Class 12 – Page 29 – Maharashtra Board Solutions

Maharashtra Board Class 12 Physics Solutions Chapter 11 Magnetic Materials

August 31, 2021August 19, 2021 by Prasanna

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 11 Magnetic Materials Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 11 Magnetic Materials

1. Choose the correct option.

i) Intensity of magnetic field of the earth at the point inside a hollow iron box is.
(A) less than that outside
(B) more than that outside
(C) same as that outside
(D) zero
Answer:
(D) zero

ii) Soft iron is used to make the core of transformer because of its
(A) low coercivity and low retentivity
(B) low coercivity and high retentivity
(C) high coercivity and high retentivity
(D) high coercivity and low retentivity
Answer:
(A) low coercivity and low retentivity

iii) Which of the following statements is correct for diamagnetic materials?
(A) µ_r < 1
(B) χ is negative and low
(C) χ does not depend on temperature
(D) All of above
Answer:
(D) All of above

iv) A rectangular magnet suspended freely has a period of oscillation equal to T. Now it is broken into two equal halves ( each having half of the original length) and one piece is made to oscillate freely. Its period of oscillation is T′, the ratio of T′ / T is.
(A) \(\frac{1}2 \sqrt{2}\)
(B) \(\frac{1}{2}\)
(C) 2
(D) \(\frac{1}{4}\)
Answer:
(B) \(\frac{1}{2}\)

v) A magnetising field of 360 Am -1 produces a magnetic flux density (B ) = 0.6 T in a ferromagnetic material. What is its permeability in Tm A^-1 ?
(A) \(\frac{1}{300}\)
(B) 300
(C) \(\frac{1}{600}\)
(D) 600
Answer:
(C) \(\frac{1}{600}\)

2 Answer in brief.

i) Which property of soft iron makes it useful for preparing electromagnet?
Answer:
An electromagnet should become magnetic when a current is passed through its coil but should lose its magnetism once the current is switched off. Hence, the ferromagnetic core (usually iron-based) used for an electromagnet should have high permeability and low retentivity, i.e., it should be magnetically ‘soft’.

ii) What happens to a ferromagnetic material when its temperature increases above curie temperature?
Answer:
A ferromagnetic material is composed of small regions called domains. Within each domain, the atomic magnetic moments of nearest-neighbour atoms interact strongly through exchange interaction, a quantum mechanical phenomenon, and align themselves parallel to each other even in the absence of an external magnetic field. A domain is, therefore, spontaneously magnetized to saturation.

The material retains its domain structure only up to a certain temperature. On heating, the increased thermal agitation works against the spontaneous domain magnetization. Finally, at a certain critical temperature, called the Curie point or Curie temperature, thermal agitation overcomes the exchange forces and keeps the atomic magnetic moments randomly oriented. Thus, above the Curie point, the material becomes paramagnetic. The ferromagnetic to paramagnetic transition is an order to disorder transition. When cooled below the Curie point, the material becomes ferromagnetic again.
Maharashtra Board Class 12 Physics Solutions Chapter 11 Magnetic Materials 2

iii) What should be retentivity and coercivity of permanent magnet?
Answer:
A permanent magnet should have a large zero-field magnetization and should need a very large reverse field to demagnetize. In other words, it should have a very broad hysteresis loop with high retentivity and very high coercivity.

iv) Discuss the Curie law for paramagnetic material.
Answer:
Curie’s law : The magnetization of a paramagnetic material is directly proportional to the external magnetic field and inversely proportional to the absolute temperature of the material.

If a paramagnetic material at an absolute temperature T is placed in an external magnetic field of induction \(\), the magnitude of its magnetization
M_z ∝ \(\frac{B_{\text {ext }}}{T}\) ∴ M_z = C\(\frac{B_{\text {ext }}}{T}\)
where the proportionality constant C is called the Curie constant.
[Notes : (1) The above law, discovered experimentally in 1895 by Pierre Curie (1859-1906) French physcist, is true only for values of Bext/ T below about 0.5 tesla per kelvin.
(2) [C] = [M_z ∙ T] / [B_ext] = [L^-1I ∙ ] /[MT^-2I^-1]
= [M^-1L^-1T²I²],
where denotes the dimension of temperature.]

v) Obtain and expression for orbital magnetic moment of an electron rotating about the nucleus in an atom.
Answer:
In the Bohr model of a hydrogen atom, the electron of charge – e performs a uniform circular motion around the positively charged nucleus. Let r, v and T be the orbital radius, speed and period of motion of the electron. Then,
T = \(\frac{2 \pi r}{v}\) …………….. (1)
Therefore, the orbital magnetic moment asso-ciated with this orbital current loop has a magnitude,
I = \(\frac{e}{T}=\frac{e v}{2 \pi r}\) …………… (2)
Therefore, the magnetic dipole moment associated with this electronic current loop has a magnitude
M₀ = current × area of the loop
= I(πr²) = \(\frac{e v}{2 \pi r}\) × πr² = \(\frac{1}{2}\) evr ……………… (3)
Multiplying and dividing the right hand side of the above expression by the electron mass m_e,
M₀ = \(\frac{e}{2 m_{\mathrm{e}}}\) (m_evr) = \(\frac{e}{2 m_{\mathrm{e}}}\) L₀ ……………. (4)
where L₀ = m_evr is the magnitude of the orbital angular momentum of the electron. \(\vec{M}_{0}\) is opposite to \(\vec{L}_{0}\).
∴ \(\vec{M}_{0}=-\frac{e}{2 m_{e}} \overrightarrow{L_{0}}\) ……………. (5)
which is the required expression.

According to Bohr’s second postulate of stationary orbits in his theory of hydrogen atom, the angular momentum of the electron in the nth stationary orbit is equal to n \(\frac{h}{2 \pi}\) , where h is the Planck constant and n is a positive integer. Thus, for an orbital electron,
Maharashtra Board Class 12 Physics Solutions Chapter 11 Magnetic Materials 1
L₀ = m_evr = \(\frac{nh}{2 \pi}\) …………… (6)
Substituting for L₀ in Eq. (4),
M₀ = \(\frac{e n h}{4 \pi m_{\mathrm{e}}}\)
For n = 1, M₀ = \(\frac{e n h}{4 \pi m_{\mathrm{e}}}\)
The quantity \(\frac{e n h}{4 \pi m_{\mathrm{e}}}\) is a fundamental constant called the Bohr magneton,
µ_B ∙ µ_B = 9.274 × 10^-24 J/T (or A∙m²) = 5.788 × 10^-5 eV/T.
[ Notes : (1) Magnetic dipole moment is conventionally denoted by µ. (2) The magnetic moment of an atom is expressed in terms of Bohr magneton (vµ_B). (3) According to quantum mechanics, an atomic electron also has an intrinsic spin angular momentum and an associated spin magnetic moment of magnitude µ₅. It is this spin magnetic moment that gives rise to magnetism in matter. (4) The total magnetic moment of the atom is the vector sum of its orbital magnetic moment and spin magnetic moment.]

vi) What does the hysteresis loop represents?
Answer:
A magnetic hysteresis loop is a closed curve obtained by plotting the magnetic flux density B of a ferromagnetic material against the corresponding magnetizing field H when the material is taken through a complete magnetizing cycle. The area enclosed by the loop represents the hysteresis loss per unit volume in taking the material through the magnetizing cycle.

vii) Explain one application of electromagnet.
Answer:
Applications of an electromagnet:

Electromagnets are used in electric bells, loud speakers and circuit breakers.
Large electromagnets are used in junkyard cranes and industrial cranes to lift iron scraps.
Superconducting electromagnets are used in MRI and NMR machines, as well as in particle accelerators of cyclotron family.
Electromagnets are used in data storage devices such as computer hard disks and magnetic tapes.

Question 3.
When a plate of magnetic material of size 10 cm × 0.5 cm × 0.2 cm (length , breadth and thickness respectively) is located in magnetising field of 0.5 × 10⁴ Am^-1 then a magnetic moment of 0.5 A∙m² is induced in it. Find out magnetic induction in plate.
Answer:
Data : l = 10 cm, b = 0.5 cm, h = 0.2 cm,
H = 0.5 × 10⁴ Am^-1, M = 5 A∙m²
The volume of the plate,
V = 10 × 0.5 × 0.2 = 1 cm² = 10^-6 m²
B = μ₀ (H + M_z) = μ₀ (H + \(\frac{M}{V}\))
The magnetic induction in the plate,
∴ B = 4π × 10^-7 (0.5 × 10⁴ + \(\frac{5}{10^{-6}}\))
= 6.290 T

Question 4.
A rod of magnetic material of cross section 0.25 cm² is located in 4000 Am^-1 magnetising field. Magnetic flux passing through the rod is 25 × 10^-6 Wb. Find out (a) relative permeability (b) magnetic susceptibility and (c) magnetisation of the rod.
Answer:
Data: A = 0.25 cm² = 25 × 10^-6 m²,
H = 4000 A∙m^-1, Φ = 25 × 10^-6 Wb
Magnetic induction is
B = \(\frac{\phi}{A}=\frac{25 \times 10^{-6}}{25 \times 10^{-6}}\) = 1 Wb/m²
(a) B = µ₀µ_rK
∴ The relative permeability of the material,
µ_r = \(\frac{B}{\mu_{0} H}=\frac{1}{4 \times 3.142 \times 10^{-7} \times 4000}\)
= \(\frac{10000}{50.272}\) = 198.91 = 199

(b) µ_r = 1 + χ_m
∴ The magnetic susceptibility of the material,
χ_m = µ_r – 1 = 199 – 1 = 198

Question 5.
The work done for rotating a magnet with magnetic dipole momentm, through 90° from its magnetic meridian is n times the work done to rotate it through 60°. Find the value of n.
Answer:
Data : θ₀ = 0°, θ₁ = 90°, θ₂ = 60°, W₁ = nW₂
The work done by an external agent to rotate the magnet from θ₀ to θ is
W = MB (cos θ₀ – cos θ)
∴ W1 = MB(cos θ₀ – cosθ₁)
= MB (cos 0° – cos 90°)
= MB (1 – 0)
= MB

∴ W2 = MB (cos 0°- cos 60°)
= MB(1 – \(\frac{1}{2}\))
= 0.5MB
∴ W₁ = 2W₂ = MB
Given W₁ = nW₂. Therefore n = 2.

Question 6.
An electron in an atom is revolving round the nucleus in a circular orbit of radius 5.3 × 10^-11 m, with a speed of 2 × 10⁶ ms^-1 Find the resultant orbital magnetic moment and angular momentum of electron. (charge on electron e = 1.6 × 10^-19 C, mass of electron m_e = 9.1 × 10^-31 kg.)
Answer:
Data: r = 5.3 × 10^-11 m, v = 2 × 10⁶ m/s,
e = 1.6 × 10^-19 C, m_e = 9.1 × 10^-31 kg
The orbital magnetic moment of the electron is
M₀ = \(\frac{1}{2}\) evr
= \(\frac{1}{2}\) (1.6 × 10^-19) (2 × 10⁶) (5.3 × 10^-11)
= 8.48 × 10^-24 A∙m²
The angular momentum of the electron is
L₀ = m_evr
=(9.1 × 10^-31) (2 × 10⁶) (5.3 × 10^-11)
= 96.46 × 10^-36 = 9.646 × 10^-35 kg∙m²/s

Question 7.
A paramagnetic gas has 2.0 × 10²⁶ atoms/m with atomic magnetic dipole moment of 1.5 × 10^-23 A m² each. The gas is at 27° C. (a) Find the maximum magnetization intensity of this sample. (b) If the gas in this problem is kept in a uniform magnetic field of 3 T, is it possible to achieve saturation magnetization? Why? (k_B = 1.38 × 10^-23 JK^-1)[Answer: 3.0× 103 A m-1, No]
(Hint: Find the ratio of Thermal energy of atom of a gas ( 3/2 kBT) and maximum potential energy of the atom (mB) and draw your conclusion)
Answer:
Data: \(\frac{N}{V}\) = 2.0 × 10²⁶ atoms/m³,
μ = 1.5 × 10^-23 Am2, T = 27 + 273 = 300 K,
B = 3T, k^B = 1.38 × 10^-23 J/K, 1 eV = 1.6 × 10^-19 J
(a) The maximum magnetization of the material,
M_z = \(\frac{N}{V}\)μ =(2.0 × 10²⁶) (1.5 ×10^-23)
= 3 × 10³ A/m

(b) The maximum orientation energy per atom is
U_m = -μB cos 180° = μB
= (1.5 × 10^-23) (3) = \(\frac{4.5 \times 10^{-23}}{1.6 \times 10^{-19}}\)
= 2.8 × 10^-4 eV

The average thermal energy of each atom,
E = \(\frac{3}{2}\) k_BT
where k_B is the Botzmann constant.
∴ E = 1.5(1.38 × 10^-23)(300)
= 6.21 × 10^-21 J = \(\frac{6.21 \times 10^{-21}}{1.6 \times 10^{-19}}\)
= 3.9 × 10^-2 eV
Since the thermal energy of randomization is about two orders of magnitude greater than the magnetic potential energy of orientation, saturation magnetization will not be achieved at 300 K.

Question 8.
A magnetic needle placed in uniform magnetic field has magnetic moment of 2 × 10^-2 A m², and moment of inertia of 7.2 × 10^-7 kg m². It performs 10 complete oscillations in 6 s. What is the magnitude of the magnetic field ?
Answer:
Data: M = 2 × 10^-2 A∙m², I = 7.2 × 10^-7 kg∙m²,
T = \(\frac{6}{10}\) = 0.6 S
T = 2π\(\sqrt{\frac{I}{M B}}\)
The magnitude of the magnetic field is
B = \(\frac{4 \pi^{2} I}{M T^{2}}\)
= \(\frac{(4)(3.14)^{2}\left(7.2 \times 10^{-7}\right)}{\left(2 \times 10^{-2}\right)(0.6)^{2}}\)
= 3.943 × 10^-3 T = 3.943 mT

Question 9.
A short bar magnet is placed in an external magnetic field of 700 guass. When its axis makes an angle of 30° with the external magnetic field, it experiences a torque of 0.014 Nm. Find the magnetic moment of the magnet, and the work done in moving it from its most stable to most unstable position.
Answer:
Data : B = 700 gauss = 0.07 tesla, θ = 30°,
τ = 0.014 N∙m
τ = MB sin θ
The magnetic moment of the magnet is
M = \(\frac{\tau}{B \sin \theta}=\frac{(0.014)}{(0.07)\left(\sin 30^{\circ}\right)}\) = 0.4 A∙m²
The most stable state of the bar magnet is for θ = 0°.
It is in the most unstable state when θ = 180°. Thus, the work done in moving the bar magnet from 0° to 180° is
W = MB(cos θ₀ – cos θ)
= MB (cos 0° – cos 180°)
= MB [1 – (-1)]
= 2 MB = (2) (0.4) (0.07)
= 0.056 J
This the required work done.

Question 10.
A magnetic needle is suspended freely so that it can rotate freely in the magnetic meridian. In order to keep it in the horizontal position, a weight of 0.1 g is kept on one end of the needle. If the pole strength of this needle is 20 Am , find the value of the vertical component of the earth’s magnetic field. (g = 9.8 m s^-2)
Answer:
Data: M = 0.2g = 2 × 10^-4kg, q_m = 20 A∙m, g =9.8 m/s²

Without the added weight at one end, the needle will dip in the direction of the resultant magnetic field inclined with the horizontal. The torque due to the added weight about the vertical axis through the centre balances the torque of the couple due to the vertical component of the Earth’s magnetic field.
∴ (Mg)\(\left(\frac{L}{2}\right)\) = (q_m B_v) L
The vertical component of the Earth’s magnetic field,
B_v = \(\frac{M g}{2 q_{\mathrm{m}}}=\frac{\left(2 \times 10^{-4}\right)(9.8)}{2(20)}\) = 4.9 × 10^-5 T

Question 11.
The susceptibility of a paramagnetic material is χ at 27° C. At what temperature its susceptibility be \(\frac{\chi}{3}\) ?
Answer:
Data: χ_m1 = χ, T₁ = 27°C = 300 K, χ_m2 = \(\frac{\chi}{3}\)
By Curie’s law,
M_z = C\(\frac{B_{0}}{T}\)
Since M_z = χ_mH = B₀ = μ₀H
Maharashtra Board Class 12 Physics Solutions Chapter 11 Magnetic Materials 3
This gives the required temperature.

12th Physics Digest Chapter 11 Magnetic Materials Intext Questions and Answers

Activity (Textbook Page No. 251)

Question 1.
You have already studied in earlier classes that a short bar magnet suspended freely always aligns in North South direction. Now if you try to forcefully move and bring it in the direction along East West and leave it free, you will observe that the magnet starts turning about the axis of suspension. Do you know from where does the torque which is necessary for rotational motion come from? (as studied in rotational dynamics a torque is necessary for rotational motion).
Answer:
Suspend a short bar magnet such that it can rotate freely in a horizontal plane. Let it come to rest along the magnetic meridian. Rotate the magnet through some angle and release it. You will see that the magnet turns about the vertical axis in trying to return back to its equilibrium position along the magnetic meridian. Where does the torque for the rotational motion come from?

Take another bar magnet and bring it near the suspended magnet resting in the magnetic meridian. Observe the interaction between the like and unlike poles of the two magnets facing each other. Does the suspended magnet rotate continuously or rotate through certain angle and remain stable? Note down your observations and conclusions.

Do you know (Textbook Page No. 255)

Effective magneton numbers for iron group ions (No. of Bohr magnetons)

Ion	Electron configuration	Magnetic moment (in terms of /i_B)
Fe³ +	[Ar] 3s²3p⁶3d⁵	5.9
Fe²⁺	[Ar] 3s²3p⁶3d⁶	5.4
Co²⁺	[Ar] 3s²3p⁶3d⁷	4.8
n²⁺	[Ar] 3s²3p⁶3d⁸	3.2

(Courtsey: Introduction to solid state physics by Charles Kittel, pg. 306 )
These magnetic moments are calculated from the experimental value of magnetic susceptibility. In several ions the magnetic moment is due to both orbital and spin angular momenta.
Answer:
In terms of Bohr magneton (µ_B), the effective magnetic moments of some iron group ions are as follows. In several cases, the magnetic moment is due to both orbital and spin angular momenta.

Ion	Configuration	Effective magnetic moment in terms of Bohr magneton (B.M) (Expreimental values)
Fe³ +	3d⁵	5.9
Fe²⁺	3d⁶	5.4
Co²⁺	3d⁷	4.8
n²⁺	3d⁸	3.2

Remember this (Textbook Page No. 256)

Question 1.
Permeability and Permittivity:
Magnetic Permeability is a term analogous to permittivity in electrostatics. It basically tells us about the number of magnetic lines of force that are passing through a given substance when it is kept in an external magnetic field. The number is the indicator of the behaviour of the material in magnetic field. For superconductors χ = – 1. If you substitute in the Eq. (11.18), it is observed that permeability of material µ = 0. This means no magnetic lines will pass through the superconductor.

Magnetic Susceptibility (χ) is the indicator of measure of the response of a given material to the external applied magnetic field. In other words it indicates as to how much magnetization will be produced in a given substance when kept in an external magnetic field. Again it is analogous to electrical susceptibility. This means when the substance is kept in a magnetic field, the atomic dipole moments either align or oppose the external magnetic field. If the atomic dipole moments of the substance are opposing the field, χ is observed to be negative, and if the atomic dipole moments align themselves in the direction of field, χ is observed to be positive. The number of atomic dipole moments of getting aligned in the direction of the applied magnetic field is proportional to χ. It is large for soft iron (χ >1000).
Answer:
Magnetic permeability is analogous to electric permittivity, both indicating the extent to which a material permits a field to pass through or permeate into the material. For a superconductor, χ = -1 which makes µ = 0, so that a superconductor does not allow magnetic field lines to pass through it.

Magnetic susceptibility (χ). analogous to electrical susceptibility, is a measure of the response of a given material to an applied magnetic field. That is, it indicates the extent of the magnetization produced in the material when it is placed in an external magnetic field. χ is positive when the atomic dipole moments align themselves in the direction of the applied field; χ is negative when the atomic dipole moments align antiparallel to the field. χ is large for soft iron (χ > 1000).

Use your brain power (Textbook Page No. 259)

Question 1.
Classify the following atoms as diamagnetic or paramagnetic.
H, O, Zn, Fe, F, Ar, He
(Hint : Write down their electronic configurations)
Is it true that all substances with even number of electrons are diamagnetic?
Answer:

Atoms	Electronic configuration	No. of Electrons	Diamagnetic/Paramagnetic
H	1s¹„	1	Diamagnetic
0	1s²2s²2p⁴	8	Paramagnetic
Zn	1s²2s²2p⁶3s²3p⁶3d¹⁰4s²	30	Diamagnetic
Fe	1s²2s²2p⁶3s²3p⁶4s²3d⁶	26	Neither diamagnetic nor paramagnetic (ferromagnetic)
F	1s²2s²2p⁵	9	Paramagnetic
Ar	1s²2s²2p⁶3s²3p⁶	18	Diamagnetic
He	Is²	2	Diamagnetic

It can be seen that all substances with an even number of electrons are not necessarily diamagnetic.

Do you know (Textbook Page No. 260)

Question 1.
Exchange Interaction: This exchange interaction in stronger than usual dipole-dipole interaction by an order of magnitude. Due to this exchange interaction, all the atomic dipole moments in a domain get aligned with each other. Find out more about the origin of exchange interaction.
Answer:
Exchange Interaction :
Quantum mechanical exchange interaction be-tween two neighbouring spin magnetic moments in a ferromagnetic material arises as a consequence of the overlap between the magnetic orbitals of two adjacent atoms. The exchange interaction in particular for 3d metals is stronger than the dipole-dipole interaction by an order of magnitude. Due to this, all the atomic dipole moments in a domain get aligned with each other and each domain is spontaneously magnetized to saturation. (Quantum mechanics and exchange interaction are beyond the scope of the syllabus.)

Use your brain power (Textbook Page No. 262)

Question 1.
What does the area inside the curve B – H (hysteresis curve) indicate?
Answer:
A magnetic hysteresis loop is a closed curve obtained by plotting the magnetic flux density B of a ferromagnetic material against the corresponding magnetizing field H when the material is taken through a complete magnetizing cycle. The area enclosed by the loop represents the hysteresis loss per unit volume in taking the material through the magnetizing cycle.

Do you know (Textbook Page No. 262)

Question 1.
What is soft magnetic material?
Soft ferromagntic materials can be easily magnetized and demagnetized.
Maharashtra Board Class 12 Physics Solutions Chapter 11 Magnetic Materials 4
Hysteresis loop for hard and soft ferramagnetic materials.
Answer:
A soft magnetic material, usually iron-based, has high permeability, low retentivity and low coercivity. In other words, it does not have appreciable hysteresis, i.e., its hysteresis loop is very narrow. Such a material magnetizes and demagnetizes more easily, by small external fields.
Maharashtra Board Class 12 Physics Solutions Chapter 11 Magnetic Materials 5

Do you know (Textbook Page No. 263)

Question 1.
There are different types of shielding available like electrical and accoustic shielding apart from magnetic shielding discussed above. Electrical insulator functions as an electrical barrier or shield and comes in a wide array of materials. Normally the electrical wires used in our households are also shielded. In case of audio recording it is necessary to reduce other stray sound which may interfere with the sound to be recorded. So the recording studios are sound insulated using acoustic material.
Answer:
There are different types of shielding, such as electrical, electromagnetic. magnetic, RF (radio fre quency) and acoustic, to shield a given space or sensitive instrument from unwanted fields of each type.

