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In This Post we are  providing Chapter-4 CHEMICAL BONDING AND MOLECULAR STRUCTURE NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON CHEMICAL BONDING AND MOLECULAR STRUCTURE

1.Give the main feature of Kossel’s explanation of chemical bonding.

Ans. Kossel in relation to chemical bonding drew attention to the following facts –

(i) In the periodic table, the highly electronegative halogens and the highly electropositive alkali metals are separated by the noble gases.

(ii) In the formation of a negative ion from a halogen atom and a positive ion from an alkali metal, atom is associated with a gain and loss of an electron by the respective atoms.

(iii) The negative and positive ions so formed attain stable noble gas electronic configurations. The noble gases have particularly eight electrons, ns² np⁶.

The –ve and +ve ions are stabilized by electrostatic attraction.

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2.How can you explain the formation of NaCl according to kossel concept?

Ans. The formation of NaCl from sodium and chlorine can be explained as

Na ® Na⁺ + e^–

[Ne] 3s¹ ® [Ne]

Cl + e^– ® Cl^–

[Ne] 3s² 3p⁵ . [Ne] 3s² 3p⁶ or [Ar]

Na⁺ + Cl^– ® Na⁺ Cl^– or NaCl.

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3.Write the significance of octet rule.

Ans. Octet rule signifies –

(i) It is useful for understanding the structures of most of the organic compounds.

It mainly applies to the second period elements of the periodic table.

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4.Write the Lewis structure for CO molecule

Ans. (i) The outer (valence) shell configurations of carbon and oxygen atoms are

Carbon : (6) – 1s² 2s² 2p²

Oxygen : (8) – 1s² 2s² 2p⁴.

The valence electrons (4 + 6 = 10)

But it does not complete octet, thus multiple bond is exhibited.

Thus,

(ii) N (2s² 2p³), O (2s² 2p⁴)

5 + (2 x 6) + 1 = 18 electrons.

Thus,

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5.Give the Lewis dot structure of HNO₃

Ans. HNO₃ ®

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6.What changes are observed in atoms undergoing ionic bonding?

Ans. Due to the electron transfer the following changes occurs –

(i) Both the atoms acquire stable noble gas configuration.

(ii) The atom that loses electrons becomes +vely charged called cation whereas that gains electrons becomes –vely charged called anion.

(iii) Cation and anion are held together by the coulombic forces of attraction to form an ionic bond.

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7.Mention the factors that influence the formation of an Ionic bond.

Ans.Ionic bond formation mainly depends upon three factors –

(i) Low ionization energy – elements with low ionization enthalpy have greater tendency to form an ionic bonds.

(ii) High electron gain enthalpy – high negative value of electron gain enthalpy favours ionic bond.

(iii) Lattice energy – high lattice energy value favours ionic bond formation.

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8.Give reason why H₂⁺ ions are more stable than H₂^– though they have the same bond order.

Ans.In H₂^– ion, one electron is present in anti bonding orbital due to which destabilizing effect is more and thus the stability is less than that of H₂⁺ ion.

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9.How would the bond lengths vary in the following species? C₂, C₂^– C₂^2-.

Ans.The order of bond lengths in C₂ , C₂^– and C₂^2- is C₂ > C₂^– > C₂^2-.

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10.Out of covalent and hydrogen bonds, which is stronger.

Ans. Covalent bond.

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11.Define covalent radius.

Ans. The covalent radius is measured approximately as the radius of an atom’s core which is in contact with the core of an adjacent atom in a bonded situation.

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12.Why NH₃ has high dipole moment than NF₃ though both are pyramidal?

Ans. In case of NH₃ the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the N-H bonds, whereas in NF₃ the orbital dipole is in the direction opposite to the resultant dipole moment of the three N-F

bonds. The orbital dipole become of lone pair decreases, which results in the low dipole moment.

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13.Draw the resonating structure of NO₃^–

Ans.

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14.On which factor does dipole moment depend in case of polyatomic molecules.

Ans.The dipole moment of the polyatomic molecule depends on individual dipole moments of bonds and also on the spatial arrangement of various bonds in the molecule.

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15.Dipole moment of Be F₂ is zero. Give reason.

Ans. In BeF₂ the dipole moment is zero because the two equal bond dipoles point in opposite directions and cancel the effect of each other.

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In This Post we are  providing Chapter-3 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

Question 1.
Calculate the energy required to convert all atoms of magnesium to magnesium ions present in 24 mg of magnesium vapors. IE₁ and IE₂ of Mg are 737.76 and 1450.73 kJ mol-1 respectively.

Answer:
Mg (g) + IE₁ → Mg+ (g) + e^– (g)
IE₁ = 737.76 kJ mol^-1

Mg+ (g) + IE₂ → Mg²⁺ + e^– (g)
IE₂ = 1450.73 kJ mol^-1

Total amount of energy needed to convert Mg (g) into ,
Mg²⁺ ion = IE₁ + IE₂

Now 24 mg of Mg = 241000 g = 241000×24 mol [1 Mole of Mg = 24 g]
= 10^-3 mole.

Question 2.
The IE₁ and IE₂ of Mg. (g) are 740 and 1450 kJ mol^-1. Calculate the percentage of Mg⁺ (g) and Mg²⁺ (g) if 1 g of Mg (g)
absorbs 50 kJ of energy.

Answer:
No. of moies of Mg vapours present in I g = 124 = 0.0147

Energy absorbed to convèrt Mg (g) to Mg⁺ (g) = 0.0417 × 740
=30.83 kJ

Energy left unused = 50 – 30.83
= 19.17 kJ
Now 19.17 kJ will be used to e Mg⁺ (g) to Mg²⁺ (g) .
∴ No. of moles of Mg+ (g) converted to Mg2+ (g) = 19.171450 = 0.132

%age of Mg⁺ (g) = 0.02850.0417 × 100 = 68.35
and %age of Mg²⁺ (g) = 100 – 68.35 = 31.65

Question 3.
Which of the elements Na, Mg, Si, P would have the greatest difference between the first and the second ionization enthalpies. Briefly explain your answer

Answer:
Among Na, Mg, Si, P, Na is an alkali metal. It has only one electron in the valence shell. Therefore, its IE₁ is low: However, after the removal of the first electron, it acquires a Neon gas configuration i.e., Na+ (1s², 2s² 2p⁶). Therefore its IE₂ is expected to be very high. Consequently, the difference in IE₁ and IE₂ comes to be greatest in the case of Na.

Question 4.
The IE₂ of Mg is higher than that of Na. On the other hand, the IE₂ of Na is much higher than that of Mg. Explain,

Answer:
The first electron in both cases has to be removed from the 3s orbital, but the nuclear charge of Na is less than that of Mg. After the removal of the first electron from Na, the electronic configuration of Na⁺ is 1s², 2s² 2p⁶, i.e., that of noble gas which is very stable and the removal of the 2nd electron is very difficult. In the case of Mg after the removal of the first electron, the electronic configuration of Mg⁺ is 1s², 2s² 2p⁶ 3s. The 2nd electron to be removed is again from 3s orbital which is easier.

Question 5.
The amount of energy released when 1 × 10¹⁰ atoms of chlorine in vapor state are converted to Cl^– ions according to the equation.
Cl (g) + e^– → Cl^– (g) is 57.86 × 10^-10 J
Calculate the electron gain enthalpy of the chlorine atom in terms of kJ mol^-1 and eV pet atom.
Answer:
The electron gain enthalpy of chlorine

Question 6.
Electronic configuration of the four elements are given below:
Arrange these elements in increasing order of their metallic character. Give reasons for your answer.

(i) [Ar]4s²
Answer:
[Ar]4s² is Calcium metal with At. no. = 20.

