralized due to coagulation.
b. The precipitation of colloids by removal of charge associated with colloidal particles is called coagulation.
ii. Alum acts as a reagent that helps in coagulation of the suspended particles by the removal of the charge associated with these particles.

Multiple Choice Questions

1. Which of the following is responsible for adsorption phenomenon?
(A) Hydrogen bonding
(R) Dipole-dipole forces
(C) Ion-dipole forces
(D) Dispersion forces
Answer:
(D) Dispersion forces

2. A substance which adsorbs another substance on its surface is called ……………..
(A) adsorbate
(B) absorbate
(C) adsorbent
(D) absorbent
Answer:
(C) adsorbent

3. During adsorption, the molecules of the substance which gets adsorbed are termed as
(A) adsorbent
(B) adsorbate
(C) absorbent
(D) absorbate
Answer:
(B) adsorbate

4. in adsorption of acetic acid on charcoal, acetic acid is ……………
(A) adsorhate
(B) adsorbent
(C) absorbent
(D) absorbate
Answer:
(A) adsorhate

5. The process of removal of an adsorbed substance from the surface is known as
(A) sorption
(B) oxidation
(C) reduction
(D) desorption
Answer:
(D) desorption

6. ………….. is the process in which adsorbate molecules are held on the surface of the adsorbent by weak van der Waals forces.
(A) Chemisorption
(B) Absorption
(C) Physisorption
(D) Biosorption
Answer:
(C) Physisorption

7. Which of the following is an example of physical adsorption?
(A) Adsorption of acetic acid in solution by charcoal
(B) Adsorption of O₂ on tungsten
(C) Adsorption of N₂ on Fe
(D) Adsorption of H₂ on Ni
Answer:
(A) Adsorption of acetic acid in solution by charcoal

8. Chemisorption is a slow process because …………….
(A) it forms multimolecular layer
(B) it is reversible
(C) it takes place at normal temperature
(D) it requires high activation energy
Answer:
(D) it requires high activation energy

9. The number of layer(s) formed on adsorbent in chemical adsorption is …………….
(A) one
(B) two
(C) three
(D) many
Answer:
(A) one

10. Which of the following statements is CORRECT regarding chemical adsorption?
(A) It is highly specific in nature.
(B) It is relatively strong.
(C) It involves the formation of monolayer of adsorbed particles.
(D) All of these.
Answer:
(D) All of these.

11. Which of the following is adsorbed to maximum extent on charcoal?
(A) H₂
(B) N₂
(C) Cl₂
(D) O₂
Answer:
(C) Cl₂

12. The relation between the amount of substance adsorbed by an adsorbent and the equilibrium pressure or …………. at any constant temperature is called adsorption isotherm.
(A) surface area
(B) volume
(C) circumference
(D) concentration
Answer:
(D) concentration

13. For equilibrium pressure (P), the mass of gas adsorbed (x) and mass of adsorbent (m) may be expressed as Freundlich adsorption isotherm as ……………

Answer:
(B) \(\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kP}^{\frac{1}{\mathrm{n}}}\)

14. When log x/m is plotted against log P, the intercept obtained …………..
(A) on Y axis is equal to log K
(B) on Y axis is equal to K
(C) on X axis is equal to log K
(D) on X axis is equal to K
Answer:
(A) on Y axis is equal to log K

15. The adsorption isotherm tends to saturate at ………….. pressure.
(A) low
(B) moderate
(C) all of these
(D) high
Answer:
(D) high

16. In Haber process for manufacture of NH₃, the catalyst used is ……………
(A) iron
(B) copper
(C) vanadium pentoxide
(D) nickel
Answer:
(A) iron

17. A substance that decreases the rate of a chemical reaction is called ……………
(A) inhibitor
(B) prohibitor
(C) promoter
(D) reactor
Answer:
(A) inhibitor

18. Whether a given mixture forms a true solution or a colloidal dispersion depends on the …………….
(A) charge of solute particles
(B) size of solvent particles
(C) size of solute particles
(D) charge of solvent particles
Answer:
(C) size of solute particles

19. An aerosol is a dispersion of a ……………
(A) gas in a solid
(B) liquid in a gas
(C) solid in a gas
(D) both (B) and (C)
Answer:
(D) both (B) and (C)

20. The dispersed phase in Pumice stone is ……………
(A) solid
(B) liquid
(C) gas
(D) none of these
Answer:
(C) gas

21. Colloidal solution in which the dispersed phase has little affinity for the dispersion medium is called ………………
(A) lyophobic colloids
(B) lyophilic colloids
(C) hydrophilic colloids
(D) emulsions
Answer:
(A) lyophobic colloids

22. Which of the following is NOT an example of macromolecular colloid?
(A) Starch
(B) Proteins
(C) S₈ molecules
(D) Nylon
Answer:
(C) S₈ molecules

23. Tyndall effect is useful ……………….
(A) to identify colloidal dispersions
(B) to count number of particles in colloidal dispersion.
(C) to determine the size of the colloidal particles
(D) all of these
Answer:
(D) all of these

24. Brownian movement is a ……………… type of property of the colloidal sol.
(A) electrical
(B) optical
(C) kinetic
(D) colligative
Answer:
(C) kinetic

25. The migration of colloidal particles under the influence of an electric field is called …………….
(A) catalysis
(B) Brownian movement
(C) electrophoresis
(D) Tyndall effect
Answer:
(C) electrophoresis

