Bhagya

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 1 Some Basic Concepts of Chemistry Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 1 Some Basic Concepts of Chemistry

Question 1.
Define chemistry.
Answer:
Chemistry is the study of matter, its physical and chemical properties, and the physical and chemical changes it undergoes under different conditions.

Question 2.
Why is chemistry called a central science?
Answer:

1. Knowledge of chemistry is required in the studies of physics, biological sciences, applied sciences, and earth and space sciences.
2. Chemistry is involved in every aspect of day-to-day life, i.e. the air we breathe, the food we eat, the fluids we drink, our clothing, transportation and fuel supplies, etc.

Hence, chemistry is called a central science.

Question 3.
Give reason: Although chemistry has ancient roots, it has developed as modern science.
Answer:
Technological development in sophisticated instruments has expanded knowledge of chemistry which, now, has been used in applied sciences such as medicine, dentistry, engineering, agriculture, and daily home use products. Hence, due to development and advancement in science and technology, chemistry has developed as modem science.

Question 4.
How is chemistry traditionally classified?
Answer:
Chemistry is traditionally classified into five branches:

• Organic chemistry
• Inorganic chemistry
• Physical chemistry
• Biochemistry
• Analytical chemistry

Question 5.
Explain the following terms:
i. Organic chemistry
ii. Inorganic chemistry
iii. Physical chemistry
Answer:
i. Organic chemistry: It deals with properties and reactions of compounds of carbon.
ii. Inorganic chemistry: It deals with the study of all the compounds which are not organic.
iii. Physical chemistry: It deals with the study of properties of matter, the energy changes and the theories, laws and principles that explain the transformation of matter from one form to another. It also provides basic framework for all the other branches of chemistry.

Question 6.
Distinguish between
i. Mixtures and pure substances
ii. Mixtures and compounds
Answer:
i.

ii.

Question 7.
What is the difference between element and compound?
Answer:
Elements cannot be broken down into simpler substances while compounds can be broken down into simpler substances by chemical changes.

Question 8.
Explain: States of matter
Answer:
There are three different states of matter as follows:

1. Solid: Particles are held tightly in perfect order. They have definite shape and volume.
2. Liquid: Particles are close to each other but can move around within the liquid.
3. Gas: Particles are far apart as compared to that of solid and liquid.

These three states of matter can be interconverted by changing the conditions of temperature and pressure.

Question 9.
Explain: Physical and chemical properties
Answer:
i. Physical properties: These are properties which can be measured or observed without changing the identity or the composition of the substance. e.g. Colour, odour, melting point, boiling point, density, etc.

ii. Chemical properties: These are properties in which substances undergo change in chemical composition. e.g. Coal bums in air to produce carbon dioxide, magnesium wire bums in air in the presence of oxygen to form magnesium oxide, etc.

Question 10.
How are properties of matter measured?
Answer:

• Measurement involves comparing a property of matter with some fixed standard which is reproducible and unchanging.
• Properties such as mass, length, area, volume, time, etc. are quantitative in nature and can be measured.
• A quantitative measurement is represented by a number followed by units in which it is measured.
• These units are arbitrarily chosen on the basis of universally accepted standards. e.g. Length of class room can be expressed as 10 m. Here, 10 is the number and ‘m’ is the unit ‘metre’ in which the length is measured.

Question 11.
Define: Units
Answer:
The arbitrarily decided and universally accepted standards are called units.
e.g. Metre (m), kilogram (kg).

Question 12.
What are the various systems in which units are expressed?
Answer:
Units are expressed in various systems like CGS (centimetre for length, gram for mass and second for time), FPS (foot, pound, second) and MKS (metre, kilogram, second) systems, etc.

Question 13.
What are SI units? Name the fundamental SI units.
Answer:
SI Units: In 1960, the general conference of weights and measures proposed revised metric system, called International system of Units i.e. SI system (abbreviated from its French name).
The seven fundamental SI units are as given below:

[Note: Units for other quantities such as speed, volume, density, etc. can be derived from fundamental SI units.]

Question 14.
What is the basic unit of mass in the SI system?
Answer:
The basic unit of mass in the SI system is kilogram (kg).

Question 15.
Name the following:
i. Full form of CGS unit system
ii. Full form of FPS unit system
iii. The SI unit of length
iv. Symbol used for Candela unit
v. SI unit of electric current
vi. SI unit of electric current
Answer:
i. Centimetre Gram Second
ii. Foot Pound Second
iii. Metre (m)
iv. Cd
v. Kelvin (K)
vi. Ampere (A)

Question 16.
Give reason: The mass of a body is more fundamental property than its weight.
Answer:

• Mass is an inherent property of matter and is the measure of the quantity of matter of a body.
• The mass of a body does not vary with respect to its position.
• On the other hand, the weight of a body is a result of the mass and gravitational attraction
• Weight varies because the gravitational attraction of the earth for a body varies with the distance from the centre of the earth.

Hence, the mass of a body is more fundamental property than its weight.

Question 17.
How is gram related to the SI unit kilogram?
Answer:
The SI unit kilogram (kg) is related to gram (g) as 1 kg = 1000 g= 10³ g.
[Note: ‘Gram’ is used for weighing small quantities of chemicals in the laboratories.
Other commonly used quantity is ‘milligram’. 1 kg = 1000 g = 10⁶ mg]

Question 18.
Why are fractional units of the SI units of length often used? Give two examples of the fractional units of length. How are they related to the SI unit of length?
Answer:
i. Some properties such as the atomic radius, bond length, wavelength of electromagnetic radiation, etc. are very small and therefore, fractional units of the SI unit of length are often used to express these properties.
ii. Fractional units of length: Nanometre (nm), picometre (pm), etc.
iii. Nanometre (nm) and picometre (pm) are related to the SI unit of length (m) as follows:
1 nm = 10^-9 m, 1 pm = 10^-12 m

Question 19.
Define: Volume
Answer:
Volume is the amount of space occupied by a three-dimensional object. It does not depend on shape.

Question 20.
State the common unit used for the measurement of volume of liquids and gases.
Answer:
The common unit used for the measurement of volume of liquids and gases is litre (L).

Question 21.
How is the SI unit of volume expressed?
Answer:
The SI unit of volume is expressed as (metre)³ or m³.

Question 22.
Name some glassware that are used to measure the volume of liquids and solutions.
Answer:

• Graduated cylinder
• Burette
• Pipette

Question 23.
What is a volumetric flask used for in laboratory?
Answer:
A volumetric flask is used to prepare a known volume of a solution in laboratory.

Question 24.
What is density of a substance? How is it measured?
Answer:
Density:

• Density of a substance is its mass per unit volume. It is the characteristic property of any substance.
• It is determined in the laboratory by measuring both the mass and the volume of a sample.
• The density is calculated by dividing mass by volume.

Question 25.
How is the SI unit of density derived? State CGS unit of density.
Answer:
i. The SI unit of density is derived as follows:
Density = \(\frac{\text { SI unit mass }}{\text { SI unit volume }}\)
= \(\frac{\mathrm{kg}}{\mathrm{m}^{3}}\)
= kg m^-3

ii. CGS unit of density: g cm^-3
[Note: The CGS unit, g cm^-3 is equivalent to \(\frac{\mathrm{g}}{\mathrm{mL}}\) or g mL^-1.]

Question 26.
State three common scales of temperature measurement.
Answer:

1. Degree Celsius (°C)
2. Degree Fahrenheit (°F)
3. Kelvin (K)

Question 27.
State the temperatures in Fahrenheit scale that corresponds to 0 °C and 100 °C.
Answer:
The temperature that corresponds to 0 °C is 32 °F and the temperature that corresponds to 100 °C is 212 °F.

Question 28.
Write the expression showing the relationship between:
i. Degree Fahrenheit and Degree Celsius
ii. Kelvin and Degree Celsius
Answer:
i. The relationship between degree Fahrenheit and degree Celsius is expressed as,
°F = \(\frac {9}{5}\) (°C) + 32
ii. The relationship between Kelvin and degree Celsius is expressed as,
K = °C + 273.15

Question 29.
Convert the following degree Fahrenheit temperature to degree Celsius.
i. 50 °F ii. 10 °F
Answer:
Given: Temperature in degree Fahrenheit = 50 °F
To find: Temperature in degree Celsius
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 50 °F in the formula,
°F = \(\frac {9}{5}\) (°C) + 32
50 = \(\frac {9}{5}\) (°C) + 32
°C = \(\frac{(50-32) \times 5}{9}\)
= 10 °C

ii. Given: Temperature in degree Fahrenheit = 10 °F
To find: Temperature in degree Celsius
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 10 °F in the formula,
°F = \(\frac {9}{5}\) (°C) + 32
10 = \(\frac {9}{5}\) (°C) + 32
°C = \(\frac{(10-32) \times 5}{9}\)
= -12.2 °C
Ans: i. The temperature 50 °F corresponds to 10 °C.
ii. The temperature 10 °F corresponds to -12.2 °C.

Question 30.
What is a chemical combination?
Answer:

• The process in which the elements combine with each other to form compounds is called chemical combination.
• The process of chemical combination is governed by five basic laws which were discovered before the knowledge of molecular formulae.

Question 31.
State and explain the law of definite proportions.
Answer:
Law of definite proportions:
i. The law states that “A given compound always contains exactly the same proportion of elements by weight”.
ii. French chemist, Joseph Proust worked with two samples of cupric carbonate; one of which was naturally occurring cupric carbonate and other was synthetic sample. He found the composition of elements present in both the samples was same as shown below:

iii. Thus, irrespective of the source, a given compound always contains same elements in the same proportion.

Question 32.
State and explain the law of multiple proportions.
Answer:
Law of multiple proportions:
i. John Dalton (British scientist) proposed the law of multiple proportions in 1803.
ii. It has been observed that two or more elements may combine to form more than one compound.
iii. The law states that, “ When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers”.
e.g. Hydrogen and oxygen combine to form two compounds, water and hydrogen peroxide.

Here, the two masses of oxygen (16 g and 32 g) which combine with the fixed mass of hydrogen (2 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 16 : 32 or 1 : 2.

Question 33.
Show that NO and NO₂ satisfy the law of multiple proportions.
Answer:
Nitrogen and oxygen combine to form two compounds, nitric oxide (NO) and nitrogen dioxide (NO₂).

Here, the two masses of oxygen (16 g and 32 g) which combine with the fixed mass of nitrogen (14 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 16 : 32 or 1 : 2.
This is in accordance with the law of multiple proportions.

Question 34.
Show that carbon monoxide and carbon dioxide satisfy the law of multiple proportions.
Answer:
Chemical reaction of carbon with oxygen gives two compounds, carbon monoxide (CO) and carbon dioxide (CO₂).

Here, the two masses of oxygen (16 g and 32 g) which combine with the fixed mass of carbon (12 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 16 : 32 or 1 : 2.
This is in accordance with the law of multiple proportions.

Question 35.
Show that SO₂ and SO₃ satisfy the law of multiple proportions.
Answer:
Chemical reaction of sulphur with oxygen gives two compounds, sulphur dioxide (SO₂) and sulphur trioxide (SO₃).

Here, the two masses of oxygen (32 g and 48 g) which combine with the fixed mass of sulphur (32 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 32 : 48 or 2 : 3.
This is in accordance with the law of multiple proportions.

Question 36.
State and explain Gay Lussac’s law of gaseous volume with two examples.
Answer:
Gay Lussac’s law:
i. Gay Lussac proposed the law of gaseous volume in 1808.
ii. Gay Lussac’s law states that, “ When gases combine or are produced in a chemical reaction, they do so in a simple ratio by volume, provided all gases are at same temperature and pressure
e.g. a. Under identical conditions of temperature and pressure, 100 mL of hydrogen gas combine with 50 mL of oxygen gas to produce 100 mL of water vapour.

Thus, the simple ratio of volumes is 2 : 1 : 2.

b. Under identical conditions of temperature and pressure, 1 L of nitrogen gas combine with 3 L of hydrogen gas to produce 2 L of ammonia gas.

Thus, the simple ratio of volumes is 1 : 3 : 2.

Question 37.
Give two examples which support the Gay Lussac’s law of gaseous volume.
Answer:
i. Under identical conditions of temperature and pressure, 1 L of hydrogen gas reacts with 1 L of chlorine gas to produce 2 L of hydrogen chloride gas.

Thus, the ratio of volumes is 1 : 1 : 2
This is in accordance with Gay Lussac’s law.

ii. Under identical conditions of temperature and pressure, 200 mL sulphur dioxide combine with 100 mL oxygen to form 200 mL sulphur trioxide.

Thus, the ratio of volumes is 2 : 1 : 2.
This is in accordance with Gay Lussac’s law.

Question 38.
Match the following:

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

Question 39.
32 g of oxygen reacts with some carbon to make 56 grams of carbon monoxide. Find out how much mass must have been used.
Answer:
Given: Mass of oxygen (reactant) = 32 g, mass of carbon monoxide (product) = 56 g
To find: Mass of oxygen (reactant)
Calculation: 12 g of carbon combine with 16 g oxygen to form 28 g of carbon monoxide as follows:

Hence, (2 × 12 = 24 g) of carbon will combine with (2 × 16 = 32 g) of oxygen to give (2 × 28 = 56 g) carbon monoxide.
Ans: Mass of carbon used = 24 g

Question 40.
Calculate the mass of sulphur trioxide produced by burning 64 g of sulphur in excess of oxygen. (Average atomic mass: S = 32 u, O = 16 u).
Solution:
Given: Mass of sulphur (reactant) = 64 g
To find: Mass of sulphur dioxide (product)
Calculation: 32 g of sulphur combine with 48 g oxygen to form 80 g of sulphur trioxide as follows:

Hence, (2 × 32 = 64 g) of sulphur will combine with (2 × 48 = 96 g) of oxygen to give (2 × 80 = 160 g) sulphur trioxide.
Ans: Mass of sulphur trioxide produced = 160 g

Question 41.
Explain Dalton’s atomic theory.
Answer:
John Dalton published “A New System of chemical philosophy” in the year of 1808. He proposed the following features, which later became famous as Dalton’s atomic theory.

• Matter consists of tiny, indivisible particles called atoms.
• All the atoms of a given elements have identical properties including mass. Atoms of different elements differ in mass.
• Compounds are formed when atoms of different elements combine in a fixed ratio.
• Chemical reactions involve only the reorganization of atoms. Atoms are neither created nor destroyed in a chemical reaction.

Dalton’s atomic theory could explain all the laws of chemical combination.

Question 42.
Give reason: Dalton’s atomic theory explains the law of conservation of mass.
Answer:

• The law of conservation of mass states that, “Mass can neither be created nor destroyed” during chemical combination of matter.
• According to Dalton’s atomic theory, chemical reactions involve only the reorganization of atoms. Therefore, the total number of atoms in the reactants and products should be same and mass is conserved during a reaction.

Hence, Dalton’s atomic theory explains the law of conservation of mass.

Question 43.
Give reason: Dalton’s atomic theory explains the law of multiple proportion.
Answer:

• The law of multiple proportion states that, “When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers
• According to Dalton’s atomic theory, compounds are formed when atoms of different elements combine in fixed ratio.

Hence, Dalton’s atomic theory explains the law of multiple proportion.

Question 44.
Define: Atomic mass unit (amu).
Answer:
Atomic mass unit or amu is defined as a mass exactly equal to one twelth of the mass of one carbon-12 atom.

Question 45.
How is relative atomic mass of an atom measured?
Answer:

• The mass of a single atom is extremely small, i.e. the mass of a hydrogen atom is 1.6736 × 10^-24 g. Hence, it is not possible to weigh a single atom.
• In the present system, mass of an atom is determined relative to the mass of an atom of carbon-12 as the standard. This was decided in 1961 by international agreement.
• The atomic mass of carbon-12 is assigned as 12.00000 atomic mass unit (amu).
• The masses of all other elements are determined relative to the mass of an atom of carbon-12 (C-12).
• The atomic masses are expressed in amu which is exactly equal to one twelth of the mass of one carbon-12 atom.
• The value of 1 amu is equal to 1.6605 × 10^-24 g.

Question 46.
What is meant by Unified Mass unit?
Answer:

• Presently, instead of amu, Unified Mass has now been accepted as the unit of atomic mass.
• It is called Dalton and its symbol is ‘u’ or ‘Da’.

Question 47.
What is average atomic mass?
Answer:
The atomic mass of an element which exists as mixture of two or more isotopes is the average of atomic masses of its isotopes. This is called average atomic mass.

Question 48.
Define: Molecular mass
Answer:
Molecular mass of a substance is the sum of average atomic masses of the atoms of the elements which constitute the molecule.
OR
Molecular mass of a substance is the mass of one molecule of that substance relative to the mass of one carbon-12 atom.

Question 49.
How is molecular mass of a substance calculated? Give example.
Answer:
Molecular mass is calculated by multiplying average atomic mass of each element by the number of its atoms and adding them together.
e.g. Molecular mass of carbon dioxide (CO₂) is calculated as follows:
Molecular mass of CO₂ = (1 × average atomic mass of C) + (2 × average atomic mass of O)
= (1 × 12.0 u) + (2 × 16.0 u)
= 44.0 u

Question 50.
Mass of an atom of hydrogen in gram is 1.6736 × 10^-24 g. What is the atomic mass of hydrogen in u?
Solution:
Given: Mass of an atom of hydrogen in gram is 1.6736 × 10^-24 g.
To find: Atomic mass of hydrogen in u
Calculation: 1.66056 × 10^-24 g = 1 u
∴ 1.6736 × 10^-24 g = x
x = \(\frac{1.6736 \times 10^{-24} \mathrm{~g}}{1.66056 \times 10^{-24} \mathrm{~g} / \mathrm{u}}\) = 1.008u
Ans: The atomic mass of hydrogen in u = 1.008 u

Question 51.
The mass of an atom of one carbon atom is 12.011 u. What is the mass of 20 atoms of the same isotope?
Solution:
Mass of l atom of carbon = 12.011 u
∴ Mass of 20 atoms of same carbon isotope = 20 × 12.011 u = 240.220 u
Ans: The mass of 20 atoms of same carbon isotope = 240.220 u

Question 52.
Calculate the average atomic mass of neon using the following data:

Solution:
Average atomic mass of Neon (Ne)

Ans: Average atomic mass of neon = 20.1707 u

Question 53.
Calculate the average atomic mass of argon from the following data:

Solution:
Average atomic mass of argon (Ar)

Ans: Average atomic mass of argon = 39.974 g mol^-1

Question 54.
Calculate the molecular mass of the following in u:
i. H₂O ii. C₆H₅Cl iii. H₂SO₄
Solution:
i. Molecular mass of H₂O = (2 × Average atomic mass of H) + (1 × Average atomic mass of O)
= (2 × 1.0u) + (1 × 16.0 u)
= 18 u

ii. Molecular mass of C₆H₅Cl = (6 × Average atomic mass of C) + (5 × Average atomic mass of H) + (1 × Average atomic mass of Cl)
= (6 × 12.0 u) + (5 × 1.0 u) + (1 × 35.5 u)
= 112.5 u

iii. Molecular mass of H₂SO₄ = (2 × Average atomic mass of H) + (1 × Average atomic mass of S) + (4 × Average atomic mass of O)
= (2 × 1.0 u) + (1 × 32.0 u) + (1 × 16.0 u)
= 98 u
Ans: i. The molecular mass of H₂O = 18 u
ii. The molecular mass of C₆H₅Cl = 112.5 u
iii. The molecular mass of H₂SO₄ = 98 u

Question 55.
Find the mass of 1 molecule of oxygen (O₂) in amu (u) and in grams.
Solution:
Molecular mass of O₂ = 2 × 16 u
∴ Mass of 1 molecule = 32 u
∴ Mass of 1 molecule of O₂= 32 × 1.66056 × 10^-24 g = 53.1379 × 10^-24 g
Ans: Mass of 1 molecule in amu = 32 u
Mass of 1 molecule in grams = 53.1379 × 10^-24 g

Question 56.
Find the formula mass of
i. NaCl ii. Cu(NO₃)₂
Solution:
i. Formula mass of NaCl
= Average atomic mass of Na + Average atomic mass of Cl
= 23.0 u + 35.5 u = 58.5 u

ii. Formula mass of Cu(NO₃)₂
= Average atomic mass of Cu + 2 × (Average atomic mass of N + Average atomic mass of three O)
= 63.5 + 2 × [14 + (3 × 16)] = 187.5 u
Ans: i. Formula mass of NaCl = 58.5 u
ii. Formula mass of Cu(NO₃)₂ = 187.5 u

Question 57.
Find the formula mass of
i. KCl
ii. AgCl
Atomic mass of K = 39 u, Ag =108 u and Cl = 35.5 u.
Solution:
i. Formula mass of KCl
= Average atomic mass of K + Average atomic mass of Cl
= 39 u + 35.5 u = 74.5 u

ii. Formula mass of AgCl
= Average atomic mass of Ag + Average atomic mass of Cl
= 108 + 35.5 = 143.5 u
Ans: i. Formula mass of KCl = 74.5 u
ii. Formula mass of AgCl = 143.5 u

Question 58.
Calculate the number of moles and molecules of urea present in 5.6 g of urea.
Solution:
Given: Mass of urea = 5.6 g
To find: The number of moles and molecules of urea
Formulae: i. Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
ii. Number of molecules = Number of moles × Avogadro’s constant
Mass of urea = 5.6 g
Molecular mass of urea, NH₂CONH₂
= (2 × Average atomic mass of N) + (4 × Average atomic mass of H) + (1 × Average atomic mass of C) + (1 × average atomic mass of O)
= (2 × 14 u) + (4 × 1 u) + (1 × 12 u) + (1 × 16 u) = 60 u
∴ Molar mass of urea = 60 g mol^-1
∴ Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{5.6 \mathrm{~g}}{60 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 0.09333 mol

[Calculation using log table:
\(\frac{5.6}{60}\)
= Antilog₁₀ [log₁₀ (5.6) – log₁₀ (60)]
= Antilog₁₀ [0.7482 – 1.7782]
= Antilog₁₀ [latex]\overline{2} .9700[/latex]
= 0.09333]

Now,
Number of molecules of urea
= Number of moles × Avogadro’s constant
= 0.09333 mol × 6.022 × 10²³ molecules/mol
= 0.5616 × 10²³ molecules (by using log table)
= 5.616 × 10²² molecules
Ans: Number of moles of urea = 0.0933 mol
Number of molecules of urea = 5.616 × 10²² molecules

[Calculation using log table:
0.09333 × 6.022
= Antilog₁₀ [log₁₀ (0.09333) + log₁₀ (6.022)]
= Antilog₁₀ [\(\overline{2} .9698\) + 0.7797]
= Antilog₁₀ [latex]\overline{1} .7495[/latex]
= 0.5616]

Question 59.
Calculate the number of atoms in each of the following:
i. 64 u of oxygen (O)
ii. 42 g of nitrogen (N)
Solution:
i. 64 u of oxygen (O) = x atoms
Atomic mass of oxygen (O) = 16 u
∴ Mass of one oxygen atom = 16 u
∴ x = \(\frac{64 \mathrm{u}}{16 \mathrm{u}}\) = 4 atoms

ii. 42 g of nitrogen (N)
Atomic mass of nitrogen = 14 u
∴ Molar mass of nitrogen = 14 g mol^-1
Now,
Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{42 \mathrm{~g}}{14 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 3 mol
Now,
Number of atoms = Number of moles × Avogadro’s constant
= 3 mol × 6.022 × 10²³ atoms/mol
= 18.07 × 10²³ atoms
= 1.807 × 10²⁴ atoms
Ans: i. Number of oxygen atoms in 64 u = 4 atoms
ii. Number of nitrogen atoms in 42 g = 1.807 × 10²⁴ atoms

Question 60.
Calculate the number of atoms in each of the following.
i. 52 moles of Argon (Ar)
ii. 52 u of Helium (He)
iii. 52 g of Helium (He)
Solution:
i. 52 moles of Argon
1 mole Argon atoms = 6.022 × 10²³ atoms of Ar
∴ Number of atoms = 52 mol × 6.022 × 10²³ atoms/mol
= 313.144 × 10²³ atoms of Argon

ii. 52 g of He
Molar mass of He = mass of 1 atom of He = 4.0 u
4.0 u = 1 He
∴ 52 u = x
∴ x = 52 u × \(\frac{1 \text { atom of He }}{4.0 \mathrm{u}}\) = 13 atoms of He

iii. 52 g of He
Molar mass of He = 4.0 g mol^-1
Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{52 \mathrm{~g}}{4.0 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 13 mol
Number of atoms of He = Number of moles × Avogadro’s constant
= 13 mol × 6.022 × 10²³ atoms/mol
= 78.286 × 10²³ atoms of He
Ans. i. Number of argon atoms in 52 moles = 313.144 × 10²³ atoms of Argon
ii. Number of helium atoms in 52 u = 13 atoms of He
iii. Number of helium atoms in 52 g = 78.286 × 10²³ atoms of He

Question 61.
Calculate the number of atoms of ‘C’, ‘H’ and ‘O’ in 72.5 g of isopropanol, C₃H₇OH (molar mass 60 g mol^-1).
Solution:
Mass of isopropanol(C₃H₇OH) = 72.5 g
The number of atoms of C, H, O
Calculation: Molecular formula of isopropanol, is C₃H₇OH.
Molar mass of C₃H₇OH = 60 g mol^-1
Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{72.5 \mathrm{~g}}{60 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 1.208 mol
∴ Moles of isopropanol = 1.21 mol
Number of atoms = Number of moles × Avogadro’s constant
Now, 1 molecule of isopropanol has total 12 atoms, out of which 8 atoms are of H, 3 of C and 1 of O.
∴ Number of C atoms in 72.5 g isopropanol = (3 × 1.208) mol × 6.022 × 10²³ atoms/mol
= 2.182 × 10²⁴ atoms of carbon
∴ Number of ‘H’ atoms in 72.5 g isopropanol = (8 × 1.208) mol × 6.022 × 10²³ atoms/mol
= 5.819 × 10²⁴ atoms of hydrogen
∴ Number of ‘O’ atoms in 72.5 g isopropanol = (1 × 1.208) mol × 6.022 × 10²³ atoms/mol
= 7.274 × 10²³ atoms of oxygen
Ans. 72.5g of isopropanol contain 2.182 ×10²⁴ atoms of H and 7.274 × 10²³ atoms of O.

