Prasanna

Students get through the MP Board Class 11th Chemistry Important Questions Chapter 1 Some Basic Concepts of Chemistry which are most likely to be asked in the exam.

MP Board Class 11th Chemistry Important Questions Chapter 1 Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry Class 11 Important Questions Very Short Answer Type

Question 1.
How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?
Solution:
Molar mass of Na₂CO₃ = (2 × 23) + 12+ (3 × 16)
= 106 g mol^-1
Mass of 0.50 mol Na₂CO₃ = Number of moles × Molar mass
= 0.50 × 106 = 53g Na₂CO₃
Thus, 0.50 M Na₂CO₃ means, 53 gm of Na₂CO₃ is present in 1 litre solution.

Question 2.
What will be the mass of one ¹²C atom in g?
Solution:
Mass of 1 atom of ¹²C = \(\frac{\text { Atomic mass of } \mathrm{C}}{\text { Avogadro number }}\)
\(=\frac{12}{6 \cdot 022 \times 10^{23}}\)
= 1.9927 × 10^-23g .

Question 3.
What is the law of constant proportion?
Answer:
The law states that, “A chemical compound is always found to be made up of the same elements combined together in the same proportion by weight.”
Example : Water from different sources is collected and decomposed. After decomposition in each sample of water, it is found that the weight of hydrogen and oxygen is according to the ratio 2 : 16 or 1 : 8.

Question 4.
N₂ and H₂ combine to form NH₃. Under same condition of temperature and pressure ratio in the volume is 1 : 3 : 2. Write the name of the law which proves it.
Or,
Write Gay Lussac’s Law of gaseous volume.
Answer:
Gay Lussac’s law of combining gaseous volume :
When gases react together, they do so in volumes which bear a simple ratio to one another and to the volumes of the gaseous products provided all measurements were made under similar condition of tem¬perature and pressure.
For example,

The simple ratio in their volume is 1 : 3 : 2.

Question 5.
Write the definition of the following :
(i) Gram atomic mass,
(ii) Gram molecular mass.
Answer:
(i) Gram atomic mass.:
Gram atomic mass of an element may be defined as the mass, which is numerically equal to its atomic mass, expressed in atomic mass units (a.m.u.).

(ii) Gram molecular mass :
The gram molecular mass of an element is the mass of the substance in gram, which is numerically equal to the molecular mass of substance in atomic mass unit. It is also called one gram molecule of the substance.

Question 6.
What is empirical formula?
Answer:
Empirical formula is the simple ratio of the atoms of various elements present in one molecule of the compound.
Significance : It represents the ratio between the different atoms of the compound.
Example : Molecular formula of glucose is C₆H₁₂O₆ ratio between its elements is 1:2:1. Therefore its empirical formula is CH₂O.
Molecular formula = (Empirical formula)_n
Where, n is an integer i.e.,n= 1,2, 3 etc.

Question 7.
Define Molality of a solution.
Answer:
Molality of a solution is the number of moles of solute present in 1000 gm of solvent. It is represented by m.

Question 8.
What is mole?
Answer:
The number of elementary particles present in 12 grams of C-12 is called mole. It is equal to 6.023 x 10²³ particles. This is known as Avogadro number.
6.023 x 10²³ atoms, molecules or ions are called one mole atom, one mole molecule or one mole ions respectively.

Question 9.
What is Limiting Reagent?
Answer:
The reactant which is present in lesser amount gets consumed first. Such a reactant is called limiting reagent because it limits the product formed.

Question 10.
Write the following in scientific notation :
(i) 0.0048
(ii) 234000
(iii) 8008
(iv) 500.0
(v) 6.0012.
Answer:
(i) 4.8 × 10^-3
(ii) 2.34 × 10⁵
(iii) 8.008 × 10³
(iv) 5.0 × 10²
(v) 6.0012 × 10°

Question 11.
Calculate the molecular mass of the following :
(i) H₂O
(ii) CO₂
(iii) CH₄.
Answer:
(i) Molecular mass of H₂O = 2 × Atomic mass of Hydrogen + 1 × Atomic mass of Oxygen
= 2 × 1 + 1 × 16 = 18

(ii) Molecular mass of CO₂ = 1 × Atomic mass of Carbon + 2 × Atomic mass of Oxygen
= 1 × 12 + 2 × 16
= 44

(iii) Molecular mass of CH₄ = 1 × Atomic mass of Carbon + 4 × Atomic mass of Hydrogen
= 1 × 12 + 4 × 1
= 16.

Question 12.
What do you understand by Significant figures?
Answer:
Number of significant figures in a certain value are all the certain digits plus one uncertain digits, For example, number of significant figures in 4.028 are 4.

Question 13.
Tell the number of significant figures in the following :
(i) 0.0025,
(ii) 208,
(iii) 5005,
(iv) 126.000,
(v) 500.00,
(vi) 2.0034.
Answer:
(i) 2,
(ii) 3,
(iii) 4,
(iv) 3,
(v) 4,
(vi) 5.

Question 14.
How are significant figures rounded up?
Answer:
For rounding up a number to three significant figures, if fourth number is smaller than 5 then it is left but if the fourth number is 5 or more than 5 then 1 is added to the third significant figure.

Question 15.
Round up the following into three significant figures :
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808.
Answer:
(i) 34.2
(ii) 10.4
(iii) 0.0460
(iv) 2810.

Question 16.
What are the S.I. Units of mass, Length, Time, Temperature and amount of substance?
Answer:
Unit of mass is kilogram (kg), length is metre (m), time-second (s), temperature is kelvin (K) and amount of substance is mole (mol).

Question 17.
Convert the following into basic units :
(i) 28.7 pm
(ii) 15.15 μs
(iii) 25365 mg.
Solution:
(i) Basic unit of length is metre (m)
2.87 pm × \(\frac{10^{-12} \mathrm{~m}}{1 \mathrm{pm}}\) = 2.87 × 10^-11m

(ii) Basic unit of Time is second (s)
15.15,_μs × \(\frac{10^{6} \mathrm{~s}}{1 \mu \mathrm{s}}\)= 1.515 × 10s

(iii) Basic unit of mass is kilogram (kg)
25365 mg × \(\frac{1 \mathrm{~g}}{1000 \mathrm{mg}} \times \frac{\mathrm{lkg}}{1000 \mathrm{~g}}\) = 2.5365 × 10^-2 kg.

Question 18.
If 10 volume of dihydrogen reacts with 5 volume of dioxygen gas then how many volume of water vapour will be obtained?
Solution:

According to Gay Lussac’s law of gaseous volume
Volume ratio 2 : 1 : 2
According to the question 10 : 5 : 10
Thus, 10 Volume H₂ will react with 5 volume O₂ to form 10 volume water vapour.

Question 19.
What is the law of conservation of matter? What is the present position of this theory?
Answer:
Law of conservation of matter :
In 1744, Lavoisier gave the law of conservation of matter based on the experiments.
This law states that when a chemical reaction takes place, the total mass of the system remains the same before and after the reaction. This law is also referred to as the law of indestructibility of matters, which implies that matter can neither be created nor destroyed by any physical or chemical means.
This law can be proved by experiments.

In the light of modern investigations, law of conservation of matter :
From modern investigations, it is proved that matter can be changed to energy. In some reactions, some energy is produced which causes loss in mass.
According to Einstein’s famous equation of energy-mass relation,
E = mc²
Where, E = energy (in erg), m = mass of the substance (in gm), c = velocity of light (in cm/sec).
In reactions, which we perform in laboratories, very less quantity of energy is absorbed or produced, so the change in mass is very less and is negligible. Such types of changes can be seen clearly in nuclear reaction (atomic bomb). For ordinary reactions, the law is quite valid.

Question 20.
What is the Law of Reciprocal proportion? Explain with example.
Answer:
Law of reciprocal proportion or Law of Equivalent proportion: This law was put forward by Richter in 1792. It states that:
When two different elements unite with the same quantity of a third element the proportions m which they do so will be the same as, or some simple multiple of the proportions in which they unite with each other.
Or
Weights of two elements (say A and B) which separately combine with a fixed weight of a third element (C) are either the same or simple multiples of the weights (of A and B) in which they combine with each other.

This law can be explained with the help of the following examples:
Consider three elements carbon, oxygen and hydrogen. Element carbon and oxygen combines with a definite mass of hydrogen in the ratio given below :
In methane (CH₄), H : C = 4 : 12
and in water (H₂O),H : O = 2 : 16 or 4 : 32
According to this law if carbon and oxygen combines with each other then their ratio should be 12 : 32 or some simple multiple of it.
We know that in carbon dioxide, CO₂, ratio of C : O = 12 : 32 which is according to this law. Hence, it illustrates law of reciprocal proportion.

Question 21.
In which of the following number of atoms is maximum :
(i) 1gAu_(s)
(ii) 1g Na_(s)
(iii) 1g Li_(s)
(iv) 1g Cl_2(g).
Answer:
Number of moles of substance = \(\frac{\text { Mass of substance }(\mathrm{g})}{\text { Molar mass }}\)
Number of Atoms = Number of moles × Avogadro number
(i) 1gAu_(s) = \(=\frac{1}{197}\) Au mole atoms =\(=\frac{1}{197}\) × 6.022 × 10²³ Au atoms
= 3.057 × 10²¹ atoms

(ii) 1g Na_(s) = \(\frac{6 \cdot 022 \times 10^{23}}{23}\) = 2.618 × 10²² atoms

(iii) 1g Li_(s) = \(\frac{6 \cdot 023 \times 10^{23}}{7}\) = 8.604 × 10²² atoms

(iv) 1g Cl_2(g) = \(\frac{1}{71}\)Cl₂ mole molecule = \(\frac{1}{71}\) × 6.022 × 10²³ Cl₂ molecule
∴ (1 mole of chlorine contains 2 atoms)

= \(\frac{2}{71}\) × 6.022 × 10²³ Cl atoms
= 1.697 × 10²² atoms
Thus, maximum number of atoms are in 1g Li_(s)

Question 22.
How is molecular formula of a substance related to its Empirical formula?
Answer:
Molecular formula is represents the actual number of atoms of various elements present in one mole of a compound.
It is either similar to the empirical formula or a simple multiple of it.
Molecular formula = n (Empirical formula)
n = \(\frac{\text { Molecular mass }}{\text { Empirical formula mass }}\)
Example: The empirical formula of a compound is CH₂O and its molecular mass is 180.
n = \(\frac{\text { Molecular mass }}{\text { Empirical formula mass }}\)
\(=\frac{180}{12+12+16}=\frac{180}{30}\) = 6
Thus, Molecular formula = n x Empirical formula
= 6(CH₂O ) = C₆H₁₂O₆.

Question 23.
(i) What is law of multiple proportion? Explain with an example.
(ii) Why is it necessary to balance chemical equations?
Answer:
(i) Law of multiple proportion :
When two elements combine to form two or more compounds, different weights of a particular element which combines with definite amount of other element, bear a simple ratio to one another.
Example : Carbon monoxide (CO) and carbon dioxide (CO₂) are two compounds of elements carbon and oxygen. The weight of oxygen which combines with a definite weight of carbon (i.e., 12) are 16 and 32 respectively. Thus, the simple ratio of weight of oxygen is 1 : 2, which supports law of multiple proportion.

(ii) It is necessary to balance a chemical equation in order to satisfy the law of conservation of mass according to which the total mass of reactants is equal to the total mass of products.

Question 24.
What is the law of constant composition? How has this law been affected by the discovery of isotopes?
Answer:
This law was stated by French chemist Louis Proust (1799). It states that ‘A pure chemical compound, regardless of its source, is always made up of the same elements combined together in the fixed proportion by weight.’
Example: Regardless of its source, water is always composed of hydrogen and oxygen combined together in the ratio of 1 : 8 by weight.

Present position of the law :
Discovery of isotopes has proved that an element can have different atoms with different atomic masses but identical properties. Thus there cannot be fixed composition of any compound.
Example : The composition of water is given below :

These figures are contradictory to the law, however the law is valid as the distribution of isotopes in an element is uniform in the nature.

Question 25.
Use the data given in the following table for calculating the molar mass of Argon available in nature :

Solution:
Average molar mass of Ar = Σf_i × A_i
= (0.00337 × 35.96755) + (0.00063 × 37.96272) + (0.99600 × 39.9624)
= 0.121 + 0.024 + 39.803
= 39.948 g mol^-1.

Question 26.
How many significant figures siinuid he in the following calculations.
(i) \(\frac{0 \cdot 02850 \times 298 \cdot 15 \times 0 \cdot 112}{0 \cdot 5785}\)
(ii) 5 × 5.364
(iii) 0.0125+0.7864+0.0215.

Solution:
(i) In multiplication and di ision in minimum exact number 0.112 is 3 significan
figures. Thus, three (3) significant figures should he in tLe tesult.
\(\frac{0 \cdot 02856 \times 298 \cdot 15 \times 0 \cdot 112}{0 \cdot 5785}\) = \(\frac{0 \cdot 953698}{0 \cdot 5785}\)
= 1.64857
Three significant figures = 1.65

(ii) In multiplication exact number is 5 and in the second number exact number is four. Thus, there should be 4 significant figures in the result.
5 × 5.364 = 26.82 (4 significant figures)

(iii) In addition and subtraction, in the result should not be more than the number beyond the decimal after addition (or subtraction). Thus, there should be 4 significant figures in the result.
0.0125 + 0.7864 + 0.0215 = 0.8204 Four significant figures.

Question 27.
Explain modern concept of Dalton’s atomic theory.
Answer:
Modem concept of Dalton’s atomic theory are

• According to Dalton, atom is indivisible, but now it is established that atom is divisible. ,
• It is not necessary that weights of atoms of the same element may be same, their weights may be different Amins of the same element havina different weights are called isotopes.
• It is not necessary that atomic massof different elements may be different, e.g., argon and calcium have same atomic mass i,. e., 40 a.m.u. .
• Atoms always do not combine in small numbers and simple ratio. For example, the ratio of C, H and O in sugar (C₁₂H₂₂O₁₁) molecule is 12 : 22 : 11 which is not a simple ratio.
• Atom is the smallest unit which can participate in chemical reaction.
• According to Dalton’s idea, atom is indivisible and it cannot be created: Nowadays breaking of atoms and formation of new atoms is possible by means of nuclear transmutation.

Question 28.
Write the difference between Atom and Molecule.
Answer:
Differences between Atom and Molecule

Question 29.
Differentiate between Molarity and Molality.
Answer:
Differences between Molarity and Molality

Some Basic Concepts of Chemistry Class 11 Important Questions Long Answer Type

Question 1.
What is chemical equation? Write the ways to balance it with example.
Answer:
Method of representing a chemical reaction with the help of their chemical formulae and symbols, is called chemical equation.

Balancing of a Chemical Equation :
In general, two methods are used for balancing chemical equations :
1. Hit and trial method :
(i) First of all atoms of that element which covers smallest space are balanced on both sides of the equation.
(ii) Those equations in which oxygen, nitrogen, hydrogen, etc. gases are formed, first of all they are written in atomic state. Such an equation obtained is called atomic equation. Atomic equation is multiplied by 2 to get molecular equation.
Example 1.
Potassium chlorate on heating gives potassium chloride and oxygen.
(a) KClO₃ → KCl + O (skeleton equation)

(b) In this equation, right side oxygen atom is multiplied by 3 to get same number of
oxygen atoms on both sides.
KClO₃ → KCl + 3O

(c) But in this equation, oxygen is in atomic form. To convert it into molecular form,
the equation is multiplied by 2.
Thus, 2KClO₃ → 2KCl + 3O₂ (balanced equation)

Example 2.
Copper reacts with sulphuric acid to give copper sulphate, sulphur dioxide
and water.
(a) Cu + H₂SO₄ → CuS₄O + SO₂ + H₂O (skeleton equation) .

(b) To balance sulphur atoms on both sides, H₂SO₄ is multiplied by 2.
Cu + 2H₂SO₄ → CuSO₄ + SO₂ + H₂O

(c) Now to balance hydrogen atoms, H₂O is multiplied by 2.
Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O (balanced equation)

2. Partial equation method:
It is generally difficult to balance the complex equations by Hit and Trial method. It is observed that complex equations gets completed in two or more steps. These equations are balanced. If necessary, then partial equation is multiplied
by a whole number. Then, all the partial equations are added in such a way that the intermediate product which is not obtained in the last step gets cancelled. This way, the last equation obtained is correct and balanced.

Example : On heating Cu with cone. H₂SO₄, copper sulphur, sulphur dioxide and water is formed. This equation can be represented by the following partial equations.

Intermediate products 2H and O being present on both the sides of the equations are cancelled.

Question 2.
What is the law of multiple proportion? Explain its situation in relation to modern research.
Answer:
Law of multiple proportion :
This was stated by Dalton in 1803, and was experimentally proved by Berzelius and Stass. It states that, “When two elements combine with each other to form two or more compounds the masses of one of the elements which combine with a fixed mass of the other bear a simple whole number ratio to each other”. For example, Nitrogen forms 5 oxides.

Various masses of oxygen 8,16,24,32,40 which combine with a definite mass 14 of nitrogen is in the form of a simple ratio.

Law of multiple proportion in relation to Modem Research:

• Law of multiple proportion can be applicable in that situation when a mass of isotope of an element combines with various masses of a definite isotope of a second element.
• This law is not true for organic compounds.

Question 3.
Explain mole and Avogadro’s number. Write the definition of Atomic mass and molecular mass on the basis of mole concept.
Answer:
Mole concept:
Atom and molecule are extremely minute particles. In a very small amount of a substance also number of atoms and molecules is very large. For example One milligram (0.001 gram) of carbon contains total 5.019 × 10¹⁹ atoms. Generally in chemical studies large amount of substances are used in which number of atoms or molecules is extremely high. It is impossible to count these molecules but its knowledge is essential. Just as 12 pen is a dozen pen, 1000 metre is a kilometre, similarly group of 6.023 × 10²³ particles is called one mole. These particles can be molecules atoms or ion.

Thus mole is a unit which represents 6.023 × 10²³ particles (molecule, atoms or ion)
Number of molecules in one mole molecule = 6.023 × 10²³
Number of atoms in one mole atom = 6.023 × 10²³
Number of ions in one mole ion = 6.023 × 10²³
It also represents molar mass :
One gram mole oxygen i.e., 32 gram oxygen contain total 6.023 × 10²³ molecules. One gram mole sodium i.e., 23 gram sodium contain 6.023 × 10²³ atoms.
Formula mass of sodium chloride is 58.5. Thus 58.5 gram sodium chloride contain one mole sodium ion and one mole chloride ion.
“At same temperature and same pressure equal volume of gases contains equal number of atoms.” According to this law, at N.T.P. one gram of all gases contain 6.023 × 10²³ atoms. This number is known as Avogadro’s number and it is denoted by N.

Question 4.
What is Stoichiometry? How are the problems based on chemical equations solved by it?
Answer:
Stoichiometry is derived from the Greek word ‘stoicheion ’ which means element and ‘ metron ’ meaning measure. The word stoichiometry means measurement of amount of element or compound.

In a balanced chemical equation, a quantitative relation exists between the molecules, masses, moles and volume of the reactants and products participating a chemical equation. Problems related to these calculations are of three types :

(a) Involving Mass-Mass relationship :
In such type of problems mass of a product or reactant is given and that of the other is calculated.

(b) Involving Mass-Volume relationship:
In such type of problems mass/volume of a reactant or product is given and that of the other is calculated.

(c) Involving Volume-Volume relationship :
In such type of problems volume of
one reactant or product is given and that of the other is calculated.
To solve the above type of questions (Problems) the following procedure is adopted:

• Balanced chemical equation is written in its molecular form.
• Select the symbols and formulae of those species (atoms or molecules) whose mass/volume are either given or are known by calculation.
• Atomic mass/molecular mass/mole/molar volume of molecules and atoms included in calculation are written.
• By common mathematical calculations amount of the selected substance is calculated.
  Think about the following balanced equation :

Number of molecules (mole) is known as stoichiometric coefficient.

Question 5.
What is a chemical equation? What its limitations? How can it be made more informative?
Answer:
Chemical equation :
Method of representing a chemical reaction with the help of chemical formula and symbol of species involved in the reaction called chemical equation. For example, zinc reacts with dilute hydrochloric acid give zinc chloride and hydrogen can
be represented as
Zn + 2HCl → ZnCl₂ + H2↑

Limitations Of a chemical equation :
Following informations cannot be obtained from a chemical equation:

• Physical state of reactants and products.
• Concentration of reactants and products cannot be furnished.
• Does not give information about the condition of the reaction,
• No knowledge is obtained about the velocity of the reaction.
• No knowledge about the absorption or evolution of heat of the reaction is obtained.
• Chemical equation does not give any information whether a gas is evolved or a precipitate is formed.
• Chemical equation does not furnish any information about what precautions are necessary for reaction.
• Chemical equation does not furnish any idea about the method of the chemical reaction.
• No information regarding the nature of reaction, i.e., reversible or irreversible.
• It does not provide information about phosphorescence or explosion.
• How much time it will take in completion, this type of information is not obtained.

A chemical equation can be made more clear and informative and useful by balancing
the chemical equations and using the suitable signs. The following are the advantages of it:
(a) Number of reactants and products participating in the reaction.
(b) If atomic mass of the elements are known, then molecular mass of reactants and products can be calculated.
(c) If reactant and products are gaseous, then their volume can be calculated.
(d) It proves the law of conservation of matter.
(e) Valency and Equivalent mass of the elements can be calculated.

Some Basic Concepts of Chemistry Class 11 Important Numerical Questions

Question 1.
Calculate the mass percent of different elements present in sodium sulphate (Na₂SO₄).
Solution:
Mass percent of Element \(=\frac{\text { Mass of the element in the compound }}{\text { Molar mass of the compound }}\) × 100
Molar mass of Na₂SO₄ = (2 × 22.99) + 32.06 + (4 × 16.00)
= 142.04g
Mass percent of Sodium = \(\frac{45 \cdot 98 \times 100}{142 \cdot 04}\)
= 32.37%
Mass percent of Sulphur = \(\frac{32 \cdot 06 \times 100}{142 \cdot 04}\)
= 22.57%
Mass percent of Oxygen \(=\frac{64 \times 100}{142 \cdot 04}\)
= 45.06%.

Question 2.
Determine the Empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
Solution:

Question 3.
Calculate the amount of carbon dioxide that could be produced when :
(i) 1 Mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
Solution:

(i) According to the above equation on burning 1 mole of carbon completely 44g of CO₂ is produced.
(ii) According to the above equation, for burning 1 mole of carbon 32g of oxygen is required. But only 16gm of oxygen is available. Thus, oxygen is the limiting reagent.
Thus, CO₂ obtained from 16g of oxygen = \(\frac{44 \times 16}{32}\)
= 22 gm.

Question 4.
Calculate the mass of sodium acetate (CH₃COONa) required to make 500 ml of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^-1.
solution:
Molarity M = \(\frac{w \times 1000}{M \times \text { Volume of solution (in ml) }}\)
Where w = Mass of solute
M = Molar mass of solute
Given
Molarity of solution = 0.375 M
Molar mass of solute = m = 82.0245g mol^-1
Volume of solution = 500 ml
Mass of solute w = ?
Mass of solute w = \(\frac{m \times M \times V(\text { in } m l)}{1000}\)
\(\frac{m \times M \times V(\text { in } m l)}{1000}\)
=15.379g
= 15.38g.

Question 5.
How much copper can be obtained from 100g of copper sulphate (CuSO₄)?
Solution:
Molar mass of CuSO₄ = 63.54 + 32.06 + (4 × 16)
= 159.6g mol^-1
159.6g of CuSO₄ contains Cu = 63.54g
∴ 1 g of CuSO₄ will contain Cu = \(\frac{63 \cdot 54}{159 \cdot 6}\) g
∴ 100g of CuSO₄ will contain Cu = \(\frac{63 \cdot 54}{159 \cdot 6}\) × 100 = 39.81g

Question 6.
Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively.
Solution:
First of all Empirical formula is determined,

The values given are same.
Empirical formula of iron oxide = Fe₂O₃
Calculation of Empirical formula mass = (2 × 55.85) + (3 × 16)
= 159.7gmol^-1
n = \([latex]\begin{array}{c}
\text { Molar mass } \\
\hline \text { Empirical formula mass }
\end{array}\)[/latex]
= \(\frac{159 \cdot 7}{159 \cdot 7}\) = 1
∴ Molecular formula = (Empirical formula),,
= (Fe₂O₃ )₁
= Fe₂O₃.

Question 7.
Calculate the concentration of nitric acid in moles per litre in a sample which has a density 1.41gm^-1 and the mass percent of nitric acid in it being 69%.
Solution:
Molarity = \(\frac{w \times 1000}{\mathrm{M} \times \mathrm{V}_{(\mathrm{ml})}}\)
Given, d = 1.41 gml^-1 mass percent of HNO₃ = 69%
69% HNO₃ means, In 100g solution of HNO₃, 69g of HNO₃ is present.
Thus, mass of HNO₃ (solute) w = 69g
Molar mass of HNO₃ = 1.0079 + 14.0067 + (3 × 16.00)
= 63.0146 gmol^-1
Density d \(=\frac{m}{V}\)
or
V = \(\frac{m}{d}=\frac{100 \mathrm{~g}}{1 \cdot 41 \mathrm{gml}^{-1}}\)
Molarity = \(\frac{69 \times 1000 \times 1 \cdot 41}{63 \cdot 0146 \times 100}\) =15.439 M

Question 8.
If the speed of light is 3.0 × 10⁸ ms^-1, calculate the distance travelled by light in 2.00ns.
Solution:
2.00ns = 2.00 × 10^-9s (1 ns=10^-9s)
Distance covered = Velocity × Time
=3.0 × 10⁸ms^-1 × 2.0 × 10^-9S
= 6.0 × 10^-1 = 0.6m.

Question 9.
Calculate the atomic mass (average) of chlorine using the following data :

Solution:
Average atomic mass \(\overline{\mathrm{A}}\), is equal to the product of relative abundance F_i and their corresponding molar mass A_i.
Average atomic mass \(\overline{\mathrm{A}}\) = Σ f_i A_i = F₁ × A₁ + F₂ × A₂ + ……………
Thus, average atomic mass of chlorine \(\overline{\mathrm{A}}\) = 0.7577 × 34.9689 + 0.2423 × 36.9659 g
= 26.4959 + 8.9568
= 35.4527u ≈ 35.5u.

Question 10.
What is the concentration of sugar (C₁₂H₂₂O₁₁) in moT1 if its 20g are dissolved in enough water to make a final volume up to 2L?
Solution:
Molar mass ofsugar(C₁₂H₂₂O₁₁)(M) = (12 x 12) + (22 x 1 .0079) + (11 x 16)
= 342 g mol^-1
Given, w = 20g and volume = 2L
Molarity (M) = \(\frac{w}{M \times \text { Volume of solution in litre }}\)
\(=\frac{20}{342 \times 2}\)
= 0.0292 mol L^-1

Question 11.
If the density of methanol is 0.793 kgL^-1, what is the volume needed for making 2.5 L of its 0.25 M solution?
Solution:
Given, d = 0.793 kgL^-1 = 0.793 × 10³ gL^-1
Final volume V₂ = 2.5 L
Final Molarity M₂ 0.25 M
Initial molarity of solution = M₁ = ?
Initial volume = V₁ = ?
Molar mass of methanol (CH₃OH) = (1 × 12) + (4 × 1.0079) + (1 × 16.00)
= 32.0416 ≈ 32gmol^-1
Molarity M₁ = \(\frac{0.793 \times 10^{3} g \mathrm{~L}^{-1}}{32 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 24.781 mol L^-1
∵ Thus, M₁V₁ ≡ M₂V₂
24.781 × V₁ = 0.25 × 2.5
V₁ = \(\frac{0 \cdot 25 \times 2 \cdot 5}{24 \cdot 781}\)= 0.02522L = 25.22 ml

Question 12.
Pressure is determined as force per unit area of the surface. The S.I. unit of pressure, pascal is shown below :
1 Pa = 1 Nm^-2
If mass of air at sea level is 1034 gcm^-2. Calculate the pressure in Pascal.
Solution:
Pressure is force or mass per unit area.
Pressure = \(\frac{\text { Force }}{\text { Area }}\)

= 101332.0 Nm^-2[IN = 1kg ms^-2]
= 1.01332 × 10⁵ Pa [1Nm^-2= 1 pa]

Question 13.
A sample of drinking water was found to be severely contaminated with chloroform CHC13, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in the water sample.
Solution:
(i) 15 ppm means :
∵ 10⁶ gram solution contains 15 gm CHCl3
∴ 1 gram solution contains = \(\frac{15}{10^{6}}\)
∴ 100 gram solution contains = \(\frac{15 \times 100}{10^{6}}\) = 15 x 10^-14gm ≈ 1.5 x 10^-3%

(ii) Molar mass of CHCl₃ = 12 + 1 + (3 × 35.5)
= 119g mol^-1
1.5 × 10^-3% means 100g of sample contains
1.5 × 10^-3g CHCl₃
Molarity M = \(\frac{w \times 1000}{m \times \text { Volume of sample }}\)
For water density = \(\frac{1 \cdot 5 \times 10^{-3} \times 1000}{119 \times 100}\) = 0.000126 ,
= 1.26 × 10^-4M.

Question 14.
The following data are obtained when dinitrogen and dioxygen react
together to form different compounds :

(a) Which law of chemical combination is obeyed by the above experimental
data. Give its statement.
(b) Fill in the blanks in the following conversions :
(i) 1km = ………………. mm = …………….. pm
(ii) 1mg = ………………… kg = ……………….. ng
(iii) 1mL = ………………. L = ……………… dm³.
Answer:
(a) Law of multiple proportion :
According to this law,’ “If two elements combine together to form more than one compound then the masses of one of the element which combine with a fixed mass of the other element is in the form of simple multiple ratio”.
In the given example, in the four oxides if mass of nitrogen is fixed 28g then the ratio of oxygen which combine is 32, 64, 32, 80 which is a simple ratio is 2 : 4 : 2 : 5. Thus, the given data follows the law of multiple proportion.
1km = 1km × \(\frac{1000 \mathrm{~m}}{\mathrm{lkm}}\) × \(\frac{100 \mathrm{~cm}}{1 \mathrm{~m}}\) × \(\frac{10 \mathrm{~mm}}{\mathrm{lcm}}\) 10⁶mm
1km = 1km × \(\frac{1000 \mathrm{~m}}{\mathrm{lkm}}\) ×\(\frac{1 \mathrm{pm}}{10^{-12} \mathrm{~m}}\) = 10^-15
∴ 1km = 10⁶mm = 10¹⁵pm

(ii) 1mg = 1mg × \(\frac{\lg }{1000 \mathrm{mg}}\) × \(\frac{1 \mathrm{~kg}}{1000 \mathrm{~g}}\) = 10^-610kg
1mg = 1mg × \(\frac{\lg }{1000 \mathrm{mg}}\) × \(\frac{\ln g}{10^{-9} \mathrm{~g}}\) = 10⁶10ng
∴ 1mg = 10^-610kg = 10⁶10ng

(iii) 1mL = 1mL × \(\frac{1 \mathrm{~L}}{1000 \mathrm{~mL}}\) = 10^-3L
1mL = 1cm³ = 1cm³ × \(\frac{1 \mathrm{dm} \times 1 \mathrm{dm} \times 1 \mathrm{dm}}{10 \mathrm{~cm} \times 10 \mathrm{~cm} \times 10 \mathrm{~cm}}\) = 10³dm³
∴ 1mL = 10^-3L = 10^-3dm³

Question 15.
In a reaction
A + B₂ → AB₂
Identify the limiting reagent, if any, in the following reaction mixtures :
(1) 300 atoms of A + 200 molecules of B
(ii) 2 mole A +3 mole B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mole A + 25 mole B
(v) 2.5 mole A +5 mole B.
Solution:
Limiting reagent first gets consumed (finishes) in a reaction.. Thus, to determine it
amount of A and 13 are compared.
A + B₂ → AB₂
(i) According to the above equation 1 atom of A reacts with I molecule of B. Thus,
200 atoms of Awil1 react with 200 molecules of B.
Thus, B is the limiting reagert and A will be in excess.

(ii) According to the above equation 1 mole of A reacts with I mole of B. Thus, 2 moles
of A will react with 2 moles of B.
In this condition, A will be the limiting reagent, and B will be in excess.

(iii) According to the above equation I atom of A reacts with 1 molecule of B.
Thus, 100 atoms of A will react with 100 molecules of A.
Thus, it is a stoichiometric mixture.
Thus, there is no limiting reagent neither A nor B.

(iv) In this condition 2.5 mole of A will react with 2.5 mole of B.
Thus, B is limiting reagent and A is in excess.
(v) In this condition A will be the limiting reagent and B will be in excess.

Question 16.
Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation :
N_2(g) + 3H_2(g) → 2NH_3(g)
(i) Calculate the mass of ammonia produced if 2.00 × 10³g dinitrogen reacts with
1.00 × 10³g of dihydrogen.
(ii) Will any of the two reactants remain unreacted.
(iii) If yes, which one and what would be its mass.
Solution:

∵ 28g of N₂ reacts with = 6gH₂
∴ 1g N₂ will reacts with = \(\frac{6}{28}\) gH₂
∴ 2000g N₂ will reacts with = \(\frac{2000 \times 6}{28}\) = 428.57gH₂
Thus, N₂ will be the limiting reagent.N₂ limits the amount of ammonia produced,
∴ 28g N₂ produces = 34g NH₃
∴ 1 gN₂ produces = \(\frac{34}{28}\)gNH₃
∴ 2000 g N₂ produces = \(\frac{34}{28}\) × 2000 = 2428.57g NH₃
= 2429g NH₃
(ii) H₂ will be left. It is in excess.
(iii) Remaining amount of H₂= 1000 – 428.57
= 571.43g.

Question 17.
A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38g carbon dioxide, 0.690g of water and no other products. A volume of 10.0L (measured at STP) of this welding gas is found to weigh 11.6g. Calculate
(i) Empirical formula,
(ii) Molar mass of the gas and
(iii) Molecular formula.
Solution:
(i) ∵ 44g CO₂ = 12g Carbon
∴ 3.38g CO₂ = \(\frac{12}{44}\) × 3.38g = 0.9218g Carbon
∵ 18g H₂O = 2g Hydrogen
∴ 0.690g H₂O = \(\frac{2}{18}\) × 0.690g = 0.0767g Hydrogen
Mass of compound = 0.9218 + 0.0767 = 0.9985g
Since, only carbon and hydrogen are present in the compound.
% Amount of C in the compound = \(\frac{0 \cdot 9218}{0 \cdot 9985}\) × 100 = 92.32%
% Amount of H in the compound = \(\frac{0 \cdot 0767}{0 \cdot 9985}\) × 100 = 7.68% .