(ii) [Ar]3d¹⁰ 4s²
Answer:
[Ar]3d¹⁰ 4s² is Zinc metal with At. no. = 30.

(iii) [Ar]3d¹⁰ 4s² 4p⁶ 5s²
Answer:
[Ar]3d¹⁰ 4s² 4p⁶ 5s² is Strontium metal with At. no. = 38.

(iv) [Arl 3d¹⁰ 4s² 4p⁶ 5s¹
Answer:
[Ar] 3d¹⁰ 4s² 4p⁶, 5s¹ is*Rubidium metal with At. no. = 37.

Alkali metals are the most metallic, followed by alkaline earth metals and transition metals. Among alkali metals – Rubidium (37) is the most metallic. Among alkaline earth metals (Ca, Sr) Sr (Strontium) is more metallic than Calcium (Ca) as the metallic character increases from top to bottom in a group. Zinc – the transition metal is the least metallic. Thus metallic character increases from
Zn < Ca < Sr < Rb or (ii) < (i) < (iii) < (iv)

Question 7.
The formulation of F (g) from F (g) is exothermic whereas that of O^2- (g) from O (g) is endothermic. Explain.

Answer:
F (g) + e^– (g) → F (g); ΔH = Negative
Energy is released when an extra electron from outside is added to a neutral isolated gaseous atom of an element.
To convert, O (g) to O^2- (g) two steps are required
(i) O (g) + e^– (g) → O^– (g); ΔH₁ = – 141 kJ mol^-1
(ii) O^– (g) + e^– (g) → O^2- (g); ΔH₂ = + 780 kJ mol^-1

Hence the over all processes endothermic (+ 780 – 141 = + 639 kJ mol^-1) whereas F (g) to F^– (g) is exothermic.

Question 8.
Explain the important general characteristics of groups in the modem periodic table in brief.

Answer:
The elements of a group show the following important similar characteristics.
(0 Electronic configuration. All elements in a particular group have similar outer electronic configuration e.g., all elements of group I’, i.e., alkali metals have ns1 configuration in their valency shell. Similarly, group 2 elements (alkaline Earths) haye ns² outer configuration and halogens (group 17) have ns² np⁵ configuration (where n is the outermost shell).

(it) Valency. The valency of an element depends upon the number of electrons in the outermost shell. So elements of a group show the same valency, e.g., elements of group 1 show + 1 valency and group 2 show + 2 valencies i.e. valency i.e., NaCl > MgCl₂ etc.

(iii) Chemical properties. The chemical properties of the elements are related to the number of electrons in the outermost shell of their atoms. Hence all elements belonging to the same group show similar chemical properties. But the degree of reactivity varies gradually from top to bottom in a group. For example, in group 1 all the elements are highly reactive metals but the degree of reactivity increases from Li to Cs. Similarly, elements of group 17, i.e., halogens: F, Cl, Br, I are all non-metals and they’re- reactivity goes on decreasing from top to bottom.

Question 9.
Explain the electronic configuration in periods in the periodic table.

Answer:
Each successive period in the periodic table is associated with the filling Up of the next higher principal energy level (n – 1, n – 2, etc.). It can be readily seen that the number of elements in each period is twice the number of atomic orbitals available in the energy level that is being filled. The first period starts with the filling of the lowest level (1s) and has thus the two elements – hydrogen (1s¹) and helium (1s²) when the first shell (K) is completed. The second period starts with lithium and the third electron enters the 2s orbital.

The next element, beryllium has four electrons and has the electronic configuration 1s² 2s². Starting from the next element boron, the 2p orbitals are filled with electrons when the L shell is completed’ at neon (2s² 2p⁶). Thus there are 8 elements in the second period. The third period (n = 3) being at sodium, and the added electron enters a 3s orbital. Successive filling of 3s and 3p orbitals give rise to the third period of 8 elements from sodium to argon.

The fourth period (n = 4) starts at potassium with the filling up of 4p of 4s orbital. Before the 4p orbital is filled, the filling up of 3d orbitals becomes energetically favorable and we come across the so-called 3d transition series of elements. The fourth period ends at krypton with the filling up of the 4p orbitals. Altogether we have 18 elements in the fourth period. The fifth period (n = 5) beginning with rubidium is similar to the fourth period and contains the 4d transition series starting at yttrium (Z = 39).

This period ends at xenon with the filling up of the 5p orbitals. The sixth period (n = 6) contains 32 elements and successive electrons enter 6s, 4/, 5d, and 6p orbitals, in that order. Filling up of the 4/ orbitals being with cerium, (Z = 58) and ends at lutetium (Z = 71) to give the 4/-inner transition series which is called the lanthanide series. The seventh period (n = 7) is similar to the sixth period with the successive filling up of the 7s, 5f, 6d, and 7p orbitals and includes most of the man-made radioactive elements.

This period will end at the element with atomic number 118 which would belong to the noble gas family. Filling up of the 5f orbitals after actinium (Z = 89) gives the 5f-inner transition series known as the actinide series. The 4f and 5f transition series of elements are placed separately in the periodic table to maintain its structure and to preserve the principle of classification by keeping elements with similar properties in a single column.

Question 10.
Explain the variation of valence in the periodic table.

Answer:
Variation of valence in a group as well as across a period in the periodic table occurs as follows:
1. In a group: All elements in a group show the same valency. For example, all alkali metals (group 1) show a valency of 1+. Alkaline earth metals (group 2) show a valency of 2+.

However, the heavier elements of p-block elements (except noble gases) show two valences: one equal to the number of valence electrons or 8-No. of valence electron# and the other two less. For example, thallium (Tl) belongs to group 13. It shows valence of 3+ and 1+.

Lead (Pb) belongs to group 14. If shows valance of 4+ and 2+.
Antimony (Sb) and Bismuth (Bi) belong to group 15. They show valence of 5+ and 3+ being more stable.

This happens due to the non-participation of tie two s-electrons present in the valence shell of these elements. This non-participation of one pair of s-electrons in bonding is called the inert-pair effect.

2. In a period: The number of the valence electrons increases – in going from left to right in a period of the periodic table. Therefore the valency of the elements in a period first increases, and then decreases.

In This Post we are  providing Chapter-2 STRUCTURE OF ATOM NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON STRUCTURE OF ATOM

1.Which experiment led to the discovery of electrons and how?  

Ans:The cathode ray discharge tube experiment performed by J.J. Thomson led to the discovery of negatively charged particles called electron.

A cathode ray tube consists of two thin pieces of metals called electrodes sealed inside a glass tube with sealed ends. The glass tube is attached to a vacuum pump and the pressure inside the tube is reduced to 0.01mm. When fairly high voltage (10, 000V) is applied across the electrodes, invisible rays are emitted from the cathode called cathode rays. Analysis of this rays led to the discovery electrons.

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2.Give the main properties of canal ray experiment.

Ans:The canal ray experiment led to the discovery of –

(i)The anode rays, travel in straight line

(ii)They are positively charged as they get deflected towards the –ve end when subjected to an electric and magnetic field.

(iii)They depend upon the nature of gas present in the cathode tube.

(iv)The charge to mass ration (e/m) of the particle is found to depend on the gas from which they originate.

(v)They are also material particles

The analysis of these proportions led to the discovery of positively charged proton.

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3.Find out atomic number, mass number, number of electron and neutron in an element?

Ans: The mass no. of 

The atomic no. of  

No. of proton is = Z – A = 40 – 20 = 20

No. of electron its (A) = 20

No. of proton is (A) = 20

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4.Give the main features of Thomson’s Model for an atom.

Ans: J.J. Thomson proposed that an atom consists of a spherical sphere (radius of about 10^-10m)in which the positive charges are uniformly distributed the electrons are embedded into it in such a manner so as to give stable electrostatic arrangement.

This model is also called raisin pudding model.