Question 62.
Calculate the number of moles and molecules of ammonia (NH₃) gas in a volume 67.2 dm³ of it measured at STP.
Solution:
Given: Volume of ammonia at STP = 67.2 dm³
To find: Number of moles and molecules of ammonia
Formulae: i. Number of moles of a gas (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
ii. Number of molecules = Number of moles × 6.022 × 10²³ molecules mol^-1
Calculation: Molar volume of a gas = 22.4 dm³ mol^-1 at STP.
Number of moles (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
Number of moles of NH₃ = \(\frac{67.2 \mathrm{dm}^{3}}{22.4 \mathrm{dm}^{3} \mathrm{~mol}^{-1}}\)
Number of molecules = Number of moles × 6.022 × 10²³ molecules mol^-1
3.0 mol × 6022 × 10²³ molecules mol^-1
= 18.066 × 10²³ molecules
Ans: Number of moles of ammonia = 3.0 mol
Number of molecules of ammonia = 18.066 × 10²³ molecules

Question 63.
3.40 g of ammonia at STP occupies volume of 4.48 dm³. Calculate molar mass of ammonia.
Solution:
Given: Mass of ammonia = 3.40 g
Volume at STP = 4.48 dm³
To Find: Molar mass of ammonia
Calculation: Let ‘x’ grams be the molar mass of NH₃.
Molar volume of a gas = 22.4 dm³ mol^-1 at STP.
Volume occupied by 3.40 g of NH₃ at S.T.P = 4.48 dm³
Volume occupied by ‘x’ g of NH3 at S.T.P = 22.4 dm³
∴ x = \(\frac{22.4 \times 3.40}{4.48}\) = 17.0 g mol^-1
Ans: Molar mass of ammonia is 17.0 g mol^-1.

Question 64.
Veg puffs from a particular bakery have an average mass of 27.0 g, whereas egg puffs from the same bakery have an average mass of 40 g.
i. Suppose a person buys 1 kg of veg puff from the bakery. Calculate the number of veg puffs he receives.
ii. Determine the mass of egg puffs (in kg) that will contain the same number of eggs puffs as in one kilogram of veg puffs.
Solution:
i. Mass of a veg puff = 27.0 g = 0.027 kg
∴ Number of veg puffs in 1 kg = 1 / 0.027 = 37
ii. One kilogram of veg puffs contains 37 veg puffs.
Mass of 37 egg puffs = 37 × 0.040 = 1.48 kg
Ans: i. 37 veg puffs in 1 kg of puff.
ii. Mass of 37 egg puffs is 1.48 kg

Multiple Choice Questions:

1. The branch of chemistry which deals with carbon compounds is called ……………. chemistry.
(A) organic
(B) inorganic
(C) carbon
(D) bio
Answer:
(A) organic

2. A/an is a simple combination of two or more substances in which the constituent substances retain their separate identities.
(A) compound
(B) mixture
(C) element
(D) All of these
Answer:
(B) mixture

3. Which one of the following is NOT a mixture?
(A) Paint
(B) Gasoline
(C) Liquefied Petroleum Gas (LPG)
(D) Distilled water
Answer:
(D) Distilled water

4. The sum of the masses of reactants and products is equal in any physical or chemical reaction. This is in accordance with ………………
(A) law of multiple proportion
(B) law of definite composition
(C) law of conservation of mass
(D) law of reciprocal proportion
Answer:
(C) law of conservation of mass

5. A sample of calcium carbonate (CaCO₃) has the following percentage composition: Ca = 40 %; C = 12 %; O = 48 %. If the law of definite proportions is true, then the weight of calcium in 4 g of a sample of calcium
carbonate from another source will be ……………..
(A) 0.016 g
(B) 0.16 g
(C) 1.6 g
(D) 16 g
Answer:
(C) 1.6 g

6. Two elements, A and B, combine to form two compounds in which ‘a’ g of A combines with ‘b₁’ and ‘b₂’g of B respectively. According to law of multiple proportion ………………
(A) b₁ = b₂
(B) b₁ and b₂ bear a simple whole number ratio
(C) a and b₁ bear a whole number ratio
(D) no relation exists between b₁ and b₂
Answer:
(B) b₁ and b₂ bear a simple whole number ratio

7. At constant temperature and pressure, two litres of hydrogen gas react with one litre of oxygen gas to produce two litres of water vapour. This is in accordance with ……………….
(A) law of multiple proportion
(B) law of definite composition
(C) law of conservation of mass
(D) law of gaseous volumes
Answer:
(D) law of gaseous volumes

8. One mole of oxygen molecule weighs …………….
(A) 8 g
(B) 32 g
(C) 16 g
(D) 6.022 × 10²³ g
Answer:
(B) 32 g

9. The mass of 0.002 mol of glucose (C₆H₁₂O₆) is ………………
(A) 0.20 g
(B) 0.36 g
(C) 0.50 g
(D) 1.80 g
Answer:
(B) 0.36 g

10. Which of the following is CORRECT?
(A) 1 mole of oxygen atoms contains 6.0221367 × 10²³ atoms of oxygen.
(B) 1 mole of water molecules contains 6.0221367 × 10²³ molecules of water.
(C) 1 mole of sodium chloride contains 6.0221367 × 10²³ formula units of NaCl.
(D) All of these
Answer:
(D) All of these

11. 180 g of glucose (C₆H₁₂O₆) contains ……………. carbon atoms.
(A) 1.8 × 10²³
(B) 1.8 × 10²⁴
(C) 3.6 × 10²³
(D) 3.6 × 10²⁴
Answer:
(C) 3.6 × 10²³

12. The number of molecules present in 8 g of oxygen gas is …………….
(A) 6.022 × 10²³
(B) 3.011 × 10²³
(C) 12.044 × 10²³
(D) 1.505 × 10²³
Answer:
(D) 1.505 × 10²³

13. The number of molecules in 22.4 cm³of ozone gas at STP is ……………….
(A) 6.022 × 10²⁰
(B) 6.022 × 10²³
(C) 22.4 × 10²⁰
(D) 22.4 × 10²³
Answer:
(A) 6.022 × 10²⁰

14. 11.2 cm³ of hydrogen gas at STP, contains …………….. moles.
(A) 0.0005
(B) 0.01
(C) 0.029
(D) 0.5
Answer:
(A) 0.0005

15. The mass of 224 mL of hydrogen gas at STP is
(A) 0.02 g
(B) 0.224 g
(C) 2.24 g
(D) 20.0 g
Answer:
(A) 0.02 g

16. 4.4 g of an unknown gas occupies 2.24 L of volume under STP conditions. The gas may be ………………
(A) CO₂
(B) CO
(C) O₂
(D) SO₂
Answer:
(A) CO₂

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 9 Optics Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 9 Optics

Question 1.
Express the speed of EM waves in terms of permittivity and permeability of the medium.
Answer:
In a material medium, the speed of EM waves is given by, c = \(\sqrt{\frac {1}{εµ}}\)
Where, ε = Permittivity and µ = permeability.
These constants depend on the electric and magnetic properties of the medium.

Question 2.
How can one classify commonly observed phenomena of light on the basis of the nature of light?
Answer:
Commonly observed phenomena concerning light can be broadly split into three categories:

1. Ray optics or geometrical optics: Ray optics can be used for understanding phenomena like reflection, refraction, double refraction, total internal reflection, etc.
2. Wave optics or physical optics: Wave optics explains the phenomena of light such as, interference, diffraction, polarisation, Doppler effect etc.
3. Particle nature of light: Particle nature of light can be used to explain phenomena like photoelectric effect, emission of spectral lines, Compton effect etc.

Question 3.
State the fundamental laws on which ray optics is based.
Answer:
Ray optics is based on the following fundamental laws:
i. Light travels in a straight line in a homogeneous and isotropic medium.

ii. Two or more rays can intersect at a point without affecting their paths beyond that point.

iii. Laws of reflection:
a. Reflected ray lies in the plane formed by incident ray and the normal drawn at the point of incidence and the two rays are on either side of the normal.
b. Angles of incidence and reflection are equal (i = r).

iv. Laws of refraction:
a. Refracted ray lies in the plane formed by incident ray and the normal drawn at the point of incidence; and the two rays are on either side of the normal.

b. Angle of incidence (90 and angle of refraction (62) are related by Snell’s law, given by,
n₁ sin θ₁ = n₂ sin θ₂
where, n₁, n₂ = refractive indices of medium 1 and medium 2 respectively.

Question 4.
Explain Cartesian sign conventions using a graph.
Answer:
According to Cartesian sign conventions:
i. All distances are measured from the optical centre or pole.

ii. Figures should be drawn in such a way that the incident rays travel from left to right. Thus, a real object should be shown to the left and virtual object or image to the right of pole (or optical centre).

iii. X-axis can be conveniently chosen as the principal axis with origin at the pole.

iv. Distances to the left of the pole are negative and those to the right of the pole are positive.

v. Distances above the principal axis (X-axis) are positive while those below it are negative.

Question 5.
Define and represent in a neat diagram the following terms:
i. Diverging beam
ii. Converging beam
Answer:
i. A diverging beam of light corresponds to rays of light coming from real point object.

ii. A converging beam corresponds to rays of light directed to a virtual point object or image.

Question 6.
Thickness of the glass of a spectacle is 2 mm and refractive index of its glass is 1.5. Calculate time taken by light to cross this thickness. Express your answer with most convenient prefix attached to the unit ‘second’.
Answer:
Speed of light in vacuum, c = 3 × 10⁸ m/s
Given that:
Refractive index, (n_glass) = 1.5
Thickness of the glass = 2 mm
= 2 × 10sup>-3 m
∵ Re fractive index (n_glass) = \(\frac{\text { speed of light in vacuum }(\mathrm{c})}{\text { speed of light in glass (v) }}\)
∵ v = \(\frac{\mathrm{c}}{\mathrm{n}_{\text {glass }}}=\frac{3 \times 10^{8}}{1.5}\) = 2 × 10⁸ m/s
As v = \(\frac {s}{t}\)
time taken (t) to cross the thickness (s),
t = \(\frac{\mathrm{s}}{\mathrm{v}}=\frac{2 \times 10^{-3}}{2 \times 10^{8}}\) = 1 × 10^-11 s
Most convenient unit to express this small time is nano second. (1 ns = 10^-9 s)
∴ t = 0.01 × 10^-9 s = 0.01 ns

Question 7.
Explain the properties of the image formed after reflection of light from a plane surface.
Answer:

1. The image of an object kept in front of a plane reflecting surface is virtual and laterally inverted.
2. Image is of the same size as that of the object.
3. It is situated at the same distance as that of object but on the other side of the reflecting surface.

Question 8.
Explain the formula to find the number of images formed when an object is placed in between two plane mirrors inclined at an angle θ.
Answer:

1. If an object is kept between two plane mirrors inclined at an angle θ, multiple images (N) are formed due to multiple reflections from both the mirrors.
2. The number of images can be calculated using formula n = \(\frac {360}{θ}\)
3. Exact number of images seen (N) depends upon the angle between the mirrors and position of the object.
4. When n is an even integer, for all positions of the object the number of images formed are N = n – 1.
5. When n is an odd integer:
   a. For an object placed at the angle bisector of the mirrors: N = n- 1
   b. For an object placed off the angle bisector of the mirrors: N = n
6. If n is not an integer, N = m, where m is integral part of n.

Question 9.
Define radius of curvature of a spherical mirror.
Answer:
Radius of the sphere of which a mirror is a part is called as radius of curvature of the mirror.

Question 10.
What is the focal length of a spherical mirror? Give its relation with the radius of curvature.
Answer:
i. For a concave mirror focal length is the distance at which parallel incident rays converge. For a convex mirror, it is the distance from where parallel rays appear to be diverging after reflection.

ii. In case of spherical mirrors, half of radius of curvature is focal length of the mirror,
f = \(\frac {R}{2}\)

Question 11.
Show with the help of a ray diagram that focal length of convex mirror is positive while that of concave mirror is negative.
Answer:
i. According to sign conventions, the incident rays are drawn from left of the mirror to the right as shown in the ray diagrams below.

ii. As the rays incident on convex mirror appear to converge at a point on the positive side of the origin, the focal length of the convex mirror is positive.

iii. However, in case of concave mirror, the rays converge at a point on negative side of the origin, the focal length of the concave mirror is negative.

Question 12.
Give relation between focal length, object distance and image distance for a small spherical mirror.
Answer:
For a point object or for a small finite object, the focal length of a small spherical mirror is related to object distance and image distance as,
\(\frac {1}{f}\) = \(\frac {1}{v}\) + \(\frac {1}{u}\)

Question 13.
What is lateral magnification? How does it vary in different types of spherical mirrors?
Answer:
i. Ratio of linear size of image to that of the object, measured perpendicular to the principal axis, is defined as the lateral magnification.
∴ m = \(\frac {h_2}{h_1}\) = \(\frac {v}{u}\) (for spherical lenses)
m = –\(\frac {v}{u}\) (for spherical mirrors)

ii. For any position of the object, a convex mirror always forms virtual, erect and diminished image. Thus, lateral magnification for convex lens is always m < 1.

iii. In the case of a concave mirror, it depends upon the position of the object.

Question 14.
Complete the following table for a concave mirror?

Answer:

Question 15.
Explain with proper diagram why parabolic mirrors are preferred over spherical one.
Answer:
i. Unlike spherical shape, every point on a parabola is equidistant from a straight line and a point.

ii. Consider given parabola having RS as directrix and F as the focus. Points A, B, C on it are equidistant from line RS and point F.

iii. Hence A’A = AF, B’B = BF, C’C = CF, and so on.

iv. If rays of equal optical path converge at a point, that point is the location of real image corresponding to that beam of rays.

v. From figure, the paths A”AA’, B”BB’. C”CC’, etc., are equal paths when mirror is neglected.

vi. If the parabola ABC is a mirror then by definition of parabola the respective optical paths,
A”AF = B”BF = C”CF

vii. Thus, F is the single point focus for entire beam of rays parallel to the axis and there is no spherical aberration.
Hence, parabolic mirrors are preferred over spherical one as there is no spherical aberration.

Question 16.
A small object is kept symmetrically between two plane mirrors inclined at 38°. This angle is now gradually increased to 41°, the object being symmetrical all the time. Determine the number of images visible during the process.
Answer:
The object is kept symmetrically between two plane mirrors. This implies the object is placed at angle bisector.
Thus, for θ = 38°,
n = \(\frac {360}{38}\) = 9.47
As it is not integral, N = 9 (the integral part of n)
∴ For going from 38° to 41°, the mirrors go through angles 39° and 40°. Number of images formed will remain 9 for all angles between 38° and 40°.
For angles > 40°, the n goes on decreasing and when θ = 41°,
n = \(\frac {360}{41}\) = 8.78 i.e., N = 8

Question 17.
A thin pencil of length 20 cm is kept along the principal axis of a concave mirror of curvature 30 cm. Nearest end of the pencil is 20 cm from the pole of the mirror. What will be the size of image of the pencil?
Answer:
For the pencil kept along the principal axis and the end of the pencil closest to pole is at 20 cm,
say, u₁ = -20 cm
Flence, the other end of the stick is at distance, u₂ = (u₁ + 20) = -40 cm from pole of the mirror.
As R = -30 cm, F = \(\frac {R}{2}\) = -15 cm

∴ v₂ = -24 cm
Here, negative signs indicate that images are formed on the left of the mirror.

The length of the image formed is given by,
v = v₂ – v₁ = -24 – (-60) = 36 cm.

Question 18.
An object is placed at 15 cm from a convex mirror having radius of curvature 20 cm. Find the position and kind of image formed by it.
Answer:
Given: u = – 15 cm,
f = \(\frac {R}{2}\) = + \(\frac {20}{2}\) = + 10cm
To find: Nature and position of image (v)
Formula: \(\frac {1}{v}\) + \(\frac {1}{u}\) = \(\frac {1}{f}\)
Calculation:
From formula,
∴ \(\frac {1}{v}\) = \(\frac {1}{f}\) – \(\frac {1}{u}\)
= \(\frac {1}{+10}\) – \(\frac {1}{-15}\) = \(\frac {1}{10}\) + \(\frac {1}{15}\)
= \(\frac {2+3}{30}\) = \(\frac {5}{30}\) = \(\frac {1}{6}\)
∴ v = 6 cm

Question 19.
Prove that refractive index of a glass slab is given by the formula,
n = \(\frac {Real depth}{Apparent depth}\)
Answer:
i. Consider a plane parallel slab of a transparent medium of refractive index n.

ii. A point object O at real depth R appears to be at I at apparent depth A, when seen from outside (air).

iii. Consider incident ray OA and OB as shown in figure.

iv. Assuming i and r to be very small, we can write,
tan r = \(\frac {x}{A}\) ≈ sin r and tan i = \(\frac {x}{R}\) ≈ sin i

v. According to Snell’s law, for a ray travelling from denser medium to rarer medium,
n = \(\frac{\sin \mathrm{r}}{\sin \mathrm{i}} \approx \frac{\left(\frac{\mathrm{x}}{\mathrm{A}}\right)}{\left(\frac{\mathrm{x}}{\mathrm{D}}\right)}=\frac{\mathrm{R}}{\mathrm{A}}=\frac{\text { Real depth }}{\text { Apparent depth }}\)

Question 20.
The depth of a pond is 10 m. What is the apparent depth for a person looking normally to the water surface? n_water = 4/3.
Answer:
Given: Real depth of pond, d_real = 10 m,
n_w = \(\frac {4}{3}\)
To find: Apparent depth
Formula: n = \(\frac {Realdepth}{Apparent depth}\)
Calculation: From formula,
∴ Apparent depth = \(\frac {Realdepth}{n}\) = \(\frac {10}{(\frac{4}{3})}\)
= \(\frac {10×3}{4}\) = 7.5 m

Question 21.
A crane flying 6 m above a still, clear water lake sees a fish underwater. For the crane, the fish appears to be 6 cm below the water surface. How much deep should the crane immerse its beak to pick that fish?
For the fish, how much above the water surface does the crane appear? Refractive index of water = 4/3.
Answer:
For crane, apparent depth of fish = 6 cm,
Given that refractive index (n_w) = \(\frac {4}{3}\)
n_w = \(\frac {Realdepth}{Apparent depth}\)
∴ Apparent depth = \(\frac {4}{3}\) × 6 = 8 cm
Similarly, for fish, real height of crane = 6 m and
\(\frac{\mathrm{n}_{\mathrm{air}}}{\mathrm{n}_{\mathrm{w}}}=\frac{1}{\mathrm{n}_{\mathrm{w}}}=\frac{\text { Real height }}{\text { Apparent height }}\)
\(\frac {3}{4}\) = \(\frac {6}{Apparent height}\)
i.e., Apparent height = \(\frac {4×6}{3}\) = 8 m

Question 22.
Write a short note on Periscope.
Answer:
i. Instrument used to see the objects on the surface of a water body from inside of water is called periscope.
ii. It consists of two right angled prisms. The incident rays of light are reflected twice through these prisms.
iii. Total internal reflections occur inside these prisms and a clear view of the surface of water is obtained.

Question 23.
A ray of light passes from glass (n_g = 1.52) to water (n_w = 1.33). What is the critical angle of incidence?
Answer:
Given: n_g = 1.52, n_w = 1.33
To find: Critical angle (i_c)
formula: sin i_c = \(\frac {n_2}{n_1}\) = \(\frac {n_w}{n_g}\)
Calculation:
From formula,
i_c = sin^-1 (\(\frac {1.33}{1.52}\)) = sin^-1 (0.875) = 61°2′

Question 24.
There is a tiny LED bulb at the center of the bottom of a cylindrical vessel of diameter 6 cm. Height of the vessel is 4 cm. The beaker is filled completely with an optically dense liquid. The bulb is visible from any inclined position but just visible if seen along the edge of the beaker. Determine refractive index of the liquid.
Answer:

As the bulb is just visible from the edge, the angle of incidence formed by ray OP must be equal to critical angle.
∴ refractive index (n) = \(\frac {1}{sin i_c}\)
From Figure,
tan i_c = \(\frac {PQ}{OQ}\) = \(\frac {4}{3}\)
∴ sin i_c = \(\frac {OQ}{OP}\) = \(\frac {3}{5}\)
∴ n_liquid = \(\frac {5}{3}\)

Question 25.
What are convex and concave lenses? For which condition, convex lens will have negative focal length?
Answer:

1. A lens is said to be convex if it is thicker in the middle and narrowing towards the periphery. According to Cartesian sign convention, its focal length is positive.
2. Convex lens is visualized to be internal cross section of two spheres (or one sphere or a plane surface).
3. A lens is concave if it is thicker at periphery and narrows down towards centre and has negative focal length.
4. Concave lens is visualized to be external cross section of two spheres.
5. For lenses of material optically rarer than the medium in which those are kept, convex lenses will have negative focal length and they will diverge the incident rays.

Question 26.
Which lenses can be considered as thin lenses?
Answer:
Lenses for which the maximum thickness is at least 50 times smaller than all the other distances are considered as thin lenses.