Thus, Empirical formula = CH

(ii) Calculation of molecular mass of gas:
∵ At STP, mass of 10.L of gas = 11.6g
∴ At STP, mass of 22.4L of gas =\(\frac{11 \cdot 6 \times 22 \cdot 4}{10 \cdot 0}\) = 25.984g ≈ 26g mol^-1.

(iii) Calculation of Molecular formula:
Empirical formula mass (CH) = 12 + 1 = 13
n = \(\frac{\text { Molecular mass }}{\text { Empirical formula mass }}\) = \(\frac{26}{13}\) =2
Molecular formula = (Empirical formula)_n
=(CH)₂
=C₂H₂. [n = 2]

Question 18.
Calculate the molarity of a solution of ethanol in water in which mole fraction of ethanol is 0.040 (assume the density of water to be 1).
Solution:
Number of moles of solute (Ethanol) in 1 litre solution will be molarity.
1L ethanol solution (dilute solution) = 1L water.
Number of moles of H₂O in 1L water = \(\frac{1000 \mathrm{~g}}{18}\) = 55.55mol
In Binary solution, two components are present.
X₁+ X₂ = 1
X_H2O = 1 – X_C2H5OH
X_H2O = 1 – 0.040 = 0.96
X_H2O = \(\frac{\cdot n_{\mathrm{H}_{2} \mathrm{O}}}{n_{\mathrm{H}_{2} \mathrm{O}}+n_{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}}\)
0.96 = \(\frac{55 \cdot 55}{55 \cdot 55+n_{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}}\)
On cross multiplication,
(0.96 x 55.55 + 0.96n_C2H5OH = 55.55
53.328 + 0.96n_C2H5OH = 55.55
0.96n_C2H5OH = 55.55 – 53.328 = 2.222
n_C2H5OH = \(\frac{2 \cdot 222}{0 \cdot 96}\)
2.3145 mol.

Question 19.
Calcium carbonate reacts with aqueous HCl to give CaCl2 and C02 according to the reaction :
CaCO_3(s) +2HCL_(aq) → CaCl_2(aq) + CO_2(g) + H₂O_(l)
What mass of CaCO₃ is required to react completely with 25ml of 0.75 M HCl?
Solution:
(i) Mass of 25 ml of 0.75 M HCl
Molarity = \(\frac{w \times 1000}{m \times V(m l)}\)
0.75 = \(\frac{w \times 1000}{36 \cdot 5 \times 25}\)
Molecular mass of HCl = 1 + 35.5 36.5 g/mol
w = 0.75 × 36.5 \(\frac{25}{1000}\) = 0.6844g.

(ii) CaCO_3(s) + 2HCl_(aq) ) → CaCl_2(aq) + CO_2(g) + H₂O₍₁₎
100g 2 × 36.5 = 73g
∵ 73g HC1 completely reacts with = 100g CaCO₃
∴ 1 g HCl completely reacts with = \(\frac{100}{73}\) g CaCO₃
∴ 0.6844g HCl will completely reacts with = \(\frac{100 \times 0.6844}{73}\)
= 0.9375g CaCO₃.

Question 20.
Chlorine is prepared in the laboratory by treating manganese dioxide (MnO) with aqueous hydrochloric acid according to the reaction:
4HCl_(aq) + MnO_(2s) → 2H₂O_(l) + MnCl_2(aq) + Cl_2(g)
How many grams of HCl reacts with 5.0g of manganese dioxide?
Solution:
MnO₂ + 4HCl → MnCl₂ +2H₂O+Cl₂
1 mole 4 mole
87g 36.5 × 4
= 146g
∵ 87g MnO₂ reacts with 146 g HCl
∴ 1 g MnO₂ will reacts with = \(\frac{146}{87}\) g HCl
∴ 5g MnO₂will react with = \(\frac{146 \times 5}{87}\) = 8.39g HCl.

Question 21.
Calculate the number of atoms in each of the following:
(i) 52 moles of Ar
(ii) 52u of He
(iii) 52g of He.
Solution:
(i) ∵ 1 mole Ar = 6.022 × 1o²³ atoms
∴ 52 moIes Ar = 52 × 6.022 × 10²³ atoms

(ii) ∵ 4u of He = I He atom
∴ 52u of He = \(\frac{52}{4}\) He atom = 13 He atoms.

(iii) 1 mole atom of He 4g = 6.022 × atoms²³
∴ 52g of He \(=\frac{52 \times 6 \cdot 022 \times 10^{23}}{4}\) = 78.286 × 10²³ atoms
= 7.82886 × 10²²⁴ atoms.

Some Basic Concepts of Chemistry Class 11 Important Questions Objective Type

1. Choose the correct answer:

Question 1.
Volume of 2.0 gram of hydrogen gas at N.T.P. is :
(a) 224 litre
(b) 22.4 litre
(c) 2.24 litre
(d) 112 litre.
Answer:
(b) 22.4 litre

Question 2.
Molarity of pure water is :
(a) 18
(b) 50
(c) 55.6
(d) 100.
Answer:
(c) 55.6

Question 3.
Number of atoms in 12 gm of ₆C¹² is :
(a) 6
(b) 12
(c) 6.02 × 10²³
(d) 12 × 6.02 × 10²³.
Answer:
(c) 6.02 × 10²³

Question 4.
Total mass of the product at the end of chemical reaction :
(a) Always increases
(b) Always decreases
(c) Does not change
(d) Always increases or decreases.
Answer:
(c) Does not change

Question 5.
Five oxides of nitrogen N₂O, NO, N₂O₃, N₂O₄ and N₂O₅ follow which law of chemical combination:
(a) Constant proportion law
(b) Reciprocal proportion law
(c) Gay Lussac’s law of gaseous volume
(d) Multiple proportion law.
Answer:
(d) Multiple proportion law.

Question 6.
Empirical formula of a compound is CH₂. Mass of one mole of a compound is 42 grams. Its molecular formula will be :
(a) CH₂
(b) C₃H₆
(C) C₂H₂
(d) C₃H₈.
Answer:
(b) C₃H₆

Question 7.
Who proposed law of definite proportion :
(a) Lavoisier 1774
(b) Dalton 1804
(c) Ritcher 1792
(d) Proust 1799.
Answer:
(c) Ritcher 1792

Question 8.
Percentage amount of oxygen in ZnSO₄.7H₂O will be :
(a) 22.65%
(b) 11.15%
(c) 22.30%
(d) 43.90%.
Answer:
(d) 43.90%.

Question 9.
One mola solution is that in which one mole solute is present in :
(a) 1000 gram solvent
(b) 1 litre solution
(c) 1 litre solvent
(d) 22.4 litre solution.
Answer:
(a) 1000 gram solvent

Question 10.
Mass of 22.4 litre SO₂ at N.T.P. will be :
(a) 44 gram
(b) 0.44 gram
(c) 64 gram
(d) 3 gram
Answer:
(c) 64 gram

Question 11.
What will be the percent mass of copper in cupric oxide :
(a) 22.2%
(b) 79.8%
(c) 63.5%
(d) 16%
Answer:
(b) 79.8%

Question 12.
Who is considered to be the father of modern chemistry :
(a) Priestly
(b) Lavoisier
(c) Robert Boyle
(d) Rutherford.
Answer:
(b) Lavoisier

Question 13.
Mass of 112 cm³ CH₄ at S.T.P. will be :
(a) 016 gram
(b) 0.8 gram
(c) 0.08 gram
(d) 1.6 gram
Answer:
(d) 1.6 gram

Question 14.
Sometimes it is observed that there is a decrease in mass in a chemical reaction. This is observed to be against which of the following laws :
(a) Constant proportion
(b) Multiple proportion
(c) Conservation of mass
(d) Reciprocal proportion.
Answer:
(c) Conservation of mass

Question 15.
Separation of two substances by fractional distillation method is based on the difference of their:
(a) Density
(b) Volatility
(c) Solubility
(d) Crystalline shape.
Answer:
(c) Solubility

Question 16.
Maximum number of molecules is :
(a) In 54 gram nitrogen peroxide
(b) In 28 gram carbon monoxide
(c) In 36 gram water
(d) In 46 gram carbon monoxide.
Answer:
(a) In 54 gram nitrogen peroxide

Question 17.
Molecular mass of O₂ and SO₂ are 32 and 64 respectively. If one litre O₂ at 13°C and 750 mm, pressure contain N molecules, then what will be the number of molecules in two litre SO₂ at the same temperature and pressure :
molecules in tw
(a) \(\frac{\mathrm{N}}{2}\)
(b) N
(c) 2N
(d) 4N.
Answer:
(c) 2N

Question 18.
Number of moles in 180 gram water is:
(a) 1 mole
(b) 10 mole
(c) 18 mole
(d) 100 mole.
Answer:
(b) 10 mole

Question 19.
If at normal temperature and pressure, two gases are placed in two containers of same volume, then in them :
(a) Number of molecules will be same
(b) Number of atoms will be same
(c) Their masses will be same
(d) Their densities will be same.
Answer:
(a) Number of molecules will be same

Question 20.
By the decomposition of 3.4 gram H₂O₂ at N.T.P., the following volume of O₂ will be obtained:
(a) 0.56 litre
(b) 1.12 litre
(c) 2.24 litre
(d) 3.36 litre.
Answer:
(b) 1.12 litre

Question 21.
By the reaction of 2 litres N₂ and 2 litres H₂ at N.T.P, volume of ammonia obtained will be:
(a) 0.665 litre
(b) 1.0 litre
(c) 1.33 litres
(d) 2.66 litre.
Answer:
(c) 1.33 litres

Question 22.
On burning 2 moles C₂H₅OH in excess of air, C0₂ obtained will be :
(a) 88 gram
(b) 176 gram
(c) 44 gram
(d) 22 gram.
Answer:
(b) 176 gram

Question 23.
12 gram Mg (atomic mass = 24) on reacting completely with acid produce H₂ whose volume at N.T.P. will be :
(a) 22.4 litre
(b) 11.2 litre
(c) 44.8 litre
(d) 6.4 litre.
Answer:
(c) 44.8 litre

Question 24.
12 gram At (atomic mass = 27) reacts with ………… gm oxygen completely :
(a) 8 gram
(b) 16 gram
(c) 32 gram
(d) 24 gram.
Answer:
(d) 24 gram.

Question 25.
2.76 gram silver carbonate (Atomic mass of Ag = 208) is heated vigorously. The
weight of the residue obtained will be:
(a) 2.48 gram
(b) 2.16 gram
(c) 2.32 gram
(d) 2.64 gram.
Answer:

Question 26.
Number of gram atoms of sulphur (atomic mass of S = 32) in 80 gram Sulphur is:
(a) 2.5
(b) 3.2
(c) 5
(d) 10.
Answer:
(a) 2.5

Question 27.
Ammonia gas is formed by the equation N₂ + 3H₂ → 2NH₃. Nitrogen gas required to prepare 10 litre ammonia will be:
(a) 5 litre
(b) 15 litre
(c) 75 litre
(d) 10 litre.
(a) 5 litre

Question 28.
Volume of 0.1 mole gas at N.TP. will be:
(a) 22.4 litre
(b) 2.24 litre
(c) 0.224 litre
(d) 22400 litre.
Answer:
(b) 2.24 litre

Question 29.
10 mole of water is:
(a) 10 gram
(b) 1oo gram
(c) 18 gram
(d) 180 gram.
Answer:
(d) 180 gram.

Question 30.
Information not obtained by chemical equation is :
Zn_(s) + 2HCl_(l) → ZnCl_2(l) + H_2(g)
(a) Physical state of reactants and products
(b) Concentration of reactants and products
(c) Heat change in the reaction
(d) Direction of reaction.
Answer:
(c) Heat change in the reaction

Question 31.
Atomic number olDeuterium is:
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(a) 1

Question 32.
Molarity of a solution formed by 7.1 gram Na₂SO₄ in 100 ml water is (Atomic mass of Na = 23,S = 32,O = 6):
(a) 2.0 M
(b) 1.0 M
(c) 0.5 M
(d) 0.05 M.
Answer:
(c) 0.5 M

Question 33.
A compound contains 50% C, 50% 0 and its molecular mass is about 290. Its
molecular formula is:
(a) CO
(b) C
(c) C₁₂ O₉
(d) C₃O₃.
Answer:
(c) C₁₂ O₉

Question 34.
A solution contains 20 moles of solute and total number of moles Is 80. Mole
fraction of the solute will be:
(a) 2.5
(b) 0.25
(c) 1
(d) 0.75.
Answer:
(b) 0.25

Question 35.
Number of molecules in 16 gram Methane is: ,
(a) 3.0 × 10²³
(b) 6.02 × 10²³
(c) \(\frac{16}{6 \cdot 02}\) × 10²³
(d) \(\frac{16}{3 \cdot 0}\) × 10²³
Answer:
(b) 6.02 × 10²³

Question 36.
Number of molecules (approx.) in 4.25 gm NH₃ is :
(a) 1 × 10²³
(b) 2 × 10²³
(c) 4 × 10²³
(d) 6 × 10²³
Answer:
(d) 6 × 10²³

Question 37.
Aqueous solution of 6.3 gm Oxalic acid (dihydrate) was diluted to 250 ml. To neutralize 10 ml of this solution completely, volume of 0.1 N NaOH required will be:
(a) 40 ml
(b) 20ml
(c) 10ml
(d) 4ml.
Answer:
(a) 40 ml

Question 38.
Weight of how many mole electrons will be 1 kilogram :
(a) 6.023 × 10²³
(b) \(\frac{1}{9 \cdot 108}\) × 10³¹
(c) \(\frac{6 \cdot 023}{9 \cdot 108}\) × 10⁵⁴
(d) \(\frac{1}{9 \cdot 108 \times 6 \cdot 023}\) × 10⁸
Answer:
(d) \(\frac{1}{9 \cdot 108 \times 6 \cdot 023}\) × 10⁸

Question 39.
Number of significant figures in 0.0500 is :
(a) One
(b) Three
(c) Two
(d) Four.
Answer:
(b) Three

Question 40.
Number of significant figures in a correctly written answer of sum of 29.4406, 3.2 and 2.25 will be :
(a) Three
(b) Four
(c) Two
(d) Five.
Answer:
(a) Three

Question 41.
Density of a solution prepared by dissolving 120 g. Urea (Molecular mass = 604)
in 1000 gm water is 1.15g/ml. Molarity of the solution is :
(a) 1.78 M
(b) 1.02 M
(c) 2.05 M
(d) 0.50 M.
Answer:
(c) 2.05 M

Question 42.
What will be the density (in gm L^-1) of 3.60 M sulphuric acid which according to mass is 29% ? (Given Molar mass of sulphuric acid = 98 gm mol^-1):
(a) 1.45
(b) 1.64
(c) 1.88
(d) 1.22.
Answer:
(d) 1.22.

Question 43.
Percentage of water of crystallization in a pure sample of green vitriol (FeS0₄.7H₂0) is:
(a) 45.00
(b) 54.23
(c) 45.32
(d) 44.22.
Answer:
(c) 45.32

Question 44.
Mass of electron is 9.108 × 10^-31 kg. Its significant figure is :
(a) One
(b) Two
(c) Three
(d) Four.
Answer:
(d) Four.

Question 45.
Which of the following number has maximum significant figures :
(a) 0.00453
(b) 4.8046
(c) 5.643
(d) 5.6 × 10³ cm.
Answer:
(b) 4.8046

Question 46.
Volume occupied by one molecule of water (Density = 1g cm^-3) is :
(a) 3.0 × 10^-23cm³
(b) 5.5 × 10²³cm³
(c) 9.0 × 10^-23cm³
(d) 6.02 × 10^-23cm³.
Answer:
(a) 3.0 × 10^-23cm³

2. Fill in the blanks:

1. Deuterium has …………………. atom.
Answer:
one

2. Empirical formula of benzene is ………………..
Answer:
CH

3. Under similar conditions of temperature and pressure equal volume of all gases contain equal number of molecules. This was proposed by …………………….
Answer:
Avogadro

4. Number of moles in 180 gram water will be ………………….
Answer:
10 mol

5. S.I. unit of atomic mass is ……………………..
Answer:
a.m.u(u)

6. Kilogram/metre³ is the unit of …………………
Answer:
Density

7. Five oxides of nitrogen N₂O, NO, N₂O₃, N₂O₄, N₂O₅ obey ………………….. law of chemical combination.
Answer:
Multiple proportion

8. 1 mole CO₂ contain ……………………. carbon atom.
Answer:
6.023 × 10²³

9. Mass of 22.4 litre SO₂ at N.T.R is …………………..
Answer:
64 gram

10. Two gases placed if in two containers of equal volume, under normal temperature
and pressure then the number of molecules in them will be ………………….
Answer:
same

11. On burning 2 mole C₂H₅OH in excess of air …………………… CO is obtained.
Answer:
176 gram

12. Volume of 2 gram H₂ at N.T.P. is …………………
Answer:
22.4 litre

13. At the end of a chemical reaction, the total mass of the reactant remains ………………….
Answer:
Unchanged

14. Simplest formula of a compound with molecular formula of benzene is …………………..
Answer:
CH

15. A compound contains C = 40.6%, H = 6.5%, O = 52.8%. Its empirical formula will
be …………………
Answer:
CH2O

3. Match the following:
I.

Answer:
1. (c) 10^-6
2. (e) 10
3. (a) 10⁶
4. (b) 10⁹
5. (d) 10^-15

II.

Answer:
1. (d) 22.4
2. (a) Empirical formula × η
3. (b) Molality
4. (e) Element.
5. (c) Molarity

4. Answer in one word/sentence:

1. Standard unit of atomic mass is
Answer:
Atomic mass unit (a.m.u.)

2. Gram molecular mass of O₂ is
Answer:
32 gram

3. The ratio by weight of Na and Cl elements in NaCl obtained from various sources is 23 : 35.5. This is proved by which law?
Answer:
Law of constant or definite proportion

4. 16 and 32 gram by weight of oxygen combine separately with 28 gram N₂ to form two oxides NO and N₂O₂. This is proved by which law.
Answer:
Law of multiple proportion

5. Number of molecules 6 023 × 10²³ present in one gram molecular mass of any substance is called.
Answer:
Mole or Avogadro number

Students get through the MP Board Class 11th Chemistry Important Questions Chapter 14 Environmental Chemistry which are most likely to be asked in the exam.

MP Board Class 11th Chemistry Important Questions Chapter 14 Environmental Chemistry

Environmental Chemistry Class 11 Important Questions Very Short Answer Type

Question 1.
Define Environmental chemistry.
Answer:
Environmental chemistry is a branch of science which deals with the study of effects of chemicals on our surroundings (like origin, transport, reactions and effects and fates of chemical species etc.).

Question 2.
Explain Tropospheric pollution.
Answer:
The tropospheric pollution occurs because of the presence of undesirable gaseous and solid particles in the air. In the troposphere mainly the following gaseous and particulate pollutants are present.

• Gaseous air pollutants are oxides of sulphur, nitrogen and carbon dioxide, hydrogen sulphide, ozone and other oxidants.
• Particulate pollutants are dust, fumes, mist, spray, smoke etc.

Question 3.
Make a list of the gases responsible for greenhouse effect.
Answer:
For greenhouse effect mainly CO₂ gas is responsible. Apart from CO₂, other greenhouse gases are methane, water vapour, nitrous oxide, chlorofluorocarbon (CFC) and ozone.

Question 4.
What do you mean by Biochemical oxygen demand BOD?
Answer:
The amount of oxygen required by bacteria to break down the organic matter present in a certain volume of a sample of water is called Biochemical oxygen demand (BOD). Amount of BOD in the water is a measure of the amount of organic material in the water, in terms of how much oxygen will be required to break it down biologically. For pure and clean water BOD value is less than 5 ppm.

Question 5.
Due to greenhouse effect, earth’s temperature is increasing. Which substances are responsible for greenhouse effect?
Answer:
Greenhouse gases like carbon monoxide, methane, nitrous oxide, ozone and chlorofluorocarbons (CFCs) are responsible for greenhouse effect. These gases absorb the radiations passing through the earth’s surface and increase the earth’s temperature.

Question 6.
Ozone is a poisonous and a strong oxidizing reagent still its presence in the upper part of stratosphere is very important. What will happen if ozone is completely removed from this region?
Answer:
Ozone prevents the harmful ultraviolet rays of the sun to reach the earth’s surface due to which human life is protected by the harmful effects of these ultraviolet rays.
If ozone is completely removed from the upper region of stratosphere, then ultraviolet rays will reach the earth and cause various diseases like skin cancer etc.

Question 7.
What are the sources of oxygen dissolved in air?
Answer:
The sources of oxygen dissolved in water are :

• Photosynthesis,
• Natural Aeration,
• Mechanical Aeration.

Question 8.
Oxygen dissolved in water is very important for aquatic life. Which factor is responsible for the reduction of dissolved oxygen in water?
Answer:
For the reduction of dissolved oxygen in water, discharge of phosphate and nitrate fertilizers, detergents, animal excreta, food, paper and pulp industry, organic waste from industries are responsible. Micro-organisms which oxidize organic substances also use dissolved oxygen. Along with this, photosynthesis stops at night but aquatic plants and animals continuously breathe due to which deficiency of dissolved oxygen in water is caused.

Question 9.
What are Biodegradable and non-biodegradable pollutants?
Answer:
Biodegradable pollutants are substances which are decomposed by organisms.
Example: Excreta, cow dung, fruits and vegetable peels.

Non-biodegradable pollutants are substances which are not decomposed by organisms.
Example: Mercury, Lead, DDT, glass, plastic etc.

Question 10.
What is pollution?
Answer:
In a balanced environment, every component is present in a definite amount and ratio. Sometimes, in the environment amount of one or more component either increases than required or harmful components enter into the environment which are harmful for living organism. This is known as pollution.

Question 11.
What are pollutants?
Answer:
Such substances whose amount increases up to a certain limit in the atmosphere due to which it becomes harmful for plant and animal kingdom are called pollutants. They are as follows:
CO, CO₂, NO, N₂, S₂ etc.

Question 12.
What are contaminants?
Answer:
Some substances do not exist in the environment but are made in artificial form by man. These substances pollute the environment without getting destroyed. They are called contaminants.

Question 13.
Write the names of gases or chemical responsible for the depletion of ozone layer.
Answer:
Nitric oxide, atomic oxygen and chlorofluorocarbon present in the atmosphere is responsible for the depletion of ozone layer.

Question 14.
What are greenhouse gases?
Answer:
CO₂, ozone and water vapour are known as greenhouse gases. They possess the property of absorbing infrared radiations. Thus, they are known as greenhouse gases.

Question 15.
What is polluted air?
Answer:
If air contains such substances which are harmful for health of organisms, then such air is called polluted air. Polluted air mainly contains suspended molecules of CO₂, SO₂, SO₃, CO.

Question 16.
Why is depletion of ozone observed Antarctica?
Answer:
Because in the other part of stratosphere, free chlorine gets converted to chlorine molecules, whereas at Antarctica compounds produced get surrounded by chlorine free radical which cause depletion of ozone.

Question 17.
What is the utility of BOD measurement in a water sample?
Answer:
BOD is measure of pollution caused by organic biodegradable substances present in a given sample of water. Lower value of BOD indicates that less amount of organic waste is present in water.

Question 18.
Oxidation of sulphur dioxide to sulphur trioxide is a slow change in the absence of a catalyst but in presence of atmosphere it oxidizes readily. Why is it so? Also give chemical equation for the conversion of SO₂ to SO₃.
Answer:
In the oxidation of SO₂ to SO₃, particulate matter present in polluted air acts as a catalyst.

Question 19.
How is ozone produced in stratosphere?
Answer:
In stratosphere, ozone is produced by the action of ultraviolet radiations on dioxygen (O₂) molecule. Ultraviolet rays decompose oxygen molecule to oxygen atom. Oxygen atom combine with oxygen molecule to form ozone molecule.

Question 20.
What is chlorosis?
Answer:
In plants, chlorophyll is formed slowly This is due to presence of SO₂. This pollution is called chlorosis.

Question 21.
What is Metathesis?
Answer:
Metathesis is the name of that science, in which utility of scientific methods for a common man is studied.

Question 22.
What are Primary and Secondary Pollutants?
Answer:
Primary Air Pollutants: Air pollutants which directly pollute the air by natural or human activities are called primary pollutants. Like CO, NO, CO_2, SO₂.
Secondary Air Pollutants: Such air pollutants which are prepared by the chemical reactions between pollutants or by the reaction of gases and pollutants in the atmosphere are called secondary pollutants.
Like: SO₃+H₂O → H₂SO₄.

Question 23.
What is photochemical smog?
Answer:
Mixture of smoke and fog is known as smog. When higher concentration of oxidizing agents are present in this smog, it is called photochemical smog. Smoke is an air pollutant, in which various solid particles are suspended in the atmosphere due to incomplete combustion of fuel substances.

Question 24.
What is Acid rain?
Answer:
Due to air pollution, concentration of various gases like CO₂, SO₂, SO₃, NO₂ etc. increases in the atmosphere. These gases dissolve in raindrops according to their solubilities and form acid. These drops fall as rain and are known as acid rain.
CO₂+H₂O → H₂CO₃
SO₃+H₂O → H₂SO₄.

Question 25.
When is CO₂ known as pollutant?
Answer:
Normal amount of CO₂ in the atmosphere is never harmful, but plants prepare their food by it. But, when due to various activity its concentration becomes more than required, then it disturbs the natural balance and become harmful.

Question 26.
Carbon dioxide is considered to be more responsible for greenhouse effect. Why?
Answer:
Water vapour is found to be very close to the earth, and ozone is found at the upper part of the atmosphere. Compared to this CO₂ is evenly found in all places of the atmosphere. Therefore, for greenhouse effect, CO₂ is found to be more responsible because CO₂ possess the tendency to absorb the infrared radiations of the sun and produce greenhouse effect.

Question 27.
Why is acid rain considered as a threat to Tajmahal?
Answer:
Tajmahal is made of marbles. The acid rain contains H2S04 which attacks the marble (CaCO₃).
CaCO₃(s) + H₂SO₄(aq) → CaSO₄(aq) + CO₂+ H₂O(l)
As a result, the marble becomes pitted and weakened mechanically. Therefore, the monument is being slowly eaten away and marble is getting discoloured and lustreless.

Question 28.
A person uses water supplied by municipality. Due to shortage of water, he started using underground water. He felt laxative effect. What could be the cause?
Answer:
Laxative effect can be felt only when the concentration of sulphate present in water is greater than 500 ppm otherwise at moderate concentration sulphate is harmless.

Question 29.
What do you understand by Green chemistry? How is it helpful to control environmental pollution?
Answer:
Green chemistry is a technique to utilize the known knowledge and principles of chemistry and other sciences by which harmful effects of environment can be reduced. Green chemistry is a process of production which can bring minimum pollution or harm to the environment. Co-products produced in a process if not removed in a beneficial way increase pollution in the environment. Such processes are not only harmful for environmental view but are also costly. Along with modified work, use of present knowledge to reduce chemical harm is the basis of Green chemistry.

Question 30.
Why is carbon monoxide more dangerous than carbon dioxide? Explain.
Answer:
Carbon monoxide is a poisonous gas. It combines with haemoglobin present in blood to form carboxyhaemoglobin. It is 300 times more stable than oxyhaemoglobin. When the percentage of carboxyhaemoglobin in the blood becomes 3-4% then ability of the blood to carry oxygen becomes very less.

Due to minimum oxygen, headache, weak eyesight, reduction in awareness and judgement, cardiovascular diseases etc. are produced. Carbon dioxide gas does not combine with haemoglobin. Therefore, it is a less harmful pollutant. Increase in its amount in the environment leads to global warning (increase in earth’s temperature).

Question 31.
What is Fly ash pollutant?
Answer:
Fly ash are small ash particles formed by the combustion of fossil fuels which are carried into the air by gases produced during combustion.

Question 32.
What is contaminated water? Write name of diseases caused due to contaminated water.
Answer:
Water in which due to pollution various harmful products of living beings and micro-organisms are present is called contaminated water. Diseases caused due to contaminated water are :

• Cholera,
• Jaundice,
•  Diarrhoea,
• Skin disease.

Question 33.
What do you understand by Global warming?
Answer:
Increase in earth’s temperature, due to absorption of sun’s energy by various gases like CO₂, methane, chlorofluorocarbon etc. present in large amount in the atmosphere is called Global warming.

Question 34.
What is Ionosphere?
Answer:
A thick layer about 40 km at a height of about 50 to 90 km from the earth’s surface is called Mesosphere. All the gases in this region exist in ionized state, therefore it is called Ionosphere.
O₂ + O⁺ → O₂ + O
O⁺ + N₂ → NO⁺ + N
N₂+O₂ → N₂ + O₂.

Environmental Chemistry Class 11 Important Questions Short Answer Type 

Question 1.
How is the poisonous effect of CO produced on man and animals?
Answer:
CO has poisonous effect on man and plants.
(i) It combines with haemoglobin of the blood more strongly than oxygen. CO + Hb → CO – Hb (Carboxy haemoglobin).
As a result of this amount of haemoglobin available in the blood for the transport of oxygen to the body cells decreases. The normal metabolism is thus, impaired due to less O₂ level. This will cause suffocation and will ultimately lead to death. Carbon monoxide if present in air can cause mental impairment, respiratory problems, muscular weakness and dizziness.

(ii) A high concentration of CO (100 PPM or more) will harmfully affect the plants causing leaf drop; reduction of leaf size and premature aging etc.

Question 2.
Write the harmful effects of sulphur dioxide.
Answer:
Harmful effects of SO₂ :

• SO₂ harms the respiratory tube and lungs due to which various lung diseases, cancer etc. can be caused.
• It stops the growth of plants. Due to this, change of green leaves to yellow is called chlorosis.
• Acid rain due to SO₂ harms the monuments and stones.
• SO₂ gas enhances corrosion.

Question 3.
Write harmful effects of nitrogen oxide.
Answer:
Harmful Effects of Nitrogen oxide :

• Presence of gases like NO, NO₂ etc. more than required cause irritation in eyes.
• Its excess amount may lead to lung and heart diseases.
• Oxides of nitrogen enhance corrosion in metals.
• Reduce rate of photosynthesis.
• Reduce the brightness of dyes.

Question 4.
A farmer was using pesticides on his farm. He used the produce of his farm as food for rearing fishes. He was told that fishes were not fit for human consumption because large amount of pesticides had accumulated in the tissues of fishes. Explain how did it happen?
Answer:
Pesticides transfer from the crops to fish as their food. This way, pesticides through the food chain reach the higher level from the lower level. With time, concentration of pesticides in fish increased to such a level that serious metabolic and body active states were produced. That is why, fishes were not suitable as food for man.

Question 5.
For dry cleaning, in place of tetrachloroethene along with suitable deter-gent liquified carbon dioxide is an alternative. Which loss of the environment can be stopped by the use of tetrachloroethene? Is the use of liquified carbon dioxide with detergent a completely prevention from the view of pollution?
Answer:

• Tetrachloroethene (Cl₂C = CCl₂) is a possible cause of cancer, it also pollutes ground water. Its harmful effects can be controlled by the use of liquified carbon dioxide and suitable detergent.
• Use of liquified CO₂ with suitable detergent is not completely preventive because maximum detergents are non-biodegradable and they also pollute the water. Along with this, liquified CO₂ enter the environment at the end and increase green house effect.

Question 6.
What are the ill effects of greenhouse effect?
Answer:
Main cause of greenhouse effect is the presence of large amount of CO₂ in the atmosphere which absorbs the infrared radiations and increase the temperature of the environment. This way, due to increase in temperature of the environment weather will change. Sun rays gives birth to serious disease like cancer.

Some places will suffer from drought, hot air will flow in some places, some places will have strong storm and somewhere floods may come. Huge icecaps of Arctica and Antarctica will melt, by which sea level will rise and places near sea shores will submerge.

Question 7.
What is smog? Explain its mechanism with its types.
Answer:
Mixture of smoke and fog is known as smog.
Types of Smog:

1. Classical smog: SO₂ gas mixed with fog combines with smoke to form classical smog.
2. Photochemical smog: In a Sunny place, primary air pollutant form secondary pollutant by photochemical reaction. They are oxidizing in nature. Where pollutant gas is more, wind does not blow, there with smoke, smog is formed.

Mechanism :
1. If automobile fuel is used before sunrise, then mainly CO and NO gas are released. With contact of air it forms NO₂ gas.
2NO + O₂ → 2NO₂

2.By the effect of ultraviolet rays active oxygen is released by NO₂.
NO₂ → NO + O
3. This active oxygen converts the hydrocarbon released from the fuel to free radical.
4. These organic free radicals form undesirable compounds by chain reactions and these pollutants form smog.

Question 8.
Discuss the causes of water pollution.
Answer:
Following are the causes of water pollution :

• Waste products from big factories are dumped in sea, rivers or lakes.
• Various fertilizers, manure and insecticides are used in fields.
• All these products mix with the soil of fields, dissolve in rainwater and fall in rivers and pools.
• Atomic electric centre pollute the water.
• Due to increase in population, colonies are formed at the river banks, and waste from the houses pollute the water.
• In rural areas, people take bath, wash clothes in the rivers which pollute water.

Question 9.
How is synthetic greenhouse prepared?
Answer:
Synthetic greenhouse: In nature, coating of CO₂ is forming greenhouse, but synthetic greenhouse can be synthesized by studying its mechanism.

Actually the transparent glass roof and wall of the glasshouse allow sun rays to pass through and strike the surface of the house. The reflected radiation is of longer wavelength than the incident radiation. A significant portion of reflected radiation absorbed by glass. As radiation of longer wavelength (Infrared radiation) generates heat, this causes rise in tem¬perature inside the glasshouse.

Question 10.
You have dug pits in your agricultural area or garden for producing compost. Discuss the process of production of best compost in the light of bad odour, Hies and recycling of waste.
Answer:
For producing compost, pits should be dug in suitable place by which we can be prevented from bad odours and flies. Biodegradable domestic waste like tea leaves, fruits and vegetable waste etc. should be put in the compost pit and should be covered by thin layer of sand. After some time, these waste materials, by the effect of heat and bacteria get converted to compost.

