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In this article, we share MP Board Class 12th Maths Book Solutions Chapter 7 Integrals Ex 7.9 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 7 Integrals Ex 7.9

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 9 Differential Equations Miscellaneous Exercise Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Miscellaneous Exercise

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 8 Application of Integrals Miscellaneous Exercise Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 8 Application of Integrals Miscellaneous Exercise

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 10 Vector Algebra Ex 10.1 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 10 Vector Algebra Ex 10.1

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 1 Relations and Functions Miscellaneous Exercise Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise

MP Board Class 12th Chemistry Important Questions Chapter 2 Solutions

Solutions Important Questions

Very Short Answer Type Questions

Question 1.
Write formula of van’t Hoff factor ‘i’
Answer:
van’t Hoff factor, i = \(\frac {Observed value of colligative property}{Calculated value of colligative property}\)

Question 2.
Write down van’t Hoff equation. Give formula used for calculating molecular mass with its help.
Answer:
van’t Hoff equation, πV = nRT
or π = \(\frac { nRT}{V}\)
or π = \(\frac { WRT}{MV}\)
or π = \(\frac { WRT}{πV}\)
Where, W is mass of solute, R is Solution constant, T is temperature, n is osmotic pressure and V is volume of Solution.

The normality formula of a solution is the gram equivalent weight of a solute per liter of solution.

Question 3.
Define Normality.
Answer:
Number of gram equivalent of solute present in one litre of Solution: is called normality. It is represented by N.
Normality (N) = \(\frac {Number of gram equivalent of solute}{Volume of Solution: in litres}\)
or N = \(\frac {Mass of solute in grams }{Equivalent mass of solute}\) x \(\frac { 1000}{ Volume of Solution: in ml }\)
∵ [No. of gram equivalent =\(\frac {Mass of solute in grams}{Equivalent mass of solute}\)]
Normality of a Solution changes with temperature as it is based on mass – volume relationship and volume changes with change in temperature.

Question 4.
Differentiate between Molarity and Molality.
Answer:
Differences between Molarity and Molality :
Molarity (M):

• Molarity involves the total volume of Solution.
• In molarity, gram moles of solute are dissolved in 1 litre of Solution:.
• Molarity changes with temperature because volume changes with temperature.

Molality (m):

• Molality involves the mass of solvent.
• In molality, gram moles of solute are dissolved in 1 kg of solvent. Here volume of Solution is not considered.
• Molality is independent of temperature as it takes mass into consideration.

Question 5.
Explain the following term : Parts per million.
Answer:
Parts per million:
When a solute is present in very minute amounts (in traces), the concentration is expressed in parts per million abbreviated as ppm. The parts may be of mass or volume. It is the parts of a component per million parts of the Solution.
i.e., ppm = \(\frac {Mass of component A}{Total mass of Solution:}\) x 10⁶
or ppm = \(\frac {Volume of component A}{Total volume of Solution:}\) x 10⁶
Where, ppm. is the concentration of component A in parts per million.

Question 6.
What role does the molecular interaction play in a Solution: of alcohol and water? (NCERT)
Answer:
Alcohols dissolve in water due to formation of inter – molecular H – bonding with water.

Question 7.
Why do gases always tend to be less soluble in liquids as the temperature is raised? (NCERT)
Answer:
Gas + Liquid ⇌ Dissolved gas   ∆ H = – ve
The dissolution of a gas in a liquid is exothermic process. Therefore in accordance with Le – Chatelier’s principle with increase in temperature, the equilibrium shifts in back-ward direction. Therefore, the solubility of gas in Solution decreases with the rise in temperature.

Question 8.
State Henry’s. (NCERT)
Answer:
Henry’s law:
According to this law, ‘The mass of a gas dissolved per unit volume of a solvent at constant temperature, is proportional to the pressure of the gas with which the solvent is in equilibrium’. Let in unit volume of solvent, mass of the gas dissolved is m and equilibrium pressure be P, then m ∝P or m = KP, where K is a constant.

Question 9.
What are colligative properties?
Answer:
The physical properties of Solution which depend on the number of total particles present in Solution and the ratio of number of particles of solute not on the nature of solute particles are known as colligative properties.

Question 10.
Give an example of Solution: of solid in solid.
Answer:
Mixture of copper and gold is an example of solid in solid.

Question 11.
On dissolving ethanoic acid in benzene, experimental molecular mass of ethanoic acid is generally found to be double. Why?
Answer:
On dissolving ethanoic acid in benzene, it dimerizes due to the formation of hydrogen bond. Thus, its experimental value is generally found to be double.

Question 12.
What is vapour pressure? What is the effect of temperature on it?
Answer:
Pressure exerted by the vapours on the surface of a liquid at the state of equilibrium is known as vapour pressure. On increasing the temperature vapour pressure increases.

Question 13.
What type of deviation is represented by acetone and CS₂ Solution?
Answer:
Positive deviation.

Osmotic pressure calculator … The factor i is also called the dissociation factor or the van’t Hoff factor.

Question 14.
Why is CaCl₂ used for clearing off snow from the roads?
Answer:
On adding CaCl₂, freezing point of water decreases. Therefore, CaCl₂ is used for clearing off snow from the roads.

Question 15.
Based on solute – solvent interactions, arrange the following in order of increasing solubility in n – octane and explain :
Cyclohexane, KCl, CH₃OH, CH₃CN.
Answer:
For solubility we know Tike dissolves like’, n – octane is a non – polar solvent, hence non-polar compounds will be more soluble.
KCl < CH₃OH < CH₃CN < Cyclohexane.

Question 16.
What is transition temperature?
Answer:
The temperature at which the nature of solubility changes (i.e., first it increases, then decreases) is known as transition temperature. Solubility of sodium sulphate in water increases upto 32.4, then it starts decreasing. Thus, 32.4°C is the transition temperature of sodium sulphate.