26. The capacity of an ion to coagulate a colloidal solution depends on ……………….
(A) its shape
(B) its valency
(C) the sign of charge
(D) both (B) and (C)
Answer:
(D) both (B) and (C)

27. ……………… is an example of water in oil type of emulsion.
(A) Milk
(B) Cod liver oil
(C) Vanishing cream
(D) Paint
Answer:
(B) Cod liver oil

28. Which of the following has highest precipitation power to precipitate negative sol?
(A) Al³⁺
(B) Mg²⁺
(C) Na⁺
(D) K⁺
Answer:
(A) Al³⁺

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 12 Chemical Equilibrium Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 12 Chemical Equilibrium

Question 1.
Explain irreversible reaction.
Answer:
Irreversible reaction:
i. Reactions which occur only in one direction, namely, from reactant to products are called irreversible reactions.
ii. They proceed in only a single direction until one of the reactants is exhausted.
iii. The direction in which an irreversible reaction occurs is indicated by an arrow (→) pointing towards the products in the chemical equation.
e.g. a. \(\mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{~g})} \stackrel{\text { Burn }}{\longrightarrow} \mathrm{CO}_{2(\mathrm{~g})}\)
b. \(2 \mathrm{KClO}_{3(\mathrm{~s})} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{KCl}_{(\mathrm{s})}+3 \mathrm{O}_{2(\mathrm{~g})}\)

Question 2.
What is a closed system?
Answer:
A system in which there is no exchange of matter with the surroundings is called a closed system.

Question 3.
What is an open system?
Answer:
A system in which exchange of both matter and heat occurs with the surroundings is called an open system.

Question 4.
Why was calcium oxide used in theatre lighting?
Answer:
Calcium oxide (CaO) on strong heating glows with a bright white light. Hence, CaO was used in theatre lighting, which gave rise to the phrase ‘in the limelight’.

Question 5.
Explain liquid-vapour equilibrium with an example.
Answer:
Liquid-vapour equilibrium:
i. Consider reversible physical process of evaporation of liquid water into water vapour in a closed vessel. Initially, there is practically no water vapour in the vessel.

ii. When the liquid evaporates in the closed container, the liquid molecules escape from the liquid surface into vapour phase building up vapour pressure. They also condense back into liquid state because the container is closed.

iii. In the beginning the rate of evaporation is high and the rate of condensation is low. But with time, as more and more vapour is formed, the rate of evaporation goes down and the rate of condensation increases. Eventually the two rates become equal. This gives rise to a constant vapour pressure. This state is known as an ‘equilibrium state’.
In this state, the rate of evaporation is equal to the rate of condensation.
It may be represented as: H₂O_(l) ⇌ H₂O_(Vapour)

iv. At equilibrium, the pressure exerted by the gaseous water molecules at a given temperature remains constant, known as the equilibrium vapour pressure of water (or saturated vapour pressure of water or aqueous tension). The saturated vapour pressure increases with increase of temperature.

[Note: The saturated vapour pressure of water at 100 °C is 1 atm (1.013 bar). Hence, water boils at 100 °C when pressure is 1 atm.]

Question 6.
What is meant by the term ‘normal boiling point’ of a liquid?
Answer:
For any pure liquid at 1 atm pressure, the temperature at which its saturated vapour pressure equals to atmospheric pressure is called the normal boiling point of that liquid.
e.g. The boiling point of ethyl alcohol is 78 °C i.e., the saturated vapour pressure of ethyl alcohol at 78 °C is 1 atm (1.013 bar).

Question 7.
Give an example of solid-liquid equilibrium.
Answer:
A mixture of ice and water in a perfectly insulated thermos flask at 273 K is an example of solid-liquid equilibrium.
H₂O_(s) ⇌ H₂O_(l)

Question 8.
Identify the type of equilibrium in the following physical processes:
i. Camphor_(s) ⇌ Camphor_(g)
ii. Ammonium chloride_(s) ⇌ Ammonium chloride_(g)
iii. Carbon dioxide gas ⇌ Dry ice
iv. Water ⇌ Ice
Answer:
i. Solid – vapour equilibrium
ii. Solid – vapour equilibrium Solid
iii. Solid – vapour equilibrium
iv. Solid – liquid equilibrium

Question 9.
Name two substances that undergoes sublimation.
Answer:
Camphor, ammonium chloride.

Question 10.
Write a short note on chemical equilibrium.
Answer:
Chemical equilibrium:

• If a reaction takes place in a closed system so that the products and reactants cannot escape, we often find that reaction does not give a 100% yield of products. Instead some reactants remain after the concentrations stop changing.
• When there is no further change in concentration of reactant and product, the chemical reaction has attained equilibrium, with the rates of forward and reverse reactions being equal.
• Chemical equilibrium at a given temperature is characterized by constancy of measurable properties such as pressure, concentration, density, etc.
• Chemical equilibrium can be approached from either side of the chemical reaction.

Question 11.
Explain the law of mass action and give its mathematical representation.
Answer:
Statement: The law of mass action states that the rate of a chemical reaction at each instant is proportional to the product of concentrations of all the reactants.
Explanation: A rate equation can be written for a reaction by applying the law of mass action as follows: Consider a reaction, A + B → C
Here A and B are the reactants and C is the product. The concentrations of chemical species are expressed in mol L^-1 and denoted by putting the formula in square brackets. On applying the law of mass action to this
reaction, a proportionality expression can be written as: Rate ∝ [A] [B]
This proportionality expression is transformed into an equation by introducing a proportionality constant, k, as follows:
Rate = k [A] [B]
This equation is called the rate equation and the proportionality constant, k, is called the rate constant of the reaction.