Compost pits should be covered by sand by which flies cannot enter and bad odour is minimum.
Non-biodegradable waste like plastic, glass, metal shavings, polythene bags are sent for recycling. By recycling useful products are obtained from waste products.

Question 11.
How are domestic waste used as manure?
Answer:
Domestic wastes are collected in small containers and sent to disposal area. Here biodegradable waste and non-biodegradable waste are separated from the garbage. Biodegradable waste like vegetables and fruit waste, animal waste etc. are in open ground beneath the soil where it degrades and after some days get converted to manure.

Question 12.
Write the ill effects of water pollution and how is it controlled.
Answer:
III effects of water pollution:

• By the use of polluted water, cholera, jaundice, typhoid like contaminated diseases are spread.
• Due to discharged excreta, amounf of oxygen dissolved in water decreases due to which aquatic organisms die.
• Due to soap, detergent waste, polluted water becomes a danger for the life of fish.

Control:

1. Discharged excreta should not be thrown in rivers, pools etc. by chemical reaction should be converted to manure.
2. Atomic identification below the sea should be banned.
3. By the process of recycling cow dung and excreta should be converted to useful material.

Question 13.
What is soil pollution? How can soil pollution be controlled?
Answer:
Soil Pollution: Any undesirable change in physical, chemical or biological properties of soil which has harmful effect on human beings and other organisms or the natural character and utility of soil or earth is destroyed is called soil pollution.
Control:

• Industrial waste should be treated in the factories only and then dumped on earth.
• Use of fertilizers, insecticides and fungicidal medicines should be minimized.
• Cutting of trees should be stopped to prevent soil erosion and destruction of the layer of earth.

Question 14.
Write the ill effects of depletion of ozone layer.
Answer:
Following is the ill effects of ozone layer :

• Entrance of ultraviolet rays directly on earth without obstruction can lead to various diseases like cancer.
• Ultraviolet rays decrease the immunity of body.
• CFC increases greenhouse effect, as a result temperature of earth rises.

Control:

1. Production of CFC should be reduced.
2. Search the option of CFC.

Question 15.
Explain the mechanism of depletion of ozone layer.
Answer:
The mechanism of depletion of ozone layer are the following :
1. Depletion by Nitrogen oxide: Some amount of N₂O is present in the stratosphere. It combines with atomic oxygen to form NO.
N₂O+O → 2NO
NO gas combines with O₃ to form NO2
NO + O₃ → NO₂ + O₂.
NO₂ breaks by atomic oxygen to form NO again
NO₂+O → N0 + O₂
This cycle of NO is the cause of depletion of O₃.

2. Depletion by C.F.C.: It produces free radical by photochemical decomposition.

Formation of free radicals is continuously depleting ozone layer.

Environmental Chemistry Class 11 Important Questions Long Answer Type

Question 1.
A large number of fish are suddenly found floating dead on a lake. There is no evidence of toxic dumping but you find an abundance of phytoplankton. Suggest a reason for fish kill.
Answer:
Large amount of phosphate and nitrate salts increases the number of phytoplanktons in water. These phytoplanktons use the dissolved oxygen in water to such a large extent, that it is not available in sufficient amount for the respiration of other organisms. Along with it excessive population of bacteria start the decomposition of organic substance in water like leaves, grass etc. It uses the dissolved oxygen in water. Micro-organisms (like algae) are in such a larger amount that they cover the water level as a sheet and prevents the penetration of sunlight in the water level as a result photosynthesis is obstructed. By all these processes, amount of dissolved oxygen decreases 6 ppm. Thus, growth of fish is stopped and they do not remain alive.

Question 2.
What would have happened, if there would have been no greenhouse gases in the atmosphere?
Answer:
Carbon dioxide, methane, water vapour, CFCs and ozone are green house gases. These gases absorb the developed heat by the objects situated near the earth’s surface and heat it up. This is known as natural green house effect because it controls temperature and makes life better on earth. If there would have been no green gases in the atmosphere, then there would have been no vegetation or life present on earth (because earth would have converted to a cool house).

Question 3.
Into how many parts is the atmosphere divided. Explain the chemical reactions occuring in every region.
Answer:
Atmosphere is divided into four parts :

1. Troposphere,
2. Stratosphere,
3. Mesosphere,
4. Thermosphere.

1.Troposphere: The region upto a height of 8 to 12 km from the earth’s surface is called Troposphere. Its main components are N₂, O₂, CO₂ and H₂O. Fossil fuels bum in oxygen to form CO₂.
CH₄ +2O₂ → CO₂ +2HzO
In the troposphere, compounds are decomposed by various bacteria.
Bacteria compound + O₂ → CO₂ + H₂O
Photosynthesis also occur in this layer.

2. Stratosphere: About 10 to 50 km, layer of 40 km thickness is known as Stratosphere.
(a) Photochemical decomposition of oxygen.

Excited oxygen molecule 02 liberate radiation of 636, 630, 558nm wavelength in visible region. This is known as airglow.

Oxygen molecule gets excited and lose electron by which positively charged oxygen ion is formed.
O₂ +O → O₃
O₃ molecule again decomposes by ultraviolet rays.

O₃ + O → O₂ + O₂

To provide stability to ozone, others like N₂, O₂ are required which absorb the extra energy and stop the decomposition of O₃.
O + O₂+M → O₃ + M
Alternatively, if species like NO or OH are present in stratosphere, then they decompose o3.
O3 + OH → O₂ + HOO
O3 + NO →NO₂ + O₂
NO₂ +O → NO + O₂
N₂ molecule undergo self decomposition by radiation and combine with O₂ to form NO.

O₂ + N → NO + O

3. Mesosphere: About 50 to 90 km height above the earth’s surface a 40 km thick layer is called Mesosphere.
O₂ + O → O₂ + O
O₊ + N₂ → NO⁺ + N
N2⁺+O₂ → N₂ + O₂

4. Thermosphere: This layer of atmosphere starts from a height of 90 km from the earth’s surface. This region till 500 km height is known as Thermosphere.

Question 4.
How is acid rain damaging statues and monuments in India?
Answer:
In India, statues and monuments are made up of marble i.e. CaCO₃. The surrounding air mainly contain oxides of sulphur and nitrogen in large amount. The main reason is the presence of large number of industrial units and energy plants around it. Oxides of sulphur and nitrogen are acidic. SO₂ and NO₂ oxidize with water to form mineral acids which is the main source of acid rain.
2SO_2(g) + O_2(g) + 2H₂O_(l) → 2H₂SO _4(aq)
4NO_2(g) + O_2(g) + 2H₂O_(l) → 4HNO_3(aq)
This acid rain reacts with the marble of statues and monuments and destroy them.
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂

Question 5.
What is smog? How is classical smog different from photochemical smog?
Answer:
The word smog is derived from smoke and fog. This is the most common example of air pollution that occurs in many cities throughout the world. There are two types of smog.

1. Classical smog: It occurs in cool humid climate. It is a mixture of smoke, fog and sulphur dioxide. Chemically it is a reducing mixture.
2. Photochemical smog: Photochemical smog occurs in warm, dry and sunny climate. The main components of photochemical smog result from the action of sunlight on unsaturated hydrocarbons and nitrogen oxides produced by automobiles and factories. Chemically it is oxidizing smog as it contains high concentration of oxidizing agents.

Question 6.
How many types of particulate are there? Explain.
Answer:
Particulate: Finely divided solid or liquid particles suspended in air are the cause of pollution. Size of these particles can be from 0.02 Å to 50000Å. Our eyes can see particles of the size about 10000Å.
Various types of particles and their source :

1.Soot: Due to incomplete combustion of carbon-containing fuels minute particles of carbon spread in the environment. This type of combustion generally occur at all places because from wood, coal etc. used as domestic fuels carbon-containing fuels are used in every field.

2. Dust: Sources of dust are present at all places on the earth. These are released in the environment due to natural or human sources during volcanic eruption, storm, during certain industrial processes like crushing, grinding of limestone, cement as fly ash and silica dust. Apart from this, dust is also spread due to running of vehicles.

3. Metal particles: Particles of metals like lead, mercury, chromium, arsenic, zinc, iron, nickel, cadmium etc. spread in the environment by various types of metal treatment.
Though their quantity is very less, still they produce harmful effect even if present in minute quantity.

4. Metal oxide: Generates by combustion of fuels containing metallic compounds.
For example when coal containing iron pyrites is burnt, particulates of Fe₃O₄ are introduced into the air.
3FeS₂+8O₂ → Fe3O₄ + 6SO₂

5. Particles of metal salts and fly ash: Limestone present in coal is left in ash in the form of CaO and can fly with ash in dust. Minute drops of H₂SO₄ situated in the environment react with CaO to form particles of CaSO₄. Ammonia present in air react with H₂SO₄ and form particles of (NH₄)₂SO₄. The flying of ash with air in the form of minute particles producing pollution is called fly ash pollution.

6. Asbestos particles: Asbestos is used in the factories and industries in various forms. Therefore asbestos particles spread in the environment.

7. Mist: We know that water vapour condense into minute droplets in the atmosphere which appear as mist to us. Similarly acids like H₂SO₄ and HNO₃ formed by various activities in the atmosphere also produce mist in the form of minute particles and spread pollution.

8. PAH and TEL: PAH means Polycyclic Aromatic hydrocarbon. TEL means Tetra Ethyl Lead. Due to incomplete combustion of fossil fuels and petroleum, PAH pollutes the environment in the form of minute particles. Tetraethyl lead, Pb (C₂H₅)₄ is added to gasoline as antiknock agent.

During combustion in the engine, it is oxidized to PbO which deposits in different parts of the engine and may cause damage. To avoid this damage, Pb(C₂H₅)₄ is mixed with dichloroethane and dibromo ethane which converts PbO into PbCl₂ and PbBr₂ which are volatile and thus come along with the exhaust gases and introduce into the atmosphere.
Pb(C₂H₅)₄ + O₂ + C₂H₄C₁₂ + C₂H₄Br₂ → CO₂ + H₂O + PbCl₂ + PbBr₂
Thus, engine is protected but air gets polluted.

Question 7.
What is Acid rain. Explain in detail its effect on the environment.
Answer:
All pollutant gases spread as pollutant particles in the form of smoke by the burning of fossil fuels and other fuels. Due to high temperature of industries and other engines, oxides of nitrogen spread in the atmosphere by the combination of N₂ and O₂. These gases mix in raindrops form acid and then fall on the earth as acid raindrops. This is called acid rain.
CO₂ + H₂O → H₂CO₃
SO₂ + H₂O → H₂SO₃

Effects of Acid rain :

• Acid rain neutralizes the base present in the soil, as a result of which soil becomes acidic. This affects the bacteria, plants and nitrogen fixation.
• Due to acid rain, water of rivers, lakes become acidic which has harmful effect on organisms.
• Acid rain leads to depleting forest resources and vegetation.
• Statues and monuments are losing their brightness because the metals present in them get corroded. Due to acid rain marble of Tajmahal is depleting.
• Acid rain harms leaves of aquatic plants and reduces the rate of photosynthesis.
• Acid rain has harmful effect on skin, lungs and neck.

Control of Acid rain :

1. Rivers and lakes cannot be prevented from acid rain but by mixing lime, its effect can be reduced.
2. By pouring lime in the fields, effect of acid rain can be reduced.
3. By using coal with low content of sulphur, unleaded petrol process of combustion of fuel in plants can be controlled and effect of acid rain can be destroyed.
4. Measures to control the effect of gases causing acid pollution and converting unburnt hydrocarbons to ineffective compound are being tried.

Question 8.
Explain the process of pollution of oxides of sulphur and nitrogen. How can they be controlled?
Answer:
Oxide of sulphur (SO_x): Among the oxides of sulphur, SO₂ is the main pollutant. It gets partially oxidised to SO₃ by air. SO₃ forms H₂SO₄ in presence of moisture this way SO₂ is a primary pollutant and SO₃ and H₂SO₄ is a secondary pollutant.
Process of pollution : (i) Fossil fuels and petroleum contain 0-3% to 3% of S which form S02 on burning.
S+O₂ → SO₂

(ii) SO₂ gas is released by industrial extraction of metals.
2FeS₂ + 5O₂ → 2FeO + 4SO₂
2PbS + 3O₂ → 2PbO + 2SO₂
2ZnS + 3O₂ → 2Zn0 + 2SO₂

(iii) During biological activities of various bacteria H₂S gas is released which oxidizes by air to form SO₂.
H₂S+\(\frac{3}{2}\) O₂ → SO₂+H₂O

(iii) Volcanic gases contain sufficient amount of SO₂

Control:

• To control SO₂, impurity of S should be removed from fuels.
• To remove sulphur from the engine by physical methods it is washed properly by making powder.
• In chemical methods, fossil fuels are carbonated.
• Liquation and vaporization are other methods of sulphur control.
• Environmental SO₂ can be controlled by scrubbing.

Oxides of nitrogen (NO_x): Nitric oxide is primary pollutant whereas N₂O, NO₂ or other oxides are secondary pollutant which are formed in the environment by various chemical activities.
Source:

• Due to high temperature produced by lightning in the atmosphere N₂ and O₂ combine to form NO.
• Production of NO due to high temperature in automobile engine.
• Formation of NO and other oxides in various industries.
• Formation of NO by natural and biological activities and due to fire in jungle.

Control:

• In the combustion of fuel, on keeping the amount of airless, the equilibrium
  for the manufacture of NO can be shifted to backward which can control the formation of NO.
  N₂ + O₂ ⇌ double arow 2NO
• Oxides of nitrogen released from automobile engines can be reduced by catalyst like Pt to N₃ or NH₃.
• Oxides of N₂ are acidic, therefore can be removed by basic solution.
• Scrubbing of nitrogen oxides can be done by acids.

Question 9.
Have you seen soil pollution in the nearby area? What measures will you take to control soil pollution ?
Answer:
Yes, it can be controlled by the following methods :
1. insecticides and pesticides which are used to protect our crops spread soil pollution. Therefore its use should be limited.

2. After the second world war, DDT was used in agriculture to control insects, pests, weeds and various crop diseases. Due to its adverse effect, its use was banned in India. Pesticides like Aldrin and Dieldrin are actually toxic. These are insoluble and non-biodegradable in water due to which they cause serious metabolic and physiological disorders.

Nowadays organophosphates and carbamates are used as pesticides. They are more biodegradable compounds but they are serious toxins and are harmful to human beings. Thus, fertilizers, detergents, pesticides, polymers like chemicals are used only under extreme necessity.

3. Biodegradable domestic waste should be dumped in pits.
4. Non-biodegradable waste should be recycled.
5. Use of polythene should be avoided.
6. Domestic waste, biological waste and chemical waste should be burnt. As a result of incineration volume of waste products is reduced.

Question 10.
Sometime ago formation of polar stratosphere clouds was reported over Antarctica. Why these formed? What happens when such clouds break up by warmth of sunlight?
Answer:
In summer season, nitrogen dioxide and methane react with chlorine monoxide and chlorine atoms and form chlorine sink which prevent ozone depletion to a large extent.
In winter season, special type of clouds, polar stratosphere clouds are formed over Antarctica. These clouds provide a type of surface by which chlorine nitrate hydrolysis to form hypochlorous acid. It reacts with hydrogen chloride to form molecular chlorine.

In spring, when sunlight return to Antarctica, heat of sun disperse the clouds and HOCl and Cl₂ decompose by sunlight.

This way, the chloride radical produced start the chain reaction for the depletion of ozone.

Question 11.
Describe at least 7 medicines obtained from plants.
Answer:
1. Neem: It is the plant of family Malice.
Uses :

• It keeps the environment pure,
• Neem leaves are insecticides,
• Neem oil is used in the treatment of wound.

2. Amla: Botanical name: Emblica Officinalis.
Uses :

• Used to prepare Triphala powder,
• Powder of seeds of Amla are used for washing hairs.

3. Tulsi: Botanical name: Ocimum sanctum.
Uses :

• It is grown in houses and temples to keep the environment clean,
• Its leaves are used for the treatment of fever and cough,
• It has antiseptic property.

4. Mahua: Botanical name: Madhuca indica.
Uses :

• Leaves, seeds and flowers are used for the treatment of skin disease,
• Its flowers are used to prepare wine.

5. Arandi: Botanical name: Ricinus communis.
Uses :

• Its oil is used to prepare soap, candles etc.
• Oil is used as medicine.

6. Harra : Botanical name : Terminalia chebula.
Uses :

• In dyeing leather,
• In the manufacture of medicines,
• Powder of fruit is used for the treatment of asthma.

7. Khus: Botanical name: Velivera ziga includes.
Uses :

• Oil is used in cosmetics, perfume and making beverage,
• It is insect repellant.
• It is used in the treatment of arthritis, backache, sprain etc.

Question 12.
How will you use Green chemistry for the following :
1. To control photochemical smog.
2. To avoid use of halogenated solvents in dry-cleaning and that of chlorine in bleaching.
3. To reduce use of synthetic detergents.
4. To reduce the consumption of petrol and diesel.
Answer:
1. Some plants like Pinus, Pyrus can metabolise nitric oxide (NO). Thus, their growth can control photochemical smog.

2. In dry-cleaning, use of halogenated solvent is replaced by a process, where liquified CO₂ with a suitable detergent is used. Use of H₂O₂ in place of Cl₂ for bleaching clothes gives better results and makes use of lesser amount of water.

3. Soap is 100% biodegradable. Thus, it can be used in place of detergents. Nowadays biodegradable detergents are also available. Thus, these are used in place of non-biodegradable detergents.

4. CNG (Compressed Natural Gas) should be used because it causes least pollution. Along with this, use of electrical automobiles can also reduce the consumption of petrol and diesel.

Question 13.
Differentiate between London classical smog and photochemical smog.
Answer:
Differentiate between London classical smog and Photochemical smog

Question 14.
(a) What are the main sources of soil pollution?
(b) How can Environmental pollution be controlled?
Answer:
(a) Main sources of soil pollution are :

• Industrial waste.
• Insecticides, pesticides and herbicides.
• Fertilizer, DDT, BHC, NaClO₃, Na₃ASO₃ etc.
• Radioactive products.

(b) Environmental pollution includes domestic waste, industrial waste etc. which can be controlled by the following methods :

1. By the recycling of glass, plastic, iron, polythene, paper etc.
2. By burning.
3. By management of waste.
4. By the use of Green chemistry.
5. By sewage treatment.
6. By awareness.
7. By planting more trees.
8. By adopting control tools and techniques.
9. By installing high chimneys fitted with equipments to discharge pollutants.

Mechanical Properties of Fluids Class 11 Important Questions Objective Type

1. Choose the correct answer:

Question 1.
Source of non-pollution energy :
(a) Fossil fuel
(b) Sun
(c) Gasoline
(d) Nuclear energy.
Answer:
(b) Sun

Question 2.
Which radiations manufacture O₃ :
(a) Ultraviolet
(b) Visible
(c) Infrared
(d) Radio waves.
Answer:
(a) Ultraviolet

Question 3.
Which radiations provide greenhouse effect:
(a) Infrared
(b) Visible
(c) Ultraviolet
(d) X-rays.
Answer:
(a) Infrared

Question 4.
PAN is responsible for :
(a) Depletion of ozone layer
(b) Smog
(c) Acid rain
(d) Poisonous food.
Answer:
(b) Smog

Question 5.
Which is not an air pollutant:
(a)H₂
(b) H₂S
(c) NO_x
(d) O₃.
Answer:
(a) H₂

Question 6.
Air pollutant released from Jet aeroplanes in the form of an air pollutant:
(a) Photochemical oxidant
(b) Photochemical reductant
(c) Aerosol
(d) Physical pollutant.
Answer:
(c) Aerosol

Question 7.
Is not present in acid rain :
(a) H₂SO₄
(b) HNO₃
(c) H₂SO₃
(d) CH₃COOH.
Answer:
(d) CH₃COOH.

Question 8.
Is responsible for disease of lungs :
(a) O₂
(b) N₂
(c) CO₂
(d) SO₂.
Answer:
(d) SO₂.

Question 9.
03 is manufactured in :
(a) Troposphere
(b) Stratosphere
(c) Mesosphere
(d) Thermosphere.
Answer:
(b) Stratosphere

Question 10.
For acid rain ‘sink’ is :
(a) Leaves
(b) Reservoir
(c) Limestone
(d) CO₂.
Answer:
(c) Limestone

Question 11.
Primary pollutant is :
(a) SO₂
(b) NO₂
(c) N₂O
(d) NO.
Answer:
(d) NO.

Question 12.
Most dangerous is :
(a) Smoke
(b) Dust
(c) Smog
(d) NO.
Answer:
(c) Smog

Question 13.
Which of the following easily combine with haemoglobin:
(a) CO
(b) NO
(c) O₂,
(d) CO₂.
Answer:
(b) NO

Question 14.
Aromatic compound which is found in the form of particles :
(a) Benzene
(b) Toluene
(c) Nitrobenzene
(d) Polyring hydrocarbon.
Answer:
(d) Polyring hydrocarbon.

Question 15.
Following disease is caused due to depletion of ozone :
(a) Blood cancer
(b) Lung cancer
(c) Skin cancer
(d) Chest cancer.
Answer:
(c) Skin cancer

Question 16.
Photochemical smog is formed :
(a) In summer mornings
(b) In winter days
(c) In rainy days in the morning
(d) In day time in rains.
Answer:
(b) In winter days

Question 17.
Which of the following compound does not form photochemical smog:
(a) NO
(b) O₃
(c) C_xH_y
(d) SO₂.
Answer:
(d) SO₂.

Question 18.
Main pollutant hydrocarbon is :
(a) Methane
(b) Ethane
(c) Propane
(d) Butane.
Answer:
(a) Methane

Question 19.
Which of the following acid is in abundance in acid rain :
(a) HNO₃
(b) H₂SO₄
(c) HCl
(d) H₂CO₃.
Answer:
(b) H₂SO₄

Question 20.
Which of the following is not a pollutant:
(a) NO₂
(b) CO₂
(c) O₃
(d) (C_xH_y).
Answer:
(b) C0₂

Question 21.
Which of the following is not a greenhouse gas :
(a) CO₂
(b) CH₄
(e) CFCs
(d) O₂.
Answer:
(d) O₂.

Question 22.
Region near earth’s surface is :
(a) Thermosphere
(b) Mesosphere
(c) Stratosphere
(d) Troposphere.
Answer:
(c) Stratosphere

Question 23.
Main source of CO pollutant is :
(a) Industrial process
(b) Means of transport
(c) Earthquake
(d) Volcano.
Answer:
(b) Means of transport

Question 24.
Marble is the main sink of:
(a) Metallic pollutant
(b) NH₃ pollutant
(c) Acid pollutant
(d) None.
Answer:
(c) Acid pollutant

Question 25.
Which of the following is not a Nobel Prize winner in 2005 :
(a) Y. Chadwin
(b) R.H. Grooms
(c) R.R. Shark
(d) Polythene.
Answer:
(d) Polythene.

Question 26.
Is correct for photochemical smog :
(a) It is reducing in nature
(b) Is formed in winter season
(c) Is a mixture of smog and fog
(d) Cause irritation in eyes.
Answer:
(d) Cause irritation in eyes.

2. Fill in the blanks:

1. The air pollutant released by jet aeroplanes in the form of fluro carbon is …………… .
Answer:
aerosol

2. D.D.T. is …………… poisonous pollutant as compared to B.H.C.
Answer:
more
3. 03 is formed in the …………… level of atmosphere.
Answer:
stratosphere

4. A definite tolerable level of pollutants in the environment is expressed by …………… .
Answer:
TLV

5. Maximum air pollutants are present in …………… level of the atmosphere.
Answer:
troposphere

6. Ozone layer prevent us from …………… rays.
Answer:
ultraviolet

7. Oxides of …………… and …………… cause acid rain.
Answer:
nitrogen, sulphur

8. …………… is the main cause of ozone layer depletion.
Answer:
C.F.C
9. …………… gas is responsible for green house effect.
Answer:
CO₂
10. Components which produce pollution are called …………… .
Answer:
pollutant

11. …………… causes harmful effect on lungs.
Answer:
Photochemical smog

12. SO₂ pollutant is responsible for the disease of …………… .
Answer:
Lungs

13. …………… is responsible for ultraviolet light.
Answer:
Skin cancer

14. …………… also cause harm to ozone layer.
Answer:
Methane.

3. Match the following:

I.

Answer:
1. (b) CFC
2. (a) Fly ash
3. (d) Infrared radiations
4. (c) Smog
5. (e) Bone disease.

II.

Answer:
1. (c) Aerosol
2. (d) SO₂
3.  (a) Water pollution
4. (b) Depletion of ozone layer.

4. Answer in one word/sentence :

1. Write the definition of pollutants.
Answer:
Chemicals which affect man, animal and plant kingdom are called pollutants.

2. Write the names of two air pollutants.
Answer:
SO₂, SO₃.

3. Name two pollutants which depletes ozone layer.
Answer:

1. Cycle of nitric oxide (NO) and
2. C.F.C. (Chlorofluorocarbon) in which CFC is main.

4. In which region ozone is found?
Answer:
Ozone is found in stratosphere region.

5. Tell two health problems caused by SO₂.
Answer:

1. SO₂ affects the respiratory canal and lungs due to which various health diseases are caused like cancer also.
2. Acid rain caused due to SO₂ produces boils on the skin.

6. What is Acid rain?
Answer:
Various gaseous pollutant present in the atmosphere like SO₂, SO₃, NO₂, NO dissolve in raindrop.
These drops fall with rain and are called acid rain.
SO₂ + H₂O → H₂SO₃
SO₃+H₂O → H₂SO₄.

7. Name any two greenhouse gases.
Answer:
CFC and CO₂.

8. What is PAN?
Answer:
Peroxy acyl nitrate which is photochemical smog.

9. What is CFC?
Answer:
It is chlorofluorocarbon which is the main cause of depletion of ozone layer.

10. What is green chemistry?
Answer:
A technique to check pollution in which such chemical reactions are suggested which do not cause pollution and if pollution spreads then it can be destroyed. This is called green chemistry.

11. What is TLV?
Answer:
A definite value of pollutants can be tolerated. This is expressed by TLV. TLV means ‘Threshold Limit Value’.

12. What are particulate pollutants?
Answer:
The pollutants mixed up with air in liquid or solid state in such a manner that the remain suspended for a long time are called particulates.

13. What is specimasion?
Answer:
Many pollutants can be made from an element. The method of determining which of the product is more dangerous is called specimasion.

14. What is sink?
Answer:
Sink is that in which the substance totally gets consumed and then also there is no effect on sink. ‘

15. Name the biggest sink of the earth.
Answer:
The biggest sink of the earth is the sea.

16. What is greenhouse effect?
Answer:
Greenhouse effect: The heating of atmosphere due to absorption of infrared radiation by carbon dioxide and other gases is called greenhouse effect.

17. Explain the mechanism of acid rain.
Answer:
Due to combustion of fuels various pollutant particles of all gas pollutants in the form of smoke spread. Due to very high temperature of industries and other fuels N2 and 02 combine and spread in the atmosphere as oxides of nitrogen. These various gases fall on the earth as rain drops which is called acid rain.

18. Gas leaked in 1984 in Bhopal gas tragedy is.
Answer:
CH₃-N = C= O Methyl isocyanate

19. Name the main component responsible for depletion of ozone layer.
Answer:
Chlorofluorocarbon.

20. Name the person who started “Chipko Andolan” for forest conservation.
Answer:
Sunderlal Bahuguna.

21. Write name of disease caused by water pollution.
Answer:
Water pollution causes gastrointestinal diseases such as cholera, typhoid, dysentry, gastroenteritis, polio, hepatitis etc.

22. Name the important medicinal plant used to remove environmental pollution and also useful in skin cancer.
Answer:
Neem (Azaderecta indica).

23. Which gas is the cause of greenhouse effect?
Answer:
Carbon dioxide (CO₂).

24. Which is the main sink of CO pollutant?
Answer:
Organisms present on earth.

25. Write the name of four methods used in Green chemistry.
Answer:

1. Use of sunlight,
2. Micro-oven,
3. Micro-waves,
4. Use of Sound waves and Enzyme.

Students get through the MP Board Class 11th Physics Important Questions Chapter 9 Mechanical Properties of Solids which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 9 Mechanical Properties of Solids

Mechanical Properties of Solids Class 11 Important Questions Very Short Answer Type

Question 1.
What do you mean by deforming force?
Answer:
External force applied on a body is known as deforming force.

Question 2.
What is deformation?
Answer:
If an application of deforming force, a change in the shape or size of a body takes place, then it is said to be deformed.

Question 3.
What is meant by elasticity?
Answer:
It is the property of a body by virtue of which, it regains its original length, volume and shape when the deforming force is removed.

Question 4.
What is stress and its unit?
Answer:
The internal force of restitution per unit area of a deformed body is called stress.
Stress = \(\frac{\text { Force }}{\text { Area }}\)
Unit: SI unit of stress is Nm^-2.

Question 5.
Define plasticity.
Answer:
Plasticity is the property of remaining deformed even after the removal of deforming forces.

Question 6.
What do you mean by elastic limit?
Answer:
The maximum value of deforming force, by which if the force is applied on a body, the body looses its elastic property, is called elastic limit.

Question 7.
What do you mean by brittleness?
Answer:
Some substances are neither elastic nor plastic but they break easily into pieces, this property of a body is called brittleness and the bodies are called brittle substances.

Question 8.
What is difference between stress and pressure?
Answer:
Stress is internal force while pressure is external force.

Question 9.
How many types of stress are there?
Answer:
There are three types of stress :

1. Tensile stress or longitudinal stress,
2. Normal stress,
3. Tangential or shearing stress

Question 10.
What is strain? Write its unit.
Answer:
The ratio of change in length, volume or shape and its original length, volume or shape is called strain.
As it is the ratio of same quantities, therefore it has no unit.

Question 11.
What is longitudinal stress?
Answer:
It is the restoring force developed per unit cross-sectional area of a body when the length of the body increases in the direction of deforming force.

Question 12.
Define tangential stress.
Answer:
When the deforming force is applied along the plane of the area of cross-section of the body then the stress produced is known as tangential stress.

Question 13.
State the types of strain.
Answer:
There are three types of strain :

1. Longitudinal strain,
2. Shearing strain and
3. Volume strain.

Question 14.
What is longitudinal strain?
Answer:
Within elastic limit, it is the ratio of change in length to original length.
i.e., Longitudinal strain = \(\frac{\text { Change in length }}{\text { Initial length }}\)

Question 15.
Define volume strain.
Answer:
The change in unit volume due to deforming force is called volume strain.
Or
Within elastic limit, the ratio of change in volume to the initial volume is called volume strain, i.e.,
Volume strain = \(\frac{\text { Change in volume }}{\text { Initial volume }}\)

Question 16.
Define shearing strain.
Answer:
When a deforming force is applied on a body along its surface in such a way that the volume of the body remains constant, but the shape changes, then the body is said to be sheared and the strain produced is called shearing strain. Within elastic limit, it is measured by the ratio of the relative displacement of one plane to its distance from fixed plane, i.e.,
Shearing strain, θ = \(\frac{\Delta l}{l}\)
Δ l is relative displacement between two layers and l is the distance between the layers.

Question 17.
State Hooke’s law.
Answer:
Within elastic limit the stress is directly proportional to the strain. i.e.,
Stress ∝ Strain.

Question 18.
What are the types of modulus of elasticity?
Answer:
There are three types of modulus of elasticity :

1. Young’s modulus,
2. Bulk modulus,
3. Modulus of rigidity.

Question 19.
Define Young’s modulus and give its unit.
Answer:
Within elastic limit the ratio of longitudinal stress to the longitudinal strain is called Young’s moduluSIe.,
Young s modulus = \(\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\)
Unit: Its SI unit is Nm^-2.

Question 20.
What is compressibility?
Answer:
‘The reciprocal of bulk modulus is called compressibility.

Question 21.
Define coefficient of rigidity.
Answer:
Within elastic limit, the ratio of tangential stress to the shearing strain is called coefficient of regidity.

Question 22.
Write the following in form of increasing elasticity of copper, iron, glass and rubber.
Answer:
Rubber < Glass < Copper < Iron.

Question 23.
If an ivory ball and an iron ball are dropped from the same height on a hard floor. Which one will rebounce higher?
Answer:
The ivory ball will rebounce higher because its elasticity is greater than that of iron ball.

Question 24.
Springs are made of steel and not of copper. Why?
Answer:
The Young modulus of steel is more than that of copper. If the same deforming force is applied on the spring of steel and spring of copper then the steel regain its initial position faster than that of copper spring. So the spring are made of steel.

Question 25.
What do you mean by restoring force?
Answer:
The force due to which the body regain its original shape and size is called restoring force.

Question 26.
State the condition for deforming force and restoring force to be equal and opposite.
Answer:
When the strain produced in the body is within the elastic limit then the deform¬ing force and restoring force are equal and opposite to each other.

Question 27.
On increasing temperature, Young modulus of elasticity increases or de¬creases.
Answer:
Decreases.

Question 28.
Why the work is done in stretching a wire?
Answer:
When a wire is stretched, the inter reaction force is developed within the wire and hence work is done against this force.

Question 29.
What happens with work to stretch a wire?
Answer:
The work done to stretch a wire is stored as elastic potential energy within wire.

Question 30.
If the length of the wire is halved, then what will be the effect in its Young modulus of elasticity?
Answer:
No effect on Young modulus of elasticity as it depends on the mass of the mate¬rial of the body not in its length.

Question 31.
If the length of wire is cut to its half then :
(i) What will be the effect on increase in length?
Answer:
Increase in length will get halved with respect to initial length.

(ii) Effect on maximum weight it can resist?
Answer:
No effect.

Question 32.
If a wire is replaced by another wire of same length and material but of double diameter, then :
(i) What will be increase in length for same given weight?
Answer:
Increase in length will remain one-fourth.

(ii) Effect on maximum weight that it can resist?
Answer:
Maximum weight it can resist will become four times.

Question 33.
Why the suspension wires are made up of quartz or phospher bronze?
Answer:
This is because the elastic after effect is negligible in quartz and phospher bronze.

Question 34.
In Searl’s apparatus, two similar wires of same material are taken, why?
Answer:
The temperature of room changes, hence the effect of temperature will be same on both the wires. Thus, the change in temperature does not cause any error.

Question 35.
In Searl’s experiment the readings are taken both by loading and unload¬ing the weights, why?
Answer:
There are two reasons :

1. It ensures that the experiment is performed within elastic limit.
2. The errors produced due to elastic fatigue, torsion and backlash ‘are removed.

Question 36.
Why the readings are taken after some time in the experiments of elasticity?
Answer:
To remove the error caused by elastic after effect.