Question 17.
Define molarity and molality. (NCERT)
Answer:
Molarity “is defined as number pf gram mole§ of solute dissolved in a litre of Solution. It is denoted by M.
Molality (M) = \(\frac {Mass of solute in gram per litre}{ Molecular mass of solute}\)

Molality is defined as number of moles of solute present in a kilogram (1000 gram) of solvent. It is denoted by m.
Molality (m) = \(\frac {Mass of solute in kg of solvent}{Gram molecular mass of solute}\)

Solutions Short Answer Type Questions

Question 1.
Write down Raoult’s law.
Answer:
The vapour pressure of a solution containing non – volatile solute is directly pro – portional to the mole fraction of the solute.
Mathematically,
\(\frac { { X }_{ A } }{ { X }_{ A }+{ X }_{ B } } \) = \(\frac { P{ A }^{ 0 } }{ { P }_{ A }+{ P }_{ B } } \)
Where, P⁰_A = Vapour pressure of pure solvent, P_A = Vapour pressure of solvent in solution, X_B = Mole fraction of solute.

Question 2.
What is Azeotropic mixture? They are of how many types? (MP 2018)
Answer:
Azeotropic mixture is the mixture of liquids which boil at one temperature with-out any change in composition. For example, at the composition of 95 6% alcohol and 4.4% water. It form an azeotropic mixture which boils at 78.13°C. Components of this mixture cannot be separated fully by fractional distillation.

They are of two types :
1. Low boiling azeotropic mixture:
Such Solutions which represent positive deviation towards Raoulfs law i.e. their vapour pressure is high thus their boiling point is low are known as low boiling azeotropic mixture.

Example:

• CS₂ + Acetone
• C₂ H₅ OH + n – hexane.

2. High boiling azeotropic mixture:
Such Solutions which represent negative deviation towards Raoult’s law i.e. their vapour pressure is low thus their boiling point is high are known as high boiling azeotropic mixture.

Example:

• Acetone + Chloroform
• Ether + Chloroform.

Question 3.
What are ideal and non – ideal Solutions? Explain with example.
Answer:
Ideal Solutions:
Ideal Solutions are those Solutions in which Raoult’s law can be applied completely for all concentrations of the Solutions and at all temperatures.

Condition for ideal Solutions are following :

1. P_A = P_A⁰ X _A and P_B= P⁰_B X B
2. ∆V_mixing = 0 and
3. ∆H_mixing = 0 and

Example:
C₂H₅Br + C₂H₅Cl, C₆H₆ + C₆H₅CH₃, CCl₄ + SiCl₄etc.

Non – ideal Solutions:
Solutions in which Raoult’s law cannot be applied completely for all concentrations and temperatures are called non – ideal Solutions.
For these Solutions:

1. P_A ≠ P⁰_A X_A and p_B ≠ P⁰_BX_B 
2. ∆V_mixing ≠ 0 and
3. ∆H_mixing ≠ 0 and

Example:
Benzene + Acetone, CHCl₃ + HNO₃ etc.

Question 4.
Establish van’t Hoff Solution: equation. (MP 2018)
Answer:
Osmotic pressure of dilute Solution of a non – volatile solute is proportional to absolute temperature of the Solution at constant concentration. This is known as van’t Hoff law.
π ∝ T .. (1)
Derivation : Osmotic pressure n of a Solution: is directly proportional to a molar concentration.
π ∝ C
By eqn.(l) and eqn.(2) π ∝ C (T is constant) .. (2)
or π = RCT .. (3)
Where, R = Gas constant
or
C = \(\frac {1}{V}\)
π = \(\frac {RT}{V}\) .. (4)
πV = RT .. (5)
This is known as van’t Hoff Ideal Solution equation.

Question 5.
Define the following:

1. Molal elevation boiling point constant.
2. Molal freezing point depression constant. (MP 2018)

Answer:
1. Molal elevation boiling point constant : Molal elevation constant can be defined as “The elevation in boiling point of the Solution in which 1 gm of solute is dissolved in 1000 gm of solvent.”
∴ Elevation in boiling point
∆ T_b ∝ m
∆ T_b = K_bm
m= 1, ∆ T_b = K_b
Where K_b = Molal boiling point elevation constant.

2. Molal freezing point depression constant : Molal depression constant may be defined as “The depression in freezing point for 1 molal Solution: i.e., Solution: in which 1gm mole of solute is dissolved in 1000 gm of solvent.”
∴ Depression in freezing point
∆ T_f ∝ m
∆ T_f = k_f
If m = 1, ∆ T_f = k_f
Where K_f = Molal freezing point depression constant.

Question 6.
What is Raoult’s law? Establish its mathematical expression.
Or,
What is Raoult’s law? How can molar mass of a non – volatile solute be determined with its help?
Answer:
Raoult’s law:
For a Solution in which solute is non – volatile, the Raoulf s law may be stated as following :
“At any constant temperature, vapour pressure of solvent collected above the Solution of non – volatile solute, is directly proportional to the mole fraction of solute.” If a non – volatile solute is added to a volatile solvent, the vapour pressure of the solvent decreases. The vapour pressure of the solvent is directly proportional to its mole fraction. As the solute is non – volatile, the vapour pressure of the Solution (P) will be equal to the vapour pressure of the solvent (PA).
P = P_A ∝ X_A
or P_A=KX_A .. (1)
Where, K = Proportionality constant.
Apply eqn. (1) for pure solvent if X_A= 1 and P_A = P⁰_A
Then, P⁰_A = K x 1 .. (2)
Where, p⁰_A= Vapour pressure of pure solvent.
Putting the value of K from eqn. (2) in eqn. (1),
P_A=P⁰_AX_A .. (3)
Where, X_A is the mole fraction of the solvent in the Solution.
The mole fraction of solute is represented as XB.
So, X_A+X_B=1
or X_A= 1 – X_B .. (4)
From eqns. (3) and (4),
P_A= P⁰_A(1-X_B) = P⁰ – P⁰_AX_B
X_B = \(\frac { { P }_{ A\quad }^{ 0 }-\quad { P }_{ A } }{ { P }_{ A }^{ 0 } } \)
P⁰_A is lowering in vapour pressure and \(\frac { { P }_{ A\quad }^{ 0 }-\quad { P }_{ A } }{ { P }_{ A }^{ 0 } } \) is relative lowering in vapour pressure.

On the basis of equation (5) Raoult’s law can be defined as “The relative lowering in vapour pressure of a Solution containing non – volatile solute is equal to mole fraction of solute”.