Question 12.
Write the rate equation for the following reactions:
i. C + O₂ → CO₂
ii. 2KClO₃ → 2KCl + 3O₂
Answer:
The rate equation is written by applying the law of mass action.
i. The reactants are C and O₂
Rate ∝ [C] [O₂]
∴ Rate = k [C] [O₂]
ii. The reactant is KClO₃ and its 2 molecules appear in the balanced equation.
∴ Rate ∝ [KClO₃]²
∴ Rate = k [KClO₃]²

Question 13.
Derive the expression of equilibrium constant, K_C for the reaction:
A + B ⇌ C + D
Answer:
Consider a hypothetical reversible reaction A + B ⇌ C + D.
Two reactions, namely, forward and reverse reactions occur simultaneously in a reversible chemical reaction. The rate equations for the forward and reverse reactions are:
Rate_forward ∝ [A][B]
∴ Rate_forward = k_f [A] [B] …… (1)
∴ Rate_reverse ∝ [C] [D]
∴ Rate_reverse = k_r [C] [D] …. (2)
At equilibrium, the rates of forward and reverse reactions are equal. Thus,
Rate_forward = Rate_reverse
∴ k_f [A] [B] = k_r [C] [D]
∴ \(\frac{\mathrm{k}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{r}}}=\mathrm{K}_{\mathrm{C}}=\frac{[\mathrm{C}][\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]}\) …….. (3)
K_C is called the equilibrium constant.

Question 14.
Show that the equilibrium constant of the reverse chemical reaction (K_C) is the reciprocal of the equilibrium constant (K_C).
Answer:
Consider a reversible chemical reaction:
aA + bB ⇌ cC + dD
The equilibrium constant, K_C = \(\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\)
Consider the reverse reaction:
cC + dD ⇌ aA + bB.
The equilibrium constant, K_C is:
K_C = \(\frac{[\mathrm{A}]^{a}[\mathrm{~B}]^{\mathrm{b}}}{[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}=\frac{1}{\mathrm{~K}_{\mathrm{C}}}\)
Thus, equilibrium constant of the reverse chemical reaction (K_C) is the reciprocal of the equilibrium constant K_C.

Question 15.
Write equilibrium constant expressions for both forward and reverse reaction for the synthesis of ammonia by the Haber process.
Answer:
Synthesis of ammonia by Haber process:
N_2(g) + 3H_2(g) ⇌ 2NH_3(g)

Question 16.
How are the equilibrium constants of the following pair of equilibrium reactions related?

Answer:

ii. K_C = \(\frac{\left[\mathrm{CO}_{2}\right]\left[\mathrm{N}_{2}\right]}{[\mathrm{CO}]\left[\mathrm{N}_{2} \mathrm{O}\right]}\)

Question 17.
Write KP expression for the reaction:
aA_(g) + bB_(g) ⇌ cC_(g) + dD_(g)
Answer:
For the given reaction,
K_P = \(\frac{\left(P_{c}\right)^{c}\left(P_{D}\right)^{d}}{\left(P_{A}\right)^{a}\left(P_{B}\right)^{b}}\)

Question 18.
N_2(g) + 3H_2(g) ⇌ 2NH_3(g)
Write expressions for K_P and substitute expressions for P_N2, P_H2 and P_NH3 using ideal gas equation.
Answer:
For the given reaction, K_P = \(\frac{\left(P_{N H_{3}}\right)^{2}}{\left(P_{N_{2}}\right)\left(P_{H_{2}}\right)^{3}}\)

[Note: The above question is modified to apply appropriate textual context, i. e., to indicate that students need to use ideal gas equation to derive expressions for P_N2, P_H2 and P_NH3]

Question 19.
For a chemical equilibrium reaction
H_2(g) + I_2(g) ⇌ 2HI_(g),
write an expression for K_P (and relate it to K_C).
Answer:

Question 20.
Write the relationship between K_C and K_P for the following equilibria:

Answer:

Question 21.
Write the expressions for K_C and K_P and the relationship between them for the equilibrium reaction,
2A_(g) + B_(g) ⇌ 3C_(g) + 2D_(g)
Answer:

Question 22.
Explain in short homogeneous equilibrium and heterogeneous equilibrium.
Answer:
i. In a homogeneous equilibrium, the reactants and products are in the same phase.
e.g. Dissociation of HI:
2HI_(g) ⇌ H_2(g) + I_2(g)
ii. In a heterogeneous equilibrium, the reactants and products exist in different phases, e.g. Formation of NH₄Cl:
NH_3(g) + HCl_(g) ⇌ NH₄Cl_(s)

Question 23.
The unit of K_C is different for different reactions. Explain this statement with suitable examples.
Answer:
Unit of equilibrium constant:
i. The unit of equilibrium constant depends upon the expression of K_C which is different for different equilibria. Therefore, the unit of K_C is also different for different reactions.
ii. Consider the following equilibrium reaction:

iii. Consider the following equilibrium reaction:

Question 24.
Write the equilibrium constant expression for the decomposition of baking soda. Deduce the unit of K_C from the above expression.
Answer:

[Note: Considering gaseous reactants and products, in this reaction, Δn = 2 – 0 = 2
∴ Units of K_C = (mol dm^-3)^Δn
= (mol dm^-3)²
= mol² dm^-6
Thus, the units of the above reaction is mol² dm^-6.]