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5.What did Rutherford conclude from the observations of   scattering experiment?

Ans: Rutherford proposed the nuclear model of an atom as

(i) The positive charge and most of the mass of an atom was concentrated in an extremely small region. He called it nucleus.

(ii) The nucleus is surrounded by electrons that move around the nucleus with a very high speed in orbits.

(iii) Electron and nucleus are held together by electrostatic forces of attraction.

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6.What is the relation between kinetic energy and frequency of the  photoelectrons?

Ans: Kinetic energy of the ejected electron is proportional to the frequency of the electromagnetic radiation.

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7.What transition in the hydrogen spectrum would have the same  wavelength as the Balmer transition, n = 4 to n = 2 of He⁺ spectrum?

Ans: For the Balmer transition, n = 4, to n = 2 in a He⁺ ion, we can write.

For a hydrogen atom

Equating equation (ii) and (i), we get

This equation gives n₁ = 1 and n = 2. Thus the transition n = 2 to n = 1 in hydrogen atom will have same wavelength as transition, n = 4 to n = 2 in He⁺

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8.Spectral lines are regarded as the finger prints of the elements. Why?

Ans: Spectral lines are regarded as the finger prints of the elements because the elements can be identified from these lines. Just like finger prints, the spectral lines of no two elements resemble each other.

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9.Why cannot the motion of an electron around the nucleus be determined

accurately?

Ans: Because there is an uncertainty in the velocity of moving electron around the nucleus (Heisenberg’s Uncertainty Principle).

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10.Calculate the uncertainty in the momentum of an electron if it is confined to a linear region of length

Ans: According to uncertainty Principle

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11.Give the mathematical expression of uncertainty principle.    

Ans:Mathematically, it can be given as 

Where is the uncertainty in position and  is the uncertainty in momentum (or velocity) of the particle.

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12.Which quantum number determines

(i)  energy of electron      

(ii) Orientation of orbitals.

Ans. (i) Principal quantum number (n), and

(ii) Magnetic quantum number (m).

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13.Arrange the electrons represented by the following sets of quantum number in decreasing order of energy. 

1.   n = 4,  l = 0, m = 0, s = +1/2

2.   n = 3,  l = 1, m = 1,  s = -1/2

3.   n = 3,  l = 2, m = 0, s = +1/2

Ans.(i)Represents 4s orbital

(ii) Represents 3p orbital

(iii)Represents 3d orbital

(iv)Represents 3s orbital

The decreasing order of energy   3d > 4s > 3p > 3s

n = 3,  l = 0, m = 0, s = -1/2

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14.What designations are given to the orbitals having

(i)    n = 2, l = 1     (ii)   n = 2, l = 0     (iii) n = 4, l = 3

(iv)   n = 4, l = 2      (v)    n = 4, l = 1? 

Ans.    (i)           Here, n = 2, and l = 1

Since l = 1 it means a p-orbital, hence the given orbital is designated as 2p.

(ii) Here, n = 2 and l = 0

Since l = 0 means s – orbital, hence the given orbital is 2s.

(iii) Here, n = 4 and  l = 3

Since, l = 3 represents f – orbital, hence the given orbital is a 4f orbital.

(iv) Here, n = 4 and l = 2

Since, l = 2 represents d – orbital, hence the given orbital is a 4d – orbital.

(v) n = 4  and l = 1

 since, l = 1 means it is a p – orbital, hence the given orbital can be designated as – 4p orbital.

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15.Write the electronic configuration of (i) Mn⁴⁺, (ii) Fe³⁺ (iii)  Cr²⁺ and Zn²⁺ Mention the number of unpaired electrons in each case.

Ans.(i)       Mn  (z = 25), Mn⁴⁺  (z = 21)

The electronic configuration of Mn⁴⁺ to Given by

1s² 2s² 2p⁶ 3s² 3p⁶ 3d³

As the outermost shell 3d has 3 electrons, thus the number of unpaired

electrons is 3.

 (ii) Fe   (z = 26), Fe³⁺ (z = 23)

The electronic configuration of Fe³⁺ is given lay

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

The number of unpaired electron is 5.

(iii) Cr (z = 24),  Cr²⁺ (z = 22)

The electronic configuration of Cr²⁺ is

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴

The number of unpaired electron is 4.

(iv) Zn (z = 30), Zn²⁺  (z = 28)

The electronic configuration of Zn²⁺ is

1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰

The number of unpaired electron is 0.

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In This Post we are  providing Chapter-1 SOME BASIC CONCEPTS OF CHEMISTRY NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON SOME BASIC CONCEPTS OF CHEMISTRY

Question 1.
Define Mole. What is its numerical value?

Answer:
A mole is the amount of a substance that contains as many entities (atoms, molecules, or other particles) as there are atoms in exactly 0.012 kg or 12 g of the carbon-12 isotope.

Its numerical value is 6.023 × 10²³.

Question 2.
Define molarity. Is it affected by a change in temperature?

Answer:
The molarity of a solution is defined as the number of moles of the solute present per liter of the solution. It is represented by the symbol M. Its value changes with the change in temperature.

Question 3.
What do you mean by Precision and accuracy?

Answer:
Precision and accuracy: The term precision refers to the closeness of the set of values obtained from identical measurements of a quantity.

Accuracy refers to the closeness of a single measurement to its true value.

Question 4.
Distinguish between fundamental and the derived units.

Answer:
Fundamental units: Fundamental units are those units by which other physical units can be derived. These are mass (M), Length (L), time (T), temperature (°).

Derived units: The units which are obtained by the combination of the fundamental units are called derived units.

Question 5.
Define molality and write its temperature dependence.

Answer:
Molality is defined as the number of g moles of the solute dissolved per kilogram of the solvent.
Molality (m) =  Mole of solute  Mass of the solvent in kg

The molality of the solution does not depend upon the temperature.

Question 6.
Two containers of equal capacity A₁ and A₂ contain 10 g of oxygen (O₂) and ozone (O₃) respectively. Which of the two will have greater no. of O-atoms and which will give greater no. of molecules?

Answer:
10 g of O₂ = 1032 mol = 1032 × 6.02 × 10²³ molecules
= 1.88 × 10²³ molecules
= 3.76 × 10²³ atoms.

10 g of O₂ = 1048 mole = 1048 × 6.02 × 10²³ molecules
= 1.254 × 10²³ molecules
= 3.76 × 10²³ atoms

Thus both A₁ and A₂ contain the same no. of atoms, but A₁ contains more numbers of molecules.

Question 7.
Assuming the density of water to be 1 g/cm³, calculate the volume occupied by one molecule of water.

Answer:
1 Mole of H₂O = 18 g = 18 cm³[∵ density of H₂O = 1 g/cm³]
= 6.022 × 10²³ molecules of H₂O
1 Molecule will have a volume
= 186.022×1023 cm- = 2.989 × 10^-23 cm³.

Question 8.
State the law of Multiple Proportions. Explain with two examples.

Answer:
The Law of Multiple Proportions states:
“When two elements combine to form two or more than two chemical compounds than the weights of one of elements which combine with a fixed weight of the other, bear a simple ratio to one another.

Examples:
1. Compound of Carbon and Oxygen: C and O combine to form two compounds CO and CO₂.
In CO₂ 12 parts of wt. of C combined with 16 parts by wt. O.
In CO₂ 12 parts of wt. of C combined with 32 parts by wt. of O.
If the weight of C is fixed at 12 parts by wt. then the ratio in the weights of oxygen which combine with the fixed wt. of C (= 12) is 16: 32 or 1: 2.
Thus the weight of oxygen bears a simple ratio of 1: 2 to each other.