Question 27.
Give the expression for the focal length of combination of lenses when
i. Lenses are kept in contact with each other
ii. Two lenses kept at a distance d apart from each other.
Answer:
i. For thin lenses kept in contact:
\(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_{1}}+\frac{1}{\mathrm{f}_{2}}+\frac{1}{\mathrm{f}_{3}}\) + ………

ii. For two lenses kept distance d apart:
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}-\frac{d}{f_{1} f_{2}}\)

Question 28.
An object is placed infront of a convex surface separating two media of refractive index 1.1 and 1.5. The radius of curvature is 40 cm. Where is the image formed when an object is placed at 220 cm from the refracting surface?
Solution:
Given: n₁ = 1.1, n2 = 1.5, R = + 40 cm,
u = -220 cm
To find: Position of image (v)
Formula: \(\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{\left(n_{2}-n_{1}\right)}{R}\)
Calculation: From formula,

Question 29.
A glass paper-weight (n = 1.5) of radius 3 cm has a tiny air bubble trapped inside it. Closest distance of the bubble from the surface is 2 cm. Where will it appear when seen from the other end (from where it is farthest)?
Answer:
From figure, distance OR = 2 cm
∴ Distance OP = 4 cm

According to sign conventions,
OP = u = -4 cm and CP = R = -3 cm
For refraction at curved surface,

Question 30.
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?
Answer:
Given: n = 1.55, f = 20 cm,
R₁ = R and R₂ = – R
(By sign convention)
To Find: Radius of curvature (R)
Formula: \(\frac{1}{\mathrm{f}}=(\mathrm{n}-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
Calculation: From formula,
\(\frac {1}{20}\) = (1.55 – 1) \(\left[\frac{1}{R}-\left(-\frac{1}{R}\right)\right]=0.55 \times \frac{2}{R}\)
∴ R = \(\frac {1.10}{1}\) × 20 = 22 cm

Question 31.
A dense glass double convex lens (n = 2) designed to reduce spherical aberration has |R₁| : |R₂| = 1:5. If a point object is kept 15 cm in front of this lens, it produces its real image at 7.5 cm. Determine R₁ and R₂.
Answer:
Given: |R₁| : |R₂| = 1 : 5, u = -15 cm,
v = +7.5 cm, n = 2
To Find: Radii of curvature of double convex lens (R₁) and (R₂)
Formula:
i. \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
ii. \(\frac {1}{f}\) = (n – 1) \(\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
Calculation: From formula (i),
\(\frac{1}{f}=\frac{1}{7.5}-\frac{1}{(-15)}=\frac{1}{5}\)
∴ f = +5 cm
Substituting this value in formula (ii), we get,
\(\frac{1}{5}=(2-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
∴ \(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}=\frac{1}{5}\)
By sign conventions,
\(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\left(-\mathrm{R}_{2}\right)}=\frac{1}{5}\) ………….. (1)
Also \(\frac{\left|R_{1}\right|}{\left|R_{2}\right|}=\frac{1}{5}\)
∴ |R₂| = 5 |R₁| …………… (2)
Substituting in equation (1),
∴ \(\frac{1}{R_{1}}-\frac{1}{\left(-5 R_{1}\right)}=\frac{1}{5}\)
∴ \(\frac {6}{5R_1}\) = \(\frac {1}{5}\)
∴ R₁ = 6 cm
Using in equation (2),
R₂ = 5 × 6 = 30 cm

Question 32.
Why are prism preferred for dispersion over two parallel surfaces? Explain its construction in brief.
Answer:
i. In case of two parallel surfaces, for dispersion to be easily detectable, they must be separated over a large distance.
ii. In order to have appreciable and observable dispersion, two parallel surfaces are not useful. In such case we use prisms, in which two refracting surfaces inclined at an angle.
iii. Commonly used prisms have three rectangular surfaces forming a triangle.
iv. Two of which take part in refraction at a time. The one, not involved in refraction is called base of the prism.
v. Any section of prism perpendicular to the base is called principal section of the prism. Commonly all the rays considered during refraction lie in this plane.

Question 33.
Draw neat labelled diagrams showing refraction of a monochromatic light and white light through a prism.
Answer:

Question 34.
For a prism prove that i + e = A + δ where the symbols have their usual meanings.
Answer:
i. Consider a principal section ABC of a prism of absolute refractive index n kept in air as shown in figure.

ii. Let A be the refracting angle of prism and surface BC be the base.

iii. A monochromatic ray PQ obliquely strikes first reflecting surface AB such that, angle of incidence ∠PQM at Q is i.

iv. After refraction at Q, the ray deviates towards the normal and strikes second refracting surface AC at R which is the point of emergence.

v. Angles of refraction at Q (∠NQR) and at R (∠QRN) are r₁ and r₂ respectively.

vi. After R. the ray deviates away from normal and finally emerges along RS making e as the angle of emergence.

vii. Emergent ray RS meets an extended incident ray QT at X if traced backward. In this case, ∠TXS gives the angle of deviation.

viii. From figure,
∠AQN = ∠ARN = 90°
∴ From quadrilateral AQNR,
A + ∠QNR = 180° ………. (1)
From ∆ QNR,
r₁ + r₂ + ∠QNR = 180° ………. (2)
∴ A = r₁ + r₂ ……… (3)

ix. Angle δ forms an exterior angle for ∆ XQR.
∴ ∠XQR + ∠XRQ = δ
∴ (i – r₁) + (e – r₂) = δ
∴ (i + e) – (r₁ + r₂) = δ
From equation (3),
δ = i + e – A
∴ i + .e = A + δ

Question 35.
Explain δ versus i curve for refraction of light through a prism.
Answer:
i. Variation of angle of incidence i with angle of deviation δ is as shown in figure.

ii. It shows that, with increasing values of i, the angle of deviation δ decreases initially to a certain minimum (δ_m) value and then increases.

iii. The curve shown in the figure is not a symmetric parabola, but the slope in the part towards right is less.

iv. Except at δ = δ_m there are two values of i for any given δ. From principle of reversibility of light, we can conclude that if one of these values is i, the other must be e and vice versa. Thus at δ = δ_m, we have i = e.

Question 36.
Show that, at condition of minimum deviation, n = \(\frac{\sin \left(\frac{\mathbf{A}+\boldsymbol{\delta}_{\mathrm{m}}}{\mathbf{2}}\right)}{\sin \left(\frac{\mathbf{A}}{\mathbf{2}}\right)}\)

i. For every angle of deviation except angle of minimum deviation, there are two values of angle of incidence.

ii. However, at angle of minimum deviation there is only one corressponding angle of incidence.

iii. From principle of reversibility in path PQRS, the values of i and e are interchangeable for every δ. Thus, at minimum deviation, i = e.

iv. This implies the angles of refraction r₁ and r₂ are also equal. Also, A = r₁ + r₂
∴ A = 2 r i.e., r = \(\frac {A}{2}\) ……. (1)

v. In case of minimum deviation, QR is parallel to base BC and the figure is symmetric.

vi. Using i + e = A + δ,
i + i = A + δ_m
∴ i = \(\frac {A+δ_m}{2}\) …………….(2)

vii. According to Snell’s law,
n = \(\frac {sin i}{sin r}\)
Thus, using equations (1) and (2),
n = \(\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}\)
This is the prism formula.

Question 37.
For grazing emergence of a ray in a prism, find out minimum possible values for angle of incidence (i) and angle of refraction (r₁) for commonly used glass prism.
Answer:

i. At the point of emergence in prism, the ray travels from a denser medium into rarer.
Thus, if r₂ = sin^-1 (\(\frac {1}{n}\)) is the critical angle, the angle of emergence e = 90°. This is called grazing emergence.

ii. Angle of prism A is constant for a given prism and A = r₁ + r₂. Hence the corresponding r₁ and i will have their minimum possible values.

iii. For commonly used glass prisms,
n = 1.5 (r₂)_max = sin^-1 (\(\frac {1}{n}\)) = sin^-1 (\(\frac {1}{1.5}\)) = 41.49°

iv. If, prism is symmetric (equilateral),
A = 60°
∴ r₁ = 60° – 41°49′ = 18°11′
∴ n = 1.5 = \(\frac{\sin \left(\mathrm{i}_{\min }\right)}{\sin \left(18^{\circ} 11^{\prime}\right)}\)
sin (i_min) = 1.5 × sin (18°11′)
∴ ii_min = 27°55′ ≅ 28°.

Question 38.
Derive the formula for angle of deviation for thin prisms.
OR
Show that in a thin prism, for small angles of incidence, angle of deviation is constant (independent of angle of incidence).
Answer:
For thin prisms (refracting angle < sin 10°). sin θ ≈ θ
∴ Refractive index, n = \(\frac{\sin \mathrm{i}}{\sin \mathrm{r}_{1}} \approx \frac{\mathrm{i}}{\mathrm{r}_{1}}\)
Also n = \(\frac{\sin e}{\sin \mathrm{r}_{2}} \approx \frac{\mathrm{e}}{\mathrm{r}_{2}}\)
∴ i ≈ n r₁ and e ≈ nr₂

ii. Substituting this in, i + e = A + δ, we get,
i + e = n (r₁ + r₂) = nA = A + δ
∴ S = A(n – 1)
A and n are constant for a given prism. Thus, for a thin prism, for small angles of incidence, angle of deviation is constant (independent of angle of incidence).

Question 39.
Give the expression for mean deviation for a beam of white light.
Answer:
For a beam of white light, yellow colour is practically chosen to be the mean colour for violet and red.
This gives mean deviation as,
δ_VR = \(\frac{\delta_{\mathrm{V}}+\delta_{\mathrm{R}}}{2}\) ≈ δ_Y = A(n_Y – 1)
Where, n_Y = refractive index for yellow colour.

Question 40.
A fine beam of white light is incident upon the longer side of a plane parallel glass slab of breadth 5 cm at angle of incidence 60°. Calculate lateral deviation of red and violet rays and lateral dispersion between them as they emerge from the opposite side. Refractive indices of the glass for red and violet are 1.51 and 1.53 respectively.
Answer:
As shown in figure,
VM = L_V = lateral deviation for violet colour,
RT = L_R = lateral deviation for red colour,
∴ Lateral dispersion between these colours, L_VR = L_V – L_R

∴ r_R = sin^-1 (0.5735) ≈ 35°
Similarly,
sin r_V = \(\frac{\sin 60^{\circ}}{1.53}=\frac{\sqrt{3}}{2 \times 1.53}=\frac{1.732}{3.06}\)
= antilog {log (1.732) – log (3.06)}
= antilog {0.2385 – 0.4857}
= antilog 11.7528}
= 0.5659
∴ sin r_V = 0.566
∴ r_V = sin^-1 (0.566) = 34°28′
∴ Angle of deviation for red colour
= i – r_R = 60° – 35° = 25°
and that for violet colur = i – r_V = 60° – 34°28′
= 25°32′
From figure, in ∆ANR,
AR = \(\frac{\mathrm{AN}}{\cos \mathrm{r}_{\mathrm{R}}}=\frac{5}{\cos \left(35^{\circ}\right)}=\frac{5}{0.8192}\) = 6.104 cm
Similarly ∆ANV,
AV = \(\frac{\mathrm{AN}}{\cos \mathrm{r}_{\mathrm{V}}}=\frac{5}{\cos \left(34^{\circ} 28^{\prime}\right)}=\frac{5}{0.8244}\) = 6.065 cm
∴ For red colour, L_R = RT = AR [sin(i – r_R)]
= AR [sin (25°)]
= 6.104 × 0.4226
= 2.58 cm
For violet colour, L_V = VM
= AV [sin (i – r_V)]
= AV × sin (25° 32′)
= 6.065 × 0.431
= 2.61 cm
∴ L_VR = L_V – L_R = 2.61 – 2.58 = 0.03 cm
= 0.3 mm

Question 41.
For a glass (n = 1.5) prism having refracting angle 60°, determine the range of angle of incidence for which emergent ray is possible from the opposite surface and the corresponding angles of emergence. Also calculate the angle of incidence for which i = e. How much is the corresponding angle of minimum deviation?
Answer:
For an equilateral prism of glass, the minimum angle of incidence for which the emergent ray just emerges is i_min = 27° 55′. Corresponding angle of emergence is, e_max = 90°.
From the principle of reversibility of light, i_max = 90° and e_min = 27°55′
Also, for equilateral glass prism at minimum deviation,

Also, from prism formula,
i + e = A + δ
At minimum deviation,
∴ i + i = 60 + 37°10′ = 97°10′
∴ i = 48°35′

Question 42.
For a dense flint glass prism of refracting angle 10°, obtain angular deviation for extreme colours and dispersive power of dense flint glass. (n_red = 1.712, n_violet = 1.792)
Answer:
Given: A = 10°, n_R = 1.712, n_V = 1.792
To find:
i. Angular deviation for extreme colours (δ_V and δ_R)
ii. Dispersive power of flint glass (ω)
Formulae:
i. δ = A(n – 1)
ii. ω = \(\frac{\delta_{\mathrm{V}}-\delta_{\mathrm{R}}}{\left(\frac{\delta_{\mathrm{V}}+\delta_{\mathrm{R}}}{2}\right)}\)

Question 43.
The refractive indices of the material of the prism for red and yellow colour are 1.620 and 1.635 respectively. Calculate the angular dispersion and dispersive power, if refracting angle is 8°.
Solution:
Given: n_R = 1.620, n_Y = 1.635, A = 8°
To find:
i. Angular dispersion (δ_V – δ_R)
ii. Dispersive power (ω)
Formulae:
i. δ_v – δ_r = A(n_V – n_R)
ii. ω = \(\frac{\mathrm{n}_{\mathrm{V}}-\mathrm{n}_{\mathrm{R}}}{\mathrm{n}_{\mathrm{Y}}-1}\)
Calculation: Since, n_Y = \(\frac {n_V+n_R}{2}\)
∴ n_V = 2n_Y – n_R
n_V = 2 × 1.635 – 1.620 = 3.27 – 1.620
∴ n_V = 1.65
From formula (i),
δ_V – δ_R = 8(1.65 – 1.620)
= 8 × 0.03 = 0.24°
∴ δ_V – δ_R = 0.24°
From formula (ii),
ω = \(\frac{1.65-1.620}{1.635-1}=\frac{0.03}{0.635}\) = 0.0472

Question 44.
What could be the possible reasons for the upward bending of the light ray during hot days?
Answer:
Possible reasons for the upward bending at the road could be:
i. Angle of incidence at the road is glancing. At glancing incidence, the reflection coefficient is very large which causes reflection.
ii. Air almost in contact with the road is not steady. The non-uniform motion of the air bends the ray upwards and once it has bent upwards, it continues to do so.

Question 45.
State some properties of rainbow.
Answer:

1. It is seen during rains and on the opposite side of the Sun.
2. It is seen only during mornings and evenings and not throughout the day.
3. In the commonly seen rainbow red arch is outside and violet is inside.
4. In the rarely occurring concentric secondary rainbow, violet arch is outside and red is inside.
5. It is in the form of arc of a circle.
6. Complete circle can be seen from a higher altitude, i.e. from an aeroplane.
7. Total internal reflection is not possible in this case.

Question 46.
Why is total internal reflection not possible during formation of a rainbow ?
Answer:
i. For total internal reflection, the angle of incidence in the denser medium must be greater than critical angle for the given pair of media.

ii. The relative refractive index of air with the water drop is just less than 1 and hence the critical angle is almost equal to 90°.

iii. Angle of incidence i in air, at the water drop, can’t be greater than 90°.

iv. As a result, angle of refraction r in water will always be less than the critical angle.

v. The figures shown indicates that this angle r itself acts as an angle of incidence at any point for one or more internal reflections. But this does not indicate the total internal reflection.

Question 47.
Rainbow is seen only for a definite angle range with respect to the ground. Justify.
Answer:
i. For clear visibility of rainbow, a beam must have enough intensity.

ii. The curve for angle of deviation and angle of incidence is almost parallel to X-axis near minimum deviation i.e., for majority of angles of incidence in this range, the angle of deviation is nearly the same and those rays form a beam of enough intensity.

iii. Rays beyond this range suffer wide angular dispersion and thus will not have enough intensity for visibility. Hence, the rainbow is seen only for a definite angle range with respect to the ground for which the intensity of the beam is enough for the visibility.

Question 48.
How is the range of angles for which rainbows can be observed calculated?
Answer:
i. Angle of deviation for the final emergent ray, can be shown to be equal to δ = π + 2i – 4r for primary rainbow and δ = 2π + 2i – 6r for the secondary rainbow.
ii. Using these relations along with Snell’s law, sin i = n sin r, derivatives of angle of deviation (δ) is obtained.
iii. Second derivative of δ comes out to be negative, which shows that it is the minima condition.
iv. Equating first derivative to zero corresponding values of i and r are obtained. Thus, from the figures shown, the corresponding angles θ_R and θ_V at the horizontal are obtained. These angles give the visible angular position for the rainbow.

Question 49.
When can one see complete circle of a rainbow? Explain in detail.
Answer:
i. Figure given below illustrates formation of primary and secondary rainbows with their common centre O. It is the point where the line joining the sun and the observer meets the Earth when extended.

ii. P is location of the observer. Different colours of rainbows are seen on arches of cones of respective angles.

iii. Smallest half angle refers to the cone of violet colour of primary rainbow, which is 41°.

iv. As the Sun rises, the relative position of common centre of the rainbows with respect to observer shifts down. Hence as the Sun comes up, smaller and smaller part of the rainbows will be seen. If the Sun is above 41°, violet arch of primary rainbow cannot be seen.

v. Beyond 53°, no rainbows will be visible. That is why rainbows are visible only during mornings and evenings and in the shape of a bow.

vi. However, if observer moves up (may be in an aeroplane), the line PO itself moves up making lower part of the arches visible. After a certain minimum elevation, entire circle for all the cones can be visible.

Question 50.
Define following terms:
i. Longitudinal chromatic aberration
ii. Circle of least confusion
iii. Transverse chromatic aberration
Answer:
i. Longitudinal chromatic aberration:
Due to different refractive indices and angle of deviations, violet and red colours of a white light converge at different focal points, f_V and f_R. The distance between f_V and f_R is measured as the longitudinal chromatic aberration.

ii. Circle of least confusion:
In presence of aberration the image is not a single point but always a circle. At particular location on the screen, this circle has minimum diameter. This is called circle of least confusion.

iii. Transverse chromatic aberration:
Radius of the circle of least confusion is called the transverse chromatic aberration.

Question 51.
After Cataract operation, a person is recommended with concavo-convex spectacles of curvatures 10 cm and 50 cm. Crown glass of refractive indices 1.51 for red and 1.53 for violet colours is used for this. Calculate the lateral chromatic aberration occurring due to these glasses.
Answer:
For a concavo-convex lens, with convex shape facing the object, both the radii of curvature are positive as shown in the figure.

= (1.53 – 1) × 0.08 = 0.0424
∴ f_v = 23.58 cm
∴ Longitudinal (lateral) chromatic aberration
= f_V – f_R = 24.51 – 23.58 = 0.93 cm

Question 52.
Why do we need optical instruments for?
Answer:
i. Due to the limitation for focusing the eye lens it is not possible to take an object closer than a certain distance. This distance is called least distance of distance vision D. For a normal, unaided human eye D = 25cm.
ii. If an object is brought closer than this, we cannot see it clearly.
iii. If an object is too small the corresponding visual angle from 25 cm is not enough to see it and if we bring it closer than that, its image on the retina is blurred.
iv. Also, the visual angle made by cosmic objects far away from us such as stars is too small to make out minor details and we cannot bring those closer.
v. In such cases we need optical instruments like microscope and telescopes to observe these things clearly.

Question 53.
A convex lens has focal length of 2.0 cm. Find its magnifying power if image is formed at DDV.
Answer:
Given: f = 2 cm, v = D = 25 cm
To find: Magnifying power (M.P.)
Formula: M.P = 1 + \(\frac {D}{f}\)
Calculation:
From formula,
M.P = 1 + \(\frac {25}{2}\)
M.P. = 1 + 12.5 = 13.5

Question 54.
A magnifying glass of focal length 10 cm is used to read letters of thickness 0.5 mm held 8 cm away from the lens. Calculate the image size. How big will the letters appear? Can you read the letters if held 5 cm away from the lens? If yes, of what size would the letters appear? If no, why not?
Answer:
Given that, f = +10 cm, u = -8 cm,
From thin lens formula,
\(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
∴ \(\frac{1}{10}=\frac{1}{\mathrm{v}}-\frac{1}{-8}\)
∴ v = 40 cm
Magnification of a lens is,
m = \(\frac {v}{u}\) = \(\frac {Object size h-i}{Object size h-0}\)
∴ \(\frac {40}{8}\) = \(\frac {h_1}{0.5}\)
∴ h₁ = 2.5 cm
This implies the height of the image is 5 times that of the object.
Magnifying power,
M = \(\frac {D}{u}\) = \(\frac {25}{8}\) = 3.125
∴ Image will appear to be 3.125 times bigger,
i.e., 3.125 × 0.5 = 1.5625 cm
For u = -5 cm, v will be -10 cm
For an average human being to see clearly, the image must be at or beyond 25 cm. Thus it will not possible to read the letters if held 5 cm away from the lens.

Question 55.
A compound microscope has a magnification of 15. If the object subtends an angle of 0.5° to eye, what will be the angle subtended by the image at the eye?
Answer:
Given: M.P = 15, α = 0.5°
To Find: Angle(β)
Formula: M.P = \(\frac {β}{α}\)
Calculation:
From formula,
β = M.P × α = 15 × 0.5 = 7.5°

Question 56.
A compound microscope has a magnifying power of 40. Assume that the final image is formed at DDV(25 cm). If the focal length of eyepiece 10 cm, calculate the magnification produced by objective.
Answer:
M.P = 40, D = 25 cm, f_e = 10 cm
To Find: Magnification (m₀)
Formula: M.P = m₀ × M_e
Calculation:
From the formula,

Question 57.
The pocket microscope used by a student consists of eye lens of focal length 6.25 cm and objective of focal length 2 cm. At microscope length 15 cm, the final image appears biggest. Estimate distance of the object from the objective and magnifying power of the microscope.
Answer:
Given: f_e = 6.25 cm, f₀ = 2 cm, L = 15 cm
As image appears biggest, V_e = -25 cm.
To find:
i. Distance of object from objective (u₀)
ii. Magnifying power (M)
Formula:
i. \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
ii. L = |v₀| + |u_e|
iii. M = \(\left(\frac{v_{o}}{u_{o}}\right)\left(\frac{D}{u_{e}}\right)\)
Calculation: For eyelens, using formula (i),

∴ u_e = 5 cm
From formula (ii),
|v₀| = L – |u_e|
= 15 – 5 = 10 cm
Using formula (i) for objective,

Question 58.
Focal length of the objective of an astronomical telescope is 1 m. Under normal adjustment, length of the telescope is 1.05 m. Calculate focal length of the eyepiece and magnifying power under normal adjustment.
Answer:
Given: f₀ = 1 m, L = 1.05 m
To find:
i. Focal length of eyepiece (f_e)
Magnifying power under normal adjustment (M)
Formula:
i. L = f₀ + f_e
ii. M = \(\frac {f_0}{f_e}\)
Calculation. From formula (i),
f_e = L – f₀ = 1.05 – 1 = 0.05 m
From formula (ii),
M = \(\frac {1}{0.05}\) = 20

Question 59.
Magnifying power of an astronomical telescope is 12 and the image is formed at D.D.V. If the focal length of the objective is 90 cm, what is the focal length of the eyepiece?
Answer:
Given: M.P = 12, v = D, f₀ = 90 cm,
To find: Focal length of eye piece (f_e)
Formula: M.P = \(\frac {f_0}{f_e}\) (1 + \(\frac {f_e}{D}\))
Calculation:
From formula.
12 = \(\frac {90}{f_e}\) (1 + \(\frac {f_e}{25}\))
∴ f_e = 10.71 cm

Question 60.
Two convex lenses of an astronomical telescope have focal length 1.3 m and 0.05 m respectively. Find the magnifying power and the length of the telescope.
Answer:
Given: f₀ = 1.3 m, f_e = 0.05 m
To find:
i. Magnifying power of telescope (M.P.)
ii. Length of telescope (L)
Formulae:
i. M.P = \(\frac {f_0}{f_e}\)
ii. L = f₀ + f_e
Calculation: From formula (i),
M.P = \(\frac {1.3}{0.05}\) = 26
From formula (ii),
L = 1.3 + 0.05 = 1.35 m

Question 61.
What is the angle of deviation of reflected ray if ray of light is incident on a plane mirror at an incident angle θ?
Answer: When a ray of light is incident on a plane mirror at an angle θ, the reflected ray gets deviated by an angle of (π – 2θ).