Question 37.
What is elastic hysteresis?
Answer:
On applying deforming force on a body the strain is produced in it on removal of deforming force some time body do not regain its original position completely. Some strain remains left in that body. It is called elastic hysteresis.

Question 38.
Write down the effect of impurities on elasticity.
Answer:
In the presence of impurities the elasticity of the bodies is increased.

Question 39.
What is the effect of temperature on elasticity?
Answer:
With increase in temperature, usually the elasticity of the bodies decreases and with decrease in temperature, the elasticity increases. But the effect of temperature is negligible in the elasticity of’Invar steel’.

Mechanical Properties of Solids Class 11 Important Questions Short Answer Type

Question 1.
State the difference between elastic and plastic bodies.
Answer:
Difference between Elastic and Plastic bodies :

Question 2.
Steel is more elastic than rubber. Explain with reason.
or
Why steel is more elastic than rubber?
Answer:
Let two wires of steel and rubber of equal length L and equal area of cross-section A are taken. Both are subjected by a force F, so that the elongation produced in steel is l_s and in rubber l_r.
∴ Modulus of elasticity of steel, Y_s = \(\frac{F \cdot L}{A \cdot l_{s}}\) …(1)
and Modulus of elasticity of rubber, Y_r = \(\frac{F \cdot L}{A \cdot l_{r}}\) …(2)
Now, dividing eqn. (1) by eqn. (2),
\(\frac{Y_{s}}{Y_{r}}=\frac{l_{r}}{l_{s}}\)
∵ l_r > l_s
∴ Y_s > Y_r
Hence, it is clear that steel is more elastic than rubber.

Question 3.
What do you mean by Young’s modulus of elasticity. Write its unit and dimensional formula. Derive the formula for Young’s modulus and hence define it.
Answer:
Within elastic limit, the ratio of longitudinal stress to the longitudinal strain is called Young’s modulus.
i.e., Y = \(\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\)

SI unit is Nm^-2 and dimensional formula is [ML^-1T^-2].
Formula: Let a mass M is suspended by a wire of length L and hence the elongation produced is l. If the radius of the wire is r, then
Stress = \(\frac{\text { Force }}{\text { Area }}=\frac{M g}{\pi r^{2}}\)
and strain = \(\frac{l}{L}\)
But Y = \(\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\)
∴ Y = \(\frac{M g / \pi r^{2}}{l / L}\)
Or
Y = \(=\frac{M g L}{\pi r^{2} l}\)
Within elastic limit, Young’s modulus of elasticity is numerically equal to that force, which extends the length of wire of unit cross-sectional area by the original length i.e., the final length becomes double the initial length.

Question 4.
What do you mean by bulk modulus? Establish the formula for bulk modulus.
Answer:

Bulk modulus : Within elastic limit, the normal stress is directly proportional to the volume strain.
∴ K = \(\frac{\text { Normal stress }}{\text { Volume strain }}\)
Formula : Let initial volume of a gas is V and pressure P. If the pressure is increased to P + p, so that its volume becomes V- v, hence the decrease in volume is v.
Since, force acting per unit area is called pressure. Hence,
Normal stress = Increase in pressure = (P + p)- P = p
Volume strain = – \(\frac{v}{V}\)
By Hooke’s law, within elastic limit,
Bulk modulus of elasticity = \(\frac{\text { Normal stress }}{\text { Volume strain }}\)
K = \(\frac{p}{-\frac{v}{V}}=\frac{-p V}{v}\)
-ve sign shows that, with increase in pressure, volume decreases.

Question 5.
What do you mean by coefficient of rigidity? Derive formula for it.
Answer:
Coefficient of rigidity : Within elastic limit the ratio of tangential stress and shearing strain is called modulus of rigidity.
i.e., η = \(\frac{\text { Tangential stress }}{\text { Shearing strain }}\)

Formula derivation : Let ABCDEFGH be a cuboid (of side l). A tangential force F is applied on the surface ADEF, If the area of surface is A and angular displacement due to force F is θ, then
Tangential stress = \(\frac{F}{A}\)
If the relative displacement between the layers is
AA’ = X, then
Shearingstrain, θ = \(\frac{A A^{\prime}}{A B}=\frac{x}{l}\)
Modulus of rigidity = \(\frac{\text { Tangential stress }}{\text { Shearing strain }}\)
∴ η = \(\frac{F / A}{x / l}=\frac{F l}{x A}\)

Question 6.
Why are girders I shaped?
Answer:

Girders of iron are made of I shapped.
Let l be the length of a rod, b be the breadth and d be its thickness. If weight W be suspended at its middle point the bending (δ) is given by
δ = \(\frac{\mathrm{W} l^{3}}{b d^{3} \mathrm{Y}}\) …(1)
For the strength of girder, should be smaller, δ will be smaller if
(a) Y is large.
(b) From eqn. (1) it is clear that δ∝\(\frac{1}{b}\) and δ∝\(\frac{1}{d^{3}}\) Therefore depth d should be greater than breadth b.
Therefore, girders of made of I shape.

Question 7.
Define Poisson’s ratio. What is its unit? What is its value?
Answer:
When a deforming force F is applied to a wire, then its length increases, but its radius or breadth decreases. This strain which occurs along the width of the body is called lateral strain. Within elastic limit, the lateral strain is proportional to longitudinal strain.
i.e., Lateral strain ∝ Longitudinal strain
or
Lateral strain = σ × Longitudinal strain
or
Poisson’s ratio,
Where D = Initial diameter of wire, d = Change in diameter, L = Original length of wire, l = Change in length.
It has no unit. Theoretical values lies from – 1 to 0.5.

Question 8.
Derive an equation for the force acting on the clamps, when the wire gets cooled. f’
Answer:

Let a wire of length L and area of cross-section A be clamped between the two ends. If it is being cooled, then its temperature decreases by Δ t°C, as a result it gets strinked and so it tries to contract the clamps. This force has to be found out. If Y is the Young’s moduliis of elasticity and a is the coefficient of linear expansion, then
α = \(\frac{\text { Change in length }(l)}{\text { Original length }(L) \times \text { Change in temp. }(\Delta t)}\)
Hence, αLΔt = l or \(\frac{l}{L}\) = αΔt
∴ Strain = αΔt …(1)
But, Young s modulus, Y = \(\frac{\text { Stress }}{\text { Strain }}\)
∴ Stress = Y × Strain= YαΔt …(2)
Moreover, Stress = \(\frac{\text { Force }}{\text { Area }}\)
or
Force = Stress × Area = YαΔt × A
∴ F =YAαΔt
This is the required relation.

Question 9.
Figure shows the strain-stress curve for a given material. What are:
(a) Young’s modulus and
(b) Approximate yield strength for his material?

Answer:
From the given graph for a stress of 150 × 10⁶ Nm^-2, the strain is 0.002.
(a) Young’s modulus of the material (Y) is given by
∴ Y = \(\frac{\text { Stress }}{\text { Strain }}=\frac{150 \times 10^{6}}{0.002}=\frac{150 \times 10^{6}}{2 \times 10^{-3}}\)
= 75 ×10⁹ Nm^-2
= 7.5 × 10¹⁰Nm^-2.

(b) Yield strength of a material is defined as the maximum stress it can sustain.
∴ From graph, the approximate yield strength of the given material = 300 × 10⁶ Nm2 = 3 × 10⁸ Nm^-2. Actually, it is slightly less than 3 × 10⁹ Nm^-2.

Question 10.
The stress-strain graphs for materials A and if are shown in figure (a) and (b):

The graphs are drawn to the same scale :
(a) Which of the materials has greater Young’s modulus?
(b) Which of the two is the stronger material?
Answer:
(a) From the two graphs we note that for a given strain, stress for A is more than that of B. Hence Young’s modulus = \(\left(\frac{\text { Stress }}{\text { Strain }}\right)\) is greater forA than that of B.

(b) Strength of a material is determined by the amount of stress (load) required to cause breaking or fracture of material corresponding the breaking point.
∴ Material A is stronger than B as it can withstand more load without breaking than the material B corresponding to point D.

Question 11.
Prove that energy stored per unit volume in a stretched string is, given by:
U = \(\frac{1}{2}\) × Stress × Strain
= \(\frac{1}{2}\) × Young’s modulus × (Strain )²
Derive the formula for the potential energy of a stretched string.
Answer:
Consider a wire of length L and it is extended by a force F due to which its extension takes place by l.
If A is the area of cross-section of the wire and Fis the Young’s modulus of elasticity,
then
Y = \(\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\) = \(\frac{F / A}{l / L}\) = \(\frac{F L}{A l} .\)
Or
YAl = Fl or F = \(\frac{Y A l}{L}\) …(1)

When, the wire is stretched by a force F, then work is done by the deforming force
which is stored in the wire in the form of potential energy.
Small amount of work done to stretch by dl i.e.,
dW = Fdl= \(\frac{Y A l}{L}\)dl
∴ The total amount of work done will be

Where, \(\frac{l}{L}\)is the longitudinal strain
Work done per unit volume is called elastic potential energy per unit volume (U).
∴ U= \(\frac{1}{2}\) × Y × (Strain)², [from eqn. (3)]
But, Y = \(\frac{\text { Stress }}{\text { Strain }}\)
∴ U= \(\frac{1}{2}\) × \(\frac{\text { Stress }}{\text { Strain }}\) × (Strain)²
or
U= \(\frac{1}{2}\) × Stress × Strain. Proved.

Mechanical Properties of Solids Class 11 Important Numerical Questions

Question 1.
A steel wire of length 4.7 m and cross-sectional area 3.0 × 10^-5 m²stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area 4.0 × 10^-5
m² under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Solution:
Here, for steel wire,
Length of wire, l₁ = 4.7 m
Area of cross-section, a₁ = 3.0 × 10^-5 m²
Stretching, Δl₁, = Δl(say)
Stretching force on steel, F₁ = F
For copper wire,
Length of wire,l₂ = 3.5 m
Area of cross-section, a₂ = 4.0 × 10^-5 m²
Stretching, Δl₂ = Δl(given);
Stretching force on copper, F₂ = F
Let Y₁ and Y₂ be the Young’s modulus of steel and copper wire respectively.

Dividing eqn. (1) by eqn. (2), we get

= \(\frac{18 \cdot 8}{10 \cdot 5}\) = 1.79 = 1.8:1

Question 2.
Two wires of diameter 0.25 cm. One made of steel and the other made of brass are loaded as shown in fig. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. If Young’s modulus of steel and brass are 2.0 × 10¹¹ Pa and 0.91 × 10¹¹ Pa, then compare the elongation of the steel and the brass wires.
Solution:

Using, Y = \(\frac{M g L}{\pi r^{2} l}\)
l = \(\frac{M g L}{\pi r^{2} Y}\)
For brass wire,
M= 6 kg, L = 1.0 m, Y = 0.91 × 10¹¹N/m²
and 2r = 0.25 cm = 0.25 × 10^-2m
r = 0.125 × 10^-2m
Putting the values in formula,
l = \(\frac{6 \times 9 \cdot 8 \times 1 \cdot 0}{3 \cdot 14 \times\left(0 \cdot 125 \times 10^{-2}\right)^{2} \times 0 \cdot 91 \times 10^{1}}\)
= 1.3 × 10^-4 m.
For steel wire,
M = 6 + 4 = 10 kg,L = 1.5 m, Y = 2.0 × 10¹¹ N/m²and 2r = 0.25cm
∴ r = 0.125 × 10^-2m
Putting the values in formula,
l = \(\frac{10 \times 9 \cdot 8 \times 1 \cdot 5}{3 \cdot 14 \times\left(0 \cdot 125 \times 10^{-2}\right)^{2} \times 2 \cdot 0 \times 10^{1}}\)
= 1.5 × 10^-4m

Question 3.
A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44500 N. force, producing only elastic deformation. Calculate the resulting strain. Y for copper = 1.1 × 10¹¹ Nm^-2.
Solution:
Here, Y = 1.1 × 10¹¹Nm^-2
A = Area of cross-section
= 15.2 mm × 19.1 mm
= 15.2 × 10^-3m × 19.1 × 10^-3 m
Force, F = 44500 N
Resulting Strain = Longitudinal Strain = ?
∴ Y = \(\frac{\text { Stress }}{\text { Strain }}\)
or
Strain \(=\frac{\text { Stress }}{Y}=\frac{F}{A Y}\)
or
Longitudinal Strain = \(\frac{44500}{15 \cdot 2 \times 19 \cdot 1 \times 10^{-6} \times 1 \cdot 1 \times 10^{1}}\)
= 139.34 × 10^-3 m = 0.139.

Question 4.
A steel cable with a radius of 1.5 cm supports a chair lift at a ski area. If the maximum stress is not to exceed 10⁸ Nm^-2, what is the maximum load the cable can support?
Solution:
Here, radius of steel cable r = 1.5cm = 1.5 × 10^-2 m
Max. Stress = 10⁸ Nm^-2
∴ Area of cross-section of cable A = πr^-2 = π(1.5 × 10^-2)²
Maximum load the cable can withstand = Maximum force ? We know that,
Maximum stress = \(\frac{\text { Maximum force }}{\text { Area of cross-section }}\)
or
Maximum force = Maximum stress × Area of cross-section
= 10⁸ × π × (1.5 × 10^-2)²
= 3.142 × 2.25 × 10⁸ × 10^-4 N
or
Maximum load the cable can withstand = 7.1 × 10⁴N.

Question 5.
A 4.0 metre long copper wire of area of cross-section 1.2 cm² is stretched by a force of 4.8 × 10³N. If the Young’s modulus of copper is Y = 1.2 × 10¹¹ N/m², then calculate:
(i) Stress,
(ii) Strain,
(iii) Increase in length of wire.
Solution:
Given : L=4.0 metre, A = 1.2 cm² = 1.2 × 10^-4m², F = 4.8 × 10³ N, Y= 1.2 × 10¹¹ N/m²

(i) Stress : The stress on wire will be = \(\frac{F}{A}=\frac{4 \cdot 8 \times 10^{3}}{1 \cdot 2 \times 10^{-4}}\)
∴ Stress = 4 × 10⁷ N/m².

(ii) Strain: The Young’s modulus of wire is Y = \(\frac{\text { Stress }}{\text { Strain }}\)
∴ Strain = \(\frac{\text { Stress }}{\mathrm{Y}}=\frac{4 \times 10^{7}}{1 \cdot 2 \times 10^{11}}\)
∴ Strain = 3.3 × 10^-4

(iii) Increase in Length of wire : From the formula of Young’s modulus
Y = \(\frac{\mathrm{FL}}{\mathrm{A} . l}\)
∴ l = \(\frac{\mathrm{FL}}{\mathrm{AY}}\)
\(=\frac{4 \cdot 8 \times 10^{3} \times 4}{1 \cdot 2 \times 10^{11} \times 1 \cdot 2 \times 10^{-4}}\)
= 13.2 × 10^-4 = 1.32 mm.

Question 6.
Two wires made of same material, length of first wire is half the length of second wire and its diameter is double the diameter of second wire. If same weight is suspended in both wires then find the ratio of increase in length.
Solution:

Question 7.
On stretching a gold wire, Its length Increases by 1%. Find out linear strain of it, if its coefficient of Young modulus is 8 × 10¹⁰ N/m². Find out the stress produced.
Solution:
According to question,
l = 1%, Y = 8 × 10¹⁰ N/m²
∴ Linear strain = \(\frac{\text { Increase in length }}{\text { Original length }}\)
= \(\frac{1}{100}\)
= 0.01
Stress = Y × Strain = (8 × 10¹⁰) × (0.01)
=8 ×10⁸N/m².

Question 8.
2 kg weight elongates a wire of 3 m length by 1 mm. The diameter of wire is 1 mm. Determine Young’s modulus of the material of the wire.
Solution:
Given : M = 2 kg, L = 3 m, l = 1 mm = 10^-3 m, r= \(\frac{1}{2}\)mm= 0.5 x 10^-3m, g = 9.8 ms^-2. ,
We have, Y = \(\frac{M g L}{\pi r^{2} l}\)
or
Y = \(\frac{2 \times 9 \cdot 8 \times 3}{3 \cdot 14 \times\left(0 \cdot 5 \times 10^{-3}\right)^{2} \times 10^{-3}}\)
=  \(\frac{58 \cdot 8 \times 10^{6} \times 10^{3}}{3 \cdot 14 \times 0 \cdot 25}\)
∴ Y = \(\frac{588 \times 10^{12}}{314 \times 25}\)
= 7.49 × 10¹⁰Nm^-2

Question 9.
The diameter of a wire is \(\frac{2}{\sqrt{\pi}}\) cm. The force required to double the length is 2 × 10¹²dyne. Calculate Young’s modulus of the material of the wire.
Solution:
Given: r = \(\frac{1}{2}\) × \(\frac{2}{\sqrt{\pi}}\)cm = \(\frac{1}{\sqrt{\pi}}\) 10^-2 m, L = l
and Mg = 2 × 10¹²dyne = 2 × 10¹² × 10^-5N = 2 × 10⁷N
Now, we have Y = \(\frac{M g L}{\pi r^{2} l}\)

∴ Y = \(\frac{2 \times 10^{7}}{10^{-4}}\) = 2 × 10¹¹ Nm^-2

Question 10.
Calculate the force required to elongate 1 m long the 2 mm thick wire by 1 mm. (Y = 2 x 10¹¹ Nm^-2)
Solution:
Given: L = 1m,r = \(\frac{2}{2}\) = 1mm =10^-3m = 10^-3m, l = 1mm 10^-3m Y = 2 × 10¹¹Nm^-2
Frmula, Y = \(\frac{M g L}{\pi r^{2} l}\)
or
Mg = \(\frac{Y \pi r^{2} l}{L}\)
\(=\frac{2 \times 10^{11} \times 3 \cdot 14 \times\left(10^{-3}\right)^{2} \times 10^{-3}}{1}\)
= 6.28 × 10¹¹ × 10^-9
= 6.28 × 10² = 628 N.

Question 11.
Calculate the work done in extending a wire of length lm and of cross-section 1mm2 through 2mm. (Given y = 2 × 10¹¹Nm^-2)
Solution:
Given: L= 1m, A = nr² = 1mm² = (1 × 10^-3m)²= 10^-6m²
l = 2mm = 2 × 10^-3m and Y = 2 × 10^-11 Nm^-2
Strain = \(\frac{l}{L}=\frac{2 \times 10^{-3}}{1}\) = 2 × 10^-3
∴ Volume of wire = A.L = 10^-6 × 1 = 10^-6m³
Work done = \(\frac{1}{2}\) × Y × (Strain)² × Volume
= \(\frac{1}{2}\) × 2 × 10¹¹ × (2 ×10^-3)¹¹× 10^-6
= 0.4 J.

Question12.
A rubber ball while taking to the bottom of a lake 200 m deep its volume de-creases by 0.1%. Calculate the Bulk modulus of rubber. Density of water is 10³ kg/ m³ and g = 10 m/sec².
Solution:
Given : Depth h = 200m, Density of water = 10³ kg/m³ Acceleration due to gravity g = 10 m/sec²,
Volume strain = 0.1% = –\(\frac{0 \cdot 1}{100}\)
∴ Bulk modulus K = \(\frac{\frac{-p}{\Delta v}}{\frac{V}{V}}\)
or
K = \(\frac{-200 \times 10^{3} \times 10}{\frac{-0 \cdot 1}{100}}\)
= 2 × 10⁹N/m²

Question 13.
A weight of 20 kg is suspended in a wire. Area of cross-section of it is 1mm². The length of the wire in extended state is 6 m. On withdrawing the weight its length become 5.995 m. Find out Young modulus of elasticity.
Solution:
Given : M = 20 kg, A = 1mm² = 10^-6m², L = 6m,
Change in length = 6 – 5.995 = 0.005 m
∴ Y = \(\frac{M g L}{A l}\)
or
Y \(=\frac{20 \times 9 \cdot 8 \times 6}{10^{-6} \times 0-005}\)
= 2.35 × 10¹¹N/m².

Question 14.
On iron wire of radius 0.5 mm is heated up to 300°C and then it is fixed between two clamp. When the temperature reduces to 30°. What will be the force acting upon the clamp. (α = 1.2 × 10^-5/°C and Y= 1.4 × 10¹² dyne/cm²)
Solution:
Given : r = 0.5mm = 0.05 cm, α = 1.2 × 10×^-5/°C, Y = 1.4 × 10¹²dyne/cm²
ΔT = 300 – 30 = 270° C
From F = YΔ α. AT, we get
F = Y πr² α ΔT
∴ F = 1.4 × 10¹² × 314 × (0.05)² × 1.2 × 10^-5 × 270
F = 3.56 × 10⁷ dyne.

Mechanical Properties of Solids Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
A wire of radius r and length L on which a mass M is suspended so that its increase in length is T. Young modulus will be:
(a) \(\frac{\mathrm{MgL}}{\pi r^{2} l}\)
(b) \(\frac{\mathrm{Mgl}}{\pi r^{2} \mathrm{~L}}\)
(c) \(\frac{\pi r^{2} L}{M g l}\)
(d) \(\frac{\pi r^{2} l}{\mathrm{MgL}}\)
Answer:
(a) \(\frac{\mathrm{MgL}}{\pi r^{2} l}\)

Question 2.
Which substance is more elastic:
(a) Glass
(b) Steel
(c) Plastic
(d) Rubber.
Answer:
(b) Steel

Question 3.
Magnitude of Poisson’s ratio is in between :
(a) – 1 and \(\frac{1}{2}\)
(b) –\(\frac{3}{4}\) and –\(\frac{1}{2}\)
(c) \(\frac{1}{2}\) and 1
(d) 1 and 2
Answer:
(a) – 1 and \(\frac{1}{2}\)

Question 4.
Work done per unit volume for a wire is :
(a) Stress × Strain
(b) \(\frac{1}{2}\)× Stress × Strain
(e) 2(Stress × Strain)
(d) Stress/Strain.
Answer:
(b) \(\frac{1}{2}\)× Stress × Strain

Question 5.
Four wire made of same material, equal weight is suspended on them, increase in length will be maximum for wire :
(a) Length 50 cm and diameter 0.5 mm
(b) Length 100 cm and diameter 1 mm
(c) Length 200 cm and diameter 2 mm
(d) Length 300 cm and diameter 3 mm.
Answer:
(a) Length 50 cm and diameter 0.5 mm

Question 6.
10 newton force is needed to break a wire of radius one millimetre. What is force needed to break a wire of radius 3 mm :
(a) 90N
(b) \(\frac{10}{3}\)N
(c) \(\frac{10}{9}\)N
(d) 30N
Answer:
(a) 90N

Question 7.
In general modulus of rigidity is than modulus of elasticity:
(a) Less
(b) More
(c) Equal
(d) None of these.
Answer:
(a) Less

Question 8.
A wire of length L, Area of cross-section A, Young modulus of elasticity Y and coefficient of linear expansion a is heated up to t°C. Force experience on wire will be :
(a) γAαt
(b) γAαLt
(c) tALα
(d) \(\frac{t \alpha \mathrm{L}}{\mathrm{A}}\)
Answer:
(a) γAαt

2. Fill in the blanks:

1. SI unit of stress is …………………
Answer:
Newton/metre²

2. On increasing temperature, coefficient of elasticity ………………….
Answer:
decreases

3. Within elastic limit ……………….. is directly proportional to strain.
Answer:
stress

4. Rubber is ………………. elastic than steel.
Answer:
less

5. For an ideal rigid body Young modulus is ………………..
Answer:
infinity

3. Match the following:

Answer:
1. (d) Applicable within elastic limit
2. (e) Volume strain
3. (a) Proportional to area of cross-section
4. (b) Zero
5. (c) Shearing strain

4. Write true or false:

1. Hooke’s law is applicable within elastic limit.
Answer:
True

2. Diamond can be considered as rigid body.
Answer:
True

3. Product of strain and stress is equal to stored energy.
Answer:
False

4. Young modulus of elasticity is dimensionless.
Answer:
False

5. Breaking stress depends on body of material.
Answer:
True

6. Young modulus of elasticity is valid for solid only.
Answer:
True

7. Rubber is more elastic than steel.
Answer:
False

8. Hookes law is defined within elastic limit.
Answer:
False

9. Quartz does not exhibit elastic after effect practically.
Answer:
True

10. It is difficult to twist a small rod in comparison to long rod.
Answer:
True

11. A hollow rod of same length and mass is more strong than a solid rod.
Answer:
True

Students get through the MP Board Class 11th Physics Important Questions Chapter 10 Mechanical Properties of Fluids which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 10 Mechanical Properties of Fluids

Mechanical Properties of Fluids Class 11 Important Questions Very Short Answer Type

Question 1.
What do you mean by pressure? Is it scalar or vector? Write its SI unit.
Answer:
The force acting normally on unit area of the surface is known as pressure. It is a scalar quantity. The SI unit of pressure is N/m².

Question 2.
What is thrust? Write its SI unit.
Answer:
The total force acting normally on a surface, by the liquid pressure is called thrust.
Unit: Its SI unit is N.

Question 3.
What is Pascal’s law? State its two applications.
Answer:
The pressure in a fluid in equilibrium is the same everywhere if the effect of gravity can be neglected.
Or
Pressure applied to any part of an enclosed fluid at rest is transmitted equally to every portion of the fluid and the walls of the containing vessel.
Applications :

• Hydraulic lift,
• Hydraulic brakes.

Question 4.
Why is the dam of water reservoir thick at the bottom?
Answer:
It is so because the pressure of water in the reservoir increases with depth.

Question 5.
What do you understand by atmospheric pressure? What are its various units?
Answer:
The force exerted by the air of atmosphere, per unit area on the surface of earth is called atmospheric pressure.
latm = 1.013 × 10⁵Nm^-2 = 1 Pascal.

Question 6.
Explain 1 bar and 1 tor.
Answer:
1 bar: It is the unit of pressure.
1 bar = 1.0 × 10⁵Nm^-2 =10⁵Pa.
1 tor: It is also unit of pressure named after Torricelli’s.

Question 7.
What is the value of atmospheric pressure?
Answer:
1.013 ×10⁵ Newton/metre².

Question 8.
“Atmospheric pressure is 760 mm of Hg”. What it means?
Answer:
It means that it equilibrium the 760 mm height of mercury column pressure.

Question 9.
What is known as Torricellian vacuum?
Answer:
The vacuum is the simple barometer above the mercury column is known as Torricellian vacuum.

Question 10.
What will be the effect on mercury level of barometer if a small hole is made in the upper portion?
Answer:
The level of mercury will come down.

Question 11.
What will be the effect on the level of mercury in a barometer if a water droplet is inserted inside it.
Answer:
The level of mercury will fall due to the water vapour formation when a water droplet is inserted inside the barometer.

Question 12.
What will be the effect on level of mercury of the barometer is baught to hill?
Answer:
The level of mercury will fall as atmospheric pressure at hill is less.

Question 13.
How atmospheric pressure is change with altitude?
Answer:
Atmospheric pressure will decrease with altitude.

Question 14.
Which liquid is used in barometer?
Answer:
Mercury is used in barometer.

Question 15.
The surface tension of a liquid does not depend upon the area of the surface. Why?
Answer:
Because the surface tension is defined as the force acting on unit length of the line drawn on the liquid surface.

Question 16.
What is difference between cohesive force and adhesive force?
Or
What do you mean by cohesive force and adhesive force?
Answer:
Cohesive force: The force of attraction between the molecules of same sub-stance is called cohesive force.
Adhesive force: The force of attraction between the molecules of different sub-stances is called adhesive force.

Question 17.
What is surface tension? Give its SI unit.
Answer:
In equilibrium, the force acting per unit length on an imaginary line drawn on the liquid surface, perpendicular to the line and tangential to the surface is called surface tension.
Its SI unit is Nm^-1.

Question 18.
How can be the surface tension of a liquid be reduced?
Answer:
Surface tension of a liquid can be reduced by increasing temperature.

Question 19.
Soaps and detergents are helpful in cleaning the clothes. Why?
Answer:
The soap or detergent reduces the surface tension of water. Therefore, the soap solution can penetrate deep into the pores, where water cannot normally reach and thus dirt and dust are removed.

Question 20.
Why hot soup tastes better than cold soup?
Answer:
Soup is a liquid. Whenever a liquid is heated, its surface tension decreases and it gets more spread up in the tongue and it reaches to the sensitive layers rather than cold soup. Hence, hot soup tastes better than cold soup.

Question 21.
At which surface the pressure is excess, on what factors it depends?
Answer:
The excess of pressure is at concave surface. This is because of surface tension. The excess pressure depends upon :

• Surface tension of liquid and
• Radius of curvature of liquid surface.

Question 22.
What is angle of contact? How do the angles of contact differ for the liquids which (i) wet the solid surface and (ii) does not wet the solid surface?
Answer:
The angle that the tangent to the liquid surface at the point of contact makes with the solid surface inside the liquid is called the angle of contact. The angle of contact is acute for the liquids which wet the solid surface and is obtuse which do not wet.

Question 23.
What is capillarity? Give example.
Answer:
The phenomenon of ascending and descending of liquid in a capillary is called capillarity. In kerosene lamp oil rises in the wick due to capillarity.

Question 24.
Towels are used to wipe and dry the body after bathe. Why?
Answer:
A towel has number of pores, which act as capillary, hence it soaks the water of the body.

Question 25.
If wax is polished on clothes, they become water proof. Why?
Answer:
When wax is polished, then the pores of the threads, which act as capillaries get closed and it cannot suck water. Hence it becomes water proof.

Question 26.
Why the blood pressure in humans is greater at the feet than at the brain? (NCERT)
Answer:
The height of the blood column is quite large at the feet than at the brain. There fore the blood pressure in humans is greater at the feet than at the brain.

Question 27.
What are the factor that effect air pressure at a place?
Answer:

1. Height of atmosphere
2. Density of atmosphere
3. Acceleration due to gravity.

Question 28.
On which principle hydraulic lift work?
Answer:
Hydraulic lift work on Pascal’s law.

Question 29.
Why the droplets of liquid are spherical?
Answer:
The free surface of liquid has tendency to achieve to least surface area, because of its potential energy. The sphere has the least surface area therefore droplets are spherical.

Question 30.
The oil spreads over the water surface but when the water is put into oil it becomes spherical droplets. Why?
Answer:
This is because the adhesive force between the molecules of water and oil is greater than the cohesive force of oil.

Question 31.
When a glass rod is heated at one end, that end becomes spherical, why?
Answer:
When the end of a glass rod is heated, it melts and changes into liquid, hence due to the surface tension the liquid tries to achieve the least surface area, thus it becomes spherical.

Question 32.
In which material of capillary the water will descent not ascent? What will happen if a capillary made up of silver is dipped into water?
Answer:
The water will descent in the capillary made up of paraffin wax. The water will neither rise nor descent in silver capillary.

Question 33.
When two lead pieces are pressured against each other they stick together, why?
Answer:
When two pieces of lead are pressured against each other, then the molecules come into contact and the cohesive force acts between the molecules and hence they stick together.

Question 34.
The small needle of steel floats in the water but it sinks into water with little soap solution. Why?
Answer:
The needle floats in the water because the surface tension force balance the weight of the needle. When the soap solution is added into water its surface tension decrease and hence the force which balances the needle decreases and needle sinks.

Question 35.
Why the farmers plough the fields after rain?
Answer:
When the fields are ploughed the capillaries of earth broken and the under- ground water remains there which is used by the plants.

Question 36.
What is molecular range?
Answer:
The maximum distance, within which the molecules of a liquid experience cohesive force.
Unit: It is of the order 10^-9 m.

Question 37.
What is meant by surface film of a liquid?
Answer:
Surface film is a thin film of liquid near its surface and having thickness equal to the molecular range for that liquid.

Question 38.
What is viscosity ?
Answer:
The inherent property of the fluids (liquids or gases) by virtue of which they oppose the relative motion between their layers is called viscosity.

Question 39.
What do you mean by ideal liquid?
Answer:
The liquid whose viscosity and compressibility both are zero called an ideal liquid.

Question 40.
What do you understand by streamline flow?
Answer:
When a fluid flows in such a way, that at any point the velocity of all the particles of fluid remain constant, then its flow is called streamline flow.

Question 41.
What do you mean by turbulent flow?
Answer:
The flow of the fluid in which the motion of the particles are irregular and zig-zag is called turbulent flow.
Fast-flowing rivers during the flood, storms are the examples of turbulent flow.

Question 42.
Two streamline flow do not intersect each other, why?
Answer:
If two streamline flow intersects at a point, then at that point, there will be two directions of velocity of the particles of fluid, which is not possible, hence they do not intersect.

Question 43.
State Stoke’s law.
Answer:
Stoke’s law states that, if a sphere of radius r moving with uniform velocity v, through a homogeneous incompressible fluid of viscosity p, the retarding force Fis given by
F = 6 πηrν.

Question 44.
What is terminal velocity?
Answer:
When a ball falls through a liquid then initially the ball is accelerated due to gravity but soon it achieves a constant velocity called terminal velocity.

Question 45.
Define coefficient of viscosity.
Answer:
The coefficient of viscosity of a liquid is the viscous force acting tangentially per unit area of a liquid layer having unit velocity gradient in a direction perpendicular to the direction of flow of liquid.

Question 46.
What is critical velocity?
Answer:
The maximum velocity of a fluid, below which the flow remains streamline flow, is called critical velocity.

Question 47.
State the principle of continuity.
Answer:
When a non-viscous and incompressible liquid flows through a tube of non-uniform bore and its flow is streamlined, then the product of velocity of flow and the area of cross-section of the tube at every transverse cross-section remains constant.
i.e., A.v = a constant
or A₁v₁=A₂v₂.

Question 48.
How many types of energy does a flowing liquid possess?
Answer:
The flowing liquid possesses three types of energies :

1. Potential energy,
2. Kinetic energy and
3. Pressure energy.

Question 49.
Two small light balls are hanging near to each other by two threads. If air is blown between them, they come near to each other, why?
Answer:
When air is blown the speed of air increases, hence the pressure between the balls decreases. Thus due to outer pressures, the balls come closer.

Question 50.
Air is blown below the pan of a balance in equilibrium, what will happen?
Answer:
When the air is blown below the pan, the velocity of air is increased and hence the pressure below the pan will decrease. Hence the air above the pan, presses the pan downward, thus that pan will be lowered down.

Question 51.
When a fast-moving train passes through a platform, all the dust particle and others attract toward train, why?
Answer:
When a fast-moving train passes through a platform according, Bernoulli’s theorem pressure of that place get dropped at once, so the dust particle and other’s get attracted toward the train.