Question 7.
Give relation between elevation in boiling point and molecular mass of solute.
Answer:
Relation between elevation in boiling point and molecular mass of solute:
Suppose, W_B gram of non – volatile solute dissolve in WA gram of solvent and the molecular mass of non – volatile solute is M_B gram. Then, molality, m will be

Question 8.
What is van’t Hoff factor? How does it equation?
Answer:
van’t Hoff Equation : van’t Hoff equation

Question 9.
What are constant boiling mixture? Write three differences in Ideal solution and Non – ideal Solutions. (MP 2011,15,16)
Answer:
Constant boiling mixtures or azeotropic mixture. A Solution which distils with out change in composition is called azeotropic mixture.
Differences between Ideal and Non – ideal Solution:

Question 10.
Define the following :

1. Reverse osmosis
2. Isotonic Solution
3. Semi – permeable membrane.

Answer:
1. Reverse osmosis:
It may be noted that if a pressure higher than the osmotic pressure is applied on the Solution the solvent will flow from the Solution into the pure solvent through the SPM. The process is called reverse osmosis.

2. Isotonic Solutions:
The Solution:s which have same osmotic pressure are called as isotonic Solution:s. For example, 0.91% Solution of NaCl called as saline water is isotonic with human blood corpuscles. That is why medicines are mixed with saline water before intravenous injections.

Osmotic pressure is a colligative property and depends on the number of solute particles in a Solution So, the isotonic Solution:s must have same number of solute particles in a given volume of Solution Consequently, the isotonic Solution:s are those which have same molality, if that the solute neither associate nor dissociate in the Solution.

3. Semipermeable membrane:
These are the membrane which allow the movement of the solvent molecules through them. The membrane appear to be continuous sheets or films. But they have very tiny holes or pores which are semimicro in nature. Through the holes, only the molecules of solvent can pass while those of bigger solute molecules cannot pass through.

Solutions Long Answer Type Questions

Question 1.
Write five differences in Solution: having Positive deviation and Negative deviation.
Answer:
Differences between Positive deviation and Negative deviation :

Question 2.
Explain in brief Berkeley and Hartley’s method of osmotic pressure measurement and state its uses.
Answer:
Berkeley and Hartley’s method:
In this method, pressure is applied over the Solution to stop the flow of solvent. This pressure is equivalent to osmotic pressure.
In this method, the apparatus consists of a strong vessel made up of steel in which porous pot is fitted. In the porous pot, copper ferrocyanide semipermeable membrane is deposited. The porous pot is fitted with a capillary tube on one side and a water reservoir on the other side. A piston and pressure gauge are fitted to the steel vessel.

The porous pot and steel vessel are filled with water and Solution respectively. Osmosis takes place and water moves into the steel vessel from the porous pot through the semipermeable membrane. This is shown by fall in water level in the capillary tube. This flow of water is stopped by applying external pressure on the Solution with the help of piston.

This method has the following advantages :

1. It takes comparatively lesser time to determine osmotic pressure.
2. Concentration of Solution: does not change, hence better results are obtained.
3. As high pressure is not exerted over semipermeable membrane, it does not break.
4. High osmotic pressure can be measured.

Question 3.
What is molal freezing point depression constant? Derive the formula to establish relation between molal freezing point depression constant and molecular mass of solute.
Or,
What is molal freezing point depression constant? Show that depression in freezing point is a colligative property. How can molecular mass of solute be determined from depression in freezing point?
Answer:
Molal freezing point depression constant is equal to depression in freezing point of the Solution when 1 gm mole is dissolved in 1000 gm of solvent. It is represented by i.e.,
K_f = \(\frac {depression in freezing point}{Number of moles}\)
or ∆ T_f = K_f x m
As K_f is molal freezing point depression constant, then
∆ T_f ∝m
Where, m = Number of moles dissolved in 1000 gm solvent.
Thus, depression in freezing point is proportional to molality of Solution. Molality is directly proportional to number or molecules of solute substance. Therefore, depression in freezing point is a colligative property.
Calculation of molecular mass of solute:
By determination of depression in freezing point, the Molecular mass of non – volatile solute can be determined.
For a Solution: of non – volatile solute,
∆T_f = K_f x m .. (1)
Let WB gram non – volatile solute is dissolved in w_A gram solvent and molecular mass of solute is M_B.

Question 4.
What is elevation in boiling points? How addition of a non – volatile solute elevates the boiling point of a solvent? Explain it with the help of graph diagram.
Answer:
The vapour pressure of the Solution:
containing a non – volatile solute is always less than that of pure solvent. Therefore, the Solution has to be heated to higher temperature so that its vapour pressure become equal to the atmospheric pressure. Thus, the boiling point of Solution (T_b)is always higher than the boiling point of solvent (T_b⁰). The difference T_b – T_b⁰ is called elevation in boiling point.

If we plot graph between temperature and vapour pressure of a pure solvent and its Solution, then following curve is obtained. Curve AB gives the vapour pressure for the pure solvent and the curve CD gives the vapour pressure of the Solution: at different temperature. At temperature T_b⁰ the vapour pressure of the solvent becomes equal to the atmospheric pressure hence it boils at T_b⁰.

Now, by the addition of non – volatile solute, lowering of vapour pressure of the Solution takes place. And to increase the vapour pressure of the Solution to become equal to atmospheric pressure, the temperature rises. Hence, at T_b the Solution boils. Thus, the boiling point is now elevated from T_b⁰ to T_b. The rise in temperature that results by the addition of a non – volatile solute in a solvent is termed as elevation in boiling point. It is represented by ∆T_b.
So, elevation in boiling point (∆T_b) = T_b – T_b⁰.