Maharashtra Board OCM 12th Commerce Solutions Chapter 1 Principles of Management

August 31, 2021August 18, 2021 by sachin

Balbharti Maharashtra State Board Organisation of Commerce and Management 12th Textbook Solutions Chapter 1 Principles of Management Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Organisation of Commerce and Management Solutions Chapter 1 Principles of Management

1. (A) Select the correct option and rewrite the sentence

Question 1.
………………… was regarded as Father of Scientific Management.
(a) Henry Fayol
(b) F. W. Taylor
(c) Philip Kotler
Answer:
(b) F. W. Taylor

Question 2.
Principle of ……………… is based on ‘A place for everything and everything in its place’.
(a) Discipline
(b) Order
(c) Equity
Answer:
(b) Order

Question 3.
Member of organization should receive orders from ………………..
(a) many superior
(b) one superior
(c) all superiors
Answer:
(b) one superior

Question 4.
Scalar chain means the hierarchy of …………….. from the top level to the lower level for the purpose of communication.
(a) discipline
(b) unity
(c) authority
Answer:
(c) authority

Question 5.
Taylor recommended total ……………….. foremen to control the various aspects of production.
(a) eight
(b) three
(c) two
Answer:
(a) eight

1. (B) Match the pairs

Group A	Group B
A) Henry Fayol	1) Eight Foremen
B) Principle of Unity of Direction	2) F.W. Taylor
C) Principles of Management	3) Proper division of all activities
D) Scientific Management Theory	4) general guidelines
E) Functional Organization	5) A place for everything and everything in its place
	6) One head-one plan
	7) Low wage rate
	8) Harmony between the employees and management
	9) German engineer
	10) Modern Management

Answer:

Group A	Group B
A) Henry Fayol	1) Modern Management
B) Principle of Unity of Direction	2) One head-one plan
C) Principles of Management	3) general guidelines
D) Scientific Management Theory	4) F.W. Taylor
E) Functional Organization	5) Eight Foremen

1. (C) Give one word/phrase/term for the following statements:

Question 1.
The study of movement of an employee as well as machine while completing particular task.
Answer:
Motion Study

Question 2.
The technique of observing and recording the time required by an employee to complete a given task.
Answer:
Time Study

Question 3.
Study consists of an organised, systematic and critical assessment of various activities.
Answer:
Work Study

Question 4.
The principle which deals with ‘to do work with innovative way’.
Answer:
Principle of Initiative

Question 5.
The principle which is based on ‘a place for everything and everything in its place’.
Answer:
Principle of Order.

1. (D) State whether the following statements are True or False

Question 1.
The principles of management are universal in nature.
Answer:
True

Question 2.
Management principles are applied differently under different situations.
Answer:
True

Question 3.
Only some principles of management are important.
Answer:
False

Question 4.
Henry Fayol has given different techniques of management.
Answer:
False

Question 5.
F. W. Taylor has proposed 14 principles of management.
Answer:
False

Question 6.
Each member of organisation should receive orders only from one superior.
Answer:
True

1. (E) Find the odd one.

Question 1.
Principle of Authority and Responsibility, Motion Study, Principle of Division of Work, Principle of Discipline.
Answer:
Motion Study

Question 2.
Fatigue Study, Principle of Unity of Command, Work Study, Motion Study.
Answer:
Principle of unity of Command

1. (F) Answer in one sentence

Question 1.
What is principle of unity of command?
Answer:
Principal of unity of command implies that every employee should receive orders and instructions from one boss only and he should be responsible and accountable to him only.

Question 2.
What is standardization of tools and equipment?
Answer:
Standardisation of tools and equipment refers to providing the standard tools and equipment for production and maintaining standard working conditions and environment at the place of work.

Question 3.
What is differential wage rate?
Answer:
Differential wage rate means offering higher rate of wages to those employees who complete the work more than the standard quantity and lower rate of wages to those employees who perform below the standard fixed.

Question 4.
What is Subordination of Individual Interest into Organisational Interest?
Answer:
Subordination of individual interest into organisational interest means the interest of an individual must be given less importance than the interest of the organisation.

Question 5.
What is the meaning of principle?
Answer:
Principle means a fundamental truth or proposition that serves as the foundation for a system of belief or behaviour or for a chain of reasoning.

1. (G) Correct the underlined word and rewrite the following sentences.

Question 1.
Security in job always affects adversely on the efficiency of employees.
Answer:
Insecurity in job always affects adversely on the efficiency of employees.

Question 2.
Esprit de corps means ‘division is strength’.
Answer:
Esprit de corps means ‘unity is strength’.

Question 3.
Each member of organisation should receive orders from only one subordinate.
Answer:
Each member of organisation should receive orders from only one superior.

Question 4.
Decentralisation means concentration of powers and authorities at a specific position.
Answer:
Centralisation means concentration of powers and authorities at a specific position.

Question 5.
Management techniques are applied differently under different conditions.
Answer:
Management principles are applied differently under different conditions.

2. Explain the following terms/concepts

Question 1.
Motion Study
Answer:
(1) This is one of the important techniques f of scientific management developed by E W. Taylor. Motion study refers to the movements of employees, as well as movement of machine while completing . a particular task. The motion study helps the manager to know the movements required for a person to do a particular job.

(2) This study is useful to manager to eliminate the unnecessary movements or their sequence for doing the job. It also helps the manager to combine some actions or movements in the process. It increases efficiency and productivity of the employees and helps in reducing the wastage of time, raw material and improving the usefulness of resources.

Question 2.
Differential Piece Rate System.
Answer:
(1) According to F. W. Taylor, the differential piece wage rate plan is necessary to motivate the employees to attain higher standard performance and to earn remuneration at a higher rate. Differential piece wage rate plan suggests that remuneration should be fixed and paid in such a way that average worker is motivated to attain standard output.

(2) In differential piece wage rate plan, the. standard is determined for production by the management. The workers who produce more them the standard are to be paid more in the form of incentives and those who produce less than the standard are to be paid less by way of penalty. This technique encourages those who perform well and motivates those who have performed less than the standard required to improve their performance.

(3) Taylor suggested the differential piece wage system and further stated that the discrimination should be made between the efficient and inefficient workers. This technique explains that, efficient workers should be paid more remuneration in comparison to inefficient workers.

(4) Thus, the differential piece-rate wage plan technique motivates the able employees to attain higher performance and earn wages at higher rate.

Question 3.
Fatigue Study
Answer:
The term ‘fatigue’ implies physical or mental exhaustion. Long working hours with insufficient breaks, heavy working tools, target pressure and poor working conditions lead to fatigue. It reduces efficiency of the employees and creates adverse effect on their health. To reduce fatigue it is very important to keep and maintain the operational efficiency of the workers.

Question 4.
Time Study.
Answer:
(1) Time study is one of the important techniques of scientific management developed by E W. Taylor. Time study is useful to the manager to determine and record the time needed to complete a particular job or task. It is based on the speed of average worker.

(2) Under this technique of scientific management, every part of the entire work is considered in detail and the time required to complete each element of job or task is ascertained. On the basis of the time study, the manager determines the standard time required to complete a particular job. This also facilitates the manager to decide the remuneration to be paid and efficiency of the worker and to control the cost of work.

3. Study the following case/situation and express your opinion.

Question 1.
Mr. Harshad is an entrepreneur and engaged in production of eco-friendly utensils. Both male and female workers are working in his factory. All male employees are directly working on machines whereas female employees are working in Packaging Department. Mr. Sharath is working as Finance Manager while Mrs. Naina is working as HR Manager who is responsible for recruiting employees in the factory. On this basis:
(i) Identify any one principle of management in above case.
(ii) What is the designation of Mrs. Naina in this organisations?
(iii) Who is responsible for overall planning of the organisation?
Answer:
(i) In the above case, principle of Division of work is identified.
(ii) The designation of Mrs. Naina is Human Resource Manager.
(iii) Mr. Harshad, an entrepreneur of eco-friendly utensils, is responsible for overall planning ; of the organisation.

Question 2.
In ‘Fine Diamonds Ltd.’ 200 employees are working in three shifts. In first shift 60 employees, in second shift 60 employees and in third shift 80 employees are working without sufficient breaks except lunch break and shift change break. No employee is able to complete the work in designated time due to inappropriate time management which results into delay for next shift employees.
(i) Identify which scientific principle needs to be followed by the company.
(ii) Suggest two scientific techniques which can be used for smooth flow of work in ‘Fine Diamonds Ltd.
(iii) Why the work is not being completed in time?
Answer:
(i) In the above case, Principle of scientific management called Science, Not Rule of Thumb needs to be followed.
(ii) Work study based on the technique of fatigue study and Scientific task setting can be used for smooth flow of work in ‘Fine Diamonds Ltd.’
(iii) Reasons for non-completion of work in time are:

Long working hours without sufficient breaks reduces the efficiency of labour.
Inappropriate time management (planning) done by the departmental head (Functional Manager) result into delay for next shift employees.
Standard required time and standard output are not defined by the manager or management of “Fine Diamonds Ltd.”

4. Answer in brief

Question 1.
Explain any five principles of management of Henry Fayol.
Answer:
(1) Principle of Division of Work : According to this principle, the entire work such as technical, financial, commercial, accounting, management and security operations should be assigned to different employees as per their qualification, qualities, capabilities and experience. It gives benefits of specialisation and improves efficiency and expertise of employees. This in turn helps to attain expected productivity level.

(2) Principle of Authority and Responsibility:
Authority implies right or freedom to take decisions. The manager should be given authority to get the work done systematically from the subordinates. Authority should always go with corresponding responsibility, e.g. if manager is given authority to complete a specific task within a given time, he would be held responsible for the same. Manager should have adequate authority to take managerial decisions on his own to achieve the goal.

(3) Principle of Discipline: Fayol insists that discipline is essential for smooth working of an organisation. It helps to achieve the goals set in the organisation. In relation to organisation, discipline means strictly obeying instructions of the superiors. There should be clear and utmost understanding between management and employees in respect to organisation’s rules. It is necessary to observe basic discipline at all levels of management.

(4) Principle of Unity of Command: According to this principle, every employee should receive orders and instructions from one boss only and he should be accountable to him only. If he receives orders from more them one boss at a time, he will not understand whose orders should be executed first. To avoid this, organisational hierarchy should be well defined and each employee in the organisation should know his immediate boss. He should receive orders and instructions from him only and should report to him only.
Maharashtra Board OCM 12th Commerce Solutions Chapter 1 Principles of Management 1

(5) Principle of Unity of Direction: Fayol advocates “One head for One plan”. It means that a group of employees working on similar activities should have common objectives and must work under one head (senior). All the objectives of the different groups should be in line with the objectives of the organisation. If this principle is followed there will be an effective co-ordination of the individual efforts and energies.

(6) Principle of Subordination of Individual Interest to Organisational Interest: According to this principle, the interest of the organisation must always be given greater importance than the interest of an employee or a group of employees. While taking decision, the manager should always consider the interest of whole group rather than the interest of an individual employee. Similarly, employees should surrender their personal goals or interests before that of an organisation e.g. while playing a game a team players should always play to win the game and not for his individual records.

Question 2.
Describe any four techniques of scientific management.
Answer:
(1) Work Study : Before allocating the work among the workers, systematic work study should be done by the management. Work study includes Em organised systematic and critical assessment of different activities or functions. It is based on the different techniques like time study, method study, motion study and fatigue study.
(i) Time Study : Under this method, the manager observes and record the time an employee takes to complete a particular job or task. This technique is useful to fix standard time needed to complete a specific task under given conditions. It measures the efficiency of an employee and helps to control the cost of work.

(ii) Method Study : In order to get best quality with cost effectiveness, it is important and challenging for a manager to identify the best method from various available methods to complete the specific job. This method is useful to reduce the wastage of time, raw material and to improve usefulness of the resources to achieve defined objectives. It is also useful to determine the methods to handle the raw materials, storage, inspection and transportation.

(iii) Motion Study : Motion study refers to the close study of the movements of employees as well as machines in completing a particular task. This technique is useful to manager to eliminate the unnecessary movements and to find out the best method of completing a specific task. It improves efficiency and productivity of the employees. This method is also useful to understand and decide about the elimination of some elements of a job or changing their sequence for smooth flow of work.

(iv) Fatigue Study : The term ‘fatigue’ implies physical or mental exhaustion. Long working hours with insufficient breaks, heavy working tools, target pressure and poor working conditions lead to fatigue. It reduces efficiency of the employees and creates adverse effect on their health. To reduce fatigue it is very important to keep and maintain the operational efficiency of the workers.

(2) Standardisation of Tools and Equipment:
On the basis of experiments conducted at work place, Taylor insisted to provide standard tools and equipment, standard working environment and standard methods of production. It helps to reduce spoilage and wastage of materials, cost of production, fatigue among the employees on the one hand and improves quality of work on the other hand.

(3) Scientific Task Setting : Taylor laid stress on the need for fixing a fair day’s work. The technique of scientific task setting is useful to restrain the employees from performing the task much below their capacity. As a result, they will complete their task according to the standards given and management will be able to keep proper control on the optimum use of available workforce.

(4) Scientific Selection and Training: By using scientific selection procedures, management easily select right persons for the right jobs. According to this technique, job specifications required to be fixed and employees are selected as per predetermined standards in an impartial way. After their selection, the management should arrange proper training programmes to increase efficiency.

(5) Functional Organisation : Taylor suggested that planning of the work is to be done by different people and actual work is to be supervised by different set of people. Every worker in the factory is to be supervised by two different sets of supervisors. He further suggested total eight foremen to control the various parts of the production. They are categorised as follows:
(A) At planning level:

Route clerk : Explains the movement of work from one machine to other.
Instruction clerk : Gives and records instructions to complete the work.
Time and cost clerk : Decides the time to complete the work and work out the cost.
Discipline : To see to it that workers work as per factory rules.

(B) At implementation level:

Gang Boss : Gets the actual work done from employees.
Speed Boss : Takes care that work is done in specified time.
Repair Boss : Manages security and maintenance of mechanism.
Inspector : Makes sure that work is completed as per specified standards.

5. Justify the following statements

Question 1.
Principles of management are flexible in nature.
Answer:
(1) Principles of management are the statements of fundamental truth which act as guidelines for managerial decision-making and action. They establish cause and effect relationship. They are evolved through observation, analysis and experiments.

(2) Principles of management although fundamental, are not rigid. They are flexible in nature in the sense they can be changed or modified according to the situation and requirements of the organisation. Managers can change these principles to suit the requirements of the organisation.

(3) Principles of management are flexible guidelines providing ample scope for making changes according to the nature of enterprise, its size, competitive situation, etc. For instance, in the context of present business scenario, at many places the ‘family management has been substituted by professional management’.

(4) Modern business world is dynamic. The situations in a business enterprise keep on changing continuously. No two circumstances are identical. Principles of management can be changed, adjusted or modified and used in the enterprise as per its changing needs and requirements. By identifying problems of business changes will be accepted.

Question 2.
Management principles are helpful in optimum utilization of resources.
Answer:
(1) Materials or abstract qualities that a person or organisation uses to perform the work is called resources, e.g. tools, stocks, time, employees, etc. In every organisation, two types of resources are used and they are: (i) Physical resources such as materials, machines, money, etc. and (ii) Human resources i.e. manpower.

(2) The different types of resources are used in the organisation to manufacture or produce different types of goods and services. The resources are scarce in relation to their demand and therefore resources should not be wasted and misused. They should be used carefully and up to their optimum capacity.

(3) The basic aim and function of management are to make and maintain proper balance and allocate these resources by putting them to maximum possible use and control on wastage of resources. Through the use of different techniques and management principles, management maintains discipline and healthy working environment to establish cordial relationship with the employees.

(4) It helps to increase the efficiency level of employees and to manage the administration effectively, e.g. use of modern and standard tools and machineries. It also helps to increase quality, productivity and level of efficiency of human resources.

Question 3.
Principle of equity is important.
Answer:
(1) The principle of equity suggests that employer should give kind, fair, just and equal treatment to the employees. Managers should be kind, impartial and fair to their subordinates.

(2) The principle of equity further states that there should not be any discrimination between the employees while making the payment of wages. The employees working on the same level but in different departments should be paid same wages.

(3) The wages payable to employees should not depend on the departments but the level at which they are working, e.g. Foremen should be paid higher wages than that is paid to employees working under them.

(4) The principle of equity also states that there should not be any discrimination between the employees while distributing work between them. As far as possible, there should be equal distribution of work. Thus, equality in treatment of employees boosts the morale and develops a sense of belongingness among the employees. It helps to develop loyalty of employees towards organisation and avoid conflicts.

Question 4.
Taylor emphasized on standardization of tools and equipment.
Answer:
(1) Fredrick Winslow Taylor published Principles of Scientific Management. His primary objective was to increase efficiency of employees by scientifically designing jobs. According to his views, management problems should be solved through experiments and use of scientific techniques rather than rules of thumb and triad and error approach.

(2) Taylor had conducted many experiments at workplace and as a result of those experiments, he advocated standardisation of tools and equipment. Standard here, means a level of quality or achievement, especially a level that is acceptable.

(3) According to him, standardised working environment and standardised methods of production tools and equipment help to reduce spoilage and wastage of materials. This in turn reduces the overall cost of production.

(4) Similarly use of standardised tools and equipment increases efficiency of employees and also helps to reduce fatigue among the workers. This improves the quality of work.

Question 5.
Differential piece wage rate plan is necessary.
Answer:
(1) According to F. W. Taylor, the differential piece wage rate plan is necessary to motivate the employees to attain higher standard performance and to earn remuneration at a higher rate. Differential piece wage rate plan suggests that remuneration should be fixed and paid in such a way that average worker is motivated to attain standard output.

(4) Thus, the differential piece-rate wage plan technique motivates the able employees to attain higher performance and earn wages at higher rate.

6. Attempt the Following

Question 1.
Explain in detail any five Principles of Management given by Henry Fayol’s?
Answer:
(1) Principle of Division of Work : According to this principle, the entire work such as technical, financial, commercial, accounting, management and security operations should be assigned to different employees as per their qualification, qualities, capabilities and experience. It gives benefits of specialisation and improves efficiency and expertise of employees. This in turn helps to attain expected productivity level.

Question 2.
Describe different techniques of scientific management.
Answer:
(1) Work Study : Before allocating the work among the workers, systematic work study should be done by the management. Work study includes Em organised systematic and critical assessment of different activities or functions. It is based on the different techniques like time study, method study, motion study and fatigue study.
(i) Time Study : Under this method, the manager observes and record the time an employee takes to complete a particular job or task. This technique is useful to fix standard time needed to complete a specific task under given conditions. It measures the efficiency of an employee and helps to control the cost of work.

Route clerk : Explains the movement of work from one machine to other.
Instruction clerk : Gives and records instructions to complete the work.
Time and cost clerk : Decides the time to complete the work and work out the cost.
Discipline : To see to it that workers work as per factory rules.

(B) At implementation level:

Gang Boss : Gets the actual work done from employees.
Speed Boss : Takes care that work is done in specified time.
Repair Boss : Manages security and maintenance of mechanism.
Inspector : Makes sure that work is completed as per specified standards.

Question 3.
Elaborate Principles of Scientific Management.
Answer:
The principles of scientific management are as follows:

Science, Not Rule of Thumb
Harmony, Not Discord
Mental Revolution
Co-operation, Not Individualism
Division of Responsibility
Development of employer and employees for greater efficiency and maximum prosperity

1. Science, Not Rule of Thumb : Rule of thumb method is based on personal judgements of the manager which should be substituted with the methods developed through scientific analysis of work. Taylor emphasised more on the use of scientific method for every small job. This principle related with selecting the best way of doing a work after scientific analysis. Under this method, standard required time and standard output are defined by the manager. This method is useful to save time and human energy, to get expected standard output and to increase organisational efficiency.

2. Harmony, Not Discord : This principle states that, in every organisation these should be proper co-ordination and harmonious relations between the management and employees. This will help in minimising conflicts between them and in achieving the goals of the organisation. The perfect understanding between employees and management is also helpful in creating healthy work environment. Organisation should also think about the maximum prosperity of the employees.

3. Mental Revolution : The concept of ‘mental revolution’ was introduced by Taylor. This principle highlights on the complete change in the attitude of the management and employees toward each other. Both should recognise their equal importance in the organisation. They should co-operate with each other to achieve goals or objectives of the organisation. This in turn will increase productivity and profits.

4. Co-operation, Not Individualism : This principle states that there should be mutual co-operation between employees and management. Co-operation, trust, team spirit, etc. are important to avoid internal competition and to create healthy working environment. Management should always appreciate and consider the suggestions given by the employees in decision-making process. The management should treat the employees as an integral part of the organisation in all respects. Employees should also resist themselves from going on strikes and making unacceptable or unnecessary demands from the management. Thus, they should see each other as two pillars of the organisation.

5. Division of Responsibility : This principle states that while dividing the work there should be’ corresponding division of responsibility between the managers and employees. Major planning should be done by the top and middle level management and employees should concentrate on its execution. The reporting of the jobs should be done by the subordinates as per the instructions given by their superiors. For the best performances, the management should always help, encourage and guide the employees.

6. Development of employer and employees for greater efficiency and maximum prosperity : Profitability and best performance of any organisation mostly depend on the skills, intelligence and capabilities of its employees. Arranging and providing training and development programmes for the employees at regular interval or whenever required are absolutely important. It helps to increase profitability of the organisation. Proper opportunity should be given to each ; employee to attain his highest efficiency and ; maximum prosperity.

Question 4.
Explain nature (characteristics) of principles of management.
Answer:
The nature (characteristics) of principles of management is (are) explained as follows:
(1) Universal application : Management principles are universal in nature. They are applicable to all types of organisations irrespective of the type, size or nature, e.g. government, college, hospital, bank. etc. Their application may have to be modified, but they are suitable for all kinds of organisations, whether in private sector or public sector. Similarly, principles of management are applicable to all levels of management. For instance, the principle of division of labour is applicable to all types of organisations.

(2) General guidelines : Management principles give general guidelines to tackle the organisational situations wisely and to solve the problems systematically. They are not rigid. Application of management principles depend upon the situation, size and nature of organisation, e.g., when we say according to principle of remuneration, the employees must be paid fair remuneration. The term ‘fair’ may vary as per nature, size and financial ability of the organisation.

(3) Principles are formed by practice and experiments : The management principles are developed gradually with thorough research work, experiments and systematic observations. The results of such observations and experiments are developed after their practice in different organisations.

(4) Flexibility : Management principles, although fundamental are not rigid statements. They have to be applied differently under various conditions. It is possible to make suitable changes in their application according to the requirement of the organisation. Thus, Management principles are flexible guidelines providing ample scope for making changes according to the nature of enterprise, its size, competitive situation, etc.

(5) Behavioural in nature : Management is a teamwork or a group activity. Management principles aim at influencing individual efforts and directing them to achieve various objectives of the organisation. They are directed towards regulating human behaviour so that people give their best to the organisation. Thus, principles of management are designed to influence human behaviour.

(6) Cause and effect relationship : Management principles indicate cause and effect relationship. Each principle has a definite effect on the efficiency or working of management. For example, payment of good remuneration and incentives increases output. Similarly, effective advertisement given by the organisation increases the sale of a product.

(7) All principles are of equal importance : All the principles of management have equal importance and they also carry equal weightage with reference to their applicability in the organisation. For example, it cannot be said that the principle of division of labour is more important than the principle of unity of command or vice versa. Management principles are not static in nature. They are not absolute like principles of pure sciences like Chemistry, Mathematics, etc. They are the principles of social science. They are to be modified and applied according to the size and nature of the organisation, keeping in mind the requirements.

7. Answer the following questions

Question 1.
What are the techniques of scientific management? Explain in detail.
Answer:
The techniques of scientific management given by F. W. Taylor are explained as follows:
(1) Work Study : Before allocating the work among the workers, systematic work study should be done by the management. Work study includes Em organised systematic and critical assessment of different activities or functions. It is based on the different techniques like time study, method study, motion study and fatigue study.

(i) Time Study : Under this method, the manager observes and record the time an employee takes to complete a particular job or task. This technique is useful to fix standard time needed to complete a specific task under given conditions. It measures the efficiency of an employee and helps to control the cost of work.

Route clerk : Explains the movement of work from one machine to other.
Instruction clerk : Gives and records instructions to complete the work.
Time and cost clerk : Decides the time to complete the work and work out the cost.
Discipline : To see to it that workers work as per factory rules.

(B) At implementation level:

Gang Boss : Gets the actual work done from employees.
Speed Boss : Takes care that work is done in specified time.
Repair Boss : Manages security and maintenance of mechanism.
Inspector : Makes sure that work is completed as per specified standards.

(6) Differential Piece – Rate Wage Plan : Taylor suggested that discrimination should be made between efficient and less efficient workers. In this technique, the standard is determined (fixed) for production. The workers who produce more than the standard output are to be paid remuneration at higher rates and those who produce less them the standard quantity are to be paid at lower rate of wages. This technique encourages the employees to attain higher standard performance to earn higher wages.

Question 2.
Explain 14 principles of Henry Fayol in detail.
Answer:
Henry Fayol, the Father of Modern Management developed the following 14 principles:
principles of management:

Principle of Division of Work
Principle of Authority and Responsibility
Principle of Discipline
Principle of Unity of Command
Principle of Unity of Direction
Principle of Subordination of Individual Interest to Organisational Interest
Principle of Centralisation
Principle of Remuneration
Principle of Scalar Chain
Principle of Order
Principle of Equity
Principle of Stability of Tenure
Principle of Initiative
Principle of Esprit de corpse (Team Work)

1. Principle of Division of Work: According to this principle, the entire work such as technical, financial, commercial, accounting, management and security operations should be assigned to different employees as per their qualification, qualities, capabilities and experience. It gives benefits of specialisation and improves efficiency and expertise of employees. This in turn helps to attain expected productivity level.