2. Compounds of Sulphate (S) and Oxygen (O):
S forms two oxides with O, viz., SO₂ and SO₃
In SO₂, 32 parts of wt. of S combine with 32 parts by wt. of O.
In SO₃, 32 parts of wt. of S combine with 48 parts by wt. of O.
If the wt. of S is fixed at 32 parts, then’ the ratio in the weights of oxygen which combine with the fixed wt. of S is 32: 48 or 2: 3.

Thus the weights of oxygen bear a simple ratio of 2: 3 to each other.

Question 9.
State the law of Constant Composition. Illustrate with two examples.

Answer:
Law of Constant Composition of Definite Proportions states: “A chemical compound is always found, to be made up of the same elements combined together in the same fixed proportion by weight”.

Examples:

1. CO₂ may be prepared in the laboratory as follows:

In all the above examples, CO₂ is made up of the same elements i. e., Carbon (C) and Oxygen (O) combined together in the same fixed proportion by weight of 12: 32 or 3: 8 by weight.

Question 10.
Define empirical formula and molecular formula. How will you establish a relationship between the two? Give examples.

Answer:
The empirical formula of a compound expresses the simplest whole-number ratio of the atoms of the various elements present in one molecule of the compound.

For example, the empirical formula of benzene is CH and that of glucose is CH₂O. This suggests that in the molecule of benzene one atom of Carbon (C) is present for every atom of Hydrogen (H). Similarly in the molecule of glucose (CH₂O), for every one atom of C, there are two atoms of H and one atom of O present in its molecule. Thus, the empirical formula of a compound represents only the atomic ratio of various elements present in its molecule.

The molecular formula of a compound represents the true formula of its molecule. It expresses the actual number of atoms of various elements present in one molecule of a compound. For example, the molecular formula of benzene is C₆H₆ and that of glucose is C₆H₁₂O₆. This suggests that in one molecule of benzene, six atoms of C and 6 atoms of H are present. Similarly, one molecule of glucose (C₆H₁₂O₆) actually contains 6 atoms of C, 12 atoms of H, and 6 atoms of O.

Relation between the empirical and molecular formula
Molecular formula = n × Empirical formula where n is an integer such as 1, 2, 3…
When n = 1; Molecular formula = Empirical formula
When n = 2; Molecular formula = 2 × Empirical formula.
The value of n can be obtained from the relation.
n =  Molecular mass  Empirical formula mass 

The molecular mass of a volatile substance can be determined by Victor Meyer’s method or by employing the relation.
Molecular mass = 2 × vapour density .

Empirical formula mass can however be obtained from its empirical formula simply by adding the atomic masses of the various atoms present in it.

Thus the empirical formula mass of glucose CH20
= 1 × 12 + 2 × 1 + 1 × 16 = 30.0 u.

In This Post we are  providing Chapter-15 WAVES NCERT MOST IMPORTANT QUESTIONS for Class 11 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON WAVES

Question 1.
A sitar wife and a tabla when sounded together give 4 beats/ sec. What do we conclude from this? As the tabla membrane is tightened, the beat rate increases or decreases, explain.

Answer:
When sitar and tabla are sounded together, they give 4 seats/ sec. From this, we conclude that the frequencies of the two sounds differ by 4. If the frequency of tabla is greater than that of a sitar, then on tightening the tabla membrane, the frequency of tabla will further increase and hence the difference in frequencies will increase.

Thus beat rate will increase. If the frequency of tabla is less than that of sitar, then on tightening the tabla membrane, the frequency of tabla will increase and the difference in frequencies will decrease. So beat rate will decrease.

Question 2.
When we start filling an empty bucket with water, the pitch of sound produced goes on changing. Why?

Answer:
An empty bucket behaves as a closed organ pipe. The frequency of fundamental note produced by it is given by
v = v4l.

As the bucket starts filling, the length (l) of the resonating air column decreases, and hence frequency increases. Since the pitch of a sound depends upon the frequency. So it changes with the change in frequency.

Question 3.
Two loud-speakers have been installed in an open space to listen to a speech. When both are operational, a listener sitting at a .particular- place receives a very faint sound. Why? What will happen if one loud-speaker is kept off?

Answer:
When the distance between two loud-speakers from the position of listener is an odd multiple of λ2, then due to destructive interference between sound waves from two loud-speakers, a feeble sound is heard by the listener.

When one loud-speaker is kept off, no interference will take place and the listener will hear the full sound of the operating loud-speaker.

Question 4.
Distinguish between progressive waves and stationary waves.

Answer:
Progressive waves:

1. The disturbance travels onward. It is1 handed over from one particle to the next.
2. Energy is transported in the medium along with the propagation of waves.
3. Each particle of the medium executes S.H.M. with the same amplitude.
4. No particle of the medium is permanently at rest.
5. Changes in pressure and density are the same at all points of the medium.

Stationary waves:

1. The disturbance is confined to a particular region and there is no onward motion.
2. No energy is transported in the medium.
3. All the particles of the medium except at nodes execute S.H.M. with different amplitude.
4. The particles of the medium at nodes are at rest.
5. The changes of pressure and density are maximum at nodes and minimum at antinodes.

Question 5.
Distinguish between musical sound and noise.

Answer:
Musical sound:

1. It produces a pleasant effect on the ear.
2. It has a high frequency.
3. There are no sudden changes in the amplitude of the musical sound waves.
4. It is a desirable sound.

Noise:

1. It produces an unpleasant effect on the ear.
2. It has a low frequency.
3. There are sudden changes in the amplitude of noise waves.
4. It is an undesirable sound.

Question 6.
What are the characteristics of wave motion?

Answer:

1. Wave motion is a form of disturbance that travels in a medium due to repeated periodic motion of the particles of the medium.
2. The wave velocity is different from the particle velocity.
3. The vibrating particles of the medium possess both K.E. and P.E.
4. The particle velocity is different at different positions of its vibrations whereas wave velocity is constant throughout a given medium.
5. Waves can undergo reflection, refraction, diffraction, dispersion, and interference.

Question 7.
Show that for 1°C change in temperature, the velocity of sound changes by 0.61 ms-1.

Answer:
We know that v ∝ T−−√ .
If v_t and v_o be the velocity of sound at T°C and 0°C respectively,

where α = vt−v0t is called temp. coefficient of the velocity of sound.

Putting v_o = 332 ms^-1 at T0 i.e. 0°C, we get
α = 332546 = 0.61 ms^-1 °C^-1

Question 8.
An electric bell is put in an evacuated room (a) near the center (b) close to the glass window, in which case the sound is heard (i) inside the room, (ii) out of the room.

Answer:

1. Sound is not heard in cases (a) and (b) inside the room as the medium is not there for the propagation of sound.
2. In case (a) sound cannot be heard outside for the reason given in (i) above.

In case (b) since the bell is very close to the window, the glass pane picks up its vibrations which are conveyed to the eardrum through the air outside the room. So, the sound can be heard in condition (b).

Question 9.
A progressive wave is given by
y = 12 sin (5t – 4x)
On this wave, how far away are the two points having a phase difference of π2?

Answer:
Here, ΔΦ = phase difference = π2

Let Δx be the corresponding path difference,

Question 10.
The figure here shows the wave, y = A sin (ωt – kx) at any instant traveling in the +x direction. What is the slope of the curve at B?

Answer:
Here, the particle velocity is maximum at B and is given by
v_o = ωA.
Also, wave velocity is given by
C = ωk
∴ So the slope v0C=ωAω/k = kA

Question 25.
Two sound waves
y₁ = A₁ sin 1000 π(t – x220 )
and y₂ = A₂ sin 1010 π(t – x220 ) are superposed. What is the frequency with which the amplitude varies?
Answer:
Rate of variation of amplitude is equal to the beat frequency.
Here, 2πν₁ = 1000 π
or
ν₁ = 500
and 2πν₂ = 1010 π
or
ν₂ = 505

∴ beat frequency = ν₂ – ν₁
= 505 – 500
= 5.