Question 5.
What are Non – Ideal Solution? How many types of are they? Explain with giving examples.
Answer:
Non – ideal Solutions:
Solutions in which Raoult’s law cannot be applied completely for all concentrations and temperatures are called non – ideal Solutions.
For these Solutions:

1. P_A ≠ P⁰_A X_A and p_B ≠ P⁰_BX_B,
2. ∆V_mixmg ≠ 0 and
3. ∆H_mixmg ≠ 0 and

Non – ideal Solution are of two types :

1. Solutions showing positive deviations:
For such Solutions the total vapour pressure will be greater than the corresponding vapour pressure according to the Raoult’s law. The boiling point of such Solutions are lowered. Because interactions between the mol-ecules of Solution: is less than the interactions between pure solvents and solutes. Thus the vapour pressure greater than expected. In these type of Solution formed enthalpy and volume increases.
Characteristics:

• P_A > P⁰_A X_A and P_B > P⁰_B X_B
• ∆V_mixlng > 0
• ∆H_mixlng > 0

Example:
Solution of Cyclohexane and Ethanol:
In Ethanol, its molecules are held together by hydrogen bond

On adding cyclohexane, the molecules tend to occupy the space between ethyl alcohol molecules. As a result the attractive forces between alcohol molecules becomes less. Thus, formation of such Solution: is slight increase in vapour pressure, endothermic and an increase in volume.

2. Solutions showing negative deviations:
For such Solutions the total vapour pressure becomes less than expected according to the Raoult’s law. Because interaction between molecules of Solution is greater than interaction between the pure solvent or solute molecules. Thus the vapour pressure of Solution is less. In these type of Solutions formed enthalpy and volume decreases.
Characteristics:

• P_A < P⁰_A X_A and P_B < P⁰_B X_B
• ∆V_mixlng < 0
• ∆H_mixlng < 0

Example:
Solution of Acetone and Chloroform:
When acetone and chloroform are mixed, there are new attractive forces due to intermolecular hydrogen bonding. These force become stronger. Thus the formation of such Solution:s is an exothermic and the vapour pressure of Solution is less and decrease in volume.

Solutions Numerical Questions

Question 1.
Calculate the mass of urea (NH₂ CONH₂ ) required in making 2.5 kg of 0.25 molal aqueous Solution. (NCERT)
Solution:
0.25 molal aqueous Solution means that
Moles of urea = 0.25 mole
Mass of solvent (water) = 1 kg = 1000 g
Molar mass of urea = 14+ 2 +12+ 16 +14+ 2 = 60 g mol^-1
∴ 0 .25 mole of urea = 60 x 0.25 mole = 15 g
Total mass of the Solution = 1000 + 15 g
= 1015 g = 1.015 g
∵ 1.015 kg of Solution contain urea = 15 g
∴ 2.5 kg of Solution will require urea = \(\frac {1.5}{1.015}\) x 2.5 kg = 37 g.

Question 2.
Henry’s law constant for CO₂ in water is 1.67 x 10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 ml of soda water when packed under 2.5 atm CO₂ pressure at 298 K. (NCERT)
Solution:
According to Henry’s law,
P = K_HX .. (1)
P = 2.5 atm = 2.5 x 101325 Pa, KH= 1.67 x 10⁸ Pa
Putting these values in equation (1), we get

Question 3.
Calculate the mass of ascorbic acid (Vitamin C, C₆H₈O₆) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. K_fK kg mol^-1. (NCERT)
Solution:
We know that ∆T_f = K_f x \(\frac{\mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}}\) .. (1)
Given ∆T_f = 1.5, K_f = 3.9 K kg mol^-1, W_A = 75g, MB (ascorbic acid, C₆H₈O6)
= 6 x 12 + 8 x l + 16 x 6 = 176
Putting, these values in equation (1),
1.5 = 3.9 x \(\frac{\mathrm{W}_{\mathrm{B}} \times 1000}{176 \times 75}\)
W_B = 5.077g.

Question 4.
Calculate the osmotic pressure in pascals exerted by a Solution prepared by dissolving 1.0 g of polymer of molar mass 1,85,000 in 450 ml of water at 37°C. (NCERT)
Solution:
π = CRT
= \(\frac {n}{V}\) RT
Here, number of moles of solute dissolved (n)
= \(\frac {1.0}{185,000}\) mol^-1
= \(\frac {1}{185,000}\) g mol^-1
v = 450mL = 0.450 L
T = 37°C = 37 + 273 = 310 K
R = 8.314 K Pa LK^-1 mol^-1
= 8.314 x 103 Pa LK^-1 mol^-1
Substituting these values we get,

Question 5.
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous Solution. What should be the molarity of such a sample of the acid if the density of the Solution is 1.504 g ml^-1? (NCERT)
Solution:
68% nitric acid by mass means that
Mass of nitric acid = 68 g
Mass of Solution = 100 g
Molar mass of HNO₃ = 63 g mol^-1
∴ 68 g HNO₃ = \(\frac {68}{63}\) mole = 1.079 mole
Density of solution = 1.504 g mL^-1
∴ Volume of solution = \(\frac {100}{1.504}\)mL = 66.5 mL = 0 0665 L
Molarity of the solution = \(\frac {Moles of the solute}{Volume of Solution in L}\)
= \(\frac {1.079}{0.0665}\) M = 16.23 M.

Question 6.
Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293K when 25g of glucose is dissolved in 450g of water. (NCERT)
Solution:
\(\frac { { p }^{ 0 }-{ p }^{ 0 } }{ { p }^{ 0 } } \) = X₂ = \(\frac { { W }_{ B }{ M }_{ A } }{ { M }_{ B }{ W }_{ A } } \)
or \(\frac { 17.535-{ P }_{ s } }{ 17.535 }\) = \(\frac { 25×18 }{ 180×450 }\)
or \(\frac { 17.535-{ P }_{ s } }{ 17.535 }\) = 5.56×10^-3
17.535 – p_s = 0.0975
P_s = 17.438 mm Hg.