2. Principle of Authority and Responsibility: Authority implies right or freedom to take decisions. The manager should be given authority to get the work done systematically from the subordinates. Authority should always go with corresponding responsibility, e.g. if manager is given authority to complete a specific task w12. Principle of Stability of Tenure:ithin a given time, he would be held responsible for the same. Manager should have adequate authority to take managerial decisions on his own to achieve the goal.

3. Principle of Discipline: Fayol insists that discipline is essential for smooth working of an organisation. It helps to achieve the goals set in the organisation. In relation to organisation, discipline means strictly obeying instructions of the superiors. There should be clear and utmost understanding between management and employees in respect to organisation’s rules. It is necessary to observe basic discipline at all levels of management.

4. Principle of Unity of Command: According to this principle, every employee should receive orders and instructions from one boss only and he should be accountable to him only. If he receives orders from more them one boss at a time, he will not understand whose orders should be executed first. To avoid this, organisational hierarchy should be well defined and each employee in the organisation should know his immediate boss. He should receive orders and instructions from him only and should report to him only.
Maharashtra Board OCM 12th Commerce Solutions Chapter 1 Principles of Management 3

5. Principle of Unity of Direction: Fayol advocates “One head for One plan”. It means that a group of employees working on similar activities should have common objectives and must work under one head (senior). All the objectives of the different groups should be in line with the objectives of the organisation. If this principle is followed there will be an effective co-ordination of the individual efforts and energies.

6. Principle of Subordination of Individual Interest to Organisational Interest: According to this principle, the interest of the organisation must always be given greater importance than the interest of an employee or a group of employees. While taking decision, the manager should always consider the interest of whole group rather than the interest of an individual employee. Similarly, employees should surrender their personal goals or interests before that of an organisation e.g. while playing a game a team players should always play to win the game and not for his individual records.

7. Principle of Centralisation: Centralisation means concentration of authority or power in a few hands at the top level. As number of employees is less in a smaller organisation there is centralisation of authority. Decentralisation means even distribution of authority or power at every level of management. As number of employees and levels of management are more in a larger organisation, there must be decentralisation of some authorities for its smooth functioning. According to Fayol, there must be a proper balance between centralisation and decentralisation, depending upon the nature and size of an organisation.

8. Principle of Remuneration: According to this principle, the employees must be paid fair and appropriate remuneration to keep them satisfied financially and to retain them within the organisation for longer period of time. While fixing remuneration various factors such as the skill, knowledge, expertise, tenure, cost of living, market trend, profitability of organisation, etc. should be considered. It boosts the morale of employees and increases efficiency and productivity.

9. Principle of Scalar Chain: According to Fayol, in the organisation decisions, orders, instructions, messages, etc. must be passed through a chain, i.e. from the general manager to the respective functional manager, then to the supervisor, then to the foreman and then ultimately, to the workers. Similarly, suggestions, information, grievances, etc., must flow from the worker in the upward direction. This is called Scalar Chain. Sometimes, following a scalar chain becomes a lengthy process. In such cases, ‘Gang Plank’ is followed which permits speedy and direct communication between the employees working at the same level of authority. However, for this, permission of the proper authority is necessary.

10. Principle of Order: The principle of order is based on ‘a place for everything and everything in its place’. According to this principle, in every organisation there should be proper, systematic and orderly arrangement of men and materials. There should be a fixed place to keep every material and thing used in the organisation and a fixed place or a seat for every employee. The purpose of this principle is to reduce wastage of time and energy. This principle emphasises more I on the proper and optimum utilisation of physical and human resources.

11. Principle of Equity: This principle states that the management should be fair as well as friendly to the subordinate staff. There should be no discrimination of employees in regard to division of work, delegation of the authorities, deciding the monetary terms, etc. This principle also states that the remuneration should depend not on the department but at the level at which employees are working, It means the employees working on the same level but in separate departments must be paid equal wages. It will also help in avoiding conflicts in an organisation.

12. Principle of Stability of Tenure: According to this principle, at the time of recruitment of employees, the management should assure them about the stability of tenure (i.e. job security). This creates a sense of belonging among the employees. Job security increases efficiency of the employees and minimises employees turnover ratio.

13. Principle of Initiative: Initiative means to do the work in an innovative way in his or her personal capacity. According to this principle, managers should give freedom, opportunity or encourage the subordinates to take initiative while working on given job. Their suggestions and ideas should be invited before framing the plan. This can work as a morale booster for the employees and leads to timely achievement of organisational goals.

14. Principle of Esprit de corpse (Team Work): ‘Esprit de corpse’ means union is strength. This principle integrates and co-ordinates the individual and group efforts. It emphasizes the spirit of teamwork. The manager as a leader should create the feeling of team spirit and understanding among the various groups. When entire group of employees works as a team, their efforts get directed towards realising the goals of the organisation.

Maharashtra Board Class 12 Physics Solutions Chapter 6 Superposition of Waves

August 31, 2021August 18, 2021 by Prasanna

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 6 Superposition of Waves Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 6 Superposition of Waves

1. Choose the correct option.

i) When an air column in a pipe closed at one end vibrates such that three nodes are formed in it, the frequency of its
vibrations is …….times the fundamental frequency.
(A) 2
(B) 3
(C) 4
(D) 5
Answer:
(D) 5

ii) If two open organ pipes of length 50 cm and 51 cm sounded together produce 7 beats per second, the speed of sound is.
(A) 307 m/s
(B) 327m/s
(C) 350m/s
(D) 357m/s
Answer:
(D) 357m/s

iii) The tension in a piano wire is increased by 25%. Its frequency becomes ….. times the original frequency.
(A) 0.8
(B) 1.12
(C) 1.25
(D) 1.56
Answer:
(B) 1.12

iv) Which of the following equations represents a wave travelling along the y-axis?
(A) x = A sin(ky – ωt)
(B) y = A sin(kx – ωt)
(C) y = A sin(ky) cos(ωt)
(D) y = A cos(ky)sin(ωt)
Answer:
(A) x = A sin(ky – ωt)

v) A standing wave is produced on a string fixed at one end with the other end free. The length of the string
(A) must be an odd integral multiple of λ/4.
(B) must be an odd integral multiple of λ/2.
(C) must be an odd integral multiple of λ.
(D) must be an even integral multiple ofλ.
Answer:
(A) must be an odd integral multiple of λ/4.

2. Answer in brief.

i) A wave is represented by an equation y = A sin (Bx + Ct). Given that the constants A, B and C are positive, can you tell in which direction the wave is moving?
Answer:
The wave is travelling along the negative x-direction.

ii) A string is fixed at the two ends and is vibrating in its fundamental mode. It is known that the two ends will be at rest. Apart from these, is there any position on the string which can be touched so as not to disturb the motion of the string? What will be the answer to this question if the string is vibrating in its first and second overtones?
Answer:
Nodes are the points where the vibrating string can be touched without disturbing its motion.
When the string vibrates in its fundamental mode, the string vibrates in one loop. There are no nodes formed between the fixed ends. Hence, there are no point on the string which can be touched without disturbing its motion.

When the string vibrates in its first overtone (second harmonic), there are two loops of the stationary wave on the string. Apart from the two nodes at the two ends, there is now a third node at its centre. Hence, the string can be touched at its centre without disturbing the stationary wave pattern.

When the string vibrates in its second overtone (third harmonic), there are three loops of the stationary wave on the string. So, apart from the two end nodes, there are two additional nodes in between, at distances one-third of the string length from each end. Thus, now the string can be touched at these two nodes.

iii) What are harmonics and overtones?
Answer:
A stationary wave is set up in a bounded medium in which the boundary could be a rigid support (i.e., a fixed end, as for instance a string stretched between two rigid supports) or a free end (as for instance an air column in a cylindrical tube with one or both ends open). The boundary conditions limit the possible stationary waves and only a discrete set of frequencies is allowed.

The lowest allowed frequency, n₁, is called the fundamental frequency of vibration. Integral multiples of the fundamental frequency are called the harmonics, the fundamental frequency being the fundamental or 2n₁, the third harmonic is 3n₁, and so on.

The higher allowed frequencies are called the overtones. Above the fundamental, the first allowed frequency is called the first overtone, the next higher frequency is the second overtone, and ‘so on. The relation between overtones and allowed harmonics depends on the system under consideration.

iv) For a stationary wave set up in a string having both ends fixed, what is the ratio of the fundamental frequency to the
second harmonic?
Answer:
The fundamental is the first harmonic. Therefore, the ratio of the fundamental frequency (n) to the second harmonic (n₁) is 1 : 2.

v) The amplitude of a wave is represented by y = 0.2 sin 4π[\(\frac{t}{0.08}\) – \(\frac{x}{0.8}\)] in SI units. Find
(a) wavelength,
(b) frequency and
(c) amplitude of the wave. [(a) 0.4 m (b) 25 Hz (c) 0.2 m]
Answer:
Maharashtra Board Class 12 Physics Solutions Chapter 6 Superposition of Waves 10
y = 0.2 sin 2π[\(\frac{t}{0.04}\) – \(\frac{x}{0.4}\)]
Let us compare above equation with the equation of a simple harmonic progressive wave:
y = A sin 2π[\(\frac{t}{T}\) – \(\frac{x}{\lambda}\)] = 0.2 sin 2π[\(\frac{t}{0.04}\) – \(\frac{x}{0.4}\)]
Comparing the quantities on both sides, we get,
A = 0.2 m, T = 0.04 s, λ = 0.4 m
∴ (a) Wavelength (λ) = 0.4 m
(b) Frequency (n) = \(\frac{1}{T}\) = \(\frac{1}{0.04}\) = 25 Hz
(c) Amplitude (A) = 0.2 m

Question 3.
State the characteristics of progressive waves.
Answer:
Characteristics of a progressive wave :

Energy is transmitted from particle to particle without the physical transfer of matter.
The particles of the medium vibrate periodically about their equilibrium positions.
In the absence of dissipative forces, every particle vibrates with the same amplitude and frequency, but differs in phase from its adjacent particles. Every particle lags behind in its state of motion compared to the one before it.
A wave motion is doubly periodic, i.e., it is periodic in time and periodic in space.
The velocity of propagation through a medium depends upon the properties of the medium.
Progressive waves are of two types : transverse and longitudinal. In a transverse mechanical wave, the individual particles of the medium vibrate perpendicular to the direction of propagation of the wave. The progressively changing phase of the successive particles results in the formation of alternate crests and troughs that are periodic in space and time. In an em wave, the electric and magnetic fields oscillate in mutually perpendicular directions, perpendicular to the direction of propagation.
In a longitudinal mechanical wave, the individual particles of the medium vibrate along the line of propagation of the wave. The progressively changing phase of the successive particles results in the formation of typical alternate regions of compressions and rarefactions that are periodic in space and time. Periodic compressions and rarefactions result in periodic pressure and density variations in the medium. There are no longitudinal em wave.
A transverse wave can propagate only through solids, but not through liquids and gases while a longitudinal wave can propagate through any material medium.

Question 4.
State the characteristics of stationary waves.
Answer:
Characteristics of stationary waves :

Stationary waves are produced by the interference of two identical progressive waves travelling in opposite directions, under certain conditions.
The overall appearance of a standing wave is of alternate intensity maximum (displacement antinode) and minimum (displacement node).
The distance between adjacent nodes (or antinodes) is λ/2.
The distance between successive node and antinode is λ/4.
There is no progressive change of phase from particle to particle. All the particles in one loop, between two adjacent nodes, vibrate in the same phase, while the particles in adjacent loops are in opposite phase.
A stationary wave does not propagate in any direction and hence does not transport energy through the medium.
In a region where a stationary wave is formed, the particles of the medium (except at the nodes) perform SHM of the same period, but the amplitudes of the vibrations vary periodically in space from particle to particle.

[Note : Since the nodes are points where the particles are always at rest, energy cannot be transmitted across a node. The energy of the particles within a loop remains localized, but alternates twice between kinetic and potential energy during each complete vibration. When all the particles are in the mean position, the energy is entirely kinetic. When they are in their extreme positions, the energy is entirely potential.]

Question 5.
Derive an expression for equation of stationary wave on a stretched string.
Answer:
When two progressive waves having the same amplitude, wavelength and speed propagate in opposite directions through the same region of a medium, their superposition under certain conditions creates a stationary interference pattern called a stationary wave.

Consider two simple harmonic progressive waves, of the same amplitude A, wavelength A and frequency n = ω/2π, travelling on a string stretched along the x-axis in opposite directions. They may be represented by
y₁ = A sin (ωt – kx) (along the + x-axis) and … (1)
y₂ = A sin (ωt + kx) (along the – x-axis) …. (2)
where k = 2π/λ is the propagation constant.

By the superposition principle, the resultant displacement of the particle of the medium at the point at which the two waves arrive simultaneously is the algebraic sum
y = y₁ + y₂ = A [sin (ωt – kx) + sin (ωt + kx)]
Using the trigonometrical identity,
sin C + sin D = 2 sin \(\left(\frac{C+D}{2}\right)\) cos \(\left(\frac{C-D}{2}\right)\),
y = 2A sin ωt cos (- kx)
= 2A sin ωt cos kx [∵ cos(- kx) = cos(kx)]
= 2A cos kx sin ωt … (3)
∴ y = R sin ωt, … (4)
where R = 2A cos kx. … (5)
Equation (4) is the equation of a stationary wave.

Question 6.
Find the amplitude of the resultant wave produced due to interference of two waves given as y₁ = A₁ sin ωt y₂ = A₂ sin (ωt + φ)
Answer:
The amplitude of the resultant wave produced due to the interference of the two waves is
A = \(\sqrt{A_{1}^{2}+2 A_{1} A_{2} \cos \varphi+A_{2}^{2}}\).

Question 7.
State the laws of vibrating strings and explain how they can be verified using a sonometer.
Answer:
The fundamental is the first harmonic. Therefore, the ratio of the fundamental frequency (n) to the second harmonic (n₁) is 1 : 2.

Here, L = 3\(\frac{\lambda}{2}\)
∴ Wavelength, λ = \(\frac{2 L}{3}\) = \(\frac{2 \times 30}{3}\) = 20 cm.

Question 8.
Show that only odd harmonics are present in the vibrations of air column in a pipe closed at one end.
Answer:
Consider a narrow cylindrical pipe of length l closed at one end. When sound waves are sent down the air column in a cylindrical pipe closed at one end, they are reflected at the closed end with a phase reversal and at the open end without phase reversal. Interference between the incident and reflected waves under appropriate conditions sets up stationary waves in the air column.

The stationary waves in the air column in this case are subject to two boundary conditions that there must be a node at the closed end and an antinode at the open end.

Taking into account the end correction e at the open end, the resonating length of the air column is L = l + e.
Maharashtra Board Class 12 Physics Solutions Chapter 6 Superposition of Waves 20

Let v be the speed of sound in air. In the simplest mode of vibration, there is a node at the closed end and an antinode at the open end. The distance between a node and a consecutive anti-node is \(\frac{\lambda}{4}\), where λ is the wavelength of sound. The corresponding wavelength λ and frequency n are
λ = 4L and n = \(\frac{v}{\lambda}\) = \(\frac{v}{4 L}\) = \(\frac{v}{4(l+e)}\) …… (1)
This gives the fundamental frequency of vibration and the mode of vibration is called the fundamental mode or first harmonic.

In the next higher mode of vibration, the first overtone, two nodes and two antinodes are formed. The corresponding wavelength λ₁ and frequency n₁ are
λ₁ = \(\frac{4 L}{3}\) and n₁ = \(\frac{v}{\lambda_{1}}\) = \(\frac{3 v}{4 L}\) = \(\frac{3 v}{4(l+e)}\) = 3n … (2)
Therefore, the frequency in the first overtone is three times the fundamental frequency, i.e., the first overtone is the third harmonic.

In the second overtone, three nodes and three antinodes are formed. The corresponding wavelength λ₂ and frequency n₂ are
Maharashtra Board Class 12 Physics Solutions Chapter 6 Superposition of Waves 30
which is the fifth harmonic.
Therefore, in general, the frequency of the pth overtone (p = 1, 2, 3, ,..) is
n_p = (2p + 1)n … (4)
i.e., the pth overtone is the (2p + 1)th harmonic.

Equations (1), (2) and (3) show that allowed frequencies in an air column in a pipe closed at one end are n, 3n, 5n, …. That is, only odd harmonics are present as overtones.

Question 9.
Prove that all harmonics are present in the vibrations of the air column in a pipe open at both ends.
Answer:
Consider a cylindrical pipe of length l open at both the ends. When sound waves are sent down the air column in a cylindrical open pipe, they are reflected at the open ends without a change of phase. Interference between the incident and reflected waves under appropriate conditions sets up stationary waves in the air column.

The stationary waves in the air column in this case are subject to the two boundary conditions that there must be an antinode at each open end.
Maharashtra Board Class 12 Physics Solutions Chapter 6 Superposition of Waves 35
Taking into account the end correction e at each of the open ends, the resonating length of the air column is L = l + 2e.

Let v be the speed of sound in air. In the simplest mode of vibration, the fundamental mode or first harmonic, there is a node midway between the two antinodes at the open ends. The distance between two consecutive antinodes is λ/2, where λ is the wavelength of sound. The corresponding wavelength λ and the fundamental frequency n are
λ = 2L and n = \(\frac{v}{\lambda}\) = \(\frac{v}{2 L}\) = \(\frac{v}{2(l+2 e)}\) …. (1)

In the next higher mode, the first overtone, there are two nodes and three antinodes. The corresponding wavelength λ₁ and frequency n₁
λ₁ = L and n₁ = \(\frac{v}{\lambda_{1}}\) = \(\frac{v}{L}\) = \(\frac{v}{(l+2 e)}\) = 2n …. (2)
i.e., twice the fundamental. Therefore, the first overtone is the second harmonic.

In the second overtone, there are three nodes and four antinodes. The corresponding wavelength λ₂ and frequency n₂ are
λ₂ = \(\frac{2 L}{3}\) and n₂ = \(\frac{v}{\lambda_{2}}\) = \(\frac{3v}{2L}\) = \(\frac{3 v}{2(l+2 e)}\) = 3n …. (3)
or thrice the fundamental. Therefore, the second overtone is the third harmonic.

Therefore, in general, the frequency of the pth overtone (p = 1, 2, 3, …) is
n_p = (p + 1)n … (4)
i.e., the pth overtone is the (p + 1)th harmonic.
Equations (1), (2) and (3) show that allowed frequencies in an air column in a pipe open at both ends are n, 2n, 3n, …. That is, all the harmonics are present as overtones.

Question 10.
A wave of frequency 500 Hz is travelling with a speed of 350 m/s.
(a) What is the phase difference between two displacements at a certain point at times 1.0 ms apart?
(b) what will be the smallest distance between two points which are 45º out of phase at an instant of time?
[Ans : π, 8.75 cm ]
Answer:
Data : n = 500 Hz, v = 350 m/s
D = n × λ
∴ λ = \(\frac{350}{500}\) = 0.7 m
(a) in t = 1.0 ms = 0.001 s, the path difference is the distance covered v × t = 350 × 0.001 = 0.35 m
∴ Phase difference = \(\frac{2 \pi}{\lambda}\) × Path difference
= \(\frac{2 \pi}{0.7}\) × 0.35 = π rad

(b) Phase difference = 45° = \(\frac{\pi}{4}\) rad
∴ Path difference = \(\frac{\lambda}{2 \pi}\) × Phase difference
= \(\frac{0.7}{2 \pi}\) × \(\frac{\pi}{4}\) = 0.0875 m

Question 11.
A sound wave in a certain fluid medium is reflected at an obstacle to form a standing wave. The distance between two successive nodes is 3.75 cm. If the velocity of sound is 1500 m/s, find the frequency. [Ans : 20 kHz]
Answer:
Data : Distance between two successive nodes =
\(\frac{\lambda}{2}\) = 3.75 × 10^-2 m, v = 1500 m/s
∴ λ = 7.5 × 10^-2m
v = n × λ
∴ n = \(\frac{1500}{7.5 \times 10^{-2}}\) = 20 kHz

Question 12.
Two sources of sound are separated by a distance 4 m. They both emit sound with the same amplitude and frequency (330 Hz), but they are 180º out of phase. At what points between the two sources, will the sound intensity be maximum? (Take velocity of sound to be 330 m/s) [Ans: ± 0.25, ± 0.75, ± 1.25 and ± 1.75 m from the point at the center]
Answer:
∴ λ = \(\frac{v}{n}\) = \(\frac{330}{330}\) = 1 m
Directly at the cenre of two sources of sound, path difference is zero. But since the waves are 180° out of phase, two maxima on either sides should be at a distance of \(\frac{\lambda}{4}\) from the point at the centre. Other
maxima will be located each \(\frac{\lambda}{2}\) further along.
Thus, the sound intensity will be maximum at ± 0.25, ± 0.75, ± 1.25, ± 1.75 m from the point at the centre.

Question 13.
Two sound waves travel at a speed of 330 m/s. If their frequencies are also identical and are equal to 540 Hz, what will be the phase difference between the waves at points 3.5 m from one source and 3 m from the other if the sources are in phase? [Ans : 1.636 π]
Answer:
Data : v = 330 m/s, n₁ = n₂ = 540 Hz
v = n × λ
∴ λ = \(\frac{330}{540}\) = 0.61 m
Here, the path difference = 3.5 – 3 m = 0.5 m
Phase difference = \(\frac{2 \pi}{\lambda}\) × Path difference
= \(\frac{2 \pi}{0.61}\) × 0.5 = 1.64π rad

Question 14.
Two wires of the same material and same cross-section are stretched on a sonometer. One wire is loaded with 1.5 kg and another is loaded with 6 kg. The vibrating length of first wire is 60 cm and its fundamental frequency of vibration is the same as that of the second wire. Calculate vibrating length of the other wire. [Ans: 1.2 m]
Answer:
Data : m₁ = m₂ = m, L₁ = 60 cm = 0.6 m, T₁ = 1.5 kg = 14.7 N, T₂ = 6 kg = 58.8 N
Maharashtra Board Class 12 Physics Solutions Chapter 6 Superposition of Waves 39
The vibrating length of the second wire is 1.2 m.

Question 15.
A pipe closed at one end can produce overtones at frequencies 640 Hz, 896 Hz and 1152 Hz. Calculate the fundamental
frequency. [Ans: 128 Hz]
Answer:
The difference between the given frequencies of the overtones is 256 Hz. This implies that they are consecutive overtones. Let n_C be the fundamental frequency of the closed pipe and n_q, n_q-1, n_q-1 = the frequencies of the q^th, (q + 1)^th and (q + 2)^th consecutive overtones, where q is an integer.

Data : n_q = 640 Hz, n_q-1 = 896 Hz, n_q+2 = 1152 Hz
Since only odd harmonics are present as overtones, n_q = (2q +1) n_C
and n_q+1 = [2(q + 1) + 1] n_C = (2q + 3) n_C
Maharashtra Board Class 12 Physics Solutions Chapter 6 Superposition of Waves 40
∴ 14q + 7 = 10q + 15 ∴ 4q = 8 ∴ q = 2
Therefore, the three given frequencies correspond to the second, third and fourth overtones, i.e., the fifth, seventh and ninth harmonics, respectively.
∴ 5n_C = 640 ∴ b_C = 128Hz

Question 16.
A standing wave is produced in a tube open at both ends. The fundamental frequency is 300 Hz. What is the length
of tube in the fundamental mode? (speed of the sound = 340 m s^-1). [Ans: 0.5666 m]
Answer:
Data : For the tube open at both the ends, n = 300 Hz and v = 340 m / s Igonoring end correction, the fundamental frequency of the tube is
n = \(\frac{v}{2 L}\) ∴ L = \(\frac{v}{2 n}\) = \(\frac{340}{2 \times 300}\) = 0.566m
The length of the tube open at both the ends is 0.5667 m.