Question 62.
Does nature of the image depend upon size of the mirror?
Answer:
No, nature of the image is independent of size of the mirror.

Question 63.
If a convex mirror is held in air and then dipped in oil, then what will be the change in its focal length?
Answer:
Focal length of spherical mirrors are independent of the medium.

Question 64.
When ray of light falls normally on a mirror, its angle of incidence is 90°. True or false? Justify your answer.
Answer:
False, when light falls normally on a mirror, its angle of incidence is zero degree.

Question 65.
In one of the performances, a magician keeps a gold ring beneath a thick glass slab (µ = \(\frac {3}{2}\)) Then he keeps a flask filled with water (µ = \(\frac {4}{3}\)), over the slab. Now when spectators one by one observe from the open end of the flask, the ring disappears at a certain angle of viewing.
i. What could be the reason behind the disappearance?
ii. At what angle of viewing the ring vanishes?
Answer:
i. The ring disappears due to total internal reflection of the light at water-air interface.

ii. _aµ_w = \(\frac {4}{3}\)
sin i_c = \(\frac {1}{µ}\)
∴ i_c = sin^-1 (\(\frac {1}{_aµ_w}\)) = sin^-1 (\(\frac {3}{4}\)) = 48.6°
Hence, for angle of viewing for which angle of incidence of ring from water to air is greater that 48.6°, the ring will vanish.

Question 66.
Why dispersion of light is not observed in glass slab but it is observed in prism?
Answer:
When a light passes from one medium to another, at one interface, it changes its speed. The glass slab and prism both have two glass-air interfaces. Hence, the light undergoes refraction twice in both the cases. When the two interfaces are parallel to each other, although the colours are separated at first interface, they all travel the same path after refracting from second interface. However, in prism, the two interfaces are not parallel. Therefore, the colours separated at first interface do not travel the same path after second refraction but emerge out at different wavelengths producing spectrum.

Question 67.
A prism manufacturer is planning to build a dispersive prism out of following materials with the refracting angles as given.
i. Glass (µ = 1.5), A = 60°
ii. Plastic (µ = 1.4), A = 90°
iii. Fluorite (µ = 1.45), A = 64°
If he desires to give the prism following relations of i and δ, then which of the above combinations can be used to construct the prism?

Answer:
From the given values of i and δ, δm = 37°
From prism formula, µ = \(\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}\)

i. For Glass (µ = 1.5), A = 60°;
µ = \(\frac{\sin \left(\frac{60^{\circ}+37^{\circ}}{2}\right)}{\sin \left(30^{\circ}\right)}\)
= 1.5
Hence, this combination can be used for fabricating the desired prism.

ii. For Plastic (µ = 1.4), A = 90°;
µ = \(\frac{\sin \left(\frac{90^{\circ}+37^{\circ}}{2}\right)}{\sin \left(45^{\circ}\right)}\)
= 1.26
As P_piastic = 1-4, this combination cannot be used.

iii. For Fluorite (µ = 1.45), A = 64°
µ = \(\frac{\sin \left(\frac{64^{\circ}+37^{\circ}}{2}\right)}{\sin \left(32^{\circ}\right)}\)
= 1.45
Hence, this combination can also be used for fabrication of prism.

Question 68.
Find the refractive index of material of following prism if the ray of light incident at angle 45° suffers minimum deviation through the prism.

Answer:

A = 60°
Also, ray of light suffers minimum deviation.
∴ 2i = A + δ_m
∴ δ_m = 2i – A = 90° – 60° = 30°
From prism formula,
µ = \(\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}\)
= \(\frac{\sin \left(\frac{60^{\circ}+30^{\circ}}{2}\right)}{\sin \left(30^{\circ}\right)}\)
= √2
Hence, refractive index of material of prism is √2.

Multiple Choice Questions

Question 1.
Time taken by light to cross a glass slab of thickness 4 mm and refractive index 3 is
(A) 4 × 10^-11 s
(B) 2 × 10^-11 ns
(C) 16 × 10^-11 s
(D) 8 × 10^-10 s
Answer:
(A) 4 × 10^-11 s

Question 2.
If mirrors are inclined to each other at an angle of 90°, the total number of images seen for a symmetric position of an object will be
(A) 3
(B) 4
(C) 5
(D) 3 or 4
Answer:
(A) 3

Question 3.
In case of a convex mirror, the image formed is
(A) always on opposite side, virtual, erect.
(B) always on the same side, virtual, erect.
(C) always on opposite side, real, inverted.
(D) dependent on object distance.
Answer:
(A) always on opposite side, virtual, erect.

Question 4.
A glass slab is placed in the path of a beam of convergent light. The point of convergence of light
(A) moves towards the glass slab.
(B) moves away from the glass slab.
(C) remains at the same point.
(D) undergoes a lateral shift.
Answer:
(A) moves towards the glass slab.

Question 5.
For a person seeing an object placed in optically rarer medium,
(A) apparent depth of the object is more than real depth
(B) apparent depth is smaller than the real depth.
(C) apparent depthe might be smaller or greater depending on the position of the person.
(D) nothing can be concluded about the depth of object from given data.
Answer:
(A) apparent depth of the object is more than real depth

Question 6.
Light travels from a medium of refractive index µ₁ to another of refractive index µ₂ (µ₁ > µ₂). For total internal reflection of light, which is NOT true?
(A) Light must travel from medium of refractive index µ₁ to µ₂.
(B) Angle of incidence must be greater than the critical angle.
(C) There is no refraction of light.
(D) Light must travel from the medium of refractive index µ₂ to µ₁.
Answer:
(D) Light must travel from the medium of refractive index µ₂ to µ₁.

Question 7.
Optical fibre is based on which of the following phenomenon?
(A) Reflection.
(B) Refraction.
(C) Total internal reflection.
(D) Dispersion.
Answer:
(C) Total internal reflection.

Question 8.
Commonly used glass have refractive index of 1.5. What is the critical angle for such glass?
(A) 49°
(B) 42°
(C) 45°
(D) 40°
Answer:
(B) 42°

Question 9.
If the refractive index of water is 4/3 and that of glass slab is 5/3. Then the critical angle of incidence for which a light ray tending to go from glass to water is totally reflected, is
(A) sin^-1 (\(\frac {3}{4}\))
(B) sin^-1 (\(\frac {3}{5}\))
(C) sin^-1 (\(\frac {2}{3}\))
(D) sin^-1 (\(\frac {4}{5}\))
Answer:
(D) sin^-1 (\(\frac {4}{5}\))

Question 10.
While deriving prism formula, which of the following condition is NOT satisfied?
(A) I = e
(B) r₁ = r₂
(C) r = \(\frac {A}{2}\)
(D) δ_m = i + e + r
Answer:
(D) δ_m = i + e + r

Question 11.
If the critical angle for the material of a prism is C and the angle of the prism is A, then there will be no emergent ray when
(A) A < 2 C
(B) A = 2 C
(C) A > 2 C
(D) A < \(\frac {C}{2}\)
Answer:
(C) A > 2 C

Question 12.
Chromatic aberrations is caused due to
(A) spherical shape of lens
(B) spherical shape of mirrors
(C) angle of deviation for violet light being more than that for red light.
(D) refractive index for violet light being less than that for red light
Answer:
(C) angle of deviation for violet light being more than that for red light.

Question 13.
In normal adjustment, magnifying powser of a astronomical telescope is given by
(A) \(\frac {D}{f_0}\) \(\frac {L}{f_e}\)
(B) \(\frac {L}{D}\) \(\frac {f_e}{f_0}\)
(C) \(\frac {f_0}{f_e}\)
(D) \(\frac {f_e}{f_0}\)
Answer:
(C) \(\frac {f_0}{f_e}\)

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 8 Sound Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 8 Sound

Question 1.
State the different types of waves.
Answer:

1. Waves that require a material medium for their propagation are called mechanical waves. Example: Sound waves, string waves, seismic waves, etc.
2. Waves that do not require a material medium for their propagation are called electromagnetic waves. Example: Light waves, radio waves, y-rays, etc.
3. The wave associated with any object when it is in motion is called a matter-wave.
4. Waves in which a disturbance created at one place travel to distant points and keeps travelling unless stopped by an external force are known as travelling or progressive waves.
5. Waves are also of stationary type.

Question 2.
Define the following terms. Give their SI units.
i. Period
ii. Frequency
iii. Velocity
Answer:
i. Period (T):
The time taken by the particle of a medium to complete one vibration is called period of the wave.
SI unit: second (s)

ii. Frequency (n):
The number of vibrations performed by a particle per second is called frequency of a wave.
SI unit: hertz (Hz)

iii. Velocity (v):
The distance covered by a wave per unit time is called the velocity of the wave.
SI unit: m/s

Question 3.
State the properties that should be possessed by a medium for a mechanical wave to propagate through it.
Answer:

1. The medium must be continuous and elastic so that it can regain its original state as soon as the deforming forces are removed.
2. The medium should possess inertia. It must be capable of storing energy and transferring it in the form of waves.
3. The frictional resistance of the medium should be negligible to avoid damped oscillations.

Question 4.
What are two types of progressive waves? State two characteristics of progressive waves.
Answer:
Progressive waves are classified into two types:
a. Transverse progressive waves
b. Longitudinal progressive waves.

Characteristics of progressive waves:
1. All the vibrating particles of medium have same amplitude, period and frequency.
2.. State of oscillation i.e., phase changes from particle to particle.

Question 5.
A violin string emits sound of frequency 510 Hz. How far will the sound waves reach when string completes 250 vibrations? The velocity of sound is 340 m/s.
Answer:
Given: n = 510 Hz, v = 340 m/s,
number of vibrations = 250
To find: Distance
Formula: v = nλ
Calculation:
From formula,
λ = \(\frac {v}{n}\) = \(\frac {340}{510}\) = \(\frac {2}{3}\) m
Distance covered in 1 vibration = \(\frac {2}{3}\) m
∴ Distance covered in 250 vibration
= \(\frac {2}{3}\) × 250 = 166.7 m 3
Answer: The distance covered by sound waves is 166.7 m

Question 6.
The speed of sound in air is 330 m/s and that in glass is 4500 m/s. What is the ratio of the wavelength of sound of a given frequency in the two media?
Answer:
Given: v_air = 330 m/s, v_glass = 4500 m/s
To find: \(\frac {λ_{air}}{λ_{glass}}\)
Formula: v = nλ
Calculation: From formula,
v_air = n λ_air
v_glass = n λ_glass
∴ \(\frac {λ_{air}}{λ_{glass}}\) = \(\frac {v_{air}}{v_{glass}}\) = \(\frac {330}{4500}\) = 7.33 × 10^-2

Question 7.
The velocity of sound in gas is 498 m/s and in air is 332 m/s. What is the ratio of wavelength of sound waves in gas to air?
Answer:
v_g = 498 m/s, v_a = 332 m/s
To find: Ratio of wavelengths (\(\frac {λ_g}{λ_a}\))
Formula: v = nλ
Calculation:
Frequency of sound wave remains same.
From formula,
For gas λ_g = \(\frac {v_g}{n}\) and for air λ_ag = \(\frac {v_a}{n}\)
∴ \(\frac {v_g}{v_a}\) = \(\frac {v_g}{v_a}\) = \(\frac {498}{332}\) = \(\frac {3}{2}\)
∴ \(\frac {v_g}{v_a}\) = 3 : 2

Question 8.
A human ear responds to sound waves of frequencies in the range of 20 Hz to 20 kHz. What are the corresponding wavelengths, if the speed of sound in air is 330 m/s? Answer:
Given: v₁ = v_g = 330 m/s, n₁ = 20 Hz,
n₂ = 20 kHz = 20 × 10³ Hz
To find: Wavelength (λ₁ and λ₂)
Formula: v = nλ
Calculation:
From formula,
λ₁ = \(\frac {v_1}{n_1}\) = \(\frac {330}{20}\) = 16.5 m
λ₂ = \(\frac {v_2}{n_2}\) = \(\frac {330}{20×10^3}\) = 16.5 × 10^-3 = 0.0165 m

Question 9.
A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (i) the reflected sound, (ii) the transmitted sound? Speed of sound in air is 340 m s^-1 and in water is 1486 m s^-1
Answer:
Given: n = 1000 kHz = 10⁶ Hz,
v_a = 340 m/s,
v_w = 1486 m/s
To find: Reflected wavelength (λ_R),
Transmitted wavelength (λ_T),
Formula: v = nλ
Calculation:
In different medium, frequency of sound wave remains same.
From formula,
Sound is reflected in air,
i. ∴ λ_R = \(\frac {v_a}{n}\) = \(\frac {330}{10^6}\) = 3.4 × 10^-4 m
Sound is transmitted in water,
ii. ∴ λ_T = \(\frac {v_w}{n}\) = \(\frac {1486}{10^6}\) = 1.486 × 10^-3 m

Question 10.
The wavelength of a sound note is 1 m in air and 2.5 m in a liquid. Find the speed of sound in the liquid, if the speed of the sound in air is 330 m/s.
Answer:
Given: λ_a = 1 m, λ_l = 2.5 m, v_a = 330 m/s,
To find: Speed of sound (v_l)
Formula: v = nλ
Calculation:
From formula,
Since the frequency n remains the same,
v_a = nλ_a and v_l = nλ_l
∴ \(\frac {v_l}{v_a}\) = \(\frac {λ_l}{λ_a}\)
∴ v_l = v_a \(\frac {λ_l}{λ_a}\) = 330 × \(\frac {2.5}{1}\) = 825 m/s

Question 11.
Define a polarised wave.
Answer:
A wave in which the vibrations of all the particles along the path of a wave are constrained to a single plane is called a polarised wave.

Question 12.
Write down the main characteristics of longitudinal waves.
Answer:
Characteristics of longitudinal waves:

1. All the particles of medium in the path of wave vibrate in a direction parallel to the direction of propagation of wave with same period and amplitude.
2. When longitudinal wave passes through a medium, the medium is divided into alternate compressions (high pressure zone) and rarefactions (low pressure zone).
3. A compression and adjacent rarefaction form one cycle of a longitudinal wave. The distance between any two consecutive points having same phase (successive compression or rarefactions) is called wavelength of the wave.
4. For propagation of longitudinal waves, the medium should possess the property of elasticity of volume (bulk modulus). Thus, longitudinal waves can travel through solids, liquids and gases. Longitudinal wave cannot travel through vacuum or empty space.
5. The compressions and rarefactions advance in the medium and are responsible for transfer of energy.
6. When longitudinal wave advances through medium, there is periodic variations in pressure and density along the path of wave and with time.
7. Since the direction of vibration of particles and direction of propagation of wave are same or parallel, longitudinal waves cannot be polarised.

Question 13.
State the mathematical expression for a transverse progressive wave travelling along the positive and negative x-axis.
Answer:
i. Consider a transverse progressive wave whose particle position is described by x and displacement from equilibrium position is described by y.
Such a sinusoidal wave can be written as follows:
∴ y (x, t) = a sin (kx – ωt + ø) ……… (1)
where a, k, ω and ø are constants,
y (x, t) = displacement as a function of position (x) and time (t)
a = amplitude of the wave,
ω = angular frequency of the wave
(kx₀ – ωt + ø) = argument of the sinusoidal wave and is the phase of the particle at x at time t.

ii. At a particular instant, t = t₀,
y (x, t₀) = a sin (kx – ωt₀ + ø)
= a sin (kx + constant)
Thus at t = t₀, shape of wave as a function of x is a sine wave.

iii. At a fixed location x = x₀
y(x₀, t) = a sin (kx₀ – ωt + ø)
= a sin (constant – ωt)
Hence the displacement y, at x = x₀ varies as a sine function.

iv. This means that the particles of the medium, through which the wave travels, execute simple harmonic motion around their equilibrium position.

v. For (kx – ωt + ø) to remain constant, x must increase in the positive direction as time t increases. Thus, the equation (1) represents a wave travelling along the positive x axis.

vi. Similarly, a wave travelling in the direction of the negative x axis is represented by,
y(x, t) = a sin (kx + ωt +ø) …….(2)

Question 14.
Explain the Laplace’s correction to the Newton’s formula for the velocity of sound in air.
Answer:
Laplace’s correction:
Laplace suggested that the compression or rarefaction takes place too rapidly. Heat produced during compression and lost during rarefaction does not get sufficient time for dissipation. Due to this, the whole heat content remains same. Thus, it is an adiabatic process.

According to Laplace, E is the adiabatic modulus of elasticity of air medium which is given by,
E = γP ….(1)
where P = pressure of the air medium γ = ratio of specific heat of air at constant pressure (c_p) and specific heat of air at constant volume (c_v). i.e., γ = c_p/c_v.

iii. Using equation, v = \(\sqrt{\frac {E}{ρ}}\), we have velocity of sound in air,
v = \(\sqrt{\frac {γP}{ρ}}\), …. [From equation (1)]
For air, γ = 1.41
At NTP, P = 0.76 × 13600 × 9.8 N/m²
ρ = 1.293 kg/m³.
∴ v = \(\sqrt{\frac {1.41×0.76×13600×9.8}{1.293}}\) = 332.35 m/s
This value is in close agreement with the experimental value.

Question 15.
What is the effect of temperature on the velocity of sound in air?
Answer:
Effect of temperature on velocity of sound:
i. Let v₀ and v be the velocity of sound in air at T₀ and T Kelvin respectively. Let ρ₀ and p be the densities of gas at temperature T0 and T respectively.

ii. Considering the number of moles n = 1 for the gas, we have,

iii. From above formula, we can conclude that velocity of sound in air increases with increase in temperature.

Question 16.
Show that for 1 °C rise in temperature, the velocity of sound in air increases by 0.61 m/s.
Answer:
Let v₀ = velocity of sound at 0 °C or 273 K
v = velocity of sound at t °C or (273 + 1) K
we have, v ∝ √T

Hence, velocity of sound increases by 0.61 m/s when temperature increases by 1 °C.

Question 17.
Suppose you are listening to an out-door live concert sitting at a distance of 150 m from the speakers. Your friend is listening to the live broadcast of the concert in another country and the radio signal has to travel 3000 km to reach him. Who will hear the music first and what will be the time difference between the two? Velocity of light = 3 × 10⁸ m/s and that of sound is 330 m/s.
Answer:
Distance between me and the speakers
(s₁) = 150 m, distance radio signal has to travel (S₂) = 3000 km, v_sound 330 m/s, v_light = 3 × 10⁸ m/s
Time taken by sound to reach me,
= \(\frac {s_1}{v_sound}\) = \(\frac {150}{330}\) = 0.4546 s
Time taken by the broadcasted sound (done by
EM waves = \(\frac {s_2}{v_light}\) = \(\frac {3000km}{30×10^5km/s}\) = \(\frac {3×1^30}{3×10^5}\) = 10^-2 s
∴My friend will hear the sound first.
The time difference will be = 0.4546 – 0.01
= 0.4446 s.

Question 18.
Consider a closed box of rigid walls so that the density’ of the air inside it is constant. On heating, the pressure of this enclosed air is increased from P₀ to P. It is now observed that sound travels 1.5 times faster than at pressure P₀. Calculate P/P₀.
Answer:
Given: v_P = 1.5 v_P0
To find: Ratio of pressure (\(\frac {p}{p_0}\))

Question 19.
The densities of nitrogen and oxygen at NTP are 1.25 kg/m³ and 1.43 kg/m³ respectively. If the speed of sound in oxygen at NTP is 320 m/s, calculate the speed in nitrogen, under the same conditions of temperature and pressure, (γ for both the gases is 1.4)
Answer:
Given: ρ_N = 1.25 kg/m³, ρ = 1.43 kg/m³,
v₀ = 320 m/s,
To find: Speed in nitrogen (v_N)
Formula: v = (\(\sqrt{\frac {γP}{ρ}}\) )
Calculation: From formula,

Question 20.
Find the temperature at which the velocity of sound in air will be 1.5 times its velocity at 0 °C
Answer:
Given: \(\frac {p}{p_0}\) = 1.5, T₀ = 0 °C = 273 K
To find: Temperature (T)
Formula: \(\frac {v}{v_0}\) = \(\sqrt{\frac {T}{T_0}}\)
Calculation:
From formula,
\(\frac {T}{T_0}\) = (\(\frac {v}{v_0}\))²
∴ T = T₀ (\(\frac {v}{v_0}\))²
∴ T = 273 (1.5)² = 614.25 K = 341.25 °C

Question 21.
The velocity of sound in air at 27 °C is 340 m/s. Calculate the velocity of sound in air at 127 °C.
Answer:
Given: T₁ = 27 °C = 27 + 273 = 300 K,
v₁ = 340 m/s,
T₂ = 127 °C = 127 + 273 = 400 K
To find: Velocity (v₂)
Formula: \(\frac {v_1}{v_2}\) = \(\sqrt{\frac {T_1}{T_2}}\)
Calculation: From formula,
v₂ = v₁ \(\sqrt{\frac {T_2}{T_1}}\) = 340, \(\sqrt{\frac {400}{300}}\)
= 340 × 1.1547
∴ v₂ = 392.6 m/s

Question 22.
The wavelength of a note is 27 m in air when the temperature is 27 °C. What is the wavelength when the temperature is increased to 37 °C?
Answer:
Given: λ₁ = 27 m,
T₁ = 27 °C = 273 + 27 = 300 K,
T₂ = 37 °C = 273 + 37 = 310 K
To find: Wavelength (λ₂)

Question 23.
We cannot hear an echo at every place. Give reason.
Answer:

1. Echo of sound depends upon the temperature of the surrounding and distance between source and reflecting surface.
2. To hear a distinct echo at 22 °C, the minimum distance required between the source of sound and reflecting surface should be 17.2 metre.
3. The velocity of sound depends on the temperature of air. Thus, the minimum distance will change with temperature. Hence, we cannot hear an echo at every place.

Question 24.
Write a short note on reverberation.
Answer:

1. Reverberation is the phenomenon in which sound waves are reflected multiple times causing a single sound to be heard more than once.
2. Sound wave gets reflected multiple times if the distance between reflecting surface and source of sound is less than 15 m.
3. During reverberation, the time interval between the successive reflections of a sound is small.
4. As a result, the reflected sound waves overlap and produce a continuously increasing loud sound which is at times difficult to understand. Measures to decrease reverberation:
5. Reverberation can be decreased by making the walls and roofs rough and by using curtains in the hall to avoid reflection of sound.
6. Chairs and wall surfaces should be covered with sound absorbing materials.
7. Porous cardboard sheets, perforated acoustic tiles, gypsum boards, thick curtains etc. should be used on the ceilings and walls.

Question 25.
Define acoustics.
Answer:
The branch of physics which deals with the study of production, transmission and reception of sound is called acoustics.