Question 25.
What are the characteristics of equilibrium constant?
Answer:
Characteristics of equilibrium constant:

• The value of equilibrium constant is independent of initial concentrations of either the reactants or products.
• Equilibrium constant is temperature dependent. Hence, K_C and K_P change with change in temperature.
• Equilibrium constant has a characteristic value for a particular reversible reaction represented by a balanced equation at a given temperature.
• Higher value of K_C or K_P means more concentration of products is formed and the equilibrium point is more towards right hand side and vice versa.

Question 26.
Explain how equilibrium constant helps in predicting the direction of the reaction.
Answer:
Prediction of the direction of the reaction:
i. For the reaction, aA + bB ⇌ cC + dD,
The equilibrium constant (K_C) is given as:
K_C = \(\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\)
where, all the concentrations are equilibrium concentrations.
ii. When the reaction is not necessarily at equilibrium, the concentration ratio is called Q_C i.e.,
Q_C = \(\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\)
iii. By comparing Q_C with K_C for a reaction under given conditions, we can decide whether the forward or the reverse reaction should occur to establish the equilibrium.
a. Q_C < K_C: The reaction will proceed from left to right, in forward direction, generating more product to attain the equilibrium.
b. Q_C > K_C: The reaction will proceed from right to left, removing product to attain the equilibrium.
c. Q_C = K_C: The reaction is at equilibrium and no net reaction occurs.

[Note: The prediction of the direction of the reaction on the basis of Q_C and K_C values makes no comment on the time required for attaining the equilibrium.]

Question 27.
Explain how K_C can be used to know the extent of the reaction?
Answer:
Extent of the reaction: The equilibrium constant expression indicates that the magnitude of K_C is:
i. directly proportional to the concentrations of the products.
ii. inversely proportional to the concentrations of the reactants.
a. Value of K_C is very high (K_C > 10³):
At equilibrium, there is a high proportion of products compared to reactants.
Forward reaction is favoured.
Reaction is in favour of products and nearly goes to completion.

b. Value of K_C is very low (K_C < 10^-3):
At equilibrium, only a small fraction of the reactants is converted into products.
Reverse reaction is favoured.
Reaction hardly proceeds towards the products.

c. Value of K_C is in the range of 10^-3 to 10³:
Appreciable concentrations of both reactants and products are present at equilibrium.

Question 28.
For the following reactions, write K_C expressions and predict direction of the reactions based on the magnitude of their equilibrium constants.
i. 2H_2(g) + O_2(g) ⇌ 2H2O_(g), K_C = 2.4 × 10⁴⁷ at 500 K
ii. 2H₂O_(g) ⇌ 2H_2(g) + O_2(g), K_C = 4.2 × 10^-48 at 500 K
Answer:
i. a. K_C expression:
K_C = \(\frac{\left[\mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\right]^{2}}{\left[\left[\mathrm{H}_{2(\mathrm{~g})}\right]\right]^{2}\left[\mathrm{O}_{2(\mathrm{~g})}\right]}\)
b. For the reaction, K_C = 2.4 × 10⁴⁷ at 500 K
If the value of K_C >>> 10³, forward reaction is favoured.
Hence, the given reaction will proceed in the forward direction and will nearly go to completion.

ii. a. K_C expression:
K_C = \(\frac{\left[\mathrm{H}_{2(\mathrm{~g})}\right]^{2}\left[\mathrm{O}_{2(\mathrm{~g})}\right]}{\left[\mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\right]^{2}}\)
b. For the reaction, K_C = 4.2 × 10^-48 at 500 K
If the value of K_C <<< 10^-3, reverse reaction is favoured.
Hence, the given reaction will proceed in the backward direction and will nearly go to completion.

Question 29.
Describe how equilibrium constant can be used to calculate the composition of an equilibrium mixture.
Answer:
An equilibrium constant can be used to calculate the composition of an equilibrium mixture.
Consider an equilibrium reaction, A_(aq) + B_(aq) ⇌ C_(aq) + D_(aq)
The equilibrium constant is 4.0 at a certain temperature.
Let the initial amount of A and B be 2.0 mol in ‘V’ litres. Let x mol be the equilibrium amount of C.
Hence, we can construct a table as shown below:

The expression for equilibrium constant can be written as:

Substituting the value of equilibrium concentration, we get

Therefore, equilibrium concentrations are 0.67 mol of A, 0.67 mol of B, 1.33 mol of C and 1.33 mol of D in V litres.

Question 30.
Explain the link between chemical equilibrium and chemical kinetics:
Answer:
Equilibrium constant (K_C) is related to rate or velocity constants of forward reaction (k_f) and reverse reaction (k_r) as:
K_C = \(\frac{\mathrm{k}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{r}}}\)
This equation can be used to determine the composition of the reaction mixture

[Note: The equilibrium refers to the relative amounts of reactants and products and thus a shift in equilibrium in a particular direction will imply the reaction in that direction will be favoured.]

Question 31.
Equal concentrations of hydrogen and iodine are mixed together in a closed container at 700 K and allowed to come to equilibrium. If the concentration of HI at equilibrium is 0.85 mol dm^-3, what are the equilibrium concentrations of H₂ and I₂ if K_C = 54 at this temperature?
Solution:
Given: [HI_(g)] = 0.85 mol dm^-3
K_C = 54 at 700 K
Equilibrium concentrations of H₂ and I₂
Formula: K_C = \(\frac{[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}{[\mathrm{A}]^{\mathrm{a}}[\mathrm{B}]^{\mathrm{b}}}\)
Balanced chemical reaction: 2HI_(g)

Equilibrium concentration of I_2(g) = Equilibrium concentration of H_2(g)

Ans: Equilibrium concentrations of H₂ and I₂ are equal to 0.12 mol dm^-3.