Question 7.
Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 x 10⁵mm Hg. Calculate the solubility of methane in benzene at 298K under 760 mm Hg. (NCERT)
Solution:
Here, KH = 4.27 x 10⁵ mm, P = 760 mm
Applying Hemy’s law, P = K_HX
\(\mathrm{X}=\frac{\mathrm{P}}{\mathrm{K}_{\mathrm{H}}}=\frac{760}{4 \cdot 27 \times 10^{5}}\) =1.78 x 10^-3
∴ Mole fraction of methane in benzene = 1.78 x 10^-3

Question 8.
Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene. (NCERT)
Solution:
A → Benzene (C₆H₆); B —> Toluene (C₇H₈)
Number of moles of benzene n_A = \(\frac {80}{78}\)
Number of moles of toluene = n_B = \(\frac {100}{92}\)
X_A = \(\frac { { n }_{ A } }{ { n }_{ A }+{ n }_{ B } } \quad\) = \(\frac {1.026}{1.026+1.087}\) = 0.486
X_B = 1 – X_A = 1 – 0.486 = 0.514
Given P°_A = 50.71 mm Hg and P°_B = 32.06 mm Hg
We know, P = P_A + P_B
= P°_AX_A + P°_AX_B
= (50.71 x 0.486) + (32.06 x 0.514)
= 24.65+ 16.48 = 41-13
Mole fraction of components in vapour phase may be calculated by using Dalton’s law,
P_A = y_A x P
y_A = \(\frac { { P }_{ A } }{ p } \quad \) = \(\frac {24.65}{41.13}\) = 0.60

Question 9.
Determine the osmotic pressure of 5% glucose Solution at 25°C. Molecular mass of glucose = 180, R = 0.0821 litre atmosphere. (MP 2013,16)
Solution:
∴5 gm glucose is dissolved in 100 ml.
∴ 180 gm glucose will be dissolved in \(\frac {100}{5}\) x 180
= 3600 ml = 3.6 litre.
We know that,
PV = RT
P x 3.6 = 0.0821 x (25 + 273) = 298
or P = \(\frac {0.52×12.5×1000}{0.63×170}\)
= 6.80 atmospheres.

Question 10.
12.5 gm of urea dissolved in 170 gm of water. The elevation in boiling point was found to be 0.63 K. If KA for water = 0.52 Km^–1, calculate the molecular mass of urea. (MP 2010)
Solution:
Formula: M_B\(\frac { { K }_{ b }x{ W }_{ B }x1000 }{ ∆{ T }_{ b }x{ W }_{ A } }\)
Given: W_A = 170gm,W_B= 12.5gm, ∆T_B=0.63K, K_b=0.52Km^-1
M_B = \(\frac {0.52×12.5×1000}{0.63×170}\)
M_B = 60.7 gm mol^-1

Question 11.
If 6.84 gram sucrose is dissolved in 100 ml. Solution then what will be its osmotic pressure at 20°C? (R = 0.082 litre atmosphere K^-1mol^-1) (Molecular mass of sucrose M_B = 342)
Solution:
Given : Weight of solute (W_B) = 6.84 gm
Molecular mass of solute (M_B) = 342
Volume of Solution: (V) =\(\frac {100}{1000}\) =0.1 litre
Temperature (T) = 20° C = 273 + 20 = 293 K
Solution constant R = 0.082 litre atmosphere K^-1 mol^-1
Osmotic pressure π = \(\frac { { W }_{ B }RT }{ { M }_{ B }V } \)
= \(\frac {6.84×0.082×293}{342×0.1}\)
= 4.8 atmosphere.

MP Board Class 12th Chemistry Important Questions

MP Board Class 9th Social Science Solutions Chapter 18 Food Security in India

MP Board Class 9th Social Science Chapter 18 Text Book Questions

Choose the Correct Answer:

Question 1.
Kharif crop is –
(a) Wheat
(b) Gram
(c) Paddy
(d) Oat.
Answer:
(c) Paddy

Question 2.
The part of Public Distribution System is –
(a) Shoe shop
(b) Gold & Silver shop
(c) Ration shop
(d) Grocery shop.
Answer:
(c) Ration shop

Question 3.
Target-ted public distribution is related to –
(a) Women
(b) Gents
(c) People living below the poverty line
(d) None of these.
Answer:
(c) People living below the poverty line

Question 4.
How much cereal is given under Antyodaya Anna Yojna –
(a) 5 kg
(b) 10 kg
(c) 15 kg
(d) 25 kg.
Answer:
(d) 25 kg.

MP Board Class 9th Social Science Chapter 18 Very Short Answer Type Questions

Question 1.
Write the names of coarse cereal.
Answer:
Sorghum (Jowar), bajra (pearl millet) and maize.

Question 2.
In which years did India face famine?
Answer:
India had to face famine is the years 1835, 1877 and 1943.

Question 3.
What is Rojgar Ashwasan Yojna?
Answer:
Under Rojgar Ashwasan Yojna 100 days employment can be provided to 18 – 60 year old unskilled labor so that they are able to earn and purchase food grains through revamped public distribution system.

Question 4.
What is meant by minimum support price?
Answer:
Declaration of support price of agricultural crops means to give the guarantee of a minimum price to farmers for their crops.

Question 5.
Write the names of any two schemes launched for food security.
Answer:

• Target-ted public distribution system
• Kam Ke Badle Anaj Yojna.

MP Board Class 9th Social Science Chapter 18 Short Answer Type Questions

Question 1.
What are the main factors of food security? Write about them.
Answer:
The main factors of food security are –

• Availability of food to the whole population of the country.
• Availability of enough money (purchasing power) to purchase the available food.
• Food should be available to all at affordable prices.
• The quality of the available food should be good.

To calculate the pH of a buffer, go to the buffer pH calculator.

Question 2.
What is Buffer Stock? Explain.
Answer:
if the production of food grains is less then to face such crisis of shortage and to distribute them through Public Distribution System, the stock of food grains kept by the government is known as Buffer Stock. Buffer Stock is the stock of food grains namely wheat and rice procured by government through Food Corporation of India (FCI).

The FCI purchases wheat and rice from farmers in states where there is surplus production. These food grains are stored in large granaries. It helps in resolving the problem of shortage of food grains during emergencies.

Question 3.
Explain Target-ted Public Distribution System.
Answer:
In 1997 target-ted Public Distribution System was introduced to ensure the availability of minimum quantity of food grains to families living below the poverty line. In this system food grains are provided to the poor on special low prices by issuing specific ration cards.

This is the largest food security plan of the world. In this system 3.5 kg. of food grains per month per family is being provided from 1st April, 2006. In the same way under the “Antyodaya Anna Yojna” 25 kg. of wheat at. Rs. 2/- per kg. and rice at Rs. 3/- per kg. is being provided to very poor families through Public Distribution System.