Question 17.
Find the fundamental, first overtone and second overtone frequencies of a pipe, open at both the ends, of length 25 cm if the speed of sound in air is 330 m/s. [Ans: 660 Hz, 1320 Hz, 1980 Hz]
Answer:
Data : Open pipe, ∠25 cm = 0.25 m, v = 330 m / s
The fundamental frequency of an open pipe ignoring end correction,
Maharashtra Board Class 12 Physics Solutions Chapter 6 Superposition of Waves 55
Since all harmonics are present as overtones, the first overtone is, n₁ = 2n₀ = 2 × 660 = 1320 Hz
The second overtone is n₂ = 3n = 3 × 660 = 1980 Hz

Question 18.
A pipe open at both the ends has a fundamental frequency of 600 Hz. The first overtone of a pipe closed at one end has the same frequency as the first overtone of the open pipe. How long are the two pipes? (Take velocity of sound to be 330 m/s) [Ans : 27.5 cm, 20.625 cm]
Answer:
Data : Open pipe, n_o = 600 Hz, n_{C, 1} = n_{o, 1} (first overtones)
For an open pipe, the fundamental frequency,
n_o = \(\frac{v}{2 L_{O}}\)
∴ The length of the open pipe is
L₀ = \(\frac{v}{2 n_{O}}\) = \(\frac{330}{2 \times 600}\) = 0.275 m
For the open pipe, the frequency of the first overtone is
2n₀ = 2 × 600 = 1200 Hz
For the pipe closed at one end, the frequency of the first overtone is \(\frac{3 v}{L_{O}}\).
By the data, \(\frac{3 v}{4 L}\) = 1200
∴ L_C = \(\frac{3 \times 330}{4 \times 1200}\) = 0.206 m
The length of the pipe open at both ends is 27.5 cm

Question 19.
A string 1m long is fixed at one end. Transverse vibrations of frequency 15 Hz are imposed at the free end. Due to this, a stationary wave with four complete loops, is produced on the string. Find the speed of the progressive wave which produces the stationary wave.[Hint: Remember that the free end is an antinode.] [Ans: 6.67 m s^-1]
Answer:
Data : L = 1 m, n = 15 Hz.
The string is fixed only at one end. Hence, an antinode will be formed at the free end. Thus, with four and half loops on the string, the length of the string is
L = \(\frac{\lambda}{4}\) + 4\(\left(\frac{\lambda}{2}\right)\) = \(\frac{9}{4}\)λ
∴ λ = \(\frac{4 L}{9}\) = \(\frac{4}{9}\) × 1 = \(\frac{4}{9}\) m
v = nλ
∴ Speed of the progressive wave
v = 15 × \(\frac{4}{9}\) = \(\frac{60}{9}\) =6.667m/s

Question 20.
A violin string vibrates with fundamental frequency of 440Hz. What are the frequencies of first and second overtones? [Ans: 880 Hz, 1320 Hz]
Answer:
Data: n =440Hz
The first overtone, n₁ = 2n =2 × 400 = 880 Hz
The second overtone, n₁ = 3n = 3 × 400 = 1320 Hz

Question 21.
A set of 8 tuning forks is arranged in a series of increasing order of frequencies. Each fork gives 4 beats per second with the next one and the frequency of last for k is twice that of the first. Calculate the frequencies of the first and the last for k. [Ans: 28 Hz, 56 Hz]
Answer:
Data : n₈ = 2n₁, beat frequency = 4 Hz
The set of tuning fork is arranged in the increasing order of their frequencies.
∴ n₂ = n₁ + 4
n₃ = n₂ + 4 = n₁ + 2 × 4
n₄ = n₃ + 4 = n₁ + 3 × 4
∴ n₈ = n₇ + 4 = n₁ + 7 × 4 = n₁ + 28
Since n₈ = 2n₁,
2n₁ = n₁ + 28
∴ The frequency of the first fork, n₁ = 28 Hz
∴ The frequency of the last fork,
n₈ = n₁ + 28 = 28 + 28 = 56 Hz

Question 22.
A sonometer wire is stretched by tension of 40 N. It vibrates in unison with a tuning fork of frequency 384 Hz. How
many numbers of beats get produced in two seconds if the tension in the wire is decreased by 1.24 N? [Ans: 12 beats]
Answer:
Data : T₁ =40N, n₁ = 384 Hz, T₂ = 40 – 1.24 = 38.76 N
Maharashtra Board Class 12 Physics Solutions Chapter 6 Superposition of Waves 60
∴ The number of beats produced in two seconds = 2 × 6 = 12

Question 23.
A sonometer wire of length 0.5 m is stretched by a weight of 5 kg. The fundamental frequency of vibration is 100 Hz. Calculate linear density of wire. [Ans: 4.9 × 10^-3 kg/m]
Answer:
Data : L = 0.5 m, T = 5 kg = 5 × 9.8 = 49 N, n = 100 Hz
n = \(\frac{1}{2 L} \sqrt{\frac{T}{m}}\)
∴ Linear density, m = \(\frac{T}{4 L^{2} n^{2}}\)
= \(\frac{49}{4(0.5)^{2}(100)^{2}}\)
= 4.9 × 10^-3 kg/m

Question 24.
The string of a guitar is 80 cm long and has a fundamental frequency of 112 Hz. If a guitarist wishes to produce a frequency
of 160 Hz, where should the person press the string? [Ans : 56 cm from one end]
Answer:
Data : L₁ = 80 cm n₁ = 112 Hz, n₂ = 160 Hz
According to the law of length, n₁L₁ = n₂L₂.
∴ The vibrating length to produce the fundamental frequency of 160 Hz,
L₂ = \(\frac{n_{1} L_{1}}{n_{2}}\) = \(\frac{112(80)}{160}\) = 56 cm

12th Physics Digest Chapter 6 Superposition of Waves Intext Questions and Answers

Can you tell? (Textbook Page No. 132)

Question 1.
What is the minimum distance between any two particles of a medium which always have the same speed when a sinusoidal wave travels through the medium ?
Answer:
When a sinusoidal wave travels through a medium the minimum distance between any two particles of the medium which always have the same speed is \(\frac{\lambda}{2}\).
Such particles are opposite in phase, i.e., their instantaneous velocities are opposite in direction.
[Note : The minimum distance between any two particles which have the same velocity is λ]

Do you know? (Textbook Page No. 140)

Question 1.
What happens if a simple pendulum is pulled aside and released ?
Answer:
If a simple pendulum is pulled aside and released, it oscillates freely about its equilibrium position at its natural frequency which is inversely proportional to the square root of its length and directly proportional to the square root of the acceleration of gravity at the place. These oscillations, called as free oscillations, are periodic and tautochronous if the displacement of its bob is small and the dissipative forces can be ignored.

Question 2.
What happens when a guitar string is plucked ?
Answer:
When a guitar string is plucked, two wave pulses of the same amplitude, frequency and phase move out from that point towards the fixed ends of the string where they get reflected. For certain ratios of wavelength to length of the string, these reflected pulses moving towards each other will meet in phase to form standing waves on the string. The vibrations of the string cause the air molecules to oscillate, forming sound waves that radiate away from the string. The frequency of the sound waves is equal to the frequency of the vibrating string. In general, the wavelengths of the sound waves and the waves on the string are different because their speeds in the two mediums are not the same.

Question 3.
Have you noticed vibrations in a drill machine or in a washing machine ? How do they differ from vibrations in the above two cases ?
Answer:
Vibrations in the body of a drill machine or that of a washing machine are forced vibrations induced by the vibrations of the motors of these machines. On the other hand, the oscillations of a simple pendulum or a guitar string are free oscillations, produced when they are disturbed from their equilibrium position and released.

Question 4.
A vibrating tuning fork of certain frequency is held in contact with a tabletop and its vibrations are noticed and then another vibrating tuning fork of different frequency is held on the tabletop. Are the vibrations produced in the tabletop the same for both the tuning forks ? Why ?
Answer:
No. Because the tuning forks have different frequencies, the forced vibrations in the tabletop differ both in frequency and amplitude. The tuning fork whose frequency is closer to a natural frequency of the tabletop induces forced vibrations of a larger amplitude.

Maharashtra Board Class 12 Physics Solutions Chapter 12 Electromagnetic Induction

August 31, 2021August 18, 2021 by Prasanna

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 12 Electromagnetic Induction Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 12 Electromagnetic Induction

1. Choose the correct option.

i) A circular coil of 100 turns with a cross-sectional area (A) of 1 m² is kept with its plane perpendicular to the magnetic field (B) of 1 T. What is the magnetic flux linkage with the coil?
(A) 1 Wb
(B) 100 Wb
(C) 50 Wb
(D) 200 Wb
Answer:
(B) 100 Wb

ii) A conductor rod of length (l) is moving with velocity (v) in a direction normal to a uniform magnetic field (B). What will be the magnitude of induced emf produced between the ends of the moving conductor?
(A) BLv
(B) BLv²
(C) \(\frac{1}{2}\)Blv
(D) \(\frac{2 B l}{\mathrm{v}}\)
Answer:
(A) BLv

iii) Two inductor coils with inductance 10 mH and 20 mH are connected in series. What is the resultant inductance of the combination of the two coils?
(A) 20 mH
(B) 30 mH
(C) 10 mH
(D) \(\frac{20}{3}\) mH
Answer:
(A) 20 mH

iv) A current through a coil of self inductance 10 mH increases from 0 to 1 A in 0.1 s. What is the induced emf in the coil?
(A) 0.1 V
(B) 1 V
(C) 10 V
(D) 0.01 V
Answer:
(A) 0.1 V

v) What is the energy required to build up a current of 1 A in an inductor of 20 mH?
(A) 10 mJ
(B) 20 mJ
(C) 20 J
(D) 10 J
Answer:
(A) 10 mJ

2. Answer in brief.

i) What do you mean by electromagnetic induction? State Faraday’s law of induction.
Answer:
The phenomenon of production of emf in a conductor or circuit by a changing magnetic flux through the circuit is called electromagnetic induction.

Faraday’s laws of electromagnetic induction :
(1) First law ; Whenever there is a change in the magnetic flux associated with a circuit, an emf is induced in the circuit.
(2) Second law : The magnitude of the induced emf is directly proportional to the time rate of change of magnetic flux through the circuit.
[Note : The phenomenon was discovered in 1830 by Joseph Henry (1797-1878), US physicist, and independently in 1832 by Michael Faraday (1791 -1867), British chemchemist and physicist.]

ii) State and explain Lenz’s law in the light of principle of conservation of energy.
Answer:
Lenz’s law : The direction of the induced current is such as to oppose the change that produces it.

The change that induces a current may be (i) the motion of a conductor in a magnetic field or (ii) the change of the magnetic flux through a stationary circuit.

Explanation : Consider Faraday’s magnet-and- coil experiment. If the bar magnet is moved towards the coil with its N-pole facing the coil, as in Fig., the number of magnetic lines of induction (pointing to the left) through the coil increases. The induced current in the coil sets up a magnetic field of its own pointing to the right (as given by Amperes right-hand rule) to oppose the growing flux due to the magnet. Hence, to move the magnet towards the coil against this repulsive flux of the induced current, we must do work. The work done shows up as electric energy in the coil.
Maharashtra Board Class 12 Physics Solutions Chapter 12 Electromagnetic induction 1
When the magnet is withdrawn, with its N-pole still facing the coil, the number of magnetic lines of induction (pointing left) through the coil decreases. The induced current reverses its direction to supplement the decreasing flux with its own, as shown in Fig.. Facing the coil along the magnet, the induced current is in the clockwise sense. The electric energy in the coil comes from the work done to withdraw the magnet, now against an attractive force. Thus, we see that Lenz’s law is a consequence of the law of conservation of energy.

[Note : The above law was discovered by Heinrich Friedrich Emil Lenz (1804-65), Russian physicist.]

iii) What are eddy currents? State applications of eddy currents.
Answer:
Whenever a conductor or a part of it is moved in a magnetic field “cutting” magnetic field lines, or placed in a changing magnetic field, the free electrons in the bulk of the metal start circulating in closed paths equivalent to current-carrying loops. These loop currents resemble eddies in a fluid stream and are hence called eddy or Foucault currents [after Jean Bernard Leon Foucault (1819-68), French physicist, who first detected them].

Applications :
(1) Dead-beat galvanometer : A pivoted moving-coil galvanometer used for measuring current has the coil wound on a light aluminium frame. The rotation of the metal frame in magnetic field produces eddy currents in the frame which opposes the rotation and the coil is brought to rest quickly. This makes the galvanometer dead-beat.

(2) Electric brakes : When a conducting plate is pushed into a magnetic field, or pulled out, very quickly, the interaction between the eddy currents in the moving conductor and the field retards the motion. This property of eddy currents is used as a method of braking in vehicles.

iv) If the copper disc of a pendulum swings between the poles of a magnet, the pendulum comes to rest very quickly. Explain the reason. What happens to the mechanical energy of the pendulum?
Answer:
As the copper disc enters and leaves the magnetic field, the changing magnetic flux through it induces eddy current in the disc. In both cases, Fleming’s right hand rule shows that opposing magnetic force damps the motion. After a few swings, the mechanical energy becomes zero and the motion comes to a stop.

Joule heating due to the eddy current warms up the disc. Thus, the mechanical energy of the pendulum is transformed into thermal energy.

v) Explain why the inductance of two coils connected in parallel is less than the inductance of either coil.
Answer:
Assuming that their mutual inductance can be ignored, the equivalent inductance of a parallel combination of two coils is given by
\(\frac{1}{L_{\text {parallel }}}=\frac{1}{L_{1}}+\frac{1}{L_{2}}\) or L_parallel = \(\frac{L_{1} L_{2}}{L_{1}+L_{2}}\)
Hence, the equivalent inductance is less than the inductance of either coil.

Question 3.
In a Faraday disc dynamo, a metal disc of radius R rotates with an angular velocity ω about an axis perpendicular to the plane of the disc and passing through its centre. The disc is placed in a magnetic field B acting perpendicular to the plane of the disc. Determine the induced emf between the rim and the axis of the disc.
Answer:
Suppose a thin conducting disc of radius R is rotated anticlockwise, about its axis, in a plane perpendicular to a uniform magnetic field of induction \(\vec{B}\) (see the figure in the above Note for reference). \(\vec{B}\) points downwards. Let the constant angular speed of the disc be ω.

Consider an infinitesimal element of radial thickness dr at a distance r from the rotation axis. In one rotation, the area traced by the element is dA = 2πrdr. Therefore, the time rate at which the element traces out the area is
\(\frac{d A}{d t}\) = frequency of rotation x dA = fdA
where f = \(\frac{\omega}{2 \pi}\) is the frequency of rotation.
.’. \(\frac{d A}{d t}=\frac{\omega}{2 \pi}\) (2 πr dr) = ωr dr
The total emf induced between the axle and the rim of the rotating disc is
\(|e|=\int B \frac{d A}{d t}=\int_{0}^{R} B \omega r d r=B \omega \int_{0}^{R} r d r=B \omega \frac{R^{2}}{2}\)
For anticlockwise rotation in \(\vec{B}\) pointing down, the axle is at a higher potential.

Question 4.
A horizontal wire 20 m long extending from east to west is falling with a velocity of 10 m/s normal to the Earth’s magnetic field of 0.5 × 10^-4 T. What is the value of induced emf in the wire?
Answer:
Data : l = 20 m, v = 10 m/s. B = 5 × 10^-5 T
The magnitude of the induced emf,
|e| = Blv = (5 × 10^-5)(20)(10) = 10^-2V = 10 mV

Question 5.
A metal disc is made to spin at 20 revolutions per second about an axis passing through its centre and normal to its plane. The disc has a radius of 30 cm and spins in a uniform magnetic field of 0.20 T, which is parallel to the axis of rotation. Calculate
(a) The area swept out per second by the radius of the disc,
(b) The flux cut per second by a radius of the disc,
(c) The induced emf in the disc.
Answer:
Data: R = 0.3m, f = 20 rps, B = 0.2T
(a) The area swept out per unit time by a given radius = (the frequency of rotations) × (the area swept out per rotation) = f(πr²)
= (20)(3.142 × 0.09) = 5.656 m²

(b) The time rate at which a given radius cuts magnetic flux
= \(\frac{d \Phi_{\mathrm{m}}}{d t}\) = B f(πr²)
= (0.2)(5.656) = 1.131 Wb/s

Question 6.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 10 A in 0.2 s, what is the change of flux linkage with the other coil?
Answer:
Data: M = 1.5 H, I_1i = 0, I_1f = 10A, ∆f = 0.2s
The flux linked per unit turn with the second coil due to current I₁ in the first coil is
Φ₂₁ = MI₁
Therefore, the change in the flux due to change in I₁ is
∆₂₁ =M(∆I₁) = M(I_1f – I_1i) = 1.5 (10 – 0)
= 15 Wb
[Note: The rate of change of flux linkage is M(∆I₁/∆t) = 15/0.2 = 75 Wb/s] .

Question 7.
A long solenoid has 1500 turns/m. A coil C having cross sectional area 25 cm2 and 150 turns (N_c) is wound tightly around the centre of the solenoid. If a current of 3.0A flows through the solenoid, calculate :
(a) the magnetic flux density at the centre of the solenoid,
(b) the flux linkage in the coil C,
(c) the average emf induced in coil C if the direction of the current in the solenoid is reversed in a time of 0.5 s. (µ₀ = 4π × 10^-7 H/m)
Answer:
Data: n = 1.5 × 10³ m , A = 25 × 10^-4 m²,
N_c = 150, I = 3A, ∆t = 0.5s,
µ₀ = 4π × 10^-7 H/m
(a) Magnetic flux density inside the solenoid,
B = u₀ nI = (4π × 10^-7)(1500)(3)
= 5.656 × 10^-3 T = 5.656 mT

(b) Flux per unit turn through the coils of the solenoid, Φ_m = BA
Since the coil C is wound tightly over the solenoid, the flux linkage of C is
N_CΦ_m = N_CBA = (150)(5.656 × 10^-3)(25 × 10^-4)
= 2.121 × 10^-3 Wb = 2.121 mWb

(c) Initial flux through coil C,
Φ_i = N_CΦ_m = 2.121 × 10^-3 Wb
Reversing the current in the solenoid reverses the flux through coil C, the magnitude remaining the same. But since the flux enters through the other face of the coil, the final flux through C is Φ_f = -2.121 × 10^-3 Wb
Therefore, the average emf induced in coil C,
e = \(-\frac{\Phi_{\mathrm{f}}-\Phi_{1}}{\Delta t}=-\frac{(-2.121-2.121) \times 10^{-3}}{0.5}\)
= 2 × 4.242 × 10^-3 = 8.484 × 10^-3 V = 8.484 mV

Question 8.
A search coil having 2000 turns with area 1.5 cm² is placed in a magnetic field of 0.60T. The coil is moved rapidly out of the field in a time of 0.2 second. Calculate the induced emf across the search coil.
Answer:
Data: N = 2000, A_i = 1.5 × 10^-4 m², A_f = 0,
B = 0.6T, ∆t = 0.2s
Initial flux, NΦ_f = NBA_i = 2000(0.6)(1.5 × 10^-4)
= 0.18 Wb
Final flux, NΦ_f = 0, since the coil is withdrawn out of the field.
Induced emf,e = \(-N \frac{\Delta \Phi_{\mathrm{m}}}{\Delta t}=-N \frac{\Phi_{\mathrm{f}}-\Phi_{1}}{\Delta t}\)
∴ e = \(-\frac{0-0.18}{0.2}\) = 0.9V

Question 9.
An aircraft of wing span of 50 m flies horizontally in earth’s magnetic field of 6 × 10^-5 T at a speed of 400 m/s. Calculate the emf generated between the tips of the wings of the aircraft.
Answer:
Data : l = 50 m, B = 6 × 10^-5T, v = 400 m/s
The magnitude of the induced emf,
|e| = Blv = (6 × 10^-5)(400)(50) = 1.2V

Question 10.
A stiff semi-circular wire of radius R is rotated in a uniform magnetic field B about an axis passing through its ends. If the frequency of rotation of the wire is f, calculate the amplitude of the alternating emf induced in the wire.
Answer:
In one rotation, the wire traces out a circle of radius R, i.e., an area A = πR².
Therefore, the rate at which the wire traces out the area is
\(\frac{d A}{d t}\) = frequency or rotation × A = fA
If the angle between the uniform magnetic field \(\vec{B}\) and the rotation axis is θ, the magnitude of the induced emf is
|e|= B\(\frac{d A}{d t}\) cosθ = BfA cosθ = Bf(πR²)cosθ
so that the required amplitude is equal to Bf(πR²).

Question 11.
Calculate the value of induced emf between the ends of an axle of a railway carriage 1.75 m long traveling on level ground with a uniform velocity of 50 km per hour. The vertical component of Earth’s magnetic field (B_v) is given to be 5 × 10^-5T.
Answer:
Data: l = 1.75 m, v = 50 km/h = 50 × \(\frac{5}{18}\) m/s.
B_v = 5 × 10^-5 T
The area swept out by the wing per unit time = 1v.
∴ The magnetic flux cut by the wing per unit time
= \(\frac{d \Phi_{\mathrm{m}}}{d t}\) = B_v(lv)
=(5 × 10^-5)(1.75)(50 × \(\frac{5}{18}\))= 121.5 × 10^-5
= 1.215 mWb/s
Therefore, the magnitude of the induced emf,
|e| =1.215 mV
[Note: In the northern hemisphere, the vertical com ponent of the Earth’s magnetic induction is downwards. Using Fleming’s right hand rule, the port (left) wing-tip would be positive.]

Question 12.
The value of mutual inductance of two coils is 10 mH. If the current in one of the coil changes from 5A to 1A in 0.2 s, calculate the value of emf induced in the other coil.
Answer:
Data: M = 10 mH = 10^-2 H, I_1i = 5A, I_1f = 1 A,
∆t = 0.2s
The mutually induced emf in coil 2 due to the changing current in coil 1,
e₂₁ = \(-M \frac{\Delta I_{1}}{\Delta t}=-M \frac{I_{1 \mathrm{f}}-I_{1 \mathrm{i}}}{\Delta t}\)
= -(10^-2) \(\left(\frac{1-5}{0.2}\right)\) = 0.2 V

Question 13.
An emf of 96.0 mV is induced in the windings of a coil when the current in a nearby coil is increasing at the rate of 1.20 A/s. What is the mutual inductance (M) of the two coils?
Answer:
Data: |e₂| = 9.6 × 10^-2 V, dI₁/dt = 1.2 A/s
|e₂| = M\(\frac{d I_{1}}{d t}\)
Mutual inductance,
M = \(\frac{\left|e_{2}\right|}{d I_{1} / d t}=\frac{9.6 \times 10^{-2}}{1.2}\)
= 8 × 10^-2 H
= 80 mH

Question 14.
A long solenoid of length l, cross-sectional area A and having N₁ turns (primary coil) has a small coil of N₂ turns (secondary coil) wound about its centre. Determine the Mutual inductance (M) of the two coils.
Answer:
We assume the solenoid to be ideal and that all the flux from the solenoid passes through the outer coil C. For a steady current Is through the solenoid, the uniform magnetic field inside the solenoid is
B = µ₀nI_s ……………….. (1)
Then, the magnetic flux through each turn of the coil due to the current in the solenoid is
Φ_CS = BA = (µ₀nI_s)(πR²) ……………… (2)
Thus, their mutual inductance is
M = \(\frac{N \Phi_{\mathrm{CS}}}{I_{\mathrm{S}}}\) = µ₀πR²nN ……………. (3)
Equation (2) is true as long as the magnetic field of the solenoid is entirely contained within the cross section of the coil C. Hence, M does not depend on the shape, size, or possible lack of close packing of the coil.
replacing n with N₁/l and N with N₂. M = µ₀A = \(\frac{N_{1} N_{2}}{l}\)
[Note: The answer given in the textbook misses out the factor of 1.] .

Question 15.
The primary and secondary coil of a transformer each have an inductance of 200 × 10^-6H. The mutual inductance (M) between the windings is 4 × 10^-6H. What percentage of the flux from one coil reaches the other?
Answer:
Data: L_P = L_S = 2 × 10^-4 H, M = 4 × 10^-6 H
M = K\(\sqrt{L_{\mathrm{P}} L_{\mathrm{S}}}\)
The coupling coefficient is
K = \(\frac{M}{\sqrt{L_{\mathrm{P}} L_{\mathrm{S}}}}=\frac{4 \times 10^{-6}}{\sqrt{\left(2 \times 10^{-4}\right)^{2}}}=\frac{4 \times 10^{-6}}{2 \times 10^{-4}}\)
= 2 × 10^-2
Therefore, the percentage of flux of the primary reaching the secondary is
0.02 × 100% = 2%

Question 16.
A toroidal ring, having 100 turns per cm of a thin wire is wound on a nonmagnetic metal rod of length 1 m and diameter 1 cm. If the permeability of bar is equal to that of free space (µ₀), calculate the magnetic field inside the bar (B) when the current (i) circulating through the turns is 1 A. Also determine the self-inductance (L) of the coil.
Answer:
Data : l = 1 m, d = 1 cm, n = 100 cm^-1 = 10⁴ m^-1,
I = 100 A, µ₀ = 4π × 10^-7 H/m
The radius of cross section, r = \(\frac{d}{2}\) = 0.5 cm
= 5 × 10^-3 m
(a) Magnetic field inside the toroid,
B = µ₀nI = (4π × 10^-7)(10⁴)(100)
= 0.4 × 3.142 = 1.257 T

(b) Self inductance of the toroid,
L = µ₀2πRn²A = µ₀n²lA = µ₀n²l(πr²)
= (4π × 10^-7)(10⁴)²(1) [π(5 × 10^-3)²]
= π² × 10^-3 = 9.87 × 10^-3 H = 9.87 mH

Question 17.
A uniform magnetic field B(t), pointing upward fills a circular region of radius, s in horizontal plane. If B is changing with time, find the induced electric field.
[Hint : Part of Maxwell’s equation, applied to a time varying magnetic flux, leads us to the equation \(\oint \vec{E} \cdot \overline{\mathrm{d} l}=\frac{-d \phi_{m}}{d t}\), where \(\vec{E}\) is the electric field induced when the magnetic flux changes at the rate of \(\frac{d \phi_{m}}{d t}\)]
Answer:
The area of the region, A = πs², remains constant while B = B(f) is a function of time. Therefore, the induced emf,
e = \(-\frac{d \Phi_{\mathrm{m}}}{d t}=-\frac{d}{d t}(B A)=-A \frac{d B(t)}{d t}=-\pi s^{2} \frac{d B(t)}{d t}\)
[Note : Emf and electric field are different physical quantities, whose respective SI units are the volt and the volt per metre. The question has accordingly been corrected.]