Question 26.
State the conditions that must be satisfied for proper acoustics in an auditorium along-with their remedies.
Answer:
i. Acoustics of an auditorium should be such that the sound is heard sufficiently loudly at all the points in the auditorium. The surface behind the speaker should be parabolic with the speaker at its focus for uniform distribution of sound in the auditorium. Reflection of sound helps to maintain good loudness through the entire auditorium.

ii. Echoes and reverberations should be reduced. More absorptive reflecting surfaces and full auditoriums help in reducing echoes.

iii. Unnecessary focusing of sound, poor audibility zone or region of silence should be avoided. Curved surface of the wall or ceiling should be avoided for this purpose.

iv. Echelon effect which arises due to the mixing of sound produced in the hall by the echoes of sound produced in front of regular structure like stairs should be reduced. Stair type construction in the hall must be avoided for this purpose.

v. To avoid outside stray sound from entering, the auditorium should be sound-proof when closed.

vi. Inside fittings, seats, etc. should not produce any sound for proper acoustics. Air conditioners instead of fans and soft action door closers should be used.

Question 27.
State the applications of acoustics observed in nature.
Answer:
Application of acoustics in nature:
i. Bats apply the principle of acoustics to locate objects. They emit short ultrasonic pulses of frequency 30 kHz to 150 kHz. The resulting echoes give them information about location of the obstacle. This helps the bats to fly in even in total darkness of caves.

ii. Dolphins navigate underwater with the help of an analogous system. They emit subsonic frequencies which can be about 100 Hz. They can sense an object about 1.4 m or larger.

Question 28.
State the medical applications of acoustics.
Answer:
i. High pressure and high amplitude shock waves are used to split kidney stones into smaller pieces without invasive surgery. A reflector or acoustic lens is used to focus a produced shock wave so that as much of its energy as possible converges on the stone. The resulting stresses in the stone causes the stone to break into small pieces which can then be removed easily.

ii. Ultrasonic imaging uses reflection of ultrasonic waves from regions in the interior of body. It is used for prenatal (before the birth) examination, detection of anomalous conditions like tumour etc. and the study of heart valve action.

iii. Ultrasound at a very high-power level, destroys selective pathological tissues which is helpful in treatment of arthritis and certain type of cancer.

Question 29.
State the underwater applications of acoustics.
Answer:

1. SONAR (Sound Navigational Ranging) is a technique for locating objects underwater by transmitting a pulse of ultrasonic sound and detecting the reflected pulse.
2. The time delay between transmission of a pulse and the reception of reflected pulse indicates the depth of the object.
3. Motion and position of submerged objects like submarine can be measured with the help of this system.

Question 30.
State the applications of acoustics in environmental and geological studies.
Answer:
i. Acoustic principle has important application to environmental problems like noise control. The quiet mass transit vehicle is designed by studying the generation and propagation of sound in the motor’s wheels and supporting structures.

ii. Reflected and refracted elastic waves passing through the Earth’s interior can be measured by applying the principles of acoustics.

iii. This is useful in studying the properties of the Earth. Principles of acoustics are applied to detect local anomalies like oil deposits etc. making it useful for geological studies.

Question 31.
A man shouts loudly close to a high wall. He hears an echo. If the man is at 40 m from the wall, how long after the shout will the echo be heard? (speed of sound in air = 330 m/s)
Answer:
Given s = 40m, v = 330 m/s
To Find: time (t)
Formula: Time = distance \(\frac {distence}{speed}\)
Calculation:
The distance travelled by the sound wave
= 2 × distance from man to wall.
= 2 × 40 = 80 m.
From formula,
∴ Time taken to travel the distance
\(\frac {distence}{speed}\) = \(\frac {80}{30}\) = 0.24 s

Question 32.
Write a short note on pitch of sound note.
Answer:

1. Pitch refers to the sharpness or shrillness of sound.
2. Increase in frequency of sound results in increase in the pitch and the sound is said to be sharper.
3. Tone refers to a single frequency of a wave.
4. A note may contain single or multiple tones.
5. High frequency is generally referred as high pitch or high tone.
6. Generally, speech of the men is of low pitch (shrill) and that of the women is of high pitch (sharp). Tones of an acoustic guitar are sharper than that of a base guitar. Sound of table is sharper than that of a dagga.

Question 33.
Write a short note on quality (timbre) of sound note.
Answer:
i. Timbre of a sound refers to the quality of the sound which depends upon the mixture of tones and overtones in the sound. Same sound played on different musical instruments feels significantly different and the musical instrument from which the sound generated can be easily identified.

Question 34.
Write a short note on loudness of sound.
OR
Explain how loudness affects the characteristics of sound.
Answer:
Loudness:
i. Loudness depends upon the intensity of vibration.

ii. Intensity of a wave is proportional to square of the amplitude (I ∝ A²) and is measured in the (SI) unit ofW/m²

iii. The human response to intensity is not linear, i.e., a sound of double intensity is louder but not doubly loud.

iv. Under ideal conditions, for a perfectly healthy human ear, the least audible intensity is I₀ = 10^-12 W/m².

v. Loudness of a sound of intensity I (measured in unit bel) is given by,
L₂ = log₁₀ (\(\frac {I}{I_0}\)) ………….. (1)

vi. Decibel is the commonly used unit for loudness.

vii. As, 1 decibel or 1 dB = 0.1 bel.
∴ 1 bel = 10 dB. Thus, loudness in dB is 10 times loudness in bel.
∴ L_dB = 10L_bel = 10 log₁₀ (\(\frac {I}{I_0}\))
For sound of least audible intensity I₀
L_dB = 10 log₁₀ (\(\frac {I_0}{I_0}\)) = 10 log₁₀ (1) = 0 ………… (2)
This corresponds to threshold of hearing.

viii. For sound of 10 dB,
10 = 10 log₁₀ (\(\frac {I}{I_0}\))
∴ (\(\frac {I}{I_0}\)) = 10 ¹ or I = 10 I₀
For sound of 20 dB,
20 = 10 log₁₀ (\(\frac {I}{I_0}\))
= (\(\frac {I}{I_0}\)) = 10² or I = 100 I₀ and so on.

ix. This implies, loudness of 20 dB sound is felt double that of 10 dB, but its intensity is 10 times that of the 10 dB sound. Similarly, 40 dB sound is left twice as loud as 20 dB sound but its intensity is 100 times as that of 20 dB sound and 10000 times that of 10 dB sound. This is the power of logarithmic or exponential scale. If we move away from a (practically) point source, the intensity of its sound varies inversely with square of the distance, i.e., I ∝ \(\frac {1}{r^2}\).

Question 35.
When heard independently, two sound waves produce sensations of 60 dB and 55 dB respectively. How much will the sensation be if those are sounded together, perfectly in phase?
Answer:
L₁ = 60 dB = 10 log₁₀ \(\frac {I_1}{I_0}\)
∴ \(\frac {I_1}{I_0}\) = 10⁶
∴ I₁ = 10⁶I₀
Similarly, I₂ = 10^5.5 I₀
As the waves combine perfectly in phase, the vector addition of their amplitudes will be given by A² = (A₁ + A₂)² = A\(_1^2\) + A\(_2^2\) + 2A₁, A₂ As intensity is proportional to square of the amplitude.
∴ I = I₁ + I₂ + 2\(\sqrt {I_1I_2}\) = 10⁵ I₀ (10¹ +10^0.5 + 2\(\sqrt {10^{1.5}}\))
= 10⁵I₀(10 + 3.1623 +2 × 10^0.75)
= 24.41 × 10⁵I₀ = 2.441 × 10⁶I₀
∴ L = 10 log₁₀ (\(\frac {I}{I_0}\)) = 10 log₁₀ (2.441 × 10⁶)
= 10[log₁₀ (2.441) + log₁₀(10⁶)]
= 10(0.3876 + 6)
L = 63.876 dB ~ 64 dB

Question 36.
The noise level in a class-room in absence of the teacher is 50 dB when 50 students are present. Assuming that on the average each student outputs same sound energy per second, what will be the noise level if the number of students is increases to 100?
Answer:
Loudness of sound is given as,

∴ L_B – L_A = 0.301 × 10 = 3.01
∴ L_B = L_A + 3.01 = 53.01 dB

Question 37.
Calculate the decibel increase if there is a two-fold increase in the intensity of a wave. (Given: log₁₀ 2 = 0.3010)
Answer:
L = 10 log₁₀ \(\frac {I}{I_0}\) decibel
L’ = 10 log₁₀ \(\frac {2I}{I_0}\) decibel
L’ – L = 10 (log₁₀ \(\frac {2I}{I_0}\) – log₁₀ (\(\frac {I}{I_0}\))
= 10 log₁₀ 2
= 10 × 0.3010
∴ L’ – L = 3.01 dB

Question 38.
Derive the expression for apparent frequency when listener is stationary and source is moving away from the listener.
Answer:

i. Consider a source of sound S moving away from a stationary listener L with velocity v_s. Let the speed of sound with respect to the medium be v (always positive). The listener uses a detector for counting each wave crest that reaches it.

ii. Let at t = 0, the source at point Si which is at a distance d from the listener, emit a crest. This crest reaches the listener at time t₁, given as, t₁ = d/v. …………(1)

iii. Let T₀ be the time period at which the waves are emitted.
At t = T₀, distance travelled by the source away from the stationary listener to reach point S₂ = v_sT₀.
∴ Distance of point S₂ from the listener = d + v_sT₀.
At S₂, The source emits second crest. This crest reaches the listener at t₂, given as,
t₂ = T₀ + (\(\frac {d+v_sT_0}{v}\)) …………. (2)

iv. Similarly, the time taken by the (p+1)^th crest (where, p is an integer, p = 1, 2, 3,…), emitted by the source at time pT₀, to reach the listener is given as,
t_p+1 = pT₀ + (\(\frac {d+pv_sT_0}{v}\)) …………. (3)
∴ the listener’s detector counts p crests in the time interval,
t_p+1 – t₁ = pT₀ + (\(\frac {d+pv_sT_0}{v}\)) – \(\frac {d}{v}\)
The period of wave as recorded by the listener is,

Where, n = frequency recorded by the listener (apparent frequency)
n₀ = frequency emitted by the source (actual frequency).
This is the expression for apparent frequency when the listener is stationary and the source is moving away from the listener.

Question 39.
Derive an expression for apparent frequency when listener is stationary and source is moving towards the listener.
Answer:

i. Consider a source of sound S moving towards a stationary listener L with velocity v_s. Let the speed of sound with respect to the medium be v (always positive). The listener use a detector for counting each wave crest that reaches it.

ii. Let at t = 0, the source at point S₁ which is at a distance d from the listener, emit a Crest. This crest reaches the listener at time t₁, given as,
∴ t₁ = d/v. ……….(1)

iii. Let T₀ be the time period at which the waves are emitted.
At t = T₀, distance travelled by the source away from the stationary listener to reach point S₂ = v_sT₀.
Distance of S₂ from the listener = d – v_sT₀.
At S₂, The source emits second crest. This crest reaches the listener at
t₂ = T₀ + (\(\frac {d-v_sT_0}{v}\)) ………….. (2)

iv. Similarly, the time taken by the (p+1)^th crest (where, p = 1,2,3,…), emitted by the source at time pT₀, to reach the listener is given as,
t_p+1 = pT₀ + (\(\frac {d-pv_sT_0}{v}\)) ……………. (3)
∴ the listener’s detector counts p crests in the time interval,
t_p+1 – t₁ = pT₀ + (\(\frac {d-pv_sT_0}{v}\)) – \(\frac {d}{v}\)
∴ the period of wave as recorded by the listener is,

Where, n = frequency recorded by the listener (apparent frequency)
n₀ = frequency emitted by the source (actual frequency).
This is the expression for apparent frequency when the listener is stationary and the source is moving towards the listener.

Question 40.
Derive the expression for apparent frequency when the source is stationary and the listener is moving towards the source.
Answer:

i. Consider a listener approaching a stationary source S with velocity v_L as shown in figure. Let the speed of sound with respect to the medium be v (always positive).

ii. Let at time t = 0, the source emits the first wave when the listener L₁ is at an initial distance d from the source.
At time t = t₁ the listener receives the first wave at the position L₂.
Distance travelled by the listener towards the stationary source during time t₁ = v_Lt₁.
Distance travelled by the sound wave during time t₁ = d – v_Lt₁
∴ time taken by the sound wave to travel this distance, t₁ = \(\frac {d-v_Lt_1}{v}\)
∴ t₁ = \(\frac {d}{v+v_L}\) ………….. (1)

iii. Let at time t = T₀ (time period of the waves emitted by the source), the source emits a second wave.
At t = t₂, the listener receives the second wave. Distance travelled by the listener towards the stationary source during time t₂ = v_Lt₂.
Distance travelled by the sound wave during time t₂ = d – v_Lt₂
∴ time taken by the sound wave to travel this distance = \(\frac {d-v_Lt_2}{v}\)
However, this time should be counted after T₀, as the second wave was emitted at t = T₀.
∴ t₂ = T₀ + \(\frac {d-v_Lt_2}{v}\)
∴ t₂ = \(\frac {vT_0+d}{v+v_L}\) …………. (2)

iv. Similarly, for the third wave, we get,
t₃ = 2T₀ + \(\frac {d-v_Lt_3}{v}\)
∴ t₃ = \(\frac {2vT_0+d}{v+v_L}\) …………. (3)

v. Extending this argument to the (p + 1)^th wave, we get,
t_p+1 = pT₀ + \(\frac {d-v_Lt_{p+1}}{v}\)
∴ t_p+1 = \(\frac {pvT_0+d}{v+v_L}\) …………. (4)

vi. The observed or recorded period T is the time duration between instances of receiving successive waves.

This is the expression for apparent frequency when the source is stationary and the listener is moving towards the source.

Question 41.
Derive the expression for apparent frequency when the source is stationary and the listener is moving away from the source.
Answer:

i. Consider a listener moving away a stationary source S with velocity V_L. Let the speed of sound with respect to the medium be y (always positive).

ii. Let at time t = 0, the source emits the first wave when the listener L₁ is at an initial distance d from the source.
At time t = t₁ the listener receives the first wave at the position L₂.
Distance travelled by the listener away from the stationary source during time t₁ = v_Lt₁.
Distance travelled by the sound wave during time t₁ = d + v_Lt₁
∴ time taken by the sound wave to travel this distance,
t₁ = \(\frac {d+v_Lt_1}{v}\)
∴ t₁ = \(\frac {d}{v-v_L}\) ………….. (1)

iii. Let at time t = T₀ (time period of the waves emitted by the source), the source emits a second wave.
At t = t₂, the listener receives the second wave. Distance travelled by the listener away from the stationary source during time t₂ = v_Lt₂.
∴ Distance travelled by the sound wave during time t₂ = d + v_Lt₂.
∴ time taken by the sound wave to travel this distance = \(\frac {d+v_Lt_2}{v}\)
However, this time should be counted after T₀, as the second wave was emitted at t = T₀.
∴ t₂ = T₀ \(\frac {d+v_Lt_2}{v}\) ………….. (2)
∴ t₂ = \(\frac {vT_0+d}{v-v_L}\)

iv. Similarly for the third wave, we get
t₃ = 2T₀ \(\frac {d+v_Lt_3}{v}\)
∴ t₃ = \(\frac {2vT_0+d}{v-v_L}\) …………..(3)

v. Extending this argument to the (p + 1)^th wave, we get,
t_p+1 = pT₀ + \(\frac {d+v_Lt_{p+1}}{v}\)
∴ t_p+1 = \(\frac {pvT_0+d}{v-v_L}\) …………..(4)

vi. The observed or recorded period T is the time duration between instances of receiving successive waves.

This is the expression for apparent frequency when the source is stationary and the listener is moving away from the source.

Question 42.
State the common properties between Doppler effect of sound and light.
Answer:
i. The recorded frequency is different than the emitted frequency in case of relative motion between listener (or observer) and source (of sound or light waves).

ii. In case of relative approach, recorded frequency > emitted frequency.

iii. In case of relative recede, recorded frequency < emitted frequency.

iv. For values of listener velocity (v_L) or source velocity (v_s) much smaller then wave speed (speed of sound or light).
n = n₀ (1±\(\frac {v_r}{v}\))
Where, vr = relative velocity
n = actual frequency of the source
n₀ = apparent frequency of the source
v = velocity of sound in air.
(upper sign is used during relative approach and lower sign is during relative recede.)

v. If velocities of source and observer (listener) are along different lines, their respective components along the line joining them should be chosen for longitudinal Doppler effect and the same mathematical treatment is applicable.

Question 43.
State the major difference between Doppler effects of sound and light.
Answer:

1. Speed of light being absolute, only relative velocity between the observer and the source matter irrespective of who is in motion. However, for obtaining exact Doppler shift for sound waves, it is absolutely important to know who is in motion.
2. In case of light, classical and relativistic Doppler effects are different while sound only has classical doppler effect.
3. Presence of wind affects the velocity of sound which affects the Doppler shift in sound.

Question 44.
A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. (i) What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of 10 m s^-1 (b) recedes from the platform with a speed of 10 m s^-1? (ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 m s^-1.
Answer:
Given: v_s = 10 m/s, v = 340 m/s, n₀ = 400 Hz
Apparent frequency (n), velocity of sound (v_s) in each case
Formulae:
i. n = n₀ (\(\frac {v}{v-v_s}\))
ii. n = n₀ \(\frac {v}{v+v_s}\)
Calculation:
a. As the train approaches the platform, using formula (i),
n = 400 (\(\frac {340}{340-10}\)) = 421.12 Hz

b. As the train recedes from the platform, using formula (ii),
n = 400 (\(\frac {340}{340+10}\)) = 388.57 Hz

ii. The relative motion of source and observer results in the apparent change in the frequency but has no effect on the speed of sound. Hence, the speed of sound remains unchanged in both the cases.

Question 45.
A train blows a whistle of frequency 640 Hz in air. Find the difference in apparent frequencies of the whistle for a stationary observer, when the train moves towards and away from the observer with the speed of 72 km/hour. (Speed of sound in air = 340 m/s)
Answer:
Given: v_s = 72 km/ hr = 20 m/s, n₀ = 640 Hz,
v = 340 m/s
To find: Difference in apparent frequencies
(n_A – n’_A)
Formulae:
i. When the train moves towards the stationary observer then,
n_A = n₀ (\(\frac {v}{v-v_s}\))
ii. When the train moves away the stationary observer then,
n’_A = n₀ (\(\frac {v}{v+v_s}\))
Calculation: From formula (i),
n_A = 640 (\(\frac {340}{340-20}\))
∴ n_A = 680 Hz
From formula (ii),
n’_A = 640 (\(\frac {340}{340+20}\))
∴ n’_A = 604.4 Hz
Difference between n_A and n’_A
= n_A – n’_A = 75.6 Hz

Question 46.
The speed limit for a vehicle on road is 120 km/hr. A policeman detects a drop of 10% in the pitch of horn of a car as it passes him. Is the policeman justified in punishing the car driver for crossing the speed limit? (Given: Velocity of sound=340 m/s).
Answer:
Given: Speed limit, v_L = 120 km/hr
n’_A = n_A – \(\frac {10}{100}\) n_A = 0.9 n_A
Velocity of sound, v = 340 m/s
To Find: Velocity of source (v_s)
i. n_A = (\(\frac {v}{v-v_s}\))n
ii. n’_A = (\(\frac {v}{v+v_s}\))n
Calculation:
Dividing formula (i) by (ii),

Question 47.
A stationary source produces a note of frequency 350 Hz. An observer in a car moving towards the source measures the frequency of sound as 370 Hz. Find the speed of the car. What will be the frequency of sound as measured by the observer in the car if the car moves away from the source at the same speed? (Assume speed of sound = 340 m/s)
Answer:
Given: n₀ = 350 Hz, v = 340 m/s,
n_A = 370 Hz
To find: Speed (v_L), Frequency (n_A)
Formulae:
i. When the car moves towards the stationary source then,
n_A = n₀ (\(\frac {v+v_s}{v}\))

ii. When the car moves away from the stationary source then,
n_A = n₀ (\(\frac {v-v_L}{v}\))
Calculation: From formula (i),
370 = 35o (\(\frac {340+v_L}{340}\))
∴ 359.43 =340 +v_L
∴ v_L = 19.43 m/s
From formula (ii),
∴ n_A = 35o (\(\frac {340-20}{340}\)) = \(\frac {35}{34}\) × 320
∴ n_A = 329.41 Hz

Question 48.
A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 m s^-1. What are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m s^-1? The speed of sound in still air can be taken as 340 m s^-1.
Answer:
Blowing of wind changes the velocity of sound. As the wind is blowing in the direction of sound, effective speed of sound v_e = v + v_w = 340 + 10 = 350 m/s
As the source and listener both are at rest, frequency is unchanged, i.e., n = 400 Hz.
∴ wavelength, λ = \(\frac {v_e}{n}\) = \(\frac {350}{400}\) = 0.875 m
For still air, v_w = 0 and v_e = v = 340 m/s
Also, as observer runs towards the stationary train v_L = 10 m/s and v_s = 0
The frequency now heard by the observer,
n = n₀ (\(\frac {v+v_L}{v}\)) = 400 (\(\frac {340+10}{340}\))
= 411.76 Hz
As the source is at rest, wavelength does not change i.e., λ’ = λ = 0.875 m
Comparing the answers, it can be stated that, the situations in two cases are different.

Question 49.
A SONAR system fixed in a submarine operates at a frequency 40 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h^-1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m s^-1.
Answer:
Frequency of SONAR (source)
n = 40 kHz = 40 × 10³ Hz
Speed of sound waves, v = 1450 m s^-1
Speed of the listener, v_L = 360 km h^-1
= 360 × \(\frac {5}{18}\) ms^-1
= 100 m s^-1
Since, the source is at rest and the observer moves towards the source (SONAR).
We have,
n = n₀ (\(\frac {v+v_L}{v}\)) = 40 × 10³ × (\(\frac {1450+100}{14540}\))
∴ n = 4.276 × 10⁴ Hz
This frequency n’ is reflected by the enemy ship and is observed by the SONAR (which now acts as observer). Therefore, in this case v_s = 100 m s^-1
Apparent frequency,
n = n₀ (\(\frac {v}{v-v_s}\))
= 4.276 × 10⁴ × (\(\frac {1450}{1450-100}\)) = 4.59 × 10⁴ Hz
∴ n = 45.9 kHz

Question 50.
A rocket is moving at a speed of 220 m s^-1 towards a stationary target. While moving, it emits a wave of frequency 1200 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (i) the frequency of the sound as detected by the target and (ii) the frequency of the echo as detected by the rocket (velocity of sound = 330 m/s)
Answer:
Given: v_s = 220 m/s, v_L = 0 m/s, n = 1200 Hz
To find: Apparent frequency (n)
i. n = n₀ (\(\frac {v}{v-v_s}\))
ii. n = n₀ \(\frac {v+v_L}{v}\)
Calculation: For first case, observer is stationary and source i.e., rocket is moving towards the target.
Hence, using formula (i),
frequency of sound as detected by the target,
n = 1200 (\(\frac {330}{330-220}\)) = 3600 Hz
For second case, target acts as a source with frequency 3600 Hz as it is the source of echo. While rocket detector acts as an observer. This means, v_s = 0 and V_L = 220 m/s
Using formula (ii),
frequency of echo as detected by the rocket,
n = 3600 (\(\frac {330+220}{330}\)) = 600 Hz

Question 51.
A bat is flying about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly towards a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?
Answer:
Here, frequency of sound emitted by bat,
n = 40 kHz
Velocity of bat, v_s = 0.03 v
where v is velocity of sound.
The bat is moving towards the flat wall. This is the case of source in motion and the observer at rest.
Therefore, the frequency of sound reflected at the wall is,
n = n_s (\(\frac {v}{v-v_s}\)) = n × (\(\frac {v}{v-0.03v}\))
= n × \(\frac {1}{0.97}\) = \(\frac {n}{0.97}\)
The frequency n’ is reflected by the wall and is again received by the bat moving towards the wall. This is the case of an observer moving towards the source with velocity v_L = 0.03 v.
The frequency observed by bat,

Question 52.
A bat, flying at velocity V_B = 12.5 m/s, is followed by a car running at velocity V_c = 50 m/s. Actual directions of the velocities of the car and the bat are as shown in the figure below, both being in the same horizontal plane (the plane of the figure). To detect the car, the bat radiates ultrasonic waves of frequency 36 kHz. Speed of sound at surrounding temperature is 350 m/s.