Question 32.
Calculate Kc at 500 K for the reaction,
2HI_(g) ⇌ H_2(g) + I_2(g) if the equilibrium concentrations are [HI] = 0.5 M, [H₂] = 0.08 M and [I₂] = 0.062 M.
Solution:
Given: T = 500 K,
At equilibrium, [HI] = 0.5 M, [H₂] = 0.08 M, [I₂] = 0.062 M.
To find: Equilibrium constant K_C
Formula: K_C = \(\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\)
Calculation: The above equilibrium reaction is given as 2HI_(g) ⇌ H_2(g) + I_2(g)
The expression of K_C is

Ans: K_C at 500 K for the given reaction is 0.0198.

Question 33.
Calculate K_C and K_P for the reaction at 295 K, N₂O₄ ⇌ 2NO_2(g) if the equilibrium concentrations are [N₂O₄] = 0.75 M and [NO₂] = 0.062 M, R = 0.08206 L atm K^-1 mol^-1.
Solution:
Given: R = 0.08206 L atm K^-1 mol^-1, T = 295 K
At equilibrium , [N₂O₄] = 0.75 M, [NO₂] = 0.062 M
To find: Equilibrium constants, K_P and K_C
Formulae: i. K_C = \(\frac{[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}{[\mathrm{A}]^{\mathrm{a}}[\mathrm{B}]^{\mathrm{b}}}\)
ii. K_P = K_C (RT)^Δn
Calculation : The equilibrium reaction is given as N₂O_4(g) ⇌ 2NO_2(g)
The expression of K_C is

K_P is related to K_C by expression: K_P = K_C (RT)^Δn
where, Δn = numbers of moles of gaseous products – number of moles of gaseous reactants
= 2 – 1 = 1
∴ K_P = K_C(RT)¹
∴ K_P = 5.13 × 10^-3 × 0.08206 × 295
∴ K_P= 123.9 × 10^-3 = 0.124
Ans: K_C and K_P for the reaction at 295 K are 5.13 × 10^-3 and 0.124 respectively.

Question 34.
The equilibrium constant K_C for the reaction of hydrogen with iodine is 54.0 at 700 K.

K_C = 54.0 at 700 K
If k_f is the rate constant for the formation of HI and k_r is the rate constant for the decomposition of HI, deduce whether k_r is larger or smaller than k_r.
ii. If the value of k_r at 700 K is 1.16 × 10^-3, what is the value of k_f ?
Solution:
Given: i. K_C = 54.0 at 700 K
ii. k_r = 1.16 × 10^-3 at 700 K
To find: i. Whether k_f is larger or smaller than k_r.
ii. Value of k_f.

Question 35.
Given the equilibrium reaction, H₂O_(g) + CH_4(g) ⇌ CO_(g) + 3H_2(g)
Using Le Chatelier’s principle, predict how concentration of CO will change when the equilibrium is disturbed by
i. adding CH₄
ii. adding H₂
iii. removing H₂O
iv. removing H₂
Answer:
i. Adding CH₄: Adding CH₄ will favour the forward reaction and the yield of CO and H₂ will increase.
ii. Adding H₂: Adding H₂ will favour the reverse reaction and the yield of CO and H₂ will decrease.
iii. Removing H₂O: Removing H₂O will favour the reverse reaction and the yield of CO and H₂ will decrease.
iv. Removing H₂: Removing H₂ will favour the forward reaction and the yield of CO and H₂ will increase.

Question 36.
By using Le Chatelier’s principle, explain the effect of change in pressure (due to volume change) on the composition of equilibrium mixture.
Answer:
Change in pressure:
i. The change in pressure has no effect on the concentrations of solids and liquids. However, it appreciably affects the concentrations of gases.
From the ideal gas equation, PV = nRT or P = \(\frac{\mathrm{n}}{\mathrm{V}}\)RT
∴ P ∝ \(\frac{\mathrm{n}}{\mathrm{V}}\)
where, the ratio n/V is an expression for the concentration of the gas in mol dm^-3.
ii. According to Le Chatelier’s principle at constant temperature, when pressure is increased, the equilibrium will shift in a direction in which the number of molecules decreases and when the pressure is decreased the equilibrium will shift in a direction in which the number of molecules increases.

[Note: For a reaction in which decrease in volume takes place, the reaction will be favoured by increasing pressure and for a reaction in which increase in volume takes place, the reaction will be favoured with lowering pressure, temperature being constant.]

Question 37.
An equilibrium mixture of dinitrogen tetroxide (colourless gas) and nitrogen dioxide (brown gas) is set up in a sealed flask at a particular temperature. Observe the effect of change of pressure on the gaseous equilibrium and complete the following table:

Answer:

Question 38.
By using Le Chatelier’s principle, explain the effect of change in pressure for the following equilibrium:
H_2(g) + I_2(g) ⇌ 2HI_(g)
Answer:
As there is the same number of molecules of gas on both sides, change of pressure has no effect on the equilibrium.