In This Post we are  providing Chapter-14 OSCILLATIONS NCERT MOST IMPORTANT QUESTIONS for Class 11 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON OSCILLATIONS

1.The springs of spring factor k, 2k, k respectively are connected in parallel to a mass m. If the mass = 0.08kg m and k = 2N|m, then find the new time period?

Ans. Total spring constant, K¹ = K₁ + K₂ + K₃ (In parallel)

= K + 2K + K

= 4K

= 4 × 2 (k = 2 N | m)

= 8 N | m

Time period,

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2.The bob of a vibrating simple pendulum is made of ice. How will the period of swing will change when the ice starts melting?

Ans .The period of swing of simple pendulum will remain unchanged till the location of centre of gravity of the bob left after melting of the ice remains at the fixed position from the point of suspension. If centre of gravity of ice bob after melting is raised upwards, then effective length of pendulum decreases and hence time period of swing decreases. Similarly, if centre of gravity shifts downward, time period increases.

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3.An 8 kg body performs S.H.M. of amplitude 30 cm. The restoring force is 60N, when the displacement is 30cm. Find: – a) Time period b) the acceleration c) potential and kinetic energy when the displacement is 12cm?

Ans.Here m = 8 kg

m = Mass, a = amplitude

a = 30cm = 0.30m

a) f = 60 N, Y = displacement = 0.30m

K = spring constant

Since, F = Ky

K = = 

As, Angular velocity = w = 

Time period, T = 

b) Y = displacement = 0.12m

Acceleration, A = w² y

A = (5)² × 0.12

A = 3.0m |s²

P.E. = Potential energy = 

Kinetic energy = K.E = 

=  

Kinetic energy = K. E. = 7.56J

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4.A particle executing SH.M has a maximum displacement of 4 cm and its acceleration at a distance of 1 cm from its mean position is 3 cm/s². What will be its velocity when it is at a distance of 2cm from its mean position?

Ans.The acceleration of a particle executing S.H.M is –

A = w² Y

w = Angular frequency ; Y = Displacement

A = Acceleration

Given A = 3cm / s² ; Y = 1cm 

So, 3 = w² × 1

w = 

The velocity of a particle executing S.H.M is :-

a = amplitude

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5.What is ratio of frequencies of the vertical oscillations when two springs of spring constant K are connected in series and then in parallel?

Ans .If two spring of spring constant K are connected in parallel, then effective resistance in parallel = K_P = K + K = 2K

Let f_P = frequency in parallel combination.

In Series combination, effective spring constant for 2 sprigs of spring constant K is :-

Let f_S = frequency in series combination 

Divide equation 2) by 1)

6.   A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation J = -α ¸, where J is the restoring couple and ¸ the angle of twist).

Ans. Mass of the circular disc, m = 10 kg

Radius of the disc, r = 15 cm = 0.15 m

The torsional oscillations of the disc has a time period, T = 1.5 s

The moment of inertia of the disc is:

I 

= 

= 0.1125 kg 

Time period, 

α is the torsional constant.

= 1.972 Nm/rad

Hence, the torsional spring constant of the wire is 1.972 Nm rad^–1.

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7. A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?

Ans.The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car.

Acceleration due to gravity = g

Centripetal acceleration 

Where,

v is the uniform speed of the car

R is the radius of the track

Effective acceleration () is given as:

Time period, 

Where,l is the length of the pendulum

∴Time period, T 

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9. Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?

(a) the rotation of earth about its axis.

(b) motion of an oscillating mercury column in a U-tube.

(c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.

(d) general vibrations of a polyatomic molecule about its equilibrium position.

Ans.(b) and (c) are SHMs

(a) and (d) are periodic, but not SHMs

(a) During its rotation about its axis, earth comes to the same position again and again in equal intervals of time. Hence, it is a periodic motion. However, this motion is not simple harmonic. This is because earth does not have a to and fro motion about its axis.

(b) An oscillating mercury column in a U-tube is simple harmonic. This is because the mercury moves to and fro on the same path, about the fixed position, with a certain period of time.

(c) The ball moves to and fro about the lowermost point of the bowl when released. Also, the ball comes back to its initial position in the same period of time, again and again. Hence, its motion is periodic as well as simple harmonic.

(d) A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the superposition of individual simple harmonic motions of a number of different molecules. Hence, it is not simple harmonic, but periodic.

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10.  Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion ( is any positive constant):

(a)  

(b) 

(c) 3 cos (π/4 – 2t)

(d) cos t + cos 3t + cos 5t

(e) exp 

(f) 1 + t + 

Ans.(a) SHM

The given function is:

This function represents SHM as it can be written in the form:

Its period is: 

(b) Periodic, but not SHM

The given function is:

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In This Post we are  providing Chapter- 13 KINETIC ENERGY NCERT MOST IMPORTANT QUESTIONS for Class 11 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON KINETIC ENERGY

Question 1.
Two identical cylinders contain helium at 2 atmospheres and argon at 1 atmosphere respectively. If both the gases are filled in one of the cylinders, then:
(a) What would be the pressure?

Answer:
(2 + 1) = 3 atmosphere.

(b) Will the average translational K.E. per molecule of both gases be equal?

Answer:
Yes, because the average translational K.E./molecule (32kT) depends only upon the temperature.

(c) Will the r.m.s. velocities are different?

Answer:
Yes, because of the r.m.s. velocity depends not only upon temperature but also upon the mass.

Question 2.
Why hydrogen escapes more rapidly than oxygen from the earth’s surface?

Answer:
We know that C_rms ∝ 1ρ√

Also ρ₀ = 16 ρ_H. So C_rms of hydrogen is four times that of oxygen at a given temperature. So the number of hydrogen molecules whose velocity exceeds the escape velocity from earth (11.2 km s^-1) is greater than the no. of oxygen molecules. Thus hydrogen escapes from the earth’s surface more rapidly than oxygen.

Question 3.
Distinguish between the terms evaporation, boiling and vaporization.

Answer:
Evaporation: It is defined as the process of conversion of the liquid to a vapor state at all temperatures and occurs only at the surface of the liquid.

Boiling: It is the process of rapid conversion of the liquid to a vapour state at a definite temperature and occurs throughout the liquid.

Vaporization: It is the general term for the conversion of liquid to vapor state. It includes both evaporation and boiling.

Question 4 .
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapor and other constituents) in a room of capacity 125.0 m³ at a temperature of 127°C and 2 atm pressure, k = 1.38 × 10^-23 JK^-1.

Answer:
Here, T = 127°C + 273 = 400 K
k = 1.38 × 10^-23 JK^-1 P = 2 atmosphere
= 2 × 1.01 × 10⁵ Nm^-2
= 2.02 × 10⁵ Nm^-2

V = volume of room = 125 m³
N’ = no. of molecules in the room =?
∴ R = Nk = 6.023 × 10²³ × 1.38 × 10^-23
= 8.31 JK^-1 mol^-1

Let n = no. of moles of the air in the given volume.
∴ Using gas equation,
PV = nRT, we get
n = PVRT=2.02×105×1258.31×400
= 7.60 × 103 moles

∴ N’ = Nn = 6.023 × 10²³ × 7.60 × 10³
= 45.77 × 10²⁶.

Question 5.
Calculate the temperature at which the oxygen molecules will have the same r.m.s. velocity as the hydrogen molecules at 150°C. The molecular weight of oxygen is 32 and that of hydrogen is 2.