Under this system a differential price system was adopted including people below the poverty line (BPL) and also for people above the poverty line (APL)- in which prices of wheat and rice are fixed.

Question 4.
What is Revamped Public Distribution System? Explain.
Answer:
Revamped Public Distribution System In January 1992, the Public Distribution System was amended and a revised Public Distribution System was introduced to supply essential goods to consumers of remote area, schedule tribes, backward classes, drought – affected and mountainous areas of the country. Its characteristics are as follows:

1. Preference has been given to the people of drought – affected areas, desert areas, mountainous area and slums in urban areas.

2. It is aimed at providing more quantity of food at comparatively low prices. Other than six chief essential commodities goods like tea, soap, pulse, iodized salt are included in it.

3. Rojgar Aswan Plan has been started in the development blocks included under this plan, in which 100 days employment can be provided tr 18 – 60 – year – old unskilled labor so that they are able to earn and purchase food-grains through Revamped Public Distribution System.

Question 5.
What is the role of cooperative in food security? Explain.
Answer:
In India the role of cooperative is very important in providing food security. This work is done by the consumer cooperative societies through the ration shops for the sale of food grains for the poor. In India there are different system of consumer cooperatives at national, state, district and village levels.

Out of these National Consumers. Cooperative (Federation) Ltd. is an organization at national level. State Cooperative Consumers’ Organization is affiliated to this forum (federation). There are 794 consumer cooperative stores at central level (wholesale) and 24,078 stores’at primary level.

In rural areas nearly 44,418 village level primary agricultural credit society are distributing essential goods along with their ordinary business. To fulfill the needs of consumers, consumer cooperative societies are running nearly 37,226 retail selling centers in urban and semi – urban areas.

Question 6.
Differentiate between Kharif and Rabi Crops.
Answer:
Kharif:

• Kharif crops are shown in June or early July and harvested at tire end of Monsoon. This means sowing is done at the onset of monsoon (June and July) while harvesting is done at the end of Monsoon (October and November).
• Kharif crops are rice, jowar, maize, millet, groundnut, jute and cotton pulses are also grown in this session.
• Kharif crops are of tropical climate. (Some changes in this pattern take place, see the note in bracketed column of rabi crop.)

Rabi:

• Rabi crop season starts in November-December and hares – ting .takes place in the months of April and May.
• the major rabi crops are – wheat, barley, gram,, linseed, rasped and mustard.
• in the southern half of the Indian peninsula where temperatures are sufficiently high and rain occurs in winters this classification become blurred. Crops like Jowar are grown in.

MP Board Class 9th Social Science Chapter 18 Long Answer Type Questions

Question 1.
What are the major cereals of India? Describe.
Answer:
India is an agricultural country and the chief occupations is farming. Cereals are grown is about 70 to 80% of the total agricultural land. rice, wheat millers and maize are the major cereals.

1. Rice:
Rice is the staple food crap of India. It thrives well in hot and humid climate. It requires temperature of 25°C, and rainfall of 100 cm and above and loamy fertile soil. The chief rice producing areas in India are the coastal strips, the Gainga Plains, the Brahmaputra valley, the deltas of Godavari, Krishna and Kaveri.

High yielding varieties, improved planting techniques, irrigation facilities and use of fertilizers have enabled Sadie to increase the production of rice in 1985-86 to 61 million tonnes and average yield per hectare rose to 15.7 Quintilian (2 \(\frac { 1}{2}\) tonnes, increase)

2. Wheat:
Wheat is the second important food crop of India. It requires a cool and moist climate during growing season and a dry climate at the time of ripening. Annual rainfall of 50 to 75 cm and fertile and sandy soil is very suitable – The chief wheat growing areas are Punjab, Haryana IIP, Bihar, MP., Rajashtan Maharashtra, etc.

because all the conditions far wheat production are found here. The use of high yielding varieties,, fertilizers and irrigation facilities helped to raise fire average yield per acre and increased the production of wheat 51 million mark in 1985 – 89. This breakthrough in the production of wheat is called the Green Resolution.

3. Millets:
Sorghum (Jo-war) and bajra are important millets in India. Sorghum (Jowar) In India Jowar has been grown from ancient period. It is used as fodder for cattle and as a food for human beings. In India it is food of the poor, ha foreign countries it is used to prepare starch and glucose.

In northern India it is a Kharif crop but in southern India it is a crop of Kharif and Rabi both. About 87 percent of the total production of Jowar (sorghum) in the country is produced in Madhya Pradesh, Maharashtra, Karnataka and Andhra Pradesh.

4. Bajra:
It is a Kharif crop in northern India. In southern India it is a crop of Rabi and Kharif both. It is used as fodder for cattle. India is the largest producer of Bajra in the world, hi India main Bajra producing states are Rajasthan, Gujarat, Maharashtra, Uttar Pradesh, Tamilnadu, Haryana, Karnataka, Andhra Pradesh and Punjab. Of the total production of Bajra in the country 96 per cent is grown here.

5. Maize:
Maize is the crop of plains and mountainous region. It is used as fodder for cattle and as food to eat. Man uses its different varieties for food products. In foreign countries starch and glucose are prepared from this. In India it is grown in about all states but mainly it is grown in Uttar Pradesh, Punjab, Bihar, Madhya Pradesh, Rajasthan, Gujarat and Karnataka.

Question 2.
What is food security and why is it necessary? Explain.
Answer:
Food security related to the food – related needs of man. In simple words food security means availability of nutritive food to all. Also people should have purchasing power (money) for the arrangement and availability of food at reasonable prices. According to the World Development Report, 1986 “Food security is the availability of adequate food at all time for active and healthy life for all”.

Necessity of Food Security:
In the present Indian situation food security has become very important. Our economy is developing but the population is also increasing rapidly. So to meet the increasing demand, food security has become necessary. The causes responsible for this can be divided into two parts internal causes and external causes.

I. Internal causes:
Internal causes include those which are related to the internal conditions of the country. Following factors are included in these.

1. Basis of life:
India is a country of large population and the birth rate is also very high. Therefore food security is necessary.