Mechanical Properties of Fluids Class 11 Important Questions Short Answer Type 

Question 1.
Explain why :
(a) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km.
(b) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area. (NCERT)
Answer:
(a) Atmospheric pressure at height of 6km
h = 6km = 6× 10₃m, d = 1.30kg m^-3, g = 9.8 ms^-2
∴ P = hdg = 6 × 10³ × 1.30 × 9.8 = 0.7644 × 10⁵Pa
Atmospheric pressure at sea level = 1.013 × 10⁵Pa
Thus, it is clear that the atmospheric pressure at height of 6 km is nearly half the atmospheric pressure. But the atmosphere is extended nearly up to 100km from the surface of earth. It is so because the density of air decreases as we go up from the surface of earth.

(b) When the force is exerted at a point on the liquid, then the pressure is equally transmitted in all directions. Thus, the pressure is not concerned with direction. Hence, the hydrostatic pressure is a scalar quantity.

Question 2.
Explain why :
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.
(b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.) (NCERT)
Answer:
(a) When a drop of liquid is put on the solid surface then three interfaces are formed:

1. Solid-liquid interface,
2. Solid-vapour interface,
3. Liquid-vapour interface.

Let the surface tension in these interfaces are T_SLT_SV and T_LV respectively. Then at point 0 for the equilibrium of liquid
T_SV = T_SL+T_LV cos θ,
where θ is angle of contact.
∴ cos θ = \(\frac{T_{S V}-T_{S L}}{T_{L V}}\)
For mercury – glass pair Tsv < TSL ∴ cos θ is negative i.e., θ is obtuse angle.
For water-glass Tsv >T_SL
cos θ is positive i.e., θ is acute angle.

(b) The angle of contact for water glass pair is acute angle. Hence, the water spreads over the surface of glass. The angle of contact for mercury-glass pair the angle of contact is obtuse angle. Hence, the mercury tries to contract and acquire the shape of drop so that the angle of contact becomes obtuse angle.

In the other words, the adhesive force between molecules of water and glass is greater than the cohesive forces between the molecules of water. Hence, water wets the glass. While the adhesive force between the molecules of mercury and glass is less than cohesive forces between the molecules of mercury, hence the mercury does not wet the glass.

Question 3.
Explain why :
(a) Surface tension of a liquid is independent of the area of the surface.
(b) A drop of liquid under no external forces is always spherical in shape. (NCERT)
Answer:
(a) The force acting per unit length on either side of an imaginary line at right angles on the surface of liquid is called surface tension. This force does not depend on the area of surface.

(b) In the absence of external force, only surface tension acts on the liquid surface. The surface tension tries to minimize the surface area of liquid. For the given volume the surface area of sphere is least. Hence, in the absence of external forces the liquid drop acquires the spherical shape.

Question 4.
Explain why :
(a) To keep a piece of paper horizontal, you should blow over, not under it.
(b) When we try to close a water tap with our Angers, fast jets of water gush through the openings between our Angers. (NCERT)
Answer:
(a) When a piece of paper (paper strip) is blown over, then the velocity of air in that region increases, hence the kinetic energy of air increases, then according to Bernoulli’s theorem the pressure in that region decreases than the pressure in below region. Hence, due to decrease in pressure in upper region, the paper strip becomes horizontal.

(b) When we try to close the water tap with our fingers. The area for the flow of liquid decreases.
According to principle of continuity Av = constant.

Obviously if A decreases then v increases. Thus, the velocity of flow of liquid increases and fast jets of water gush through the opening between our finger.

Question 5.
Explain the construction and working of hydraulic lift.
Answer:
Construction: The construction of a hydraulic lift is shown in the Fig. In it there are two cylinders C₁ and C₂ of different area of cross-sections. These are interconnected by means of a narrow tube. Two frictionless and watertight pistons A and B are fitted in the cylinders C₁ and C₂ respectively. The area of cross-section of the piston A is small while that of the piston B is quite large. The body to be lifted up is placed on the platform attached to the piston B. The force is applied on the piston A. The two cylinders are filled with water.

Principle: The hydraulic lift consists of two cylinders C₁ and C₂ and they are connected by a horizontal tube T. There are two air resistance pistons P₁ and P₂ and their area of cross-section is A₁ and A₂ respectively. Such that A₁ solid A₂ i. e., area of container C₁ is less than that of C₂ On piston P₁ if F, is the force applied, then applied pressure will be
P = \(\frac{F_{1}}{A_{1}}\) …….. (1)
This pressure now gets transmitted to the container C₂ and so the liquid gives an upward force to the piston P₂. By Pascal’s law,
Pressure on piston P₂ = Pressure on piston P1 = \(\frac{F_{1}}{A_{1}}\)
Force F₂ on piston P₂ = Pressure × Area
or F₂ = \(\frac{F_{1}}{A_{1}} \) × A₂ ……….. (2)
As A₁ < A₂, hence F₂>F₁ So by applying lesser force on small area, we get large force on larger areas. This is the principle of hydraulic lift.

Question 6.
Show that a body when immersed in a liquid looses its weight equal to the weight of the liquid displaced by the body.
Answer:
According to this principle, when a body is immersed wholly or partially in a liquid at rest, it loses some of its weight which is equal to the weight of liquid displaced by immersed part of the body.

Consider a rectangular body of mass M and height h is completely immersed in a liquid of density p. Let A be the area of the lower and upper face of the body and l is the depth of the upper face of the body from the free surface of liquid.

Volume of liquid displaced = Volume of the body ( V) = Ah.
Mass of the liquid displaced, m = Vρ= Ahρ ………… (1)
Pressure acting on the upper face of the body due to
liquid, P₁ = lρg ……………… (2)
and Pressure acting on the lower face of body, P₂ = (l+ h)ρg ………….. (3)
Downward force on the upper face of the body is Fx = P1A = IρgA …………… (4)
Upward force on the lower face of the body is F₂ = P₂A = A(l + h)ρg …………… (5)
Net upward thrust on the body
F = F₂-F₁ =(l + h)ρgA – lρgA
or
F = Ahρg …………….. (6)
From eqns. (1) and (6), we get
F = mg = Weight of liquid displaced Actual weight of the body W = Mg
Upward thrust F = mg
Hence, the apparent weight of the body in liquid = W-F= Mg – mg.

Question 7.
What is Pascal’s law? State and prove it.
Answer:
In a closed liquid the pressure applied at any part is equally transmitted in all direction and in the same amount.
Or
If a liquid is in equilibrium, then the pressure in every part of it is equal.
Deduction of Pascal’s law: Consider two points O₁ and O₂ inside a liquid. Now, imagine a right circular cylinder with O₁, O₂ as the axis. Its end faces will be circular with O₁ and O₂ as centres.

The liquid inside the cylinder is in equilibrium, therefore the forces exerted by the outside liquid on the inner liquid will be perpendicular to the surface.
In accordance with it the forces acting on the end faces of the cylinder with O₁ and O₂ as centre will be perpendicular to them.

If the force acting on the face with O₁ as centre is F₁ then
Pressure at O₁ P = \(\frac{F_{1}}{A}\)
Where A is the area of the end face.
∴ F₁ = P₁A
Similarly, if F₂ is the force acting on the end face with O₂ as centre, then
Pressure at O₂ is P₂ = \(\frac{F_{2}}{A}\)
or F₂ = P₂ A
As the liquid is in equilibrium.
∴ F₁=F₂
P₁ A = P₂A
or P₁ = P₂
Therefore, the pressure on O₁ and O₂ are equal.

Question 8.
State Pascal’s law. Explain the effect of gravity on Pascal’s law.
Answer:
Pascal’s law:

In a closed liquid the pressure applied at any part is equally transmitted in all direction and in the same amount.
Or
If a liquid is in equilibrium, then the pressure in every part of it is equal.
Deduction of Pascal’s law: Consider two points O₁ and O₂ inside a liquid. Now, imagine a right circular cylinder with O₁, O₂ as the axis. Its end faces will be circular with O₁ and O₂ as centres.

The liquid inside the cylinder is in equilibrium, therefore the forces exerted by the outside liquid on the inner liquid will be perpendicular to the surface.
In accordance with it the forces acting on the end faces of the cylinder with O₁ and O₂ as centre will be perpendicular to them.

If the force acting on the face with O₁ as centre is F₁ then
Pressure at O₁ P = \(\frac{F_{1}}{A}\)
Where A is the area of the end face.
∴ F₁ = P₁A
Similarly, if F₂ is the force acting on the end face with O₂ as centre, then
Pressure at O₂ is P₂ = \(\frac{F_{2}}{A}\)
or F₂ = P₂ A
As the liquid is in equilibrium.
∴ F₁=F₂
P₁ A = P₂A
or P₁ = P₂
Therefore, the pressure on O₁ and O₂ are equal.

Suppose a trough is filled with a liquid of density ρ. A and B are two points on the same vertical line and separated by a distance of h. A is above the point B shown in Fig.
Imagine a right cylinder of height h and area of cross-section A with the vertical line AB as axis, Weight of the cylinder = Mass × g
= Volume × Density × g
= Ahpg, (∵ Volume=Area × Height)

This weight will be acting vertically downwards. If the pressures at the points A and B are P₁ and P₂ respectively, then the force acting on the upper face of the cylinder in the downward direction is
F₁ = Pressure x Area
or
F₁ = P₁A
Similarly, the force acting on the lower face of the cylinder in the upward direction is F₂ = P₂A
But, the liquid is in equilibrium, therefore F₁ + Ahρg – F₂ = 0
∴ P₁A + Ahρg- P₂A = 0
or (P₂ – P₁ )A = Ahρg
or P₂ – P₁ = hρg …………… (1)
From eqn. (1), it is clear that there is a difference of pressure between any two points in the same vertical line in a liquid.

Question 9.
Explain why: The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection. (NCERT)
Answer:
The area of needle of a syringe is very small. According to principle of continuity Aν = constant. Since area is small therefore ν will be large. The doctor exerts pressure P with his thumb. According to Bernoulli’s theorem.
P + \(\frac{1}{2}\) dv²+ dgh = Constant

In this expression, the power of P is 1 and that of ν is 2. Obviously, ν is more effective than P. This is reason due to which the size of needle of syringe controls flow rate better than the thumb pressure exerted by doctor while administering an injection.

Question 10.
The farmer plough their field after rain. Why?
Answer:
In a dry clay, number of pores are there, which acts as capillaries. After rain if farmers plough their field, then the capillaries of earth broken and the underground water remain there, which is used by the plant.

Question 11.
Explain why: A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel. (NCERT)
Answer:
The velocity of liquid emerging out of a small hole is high (∵ Aν = constant). Hence, its linear momentum will be large. Since no external force is acting on the system, so the linear momentum of the system remains conserved. Thus, it is clear that a reaction will act on the vessel and it will experience a thrust.

Question 12.
Explain why: A spinning cricket ball in air does not follow a parabolic trajectory. (NCERT)
Answer:
When a spinning cricket ball is thrown in air then at one side of ball the resultant velocity of air increases and at other side, the velocity decreases. According to Bernoulli’s theorem, a pressure difference is created between the two sides and a resultant force acts on the ball. Hence, the ball follows a curved path, it is called Magnus effect. Due to this effect, the ball does not follow a parabolic trajectory.

Question 13.
Describe Torrecelli’s experiment of atmospheric pressure. How does it measure the pressure?
Answer:
Torricelli’s Experiment: Torricelli performed an experiment to measure the atmospheric pressure. This experiment is called the
Torricelli’s experiment. The device used for the mea¬surement of atmospheric pressure is called a Barometer.
A barometer consists of a one-metre long glass tube of nearly 1.5 cm internal diameter.

One end of this tube is closed. This tube is completely filled with mercury. Now, closing the open end with thumb, we invert the tube and place the open end in a trough containing mercury such that the tube is vertical and the open end is well inside the mercury. Care is taken that no air bubble enters the tube. On removing the thumb the level of mercury in the tube starts falling and at a certain definite height the fall in level stops.

In this state the height of the level of mercury in the tube above the (fee surface of mercury in the trough i. e., the height of the mercury column is nearly 76 cm.
In the space inside the tube above the mercury, column is fully vacuumed, this empty space is called the Torricellian vacuum.

Explanation: Torricelli explained the above phenomenon as follows: The column of mercury in the tube tries to fall due to its own weight while the atmospheric pressure tries to push it upwards. When the pressure of the mercury column equals the atmospheric pressure an equilibrium is achieved and the fall of mercury level stops. In this way by measuring the height of mercury column in the tube the atmospheric pressure can be measured.

Question 14.
What are the factors which effect surface tension? How they effect surface tension?
Answer:
Factor affecting surface tension are :

• Substance dissolved in the liquid: The substance which are highly soluble in the liquid increases the surface tension, while the insoluble substance decreases surface tension.
• Contamination of the surface: Dust particles, oil, grease etc. decrease the surface tension of liquid.
• Temperature of the liquid: Generally on increasing the temperature, the surface tension decreases. According to the formula,
  T = T₀ (1 – αt)
  Where, T= S.T. at t°C, TQ = S.T. at 0°C and α = Temperature coefficient of surface tension.

Question 15.
Explain the phenomenon of capillary rise.
Answer:
Capillary rise: Let a capillary be dipped vertically in a liquid which wets the glass surface. The me¬niscus of the liquid in the capillary will be concave. Suppose the liquid does not rise in the capillary [figure (a)]. Imagine two points A and B at the same horizontal level below the surface of the liquid. Another point C is above the meniscus of the liquid

The pressure on the concave side of the meniscus is more than that on the convex side. This excess of pressure on the concave side is 2T/R, where, T is the surface tension of the liquid and R is the radius of the meniscus. The pressure on the point C is equal to the atmospheric pressure P. Therefore, the net pressure at the point
B = P – \(\frac{2 T}{R}\)

At point A the pressure is equal to the atmospheric pressure P. Points A and B are in the same horizontal level. Therefore, the pressure on both these points must be equal. Clearly, the liquid will rise in the capillary to a height such that the pressure of the liquid column at the point B equals 2 T/R. Thus, the total pressure at the point
B = P – \(\frac{2 T}{R}+\frac{2 T}{R}\) = P.

Question 16.
In strong storm, the roofs of houses fly away, High-velocity Low pressure why?
Answer: At the time of strong wind or storms the air above the roof moves with very high velocity, creating low pressure above the roof. Hence P₂ becomes greater than P₁. Due to the pressure difference P₁ -P₂, a large upthrust acts on the roof and therefore it fly away.

Question 17.
What is critical velocity? What is its relationship with Reynold’s number?
Answer:
The maximum velocity of a fluid, till the flow is streamline is called critical velocity. If the critical velocity is V_c, coefficient of viscosity is η, density of liquid is ρ and diameter of tube is D, then
V_c = \(\frac{K \eta}{\rho D}\)
Where, K is called Reynold’s number.
K is a dimensionless quantity.

Question 18.
Differentiate between the streamlined flow and turbulent flow of liquids.
Answer:
Difference between Streamlined flow and Ttirbulent flow :

Question 19.
State Newton’s law of viscosity. Define coefficient of viscosity and write its unit and dimensional formula.
Answer:
Newton’s law of viscosity: According to this law, the friction force F between the two layers is :
(i) Directly proportional to the area A of the layers, i.e.,
F ∝ A

(ii) Directly proportional to the velocity gradient between the layers, i.e.,
F ∝\(\frac{d v}{d x}\)
Combining both the laws, we get
F ∝ A .\(\frac{d v}{d x} \)
F = – η A. \(\frac{d v}{d x}\)

Where, η is a constant, called coefficient of viscosity.
Negative sign shows that viscous force acts in the opposite direction of flow.
Now, if A = 1 and \(\frac{d v}{d x}\) = 1, then F = – η
Thus, the coefficient of viscosity of a liquid is the tangential force required to maintain a unit velocity gradient between two layers of unit area separated by unit distance. Unit: MKS unit of η is N-s/m².

Question 20.
Find out dimensional formula of coefficient of viscosity.
Answer:
We know that fricition force
F = – ηA \(\frac{d v}{d x}\)
∴ Coefficient of viscosity η = – \(\frac{F}{A \frac{d v}{d x}}\)
or [η] = \(\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^{2}\right]\left[\frac{\mathrm{LT}^{-1}}{\mathrm{~L}}\right]}\) = [ML^-1T^-1].

Question 21.
What is terminal velocity? Calculate the terminal velocity of a sphere falling through a viscous liquid.
Answer:
Terminal velocity: When a ball falls through a liquid then initially the ball is accelerated due to gravity but soon it achieves a constant velocity called terminal velocity.
Formula for terminal velocity: Let a sphere of radius r, density d is falling through a liquid of density ρ and coefficient of viscosity η.
Now, the forces acting on falling body are:

• Weight of the sphere acting vertically downwards.
• Force of buoyancy liquid, acting vertically upward.
• Viscous force acting vertically upwards.

Now, Weight of sphere = Volume × Density × g
= \(\frac{4}{3}\) πr³dg,
Where, r is the radius of ball.
Force of buoyancy = Weight of liquid displaced
= \(\frac{4}{3}\) πr³ρg
∴ Resultant weight of body =\(\frac{4}{3}\) πr³ dg – \(\frac{4}{3}\) πr³ρg
= \(\frac{4}{3}\) πr³ (d – ρ) g

Again, Viscous force = 6 πηrν
When the body is falling with terminal velocity, then Viscous force = Resultant weight
∴ 6 πηrν = \(\frac{4}{3}\) πr³(d – ρ) g
or ν = \(\frac{2}{9} \frac{r^{2}(d-\rho) g}{\eta}\).

Question 22.
What is terminal velocity? On what factors does it depend?
Answer:
Terminal velocity:

Terminal velocity: When a ball falls through a liquid then initially the ball is accelerated due to gravity but soon it achieves a constant velocity called terminal velocity.
Formula for terminal velocity: Let a sphere of radius r, density d is falling through a liquid of density ρ and coefficient of viscosity η.
Now, the forces acting on falling body are:

• Weight of the sphere acting vertically downwards.
• Force of buoyancy liquid, acting vertically upward.
• Viscous force acting vertically upwards.

Now, Weight of sphere = Volume × Density × g
= \(\frac{4}{3}\) πr³dg,
Where, r is the radius of ball.
Force of buoyancy = Weight of liquid displaced
= \(\frac{4}{3}\) πr³ρg
∴ Resultant weight of body =\(\frac{4}{3}\) πr³ dg – \(\frac{4}{3}\) πr³ρg
= \(\frac{4}{3}\) πr³ (d – ρ) g

Again, Viscous force = 6 πηrν
When the body is falling with terminal velocity, then Viscous force = Resultant weight
∴ 6 πηrν = \(\frac{4}{3}\) πr³(d – ρ) g
or ν = \(\frac{2}{9} \frac{r^{2}(d-\rho) g}{\eta}\).

Factors on which it depends :

• It is directly proportional to the square of the radius of the body.
• Directly proportional to the difference of the densities of the liquid and body.
• Inversely proportional to the coefficient of viscosity of the liquid.
• Directly proportional to acceleration due.to gravity.

Question 23.
State and prove principle of continuity.
Answer:
Principle of continuity: When a non-viscous and incompressible liquid flows through a tube of non-uniform bore and its flow is streamline, then the product of velocity of flow and the area of cross-section of the tube at every transverse cross-section remains constant.
i. e., A.ν = a constant
or A₁ν₁ = A₂V₂ .
Proof: Let an incompressible and non-viscous liquid be flowing through a tube XY of non-uniform bore.
A and B are two transverse sections of the tube having areas A₁ and A₂ respectively. Also let the velocity of flow of the liquid at these transverse sections be V₁, and v₂ respectively. Let the density of the liquid bed.
The liquid flowing through the transverse section of A travels a distance v₁ in 1 second, therefore the volume of the liquid flowing per second through section at A is A₁v₁. Hence, the mass of the liquid flowing per second through this section is A₁V₁d.

Similarly, the mass of the liquid flowing through the transverse section at B is A₂v₂d. Since the liquid is incompressible, it cannot accumulate in any portion of the tube. Therefore, whatever amount of liquid enters the segment AB of the tube at section A, the same amount of liquid will flow out of the section at B.
A₁V₁d = A₂v₂d
or A₁v₁ = A₂V₂
or A.v = Constant
This is the principle of continuity.

Question 24.
How many types of energy are there in a flowing liquid? Write the expressions for them.
Answer:
A non-compressible flowing liquid possesses three types of energies :
1. Kinetic energy: If the mass of liquid flowing is m, then its kinetic energy,
K.E = \(\frac{1}{2}\) mv²
If the volume of liquid be V, then K.E. of unit volume of liquid = \(\frac{1}{2}\) \(\frac{m v^{2}}{V}\)
But, \(\frac{m}{V}\) = d. (density)
∴ K.E. of unit volume of liquid =\(\frac{1}{2}\) dv².

2. Potential energy : If a liquid of mass m is at a height h, then its
P.E.= mgh
Where, g is acceleration due to gravity.
∴ P.E of unit volume of liquid = \(\frac{m g h}{V}\) = dgh, (∵ \(\frac{m}{V}\) = d)

3. Pressure energy: For the flow of liquid pressure is applied, with result in pressure energy.
Let the pressure P is applied on a liquid of area of cross-section A, so that liquid flows through a distance of x.
∴ Pressure energy = Work done
= Force × Distance
= Pressure × Area × Distance
= P × A × x =PV, (∵ Ax = V)
∴ Pressure energy per unit volume =\(\frac{P V}{V}\) = P
and pressure energy per unit mass = \(\frac{P V}{V}\) = \(\frac{P}{\frac{m}{V}}\) = \(\frac{P}{d}\).

Question 25.
What is Stoke’s law? What are the conditions of Stoke’s law?
Answer:
According to Stoke, if a spherical body moves through a homogeneous viscous liquid, the viscous force F is :
(i) Directly proportional to the coefficient of viscosity
i.e., F ∝η
(ii) Directly proportional to the radius of body
i.e., F ∝ r
(iii) Directly proportional to the velocity of the body
i.e., F ∝ v
Combining all the laws, we get
F ∝ ηrv
or
F = Kηrv
Where K is constant of proportionality. For spherical ball, K is equal to 6π.
∴ F=6πηrv

Conditions of Stoke’s law :

• The sphere should be very small.
• The sphere should be rigid and smooth.
• The liquid should be extended infinitely.
• The liquid should be viscous and homogeneous.

Mechanical Properties of Fluids Class 11 Important Questions Long Answer Type

Question 1.
State-Bernoulli’s theorem and prove it.
Or
State Bernoulli’s theorem and prove that:
P +\(\frac{1}{2}\) ρv₂ + ρgh = A constant.
Answer:
Bernoulli’s theorem: When a non-viscous and incompressible liquid flows in streamlined condition through a tube then the total energy of its unit mass or unit volume remains constant, i.e.,
For unit volume,
P +\(\frac{1}{2}\) ρv + ρgh = a constant
and For unit mass,
\(\frac{P}{\rho}+\frac{1}{2}\)v₂ +gh = a constant
Where, P – Pressure, p = Density, v = Velocity of liquid flow, g = Acceleration due to gravity, h – Height of liquid above ground level.

Proof: Suppose an incompressible and non-viscous liquid is flowing through a tube XY of non-uniform bore. Its flow is streamlined [See fig.].
A and B are two transverse sections of the tube. The areas of these sections are A₁ and A₂ respectively. At the section A let the velocity of flow of the liquid be v, and its pressure P₁. At the section B let the corresponding quantities be v₂ and P₂ respectively.

Also let the heights of A and B above the ground be h₁ and h₂ respectively and p the density of the liquid.
The force, acting on the liquid at the section A is
F₁=P₁A₁
Therefore, the work done on the liquid entering the cross-section at A in each second
W₁ = F₁ × V₁ = P₁A₁v₁
Similarly, the amount of work done by the liquid coming out of the section at B in each second
W₂ = P₂A₂V₂
∴ Net amount of work done on the liquid
W = W₁ – W₂
= P × A × v₁ – P₂A₂V₂ ……………… (1)
But, from the principle of continuity,
A₁ V₁ = A₂V₂

Where, A₁v₁, or A₂V₂ is the volume of the liquid flowing in each second through any cross-section.
∴ Mass of the liquid flowing any cross-section in one-second
m = A₁v₁ρ= A₂v₂ρ
or
A₁v₁ = \(\frac{m}{\rho}\)
Putting this value in eqn. (1), we get
W = P₁ \(\frac{m}{\rho}\) – P₂\(\frac{m}{\rho}\)
= (P₁ – P₂)\(\frac{m}{\rho}\)
The potential energies of the liquid flowing per second at A and B are mgh₁ and mgh₂ respectively.
∴ Increases in the potential energy of the liquid
= mgh₂ – mgh₁ = mg(h₂-h₁)

The K.E. of the liquid flowing per second at A is \(\frac{1}{2}\)mv₁ ² and at B is \(\frac{1}{2}\) mv₂ ².
∴ Increase in the kinetic energy of the liquid
= \(\frac{1}{2}\) mv₂ ² – \(\frac{1}{2}\)mv₁ ²
= \(\frac{1}{2}\) m(v₂ ² – v₁ ²)

From the principle of conservation of energy,
Work done on the liquid = Increase in potential energy of the liquid + Increase in the kinetic energy of the liquid
∴ (P₁ – P₂)\(\frac{m}{\rho}\) = mg (h₂ – h₁) + \(\frac{1}{2}\)m (v₂² – v₁²)
or
P₁ – P₂ = ρg (h₂ – h₁) + \(\frac{1}{2}\)ρ(v₂² – v₁²)
or
P₁ – P₂ = ρgh₂ – ρgh₁ + \(\frac{1}{2}\)ρ(v₂² – \(\frac{1}{2}\)ρ(v₁²
or
P₁ + ρgh₁+\(\frac{1}{2}\)ρ(v₁² = P₂ + ρgh₂ + \(\frac{1}{2}\)ρ(v₂² ………… (2)
or
P + ρgh + \(\frac{1}{2}\)ρv² ……………… (3)

∴ Therefore, the total energy of unit volume of the liquid remains constant at every point during flow.
From eqn. (3),
\(\frac{P}{\rho}\) + \(\frac{1}{2}\)v² + gh = constant
The total energy of unit mass of the liquid remains constant. From eqn. (2),
P₁ + \(\frac{1}{2}\)ρv₁²+ dgh₁ = P₂ + \(\frac{1}{2}\)ρv₂² + dgh₂
For horizontal tubes : h1 = h2.
∴ P₁ + \(\frac{1}{2}\)ρv₁² = P₂ + \(\frac{1}{2}\)ρv₂²
P + \(\frac{1}{2}\)ρv² = Constant.

Question 2.
In venturi meter the area of cross-sections at two points are A₁ and A₂ and pressure difference is h. Derive an expression for the volume of liquid flowing per second.
Or
Establish the formula for the flow of liquid by venturi meter.
Or
Explain an application of Bernoulli’s theorem.
Answer:
Venturimeter: It is a device based upon the Bernoulli’s theorem. It is used to measure the rate of flow of a liquid through a given pipe.
The construction of a venturi meter is shown in the fig., AB is a horizontal pipe, the middle portion R of it, is constricted. Two vertical tubes M and N are attached to it as shown in the figure. They constitute the manometer arrangement.

When the liquid (say, water) flows through the pipe AB, then in accordance with the principle of continuity, the velocity of flow at the points is smaller and that at/? it is larger. Therefore, according to the Bernoulli’s theorem, the pressure at A will be larger than the pressure at R. This causes the liquid to rise more in-tube Ethan in the tube N. By measuring the difference in the heights of liquid column M and N we can calculate the rate of flow of liquid through pipe.

Let at A the area of cross-section of the pipe be A₁, the velocity of flow V₁ and pressure P₁. At R the area of cross-section of the tube is A₂, the velocity of flow v₂ and the pressure is P₂.
As the pipe is horizontal, according to Bernoulli’s theorem,
P₁ + \(\frac{1}{2}\)ρv₁² = P₂\(\frac{1}{2}\)ρv₂²
or
P₁ – P₂ = \(\frac{1}{2}\) ρ (v₂² – v₁²) ……………… (1)
But, from the principle of continuity,
A₁v₁=A₂V₂ =Q

Where, Q is the volume of the liquid flowing per second i.e., the rate of flow of the liquid.
Therefore,
v₁ = \(\frac{Q}{A_{1}}\) and V₂ =\(\frac{Q}{A_{2}}\)
Putting these values in the eqn. (1), we have
or
P₁ – P₂ = \(\frac{1}{2}\)ρ (\(\frac{Q^{2}}{A_{2}^{2}}-\frac{Q^{2}}{A_{1}^{2}}\) )
or
P₁ – P₂ = \(\frac{1}{2}\) \(\frac{\rho Q^{2}}{A_{1}^{2} A_{2}^{2}}\) (A₁² – A ₂²)

If the difference in levels of the liquid in tubes M and N is h, then
P₁ -P₂ = hpg
Therefore, from eqn. (2),
hpg = \(\frac{1}{2}\) [/latex] \(\frac{\rho Q^{2}}{A_{1}^{2} A_{2}^{2}}\) (A₁² – A ₂²).

Question 3.
State and prove Torricelli’s theorem.
Answer:
Torricelli’s theorem: The velocity of efflux is equal to the velocity that a body acquires when allowed to fall freely from rest through a height equal to the depth of the hole below the free surface of liquid.
Let liquid is filled in a jar, up to height H and from the free surface there is a hole below depth h.
The velocity by the liquid emerges from the hole is called the velocity of efflux.

Now, the K.E. of free surface is zero and P.E. is maximum.
∴ Energy of free surface, per unit volume is
= P+0+ ρgH = P+ ρgH
Where, P is atmospheric pressure and ρ is the density of liquid.

Similarly, the energy per unit volume of liquid at the hole
= P + \(\frac{1}{2}\)ρv²+ ρg (H -h)
Again, by the Bernoulli’s theorem,
P +ρgH=P +\(\frac{1}{2}\)ρv² +pg(H-h)
or
pgH = \(\frac{1}{2}\)ρv² +pgH-pgh
or
\(\frac{1}{2}\)v² =gh
or
v² = 2 gh,
or
v = \(\sqrt{2 g h}\)

Again, if a body is allowed to fall from a height h freely, then u = 0.
∴ v² = u² + 2 gs
or
v² = 0+2gh = 2gh
or v = \(\sqrt{2 g h}\).

Question 4.
What is meant by surface energy of a liquid? Prove that surface energy per unit area is equal to the surface tension.
Or
What is surface energy? Derive an expression for it.
Answer:
Surface energy: The potential energy stored in unit area of free surface of liquid is called its surface energy.
Let PQRS be a rectangular frame of wire. AB is another straight wire which can slide over the arms PQ and RS of the frame as shown in the Fig.

Let the frame be dipped in soap solution and taken out so that a soap film ABRQ is formed. Due to the force of surface tension, the wire AB starts sliding towards QR. If the wire is to be prevented from sliding, a force F must be applied on the wire AB. In equilibrium this force will be equal and opposite to the force of surface tension on the wire AB.
If the length of the wire AB is l and the surface tension of the soap film is T, then F = 2Tl

The factor of 2 appears because soap film has two surfaces.
If the wire AB is displaced through a distance x against the force of surface tension to the position A ‘B’, this will increase the area of the film.
Work done in increasing the area of the film,
W=Fx = 2Tlx ,
or W=T2lx
But, 2lx= A, the total increase in the area of the two surfaces of the film.
∴ W=TA
or
T = \(\frac{W}{A}\)

In the above expression if A = 1, then T= W
Hence, the surface tension of a liquid is numerically equal to the amount of work done in increasing the surface area of the film by unity under isothermal conditions. Thus, surface tension is equal to the surface energy per unit area.

Question 5.
Derive an expression for the excess of pressure inside a drop.
Answer:
Excess of pressure inside a drop: A liquid drop is spherical in shape and its outer surface is convex. Therefore, due to surface tension a net inward force acts on every molecule situated on the surface of the drop. As a result the pressure inside the drop is more than that outside. Let the radius of the drop be R and the excess of pressure inside it be P. Due to this excess of pressure an outward force acts on the surface of the drop.

Suppose due to this excess of pressure, the radius of the drop increases from R to R + ΔR.

Therefore, the work done by the excess of pressure in expansion of the drop,
W = Force × Displacement
= Pressure × Area × Displacement
= P × 4πR² × A R
Increase in the surface area of the drop = Final surface area – Initial surface area
= 4π(R + ΔR)² – 4πR² =8 πRΔR
∵ 4π( ΔR)² will be very small and can be neglected.
∴ Increase in the surface energy
= Surface tension x Increase in the surface area = T ×8 πRΔR
This increase in surface energy is due to the excess of pressure.
∴ P × 4πR² × ΔR = T × 8 πR.ΔR
or P = \(\frac{2 T}{R}\).

Question 6.
Derive an expression for excess pressure inside a bubble.
Answer:
Excess of pressure inside a bubble: Due to air inside it, the bubble has two free surface. One inside and another outside. Let the radius of the bubble be R. Due to excess of pressure P in it, an outward force acts on the surface of the bubble. This is balanced by the force due to the surface tension.
Suppose due to excess of pressure P, let the radius of the bubble increase from R to R +ΔR.
∴ Work done by the excess of pressure
= Force × Distance = Pressure × Area × Distance
= P × 4πR² × ΔR
Since, there are two surfaces in a bubble, the increase in the surface area of the bubble
= 2[4π(R + ΔR)² – 4 πR² ]
= 16πR.ΔR
∴ Increase in surface energy = T×16πR.ΔR
The increase in surface energy is due to the excess of pressure.
∴ P × 4πR² × ΔR = T × 16 πR.ΔR
or P =\(\frac{2 T}{R}\).

Question 7.
What is meant by capillarity? Give two examples of capillarity in daily life. Establish the formula to determine surface tension of water by capillary rise.
Or
Derive the formula to determine the surface tension in the laboratory.
Answer:
Capillarity: The phenomenon of ascenting and descenting of liquid in a capillary is called capillarity. In kerosene lamp oil rises in the wick due to capillarity.
Formula derivation: Let a uniform capillary tube of radius r be immersed into water and water rises up to height h. The meniscus is concave in the tube and its ra¬dius is nearly equal to the radius of capillary. The surface tension of liquid acts at the point of contact of meniscus along the direction of tangent, making an angle θ with the wall of tube.