Answer:
Here, Molecular weight of oxygen, M0 = 32
Molecular weight of hydrogen. MH = 2

Let T₀ = temp. of oxygen = ?
T_H = temp. of hydrogen
= 150°C = 150 + 273 = 423 K
C₀ = C_H

Question 6.
Calculate the r.m.s. the velocity of molecules of gas for which the specific heat at constant pressure is 6.84 cal per g mol per °C. The velocity of sound in the gas being 1300 ms^-1. R = 8.31 × 10⁷ erg per g mol per °C. J = 4.2 × 10⁷ erg cal^-1.

Answer:
Here, C_p = 6.84 cal/g mol/°C
R = 8.31 × 10 erg/g mol/°C
J = 4.2 × 10 erg/cal
v = velocity = 1300 ms^-1
= 1300 × 100 cm s^-1
C_rms =?

Using the relation,


Now using the relation,

Question 7.
Calculate the molecular K.E. of I g of an oxygen molecule at 127°C. Given R = 8.31 JK^-1 mol^-1. The molecular weight of oxygen = 32.

Answer:
Here, M = 32 g
T = 127 + 273 = 400 K

∴ Molecular K.E. of oxygen is given by
12 MC² = 32 RT
Now K.E. of 32 g of O₂ RT = 32RT

∴ K.E.of 1 g of O₂ = 32⋅RT32
or
E = 364 × 8.31 × 400 J
= 155.81 J.

Question 8.
Calculate the intermolecular B.E. in eV of water molecules from the following data:
N = 6 × 10²³ per mole
1 eV= 1.6 × 10^-19 J
L = latent heat of vaporization of water = 22.6 × 10⁵ J/kg.

Answer:
Here, molecular weight of water, M = 2 + 16 = 18g
∴ No. of molecules in 1 kg of water = 6×102318 × 1000 = 10263

L = 22.6 × 10⁵ J kg
∴ B.E .per molecule = 22.6 × 10⁵ J = B.E of 6×102318 molecule

Thus B.E. per molecule

Question 9.
Two perfect gases at absolute temperatures T₁ and T₂ are mixed. There is no loss of energy. Find the temperature of mixture if masses of molecules are m₁ and m₂ and the no. of molecules in the gases are n₁ and n₂ respectively.

Answer:
Let E₁ and E₂ be the K.E. of the two gases,
∴ E₁ = 32 kT₁ × n₁
and E₂ = 32 kT₂ × n₂

Let E be the total energy of the two gases before mixing
∴ E = E₁ + E₂ = 32K(n₁T₁ + n₂T₂) ….(1)

After mixing the gases, let T be the temperature of the mixture of the two gases
∴ E’ = 32kT(n₁ + n₂) …(2)

As there is no loss of energy,

Question 1.
Ram has to attend an interview. He was not well. He took the help of his friend Raman. On the way office, Ram felt giddy, He vomited on his dress. Raman washed his shirt. He made Ram drink enough amount of water. In spite of doing, a foul smell was coming from the shirt. Then Raman purchased a scent bottle from the nearby cosmetics shop and applied to Ram. Ram attended the interview, Performed well. Finally, he was selected.
(a) What values do you find in Raman?

Answer:
He has the presence of mind, serves others in need.

(b) The velocity of air is nearly 500m/s. But the smell of scent spreads very slowly, Why?

Answer:
This is because the air molecules can travel only along a zig-zag path due to frequent collisions. Consequently, the displacement per unit time is considerably small.

In This Post we are  providing Chapter-12 THERMODYNAMICS NCERT MOST IMPORTANT QUESTIONS for Class 11 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON THERMODYNAMICS

1.Change in internal energy is a state function while work is not, why?

Ans. The change in internal energy during a process depends only upon the initial and final state of the system. Therefore it is a state function. But the work is related the path followed. Therefore, it is not a state function.

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2.With the help of first law of thermodynamics and H = U + pv, prove = q_p

Ans.The enthalpy is defined as

H = U + pv

For a change in the stales of system,

= 

= 

= …………(i)

The first law of thermodynamics states that –

= …………………….(ii)

From (i) and (ii),

= 

When the pressure is constant,

3.Show that for an ideal gas, the molar heat capacity under constant volume conditions is equal to 3/2 R.

Ans.For an ideal gas, from kinetic theory of gases, the average kinetic energy per mole (Ek) of the gas at any temperature T_k is given by 

At (T+1)k, the kinetic energy per mole (Ek¹) is Ek¹ = 

Therefore increase in the average kinetic energy of the gas for 1⁰C (or 1K) rise in temperature is 

by definition is to the molar heat capacity of a gas at constant volume, Cv.

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4.A 1.25g sample of octane (C₁₈ H₁₈) is burnt in excess of oxygen in a bomb calorimeter. The temperature of the calorimeter rises from 294.05 to 300.78K. If heat capacity of the calorimeter is 8.93 KJ/K. find the heat transferred to calorimeter.

Ans .Mass of octane,

M = 1.250g.

= 0.00125.

Heat capacity, c = 8.93 kJ/k

Rise in temp, 

= 6.73K

Heat transferred to calorimeter

= 0.00125 x 8.93 x 6.73

= 0.075 kJ

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5.Calculate the heat of combustion of ethylene (gas) to from CO₂ (gas) and H₂O (gas) at 298k and 1 atmospheric pressure. The heats of formation of CO_2, H₂O and C₂H₄ are – 393.7, – 241.8, + 52.3 kJ per mole respectively.

Ans. C₂H₄ (g) + 30₂(g) 2CO₂(g) + 2H₂O (g)

reactants

= [2 x (CO₂) + 2 x ] – 

= 2 x[(-393.7)m+2x (-241.8)] – [(523.0) + 0)]

= [-787.4 – 483.6 ] -53.3

= – 1323.3 kJ.

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6.Give two examples of reactions which are driven by enthalpy change.

Ans. Examples of reactions driven by enthalpy change:

The process which is highly exothermic, i.e. enthalpy change is negative and has large value but entropy change is negative is said to be driven by enthalpy change, eg.

(i)

(ii)

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7.Will the heat released in the following two reactions be equal? Give reasons in support of your answer.

(i)H₂ (g) + 

Ans. No, the heats released in the two reactions are not equal. The heat released in any reaction depends upon the reactants, products and their physical states. Here in reaction (i), the water produced is in the gaseous state whereas in reaction (ii) liquid is formed. As we know, that when water vapors condensed to from water, heat equal to the latent heat of vaporization is released. Thus, more heat is released in reaction (ii).

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8.What is the relation between the enthalpy of reaction and bond enthalpy?

Ans .A chemical reaction involves the breaking of bonds in reactants and formation of new bonds in products. The heat of reaction (enthalpy change) depends on the values of the heat needed to break the bond formation .Thus

(Heat of reaction = (Heat needed to break the bonds in reactants – Heat liberated to from bonds in products).

= Bond energy in (to break the bonds) – Bond energy out (to form the bonds)

= Bond energy of reactants – Bond energy of products.

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9.The reaction C (graphite) + O₂ (g) CO₂(g) + 393.5 kJ mol^-1 represents the formation of CO₂ and also combustion of carbon. Write the values of the two processes.

Ans.(i) The standard enthalpy of formation of CO₂ is -393.5 kJ per mole of CO₂.

That is 

(ii) The staard enthalpy of combustion of carbon is – 393.5 kJ per mole of carbon i.e. 

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10.Explain how is enthalpy related to spontaneity of a reaction?

Ans.Majority of the exothermic reactions are spontaneous because there is decrease in energy.

Burning of a substance is a spontaneous process.

C(s) +O₂(g) 

Neutralisation of an acid with a base is a spontaneous reaction.

Many spontaneous reactions proceed with the absorption of heat. Conversion of water into water vapour is an endothermic spontaneous change. Therefore change in enthalpy is not the only criterion for deciding the spontaneity of a reaction.

