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MP Board Class 12th Chemistry Solutions Chapter 5 Surface Chemistry

Surface Chemistry NCERT Intext Exercises

Question 1.
Why are substance like platinum and palladium used for carrying electrolysis of aqueous solutions ?
Answer:
These metals are not attacked by the ions of the electrolytes or the products of electrolysis. Electrodes of such metals are called inert electrodes.

Question 2.
Why does physisorption decreases with the increase of temperature ?
Answer:
It is an exothermic process

Thus, when temperature is raised, reaction proceeds in backward direction and gas adsorbed gets released. It is accordance to Le-Chatelier’s principle.

Question 3.
Why are powdered substances more effective adsorbent than their crystal line forms ?
Answer:
Powdered form has greater surface area. Greater the surface area, greater is the adsorption.

Question 4.
Why is it necessary to remove CO when ammonia is obtained by Haber’s process ?
Answer:
For the preparation of ammonia in Haber’s process solid catalyst is used. It is necessary to remove the CO produced because the gas reacts with iron and forms Fe(CO)₅ which exist in liquid state at the temperature of the chamber and obstructs in the production of NH₃ because at high temperature CO and H₂ react by which production of ammonia decreases.
In short CO acts as poison for the iron catalyst used in the process. It is therefore, necessary to remove CO.

Question 5.
Why is the ester hydrolysis slow in the beginning and becomes faster after sometime ?
Answer:
CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH
Infact, this reaction is catalysed by an acid (H+). Acetic acid formed during the reac¬tion catalyses the hydrolysis of ester. Therefore, reaction becomes fast. It is called autocatalyst.

Question 6.
What is the role of desorption in the process of catalysis ?
Answer:
The product formed leaves the surface of catalyst and makes it available for adsorption of mole reactants molecules.

Question 7.
What modifications can you suggest in the Hardy-Schulze law ?
Answer:
The coagulating ion has charge opposite to that on the colloidal particles. Thus, when coagulating ion is added to colloidal solution, the change on colloidal particle is neutralised and coagulation occurs. According to Hardy-Schulze rule, greater the volume of the oppositely charged ions of the electrolyte faster is the coagulation the Hardy-Schulze law, consider the coagulation of colloidal sols because of neutralisation of charge on colloidal particles. We know that neutralisation and coagulation also takes place by mixing two oppositely charged colloidal solutions.

Thus, Hardy-Schulze law should also include the following when oppositely charged sols are mixed in right proportions, they neutralize charges of each other and undergo coagulation.

Question 8.
Why is it essential to wash the precipitate with water before estimating it quantitatively ?
Answer:
Some amount of impurities remains adsorbed on the surface of the precipitate. It is essential to remove the sticking impurities by washing with water.

Surface Chemistry NCERT TextBook Exercises

Question 1.
Distinguish between the meaning of the terms adsorption and absorption.
Answer:
Differences between Adsorption and Absorption :

Adsorption:

1. It is a process as a result of which one substance get concentrated only on the surface of the other.
2. Concentration of adsorbate on the surface of adsorbent is different than in the bulk.
3. It is a surface phenomenon.
   Example : Adsorption of water vapour on silica gel.

Absorption:

1. It is a process as a result of which one substance gets uniformly distributed in the volume of the other.
2. Concentration is uniform in the entire solid system.
3. It is a bulk phenomenon.
   Example : Adsorption of water vapour by dry calcium chloride.

Question 2.
What is the difference between physisorption and chemisorption ?
Answer:
Differences between Physical adsorption and Chemical adsorption :

Physical adsorption:

1. Low value of enthalpy of adsorption (20-40 kJ mol^-1)
2. This type of adsorption involves weak vanderWaals’ forces between adsorbent and adsorbate.
3. Usually takes place at low temperature and decreases with increase in temperature.
4. It is reversible in nature.
5. The extent of adsorption is approximately related to the case of liquification of the gas.
6. Forms multimolecular layers.
7. On increasing pressure, adsorption also increases.

Chemical adsorption:

1. High value of enthalpy of adsorption (40-400kJmol^-1 ).
2. This type of adsorption involves strong forces of attraction due to chemical bond formation.
3. Takes place at high temperature.
4. It is irreversible.
5. No such correlation.
6. Forms monomolecular layers.
7. No any effect of pressure.

Question 3.
Give reason why a finely divided substance is more effective as an adsorbent ?
Answer:
With the increase in surface area of adsorbent adsorption increases. Thus, in powdered state (finely divided substance) or in porous state surface area of metals is more. Therefore, adsorption is more in these states.

Question 4.
What are the factors which influence the adsorption of a gas on a solid ?
Answer:
Factors which affect the adsorption of gas on a solid are as follows :

1. Nature of gas
2. Surface area of adsorbent
3. Pressure
4. Temperature
5. Activity of adsorbent.

Question 5.
What is an adsorption isotherm ? Describe Freundlich adsorption isotherm.
Answer:
The graph showing the extent of adsorption versus pressure at constant tempera¬ture is known as adsorption isotherm. Adsorption isotherm graphs are of two types :
(a) Freundlich isotherm graph.
(b) Langmuir isotherm graph.

Freundlich isotherm: Freundlich gave an experimental relation between the amount of gas adsorbed by unit mass of adsorbent with pressure at a definite temperature. This relation can be represented by the following equation :
\(\frac{x}{m}=\mathrm{K.P}^{1 / n}(n>1)\) …..(1)
Where m is the mass of gas adsorbed on adsorbent x at pressure P, K and n are con¬stants which depend on the nature of adsorbent and gas at a definite temperature.
These graphs indicate at the definite pressure, physical adsorption decreases with the increase in temperature.
Taking log of equation (1)
log \(\frac { x }{ m } \) = log K + \(\frac { 1 }{ n } \) log P …..(2)
When a graph is drawn by taking log \(\frac { x }{ m } \) on y-axis and log P on x-axis, a straight line is obtained. This verifies Freundlich isotherm.
Slope = \(\frac { 1 }{ n } \) and intercept on y-axis = log K
value of \(\frac { 1 }{ n } \) can be between 0 and 1, thus eqn. (2)

can be applicable only upto a limited extention of pressure.
(i) When \(\frac { 1 }{ n } \) = 0, \(\frac { x }{ m } \) = constant, then adsorption is independent of pressure.
(ii) When \(\frac { 1 }{ n } \) = 1, \(\frac { x }{ m } \) = Kp, \(\frac { x }{ m } \) α P, then change in adsorption is proportional to pressure.

Question 6.
What do you understand by activation of adsorbent ? How is it achieved ?
Answer:
It means increasing the adsorption power of an adsorbent; it is done (i) by increasing the surface area, (ii) by making the surface rough, (iii) by removing the gas already adsorbed.

Question 7.
What role does adsorption play in heterogeneous catalysis ?
Answer:
In heterogeneous catalysis, the reactants are generally gases while catalysts are solids. The reactant molecules are adsorbed on the surface of solid catalyst by physical adsorption or chemical adsorption. As a result, the concentration of the reactant molecules on the surface of the catalyst increases and hence the rate of reaction also increases.

Question 8.
Why is adsorption always exothermic ?
Answer:
Adsorption leads to decrease in disorder, thus ∆S = +ve. For the process to be performed value of ∆G should be -ve. Thus, in equation ∆G = ∆H – T∆S, ∆G will be negative when ∆H will be negative i.e., exothermic. Thus, adsorption is an exothermic process.

Question 9.
How are the colloidal solutions classified on the basis of physical states of the dispersed phase and dispersion medium ?
Answer:
Classification on the basis of physical state of dispersed phase and dispersion medium : Depending upon the physical state of dispersed phase and dispersion medium whether these are solids, liquids or gases, eight types of colloidal systems are possible. The examples of the various types of colloids and their typical names are given in Table.

It may be noted that a gas mixed with another gas forms a homogeneous mixture and therefore it is not a colloidal system.

Different Types of Colloidal Systems

Out of the different types of colloids, the most common are sols (solid in liquids), gels (liquid in solids) and emulsions (liquid in liquid).

Question 10.
Discuss the effect of pressure and temperature on the adsorption of gases on solids.
Answer:
Effect of pressure: Adsorption is a reversible process and is accompanied by a decrease in pressure. Therefore, adsorption increases with an increase in pressure.
Effect of temperature: Adsorption is an exothermic process. Thus, in accordance with Le-Chatelier’s principle the mangnitude of adsorption decreases with an increase in temperature.

Question 11.
What are lyophilic and lyophobic sols ? Give one example of each type. Why are hydrophobic sols easily coagulated ?
Answer:
Lyophilic sols : Those sols in which there are forces of attraction between dispersed phase and dispersion medium, are called lyophilic sols. Lyophilic means liquid loving. They are directly prepared by mixing substances like gum, gelatine, starch with suitable dispersion medium like water, e.g., starch in water, albumin in water. They are reversible sols.

Lyophobic sols : The word ‘lyophobic’ means liquid hating. Substances like metals, their sulphides etc. when mixed with dispersion medium, do not form sols. They are prepared by special methods, e.g., Gold sol, As2S3 sol etc. In these sols, the particles of disperesed r phase have no affinity for dispersion medium and they are irreversible sols.

Reasons for coagulation of lyophobic sols : Lyophobic sols are easily precipited or coagulated on addition of small amounts of electrolytes, by heating or by shaking because they are not stable due to less force of attraction between dispersed phase and dispersion medium.

Question 12.
What is the difference between multimolecular and macromolecular col-loids ? Give one example of each. How are associated colloids different from these two types of colloids ?
Answer:
(i) Multimolecular colloids: In this type of colloids the colloidal particles consist of aggregates of atoms or small molecules with diameter of less than 1 nm. For example, a gold sol may contain particles of different sizes having several atoms of gold. Similarly sulphur sol contains about one thousand of S₈ molecules. These are held together by van der Waals’ forces.

(ii) Macromolecular colloids : In macromolecular colloids the dispersed particles themselves are big molecules, usually polymers. The molecular masses of these macro molecules range from thousands to millions. Synthetic compounds e.g. polyethylene, polystrene, nylon, rubber etc. are macromolecules. Since these molecules have dimensions comparable to those of colloidal particles, there dispersions are known as macromolecular colloids. Most lyophillic sols are of this category.

(iii) Associated colloids (Micelles): There are certain substances which behave as normal, strong electrolytes at low concentration, but exhibit colloidal properties at higher concentrations due to the formation of aggregated particles. These are called Micelles. These substances are also called Associated Colloids. Surface active agepts like soaps and detergents are of this class.

Question 13.
What are enzymes ? Write in brief the mechanism of anzyme catalysis.
Answer:
Enzymes are complex nitrogeneous organic compounds present in living beings. Enzymes catalyse many biochemical reactions taking place inside the human body like digestion, i.e., why enzymes are also called biocatalysts. For example,

Inversion of cane sugar is catalysed by invertase.

Mechanism of enzyme catalysis : There are present a number of cavities of characteristic shape on the surface of enzymes. The molecules of the reactant (substrate) which have complementary shape, fit into these cavities just as a key fits into a lock. An activated enzyme-substrate complex is formed which then decomposes to yield products. It involves following steps :

Step-I → Binding enzyme to substrate (reactant) to form a complex.

Step-II → Product formation from the complex

Step-III → Release of the product from the enzyme product complex.

Question 14.
How are colloids classified on the basis of:
(i) Physical states of components
(ii) Nature of dispersion medium and
(iii) Interaction between dispersed phase and dispersion medium ?
Solution:
Colloids can be classified on various bases as :
(i) On the basis of the physical state of the components (by components we mean the dispersed phase and dispersion medium). Depending on whether the components are solids, liquids, or gases, we can have eight types of colloids.
(ii) On the basis of the dispersion medium, sols can be divided as :

(iii) On the basis of the nature of the interaction between the dispersed phase and dispersion medium, the colloids can be classified as lyophilic (solvent attracting) and lyophobic (solvent repelling).

Question 15.
Explain, what is observed :
(i) When a beam of light is passed through a colloidal sol ?
(ii) An electrolyte, NaCl is added to hydrated ferric oxide sol ?
(iii) Electric current is passed through a colloidal sol ?
Answer:
(i) Scattering of light by colloidal particles takes place and path of the light becomes visible (Tyndall effect).
(ii) The positively charged colloidal particles of Fe(OH)₃ get coagulated by the oppositely charged Cl^– ions provided by NaCl.
(iii) On passing.electric current, the colloidal particles move towards the oppositely charged electrode where, they lose their charge and get coagulated. This is electrophoresis process.

Question 16.
What are emulsions ? What are their different types ? Give examples of each type.
Answer:
The colloidal system in which dispersed phase and dispersion medium both are liquid, is called emulsion.

Types of emulsions:
1. Oil in Water type (O/W) : In this type of emulsion small droplets of oil are dispersed in the water or dispersion medium, e.g., Milk, vanishing cream.

2. Water in Oil (W/O): In this type of emulsion, small quantity of water is dispersed in the form of droplets in oil (dispersion medium), e.g., Butter, Cod-liver.

Question 17.
What is demulsification ? Name two demulsifiers.
Answer:
The process of separation of constituent liquids of an emulsion is called demulsification. Demulsification can be done by centrifuging or boiling.

Question 18.
Action of soap is due to Emulsification or formation of micelle formation. Comment
Answer:
Cleaning action of soaps: Cleaning action of soap and detergents are based on miceller activity. Soaps are sodium salt of higher fatty acids e.g., sodium stearate which ionises as:

The anionic head of stearate ion (-COO^–) is hydrophilic and hence has great affinity for water. The hydrocarbon part is hydrophobic and has great affinity for oil, grease etc.

When soap dissolves in water the anions (C₁₇H₃₅COO^–) form micelle encapsulating oil or grease inside. These micelles are removed by rinsing with water. Thus the main function of soap is to convert oily and greasy dirt to colloidal particles by forming an emulsion. Soaps therefore, act as emulsifying agents. This mechanism of cleaning is also applicable to synthetic detergents like sodium lauryl sulphate CH₃(CH₂)₁₁ SO^–₄ Na⁺. Repulsion of similar charge -COO^– or -SO^–₄ covering each oil (grease) micelle prevents them to come together. Thus oil or grease remains in the suspension.

Question 19.
Give four examples of heterogeneous catalysis.
Answer:
Four examples of heterogeneous catalysis: (i) Oxidation of sulphur dioxide to form sulphur trioxide. In this reaction, Pt acts as catalyst.

(ii) Formation of ammonia by the combination of dinitrogen and dihydrogen in the presence of finely divided iron.

This process is called the Haber’s process.
(iii) Ostwald’s process : Oxidation of ammonia to nitric oxide in the presence of platinum.

(iv) Hydrogenation of vegetable is in the presence of Ni.

Question 20.
What do you mean by activity and selectivity of catalysts ?
Answer:
(a) Activity of a catalyst: The activity of a catalyst is its ability to increase the rate of a particular reaction. Chemisorptions is the main factor in deciding the activity of a catalyst. The adsorption of reactants on the catalyst surface should be neither too strong nor too weak. It should just be strong enough to make the catalyst active.
(b) Selectivity of the catalyst: The abilitiy of the catalyst to direct a reaction to yield a particular product is referred to as the selectivity of the catalyst. For example, by using different catalyst, we can get different products for the reaction between H₂ and CO.

Question 21.
Describe some features of catalysis by zeolites.
Answer:

1. Zeolites are hydrated alumino-silicates. They have three-dimensional network structure. They contain water molecules in their pores,
2. Zeolites are heated to remove water of hydration. The pores become vacant and zeolites are ready to act as catalyst.
3. The size of the pores varies from 260 pm to 760 pm. This shows that only those molecules can be adsorbed in these pores whose size is small enough to enter these pores. Thus, zeolites as molecular sieve and the shape selective catalysts.

Question 22.
What is shape selective catalysis ?
Answer:
A catalyst reaction is very specific, which depends upon the pore structure of the catalyst and on the size of the reactant and the product. Molecules is called shape-selective catalysis. For example: catalyst by zeolite is shape-selective catalysis.The pore size present in the zeolite ranges from 260-740 pm. Molecules having a pore size more than this cannot enter the zeolite and undergo the reaction.

Question 23.
Explain the following terms :
(i) Electrophoresis
(ii) Coagulation
(iii) Dialysis
(iv) Tyndall effect.
Answer:
(i) Electrophoresis : Colloidal particles are either positively charged or negatively charged. When this colloidal solution is put in electric field, colloidal particles move towards the electrodes of opposite charge and on reaching there deposited and get precipitated As, AS2S3 particles being negatively charged move towards anode and precipitated there.

The movement of colloidal particles towards the electrodes of opposite charge is called electrophoresis.This electrophoresis is also known as cataphoresis and anaphoresis depending upon movement towards cathode and anode respectively.

(ii) Coagulation : Colloidal particles are either positively charged or negatively charged. It is seen that on adding some oppositely charged ions or electrolyte, which cancels the charge of colloidal particles and these particles aggregate and change into precipitates. Precipitation of colloidal particles by electrolytes is called coagulation.

(iii) Dialysis: The process of separating impurity of crystalloids from colloidal sol by means of diffusion of the former through an animal or vegetable membrane, placed in running water, is called dialysis and the apparatus used for effecting such separation is called dialyser.

Graham’s dialyser consist [See Fig] of a parchment (or cellophone or cellulose acetate) bag, half immersed in water. The impure colloidal solution (i.e., a mixture of colloidal solution and crystalloids) is placed in the dialyser and the outside water is continuously renewed. The crystalloids slowly pass freely through the membrane into outside water, while pure colloidal solution is left inside the dialyser. This is, however, a very slow process. By allowing the process to continue over a period and pumping fresh water continually into the container, most the ions are eventually removed and the sols get purified. The membrane used is called semipermeable membrane, because it allows selectively the passage of only crystalloids particle and not colloidal particles.

(iv) Tyndall effect: In 1869 Tyndall noted that when a beam of light passes through colloidal solution, in a dark place, path of light is illu¬minated by bluish light. Colloidal particles scatter light and due to this scattering the light path, way is seen like a bright cone.

This effect is known as Tyndall effect. The same phenomenon is observed when a beam of sunlight passes through a small slit or hole in a dark room. This cone is perpendicular to light path.

Question 24.
Give four uses of emulsions.
Answer:
Applications or uses of emulsion :

1. Cleaning action of soaps and detergents are based on emulsification.
2. Emulsions are used as medicines e.g., codliver oil.
3. Many disinfectants like phenyl, dettol, lysol etc., are when mixed with water, form emulsions like oil.
4. Froth floatation process for concentration of ore is based on the principle of emulsification.
5. Digestion of fat by intestine also involves emulsification.

Question 25.
Explain the terms with suitable examples :
(i) Alcosol
(ii) Aerosol
(iii) Hydrosol.
Answer:
(i) Alcosol: It is a colloidal solution in which alcohol is the dispersion medium. For example, colloids which has cellulose nitrate as dispersed phase and ethyl alcohol as dispersion medium.
(ii) Aerosol: It is a colloidal solution in which liquid is a dispersed phase and gas is a dispersion medium e.g., fog, mist, cloud etc..
(iii) Hydrosol: It is a colloidal solution in which solid is a dispersed phase and water is a dispersion, e.g., gold sol, arsenious sulphide sol, ferric oxide sol etc.

Question 27.
Comment on the statement that “Colloid is not a substance but a state of substance”.
Answer:
The given statement is true. It is because colloidal is not the property of substance. Colloidal is a state in which all the substance can be brought by suitable methods. If the size of the particle lies in the range 1 nm to 1000 nm, it is in the colloidal state.

Surface Chemistry Other Important Questions and Answers

Surface Chemistry Objective Type Questions

Question 1.
Choose the correct answer :

Question 1.
In the adsorption process of acetic acid on activated charcoal, acetic acid is :
(a) Adsorber
(b) Absorber
(c) Adsorbent
(d) Adsorbate.

Question 2.
Cause of stability of lyophobic sol is :
(a) Brownian movement
(b) Tyndall effect
(c) Electric charge
(d) Brownian movement and Electric charge.

Question 3.
In the coagulation of As₂S₃ colloidal solution value of whose coagulating power will be minimum:
(a) NaCl
(b) KCl
(c) BaCl₂
(d) AlCl₃.

Question 4.
Adsorption process is :
(a) Endothermic
(b) Exothermic
(c) No heat change
(d) None of these.

Question 5.
Size of colloidal particles in the range of:
(a) 10^-7 to 10^-9 cm
(b) 10^-9 to 10^-1 cm
(c) 10^-5 to 10^-7 cm
(d) 10^-2 to 10^-8 cm.

Question 6.
Which of the following is not used for the preparation of lyophilic colloid :
(a) Starch
(b) Gum
(c) Gelatin
(d) Metal sulphide.

Question 7.
Sol which can act like a protective colloid :
(a) As₂S₃
(b) Gelatin
(c) Au
(d) Fe(OH)₃

Question 8.
Fog is an example of colloidal system of:
(a) Liquid dispersed in gas
(b) Gas dispersed in gas
(c) Solid dispersed in gas
(d) Solid dispersed in liquid.

Question 9.
Gelatin is mostly used in making ice creams in order to :
(a) Prevent forming a colloidal sol.
(b) Enrich the fragrance
(c) Prevent crystallization and stabilize the mixture
(d) Modify the taste.

Question 10.
In the reaction

(a) Auto catalyst
(b) Poison
(c) Negative catalyst
(d) Positive catalyst.

Question 11.
The migration of colloidal particles under the influence of electric fields is :
(a) Cataphoresis
(b) Electrodialysis
(c) Electrophoresis
(d) Electrical dispersion.

Question 12.
Hardy-Schulze law is related with :
(a) Solution
(b) Coagulation
(c) Solids
(d) Gases.

Question 13.
How many phases are present in a colloidal solution :
(a) 1
(b) 2
(c) 3
(d) 4.

Question 14.
Which is an emulsion among the following :
(a) Air
(b)Wood
(c) Butter
(d) Milk.

Question 15.
Butter is:
(a) A gel
(b) An emulsion
(c) A sol
(d) Not a form of colloid.

Question 16.
Sol which acts as protective colloid is :
(a) Gelatin
(b) Au
(c) As₂S₃
(d) Fe(OH)₂

Question 17.
Which is not correct for physisorption :
(a) A reversible process
(b) Needs low heat of adsorption
(c) Needs activation energy
(d) Needs low temperature.

Question 18.
In which among the following is Tyndall effect not expected :
(a) Suspension
(b) Emulsion
(c) Sugar solution
(d) Gold sol.

Answers:

1. (d), 2. (d), 3. (d), 4. (b), 5. (c), 6. (d), 7. (b), 8. (a), 9. (c), 10. (b), 11. (c), 12. (b), 13. (b), 14. (d), 15. (a), 16. (a), 17. (c), 18. (c).

Question 2.
Fill in the blanks :

1. Rate of physical adsorption ………………………… with the increase in temperature.
2. Particles of As₂S₃ sol are …………………………
3. In the contact process of manufacture of H₂SO₄ for Pt catalyst ………………………… act as …………………………
4. Oxidation of oxalic acid by KMnO₄ is an example of …………………………
5. Biological catalysts are necessarily …………………………
6. Movement of colloidal particles under the effect of electric field is called …………………………
7. Scattering of light by colloidal particles is called ………………………… effect|
8. Intermediate compound theory is applicable to ………………………… catalyst
9. The substance on whose surface adsorption takes place is called an …………………………
10. Milk is an example of …………………………
11. Blood is a ………………………… charged colloid.
12. Adsorption is an ………………………… process.
13. Colloidal solution of solid in liquid is called …………………………
14. Catalytic promoter substance is …………………………
15. On adding electrolyte, precipitation of colloidal particles is known as …………………………
16. According to Hardy Schulze law, coagulating capacity of ions depend on the ………………………… of ions

Answers:

1. Decreases
2. Negatively charged
3. As₂O₃, poison
4. Auto catalysis
5. Enzyme
6. Electrophoresis
7. Tyndall effect
8. Homogeneous
9. Adsorbent
10. Emulsion
11. Negatively
12. Exothermic
13. Gel
14. Molybdenum
15. Coagulation
16. Charge.

Question 3.
Match the following :

I.

Answers:

1. (f)
2. (d)
3. (a)
4. (c)
5. (g)
6. (e)
7. (b).

II.

Answers:

1. (d)
2. (a)
3. (b)
4. (e)
5. (c)
6. (f)
7. (g).

Question 4.
Answer in one word/sentence :

1. What happen when gelatin is mixed into gold sol ?
2. The substance which increases the strength of the catalyst but do not work as catalyst.
3. Dissociation of emulsion into its constituent liquid is known as :
4. The catalyst which is used for hydrogenation of oils ?
5. The decomposition of hydrogen peroxide by phosphoric acid represents which kinds of catalysis ?
6. Which catalyst converts glucose into alcohol ?
7. Among Na⁺, Ba⁺², Al⁺³, Sn⁺⁴ ions which one has the strongest coagulation power ?
8. Cl₂ gas mask works on which principle ?
9. Conversion of a precipitate into colloidal particles is known as:
10. Cleansing action of soap based on which principle ?
11. Who used the word catalyst for the first time ?
12. Write the size of colloidal particles.
13. Write nature of catalyst used in oxidation of oxalic acid by KMnO₄.
14. Give an example of catalytic poisoning.
15. Movement of particles of colloids is called.
16. Which one catalyst is used for hydrolysis of cane sugar.

Answer:

1. Protection
2. Catalyst promotor
3. Demulsification
4. Ni
5. Negative catalyst
6. Zymase
7. Sn⁺⁴
8. Adsorption
9. Peptization
10. Emulsification
11. Bergellius
12. 10^-5 to 10^-7 cm
13. Auto catalyst
14. As₂O₃ in contact process
15. Brownian motion
16. Invertase enzyme.

Surface Chemistry Very Short Answer Type Questions

Question 1.
Which inert gas is adsorbed in (i) Minimum amount and (ii) Maximum amount on the surface of gas charcoal ?
Answer:

1. Helium because of its lower atomic mass and the weakest van der Waals’ forces of attraction is adsorbed on gas charcoal in minimum amount.
2. Xenon because of its high atomic mass and the strongest van der Waals’ forces of attraction is adsorbed on gas charcoal in maximum amount.

Question 2.
Why is sky blue in colour ?
Answer:
Dust particles present in air forms colloidal solution whose particles shows Tyndall effect and hence sky appears blue in colour.

Question 3.
KMnO₄ decolourises slowly initially and then rapidly when added to oxalic acid. Why ?
Answer:
Addition of KMnO₄ to oxalic acid oxidises it. Mn⁺² ion produced in this reaction acts as autocatalyst and hence decolourisation occur rapidly.
2MnO^–₄ + 5C₂O^-2₄ + 16H^– → 2Mn⁺² + 10CO₂ + 8H₂O

Question 4.
What is Colloidal solution ?
Answer:
The heterogeneous solution in which solute particles of 10^-7 cm to 10^-4 cm diameter and solvent particles are within 10^-8 cm to 10^-7 cm are called colloidal solution.
e.g. Milk, blood, cloud, smoke, etc.

Question 5.
What is adsorbent ?
Answer:
Solid substances which adsorb the gases or solution are called adsorbent. Like : animal charcoal, silica etc.

Question 6.
What are adsorbate ?
Answer:
Gas molecules, vapours or ions which are adsorbed on the surface of solids are called adsorbate.

Question 7.
Delta is formed when river water meets sea water. Explain.
Answer:
In the river water negatively charged particles of soil and sand are present. When river water meets sea-water, various ions Na⁺, K⁺ or Mg⁺⁺ present in sea water causes coagulation and soil particles etc. are settle down. Thus a delta is formed.

Question 8.
How is rain possible by spraying silver iodide on clouds ?
Answer:
Clouds are charged due to colloidal nature. Silver iodide is an electrolyte. Spray¬ing it on clouds lead to coagulation by which it rains.

Question 9.
How is the adsorption of a gas related to its critical temperature ?
Answer:
Higher is the critical temperature of a gas, greater is the ease of liquification of gas i.e., larger are the van der Waals’ forces of attraction. Therefore, greater is the adsorption.