12th Physics Digest Chapter 12 Electromagnetic induction Intext Questions and Answers

Do you know (Textbook Page No. 274)

Question 1.
If a wire without any current is kept in a magnetic field, then it experiences no force as shown in figure (a). But when the wire is carrying a current into the plane of the paper in the magnetic field, a force will be exerted on the wire towards the left as shown in the figure (b). The field will be strengthened on the right side of the wire where the lines of force are in the same direction as that of the magnetic field and weakened on the left side where the field lines are in opposite direction to that of the applied magnetic field. For a wire carrying a current out of the plane of the paper, the force will act to the right as shown in figure (c).
Maharashtra Board Class 12 Physics Solutions Chapter 12 Electromagnetic induction 2
Answer:
Force on a current-carrying conductor in a magnetic field, \(\vec{F}=I \vec{L} \times \vec{B}\) (Refer unit 10.5). The field due to acurrent-carrying straight conductor is given by right- hand grip rule. As shown in the figure below, the combined field due to a permanent magnet and a current-carrying conductor force the conductor out of the field. The field is strengthened where the two fields are in the same direction and add constructively while the field is weakened where the two fields are opposite in direction.
Maharashtra Board Class 12 Physics Solutions Chapter 12 Electromagnetic induction 3

Use your brain power (Textbook Page No. 282)

Question 1.
It can be shown that the mutual potential energy of two circuits is W = MI₁I₂. Therefore, the mutual inductance (M) may also be defined as the mutual potential energy (W) of two circuits corresponding to unit current flowing in each circuit.
M = \(\frac{W}{I_{1} I_{2}}\)
M = W[I₁ = I₂ = 1]
Answer:
Mutual inductance of two magnetically linked coils equals the potential energy for unit currents in the coils.
1 H = 1 T∙m²/A (= 1 V∙s/A = 1 Ω ∙ s = 1 J/A²)

Use your brain power (Textbook Page No. 284)

Question 1.
Prove that the inductance of parallel wires of length l in the same circuit is given by L = \(\left(\frac{\mu_{0} l}{\pi}\right)\) ln (d / a), where a is the radius of wire and d is separation between wire axes.
Answer:
If l is the current in each wire, from Ampe’re’s law the magnitude of the magnetic field outside each wire is
B = \(\frac{\mu_{0} I}{2 \pi r}\)
By right hand grip rule, the direction of the magnetic field due to both the wires are in the same direction at the point shown. Hence, by the symmetry of the setup, the total magnetic flux through an area dA = l dr shown is two times that due to one wire.
Maharashtra Board Class 12 Physics Solutions Chapter 12 Electromagnetic induction 4

Do you know (Textbook Page No. 285)

Question 1.
The flux rule is the terminology that Feynman used to refer to the law relating magnetic flux to emf. (RP Feynman, Feynman lectures on Physics, Vol II)
Answer:
Modern applications of Faraday’s law of induction :

Electric generators and motors
Dynamos in vehicles
Transformers
Induction furnaces (industrial), induction cooking stoves (domestic)
Radio communication
Magnetic flow meters and energy meters
Metal detectors at security checks .
Magnetic hard disk and tape, storage and retrieval
Graphics tablets
ATM Credit/debit cards, ATM and point-of-sale (POS) machines
Pacemakers

Faraday’s second law of electromagnetic induction is referred by some as the “flux rule”.

Maharashtra State Board Class 12 Biology Solutions Digest

October 5, 2021August 18, 2021 by Bhagya

Maharashtra State Board 12th Std Biology Textbook Solutions Digest

Maharashtra State Board Class 12 Textbook Solutions

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State

August 31, 2021August 18, 2021 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 1 Solid State Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Chemistry Solutions Chapter 1 Solid State

1. Choose the most correct answer.

Question i.
Molecular solids are
(a) crystalline solids
(b) amorphous solids
(c) ionic solids
(d) metallic solids
Answer:
(b) amorphous solids

Question ii.
Which of the following is n-type semiconductor?
(a) Pure Si
(b) Si doped with As
(c) Si doped with Ga
(d) Ge doped with In
Answer:
(b) Si doped with As

Question iii.
In Frenkel defect
(a) electrical neutrality of the substance is changed.
(b) density of the substance is changed.
(c) both cation and anion are missing
(d) overall electrical neutrality is preserved
Answer:
(d) overall electrical neutrality is preserved

Question iv.
In crystal lattice formed by bcc unit cell the void volume is
(a) 68%
(b) 74%
(c) 32%
(d) 26%
Answer:
(c) 32%

Question v.
The coordination number of atoms in bcc crystal lattice is
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(d) 8

Question vi.
Which of the following is not correct ?
(a) Four spheres are involved in the formation of tetrahedral void.
(b) The centres of spheres in octahedral voids are at the a pices of a regular tetrahedron.
(c) If the number of atoms is N the number of octahedral voids is 2N.
(d) If the number of atoms is N/2, the number of tetrahedral voids is N.
Answer:
(c) If the number of atoms is N the number of octahedral voids is 2N.

Question vii.
A compound forms hcp structure. Number of octahedral and tetrahedral voids in 0.5 mole of substance is respectively
(a) 3.011 × 10²³, 6.022 × 10²³
(b) 6.022 × 10²³, 3.011 × 10²³
(c) 4.011 × 10²³, 2.011 × 10²³
(d) 6.011 × 10²³, 12.022 × 10²³
Answer:
(a) 3.011 × 10²³, 6.022 × 10²³

Question viii.
Pb has fcc structure with edge length of unit cell 495 pm. Radius of Pb atom is
(a) 205 pm
(b) 185 pm
(c) 260 pm
(d) 175 pm
Answer:
(d) 175 pm

2. Answer the following in one or two sentences.

Question i.
What are the types of particles in each of the four main classes of crystalline solids?
Answer:
The smallest constituents or particles of various solids are atoms, ions or molecules.

Question ii.
Which of the three types of packing used by metals makes the most efficient use of space and which makes the least efficient use ?
Answer:
fcc has the most efficient packing of particles while scc has the least efficient packing.

Question iii.
The following pictures show population of bands for materials having different electrical properties. Classify them as insulator, semiconductor or a metal.
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 1a
Answer:
Picture A represents metal conductor,
Picture B represents insulator,
Picture C represents semiconductor.

Question iv.
What is a unit cell?
Answer:

Unit cell : It is the smallest repeating structural unit of a crystalline solid (or crystal lattice) which when repeated in different directions produces the crystalline solid (lattice).
The crystal is considered to consist of an infinite number of unit cells.
The unit cell possesses all the characteristics of the crystalline solid.

Question v.
How does electrical conductivity of a semiconductor change with temperature ? Why?
Answer:

Since the energy difference between valence band and conduction band in semiconductor is not large, the electrons from valence band can be promoted to conduction by heating.
Hence electrical conductivity of a semiconductor increases with temperature.

Question vi.
The picture represents bands of MOs for Si. Label valence band, conduction band and band gap.
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 2
Answer:

Question vii.
A solid is hard, brittle and electrically non-conductor. Its melt conducts electricity. What type of solid is it?
Answer:
A solid crystalline electrolyte like NaCl is hard, brittle and electrically nonconductor. But its melt conducts electricity.

Question viii.
Mention two properties that are common to both hep and ccp lattices.
Answer:
In hcp and ccp crystal lattices coordination number is 12 and packing efficiency is 74%.

Question ix.
Sketch a tetrahedral void.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 4

Question x.
What are ferromagnetic substances?
Answer:

The substances which possess unpaired electrons and high paramagnetic character and when placed in a magnetic field are strongly attracted and show permanent magnetic moment even when the external magnetic field is removed are said to be ferromagnetic. They can be permanently magnetised.
In the solid state, the metal ions of ferromagnetic substance are grouped together into small regions called domains, where each domain acts as a tiny magnet.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 5
For example : Fe, Co, Gd, Ni, CrO₂, etc.

3. Answer the following in brief.

Question i.
What are valence band and conduction band?
Answer:
There are two types of bands of molecular orbitals as follows :

Valence band : The atomic orbitals with filled electrons from the inner shells form valence bands, where there are no free mobile electrons since they are involved in bonding.
Conduction band : Atomic orbitals which are partially filled or empty on overlapping form closely placed molecular orbitals giving conduction bands where electrons are delocalised and can conduct, heat and electricity.

Question ii.
Distinguish between ionic solids and molecular solids.
Answer:

Type/ Property	Ionic solids	Molecular solids
1. Particles of unit cell	Cations and anions	Monoatomic or polyatomic molecules
2. Interparticle forces	Electrostatic	London, dipole-dipole forces and/or hydrogen bonds
3. Hardness	Hard and brittle	Soft
4. Melting points	High 600 °C to 3000 °C	Low (-272 °C to 400 °C)
5. Thermal and electrical conductivity	Poor electrical conductors in solid state. Good conductors when melted or dissolved in water.	Poor conductor of heat and electricity
6. Examples	NaCl, CaF₂	ice, benzoic acid

Question iii.
Calculate the number of atoms in fcc unit cell.
Answer:
Number of atoms in face-centred cubic (fcc) unit cell :
In this unit cell, there are 8 atoms at 8 corners and 6 atoms at 6 face centres. Each corner contributes 1/8th atom to the unit cell, hence due to 8 corners,
Number of atoms = \(\frac {1}{8}\) × 8 = 1.
Each face centre contributes half of the atom to the unit cell, hence due to 6 face centres,
Number of atoms = \(\frac {1}{2}\) × 6 = 3.
∴ Total number of atoms present in fee unit cell = 1 + 3 = 4.
Hence the volume of the unit cell is equal to the volume of four atoms.
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 6
Face centered unit cell

Question iv.
How are the spheres arranged in first layer of simple cubic close-packed structures? How are the successive layers of spheres placed above this layer ?
Answer:
(i) Stacking of square close packed layers :
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 7
Stacking of square close packed layers

In this arrangement, the two dimensional AAAA type square closed packed layers are placed one over the other in such a way that the crests of all spheres are in contact with successive layers in all directions. All spheres of different layers are perfectly aligned horizontally and vertically forming unit cells having primitive or simple cubic structure. Since all the layers are identical and if each layer is labelled as layer A, then whole three dimensional crystal lattice will be of AAAA… type.

Each sphere is in contact with six surrounded spheres, hence the coordination number of each sphere is six.

(ii) Stacking of two hexagonal close packed layers :
A close packed three dimensional structure can be generated by arranging hexagonal close packed layers in a particular manner.

In this the spheres of second layer are placed in the depression of the first layer.
In this if first layer is labelled as A then second layer is labelled as B since they are aligned differently.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 8
Two layers of closed packed spheres

In this, all triangular voids of the first layers are not covered by the spheres of the second layer. The triangular voids which are covered by second layer spheres generate tetrahedral void which is surrounded by four spheres. The triangular voids in one layer have above them triangular voids of successive layers.

The overlapping triangular voids from two layers together form an octahedral void which is surrounded by six spheres.

Question v.
Calculate the packing efficiency of metal crystal that has simple cubic structure.
Answer:
Step 1 : Radius of sphere : In simple cubic lattice, the atoms (spheres) are present at eight corners and in contact along the edge in the unit cell.
If ‘a’ is the edge length of the unit cell and ‘r’ is the radius of the atom, then
a = 2r or r = a/2
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 9
scc structure

Step 2 : Volume of sphere :
Volume of one particle = \(\frac{4 \pi}{3}\) × r³
= \(\frac{4 \pi}{3}\) × (a/2)³ = \(\frac{\pi a^{3}}{6}\)

Step 3 : Total volume of particles : Since the unit cell contains one particle. Volume occupied by one particle in unit cell = \(\frac{\pi a^{3}}{6}\)

Step 4 : Packing efficiency :
Packing efficiency
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 10
∴ Packing efficiency = 52.36%
Percentage of void space = 100 – 52.36
= 47.64%

Question vi.
What are paramagnetic substances? Give examples.
Answer:
(1) The magnetic properties of a substance arise due to the presence of electrons.
(2) An electron while revolving around the nucleus, also spins around its own axis and generates a magnetic moment and magnetic properties.
(3) If an atom or a molecule contains one or more unpaired electrons spinning in same direction, clockwise or anticlockwise, then the substance is associated with net magnetic moment and magnetic properties. They experience a net force of attraction when placed in the magnetic field. This phenomenon is called paramagnetism and the substance is said to be paramagnetic.
For example, O₂, Cu²⁺, Fe³⁺ , Cr³⁺ , NO, etc.

Question vii.
What are the consequences of Schottky defect?
Answer:
Consequences of Schottky defect :

Since the number of ions (cations and anions) decreases but volume remains unchanged, the density of a substance decreases.
As the number of missing cations and anions is equal, the electrical neutrality of the compound remains same.
This defect arises in ionic crystals like NaCl, AgBr, KCl, etc.

Question viii.
Cesium chloride crystallizes in cubic unit cell with Cl^– ions at the corners and Cs⁺ ion in the centre of the cube. How many CsCl molecules are there in the unit cell ?
Answer:
Number of Cs⁺ ion at body centre = 1
Number of Cl^– ions due to 8 comers = 8 × \(\frac {1}{8}\) = 1
Hence unit cell contains 1 CsCl molecule.

Question ix.
Cu crystallizes in fee unit cell with edge length of 495 pm. What is the radius of Cu atom ?
Answer:
Given : a = 495 pm
Radius, r = ?
For fee structure,
radius = r = \(\frac{a}{2 \sqrt{2}}=\frac{495}{2 \times \sqrt{2}}\) = 175 cm.
Radius of Cu atom = 175 pm

Question x.
Obtain the relationship between density of a substance and the edge length of unit cell.
Answer:
(1) Consider a cubic unit cell of edge length ‘a’.
The volume of unit cell = a³

(2) If there are ‘n’ particles per unit cell and the mass of particle is ‘m’, then
Mass of unit cell = m × n.

(3) If the density of the unit cell of the substance is p then
Density of unit cell = \(\frac{\text { Mass of unit cell }}{\text { Volume of unit cell }}\)
ρ = \(\frac{m \times n}{a^{3}}\)

Question 4.
The density of iridium is 22.4 g/cm³. The unit cell of iridium is fcc. Calculate the radius of iridium atom. Molar mass of iridium is 192.2 g/mol.
Answer:
Given : Crystal structure of iridium = fcc
Molar mass of iridium = 192.2 gmol^-1
Density = ρ = 22.4 gcm^-3
Radius of iridium = ?
In fcc structure, there are 8 Ir atoms at 8 comers and 6 Ir atoms at 6 face centres.
∴ Total number of Ir atoms = \(\frac {1}{8}\) × 8 + \(\frac {1}{2}\) × 6
= 1 + 3
= 4
Mass of Ir atom = \(\frac{192.2}{6.022 \times 10^{23}}\)
= 31.92 × 10^-23 g
∴ Mass of 4 Ir atoms = 4 × 31.92 × 10^-23
= 1.277 × 10^-21 g
∴ Mass of unit cell = 1.277 × 10^-21 g
Density of unit cell = \(\frac{\text { Mass of unit cell }}{\text { Volume of unit cell }}\)
22.4 = \(\frac{1.277 \times 10^{-21}}{a^{3}}\)
∴ a³ = \(\frac{1.277 \times 10^{-21}}{22.4}\)
= 57 × 10^-24 cm³
∴ a = (57 × 10^-24)³ = 3.848 × 10^-8 cm
If r is the radius of iridium atom, then for fcc structure,
r = \(\frac{a}{2 \sqrt{2}}\)
= \(\frac{3.848 \times 10^{-8}}{2 \times 1.414}\)
= 1.36 × 10^-8 cm
= 136 pm
Radius of iridium atom = 136 pm

Question 5.
Aluminium crystallizes in cubic close packed structure with unit cell edge length of 353.6 pm. What is the radius of Al atom ? How many unit cells are there in 1.00 cm³ of Al ?
Answer:
Given : Structure of Al
= Cubic close packed structure
= ccp structure
Edge length of unit cell = a = 353.6 pm
= 3.536 × 10^-8 cm
r = ?
Number of unit cells in 1.00 cm³ of Al = ?
Radius of Al atom = r = \(\frac{a}{2 \sqrt{2}}=\frac{353.6}{2 \sqrt{2}}\)
= \(\frac{353.6}{2 \times 1.414}\) = 125 pm
Volume of one unit cell = a³ = (3.536 × 10^-8)³
= 4.421 × 10^-23 cm³
Number of unit cells = \(\frac{1.00}{4.421 \times 10^{-23}}\)
= 2.26 × 10²²
Radius of Al atom = 125 pm
Number of unit cells = 2.26 × 10²²

Question 6.
In an ionic crystalline solid atoms of element Y form hcp lattice. The atoms of element X occupy one third of tetrahedral voids. What is the formula of the compound?
Answer:
In the given hcp lattice, Y atoms are present at 12 corners and 2 face centres.
∴ Number of Y atoms = \(\frac {1}{2}\) × 12 + 2 × \(\frac {1}{2}\) = 3
There are 6 tetrahedral voids, the number of X atoms = \(\frac {1}{3}\) × 6 = 2
∴ Formula of the compound is X₂Y₃.

Question 7.
How are tetrahedral and octahedral voids formed?
Answer:
Tetrahedral void : The vacant space or void among four constituent particles having tetrahedral arrangement in the crystal lattice is called tetrahedral void.
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 11
The arrangement of four spheres around the void is tetrahedral. A tetrahedral void is formed when a triangular void made by three coplanar spheres is in contact with fourth sphere above or below it.

Octahedral void : The vacant space or void at the centre of six spheres (or atoms) which are placed octahedrally is called octahedral void.
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 12

Question 8.
Third layer of spheres is added to second layer so as to form hcp or ccp structure. What is the difference between the addition of third layer to form these hexagonal close-packed structures?
Answer:

In the formation of hexagonal closed-packed (hcp) structure, the first one dimensional row shows depressions between neighbouring atoms.
When a second row is arranged so that spheres fit in these depressions then a staggered arrangement is obtained. If the first row is A then the second row is B.
When third row is placed in staggered manner in contact with second row then A type arrangement is obtained.
Similarly, the spheres in fourth row can be arranged as B type layer. This results in ABAB… type setting of the layers. This gives hexagonal close packing (hcp) structure.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 13
Hexagonal close packing (hcp)

Question 9.
An element with molar mass 27 g/mol forms cubic unit cell with edge length of 405 pm. If density of the element is 2.7 g/cm³, what is the nature of cubic unit cell ? (fcc or ccp)
Answer:
Given : Molar mass = M = 27 g mol^-1
Nature of crystal = cubic unit cell
Edge length = a = 405 pm = 4.05 × 10^-8 cm
Density = ρ = 2.7 g cm^-3
Nature of unit cell = ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 14
= 3.997
≅ 4
Hence the nature of unit cell = face-centred cubic unit cell
Radius of Al atom = 125 pm
The nature of cubic unit cell is fcc.

Question 10.
An element has a bcc structure with unit cell edge length of 288 pm. How many unit cells and number of atoms are present in 200 g of the element? (1.16 × 10²⁴, 2.32 × 10²⁴)

Question 11.
Distinguish with the help of diagrams metal conductors, insulators and semiconductors from each other.
Answer:
Conductor:

A substance which conducts heat and electricity to a greater extent is called conductor.
In this, conduction bands and valence bands overlap or are very closely spaced.
There is no energy difference or very less energy difference between valence bands and conduction bands.
There are free electrons in the conduction bands.
The conductance decreases with the increase in temperature.
E.g., Metals, alloys.
The conducting properties can’t be improved by adding third substance.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 15 b

Insulator:

A substance which cannot conduct heat and electricity under any conditions is called insulator.
In this, conduction bands and valence bands are far apart.
The energy difference between conduction bands and valence bands is very large.
There are no free electrons in the conduction bands and electrons can’t be excited from valence bands to conduction bands due to large energy difference.
No effect of temperature on conducting properties.
E.g., Wood, rubber, plastics.
No effect of addition of any substance.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 24

Semiconductor:

A substance that has poor electrical conductance at low temperature but higher conductance at higher temperature is called semiconductor.
In this, conduction bands and valence bands are spaced closely.
The energy difference between conduction bands and valence bands is small.
The electrons can be easily excited from valence bands to conduction bands by heating.
Conductance increases with the increase in temperature.
E.g., Si, Ge
By doping, conducting properties improve. E.g. n-type, p-type semiconductors.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 25

Question 12.
What are n-type semiconductors? Why is the conductivity of doped n-type semiconductor higher than that of pure semiconductor ? Explain with diagram.
Answer:
n-type semiconductor:

n-type semiconductor contains increased number of electrons in the conduction band.
When Si semiconductor is doped with 15^th group element phosphorus, P, the new atoms occupy some vacant sites in the lattice in place of Si atoms.
P has five valence electrons, out of which four are involved in covalent bonding with neighboring Si atoms while one electrons remains free and delocalised.
These free electrons increase the electrical conductivity of the semiconductor.
The semiconductors with extra non-bonding free electrons are called n-type semiconductors.

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 16
P atom occupying regular site of Si atom

Question 13.
Explain with diagram. Frenkel defect. What are the conditions for its formation? What is its effect on density and electrical neutrality of the crystal?
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 17

Frenkel defect : This defect arises when an ion of an ionic compound is missing from its regular site and occupies interstitial vacant position between lattice points.
Cations have smaller size than anions, hence generally cations occupy the interstitial sites.
This creates a vacancy defect at its original position and interstitial defect at new position.
Frenkel defect is regarded as the combination of interstitial defect and vacancy defect.

Conditions for the formation of Frenkel defect :

This defect arises in ionic compounds with a large difference between the sizes of cation and anion.
The ionic compounds must have ions with low coordination number.

Consequences of Frenkel defect :

Since there is no loss of ions from the crystal lattice, the density of the solid remains unchanged.
The crystal remains electrically neutral.
This defect is observed in ZnS, AgCl, AgBr, Agl, CaF₂, etc.

Question 14.
What is an impurity defect? What are its types? Explain the formation of vacancies through aliovalent impurity with example.
Answer:
Impurity defect : This defect arises when foreign atoms, that is, atoms different from the host atoms are present in the crystal lattice.

There are two types of impurity defects namely

Substitutional defects and
Interstitial defects.

(1) Substitutional defects : These defects arises when foreign atoms occupy the lattice sites in place of host atoms, due to their displacements.
Examples : Solid solutions of metals (alloys). For example. Brass in which host atoms are of Cu which are replaced by impurity of Zn atoms. In this Zn atoms occupy regular sites while Cu atoms occupy substituted sites.
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 18
Brass

Vacancy through aliovalent impurity :
By addition of impurities of aliovalent ions :
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 19
Vacancy through aliovalent ion

When aliovalent ion like Sr²⁺ in small amount is added by additing SrCl₂ to NaCl during its crystallisation, each Sr²⁺ion (oxidation state 2+) removes 2 Na⁺ ions from their lattice points, to maintain electrical neutrality. Hence one of vacant lattice site is occupied by Sr²⁺ ion while other site remains vacant.

Interstitial impurity defect :
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 20
Stainless steel

A defect in solid in which the impurity atoms occupy interstitial vacant spaces of lattice structure is called interstitial impurity defect.

For example, in steel, normal lattice sites are occupied by Fe atoms but interstitial spaces are occupied by carbon atoms.

12th Chemistry Digest Chapter 1 Solid State Intext Questions and Answers

Try this… (Textbook Page No. 1)

Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 21
Observe the above figure carefully. The two types of circles in this figure represent two types of constituent particles of a solid.