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In This Post we are  providing Chapter-11 THERMAL PROPERTIES OF MATTER NCERT MOST IMPORTANT QUESTIONS for Class 11 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON THERMAL PROPERTIES OF MATTER

Question 1.
Why gas thermometers are more sensitive than mercury thermometers?

Answer:
This is because the coefficient of expansion of a gas is very large as compared to the coefficient of expansion of mercury. For the same temperature change, the gas would undergo a much larger change in volume as compared to mercury.

Question 2.
Why the brake drum of an automobile gets heated up when the automobile moves down a hill at constant speed?

Answer:
Since the speed is constant so there is no change of kinetic energy. The loss in gravitational potential energy is partially the gain in the heat energy of the brake drum.

Question 3.
A solid is heated at a constant rate. The variation of temperature with heat input is shown in the figure here:

(а) What is represented by AB and CD?

Answer:
The portions AB and CD represent a change of state. This is because the supplied heat is unable to change the temperature. While AB represents a change of state from solid to liquid, the CD represents a change of state from liquid to vapour state.

(b) What conclusion would you draw1 if CD = 2AB?

Answer:
It indicates that the latent heat of vaporization is twice the latent heat of fusion.

(c) What is represented by the slope of DE?

Answer:
Slope of DE represents the reciprocal of the thermal or heat capacity of the substance in vapour state i.e. slope 0f DE = dTdQ=1mC(∴ dQ = mCΔT).

(d) What conclusion would you draw from the fact that the slope of OA is greater than the slope of BC?
Answer:
Specific heat of the substance in the liquid state is greater than that in the solid-state as the slope of OA is more than that of BC i.e. 1mC1 > 1mC2 where C₁, C₂ are specific heats mC₁ mC₂ of the material in solid and liquid state respectively.

Question 4.
Define:
(a) Thermal conduction.

Answer:
It h defined as the process of the transfer of heat energy from one part of a solid. to another part at a lower temperature without the actual motion of the molecules. It is also called the conduction of heat.

(b) Coefficient of thermal conductivity of a material.

Answer:
It is defined as the quantity of heat flowing per second across the opposite faces of a unit cube made of that material when the opposite faces are maintained at a temperature difference of 1K or 1°C.

Question 5.
On what factors does the amount of heat flowing from the hot face to the cold face depend? How?

Answer:
If Q is the amount of heat flowing from hot to the cold face, then it is found to be:

1. directly proportional to the cross-sectional area (A) of the face
   i. e. Q ∝ A …(1)
2. directly proportional to the temperature difference between the two faces, i.e. Q ∝ Δθ ….(2)
3. directly proportional to the time t for which the heat flows i.e. Q ∝ t …. (3)
4. inversely proportional to the distance ‘d’ between the two faces
   i.e. Q ∝ 1Δx …(4)

Combining factors (1) to (4), we get
Q ∝ AΔθΔxt
or
Q ∝ K A ΔθΔxt
where K is the proportionality constant known as the coefficient – of thermal conductivity.

Question 6.
State Newton’s law of cooling and define the cooling curve. What is its importance?

Answer:
Newton’s law of cooling: States that the rate of loss of heat per unit surface area of a body is directly proportional to the temperature difference between the body and the surroundings provided the difference is not too large.

Cooling Curve: It is defined as a graph between the temperature of a body and the time. It is as shown in the figure here.

The slope of the tangent to the curve at any point gives the rate of fall of temperature.

Question 7.
Explain why heat is generated continuously in an electric heater but its temperature becomes constant after some time?

Answer:
When the electric heater is switched on, a stage is quickly reached when the rate at which heat is generated by an electric current becomes equal to the rate at which heat is lost by conduction, convection and radiation and hence a thermal equilibrium is established. Thus temperature becomes constant.

Question 8.
Specific heats of argon at constant pressure and volume are 0.125 cal g^-1 and 0.075 cal g^-1 respectively. Calculate the density of argon at N.T.P. (J = 4.18 × 10⁷ ergs/cal and normal pressure = 1.01 × 10⁶ dynes cm^-2.)

Answer:
Here, CP = 0.125 cal g^-1
Cv = 0.075 cal g^-1 J
J = 4.18 × 107 ergs cal^-1
P = 1.01 × 106 dyne cm^-2
d = density at NTP = ?
m = 1 g
T = 273 K

Using the relation,
C_p – C_v = rJ=PVTJ=PmdTJ (∵ V = md)
d = Pm TJ(Cr−Cv)

Question 9.
A piece of metal weighs 46 g in air. When it is immersed ¡n a liquid of specific gravity 124 at 27°C, it weighs 30g. When the temperature of the liquid is raised to42°C, the metal piece weighs 30.5 g. The specific gravity of the liquid at 42°C ¡s 1.20. Calculate the coefficient of linear expansion of the metal.

Answer:
Here, the Weight of the metal piece at 27°C in air 46 g
Weight of metal piece at 27°C in liquid =30 g
Weight of metal piece at 42°C in liquid = 30.5 g
α =?
Loss in weight of the metal = weight of liqiid displaced = 46 – 30
= 16 g.

The volume of metal at 27°C = Volume of liquid displaced at 27°C
or
V1 = 16g specific gravity of liquid 
= 16 g1.24gcm−3
= 12.903 cm3

Similarly volume of metal piece at 42°C = V₂ = (46−30.5)1.2gcm−3
= 12.917 cm³

∴ Coefficient of cubical expansion of the metal

Since γ = 3α
∴ α = 13 γ = 13 × 2.41 × 10⁵ °C^-1
= 0.803 × 10⁵ °C^-1

Question 10.
In an industrial process, 10 kg of water per hour is to be heated from 20°C to 80°C. To do so, steam at 150°C is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at 90°C. How many kg of steam is required per hour? Specific heat of steam = 1 Kcal kg^-1 °C^-1 and latent heat of steam = 540 Kcal kg^-1.

Answer:
C = sp. heat of steam
= 1 Kcal kg^-1 °c^-1

L = latent heat of steam
= 540 Kcal kg^-1

Let m (kg) = mass of steam required per hour.

Heat is given by steam first from 150°C to steam at 100°C = mCΔθ
= m × (150 – 100)Kcal = 50 m Kcal.

Then steam changes from steam at 100°C to water at 100°C and gives out heat = mL = 540 m Kcal.

After this water at 100°C gives heat is going to temperature 90°C = m (100 – 90) = 10m Kcal.

Total amount of heat given by the steam = 50 m + 540 m + 10 m = 600 m Kcal.
∴ 600 m K cal = 600 K cal
∴ m = 1 kilogram.

In This Post we are  providing Chapter-10 MECHANICAL PROPERTIES OF FLUIDS NCERT MOST IMPORTANT QUESTIONS for Class 11 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON MECHANICAL PROPERTIES OF FLUIDS

Question 1.
Prove Archimedes’ Principle mathematically.

Answer:
Let W₁ and W₂ be the weights of the body in the air and when completely immersed in a liquid respectively.

∴ Loss in weight of body inside the liquid = W₁ – W₂.

Proof: Let h = height of a body lying at a depth X below the free surface of a liquid of density p.
Let a = area of the face of the body parallel to the horizontal.

If P₁ and P₂ be the pressures at the upper and lower face of the
P₁ = x ρg ….(i)
P₂ = (x + h) ρg …(ii)

If F₁ and F₂ be the thrust on the upper and lower face of the body, then
F₁ = P₁a = xρag …(iii)
and acts vertically downward.

and F₂ = P₂a = (x + h) ρag …. (iv)
and acts vertically upward.

As F₂ > F₁, so net thrust acts on the body in the upward direction and is called upthrust (U)

As V = volume of the body = volume of liquid displaced by the body, so Vρ is the mass of the liquid displaced
∴ Vρg = weight of the displaced liquid

Thus loss in weight of the body when sunk in the liquid = weight of the liquid displaced.