Question 10.
What happen when a freshly precipitated Fe(OH)₃ is shaken with little amount of dilute solution of FeCl₃ ?
Answer:
A reddish brown colloidal solution of Fe(OH)₃ is obtained. This process is called peptization. The Fe⁺³ ions from FeCl₃ are absorbed on the surface of the precipitate and form positively charged colloidal solution.
Fe(OH)₃ + Fe⁺³ → [Fe(OH)₃]Fe⁺³

Surface Chemistry Short Answer Type Questions

Question 1.
What is meant by colloidal solution ? What is dialysis ? What is its principle ?
Answer:
Colloidal solution: The solutions in which solute particles are of 10^-7 cm. to 10^-4 cm. iameter and solvent particles are within 10^-8 cm. to 10^-7 cm., are called colloidal solution. Colloidal solutions are heterogeneous in nature.
Dialysis : The process by which dissolved impurities are removed with the help of parchment paper or membrane, is called dialysis.
Colloidal solution is filled in a bag, made of parchment paper and suspended in distilled water. Particles of dissolved crystalloids come out of the bag and are diffused in the water. Thus, the colloidal solution is purified.

Question 2.
Define Sol, Gel and Emulsion with example.
Answer:
When the dispersed phase of colloidal solution is solid and dispersion medium is liquid, it is called sol.
e.g., Colloidal solutions of ferric hydroxide, gold etc. in water.
Gel: When any liquid is dispersed in solid and form colloidal solution, is called gel.
e.g., Paneer, butter etc.
Emulsion : When dispersed state and dispersion medium both are liquid, it is called emulsion.
e.g., Milk, Cream etc.

Question 3.
What is homogeneous and heterogeneous catalysis ? Explain with example.
Answer:
Homogeneous catalysis : The chemical reaction in which reactants and catalysts are in same physical state, are example of homogeneous catalysis.
Example : In lead chamber process of manufacturing of sulphuric acid reactant and catalyst NO are in same physical state – gaseous.

Heterogeneous catalysis : The chemical reactions in which reactants and catalysts are in different physical state, are examples of heterogeneous catalysis.
Example: In Haber process manufacture of Ammonia.

Question 4.
What is ‘Gold number’ ? Explain with example.
Answer:
Gold Number: Protective actions of different lyophilic colloids are compared with gold number. Gold number can be defined as :

“Gold number is the number of miligrams of protective colloid (Lyophilic) which on adding prevents the coagulation of 10 ml. standard gold sol by addition of 1 ml. of 10% NaCl solution.”

Thus smaller the value of gold number, higher is it’s protecting action.

For example, gold number of gelatin and gum arobic are 0.005 and 015 respectively. It means gelatin is better protective colloid because it’s only 0-005 mg quantity is required to prevent coagulation if 10 ml Au sol by addition of 1 ml. 10% NaCl.

Question 5.
What is the importance of emulsifying agents in emulsifications ?
Answer:
The process of making emulsions is called emulsification. Emulsions are formed when suitable liquids are mixed and shaked, but the emulsions are thus formed are not stable. To make these emulsions stable some other substances are added, which are known as emulsifying agents. Soap, gum, starch etc. act as emulsifying agents. In absence of emulsifying agents, disperse drops of liquid meet each other to distroy emulsion state. It is supposed that emulsifying agents form membrane at interphase of oil and water which checks the union of droplets.

Question 6.
What is adsorption ? Give its two example and explain its mechanism.
Answer:
Adsorption is a surface phenomenon in which particles of different substances or gases are temporary linked with the surface of liquid or solids to satisfy the free valencies of solid surfaces.
Examples of adsorption :
(i) In sugar industry, animal charcoal adsorbs coloured materials.
(ii) In permutit process, Ca²⁺, Mg²⁺ ions are adsorbed and Na+ ion are released by which hardness of water is removed.

Mechanism of adsorption : Atoms or molecules present on the surface of solid con¬tain free valencies. These free valencies attract the molecules of adsorbate.

Question 7.
What do you understand by Electro-dialysis ?
Answer:
Electro-dialysis : Particles of true solutions pass through parchment paper or cellophane but sol particles cannot pass through such membranes. In dialysis, the sol filled in a bag of parchment or cellophane is suspended in pure water. This process is very slow and takes a long time for completion. However, it can be quickened under the influence of an electric field and the process is known as Electro-dialysis.

Artificial kidney used in medical science works on the principle of dialysis.

Question 8.
What do you understand by protective colloids ?
Answer:
Protective colloid: On adding small quantity of electrolyte, Lyophobic colloids easily coagulated but if some quantity of hydrophillic colloid is added in Lyophobic colloids, it minimize the effect of addition of electrolyte and coagulation is prevented or very slowly coagulated. This process is called protection of colloid. The hydrophillic colloid which is responsible for protection against coagulation is called protective colloid.

Example: If a small amount of gum, gelatin or starch is added to As₂S₃ sol, its coagulation of NaCl solution is prevented. Thus starch, gum or gelatin protects the As₂S₃ sol.

Question 9.
Give the method of preparation of colloidal solution of ferric hydroxide and sulphur in water.
Answer:
(i) To prepare sol of ferric hydroxide, FeCl₃ solution is added in the boiling water dropwise with shaking. Excess of FeCl₃ and HCl are separated by electrodialysis and sol is also stabilized by this.
FeCl₃ + 3H₂O → Fe(OH)₃ + 3HCl

(ii) H₂S gas is passed in the solution of nitric acid (oxidising agent) and colloidal solution of sulphur is obtained.
2HNO₃ + H₂S → S + 2H₂O + 2NO₂.

Question 10.
Explain physical and chemical adsorption.
Answer:
In physical adsorption the adsorbate molecules are all reacted by van der Waals’ forces on the surface of adsorbent. These van der Waals’ forces are weak in nature. The physical adsorption is also known as van der Waals’ adsorption or physical adsorption.
In chemical adsorption, the adsorbate molecules are attached with adsorbent through some chemical bonds, because some unsatisfied free valencies exist m the surface of adsorbent. This adsorption is also known as chemisorption.

Question 11.
What do you understand with enthalpy of adsorption ?
Answer:
In adsorption process, when one mole of adsorbate is adsorbed on the surface of adsorbent, the enthalpy change in this is called enthalpy of adsorption or heat of adsorption. The enthalpy of adsorption for chemisorption as about 400 kJ/mol while in physical adsorption it is about 40 kJ/mol.

Question 12.
What is catalysis ? Explain induced catalysis with an example.
Answer:
Catalysis : A catalyst is a substance which alters the rate of chemical reaction without being used in the reaction and the phenomenon is known as catalysis.
Induced Catalysis : In induced catalysis, one reaction already taking place, catalyse the other reaction also which does not occur separately.
Example : Sodium sulphite (Na₂SO₃) oxidise in the atmosphere easily while sodium arsenite (Na₃AsO₃) does not oxidise in atmosphere separately. When (Na₂SO₃) and Na₃AsO₃ both are kept together, both are oxidised.

Question 13.
Explain positive and negative catalysis with example.
Answer:
Positive Catalysis : When the velocity of any chemical reaction increases with the presence of catalyst, this type of catalysis is called positive catalysis.

Negative Catalysis : Catalysts when decrease the velocity of chemical reaction are called negative catalysts or retarders and the process is known as negative catalysis.

Question 14.
What is peptization ? Explain.
Answer:
Peptization : Peptization is a good method of preparing colloidal sols from precipitates. A fresh precipitate is taken for the purpose and a suitable reason or peptizing agents are added. These peptizing agents are generally dilute solutions of electrolytes of common ion. Peptization process is op-posite of the process of coagulation.

Example : When fresh precipitate of aluminium hydroxide is boiled with dilute HCl mixed water, colloidal solution of Al(OH)₃ is obtained.

When an electrolyte is added to a fresh precipitate then the particles of the electrolyte preferentially adsorbs an ion and due to electrostatic repulsion undergo in colloidal state. On adding electrolyte ferric chloride to ferric hydroxide precipitate, sol of ferric hydroxide is obtained.

Question 15.
Deduce the origin of electrical charge on colloidal particles.
Answer:
Colloidal particles have electrical charge and all colloidal particles of any colloidal solution have similar positive or negative charge. To explain origin of charge, following theories are given :
(i) Due to internal friction : According to this theory charge on colloidal particles originates due to mutual friction of dispersion medium and dispersed phase.
(ii) By ionisation of groups present in surface: Colloidal particles may also acquire charge by direct ionisation of different groups present in the surface. These groups may be either colloidal particles or any electrolyte added to prevent coagulation.
(iii) Electrical double layer theory : According to this theory electrical charge on colloidal particles is double layered. One layer is consist of absorbed ions and other layer is diffused layer which is consist of oppositely charged ions in dispersion medium.

Question 16.
Give reason :
(i) Milk turns sour on adding acid to it.
(ii) Alum is added to purify water.
(iii) Delta is formed where river water meets sea.
Answer:
(i) Milk is an emulsion of fats dispersed in water. Albumin and caesin are emulsifiers. Addition of acid destroys emulsifiers and hence milk gets coagulated.
(ii) In impure water, soil particles, bacteria and other soluble impurities are dissolved.
When alum is added, the Al⁺³ ion present in alum destroys the negative charge of impure water. Due to neutralization of charge the impurities are coagulated and settle down.
(iii) In the river water negatively charged particles of soil and sand are present. When river water meets sea-water, various salts present in sea water causes coagulation and soil particles etc. are settle down. Thus a delta is formed.

Question 17.
Write characteristics of catalysts.
Answer:
Characteristics of catalysts :

1. No participation : Catalysts do not participate in the chemical reaction. Only it’s physical state may be changed.
2. Small quantity : A small quantity of catalyst is always required for catalysing reaction.
3. No effect in equilibrium: Catalysts do not affect equilibrium state because catalyse both forward and backward reactions.
4. Specific nature: Catalysts are of specific nature and catalyse only definite reaction not all.
5. Effective temperature : There is a optimum temperature in which catalyst’s activity is maximum
6. Promotors and Catalytic poisons : There are some substances which increase the activity of catalyst, by their presence. These substances are known as promotors. Catalytic poisons are substances which destroy the catalytic activity.

Question 18.
Coloured glass, smoke, milk, cream, fog, jelly, pumic stone and soap lather belongs to which type of colloidal system? Name them.
Answer:

Question 19.
Compare the properties of true solution, colloidal solution and suspension.
Answer:
Distinction between True solution, Colloidal solution and Suspension :

Question 20.
What do you understand by activation of the adsorbent ? How is it obtained ?
Answer:
Activation of adsorbent means, increasing the activation power of adsorbent. It can be done by increasing the surface area of adsorbent which can be obtained as follows :

1. By removing the adsorbed gases or by heating charcoal in vacuum to extremely high temperature in steam at 650K to 1330K temperature.
2. By breaking the adsorbate into small pieces.
3. By making the surface of adsorbent rough.

Question 21.
What is the role of adsorbent in heterogenous catalysis ?
Answer:
Generally in heterogenous catalysis, adsorbate is in gaseous where as catalyst is in solid state. Adsorbate molecules are adsorbed on the surface of solid catalyst by physical or chemical adsorption. Due to increase in concentration of adsorbed molecules or by the formation of active species due to breaking up of adsorbed molecules due to reaction occurs fast. Product molecules are desorbed and catalytic surface again becomes available for the adsorption of reactant molecules. This principle is known as adsorption of heterogenous catalysis.

Surface Chemistry Long Answer Type Questions

Question 1.
Explain the following :
(i) Brownian movement
(ii) Autoca-talysis.
Answer:
(i) Brownian movement : It is observed by an ultra-microscope that colloidal particles move in a zig-zag path in all possible directions. Thus, zig-zag or random motion of sol particles is called Brownian movement. It is caused due to unequal bombardment of Small particle the molecules of dispersion medium on the sol particles. First of all this type of movement is reported by Robert Brown so it is known as Brownian movement. On increasing the size of particles, their movements also decreases and in suspension it is finished.

(ii) Autocatalysis: In autocatalysis, one of the product behave as catalyst and increases the velocity of reaction.
Example : Hydrolysis of Ethyl acetate forms Acetic acid which act as autocatalyst.
CH₃COOC₂H₅ + H₂O → [CH₃COOH] + C₂H₅OH
Acetic acid

Question 2.
Write differences between lyophilic and lyophobic colloids.
Answer:
Differences between Lyophilic and Lyophobic colloids :

Question 3.
Write Hardy Schulze’s law.
Answer:
Hardy-Schulze’s law : Coagulating power of an ion is governed by this law according to which, “Greater the valency of an ion greater will be its coagulating power.”

Thus, to coagulate negative sol, the coagulating power of different cations are found to be in the order :
\(\mathrm{Sn}^{4+}>\mathrm{Al}^{3+}>\mathrm{Ba}^{2+}>\mathrm{Na}^{+}\)
Similarly for the coagulation of a positive sol of Fe(OH)3, the coagulating power of different anions is found to be in the order :
\(\mathrm{PO}_{4}^{3-}>\mathrm{SO}_{4}^{2-}>\mathrm{Cl}^{-}\)

Question 4.
Explain intermediate compound theory of catalysis.
Answer:
Intermediate compound theory of catalysis : Various theories are given to explain the mechanism of catalysis. This theory explains the mechanism of liquid and gaseous catalysts. According to this theory catalyst combines with one of the reactant and form an intermediate compound. This intermediate compound then reacts with other reactant and form aspected compound and catalyst is released which again acts as catalyst.

For example, if A and B combine and form a compound AB where X is catalyst then AX is intermediate compound.
A + X → AX (intermediatecompound)
AX + B → AB + X
The time in this reaction is less than time required when A and B combine directly.

Example: In lead chamber process of manufacturing of H₂SO₄, NO acts as catalyst. NO combines with oxygen to form NO₂ as intermediate compound. NO₂ now reacts with other reagent SO₂ and form SO₃, the catalyst NO released which further enhance the chain.

Question 5.
Write any five differences between enzyme catalyst and normal catalyst.
Answer:
Differences between enzyme catalyst and normal catalyst:

Enzyme catalyst:

1. Is the catalyst for biochemical reactions.
2. These are active at normal body temperature and pH.
3. These are produced in living cells and are protein in normal form.
4. Their molecular mass are high.
5. These become inactive at low temperature.

Normal catalyst

1. Is the catalyst for normal chemical reactions.
2. These function at a definite favourable temperature.
3. These can be metals, non-metals or compounds of non-metals.
4. Their molecular mass are low. These are also ineffective at low temperature.

Question 6.
What is adsorption ? Explain the factors affecting it
(a) On what factors does the extent of adsorption of gases on solid depend ?
(b) Write five applications of adsorption.
Answer:
Adsorption: Refer Short Answer Type Q. No. 6.
(a) The extent of adsorption of a gas on a solid adsorbent is affected by the following factors:

1. Nature of the gas: Under given conditions of temperature and pressure, the easily liquefiable gases like NH3, C02, HC1, etc. are adsorbed to a greater extent than permanent gas.
2. Nature of the adsorbent: Activated charcoal is the most common adsorbent for the gases which are easily liquefied.
3. Specific area of the solid : Greater the specific area of the solid, greater will be its adsorption power.
4. Pressure of the gas : The extent of adsorption increases with increase in pres¬sure.
5. Effect of temperature: Adsorption at a surface initially increases till a saturation point is achieved and to this point equilibrium is developed. The magnitude of adsorption decreases with rise in temperature.
6. Activation of adsorbent: The adsorption power can be increase by making the surface of adsorbent rough or by breaking it into small pieces. Adsorption power decreases if the particles are made very small because the interparticle space will be two small to allow penetration.

(b) Various applications of adsorption are in our daily life. Some important applications are:

1. In Catalysis : In industries various catalysts are used to promote the reaction. These catalysts are generally based on adsorption principle. e.g., Haber process of ammonia, preparation of sulphuric acid by contact process etc.
2. In gas masks : Activated charcoal is used to adsorb harmful gases as CO,CH₄ etc. Activated charcoal or any other suitable adsorbent are kept in gas masks.
3. In creating vacuum : High vacuum can be created by adsorbing gases by adsorbents.
4. For decolourising and de-odourising sugar : Animal charcoal is used for decolourising sugar and in de-odourising process.
5. In chromatography: Purification of compounds by chromatography is also based on adsorption theory.

Question 7.
Write five applications of colloidal chemistry.
Answer:
The five applications of colloidal chemistry are as follows :
1. In medicines : Colloidal medicines are easily adsorbed by body tissue and are more effective, e.g., Argyrol, colloidal antimony, colloidal gold etc.

2. Smoke precipitation : Smoke and dust are also colloidal in nature and smoke precipitation is based on the principle of electrophoresis. Cottrell precipitator is nowadays commonly used in various industries.

3. Colloids in nature : Different forms of colloidal solutions can be seen in nature. Delta formation, fertile soil, rain, blue colour of sky are some examples of colloidal solutions.

4. Action of Soap: Cleaning action of soap is explained on the basis that soap solution is colloidal. Soap emulsifies greasy and oily materials sticking on the surface of body or clothes.

5. In leather industry: The colloidal particles in hides are positively charged and on soaking the hide in tannin solution (a negative solution) mutual coagulation takes place. This process is called tanning and it makes the hide hard. Solutions of chromium salts are used in place of tannin with advantage and the process is called chrome-tanning.

MP Board Class 12th Chemistry Solutions

MP Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.7

Assume π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.
Find the volume of the right circular cone with

1. radius 6 cm, height 7 cm
2. radius 3.5 cm, height 12 cm.

Solution:
1. Here, radius of the cone r = 6 cm
height (h) = 7 cm
Volume = \(\frac{1}{3}\) x πr²h
= \(\frac{1}{3}\) x \(\frac{22}{7}\) x 6 x 6 x 7 cm³
= 22 x 2 x 6 cm³
= 264 cm³

2. Here, radius of the cone (r) ;
= 3.5 cm = \(\frac{35}{10}\) cm
Height (h) = 12 m
Volume of the cone

Question 2.
Find the capacity in litres of a conical vessel with

1. radius 7 cm, slant height 25 cm
2. height 12 cm, slant height 13 cm

Solution:
1. Here, r = 7 and l = 25 cm

Thus, the required capacity of the conical vessel is 1.232 l.

2. Here, height (h) – 12 cm and l = 13 cm

Thus, the required capacity of the conical vessel is \(\frac{11}{35}\) l.

Question 3.
The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use 71 = 3.14)
Solution:
Here, height of the cone (h) = 15 cm
Volume of the cone (v) = 1570 cm³
Let the radius of the base be ‘r’ cm.

Question 4.
If the volume of a right circular cone of height 9 cm is 48 JI cm3, find the diameter of its base.
Solution:
Volume of cone = \(\frac{1}{3}\) πr²h
\(\frac{1}{3}\) x πr² x 9 = 48π
r² = \(\frac{48π}{9π}\) x 3
r² = 16
r = 4 cm
Diameter = 2 x 4 = 8 cm

Question 5.
A conical pit of top diameter 3,5 m is 12 m deep. What is its capacity in kilolitres?
Solution:
d = 3.5 m
r = 1.75 m
h = 12m
Volume of the pit = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 1. 75 x 1.75 x 12
= 38.5 m³ = 38.5 kl (1 kl= 1 m³)

Question 6.
The volume of a right circular cone is 9856 cm³. If the diameter of the base 28 cm, find.

(i) height of the cone.
(ii) slant height of the cone.
(iii) curved surface area of the cone.

Solution:
V = 9856 cm²
d =28 cm
r = 14 cm

Question 7.
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the sides 12 cm. Find the volume of the solid 30 obtained.
Solution:
h = 12cm
r = 5cm

Volume of solid, V₁ = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) x π x 5 x 5 x 12
= 100π cm³

Question 8.
If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
Solution:
h = 5 cm
r = 12 cm
Volume of solid, V₂ = \(\frac{1}{3}\) x π x 12 x 12 x 5
= 240π

Question 9.
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be by covered canvas to protect it from rain. Find the area of the canvas required.
Solution:
d = 10.5 m ⇒ r = 5.25 m
h = 3m
Volume of heap of wheat = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 5.25 x 5.25 x 3
= 86.625 m³
l² = 3² + (5.25)²
l = \(\sqrt{9+27.56}\) = \(\sqrt{36.56}\) = 6.04 cm
Area of canvas required = CSA of cone –
= \(\frac{22}{7}\) x 5.25 x 6.04 = 99.66 m²

MP Board Class 9th Maths Solutions

MP Board Class 6th Science Solutions Chapter 13 Fun with Magnets

Fun with Magnets Textbook Exercises

Question 1.
Fill in the blanks in the following:

1. Artificial magnets are made in different shapes such as …………….. , …………….. and ……………..
2. The materials which are attracted towards a magnet are called ……………..
3. Paper is not a …………….. material.
4. In olden days, sailors used to find direction by suspending a piece of ……………..
5. A magnet always has …………….. poles.

Answer:

1. Bar magnet, horse – shoe magnet, cylindrical.
2. Magnetic material
3. Plastic,
4. Suspended magnet
5. Two.

Question 2.
State whether the following statements are True (T) or False (F):

1. A cylindrical magnet has only one pole.
2. Artificial magnets were discovered in Greece.
3. Similar poles of magnet repel each other.
4. Maximum iron filings stick in the middle of a bar magnet when it is brought near them.
5. Bar magnets always point towards North – South direction.
6. A compass can be used to find East – West direction at any place.
7. Rubber is a magnetic material.

Answer:

1. False
2. False
3. True
4. False
5. True
6. False
7. False.

Question 3.
It was observed that a pencil sharpener gets attracted by both the poles of a magnet although its body is made of plastic. Name a material that might have been used to make some part of it.
Answer:
Iron.

Question 4.
Column I shows different positions in which one pole of a magnet is placed near that of the other. Column II indicates the resulting action between them for each situation. Fill in the blanks.

Answer:

Question 5.
Write any two properties of a magnet.
Answer:
The properties of a magnet are:

1. It attracts the objects made of iron, cobalt and nickle.
2. When it is suspended freely then it stands always in North – South direction.

Question 6.
Where are poles of a bar magnet located?
Answer:
On the two ends of the bar magnet.

Question 7.
A bar magnet has no markings to indicate its poles. How would you find out near which end is its north pole located?
Answer:
Take a bar magnet. Put a mark on one of its ends for identification. Now, tie a thread at the middle of the magnet so that you may suspend it from a wooden stand [Fig.]. Make sure that the magnet can rotate freely. Let it come to rest. Mark two points on the ground to show the position of the ends of the magnet when it comes to rest. Draw a line joining the two points. This line shows the direction in which the magnet Was pointing in its position of rest. Now, rotate the magnet by gently pushing one end in any direction and let it come to rest. Again, mark the position of the two ends in its position of rest.

Rotate  the magnet in other directions and magnet always comes to rest note the final direction in which it in the same direction comes to rest. We find that a freely suspended bar magnet always comes to rest in a North –  South direction. The end of the magnet that points towards North is called its North seeking end or the North pole of the magnet.

Question 8.
You are given an iron strip. How will you make it into a magnet?
Answer:
Take the iron strip and place it on the table. Now take a bar magnet and place one of its poles near one edge of the bar of iron. Without lifting the bar magnet, move it along the length of the iron bar till you reach the other end. Now, lift the magnet and bring the pole (the same pole you started with) to the same point of the iron bar from which you began (fig.) Move the magnet again along the iron bar in the same direction as you did before.

Repeat this process about 30 – 40 times. Bring a pin or some iron filings near the iron bar to check whether it has become a magnet. If not, continue the process for some more time. Remember that, the pole of the magnet and the direction of its movement should not change. In this way, an iron strip can be converted into a magnet.

Question 9.
How is a compass used to find directions?
Answer:
Magnetic Compass.
Magnetic compass consists of a small magnetized needle which is enclosed in a small round box made of aluminium. The box consists of a small pointed vertical axis (pivot) on which the magnetized needle rotates freely. The aluminium box is covered by thin glass cover.

The magnetized needle rotates freely and points towards geographic north from which true north can be calculated. This magnetic compass is used to find the direction at any place. Sailors and navigators also use this compass to find the direction during their journey.

Question 10.
A magnet was brought from different directions towards a toy boat that has been floating in water in a tub. Affect observed in each case is stated in Column I. Possible reasons for the observed affects are mentioned in Column II. Match the statements given in Column I with those in the Column II

Answer:

(i) (d)
(ii) (e)
(iii) (b)
(iv) (a)
(v) (b).

Projects and Activities

Activity 1.
collect various objects of day – to – day use from your sourroundings. Test these with the “Magnes stick”. You can also take a magnet, touch these objects with it and observe which objects stick to the magnet. Prepare a table and record your observations.
Answer:
Table: Finding the objects attracted by magnet

Fun with Magnets Text Book Questions

Question 1.
Boojho has this question for you. A tailor n was stitching buttons on his Shirt. The needle has slipped from his hand on to the floor. Can you help the tailor to find the needle?
Answer:
Yes, with the help of magnet.

Question 2.
Paheli has this puzzle for you. You are given tow indentical bars which look as if they might be made of iron. One of them is a magnet, while the other is a simple iron bar. How will you find out, which one is a magnet?
Answer:
The magnet has two poles where attraction is maximum. But in iron bar there is no pole, so attraction is same everywhere.

Fun with Magnets Additional Important Questions

Fun with Magnets Objective Type Questions

Question 1.
Choose the correct answer:
Question (i)
North pole of a magnet –
(a) Attracts north pole
(b) Remains inactive
(c) May attract or repel
(d) Repels north pole.
Answer:
(d) Repels north pole.

Question (ii)
Which one is natural magnet –
(a) Ferrite
(b) Carbon
(c) Iron
(d) Magnetite.
Answer:
(b) Carbon

Question (iii)
Which of the following is non – magnetic –
(q) Iron
(b) Nickel
(c) Cobalt
(d) Aluminium.
Answer:
(d) Aluminium.

Question (iv)
An electrolyte is a –
(a) Solid that conducts electricity
(b) Liquid that does not conduct electricity
(c) Solid that does not conduct electricity
(d) Liquid that conducts electricity and breaks up chemically in the process.
Answer:
(d) Liquid that conducts electricity and breaks up chemically in the process.

Question (v)
Strength of a magnet is –
(а) Maximum at centre
(b) Maximum at poles
(c) Minimum at poles
(d) None of these.
Answer:
(а) Maximum at centre

Question 2.
Fill in the blanks:

1. Iron and nickle are …………….. materials whereas copper and aluminium are …………….. materials.
2. The regions of strongest magnetism in a magnet are known as the ………………
3. Like poles ………………. whereas unlike poles
4. The molecular magnets in a magnet are arranged in groups called ……………..
5. Steel is used to make …………….. magnets.
6. The shaving blade is an inexpensive ………………

Answer:

1. Magnetic, non – magnetic
2. Ends
3. Repel, attract
4. Domains
5. Permanent
6. Magnet.

Question 3.
Which of the following statements are true (T) false (F):

1. The magnetite is mainly composed of oxides of iron.
2. Like poles of a magnet attract each other,
3. Plastic is a non – magnetic material.
4. A magnet looses its magnetism on striking voilently with hammer.
5. Magnets point towards East – West when freely suspended.
6. Earth behaves like a huge magnet.
7. Orested a Danish scientist noticed that a compass needle was affected by the electric current flowing in the neighbouring coil of wire.
8. The most important property of magnet is to attract iron pieces.
9. A horse – shoe magnet.can have only one pole.
10. A magnet can have two north poles.

Answer:

1. True
2. False
3. True
4. False
5. True
6. True
7. True
8. Ture
9. False
10. False.

Question 4.
Match the items of Column A with Column B:

Answer:

(i) (b)
(ii) (c)
(iii) (d)
(iv) (a).

Fun with Magnets Very Short Answer Type Questions

Question 1.
When was magnetite discovered?
Answer:
Magnetite was discovered in 800 B. C.

Question 2.
From where a mineral magnetite was discovered first.
Answer:
It was discovered in the town of Magnesia.

Question 3.
Why was magnetite named Londestone?
Answer:
It was named so because this mineral can align itself in the same direction it left to rotate freely.

Question 4.
Which is the North Pole of a bar magnet?
Answer:
It is the tip of a bar magnet which points towards the north direction.

Question 5.
Which is the South Pole of a bar magnet?
Answer:
It is the tip of a bar magnet which points towards the south direction.