Now, according to Newton’s third law, reaction force R is equal to T.
Resolving R we get horizontal components R sin θ = T sin θ acting diametrically opposite, around the meniscus, hence their resultant will be zeros.
And vertical component R cos θ= T cos θ acting upwards around the circumference of meniscus, due to which the water rises.
The total force which rises the water
= Tcosθ × 2 πr …………. (1)
(2 πr has been multiplied because it is the total circumference on which T cosθ acts)

Now, the upward force is balanced by the weight of the water.
Now, volume of water raised = Volume of water to height h + Volume of water in meniscus.
Where, volume of water in meniscus = Volume of cylinder of radius and height r -Volume of hemisphere filled with air
= πr²r – \(\frac{2}{3}\) πr³
∴ Volume of water raised = πr² h + (πr² r – \(\frac{2}{3}\) πr³)
= πr²h + \(\frac{1}{3}\)πr³

If the density of water is ρ then weight of water acting vertically downwards.
W = mg – Volume × Density × g
=( πr²h + \(\frac{1}{3}\)πr³)ρg
= πr²(h +\(\frac{1}{3}\)r) ρg

Now, in equilibrium position,
Total force upwards = Total force downwards
T cosθ.2πr = πr²(h+ \(\frac{1}{3}\)r)ρg
or T = \(\frac{r\left(h+\frac{1}{3} r\right) \rho g}{2 \cos \theta}\)
Since,\(\frac{r}{3}\) is negligible in comparison to h.
∴T = \(\frac{r h \rho g}{2 \cos \theta}\)

But, for pure water and clean glass θ = 0°
∴ cos θ = cos 0° = 1
∴ T = \(\frac{r h \rho g}{2}\)
This is the required formula.

Mechanical Properties of Fluids Class 11 Important Numerical Questions

Question 1.
A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor? (NCERT)
Solution:
Given, m = 50kg; 2r = 1.0cm ⇒ r = 0.5 cm = 0.5 × 10^-2 m
F = mg = 50 × 9.8 = 490N.
Now pressure,
P =\(\frac{F}{A}\) = \(\frac{F}{\pi r^{2}}\) = \(\frac{490}{3 \cdot 14\left(0 \cdot 5 \times 10^{-2}\right)^{2}} \)
= \(\frac{490 \times 10^{4}}{0 \cdot 785}\)
= 624.20 × 10⁴
= 6.24 × 10⁶ N/m².

Question 2.
Torricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kgm^-3. Determine the height of the wine column for normal atmospheric pressure.(NCERT)
Solution:
Given, d = 984kg/m³; g = 9.8ms ^-2, P = 1.013 × 10⁵Pa
Formula, P = hdg
∴ h = \(\frac{P}{d g}\) = \(\frac{1 \cdot 013 \times 10^{5}}{984 \times 9 \cdot 8}\) =10.5m.

Question 3.
In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 ms^-1 and 63 ms^-1 respectively. What is the lift on the wing if its area is 2.5 m²? Take the density of air to be 1.3 kg m^-3. (NCERT)
Solution:
Given, ν₁ = 70ms^-1 ;ν₂ = 63ms^-1;A = 2.5m² , d = 1.3 kgm^-3
According to Bernoulli’s theorem,
\(\frac{1}{2} d v_{1}^{2}\) + P₁ = \(\frac{1}{2} d v_{2}^{2}\) +P₂
or
P₁ – P₂ = \(\frac{1}{2} \)d (ν₂² – ν₁²)A
∴ Applied thrust force will be
F = (P₁ – P₂)A
= \(\frac{1}{2}\)d (ν₂² – ν₁²)A
= \(\frac{1}{2}\) ×1.3 [ (63)² – (70)²] × 2.5
= 0.5 × 1.3 [3969 – 4900] × 2.5
= -1.5 × 10³N.

Question 4.
What is the pressure inside the drop of mercury of radius 3-00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 × 10^-1 Nm^-1. The atmospheric pressure is 1.01 × 10⁵ Pa. Also give the excess pressure inside the drop. (NCERT)
Solution:
Given, R = 3.00mm = 3 × 10³m, T = 4.65 × 10^-1Nm –¹
Atmospheric pressure = 1.01 × 10⁵Pa
Formula: ρ = \(\frac{2 T}{R}\)
Or ρ = \(\frac{2 \times 4 \cdot 65 \times 10^{-1}}{3 \times 10^{-3}}\) = 310Pa.

Total pressure inside the drop = Atmospheric pressure + Excess pressure
= 1.01 × 10⁵ +310
= 1.01 × 10⁵ +0.00310 × 10⁵
= 1.01310 × 10⁵Pa
= 1.01 × 10⁵ Pa (UP to three significant digits).

Question 5.
At what depth in the water the pressure will be double the atmospheric pressure ?
Solution:
Let the depth = hm
∴ Pressure of Am depth water + Atmospheric pressure
= 2 × Atmospheric pressure
or Pressure of water at h m depth = Atmospheric pressure
or h × 1 × 10³ × 9.8 = 1 × 10⁵(∵ P = hρg)
or h = \(\frac{10^{5}}{10^{3} \times 9 \cdot 8}\) = 10.20m.

Question 6.
The weight of a piece of lead in water is 100 g. Find out its weight in air if relative density of lead is 11.3.
Solution:
Let die weight of lead in air be x
The weight of lead in air = 100 g.
The loss in its weight in water = x – 100

Relative density of lead = \(\frac{\text { The weight of lead in air }}{\text { The loss in its weight in water }} \)
∴ \(\frac{11 \cdot 3}{1}\) = \(\frac{x}{x-100}\)
or 11.3x – 11.30 = x
or 10.3x = 1130
x = 109.7 gm.

Question 7.
The surface tension of a liquid is 70 × 10.3 Nm^-1. A needle of 5 cm is floating on it. Calculate the lateral force acting on the pin.
Solution:
Given : T = 70 × 10³Nm^-1, l = 5 cm = 5 x 10^-2m.
Now we have T = \(\frac{F}{l}\)
F=T.l
= 70 × 10^-3 × 5 × 10 ^-2
= 3.5 × 10^-3N.

Question 8.
The radius of a soap bubble is 0.5 cm. If the surface tension of soap solution is 0.2 Nm^-1, then find the surface energy of the bubble.
Solution:
Given : R = 0.5cm = 0.5 × 10^-2m, T = 0.2Nm ^-1.
A = 4πR² = 4 × 3.14 × (0.5 × 10^-2)²
= 12.56 × 0.25 × 10^-4m²
Now, Surface energy,
W=2TA,
= 2 × 0.2 × 12.56 × 0.25 × 10^-4 = 1.256 × 10^-4 joule.

Question 9.
Calculate the work done in blowing a bubble of soap solution of radius 5cm. Surface tension of soap solution is = 3 × 10 ^-2Nm^-1
Solution:
Given : r = 5 cm =\(\frac{5}{100}\) = \(\frac{1}{20}\)m,
T = 3 × 10^-2 Nm^-1
Since, the bubble has two surfaces.
∴ A = 2 × 4πr²
or
A = 8π\(\left(\frac{1}{20}\right)^{2}\) = 628 × 10^-4m²
Now,
W = T.A
= 3 ×10^-2 × 628 x 10^-4
= 1884 x 10^-6 = 0.001884 joule.

Question 10.
Radius of a soap bubble is r and its surface tension is T. At constant temperature the radius of bubble is blown doubled, calculate the required force.
Solution:
ΔA = 2 [4πr²₂ – 4πr²₂ ]
= 8π (r²₂ – r²₂ )
Given:
r₁=r and r₂ = 2r.
ΔA = 8π (4r² -r²)
= 24πr²
∵ W=T.ΔA
∴ W=T.24 πr² = 24 πr². T.

Question 11.
A drop of diameter 2.8 mm is broken into 125 droplets of equal radii. If the surface tension of liquid is 75 dyne × cm^-1, then calculate the change in energy.
Solution:
Given :
2R = 2.8 mm
or
R = 1.4 mm
= 1.4 × 10^-3m
and, T = 75 dyne × cm^-1
= 75 × 10^-5 N × (10^-2 m)^-1
∴ T = 75 × 10^-3 Nm-1

Now, Volume of big drop = Volume of 125 droplets.
∴ \(\frac{4}{3} \pi R^{3}\) = 125 × \(\frac{4}{3} \) πr³
or R3 = 125r³
or
R = 5r
or
r = \(\frac{R}{5}\)
∴ Surface area of bigger drop = 4πR²
and Surface area of 125 droplets = 125 × 4πr²
= 125 × 4π × \(\left(\frac{R}{5}\right)^{2}\)
∴ Δ A = 20πR² – 4πR² = 16 πR²
∴ Change in energy,
ΔW = T.Δ A
= 75 × 10^-3 × 16 × πR²
= 75 × 10^-3 × 16 × \(\frac{22}{7}\) × (1.4 × 10 ^-3)²
= 7392 × 10^-9
= 7.392 × 10^-9 joule.

Question 12.
The surface tension of soap solution is l-9xl0-2 Nm”1. Calculate the excess of pressure inside the bubble of diameter 2 cm.
Solution:
Given : T = 1.9 × 10^-2 Nm^-1
R = \(\frac{2}{2}\) cm = 1cm = 1 × 10^-2m.
Now, ρ = \(\frac{4 T}{R}\)
or
ρ = \(\frac{4 \times 1 \cdot 9 \times 10^{-2}}{10^{-2}}\) = 7.6 Nm^-2.

Question 13.
Water is flowing through a tube of non-uniform cross-section. The velocity at a point A is 4 times that of another point B. Compare the diameters of tube at A and B.
Solution:
Given: ν₁ = 4v₂
We have, by the principle of continuity,

Question 14.
The diameter of a spherical ball A is half of that of B. What will be the ratio of their terminal velocities in water?
Solution:
Given: a1 = \(\frac{a_{2}}{2}\)
We have,
ν ∝ a²
or
\(\frac{v_{1}}{v_{2}}\) = \(\frac{a_{1}^{2}}{a_{2}^{2}}\)
∴ \(\frac{v_{1}}{v_{2}}\) = \(\frac{\left(a_{2} / 2\right)^{2}}{a_{2}^{2}}\) = \(\frac{1}{4}\)
∴ ν₁ :ν₂ = 1:4

Question 15.
Water is flowing through a horizontal tube of non-uniform area of cross-section. At a place where the pressure is 20 cm (of water) the velocity of flow of water is 60 cm/s. Find the velocity of flow at a place where the pressure is 15 cm (of water).
Solution:
Given, P₁ = 20 cm(of water) =20 × 1 × 980 dyne /cm², v, = 60 cm/s,
P₂ = 15 × 1 × 980 dyne/cm² and ρ= 1 gm/cm3.

By the Bernoulli’s theorem,
\(\frac{1}{2}\) ρ (ν₂² – ν₁²) = P₁ – P₂
⇒ \(\frac{1}{2}\) × 1 × (ν₂² – 60²) = 20 × 1 × 1980 – 15 × 1 × 1980
⇒ \(\frac{1}{2}\) (ν₂² – 60²) = 5 × 980
⇒ ν₂² = 9800 + 3600 = 13400
∴ ν₂ = 115.76 cm/s.

Question 16.
Calculate the terminal velocity of a spherical ball of diameter 2mm, in a viscous liquid. The density of liquid 1.0 × 10³ kg m^-3 and density of ball is 8.0 × 10³kg m³ and coefficient of viscosity of liquid is 1.0 kg m^-1s^-1.
Solution:
Given : 2a = 2mm or a = 1mm = 1 x 10^-3m,
ρ = 1.0 × 10 ³kg m^-3 ,d = 8.0 × 10³ kg m^-3, η = 1.0 kg m^-1s^-1

Now, ν = \(\frac{2}{9} \cdot \frac{a^{2}(d-\rho) g}{\eta}\)
= \(\frac{2}{9}\) . \(\frac{\left(1 \times 10^{-3}\right)^{2}\left(8 \cdot 0 \times 10^{3}-1 \cdot 0 \times 10^{3}\right) 9 \cdot 8}{1 \cdot 0}\)
= \(\frac{2}{9}\) × 10^-6 × 7 × 10³ × 9.8
= 15.2 ×10^-3 ms^-1

Question 17.
A drop of water radius 0.0015 mm is falling in the air. If the coefficient of viscosity of air is 1.8 × 10 ^-6 kgm^-1s^-1 then calculate the terminal velocity of drop. The density of air is negligible.
Solution:
Given : a = 0.0015 mm = 0.0015 × 10^-3m = 1.5 × 10⁶m,
d = 1 × 10³ kgm^-3, ρ = 1.0 × 10³ kg m^-3, η = 8.0 × 10^-6 kg m^-1 s^-1

Formula,
ν = \(\frac{2}{9} \cdot \frac{a^{2}(d-\rho) g}{\eta}\)
= \(\frac{2}{9}\) .\(\frac{\left(1 \cdot 5 \times 10^{-6}\right)^{2} \times 1 \times 10^{3} \times 9 \cdot 8}{1 \cdot 8 \times 10^{-6}}\)
= \(\frac{2}{9}\) \(\frac{2 \cdot 25 \times 10^{-12} \times 9 \cdot 8 \times 10^{3}}{1 \cdot 8 \times 10^{-6}}\)
= 2.72 × 10^-3ms^-1.

Question 18.
Two horizontal pipes of different diameters are joined together, water is flowing through them. The velocity and pressure through one pipe are 4 ms^-1 and 2.0 × 10^-3Nm^-2 respectively. The diameters of pipes are 3 cm and 6 cm. Calculate the velocity and pressure in second pipe.
Solution:
Given :
v₁ = 4 ms^-1 , P ¹ = 2.0 × 10⁴ Nm^-1
r₁cm = 1.5 × 10^-2m, r² = \(\frac{6}{2}\) = 3cm = 3 × 10^-2
Now we have
A₁ v₁ = A₂v₂
or πr₁ ²v₁ = πr₁ ²v₂
or
r₁ ²v₁ = r₁ ²v₂

∴ ν₂ = \(\frac{r_{1}^{2} \cdot v_{1}}{r_{2}^{2}} \)
= \(\frac{\left(1 \cdot 5 \times 10^{-2}\right)^{2} \times 4}{\left(3 \times 10^{-2}\right)^{2}}\)
= 1 ms^-1
Again, P₂ – P₁ = \(\frac{1}{2}\) ρ (ν₁² – ν₂²)
or P₂ – 2 × 10⁴ = \(\frac{1}{2}\) × 10³ (4² – 1²)
or
P₂ = \(\frac{1}{15}\) × 10³ + 2 × 10⁴
= 2.75 × 10 ⁴ Nm^-2

Question 19.
What will be the maximum velocity of water for streamline flow through a tube of diameter 2 cm? (η= 10³ kgm^-1s^-1).
Solution:
Given: D = 2cm = 2 × 10²m. η= 10³ kgm^-1s^-1
K=2000,
ρ= 10 kgm^-3
Now, ν_c = \(\frac{K \eta}{\rho D}\)
or ν_c = \(\frac{2000 \times 10^{-3}}{10^{3} \times 2 \times 10^{-2}}\) = 0.1ms^-1.

Mechanical Properties of Fluids Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
The sudden fall of atmospheric pressure indicates :
(a) Storm
(b) Rainfall
(c) Clear weather
(d) Coldwave.
Answer:
(a) Storm

Question 2.
On which principle the Hydraulic lift work :
(a) Pascal’s law
(b) Archimedis principle
(c) Boyle’s law
(d) Bernoulli’s theorem.
Answer:
(a) Pascal’s law

Question 3.
Systolic and diastolic blood pressure of a healthy human being is :
(a) 140 mm/80 mm of Hg
(b) 120 mm/80 mm of Hg
(c) 160 mm/90 mm of Hg
(d) 80 mm/ 120 mm of Hg.
Answer:
(b) 120 mm/80 mm of Hg

Question 4.
On which condition liquid flowing through a tube :
(a) When high density and more viscous liquid flows through less radius tube
(b) When low density and more viscous liquid flows through less radius tube
(c) When high density and less viscous liquid flows through more radius tube
(d) When less density and less viscous liquid flows through more radius tube.
Answer:
(b) When low density and more viscous liquid flows through less radius tube

Question 5.
On increasing temperature viscosity of liquid :
(a) Increases
(b) Becomes zero
(c) Decreases
(d) No effect.
Answer:
(c) Decreases

Question 6.
Bernoulli’s equation is based on which law :
(a) Conservation of momentum
(b) Conservation of energy
(c) Conservation of mass
(d) No law.
Answer:
(b) Conservation of energy

Question 7.
Rain droplet is spherical in shape, its reason is :
(a) Gravitational force
(b) Viscosity
(c) Atmospheric pressure
(d) Surface tension.
Answer:
(d) Surface tension.

Question 8.
On increasing temperature, surface tension of liquid :
(a) Increases
(b) Becomes zero
(c) Decreases
(d) No effect.
Answer:
(c) Decreases

Question 9.
At critical temperature, the surface tension of the liquid is :
(a) Zero
(b) Infinity
(c) Same as that any other temperature
(d) Cannot be determined.
Answer:
(a) Zero

Question 10.
The shape of meniscus of mercury in capillary tube is :
(a) Convex
(b) Cancave
(c) Flat
(d) Not definite.
Answer:
(a) Convex

Question 11.
Magnitude of angle of contact due to which surface of glass become wet is :
(a) 0°
(b) 90°
(c) More than 90°
(d) Less than 90°.
Answer:
(d) Less than 90°

Question 12.
On dipping glass capillary inside a mercury :
(a) Level of mercury in capillary tube increases
(b) Level of mercury in capillary tube increases and overflow
(c) Level of mercury in capillary tube falls
(d) Level of mercury in capillary tube falls and does not rise.
Answer:
(c) Level of mercury in capillary tube falls

Question 13.
Which phenomenon is not related to surface tension :
(a) Dancing of piece of camphor in water surface
(b) Spherical shape of small mercury drop
(c) The liquid become stable after jerking in a vessel
(d) None of these.
Answer:
(c) The liquid become stable after jerking in a vessel

2. Fill in the blanks:

1. Pressure exerted by one-millimeter column of mercury is called ……………………. .
Answer:
one tor

2. Blood pressure at leg of a human is ……………………. in comparison to his brain.
Answer:
more

3. In generally surface tension of a liquid ……………………. on increasing temperature.
Answer:
decreases

4. Viscosity of gas ……………………. and viscosity of liquid ……………………. on increasing temperature.
Answer:
increases, decreases

5. On mixing detergent in water, its ……………………. decreases.
Answer:
surface tension

6. When the magnitude of Reynold’s number is between 0 to 2000, the flow of liquid is ……………………. .
Answer:
streamline

7. ……………………. type of energy is associated with flowing liquid.
Answer:
three

8. Poisson’s is C.G. S. unit of ……………………. .
Answer:
coefficient of viscosity

9. The liquid which is incompressible is known as ……………………. liquid.
Answer:
ideal

10. Mercury does not wet glass because cohesive force is ……………………. than adhesive force.
Answer:
less.

3. Match the following:

I.

Answer:
1.  (c) Capillarity
2. (a) Use of detergent,
3 (b) Pressure is less
4. (e) Pressure is more
5. (d) Cohesive force.

II.

Answer:
1. (d) \( \eta A \frac{d v}{d x} \)
2. (a) \( \frac{K \cdot \eta}{\rho \cdot D} \),
3. (e) \( \frac{2}{9} \cdot \frac{r^{2}(d-\sigma) g}{\eta} \)
4. (b) A.v = constant
5. (c) P+ \( \frac{1}{2} d v^{2} \)+ dgh = constant.

4. state true or False:

1. Hydrostatic pressure is vector quantity.
Answer:
False

2. Surface tension of a liquid does not depend upon area of cross-section of the surface.
Answer:
True

3. Angle of contact of mercury with glass is more.
Answer:
True

4. The man who jump with parachute, its velocity increases initially and then become constant.
Answer:
True

5. Viscosity of gas decreases on increasing temperature.
Answer:
False

6. Surface tension is numerically equal to surface energy.
Answer:
True

7. Greece is more viscous than honey.
Answer:
False

8. In flowing liquid, streamline of layer cannot intersect each other.
Answer:
False.

MP Board Class 10th Hindi Navneet Solutions पद्य Chapter 5 प्रकृति चित्रण

प्रकृति चित्रण अभ्यास

बोध प्रश्न

प्रकृति चित्रण अति लघु उत्तरीय प्रश्न

प्रश्न 1.
कवि के अनुसार ‘वनों और बागों में किसका विस्तार है?
उत्तर:
कवि के अनुसार ‘वनों और बागों’ में बसन्त ऋतु का विस्तार है।

प्रश्न 2.
तन, मन और वन में परिवर्तन किसके प्रभाव से दिखाई दे रहा है?
उत्तर:
तन, मन और वन में परिवर्तन बसन्त ऋतु के प्रभाव से दिखाई दे रहा है।

प्रश्न 3.
कन्हैया के मुकुट की शोभा क्यों बढ़ गई है?
उत्तर:
शरद ऋतु की चाँदनी की छवि पाकर कन्हैया के मुकुट की शोभा बढ़ गई है।

प्रश्न 4.
कवि ने हिमालय की झीलों में किसको तैरते हुए देखा है?
उत्तर:
कवि ने हिमालय की झीलों में कमल नाल को खोजने वाले तैरते हुए हंसों को देखा है।

प्रश्न 5.
कवि ने बादलों को कहाँ घिरते देखा है?
उत्तर:
कवि ने मानसरोवर झील के समीप हिमालय की ऊँची चोटियों पर बादलों को घिरते देखा है।

प्रश्न 6.
कौन अपनी अलख नाभि से उठने वाले परिमल के पीछे-पीछे दौड़ता है?
उत्तर:
कस्तूरी मृग अपनी अलख नाभि से उठने वाले परिमल के पीछे-पीछे दौड़ता है।

प्रकृति चित्रण लघु उत्तरीय प्रश्न

प्रश्न 1.
बसन्त ऋतु के आगमन पर प्रकृति में कौन-कौन से परिवर्तन होते हैं?
उत्तर:
बसन्त ऋतु के आगमन पर कुंजों में भौरे गुंजार करने लगते हैं, उनके झुण्ड के झुण्ड आम के बौरों पर चक्कर लगाने लगते हैं। विहग (पक्षी) समाज में तरह-तरह की आवाजों के द्वारा बसन्त की खुशी का वर्णन होने लगता है। ऋतुराज के आने से प्रकृति में भाँति-भाँति के राग-रंग बिखरने लगते हैं।

प्रश्न 2.
यमुना तट पर किसी छटा बिखरी हुई है?
उत्तर:
यमुना तट पर अखण्ड रासमण्डल की छटा बिखरी

प्रश्न 3.
कवि के अनुसार ‘घनेरी घटाएँ’ क्या कर रही
उत्तर:
कवि के अनुसार घनेरी घटाएँ घुमड़-घुमड़ कर गर्जना करने लग जाती हैं और फिर उनकी गर्जना थमने का नाम नहीं लेती।

प्रश्न 4.
‘निशाकाल से चिर अभिशापित’ किसे कहा गया है?
उत्तर:
निशाकाल से चिर अभिशापित चकवा-चकवी को कहा गया है। कवियों की मान्यता है कि चकवा-चकवी का रात के समय वियोग हो जाता है। यह वियोग उन्हें किसी शाप के द्वारा प्राप्त हुआ है।

प्रश्न 5.
कवि ने भीषण जाड़ों में किससे ‘गरज-गरज कर भिड़ते देखा है?
उत्तर:
कवि ने भीषण जाड़ों में महादेव को झंझावात से गरज-गरज कर भिड़ते देखा है।

प्रकृति चित्रण दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
पद्माकर ने वर्षा ऋतु में प्रकृति का चित्रण किस तरह किया है? उल्लेख कीजिए।
उत्तर:
पद्माकर ने वर्षा ऋतु में चारों ओर चमकने वाली बिजली और शीतल, मन्द सुगन्धित वायु के बहने की तथा घनेरी घटाओं के घुमड़-घुमड़ कर गर्जन करने का वर्णन किया है।

प्रश्न 2.
कवि के अनुसार बसन्त का प्रभाव कहाँ-कहाँ दिखाई दे रहा है?
उत्तर:
कवि के अनुसार बसन्त का प्रभाव कूलों में, केलि में, कछारों में, कुंजों में, क्यारियों में और सुन्दर किलकती हुई कलियों में, पत्तों में, कोयल में, पलासों में, समस्त द्वीपों में और ब्रज के बाग-बगीचों तथा युवतियों में देखा जा सकता है।

प्रश्न 3.
नागार्जुन ने कवि कल्पित’ सन्दर्भ किसे माना है? स्पष्ट कीजिए।
उत्तर:
नागार्जुन ने कवि कल्पित सन्दर्भ मेघदूत को माना है। अलकापुरी से जब यक्ष को निर्वासित कर दिया गया तो वह अपनी प्रेमिका के वियोग में दुःखी रहने लगा। वर्षा ऋतु के आने पर यक्ष ने बादल को दूत बनाकर अर्थात् मेघदूत के माध्यम से ही अपनी प्रेमिका को प्रेम-सन्देश भेजा है।

प्रश्न 4.
नागार्जुन के अनुसार बसन्त ऋतु के उषाकाल का वर्णन कीजिए।
उत्तर:
नागार्जुन के अनुसार बसन्त ऋतु के उषाकाल का वर्णन इस प्रकार है-
बसन्त ऋतु का सुप्रभात था, उस समय मन्द-मन्द गति से हवा चल रही थी, सूर्य की प्रातःकालीन कोमल किरणें पर्वत शिखरों पर स्वर्णिम आभा में गिर रही थीं। रात में चकवा-चकवी का वियोग क्रन्दन सुनाई पड़ता था तथा सरवर के किनारे काई की हरी दरी पर प्रात:काल में उनको प्रणय आलाप करते देखा है।

प्रश्न 5.
निम्नलिखित काव्यांशों की सप्रसंग व्याख्या कीजिए
(अ) कौन बताए वह छायामय ………… गरज-गरज भिड़ते देखा है।
उत्तर:
कविवर नागार्जुन कहते हैं कि इस कैलाश शिखर पर धनपति कुबेर की अलका नगरी का पुराणों में उल्लेख मिलता है लेकिन आज तो इसका यहाँ कोई नामोनिशान भी नहीं है, न ही हमें कालिदास के आकाश में विचरण करने वाले उस मेघदूत का कहीं पता मिलता है। हमने बहुत प्रयत्न कर लिए पर वह ढूँढने से भी नहीं मिला। ऐसा हो सकता है कि वह छायामय मेघदूत यहीं कहीं जाकर बरस गया होगा। फिर कवि कहता है कि इन बातों को छोड़ो, मेघदूत तो कवि कालिदास की कल्पना थी।

मैंने तो भयंकर कड़कड़ाते जाड़ों में आकाश को चूमने वाले कैलाश के शिखर पर घनघोर झंझावातों में महादेव जी को उनसे गरज-गरज कर युद्ध करते देखा है। बादल को घिरते देखा है।

(ब) और भाँति कुंजन ………… और बन गए।
उत्तर:
कविवर पद्माकर कहते हैं कि बसन्त ऋतु में कुंजों में और ही प्रकार की सुन्दरता छा गई है तथा भरों की भीड़ में अनौखी गुंजार सुनाई दे रही है। डालियों, झुण्डों तथा आम के बौरों में नयी आभा आ गई है। गलियों में और ही भाँति की शोभा छा गयी है पक्षियों के समाज में भी और ही प्रकार की बोलियाँ सुनाई दे रही हैं। ऐसे श्रेष्ठ ऋतुराज बसन्त के अभी दो दिन भी नहीं चुके हैं फिर भी सर्वत्र और ही प्रकार का रस, और ही प्रकार की रीति, और ही प्रकार के राग तथा और ही प्रकार के रंग सर्वत्र छाये हुए हैं। इस ऋतु में मनुष्यों के शरीर और ही प्रकार के हो गए हैं, और ही प्रकार के मन तथा और ही प्रकार के वन और बाग हो गए हैं।

प्रकृति चित्रण महत्त्वपूर्ण वस्तुनिष्ठ प्रश्न

प्रकृति चित्रण बहु-विकल्पीय प्रश्न

प्रश्न 1.
‘कहाँ गया धनपति कुबेर वह’ पंक्ति पाठ्य-पुस्तक की किस कविता से ली गयी है? (2009)
(क) नीति-अष्टक
(ख) श्रद्धा
(ग) पथ की पहचान
(घ) बादल को घिरते देखा है।
उत्तर:
(घ) बादल को घिरते देखा है।

प्रश्न 2.
कन्हैया के मुकुट की शोभा किसके द्वारा बढ़ गई है?
(क) नगों द्वारा
(ख) शरद की चाँदनी से
(ग) तारों की आभा से
(घ) सोने का होने के कारण
उत्तर:
(ख) शरद की चाँदनी से

प्रश्न 3.
कवि ने बादलों को कहाँ घिरते देखा है? (2010, 17)
(क) नवल धवल गिरि पर
(ख) तालाबों में
(ग) चट्टानों में
(घ) बागों में।
उत्तर:
(क) नवल धवल गिरि पर

प्रश्न 4.
निशाकाल से चिर अभिशापित किसे कहा गया है?
(क) मृगों को
(ख) नायक-नायिका को
(ग) चकवा-चकवी को
(घ) भौंरों को।
उत्तर:
(ग) चकवा-चकवी को

प्रश्न 5.
हिमालय की झीलों में कवि नागार्जुन ने तैरते हुए देखा है (2015)
(क) मछली को
(ख) मृग को
(ग) हंसों को
(घ) नारी को।
उत्तर:
(ग) हंसों को

रिक्त स्थानों की पूर्ति

1. अमल धवल गिरि के शिखरों पर ……… को घिरते देखा है।
2. वर्षा ऋतु का वर्णन पद्माकर ने ….. को बढ़ाने वाली ऋतु के रूप में किया है।
3. “बीथिन में ब्रज में नवेलिन में बेलनि में बगन में बागन में बगर्यो …….. है।”
4. ……… की हरी दरी पर प्रणय-कलह छिड़ते देखा है।
5. हिमालय की झीलों में कवि नागार्जुन ने ………. को तैरते देखा है। (2011)

उत्तर:

1. बादल
2. विरह
3. बसन्त
4. शैवालों
5. हंसों।

सत्य/असत्य

1. ताप तीन प्रकार के होते हैं – दैहिक, दैविक और भौतिक।
2. ‘ऋतु वर्णन’ कविता में कविवर पद्माकर ने वर्षा ऋतु का सुन्दर वर्णन किया है।
3. हिमालय पर बादलों की स्थिति सदैव एक-सी रहती है।
4. ‘बादल को घिरते देखा है’ कविता के कवि तुलसीदास हैं। (2009)

उत्तर:

1. सत्य
2. असत्य
3. असत्य
4. असत्य

सही जोड़ी मिलाइए

उत्तर:
1. → (ख)
2. → (ग)
3. → (क)
4. → (घ)

एक शब्द/वाक्य में उत्तर

1. पद्माकर कवि को किस काल का कवि माना जाता है?
2. ‘ऋतु वर्णन’ में कवि ने प्रमुख रूप से इन पदों में किस ऋतु का वर्णन किया है।
3. पद्माकर कवि ने विरह को बढ़ाने वाली ऋतु कौन-सी मानी है?
4. चकवा और चकवी कब बेबस हो जाते हैं?
5. कवि नागार्जुन ने हिमालय की झीलों में किसको तैरते देखा? (2012)

उत्तर:

1. रीतिकाल
2. शरद ऋतु,बसन्तु ऋतु तथा वर्षा ऋतु
3. वर्षा ऋतु
4. रात्रि में
5. हंसों को।

ऋतु वर्णन पाठ सारांश

ऋतु वर्णन में पद्माकर कवि ने वसन्त ऋतु की महिमा का गुणगान किया है। यमुना के तट पर,गोपियों की रासलीला का सुन्दर वर्णन किया है। वसन्त ऋतु में कोयल की कूक का मधुर स्वर गूंजता सुनायी पड़ता है। विभिन्न प्रकार के पुष्प खिले रहते हैं तथा उन पुष्पों की सुगन्ध दूर-दूर तक फैली रहती है। बागों में पुष्पों की सुगन्ध से आकर्षित होकर भौरे गुंजार करते हैं।

आम के वृक्ष मंजरियों से भर गये हैं। प्रकृति के हरा-भरा होने से पक्षी भी मस्त होकर कलरव कर रहे हैं। कवि ने शरद ऋतु की अपूर्व शोभा का सुन्दर,वर्णन किया है। शरद ऋतु की चाँदनी चारों ओर व्याप्त है। लताओं पर तथा तमाल के वृक्षों पर चारों ओर शरद ऋतु का सौन्दर्य छाया है। शरद ऋतु में चाँदनी दिग्दिगन्त को अपनी आभा से मंडित कर रही है।

ऋतु वर्णन संदर्भ-प्रसंगसहित व्याख्या

(1) कुलन में केलि में कछारन में कंजन में,
क्यारिन में कलित कलीन किलकत है।
कहैं पद्माकर परागन में पौन में,
पातन में पिक में पलासन पतंग है।
द्वार में दिसान में दुनी में देस देसन में,
देखौ दीप दीपन में दीपत दिगंत है।
बीथिन में ब्रज में नवेलिन में बेलनि में,
बनन में बागन में बगर्यो बसंत है।

शब्दार्थ :
कूलन = नदी के किनारों में। कलित = सुन्दर। केलि क्रीड़ा में। कलीन-कलियों में। किलकंत=किलकारी मारता है। पौन = पवन, हवा। पातन = पत्तों। पिक = कोयल। दुनी = दुनिया। दीप = द्वीप। दीपत = दीप्त। दिगंत = दिशाओं में। बीथिन = गलियों में। नवेलिन = नवयुवतियों में। बगर्यो = फैला हुआ है।

सन्दर्भ :
प्रस्तुत छन्द ‘प्रकृति चित्रण’ के अन्तर्गत ‘ऋतु-वर्णन’ शीर्षक से लिया गया है। इसके रचयिता श्री पद्माकर हैं।

प्रसंग:
इस छन्द में कवि ने बसन्त की चारों ओर फैली हुई मादकता का वर्णन किया है।

व्याख्या :
कविवर पद्माकर कहते हैं कि नदी के कूलों में क्रीड़ा स्थलों में, कछारों में, कुंजों में, क्यारियों में, सुन्दर कलियों में बसन्त किलकारी मार रहा है। पुष्पों के पराग में, पवन में, पत्तों में, कोयल में और पलाशों में बसन्त पगा हुआ है। घर के द्वारों में दिशाओं में, दुनिया में, देशों में, द्वीपों में और दिशाओं में बसन्त दीप्तमान हो रहा है। गलियों में, ब्रजमण्डल में, नवयुवतियों में बेलों में, वनों में, बागों में चारों ओर बसन्त की बहार छाई हुई है।