Question 2.
Derive the condition of floatation of the body.

Answer:
When a body floats in a liquid with a part submerged in the liquid, the weight of the liquid displaced by the submerged part is always equal to the weight of the body.

Let V = volume of the body
σ = density of its material
ρ = density of the liquid in which the body floats such that its volume V ‘ is outside the liquid

Then the volume of the body inside the liquid = V – V’
Weight of the displaced liquid = (V – V’) ρg
Also weight of the body = Vσg

For the body to float,
weight of the liquid displaced by the submerged part = weight of the body.

Question 3.
(a) Why the wings of an airplane are rounded outwards (i.e. more curved) while flattened inwards? What is this shape called?

Answer:
The special design of the wings which is slightly convex upward and concave downward increases velocity at the upper surface and decreases it at the lower surface. So according to Bernoulli’s Theorem, the pressure on the upper side is less than the pressure on the lower side.

This difference of pressure provides the additional thrust on the foil called lift. This is called airfoil or aerofoil. It is a solid piece that is so shaped that it an upward vertical. force is produced on it when it moves horizontally through the air.

(b) What is an ideal liquid?

Answer:
A liquid is said to be ideal if:

1. It is incompressible.
2. It is non-viscous.
3. Its flow is steady i.e. stream-line.

Question 4 .
What is a hydrostatic paradox? Explain. Is it really a paradox?

Answer:
It is defined as the inability of a liquid to flow from a vessel having more liquid to a vessel having lesser liquid when the liquid level is the same. Consider three vessels of different shapes but the same base area as shown., The level of water is kept the same in A, B, and C. So the quantity of water is different in the vessels. However, the thrust on the bottom is the same in all of them. It may appear paradoxical

We know that

1. the pressure at a point depends on the height of the liquid column.
2. It does not depend on the quantity of the liquid and
3. thrust is the product of pressure and area of the surface. ,

In the given three cases the pressure at the base is hρg and since the area of the base is the same in all the three cases hence the thrust = hρg a where a = area of the base

So, we see that the thrust is the same in A, B, and C even though the quantity of liquid is different in them.

No. In fact, there is no paradox as such because the pressure depends on the depth of point and not on the quantity of liquid. Here O₁, O₂, and O₃ lie in the same horizontal plane, so the pressure is the same.

Question 5.
State Pascal’s law. How does it get changed in the presence of gravity?

Answer:
Pascal’s law states that for a liquid in equilibrium, the pressure is the same everywhere (provided the effect of gravity can be neglected), It may also be stated as “the pressure applied anywhere on an enclosed fluid is transmitted equally in all directions throughout the fluid”.

Effect of gravity: Consider a liquid of density p contained in a vessel. Let us find the pressure difference between the points x and y. Let us imagine a cylinder of liquid whose faces 1 and 2 are at height h. The cylinder is in equilibrium.

Let P₁ = Pressure at face 1, pressure P₂ at face 2 and the weight of the liquid in cylinder mg. Here m is the mass of the imaginary fluid cylinder. If F, and F, be the forces on the upper and lower faces of the cylinder, then F₁ = P₁ A₁, F₂ = P₂A₂. As the cylinder of liquid is in equilibrium, so the net force on it is zero.
i.e. (F₁ + mg) – F₂ = 0
or P₁A + mg – P₂A = 0
or (P₂ – P₁)A = mg

where A is the base area of the imaginary cylinder. Since mass of the liquid cylinder
m = Vρ = Ahρ (∵ V = Ah)
∴ (P₂ – P₁)A = Ahρg
or (P₂ – P₁) = hρg

If the point x lies on the surface, then P₁ = 0 and Let P₂ = P
∴ P = hρg
Equation (1) gives the expression for the pressure applied by a liquid column of height h.

Question 6.
Draw a diagram showing the construction of a hydraulic brake sIow does it work?

Answer:
The diagram showing various parts of a hydraulic brake is given here.

On applying foot pressure on the pedal, the brake fluid flows*from the master cylinder transmitting the pressure from P₁ to P₂ equally.

This expands the brake shoe and stops the wheel. When pressure is released at the foot pedals, the spring brings the brake shoe to its original position and brake fluid is forced back to the master cylinder.

Question 7.
Stake’s law deals with spherical bodies moving through a viscous fluid. Give its statement and derive it dimensionally.

Answer:
Stake’s law may be stated as “the viscous drag experienced by a spherical body of radius r moving in a fluid of viscosity η with a terminal velocity v is given by
F = 6πηrv

Derivation: Let F depends on η, r, and v, we can write
F = kη_ar_bv_c ….(1)
where k = Proportionality constant.
On writing dimensional formula on both sides, we have

∴ a = 1, b = 1, c = 1

∴ Putting values of a, b, c in equation (1), we get
F = kη₁r₁V₁
= kηrv.

For spherical bodies Stoke found k to be 6π
∴ F = 6ηπrv
Hence, derived.

Question 8.
A ring is cut from a platinum tube having 8.5 cm internal and 8.7 cm external diameter. It is supported horizontally from a pan of balance so that it comes in contact with water in a glass vessel. What is the surface tension of water if an extra 3.97 g weight is required to pull it away from water (g = 980 ms^-2)?

Answer:
Here, m = 3.97 gm
r₁ = 8.5 cm
r₂ = 8.7 cm
T = ?
The water is in contact with the inner and outer circumference of the ring. To pull it out, work has to be done against forces due to surface tension.

Question 9.
A stone of density 2.5 g cm^-3 completely immersed in seawater is allowed to sink from rest. Calculate the depth to which the stone would sink in 2s. The specific gravity of seawater is 1.025 and acceleration due to gravity is 980 cm s^-2. Neglect the effect of friction.

Answer:
Here, g = 980 cm s^-2 .
ρ₁ = density of stone = 2.5 g cm^-3.
Specific gravity of sea water = 1.025
∴ ρ₂ = density of sea water = 1.025 g cm^-3.

Now let m = mass of stone.
∴ V = volume of stone = \frac{\mathrm{m}}{\rho}=\frac{\mathrm{m}}{2.5} cm .
and this is equal to the volume of the displaced seawater.

∴ M = mass of seawater displaced.
= ρ₂ × V
= 1.025 × \frac{m}{2.5} gram

∴ Weight of sea water displaced = \frac{1.025 \times \mathrm{m}}{2.5} × g
If W₁ be the weight of the stone in seawater, then
W₁ = weight of stone – the weight of seawater displaced.

∴ Downward acceleration ‘a’ of stone in seawater is given by

Now let S = depth to which stone sinks. t = 2s, u = 0

∴ Using the relation,

Question 10.
Krishna went sightseeing to a nearby river along with his physics teacher. He noticed that the wind was blowing from the side and the sailboat still continue to move forward. He was surprised. He asked his physics teacher an explanation of this situation. The teacher has noticed his “interest explained the concept through a small example.

The physics of sailing is very interesting in that sailboats do not need the wind to push from behind in order to move. The wind can blow from the side and the sailboat can still move forward.

The answer lies in the well-known principle of aerodynamic
lift Imagine you are a passenger in a car as it’s moving along, and you place your right hand out the window, ff you tilt your hand in the clockwise sense your hand will be pushed backward and up. This is due to the force of the air which has a sideways component and upwards component (therefore your hand is pushed backward and up).
(a) What values could you find- in Krishna?
Answer:
Krishna is very interested in learning the subject; also he is interested in knowing how science helps in understanding the day-to-day experiences, observant, has the courage to ask questions.

(b) Also explain what the Magnus effect is.
Answer:
The difference in velocities of air above the ball is relatively larger than below. Hence, there is a pressure difference between the lower and upper faces and there is a net upward force on the ball. This dynamic lift due to spinning is called the Magnus effect.