Question 6.
What is a temporary magnet?
Answer:
It is piece of magnet only for a short period. This magnetism is lost as soon as the source of magnetisation is removed.

Question 7.
How should we store magnets?
Answer:
We should store magnets by joining opposite poles of two magnets preferably with a piece of iron in between.

Question 8.
Who discovered the magnetic effects of electric current?
Answer:
Danish Scientist H. C. Uersted discovered the phenomenon in 1820.

Question 9.
What are domains?
Answer:
When we study a magnet in microscope, it consists of small regions. These regions are called domains.

Question 10.
How is magnet and electric current similar in property?
Answer:
Because wires having electric current behave like a magnet. In electric current also like charges repel and unlike charges attract

.

Question 11.
What is a magnetic needle?
Answer:
This needle moves a pivot fixed at the centre of a round frame box. It is used by navigators for finding out the direction in the sea.

Question 12.
Name some non-magnetic materials.
Answer:
Plastic, cloth, paper and leather, etc. are non – magnetic materials.

Question 13.
What are the ends of a magnet called?
Answer:
Poles.

Question 14.
What is natural magnet called?
Answer:
Magnetite.

Question 15.
Which iron is used for making a permanent magnet?
Answer:
Hard steel and alloys.

Question 16.
What is the instrument in which directive property of a magnet is used?
Answer:
Mariner’s compass.

Fun with Magnets Short Answer Type Questions

Question 1.
What is a magnet? How is a temporary magnet different from a permanent magnet?
Answer:
Any material that has the property of attracting iron is called a magnet.
Difference between temporary and permanent magnet:

Temporary magnet:

• It is magnet for a short period.
• Its magnetism is lost if source of magnetisation is removed.
• It is made up of soft iron.

Permanent magnet:

• It is magnet for ever.
• It remains magnet even if the source of magnetisation is removed.
• It is made up of steel.

Question 2.
Can you make out the south and north direction with the help of bar magnet. How?
Answer:
Yes, we can make out the south and north directions with the help of bar magnet. Suspend a magnet from a string tied to its middle so that it can rotate freely. The tip of the bar magnet which point north will rest to the north direction and the tip bearing south will rest to point towards south direction.

Question 3.
Define magnetism. Also give the name of magnetic and non – magnetic substance.
Answer:
Magnetism:
Magnet has the property of attracting various substances towards itself. This property of magnet is called magnetism.

Magnetic substances:
Those substances which are attracted by a magnet are called magnetic substances. For examples-, nickel, iron steel, cobalt and alloys of these substances are magnetic in nature.

Non – magnetic substances:
Those substances which are not attracted by a magnet are called non-magnetic substances. For examples, glass, cloth, wood, cloth, etc.

Question 4.
How can you make your personal compass?
Answer:
Magnetise an iron needle using a bar magnet. Now, insert the magnetised needle through a small piece of cork or foam. Let the cork float in water in a bowl or a tub. Make sure that the needle does not touch the water (Fig.). Your compass is now ready to work.

Question 5.
What is an electromagnet? Give two applications of electromagnets?
Answer:
When an electric current flows through a coil of wire, the coil behaves like a permanent magnet. When this current carrying coil is brought near a suspended bar magnet, one end of the coil repeals the north pole of the magnet. The other end of the coil attracts the north pole of the magnet. Thus, a current carrying coil has both a north and south pole like a magnet, such a magnet is called electromagnet.

Applications of electromagnet:

1. They are used in electric bells, telephones, telegraphps, etc.
2. They are used to separate magnetic substances like nickel, cobalt and iron from non – magnetic substances like brass, copper, zinc and plastics.

Fun with Magnets Long Answer Type Questions

Question 1.
What happens when the north pole of a magnet is brought near (i) the north pole, (ii) the south pole of a freely suspended magnet?

Answer:
Suspend a bar magnet with a string. This bar magnet should be marked with north and south pole.

1. Hold another magnet in hand and bring its north pole close to each pole one by one. North pole of two magnet will repel each other as they are like poles.
2. While north of magnet, in hand comes close to south of suspended magnet, it will show attraction because they are unlike poles.

Question 2.
Can we have an isolated north pole or south pole?
Answer:
An isolated magnetic pole is not possible. Two magnetic poles of a magnet cannot be separated.
If a magnet is broken into two pieces, each piece is a magnet having north and south pole. If these pieces are further broken into more smaller pieces, each piece consists the two poles (i.e., north pole and south pole) at its ends.

Dividing a magnet infinitely would still produce tiny magnets each having its own poles at two ends. This shows that two poles of a magnet cannot be separated and an isolated magnetic pole does not exist.

Question 3.
Explain the process by which a permanent magnet can magnetise an ordinary piece of iron?
Answer:
A piece of iron can be magnetised by two methods:

1. By single touch method. Put a piece of iron on the table:
Hold a bar magnet on it vertically with north pole touching. Now move the magnet along the length. By repeating this practice many times makes that iron piece a magnet.

2. By double touch method:
Put a piece of iron on the table. Hold two bar magnets on it vertically with north and south pole of two magnets. Put them in the middle first. Move the magnets in opposite direction as shown in figure. Repeating this process converts the piece of iron into a magnet.

Question 4.
Why does a freely suspended magnet always come to rest in the north – south direction?
Answer:
This is because the earth itself behaves like a huge bar magnet with its magnetic poles near the geographical north and south poles. The south pole of this bar magnet is near the geographical north pole, whereas the north pole of the bar magnet is near the geographical south pole.

In a freely suspended magnet, therefore, the north pole points towards the geographical north pole since it is attracted by the earth’s magnetic south pole. Similarly, the south pole of the suspended magnet is attracted by the earth’s magnetic north pole and, therefore, points towards the geographical south pole.

Question 5.
Write a short note on the cautions of magnets?
Answer:
Magnets loose their properties if they are heated, hammered or dropped from some height [Fig. (a)]. Also, magnets become weak if they are not stored properly.

To keep them safe, bar magnets should be kept in pairs with their unlike poles on the same side. They must be separated by a piece of wood while two pieces of soft iron should be placed across their ends [Fig. (b)]. For horse-shoe magnet, one should keep a piece of iron across the poles.

Keep magnets away from cassettes, mobiles, television, music system, compact disks (CDs) and the computer. [See Fig. (c)]

MP Board Class 6th Science Solutions

MP Board Class 10th Hindi Navneet Solutions गद्य Chapter 6 मेरे गाँव की सुख और शांति किसने छीन ली? (संस्मरण रामनारायण उपाध्याय)

मेरे गाँव की सुख और शांति किसने छीन ली? अभ्यास

बोध प्रश्न

मेरे गाँव की सुख और शांति किसने छीन ली? अति लघु उत्तरीय प्रश्न

प्रश्न 1.
गाँधी द्वारा स्थापित आश्रम का नाम लिखिए।
उत्तर:
गाँधी द्वारा ‘सेवा ग्राम’ आश्रम की स्थापना की गयी है।

प्रश्न 2.
लोक संस्कृति का जन्म कहाँ हुआ?
उत्तर:
लोक संस्कृति का जन्म गाँवों में हुआ।

प्रश्न 3.
लेखक ने संगीत का जन्म किससे माना है?
उत्तर:
लेखक ने संगीत का जन्म श्रम से माना है।

प्रश्न 4.
ललित कलाओं का स्वभाव कैसा होता है?
उत्तर:
ललित कलाओं का स्वभाव फूल जैसा होता है।

प्रश्न 5.
ग्रामीण समूचे गाँव को किस रूप में मानता आया है?
उत्तर :
ग्रामीण समूचे गाँव को एक परिवार मानता आया है।

मेरे गाँव की सुख और शांति किसने छीन ली? लघु उत्तरीय प्रश्न

प्रश्न 1.
लेखक समाज से किन प्रश्नों को पूछना चाहता है?
उत्तर:
लेखक समाज से ये प्रश्न पूछना चाहता है कि मेरे गाँवों की सुख और शान्ति को किसने छीन लिया। गाँवों की अन्न-धन और लक्ष्मी कहाँ चली गई?

प्रश्न 2.
लोक की जीवन्त रसधारा को किसने, कहाँ सुरक्षित रखा है?
उत्तर:
लोक की जीवन्त रसधारा को गाँवों ने अपने हृदय-में सुरक्षित रखा है।

प्रश्न 3.
अँधेरे को सुहावने प्रभात में कौन, कैसे परिवर्तित करता है?
उत्तर:
गाँव की स्त्रियाँ भोर में उठकर आटे के साथ घने अँधेरे को पीसकर सुहावने प्रभात में बदल देती थीं।

प्रश्न 4.
गाँव के समृद्ध किसान की स्थिति अब कैसी हो गई है?
उत्तर:
गाँव के समृद्ध किसान की स्थिति बढ़ती हुई महँगाई और शोषणकारी समाज व्यवस्था के चलते मज़दूर के रूप में बदल गई है।

प्रश्न 5.
श्रम से किसका जन्म होना प्रतीत होता था?
उत्तर:
श्रम से संगीत का जन्म होना प्रतीत होता था और यही संगीत लोरी बनकर श्रम को हल्का करने में योगदान देता था।

प्रश्न 6.
माटी कुम्हार की हथेली के स्पर्श से किन नवीन रूपों को धारण करती थी?
उत्तर;
माटी कुंम्हार की हथेली का स्पर्श पाकर नये-नये रूप धारण करती थी। कभी वह गागर बनती, कभी खपरैल तो कभी माटी का दीप बनकर आशा की किरण जगाया करती थी।

मेरे गाँव की सुख और शांति किसने छीन ली? दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
ललित कलाओं का ग्राम्य जीवन में क्या महत्त्व था?
उत्तर:
ललित कलाओं का ग्राम्य जीवन में बड़ा महत्त्व था। ये ललित कलाएँ उनमें जीवन का रस घोलती थीं। तीज-त्यौहार, मेले-ठेले तथा चौपालों पर इनके रूप देखने को मिलते थे।

प्रश्न 2.
लोकगीत ग्रामीण जीवन में किस प्रकार रचे-बसे थे?
उत्तर:
लोकगीत ग्रामीण जीवन की रग-रग में बसे और रचे थे। ग्रामीण जनों का सम्पूर्ण जीवन इन्हीं लोकगीतों के ताने-बाने से बुना हुआ था। उनकी श्वास-प्रश्वास में ये ही गीत समाये हुए थे।

प्रश्न 3.
लेखक के अनुसार “गोकुल के सहज-सरल गाँव” का आशय स्पष्ट कीजिए।
उत्तर:
लेखक की मान्यता है कि ब्रजमण्डल के गोकुल के गाँव जिस प्रकार सहज एवं सरल थे वैसे ही इस क्षेत्र के सभी गाँव सहज और सरल थे। बनावट एवं कृत्रिमता का उनमें कोई स्थान नहीं था। वहाँ के निवासी सरल एवं भोले-भाले व्यक्ति हुआ करते थे।

प्रश्न 4.
गाँब के सुसंस्कृत आदमी की पाँच विशेषताएँ लिखिए।
उत्तर:
गाँव का आदमी निरक्षर भले हो लेकिन वह सुसंस्कृत रहा है। उसकी पाँच विशेषताएँ इस प्रकार हैं-

1. वह विश्वास पर बिक जाता है
2. वह धर्म पर झुक जाता है
3. वह थककर बैठता नहीं हैं
4. झुककर नहीं चलता है और
5. वह दुःख में भी मुस्कुराता रहता है।

प्रश्न 5.
गाँव के कुटीर उद्योगों पर आधुनिकता का क्या प्रभाव पड़ा है?
उत्तर:
गाँव के कुटीर उद्योगों पर आधुनिकता का यह प्रभाव पड़ा है कि गाँव के कुटीर उद्योग नष्ट हो गये हैं। फ्लोर मिल खुल जाने से चक्कियों का चलना बन्द हो गया है, ट्रैक्टर एवं अन्य कृषि यन्त्रों के प्रयोग से किसानों का हल आदि चलाना बन्द हो गया है।

प्रश्न 6.
गाँव का आदमी अपने समग्र जीवन से क्या-क्या देने की क्षमता रखता है?
उत्तर:
गाँव के आदमी का समग्र जीवन एक अनपढ़ी खुली किताब जैसा है। उसका रहन-सहन, खान-पान, वस्त्राभूषण, आचार-विचार, रीति-रिवाज, गीत और कथाएँ, नृत्य संगीत आदि सभी कुछ हमें कुछ न कुछ देने की क्षमता रखते हैं।

मेरे गाँव की सुख और शांति किसने छीन ली? भाषा अध्ययन

प्रश्न 1.
निम्नलिखित शब्दों का सन्धि-विच्छेद कीजिए और नाम लिखिए
उत्तर:

1. रवीन्द्र = रवि + इन्द्र = दीर्घ सन्धि।
2. निरक्षर = निः + अक्षर = विसर्ग सन्धि।
3. संग्रहालय = संग्रह + आलय = दीर्घ सन्धि।
4. सज्जन = सत् + जन = व्यंजन सन्धि।

प्रश्न 2.
निम्नलिखित शब्दों का समास-विग्रह कीजिए
उत्तर:

1. कार्य स्थल= कार्य का स्थल = तत्पुरुष समास।
2. रसधारा = रस की धारा = तत्पुरुष समास।
3. श्वास-प्रश्वास = श्वास और प्रश्वास = द्वन्द्व समास।
4. लोक संस्कृति = लोक की संस्कृति = तत्पुरुष समास।

प्रश्न 3.
निम्नलिखित शब्दों की सन्धि कीजिए
उत्तर:

1. पर + उपकार = परोपकार।
2. देव + ऋषि = देवर्षि।
3. अति + आचार = अत्याचार।
4. प्रति + एक = प्रत्येक।

प्रश्न 4.
निम्नलिखित वाक्यांशों के लिए एक-एक शब्द लिखिए.

1. ठेका लेने वाला।
2. खेती करने वाला।
3. मिट्टी के बर्तन बनाने वाला।
4. पानी भरने वाली।

उत्तर:

1. ठेकेदार
2. खेतिहर
3. कुम्हार
4. पनहारिन।

प्रश्न 5.
निम्नलिखित वाक्यों को पहचान कर अर्थ के आधार पर वाक्य प्रकार का नाम लिखिए

1. किसान के कंठ से गीत कहाँ लुप्त हो गये?
2. गाँवों में लोक संस्कृति का जन्म हुआ।
3. मेरे गाँव की शान्ति मन छीनो।
4. निर्मल चाँदनी, रात को नहीं फैली थी।
5. मैं चाहता हूँ कि आप एक बार गाँव अवश्य जायें।

उत्तर:

1. प्रश्नवाचक
2. स्वीकारात्मक
3. आदेशात्मक
4. निषेधात्मक
5. आदेशात्मक।

मेरे गाँव की सुख और शान्ति किसने छीन ली? महत्त्वपूर्ण वस्तुनिष्ठ प्रश्न

मेरे गाँव की सुख और शांति किसने छीन ली? बहु-विकल्पीय प्रश्न 

प्रश्न 1.
‘मेरे गाँव की सुख और शान्ति किसने छीन ली?’ निबन्ध के लेखक हैं-
(क) रामनारायण उपाध्याय
(ख) सरदार पूर्णसिंह
(ग) बालकृष्ण भट्ट
(घ) अज्ञेय।
उत्तर:
(क) रामनारायण उपाध्याय

प्रश्न 2.
गाँधीजी ने गाँवों को आदर्श मानकर किस गाँव की स्थापना की?
(क) सेवाग्राम
(ख) रामपुर
(ग) शान्ति निकेतन
(घ) साबरमती आश्रम।
उत्तर:
(क) सेवाग्राम

प्रश्न 3.
लोकगीतों के स्वर गलों से लुप्त होकर कहाँ कैद किये गये हैं?
(क) पुस्तकों में
(ख) कैसिटों में
(ग) संग्रहालयों में
(घ) कहीं नहीं।
उत्तर:
(ख) कैसिटों में

प्रश्न 4.
मिट्टी को नवीन आकृति प्रदान करता है
(क) मनुष्य
(ख) बढ़ई
(ग) रंगरेज
(घ) कुम्हार।
उत्तर:
(घ) कुम्हार।

प्रश्न 5.
लोक संस्कृति का जन्म हुआ (2016)
(क) शहरों में
(ख) गाँवों में
(ग) कस्बों में
(घ) विद्यालयों में।
उत्तर:
(ख) गाँवों में

रिक्त स्थानों की पूर्ति

1. रवीन्द्रनाथ टैगोर ने ………….. की स्थापना की।
2. ……….. में लोकसंस्कृति का जन्म हुआ। (2013)
3. पहले गाँव का किसान …………. माना जाता था।
4. गाँव का आदमी निरक्षर भले हो, लेकिन ………… रहा है।
5. आज गाँव में बेरोजगारी है, गाँव में नीरसता है, गाँव . हैं।

उत्तर:

1. शान्ति निकेतन
2. ग्रामों
3. समृद्ध
4. सुसंस्कृत
5. गरीब।

सत्य/असत्य

1. गाँव का आदमी निरक्षर भले ही हो लेकिन सुसंस्कृत होता है।
2. गाँव में रहने के लिए साधारण नागरिक ही नहीं कवियों का मन भी ललचाया था।
3. गाँधीजी द्वारा स्थापित आश्रम का नाम ‘सेवाग्राम’ है। (2009)
4. आज गाँव में मजदूर प्रसन्नता से गाता गुनगुनाता मिलेगा।
5. गाँव वालों का जीवन एक बिना पढ़ी खुली पुस्तक की तरह सामने बिछा है।

उत्तर:

1. सत्य
2. सत्य
3. सत्य
4. असत्य
5. सत्य।

सही जोड़ी मिलाइए

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

एक शब्द/वाक्य में उत्तर

1. लोक संस्कृति का जन्म कहाँ हुआ? (2014, 18)
2. ‘मेरे गाँव की सुख और शान्ति किसने छीन ली?’ के लेखक कौन हैं?
3. ललित कलाओं का स्वभाव कैसा होता है? (2017)
4. कहाँ के सरल और सहज गाँव धीरे-धीरे नष्ट होते जा रहे हैं?
5. अमराई में अब किसकी कूक नहीं गूंजती है?

उत्तर:

1. ग्रामों में
2. पण्डित रामनारायण उपाध्याय
3. फूल की तरह
4. गोकुल के
5. कोयल।

मेरे गाँव की सुख और शान्ति किसने छीन ली? पाठ सारांश

इस संस्मरण में पं.रामनारायण उपाध्याय जी ने भारत के गाँव की संस्कृति का अवलोकन किया है लेकिन निबन्धकार भारत के गाँव के वर्तमान को लेकर व्यथित हैं। उसका प्रमुख कारण है कि गाँव अपनी पहचान को नष्ट करते जा रहे हैं। गाँव की पहचान के साथ-साथ भारतमाता की पहचान भी धुंधली पड़ती जा रही है।

गाँधीजी और रवीन्द्रनाथ टैगोर ने भारत के गाँवों को ही आदर्श मानकर सेवाग्राम और शान्ति निकेतन की स्थापना की। हमारे देश के गाँवों की प्राकृतिक छटा अलौकिक व दर्शनीय थी लेकिन अब वही गाँव आधुनिकता के दबाव के कारण इस अनुपम सुख से पृथक् होते जा रहे हैं।

अब न तो कहीं लोकगीतों की मधुर ध्वनि सुनाई देती है न ही ढोलक की थाप। ये लोकगीत मानव जीवन का अभिन्न अंग थे तथा मानव को भरपूर स्फूर्ति प्रदान करते थे। अब इन गीतों के समाप्त होने के कारण मानव की संवेदना भी समाप्त हो गयी है।

जिस भारत को सोने की चिड़िया कहा जाता था वही भारतवासी जो कि ग्रामीण क्षेत्रों में रहते हैं, आर्थिक रूप से विपन्न हैं। आत्मनिर्भर गाँव अब छिन्न-भिन्न हो गये हैं। मजदूर व शिल्पकार रोजी-रोटी की तलाश में गाँव से पलायन कर शहर की ओर भाग रहे हैं।

पूर्व में गाँव के लोग अनपढ़ होते थे लेकिन उनमें परस्पर स्नेह. आस्था. विश्वास और परिश्रम की भावना कूट-कूटकर भरी हुई थी। वे श्रम और संघर्ष करके भी प्रसन्नतापूर्वक जीवन बिताते थे क्योंकि वे आत्मनिर्भर थे। आज भी इतिहास इस बात का प्रमाण है कि भारत के गाँव खुशहाल थे लेकिन आज गाँव में नीरसता,निर्धनता और अशान्ति है। निबन्धकार का इस निबन्ध को लिखने का उद्देश्य है कि लोकचेतना जाग्रत हो।

मेरे गाँव की सुख और शान्ति किसने छीन ली? संदर्भ-प्रसंगसहित व्याख्या

(1) जिन गाँवों ने हिन्दी साहित्य को ‘हीरो’ और ‘गोबर’ जैसे पात्र दिये, जिन गाँवों के लिए गाँधी और रवीन्द्रनाथ टैगोर ने शहरों की समस्त सुविधाओं को त्यागकर ‘शान्ति निकेतन’ और ‘सेवाग्राम’ को अपना कार्यस्थल बनाया, जिन गाँवों में रहने के लिए साधारण नागरिक ही नहीं कवियों का मन भी ललचाया था, वे ही गाँव आज अशान्ति के घर हुए जा रहे हैं और जैसे किसी भी आँख से आँसू गिरे, ऐसे गाँवों के आँचल से एक-एक घर टूटते ही चले जा रहे हैं।

कठिन शब्दार्थ :
कार्यस्थल = कामकाज करने का स्थान। आँचल = परिवेश, वातावरण।

सन्दर्भ :
प्रस्तुत गद्यांश ‘मेरे गाँव की सुख और शान्ति किसने छीन ली’ शीर्षक पाठ से लिया गया है। इसके लेखक श्री रामनारायण उपाध्याय हैं।

प्रसंग :
इसमें लेखक की चिन्ता यह है कि पहले तो गाँव सुख शान्ति के भण्डार हुआ करते थे और सभी लोग गाँवों की ओर देखा करते थे; पर आज तो वे समस्याओं के घर बनते चले जा रहे हैं।

व्याख्या :
लेखक श्री रामनारायण उपाध्याय कहते हैं कि जिन गाँवों ने प्रेमचन्द के प्रसिद्ध उपन्यास, ‘गोदान’ में ‘होरी’ और ‘गोबर’ का चित्रण किया है तथा जिन गाँवों में गाँधी और टैगोर को शहरों की सभी सुख-सुविधाएँ त्याग कर ‘सेवा ग्राम’ और ‘शान्ति निकेतन’ भाया तथा इन्हीं को इन लोगों ने अपनी कर्म भूमि बनाया। इन गाँवों के प्रति साधारण नागरिक ही नहीं अपितु पन्त जैसे महान् कवियों का मन भी ललचाया करता था, वे ही गाँव आज सुख शान्ति से रहित होकर अशान्ति के घर क्यों बनते जा रहे हैं? जिस प्रकार किसी आँख से आँसू टप-टप गिरते जाते हैं, उसी प्रकार गाँव के आँचल के एक-एक घर टूटते जा रहे हैं।

विशेष :

1. लेखक को गाँवों की वर्तमान दशा से हार्दिक दुःख होता है।
2. भाषा सहज एवं सरल है।

(2) जिन गाँवों में लोक-संस्कृति का जन्म हुआ, जिसने लोक की जीवन्त रसधारा को, अपने हृदय में सुरक्षित रखा, वे ही गाँव आज टूटते जा रहे हैं, गाँव की वह पुरानी पीढ़ी भी समाप्त होती जा रही है जिसका सम्पूर्ण जीवन श्वास-प्रश्वास की तरह गीतों के ताने-बाने पर आधारित था। अब तो वे गोकुल से सहज-सरल गाँव नष्ट होते जा रहे हैं, जहाँ स्त्रियाँ भोर में उठकर आटे के साथ घने अँधेरे को भी पीसकर सुहावने प्रभात में बदल देती थीं। जहाँ चक्की के हर फेरे के साथ गीत की नई पंक्तियाँ उठती थीं। लगता था जैसे श्रम में से संगीत का जन्म हो रहा हो और संगीत लोरी बनकर श्रम को हल्का करने में अपना योगदान दे रहा हो।

कठिन शब्दार्थ :
लोकसंसार। जीवन्त रसधारा = ऐसी धारा जो जीवन प्रदान करती हो। श्वास-प्रश्वास = साँस लेना और छोड़ना। लोक-संस्कृति = गाँव की संस्कृति।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
लेखक का मानना है कि लोक संस्कृति का जन्म गाँवों से ही हुआ था और आज वह टूटती जा रही है।

व्याख्या :
लेखक श्री रामनारायण उपाध्याय कहते हैं कि जिन गाँवों में लोक-संस्कृति का जन्म हुआ था और जिसने वहाँ के जन-जीवन में एक जीवन्त जीवन प्रवाहित किया था, वे ही गाँव आज टूटते और बिखरते जा रहे हैं। गाँव की वह पुरानी पीढ़ी आज नाश के कगार पर है जिसका सारा जीवन श्वास-प्रश्वास की तरह लोक गीतों के ताने-बाने पर टिका हुआ था। गोकुल के सहज एवं सरल गाँव जहाँ स्त्रियाँ प्रात:काल की बेला में उठकर घर की चक्की पर बैठकर लोकगीत गाते-गाते रात्रि के घने अँधेरे को पीसकर सुहावने प्रात:काल में बदल देती थीं, आज वे गाँव भी नष्ट होते जा रहे हैं। उस समय गाँव की स्त्रियाँ चक्की चलाते वक्त गीत की नई-नई पंक्तियाँ गाकर वातावरण को मधुर बना दिया करती थीं। उस समय ऐसा लगता था मानो श्रम में से संगीत का जन्म हो रहा हो और संगीत लोरी बनकर श्रम को हल्का कर दिया करता था।

विशेष :

1. पहले गाँवों में स्त्रियाँ चक्की चलाते समय, पानी भरते समय लोकगीत गाया करती थीं।।
2. भाषा सहज, सरल में भावानुकूल है।

(3) पहले जो आदमी गाँव का समृद्ध किसान माना जाता था, वही अब बढ़ती महँगाई और शोषणकारी समाज व्यवस्था के चलते अपनी जमीन से हाथ धोकर खेतिहर मजदूर में बदलता जा रहा है और सचमुच जो मजदूर था वह जमीन पर से किसान का आधिपत्य कम हो जाने से शहरों में मजदूरी करता नजर आता है।

कठिन शब्दार्थ-समृद्ध = सम्पन्न, खाता-पीता। शोषणकारी = सताने वाला, अनुचित लाभ लेने वाला। खेतिहर = मजदूरी पर खेत में काम करने वाला। आधिपत्य = अधिकार।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
लेखक यह बताना चाहता है कि गाँव के लोग पलायन कर शहरों की ओर भाग रहे हैं।

व्याख्या :
लेखक रामनारायण उपाध्याय कहते हैं कि पहले जो गाँव का सम्पन्न किसान माना जाता था वह बढ़ती महँगाई के कारण तथा शोषण करने वाली सामाजिक व्यवस्था के कारण अपनी जमीन से हाथ धोकर खेत में काम करने वाला मजदूर बनता जा रहा है और जो वास्तव में मजदूर था वह जमीन पर से किसान का अधिकार समाप्त हो जाने पर शहर में मजदूरी करता दिखाई देता है।

विशेष :

1. बढ़ती महँगाई और शोषणकारी सामाजिक व्यवस्था ने किसानों को मजदूर बना डाला है।
2. भाषा सहज एवं सरल है।

MP Board Class 10th Hindi Solutions

MP Board Class 11th Biology Solutions Chapter 6 पुष्पी पादपों का शारीर

पुष्पी पादपों का शारीर NCERT प्रश्नोत्तर

प्रश्न 1.
विभिन्न प्रकार के मेरिस्टेम की स्थिति तथा कार्य बताइए।
उत्तर:
स्थिति के अनुसार विभज्योतक निम्न प्रकार के होते हैं –
1.शीर्षस्थ विभज्योतक (Apical meristem):
ये ‘जड़ तथा तने के शीर्ष पर पाये जाते हैं इनकी कोशिकाओं के विभाजन से तने तथा जड़ें लम्बाई में बढ़ती हैं। इनसे जड़ व तनों के सिरों पर वृद्धि बिन्दु का निर्माण होता है।