विशेष:

1. कवि ने बसन्त की मादक छटा का सम्पूर्ण संसार में प्रसार दिखाया है।
2. अनुप्रास अलंकार की सुन्दर छटा।
3. भाषा-ब्रज।

(2) और भाँति कुंजन में गुजरत भीरे भौंर,
और डौर झौरन पैं बोरन के वै गए।
कहै पद्माकर सु और भाँति गलियान,
छलिया छबीले छैल औरे छ्वै छ्वै गए।
औरे भाँति बिहग समाज में अवाज होति,
ऐसे रितुराज के न आज दिन द्वै गए।
और रस औरे रीति और राग औरे रंग,
औरे तन और मन और बन है गए॥

शब्दार्थ :
भीरे = भीड़। बौरन = आम के बौर में। छैल = छवि। छ्दै छ्वै = छा-छा गए। विहग = पक्षी।

सन्दर्भ एवं प्रसंग :
पूर्ववत्।

व्याख्या :
कविवर पद्माकर कहते हैं कि बसन्त ऋतु में कुंजों में और ही प्रकार की सुन्दरता छा गई है तथा भरों की भीड़ में अनौखी गुंजार सुनाई दे रही है। डालियों, झुण्डों तथा आम के बौरों में नयी आभा आ गई है। गलियों में और ही भाँति की शोभा छा गयी है पक्षियों के समाज में भी और ही प्रकार की बोलियाँ सुनाई दे रही हैं। ऐसे श्रेष्ठ ऋतुराज बसन्त के अभी दो दिन भी नहीं चुके हैं फिर भी सर्वत्र और ही प्रकार का रस, और ही प्रकार की रीति, और ही प्रकार के राग तथा और ही प्रकार के रंग सर्वत्र छाये हुए हैं। इस ऋतु में मनुष्यों के शरीर और ही प्रकार के हो गए हैं, और ही प्रकार के मन तथा और ही प्रकार के वन और बाग हो गए हैं।

विशेष :

1. बसन्त ऋतु को इन्हीं सब विशेषताओं के कारण ऋतुराज की पदवी दी गयी है।
2. बसन्त में प्रकृति तथा पुरुष सभी में एक विचित्र प्रकार की नवीनता आ जाती है।
3. अनुप्रास की छटा।

(3) चंचला चलाकै चहूँ ओरन तें चाह भरी,
चरजि गई तो फेरि चरजन लागी री।
कहै पद्माकर लवंगन की लोनी लता,
लरजि गई तो फेरि लरजन लागी री।
कैसे धरौं धीर वीर त्रिविधि समीरै तन,
तरजि गई तो फेरि तरजन लागी री।
घुमड़ि-घुमड़ि घटा घन की घनैरों अबै,
गरजिं गई तो फेरि गरजन लागी री॥

शब्दार्थ :
चंचला = बिजली। चलाकै = चमककर। त्रिविध समीरें = शीतल, मन्द, सुगन्ध वाली हवा। घनेरी = घनी।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
इस छन्द में कवि ने पावस ऋतु का सुन्दर एवं मनोहारी वर्णन किया है।

व्याख्या :
कविवर पद्माकर कहते हैं कि पावस ऋतु में बिजली चारों ओर बड़ी चाहना से चमक रही है। एक बार वह बिजली आकाश में छा गयी तो सभी को आश्चर्यजनक लगने लगी। कवि पद्माकर कहते हैं कि लौंग (लवंग) की सुन्दर लता इस ऋतु को बहुत ही अच्छी लग रही है। एक बार यदि वह लवंग लता झुक गई तो फिर वह निरन्तर झुकती ही जाती है। वे वीर! तू ही बता ऐसे सुहावने समय में जबकि शीतल, मन्द और सुगन्धित वायु बह रही हो, तो मैं तेरे बिना कैसे धैर्य धारण करूँ। यह सुगन्धित वायु एक बार यदि तन को सुवासित करने लगी, तो फिर वह निरन्तर ही सुवासित करती रहेगी। इस ऋतु। में बादलों की घनी घटाएँ घुमड़-घुमड़ कर बार-बार आ रही हैं। यदि वे घटाएँ एक बार गर्जन करने लगी, तो फिर वे निरन्तर गर्जन करने लगेंगी।

विशेष :

1. वर्षा ऋतु का बड़ा ही मनमोहक वर्णन हुआ
2. अनुप्रास की छटा।
3. ब्रजभाषा का प्रयोग।

(4) तालने पै ताल पै तमालन पै मालन पै,
वृन्दावन बीथिन बहार बंसीवट पै। 
कहै पद्माकर अखंड रासमंडल पै, 
मण्डित उमंड महा कालिन्दी के तट पै। 
छिति पर छान पर छाजत छतान पर, 
ललित लतान पर लाड़िली की लट पै। 
छाई भली छाई यह सदर जुन्हाई जिहि, 
पाई छबि आजु ही कन्हाई के मुकुट पै।

शब्दार्थ :
तमालन = तमाल वृक्ष पर। बंसीवट = वह वट वृक्ष जहाँ श्रीकृष्ण बंसी बजाते थे। कालिन्दी = यमुना। छिति = पृथ्वी। छान = छाजन। छतान = छतों पर। लाड़िली = राधा जी। जुन्हाई = चाँदनी।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
इस छन्द में कवि ने शरद ऋतु की सुन्दर शोभा का वर्णन किया है।

व्याख्या :
कवि पद्माकर कहते हैं कि तालाबों पर, ताड़ पर, तमाल वृक्षों पर और मालाओं पर, वृन्दावन की गलियों एवं बंसीवट पर, सभी ओर शरद ऋतु की बहार छाई हुई है। कविवर पद्माकर कहते हैं कि सम्पूर्ण रासमण्डली पर तथा कालिन्दी के तट पर शरद ऋतु की महान् छटाएँ उमड़ रही हैं। पृथ्वी पर, छप्परों पर और छतों पर शरद ऋतु छाई हुई है। सुन्दर लताओं पर, लाड़िली अर्थात् राधाजी की लटों पर शरद ऋतु की यह चाँदनी भली-भाँति छाई हुई है और यही शोभा आज श्रीकृष्ण के मुकुट पर भी छाई हुई है।

विशेष :

1. शरद ऋतु की सुन्दरता का वर्णन।
2. अनुप्रास की छटा।
3. ब्रजभाषा का प्रयोग।

बादल को घिरते देखा है भाव सारांश

प्रस्तुत कविता में कवि ने हिमालय की ऊँची-ऊँची चोटियों का तथा उमड़ने-घुमड़ने वाली घटाओं का चित्ताकर्षक वर्णन किया है। हिमालय की चोटियों के नीचे अनेक प्रकार की झीलें बहती रहती हैं। उन झीलों के समीप बरसात की उमस से व्याकुल होकर हंस कमलदण्डों को खोजते हुए इधर-उधर घूम रहे हैं।

उन जल-क्रीड़ा करते हुए हंसों को देखकर मेरा मन प्रफुल्लित हो उठता है। कवि कहता है कि हिमालय पर्वत की बर्फ से ढकी अगम्य घाटियों में कस्तूरी मृग विचरण करते हुए देखे जा सकते हैं। उन मृगों की नाभि से निकलने वाली कस्तूरी की सुगन्ध को खोजते हुए वे अज्ञानतावश इधर-उधर भटक रहे हैं। मृग कस्तूरी न मिलने के कारण स्वयं से दुखी प्रतीत होते हैं।

कवि बादलों का वर्णन करते हुए कहता है कि प्राचीनकाल में कुबेर के शाप से अलकापुरी से निष्कासित यक्ष ने बादलों को दूत बनाकर अपनी प्रेमिका के समीप भेजा था। इसमें कितनी सत्यता है कौन जानता है? मैंने शीत के दिनों में बर्फ से ढकी कैलाश पर्वत की ऊँची-ऊँची चोटियों पर काले-काले बादलों को घिरते देखा है।

बादल को घिरते देखा है संदर्भ-प्रसंगसहित व्याख्या

(1) अमल धवल गिरि के शिखरों पर,
बादल को घिरते देखा है।
छोटे-छोटे मोती जैसे
उसके शीतल तुहिन कणों को
मानसरोवर के उन स्वर्णिम
कमलों पर गिरते देखा है,
बादल को घिरते देखा है।

शब्दार्थ :
अमल = निर्मल। धवल = सफेद। गिरि = पर्वत। तुहिन = ओस।

सन्दर्भ :
प्रस्तुत छन्द ‘बादल को घिरते देखा है’ शीर्षक कविता से लिया गया है। इसके कवि ‘नागार्जुन’ हैं।

प्रसंग :
घिरते हुए बादलों को देखकर कवि की कल्पना स्फुटित हुई है, उसी का यहाँ वर्णन है।

व्याख्या :
कविवर नागार्जुन कहते हैं कि निर्मल तथा श्वेत पर्वत के शिखरों पर मैंने बादल को घिरते हुए देखा है। छोटे-छोटे मोती जैसे उसके ठण्डे बर्फ के कणों को मानसरोवर के उन सुनहरे कमलों पर गिरते देखा है। बादल को घिरते देखा है।

विशेष :

1. कवि का प्रकृति चित्रण बड़ा ही मोहक है।
2. अनुप्रास की छटा।
3. भाषा-खड़ी बोली।

(2) तुंग हिमालय के कन्धों पर
छोटी-बड़ी कई झीलें हैं,
उनके श्यामल नील सलिल में
समतल देशों से आ-आकर
पावस की ऊमस से आकुल
तिक्त-मधुर विसतंतु खोजते
हंसों को तिरते देखा है।
बादल को घिरते देखा है।

शब्दार्थ :
तुंग = ऊँचे। नील सलिल में = नीले जल में। समतल = मैदानी भागों से। पावस= वर्षा। आकुल= व्याकुल। तिक्त = तीखे। मधुर =मीठे। विसतंतु = कमल नाल।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
इस छन्द में कवि ने मानसरोवर झील में तैरने वाले हंसों का वर्णन किया है।

व्याख्या :
कविवर नागार्जुन कहते हैं कि ऊँचे हिमालय पर्वत की तलहटी में छोटी और बड़ी कई झीलें हैं। उनके साँवले और नीले जल में मैदानी भागों से उड़-उड़कर आने वाले तथा
वर्षा की उमस से व्याकुल हुए एवं तीखे तथा मीठे कमल नाल को खोजते-फिरते हंसों को मैंने तैरते देखा है। बादल को घिरते देखा है।

विशेष :

1. मानसरोवर की प्राकृतिक सुषमा तथा उसमें तैरने वाले सुन्दर हंसों का मनमोहक वर्णन है।
2. अनुप्रास की छटा।
3. भाषा-खड़ी बोली।

(3) ऋतु बसंत का सुप्रभात था
मंद-मंद था अनिल बह रहा
बालारुण की मृदु किरणें थीं
अगल-बगल स्वर्णाभ शिखर थे
एक-दूसरे से विरहित हो
अलग-अलग रहकर ही जिनको
सारी रात बितानी होती,
निशा काल से चिर-अभिशापित
बेबस उन चकवा-चकई का
बंद हुआ क्रन्दन फिर उनमें
उस महान सरवर के तीरे
शैवालों की हरी दरी पर
प्रणय-कलह छिड़ते देखा है।
बादल को घिरते देखा है।

शब्दार्थ :
अनिल = वायु। बालारुण = प्रात:कालीन सूर्य। मृदु = कोमल। स्वर्णिम = सुनहरे। विरहति = अलग। होकर, वियोग की दशा में। निशाकाल = रात के समय। चिर = दीर्घ काल से। अभिशापित = अभिशाप पाये हुए। सरवर = तालाब। तीरे = किनारे। शैवालों = काई। प्रणय-कलय = प्रेम का झगड़ा।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
प्रस्तुत छन्द में कवि ने बसन्त ऋतु के सुप्रभात की मनोहर झाँकी का वर्णन किया है।

व्याख्या :
कविवर नागार्जुन कहते हैं कि बसन्त ऋतु का सुप्रभात था। उस समय वायु मंद-मंद गति से बह रही थी तथा प्रात:कालीन सूर्य की कोमल किरणे अगल-बगल के सुनहरे शिखरों पर बिखर रही थीं। उस समय एक-दूसरे से अलग होकर! सारी रात वियोग में बिता देने वाले, अनन्त काल से अलग रहने का अभिशाप पाये हुए, उन बेवस चकवा-चकवी का वियोग। दुःख से उत्पन्न क्रन्दन यकायक बन्द हो गया। प्रात:काल होने पर उस महान सरोवर के किनारे काई की हरी पट्टी पर उन चकवा-चकवी को परस्पर प्रेमालाप में झगड़ते हुए मैंने देखा है। बादल को घिरते देखा है।

विशेष :

1. चकवा-चकवी रात में एक-दूसरे से अलग होकर विरह जन्य क्रन्दन करते रहते हैं। प्रात:काल दोनों का संयोग होने पर प्रणय क्रीड़ाएँ करने लगते हैं।
2. बसन्त ऋतु की मादकता का वर्णन है।
3. भाषा-खड़ी बोली।

(4) दुर्गम बर्फानी घाटी में
शत-सहस्त्र फुट ऊँचाई पर
अलख-नाभि से उठने वाले
निज के ही उन्मादक परिमल
के पीछे धावित हो होकर
तरल तरुण कस्तूरी मृग को
अपने पर चिढ़ते देखा है,
बादल को घिरते देखा है।

शब्दार्थ :
दुर्गम = जहाँ जाना सरल न हो। बर्फानी = बर्फ से ढकी हुई। शत-सहस्र = सैकड़ों-हजारों। अलख नाभि = अदृश्य नाभि। उन्मादकं = नशीले। परिमल = पराग, सुगन्ध। धावित = दौड़कर।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
इस छन्द में कवि ने दुर्गम बर्फीली घाटियों में दौड़ते हुए कस्तूरी मृगों की चेष्टाओं का सुन्दर वर्णन किया है।

व्याख्या :
कविवर नागार्जुन कहते हैं कि दुर्गम बर्फ से ढकी हुई चोटियों में सैकड़ों-हजारों फुट की ऊँचाई पर, नाभि से उठने वाली स्वयं की ही अदृश्य उन्मादक गन्ध को ढूँढ़ते हुए तथा उसकी खोज में भटकते हुए चंचल एवं तरुण कस्तूरी मृगों को स्वयं अपने ऊपर चिढ़ते हुए मैंने देखा है, बादल को घिरते देखा है।

विशेष :

1. कस्तूरी मृग की नाभि में अदृश्य रूप से छिपी रहती है, पर कस्तूरी मृग अपने अन्दर छिपी हुई कस्तूरी को न जानकर उसकी खोज में इधर-उधर भटकता फिरता है।
2. अनुप्रास की छटा।
3. भाषा-खड़ी बोली।

(5) कहाँ गया धनपति कुबेर वह
कहाँ गई उसकी वह अलका
नहीं ठिकाना कालिदास के
व्योम-प्रवाही गंगाजल का,
ढूँढा बहुत परन्तु लगा क्या
मेघदूत का पता कहीं पर,
कौन बताए वह छायामय
बरस पड़ा होगा न यहीं पर,
जाने दो, वह कवि कल्पित था,
मैंने तो भीषण जाड़ों में
नभ-चंबी कैलाश शीर्ष पर
महादेव को झंझानिल से
गरज-गरज भिड़ते देखा है।
बादल को घिरते देखा है।

शब्दार्थ :
अलका = कुबेर की राजधानी। व्योम-प्रवाही। = आकाश में बहने वाली। नभ-चुंबी = आकाश को चूमने वाले। झंझानिल = झंझावात।।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
इस छन्द में कवि ने कैलाश पर्वत की सुन्दरता का वर्णन करते हुए पौराणिक आख्यानों की चर्चा की है।

व्याख्या :
कविवर नागार्जुन कहते हैं कि इस कैलाश शिखर पर धनपति कुबेर की अलका नगरी का पुराणों में उल्लेख मिलता है लेकिन आज तो इसका यहाँ कोई नामोनिशान भी नहीं है, न ही हमें कालिदास के आकाश में विचरण करने वाले उस मेघदूत का कहीं पता मिलता है। हमने बहुत प्रयत्न कर लिए पर वह ढूँढने से भी नहीं मिला। ऐसा हो सकता है कि वह छायामय मेघदूत यहीं कहीं जाकर बरस गया होगा। फिर कवि कहता है कि इन बातों को छोड़ो, मेघदूत तो कवि कालिदास की कल्पना थी।

मैंने तो भयंकर कड़कड़ाते जाड़ों में आकाश को चूमने वाले कैलाश के शिखर पर घनघोर झंझावातों में महादेव जी को उनसे गरज-गरज कर युद्ध करते देखा है। बादल को घिरते देखा है।

विशेष :

1. इस छन्द में कवि ने गगनचुम्बी कैलाश शिखर की प्राकृतिक छटा का सुन्दर वर्णन किया है।
2. अनुप्रास की छटा।
3. भाषा-खड़ी बोली।

MP Board Class 10th Hindi Solutions

Students get through the MP Board Class 11th Physics Important Questions Chapter 8 Gravitation which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 8 Gravitation

Gravitation Class 11 Important Questions Very Short Answer Type

Question 1.
Write the difference between gravitation and gravity.
Answer:
Gravitation : The force of attraction between any two material bodies in the universe is called force of gravitation.
Gravity: If one of the attracting bodies is earth or some other planet or natural satel¬lite, then the force of attraction is called force of gravity.

Question 2.
Explain Newton’s universal gravitational laws.
Answer:
Every body of the universe attract each other. The force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them and it acts along the line joining the two bodies.

\(\vec{F}\)AB → Force given by B on A, \(\vec{F}\)BA → Force given by A on B. Let two bodies of mass m₁ and m₂ are kept d distance apart.
∴ (i) F ∝ m₁m₂and
(ii) F ∝ \(\frac{1}{d^{2}}\)
Combining both the laws,
F ∝\(\frac{m_{1} m_{2}}{d^{2}}\)
F = G\(\frac{m_{1} m_{2}}{d^{2}}\)
Where, G is constant called universal gravitational constant.

Question 3.
Define universal gravitational constant.
Answer:
We know that, F = G\(\frac{m_{1} m_{2}}{d^{2}}\)
If m₁ = m₂= 1 and d= 1, then F= G
Hence, the universal gravitational constant is numerically equal to the force of attraction between the two bodies each of mass unity, kept at unit distance apart.

Question 4.
Find out the unit and dimensional formula of universal gravitational constant. Also, write the value.
Answer:
Since, F = \(\frac{G m_{1} m_{2}}{d^{2}}\)
or
G = \(\frac{F d^{2}}{m_{1} m_{2}}\)
Unit: S.I. unit of G will \(\frac{\mathrm{Nm}^{2}}{\mathrm{~kg} \mathrm{~kg}}\) = Nm² kg^-2 .
Dimensional formula : [G] = \(\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^{2}\right]}{[\mathrm{M}][\mathrm{M}]}\) = [M^-1L³T^-2]
Value : Value of G = 6.67 × 10^-11 Nm² kg ^-2.

Question 5.
The value of G = 6.67 × 10^-11 Nm² kg ^-2. What does it means? Why G is called universal constant?
Answer:
When two bodies, each of mass 1 kg are kept 1 m apart, then the force of attraction between them is 6.67 × 10^-11 N.
The value of G does not depend upon nature, medium, time, temperature etc. There-fore, it is called universal constant.

Question 6.
What is acceleration due to gravity ? What is its standard value?
Answer:
The acceleration produced in a body due to force of gravity is called acceleration due to gravity.
or
The rate of change of velocity in a body falling freely is called acceleration due to gravity.
It is denoted by g and its standard value is 9.8 m × s^-2.

Question 7.
Write the relation between ‘g’ and ‘G’.
Answer:
g = \(\frac{\dot{G} M}{R^{2}}\)

Question 8.
What is the value of ‘g’ at the centre of earth?
Answer:
The value of ‘g’ at the centre of earth is zero.

Question 9.
On what factors the weight of a body depends upon a planet?
Answer:

• Size of planet,
• Daily rotation of planet,
• Position of body in planet.

Question 10.
How the magnitude of ‘g’ changes on the surface of earth?
Answer:
From equator to pole value of ‘g’ increases. It is minimum at equator and maxi¬mum at pole.

Question 11.
At which place value of ‘g’ is minimum and maximum?
Answer:
Value of ‘g’ is minimum at centre of earth and maximum at pole.

Question 12.
A body is taken to moon from the centre of the earth. What change will occur in the weight and the mass of the body?
Answer:
The weight of a body at the centre of the earth is zero. When the body is moved towards the moon, the weight of the body will increase. On the surface of the earth the weight will be maximum. Further the weight will decrease and on the moon it will be \(\frac{1}{6}\) th of the weight on the surface of earth. The mass will remain same.

Question 13.
What is the value of g at the centre of the earth? What will be its weight?
Answer:
Zero, zero.

Question 14.
How does the value of g change on the surface of earth ? Where it is maximum and minimum?
Answer:
The value of g increases when the body is taken from equator to poles. The value of g is maximum at poles and minimum at the equator.

Question 15.
Can the gravity of earth be zero, at some height?
Answer:
No as F ∝ \(\frac{1}{r^{2}}\)

Question 16.
If the earth stops rotation about its axis, what will happen to the weight of the body? What change will occur in the body placed at the poles?
Answer:
If the earth stops rotation the weight of the bodies will increase. The weight of the body placed at poles will remain same.

Question 17.
A man can jump six times more on the moon, than on the earth. Why?
Answer:
Because the value of acceleration due to gravity is six times less than that of earth.

Question 18.
The gravitational mass on earth is \(\frac{g}{6}\). If a body is taken to moon from the earth, then what will happen to weight, inertial mass and gravitational mass ?
Answer:
The weight will be \(\frac{1}{6}\) time that of earth. The inertial mass and gravitational mass will not change.

Question 19.
The gravitational potential energy of a body on the surface of earth is 6.4 × 10⁶ joule. Explain.
Answer:
The work done required to throw a body away from gravitational field of earth will be 6.4 × 10⁶ joule.

Question 20.
Weight of a body at equatorial line is less than its weight at pole. Why?
Answer:
Because value of ‘g’ at equator is less than pole and since weight = Mass × ‘g’ Weight at equator is less than pole.

Question 21.
What is escape velocity? On what factors does it depend?
Answer:
It is the least velocity with which a body must be thrown vertically upwards so that it just escapes the gravitational pull of earth.
It depends upon

• Radius of planet and
• Acceleration due to gravity of the planet.

Question 22.
What are the value of escape velocities of earth and moon?
Answer:
Escape velocity of earth, = 11.2 kms^-1
and that of moon = 2.38 kms^-1

Question 23.
What is an artificial satellite?
Answer:
The man made satellite which revolves around the earth are called artificial satellite.

Question 24.
What is geo-stationary satellite or synchronous satellite ? What is its height above the earth?
Answer:
A satellite revolving around the earth in west to east direction; in the equatorial plane perpendicular to the axis of rotation with time-period equal to that of earth i.e., 24 hours, is called geo-stationary satellite.
Its height from the surface of earth is nearly 36000 km.

Question 25.
What is parking orbit? What is its radius?
Answer:
The orbit in which the geo-stationary satellite is launched is called parking orbit. Its radius is nearly 42309 km.

Question 26.
Why the orbit of artificial satellite is set out of the atmosphere?
Answer:
The orbit of satellite is set out of the atmosphere so that the satellite may not bum due to the friction with air.

Question 27.
What do you mean by weightlessness in the satellite?
Answer:
The necessary centripetal force to revolve in a circular orbit is provided by the gravity of earth. Hence, the net force on the satellite and the bodies inside it will be zero. Therefore, the weight of the body becomes zero.

Question 28.
Weightlessness is not felt on the moon, why?
Answer:
Due to gravity of moon, the bodies have weights on the moon, of course it is a satellite of earth.

Question 29.
Can a satellite be launched in an orbit which is not in the plane, passing through the centre of earth?
Answer:
No, because under this condition, the horizontal component of the force of at-traction between earth and satellite cannot balance the centripetal force. As a result, the satellite will be attracted towards the equatorial plane and its orbit will not be stable.

Question 30.
Can a satellite be launched in an orbit, so that it can be seen always above
Delhi?
Answer:
No, this is because Delhi is not on equator line.

Question 31.
What is the relation between orbital velocity and escape velocity?
Answer:
Escape velocity = √2 × Orbital velocity.

Question 32.
What is the direction of revolution of a geo-stationary satellite?
Answer:
The direction of revolution of geo-stationary satellite is same as the direction of rotation of earth about its own axis, i.e., west to east.

Question 33.
What is the angle between the plane of revolution of satellite and axis of rotation ?
Answer:
90°.

Question 34.
In an artificial satellite a bob is hanging with a string, what will be the tension in the string ?
Answer:
Zero.

Question 35.
Can the velocity of a body be equal to the velocity of light?
Answer:
No, as the mass of a body increases with the velocity. When its velocity becomes equal to that of light, the mass of the body will be infinite, which is not possible.

Question 36.
If the height of the parking orbit of a satellite in space is increased then its orbital velocity decreases, why?
Answer:
Since orbital velocity ν₀ = R\(\sqrt{\frac{g}{R+h}}\)
Here, R and g are constant.
∴ ν₀∝ \(\frac{1}{\sqrt{h}}\)
i.e., if h increases then ν₀ decreases.

Question 37.
To find out time in a satellite pendulum clock is used or spring clock.
Answer:
Spring clock is used to determine time at satellite because pendulum clock will get stop there.

Question 38.
In an artificial satellite a mass of 2 kg is kept at spring balance. What will be the reading of spring balance ?
Answer:
Zero.

Question 39.
What will happen if a glass full of water is tilt in an artificial satellite?
Answer:
Water will start swinging as a droplet inside an artificial satellite due to weightlessness.

Question 40.
Inside an artificial satellite a body of mass 1 kg is hanged with a string. What will be the tension in the string ?
Answer:
Zero.

Question 41.
Escape velocity of 5 gram object in earth is 11.2 km/s. What will be the escape velocity for 10 gram object?
Answer:
Escape velocity of 10 gram object will be 11.2 km/s because escape velocity does not depend upon the mass of the object.

Question 42.
From where a satellite obtain the centripetal force required to revolve around a planet?
Answer:
Any satellite gets the required centripetal force to revolve around the planet from the gravitation force of the planet.

Question 43.
Sun attracts earth toward itself with its gravitational force, but earth does not move toward sun, explain with reason.
Answer:
Earth is not stationary, it revolves around the sun in a circular orbit. Centripetal force required to revolve around the sun is obtained by earth from the gravitational force of sun. In this way earth remains in equilibrium.

Question 44.
On which factor the period of revolution of geo-stationary satellite depends?
Answer:
Period of revolution of geo-stationary satellite depends on distance of it from earth surface. If the distance from earth surface increases, its period of revolution also increases.

Question 45.
How the orbital velocity of a satellite depends upon the distance of its from the earth surface?
Answer:
When the distance of satellite from earth surface increases, its orbital velocity decreases.

Question 46.
Why total energy of a satellite is negative?
Answer:
Since, energy is required for satellite to place it at orbit in infinity. So total energy of a satellite is negative.

Question 47.
Gravitational potential energy of any object at earth surface is 6.4 × 10⁶ joule. Explain its meaning.
Answer:
It means energy required to send an object out of gravitational field is 6.4 × 10⁶ joule.

Gravitation Class 11 Important Questions Short Answer Type

Question 1.
Establish the relation between g and G.
Answer:
Let the mass of the earth be M and its radius be R. A body of mass m is placed on it. Therefore, the force of attraction by the earth on the body :
By Newton’s laws of gravitation,
F = \(\frac{G M m}{R^{2}}\) …(1)
But, the force with which a body is attracted towards the earth gives the weight of body.
F = mg …(2)
Equating eqns. (1) and (2), we get
mg= \(\frac{G M m}{R^{2}}\)
or
g = \(\frac{G M}{R^{2}}\)
It is the required relation.

Question 2.
Explain with mathematical calculation, how the value of g changes with altitude?
or
How does the value of ‘g’ change above the surface of the earth ? Find out the expression and explain.
Answer:
The value of g decreases with the increase of altitude.
Let the mass of earth is M and its radius is R.
∴ g = \(\frac{G M}{R^{2}}\) …(1)
If at point P at height h from the surface of earth is g’, then
g’ = \(\frac{G M}{(R+h)^{2}}\) …(2)
∴ By eqns. (1) and (2), we get

Hence, g’ < g.
Thus as height increases the value of g decreases.

Question 3.
How the value of g changes with the increase of depth? Find the value of g
at the centre of earth.
or
Prove that the value of g at the centre of earth is zero.
Answer:
Assuming the earth to be a homogeneous sphere of mass M, radius R and acceleration due to gravity on the surface is g. Let ρ be the mean density of earth.
We know that, g = \(\frac{G M}{R^{2}}\) …(1)
∵ M = Volume × Density
or
M = \(\frac{4}{3}\)πR³ × ρ …(2)
Putting the value of Min eqn. (1),
g = \(\frac{G \frac{4}{3} \pi R^{3} \times \rho}{R^{2}}\)
or
g = \(\frac{4}{3}\)πRGρ …(3)

Now, let the body be taken to depth h, below the free sur¬face hence the distance of the body from the centre of earth will be (R-h). Hence, the force of attraction on the body will be acted by the sphere of radius (R-h). Mass of earth of radius (R-h) will be M’.
∴ M’ = \(\frac{4}{3}\)π(R – h)³ ρ
If g_d be the acceleration due to gravity at depth h, then using eqn. (1),

Dividing eqn. (4) by eqn. (3), we have

Hence, from eqn. (5), the value of g decreases with the increase of depth. At the centre
ofearth, h=R
∴ g_d = g\(\left(1-\frac{R}{R}\right)\)
or
g_d= g(1 – 1)g × 0
∴ g_d= 0
Hence, value of g at the centre of earth is zero.

Question 4.
How does the value of ‘g’ change due to shape of the earth?
Answer:
The earth is not perfectly spherical in shape. It is slightly flattened at the poies
and is bulged at the equators
Let R_p and R_e be polar and equational radius respectively.,
Hence, g_p =\(\frac{G M}{R_{p}^{2}}\) and g_e =\(\frac{G M}{R_{p}^{2}}\)
Where g_p and g_e are the acceleration due to gravity at the poles and equator respectively.

Now \(\frac{g_{e}}{g_{p}}=\frac{G M}{R_{e}^{2}} \times \frac{R_{p}^{2}}{G M}=\left(\frac{R_{p}}{R_{e}}\right)^{2}\)
∵ R_p < R_e
∴ \(\left(\frac{R_{p}}{R_{e}}\right)^{2}\) < 1, hense \(\frac{g_{e}}{g_{p}}\) < 1
or
g_e < g_p

Question 5.
Show that the gravitational potential energy of an object near the earth
surface is ΔU = mgh.
Answer:
If a particle is raised from the surface of the earth to a height h above, then r_{1 =} R (earth radius) and r₂ = R + h (see figure).
So that, the change in potential energy )

Using Binomial theorem and neglecting the higher powers of h/R, we get

or
ΔU = G \(\frac{M m h}{R^{2}}\)
But, \(\frac{G M}{R_{p}^{2}}\)= g, (acceleration due to gravity at the surface of the earth).
Hence, ΔU = mgh
Thus, when a body of mass m is taken h distance above the earth’s surface, its potential energy increases by mgh.

Question 6.
What do you mean by gravitational field intensity? Establish relation be-tween intensity of gravitational field and acceleration due to gravity.
Answer:
Gravitational field :
The space around a material body in which its gravita-tional pull can be experienced is called its gravitational field.
Gravitational field intensity:
The intensity of gravitational field of a body at a point in a field is defined as the force experienced by a body of unit mass placed at that point provided the presence of unit mass does not disturb the original gravitational field.
i.e.. E = \(\frac{F}{m}\)

Let an object of mass M is at point O and at a distance of r from it at point P, we have to find out the intensity.
Let us consider a unit mass (m = 1) is placed at point P, therefore force of attraction applied by mass M on unit mass will be
F = \(\frac{G M \cdot 1}{r^{2}}=\frac{G M}{r^{2}}\)
Intensity at point P
E = \(=\frac{F}{m}=\frac{G M}{r^{2}} / 1=\frac{G M}{r^{2}}\) …(1)
and we know, acceleration due to gravity
g = \(\frac{G M}{r^{2}}\) …(2)
From eqns. (1) and (2), we get E = g
i.e., Intensity of gravitational field is equal to acceleration due to gravity of any point.

Question 7.
What is orbital velocity? Derive an expression for the orbital velocity. Obtain an expression if satellite is near to the surface of earth.
Answer:
Orbital velocity:
Orbital velocity is the velocity which is given to an artificial satellite, a few hundred kilometre above the earth’s surface so that it may start revolving round the earth.
Let the mass of satellite is m and the radius of orbit from the centre of earth is r, revolving with a velocity of ν_o
∴ Centripetal force = \(\frac{m v_{o}^{2}}{r}\) …(1)
Also, by Newton’s law of gravitation,
F = \(\sqrt{\frac{G M m}{r^{2}}}\) …(2)
Since, centripetal force is provided by gravitation force, then

\(\frac{m v_{o}^{2}}{r}=\frac{G M m}{r^{2}}\)
or
ν_o² = \(\frac{G M}{r}\)
∴ ν_o = \(\sqrt{\frac{G M}{r}}\) …(3)
If the radius of earth is R and height of satellite from the surface of earth is h, then
r = R + h
But g = \(\frac{G M}{R^{2}}\)
or
GM = gR²
Putting the values of GM = gR² and r = R + h in eqn. (3), we get
ν_o = \(\sqrt{\frac{g R^{2}}{R+h}}=R \sqrt{\frac{g}{R+h}}\)
If the satellite is near to the surface of earth,then h<<R, then R + h ≈ R
∴ ν_o = R\(\sqrt{\frac{g}{R}}=\sqrt{g R}\)

Question 8.
What is meant by the period of revolution ? Derive an expression for It.
Answer:
Period of revolution :
The time taken to complete one revolution is called period of revolution of satellite.
Let the period of revolution is Tand the radius of orbital is r.
∴ Orbital velocity = \(\frac{\text { Total path travelled }}{\text { Total time taken }}\)
ν_o = \(\frac{2 \pi r}{T}\)
or
T = \(\frac{2 \pi r}{v_{o}}\)
But r = R + h
∴ T = \(\frac{2 \pi(R+h)}{v_{o}}\)
But we have
ν_o = R \(\sqrt{\frac{g}{R+h}}\)
∴ T = \(\frac{2 \pi(R+h)}{R \sqrt{\frac{g}{R+h}}}\)
or
T = 2π(R + h)\(\frac{\sqrt{(R+h)}}{R \sqrt{g}}\)
∴ T = \(\frac{2 \pi}{R} \sqrt{\frac{(R+h)^{3}}{g}}\)

Question 9.
Determine the revolution time of a satellite, near to earth.
Answer:
We have T = \(\frac{2 \pi}{R} \sqrt{\frac{(R+h)^{3}}{g}}\)
If h<<R, then
T = \(\frac{2 \pi}{R} \sqrt{\frac{R^{3}}{g}}\)
= \(2 \pi \sqrt{\frac{R}{g}}\)
Now
R = 6.38 × 10⁶ m,g = 9.8ms^-2
∴ T = \(2 \times 3 \cdot 14 \sqrt{\frac{6 \cdot 38 \times 10^{6}}{9 \cdot 8}}\)
= 84.4 min.(approx.).