2. Dependence on Monsoon:
Majority of crops in India are dependent on monsoon for irrigation, but the monsoon is always uncertain and irregular. The distribution of rains is uneven too. Consequently droughts and famines are common features of our economy. Therefore food security is necessary.

3. Low Productivity:
In India the productivity of food grains regarding per hectare and per labor is low. From this point of view also food security is necessary.

4. Natural Calamities:
Other than the problem of monsoon, flood, insects and pets, cold waves, soil erosion etc. also destroy the food crops in some or the – other part of the country. So the problem of shortage of food crops arises. The famine in Orissa of 1835 in Punjab and Madhya Pradesh in 1877 and in West Bengal in 1943, lacks of people died of hunger. So food security is necessary to face these natural calamities.

5. Continuously Rising Prices:
The prices of food grains are increasing continuously which results in starvation. So food security is necessary to overcome this problem.

6. Progress of the Country:
No country can progress without self-sufficiency of food, and for this food security is necessary.

II. External causes:
External causes include those causes which are related to the relation of the other countries with our country. Following are the external causes:

1. Dependence on Foreign Countries:
Food is the basic need of human beings. So when this requirement is not fulfilled it becomes the primary duty of the Government to fulfill these needs of the people. If there is a shortage of food we have to depend on foreign countries. Whenever there is inadequate supply of food-grains in our country we have to import even if food graii are costly or cheap the quality is good or bad. Thus dependence on foreign countries increases.

2. Decrease in Foreign Exchange:
Whenever we import things like food grains we have to spend our foreign exchange unnecessarily. We can meet our demand for food ourselves but we cannot. This results in shortage of foreign exchange to purchase very important commodities.

3. Foreign Pressure:
Countries which supply food grains to other countries become influential and then they pursue them to follow their policies. These countries dominate those countries which import food grains from them, as a result they lose their freedom to decide their foreign policies.

In the year 1965 – 66 and 1966 – 67 due to the failure of monsoon India had to face a devastating drought and therefore wheat was imported from America. During these frequent emergencies of food grains India experienced that food security is very essential to save people from starvation, to protect self – respect that honor and sovereignty and for the development of the country.

Question 3.
How does the government provide food security to the poor? Explain.
Answer:
Government provides food security to the poor through the following measures:

1. Public Distribution System.
2. Alertness of Government
3. Measures taken in a Critical Situation.

1. Public Distribution System:
By Public Distribution System is meant that system in which different consumer goods are sold in sufficient quantity of fixed prices to the consumers specially to the poor section of society. In this system different goods such as wheat, rice, sugar, imported edible oil, coal and kerosene oil etc. are sold through ration shops or cooperative consumer stores.

The profit rate for these sellers ’are fixed and they have to sell the goods to the ration card holders on fixed price and in fixed quantities. There are three kinds of ration cards B.P.L. card, A.P.L card and Antyodaya cards. BPL cards are meant for those people who are below poverty line, APL cards are meant for those people who are above poverty line and Antyodaya cards are issued to those people who are the poorest of the poor.

The Public Distribution System is regulated by central and state government together. The central government allots food grains and other commodities to state and determines prices also. The state has the right to add transportation charges etc. to the prices fixed by Central Govt.

The transportation, collection, distribution and inspection of these goods processed under this system is done by state government. It is clear from the above table that Public Distribution System plays an important role in providing food-grains to people.

The state government can include those goods also in Public Distribution System which it can purchase if required. In India distribution of food-grains is increasing continuously through Public Distribution System as shown in the given table:

Public Distribution System and Distribution of Food Grains:

2. Alertness of Government:
Our government is very alert in providing food security to the poor. It has developed food security system to make food grains available at reasonable prices to the poor and other people.

3. Measures Taken in a Critical Situation:
The government takes several steps in order to make food grains available to the poor during the period of food crisis arising due to any other reasons or due to natural calamities. Some of these measures are given below:

I. Efforts to Increase Food Grains:
For food security it is important that the production of food grains should be enough in quantity. In this the contribution of green revolution is quite important under Green Revolution mechanization of agriculture, use of high yielding hybrid variety of seeds, fertilizers and insecticides and irrigation facilities were extended.

Also due to promotion of consolidation of land holdings, abolishing of mediators, today the country has become self-sufficient in the field of food grains. The progress of food grains in India can be explained on the basis of following table:

Food Grain Production in India (in crore tonnes):

II. Minimum Support Price:
The prices of agriculture products are very flexible, at the time of harvesting the supply increases, due to which there is enough decrease in price. As prices at this time go down below the fixed limit the producer finds it difficult to get the cost of their products.

1. Therefore government declares minimum support price for agricultural products, under which when market price of food-grains becomes less than its support price, the government starts purchasing food grains on self-declared support price. Due to which farmers get inspired to produce more and more and the government procures food grain for “Buffer Stocks”.

2. The support prices declared by the government during last years are shown here in following table:

Support of Different Food Grains (Rupees per quintal):

III. Buffer Stock:
If the production of food grains is less than to face such crisis of shortage and to distribute them through Public Distribution System, the stock of food grains kept by government is known as Buffer Stock. Buffer stock is the Stock of food grains, namely wheat and rice procured by government through Food Corporation of India (FCI).

The FCI purchases wheat and rice from farmers in states where there is surplus production. The farmers are paid a pronounced price for their crop. This price is called “Minimum Support Price”. The government declares these prices before the sowing season to provide incentives to farmers for raising production of these crops.

These food grains are stored in large granaries. It helps in resolving the problem of shortage of food grains during emergencies. The table of Buffer Stock explains that in the past years in India the stocks has been greater than fixed minimum quantity which is a sign of strong food security of India.

Status of Buffer Stock in India (in million tonnes):

Question 4.
What steps has the government taken to increase food grains?
Answer:

Government provides food security to the poor through the following measures:

1. Public Distribution System.
2. Alertness of Government
3. Measures taken in a Critical Situation.

1. Public Distribution System:
By Public Distribution System is meant that system in which different consumer goods are sold in sufficient quantity of fixed prices to the consumers specially to the poor section of society. In this system different goods such as wheat, rice, sugar, imported edible oil, coal and kerosene oil etc. are sold through ration shops or cooperative consumer stores.