2. अन्तर्विष्ट विभज्योतक (Intercalary meristem):
वास्तव में यह शीर्षस्थ विभज्योतक से पृथक् हुआ भाग है जो प्ररोह की वृद्धि के समय शीर्षस्थ भाग से अलग हो जाता है और स्थायी ऊतक में परिवर्तित नहीं होता एवं स्थायी ऊतकों के बीच में विभज्योतक ऊतक के रूप में बचा रहता है। यह ऊतक पत्ती के आधार के पास अथवा पर्व के आधार के पास स्थित रहता है। अन्तर्विष्ट विभज्योतक ऊतक सामान्यतः घासों में पर्व के आधार के पास, पुदीने (Mentha) की पर्वसन्धि के नीचे इत्यादि स्थानों पर पाये जाते हैं। ये ऊतक अस्थायी होते हैं लेकिन बाद में स्थायी ऊतकों में परिवर्तित हो जाते हैं। इस ऊतक के कारण पौधा लम्बाई में बढ़ता है।

3. पार्श्व विभज्योतक (Lateral meristem):
ये विभज्योतक ऊतक तनों तथा जड़ों के पार्यों में स्थित हैं, जैसे कि संवहन एधा और कॉर्क एधा। ये ऊतक स्थायी ऊतकों के पुनः विभेदन के कारण बनते हैं। इनके विभाजन से द्वितीयक वृद्धि होती है जिससे जड़ व तने मोटाई में वृद्धि करते हैं।

प्रश्न 2.
कार्क कैंबियम ऊतकों से बनता है, जो कार्क बनाते हैं। क्या आप इस कथन से सहमत हैं ? वर्णन कीजिए।
उत्तर:
जब द्विबीजपत्री जड़ और तने के परिधि में वृद्धि होती है तब वल्कुट तथा बाह्य त्वचा की सतहें टूटती जाती है और उन्हें नई संरक्षी कोशिका सतह की आवश्यकता होती है। इसलिए एक दूसरा मेरीस्टेमी ऊतक तैयार हो जाता है जिसे कार्क कैंबियम अथवा कागजन कहते हैं। यह प्रायः वल्कुट क्षेत्र में विकसित होता है इसकी सतह मोटी और सँकरी पतली भित्ति वाली आयताकार कोशिकाओं की बनी होती है। कागजन दोनों ओर की कोशिकाओं को बनाता है।

बाहर की ओर की कोशिकाएँ कार्क अथवा काग में बँट जाती है और अंदर की ओर की कोशिकाएँ द्वितीयक वल्कुट अथवा कागास्तर में विभेदित हो जाती है। कार्क कोशिकाओं में पानी का प्रवेश नहीं होता है क्योंकि इसकी कोशिका भित्ति पर सूबेरिन जमा रहता है। द्वितीयक वल्कुट कोशिकाएँ पैरेकाइमी होती है। कागजन, काग तथा काग मिलकर परिचर्म बनाते हैं । ये बाहरी तथा तने के भीतरी ऊतकों के बीच गैसों का आदान-प्रदान करते हैं। ये अधिकांश काष्ठीय वृक्षों में पाये जाते हैं।

प्रश्न 3.
चित्रों की सहायता से काष्ठीय एंजियोस्पर्म के तने में द्वितीयक वृद्धि के प्रक्रम का वर्णन कीजिए। इसकी क्या सार्थकता है?
उत्तर:
द्वितीयक वृद्धि (Secondary growth):
“कैम्बियम तथा कॉर्क कैम्बियम की क्रियाशीलता के फलस्वरूप क्रमशः स्टील के अन्दर तथा स्टील के बाहर द्वितीयक ऊतकों के बनने के कारण जड़ तथा तने में मोटाई में हुई वृद्धि द्वितीयक वृद्धि कहलाती है।” द्वितीयक वृद्धि के कारण ही आवृतबीजी और द्विबीजपत्री पौधे वृक्ष जैसी रचना बना पाते हैं द्वितीयक वृद्धि के अभाव के कारण ही एकबीजपत्री पादपों में सामान्यतः वृक्ष का अभाव रहता है। अत: कुछ अपवादों को छोड़कर एकबीजपत्री पादपों में द्वितीयक वृद्धि का पूर्णतः अभाव होता है। जबकि द्विबीजपत्री पादपों (जड़ एवं तना) में द्वितीयक वृद्धि पायी जाती है।

द्विबीजपत्री तने में द्वितीयक वृद्धि (Secondary growth in cambium) – एक प्रारूपी द्विबीजपत्री तने में द्वितीयक वृद्धि निम्न प्रकार से होती है –

(A) कैम्बियम की क्रियाशीलता (Activity of cambium):
आप जानते हैं कि द्विबीजपत्री तने में जाइलम तथा फ्लोएम के बीच एक पट्टी पायी जाती है जिसे पूलीय कैम्बियम (Fascicular cambium) कहते हैं। जब तना वयस्क हो जाता है अर्थात् उसमें द्वितीयक वृद्धि होनी होती है तब पूलीय कैम्बियम क्रियाशील हो जाता है तथा इसके साथ ही दो सम्वहन पूलों के बीच की कैम्बियम की सन्धि वाली कोशिकाएँ, जो मृदूतकी होती हैं, भी विभाजित हो जाती हैं अब इन्हें अन्तरपूलीय कैम्बियम (Interfascicular cambium) कहते हैं। इस प्रकार कैम्बियम का एक वलय बन जाता है जिसे कैम्बियम वलय (Cambium ring) कहते हैं।

अनुकूल परिस्थिति में कैम्बियम कोशिकाएँ विभाजन करने लगती हैं। यह विभाजन स्पर्श रेखीय (Tangential) होता है जिसके कारण वलय के बाहर द्वितीयक फ्लोएम तथा अन्दर द्वितीयक जाइलम का निर्माण होता है। सामान्यतः द्वितीयक फ्लोएम की अपेक्षा द्वितीयक जाइलम अधिक मात्रा में बनता है, इस कारण कैम्बियम का वलय परिधि की ओर खिसकता जाता है।

प्राथमिक जाइलम तथा फ्लोएम इस क्रिया के कारण दूर हो जाते हैं जबकि कैम्बियम क्रियाशीलता के पहले ये पास-पास स्थित होते हैं। इस समय प्राथमिक फ्लोएम दबाव के कारण कुचलकर अवशेष के रूप में रह जाता है जबकि प्राथमिक जाइलम केन्द्रीय मज्जा की ओर आ जाता है। कुछ कैम्बियम कोशिकाएँ द्वितीयक जाइलम तथा फ्लोएम के स्थान पर केवल मृदूतकी कोशिकाओं का निर्माण करने लगती हैं –

जिससे अक्ष के क्षैतिज द्वितीयक जाइलम से द्वितीयक फ्लोएम तक एक पट्टी दिखाई देने लगती है। जिसे द्वितीयक ऊतकों में स्थित होने के कारण द्वितीयक मेड्यूलरी रश्मियाँ (Secondary medulary rays) कहते हैं। ये संवहन ऊतक की जीवित कोशिकाओं से सम्बन्ध स्थापित करती हैं, इन्हीं से होकर फ्लोएम तथा जाइलम की जीवित कोशिकाओं को भोजन सामग्री पहुँचती है। इसके अलावा ये कोशिकाएँ भोजन संग्रह का भी कार्य करती हैं।

(B) कॉर्क कैम्बियम की सक्रियता (Activity of cork cambium):
कैम्बियम द्वारा नये ऊतकों के बनने के कारण तने के बाहरी ऊतकों पर दबाव पड़ता है जिसके कारण बाह्य त्वचा फट जाती है। इसी समय कॉर्टेक्स की बाहरी पर्त विभाजित होकर नयी कोशिकाएँ बनाने लगती हैं इस पर्त को ही कॉर्क कैम्बियम या फेलोजन (Cork cambium or Phellogen) कहते हैं। कॉर्क कैम्बियम अन्दर तथा बाहर दोनों तरफ कोशिकाओं का निर्माण करता है लेकिन बाहर की तरफ अपेक्षाकृत अधिक कोशिकाएँ बनती हैं।

कॉर्क कैम्बियम द्वारा बाहर की ओर जो कोशिकाएँ बनती हैं उन्हें कॉर्क या फेलम (Phellem) तथा अन्दर की तरफ बनी कोशिकाओं को फैलोडर्म कहते हैं। बाहर बनी कोशिकाओं में अन्तराकोशिकीय अवकाश नहीं पाया जाता है तथा ये सुबेरिन का निर्माण कर मृत हो जाती है और तने की छाल (Bark) बनाती हैं। अन्दर की ओर कोशिकाएँ द्वितीयक कॉर्टेक्स बनाती हैं जो मृदूतकी कोशिकाओं का बना होता है जिसमें हरितलवक पाया जाता है। सामान्यत: कॉर्क कैम्बियम द्वारा बनाये ऊतकों को कॉर्क (Cork or Periderm) कहा जाता है जो तने को यांत्रिक सहारा देने के साथ ही रक्षात्मक आवरण बनाता है तथा आन्तरिक ऊतकों से जल हानि को रोकता है।

प्रश्न 4.
निम्नलिखित में विभेद कीजिए –

1. वाहिका तथा ट्रैकीड
2. पैरेनकाइमा तथा कोलेनकाइमा
3. रसदारु तथा अंत:काष्ठ
4. खुला तथा बंद संवहन बंडल।

उत्तर:
1. वाहिका एवं वाहिनिका (ट्रैकीड) में अन्तर –

2. पैरेनकाइमा (मृदूतक) तथा कोलेनकाइमा (स्थूलकोण ऊतक) पैरेनकाइमा (Perenchyma):

1. इसकी कोशिकाएँ जीवित एवं पतली भित्ति वाली होती हैं।
2. इनकी कोशिका भित्ति सेल्युलोज की बनी होती है।
3. इनकी कोशिकाओं में रिक्तिकाएँ संख्या में अधिक होती हैं।
4. इनकी कोशिकाएँ गोल या बहुभुजी और अन्तराकोशिकीय अवकाश युक्त होती हैं।
5. इनका मुख्य कार्य भोज्य पदार्थों का संग्रहण करना है।
6. जब इनमें क्लोरोफिल उपस्थित होता है, तब इन्हें क्लोरेनकायमा कहते हैं। यह प्रकाश-संश्लेषण. का कार्य करता है।
7. जब इन कोशिकाओं के अन्तराकोशिकीय अवकाश बड़े-बड़े होते हैं तब इन्हें ऐरेनकायमा कहते हैं। ये जलीय पौधों में पाये जाते हैं तथा उनको तैरने में सहायता करते हैं।

स्थूलकोण ऊतक (Collenchyma):

1. यह जीवित ऊतक है, जिसकी कोशिकाएँ लम्बी होती हैं।
2. इनकी कोशिकाओं में अन्तर कोशिकीय अवकाशों का अभाव होता है।
3. इनकी कोशिकाओं की कोशिका भित्ति के कोने सेल्युलोज एवं पेक्टिन के जमाव के कारण मोटे व स्थूलित हो जाते हैं इसलिए इन्हें स्थूलकोण ऊतक कहा जाता है।
4. क्लोरोप्लास्ट उपस्थित होने पर ये प्रकाश-संश्लेषण का कार्य करते हैं।
5. यह पौधे को यांत्रिक मजबूती प्रदान करता है।
6. लचीला होने के कारण यह अंगों को तनन सामर्थ्य प्रदान करता है।
7. क्लोरोप्लास्ट उपस्थित होने पर यह प्रकाश-संश्लेषण का कार्य करता है।

3. रसदारु तथा अंतःकाष्ठ में अन्तर।

4. खुला तथा बंद संवहन बंडल

• खुला संवहन बंडल (Open vascular bundle) – इसमें जायलम तथा फ्लोएम के बीच में कैम्बियम स्थित होता है। इस प्रकार का सम्वहन पूल द्विबीजपत्री तनों में पाया जाता है।
• बन्द संवहन बंडल (Closed vascular bundle) – इस प्रकार के सम्वहन पूलों में कैम्बियम नहीं पाया जाता। यह एकबीजपत्री पौधों के तनों में पाया जाता है।

प्रश्न 5.
निम्नलिखित में शारीरिकी (Anatomy) के आधार पर अंतर बताइए

1. द्विबीजपत्री मूल तथा एकबीजपत्री मूल।
2. द्विबीजपत्री तना तथा एकबीजपत्री तना।

उत्तर:
1. एकबीजपत्री एवं द्विबीजपत्री जड़ (मूल) की आन्तरिक संरचना में अन्तर –

2. द्विबीजपत्री तथा एकबीजपत्री तने में अन्तर –

प्रश्न 6.
आप एक शैशव तने की अनुप्रस्थ काट का सूक्ष्मदर्शी अवलोकन कीजिए।आप कैसे पता करेंगे कि यह एकबीजपत्री तना अथवा द्विबीजपत्री तना है ? इसके कारण बताइए।
उत्तर:

एक शैशव तने की अनुप्रस्थ काट का सूक्ष्मदर्शी में अवलोकन करने के पश्चात् एकबीजपत्री एवं द्विबीजपत्री तना को निम्नलिखित संरचनाओं के आधार पर पहचाना जा सकता है –

द्विबीजपत्री तने की आंतरिक संरचना(Internal structure of dicot stem) – यदि हम सूर्यमुखी के तने की आंतरिक संरचना को देखें तो इसमें निम्न रचनाएँ दिखाई देती हैं –

1. बाह्यत्वचा (Epidermis):
यह सबसे बाहरी एक कोशिकीय स्तर है जिस पर क्यूटिकिल पायी जाती है। इस पर कहीं-कहीं बहुकोशिकीय रोम तथा स्टोमेटा पाये जाते हैं।

2. कॉर्टेक्स (Cortex):
यह बाह्यत्वचा के नीचे के स्तर है जो तीन स्तरों की बनी होती है –

• अधस्त्व चा (Hypodermis) – यह कोलेनकाइमेटस कोशिकाओं की 3 से 5 परतों की बनी होती है। इन कोशाओं में अन्तराकोशिकीय अवकाश अनुपस्थित तथा हरित लवक उपस्थित होता है।
• सामान्य कॉर्टेक्स (General cortex) – यह अधस्त्वचा के नीचे स्थित होता है तथा अन्तराकोशिकीय अवकाशों युक्त मृदूतकी कोशिकाओं का बना होता है।
• अन्तस्त्वचा (Endodermis) – यह कॉर्टेक्स की आन्तरिक एक कोशिकीय स्तर जो ढोलक के समान कोशिकाओं की बनी होती है। जिसमें स्टार्च कण पाये जाते हैं। इसमें कैस्पेरियन स्ट्रिप स्पष्ट दिखाई देती है।

3. पेरिसाइकिल (Pericycle):
यह परत मृदूतकी तथा दृढ़ऊतकी कोशिकाओं के एकान्तर क्रम में व्यवस्थित होने से बनती है और अन्तरस्त्वचा के नीचे स्थित होती है।

4.संवहन पूल (Vascular bundle):
इनके संवहन पूल संयुक्त (Conjoint), कोलेटरल (Collateral), खुले (Open) तथा एक घेरे में व्यवस्थित होते हैं। इनका प्रत्येक संवहन पूल जाइलम, फ्लोएम तथा कैम्बियम का बना होता है। इनका जाइलम वैसेल्स, ट्रैकीड, काष्ठ तन्तु तथा काष्ठ मृदूतक का बना होता है जबकि फ्लोएम चालनी नलिकाओं, सखि कोशिकाओं तथा मृदूतक कोशिकाओं का बना होता है। इनके जाइलम तथा फ्लोएम के बीच में पतली भित्ति वाली कोशिकाओं की एक पट्टी पायी जाती है जिसे कैम्बियम कहते हैं।

5. पिथ (Pith):
तने के मध्य में मृदूतकी कोशिकाओं का बना पिथ पाया जाता है। एकबीजपत्री तने की आंतरिक संरचना (Internal structure of monocot stem) – मक्का सामान्य रूप से पाया जाने वाला एकबीजपत्री पादप है। जिसके अनुप्रस्थ काट में निम्न संरचनाएँ दिखाई देती हैं –

1. बाह्यत्वचा (Epidermis) – इसमें क्यूटिकिल उपस्थित लेकिन रोम अनुपस्थित होते हैं।

2. हाइपोडर्मिस (Hypodermis) – बाह्यत्वचा के नीचे दृढ़ऊतकी कोशिकाओं की दो से चार परतें पायी जाती हैं, जिन्हें हाइपोडर्मिस कहते हैं।

3. भरण ऊतक (Ground tissue):
यह मृदूतकी अन्तराकोशिकीय अवकाश युक्त कोशिकाओं का बना भाग है जो अधस्त्वचा से लेकर तने के केन्द्र तक फैला होता है।

4. संवहन पूल (Vascular bundle):
इनके भरण ऊतक में बहुत – से संयुक्त (Conjoint), कोलेटरल (Collateral) तथा बन्द (Close) संवहन पूल बिखरे होते हैं अर्थात् इनमें कैम्बियम का अभाव होता है। प्रत्येक संवहन पूल के चारों तरफ स्क्लेरेनकाइमा कोशिकाएँ पायी जाती हैं। इनका संवहन पूल जाइलम तथा फ्लोएम का बना होता है। इसका जाइलम वेसेल्स, ट्रैकीड्स तथा जाइलम मृदूतक का बना होता है। जबकि फ्लोएम चालनी नलिकाओं तथा सखि कोशिकाओं का बना होता है। एकबीजपत्री तनों में फ्लोएम पैरेनकाइमा नहीं पायी जाती हैं। एकबीजपत्री तने में मज्जा (Pith) का अभाव होता है।

प्रश्न 7.
सूक्ष्मदर्शी किसी पौधे के भाग की अनुप्रस्थ काट का निम्नलिखित शरीर रचनाएँ दिखाती

1. संवहन बंडल संयुक्त, फैले हुए तथा उसके चारों ओर स्क्लेरेनकाइमी आच्छद है।
2. फ्लोएम पैरेनकाइमी है। आप इसे कैसे पहचानेंगे यह किसका है?

उत्तर:
यह एकबीजपत्री तने (Monocot stem) की अनुप्रस्थ काट की रचना है।

प्रश्न 8.
जाइलम एवं फ्लोएम को जटिल ऊतक क्यों कहते हैं ?
उत्तर:
जटिल स्थायी ऊतक (Complex permanent tissue) – पौधे के संवहनी ऊतक (Vascular tissues) ही जटिल ऊतक हैं। ये पौधे में पदार्थों को एक-स्थान से दूसरे स्थान तक ले जाने का कार्य करते हैं। इसलिए इनका नाम संवहनी ऊतक (Conductive tissue) है। जटिल ऊतक कोशिकाओं का वह समूह होता है जिसमें एक से अधिक प्रकार की कोशिकाएँ होती हैं लेकिन सब मिलकर एक इकाई की तरह कार्य करती है। जटिल ऊतक के प्रमुख उदाहरण दारू (Xylem) तथा पोषवाह (Phloem) हैं।

ऐसे ऊतक जो पानी और लवणों (Salts) को गुरुत्वाकर्षण बल के विपरीत ऊपर की ओर ले जाते हैं, जाइलम (Xylem or Wood) तथा जो ऊतक प्रकाश-संश्लेषी उत्पादों (Prepared food) के संवहन से सम्बन्धित होते हैं, उन्हें पोषवाह या बास्ट (Pholem or Bast) कहा जाता है। चूँकि कार्य सम्पादन के दृष्टिकोण से दारू और पोषवाह अलंग-अलग काम करते हैं, इसलिए उनकी रचना में भी भिन्नता पाई जाती है। चूंकि ये ऊतक एक.से अधिक ऊतकों के बने होते हैं इसलिए इन्हें जटिल ऊतक कहा जाता है।

प्रश्न 9.
रंध्रीय तंत्र क्या है? रंध्र की रचना का वर्णन कीजिए और इसका चिन्हित चित्र बनाइए।
उत्तर:
रंध्र (Stomata) ऐसी रचनाएँ हैं, जो पत्तियों की बाह्यत्वचा पर होते हैं। प्रत्येक रंध्र में दो सेम के आकार की द्वार कोशिकाएँ (Guard cells) होती हैं। द्वार कोशिका की बाहरी भित्ति पतली तथा आंतरिक भित्ति मोटी होती है। द्वार कोशिकाओं में क्लोरोप्लास्ट होता है और यह रंध्र के खुलने तथा बंद होने के क्रम को नियमित करता है। कभी-कभी कुछ बाह्यत्वचीय कोशिकाएँ, जो रंध्र के आस – पास पायी जाती हैं, इन्हें सहायक कोशिकाएँ (Subsidiary cells) कहते हैं। रंध्रीय छिद्र, द्वार कोशिका तथा सहायक कोशिकाएँ मिलकर रन्ध्रीय तंत्र (Stomatal system) का निर्माण करती हैं।

प्रश्न 10.
पुष्पीय पौधों के तीन मूलभूत ऊतक तंत्र बताइए। प्रत्येक तंत्र के ऊतक बताइए।
उत्तर:
पुष्पीय पौधों के तीन मूलभूत ऊतक तंत्र निम्नलिखित हैं –

• साधारण ऊतक
• जटिल ऊतक
• विशिष्ट ऊतक।

प्रश्न 11.
पादप शरीर (Anatomy) का अध्ययन हमारे लिए कैसे उपयोगी है ?
उत्तर:
सजीवों के आकारिकी (Morphology) का अध्ययन करने से उनके केवल बाह्य आकृति, रूप रंग में समानता अथवा भिन्नता का पता चल पाता है, लेकिन बहुत-से जीवों के बाह्य समानता होने के बावजूद उनमें विभिन्नता हो सकती है। जब हम पादपों अथवा जन्तुओं के आंतरिक संरचना का अध्ययन करते हैं तब हमें उनकी कोशिका, ऊतकों तथा अंगों में विभिन्नता दिखाई देती है। पादपों के आंतरिक संरचना का अध्ययन, विज्ञान की शरीर शाखा (Anatomy) के अन्तर्गत किया जाता है। शरीर के अध्ययन करने से पादपों की औतिकी, संरचना एवं उनके कार्यों का पता चलता है, जैसे – कि जाइलम एवं फ्लोएम ऊतकों का जल एवं भोज्य पदार्थों का संवहन करना है।

प्रश्न 12.
परिचर्म क्या है ? द्विबीजपत्री तने में परिचर्म कैसे बनता है ?
उत्तर:
कागजन कॉर्क तथा काग मिलकर परिचर्म बनाते हैं। जैसे-जैसे तने की परिधि में वृद्धि होती है वैसे ही बाहरी कॉर्टेक्स तथा बाह्य त्वचा की सतहें टूटती जाती हैं तब उन्हें संरक्षी कोशिकीय सतह की आवश्यकता पड़ती है। इसी कारण एक दूसरा प्रविभाजी ऊतक विकसित होता है जिसे कॉर्क कैम्बियम या कागजन कहते हैं। यह प्राय: कॉर्टेक्स क्षेत्र में विकसित होता है। इसकी सतहें मोटी और सँकरी भित्ति वाले आयताकार कोशिकाओं से बनी होती है।

कागजन दोनों ओर कोशिकाओं को बनाता है। बाहर की ओर की कोशिकाएँ कॉर्क अथवा काग में बँट जाते हैं। अंदर की ओर की कोशिकाएँ द्वितीयक कॉर्टेक्स अथवा काग स्तर में विभेदित हो जाती हैं। कॉर्क की कोशिकाभित्ति में सुबेरिन का जमाव हो जाता है, इस कारण इसमें पानी का प्रवेश नहीं होता है। द्वितीयक कॉर्टेक्स की कोशिकाएँ पैरेनकाइमी होती है। इस प्रकार से तीन रचनाओं कागजन, कॉर्क तथा काग मिलकर परिचर्म बनाते हैं।

प्रश्न 13.
पृष्ठाधारी पत्ती की भीतरी रचना का वर्णन चिन्हित चित्रों की सहायता से कीजिए।
उत्तर:
द्विबीजपत्री ( पृष्ठाधारी पत्ती) पत्ती की आन्तरिक संरचना (Internal structure of dicot leaf) – वे पत्तियाँ, जिनकी ऊपरी तथा निचली सतह में संरचनात्मक भिन्नता पायी जाती है पृष्ठाधारी पत्ती कहलाती है। द्विबीजपत्री पादपों में इसी प्रकार की पत्तियाँ पायी जाती हैं। आम एक सामान्य द्विबीजपत्री पौधा है, जिसकी संरचना निम्नानुसार होती है –

1. ऊपरी बाह्यत्वचा (Upper epidermis):
वे एक कोशिका स्तर मोटी पैरेनकाइमेटस कोशिकाओं की बनी होती है, जिसमें क्लोरोप्लास्ट तथा स्टेमेटा नहीं पाये जाते लेकिन उनकी बाहरी सतह पर क्यूटिकिल का एक आवरण पाया जाता है। इनकी कोशिकाएँ ढोलक के समान तथा एक-दूसरे से सटी होती हैं।

2. निचली बाह्यत्वचा (Lower epidermis):
यह एक कोशिका मोटी निचली स्तर है जिसकी संरचना बाह्य त्वचा के समान होती है लेकिन इस स्तर में कुछ विशेष प्रकार के छिद्र पाये जाते हैं जिन्हें स्टोमेटा (Stomata) कहते हैं ये वाष्पोत्सर्जन तथा वायु के आदान-प्रदान में भाग लेते हैं। स्टोमेटा एक कक्ष में खुलते हैं जिसे श्वसन कक्ष कहते हैं।

3. पर्ण मध्योतक (Mesophyll):
ऊपरी तथा निचली त्वचा के बीच के ऊतक को पर्ण मध्योतक कहते हैं। इनमें बाह्यत्वचा से लगे अन्तराकोशीय अवकाशविहीन तथा हरितलवक युक्त ऊतक पाये जाते हैं जिन्हें खम्भ मृदूतक (Palisade parenchyma) कहते हैं। ये प्रकाश-संश्लेषण में भाग लेते हैं। इन ऊतकों के नीचे के ऊतक की कोशिकाएँ अण्डाकार अन्तराकोशीय अवकाश तथा हरितलवकयुक्त होती हैं इन्हें स्पॉन्जी (Spongy parenchyma) कहते हैं।

4. संवहन पूल (Vascular bundle):
पत्तियों में संवहन पूल स्पॉन्जी मृदूतक में बिखरे रहते हैं। जिन पत्तियों में मध्यशिरा पाया जाता है उनकी मध्यशिरा का संवहन पूल सबसे बड़ा होता है। उनके सम्वहन पूल कोलेटरल या बन्द प्रकार के होते हैं तथा उनके चारों तरफ मृदूतकी कोशिकाओं का एक आवरण पाया जाता है जिसे बण्डल छाद (Bundle seath) कहते हैं। इनका जाइलम ट्रैकीड्स, वेसेल्स, काष्ठ तन्तु (Wood fire) तथा जाइलम पैरेनकाइमा का बना होता है तथा जल एवं खनिज लवणों का सम्वहन करता है।

इनका फ्लोएम, चालनी नलिकाओं, सखि कोशिकाओं तथा फ्लोएम पैरेनकाइमा का बना होता है। फ्लोएम भोज्य पदार्थों के सम्वहन का कार्य करता है। इनका जाइलम बाह्यत्वचा की ओर तथा फ्लोएम नीचे की ओर स्थित होता है जाइलम का प्रोटोजाइलम भी बाह्यत्वचा की ओर ही स्थित होता है। इसके सम्वहन पूल के ऊपर तथा नीचे पैरेनकाइमा अथवा कोलेनकाइमा की कोशिकाएँ बाह्यत्वचा तक स्थित होती हैं।

प्रश्न 14.
त्वक कोशिकाओं की रचना तथा स्थिति उन्हें किस प्रकार विशिष्ट कार्य करने में सहायता करती है ?
उत्तर:
घास में ऊपरी बाह्य त्वचा की कुछ कोशिकाएँ लंबी, खाली तथा रंगहीन होती हैं। इन कोशिकाओं को आवर्ध त्वक कोशिका कहते हैं। जब कोशिकाएँ स्फीत होती हैं, तब ये कोशिकाएँ मुड़ी हुई पत्तियों को खुलने में सहायता करती हैं। वाष्पोत्सर्जन की अधिक दर होने पर ये पत्तियाँ वाष्पोत्सर्जन की दर कम करने के लिए मुड़ जाती हैं।