Question 10.
Planet A is heavier than planet B. Which will have the greater escape velocity?
Answer:
We have, ν = \(\sqrt{\frac{2 G M}{R}}\)
If R is constant, then
ν ∝ √M
∴ \(\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{A}}{M_{B}}}\)
∵ M_A>M_B
∴ ν _A> ν_B
Hence, escape velocity of planet A will be greater than that of planet B.

Question 11.
Establish the relation between escape velocity and orbital velocity.
Answer:
Escape velocity : ν_e = \(\sqrt{2 g R}\) …(1)
Orbital velocity : ν_o = \(\sqrt{g R}\) …(2)
Dividing eqn. (1) by eqn. (2), we get
\(\frac{v_{e}}{v_{o}}=\frac{\sqrt{2 g R}}{\sqrt{g R}}=\sqrt{2}\)
∴ ν_e = \(\sqrt{2} v_{o}\)

This is the required relation between escape velocity and orbital velocity.

Question 12.
What are the characteristics of geo-stationary satellite?
Answer:

• The geo-stationary satellite revolves around the earth in equatorial plane, perpendicular to the axis of rotation of earth.
• Its direction is from west to east.
• Its time-period of revolution is 24 hours.

Question 13.
What is geo-stationary satellite ? Prove that its height from the surface of earth is 36000 km.
Answer:
Geo-stationary satellite:
A satellite revolving around the earth in west to east direction; in the equatorial plane perpendicular to the axis of rotation with time-period equal to that of earth i.e., 24 hours, is called geo-stationary satellite.
Its height from the surface of earth is nearly 36000 km.
Now, we have

= 42.22 × 10⁶ m = 42220 km (Approx.)
∴ h=(R + h) – R
= 42.22 × 10⁶ – 6.38 × 10⁶ = 35840 km
or
h = 36000 km (Approx.) Proved.

Question 14.
Why the moon has no atmosphere?
Or
Why the moon has no atmosphere while Jupiter and Saturn have dense atmosphere?
Answer:
The escape velocity for moon is 2.38 kms^-1, while the r.m.s. speed of molecules of gases at 500 K is 2.5 kms^-1, hence the molecules of gases escape from the gravity of moon. The escape velocities of Jupiter and Saturn are greater, therefore, they have atmosphere.

Question 15.
What conclusion did Newton obtain from Kepler’s law of planetary motion?
Answer:
Conclusions obtained by Newton : On the basis of Kepler’s law, Newton pro-pounded the following:
(i) The path of the planets round the sun can be approximately taken as circular, for which the necessary centripetal force can be obtained from the gravitational force of attraction between the planet and sun. The force acting on the planet is always towards the sun.

(ii) By the third law :
T² ∝ r³
For the circular path of planet, the centripetal force acting will be
F = \(\frac{m v^{2}}{r}=\frac{m(r \omega)^{2}}{r}\) = mrω²
or
F = mr \(\left(\frac{2 \pi}{T}\right)^{2}=\frac{4 \pi^{2}}{T^{2}}\) mr …(1)
Since T² ∝ r³ ⇒ T² = Kr³
Putting the value of T²in eqn (1),we get
F = \(\frac{4 \pi^{2}}{K r^{3}}\) mr = \(\frac{4 \pi^{2} m}{K r^{2}}\)
But,\(\frac{4 \pi^{2} m}{K}\) is a Constant.
∴ F = \(\frac{\text { Constant }}{r^{2}}\)
or
F ∝ \(\frac{1}{r^{2}}\)
Hfence, the force acting on the planet is inversely proportional to the square of dis-tance between them.

(iii) The force acting on the planet is directly proportional to its mass.
ie., F∝m.

Gravitation Class 11 Important Questions Long Answer Type

Question 1.
Derive an expression for the escape velocity. Does it depend upon mass?
Answer:
Escape velocity : It is the least velocity with which a body must be thrown vertically upwards so that it just escapes the gravitational pull of earth.
It depends upon

• Radius of planet and
• Acceleration due to gravity of the planet.

Let the mass of earth be M and its radius is R .At any time let the body is at a distance x from the centre of earth.
∴ Force on the body due to earth, F = \(\frac{G M m}{x^{2}}\)
∴ Work done in moving the body through dx distance, against gravity,
dW = Fdx = \(\frac{G M m}{x^{2}}\) .dx
∴ Work done in moving up to infinity,

∴ W = \(\frac{G M m}{R}\)
If the escape velocity is ν_e, then

This is the required expression.
Since, the formula is free from the mass of the body, hence the escape velocity does not deppnd upon the mass of the body.

Gravitation Class 11 Important Numerical Questions

Question 1.
Two masses, each of 1 gm, are kept 1 cm apart. Calculate the force of at-traction between them in kg wt.
Solution:
Given : m₁, = m₂ = 1 gm = 10^-3 kg, r = 1 cm = 10^-2 m and
G = 6.67 × 10^-11 Nm²kg^-2
Now F = \(\frac{G m_{1} m_{2}}{r^{2}}\)
\(=\frac{6 \cdot 67 \times 10^{-11} \times 10^{-3} \times 10^{-3}}{\left(10^{-2}\right)^{2}}\)
\(=\frac{6 \cdot 67 \times 10^{-17}}{10^{-4}}\)
6.67 × 10^-13 N
\(=\frac{6 \cdot 67 \times 10^{-13}}{10}\) kg wt
∴ F = 6.67 × 10^-14kg wt

Question 2.
The mass of a planet is 10¹⁸ kg and its radius is 10³ m. Calculate the acceleration due to gravity on that planet.
Solution:
Given : M = 10¹⁸kg, R = 10³m and G = 6.67 × 10^-11 Nm²kg^-2.
Now, g = \(\frac{G M}{R^{2}}\)
∴ g = \(\frac{6 \cdot 67 \times 10^{-11} \times 10^{18}}{\left(10^{3}\right)^{2}}\)
\(=\frac{6 \cdot 67 \times 10^{7}}{10^{6}}\)
or
g = 6.67 × 10= 66.7ms^-2.

Question 3.
Calculate the value of g at 600 km height above the surface of earth. Given the radius of earth R = 6400 km and g = 9.8 ms^-2.
Solution:
Given : R = 6400 km = 6400 × 10³ m, g = 9.8 ms ^-2,
h = 600 km = 600 × 10³ m
Formula \(\frac{g^{\prime}}{g}=\frac{R^{2}}{(R+h)^{2}}\)
0r
\(\frac{g^{\prime}}{9 \cdot 8}=\left(\frac{6400 \times 10^{3}}{(6400+600) \times 10^{3}}\right)^{2}=\left(\frac{6400}{7000}\right)^{2}\)
∴ g’ = 9.8 × \(\left(\frac{64}{70}\right)^{2}\)
= 8.192ms^-2

Question 4.
At what height the weight of a body be \(\frac{1}{3}\) times that of the surface of earth
Solution:
Given:


∴ h = R(1.732 – 1) = 0.732Rm

Question 5.
What would be the angular velocity of earth so that the weight of a body at equator becomes zero?
Solution:
Now, we have at equator,
g’ = g – Rω²
But, g’ = 0
∴ g – Rω² = 0
or
-Rω² = -g
or
ω²=g|R
or
ω = \(\sqrt{\frac{g}{R}} \text { . }\)

Question 6.
Calculate the potential energy of a body of mass 100 kg placed at a height 600 km above the surface of earth (M = 6 × 10²⁴ kg, R = 6400 km)-
Solution:
Given : M = 6 × 10²⁴ kg,R = 6400 km = 6400 × 10³m,
h = 600 km = 600 × 10³m,m = 100 kg.
Formula U = \(-\frac{G M m}{r}\)
\(=-\frac{6 \cdot 67 \times 10^{-11} \times 6 \times 10^{24} \times 100}{(6400+600) \times 10^{3}}\) (r = R +h)
\(=-\frac{6 \cdot 67 \times 6 \times 10^{15}}{7000 \times 10^{3}}\)
U = – 5.717 10⁹ joule.

Question 7.
The gravitational field intensity at a point 8000 km away from the earth surface is 6 N/kg. What is gravitational potential at that point?
Solution:
Given : E = 6 N/kg, d – 8000 km = 8000 × 10³m
∴ Formula V = -E × d
Gravitational potential V = -6 × 8000 × 10³
= -48 × 10⁶
= -4.8 × 10⁷Nm/kg.

Question 8.
An artificial satellite is revolving around the earth at height 3600 km above the surface of earth. Calculate
(i) Orbital velocity,
(ii) Period of revolution.
Solution:
Given : h = 3600 km = 3600 × 10³m, g = 9.8 m/s² and R = 6.38 × 10⁶m.
Formula: ν_o = \(\sqrt{\frac{g R^{2}}{R+h}}\)

= 10⁴ × 0.6322m/s
= 6.322 × 10³m/s = 6.322m/s

= 626744 × 1.58 = 9913.9 sec
= 2.734 hours

Question 9.
The mass of planet A Is 4 times that of planet I? and radius of A, ¡s 2 times
that of B. Compare their escape velocities.
Solution:
Given : M₁= 4M₂and R₁ = 2R₂

Question 10.
The distance of planet A is three times that of planet B. If the time of revolution of B is 6 months, then find the time of revolution of A.
Solution:By kepler’s law,
T² ∝ r³

Question 11.
The time of revolution ofaplanet revolving in a circular orbit of radius R is T. Find the time of revolution of a planet revolving in an orbit of radius 4R.
Solution:
Given: = T₁ = T, r₁ = R,r₂ = 4R.
By Kepler’s law,
\(\left(\frac{T_{1}}{T_{2}}\right)^{2}=\left(\frac{r_{1}}{r_{2}}\right)^{3}\)
or
\(\left(\frac{T}{T_{2}}\right)^{2}=\left(\frac{R}{4 R}\right)^{3}=\frac{1}{64}\)
or
\(\frac{T}{T_{2}}=\frac{1}{8}\)
∴ T₂ = 8T

Question 12.
If the radius of the moon is 1.74 × 10⁶ m and its mass is 7.36 × 10²² kg, then calculate the escape velocity from the surface of the moon.
Solution:
Given: R = 1.74 × 10⁶ m, M = 7.36 × 10²² kg,
G = 6.67 × 10^-11Nm²/kg²
∵ Escape velocity, ν_e = \(\sqrt{\frac{2 G M}{R}}\)
∴ ν_e = \(\sqrt{\frac{2 \times 6 \cdot 67 \times 10^{-11} \times 7 \cdot 36 \times 10^{22}}{1 \cdot 74 \times 10^{6}}}\)
= 2.38 × 10³ m/sec = 2.38 km/sec.

Question 13.
The escape velocity of earth is 11 kms1. If the radius of a planet is double of earth’s radius and mean density are same, then find the escape velocity of that planet.
Solution:
Escape velocity,
ν = \(\sqrt{\frac{2 G M}{R}}\)

or
ν₂ = 22 km/s.

Question 14.
The mass of a planet is 3 × 10²⁴ kg and radius is 5 × 10⁶ m. What will be the energy required to move, 5 kg body to infinity ?
Solution:
Given : M = 3 × 10²⁴kg,R = 5 × 10⁶m, m = 5kg,
G = 6.67 × 10¹¹ Nm²kg^-2.
Formula U = \(\frac{G M m}{2 R}\)
\(=\frac{6 \cdot 67 \times 10^{-11} \times 3 \times 10^{24} \times 5}{2 \times 5 \times 10^{6}}\)
U = 100.05 × 10⁶ = 1010⁸ joule.

Gravitation Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
Newton’s laws of gravitation is true for :
(a) All bodies
(b) Only for heavenly bodies
(c) Only for small bodies
(d) Only for charged objects.
Answer:
(a) All bodies

Question 2.
Value of ‘g’ at moon with comparison to earth is :
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{1}{5}\)
(d) \(\frac{1}{6}\)
Answer:
(d) \(\frac{1}{6}\)

Question 3.
value of universal gravitational constant G is :
(a) Same at all places
(b) Unequal at all places
(c) Zero
(d) Infinity.
Answer:
(b) Unequal at all places

Question 4.
Relation between g and G is :
(a) g = GM
(b) gR² = GM
(c) gM = GR²
(d) gM/GR².
Answer:
(b) gR² = GM

Question 5.
Magnitude of g outside the gravitational field is:
(a) Infinity
(b) Zero
(c) 9.8 m/s²
(d) 980 newton.
Answer:
(b) Zero

Question 6.
Magnitude of g at the centre of earth is:
(a) Zero
(b) Infinity
(c) 9.8 newton
(d) 980 dyne.
Answer:
(a) Zero

Question 7.
Magnitude of g at pole is:
(a) Less
(b) More
(c) Zero
(d) None of these.
Answer:
(b) More

Question 8.
Magnitude of g at equator:
(a) More
(b) Less
(c) Infinity
(d) Zero
Answer:
(b) Less

Question 9.
Formula for gravitational potential energy is:
(a) \(\frac{G M m}{r}\)
(b) \(\frac{G m}{r}\)
(c) \(\frac{R}{G M}\)
(d) gR² = GM.
Answer:
(a) \(\frac{G M m}{r}\)

Question 10.
Formula for escape velocity is:
(a) gR
(b) MR .
(c) \(\sqrt{2 g R}\)
(d) \(\text { (d) } \sqrt{g R} \text { . }\)
Answer:
(c) \(\sqrt{2 g R}\)

Question 11.
There is no atmosphere at moon, because :
(a) Moon is closer to earth
(b) It revolves round the earth
(c) It gets light from sun
(d) Escape velocity of gas molecules is less than their root mean-square velocity.
Answer:
(d) Escape velocity of gas molecules is less than their root mean-square velocity.

Question 12.
If a body is projected at angle of 50° then its escape velocity will be :
(a) 11.6 km/s
(b) 11.2km/s
(c) 12.8 km/s
(d) 16.2 km/s.
Answer:
(b) 11.2km/s

Question 13.
Distance of geo-stationary satellite from earth surface :
(a) 6 R
(b) 7 R
(c) 5 R
(d) 3 R.
Answer:
(a) 6 R

Question 14.
Kepler’s second law is based on :
(a) Newton’s first law
(b) Theory of Relativity
(c) Newton’s second law
(d) Laws of conservation of angular momentum.
Answer:
(d) Laws of conservation of angular momentum.

Question 15.
Period of revolution of geo-stationary satellite is :
(a) 20 hours
(b) 22 hours
(c) 24 hours
(d) 25 hours.
Answer:
(c) 24 hours

Question 16.
Formula for orbital velocity is :
(a) gR
(b) g² R²
(c) g³R³
(d) \(\sqrt{g R}\)
Answer:
(c) g³R³

Question 17.
Who invented first the laws of planetary motion :
(a) Newton
(b) Kepler
(c) Galileo
(d) Aryabhatt.
Answer:
(b) Kepler

Question 18.
Radius of parking orbit is :
(a) 36000 km
(b) 42309 km
(c) 6400 km
(d) 80000 km.
Answer:
(b) 42309 km

2. Fill in the blanks:

1. The acceleration of body due to gravity is called …………………..
Answer:
acceleration due to gravity

2. Gravitational force in nature is a …………………….. force.
Answer:
weak

3. Acceleration due to gravity does not depend upon ………………. of the body.
Answer:
mass

4. At height the value of ‘g’ is ………………………
Answer:
less

5. At centre of earth the value of ‘g’ is ………………………
Answer:
Zero

6. G is a ……………… quantity.
Answer:
scalar

7. Gravitational mass and inertial mass are …………………… to each other.
Answer:
equal

8. Gravitational intensity is equal to ……………………..
Answer:
acceleration due to gravity

9. Gravitational potential energy at infinity is ……………………..
Answer:
zero

10. Magnitude of ‘g’ at equatorial position is …………………..
Answer:
minimum

11. Magnitude of ‘g’ at pole is ……………..
Answer:
maximum

12. Orbital velocity does not depend upon ………………… of a body.
Answer:
mass

13. Escape velocity does not depend upon …………….. of the body.
Answer:
mass

14. Magnitude of escape velocity is ……………… km/s.
Answer:
11.2

15. The time-period of geo-stationary satellite is ………………. hours.
Answer:
24

3. Match the following:
I.

Answer:
1. (c) Planetary motion
2. (a) Zero
3. (e) \(\sqrt{2 g R}\)
4. (b) \(-\frac{G M m}{2 r}\)
5. (d) Time-period of revolution is 24 hours

II.

Answer:
1. (e) Vector.
2. (c) Variable
3. (b) S calar quantity
4. (a) Weak force
5. (d) E = mc²

4. Write true or false:

1. Newton’s laws is universal.
Answer:
True

2. Magnitude of G is different at different places.
Answer:
False

3. Magnitude of ‘g’ is equal at any point on surface of earth.
Answer:
False

4. Ratio of gravitational mass and inertial mass is one.
Answer:
True

5. Gravitational potential is negative at every point on earth surface.
Answer:
True

6. Principle of superposition is not applicable for gravitation.
Answer:
False

7. Value of escape velocity of a body depends upon its mass.
Answer:
False

8. Height of polar satellite from earth surface is 36000 km.
Answer:
False

9. Magnitude of gravitation potential energy at infinity is zero.
Answer:
True

10. Gravitational force is a conservative force.
Answer:
True

Students get through the MP Board Class 11th Physics Important Questions Chapter 7 System of Particles and Rotational Motion which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 7 System of Particles and Rotational Motion

System of Particles and Rotational Motion Class 11 Important Questions Very Short Answer Type

Question 1.
Define the rotational motion.
Answer:
When a force is applied on a rigid body then it revolves about its own axis. This motion of the body is called rotational motion.

Question 2.
Define the term rigid body.
Answer:
A body is said to be a rigid body if an applying external force or during its motion the distance between constituent particle remain unchanged.

Question 3.
Differentiate rotatory motion and circular motion.
Answer:
In rotatory motion the axis of motion lies within the body while in circular motion the axis lies outside the body.

Question 4.
What is torque? Write its unit and dimensional formula.
Answer:
The turning effect of a force, to rotate a body about an axis is called torque. The torque is equal to product of force and perpendicular distance of force from the axis of rotation.
Torque = Force × Moment arm
τ = F × d
Unit: Its S.I. unit is.N × m.
Dimensional formula: [ML²T^-2].

Question 5.
What do you mean by angular velocity and angular acceleration. Write its S.I. units and dimensional formula.
Answer:
Angular velocity : Rate of Change of angular displacement per unit time.
Its unit is radian/second and dimensional formula is [M⁰L⁰T^-1].
Angular acceleration : The rate of change in angular velocity of the body about the given axis of rotaion.
Its unit is radian/second² and dimensional formula is [M⁰L⁰T^-2]

Question 6.
What do you mean by angular momentum?
Answer:
It is defined on the product of linear momentum and perpendicular distance of line of action of linear vector from the axis of rotation.
Angular momentum = Linear momentum × perpendicular distance of linear momentum from the axis of rotation.
Its SI unit is J × s. and dimensional formula is [ML²T^-1 ].

Question 7.
Define moment of inertia and give its unit and dimensional formula.
Answer:
Moment of inertia of a rotating body is the property by virtue of which, it opposes the torque applied to change its position of rest or rotational motion.
If the mass of a body is m and its distance from the axis of rotation is r, then moment of inertia, I = mr²
Unit: Its S.I. unit is kg × m².
Dimensional formula: [ML²T⁰].

Question 8.
Define radius of gyration and give its unit.
Answer:
The radius of gyration of a body is equal to the perpendicular distance from the axis of rotation of a point at which if the mass of a body is assumed to be concentrated the moment of inertia is the same as the actual ‘moment of inertia of the body about that axis. It is represented by k.
UnIt: Its S.I. unit is metre.

Question 9.
On what factors does the moment of inertia of a body depend?
Answer:
(a) On mass of body (2) Distribution of mass from the axis of rotation.

Question 10.
What will be the effect on moment of inertia if the direction of motion of the body is changed?
Answer:
There will be no effect on moment of inertia.

Question 11.
Is it essential to apply moment of couple to a rotating body? Explain with reason?
Answer:
No, it is not essential to apply moment of couple to a rotating body because moment of couple only create angular acceleration.

Question 12.
It is easy to rotate a stone tied with a small string, than that of a longer. Why?
Answer:
We know that, τ = Iω. If the string is long its moment of inertia will be greater and therefore greater torque will be required to rotate the stone. But, for small string less torque is required. Therefore, it is easy to rotate the stone with smaller string.

Question 13.
Write the relation between torque and angular momentum.
Answer:
τ = \(\frac{d L}{d t}\)

Question 14.
Write the relation between angular momentum and moment of inertia.
Answer:
Angular momentum = Moment of inertia × Angular velocity.

Question 15.
Why the spokes are used in a wheel?
Answer:
The spokes increases the distance of rim from the axis of rotation, hence its moment of inertia increases, which makes the rotation of wheel uniform and maintains its rotation.

Question 16.
The handle of a door is fixed away from the hinge. Why?
Answer:
The handle is fixed away from the hinge because, it increases the perpendicular distance between the line of action of force and hinge. Hence, the torque increases.

Question 17.
The handle of a hand pump is made long, why?
Answer:
Long handle increases the distance between the piston and the force, thus the torque increases, so less force is required to pump the water.

Question 18.
What will be the velocity of a point on a rotational axis in rotational motion?
Answer:
The velocity of a point situated in rotation axis in rotational motion is zero.

Question 19.
Differentiate inertia and moment of inertia.
Answer:
Difference between Inertia and Moment of inertia :

Question 20.
What is the physical importance of moment of inertia?
Answer:
To rotate a body about an axis from the rest or to change angular velocity or angular acceleration of rotating body, a torque is applied, more the moment of inertia more will be the torque. This is the physical importance of moment of inertia.

Question 21.
What do you mean by mechanical equilibrium of a system?
Answer:
When the vector addition of total forces and vector addition of total moment of inertia is’ zero then the system is said to be in mechanical equilibrium.

Question 22.
Write the principle of moments.
Answer:
According to this principle anti clockwise moments are taken positive and clockwise moments are taken negative when the body is in rotational equilibrium.

Question 23.
What is moment of couple? Write its SI unit and dimensional formula. Write the factor on which it depends.
Answer:

The product of a force and the perpendicular distance between the line of action of forces is called moment of couple.
Moment of couple = Force × Couple arm
Unit: Its S.I. unit is N × m.
The factor it depends are :

• Magnitude of force.
• Perpendicular distance between the line of action of forces.

Question 24.
What is the physical importance of angular momentum?
Answer:
We know that,
Torque = Force × Perpendicular distance of line of action from the axis and
Angular momentum = Linear momentum × Perpendicular distance of particle from the
axis.
Hence, as the torque measures the tinning effect, similarly angular momentum measures the rotational motion.

Question 25.
What is geometrical meaning of angular momentum?
Answer:
The angular momentum is equal to twice of the product of mass and areal velocity of a particle.
i.e., Angular momentum = 2 × Mass × Areal velocity

Question 26.
A dancer is rotating with stretched hands, when the closes her hands, her speed of rotation increases. Why?
Answer:
We know that in the absence of external force the angular momentum of a rotating body is conserved and is equal to Iω. When the dancer closes her hands, her moment of inertia I decreases hence ω i.e., angular velocity increases.

Question 27.
When a rotating stone tied with a string is stoped, the string winds up in the hand fastly. Why?
Answer:
When the string is started wind up in the hand, the length of string decreases and hence the moment of inertia of stone decreases. But, we know that angular momentum Iω is constant. As I decreases ω increases, therefore the string winds up very fast.

Question 28.
Which factor depends on moment of inertia of the body that moves around a body.
Answer:
We know I = Σmr²
Therefore moment of inertia,

• Is directly proportional to the mass.
• Is directly proportional to the square of distance from the axis of rotation.

Question 29.
Is centre of mass a reality?
Answer:
No, it is only a mathematical concept.

Question 30.
It is difficult to open the door by pushing it or pulling it at the hinge. Why?
Answer:
Because at the hinge the distance of the point of application of the force from the axis of rotation is negligible, so the torque is very small.

Question 31.
Why a helicopter must necessarily have two propellors?
Answer:
If there were only one propellor in the helicopter, then it would have to turn in the opposite direction to conserve its angular momentum.

System of Particles and Rotational Class 11 Important Questions Short Answer Type

Question 1.
Define vector product of two vectors and give one example of vector product. Also write its properties.
Answer:

Let the angle between two vectors \(\vec{A} \text { and } \vec{B}\) is θ, then their cross product is defined as
\(\vec{A} \times \vec{B}=A B \sin \theta \hat{n}\)
Where, \(\hat{n}\) is unit vector perpendicular to the plane consisting vectors \(\vec{A} \text { and } \vec{B}\) .
The direction of \(\vec{A} \times \overrightarrow{\vec{B}}\) follows the screw rule, as shown in the figure.

Example : Let a force \(\vec{F}\) is acting at a point whose position vector is \(\vec{r}\) , then the torque \(\vec{τ}\) is given by the cross product of \(\vec{F}\) and \(\vec{r}\)
i.e.. \(\vec{\tau}=\vec{r} \times \vec{F}\)
Properties: I. Vector product of two vector is always a vector quantity.
2. It does not obey commutative law
\(\vec{a} \times \vec{b} \neq \vec{b} \times \vec{a}\)
3. It does not obey associative law
\(\vec{a} \times(\vec{b} \times \vec{c}) \neq(\vec{a} \times \vec{b}) \times \vec{c}\)
4. \(\vec{a} \times \vec{a}\) = 0 (zero vector)

Question 2.
Show that the area of the triangle contained between the vectors  \(\vec{a} \text { and } \vec{b}\) is one half of the magnitude of a× b.
Answer:

Let the vectors \(\vec{a} \text { and } \vec{b}\) represent the two adjacent sides of ΔAOB such that
OA =b, OB = a
Let angle between a and b be θ.
Such that∠AOB= θ
and also let h be the height of the triangle such that
h=AC
Now in right angled ΔOCA,
sinθ = \(\frac{A C}{O A}\)
or
AC=OA sinθ
or
h = b sinθ … (1)
We know that the area of the triangle AOB, is given b
= \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × OB x AC
= \(\frac{1}{2}\) × a × h
= \(\frac{1}{2}\) × a × b sinθ
= \(\frac{1}{2}\)absinθ …(2)
Also by the definition of cross-product of two vectors, we know that
a × b = ab sinθ\(\hat{n}\)
or
\(|a \times b|=|a b \sin \theta \hat{n}|\)
= ab sinθ\(|\hat{n}|\)
= ab sinθ (\(|\hat{n}|\)
= absinθ, (\(|\hat{n}|\)
= absinθ, (∵ (\(|\hat{n}|\) = 1) …(3)
∴ From eqns. (2) and (3), we get
Area of ΔAOB = \(\frac{1}{2} \mid a \times b\)
= \(\frac{1}{2}\) [magnitude of a × b].

Question 3.
Give the location of the centre of mass of a
(i) sphere
(ii) cylinder
(iii) ring and
(iv) cube each of uniform mass density. Does the centre of mass of a body neccessarily lie inside the body?
Answer:
C.M. lies at the :
(i) Centre of sphere.
(ii) Mid-point of axis of symmetry of the cylinder i.e., its geometrical centre.
(iii) Centre of ring.
(iv) Point of intersection of diagonals i.e., at its geometrical centre.

No, in some cases, C.M. of a body like ring, hollow cylinder, hollow sphere and hollow cube etc. may lie outside.

Question 4.
Define work in rotational motion.
Answer:
As work is define as product of force and displacement in direction of force in linear motion, same like that work done by a torque in rotational motion is define as product of torque and angular displacement.
Work W = τ.d.θ

Question 5.
Derive an expression for the angular momentum of a body in rotational motion.
Or
Prove that J = Iω
Answer:
Consider a body which is made up number of particles of masses m₁,m₂,m₃,…. situated at the distances r₁,r₂, r₃ from the axis of rotation. If the linear velocities are
ν₁, ν₂, ν₃,…. respectively, then
Linear momentum of m₁ = m₁ν₁ = m₁r₁ω , (∵ ν = r ω)
and Angular momentum = m₁r₁ω x r₁ = m₁r₁² ω
Similarly, angular momentum of m₂,m₃,…. will be m₂r₂² ω,m₃r₃² ω,…..
∴ Angular momentum of whole body = m₁r₁²ω + m₂r₂² ω, + m₃r₃² ω, +….
or
J = ω (m₁r₁² +m₂r₂² +m₃r₃² +….)
= ωΣ mr²
or
J = ωI
∴ Angular momentum = Angular velocity × Moment of inertia.

Question 6.
What do you understand by angular momentum? Establish the relation between angular momentum and rotational K.E.
Answer:
Angular momentum : If a particle is moving about an axis, then the moment of linear momentum is called its angular momentum about that axis.
∴ Angular momentum = Linear momentum × Perpendicular distance of particle from the axis of rotation

Relation between angular momentum and rotational K.E.: We know that rotational kinetic energy (E_k) is given by the formula
E_k = \(\frac{1}{2}\) Iω² …(1)
But, Angular momentum, J = Iω
or
ω = \(\frac{J}{I}\)
Putting the value of ω in eqn. (1), we get
E_k = \(\frac{1}{2}\) I\(\left(\frac{J}{I}\right)^{2}\) = \(\frac{1}{2}\) \(\frac{J^{2}}{I}\)
or
J² =2I E_k
or
J = \(\sqrt{2 I E_{k}}\) …(2)
Eqn. (2) is the required relation.
If angular momentum remains conserved, then \(\sqrt{2 I E_{k}}\) lEk will be constant
or
2I E_k = Constant
or
E_k = \(\frac{\text { Constant }}{2 I}\)
or
E_k ∝ \(\frac{1}{I}\)

Question 7.
What is the law of conservation of angular momentum? Prove it.
Answer:
Law of conservation of angular momentum: In the absence of external torque, the angular momentum of a system remains constant.
i.e., J = a constant
Then, Iω = a constant
Now, we know that rate of change of angular momentum is equal to the external torque
applied.
i.e., τ = \(\frac{d J}{d t}\)
If τ = 0
Then, \(\frac{d J}{d t}\) = 0
or
J = a constant
or
J = Iω
If in two different conditions the moment of inertia of a body are I₁ and I₂and angular velocities are ω₁and ω₂respectively, then
I₁ω₁ = I₂ω₂

Question 8.
Prove that the rate of change of angular momentum is equal to the torque applied.
Or
Prove that τ = \(\frac{d J}{d t}\)
Answer:
Let a torque τ is applied on a body of moment of inertia I, so that its angular
acceleration becomes α, then
τ = I.α ….(1)
But, α = \(\frac{d \omega}{d t}\)
∴ τ = I.\(\frac{d \omega}{d t}\) …(2)
Also we have J = Iω
∴ \(\frac{d J}{d t}=\frac{d}{d t}\) (Iω) = \(\frac{d \omega}{d t}\)
From eqn. (2), we get
Or
\(\frac{d J}{d t}\) = Iω = τ [ from eqn .(1)]
Hence, rate of change of angular momentum is equal to the torque applied.

Question 9.
Derive an expression for the moment of inertia of a rigid body.
Or
Prove that I = Σ mr².
Answer:

Let a body rotating about an axis XY supposing that the body is made up of number of particles of masses m₁,m₂,m₃,…. situated at the distances r₁,r₂,r₃, from the axis of rotation.
∴ The moment of inertia of the particles will be m₁r₁² , m₂r₂² , m₃r₃²,
Hence, the moment of inertia of whole body,
I = m₁r₁² + m₂r₂² + m₃r₃² +
or
I = Σmr²
Where, Σ is the sign of summation.

Question 10.
Compare the different derivatives in linear motion and rotational motion.
Answer:
Different derivatives in linear and rotational motion :

Question 11.
A planet is revolving around a sun in an elliptical orbit. How angular velocity and linear velocity get changes in the orbit?
Answer:
(1) Since angular momentum L = mr²ω is constant. Therefore when planet go
far from sun r increases and hence co decreases and when planet go near to sun r decreases
and hence co increases.
(2) Since L = mνr constant. Therefore when planet go far from sun r increases and
hence ν decreases and when planet is close to sun r decreases and hence ν increases.
[∵ ν ∝\(\frac{1}{r}\)

Question 12.
A ladder is at rest against a wall. Is the ladder more likely to slip when a person climb up why?
Answer:
It is more likely to slip when a man climb upward in the ladder because due to the fact that the weight of the man will provide an extra torque for the slipping of the ladder.

Question 13.
What do you understand by angular momentum? Derive its’s expression.
Answer:
If a particle is moving about an axis, then the moment of linear momentum is called its angular momentum about that axis.
Angular momentum = Linear momentum × Perpendicular distance of particle from the axis of rotation.
Let’ m’ be the mass of a particle P whose position vector be V with respect to origin
O and its linear momentum be \(\vec{p}(=m \vec{v})\) and angular momentum be \(\vec{J}=(\vec{p} \times \overrightarrow{O N})\) = pr sinθ
In right angled Δ OPN,

ON = OP sinθ =r sinθ
or
J=rp sinθ
Where θ is angle between \(\vec{r} \text { and } \vec{p}\)
or
\(\vec{J}=\vec{r} \times \vec{p}\)
or
\(\vec{J}=\vec{r} \times(m \cdot \vec{v})\)
or
\(\vec{J}=m(\vec{r} \times \vec{v})\)