Question 39.
Explain the effect of change in pressure due to volume change of the following equilibria:
i. 2NO_(g) + Cl_2(g) ⇌ 2NOCl_(g)
ii. 2NO_(g) ⇌ N_2(g) + O_2(g)
Answer:
i. 2NO_(g) + Cl_2(g) ⇌ 2NOCl_(g)
In the forward reaction, the number of molecules decreases (3 to 2) and in the reverse reaction the number of molecules increases (2 to 3).
a. Effect of increase in pressure: According to Le Chatelier’s principle, when pressure is increased the forward reaction is favoured as the number of molecules decreases. Thus, when the pressure of the equilibrium system is increased at constant temperature by reducing the volume, the yield of NOCl increases.
b. Effect of decrease in pressure: When the pressure is decreased the equilibrium will shift from right to left. Therefore, the yield of NOCl will decrease.

ii. 2NO_(g) ⇌ N_2(g) + O_2(g)
As both reactants and products have equal numbers of moles (or molecules), there is no effect of change in pressure (due to volume change) on the composition of the equilibrium mixture.

Question 40.
Explain the effect of change in temperature on the value of K_C.
Answer:

• The value of equilibrium constant is unaffected if temperature remains constant.
• However, a change in temperature alters the value of equilibrium constant.
• In a reversible reaction, one of the reactions is exothermic (heat is released) and the other is endothermic (heat is absorbed).
• The value of equilibrium constant for an exothermic reaction decreases with increase in the temperature and that of endothermic reaction increases with the increase in temperature.

Question 41.
Explain the effect of change in temperature on the following equilibria:
CO_(g) + 2H_2(g) ⇌ CH₃OH_(g) ; ΔH = – 90 kJ
Answer:
i. The forward reaction is exothermic and reverse reaction is endothermic. According to Le Chatelier’s principle, when the temperature of the equilibrium mixture increases, the equilibrium shifts from right to left in endothermic direction. Therefore, the yield of CH₃OH decreases at high temperature.

ii. When the temperature decreases, the forward exothermic reaction is favoured. Therefore, the yield of CH₃OH increases at low temperature.
Thus, the decomposition of CH₃OH into CO and H₂ is favoured with increase in temperature, whereas formation of CH₃OH is favoured with decrease in temperature.

Question 42.
By using Le Chatelier’s principle, explain the effect of addition of a catalyst on the composition of equilibrium mixture.
Answer:

• When a catalyst is added to the equilibrium mixture, the rates of forward and reverse reactions increases to the same extent. Hence, the position of equilibrium remains unaffected.
• A catalyst does not change the composition of equilibrium mixture. The equilibrium concentrations of reactants and products remain same and catalyst does not shift the equilibrium in favour of either reactants or products.
• The value of equilibrium constant is also not affected by the presence of a catalyst.

[Note: A catalyst does not appear in the balanced chemical equation and in the equilibrium constant expression.]

Question 43.
Consider an esterification reaction:

What will happen if H+ ions are added to the reaction mixture?
Answer:
H⁺ ions act as catalyst in the esterification reaction. Hence, the addition of H⁺ ions reduces the time for the completion of reaction.

Question 44.
Complete the following table that shows the shifts in the equilibrium position for the reaction:
N₂O_4(g) + Heat ⇌ 2NO_2(g)

Answer:

Question 45.
Summarize effects of following four factors on the position of equilibrium and value of K_C:
i. Concentration
ii. Pressure
iii. Temperature
iv. Catalyst
Answer:

Effect of	Position of equilibrium	Value of KC
Concentration	Changes	No change
Pressure	Changes if reaction involves change in number of gas molecules	No change
Temperature	Change	Change
Catalyst	No change	No change

Question 46.
State TRUE or FALSE. Correct the false statement.
i. The value of equilibrium constant depends on temperature.
ii. If Q_C < K_CC, the reaction will proceed from right to left consuming more product to attain equilibrium.
iii. Any change in the pressure of a gaseous reaction mixture at equilibrium, changes the value of K_C.
iv. In a reversible reaction, the reverse reaction has an energy change that is equal and opposite to that of the forward reaction.
Answer:
i. True
ii. False
If Q_C > K_C the reaction will proceed from right to left consuming more product to attain equilibrium.
iii. False
Any change in the pressure of a gaseous reaction mixture at equilibrium, does not change the value of K_C.
iv. True

Question 47.
Draw the flowchart showing the manufacture of NH₃ by Haber process.
Answer:

Question 48.
Explain in short: The Haber process.
Answer:
Haber process:

• The Haber process is the process of synthesis of ammonia gas by reacting together hydrogen gas and nitrogen gas in a particular stoichiometric ratio by volumes and at selected optimum temperature and pressure.
• The chemical reaction is: \(\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \stackrel{\text { Catalyst }}{\rightleftharpoons} 2 \mathrm{NH}_{3(\mathrm{~g})}+\text { Heat }\)
  The reaction proceeds with a decrease in number of moles (Δn = -2) and the forward reaction is exothermic.
• Iron (containing a small quantity of molybdenum) is used as catalyst.
• The optimum temperature is about 773 K and the optimum pressure is about 250 atm.

Question 49.
Consider the reaction P_(g) + Q_(g) ⇌ PQ_(g). Diagram ‘X’ represents the reaction at equilibrium.

i. If each molecule (sphere) represents a partial pressure of 1 atm, calculate the value of K_P.
ii. Predict the change in equilibrium, when the volume is increased by 50 percentage.
Answer:
i. For the given equilibrium mixture:

Chemical species	P	Q	PQ
Partial pressure	4	6	7

K_P = \(\frac{\mathrm{p}_{\mathrm{PQ}}}{\mathrm{p}_{\mathrm{p}} \times \mathrm{p}_{\mathrm{Q}}}=\frac{7}{4 \times 6}\) = 0.29
ii. Increasing the volume will shift the equilibrium position to the side with higher number of gaseous moles. In the given reaction, the equilibrium will shift to the left (toward reactant) resulting in an increase in the concentration of P and Q accompanied by a corresponding decrease in concentration of PQ.