पुष्पी पादपों का शारीर अन्य महत्वपूर्ण प्रश्नोत्तर

पुष्पी पादपों का शारीर वस्तुनिष्ठ प्रश्न

प्रश्न 1.
सही विकल्प चुनकर लिखिए –

1. कुछ पादप कोशिकाओं की कोशिकाभित्ति मोटी तथा सिरों पर नुकीली होती है, यह हो सकती –
(a) पैरेनकाइमा
(b) क्लोरेनकाइमा
(c) स्क्लेरेनकाइमा
(d) ऐरेनकाइमा।
उत्तर:
(c) स्क्लेरेनकाइमा

2. शीर्षस्थ ऊतक के डर्मेटोजन का प्रमुख कार्य है –
(a) उपत्वचा का निर्माण
(b) बाह्यत्वचा का निर्माण
(c) संवहन ऊतक का निर्माण
(d) हाइपोडर्मिस का निर्माण।
उत्तर:
(b) बाह्यत्वचा का निर्माण

3. लेटेक्स वाहिकाएँ पायी जाती हैं –
(a) दारु ऊतक में
(b) पोषवाह ऊतक में
(c) वल्कुट में
(d) उपर्युक्त में से किसी में नहीं।
उत्तर:
(c) वल्कुट में

4. एक ऊतक जिसकी कोशिकाएँ सक्रिय बढ़ने वाले अंगों के यान्त्रिक ऊतकों को बनाती हैं और
इनकी कोशिका भित्तियों में कोशिकाओं के कोनों पर सेल्यूलोजिक, अलिग्नीन स्थूलन पाये जाते हैं, ये हो सकती हैं
(a) स्क्लेरेनकाइमा
(b) पैरेनकाइमा
(c) कोलेनकाइमा
(d) क्लोरेनकाइमा।
उत्तर:
(c) कोलेनकाइमा

5. एकबीजपत्रियों के प्रारूपी संवहन पूल होते हैं –
(a) बहिफ्लोएमी
(b) उभय फ्लोएमी
(c) संकेन्द्री
(d) अरीय।
उत्तर:
(a) बहिफ्लोएमी

6. ऊतक किसे कहते हैं –
(a) कोशिकाओं का अलग-अलग विकसित होने वाला समूह
(b) कोशिकाओं का वह समूह जो रचना, उत्पत्ति तथा कार्य में समान हो
(c) कोशिकाओं का वह समूह जो रचना में समान लेकिन कार्य में भिन्न हो
(d) कोशिकाओं का एकसमान आयु वाला समूह।
उत्तर:
(b) कोशिकाओं का वह समूह जो रचना, उत्पत्ति तथा कार्य में समान हो

7. पैरेनकाइमा (मृदूतकी) ऊतक किसे कहते हैं –
(a) मोटी कोशिकीय भित्ति वाली कोशिकाओं का समूह
(b) मृत कोशिकाओं का समूह
(c) पतली कोशिकीय भित्ति वाली जीवित कोशिकाओं का समूह
(d) लिग्निन युक्त कोशिकाओं का समूह।
उत्तर:
(c) पतली कोशिकीय भित्ति वाली जीवित कोशिकाओं का समूह

8. एक ऊतक जिसकी कोशिकाएँ महीन भित्ति वाली, समव्यासी और अन्तराकोशिकीय स्थान वाली हैं, वह होगा –
(a) पैरेनकाइमा
(b) कोलेनकाइमा
(c) स्क्लेरेनकाइमा
(d) क्लोरेनकाइमा।
उत्तर:
(a) पैरेनकाइमा

9. डर्मेटोजन, पेरीब्लेम तथा प्लीरोम पाये जाते हैं –
(a) पूर्णाग्र में
(b) तने की गाँठों पर
(c) मूलाग्र तथा तनाग्र में
(d) जायलम तथा फ्लोएम में।
उत्तर:
(c) मूलाग्र तथा तनाग्र में

10. पादपों के जलोत्सर्जक या हाइडेथोड हैं –
(a) वाष्पोत्सर्जन करने वाली रचनाएँ
(b) रसारोहण करने वाली रचनाएँ
(c) म्यूसीलेज स्रावित करने वाली रचनाएँ
(d) जल स्रावित करने वाली रचनाएँ।
उत्तर:
(d) जल स्रावित करने वाली रचनाएँ।

11. रसकाष्ठ होता है –
(a) द्वितीयक जाइलम का बाहरी क्रियाशील भाग
(b) द्वितीयक जाइलम का आंतरिक अक्रियाशील भाग
(c) द्वितीयक जाइलम का आंतरिक तथा बाह्य भाग
(d) इनमें से कोई नहीं।
उत्तर:
(a) द्वितीयक जाइलम का बाहरी क्रियाशील भाग

12. पुराने तने में गैस का आदान-प्रदान किसके द्वारा होता है –
(a) स्टोमेटा
(b) हाइडेथोड्स
(c) वातरंध्र
(d) पैलेड कोशिका।
उत्तर:
(c) वातरंध्र

13. पेरीडर्म में होते हैं –
(a) कॉर्क कैम्बियम, कॉर्क, द्वितीयक कॉर्टेक्स
(b) कॉर्क कैम्बियम तथा कॉर्क
(c) कॉर्क
(d) कॉर्क तथा द्वितीयक फ्लोएम।
उत्तर:
(a) कॉर्क कैम्बियम, कॉर्क, द्वितीयक कॉर्टेक्स

14. जैसे-जैसे द्वितीयक वृद्धि होती है किसकी मोटाई में वृद्धि होती है –
(a) अन्त:काष्ठ
(b) रसकाष्ठ
(c) अन्तः एवं रसकाष्ठ
(d) इनमें से कोई नहीं।
उत्तर:
(a) अन्त:काष्ठ

15. वृक्ष की आयु का पता लगाने का सबसे आसान तरीका है –
(a) तने का व्यास मापना
(b) पत्तियों की संख्या की गिनती करना
(c) तने के आधार पर वार्षिक वलयों की गणना
(d) शाखाओं की गणना करना।
उत्तर:
(c) तने के आधार पर वार्षिक वलयों की गणना

16. द्विबीजपत्री तने के काष्ठ में जाइलम का सबसे नया स्तर स्थित होता है –
(a) कैम्बियम के बाहर
(b) कैम्बियम के अंदर
(c) पिथ के बाहर
(d) कॉर्टेक्स के अंदर।
उत्तर:
(b) कैम्बियम के अंदर

17. द्विबीजपत्री तने में संवहन पूल होते हैं –
(a) खुला, कोलैटरल तथा एन्डार्क
(b) बंद, कोलैटरल तथा एन्डार्क
(c) खुला, कोलैटरल तथा एक्सार्क
(d) बंद, कोलैटरल तथा एक्सॉर्क।
उत्तर:
(a) खुला, कोलैटरल तथा एन्डार्क

18. द्विबीजपत्री जड़ में पाये जाने वाले संवहन पूल होते हैं –
(a) रेडियल एक्सॉर्क
(b) संयुक्त एण्डार्क
(c) रेडियल एन्डार्क
(d) संयुक्त एक्सॉर्क।
उत्तर:
(a) रेडियल एक्सॉर्क

19. कॉर्क का निर्माण किससे होता है –
(a) कॉर्क कैम्बियम
(b) संवहन कैम्बियम
(c) फ्लोएम
(d) जाइलम।
उत्तर:
(a) कॉर्क कैम्बियम

20. कॉर्क कैम्बियम का कार्य किसका निर्माण करना होता है –
(a) द्वितीयक जाइलम तथा द्वितीयक फ्लोएम
(b) कॉर्क तथा द्वितीयक कॉर्टेक्स
(c) द्वितीयक कॉर्टेक्स तथा फ्लोएम
(d) कॉर्क।
उत्तर:
(b) कॉर्क तथा द्वितीयक कॉर्टेक्स

21. कैस्पेरियन बैन्ड्स कहाँ पाये जाते हैं:
(a) एपिडर्मिस
(b) एण्डोडर्मिस
(c) पेरीसाइकिल
(d) फ्लोएम।
उत्तर:
(a) एपिडर्मिस

22. वार्षिक वलय का निर्माण किसकी सक्रियता से होता है –
(a) कैम्बियम
(b) जाइलम
(c) फ्लोएम
(d) जाइलम तथा फ्लोएम दोनों।
उत्तर:
(a) कैम्बियम

23. टाइलोसेज किसमें पाये जाते हैं –
(a) द्वितीयक जाइलम
(b) द्वितीयक फ्लोएम
(c) थैलस ऊतक
(d) कॉर्क कोशाएँ।
उत्तर:
(a) द्वितीयक जाइलम

24. वेसेल्स के अंदर मृदूतक कोशिकाओं का बैलून के समान अतिवृद्धि कहलाती है –
(a) हिस्टोजन
(b) टाइलोसेज
(c) फैलोजेन
(d) ट्यूनिका।
उत्तर:
(b) टाइलोसेज

25. एकबीजपत्री पौधों में ग्राफ्टिग संभव नहीं है, क्योंकि –
(a) कैम्बियम नहीं पाया जाता
(b) संवहन पूल बिखरे होते हैं
(c) समान्तर शिराविन्यास पाये जाते हैं
(d) पृथक् अरीय संवहन पूल पाये जाते हैं।
उत्तर:
(a) कैम्बियम नहीं पाया जाता

26. संयुक्त, कोलैटरल, खुला तथा एण्डार्क संवहन पूल किसमें पाये जाते हैं –
(a) एकबीजपत्री तना
(b) एकबीजपत्री जड़
(c) द्विबीजपत्री जड़
(d) द्विबीजपत्री तना।
उत्तर:
(d) द्विबीजपत्री तना।

27. जड़ों में पार्श्व शाखाएँ किससे विकसित होती हैं –
(a) एपिब्लेमा
(b) पेरीसाइकिल
(c) कॉर्टेक्स
(d) एण्डोडर्मिस।
उत्तर:
(b) पेरीसाइकिल

28. कॉर्क कोशिकाएँ किसके जमाव के कारण अपारगम्य होती हैं –
(a) क्यूटिन
(b) लिग्निन
(c) सुबेरिन
(d) काइटिन।
उत्तर:
(b) लिग्निन

29. वेलामेन ऊतक किसमें भाग लेते हैं –
(a) श्वसन
(b) नमी का अवशोषण
(c) वाष्पोत्सर्जन
(d) संरक्षण।
उत्तर:
(b) नमी का अवशोषण

30. एकबीजपत्री तने में संवहन पूल होते हैं –
(a) बाइकोलैटरल तथा बंद
(b) बाइकोलैटरल तथा खुला
(c) (a) एवं (b) दोनों
(d) इनमें से कोई नहीं।
उत्तर:
(a) बाइकोलैटरल तथा बंद

प्रश्न 2.
एक शब्द में उत्तर दीजिए –

1. द्वितीयक वृद्धि से बनने वाली रचना जो वातावरण से गैसों का आदान-प्रदान करती है।
2. पुष्पीय पौधों की पत्तियों के सिरों पर पाई जाने वाली विशेष रचना जिनसे पानी की बूंदों के रूप में स्राव होता है।
3. एक पार्श्व विभज्योतक ऊतक जिसके कारण पौधों की मोटाई में वृद्धि होती है।
4. अरीय संवहन पूल किसमें पाये जाते हैं ?
5. किसी एक बीजपत्री पौधे का नाम जिसमें द्वितीयक वृद्धि होती है।
6. वृक्ष में हल्के रंग की लकड़ी को क्या कहते हैं?
7. द्विबीजपत्री जड़ में संवहन पूल की क्या संख्या होती है ?
8. दो बाह्य त्वचीय अतिवृद्धियों के नाम लिखिये।
9. रबर क्षीरी, कोशिका वाले पौधे का नाम लिखिये।
10. कैम्बियम की उपस्थिति वाले संवहन पूल को क्या कहते हैं?
11. फैलोजन, फैलोडर्म तथा फैलम का सम्मिलित नाम क्या है ?
12. द्विबीजपत्री तनों के बाह्य भाग में पाया जाने वाला काष्ठीय भाग क्या कहलाता है?
13. ऐसा संवहन पूल जिसमें जाइलम के दोनों ओर कैम्बियम तथा फ्लोएम पाये जाते हैं ?
14. तने में एपिडर्मिस के नीचे पाया जाने वाला कोलेनकाइमेटस भाग क्या कहलाता है ?

उत्तर:

1. वातरंध्र
2. जलरंध्र
3. कैम्बियम
4. जड़
5. ड्रैसीना
6. रसकाष्ठ
7. 1 से 6 तक
8. ट्राइकोम्स व सिस्टोलिथ
9. यूफोर्बिया
10. खुला संवहन पूल
11. पेरीडर्म
12. रसकाष्ठ
13. बाइकोलैटरल
14. हाइपोडर्मिस।

प्रश्न 3.
रिक्त स्थानों की पूर्ति कीजिए –

1. डर्मेटोजन, पेरीब्लेम तथा प्लीरोम …………….. में पाये जाते हैं।
2. वाहिकाओं के बीच पाई जाने वाली गुब्बारे जैसी रचना को …………. कहते हैं।
3. वार्षिक वलय …………….. से बनता है।
4. लेटेक्स वाहिकाएँ …………….. में पाई जाती हैं। ………..
5. द्विबीजपत्री तने में सबसे पुराना फ्लोएम …………….. में स्थित होता है।
6. पौधों की ……………. में वृद्धि को द्वितीयक वृद्धि कहते हैं।
7. द्वितीयक वृद्धि ………….. में सक्रियता के कारण होती है।
8. कॉर्टेक्स सामान्यतः ……………. कोशिकाओं द्वारा बना होता है।
9. द्वितीयक संवहनी ऊतकों का निर्माण ……………. में होता है।
10. तने में वार्षिक वलय ……………. तथा ……………. के कारण होता है।

उत्तर:

1. मूलाग्र व तनाग्र में
2. टायलोसिस
3. स्प्रिंग व ऑटम काष्ठ से
4. कॉर्टेक्स
5. प्राथमिक फ्लोएम के अन्दर
6. मोटाई
7. कैम्बियम
8. पैरेन्काइमेट्स
9. कैम्बियम
10. वसन्त काष्ठ, शरद काष्ठ।

प्रश्न 4.
उचित संबंध जोडिए –

उत्तर:

1. (c) संयुक्त कोलेटरल वर्षी पूल
2. (d) अरीय 6 से कम संवहन पूल
3. (a) संयुक्त कोलेटरल अवर्धी पूल
4. (e) अरीय 6 से अधिक संवहन पूल।
5. (b) दृढ़ोतकी बंडल छाद

उत्तर:

1. (d) संयुक्त कोलैटरल बंद पूल
2. (e) अंत:काष्ठ में अतिवृद्धि।
3. (a) द्विबीजपत्री पौधा
4. (b) चौड़ी व बड़ी कोशिकाएँ
5. (c) सँकरी व छोटी कोशिकाएँ

उत्तर:

1. (c) संवहन ऊतक का कोर
2. (d) कैस्पेरियन पट्टी।
3. (a) वेलामेन ऊतक
4. (b) एकबीजपत्री जड़

उत्तर:

1. (b) द्वितीयक संवहन ऊतक
2. (a) पेरीडर्म
3. (d) टाइलोसेज़।
4. (c) सखी कोशाएँ

प्रश्न 5.
सत्य / असत्य बताइए

1. विकसित प्रौढ़ पादपों में द्वितीयक वृद्धि पाई जाती है।
2. छाल पर वात विनिमय हेतु कुछ विशिष्ट छिद्र होते हैं जिसे जलरंध्र कहते हैं।
3. ऐसे ऊतक जिनकी कोशिकाओं में विभाजन क्षमता नहीं होती प्रविभाजी ऊतक कहलाते हैं।
4. किसी तने के अनुप्रस्थ काट में वार्षिक वलयों को गिनकर वृक्ष की उम्र का पता लगाया जाता है।
5. अपवादस्वरूप कुछ एकबीजपत्री पादपों के तनों में भी द्वितीयक वृद्धि पाई जाती है।
6. एकबीजपत्री पौधों में द्वितीयक वृद्धि नहीं पाई जाती।
7. कॉर्क कैम्बियम का निर्माण संवहन पूल में होता है।
8. कॉर्क कैम्बियम को फैलोजन भी कहा जाता है।
9. वार्षिक वलय की संख्या के आधार पर पौधे की आयु का पता लगाया जा सकता है।
10. पेरीडर्म तथा छाल वानस्पतिक रूप से एक समान होते हैं।
11. तने की छाल पर पाये जाने वाले छिद्र को वातरंध्र कहते हैं।
12. घावों को भरना पौधे में वैस्कुलर कैम्बियम की सक्रियता के कारण होता है।
13. पर्ण विलगन के समय विलगन परत का निर्माण होता है।
14. द्विबीजपत्री तने में अरीय संवहन पूल पाये जाते हैं।
15. व्यापारिक दृष्टि से अंत:काष्ठ अधिक महत्वपूर्ण होता है।

उत्तर:

1. सत्य
2. असत्य
3. असत्य
4. सत्य
5. सत्य
6. सत्य
7. असत्य
8. सत्य
9. सत्य
10. असत्य
11. सत्य
12. असत्य
13. सत्य
14. असत्य
15. सत्य।

पुष्पी पादपों का शारीर अति लघु उत्तरीय प्रश्न

प्रश्न 1.
ऊतक क्या है ?
उत्तर:
एकसमान उत्पत्ति एवं विकास वाली कोशिकाओं का ऐसा समूह जो एकसमान कार्य करता हो, ऊतक कहलाता है।

प्रश्न 2.
जाइलम का कार्य बताइए।
उत्तर:

• जाइलम जल एवं उसमें घुले लवणों का संवहन करता है।
• यह पौधों को यांत्रिक शक्ति (मजबूती) प्रदान करता है।

प्रश्न 3.
द्वार कोशिकाएँ कहाँ पायी जाती हैं ? इसके क्या कार्य हैं ?
उत्तर:
द्वार कोशिकाएँ (Guard cells) पत्तियों के रन्ध्रों में पायी जाती हैं, ये रन्ध्रों के खुलने एवं बन्द होने की क्रिया को नियंत्रित करती हैं। .

प्रश्न 4.
कैम्बियम क्या है? इसके कार्य बताइए।
उत्तर:
कैम्बियम एक पार्श्व विभज्योतक ऊतक है, जिसके कारण पौधे की मोटाई में वृद्धि होती है।

प्रश्न 5.
स्थायी ऊतक किसे कहते हैं?
उत्तर:
ऐसे ऊतक जिनमें विभाजन की क्षमता नहीं पायी जाती है, उन्हें स्थायी ऊतक कहते हैं।

प्रश्न 6.
बास्ट रेशा क्या है?
उत्तर:
फ्लोएम के साथ सम्बद्ध दृढ़ोतक कोशिकाएँ बास्ट रेशे कहलाती हैं।

प्रश्न 7.
जायलम ऊतक में पायी जाने वाली कोशिकाओं के नाम बताइए।
उत्तर:
जायलम में निम्नलिखित चार प्रकार की कोशिकाएँ पायी जाती हैं –

• वाहिनिका (Tracheids)
• वाहिनी (Vessel or Tracheae)
• जायलम रेशे या काष्ठ रेशे (Xylem fibres or Wood fibres)
• जायलम या काष्ठ पैरेनकायमा (Xylem or Wood parenchyma)।

प्रश्न 8.
कैम्बियम कहाँ पाया जाता है?
उत्तर:
कैम्बियम प्रायः द्विबीजपत्री पौधों के संवहन पूलों में पार्श्व विभज्योतक के रूप में संवहन पूलों (V. B.) में जायलम एवं फ्लोएम के बीच पाया जाता है।

प्रश्न 9.
फ्लोएम में पायी जाने वाली कोशिकाओं के नाम लिखिए।
उत्तर:
फ्लोएम में निम्नलिखित कोशिकाएँ पायी जाती हैं –

• चालनी नलिका (Sieve tube)
• सखि कोशिकाएँ (Companian cells)
• फ्लोएम पैरेनकायमा (Phloem parenchyma)
• बास्ट रेशे (Bast fibres)।

प्रश्न 10.
ऊतक तंत्र किसे कहते हैं?
उत्तर:
पुष्पीय पौधों में अनेक प्रकार के स्थायी ऊतक आपस में संयुक्त होकर एक या कई कार्यों का निर्वहन करते हैं। इन संयुक्त ऊतकों से निर्मित तंत्र को ही ऊतक तंत्र (Tissue system) कहते हैं।

प्रश्न 11.
काष्ठ क्या है?
उत्तर:
तनों में उपस्थित द्वितीयक जायलम को काष्ठ (Wood) कहते हैं।

पुष्पी पादपों का शारीर लघु उत्तरीय प्रश्न

प्रश्न 1.
प्रविभाजी / विभज्योतक ऊतक की चार विशेषताएँ बताइए।
उत्तर:
प्रविभाजी विभज्योतक ऐसी कोशिकाओं से मिलकर बना होता है, जिसमें बार-बार विभाजन की क्षमता होती है। इनमें निम्नलिखित विशेषताएँ पायी जाती हैं –

• इनकी आकृति बहुभुजी या चौकोर होती है। कोशिकाओं की भित्ति पतली व सेल्युलोज की बनी होती है।
• कोशिकाओं में बड़े तथा निश्चित केन्द्रक होते हैं। कोशिकाएँ आपस में सटी हुई होती हैं अर्थात् अन्तरकोशिकीय अवकाशों का अभाव होता है।
• कोशिकाओं में लवण प्राकलवक (Proplastid) अवस्था में होते हैं।
• इन कोशिकाओं में जीवद्रव्य सघन तथा प्रायः रिक्तिकाहीन होता है। ये कोशिकाएँ सक्रिय उपापचय की अवस्था में रहती हैं।

प्रश्न 2.
जड़ एवं तने के वैस्कुलर बंडल में अन्तर स्पष्ट कीजिए।
अथवा
अरीय व संयुक्त वाहिनी पूल में अन्तर स्पष्ट कीजिए।
उत्तर:
अरीय व संयुक्त वाहिनी पूल में अन्तर –

प्रश्न 3.
वातरंध्र पर संक्षिप्त टिप्पणी दीजिए।
उत्तर:
तने की छाल पर पाये जाने वाले छिद्र को वातरंध्र कहते हैं। इसकी सहायता से वातावरण से गैसों का आदान-प्रदान होता है। द्वितीयक वृद्धि के बाद बनने वाली कॉर्क की कोशाएँ जल तथा वायु के लिए अपारगम्य होती हैं। वातरंध्र बनने के पहले शिशु तने के उन्हीं स्थानों पर स्टोमेटा पाये जाते हैं। इनका निर्माण पेरिडर्म के निर्माण के साथ ही प्रारंभ हो जाता है। इन स्थानों पर फैलोजन की कोशिकाएँ बाहर की ओर कॉर्क कोशिकाओं का निर्माण नहीं कर पातीं। इनके स्थान पर ये पतली भित्ती वाली पैरेनकाइमेटस कोशाएँ बनाती हैं। इन कोशाओं को पूरक कोशाएँ कहते हैं।

इनके दबाव के कारण एपिडर्मिस की कोशिकाएँ टूट जाती हैं तथा छिद्र के समान संरचना बन जाती है इसे वातरंध्र कहते हैं। परक कोशिकाओं के बीच पर्याप्त अंतराकोशिकीय अवकाश पाया जाता है। अनेक स्थितियों में वातरंध्र स्टोमेटा के स्थान पर नये स्थान पर भी बनते हैं। नये तनों की सतह पर वातरंध्र को छोटे – छोटे उभार के रूप में आसानी से देखा जा सकता है। शीतऋतु में सामान्यतः कॉर्क कोशिकाओं द्वारा बँक जाते हैं। बसंत ऋतु आने पर फैलोजेन की सक्रियता के कारण पुरानी कॉर्क कोशिकाएँ नष्ट हो जाती हैं तथा नई कोशिकाएँ तेजी से बनने लगती हैं, इससे वातरंध्र खुल जाते हैं।

प्रश्न 4
द्विबीजपत्री एवं एकबीजपत्री पत्ती की आन्तरिक संरचना में अन्तर स्पष्ट कीजिए।
अथवा
एकबीजपत्री एवं द्विबीजपत्री में अन्तर लिखिए।
उत्तर:
द्विबीजपत्री तथा एकबीजपत्री की आन्तरिक संरचना में अन्तर –

पुष्पी पादपों का शारीर दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
विभिन्न प्रकार के संवहन बण्डलों की रचना समझाइए।
उत्तर:
संवहन पूलों का प्रकार (Types of vascular bundles):
विभिन्न अंगों या ऊतकों का बना वह तन्त्र जो संवहन का कार्य करता है, संवहन तन्त्र कहलाता है। पौधों का सम्वहन तन्त्र अनेक छोटी-छोटी समान इकाइयों का बना होता है, जिन्हें संवहन पूल (Vascular bundle) कहते हैं। प्रत्येक सम्वहन पूल जायलम तथा फ्लोएम या जायलम, फ्लोएम तथा कैम्बियम का बना होता है। जायलम तथा फ्लोएम की स्थिति तथा कैम्बियम की उपस्थिति और अनुपस्थिति के आधार पर सम्वहन पूल निम्नलिखित प्रकार के हो सकते हैं –

1. अरीय (Radial) – इसमें जायलम तथा फ्लोएम एक-दूसरे से अलग – अलग भिन्न – भिन्न त्रिज्याओं पर एकान्तरित क्रम में स्थित होते हैं। यह सम्वहन पूल मुख्यतः जड़ों में पाया जाता है।

2. संयुक्त (Conjoint) – इसमें जायलम तथा फ्लोएम एक ही त्रिज्या पर एक – दूसरे से जुड़े हुए स्थित होते हैं। यह मुख्यतः तने में पाया जाता है। यह सम्वहन पूल दो प्रकार का होता है –

(a) कोलैटरल (Collateral):
इन सम्वहन पूलों में एक ही त्रिज्या पर फ्लोएम बाहर की तरफ तथा जायलम अन्दर की तरफ स्थित होता है। यह पुनः दो प्रकार का होता है –

1. कोलैटरल खुला (Collateral open) – इसमें जायलम तथा फ्लोएम के बीच में कैम्बियम स्थित होता है। इस प्रकार का सम्वहन पूल द्विबीजपत्री तनों में पाया जाता है।
2. कोलैटरल बन्द (Collateral closed) – इस प्रकार के सम्वहन पूलों में कैम्बियम नहीं पाया जाता। यह एक बीजपत्री पौधों के तनों में पाया जाता है।

(b) बाइकोलैटरल (Bicollateral):
इस प्रकार के सम्वहन पूल में जायलम बीच में स्थित होता है और इसके दोनों तरफ कैम्बियम तथा फ्लोएम पाये जाते हैं। इस प्रकार यह एक जायलम, दो कैम्बियम तथा दो फ्लोएम का बना होता है। कुकरबिटेसी, सोलेनेसी, कान्वल्वुलेसी और एपोसाइनेसी कुल के पौधों में इस प्रकार का सम्वहन पूल पाया जाता है।

3. संकेन्द्री (Concentric):
इस प्रकार के सम्वहन बण्डल में जायलम तथा फ्लोएम में से एक-दूसरे को पूरी तरह से घेरे रहता है। यह बण्डल बन्द प्रकार का होता है, अर्थात् इसमें कभी भी कैम्बियम नहीं पाया जाता है। यह दो प्रकार का होता है –

(a) एम्फीक्राइबल (Amphicribal):
इसमें जायलम को फ्लोएम पूरी तरह से घेरे रहता है। यह सम्वहन पूल कुछ जलीय पौधों और टेरिडोफाइट्स जैसे – सिलैजिनेला (Selaginella) और लाइकोपोडियम (Lycopodium) में पाया जाता है।

(b) एम्फीवेसल (Amphivasal):
इसमें फ्लोएम को जायलम चारों तरफ से घेरे रहता है। यह कुछ एकबीजपत्री पौधों के तनों में पाया जाता है। जैसे – ड्रैसीना (Dracena), यूक्का (Yucca)।

MP Board Class 11th Biology Solutions

MP Board Class 10th Hindi Navneet Solutions गद्य Chapter 5 पहली चूक (व्यंग्य निबन्ध, श्रीलाल शुक्ल)

पहली चूक अभ्यास

बोध प्रश्न

पहली चूक अति लघु उत्तरीय प्रश्न

प्रश्न 1.
‘आपका नाम बाजरा तो नहीं है’ यह कथन किसने किससे पूछा?
उत्तर:
यह कथन लेखक ने रामचरन से पूछा।

प्रश्न 2.
रामचरन ने किसानों को पुकारते हुए क्या कहा?
उत्तर:
रामचरन ने किसानों को पुकारते हुए कहा-“यह देखो, ये भैया तो बाजरे को आदमी समझ रहे हैं।”

प्रश्न 3.
लेखक ने भाषायी चूक से बचने के लिए कौन-कौनसी तैयारी की ?
उत्तर:
लेखक ने भाषायी चूक से बचने के लिए कृषि शास्त्र की मोटी-मोटी किताबें मँगवाकर उनका अध्ययन आरम्भ कर दिया।

प्रश्न 4.
लेखक को बीज गोदाम क्यों जाना पड़ा?
उत्तर:
लेखक को बीज और खाद खरीदने के लिए बीज गोदाम जाना पड़ा।

प्रश्न 5.
लेखक के अनुसार कंद-मूल-फल खाने के कारण ऋषियों की दिलचस्पी किस व्यवसाय में थी?
उत्तर:
कंद-मंद-फल खाने के कारण ऋषियों की दिलचस्पी हॉर्टिकल्चर में थी।

पहली चूक लघु उत्तरीय प्रश्न

प्रश्न 1.
चाचा ने लेखक को खेती के बारे में क्या समझाया?
उत्तर:
चाचा ने लेखक को खेती के बारे में समझाया कि खेती का काम है तो बड़ा उत्तम, पर फारसी पढ़कर जिस प्रकार तेल नहीं बेचा जा सकता, वैसे ही अंग्रेजी पढ़कर खेत नहीं जोता जा सकता।

प्रश्न 2.
लेखक ने रामचरन के कन्धे पर हाथ रखते हुए मित्र भाव से क्या-क्या प्रश्न किये?
उत्तर:
लेखक ने रामचरन के कन्धे पर हाथ रखते हुए मित्र भाव से कहा-“तो भाई रामचरन, मुझे बताओ यह बाजरा कौन है? कहाँ रहता है? यह खड़ा कहाँ है? इसे क्यों खड़ा किया गया है?”