There is an ultrasonic frequency detector fitted in the car. Calculate the frequency recorded by this detector. The ultrasonic waves radiated by the bat are reflected by the car. The bat detects these waves and from the detected frequency, it knows about the speed of the car. Calculate the frequency of the reflected waves as detected by the bat. (sin 37° = cos 53° ~ 0.6, sin 53° = cos 37° ~ 0.8)
Answer:
The components of velocities of the bat and the car, along the line joining them, are
v_c cos 53° ~ 50 × 0.6 = 30 m s^-1 and
v_B cos 37° ~ 12.5 × 0.8 = 10 m s^-1

Doppler shifted frequency n = n₀ (\(\frac {v±v_1}{v±v_s}\))
upper signs to be used during approach, lower signs during recede.
Case I: Frequency radiated by the bat
n₀ = 36 × 10³ Hz,
The source (bat) is receding, while the listener (car) is approaching
v_S = v_B cos 37° = 10 m/s and
V_L = v_C cos 53° = 30 m/s
∴ Frequency detected by the detector in the car,
n = n₀ (\(\frac {v+v_L}{v+v_s}\))
∴ n = 36 × 10³ (\(\frac {350+30}{350+10}\)) = 36 × 10³ × \(\frac {38}{36}\)
∴ n = 38 × 10³ Hz = 38 kHz

Case II: The car is the source.
Emitted frequency by the car, is given as,
n₀ = 38 × 10³ Hz,
Car, the source, is approaching the listener (bat).
Now bat-the listener is receding while car the source is approaching,
∴ v_s = v_c cos 53° = 30 m/s
∴ v_L = v_B cos 37° = 10 m/s
∴ n = n₀ (\(\frac {v-v_L}{v-v_s}\))
∴ n = 38 × 10³ (\(\frac {350-10}{350-30}\)) = 38 × 10³ × \(\frac {34}{33}\)
= 39.15 × 10³ Hz
∴ n = 39.15 kHz

Question 53.
Source of sound is placed at one end of a copper bar of length 1 km. Two sounds are heard at the other end at an interval of 2.75 seconds, (speed of sound in air = 330 m/s)
i. Why do we hear two sounds?
ii. Find the velocity of sound in copper.
Answer:
i. Two sounds are heard because sound travels through air as well as through copper.

ii. In air, t₁ = \(\frac {distence}{time}\) = \(\frac {1000}{330}\) = 3.03 s
As the time interval is 2.75 seconds and sound travels faster in copper.
∴ In copper, t₂ = 3.03 – 2.75 = 0.28 s
∴ velocity of sound in copper = \(\frac {1000}{0.28}\) = 3571 m/s

Question 54.
If all the persons mentioned in the table below are listening to a match commentary on the same channel at their respective locations positioning at same distance from television, then will they hear the same line of the commentary at same instant of time? Justify your answer.

Answer:
As the order of humidity for the above locations is Mumbai > Bangalore > Hyderabad > Delhi.
As velocity of sound increases with increase in humidity, the order of velocity of sound at their respective locations is Mumbai > Bangalore > Hyderabad > Delhi.
Hence, the order of persons who would listen the line of commentary first to last is Nilesh, Arjun, Virendra, Vikas.

Question 55.
Speed of sound is greater during day than at night. True or False? Justify your answer.
Answer:
True. At night, the amount of CO₂ in atmosphere increases the density of atmosphere. Since, Speed of sound is inversely proportional to the square root of density. Hence, speed of sound is greater during day than in night.

Question 56.
Case I: During summer (33 °C), Prakash was waiting for a train at the platform, train arrived tt seconds after he heard train’s whistle.
Case II: During winter (19 °C), train arrived t₂ seconds after Prakash heard the sound of train’s whistle.
i. Will t₂ be equal to t₁? Justify your answer.
ii. Calculate the velocity of sound in both the cases.
(velocity of sound in air at 0 °C = 330 m/s)
Answer:
i. Velocity of sound is directly proportional to square root of absolute temperature.
Hence, whistle’s sound will be first heard by Prakash in summer than in winter.
Therefore, the time interval between sound and train reaching Prakash in summer will be more than in winter.
i.e.,t₁ > t₂

ii. When t = 33 °C
∴ v₁ = v₀ + 0.61t
= 330 + 0.61 × 33
∴ v₁ = 350.13 m/s
When t = 19 °C
v₂ = v₀ + 0.61t
= 330 + 0.61 × 19
v₂ = 341.59 m/s

Question 57.
You are at a large outdoor concert, seated 300 m from speaker system. The concert is also being broadcast live. Consider a listener 5000 km away who receives the broadcast. Who will hear the music first, you or listener and by what time difference? (Speed of light = 3 × 10⁸ m/s and speed of sound in air = 343 m/s)
Answer:
s₁ = 300 m,
v₁ = 343 m/s,
∴ t₁ = \(\frac {s_1}{v_1}\) = \(\frac {300}{343}\) = 0.8746 s
Now,
s₂ = 5000 km = 5 × 10⁶ m,
v₂ = c = 3 × 10⁸ m/s
∴ t₂ = \(\frac {s_2}{c}\) = \(\frac {5×10^6}{3×10^8}\) = 0.0167 s
∴ t₂ < t₁
∴ Listener will hear the music first.
Time difference = t₁ – t₂
= 0.8746 – 0.0167
= 0.8579 s
The listener will hear the music first, about 0.8579 s before the person present at the concert.

Question 58.
When a source of sound moves towards a stationary observer, then the pitch increases. Give reason.
Answer:
When a source of sound moves towards a stationary observer, then the increase in pitch is due to actual or apparent change in wavelength. When the source of sound moves towards an observer at rest, waves get compressed and the effective velocity of sound waves relative to source becomes less than the actual velocity. Hence the wavelength of sound waves an decreases which results into increase in pitch.

Question 59.
In the examples given below, state if the wave motion is transverse, longitudinal or a combination of both?

1. Light waves travelling from Sun to Earth.
2. ultrasonic waves in air produced by a vibrating quartz crystal.
3. Waves produced by a motor boat sailing in water.

Answer:

1. Light waves (from Sun to Earth) are electromagnetic waves which are transverse in nature.
2. Ultrasonic waves in air are basically sound waves of frequency greater than the audible frequencies. Therefore, these waves are longitudinal.
3. The water surface is cut laterally and pushed backwards by the propeller of motor boat. Therefore, the waves produced are a mixture of longitudinal and transverse waves.

Question 60.
Internet my friend
Answer:
https://hyperphysics.phys-astr.gsu.edu/ hbase/hframe.html
[Students are expected to visit the above mentioned website and collect more information about sound.]

Multiple Choice Questions

Question 1.
Water waves are …………….
(A) longitudinal
(B) transverse
(C) both longitudinal and transverse
(D) neither longitudinal nor transverse
Answer:
(C) both longitudinal and transverse

Question 2.
Sound travels fastest in ……………..
(A) water
(B) air
(C) steel
(D) kerosene oil
Answer:
(C) steel

Question 3.
At room temperature, velocity of sound in air at 10 atmospheric pressure and at 1 atmospheric pressure will be in the ratio ……………..
(A) 10 : 1
(B) 1 : 10
(C) 1 : 1
(D) cannot say
Answer:
(C) 1 : 1

Question 4.
In a gas, velocity of sound varies directly as ………………
(A) square root of isothermal elasticity.
(B) square of isothermal elasticity.
(C) square root of adiabatic elasticity.
(D) adiabatic elasticity.
Answer:
(C) square root of adiabatic elasticity.

Question 5.
At a given temperature, velocity of sound in oxygen and in hydrogen has the ratio …………………
(A) 4 : 1
(B) 1 : 4
(C) 1 : 1
(D) 2 : 1
Answer:
(B) 1 : 4

Question 6.
With decrease in water vapour content in air, velocity of sound …………………..
(A) increases
(B) decreases
(C) remains constant
(D) cannot say
Answer:
(B) decreases

Question 7.
The temperature at which speed of sound in air becomes double its value at 0 °C is ……………….
(A) 546 °C
(B) 819 °C
(C) 273 °C
(D) 1092 °C
Answer:
(B) 819 °C

Question 8.
The velocity of sound in air at NTP is 330 m/s. What will be its value when temperature is doubled and pressure is halved?
(A) 330 m/s
(B) 165 m/s
(C) 330 √2 m/s
(D) \(\frac {330}{√2}\) m/s
Answer:
(D) \(\frac {330}{√2}\) m/s

Question 9.
A series of ocean waves, each 5.0 m from crest to crest, moving past the observer at a rate of 2 waves per second have wave velocity
(A) 2.5 m/s
(B) 5.0 m/s
(C) 8.0 m/s
(D) 10.0 m/s
Answer:
(D) 10.0 m/s

Question 10.
A radio station broadcasts at 760 kHz. What is the wavelength of the station?
(A) 395 m
(B) 790 m
(C) 760 m
(D) 197.5 m
Answer:
(A) 395 m

Question 11.
If the bulk modulus of water is 2100 MPa, what is the speed of sound in water?
(A) 1450 m/s
(B) 2100 m/s
(C) 0.21 m/s
(D) 21 m/s
Answer:
(A) 1450 m/s

Question 12.
If speed of sound in air at 0°C is 331 m/s. What will be its value at 35° C?
(A) 331 m/s
(B) 366 m/s
(C) 351.6 m/s
(D) 332 m/s.
Answer:
(C) 351.6 m/s

Question 13.
For a progressive wave, in the usual notation
(A) v = λT
(B) n = \(\frac {v}{λ}\)
(C) T = λv
(D) λ = \(\frac {1}{n}\)
Answer:
(B) n = \(\frac {v}{λ}\)

Question 14.
At normal temperature, for an echo to be heard the reflecting surface should be at a minimum distance of ………………. m.
(A) 34.4
(B) 17.2
(C) 10
(D) 20
Answer:
(B) 17.2

Question 15.
In a transverse wave, are regions of negative displacement.
(A) rarefactions
(B) compressions
(C) crests
(D) troughs
Answer:
(D) troughs

Question 16.
If pressure of air gets doubled at constant temperature then velocity of sound in air ……………….
(A) gets doubled
(B) remains unchanged
(C) √2 times initial velocity
(D) becomes half
Answer:
(B) remains unchanged

Question 17.
Wave motion has ……………………
(A) single periodicity.
(B) double periodicity.
(C) only periodicity in space.
(D) only periodicity in time.
Answer:
(B) double periodicity.

Question 18.
The speed of the mechanical wave depends upon ………………
(A) elastic properties of the medium only.
(B) density of the medium only.
(C) elastic properties and density of the medium
(D) initial speed.
Answer:
(C) elastic properties and density of the medium

Question 19.
Longitudinal waves CANNOT be …………………
(A) reflected
(B) refracted
(C) scattered
(D) polarised
Answer:
(D) polarised

Question 20.
Wavelength of the transverse wave is 30 cm. If the particle at some instant has displacement 2 cm, find the displacement of the particle 15 cm away at the same instant.
(A) 2 cm
(B) 17 cm
(C) -2 cm
(D) -17 cm
Answer:
(C) -2 cm

Question 21.
The wavelength of sound in air is 1.5 m and that in liquid is 2 m. If the velocity of sound in air is 330 m/s, the velocity of sound in liquid is
(A) 330 m/s
(B) 440 m/’s
(C) 495 m/s
(D) 660 m/s
Answer:
(B) 440 m/’s

Question 22.
The velocity of sound in a gas is 340 m/s at the pressure P, what will be the velocity of the gas when only pressure is doubled and temperature same?
(A) 170 m/s
(B) 243 m/s
(C) 340 m/s
(D) 680 m/s
Answer:
(C) 340 m/s

Question 23.
Choose the correct statement.
(A) For 1 °C rise in temperature, velocity of sound increases by 0.61 m/s.
(B) For 1 °C rise in temperature, velocity of sound decreases by 0.61 m/s.
(C) For 1 °C rise in temperature, velocity of sound decreases by \(\frac {1}{273}\) m/s.
(D) For 1 °C rise in temperature, velocity of sound increases by \(\frac {1}{273}\) m/s.
Answer:
(A) For 1 °C rise in temperature, velocity of sound increases by 0.61 m/s.

Question 24.
A sound note emitted from a certain source has a velocity of 300 m/s in air and 1050 m/s in water. If the wavelength of sound note in air is 2 m, the wavelength in water is …………
(A) 2 m
(B) 6 m
(C) 7 m
(D) 12 m
Answer:
(C) 7 m

Question 25.
A thunder clap was heard 6 seconds after a lightening flash was seen. If the speed of sound in air is 340 m/s at the time of observation, the distance of the listener from the thunder clap is ………………
(A) 56.6 m
(B) 346 m
(C) 1020 m
(D) 2040 m
Answer:
(D) 2040 m

Question 26.
The speed of sound in air at NTP is 330 m/s. The period of sound wave of wavelength 66 cm is …………………
(A) 0.2 s
(B) 0.1
(C) 0.1 × 10^-2 s
(D) 0.2 × 10^-2 s
Answer:
(D) 0.2 × 10^-2 s

Question 27.
If the velocity of sound in hydrogen is 1248 m/s, the velocity of sound in oxygen is [Given: M_O = 32 and M_H = 2]
(A) 1248 m/s
(B) 624 m/s
(C) 312 m/s
(D) 300 m/s
Answer:
(C) 312 m/s

Question 28.
If the source is moving away from the observer, then the apparent frequency …………..
(A) will increase
(B) will remain the same
(C) will be zero
(D) will decrease
Answer:
(D) will decrease

Question 29.
The working of SONAR is based on …………………
(A) resonance
(B) speed of a star
(C) Doppler effect
(D) speed of rotation of sun
Answer:
(C) Doppler effect

Question 30.
The formula for speed of a transverse wave on a stretched spring is ……………… (m = linear mass density, T = tension in Spring)
(A) v = \(\sqrt{\frac {m}{T}}\)
(B) v = \(\sqrt{\frac {T}{m}}\)
(C) v = (\(\frac {m}{T}\))^{\(\frac {3}{2}\)}
(D) v = (\(\frac {T}{m}\))^{\(\frac {3}{2}\)}
Answer:
(B) v = \(\sqrt{\frac {T}{m}}\)

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 7 Thermal Properties of Matter Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 7 Thermal Properties of Matter

Question 1.
State various units of heat and relate them to SI units of heat.
Answer:

1. CGS unit of heat: erg and it is related to SI unit as 1 J = 10⁷ erg
2. Thermodynamic unit of heat: calorie (cal) and it is related as 1 cal = 4.184 J

Question 2.
What is thermal equilibrium?
Answer:

1. When two bodies at different temperatures come into the contact with each other, they exchange heat.
2. After some time, temperature of two bodies become equal and heat transfer between them stops.
3. The two bodies are then said to be in thermal equilibrium with each other.

Question 3.
State true of false. Correct the statement and rewrite if false.
i. Heat transfer takes place between the body and the surrounding medium until the body and the surrounding medium are at the same temperature.
ii. Whenever two bodies are in contact, there is a transfer of heat.
Answer:

1. True.
2. False.

Whenever two bodies at different temperature are in contact, there is a transfer of heat.

Question 4.
Give reason: Temperature is said to be a measure of average kinetic energy of the atoms/molecules of the body.
Answer:

1. Matter consists of particles which are in continuous vibrational motion and thereby possess kinetic energy.
2. When external energy is provided to these particles, internal energy of particles increases.
3. This increase in internal energy is in the form of increased kinetic energy of atoms/molecules and raises temperature of body (except at melting or boiling point of the body).
4. Greater the kinetic energy, faster the atoms/ molecules move and temperature of body becomes higher.
5. Thus, temperature of body is directly proportional to its kinetic energy.
   Hence, temperature is said to be a measure of average kinetic energy of the atoms and molecules of the body.

Question 5.
Why do solid particles possess potential energy?
Answer:
The solid particles possess potential energy due to the interatomic forces that hold the particles together at some mean fixed positions.

Question 6.
What happens when heat is supplied to a solid at its melting point?
Answer:

1. When heat is supplied to a solid at its melting point, average kinetic energy of constituent particles does not change.
2. As a result, temperature of body remains constant.
3. Supplied energy is used to weaken the bonds between constituent particles.
4. While order of magnitude of average distance between the molecules remains almost same as solid, substance melts, i.e., changes into liquid state, at melting point.

Question 7.
Why do solids have definite shape and volume?
Answer:
The solid particles possess potential energy due to the interatomic forces that hold the particles together at some mean fixed positions.
Hence, solids have definite shape and volume.

Question 8.
Why is density of liquid at melting point nearly same as density of solids at melting point?
Answer:

1. During change of state from solid to liquid, mass of substance does not change.
2. Also, mean distance between particles during change of state does not alter at melting point.
3. Density depends upon mass and volume, in turn, on mean distance between particles.
   Hence, density of liquid at melting point is nearly same as density of solids at melting point.

Question 9.
State true or false. If false correct the statement and rewrite.
Due to weakened interatomic bonds liquid do not possess definite volume but have definite shape.
Answer:
False.
Due to weakened bonds liquids do not possess definite shape but have definite volume.

Question 10.
What happens when heat is supplied to liquid its freezing point (melting point)?
Answer:

1. On heating, the atoms/molecules in liquid gain kinetic energy and temperature of the liquid increases.
2. If liquid is continued to heat further, at the boiling point, the constituents overcome the interatomic/molecular forces.
3. The mean distance between the constituents increases so that the particles are farther apart.
4. At boiling point, the liquid gets converted into gaseous state.

Question 11.
Why, according to kinetic theory of gases, gases have neither definite volume nor shape?
Answer:
As per kinetic theory of gases, for an ideal gas, there are no forces between the molecules of a gas. Hence, gases neither have a definite volume nor shape.

Question 12.
Match the pairs.

Answer:
i – b,
ii – c,
iii – a

Question 13.
Distinguish between an adiabatic wall and diathermic wall.
Answer:

Question 14.
State and explain zeroth law of thermodynamics.
Answer:
Statement: If two bodies A and B are in thermal equilibrium and also A and C are in thermal equilibrium then B and C are also in thermal equilibrium.
Explanation:

1. Consider two sections of a container separated by an adiabatic wall containing two different gases as system A and system B.
2. Systems A and B are independently brought in thermal equilibrium with a system C.
3. When the adiabatic wall separating systems A and B is removed, there will be no transfer of heat from system A to system B or vice versa.
4. This indicates that systems A and B are also in thermal equilibrium.
5. This means, if systems A and B are separately in thermal equilibrium with a system C, then A and B are also mutually in thermal equilibrium.

Question 15.
What is thermometry? What is thermometer?
Answer:
Thermometry is the science of temperature and its measurement. The device used to measure temperature is a thermometer.

Question 16.
State the principle used to measure the temperature of a system using a thermometer.
Answer:
When two or more systems/bodies are in thermal equilibrium, their temperatures are same. This principle is used to measure the temperature of a system by using a thermometer.

Question 17.
Explain how thermal equilibrium is attained between thermometer and the patient holding thermometer, in mouth.
Answer:

1. Thermometer indicating lower temperature is held in mouth by patient.
2. As body of patient is at higher temperature, heat energy is transferred from patient to thermometer.
3. When temperature of thermometer becomes same as temperature of patient, heat exchange stops and thermal equilibrium is attained between thermometer and body of patient.

Question 18.
Define the following terms.
i) Ice point
ii) Steam point
Answer:

1. Ice point: The temperature at which pure water freezes at one standard atmospheric pressure is called as ice point or freezing point.
2. Steam point: The temperature at which pure water boils into steam or steam changes to liquid water at one standard atmospheric pressure is called as steam point or boiling point.

Question 19.
Explain Celsius and fahrenheit scale of temperature. Give relation between the two scales with the help of the graph.
Answer:

1. Celsius scale:
   • The ice point (melting point of pure ice) is marked as 0 °C (lower point) and steam point (boiling point of water) is marked as 100 °C (higher point).
   • Both are taken at one atmospheric pressure.
   • The interval between these points is divided into loo equal pans. Each of these parts is called as one degree celsius and it is written as 1 oc.
2. Fahrenheit scale:
   • The ice point (melting point of pure ice) is marked as 32 °F and steam point (boiling point of water) is marked as 212 °F.
   • The interval between these two reference points is divided into 180 equal parts. Each part is called as degree fahrenheit and is written as 1 °F.
3. Relation between fahrenheit temperature and celsius temperature:
   \(\frac{\mathrm{T}_{\mathrm{F}}-32}{180}\) = \(\frac{T_{C}-0}{100}\)
   Where, T_F = temperature in fahrenheit scales.
   T_C = temperature in celsius scale.
   The graph of T_F versus T_C is as shown
   

Question 20.
What is thermometer? Explain with examples, the thermometric property used in a thermometer.
Answer:

1. An instrument designed to measure temperature is called as thermometer.
2. Any property of a substance which changes sufficiently with temperature can be used as a basis of constructing a thermometer and is known as the thermometric property.
3. There are different types of thermometers.
   • In a constant volume gas thermometer, the pressure of a fixed volume of gas (measured by the difference in height) is used as the thermometric property.
   • The liquid-in-glass thermometer depends on the change in volume of the liquid with temperature. Small change in temperature changes the volume of liquid considerably. Two such liquids are mercury and alcohol. Mercury thermometers are used for measurement of temperature range -39 °C to 357 °C while alcohol thermometers are used only to measure temperatures near ice point (melting point of pure ice).
   • The resistance thermometer uses the change of electrical resistance of a metal wire with temperature.
   • Normally in research laboratories, a thermocouple is used to measure the temperature. A thermocouple is a junction of two different metals or alloys eg.: copper and iron joined together.
   • When two such junctions at the two ends of two dissimilar metal rods are kept at two different temperatures, an electromotive force is generated that can be calibrated to measure the temperature.
4. Thermometers are calibrated so that a numerical value may be assigned to a given temperature. The standard fixed points are melting point of ice and boiling point of water.

Question 21.
State characteristics of thermometer.
Answer:

1. Thermometer must be sensitive, i.e., a noticeable change in the thermometric property should be observed for a very small change in temperature.
2. It has to be accurate.
3. It should be easily reproducible.
4. It is important that the system attains thermal equilibrium with the thermometer quickly.

Question 22.
Explain relation between unknown temperature T and thermodynamic property P_T at temperature T.
Answer:
If the values of a thermometric property are P₁ and P₂ at the ice point (0 °C) and steam point (100 °C) respectively and the value of this property is P_T at unknown temperature T, then T is given by the following equation.
T = \(\frac{100\left(P_{T}-P_{1}\right)}{P_{2}-P_{1}}\)

Question 23.
State true or false. If false correct the statement and rewrite.
Ideally, there should be no difference in temperatures recorded on two different thermometers.
Answer:
True.