Multiple Choice Questions

1. Which of the following is expression of K_C for
2NH_3(g) ⇌ N_2(g) + 3H_2(g)?

Answer:
(A) \(\frac{\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}{\left[\mathrm{NH}_{3}\right]^{2}}\)

2. For the system 3A + 2B ⇌ C, the expression for equilibrium constant is …………..

Answer:
(D) \(\frac{[\mathrm{C}]}{[\mathrm{A}]^{3}[\mathrm{~B}]^{2}}\)

3. For the reaction C_(s) + CO_2(g) ⇌ 2CO_(g) the partial pressure of CO₂ and CO are 4 and 8 atm, respectively, then K_P for the reaction is ……………
(A) 16 atm
(B) 2 atm
(C) 5 atm
(D) 4 atm
Answer:
(A) 16 atm

4. The equilibrium constant value for the reaction:
2H_2(g) + O_2(g) ⇌ 2H₂O_(g) is 2.4 × 10⁴⁷ at 500 K. What is the value of equilibrium constant for the reaction:
2H₂O_(g) ⇌ 2H_2(g) + O_2(g) ?
(A) 0.41 × 10^-46
(B) 0.41 × 10⁴⁷
(C) 0.41 × 10^-48
(D) 0.41 × 10^-47
Answer:
(D) 0.41 × 10^-47

5. For the reaction CO_(g) + Cl_2(g) ⇌ COCl_2(g), K_P/K_C is equal to ……………
(A) \(\frac{1}{\mathrm{RT}}\)
(B) RT
(C) \(\sqrt{\mathrm{RT}}\)
(D) 1.0
Answer:
(A) \(\frac{1}{\mathrm{RT}}\)

6. For which of the following reaction, K_P = K_C?
(A) PCl_5(g) ⇌ PCl_3(g) + Cl_2(g)
(B) N_2(g) + 3H_2(g) ⇌ 2NH_3(g)
(C) H_2(g) + I_2(g) ⇌ 2HI_(g)
(D) 2NO_2(g) ⇌ N₂O_4(g)
Answer:
(C) H_2(g) + I_2(g) ⇌ 2HI_(g)

7. For the equilibrium reaction
2NO_2(g) ⇌ N₂O_4(g) + 60.0 kJ, the increase in temperature ……………..
(A) favours the formation of N₂O₄
(B) favours the decomposition of N₂O₄
(C) does not affect the equilibrium
(D) stops the reaction
Answer:
(B) favours the decomposition of N₂O₄

8. The following reaction occurs in the blast furnace where iron ore is reduced to iron metal:
3Fe₂O_3(s) + 3CO_(g) ⇌ 2Fe_(l) + 3CO_2(g)
Using the Le Chatelier’s principle, predict which one of the following will NOT disturb the equilibrium?
(A) Removal of CO
(B) Removal of CO₂
(C) Addition of CO₂
(D) Addition of Fe₂O₃
Answer:
(D) Addition of Fe₂O₃

9. The reaction A + B ⇌ C + D + heat, has reached equilibrium. The reaction may be made to proceed forward by
(A) adding more C
(B) adding more D
(C) decreasing the temperature
(D) increasing the temperature
Answer:
(C) decreasing the temperature

10. Identify the CORRECT statement.
(A) Catalyst lowers activation energy for the forward and reverse reactions by exactly the same amount.
(B) The value of equilibrium constant decreases in presence of a catalyst.
(C) Catalyst affect the position of the equilibrium.
(D) Catalyst changes the equilibrium composition of a reaction mixture.
Answer:
(A) Catalyst lowers activation energy for the forward and reverse reactions by exactly the same amount.

11. The equilibrium constant for the reaction:
N_2(g) + O_2(g) ⇌ 2NO_(g) is 4 × 10^-4 at 2000 K. In presence of a catalyst, the equilibrium is attained ten times faster. Therefore, the equilibrium constant in presence of catalyst of 2000 K is …………..
(A) 40 × 10^-4
(B) 4 × 10^-2
(C) 4 × 10^-3
(D) 4 × 10^-4
Answer:
(D) 4 × 10^-4

12. The rate of formation of NH₃ can be increased by using catalyst …………….
(A) Fe + Co
(B) Mo + Fr
(C) Fe + Mo
(D) Fe + Mg
Answer:
(C) Fe + Mo

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.

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^–

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.

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 v_d.

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 (v_d)
Formula: v_d = \(\frac {I}{nAe}\)
Calculation: From formula,
v_d = \(\frac {I}{nπr^2e}\)
∴ v_d = \(\frac {100}{10^{28}×3.142×10^{-4}×1.6×10^{-19}}\)
= \(\frac {10^{-3}}{3.142×1.6}\)
= 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.

Question 14.
Draw a graph showing the I-V curve for a good conductor and ideal conductor.
Answer:

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,
v_d = a τ = \(\frac {eEτ}{m}\) ………………(From 1)
v_d = \(\frac {eEτ}{m}\) = \(\frac {J}{ne}\) ……………(2) [∵ v_d = \(\frac {J}{ne}\)]

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.

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.

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

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.

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.

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 × 10^z 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.

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 R_s
Thus, the equivalent resistance of the series circuit is, R_s = 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,
R_s = R₁ + R₂ + R₂ + ………….. + R_n = \(\sum_{i=1}^{i=n} R_{i}\)

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.