प्रश्न 3.
पंचभूत और पंचगव्य में क्या अन्तर है?
उत्तर:
मानव शरीर का पंचभूतों अर्थात्-पृथ्वी, जल, अग्नि, आकाश और वायु से मिलकर निर्माण होता है। पंचगव्य से आशय है-गाय से मिलने वाली पाँच वस्तुएँ-दूध, दही, घी, गोबर और गौ मूत्र।

प्रश्न 4.
“शरीर ही मिट्टी का बना हुआ है”-किस अर्थ में प्रयुक्त हुआ है?
उत्तर:
मानव शरीर के निर्माण में पंचभूतों का योग होता है और ये पंच भूत हैं-पृथ्वी (मिट्टी), जल, अग्नि, आकाश और वायु। अतः जब शरीर के निर्माण में मिट्टी की प्रमुख भूमिका है तब फिर हमें मिट्टी से क्यों परहेज करना चाहिए?

प्रश्न 5.
लेखक ने अन्तिम बार शहर जाने का फैसला क्यों किया?
उत्तर:
लेखक ने अन्तिम बार शहर जाने का फैसला इसलिए किया कि वह शहर में रहकर बाजरे के विषय में अपनी रिसर्च कर सकेगा तथा वह बता पायेगा कि खाद के प्रयोग से बाजरे की लताओं में मीठे और बड़े-बड़े फल लाये जा सकते हैं।

पहली चूक दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
‘पहली चूक’ पाठ का केन्द्रीय भाव लिखिए।
उत्तर:
‘पहली चूक’ शीर्षक पाठ में लेखक ने उन शिक्षित जनों को अपनी रचना का केन्द्र बिन्दु बनाया हैं जो शहर में निवास करते हैं और ग्रामीण वातावरण से पूर्णतः अनभिज्ञ होते हैं। वे पुस्तकों और सिनेमाओं से प्राप्त जानकारी के आधार पर अपने मन में काल्पनिक ग्राम का निर्माण किया करते हैं। उनकी यह कल्पना उस समय खण्ड-खण्ड होकर बिखर जाती है, जब वे ग्रामीण परिवेश में जाकर वास्तविक बातों को जान पाते हैं।

प्रश्न 2.
‘सिनेमा से खेती की शिक्षा’ पर लेखक के विचार लिखिए।
उत्तर:
सिनेमा खेती की उन्नति का एक अच्छा साधन है। सिनेमा द्वारा खेती का बड़ा प्रचार हुआ है। बड़े-बड़े हीरो खेत जोतते जाते हैं और गाते जाते हैं। हीरोइन खेत पर टोकरी में रोटी लेकर आती है। हरी-भरी फसल में आँख-मिचौली का खेल होता है। फसल काटते समय हीरोइन के साथ बहुत-सी लड़कियाँ नाचती हैं और गाती भी हैं। वे नाचती जाती हैं और फसल अपने आप कटती जाती है। ऐसे ही मधुर दृश्यों को देखकर, पढ़े-लिखे आदमी गाँवों में आने लगते हैं और खेतों के चक्कर काटने लगते हैं। इस प्रकार सिनेमा द्वारा खेती की शिक्षा मिलती है।

प्रश्न 3.
‘फारस में तेल बेचने वाले संस्कृत नहीं बोलते’ इस कथन का भाव विस्तार कीजिए।
उत्तर:
लेखक का चाचा लेखक को समझाता है कि जिस प्रकार फारस में तेल बेचने वाले संस्कृत नहीं बोलते, खेत जोतते हैं। उसी प्रकार भारत में फारसी पढ़कर तेल नहीं बेचा जाता है। उसी प्रकार अंग्रेजी पढ़कर भी खेती नहीं की जा सकती।

पहली चूक भाषा अध्ययन

प्रश्न 1.
निम्नलिखित शब्दों का सन्धि-विच्छेद करते हुए सन्धि का नाम बताइए
उत्तर:

1. वागीश = वाक् + ईश = व्यंजन सन्धि।
2. कीटाणु = कीट + अणु = दीर्घ सन्धि।
3. ग्रन्थालय = ग्रन्थ + आलय = दीर्घ सन्धि।
4. महेश = महा + ईश = गुण सन्धि।
5. उन्नति = उत् + नति = व्यंजन सन्धि।।

प्रश्न 2.
निम्नलिखित वाक्यों में संकेत के अनुसार परिवर्तन कीजिए-

1. खेत के पूरब में गन्ने का एक खेत है। (प्रश्न वाचक)
2. मैं खेती के बारे में अधिक जानता हूँ। (निषेधात्मक वाक्य)
3. यह आप क्यों पूछ रहे हो ? (विस्मयबोधक वाक्य)

उत्तर:

1. क्या खेत के पूरब में गन्ने का एक खेत है?
2. मैं खेती के बारे में अधिक नहीं जानता हूँ।
3. अरे! यह आप क्यों पूछ रहे हो?

पहली चूक महत्त्वपूर्ण वस्तुनिष्ठ प्रश्न

पहली चूक बहु-विकल्पीय प्रश्न

प्रश्न 1.
‘पहली चूक’ पाठ में लेखक ने किसे केन्द्र-बिन्दु बनाया है? (2009)
(क) ग्रामीणों को
(ख) शिक्षितों को
(ग) मूरों को
(घ) अज्ञानियों को।
उत्तर:
(क) ग्रामीणों को

प्रश्न 2.
गोबर पवित्र वस्तु है, उसका स्थान है
(क) पंचभूत में
(ख) मिट्टी में
(ग) पंचगव्य में
(घ) किसी में नहीं।
उत्तर:
(ग) पंचगव्य में

प्रश्न 3.
“बेटा! यह खेती का पेशा तुमसे नहीं होगा।” किसने कहा है?
(क) चाचा ने
(ख) लेखक ने
(ग) किसान ने
(घ) रामचरन ने।
उत्तर:
(क) चाचा ने

प्रश्न 4.
खेती करने वाला कहलाता है (2009)
(क) मजदूर
(ख) मुनीम
(ग) माली
(घ) कृषक।
उत्तर:
(घ) कृषक।

प्रश्न 5.
‘पहली चूक’ किस विधा में लिखी गई रचना है? (2018)
(क) कहानी
(ख) एकांकी
(ग) निबन्ध
(घ) व्यंग्य निबन्ध।
उत्तर:
(घ) कृषक।

रिक्त स्थानों की पूर्ति

1. मिट्टी का स्थान यदि पंचभूत में है तो गोबर का स्थान …………. में है।
2. क्योंकि फारस में तेल बेचने वाले …………. नहीं बोलते,खेत जोतते हैं।
3. हरी-भरी फसल में ……….. का खेल होता है।
4. गाँव की पुरानी पीढ़ी भी ……….. होती जा रही है।
5. पहली चूक …………. विधा की रचना है। (2012, 17)

उत्तर:

1. पंचगव्य
2. संस्कृत
3. आँख मिचौली
4. समाप्त
5. व्यंग्य

सत्य/असत्य

1. ‘पहली चूक’ का शीर्षक सटीक एवं सार्थक नहीं है।
2. लेखक ने खेत में खड़े बाजरे को आदमी समझा।
3. बीज खरीदने के लिए बीजगोदाम जाना पड़ता है।
4. ‘मैं वैज्ञानिक ढंग से खेती करूँगा।’ अन्त में लेखक ने यही निर्णय लिया।

उत्तर:

1. असत्य
2. सत्य
3. सत्य
4. सत्य

सही जोड़ी मिलाइए

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

एक शब्द/वाक्य में उत्तर

1. पहली चूक किसके द्वारा हुई थी?
2. “फारसी पढ़कर तेल नहीं बेचा जा सकता।” यह कथन किसका है?
3. लेखक ने खेत की मेंड़ पर खड़े किस किसान से पूछा-“आपका नाम बाजा तो नहीं है?
4. अन्त में लेखक ने किस व्यवसाय को श्रेष्ठ माना?
5. खेती के प्रचार व प्रसार का जीता-जागता उदाहरण क्या है?

उत्तर:

1. लेखक के द्वारा
2. लेखक के चाचा का
3. रामचरन से
4. खेती
5. सिनेमा।

पहली चूक पाठ सारांश

‘पहली चूक’ श्रीलाल शुक्ल द्वारा रचित उच्चकोटि का व्यंग्यात्मक निबन्ध है। इस निबन्ध में श्रीलाल शुक्ल जी ने शहर में रहने वालों पर तीखा प्रहार किया है क्योंकि शहर में रहने वाले व्यक्तियों का ग्रामीण वस्तुओं से तथा ग्राम के वातावरण से तनिक भी परिचय नहीं होता है। शहर में रहने वाले लोगों को गाँव की जानकारी केवल पुस्तकों के द्वारा होती है या वे सिनेमा के द्वारा ग्रामीण व्यवस्था से परिचित होते हैं।

लेखक को जब बी.ए. करने के बाद नौकरी नहीं मिली तो उसने अपने गाँव जाकर खेती करने का निर्णय किया। लेकिन उनके चाचा ने कहा कि किसी भी कार्य को करने से पूर्व उसका अनुभव आवश्यक है,अंग्रेजी पढ़ लेने मात्र से वह खेती नहीं कर सकता है। खेती के लिए अथक परिश्रम की आवश्यकता होती है। दिन-रात,मिट्टी,गोबर तथा धूप,सर्दी व बरसात में परेशानियों से जूझना पड़ता है।

लेखक को उसके चाचा ने यह भी बताया यह शरीर मिट्टी से बना है तथा गोबर का पंचगव्य में स्थान है। अतः पवित्र है। लेखक अपने चाचा की बातों से प्रभावित हुआ और बोला कि खेती जाहिलों का पेशा नहीं है। टालस्टॉय किसान था,वाल्टेयर ने भी स्वयं खेती की थी। अतः खेती को सर्वोत्तम व्यवसाय मानकर लेखक ने खेती करने का निर्णय किया। लेखक ग्रामीण वस्तुओं से अपरिचित था अतः उसे यह नहीं मालूम था कि गन्ना क्या होता है अथवा बाजरा क्या होता है।

लेखक चाचा द्वारा बताये स्थान पर पहुँचकर वहाँ पर बैठे एक किसान से पूछने लगा, “क्या तुम बाजरा हो?” वह बोला, “मेरा नाम रामचरण है।”

बाद में उसे छायादार वृक्ष समझने लगा। लेखक ने निश्चय किया कि मैं अब शहर में रहकर रिसर्च करूँगा तथा समझने की चेष्टा करूँगा कि खाद बाजरे को हरा-भरा करने का अचूक साधन है।

पहली चूक संदर्भ-प्रसंगसहित व्याख्या

(1) “उत्तम खेती मध्यम बान,
अधम चाकरी भीख निदान।”

कठिन शब्दार्थ :
उत्तम = सबसे अच्छा। मध्यम = बीच का। बान = व्यवसाय, धन्धा। अधम = गिरा हुआ, ओछा। चाकरी = नौकरी।

सन्दर्भ :
प्रस्तुत अंश ‘पहली चूक’ शीर्षक निबन्ध से लिया गया है। इसके लेखक श्रीलाल शुक्ल हैं।

प्रसंग :
इसमें लेखक ने अन्य सभी व्यापारों से कृषि को सबसे अच्छा बताया है।

व्याख्या :
लेखक कहता है कि उपर्युक्त कहावत प्रायः सुनने को मिल जाती है और इसका अर्थ यह है. कि संसार में अन्य सभी व्यापारों में खेती करना सबसे उत्तम काम है, मध्यम काम व्यापार करना है, चाकरी अर्थात् नौकरी अधम है और भीख माँगना सबसे तुच्छ है।

विशेष :
भाषा सहज एवं सरल है।

(2) शरीर ही मिट्टी का बना हुआ है और यह गोबर तो परम पवित्र वस्तु है। मिट्टी का स्थान यदि पंचभूत में है तो गोबर का स्थान पंचगण्य में है।

कठिन शब्दार्थ :
पंचभूत = मानव शरीर का निर्माण पाँच भूतों से होता है और वे हैं-पृथ्वी, जल, अग्नि, वायु और आकाश। पंचगण्य = गाय के पाँच उपादान होते हैं-दूध, दही, घी, गोबर और गौ मूत्र।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
लेखक ने बी. ए. करने के पश्चात् गाँव जाकर अपने चाचा से खेती का व्यवसाय करने की इच्छा व्यक्त की तो चाचा लेखक को खेती की बुराइयाँ एवं परेशानियाँ बताते हुए कहते हैं कि इसमें दिन-रात पानी और पसीना. मिट्टी और गोबर से खेलना पड़ता है। चाचा के इस तर्क का उत्तर देते हुए लेखक कहता है।

व्याख्या :
लेखक कहता है कि चाचाजी मनुष्य का यह शरीर ही मिट्टी का बना हुआ है और यह गोबर तो परम पवित्र वस्तु है। मिट्टी का स्थान मानव शरीर का निर्माण करने वाले पंच भूतों अर्थात् पृथ्वी, जल, अग्नि, वायु और आकाश में से एक है तथा उसी प्रकार गोबर का स्थान पंचगण्य अर्थात् दूध, दही, घी, गोबर और मूत्र में है। अतः मुझे मिट्टी और गोबर से कोई परेशानी नहीं होगी।

विशेष :

1. लेखक ने मिट्टी और गोबर का महत्त्व बताया
2. भाषा सरल एवं सहज है।

(3) सिनेमा खेती की उन्नति का एक अच्छा साधन है। सिनेमा द्वारा खेती का बड़ा प्रचार हुआ है। बड़े-बड़े हीरो खेत जोतते जाते हैं और गाते जाते हैं। हीरोइन खेत पर टोकरी में रोटी लेकर आती है। हरी-भरी फसल में आँख-मिचौंली का खेल होता है। फसल काटते समय हीरोइन के साथ बहुत-सी लड़कियाँ नाचती हैं और गाती भी हैं। वे नाचती जाती हैं और फसल अपने आप कटती जाती है। ऐसे ही मधुर दृश्यों को देखकर, पढ़े-लिखे आदमी गाँवों में आने लगते हैं और खेतों के चक्कर काटने लगते हैं। इस प्रकार सिनेमा द्वारा खेती की शिक्षा मिलती है। सच पूछिए तो खेती करने की सच्ची शिक्षा मुझे सिनेमा से ही मिली थी।

कठिन शब्दार्थ :
आँख मिचौली = लुकाछिपी का खेल। मधुर = सुन्दर, मोहक।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
लेखक बताता है कि सिनेमा देखकर ही उसे इतना पढ़-लिख जाने के बाद भी खेती करने की शिक्षा मिली है।

व्याख्या :
लेखक कहता है कि आजकल सिनेमा खेती की उन्नति का अच्छा साधन बन गया है। इन सिनेमाओं ने खेती के प्रचार में बहुत बड़ा योगदान किया है। बड़े-बड़े प्रसिद्ध हीरो खेत जोतते जाते हैं और मस्ती में गाने गाते जाते हैं। हीरोइन खेत पर टोकरी में रोटी खाने के लिए लेकर आती है। हीरो और हीरोइन हरी-भरी फसल में आँख-मिचौली का खेल खेलते हैं। फसल जब कटने को होती है तो उस समय हीरोइन के साथ अनेक सुन्दर लडकियाँ नाचती और गाती हैं। वे एक ओर नाचती जाती हैं और दूसरी ओर फसल अपने आप कटती जाती है। इसी प्रकार के मनमोहक दृश्यों को देखकर पढ़े-लिखे लोग गाँवों की ओर जाने लगते हैं तथा वे खेतों के चक्कर काटते रहते हैं। इस प्रकार हम कह सकते हैं कि सिनेमा द्वारा खेती की शिक्षा प्राप्त होती है। सच मानो तो खेती करने की सच्ची शिक्षा मुझे सिनेमा से ही मिली थी।

विशेष :

1. लेखक ने सिनेमाओं को खेती सिखाने का उचित साधन माना है।
2. भाषा सहज, सरल है।

(4) जैसे खूटे से छूटी हुई घोड़ी भूसे के ढेर पर मुँह मारती है, जैसे धूप में बँधी हुई भैंस तालाब की ओर दौड़ती है, वैसे ही तुम्हारा शहर की ओर जाना बड़ा स्वाभाविक और उचित है।

कठिन शब्दार्थ :
स्वाभाविक = स्वभावगत। सन्दर्भ-पूर्ववत्।

प्रसंग :
लेखक का मत है कि पढ़ा-लिखा व्यक्ति शहर की ओर ही भागता है।

व्याख्या :
लेखक कहता है कि जैसे छूटे से बँधी हुई घोड़ी भूसे के ढेर को देखकर उसी पर टूट पड़ती है और धूप में बँधी हुई भैंस तालाब की ओर भागती है, उसी प्रकार पढ़ा-लिखा व्यक्ति शहर की ओर भागता हुआ दिखाई देता है और उसका यह कार्य स्वाभाविक और उचित ही है।

विशेष :

1. लेखक ने पढ़े-लिखे व्यक्ति की स्वाभाविक भावना को बताया है।
2. भाषा सहज और सरल है।

MP Board Class 10th Hindi Solutions

MP Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.6

Assume = \(\frac{22}{7}\) unless stated otherwise.

Question 1.
The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm³ = 1l)
Solution:
Circumference of base, C =132 cm ,
h = 25 cm
C = 2πr
132 = 2 x \(\frac{22}{7}\) x r
r = \(\frac{132×7}{2×22}\) = 21 cm
Volume of cylinder =πr²h
= \(\frac{22}{7}\) x 21 x 21 x 25 = 34650 cm³
= \(\frac{34650}{1000}\)

Question 2.
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g.
Solution:

Inner diameter
d₁ = 24 cm ⇒ r₁ = 12 cm
Outer diameter, d₂ = 28 cm ⇒ r₂ = 14 cm
Volume of wood in the piple = πr₂²h – πr₁²h
πh(r₂² – r₁²) = \(\frac{22}{7}\) x 35(14² – 12²)
= \(\frac{22}{7}\) x 35 x 52 = 5720 cm³
Mass of pipe = 0.6 x 5720 = 3432 gm
= 3.432 kg.

Question 3.
A soft drink is available in two packs –

1. a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and
2. a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Solution:
1. Volume of cuboidal can = (5 x 4 x 15) = 300 cm³

2. Volume of cylindrical can = πr²h = \(\frac{22}{7}\) x 3.5 x 3.5 x 10
= 385 cm²
Capacity of cylindrical can is more than the cuboidal can by 85 cm2.

Question 4.
If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then find

1. radius of its base
2. its volume. (Use π = 3.14)

Solution:
1. CSA of cylinder = 94.2 cm²
2πrh = 94.2 cm²
2 x 3.14 x 5 = 94.2 cm²
r = \(\frac{94.2}{2×3.14×5}\) = 3 cm
Volume of the cylinder = πr²h
= 3.14 x 3 x 3 x 5
= 3.14 x 45 = 141.3 cm³

Question 5.
It costs ₹ 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m², find

1. inner curved surface area of the vessel
2. radius of the base
3. Capacity of the Vessel.

1. Inner curved surface area =

2. Let r be the radius of the base ICSA = 2πrh
110 = 2 x \(\frac{22}{7}\) x r x 10 r
= 1.75 m

3. Capacity of vessel = πr²h = \(\frac{22}{7}\) x 1.75 x 1.75 x 10
= 96.25 m³.

Question 6.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?
Solution:
h = 1 m
V = 15.4l
= \(\frac{15.4}{1000}\)m³ = 0.0154 m³
Volume of the vessel = πr²h
0. 0154 = \(\frac{22}{7}\) x r² x 1
∴ r² = \(\frac{0.0154×7}{22}\)
⇒ r² = 0.0049
∴ r = 0.07 m
Area of metal sheet required = 2nr (h + r)
= 2x \(\frac{22}{7}\) x 0.07(1 + 0.07)
= 0.4708 m².

Question 7.
A lead pencil consists of a cylinder ofvyood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
Solution:
Volume of the pencil = πr²h
= \(\frac{22}{7}\) x 3.5 x 3.5 x 140 = 5390 mm³
Volume of graphite = πr²h
= \(\frac{22}{7}\) x 0.5 x 0.5 x 140 = 110 mm³
Volume of wood = Volume of pencil – Volume of graphite
= 5390 – 110
= 5280 mm³
=5.28 cm³.

Question 8.
A patient in a hospital is given soup daily in a cylindrical biftvl of diameter 7 cm. If the bowl is filled with/soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Solution:
d = 7cm
r = 3.5 cm
h = 4cm
Volume of the bowl = πr²h
= \(\frac{22}{7}\) x 3.5 x 3.5 x 4 = 154 cm³
Volume of soup needed for 250 patients = 154 x 250
= 38500 cm³
= 38.5l.

MP Board Class 9th Maths Solutions

MP Board Class 6th Science Solutions Chapter 12 Electricity and Circuits

Electricity and Circuits Textbook Exercises

Question 1.
Fill in the blanks:

1. A device that is used to break an electric circuit is called ……………… terminals.
2. An electric cell has ……………… terminals.

Answer:

1. Switch
2. Two.

Question 2.
Mark ‘True’ or ‘False’ for following statements:

1. Electric current can flow through metals.
2. Instead of metal wires, a jute string can be used to make a circuit.
3. Electric current can pass through a sheet of thermo Col.

Answer:

1. True
2. False
3. False.

Question 3.
Explain why the bulb would not glow in the arrangement show in Fig?

Answer:
The bulb will not glow because the circuit is not completed due to the presence of an insulator between the circuit. Here insulator is screw driver.

Question 4.
Complete the drawing shown in figure to indicate where the free ends of the two wires should be joined to make the bulb glow.

Answer:
The complete diagram in shown below:

Question 5.
What is the purpose of using an electric switch? Name some electrical gadgets that have switches built into them.
Answer:
An electric switch is used to open or close the circuit. Electric gadgets that have switches built into them are freezer, washing machines, microwaves, toaster, heaters, electric – bulb, tube – light, electric iron, etc.

Question 6.
Would the bulb glow after completing the circuit shown in Figure of Question 4 above if instead of safety pin we use an eraser?
Answer:
No.

Question 7.
Would the bulb glow in the circuit shown in Fig.

Answer:
No, because both of the terminal of battery is connecting with one terminal of bulb.

Question 8.
Using the “conduction tester” on an object it was found that the bulb begins to glow. Is that object a conductor or an insulator? Explain.
Answer:
That object is a conductor because electricity can pass through a conductor and not through an insulator. If the object is an insulator then bulb could not glow.

Question 9.
Why should an electrician use rubber gloves while repairing an electric switch at your home? Explain.
Answer:
An electrician use rubber gloves while repairing an electric switch because rubber is a bad conductor of electricity or in other words rubber is an insulator.

Question 10.
The handles of the tools like screwdrivers and pliers used by electricians for repair work usually have plastic or rubber covers on them. Can you explain why?
Answer:
Because plastic and rubber, both are the bad conductor of electricity. Hence, they protect electricians against electric shock.

Projects and Activities

Activity 1.
Make a table to show different materials allows current to pass through it or not.
Answer:
Conductors and Insulators:

Activity 2.
Describe a simple experiment to test whether a given material is a conductor or an insulator.
Answer:
To identify given material is a conductor or an insulator.

Requirements:
Pencil, matchstick, alpin, rubber tube, cell wire, bulb.

Procedure:
Take a small bulb, a battery and a length of wire and connect them in a circuit as show in figure. This circuit is not closed and no current will pass through it. Close the gap between the points A and B by inserting the graphite of your pencil between them. Repeat this by successively replacing the graphite by rubber tube, alpin and matchstick. Record the observations in the table given below.
Table

Electricity and Circuits Additional Important Questions

Electricity and Circuits Objective Type Questions

Question 1.
Choose the correct answer:

Question (i)
S.I. unit of electic current is –
(a) Farad
(b) Volt
(c) Ampere
(d) Coulomb.
Answer:
(c) Ampere

Question (ii)
The types of charges are –
(a) Two
(b) Three
(c) Four
(d) None of these.
Answer:
(a) Two

Question (iii)
Voltaic cell was invented by –
(a) Alessandro Volta
(b) J. F. Volta
(c) Georges Leclanche
(d) None of these.
Answer:
(a) Alessandro Volta

Question (iv)
The principle of a voltaic cell was used by –
(a) J. F. Daniel
(b) Alessandro Volta
(c) Georges Leclanche
(d) None of these.
Answer:
(a) J. F. Daniel

Question (v)
The primary cells are –
(a) Voltaic cell
(b) Daniel cell
(c) Dry cell
(d) All of these.
Answer:
(d) All of these.

Question (vi)
Electric current is the flow of particles with –
(a) A negative charge
(b) A positive charge
(c) Both positive and negative charges flowing opposite to each other.
(d) Either positive or negative charge depending on the material.
Answer:
(a) A negative charge

Question (vii)
The conductors of electricity are –
(a) mica
(6) wood
(c) glass
(d) none of these.
Answer:
(d) none of these.

Question 2.
Fill in the blanks:

1. Flow of ………….. through a conductor, when a potential difference is applied across its terminals, is called electric current.
2. The actual direction of current flowing through a circuit is from negative to positive electrode, whereas the …………….. direction is from negative to positive.
3. S.I. unit of electric current is ……………..
4. …………….. is used to measure current flowing through a conductor and voltmeter is used to measure ……………. difference across the ends of a conductor.
5. The bulb with broken filament is called a …………… bulb.
6. The thin tiny wire inside the glass cover in a bulb is called ……………….
7. The bulb does not glow when the ……………. is open.