Question 24.
List an advantage and a disadvantage of constant volume gas thermometer.
Answer:
Advantage: There is no difference in temperatures recorded on two different constant volume gas thermometers. Hence, it is very accurate.
Disadvantage: Constant volume gas thermometer is bulky instrument. Hence, it is not easily portable.

Question 25.
Give short note on liquid-in-glass thermometer.
Answer:

1. Liquid-in-glass thermometer depends on the change in volume of the liquid with temperature.
2. When the bulb is heated, the liquid in a glass bulb expands upward in a capillary tube.
3. The liquid is such that it is easily seen and expands (or contracts) rapidly and by a large amount over a wide range of temperature.
4. Most commonly used liquids are mercury and alcohol as they remain in liquid state over a wide range. Mercury freezes at -39 °C and boils at 357 °C; alcohol freezes at -115 °C and boils at 78 °C.

Question 26.
What are thermochromic liquids? Give two examples.
Answer:

1. Thermochromic liquids are ones which change colour with temperature.
2. These liquids are very sensitive to temperature, especially in range of room temperature.
3. Hence, only specific liquids display distinct colour variations at normal temperature.
   Examples: Titanium dioxide and zinc oxide are white at room temperature but when heated change to yellow.

Question 27.
Write a note on resistance thermometer.
Answer:

1. Resistance thermometer uses the change of electrical resistance of a metal wire with temperature.
2. It measures temperature accurately in the range -2000 °C to 1200 °C is best for steady temperatures.
3. It is bulky and hence not easily portable.

Solved Examples

Question 28.
If the temperature in the room is 29 °C, what is its temperature in degree fahrenheit?
Solution:
Given: T_C = 29 °C
To find: Temperature in degree fahrenheit (T_F)

Answer:
Temperature in the room in degree fahrenheit is 84.2 °F.

Question 29.
Average room temperature on a normal day is 27 °C. What is the room temperature in °F?
Solution:
T_C = 27 °C
Room temperature in °F
Calculation: From formula,

Answer:
Room temperature in °F is 80.6 °F.

Question 30.
Normal human body temperature in fahrenheit is 98.4 °F. What is the body temperature in °C?
Solution:
T_F = 98.4 °F
Formula: Body temperature in °C (T_C)

Body temperature in °C is 36.89 °C.

Question 31.
The length of a mercury column in a mercury-in-glass thermometer is 25 mm at the ice point and 180 mm at the steam point. What is the temperature when the length is 60 mm?
Solution:
Here the thermometric property P is the length of the mercury column.
Using equation,
T = \(\frac{100\left(\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{1}\right)}{\left(\mathrm{P}_{2}-\mathrm{P}_{1}\right)}\)
For P₁ = 25 mm,
P₂ = 180 mm,
P_T = 60 mm
T = \(\frac{100(60-25)}{(180-25)}\)
= 22.58 °C
The temperature corresponding to the length of 60 mm is 22.58 °C.

Question 32.
A resistance thermometer has resistance 95.2 Ω at the ice point and 138.6 Ω at the steam point. What resistance would be obtained if the actual temperature is 27 °C?
Solution:
Here the thermometric property P is the resistance. If R is the resistance at 27 °C,
Using equation,
T = \(\frac{100\left(\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{1}\right)}{\left(\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{1}\right)}\)
For T = 27 °C, P₁ = 95.2 Ω, P₂ = 138.6 Ω .
∴ 27 = \(\frac{100(\mathrm{R}-95.2)}{(138.6-95.2)}\)
∴ R = \(\frac{27 \times(138.6-95.2)}{100}\) + 95.2
= 11.72 + 95.2
= 106.92Ω
The resistance obtained would be 106.92 Ω.

Question 33.
Explain the need for thermodynamic (absolute) scale.
Answer:

1. The two fixed point scale, Celsius scale and Fahrenheit scale had a practical shortcoming for calibrating the scale.
2. It was difficult to precisely control the pressure and identify the fixed points, especially for the boiling point as the boiling temperature is very sensitive to changes in pressure.
3. Hence, a one fixed point scale was adopted to define a temperature scale.
4. This scale is called the absolute scale or thermodynamic scale.

Question 34.
What is triple point of water? State its physical significance.
Answer:

1. The triple point of water is that point where water in a solid, liquid and gas state co-exists in equilibrium and this occurs only at a unique temperature and a pressure.
2. To know the triple point, one has to see that three phases coexist in equilibrium and no one phase in dominating. This occurs for each substance at a single unique combination of temperature and pressure.
3. Thus, if three phases of water solid ice, liquid water and water vapour coexist, the pressure and temperature are automatically fixed.
4. Internationally, triple point of water has been assigned as 273.16 K at pressure equal to 6.11 × 10² Pa or 6.11 × 10^-3 atmosphere, as the standard fixed point for calibration of thermometers.
5. The physical significance of triple point of water is that, it represents unique condition and it is used to define the absolute temperature.

Question 35.
Write a short note on absolute scale of temperature.
Answer:

1. The absolute scale of temperature, is so termed since ills based on the properties of an ideal gas and does not depend on the property of any particular substance.
2. The zero of this scale is ideally the lowest temperature possible although it has not been achieved in practice.
3. It is termed as Kelvin scale after Lord Kelvin with its zero at -273.15 °C and temperature intervals same as that on the Celsius scale. It is written as K (without °).

Question 36.
Draw a neat and well labelled diagram to show comparison of kelvin, Celsius and fahrenheit temperature scales.
Answer:

Question 37.
Answer the following:
i) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?
ii) The absolute temperature (Kelvin scale) T is related to temperature t_c on the Celsius scale by t_c = T – 273.15. Why do we have 273.15 in this relation and not 273.16?
Answer:
i) The triple point of water has been assigned a fixed value of 273.15 K. This number represents a unique value associated with a unique condition of temperature and pressure in which all the three phases of water co-exist. On the other hand, melting point of ice and boiling point of water do not have a unique set of values as they are subject to changes in pressure and volume. For this reason, triple point of water is a standard fixed point in modem thermometry.

ii) On Celsius scale, the melting point of ice at normal pressure has a value 0 °C. The value corresponding to this value on the absolute scale is 273.15 K. The value 273.16 K denotes the triple point of water which has a value,
273.16 – 273.15 = 0.01 °C on Celsius scale as per the given relation.

Question 38.
State Charles’ law and give its formula.
Answer:
Charles’ law:
At constant pressure, volume of a fixed mass of gas is directly proportional to its absolute temperature.
Mathematically,
V ∝ T … ( at constant pressure)
∴ V = kT
where k is constant of proportionality
∴ \(\frac{\mathrm{V}}{\mathrm{T}}\) = k = constant
For two gases, \(\frac{V_{1}}{T_{1}}\) = \(\frac{V_{2}}{T_{2}}\) = constant.

Question 39.
State Pressure law and give its formula.
Answer:
Pressure (Gay Lussac’s) law:
At constant volume, pressure of a fixed mass of gas is directly proportional to its absolute temperature.
Mathematically,
P ∝ T .. .(at constant volume)
∴ P = kT
Where, k is constant of proportionality P
∴ \(\frac{\mathrm{P}}{\mathrm{T}}\) = k = constant
For two gases, \(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\) = constant

Question 40.
State Boyle’s law and give its formula.
Answer:
Boyle’s law:
At constant temperature, the volume of a fixed mass of a gas is inversely proportional to its pressure.
Mathematically,
V ∝ \(\frac{1}{\mathrm{P}}\).. ..(at constant temperature)
V = k × \(\frac{1}{\mathrm{P}}\)
where, k is constant of proportionality.
PV = k = constant For two gases,
P₁V₁ = P₂V₂ = constant

Question 41.
Derive ideal gas equation PV = nRT.
Answer:

1. The relation between three variables of a gas i.e., pressure, volume and absolute temperature is called as ideal gas equation.
   From Boyle’s law,
   V ∝ \(\frac{\mathrm{T}}{\mathrm{P}}\), at constant temperature ….(1)
   From Charles’ law,
   V ∝ T, at constant pressure … .(2)
2. Combining equations (1) and (2) we get,
   ∴ V ∝ \(\frac{\mathrm{T}}{\mathrm{P}}\)
   ∴ \(\frac{\mathrm{PV}}{\mathrm{T}}\)= constant
3. For one mole of a gas,
   \(\frac{\mathrm{PV}}{\mathrm{T}}\) = R or PV = RT … (3)
   where R is the constant of proportionality.
4. Equation (3) is called ideal gas equation. The value of constant R is same for all gases. Therefore, R is called as universal gas constant. R = 8.31 JK^-1mol^-1.
5. For ‘n’ moles of gas, i.e. if the gas contains ‘n’ moles, equation (3) can be written as,
   PV = nRT

Solved Examples

Question 42.
Express T = 24.57 K in Celsius and fahrenheit.
Solution:
Given: T_K = 24.57 K
To find: Temperature in Celsius (T_C),
Temperature in fahrenheit (T_F)

Calculation:
From formula,
∴ T_C = T_K – 273.15
= 24.57 – 273.15
= -248.58 °C

Temperature in celsius (T_C) is – 248.58 °C
Temperature in fahrenheit (T_F) is -415.44 °F

Question 43.
Calculate the temperature which has the same value on fahrenheit scale and kelvin scale.
Solution:
Given: T_K = T_F = x
To find: Temperature at which Kelvin and Fahrenheit scales coincide (x)
Formula: \(\frac{\mathrm{T}_{\mathrm{F}}-32}{180}\) = \(\frac{\mathrm{T}_{\mathrm{K}}-273.15}{100}\)
Calculation: From formula.
∴ \(\frac{x-32}{180}\) = \(\frac{x-273.15}{100}\)
∴ 5(x – 32) = 9(x – 273.15)
∴ 5x – 160 = 9x – 2458.35
∴ 4x = 2298.35
∴ x = \(\frac{2298.35}{4}\) = 574.6
∴ x = 574.6 °F
The temperature at which kelvin and fahrenheit scales coincide is 574.6 °F.

Question 44.
The triple points of neon and carbon dioxide are 24.57 K and 216.55 K, respectively. Express these temperatures on
the celsius and fahrenheit scales. (NCERT)
Solution:
Given: For neon, T_K = 24.57 K
For carbon dioxide, T_K = 216.55 K
To find:
i) Triple point of neon on celsius (T_CN) and fahrenheit scale (T_FN)
ii) Triple point of carbon dioxide on Celsius (T_CC) and fahrenheit scale (T_FC)

Formulae:
i) T_K – 273.15 = T_C
ii) \(\frac{\mathrm{T}_{\mathrm{K}}-273.15}{100}\) = \(\frac{\mathrm{T}_{\mathrm{c}}-32}{180}\)
Calculation: From formula (i),
T_C = T_K – 273.15
For neon, T_CN = 24.57 – 273.15
∴ T_CN = -248.58 °c
For carbon dioxide.
T_CC = 216.55 – 273.15
∴ T_CC = -56.60 °c
From formula (ii),
T_F = \(\frac{9}{5}\)(T_k – 273.15) + 32
For neon, T_k = 24.57 K
∴ T_FN = \(\frac{9}{5}\)[24.57 – 273.15] + 32
∴ T_FN = -415.44 °F
For CO₂, T_K = 216.55 K
∴ T_FC = \(\frac{9}{5}\)(216.55 – 273.15) + 32
∴ T_FC = -69.88 °F
i) Triple point of neon on celsius scale is -248.58 °c and on a fahrenheit scale is -415.44 °F.
ii) Triple point of carbon-dioxide on celsius scale is -56.60 °C and on fahrenheit scale is -69.88 °F.

Question 45.
Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between T_A and T_B?
Solution:
Triple point of water is, T = 273.16 K
Since the absolute scales measure the triple point as 200 A and 350 B.
∴ 200A = 350B = 273.16 K
∴ 1A = \(\frac{273.16}{200}\)K and 1B = \(\frac{273.16}{350}\)K
If T_A and T_B are the temperatures on the two scales, then
\(\frac{273.16}{200}\)T_A = \(\frac{273.16}{350}\)T_B
∴ T_A = \(\frac{200}{350}\) T_B = \(\frac{4}{7}\)T_B
\(\frac{\mathbf{T}_{\mathbf{A}}}{\mathbf{T}_{\mathbf{B}}}\) = \(\frac{4}{7}\)
The relation between T_A and T_B is T_A : T_B = 4 : 7

Question 46.
The pressure reading in a thermometer at steam point is 1.367 × 10³ Pa. What is pressure reading at triple point knowing the linear relationship between temperature and pressure?
Solution:
Given: P = 1.367 × 10³ Pa at steam point (T) i.e., at 273.15 + 100 = 373.15 K.
Linear relationship between temperature and pressure means that.
P ∝ T ⇒ P₁T₁ = P₂T₂
To find: Pressure reading (P_triple)
Formula: P_triple = 273.16 × \(\left(\frac{\mathrm{P}}{\mathrm{T}}\right)\)
where P_triple and P are the pressures at temperature of triple point (273.16 K) and T (375.15 K) respectively.
Calculation: From formula,
∴ P_triple = 273.16 × \(\left(\frac{1.367 \times 10^{3}}{373.15}\right)\)
= 1000 × 10³ Pa
Pressure reading is 1.000 × 10³ Pa.

Question 47.
When the pressure of 0.75 litre of a gas at 27 °C is doubled, its temperature rises to 111°C. Calculate the final volume of a gas.
Solution:
Given: V₁ = 0.75 litre = 750 cm³,
T₁ = 27 + 273.15 = 300.15 K,
T₂ = 111 + 273.15 = 384.15 K,
P₂ = 2P₁
To find: Final volume (V₂)
Formula: \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}\) = \(\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
Calculation: From formula,

∴ V₂ = 480 cm³
The final volume of the gas is 480 cm³.

Question 48.
A certain mass of a gas at 20 °C is heated until both its pressure and volume are doubled. Calculate the final temperature.
Solution:
T₁ = 20 + 273.15 = 293.15 K,
P₂ = 2P₁, and V₂ = 2V₁
To find: Final temperature (T₂)
Formula: \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}\) = \(\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
Calculation: From formula,
\(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}\) = \(\frac{\left(2 \mathrm{P}_{1}\right)\left(2 \mathrm{~V}_{1}\right)}{\mathrm{T}_{2}}\)
∴ \(\frac{1}{\mathrm{~T}_{1}}\) = \(\frac{4}{\mathrm{~T}_{2}}\)
∴ T₂ = 4 × 293.15 = 1172.6 K
= 1172.6 – 273.15 = 899.45 °C
∴ T₂ = 1172.6 K or 899.45 °C
The final temperature of the gas is 1172.6 K or 899.45 °C.

Question 49.
Explain how substances expand on the basis of vibrational motion of atoms.
Answer:

1. The atoms in a solid vibrate about their mean positions.
2. When heated, they vibrate faster and force each other to move a little farther apart. This results into expansion.
3. The molecules in a liquid or gas move with certain speed.
4. When heated, they move faster and force each other to move a little farther apart. This results
   in expansion of liquids and gases on heating.
5. The expansion is more in liquids than in solids; gases expand even more.

Question 50.
What is thermal expansion? List types of thermal expansion.
Answer:

1. A change in the temperature of a body causes change in its dimensions.
2. The increase in the dimensions of a body due to an increase in its temperature is called thermal expansion.
3. There are three types of thermal expansion:
   • Linear expansion
   • Areal expansion
   • Volume expansion

Question 51.
Define linear expansion of solids.
Answer:
The expansion in length of a solid due to thermal energy is called linear expansion.

Question 52.
Define coefficient of linear expansion of solid. State its unit and dimensions.
Answer:

1. The coefficient of linear expansion of a solid is defined as increase in the length per unit original length at 0 °C per degree rise in temperature.
   It is denoted by α and given by,
   α = \(\frac{l_{2}-l_{1}}{l_{1}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)}\)
   where l₁ = initial length at temperature T₁ °C
   l₂ = final length at temperature T₂ °C
2. Unit: °C^-1 or K^-1
3. Dimensions: [L⁰M⁰T⁰K^-1]

Question 53.
Derive an expression for the coefficient of linear expansion in solid.
Answer:
i) If the substance is in the form of a long rod of length l, then for small change ∆T, in temperature, the fractional change ∆l/l, in length is directly proportional to ∆T.

[Note: Linear expansion ∆l is exaggerated for explanation.]
\(\frac{\Delta l}{l}\) ∝ ∆T
∴ \(\frac{\Delta l}{l}\) = α∆T … (1)
where, α is called the coefficient of linear expansion of solid.

ii) Rearranging terms in equation (1),
α = \(\frac{\Delta l}{l \Delta \mathrm{T}}\)
= \(\frac{l_{\mathrm{T}}-l_{0}}{l_{0}\left(\mathrm{~T}-\mathrm{T}_{0}\right)}\)
where,
l₀ = length of rod at 0 °C,
l_T = length of rod when heated to T °C,
T₀ = 0 °C (initial temperature)
T = final temperature,
∆l = l_T – T₀ = change in length,
∆T = T – T₀ = rise in temperature.

iii) If l₀ = 1 m and T – T₀ = 1 °C, then α = l_T – l₀ (numerically).

iv) As magnitude of α varies negligibly with temperature, it is assumed to be constant for a particular material.
Hence, it is not essential to take initial temperature as 0 °C.
This modifies equation (2) into,
α = \(\frac{l_{2}-l_{1}}{l_{1}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)}\)
where l₁ = initial length at temperature T₁ °C
l₂ = final length at temperature T₂ °C

Question 54.
State true or false. If false correct the statement and rewrite.
i) Coefficient of linear expansion is same for all substances.
ii) Metals have high values for the coefficient of linear expansion, than non-metals.
Answer:

1. False.
   Coefficient of linear expansion is different for different substances.
2. True.

Question 55.
Define areal expansion of solids.
Answer:
The increase in the surface area of a solid, on heating is called areal expansion or superficial expansion of solids.

Question 56.
Define coefficient of superficial (areal) expansion of a solid, state its unit and dimensions.
Answer:

1. Coefficient of superficial (areal) expansion of a solid is defined as the increase in area per unit original area at 0 °C per degree rise in temperature.
   It is denoted by β and given by,
   β = \(\frac{A_{2}-A_{1}}{A_{1}\left(T_{2}-T_{1}\right)}\)
   where, A₁ = Area of solid sheet at T₁ °C,
   A₂ = Area of solid sheet at T₂ °C
2. Unit: 0°C^-1 or K^-1
3. Dimensions: [L⁰M⁰T⁰K^-1

Question 57.
Derive expression for coefficient of areal expansion.
Answer:
i) If a substance is in the form of a plate of area A, then for small change ∆T in temperature, the fractional change in area, ∆A/A in figure given below, is directly proportional to ∆T.

[Note: Areal expansion ∆A is exaggerated for explanation]
\(\frac{\Delta \mathrm{A}}{\mathrm{A}}\) ∝ ∆T
∴ \(\frac{\Delta \mathrm{A}}{\mathrm{A}}\) = β∆T … (1)
where β is called the coefficient of areal expansion of solid.

ii) Rearranging terms in equation (1),
β = \(\frac{\Delta \mathrm{A}}{\mathrm{A} \Delta \mathrm{T}}\)
= \(\frac{A_{T}-A_{0}}{A_{0}\left(T-T_{0}\right)}\)
where,
A₀ = volume at 0 °C,
A_T = volume when heated to T °C,
T₀ = 0 °C (initial temperature),
T = final temperature,
∆A = A_T – A₀ = change in area,
∆T = T – T₀ = rise in temperature.
iii) If A₀ = 1 m², T – T_o = 1 °C, then β = A_T – A₀ (numerically).
iv. As, β does not vary significantly with temperature. Hence, if A₁ is the area of a metal plate at T₁ °C and A₂ is the area at higher temperature at T₂ °C, then
β = \(\frac{A_{2}-A_{1}}{A_{1}\left(T_{2}-T_{1}\right)}\)

Question 58.
Define volume expansion of solids.
Answer:
The increase in volume due to heating is called volume expansion or cubical expansion.

Question 59.
Define coefficient of cubical (volume) expansion of a solid. State its unit and dimensions.
Answer:
i) Coefficient of cubical (volume) expansion:
Coefficient of cubical (volume) expansion of a solid is defined as increase in volume per unit original volume at O °C per degree rise in temperature.
It is denoted by γ and is given by,
γ = \(\frac{V_{2}-V_{1}}{V_{1}\left(T_{2}-T_{1}\right)}\)
where,
V₁ = Volume of solid at T₁ °C,
V₂ = Volume of solid at T₂ °C
ii) Dimensions: [L⁰M⁰T⁰K^-1]

[Note: Units and dimension of areal expansion in solid (β) and cubical expansion in solid (γ) are same as that of linear expansion in solid (α).

Question 60.
Derive an expression for coefficient of cubical expansion.
Answer:
i) If the substance is in the form of a cube of volume V, then for small change ∆T in temperature, the fractional change, ∆/V in volume is directly proportional to ∆T.

[Note: Volume expansion ∆V is exaggerated for explanation.)
ii) γ = \(\frac{\Delta \mathrm{V}}{\mathrm{V} \Delta \mathrm{T}}\) = \(\frac{\mathrm{V}_{\mathrm{T}}-\mathrm{V}_{0}}{\mathrm{~V}_{0}\left(\mathrm{~T}-\mathrm{T}_{0}\right)}\)
where, V₀ = volume at 0 °C,
V_T = volume when heated to T °C,
T₀ = 0 °C (initial temperature),
T = final temperature.
∆V = V_T – V₀ = change in volume,
γ = V_T – T₀ = rise in temperature.

iii) If V₀ = 1 m³, T – T₀ = 1 °C, then γ = V_T – V₀ (numerically).

iv) If V₁ is the volume of a body at T₁ °C and V₂ is the volume at higher temperature T₂ °C, then
γ₁ = \(\frac{\mathrm{V}_{2}-\mathrm{V}_{1}}{\mathrm{~V}_{1}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)}\)
γ₁, is the coefficient of volume expansion at temperature T₁ °C.

Question 61.
How does coefficient of volume expansion depend upon temperature?
Answer:

1. As compared to coefficient of linear expansion (α) and coefficient of areal expansion (β), γ changes more with temperature.
2. It is constant only at high temperatures.

Question 62.
Do coefficient of areal expansion and coefficient of volume expansion depend upon nature of material?
Answer:
Yes, coefficient of areal expansion and coefficient of volume expansion depend upon nature of material.

Question 63.
Explain expansion in fluids.
Answer:

1. Since fluids possess definite volume and take the shape of the container, they exhibit only change in volume significantly.
2. Equations valid for cubical or volume expansion of fluids are:
   γ = \(\frac{\Delta V}{V \Delta T}=\frac{V_{T}-V_{0}}{V_{0}\left(T-T_{0}\right)}\)
   where, V₀ = volume at 0 °C,
   V_T = volume when heated to T °C,
   T₀ = 0 °C (initial temperature),
   T = final temperature,
   ∆V = V_T – V₀ = change in volume,
   ∆T = T – T₀ = rise in temperature.
   and γ₁ = \(\frac{V_{2}-V_{1}}{V_{1}\left(T_{2}-T_{1}\right)}\)
   where, V₁ is volume of body at T₁ °C, V₂ is volume of body at higher temperature T₂ °C and γ₁ is coefficient volume expansion at T₁ °C.
3. As fluids are kept in containers, while dealing with the volume expansion of fluids, expansion of the container also needs to be considered.
4. If expansion of fluid results in a volume greater than the volume of the container, the fluid overflows if the container is open.
5. If the container is closed, volume expansion of fluid causes additional pressure on the walls of the container.

Question 64.
Use the data given in the table below:

What conclusions can be drawn using the data?
Answer:

1. Coefficient of volume expansion (γ) is a characteristic of the substance.
2. It (γ) has higher order of magnitude for liquids than that of solids.