Question 1.
Will you call the arrangement of particles in this solid regular or irregular ?
Answer:
The arrangement of particles in this solid is regular.

Question 2.
Is the arrangement of constituent particles in directions \(\overrightarrow{\mathbf{A B}}\), \(\overrightarrow{\mathbf{C D}}\) and \(\overrightarrow{\mathbf{E F}}\) same or different?
Answer:
\(\overrightarrow{\mathbf{A B}}\) represents arrangement of identical particles of one type.
\(\overrightarrow{\mathbf{C D}}\) represents arrangement of identical particles of another type.
\(\overrightarrow{\mathbf{E F}}\) represents regular arrangement of two different particles in alternate positions.

Use your brain power ! (Textbook Page No. 2)

Question 1.
Identify the arrangements A and B as crystalline or amorphous.
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 22
Answer:
Arrangement in image A indicates the substance is crystalline.
Arrangement in image B indicates the substance is amorphous.

Try this… (Textbook Page No. 3)

Question 1.
Graphite is a covalent solid yet soft and good conductor of electricity. Explain.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 1 Solid State 23

Each carbon atom in graphite is sp² hybridised and covalently bonded to other three sp2 hybridised carbon atoms forming σ bonds and the fourth electron in 2p_z orbital of each carbon atom is used in the formation of a π bond. This results in the formation of hexagonal rings in two dimensions.
In graphite, the layers consisting of hexagonal carbon network are held together by weak van der Waal’s forces imparting softness.
The electrons in π bonds in the ring are delocalised and free to move in the delocalised molecular orbitals giving good electrical conductance.

Use your brain power ! (Textbook Page No. 13)

Question 1.
Which of the three lattices scc, bcc and fcc has the most efficient packing of particles ? Which one has the least efficient packing ?
Answer:
fcc has the most efficient packing of particles while see has the least efficient packing.

Can you think ? (Textbook Page No. 20)

Question 1.
When ZnO is heated it turns yellow and returns back to original white colour on cooling. What could be the reason ?
Answer:
When colourless ZnO is strongly heated, the metal atoms are deposited on crystal surface and anions O^2- migrate to the surface producing vacancies at anion lattice points.

These anions combine with Zn atoms forming ZnO and release electrons.
Zn + O^2- → ZnO + 2e^–

These released electrons diffuse into the crystal and occupy vacant sites of anions and produce F- centres. Due to these colour centres, ZnO turns yellow.

Can you tell ? (Textbook Page No. 23)

Let a small quantity of phosphorus be doped into pure silicon.

Question 1.
Will the resulting material contain the same number of total number of electrons as the original pure silicon ?
Answer:
Total number of electrons in doped silicon will be more than in original silicon.

Question 2.
Will the material be electrically neutral or charged ?
Answer:
Material will be electrically neutral.

Maharashtra State Board Class 12 Chemistry Solutions Digest

September 24, 2021August 18, 2021 by Bhagya

Maharashtra State Board 12th Std Chemistry Textbook Solutions Digest

Maharashtra State Board Class 12 Textbook Solutions

Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations

August 31, 2021August 18, 2021 by Prasanna

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 5 Oscillations Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 5 Oscillations

1. Choose the correct option.

i) A particle performs linear S.H.M. starting from the mean position. Its amplitude is A and time period is T. At the instance when its speed is half the maximum speed, its displacement x is
(A) \(\frac{\sqrt{3}}{2}\)A
(B) \(\frac{2}{\sqrt{3}}\)
(C) A/2
(D) \(\frac{1}{\sqrt{2}}\)
Answer:
(A) \(\frac{\sqrt{3}}{2}\)A

ii) A body of mass 1 kg is performing linear S.H.M. Its displacement x (cm) at t (second) is given by x = 6 sin (100t + π/4). Maximum kinetic energy of the body is
(A) 36 J
(B) 9 J
(C) 27 J
(D) 18 J
Answer:
(D) 18 J

iii) The length of second’s pendulum on the surface of earth is nearly 1 m. Its length on the surface of moon should be [Given: acceleration due to gravity (g) on moon is 1/6 th of that on the earth’s surface]
(A) \(\frac{1}{6}\) m
(B) 6 m
(C) \(\frac{1}{36}\) m
(D) \(\frac{1}{\sqrt{6}}\) m.
Answer:
(A) \(\frac{1}{6}\) m

iv) Two identical springs of constant k are connected, first in series and then in parallel. A metal block of mass m is suspended from their combination. The ratio of their frequencies of vertical oscillations will be in a ratio
(A) 1:4
(B) 1:2
(C) 2:1
(D) 4:1
Answer:
(B) 1:2

v) The graph shows variation of displacement of a particle performing S.H.M. with time t. Which of the following statements is correct from the graph?
(A) The acceleration is maximum at time T.
(B) The force is maximum at time 3T/4.
(C) The velocity is zero at time T/2.
(D) The kinetic energy is equal to total energy at time T/4.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 1
Answer:
(B) The force is maximum at time 3T/4.

2. Answer in brief.

i) Define linear simple harmonic motion.
Answer:
Definition: Linear simple harmonic motion (SHM) is defined as the linear periodic motion of a body, in which the force (or acceleration) is always directed towards the mean position and its magnitude is proportional to the displacement from the mean position.
OR
A particle is said to execute linear SHM if the particle undergoes oscillations about a point of stable equilibrium, subject to a linear restoring force always directed towards that point and whose magnitude is proportional to the magnitude of the displacement of the particle from that point.
Examples : The vibrations of the tines (prongs) of a tuning fork, the oscillations of the needle of a sewing machine.

ii) Using differential equation of linear S.H.M, obtain the expression for
(a) velocity in S.H.M.,
(b) acceleration in S.H.M.
Answer:
The general expression for the displacement of a particle in SHM at time t is x = A sin (ωt + α) … (1)
where A is the amplitude, ω is a constant in a particular case and α is the initial phase.
The velocity of the particle is
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 10
Equation (2) gives the velocity as a function of x.
The acceleration of the particle is
a = \(\frac{d v}{d t}\) = \(\frac{d}{d t}\) [Aω cos (ωt + α) J at at
∴ a = – ω² A sin (ωt + α)
But from Eq. (1), A sin (ωt + α) = x
∴ a = -ω²x … (3)
Equation (3) gives the acceleration as a function of x. The minus sign shows that the direction of the acceleration is opposite to that of the displacement.

iii) Obtain the expression for the period of a simple pendulum performing S.H.M.
Answer:
An ideal simple pendulum is defined as a heavy point mass suspended from a rigid support by a weightless, inextensible and twistless string, and set oscillating under gravity through a small angle in a vertical plane.

In practice, a small but heavy sphere, called the bob, is used. The distance from the point of suspension to the centre of gravity of the bob is called the length of the pendulum.

Consider a simple pendulum of length L₁ – suspended from a rigid support O. When displaced from its initial position of rest through a small angle θ in a vertical plane and released, it performs oscillations between two extremes, B and C, as shown in below figure. At B, the forces on the bob are its weight \(m \vec{g}\) and the tension \(\overrightarrow{F_{1}}\) in the string. Resolve \(m \vec{g}\) into two components : mg cos θ in the direction opposite to that of the tension and mg sin θ perpendicular to the string.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 20
mg cos θ balanced by the tension in the string. mg sin θ restores the bob to the equilibrium position.
Restoring force, F = – mg sin θ
If θ is small and expressed in radian,
sin θ \(\approx\) θ = \(\frac{\text { arc }}{\text { radius }}\) = \(\frac{\mathrm{AB}}{\mathrm{OB}}\) = \(\frac{x}{L}\)
∴ F = – mgθ = -mg\(\frac{x}{L}\) …. (1)
Since m, g and L are constant,
F ∝ (-x) …. (2)

Thus, the net force on the bob is in the direction opposite to that of displacement x of the bob from its mean position as indicated by the minus sign, and the magnitude of the force is proportional to the magnitude of the displacement. Hence, it follows that the motion of a simple pendulum is linear SHM.
Acceleration, a = \(\frac{F}{m}\) = –\(\frac{g}{L}\)x … (3)
Therefore, acceleration per unit displacement
= |\(\frac{a}{x}\)| = \(\frac{g}{L}\) ….. (4)
Period of SHM,
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 25
This gives the expression for the period of a simple pendulum.

iv) State the laws of simple pendulum.
Answer:
The period of a simple pendulum at a given place is
T = 2π\(\sqrt{\frac{L}{g}}\)
where L is the length of the simple pendulum and g is the acceleration due to gravity at that place. From the above expression, the laws of simple pendulum are as follows :

(1) Law of length : The period of a simple pendulum at a given place (g constant) is directly proportional to the square root of its length.
∴ T ∝\(\sqrt{L}\)
(2) Law of acceleration due to gravity : The period of a simple pendulum of a given length (L constant) is inversely proportional to the square root of the acceleration due to gravity.
∴ T ∝ \(\frac{1}{\sqrt{g}}\)
(3) Law of mass : The period of a simple pendulum does not depend on the mass or material of the bob of the pendulum.
(4) Law of isochronism : The period of a simple pendulum does not depend on the amplitude of oscillations, provided that the amplitude is small.

v) Prove that under certain conditions a magnet vibrating in uniform magnetic field performs angular S.H.M.
Answer:
Consider a bar magnet of magnetic moment μ, suspended horizontally by a light twistless fibre in a region where the horizontal component of the Earth’s magnetic field is B_h. The bar magnet is free to rotate in a horizontal plane. It comes to rest in approximately the North-South direction, along B_h. If it is rotated in the horizontal plane by a small
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 30
displacement θ from its rest position (θ = 0), the suspension fibre is twisted. When the magnet is released, it oscillates about the rest position in angular or torsional oscillation.

The bar magnet experiences a torque \(\tau\) due to the field B_h. Which tends to restore it to its original orientation parallel to B_h. For small θ, this restoring torque is
\(\tau\) = – μB_h sin θ = – μB_hμ …. (1)

where the minus sign indicates that the torque is opposite in direction to the angular displacement θ. Equation (1) shows that the torque (and hence the angular acceleration) is directly proportional in magnitude of the angular displacement but opposite in direction. Hence, for small angular displacement, the oscillations of the bar magnet in a uniform magnetic field is simple harmonic.

Question 3.
Obtain the expression for the period of a magnet vibrating in a uniform magnetic field and performing S.H.M.
Answer:
Definition : Angular SHM is defined as the oscillatory motion of a body in which the restoring torque responsible for angular acceleration is directly proportional to the angular displacement and its direction is opposite to that of angular displacement.
The differential equation of angular SHM is
I\(\frac{d^{2} \theta}{d t^{2}}\) + cθ = 0 ….. (1)
where I = moment of inertia of the oscillating body, \(\frac{d^{2} \theta}{d t^{2}}\) = angular acceleration of the body when its angular displacement is θ, and c = torsion constant of the suspension wire,
∴ \(\frac{d^{2} \theta}{d t^{2}}\) + \(\frac{c}{I}\)θ = 0
Let \(\frac{c}{I}\) = ω², a constant. Therefore, the angular frequency, ω = \(\sqrt{c / I}\) and the angular acceleration,
a = \(\frac{d^{2} \theta}{d t^{2}}\) = -ω²θ … (2)
The minus sign shows that the α and θ have opposite directions. The period T of angular SHM is
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 80
This is the expression for the period in terms of torque constant. Also, from Eq. (2),

Question 4.
Show that a linear S.H.M. is the projection of a U.C.M. along any of its diameter.
Answer:
Consider a particle which moves anticlockwise around a circular path of radius A with a constant angular speed ω. Let the path lie in the x-y plane with the centre at the origin O. The instantaneous position P of the particle is called the reference point and the circle in which the particle moves as the reference circle.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 35
The perpendicular projection of P onto the y-axis is Q. Then, as the particle travels around the circle, Q moves to-and-fro along the y-axis. Line OP makes an angle α with the x-axis at t = 0. At time t, this angle becomes θ = ωt + α.
The projection Q of the reference point is described by the y-coordinate,
y = OQ = OP sin ∠OPQ, Since ∠OPQ = ωt + α, y = A sin(ωt + α)
which is the equation of a linear SHM of amplitude A. The angular frequency w of a linear SHM can thus be understood as the angular velocity of the reference particle.

The tangential velocity of the reference particle is v = ωA. Its y-component at time t is v_y = ωA sin (90° – θ) = ωA cos θ
∴ v_y = ωA cos (ωt + α)
The centripetal acceleration of the reference particle is a_r = ω²A, so that its y-component at time t is a_x = a_r sin ∠OPQ
∴ a_x = – ω²A sin (ωt + α)

Question 5.
Draw graphs of displacement, velocity and acceleration against phase angle, for a particle performing linear S.H.M. from (a) the mean position
(b) the positive extreme position. Deduce your conclusions from the graph.
Answer:
Consider a particle performing SHM, with amplitude A and period T = 2π/ω starting from the mean position towards the positive extreme position where ω is the angular frequency. Its displacement from the mean position (x), velocity (v) and acceleration (a) at any instant are
x = A sin ωt = A sin\(\left(\frac{2 \pi}{T} t\right)\) (∵ω = \(\frac{2 \pi}{T}\))
v = \(\frac{d x}{d t}\) = ωA cos ωt = ωA cos\(\left(\frac{2 \pi}{T} t\right)\)
a = \(\frac{d v}{d t}\) = -ω² A sin ωt = – ω²A sin\(\left(\frac{2 \pi}{T} t\right)\) as the initial phase α = 0.
Using these expressions, the values of x, v and a at the end of every quarter of a period, starting from t = 0, are tabulated below.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 38
Using the values in the table we can plot graphs of displacement, velocity and acceleration with time.

Conclusions :

The displacement, velocity and acceleration of a particle performing linear SHM are periodic (harmonic) functions of time. For a particle starting at the mean position, the x-t and a-t graphs are sine curves. The v-t graph is a cosine curve.
There is a phase difference of \(\frac{\pi}{2}\) radians between x and v, and between v and a.
There is a phase difference of π radians between x and a.

Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 40

Consider a particle performing linear SHM with amplitude A and period T = 2π/ω, starting from the positive extreme position, where ω is the angular frequency. Its displacement from the mean position (x), velocity (y) and acceleration (a) at any instant (t) are
x = A cos ωt = A cos \(\left(\frac{2 \pi}{T} t\right)\) (∵ω = \(\frac{2 \pi}{T}\))
v = -ωA sin ωt = – ωA sin \(\left(\frac{2 \pi}{T} t\right)\)
a = -ω²A cos ωt = -ω²A cos \(\left(\frac{2 \pi}{T} t\right)\)
Using these expressions, the values of x, y and a at the end of every quarter of a period, starting from t = 0, are tabulated below.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 41
Using these values, we can plot graphs showing the variation of displacement, velocity and acceleration with time.

Conclusions :

The displacement, velocity and acceleration of a particle performing linear SHM are periodic (harmonic) functions of time. For a particle starting from an extreme position, the x-t and a-t graphs are cosine curves; the v-t graph is a sine curve.
There is a phase difference of \(\frac{\pi}{2}\) radians between x and v, and between v and a.
There is a phase difference of n radians between x and a.

Explanations :
(1) v-t graph : It is a sine curve, i.e., the velocity is a periodic (harmonic) function of time which repeats after a phase of 2π rad. There is a phase difference of π/2 rad between a and v.
v is minimum (equal to zero) at the extreme positions (i.e., at x = ± A) and v is maximum ( = ± ωA) at the mean position (x = 0).

(2) a-t graph : It is a cosine curve, i.e., the acceleration is a periodic (harmonic) function of time which repeats after a phase of 2π rad. There is a phase difference of π rad between v and a. a is minimum (equal to zero) at the mean position (x = 0) and a is maximum ( = \(\mp\)ω²A) at the extreme positions (x = ±A).
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 43

Question 6.
Deduce the expressions for the kinetic energy and potential energy of a particle executing S.H.M. Hence obtain the expression for total energy of a particle performing S.H.M and show that the total energy is conserved. State the factors on which total energy depends.
Answer:
Consider a particle of mass m performing linear SHM with amplitude A. The restoring force acting r on the particle is F = – kx, where k is the force constant and x is the displacement of the particle from its mean position.
(1) Kinetic energy : At distance x from the mean position, the velocity is
v = ω\(\sqrt{A^{2}-x^{2}}\)
where ω = \(\sqrt{k / m} .\) The kinetic energy (KE) of the particle is
KE = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\) mω² (A² – x²)
= \(\frac{1}{2}\)k(A² – x²) … (1)
If the phase of the particle at an instant t is θ = ωt + α, where α is initial phase, its velocity at that instant is
v = ωA cos (ωt + α)
and its KE at that instant is
KE = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\)mω²A² cos²(ωt + α) ….. (2)
Therefore, the KE varies with time as cos² θ.

(2) Potential energy : The potential energy of a particle in linear SHM is defined as the work done by an external agent, against the restoring force, in taking the particle from its mean position to a given point in the path, keeping the particle in equilibirum.

Suppose the particle in below figure is displaced from P₁ to P₂, through an infinitesimal distance dx against the restoring force F as shown.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 45
The corresponding work done by the external agent will be dW = ( – F)dx = kx dx. This work done is stored in the form of potential energy. The potential energy (PE) of the particle when its displacement from the mean position is x can be found by integrating the above expression from 0 to x.
∴ PE = \(\int\)dW = \(\int_{0}^{x}\) kx dx = \(\frac{1}{2}\) kx² … (3)
The displacement of the particle at an instant t being
x = A sin (wt + α)
its PE at that instant is
PE = \(\frac{1}{2}\)kx² = \(\frac{1}{2}\)kA² sin²(ωt + α) … (4)
Therefore, the PE varies with time as sin²θ.

(3) Total energy : The total energy of the particle is equal to the sum of its potential energy and kinetic energy.
From Eqs. (1) and (2), total energy is E = PE + KE
= \(\frac{1}{2}\)kx² + \(\frac{1}{2}\)k(A² – x²)
= \(\frac{1}{2}\)kx² + \(\frac{1}{2}\)kA² – \(\frac{1}{2}\)kx²
∴ E = \(\frac{1}{2}\)kA² = \(\frac{1}{2}\)mω²A² … (5)
As m is constant, ω and A are constants of the motion, the total energy of the particle remains constant (or its conserved).

Question 7.
Deduce the expression for period of simple pendulum. Hence state the factors on which its period depends.
Answer:
An ideal simple pendulum is defined as a heavy point mass suspended from a rigid support by a weightless, inextensible and twistless string, and set oscillating under gravity through a small angle in a vertical plane.

In practice, a small but heavy sphere, called the bob, is used. The distance from the point of suspension to the centre of gravity of the bob is called the length of the pendulum.

The period of a simple pendulum at a given place is
T = 2π\(\sqrt{\frac{L}{g}}\)
where L is the length of the simple pendulum and g is the acceleration due to gravity at that place. From the above expression, the laws of simple pendulum are as follows :

Question 8.
At what distance from the mean position is the speed of a particle performing S.H.M. half its maximum speed. Given path length of S.H.M. = 10 cm. [Ans: 4.33 cm]
Answer:
Data : v = \(\frac{1}{2}\)v_max, 2A = 10 cm
∴ A = 5 cm
v = ω\(\sqrt{A^{2}-x^{2}}\) and v_max = ωA
Since v = \(\frac{1}{2}\)v_max,
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 50
This gives the required displacement.

Question 9.
In SI units, the differential equation of an S.H.M. is \(\frac{d^{2} x}{d t^{2}}\) = -36x. Find its frequency and period. Find its frequency and period.
[Ans: 0.9548 Hz, 1.047 s]
Answer:
\(\frac{d^{2} x}{d t^{2}}\) = -36x
Comparing this equation with the general equation,
\(\frac{d^{2} x}{d t^{2}}\) = -ω²x
We get, ω² = 36 ∴ ω = 6 rad/s
ω = 2πf
∴ The frequency, f = \(\frac{\omega}{2 \pi}\) = \(\frac{6}{2(3.142)}\) = \(\frac{6}{6.284}\) = 0.9548 Hz
and the period, T = \(\frac{1}{f}\) = \(\frac{1}{0.9548}\) = 1.047 s

Question 10.
A needle of a sewing machine moves along a path of amplitude 4 cm with frequency 5 Hz. Find its acceleration \(\frac{1}{30}\)s after it has crossed the mean position. [Ans: 34.2 m/s²]
Answer:
Data : A = 4 cm = 4 × 10^-2 m, f = 5Hz, t = \(\frac{1}{30}\)s
ω = 2πf = 2π (5) = 10π rad/s
Therefore, the magnitude of the acceleration,
|a| = ω²x = ω²A sin ωt
= (10π)² (4 × 10²)
= 10π² sin \(\frac{\pi}{3}\) = 10 (9.872)(0.866) = 34.20 m/s²

Question 11.
Potential energy of a particle performing linear S.H.M is 0.1 π² x² joule. If mass of the particle is 20 g, find the frequency of S.H.M. [Ans: 1.581 Hz]
Answer:
Data : PE = 0.1 π² x² J, m = 20 g = 2 × 10^-2 kg
PE = \(\frac{1}{2}\)mω²x² = \(\frac{1}{2}\)m (4π²f²)x²
∴ \(\frac{1}{2}\)m(4π²f²)x² = 0.1 π² x²
∴ 2mf² = 0.1 ∴ f² = \(\frac{1}{20\left(2 \times 10^{-2}\right)}\) = 2.5
∴ The frequency of SHM is f = \(\sqrt{2.5}\) = 1.581 Hz

Question 12.
The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the centre of the path. [Ans: 6.324 cm/s]
Answer:
Data : m = 2 kg, E = 40 J
The speed of the body while crossing the centre of the path (mean position) is v_max and the total energy is entirely kinetic energy.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 60

Question 13.
A simple pendulum performs S.H.M of period 4 seconds. How much time after crossing the mean position, will the displacement of the bob be one third of its amplitude. [Ans: 0.2163 s]
Answer:
Data : T = 4 s, x = A/3
The displacement of a particle starting into SHM from the mean position is x = A sin ωt = A sin \(\frac{2 \pi}{T}\) t
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 61
∴ the displacement of the bob will be one-third of its amplitude 0.2163 s after crossing the mean position.

Question 14.
A simple pendulum of length 100 cm performs S.H.M. Find the restoring force acting on its bob of mass 50 g when the displacement from the mean position is 3 cm. [Ans: 1.48 × 10^-2 N]
Answer:
Data : L = 100 cm, m = 50 g = 5 × 10^-2 kg, x = 3 cm, g = 9.8 m/s²
Restoring force, F = mg sin θ = mgθ
= (5 × 10^-2)(9.8)\(\left(\frac{3}{100}\right)\)
= 1.47 × 10^-2 N

Question 15.
Find the change in length of a second’s pendulum, if the acceleration due to gravity at the place changes from 9.75
m/s² to 9.80 m/s². [Ans: Decreases by 5.07 mm]
Answer:
Data : g₁ = 9.75 m/s², g₂ = 9.8 m/s²
Length of a seconds pendulum, L = \(\frac{g}{\pi^{2}}\)
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 62
∴ The length of the seconds pendulum must be increased from 0.9876 m to 0.9927 m, i.e., by 0.0051 m.

Question 16.
At what distance from the mean position is the kinetic energy of a particle performing S.H.M. of amplitude 8 cm, three times its potential energy? [Ans: 4 cm]
Answer:
Data : A = 8 cm, KE = 3 PE
KE = \(\frac{1}{2}\) (A² – x²) and PE = \(\frac{1}{2}\)kx²
Given, KE = 3PE.
∴ \(\frac{1}{2}\)k(A² – x²) = 3\(\left(\frac{1}{2} k x^{2}\right)\)
∴ A² – x² = 3x² ∴ 4x² = A²
∴ the required displacement is
x = ±\(\frac{A}{2}\) = ±\(\frac{8}{2}\) = ± 4 cm

Question 17.
A particle performing linear S.H.M. of period 2π seconds about the mean position O is observed to have a speed of \(b \sqrt{3}\) m/s, when at a distance b (metre) from O. If the particle is moving away from O at that instant, find the
time required by the particle, to travel a further distance b. [Ans: π/3 s]
Answer:
Data : T = 2πs, v = b\(\sqrt{3}\) m/s at x = b
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 63
∴ Assuming the particle starts from the mean position, its displacement is given by
x = A sin ωt = 2b sin t
If the particle is at x = b at t = t₁,
b = 2b sint₁ ∴ t₁ = sin^-1 \(\frac{1}{2}\) = \(\frac{\pi}{6}\)s
Also, with period T = 2πs, on travelling a further distance b the particle will reach the positive extremity at time t₂ = \(\frac{\pi}{2}\)s.
∴ The time taken to travel a further distance b from x = b is t₂ – t₁ = \(\frac{\pi}{2}\) – \(\frac{\pi}{6}\) = \(\frac{\pi}{3}\)s.