Answer:

1. Free electrons
2. Conventional
3. Ampere,
4. Ammeter, potential difference
5. Fused
6. Filament
7. Switch.

Question 3.
Which of the following statements are true (T) or false (F):

1. Materials which do not allow an electric current to flow through them are called conductors.
2. In an electric press which effect of electric current used is heating effect.
3. Rate of flow of charge is called electric current.
4. Electric cell converts chemical energy to mechanical energy.
5. Voltaic cell was invented by Volta.
6. Dry cell is portable.
7. An electric cell cannot to be used in a wrist watch.

Answer:

1. False
2. True
3. True
4. False
5. True,
6. True
7. False.

Electricity and Circuits Very Short Answer Type Questions

Question 1.
What are the essential components of an electric circuit?
Answer:
The essential components of an electric circuit are battery or cell, conductor and a key.

Question 2.
What is an electric circuit?
Answer:
The path of an electric current is referred to as electric circuit.

Question 3.
Can we see electricity?
Answer:
No, we cannot see electricity but can observe its effects.

Question 4.
Name two sources of electric current?
Answer:
The two sources of electric current are Voltaic cell and Daniel cell.

Question 5.
What is a conductor? Give two examples.
Answer:
Substances, such as metals, that can conduct electric current are called conductors. For examples, Copper, Aluminium.

Question 6.
What is an insulator? Give three examples.
Answer:
The materials that do not allow current to pass through them are called insulators. For examples, Mica, Wood and Rubber.

Question 7.
Make a list of materials around you which conduct electricity and a list of those that do not.
Answer:
Conductors of electricity are:
All metals, acid base and salt solutions, aluminium, iron, copper and nickel.

Do not conduct electricity are:
Paper, rubber, wool, nylon, polythene and backelite.

Question 8.
Who supplies electricity to tourch bulb?
Answer:
Electric cell supplies electricity to torch bulb.

Question 9.
What are the ready sources of electric current?
Answer:
The ready sources of electric current are dry cell and battery. Every dry cell and battery had got two terminals or connection points marked (+) and (-).

Question 10.
Which scientist made earliest attempts to obtain an electric current?
Answer:
The earliest attempts to obtain electric current was made in the year 1790 by Alessandro Volta an Italian scientist.

Question 11.
What is a fused bulb?
Answer:
When the filament is broken in a bulb, it is called the fused bulb.

Question 12.
Define an open circuit?
Answer:
When there is a gap between two terminals, it is called an open circuit.

Question 13.
Define a closed circuit?
Answer:
A circuit where there is no gap between two terminals is called a closed circuit.

Question 14.
Draw symbol of (i) a cell, (ii) battery and (iii) a key.
Answer:

Question 15.
Name two insulators?
Answer:
Plastics and rubber.

Question 16.
Who discovered the dry cell?
Answer:
Georges Leclanche discovered the dry cell.

Question 17.
What is meant by battery?
Answer:
The positive terminal of one cell is kept in contact with the negative terminal of the other cell. When two or more then two cells are joined together in this way, we get a battery.

Question 18.
What energy is converted to electrical energy in an electric cell?
Answer:
The chemicals in a cell produce chemical energy. This chemical energy is converted to electrical energy in a cell.

Question 19.
What are the uses of storage batteries?
Answer:
Storage batteries are used where electric current is needed for longer time of more voltage. For examples, in cars, trucks, buses. It is also used in submarines, radars and satellites.

Question 20.
Give some uses of dry cells?
Answer:
Uses of dry cells. In radio, transistor, tourches, remote, camera and calculator.

Question 21.
What is filament of a bulb?
Answer:
The thin wire that gives off light is called the filament of the bulb.

Electricity and Circuits Short Answer Type Questions

Question 1.
What is current?
Answer:
The rate of flow of electric charge through a conductor per unit time is called electric current. In other words, we can say that the amount of electric current is the ratio of charge flowing through a conductor to the time taken in the flowing of the charge.

Question 2.
List the appliances around you that depend on electricity for their operation. List the appliances that do not use electrical energy?
Answer:
Appliances that depend on electricity are:
Electric press, geyser, washing machine, television, radio, fan, cooler etc.

Appliances that do not depend on electricity are:
Cooking gas, stove, solar cooker, sewing machine, cycle, rickshaw, etc.

Question 3.
We selectively choose certain materials to make a path and reject others. Why?
Answer:
Some materials do not allow an electric current to pass through them. Such materials are called insulators. Materials which allow electric current to pass through them are called conductors.

Question 4.
What happens when a charged electroscope is connected to a uncharged electroscope by a wire? What is that called?
Answer:
When a charged electroscope is joined to an uncharged electroscope by a wire, charges flow from the charged to the uncharged electroscope through the wire till they are equalised. This flow of charges forms an electric current. It is measured in ampere.

Question 5.
What is the nature of electric current?
Answer:
As water flows from higher level to lower level in a pipe, electric current also flows from the source of electric current to the target where it is needed through metal wires.

Question 6.
Define an electric switch?
Answer:
A switch is a simple device that either breaks the circuit or completes it. When the switch is open, the circuit is not complete and the current does not flow. In the other position, when the switch is closed, the circuit is complete and current can flow through the circuit.

Question 7.
On what principle scientist Volta made his cell?
Answer:
The principle was that when two strips of different metals are dipped in an acid solution an electric current begins to flow through them. Such a simple source of current, or a cell is called as a voltaic cell in honour of its inventor.

Question 8.
What are the draw backs in voltaic cell? Who then improved the design?
Answer:
Voltaic cell is not a good source of current as the flow through the wire in such a cell is not smooth and steady. J. F. Daniel (1790 – 1845) made an improved design of voltaic cell in the year 1836.

Electricity and Circuits Long Answer Type Questions

Question 1.
Explain an electric cell with suitable diagram and their construction.
Answer:
Electricity to the bulb in a torch is provided by the electric cells. Electric cells are also used in alarm clocks, wristwatches, transistor, radios, cameras and many other devices. In an electric cell their is a small metal cap on one side and a metal disc on the other side. The metal cap is the positive terminal of the electric cell while the metal disc is the negative terminal.

All electric cells have two terminals one positive and another negative terminal. An electric cell produces electricity from the chemical stored inside it. When the chemicals in the electric cell are used up, the electric cell stops producing electricity. The electric cell then has to be replaced with a new one.

Question 2.
What is meant by an electric circuit?
Answer:
When we connect the two ends of a cell to a bulb using copper wires, the bulb lights up. This is because we have provided a path for electrons to flow the negative terminal of the battery to the positive terminal through a bulb. Such a path of an electric current is known as a electric circuit.

Question 3.
Find the various ways in which two batteries and two bulbs can be connected in a working circuit.
Answer:
Two batteries and two bulbs can be connected in working circuit in two ways.
1. In series as shown in figure.

2. In parallel as shown in figure.

Question 4.
Draw a diagram of a torch bulb?
Answer:

Question 5.
Draw a neat diagram to show the inside view of a torch?
Answer:

MP Board Class 6th Science Solutions

MP Board Class 6th Science Solutions Chapter 11 Light, Shadows and Reflections

Light, Shadows and Reflections Textbook Exercises

Question 1.
Rearrange the boxes given below to make a sentence that helps us understand opaque objects?

Answer:

Question 2.
Classify the objects or materials given below as opaque, transparent or translucent and luminous or non -luminous? Air, water, a piece of rock, a sheet of aluminium, a mirror, a wooden board, a sheet of polythene, a CD, smoke, a sheet of plane glass, fog, a piece of red hot iron, an umbrella, a lighted fluorescent tube, a wall, a sheet of carbon paper, the flame of a gas burner, a sheet of cardboard, a lighted torch, a sheet of cellophane, a wire mesh, kerosene stove, sun, firefly, moon.
Answer:
Obaque objects:
A piece of rock, a sheet of aluminium, a mirror, a wooden board, a CD, a wall, a sheet of cardboard.

Transparent objects:
Air, water, sheet of plane glass.

Translucent objects:
A sheet of polythene, smoke, fog, a sheet of carbon paper, a sheet of cellophone.

Luminous objects:
A piece of red hot iron, a lighted fluorscent tube, the flame of a gas burner, a lighted torch, sun, firefly, kerosene stove.

Non – luminous Objects:
An umberela, a wire mesh, moon.

Question 3.
Can you think of creating a shape that would give a circular shadow if held in one way and a rectangular shadow if held in another way?
Answer:
Yes, we can think of creating a shape that would give a circular shadow if held in one way and a rectangular shadow of held in another way.

Question 4.
In a completely dark room, if you hold up a mirror in front of you, will you see a reflection of yourself in the mirror?
Answer:
No, in a completely dark room, we cannot see our right – left inverted image in the mirror, because there is no source of light.

Projects and Activities:

Activity 1.
Look around you and collect as many objects as you can an eraser, plastic scale, pen, pencil, notebook, single sheet of paper, tracing paper or a piece of cloth. Try to look at something far away, through each of these objects. Is light from a far away object able to travel to your eye, through any of the objects?
Record your observations in a table.
Answer:

Activity 2.
Describe an experiment which shows that light propagates in a straight line?
Answer:
In a homogeneous medium, light travels in a straight line path. This is called rectilinear propagation of light. This can be demonstrated with the help of three card – board pieces A, B and C with a fine hole at their centres. A candle or a bulb is placed on one side and the boards are arranged such that the holes are in straight lines as shown in figure. Looking from the other side, it is found that the fight from the candle is received only when the three holes are in a straight fine. If one of the card board pieces is displaced, the fight is no longer seen. This clearly demonstrates that the fight travels in a straight fine path.

Things to Think About:

Question 1.
Opaque objects cast shadows, isn’t it? Now, if we hold a transparent object in the Sun, do we see anything on the ground that gives us a hint that we are holding something in our hand?
Answer:
Yes, it gives us a fuzzy shadow on the ground, which gives us a hint that we are holding some thing in our hand.

Question 2.
We saw that changing colour of opaque objects does not change the colour of their shadows. What happens if we place an opaque object in coloured light? You can cover the face of a torch with a coloured transparent paper to do this. (Did you ever noticed the colours of evening shadows just as the Sun is setting?)
Answer:
Yes, the colour of shadow depends upon the colour of the objects. If we can change the colour of opaque objects, the same colour will be seen in shadow.

Light, Shadows and Reflections Additional Important Questions

Light, Shadows and Reflections Objective Type Questions

Question 1.
Choose the correct answer:

Question (i)
A pinhole camera forms an image of a building on its screen. Compared to the building, its image would be –
(а) Larger and inverted
(b) Larger and upright
(c) Smaller and inverted
(d) Smaller and upright.
Answer:
(c) Smaller and inverted

Question (ii)
The transparent object out of the following is –
(a) Water
(b) Moon
(c) Rubber sheet
(d) None of these
Answer:
(a) Water

Question (iii)
Which of the following objects allow more light to pass through –
(a) Translucent
(b) Opaque
(c) Transparent
(d) None of these.
Answer:
(a) Translucent

Question (iv)
The objects which do not allow the light to pass through them at all are called –
(a) Opaque
(b) Translucent
(c) Transparent
(d) None of these.
Answer:
(a) Opaque

Question (v)
In a room lighted by an electric bulb, it is found that the shadow of the ceiling fan hung in the centre is falling on a wall opposite to the door. What can you say about the position of the bulb –
(a) On the wall on your right as you enter the room.
(b) On the wall on your left as you enter the room.
(c) On the wall facing the door.
(d) On the wall on which the door is fixed.
Answer:
(b) On the wall on your left as you enter the room.

Question 2.
Fill in the blanks:

1. Opaque objects cast shadow.
2. The shadow is longest during and
3. The totally dark portion of a shadow is called
4. Surfaces act like mirrors.
5. Solar eclipse occurs only on moon days.
6. Lunar eclipse occurs only on moon days.
7. A pin – hole camera is based on propagation of light.
8. Surfaces act like mirrors.

Answer:

1. Dark
2. Sunset, sunrise
3. Umbra
4. Polished,
5. New
6. Full
7. Straight
8. Luminous.

Question 3.
Which of the following statements are true (T) or false (F):

1. A fire fly is a hot source of fight.
2. The sun is a natural source of fight.
3. The tourch is a man – made source of fight.
4. The partially dark portion of a shadow is called umbra.
5. Formation of a sharp shadow of an object in sunlight is an evidence of the rectilinear propagation of fight.
6. We do not see the shadow of the aeroplane flying high in the sky because the Sun is a point source of fight.
7. Fluroscent tube is an extended source of light.
8. If we keep 100 sheets of transparent paper one above the other, the heap of paper would behave as an opaque body.
9. If we keep 100 sheets of translucent paper one above the other, the heap would behave as an opaque body.
10. Regular reflection obeys only one law, namely, the angle of incidence is equal to the angle of reflection.
11. Light does not interact with the living matter.
12. Image formed in a convex mirror is shorter anti real. (xii) Concave mirror always forms a real image.

Answer:

1. False
2. True
3. True
4. False
5. True
6. False
7. True
8. True
9. True
10. False
11. False
12. True
13. False.

Light, Shadows and Reflections Very Short Answer Type Questions.

Question 1.
Name any four man – made sources of light?
Answer:
The man – made sources of light are:

1. An electric bulb
2. A torch
3. A candle
4. A lamp.

Question 2.
Why do objects in a room become visible even if sunlight does not enter it?
Answer:
The objects in a room become visible during the day because of reflection of sunlight by the several non – luminous objects around us.

Question 3.
Give the names of four different sources of light.
Answer:
The four different sources of light are:

1. Sun
2. Stars
3. Moon
4. Firefly.

Question 4.
How does light travel from one point to the other?
Answer:
Light travel from one point to the other in straight line.

Question 5.
Why is the image formed in a pinhole camera inverted?
Answer:
The image formed in a pinhole camera is inverted because light travels in straight lines.

Question 6.
What is luminous objects?
Answer:
Objects like the sum that give out or emit light of their own are called luminous objects.

Question 7.
Does the flame of a gas stove emit light?
Answer:
Yes.

Question 8.
Give one example of living thing, which emits light?
Answer:
Firefly (Jugnu) is a living source of light.

Question 9.
Why translucent objects cast feeble shadow?
Answer:
Translucent objects cast feeble shadows because they allow the partial light to pass through them. So, every light or feeble shadow is formed behind the object.

Question 10.
State the conditions when annular solar eclipse occurs?
Answer:
An annular solar eclipse occurs when only the tip of the umbra of the moon falls on the earth.

Question 11.
Tube light is a cold source of light. Is this statement correct? If yes, then why?
Answer:
Yes, this statement is correct because tube light gives light only. It does not produce any heat. This is why tube light is called a cold source of light.

Question 12.
Classify the following into transparent, opaque and translucent objects: Wax, spectacles, a heap of salt, a stone, dense smoke, wood, skin, balloon, rubber, membrane of a table, blood and milk?
Answer:
Transparent objects:
Spectacles, membrane of a tabla.

Opaque objects:
Wax, wood, a heap of salt, a stone.

Translucent objects:
Dense smoke, skin, balloon, rubber, blood and milk.

Question 13.
What is the meaning of shadow?
Answer:
When any obstacte comes in the way of light then the portion on the screen where light does not reach is called shadow.

Question 14.
Why is the shadow of edges of a blade not clear?
Answer:
Because at the edges of the blade, the light is diffracted, i.e., bent slightly in the shadow region. Hence the edges of shadow of blade are not sharp.

Question 15.
Is the medium required for the propagation of light?
Answer:
No, the light can travel in vacuum.

Light, Shadows and Reflections Short Answer Type Questions

Question 1.
What is light?
Answer:
Light is the external physical reason which makes things visible. Light makes it possible to see the objects but light itself cannot be seen. When light falls on any body, it gets reflected and reaches our eyes and we feed the presence of that body. Light is also a form of energy. It can be converted to other form of energy. Light is propagated in the form of electromagnetic waves. Its wavelength is about 4000 A⁰ to 8000 A⁰. It can travel through vacuum with velocity of 3 x 10⁸ m/see.

Question 2.
What do you mean by luminious and non – luminious bodies?
Answer:
The bodies which themselves are a source of light, and emit light they are called luminious. For examples, the sun, a burning candle, a light bulb, etc. Those bodies which do not emit light but are made visible by light falling from some luminious object on them are called non – luminious bodies. For examples, the earth, the moon, the table, etc.

Question 3.
Give the definitions of

1. Transparent bodies
2. Translucent bodies
3. Opaque bodies.

Answer:

1. Transparent bodies are those through which light can pass freely and through which things can be distinctly seen. For example, air, water, glass, etc.

2. Translucent bodies are those which allow only a part of the light to pass through them and things cannot be seen distinctly. For example, butter-paper, ground glass, etc.

3. Opaque bodies are those which do not allow any light to pass through them and so we cannot through them. For example, stone, iron, wood, etc.

Question 4.
Give the differences between transparent and opaque objects.
Answer:
The differences between the transparent and opaque objects are:
Transparent Object:

• The light passes through these objects.
• They cast no shodow.

Opaque Object:

• The light cannot pass through these objects.
• They cast dark shadows.

Question 5.
State the differences between transluscent and opaque objects.
Answer:
Difference between transparent and transluscent materials:
Transparent Materials:

• The light pass through easily.
• Through it objects can. be seen clearly.

Transluscent Materials:

• It allows a part of light to pass through it.
• Through it objects can only be seen dimly.

Question 6.
How can you see a solar eclipse?
Answer: Solar eclipse can be seen by using a glass. There is another alternative method to see the sun. Take a piece of cardboard having a hole in the centre. This cardboard forms an image of sun on the wall. Thus the solar eclipse can be seen by this image safely.

Question 7.
Mention some of the uses of plane mirror.
Answer:
Uses of plane mirror are:

1. In hair cutting saloons, shops and at home.
2. For constructing periscope.
3. For constructing kelidoscope.

Question 8.
State two properties each of umbra and penumbra.
Answer:
Properties of Umbra:

1. The darkest part of the shadow is called umbra.
2. The light in umbra does not reach from any source of light.

Properties of Penumbra:

1. In this region light reaches from one part of the source.
2. This region is not completely dark.

Question 9.
What is regular reflection?
Answer:
When the light falls on the smooth surface the scattered light rays move in a definite direction. This gives the clear image of the object. This. is called regular reflection.

Question 10.
State the difference between umbra and penumbra.
Answer:
Difference between umbra and penumbra:
Umbra:
The region of the space where the light does not reach from any source, is the darkest. This darkest part of the shadow is called umbra.

Penumbra:
The dark circular path is surrounded by a less dark portion. This shadow part is called penumbra.

Light, Shadows and Reflections Long Answer Type Question

Question 1.
Draw a diagram to show the position of the screen so that no umbra is formed on it if the source of light is bigger than the obstacle.
Answer:

Question 2.
Describe clearly the event of partial lunar eclipse and explain is cause.
Answer:
The ray diagram shows the various positions of the Moon as it passes through the penumbra and the umbra of the earth during a lunar eclipse. At position A, the moon is yet to enter the penumbra and it looks bright as usual. At position B, it has entered the penumbra. At this position, the moon looks pale, as if it has gone behind the clouds. After some time, a part of the moon enters the umbra. This part is then not visible from the earth, but the remaining part of the moon is still visible. This is called a partial lunar eclipse. Lunar eclipse, always takes place on full moon day but not on all full moon days.

Question 3.
How can you make a pin hole camera? Describe.
Answer:
Take two boxes so that one can slide into another with no gap in between them. Cut open one side of each box. On the opposite face of the larger box, make a small hole in the middle [Fig. (a)]. In the smaller box, cut out from the middle a square with a side of about 5 cm to 6 cm. Cover this open square in the box with tracing paper (translucent screen) [Fig. (b)]. Slide the smaller box inside the larger one with the hole, in such a way that the side with the tracing paper is inside [Fig. (c)]. Your pin hole camera is ready for use.

Holding the pin hole camera look through the open face of the smaller box.# You should use a piece of black cloth to cover your head and the pinhole camera. Now, try to look at some distant objects like a tree or a building through the pinhole camera. Make sure that the objects you wish to look at through your pinhole camera are in bright sun. shine. Move the smaller box forward or backward till you get a picture on the tracing paper pasted at the other end.

Question 4.
Describe the working of a its uses?
Answer:
A principle is an obtical instrument which is used to see the overhead objects. In a principls there are two mirror. A and B fixed at the two ends of a vertical tube whose reflecting surfaces are parallel and face each other.

Principle:
It is based on the principle of reflection of plane mirrors which are placed parallel. Light rays are coming through an object and entering from the apperture C. These rays strike on the mirror A at an angle of incidence 45° and is reflected along the axis of the tube striking the mirror at 45° again. This is then finally reflected parallel to their original path and reach is to the eyes of observer.

MP Board Class 6th Science Solutions

MP Board Class 12th Chemistry Solutions Chapter 4 Chemical Kinetics

Chemical Kinetics NCERT Intext Exercises

Question 1.
For the reaction R → P, the concentration of reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Answer:
For the reaction R → P

Question 2.
In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L^-1 to 0.4 mol L^-1 in 10 minutes. Calculate the rate during this interval ?
Answer:

Question 3.
For a reaction, A + B → Product; the rate law is given by, r = k [A]^1/2 [B]². What is the order of the reaction ?
Answer:
Order of reaction = sum of powers of concentration of reactants.
Order of reaction = \(\frac { 1 }{ 2 } \) + 2 = \(\frac { 5 }{ 2 } \) = 2.5.

Question 4.
The conversion of molecules A; toy follows second order kinetics. If concetration of JC is increased to three times how will it affect the rate of formation of y².
Answer:
For the reaction x → y
Rate of reaction (r) = k[x]² …(1)
If concentration of x is increased three times, now
Rate of reaction (r)¹ = K[3x]² = k[9x²]
Dividing equation (2) by eq. (1)
\(\frac{r^{1}}{r}=\frac{k\left[9 x^{2}\right]}{k\left[x^{2}\right]}=9\)
Thus, rate of reactions will become 9 times.

Question 5.
A first order reaction has a rate constant 1.15 × 10^-3 s^-1. How long will 5g of this reactant take to reduce to 3g ?
Solution:
According to question,

Applying first order kinetic equation

Question 6.
Time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Solution:
For a first order reaction,

Question 7.
What is the effect of rate constant on temperature ? How can this effect of temperature be measured quantitatively ?
Answer:
Rate of reaction increases with increase in temperature and with 10°C rise in temperature its value becomes two times.
Arrhenius equation expresses the effect of temperature on constant
\(\mathrm{K}=\mathrm{A}_{e}^{-\mathrm{E}_{a} / \mathrm{RT}}\)
Where A is frequency factor and Efl is activation energy of the reaction.

Question 8.
The rate of the chemical reaction doubles for ah increase of 10K in absolute temperature from 298K. Calculate Ea.
Solution:
Given : k₂ = 2k₁, T₁ = 298K, T₂ = 308K, R = 8.314 JK^-1 mol^-1
We know that

Question 9.
The activation energy for the reaction
2HI_(g) → H_2(g) + I_2(g)
is 209.5kJ mol^-1 at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy ?
Solution:
According to question, E_a = 209.5kJ mol^-1 = 209.5 × 103 J mol^-1, T = 581K Fraction of molecules with energy equal to or greater than activation energy is given by
[R = 8.314J]

Chemical Kinetics NCERT TextBook Exercises

Question 1.
From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

Solutions:

Question 2.
For the reaction :
2A + B → A2B
the rate= k [A][B]² with k = 2.0 × 10^-6 mol^-2 L² s^-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L^-1, [B] = 0.2 mol L^-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^-1
Solution:
Given, Rate = k [A][B]²
Initial rate = 2 × 10-6 × [0.1] × [0.2]²
= 8 × 10^-9 mol L^-1 s^-1.
Decrease in concentration of A
i.e. ∆[A] = 0.1 – 0.06 = 0.04 mol L^-1
Decrease in concentration of B
i.e. ∆[B] = \(\frac { 1 }{ 2 } \) × 0.04 = 002 mol L^-1
Remaining concentration of B
= 0.2 – 0.02 = 0.18M
Rate = k[A] [B]²
= 2 × 10-6 × 0.06 × [0.18]²
= 3.89 × 10^-9 mol L^-1 s^-1.

Question 3.
The decomposition of NH₃ on platinum surface is zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 × 10^-4 mol^-1 L s^-1 ?
Solution:
2NH₃ → N₂ + 3H₂

Question 4.
The decomposition of dimethyl ether leads to the formation of CH₄, H₂ and CO and the reaction rate is given by Rate = k [CH₃OCH₃]^3/2. The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.,
Rate = k(P_CH3OCH3)^3/2
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants ?
Solution:
The rate law of reaction is
Rate = k [P_CH3OCH3]^3/2
Rate = Pressure change/Time change
Unit of rate = bar min^-1

Question 5.
Mention the factors that affect the rate of a chemical reaction.
Answer:
Rate of reaction depends upon the following factors:

1. Concentration of reactants
2. Temperature
3. Catalyst
4. Nature of reactant etc.

Question 6.
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is :
(i) doubled
(ii) reduced to half ?
Solution:
Let the reaction A → B is a second order reaction, for A

(i) When the concentration of [A] is doubled
A = 2 a
The new rate of reaction

The rate of reaction will become four times when concentration is doubled.

(ii) When concentration of [A] is 1/2.
A = \(\frac { a }{ 2 } \)
So new rate of reaction

Therefore, the rate of the reaction could be reduced to l/4^th.

Question 7.
What is the effect of temperature on the rate constant of a reaction ? How can this temperature effect on rate constant be represented quantitatively ?
Answer:
Rate constant of the reaction increases with temperature, it is independent whether the reaction is exothermic or endothermic. Arrhenius equation shows the dependence of rate constant on temperature.

Question 8.
In a pseudo first order hydrolysis of ester in water, the following results were obtained:

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
Solution:

Where [R]0 = 0.55 M is initial concentration of ester (t = 0) and [R] is the concentration of ester at time ‘t’

Question 9.
A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled ?
Solution:

Let [A] = x and [B] = y
Rate (r₁) = kxy² ….(i)

(ii) When [B] = 3y
r₂ = k x (3y)² ….(ii)
On dividing equation (ii) by (i),
r₂ = 9r₁
i. e. rate increases 9 times.

(iii) When both [A] and [B] are doubled.
r₃ = k (2x) (2y)² ….(iii)
Dividing equation (iii) by (i),
r₃ = 8r₁
i.e. rate increases 8 times.

Question 10.
In a reaction between A and B, the initial rate of reaction (r₀) was measured for different initial concentrations of A and B as given below:

What is the order of the reaction with respect to A and B ?
Solution:
Assuming that the order of reaction w.r.t. A is x and w.r.t. B is y.
Rate = k [A]^x [B]^y
Rate₁ = 1(0.20)^x (0.30)^y = 5.07 × 10^-5 ….(i)
Rate₂ = k(0.20)^x (0.10)^y = 5.07 × 10^-5 …..(ii)

Order of reaction w.r.t A = 1.5
Order of reaction w.r.t B = 0.

Question 11.
The following results have been obtained during the kinetic studies of the reaction :
2A + B → C + D

Determine the rate law and the rate constant for the reaction.
Solution:
Comparing experiments I and IV and substituting the values.
As concentration of ‘B’ remains constant therefore we get order w.r.t. ‘A’.
(Rate)₁ = k (0.1)^x (0.1)^y = 6.0 × 10^-3 ….(i)
(Rate)₂ = k (0.4)^x (0.1)^y = 2.40 × 10^-2 ….(ii)
Dividing equation (ii) by (i),

Similarly by comparing experiments II and III, we get the order w.r.t. ‘B’.
(Rate)₂ = k (0.3)^x (0.2)^y = 7.2 × 10^-2 ….(iii)
(Rate)₃ = k (0.3)^x (0.4)^y = 2.88 × 10^-1 …(iv)
Dividing equation (iv) by (iii),

Question 12.
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:

Solution:
Given that the reaction between A and B is first order w.r.t. A and zero order w.r.t. B.
∴ Rate = it [A]¹ [B]0 but [B]⁰ = 1.
Rate = k [A]
From experiment I,
2 × 10^-2 = k(0.1) ⇒ k = 0.2 min^-1
From experiment II,
4 × 10^-2 = 0.2 [A] ⇒ [A] = 0.2 min L^-1
From experiment III,
Rate = (0.2) (0.4) = 0.08 mol L^-1 min^-1
From experiment IV,
2 × 10^-2 = 0.2[A] ⇒ [A] = 0.1 mol L^-1

Question 13.
Calculate the half-life of a first order reaction from their rate constants given below:
(i) 200 s^-1
(ii) 2 min^-1
(iii) 4 years^-1.
Solution:

Question 14.
The half-life for radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.
Solution:
Radioactive decay follows first order kinetics,

Question 15.
The experimental data for decomposition of N₂O₅
[2N₂O_s → 4NO₂ + O₂]
In gas phase at 318K are given below:

(i) Plot [N₂O₅] against t.
(ii) Find the half-life period for the reaction.
(iii) Draw a graph between log [N₂O₅] and t.
(iv) What is the rate law?
(v) Calculate the half-life period from k and compare it with (ii).
Solution:
(i) The plot of [N₂O₅] versus time is shown below :

(ii) Initial concentration of N₂O₅ = 1.63 × 10^-2 M
Half of initial concentration = 1.63 × 10^-2 \(\frac { 1 }{ 2 } \) × = 0.815 × 10^-2 M
Time corresponding to half of initial cone. (t_1/2) from the plot = 1440 sec.