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, R_p is the equivalent resistance in parallel combination.

vii. If ‘n’ number of resistors R₁, R₂, R₃, ………….. R_n 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 × 10^z Ω ± tolerance
∴ Value of resistance = 47 × 10³ Ω ± 5%
= 47 kΩ ± 5%

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 × 10^z 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 × 10^z ± 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 ×10^z ± T%) Ω
= (47 × 10° ± 10%) Ω
Value of resistance = 47 Ω ± 10%

To find: Value of resistance
Formula: Value of resistance
= (xy × 10^z ± 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 × 10^z ± 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 × 10^z ± 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 × 10^z ± 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

Answer:
i. R₁ and R₂ are connected in parallel. This combination (R_p) is connected in series with R₃.
∴ Total resistance, R_T = R_p + R₃
R_p = \(\frac {R_1R_2}{R_1+R_2}\) = \(\frac {3×6}{3+6}\) = 2 Ω
∴ R_T = 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,

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

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,

The volume of wire remains the same in two cases, we have

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:

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 (T_c), 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 (R_T)
Formula: R_T = R₀(l + α T)
Calculation: From formula,
R_T = 2.5 [1+ (0.004 × 80)]
= 2.5(1 + 0.32)
R_T = 2.5 × 1.32
R_T = 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: R_T = 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: R_T = R_o (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: R_T = 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.

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 (-)

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:

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.
   I_max = \(\frac {E}{R}\)
3. I_max 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 V_AB, V_BC and V_DC.

Answer:

For equivalent resistance (R_eq):
R_AB is given as,
\(\frac{1}{\mathrm{R}_{\mathrm{AB}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}\)
∴ R_AB = 2 Ω
R_BC = R₃ = 1 Ω
Also, R_CD is given as,
\(\frac{1}{\mathrm{R}_{\mathrm{CD}}}=\frac{1}{\mathrm{R}_{4}}+\frac{1}{\mathrm{R}_{5}}=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}\)
∴ R_CD = 3 Ω
∴ R_eq = R_AB + R_BC + R_CD
= 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.
∴ I_BC = 2 A

iii. Voltage drops across AB, BC and CD:
V_AB = IR_AB = 2 × 2 = 4 V
V_BC = IR_BC = 2 × 1 = 2 V
V_CD = IR_CD = 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):

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. R_s = 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

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 + …………. + V_n
∴ 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 + ………. + I_n
∴ 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

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.

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+ …….. + E_n

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 + ……… + r_n

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.

iii. Consider points A and B having potentials V_A and V_B, respectively.

iv. For the first cell the potential difference across its terminals is, V = V_A – V_B = 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 = V_A – V_B = E₂ – I₂r₂
∴ I₂ = \(\frac {E_2V}{r_2}\) ………. (2)

vi. Since, I = I₁ + I₂ ………….. (3)
Combining equations (1), (2) and (3),

viii. If we replace the cells by a single cell connected between points A and B with the emf E_eq and the internal resistance r_eq then,
V = E_eq– Ir_eq
From equations (4) and (5),

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

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.

Answer:
i. Here, the circuit can be represented as,

ii. Here, the circuit can be represented as,

iii. Here, the circuit can be represented as,

∴ 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.

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.

Answer:
As R = \(\frac {V}{I}\)

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,
E_eq = E + E + …….. (n times) = nE

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

iv. The terminal p.d. of the combination is
V = E_eq – Ir_eq = 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.

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) v_d
Answer:
(C) \(\frac {v_d}{2}\)

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

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

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

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

(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.

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

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 Q_in 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.

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, ø_m 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:

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.

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 (v_m) of EM waves is given by v_m = \(\frac {1}{\sqrt{µε}}\)
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. I_E = I_B.

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,
E_y = E₀ sin (kx – ωt)
B_z = B₀ sin (kx – ωt)
where, E₀ is the amplitude of the electric field (E_y) and B₀ is the amplitude of the magnetic field (B_z).

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, E_y = E₀ sin (\(\frac {2πx}{λ}\) – 2πvt)
B_z = 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

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,

………… (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
E_y = E₀ sin (kx – ωt)
E_y = E₀ sin [(\(\frac {2π}{λ}\))x – (2πv)t]
E_y = E₀ sin 2π [\(\frac {x}{λ}\) – vt]
E_y = E₀ sin 2π [\(\frac {x}{10^{-2}}\) – 3 × 10¹⁰ t]
E_y = 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

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.

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.

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:

Question 42.
Draw a diagram showing different types of EM waves.
Answer:

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.

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:

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.

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),

= 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: h_t = 68 m, h_r = 34 m,
R = 6400 km = 6.4 × 10⁶ m
To find: Maximum distance or range (d)
Formula: d = \(\sqrt{2Rh}\)
Calculation:
From formula,

= 2.086 × 10⁴
= 20.86 km
d = d_t + d_r = 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.

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.

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.

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.

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.
   

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.

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 \(\vec{E}\).

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

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

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 ‘h_t‘ and ‘h_r’ are height of transmitting and receiving antennae and ‘R’ is radius of the earth, the range of space wave is
(A) \(\sqrt {2R}\) (h_t + h_r)
(B) 2R \(\sqrt {(h_t + h_r)}\)
(C) \(\sqrt {2R(h_t + h_r)}\)
(D) \(\sqrt {2R}\) (√h_t + √h_r)
Answer:
(D) \(\sqrt {2R}\) (√h_t + √h_r)

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