Question 18.
The period of oscillation of a body of mass m₁ suspended from a light spring is T. When a body of mass m₂ is tied to the first body and the system is made to oscillate, the period is 2T. Compare the masses m₁ and m₂ [Ans: 1/3]
Answer:
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 64
This gives the required ratio of the masses.

Question 19.
The displacement of an oscillating particle is given by x = asinωt + bcosωt where a, b and ω are constants. Prove that the particle performs a linear S.H.M. with amplitude A = \(\sqrt{a^{2}+b^{2}}\)
Answer:
x = asinωt + bcosωt
Let a = A cos φ and b = A sin φ, so that
A² = a² + b² and tan φ = \(\frac{b}{a}\)
∴ x = A cos φ sin ωt + A sin φ cos ωt
∴ x = A sin (ωt + φ)
which is the equation of a linear SHM with amplitude A = \(\sqrt{a^{2}+b^{2}}\) and phase constant φ = tan^-1 \(\frac{b}{a}\), as required.

Question 20.
Two parallel S.H.M.s represented by x₁ = 5sin (4πt + \(\frac{\pi}{3}\)) cm and x₂ = 3sin(4πt + π/4) cm are superposed on a particle. Determine the amplitude and epoch of the resultant S.H.M. [Ans: 7.936 cm, 54° 23′]
Answer:
Data: x₁ = 5 sin (4πt + \(\frac{\pi}{3}\)) = A₁ sin(ωt + α),
x₂ = 3 sin (4πt + \(\frac{\pi}{4}\)) = A₂ sin(ωt + β)
∴ A₁ = 5 cm, A₂ = 3 cm, α = \(\frac{\pi}{3}\) rad, β = \(\frac{\pi}{4}\) rad
(i) Resultant amplitude,
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 65

(ii) Epoch of the resultant SHM,
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 66

Question 21.
A 20 cm wide thin circular disc of mass 200 g is suspended to a rigid support from a thin metallic string. By holding the rim of the disc, the string is twisted through 60° and released. It now performs angular oscillations of period 1 second. Calculate the maximum restoring torque generated in the string under undamped conditions. (π³ ≈ 31)
[Ans: 0.04133 N m]
Answer:
Data: R = 10cm = 0.1 m, M = 0.2 kg, θ_m = 60° = \(\frac{\pi}{3}\) rad, T = 1 s, π³ ≈ 31
The Ml of the disc about the rotation axis (perperdicular through its centre) is
I = \(\frac{1}{2}\)MR² = (0.2)(0.1)² = 10^-3 kg.m²
The period of torsional oscillation, T = 2π\(\sqrt{\frac{I}{c}}\)
∴ The torsion constant, c = 4πr²\(\frac{I}{T^{2}}\)
The magnitude of the maximum restoring torque,
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 67

Question 22.
Find the number of oscillations performed per minute by a magnet is vibrating in the plane of a uniform field of 1.6 × 10^-5 Wb/m². The magnet has moment of inertia 3 × 10^-6 kgm² and magnetic moment 3 A m². [Ans:38.19 osc/min.]
Answer:
Data : B = 1.6 × 10^-5 T, I = 3 × 10^-6kg/m²,
µ = 3 A.m²
The period of oscillation, T = 2π \(\sqrt{\frac{I}{\mu B_{\mathrm{h}}}}\)
∴ The frequency of oscillation is
f = \(\frac{1}{2 \pi}\)\(\sqrt{\frac{\mu B}{I}}\)
∴ The number of oscillations per minute
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 68
= 38.19 per minute

Question 23.
A wooden block of mass m is kept on a piston that can perform vertical vibrations of adjustable frequency and amplitude. During vibrations, we don’t want the block to leave the contact with the piston. How much maximum frequency is possible if the amplitude of vibrations is restricted to 25 cm? In this case, how much is the energy per unit mass of the block? (g ≈ π² ≈ 10 m s^-2)
[Ans: n_max = 1/s, E/m = 1.25 J/kg]
Answer:
Data : A = 0.25 m, g = π² = 10 m/s²
During vertical oscillations, the acceleration is maximum at the turning points at the top and bottom. The block will just lose contact with the piston when its apparent weight is zero at the top, i. e., when its acceleration is a_max = g, downwards.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 69
This gives the required frequency of the piston.

12th Physics Digest Chapter 5 Oscillations of Waves Intext Questions and Answers

Can you tell? (Textbook Page No. 112)

Question 1.
Why is the term angular frequency (ω) used here for a linear motion ?
Answer:
A linear SHM is the projection of a UCM on a diameter of the circle. The angular speed co of a particle moving along this reference circle is called the angular frequency of the particle executing linear SHM.

Can you tell? (Textbook Page No. 114)

Question 1.
State at which point during an oscillation the oscillator has zero velocity but positive acceleration ?
Answer:
At the left extreme, i.e., x = – A, so that a = – ω²x = – ω²(- A) = ω²A = a_max

Question 2.
During which part of the simple harmonic motion velocity is positive but the displacement is negative, and vice versa ?
Answer:
Velocity v is positive (to the right) while displacement x is negative when the particle in SHM is moving from the left extreme towards the mean position. Velocity v is negative (to the left) while displacement x is positive when the particle in SHM is moving from the right extreme towards the mean position.

Can you tell? (Textbook page 76)

Question 1.
To start a pendulum swinging, usually you pull it slightly to one side and release. What kind of energy is transferred to the mass in doing this?
Answer:
On pulling the bob of a simple pendulum slightly to one side, it is raised to a slightly higher position. Thus, it gains gravitational potential energy.

Question 2.
Describe the energy changes that occur when the mass is released.
Answer:
When released, the bob oscillates in SHM in a vertical plane and the energy oscillates back and forth between kinetic and potential, going completely from one form of energy to the other as the pendulum oscillates. In the case of undamped SHM, the motion starts with all of the energy as gravitational potential energy. As the object starts to move, the gravitational potential energy is converted into kinetic energy, becoming entirely kinetic energy at the equilibrium position. The velocity becomes zero at the other extreme as the kinetic energy is completely converted back into gravitational potential energy,
and this cycle then repeats.

Question 3.
Is/are there any other way/ways to start the oscillations of a pendulum? Which energy is supplied in this case/cases?
Answer:
The bob can be given a kinetic energy at its equilibrium position or at any other position of its path. In the first case, the motion starts with all of the energy as kinetic energy. In the second case, the motion starts with partly gravitational potential energy and partly kinetic energy.

Can you tell? (Textbook Page No. 109)

Question 1.
Is the motion of a leaf of a tree blowing in the wind periodic ?
Answer:
The leaf of a tree blowing in the wind oscillates, but the motion is not periodic. Also, its displacement from the equilibrium position is not a regular function of time.

Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

August 31, 2021August 18, 2021 by Bhagya

Balbharti Maharashtra State Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
_____________ is related to money and money management.
(a) Production
(b) Marketing
(c) Finance
Answer:
(c) Finance

Question 2.
Finance is the management of _____________ affairs of the company.
(a) monetary
(b) marketing
(c) production
Answer:
(a) monetary

Question 3.
Corporation finance deals with the acquisition and use of _____________ by business corporation.
(a) goods
(b) capital
(c) land
Answer:
(b) capital

Question 4.
Company has to pay _____________ to government.
(a) taxes
(b) dividend
(c) interest
Answer:
(a) taxes

Question 5.
_____________ refers to any kind of fixed assets.
(a) Authorised capital
(b) Issued capital
(c) Fixed capital
Answer:
(c) Fixed capital

Question 6.
_____________ refers to the excess of current assets over current liabilities.
(a) Working capital
(b) Paid-up capital
(c) Subscribed capital
Answer:
(a) Working capital

Question 7.
Manufacturing industries have to invest _____________ amount of funds to acquire fixed assets.
(a) huge
(b) less
(c) minimal
Answer:
(a) huge

Question 8.
When the population is increasing at a high rate, certain manufacturers find this as an opportunity to _____________ business.
(a) close
(b) expand
(c) contract
Answer:
(b) expand

Question 9.
The sum of all _____________ is gross working capital.
(a) expenses
(b) current assets
(c) current liabilities
Answer:
(b) current assets

Question 10.
_____________ means mix up of various sources of funds in desired proportion.
(a) Capital budgeting
(b) Capital structure
(c) Capital goods
Answer:
(b) Capital structure

1B. Match the pairs:

Question 1.
Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance 1B Q1
Answer:

Group ‘A’	Group ‘B’
(a) Capital budgeting	(6) Investment decision
(b) Fixed capital	(5) Fixed assets
(c) Working capital	(1) Sum of current assets
(d) Capital structure	(9) Mix up various sources of funds
(e) Corporate finance	(2) Deals with acquisition and use of capital

1C. Write a word or term or a phrase that can substitute each of the following statements:

Question 1.
A key determinant of the success of any business function.
Answer:
Finance

Question 2.
The decision of the finance manager ensures that the firm is well-capitalized.
Answer:
Financing decision

Question 3.
The decision of the finance manager to deploy the funds in a systematic manner.
Answer:
Investment decision

Question 4.
Capital is needed to acquire fixed assets that are used for a longer period of time.
Answer:
Fixed capital

Question 5.
The sum of current assets.
Answer:
Gross working capital

Question 6.
The excess of current assets over current liabilities.
Answer:
Networking capital

Question 7.
The process of converting raw material into finished goods.
Answer:
Production cycle

Question 8.
The boom and recession cycle in the economy.
Answer:
Economic Trend

Question 9.
The ratio of different sources of funds in the total capital.
Answer:
Capital Structure

Question 10.
The internal source of financing.
Answer:
Retained earnings

1D. State whether the following statements are True or False:

Question 1.
Finance is related to money and money management.
Answer:
True

Question 2.
The business firm gives a green signal to the project only when it is profitable.
Answer:
True

Question 3.
Corporate finance brings coordination between various business activities.
Answer:
True

Question 4.
Fixed capital is also referred as circulating capital.
Answer:
False

Question 5.
Working capital stays in the business almost permanently.
Answer:
False

Question 6.
The business will require huge funds if assets are acquired on a lease basis.
Answer:
False

Question 7.
The business dealing in luxurious products will require a huge amount of working capital.
Answer:
True

Question 8.
A firm with large-scale operations will require more working capital.
Answer:
True

Question 9.
Liberal credit policy creates a problem of bad debt.
Answer:
True

Question 10.
Financial institutions and banks cater to the working capital requirement of the business.
Answer:
True

1E. Find the odd one.

Question 1.
Land and Building, Plant and Machinery, Cash.
Answer:
Cash

Question 2.
Debenture Capital, Equity Share Capital, Preference Share Capital.
Answer:
Debenture Capital

Question 3.
Fixed Capital, Capital Structure, Working Capital.
Answer:
Capital Structure

1F. Complete the sentences.

Question 1.
Initial planning of capital requirement is made by _____________
Answer:
entrepreneur

Question 2.
When there is boom in economy, sales will _____________
Answer:
increase

Question 3.
The process of converting raw material into finished goods is called _____________
Answer:
production cycle

Question 4.
During recession period sales will _____________
Answer:
decrease

1G. Select the correct option from the bracket.

Question 1.
Maharashtra Board Class 12 Secretarial Practice Solutions Chapter 1 Introduction to Corporate Finance 1G Q1
(To have the right amount of capital, deploy funds in a systematic manner, fixed capital, working capital, capital structure, carry dividend at a fixed rate)
Answer:

Group ‘A’	Group ‘B’
(a) Financing decision	(1) To have the right amount of capital
(b) Fixed capital	(2) Longer period of time
(c) Investment decision	(3) Deploy funds in a systematic manner
(d) Working capital	(4) Circulating capital
(e) Combination of various sources of funds	(5) capital structure

1H. Answer in one sentence.

Question 1.
Define corporate finance.
Answer:
Corporate finance deals with the raising and using of finance by a corporation.

Question 2.
What is fixed capital?
Answer:
Fixed capital is the capital that is used for buying fixed assets that are used for a longer period of time in the business eg. Capital for plant and machinery etc.

Question 3.
What is working capital?/Define working capital.
Answer:
Working capital is the capital that is used to carry out day-to-day business activities and takes into consideration all current assets of the company.
Eg: for building up inventories.

Question 4.
What is the production cycle?
Answer:
The process of converting raw material into finished goods is called the production cycle.

Question 5.
Define capital structure.
Answer:
Capital structure means to mix up various sources of funds in the desired proportion. To decide capital structure means to decide upon the ratio of different types of capital.

1I. Correct the underlined word and rewrite the following sentences.

Question 1.
Finance is needed to pay dividends to debenture holders.
Answer:
Finance is needed to pay interest to debenture holders.

Question 2.
When there is a recession in the economy sales will increase.
Answer:
When there is a boom in the economy sales will increase.

Question 3.
Share is an acknowledgment of a loan raised by the company.
Answer:
A debenture is an acknowledgment of a loan raised by a company.

Question 4.
Equity shares carry dividends at a fixed rate.
Answer:
Preference shares carry dividends at a fixed rate.

2. Explain the following terms/concepts.

Question 1.
Financing decision
Answer:
A financing decision is a right decision that is made by a finance manager of any corporation ensuring that the firm is well capitalized with the right combination of debt and equity, having access to multiple choices of sources of financing.

Question 2.
Investment decision
Answer:
Investment decisions mean capital budgeting i.e. finding investments and deploying them successfully in the business for greater profits.

Question 3.
Fixed capital
Answer:
Fixed capital is the capital that is used for buying fixed assets that are used for a longer period of time in the business. These assets are not meant for. resale. Examples of fixed capital are capital used for purchasing land and building, furniture, plant, and machinery, etc.

Question 4.
Working Capital
Answer:
Working capital is the capital that is used to carry out day-to-day business activities. It takes into consideration all current assets, of the company. It also refers to ‘Gross Working Capital’.
Examples of working capital are

for building up inventories.
for financing receivables.
for covering day-to-day operating expenses.

3. Study the following case/situation and express your opinion.

1. The management of ‘Maharashtra State Road Transport Corporation’ wants to determine the size of working capital.

Question (a).
Being a public utility service provider will it need less working capital or more?
Answer:
MSRTC being a public utility service provider will need less working capital because of a continuous flow of cash from there, customers thus liabilities are taken care of.

Question (b).
Being a public utility service provider, will it need more fixed capital?
Answer:
Being a public utility service provider MSRTC will need a huge amount of funds to acquire fixed assets thus it will need more fixed capital.

Question (c).
Give one example of a public utility service that you come across on a day-to-day basis.
Answer:
The Indian Railways.

2. A company is planning to enhance its production capacity and is evaluating the possibility of purchasing new machinery whose cost is ₹ 2 crore or has alternative of machinery available on a lease basis.

Question (a).
What type of asset is machinery?
Answer:
Machinery is a Fixed Asset.
A fixed asset may be held for 5, 10 or 20 years and more. But if assets are acquired on a lease or rental basis, then less amount of funds for fixed assets will be needed for business.

Question (b).
Capital used for the purchase of machinery is fixed capital or working capital.
Answer:
Capital used for the purchase of machinery is fixed capital.

Question (c).
Does the size of a business determine the fixed capital requirement?
Answer:
Yes. Where a business firm is set up to carry on large-scale operations, its fixed capital requirements are likely to be high.

4. Distinguish between the following.

Question 1.
Fixed Capital and Working Capital
Answer:

Points	Fixed Capital	Working Capital
1. Meaning	Fixed capital refers to any kind of physical asset, a portion of total capital that is invested in fixed assets.	Working capital refers to the sum of current assets or gross working capital.
2. Nature	It stays in the business almost permanently.	Working capital is circulatory capital. It keeps changing.
3. Purpose	It is invested in fixed assets such as land, building, equipment, etc.	Working capital is invested in short-term assets such as cash, account receivable, inventory, etc.
4. Sources	Fixed capital funding can come from selling shares, debentures, bonds, long-term loans, etc.	Working capital can be funded with short-term loans, deposits, trade credit, etc.
5. Objectives of investors	Investors invest money in fixed capital hoping to make a future profit.	Investors invest money in working capital for getting immediate returns.
6. Risk	Investment in fixed capital implies more risk.	Investment in working capital is less risky. Eg. Land, building, plant and machinery
7. Decisions	Decisions relating to fixed capital investment are generally made by top-level management. Eg. Cash, bills receivable, inventories, cash at the bank	Decisions relating to working capital needs are generally made by middle-level or lower-level management.

5. Answer in brief:

Question 1.
Define capital structure and state its components.
Answer:
Definition: R.H. Wessel “The long term sources of funds employed in a business enterprise.”
John H. Hampton “A firm’s capital structure is the relation between the debt and equity securities that make up the firm’s financing of its assets.” Thus, the term capital structure means security mix. It refers to the proportion of different securities raised by a firm for long-term finance.

Components/Parts of Capital Structure:
There are four basic components of capital structure. They are as follows:
(i) Equity Share Capital:

It is the basic source of financing activities of the business. Equity share capital is provided by equity shareholders.
They buy equity shares and help a business firm to raise necessary funds. They bear the ultimate risk associated with ownership.
Equity shares carry dividends at a fluctuating rate depending upon profit.

(ii) Preference Share Capital:

Preference shares carry preferential rights as to payment of dividends and have priority over equity shares for return of capital when the company is liquidated.
These shares carry dividends at a fixed rate.
They enjoy limited voting rights.

(iii) Retained earnings:

It is an internal source of financing.
It is nothing but ploughing back of profit.

(iv) Borrowed capital: It comprises of the following:

Debentures: A debenture is an acknowledgment of a loan raised by the company. The company has to pay interest at an agreed rate.
Term Loan: Term loans are provided by the bank and other financial institutions. They carry fixed rate of interest.

Question 2.
State any four factors affecting fixed capital requirements?
Answer:
(i) Nature of business:

The nature of business certainly plays a role in determining fixed capital requirements. They need to invest a huge amount of money in fixed assets.
e.g. Rail, road, and other public utility services have large fixed investments.
Their working capital requirements are nominal because they supply services and not the product.
They deal in cash sales only.

(ii) Size of business:
The size of a business also affects fixed capital needs. A general rule applies that the bigger the business, the higher the need for fixed capital. The size of the firm, either in terms of its assets or sales, affects the need for fixed capital.

(iii) Scope of business:
Some business firms that manufacture the entire range of their production would require a huge investment in fixed capital. However, those companies that are labour intensive and who do not use the latest technology may require less fixed capital and vice versa.

(iv) Extent of lease or rent:
Companies who take their assets on a lease basis or on a rental basis will require less amount of funds for fixed assets. On the other side, firms that purchase assets will naturally require more fixed capital in the initial stages.

Question 3.
What are Corporate Finance and State’s two decisions which are basic of corporate finance?
OR
Write short note on Corporate Finance
Answer:
Corporate finance deals with the raising and using of finance by a corporation. It includes various financial activities like capital structuring and making investment decisions, financial planning, capital formation, and foreign capital, etc. Even financial organisations and banks play a vital role in corporate financing.

Henry Hoagland expresses, “Corporate Finance deals primarily with the acquisition and use of capital by the business corporation”.

Following two decisions are the basis of corporate finances:
(i) Financing decision:
Every business firm must carefully estimate its capital needs i.e. working capital and fixed capital. The firm needs to mobilize funds from the right sources also maintaining the right combination of debt capital and equity capital. For this balance, a company may go for or raise equity capital or even opt for borrowed funds by way of debentures, public deposits term loans, etc. to raise funds.

(ii) Investment decision:
Once the capital needs are accessed, the finance manager needs to take correct decisions regarding the use of the funds in a systematic manner, productively, using effective cost control measures to generate high profits. Finding investments through proper decisions and using them successfully in business is called ‘capital budgeting

6. Justify the following statements.

Question 1.
The firm has multiple choices of sources of financing.
Answer:

Business firms require finance in terms of working capital and fixed capital.
Funds are required at different stages of business.
The company can raise funds from various sources i.e. from internal and external sources.
Internal sources could be cash inflows on sales turnover, income from investments, and retained earnings.
External sources can be obtained for short-term requirements through cash credit, overdraft trade credit, discounting bills of Exchange issues of commercial paper, etc.
For long-term needs, a firm can meet its financing needs through the issue of shares, debentures, bonds, public deposits, etc. Thus, it is rightly said that the firm has multiple choices of sources of financing.

Question 2.
There are various factors affecting the requirements of fixed capital.
Answer:

Fixed capital being long-term capital is required for the development and expansion of the company.
The nature and size of a business have a great impact on fixed capital. Manufacturing businesses require huge fixed capital whereas trading organizations like retailers require less fixed capital.
Methods of acquiring assets on rentals or on a lease/installment basis will require less amount of fixed assets.
If fixed assets are available at low prices and concessional rates then it would reduce the need for investment in fixed assets.
International conditions and economic trends like a boom period will require high investment in fixed assets and a recession will lead to less requirement.
Similarly, consumer preferences, competition, and highly demanded goods and services will require a large amount of fixed capital. E.g. Mobile phones. Thus, it is rightly said that there are various factors affecting the requirements of fixed capital.

Question 3.
Fixed capital stays in the business almost permanently.
Answer:
Factors determining fixed capital requirements are:

Fixed capital refers to capital invested for acquiring fixed assets.
These assets are not meant for resale.
Fixed capital is capital used for purchasing land and building, furniture, plant, and machinery, etc.
Such cap al is usually required at the time of the establishment of a new company.
Existing companies may also need such capital for their expansion and development, replacement of equipment, etc.
Modern industrial processes require the increased use of heavy automated machinery. Thus, it is rightly said that fixed capital stays in the business almost permanently.

Question 4.
Capital structure is composed of owned funds and borrowed funds.
Answer:

Capital structure means to mix up of various sources of funds in desired proportions.
To decide capital structures means to decide upon the ratio of different types of capital.
A firm’s capital structure is the relation between the debt and equity securities that make up the firm’s financing of its assets.
The capital structure is composed of own funds which include share capital, free serves, and surplus, and borrowed funds which represent debentures, bank loans, and long-term loans provided by financial institutions.
Thus capital structure = Equity share capital + preference share capital + reserves + debentures.
Thus, it is rightly said that capital structure is composed of owned funds and borrowed funds.

Question 5.
There are various factors affecting the requirement of working capital.
Answer:

The nature and size of a business affect the requirement of working capital. Trading or merchandising firms and big retail enterprises need a large amount of capital compared to small firms which need a small amount of working capital.
If the period of the production cycle is longer then the firm needs more amount of working capital. If the manufacturing cycle is short, it requires less working capital.
During the boom period sales will increase leading to increased investment in stocks, thus requiring additional working capital and during the recession, it is vice versa.
Along with the expansion and growth of the firm or company in terms of sales and fixed assets, the requirement of working capital increases.
If there is proper coordination, communication, and co-operation between production and sales departments then the requirement of working capital is less.
A liberal credit policy increases the possibility of bad debts and in such cases, the requirement of working capital is high, whereas a firm making cash sales requires less working capital.

7. Answer the following questions.

Question 1.
Discuss the importance of Corporate Finance?
Answer:
Corporate finance deals with the raising and utilizing of finance by a corporation. It also deals with capital structuring and making investment decisions, financial planning of capital, and the money market. The finance manager should ensure that:
The firm has adequate finance and it’s being utilized effectively;
Generate minimum return for its owners.

The importance of Corporate Finance are as follows:
(i) Helps in decision making:
Most important decisions of business enterprises are made on the basis of availability of funds, as without finance any function of business enterprise is difficult to be performed independently. Obtaining the funds from the right sources at a lower cost and productive utilization of funds would lead to higher profits. Thus corporate finance plays a significant role in the decision-making process.

(ii) Helps in raising capital for a project:
A new business venture needs to raise capital. Business firms can raise funds by issuing shares, debentures, bonds or even by taking loans from the banks.

(iii) Helps in Research and Development
Research and Development need to be undertaken by firms for growth and expansion of business and to enjoy a competitive advantage. Research and development mostly involve lengthy and detailed technical work for the execution of projects. Through surveys and market analysis etc. companies may have to upgrade old products or develop new products to face competition and attract consumers. Thus the availability of adequate finance helps to generate high efficiency.

(iv) Helps in the smooth running of the business firm:
A smooth flow of corporate finance is important to pay the salaries of employees on time, pay loans, and purchase the required raw materials. At the same time finance is needed for sales promotion of existing products and more so for the launch of new products effectively.

(v) Brings co-ordination between various activities:
Corporate finance plays a significant role in the coordination and control of all activities in an organization. Production activity requires adequate finance for the purchase of raw materials and meeting other day-to-day financial requirements for the smooth running of the production unit. If the production increases, sales will also increase by contributing the income of the concern and profit to increase.