(iii) The plot of log [N₂O₅] Versus time :

(iv) Since the graph between log [N₂O₅] vs time is a straight line, the reaction is of 1st order.
Rate = k [N₂O₅]

Question 16.
The rate constant for a first order reaction is 60s^-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16^th value ?
Solution:

Question 17.
During nuclear explosion, one of the products is ⁹⁰Sr with half-life of 28.1 years. If µg of ⁹⁰Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.
Solution:

Question 18.
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Solution:

Question 19.
A first order reaction takes 40 min for 30% decomposition. Calculate t_1/2.
Solution:
30% decomposition means that x = 30% of a = 0.30 a
As reaction is of first order

Question 20.
For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained :

Calculate the rate constant.
Solution:

Question 21.
The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume.
SO₂Cl₂(g) → SO₂(g) + cl₂(g)

Calculate the rate of the reaction when total pressure is 0.65 atm.
Solution:

For the first order reaction in terms of pressure

Question 22.
The rate constant for the decomposition of N₂O₅ at various temperatures is given below:

Draw a graph between In k and 1/T and calculate the values of A and E_a. Predict the rate constant at 30°C and 50°C.
Solution:
To draw the plot of log k versus 1/T, we can rewrite the given data as follows:

Draw the graph as shown in the ahead figure :


Compare it with y = mx + c
log A = Value of intercept on y axis i.e„ on logk axis
[y₂ – y₁ = -1 – (-7.2)] = (-1 + 7.2) = 6.2
log A = 6.2
A = Antilog 6.2
= 1.585 × 10⁶ s^-1
The value of rate constant k can be found from graph as follows :

We can also calculate the value of k from the following formula :

Question 23.
The rate constant for the decomposition of hydrocarbons is 2.418 × 10^-s s^-1 at 546 K. If the energy of activation is 179-9 kj/mol, what will be the value of pre-exponential factor.
Solution:
According to the question, Rate constant (k) = 2.418 × 10^-5 s^-1, Temperature (T) = 546K, Activation energy (E_a) = 179.9 kJ mol^-1

Question 24.
Consider a certain reaction A → Products, with k = 2.0 × 10^-2 s^-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L^-1.
Solution:
Unit of k is s^-1.

Question 25.
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
Solution:

Let initial concentration of sucrose (a) = 1M
Concentration after 8hrs = (a – x) = (1 – x). Where ‘x’ is the amount of sucrose decomposed.

Question 26.
The decomposition of hydrocarbon follows the equation k = (4.5 × 10¹¹ s^-1) e^– 28000 K/T Calculate E_a.
Solution:
k = (4.5 × 10¹¹ s^-1) e^-28000K/T
Comparing the equation with Arrhenius equation
\(k=\mathrm{A} e^{-\mathrm{E}_{a} / \mathrm{RT}}\)

Question 27.
The rate constant for the first order decomposition of H₂O₂ is given by the following equation:
log k = 14.34 – 1.25 × 10⁴ K/T
Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
Solution:
(i) We know that
\(k=\mathrm{A} e^{-\mathrm{E}_{a} / \mathrm{RT}}\)
Taking logarithm on both sides of equation (i), we get

Question 28.
The decomposition of A into product has value of k as 4.5 × 10³ s^-1 at 10°C and energy of activation 60 kj mol^-1. At what temperature would k be 1.5 × 10⁴ s^-1 ?
Solution:
Given k₁ = 4.5 × 10³ s^-1, k₂ = 1.5 × 10⁴ s^-1, T₁ =10°C = 10 + 273 = 283K, E_a = 60kJ mol^-1 = 60000J mol^-1
Applying Arrhenius equation and substituting the values, we get

Question 29.
The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 10¹⁰ s^-1. Calculate k at 318 K and Ea.
Solution:
For 10% completion of the reaction

Question 30.
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Solution:

Chemical Kinetics Other Important Questions and Answers

Chemical Kinetics Objective Type Questions

Question 1.
Choose the correct answer :

Question 1.
For most of the reactions, the value of temperature coefficient lies in between:
(a) 1 and 3
(b) 2 and 3
(c) 1 and 4
(d) 2 and 4.

Question 2.
For first order reaction value of t_1/2 is :

Question 3.
A first order reaction gets completed to 75% in 32 minutes. How much time would have been required for 50% completion :
(a) 24 minute
(b) 16 minute
(c) 8 minute
(d) 4 minute.

Question 4.
For reaction H_2(g) + Br_2(g) → 2HBr_(g) the experimental values indicate that:
Rate of reaction = k [H₂] [Br₂]^1/2 Molecularity and order of reaction is :
(a) 2, \(\frac { 3 }{ 2 } \)
(b) \(\frac { 3 }{ 2 } \),\(\frac { 3 }{ 2 } \)
(c) 1,1
(d) 1,\(\frac { 1 }{ 2 } \)

Question 5.
The reaction 2FeCl₃ + SnCl₂ → 2FeCl₂ + SnCl₄ is an example of:
(a) First order reaction
(b) Second order reaction
(c) Third order reaction
(d) None of these.

Question 6.
In the reaction 2A + B → A₂B, the rate of consumption of reactant A is:
(a) Half of the consumption rate of B
(b) Equal to the consumption rate of B
(c) Twice to the consumption rate of B
(d) Equal to the rate of formation of A₂B.

Question 7.
Hydrolysis of sucrose to glucose and fructose is :
C₁₂H₂₂O₁₁ + H₂O > C₆H₁₂O₆ + C₆H₁₂O₆
sucrose  glucose  fructose
an example of:
(a) First order reaction
(b) Second order reaction
(c) Third order reaction
(d) Zero order reaction.

Question 8.
At a given temperature the rate of reaction becomes slow when :
(a) Energy of activation becomes high
(b) Energy of activation becomes low
(c) Entropy changes
(d) Initial concentration of reactants remains constant.

Question 9.
Plants prepare starch in the process of:
(a) Flash photolysis
(b) Photolysis
(c) Photosynthesis
(d) None of these.

Question 10.
For first order reaction the specific reaction constant depends upon:
(a) Concentration of reactants
(b) Concentration of products
(c) Time
(d) Temperature.

Question 11.
In the reaction between A and B to form C, A represents first order and B rep¬resents second order. Rate equation will be written as :
(a) Rate = k [A]² [B]
(b) Rate = k [A] [B]²
(c) Rate = k [A]^1/2 [B]
(d) Rate = k [A] [B]^1/2.

Question 12.
Molecularity of reaction of Inversion of sugar :
(a) 3
(b) 2
(c) 1
(d) 0.

Question 13.
Minimum energy required for molecules to react is called :
(a) Potential energy
(b) Kinetic energy
(c) Nuclear energy
(d) Activation energy.

Question 14.
2A + B → A₂B, if concentration A is doubled and concentration of B is being half, the rate of reaction will:
(a) Increase 4 times
(b) Decrease 2 times
(c) Increase 2 times
(d) Remains unchanged.

Question 15.
Half life for second order reaction :
(a) Is proportional to initial concentration
(b) Does not depend upon initial concentration
(c) Inversely proportional to initial concentration
(d) Inversely proportional to square root of intial concentration.

Question 16.
The reaction 2H₂O₂ → 2H₂O + O₂, r = k[H₂O₂] is:
(a) Zero order reaction
(b) First order reaction
(c) Second order reaction
(d) Third order reaction.

Question 17.
Thermal decomposition of a compound is first order reaction. If a sample of compound decomposes 50% in 120 minutes, how much time it will take for 90% decomposition:
(a) About 240 minutes
(b) About 480 minutes
(c) About 450 minutes
(d) About 400 minutes.

Question 18.
Arrhenius equation is :

Question 19.
When temperature is raised, rate of reaction increases. This is due to :
(a) Decrease in activation energy
(b) Increase in number of collisions
(c) Decrease in number of active molecules
(d) Decrease in number of collisions.

Question 20.
Unit of rate constant of second order reaction is :
(a) mol^-1 litre^-1 second^-1
(b) mol litre^-1 second^-1
(c) mol litre second
(d) mol^-1 litre second^-1.

Question 21.
Reaction

(a) Bimolecular and second order
(b) Unimolecular and first order
(c) Bimolecular and first order
(d) Bimolecular and zero order.

Question 22.
Unit of rate constant of first order reaction is :
(a) mol L^-1 S^-1
(b) mol^-1 LS^-1
(c) S^-1
(d) mol L^-1 S.

Question 23.
Time required for the completion of 90% reaction of first order is :
(a) 1-1 times that of half life
(b) 2-2 times that of half life
(c) 3-3 times that of half life
(d) 4-4 times that of half life.

Question 24.
The rate of chemical reaction depends upon :
(a) Active mass
(b) Atomic mass
(c) Equivalent weight
(d) Molecular mass.

Question 25.
Unit of reaction rate is :
(a) mol litre^-1 second^-1
(b) mol^-1 litre second^-1
(c) mol^-1 litre^-1 second
(d) mol litre second.

Answer:
1. (b), 2. (a), 3. (b), 4. (a), 5. (d), 6(b), 7. (a), 8. (a), 9. (c), 10. (d), 11. (b), 12. (c), 13. (d), 14. (d), 15. (c), 16. (b), 17. (d), 18. (d), 19. (b), 20. (d), 21. (c), 22. (c), 23. (c), 24. (a), 25. (a).

Question 2.
Fill in the blanks :

1. The rate of a reaction does not depends on the concentration of the reacting species, then the reaction is of …………………
2. Half life period of a radioactive element is 140 days. On taking 1 gm element initially the amount left after 560 days will be …………………
3. Fast reactions are completed in less than ………………… seconds.
4. In the mechanism of a reaction, the slowest step is called …………………
5. Study of fast reactions is done by …………………
6. Unit of rate constant for third order reaction is …………………
7. Difference in the minimum and maximum energy state of reactants is called …………………
8. Reactions which take place by the absorption of radiations are called …………………
9. The total number of molecules which participate in a reaction is called …………………
10. ………………… gives the idea about the mechanism of the reaction.
11. Hydrolysis of Ethyl acetate in acidic medium is an example of ………………… order reaction.
12. Molecularity is always a …………………
13. For the excitation of one mole reactant ………………… photons are required.
14. Rate of change in concentration of reactant at a specific instant of time is known as ………………… rate.
15. Rate of reaction is ………………… to the concentration of reactant.
16. cm-1 is the unit of …………………
Answer:
1. Zero
2. \(\frac { 1 }{ 16 } \) gm
3. 10^-9
4. Rate determining step
5. Flash photolysis
6. mol^-2 Litre² second^-1
7. Activation energy
8. Photochemical reactions
9. Molecularity
10. Order of reaction
11. Pseudo unimolecular
12. whole number
13. one mole
14. Instantaneous
15. Proportional
16. cell constant.

Question 3.
Match the following :


Answers:

1. (c)
2. (e)
3. (a)
4. (b)
5. (d)
6. (g)
7. (f).

Question 4.
Answer in one word / sentence :

1. What is the relation between threshold energy and activation energy ?
2. What is the expression of rate constant for first order reaction ?
3. Write alternative form of Arrhenius equation.
4. If reaction A + B → C is a zero order reaction. Write rate law expression for it.
5. If rate equation for a reaction is Rate = k [NO₂]² [Cl₂] then state the order of reaction with respect to Cl₂ and order with respect to NO₂ and total order of reaction.
6. Write rate equations on the basis of concentration of reactants and products for the following reactions.
2NO₂ → 2NO + O₂
7. What is rate determining step ?
8. Write an example of zero order reaction.
9. What is Quantum Efficiency ?
10. Write Arrhenius Equation.
11. Write the expression of half life period for first order reaction.
12. What is the effect of surface area of reactant on the rate of reaction ?
13. What is instantaneous rate ?
14. What is half-life period of a reaction ?
15. What is the unit of k for zero order reaction ?
16. Explain Threshold energy.
17. What are fast reactions ?
18. For a zero order reaction tm is proportional to what ?
Answer:
1. Activation energy = Threshold energy – Energy of molecules in normal state
2.

3.

4.

5. Order with respect to cl₂ = l, Order with respect to NO₂ = 2, Thus, Total order = 1 + 2 = 3
6.

7. The slowest step is the rate determining step
8.

9. Quantum Efficiency

10.

11.

12. Larger surface area increases the rate of reaction
13. Rate of change in concentration of reactant or product at a specific instant of time is known as instantaneous rate
14. Half-life period of a reaction is the time in which concentration of reactant is reduced to one half of the initial concentration
15. Litre mol^-1 sec^-1
16. The minimum amount of energy which the reactant molecules should possess for effective collision
17. Reactions which gets completed in 10-9 second or even lesser time interval are called fast reactions
18. Initial concentration of reactant [A]

Chemical Kinetics Very Short Answer Type Questions

Question 1.
What is first order reaction ?
Answer:
If the rate of reaction depends on the first power of concentration of reactant then the order of reaction is said to be first order reaction.

Question 2.
What is Energy Barrier ?
Answer:
The minimum energy achieved by the reactant only after which it can be converted to product is known as energy barrier. Reactant molecules cannot form activated complex till they reach this height (activation energy) and cannot be converted to form product.

Question 3.
Explain the rate determining step.
Answer:
Some chemical reactions complete in one or more steps. Rate of reaction is determined by the slowest step which is known as rate determining step.

Question 4.
Write the characteristics of photochemical reactions. Any three.
Answer:
Characteristics :

1. Absorption of magnetic radiations is necessary for these reactions.
2. These reactions are unaffected by temperature but intensity of radiations affects them.
3. Light is necessary for activating the reactants.

Question 5.
What are the applications of Arrhenius equation ? (Any two)
Answer:

1. In calculating activation energy.
2. In determining the rate constant at a temperature by the help of rate constant at another temperature of the reaction.

Question 6.
What is photochemical reaction ? Give an example.
Answer:
Such reactions which are induced by light or other electromagnetic radiations are known as photochemical reactions. Wavelength of these radiations is from 2000Å to 8000Å.

Question 7.
What is specific reaction rate ?
Answer:
Specific reaction rate of a reaction at a given temperature is equal to that rate of a reaction when concentration of each reactant is unity.

Rate of reaction = k where k = Specific reaction rate constant.

Question 8.
When is the average rate of the reaction equal to its instantaneous rate ?
Answer:
When value of time interval is nearly zero or when time by infinite form is minute then the average rate of reaction is comparable to its instantaneous rate.

Question 9.
What are pseudo unimolecular reactions ?
Answer:
There are some reactions in which molecularity is more than one i.e. two or more molecules are present, but in the chemical reaction concentration of only one reactant molecule is changed and it is only responsible for rate of reaction. Thus, order is one. These reactions are called pseudo unimolecular reactions.

Above reaction is bimolecular but concentration of water does not affect the rate of reaction, thus, rate of reaction is proportional to the concentration of sucrose only.
Rate = k[C₁₂H₂₂O₁₁]
Thus, inversion of sucrose is a first order reaction. It is known as pseudo unimoleciilar reaction.

Question 10.
What is temperature coefficient ?
Answer:
Generally, rate of any reaction increases due to increase in temperature. On increasing temperature by 10 C, velocity of reaction is increased up to 2-3 times. If velocity constant of a reaction is and it is increased to increasing temperature by 10 C, the ratio of these two constant is called temperature coefficient, i.e.,

Thus, at various time the ratio of rate of reaction which differ by 10°C is known as temperature coefficient.

Chemical Kinetics Short Answer Type Questions

Question 1.
Write down the expression representing the rate of reaction.
Answer:
Rate of reaction is the rate of change in concentration of reactant or concentration of product in unit time interval.

Unit of rate of reaction depend on the units of concentration and time. If concentration is represented in mole per litre and time in second, then unit of rate of reaction is mol per litre per second. If time is expressed in minute then unit of reaction rate is mole per litre per minute.

Question 2.
What is the meaning of instantaneous rate of reaction ?
Answer:
The rate of reaction does not remain constant during the whole time interval because rate of reaction depends upon the concentration of reactants. As the concentration of reactants decreases with time, the rate of reaction also decreases with time.

In order to express the reaction rate as accurately as possible, the instantaneous rate of reaction is expressed. For this the time interval (∆t) is taken as small as possible.

Where, d [A] is a change in concentration of reactant A.

Question 3.
What do you know about molecularity of any reaction ?
Answer:
Molecularity of Reaction : ‘Number of moles of reactant participating in elementary step of chemical reaction is called molecularity of the reaction.’

There are many reactions which proceed through the number of steps and each step being independent is called elementary step. The rate of reaction is determined by slowest step and it is known as rate determining step. Here molecularity can be defined as “Number of molecules/atoms or ions participating in rate determining step is called molecularity.”

Example : (i) Unimolecular reaction :
O₃ → O₂ + O
(ii) Bimolecular reaction:
NO + O₃ → NO₂ + O₂

Question 4.
Differentiate molecularity and order of reaction.
Answer:
Differences between Molecularity and Order of reaction :

Molecularity:

1. It is the total number of molecules which participate in the reaction.
2. It is a theoretical concept.
3. It is always a whole number.
4. Its value is never zero.
5. It does not provide any information about mechanism of reaction.

Order of reaction

1. It is the number of molecules which participate in reaction and whose concentration is changed.
2. Order of reaction is determined by experimentally.
3. Fractional values are also possible.
4. Zero value is possible.
5. It provides information about mechanisms of reaction.

Question 5.
Write any four factors which affects rate of a chemical reaction.
Answer:
The rate of reaction depends upon the following factors :
(i) Concentration of reactants: At constant temperature, the rate of a reaction increases by increasing the concentration of the reactants.

(ii) Temperature of the system : If the concentration of the reactants are constant then the rate of reaction increases by increasing temperature. For 10 degree rise of temperature, the reaction rate becomes double or triple.

(iii) Presence of catalyst: Positive catalyst increases the rate of reaction and negative catalyst decreases the reaction rate.

(iv) Nature of the reactants: Nature of reactants also affect the reaction rate. In any chemical reaction some old bonds are broken and new bonds are formed. Thus, in case of more simple molecules the lesser is the number of bond breaking and the rate of reaction increase whereas in case of complex molecules more bonds are broken and rate decreases.

(v) Exposure to radiations: The rate of some reactions increases due to some special radiations.

Question 6.
Write characteristics of rate constant
Answer:

1. At fixed temperature, the value of k is constant.
2. For a particular reaction, k is independent of concentration but depends on temperature.
3. The value of k is different for different reactions.
4. It is a measure of intrinsic rate of reaction i. e., larger the value of k, faster will be the reaction and viceversa.

Question 7.
What do you understand by order of reaction ? Give example.
Answer:
Order of reaction : The order of a reaction is defined as the sum of all the powers to which concentration terms in the rate law are raised to express the observed rate of the reaction. Suppose there is a general reaction,
aA + bB + cC → product
For which the rate law is
Rate = – \(\frac { dx }{ dt } \) = k[A]^p [B]^q [C]^r
Then the order of the reaction n = p + q + r, where, p, q and r are the orders with respect to individual reactants and overall order is the sum of the exponents i.e.,p + q + r.
When n = 1 the reaction is of first order, if n = 2 the reaction is of second order and so on.
For example : Decomposition of ammonium nitrite occurs as follows :
NH₄NO₂ → N₂ + 2H₂O
Rate of reaction = – \(\frac { dx }{ dt } \) = k[NH₄NO₂]
Thus, order of this reaction will be 1.

Question 8.
What do you mean by zero order reaction ? Give one example,
Answer:
In some reactions, rate does not depends, upon the concentration of reactants. This type of reactions are called zero order reaction.
Example : In contact of Au or Pt, the ammonia molecule dissociate and rate of this reaction is independent of concentration of ammonia.

Question 9.
How does rate of any reaction depend on temperature ? Explain.
Answer:
Generally, rate of any reaction increases due to increase in temperature. On increasing temperature by 10°C, velocity of reaction is increased up to 2-3 times. If velocity constant of a reaction is and it is increased to increasing temperature by 10°C, the ratio of these two constant is called temperature coefficient, i.e.,
\(\frac{k_{t+10}}{k_{t}} \approx 2 \text { to } 3\)
Arrhenius provide the following relation for showing the effect of temperature on velocity constant.
i.e., \(k=\mathbf{A} \cdot e^{-\mathrm{E}_{\alpha} / \mathrm{RT}}\)

Question 10.
What is activation energy ?
Answer:
According to Arrhenius, any chemical reaction is only possible when reacting . molecules are activated with minimum energy which is called threshold energy. Kinetic energy of most of the molecules are less than this minimum energy. The excess energy which is required to activate reactant molecules, is called activation energy.

Activation energy can be determined by the use of Arrhenius equation.

Question 11.
Write four differences between Rate of reaction and Rate constant.
Answer:
Differences between Rate of reaction and Rate constant:

Rate of reaction:

1. It is expressed in terms of consumption of reactants or formation of product per unit time.
2. It depends on concentration of reactant at particular moment.
3. It generally decreases with the progress of reaction.
4. Its unit is mol L^-1 cm^-1.

Rate constant:

1. It is proportionality constant in differential form in rate law or rate equation.
2. It is independent of concentration of reactant.
3. It does not depend on the progress of reaction.
4. It changes according to order of reaction.

Question 12.
Prove that half-life period of zero order reaction is proportional to initial concentration of reactant.
Answer:
If rate of reaction does not depend on the concentration of reactants, then it is known as zero order reaction.

For a zero order reaction the relation between rate constant and concentration is expressed by the following equation :

Thus, half-life period of zero order reaction is proportional to initial concentration of reactant.

Question 13.
Prove that half-life period is independent of the initial concentration for the first order reaction.
Answer:
Half-life period for the reaction is that period in which the initial concentration of reactant is reduced to half.
For first order reaction :

Thus, the half-life period is independent of the initial concentration for the first order reaction.

Question 14.
What is Integrated rate law method ? Write a note.
Answer:
Integrated rate equation for first order reaction is :

By this equation, order of reaction can be calculated. By knowing the initial concen¬tration (a) of reactant, concentration at a definite time (a – x) can be known.

This way, by substituting the values of t and (a – x), value of k is calculated. If value of k comes to be constant, the reaction is of first order. For, first order reaction, on plotting a graph between concentration log)0 (a – x) and t, a straight line is obtained. For other orders of reaction suitable equations are used and orders are determined.

Question 15.
A first order reaction is 90% complete in 40 minutes. Calculate its half- life period. (log 2 = 0.3010).
Solution:
Let initial cone, of reactant (a) = 100, t = 40 minutes
90% reaction gets completed in 40 minutes i.e.,

Question 16.
Show that time required for completing 99.9% of a first order reaction is 10 times of its half-life period.
Solution:
If initial concentration of reactant is a, then t = ? for x = 0.999 a
We know that for a first order reaction

Question 17.
Write unit of rate constant k for the zero order, first order and second order reaction
Solution:

Question 18.
(1) 2N₂O_s 4NO₂ + O₂
(2) H²₂ + I₂ → 2HI
Reaction (1) is first order reaction and (2) is second order reaction. Why ?
Answer:
(1) In reaction 2N₂O₅ → 4NO₂ + O₂, when graph is plotted between rate of reaction and then again graph is plotted between rate of reaction and [N₂O₅]², it is shown that in the first graph straight line is obtained i.e.
rate ∝ [N₂O⁵]
or rate = k[N₂O⁵]
Therefore, 2N₂O₅ → 4NO₂ + O₂ is first order reaction.

(2) In reaction H₂ + I₂ → 2HI when graph is plotted between rate of reaction and (H₂) (I₂), it is seen that straight line is obtained.
Therefore rate ∝ [H₂][I₂]
Hence, this reaction is of 2^nd order reaction.

Chemical Kinetics Long Answer Type Questions

Question 1.
Determine the expression for zero order reaction.
Answer:
Reactions in which rate of reaction does not depend on the concentration of the reactants are called zero order reactions. Consider a zero order reaction
A → B
Where, A and B are concentration of reactants and products. Since in this type of reaction, rate of reaction does not depend on the concentration of reactant therefore, rate of change in concentration of reactant remains constant.
Rate of reaction = Constant.
Let initial cone, of reactants be a moles /It and after time ‘t’ x moles are converted into product. Then cone, of A after time t will be (a – x) mol /It.

Thus, eqn. (7) is the velocity equation for zero order reaction.

Question 2.
Write a note on Arrhenius equation.
Answer:
Arrhenius equation : Arrhenius gave a relation between rate constant of reaction and temperature which is known as Arrhenius equation i.e.,
\(k=\mathbf{A} e^{-\Delta \mathbf{E}_{a} / \mathbf{R} \mathbf{T}}\)
Where, k = Rate constant of reaction, A = Frequency factor, Ea = Activation energy, R = Gas constant, T = Absolute temperature.

Taking logarithm both sides of above equation we get

E_a and A can be determined by measuring rate constants of the reaction at two different temperature. Let k₁ and k₂ are the rate constant for the reaction at two temperatures T₁ and T₂ respectively. Then from eqn. (1)

Application : Activation energy AE_a may be calculated easily from this equation.

Question 3.
Write the Arrhenius equation in the form of equation of straight line. What will be the slope of a graph using this equation ? Calculate the activation energy for the decomposition in a decomposition reaction in which the value of slope obtained is – 9920 when log k is plotted against \(\frac { 1 }{ T } \).
Answer:
Arrhenius equation : Arrhenius gave a relation between rate constant of reac¬tion and temperature which is known as Arrhenius equation, i.e.,
\(k=\mathrm{A} e^{-\mathrm{E}_{a} / \mathrm{RT}}\)
Where, k = Rate constant of reaction, A = Frequency factor, Ea = Activation energy, R = Gas constant and T = Absolute temperature.
Taking logarithm both sides of above equation, we get

This is a straight line equation. If log₁₀ A and \(\frac { 1 }{ T } \) is plotted at different temperatures, a straight line is obtained whose slope = \(\frac{-\mathrm{E}_{a}}{2 \cdot 303 \mathrm{R}}\) From the graph, value of slope = \(\frac{\mathrm{E}_{a}}{2 \cdot 303 \mathrm{R}}\) can be calculated and activation energy (E_a) can be calculated.

Alternatively, E_a and A can be determined by measuring rate constant of the reaction at two different temperatures. Let k₁ and k₂ are the rate constants for the reaction at two temperatures T₁ and T₂ respectively. Then, from eqn. (1),

Calculation of activation energy :
E_a = -2.303 × slope × R = -2.303 × (-9920) × 1.986
∴ E_a = 45334 cal/gm/mol.

Question 4.
Derive an expression for velocity constant of first order reaction.
Answer:
The reaction in which velocity of reaction depends upon the concentration of one mole, are called first order reaction.
Let this reaction is
A → Product
Suppose intial concentration of A is a gram mole and after t second x mole consumed and remaining concentration is (a – x) gram mole.
So after t time, the rate of reaction will be proportional to (a – x)

Putting the value of c from eqn. (4), into eqn. (3).
– ln(a – x) = kt – In a
or In a – ln(a – x) = kt

It is the desired expression for first order of reaction.

MP Board Class 12th Chemistry Solutions