Question 65.
Explain how behaviour of water is different than solids and liquids when heat is supplied.
Answer:

1. Normally solids and liquids expand on heating. Hence their volume increases on heating.
2. Since the mass is constant, it results in a decrease in the density on heating.
3. Water expand on cooling from 4 °C to 0 °C.
4. Hence its density decreases on cooling in this temperature range.

Question 66.
Derive relation between coefficient of linear expansion (α) and coefficient of areal expansion (β).
Answer:
Consider a square plate of side l₀ at 0 °C and h at T °C.

1. l_T = l_o (1 + αT) .
   If area of plate at 0 °C is A_o, A_o = \(l_{0}^{2}\)
   If area of plate at T °C is A_T,
   A_T = \(l_{\mathrm{T}}^{2}\) = \(l_{0}^{2}\)(1 + αT)²
   or A_T = A₀(1 + αT)² …. (1)
   Also,
   A_T = A₀(1 + βT) … (2)
   
2. Using Equations (1) and (2), A₀(1 + αT)² = A₀(1 + βT)
   ∴ 1 + 2αT + α²T² = 1 + βT
3. Since the values of a are very small, the term α²T² is very small and may be neglected, ∴ β = 2α
4. The result is general because any solid can be regarded as a collection of small squares.

Question 67.
Derive relation between coefficient of linear expansion (α) and coefficient of cubical expansion (γ).
Answer:

1. Consider a cube of side l_o at 0 °C and l_T at T°C.
   ∴ l_T = l_o(l + αT)
   If volume of the cube at 0 °C is V₀, V₀ = \(l_{0}^{3}\)
   If volume of the cube at T °C is
   V_T, V_T = \(l_{\mathrm{T}}^{3}\) = \(l_{0}^{3}\)(1 + αT)³
   V_T = V₀ (1 + αT)³ ….(1)
   Also,
   V_T = V₀(1 + γT) ….(2)
   
2. Using Equations (1) and (2),
   V_o(1 + αT)³ = V₀(1 + γT)
   ∴ 1 + 3αT + 3α²T² + α³T³ = 1 + γT
3. Since the values of a are very small, the terms with higher powers of a may be neglected.
   ∴ γ = 3α
4. The result is general because any solid can be regarded as a collection of small cubes.

Question 68.
State the relation between α, β and γ and write their meaning.
Answer:
Relation between α, β and γ is given by,
α = \(\frac{\beta}{2}\) = \(\frac{\gamma}{3}\)
where, α = coefficient of linear expansion.
β = coefficient of superficial expansion.
γ = coefficient of cubical expansion.

Solved Examples

Question 69.
The length of a rail on a railway line is 25 m at 10 °C. During summer, maximum temperature attained in the region is 50 °C. Find the minimum gap between the rails, (a = 1.2 × 10^-5/°C)
Solution:
Given: L₁ = 25 m, T₁ = 10 °C, T₂ = 50 °C,
α = 1.2 × 10^-5/°C
To find: Minimum gap between rails (L₂ – L₁)
Formula: L₂ – L₁ = L₁α(T₂ – T₁)
Calculation: From formula,
L₂ – L₁ = 25 × 1.2 × 10^-5 × (50 – 10)
= 25 × 1.2 × 10^-5 × 40
= 1.2 × 10^-2 m
∴ L₂ – L₁ = 1.2 cm
The minimum gap between the rails is 1.2 cm.

Question 70.
The length of a metal rod at 27 °C is 4 cm. The length increases to 4.02 cm when the metal rod is heated upto 387 °C. Determine the coefficient of linear expansion of the metal rod.
Solution:
Given: T₁ = 27 °C, T₂ = 387 °C
L₁ = 4 cm = 4 × 10^-2 m
L₁ = 4.02 cm = 4.02 × 10^-2 m
To find: Coefficient of linear expansion

Coefficient of linear expansion is 1.389 × 10^-5/°C

Question 71.
The length of a metal rod is 150 cm at 25 °C. Find its length when it is heated to 150 °C. (α_steel = 2.2 × 10⁵ /°C)
Solution:
Given. L₁ = 150 cm, T₁ = 25 °C, T₂ = 150 °C,
α_steel = 2.2 × 10⁵ /°C
To find: Length of rod (L₂)
Formula: α = \(\frac{L_{2}-L_{1}}{L_{1}\left(T_{2}-T_{1}\right)}\)
Calculation: From formula,
L₂ – L₁ = L₁ α(T₂ – T₁)
∴ L₂ = L₁[1 + α(T₂ – T₁)]
= 150[1 + 2.2 × 10⁵ × (150 – 25)]
= 150(1 + 2.2 × 10⁵ × 125)
= 150(1 + 0.00275)
= 150× 1.00275 = 150.4125
∴ L₂ = 150.4cm
Length of the rod at 150 °C) is 150.4 cm.

Question 72.
Length of a metal rod at temperature 27 °C is 4.256 m. Find the temperature at which the length of the same rod increases to 4.268 m. (α for iron = 1.2 × 10⁵ K^-1)
Solution:
Given: T₁ = 27°C, L₁ = 4.256 m, L₂ = 4.268 m,
α = 1.2 × 10⁵ K^-1
To find: Temperature (T₂)

= antilog [log 1114.912 – log 4.256]
= antilog [3.0468 – 0.6290]
= antilog [2.4 1781
= 2.617 × 10²
= 261.7°C
Required temperature is 261.7 °C.

Question 73.
A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10^-5 °C^-1.
Solution:
Given: d₁ = 4.24 cm, ∆T = 227 – 27 = 200 °C,
α = 1.70 × 10^-5 °C^-1
To find: Change in diameter (∆d)
Formula: α = \(\frac{d_{2}-d_{1}}{d_{1} \Delta T}=\frac{\Delta d}{d_{1} \Delta T}\)
Calculation: From formula.
∆d = α x d₁ x αT
= 1.70 × 10 × 4.24 × 200
∴ ∆d = 1.44 × 10^-2 cm
The change in diameter of the hole is 1.44 × 10^-2 cm.

Question 74.
A thin aluminium plate has an area 286 cm2 at 20 °C. Find its area when it is heated to 180 °C.
(β for aluminium = 4.9 × 10^-5 °C)
Solution:
Given: T₁ = 20°C, T₂ = 180°C, A₁ = 286 cm²
β = 4.9 × 10^-5 °C
To find: Final area (A₂)
Formula: β = \(\frac{A_{2}-A_{1}}{A_{1}\left(T_{2}-T_{1}\right)}\)
Calculation: From formula,
A₂ = A₁ [1 + β(T₂ – T₁)]
= 286[1 + 4.9 × 10^-5 (180 – 20)]
= 286[1 + 4.9 × 10^-5 × 160]
= 286 [1 + 784.0 × 10]
= 286 [1 + 0.00784]
= 286 [1.00784]
∴ A₂ = 288.24 cm²
Its area when its heated is 288.24 cm².

Question 75.
The surface area of the metal plate is 2.4 × 10^-2 m² at 20°C. When the plate is heated to 185 °C, its area increases by 0.8 cm². Find the coefficient of areal expansion of metal.
Solution:
Given: T₁ = 20 °C, A₁ = 2.4 × 10^-2 m²,
T₂ = 185°C,
∆A = 0.8cm² = 0.8 × 10^-4 m²
To find: Coefficient of areal expansion (β)

The coefficient of areal expansion of the metal is 2.02 × 10^-5/ °C.

Question 76.
A liquid occupies a volume of 2 × 10^-4 m³ at 0 °C. Calculate the increase in its volume If It is heated to 80 °C. [The coefficient of cubical expansion of the liquid is 4 × 10^-4 K^-1]
Solution:
Given: V₀ = 2 × 10^-4 m³, T₀ = 0°C,T = 80°C,
γ_r = 4 × 10^-4 K^-1,
T – T₀ = 80 – 0 = 80°C
To find: Increase in volume (∆V)
Formula. ∆V = V₀γ_r(T – T₀)
Calculation: From formula.
∆V = (2 × 10^-4) (4 × 10^-4) × 80
∴ ∆V = 6.4 × 10^-6 m³
Increase m volume of the liquid is 6.4 × 10^-6 m³.

Question 77.
A liquid at O °C is poured in a glass beaker of volume 600 cm3 to fill it completely. The beaker is then heated to 90 °C. How much liquid will overflow?
(γ_liquid = 1.75 × 10^-4/ °C, γ_glass = 2.75 × 10^-5/ °C)
Solution:
Given: V₁ = 600 cm³, T₁ = 0 °C, T₂ = 90 °C
γ_liquid = 1.75 × 10^-4/°C,
γ_glass = 2.75 × 10^-4/°C
To find: Volume of liquid that overflows
Formula: γ = \(\frac{\mathrm{V}_{2}-\mathrm{V}_{1}}{\mathrm{~V}_{1}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)}\)
Calculation: From formula,
Increase is volume = V₂ – V₁
= γ V₁(T₂ – T₁)
Increase in volume of beaker
= γ_glass × V₁ (T₂ – T₁)
= 2.75 × 10^-5 × 600 × (90 – 0)
= 2.75 × 10^-5 × 600 × 90
= 148500 × 10^-5 cm³
∴ Increase in volume of beaker = 1.485 cm³
Increase in volume of liquid
= γ_liquid × V₁ (T₂ – T₁)
= 1.75 × 10^-4 × 600 × (90 – 0)
= 1.75 × 10^-4 × 600 × 90
= 94500 × 10^-4 cm³
∴ Increase in volume of liquid = 9.45 cm³
∴ Volume of liquid which overflows
= (9.45 – 1.485) cm³
= 7.965 cm³
Volume of liquid that overflows is 7.965 cm³.

Question 78.
The surface area of an iron plate is 80 cm² at 20 °C. Find its surface area at 120 °C. (α_iron = 1.25 × 10^-5 / °C)
Solution:
Given: A₁ = 80 cm², T₁ = 20 °C, T₂ = 120 °C,
α_iron = 1.25 × 10^-5 / °C
To find: Surface area (A₂)
Formula: A₂ = A₁ [1 + β (T₂ – T₁)]
Calculation: β_iron = 2 × α_iron = 2.5 × 10^-5 / °C
From formula,
A₂ = 80[1 + 2.5 × 10^-5 (120 – 20)]
∴ A₂ = 80.2 cm²
Surface area of the iron plate at 120 °C is 80.2 cm².

Question 79.
A sheet of brass is 50 cm long and 8 cm broad at 0 °C. If the surface area at 100 °C is 401.57 cm² find the coefficient of linear expansion of brass.
Solution:
Given: l = 50 cm, b = 8cm.
∴ A₁ = l × b = 50 x 8 = 400 cm²,
T₁ = 0°C, T₂ = 100°C,
A₂ = 401.57 cm²
To find: coefficient of linear expansion (α)

Coefficient of linear expansion of brass is 1.962 × 10^-5/°C.

Question 80.
On heating a glass block of 10.000 cm³ from 25 °C to 40 °C, its volume increases by 4 cm3. Calculate coefficient of linear elipansion of glass.
Solution:
Given: V = 10,000 cm³, ∆V = 4 cm³,
∆T = 40 – 25 = 15°C,
To find: Coefficient of linear expansion (α)

Coefficient of linear expansion of the glass block is 8.89 × 10^-6 /°C.

Question 81.
Explain in detail what is specific heat or specific heat capacity of a substance.
Answer:

1. Specific heat capacity is defined as the amount of heat per unit mass absorbed or given out by the substance to change its temperature by one uni! (one degree) Le., 1 °C or 1 K.
2. The amount of heat (∆Q) required to change the temperature of a substance is directly proportional to:
   • the mass of the substance (m).
   • change in temperature of the substance (∆T).
     ∴ ∆Q ∝ m and ∆Q ∝∆T
     ∴ ∆Q ∝ m∆T
     ∴ ∆Q = sm∆T …………….. (1)
     where ‘s’ is specific heat or specific heat capacity of a substance.
     From equation (1).
     s = \(\frac{\Delta Q}{m \Delta T}\)
     If m = 1 kg and ∆T = 1 °C,then s = ∆Q.
3. Unit: S.I. unit of specific heat is J kg^-1 °C^-1 or J kg^-1 K^-1 and C.G.S. unit is erg g^-1 K^-1 or erg g^-1 °C^-1.
4. Example: The specific heat of water is 4.2 J kg^-1 °C^-1
   It means that 4.2 J of energy must be added to 1 kg of water to rise its temperature by 1 °C.
5. The specific heat capacity is a property of the substance.
6. Specific heat capacity weakly depends on temperature of object. Except for very low temperatures. the specific heat capacity is almost constant for all practical purposes.

Question 82.
State the heat equation.
Answer:
Heat received or given out (Q)
= mass (m) temperature change (∆T) × specific heat capacity (s).
or Q = m × ∆T × s

Question 83.
Write a note on: Molar specific heat.
Answer:

1. If the amount of substance is specified in terms of moles (μ) instead of mass (m) in kg. then the specific heat is called molar specific heat (C).
2. It is given by, C = \(\frac{1}{\mu} \frac{\Delta Q}{\Delta \mathrm{T}}\)
3. The SI unit of molar specific heat capacity is J/mol °C or J/mol K.
4. Like specific heat, molar specific heat also depends on the nature of the substance and its temperature.

Question 84.
Give reason: Water is used as a coolant in automobile radiators.
Answer:

1. Water has the highest specific heat capacity compared to other substances.
2. As a result, water requires higher amount of energy to get heated.
3. This allows water to absorb heat readily while increasing its temperature minimally.
   Hence, water is used as a coolant in automobile radiators.

Question 85.
Give reason: Water is preferred as heater in hot water hag than other liquids.
Answer:

1. Water has the highest specific heat capacity compared to other substances.
2. This means certain mass of water heated to certain temperature contains more heat than the same mass of any other liquid heated to same temperature.
3. As a result, water takes longer time to cool than any other liquid heated to same temperature.
   Hence, water is preferred as heater in hot water bag than other liquids.

Question 86.
Explain why specific heat capacity of a gas at constant pressure is greater than that at constant volume.
Answer:

1. When the gas is heated at constant volume. there is no work done against external pressure.
2. Hence, all the supplied heat is used in raising the temperature of the gas.
3. But when the gas is heated at constant pressure, volume of the gas changes.
4. Due to this, part of the supplied heat is used by the gas to expand against external pressure and remaining part of heat supplied is used to raise the temperature.
5. Because of this, for the same rise in temperature, the heat to be supplied at constant pressure is greater than that for heating at constant volume.
   Hence, specific heat capacity of a gas at constant pressure is greater than specific heat capacity at constant volume.

Question 87.
Define principal and molar specific heat of a gas at constant volunie and constant pressure.
Answer:

1. Principal specific heat:
   • Principal specific heat of a gas at constant volume (s_v):
     Principal specific heat of a gas at constant volume is defined as the quantity of heat absorbed or released for rise or fall temperature of unit mass of a gas through 1 K (or 1 °C), when its volume is kept constant.
   • Principal specific heat of a gas at constant pressure (s_p): Principal specfic heat of a gas at constant pressure is defined as the quantity of heat absorbed or released for rise or fall the temperature of unit mass of a gas through 1 K (or 1 °C), when its pressure is kept constant.
2. Molar specific heat:
   • Molar specific heat of a gas at constant volume (C_V): Molar specific heat of a gas at Constant volume is defined as the quantity of absorbed or released for rise or fall the temperature of one mole of the gas through 1 K (or 1 °C), when its volume is kept constant.
   • Molar specific heat of a gas at constant pressure (C_P): Molar specific heat of a gas at constant pressure is defined as the quantity of heat absorbed or released for rise or fall the temperature of one mole of the gas through 1 K (or 1 °C), when its pressure is kept constant.

Question 88.
State the relation between principal specific heat capacity and molar specific heat capacity for a gas.
Answer:
Molar specific heat capacity = Molecular weight × principal specific heat capacity.
i.e. C_V = μ × S_P and C_V = μ × S_V
where, μ is the molecular weight of the gas.
[Note: Symbols (S_V) and (S_P) are used as per standard convention.]

Question 89.
What is heat capacity?
Answer:

1. Heat capacity or thermal capacity of a body is the quantity of heat needed to raise or lower the temperature of the whole body by 1 o°C (or 1 K).
2. Heat capacity can be written as Heat received or given out
   = mass × 1 × specific heat capacity
   Heat capacity = Q = m × s
   Heat capacity (thermal capacity) is measured in J/°C.

Solved Examples

Question 90.
If the temperature of 4 kg mass of a material of specific heat capacity 300 J/ kg °C rises from 20 °C to 30 °C. Find the heat received.
Solution:
m = 4 kg, s = 300 J/kg °C
∆T = 30 – 20 = 10 °C
To find: Heat received (Q)
Formula: Q = ms∆T
Calculation: From formula,
Q = 4 × 300 × 10
∴ Q = 12000 J
Heat received is 12000 J.

Question 91.
How much heat is required to raise temperature of 750 g of copper pot from 20 to 50 °C?
(The specific heat of copper is 0.094 kcal/kg °C)
Solution:
Given: s = 0.094 kcal/kg°C,
m = 750 g = 0.750kg,
T₁ = 20°C and T₂ = 50°C
Rise in temperature,
∆T = T₂ – T₁ = 50 – 20 = 30°C
To find: Heat required (Q)
Formula: Q = m × s × ∆T
Calculation: From formula,
Q = 0.750 × 0.094 × 30
∴ Q = 2.115 kcal
Heat required to raise the temperature of copper pot is 2.115 kcal.

Question 92.
Calculate the difference in the temperatures between the water at the top and bottom of a water fall 200 m high. Specific heat of water is 4200 J kg^-1 °C^-1.
Solution:
Given: s = 4200 J kg^-1 °C^-1, h = 200 m
To find: Difference in temperatures (∆T)
Formulae:
i) Q = ms∆T
ii) P.E. = mgh
Calculation: From formulae (j) and (ii)
When water falls from top to bottom. assuming no loss in energy, potential energy is converted into heat energy.
∴ Q = P.E.
∴ ms∆T = mgh
∴ s∆T = gh
∴ ∆T = \(\frac{\mathrm{gh}}{\mathrm{s}}=\frac{9.8 \times 200}{4200}\)
∴ ∆T = 0.467 °C
The difference in temperatures between the water at top and bottom is 0.467 °C.

Question 93.
Find thermal capacity for a copper block of mass 0.2 kg, if specific heat capacity of copper is 290 J/kg °C.
Solution:
Given: m = 0.2 kg, s = 290 J/kg °C
To find: Thermal capacity
Formula: Thermal capacity = m × s
Calculation: From formula,
Thermal capacity = 0.2 × 290
= 58 J/ °C
Thermal capacity is 58 J/ °C.

Question 94.
What is calorimetry?
Answer:
Calorimetry is an experimental technique for quantitative measurement of heat exchange.

Question 95.
Explain construction of calorimeter with the help of a labelled diagram.
Answer:

1. A device in which heat measurement can be made is called calorimeter.
2. It consists of a cylindrical vessel and stirrer as well as lid of the same material like copper or aluminium.
3. The vessel is kept inside a wooden jacket which contains heat insulating materials like glass. wool etc. to prohibit any transfer of heat into or out of the calorimeter.
   

Question 96.
State the principle behind working of calorimeter.
Answer:
Calorimeter being isolated system works on the principle of conservation of energy, where heat gained equals heat lost.

Question 97.
Explain the technique “method of mixtures’.
Answer:

1. Method of mixtures is a technique used to determine specific heat capacity of a material using calorimeter.
2. In this technique a sample ‘A’ of the substance is heated to a high temperature which is accurately measured.
3. The sample ‘A’ is then placed quickly in the calorimeter containing water.
4. The contents are stirred constantly until the mixture attains a final common temperature.
5. The heat lost by the sample ‘A’ will be gained by the water and the calorimeter.
6. The specific heat of the sample ‘A’ of the substance can be calculated as follows:
   • Let,
     m₁ = mass of the sample ‘A’
     m2₂ = mass of the calorimeter and the stirrer
     m₃ = mass of the water in calorimeter
     s₁ = specific heat capacity of the substance of sample ‘A’
     s₂ = specific heat capacity of the material of calorimeter (and stirrer)
     s₃ = specific heat capacity of water
     T₁ = initial temperature of the sample ’A’
     T₂ = initial temperature of the calorimeter stirrer and water
     T = final temperature of the combined system
   • Using heat equation,
     Heat lost by the sample ‘A’ = m₁s₁ (T₁ – T)
     Heat gained by the calorimeter and the stirrer = m₂s₂ (T – T₂)
     Heat gained by the water = m₃s₃ (T – T₂)
   • c. Assuming no loss of heat to the surroundings, the heat lost by the sample goes into the calorimeter, stirrer and water,
     ∴ m₁s₁(T₁ – T) = m₂s₂(T – T₂) + m₃s₃(T – T₂) ………….. (1)
   • Knowing the specific heat capacity of water and copper material of the calorimeter and the stirrer, specific heat capacity (si) of material of sample ‘A’ can be calculated.
     e. Rearranging terms of equation (1),
     s₁ = \(\frac{\left(\mathrm{m}_{2} \mathrm{~s}_{2}+\mathrm{m}_{3} \mathrm{~s}_{3}\right)\left(\mathrm{T}-\mathrm{T}_{2}\right)}{\mathrm{m}_{1}\left(\mathrm{~T}_{1}-\mathrm{T}\right)}\)
7. One can find specific heat capacity of water or any liquid using the following expression, if the specific heat capacity of the material of calorimeter and sample is known
   s₃ = \(\frac{\mathrm{m}_{1} \mathrm{~s}_{1}\left(\mathrm{~T}_{1}-\mathrm{T}\right)}{\mathrm{m}_{3}\left(\mathrm{~T}-\mathrm{T}_{2}\right)}-\frac{\mathrm{m}_{2} \mathrm{~s}_{2}}{\mathrm{~m}_{3}}\)

Question 98.
In method of mixtures, why is it essential that density of solid sample be greater than the liquid in calorimeter?
Answer:

1. In method of mixtures, solid sample gives away heat to liquid, and heat exchange between, solid, liquid and calorimeter is considered as isolated.
2. If density of solid sample is lesser than liquid in calorimeter, sample will float on liquid.
3. This will cause partial heat loss to air inside the calorimeter and standard heat equations of calorimeter will no longer be applicable.
   Hence, in method of mixtures it is essential that density of solid sample be greater than liquid in calorimeter.

Solved Examples

Question 99.
A sphere of aluminium of 0.06 kg is placed for sufficient time in a vessel containing boiling water so that the sphere is at 100 °C. It is then immediately transferred to 0.12 kg copper calorimeter containing 0.30 kg of water at 25 °C. The temperature of water rises and attains a steady state at 28 °C. Calculate the specific heat capacity of aluminium.
(Specific heat capacity of water, s_w = 4.18 × 10³ J kg^-1 K^-1, specific heat capacity of copper, s_Cu = 0.387 × 10³ J kg^-1 K^-1)
Solution:
Given: Mass of aluminium sphere = m₁ = 0.06 kg
Mass of copper calorimeter = m₂ = 0.12 kg
Mass of water in calorimeter = m₃ = 0.30 kg
Specific heat capacity of copper
= S_Cu = s₂ = 0.387 × 10³ J/kg K = 387 J/kg K
Specific heat capacity of water
= S_w = s₃ = 4.18 × 10³ J/kg K = 4180 J/kg K
Initial temperature of aluminium sphere
= T₁ = 100 °C
Initial temperature of calorimeter and water
= T₂ = 25 °C
Final temperature of the mixture = T = 28 °C
To find: Specific heat capacity of aluminium (s_al)

= 903.08 J/kg K
Specific heat capacity of aluminium is 903.08 J/kg K.