(vi) Promotes expansion and diversification:
Corporate finance provides money for the purchase of modern machines and sophisticated technology. Modern machines and technology help to improve the performance of the firm in terms of profits. It also helps the firm to expand and diversify the business.

(vii) Managing risk:
Companies have to manage several risks such as sudden fall in sales, loss due to natural calamity, loss due to workers strikes, change in government policies, etc. Financial aids help in such situations to manage such risks.

(viii) Replace old assets:
Assets like plants and machinery have become old and outdated over the years. Finance is required to purchase new assets or replace the old assets with new assets having new technology and features.

(ix) Payment of dividend and interest:
Finance is needed to pay the dividend to shareholders, interest to creditors, bank, etc.

(x) Payment of taxes/fees:
The company has to pay taxes to the government such as Income tax, Goods and Service Tax (GST), and fees to the Registrar of Companies on various occasions. Finance is needed for paying these taxes and fees.

Question 2.
Discuss the factors determining working capital requirements?
Answer:
Working Capital = Current Assets – Current Liabilities.
In other words, it is also called ‘Circulating Capital’. Also, refer to ‘GROSS WORKING CAPITAL.’ Management needs to determine the size of working capital with reference to the economic environment and other aspects within the business firm.

Factors determining/influencing working capital requirements are as follows:
(i) Nature of Business:
The working capital requirements are highly influenced by the nature of the business. Trading/ merchandising forms concerned with the distribution of goods require a huge amount of working capital to maintain a large stock of the variety of goods to meet customers’ demands are extend credit facilities to attract them. Whereas public utility concerns have to maintain small working capital because of a continuous flow of cash from their customers.

(ii) Size of business:
The size of a business also affects the requirements of working capital. Size of the firm refers to the scale of operation i.e. a firm with large scale operations will require more working capital and vice versa.

(iii) Volume of Sales:
The volume of sales and the size of the working capital have a direct relationship with each other. If the volume of sales increases there is an increase in the amount of working capital.

(iv) Production cycle:
The process of converting raw material into finished goods is called the ‘production cycle’. If the production cycle period is longer, the firm needs more amount of working capital. If the manufacturing cycle is short, it requires less working capital.

(v) Business cycle:
When there is a boom in the economy, sales will increase resulting in to increase in investment in stock. This will require additional working capital. During a recession period, sales will decline and consequently, the need for working capital will also decrease.

(vi) Terms of purchases and sales:
If credit terms of purchase are favourable and terms of sales are less liberal, then the requirement of cash will be less. Thus, the working capital requirement will be reduced.
A firm that enjoys more credit facilities needs less working capital. On the other hand, if a firm does not get proper credit for purchases and adopts a liberal credit policy for sales if requires more working capital.

(vii) Credit Control:
Credit control includes the factors such as volume of credit sales, the terms of credit sales, the collection policy etc. A firm with a good credit control policy will have more cash flow reducing the working capital requirement. Whereas if the firm’s credit policy is liberal there would be more requirements of the working capital.

(viii) Growth and Expansion:
Those firms which are growing and expanding at a rapid pace need more working capital compared to those firms which are stable in their growth.

(ix) Management ability:
The requirement of working capital is reduced if there is proper coordination in the production and distribution of goods. A firm stocking on heavy inventory calls for a higher level of working capital.

(x) External factors:
If the financial institutions and banks provide funds to the firm as and when required, the need for working capital is reduced.

Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits

August 31, 2021August 18, 2021 by Prasanna

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 13 AC Circuits Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 13 AC Circuits

1. Choose the correct option.

i) If the RMS current in a 50 Hz AC circuit is 5A, the value of the current \(\frac{1}{300}\) seconds after its value becomes zero is
(A) 5\(\sqrt {2}\) A
(B) 5\(\sqrt{\frac{3}{2}}\) A
(C) \(\frac{5}{6}\) A
(D) \(\frac{5}{\sqrt{2}}\) A
Answer:
(B) 5\(\sqrt{\frac{3}{2}}\) A

ii) A resistor of 500 Ω and an inductance of 0.5 H are in series with an AC source which is given by V = 100 \(\sqrt {2}\) sin (1000t). The power factor of the combination
(A) \(\frac{1}{\sqrt{2}}\)
(B) \(\frac{1}{\sqrt{3}}\)
(C) 0.5
(D) 0.6
Answer:
(A) \(\frac{1}{\sqrt{2}}\)

iii) In a circuit L, C & R are connected in series with an alternating voltage of frequency f. the current leads the voltage by 450. The value of C is
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 1
Answer:
(B) \(\frac{1}{2 \pi f(2 \pi f L-R)}\)

iv) In an AC circuit, e and i are given by e = 150 sin (150t) V and i = 150 sin (150 t + \(\frac{\pi}{3}\)) A. the power dissipated in the circuit is
(A) 106W
(B) 150W
(C) 5625W
(D) Zero
Answer:
(C) 5625W

v) In a series LCR circuit the phase difference between the voltage and the current is 45°. Then the power factor will be
(A) 0.607
(B) 0.707
(C) 0.808
(D) 1
Answer:
(B) 0.707

2. Answer in brief.

i) An electric lamp is connected in series with a capacitor and an AC source is glowing with a certain brightness. How does the brightness of the lamp change on increasing the capacitance ?
Answer:
Impedance, Z = \(\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}}\), where R is the resistance of the lamp, w is the angular frequency of AC and C is the capacitance of the capacitor connected in series with the AC source and the lamp. When C is increased, \(\) decreases. Hence, Z increases.
Power factor, cos Φ = \(\frac{R}{Z}\)
As Z increases, the power factor decreases.
Now, the average power over one cycle,
P_av = v_rms i_rms cos Φ
= V_rms \(\left(\frac{V_{\mathrm{rms}}}{Z}\right)\) cos Φ
= \(\frac{V_{\mathrm{rms}}^{2}}{\mathrm{Z}} \cos \phi\)
∴ P_av decreases as Z increases and cos Φ decreases.
As the current through the lamp \(\left(\frac{V_{\mathrm{rms}}}{Z}\right)\) decreases, the brightness of the lamp will decrease when C is increased.

ii) The total impedance of a circuit decreases when a capacitor is added in series with L and R. Explain why ?
Answer:
For an LR circuit, the impedance,
Z_LR = \(\sqrt{R^{2}+X_{\mathrm{L}}^{2}}\), where X_L is the reactance of the inductor.
When a capacitor of capacitance C is added in series with L and R, the impedance,
Z_LCR = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\) because in the case of an inductor the current lags behind the voltage by a phase angle of \(\frac{\pi}{2}\) rad while in the case of a capacitor the current leads the voltage by a phase angle of \(\frac{\pi}{2}\) rad. The decrease in net reactance decreases the total impedance (Z_LCR < Z_LR).

iii) For very high frequency AC supply, a capacitor behaves like a pure conductor. Why ?
Answer:
The reactance of a capacitor is X_C = \(\frac{1}{2 \pi f C}\), where f is the frequency of the AC supply and C is the capacitance of the capacitor. For very high frequency, f, X_C is very small. Hence, for very high frequency AC supply, a capacitor behaves like a pure conductor.

iv) What is wattless current ?
Answer:
The current that does not lead to energy consumption, hence zero power consumption, is called wattless current.
In the case of a purely inductive circuit or a purely capacitive circuit, average power consumed over a complete cycle is zero and hence the corresponding alternating current in the circuit is called wattless current.
[Note : In this case, the power factor is zero.]

v) What is the natural frequency of L C circuit ? What is the reactance of this circuit at this frequency
Answer:
The natural frequency of LC circuit is \(\frac{1}{2 \pi \sqrt{L C}}\) ,
where L is the inductance and C is the capacitance. The reactance of this circuit at this frequency is
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 18

Question 3.
In a series LR circuit X_L = R and power factor of the circuit is P₁. When capacitor with capacitance C such that X_L = X_C is put in series, the power factor becomes P₂. Calculate P₁ / P₂ .
Answer:
For a series LR circuit, power factor,
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 17

Question 4.
When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero.
Answer:
In an AC circuit containing only an ideal inductor, the current i lags behind the emf e by a phase angle of \(\frac{\pi}{2}\) rad. Here, for e = e₀ sin ωt, we have, i = i₀ sin(ωt – \(\frac{\pi}{2}\))
Instantaneous power, P = ei
= (e₀ sin ωt) [i₀ (sin ωt cos \(\frac{\pi}{2}\) – cos ωt sin \(\frac{\pi}{2}\))]
= – e₀i₀ sin ωt cos ωt as cos \(\frac{\pi}{2}\) = 0 and sin \(\frac{\pi}{2}\) = 1.
Average power over one cycle,
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 15

= e_rms i_rms cos Φ = e_rms i_rms (\(\frac{R}{Z}\)), where the impedance Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\).
∴ P_av = 0, i.e., the circuit does not dissipate power.

Question 5.
Prove that an ideal capacitor in an AC circuit does not dissipate power
Answer:
In an AC circuit containing only an ideal inductor, the current i lags behind the emf e by a phase angle of \(\frac{\pi}{2}\) rad. Here, for e = e₀ sin ωt, we have, i = i₀ sin(ωt – \(\frac{\pi}{2}\))
Instantaneous power, P = ei
= (e₀ sin ωt) [i₀ (sin ωt cos \(\frac{\pi}{2}\) + cos ωt sin \(\frac{\pi}{2}\))]
= – e₀i₀ sin ωt cos ωt as cos \(\frac{\pi}{2}\) = 0 and sin \(\frac{\pi}{2}\) = 1.
Average power over one cycle, P_av
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 15

= e_rms i_rms cos Φ = e_rms i_rms (\(\frac{R}{Z}\)), where the impedance Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\).
∴ P_av = 0, i.e., the circuit does not dissipate power.

Question 6.
(a) An emf e = e₀ sin ωt applied to a series L – C – R circuit derives a current I = I₀ sinωt in the circuit. Deduce the expression for the average power dissipated in the circuit.
(b) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.
Answer:
(a) Instantaneous power,
P = ei
= (e₀ sin ωt) [i₀ (sin ωt ± Φ)]
= e₀i₀ sin ωt(sin ωt cos Φ ± cos ωt sin Φ)
= e₀i₀ sin² ωt ± e₀i₀ sin Φ sin ωt cos ωt
Average power over one cycle,
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 15

= e_rms i_rms cos Φ = e_rms i_rms (\(\frac{R}{Z}\)), where the impedance Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\).

(b) P_av = e_rms i_rms cos Φ
The factor cos Φ is called as power factor. For circuits used for transporting electric power, a low power factor means the power available on transportation is much less than e_rms i_rms It means there is significant loss of power during transportation.

Question 7.
A device Y is connected across an AC source of emf e = e₀ sinωt. The current through Y is given as i = i₀ sin(ωt + π/2)
a) Identify the device Y and write the expression for its reactance.
b) Draw graphs showing variation of emf and current with time over one cycle of AC for Y.
c) How does the reactance of the device Y vary with the frequency of the AC ? Show graphically
d) Draw the phasor diagram for the device Y.
Answer:
(a) The device Y is a capacitor. Its reactance is X_c = \(\frac{1}{\omega C}\),
where ω is the angular frequency of the applied emf and C is the capacitance of the capacitor.

(b)
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 6

(c) X_C = \(\frac{1}{\omega C}=\frac{1}{2 \pi f C}\). Thus X_C ∝ \(\frac{1}{f}\), where f is the frequency of AC. Suppose C = \(\left(\frac{1000}{2 \pi}\right)\) pF
For f= 100 Hz, X_C = 1 × 10⁷Ω = 10MΩ;
for f = 200 Hz, X_C = 5 MΩ;
for f = 300 Hz, X_C = \(\frac{10}{3}\) MΩ;
for f = 400 Hz, X_C = 2.5 MΩ
for f = 500 Hz, X_C = 2 MΩ and so on
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 8

(d)
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 9
The phasor representing the peak emf (e₀) makes an angle (ωt) in an anticlockwise direction with respect to the horizontal axis. As the current leads the voltage by 90°, the phasor representing the peak current (i₀) is turned 90° anticlockwise with respect to the phasor representing emf e₀. The projections of these phasors on the vertical axis give instantaneous values of e and i.

Question 8.
Derive an expression for the impedance of an LCR circuit connected to an AC power supply.
Answer:
Figure shows an inductor of inductance L, capacitor of capacitance C, resistor of resistance R, key K and source (power supply) of alternating emf (e) connected to form a closed series circuit.
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 10
We assume the inductor, capacitor and resistor to be ideal. As these are connected in series, at any instant, they carry the same current i = i₀ sin ωt. The voltage across the resistor, e_R = Ri, is in phase with the current. The voltage across the inductor, e_L = X_Li, leads the current by \(\frac{\pi}{2}\) rad and that across the capacitor, e_C = X_Ci, lags behind the current by \(\frac{\pi}{2}\) rad. This is shown in the phasor diagram.
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 11
is the effective resistance of the circuit. It is called the impedance.

Question 9.
Compare resistance and reactance.
Answer:
(1) Resistance is opposition to flow of charges (current) and appears in a DC circuit as well as in an AC circuit.
The term reactance appears only in an AC circuit. It occurs when an inductor and/or a capacitor is used.

(2) In a purely resistive circuit, current and voltage are always in phase.
When reactance is not zero, there is nonzero phase difference between current and voltage.

(3) Resistance does not depend on the frequency of AC.
Reactance depends on the frequency of AC. In case of an inductor, reactance increases linearly with frequency. In case of a capacitor, reactance decreases as frequency of AC increases; it is inversely proportional to frequency.

(4) Resistance gives rise to production of Joule heat in a component.
In a circuit with pure reactance, there is no production of heat.

Question 10.
Show that in an AC circuit containing a pure inductor, the voltage is ahead of current by π/2 in phase.
Answer:
Figure 13.8 shows an AC source, generating a voltage e = e₀ sin ωt, connected to a key K and a pure inductor of inductance L to form a closed circuit.
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 2
On closing the key K, an emf is induced in the inductor as the magnetic flux linked with it changes with time. This emf opposes the applied emf and according to the laws of electromagnetic induction by Faraday and Lenz, we have,
e’ = -L\(\frac{d i}{d t}\) ………………. (1)
where e’ is the induced emf and i is the current through the inductor. To maintain the current; e and e’ must be equal in magnitude and opposite in direction.

According to Kirchhoff’s voltage law, as the resistance of the inductor is assumed to be zero, we
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 3
where C is the constant of integration. C must be time independent and have the dimension of current. As e oscillates about zero, i also oscillates about zero and hence there cannot be any time independent component of current.
∴ C = 0. ∴ i = –\(\frac{e_{0}}{\omega L}\)cos ωt = – \(\frac{e_{0}}{\omega L}\)sin(\(\frac{\pi}{2}\) – ωt)
∴ i = \(\frac{e_{0}}{\omega L}\) sin(ωt – \(\frac{\pi}{2}\)) ……………. (3)
as sin (-θ) = – sin θ
From Eq. (3), i_peak = i₀ = \(\frac{e_{0}}{\omega L}\)
∴ i = i₀ sin(ωt – \(\frac{\pi}{2}\)) ………………. (4)
Comparison of this equation with e = e₀ sin ωt shows that e leads i by \(\frac{\pi}{2}\) rad, i.e., the voltage is ahead of current by \(\frac{\pi}{2}\) rad in phase.

Question 11.
An AC source generating a voltage e = e₀ sinωt is connected to a capacitor of capacitance C. Find the expression for the current i flowing through it. Plot a graph of e and i versus ωt.
Answer:
Figure 13.12 shows an AC source, generating a voltage e = e₀ sin ωt, connected to a capacitor of capacitance C. The plates of the capacitor get charged due to the applied voltage. As the alternating voltage is reversed in each half cycle, the
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 4
capacitor is alternately charged and discharged. If q is the charge on the capacitor, the corresponding potential difference across the plates of the capacitor is V = \(\frac{q}{C}\) ∴ q = CV. q and V are functions of time, with V = e = e₀ sin ωt. The instantaneous current in the circuit is i = \(\frac{d q}{d t}=\frac{d}{d t}\)(CV) = C \(\frac{d v}{d t}\) = C \(\frac{d}{d t}\) (e₀ sin ωt) = ωC e₀ cos ωt
∴ i = \(\frac{e_{0}}{(1 / \omega C)} \sin \left(\omega t+\frac{\pi}{2}\right)=i_{0} \sin \left(\omega t+\frac{\pi}{2}\right)\)
where i₀ = \(\frac{e_{0}}{(1 / \omega C)}\) is the peak value of the current.
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 5
Table gives the values of e and i for different values of cot and Fig shows graphs of e and i versus ωt. i leads e by phase angle of \(\frac{\pi}{2}\) rad.

Question 12.
If the effective current in a 50 cycle AC circuit is 5 A, what is the peak value of current? What is the current 1/600 sec.after if was zero ?
Answer:
Data : f = 50 Hz, i_rms = 5 A, t = \(\frac{1}{600}\) s
The peak value of the current,
i₀ = i_rms\(\sqrt {2}\) = (5)(1.414) = 7.07 A
= i₀sin (2πft)
= 7.07 sin [2π(5o) (\(\frac{1}{600}\))]
= 7.07 sin (\(\frac{\pi}{6}\)) = (7.07)(0.5)
= 3.535 A
This is the required current.

Question 13.
A light bulb is rated 100W for 220 V AC supply of 50 Hz. Calculate (a) resistance of the bulb. (b) the rms current through the bulb.
Answer:
Data: Power (V_rms i_rms) = 100 W, V_rms = 220V,
f = 50 Hz
The rms current through the bulb,
i_rms = \(\frac{\text { power }}{V_{\mathrm{rms}}}=\frac{100}{220}\) = 0.4545 A
The resistance of the bulb,
R = \(\frac{V_{\mathrm{rms}}}{i_{\mathrm{rms}}}=\frac{220}{(100 / 220)}\) = (22) (22) = 484 Ω

Question 14.
A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak)
in the circuit. If the frequency is doubled, what will happen to the capacitive reactance and the current.
Answer:
Data : C = 15 µF = 15 × 10-6 F, V_rms = 220V, f = 50 Hz,
The capacitive reactance = \(\frac{1}{2 \pi f C}\)
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 12
If the frequency is doubled, the capacitive reactance will be halved and the current will be doubled.

Question 15.
An AC circuit consists of only an inductor of inductance 2 H. If the current is represented by a sine wave of amplitude 0.25 A and frequency 60 Hz, calculate the effective potential difference across the inductor (π = 3.142)
Answer:
Data : L = 2H, i₀ = 0.25 A, f = 60 Hz, π = 3.142
ωL = 2πfL = 2(3.142)(60)(2) = 754.1 Ω
The effective potential difference across the inductor = ωLi_rms = ωL \(\frac{i_{0}}{\sqrt{2}}\)
= \(\frac{(754.1)(0.25)}{1.414}\) = 133.3 V

Question 16.
Alternating emf of e = 220 sin 100 πt is applied to a circuit containing an inductance of (1/π) henry. Write an equation for instantaneous current through the circuit. What will be the reading of the AC galvanometer connected in the circuit?
Answer:
Data: e = 220 sin 100 πt, L = (\(\frac{1}{\pi}\))H
Comparing e = 220 sin 100 πt with
e = e₀ sin ωt, we get
ω = 100 π ∴ ωL = (100 π) (\(\frac{1}{\pi}\)) = 100 Ω
∴ The instantaneous current through the circuit
= i = \(\frac{e_{0}}{\omega L}\) sin(100 πt – \(\frac{\pi}{2}\))
= \(\frac{220}{100}\) sin (100 πt – \(\frac{\pi}{2}\)) = 2.2 sin (100 πt – \(\frac{\pi}{2}\)) in ampere [assuming that e is in volt.]
i_rms = \(\frac{i_{0}}{\sqrt{2}}=\frac{2.2}{1.414}\) = 1.556 A is the reading of the AC galvanometer connected in the circuit.

Question 17.
A 25 µF capacitor, a 0.10 H inductor and a 25Ω resistor are connected in series with an AC source whose emf is given by e = 310 sin 314 t (volt). What is the frequency, reactance, impedance, current and phase angle of the circuit?
Answer:
Data: C = 25 µF = 25 × 10^-6F, L = 0.10H, R = 25 Ω ,
e = 310 sin (314 t) [volt]
Comparing e = 310 sin (314 t) with
e = e₀ sin (2πft), we get,
the frequency of the alternating emf as
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 13
cos Φ = \(\frac{R}{Z}=\frac{25}{99.2}\) = 0.2520
∴ The phase angle, Φ = cos^-1(0.2520) = 75.40° = 1.316 rad

Question 18.
A capacitor of 100 µF, a coil of resistance 50Ω and an inductance 0.5 H are connected in series with a 110 V-50Hz source. Calculate the rms value of current in the circuit.
Answer:
Data : C = 100 µF = 100 × 10^-6 F = 10^-4 F,
R = 50 Ω, L = 0.5H, f = 50 Hz, V_rms = 110 V
∴ ωL = 2πfL = 2 (3.142)(50)(0.5) = 157.1 Ω
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 14
2500 + 15700 = 18200 Ω²
∴ Impedance, Z = \(\sqrt {18200}\) Ω = 134.9 Ω
The rms value of the current in the circuit,
i_rms = \(\frac{V_{\mathrm{rms}}}{\mathrm{Z}}=\frac{110}{134.9} \mathrm{~A}\)
= 0.8154 A

Question 19.
Find the capacity of a capacitor which when put in series with a 10Ω resistor makes the power factor equal to 0.5. Assume an 80V-100Hz AC supply.
Answer:
Data : R = 10 Ω, power factor = 0.5, f = 100 Hz
Power factor = \(\frac{1}{2 \pi f C R}\)
∴ 0.5 = \(\frac{1}{2(3.142)(100) C(10)}\)
∴ C = \(\frac{1}{3.142 \times 10^{3}}\)
= \(\frac{10 \times 10^{-4}}{3.142}\)
= 3.182 × 10^-4 F
This is the capacity of the capacitor.

Question 20.
Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.
Answer:
Data : f = 50 Hz, i = \(\frac{i_{0}}{\sqrt{2}}\) ∴ \(\frac{i}{i_{0}}=\frac{1}{\sqrt{2}}\)
i = i₀ sinωt
∴ sinωt = \(\frac{i}{i_{0}}=\frac{1}{\sqrt{2}}\)
∴ ωt = \(\frac{\pi}{4}\) rad
∴ 2πft = \(\frac{\pi}{4}\)
∴ t = \(\frac{1}{8 f}=\frac{1}{8(50)}=\frac{1}{400}\)
= \(\frac{1000 \times 10^{-3}}{400}\) = 2.5 × 10^-3 s
This is the required time.

Question 21.
Calculate the value of capacitance in picofarad, which will make 101.4 micro henry inductance to oscillate with frequency of one megahertz.
Answer:
Data : f_r = 10⁶ Hz, L = 101.4 × 10^-6 H
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 19
= \(\frac{10000 \times 10^{-10}}{4(3.142)^{2}(101.4)}\) = 2.497 × 10^-10 F
= 249.7 × 10^-12 F = 249.7 picofarad
This is the value of the capacity.

Question 22.
A 10 µF capacitor is charged to a 25 volt of potential. The battery is disconnected and a pure 100 m H coil is connected across the capacitor so that LC oscillations are set up. Calculate the maximum current in the coil.
Answer:
Data: C = 10 µF = 10 × 10^-6F = 10^-5F,
L = 100mH = 100 × 10^-3 H = 10^-1 H, V = 25V
For reference, see the solved example (8) above.
\(\frac{1}{2}\)CV² = \(\frac{1}{2}\)Li²
∴i² = \(\frac{C}{L} V^{2}=\frac{10^{-5}}{10^{-1}}(25)^{2}\)
∴i = 25 × 10^-2 A = 0.25 A
This is the maximum current in the coil.

Question 23.
A 100 µF capacitor is charged with a 50 V source supply. Then source supply is removed and the capacitor is connected across an inductance, as a result of which 5A current flows through the inductance. Calculate the value of the inductance.
Answer:
Data: C = 100 µF = 100 × 10^-6 F = 10^-4 F,
V = 50V, i = 5A
The energy stored in the electric field in the capacitor
= \(\frac{1}{2}\)CV²
The energy stored in the magnetic field in the inductor = \(\frac{1}{2}\)Li²
Here, \(\frac{1}{2}\)CV² = \(\frac{1}{2}\)Li²
∴ L = C\(\frac{V^{2}}{i^{2}}\)
∴ L = C\(\left(\frac{V}{i}\right)^{2}=10^{-4}\left(\frac{50}{5}\right)^{2}\) = 10^-4 × 10²
= 10^-2H
This is the value of the inductance.