The profit rate for these sellers ’are fixed and they have to sell the goods to the ration card holders on fixed price and in fixed quantities. There are three kinds of ration cards B.P.L. card, A.P.L card and Antyodaya cards. BPL cards are meant for those people who are below poverty line, APL cards are meant for those people who are above poverty line and Antyodaya cards are issued to those people who are the poorest of the poor.

The Public Distribution System is regulated by central and state government together. The central government allots food grains and other commodities to state and determines prices also. The state has the right to add transportation charges etc. to the prices fixed by Central Govt.

The transportation, collection, distribution and inspection of these goods processed under this system is done by state government. It is clear from the above table that Public Distribution System plays an important role in providing food-grains to people.

The state government can include those goods also in Public Distribution System which it can purchase if required. In India distribution of food-grains is increasing continuously through Public Distribution System as shown in the given table:

Public Distribution System and Distribution of Food Grains:

2. Alertness of Government:
Our government is very alert in providing food security to the poor. It has developed food security system to make food grains available at reasonable prices to the poor and other people.

3. Measures Taken in a Critical Situation:
The government takes several steps in order to make food grains available to the poor during the period of food crisis arising due to any other reasons or due to natural calamities. Some of these measures are given below:

I. Efforts to Increase Food Grains:
For food security it is important that the production of food grains should be enough in quantity. In this the contribution of green revolution is quite important under Green Revolution mechanization of agriculture, use of high yielding hybrid variety of seeds, fertilizers and insecticides and irrigation facilities were extended.

Also due to promotion of consolidation of land holdings, abolishing of mediators, today the country has become self-sufficient in the field of food grains. The progress of food grains in India can be explained on the basis of following table:

Food Grain Production in India (in crore tonnes):

II. Minimum Support Price:
The prices of agriculture products are very flexible, at the time of harvesting the supply increases, due to which there is enough decrease in price. As prices at this time go down below the fixed limit the producer finds it difficult to get the cost of their products.

1. Therefore government declares minimum support price for agricultural products, under which when market price of food-grains becomes less than its support price, the government starts purchasing food grains on self-declared support price. Due to which farmers get inspired to produce more and more and the government procures food grain for “Buffer Stocks”.

2. The support prices declared by the government during last years are shown here in following table:

Support of Different Food Grains (Rupees per quintal):

III. Buffer Stock:
If the production of food grains is less than to face such crisis of shortage and to distribute them through Public Distribution System, the stock of food grains kept by government is known as Buffer Stock. Buffer stock is the Stock of food grains, namely wheat and rice procured by government through Food Corporation of India (FCI).

The FCI purchases wheat and rice from farmers in states where there is surplus production. The farmers are paid a pronounced price for their crop. This price is called “Minimum Support Price”. The government declares these prices before the sowing season to provide incentives to farmers for raising production of these crops.

These food grains are stored in large granaries. It helps in resolving the problem of shortage of food grains during emergencies. The table of Buffer Stock explains that in the past years in India the stocks has been greater than fixed minimum quantity which is a sign of strong food security of India.

Status of Buffer Stock in India (in million tonnes):

Question 5.
What is Public Distribution System and what are its main constituents? Describe.
Answer:
By Public Distribution System is meant that system in which different consumer goods are sold in sufficient quantity at fixed prices to the consumers specially to the poor section of society. In this system different goods (wheat, rice, sugar, imported edible oil, coal and kerosene oil etc.) are sold through ration shops or cooperative consumer stores.

The profit rate for these sellers are fixed and they have to sell the goods to the ration card holders on fixed price and in fixed quantities. There are three kinds of ration cards BPL Card, APL Card and Antyodaya Card. BPL cards are for people below poverty line. APL cards are for people above poverty line.

Antyodaya Cards are meant for the poorest of the poor. The Public Distribution System is regulated by central and state government together. The central government allots food grains and other commodities to state and determines prices also. The state has the right to add transportation charge etc.

to the prices fixed by Central Government. The transportation, collection, distribution and inspection of these goods processed under this system is-done by state government. The state government can include those goods also in Public Distribution System which it can purchase if required.

The main constituents of Public Distribution System. Public Distribution System includes fair-price shops. Fair Price Shop for selling cloth, soft coke depot, Super market and Kerosene shops. In India distribution of food grains is increasing continuously through Public Distribution System. The following table will show that Public Distribution system plays an important role in providing food grains to people.

Public Distribution System and Distribution of Food Grains (in million tonnes):

Question 6.
How is the Public Distribution System conducted? Describe.
Answer:

By Public Distribution System is meant that system in which different consumer goods are sold in sufficient quantity at fixed prices to the consumers specially to the poor section of society. In this system different goods (wheat, rice, sugar, imported edible oil, coal and kerosene oil etc.) are sold through ration shops or cooperative consumer stores.

The profit rate for these sellers are fixed and they have to sell the goods to the ration card holders on fixed price and in fixed quantities. There are three kinds of ration cards BPL Card, APL Card and Antyodaya Card. BPL cards are for people below poverty line. APL cards are for people above poverty line.

Antyodaya Cards are meant for the poorest of the poor. The Public Distribution System is regulated by central and state government together. The central government allots food grains and other commodities to state and determines prices also. The state has the right to add transportation charge etc.

to the prices fixed by Central Government. The transportation, collection, distribution and inspection of these goods processed under this system is-done by state government. The state government can include those goods also in Public Distribution System which it can purchase if required.

The main constituents of Public Distribution System. Public Distribution System includes fair-price shops. Fair Price Shop for selling cloth, soft coke depot, Super market and Kerosene shops. In India distribution of food grains is increasing continuously through Public Distribution System. The following table will show that Public Distribution system plays an important role in providing food grains to people.

Public Distribution System and Distribution of Food Grains (in million tonnes):

MP Board Class 9th Social Science Chapter 18 Other Important Questions

Question 1.
Choose the right answer:
(i) Kharif crops are sown in the month of –
(a) October
(b) November
(c) December
(d) January
Answer:
(a) October

Question 2.
Rabi crops are harvested in the month of –
(a) February or March
(b) March or April
(c) April or May
(d) May or June.
Answer:
(b) March or April

Question 3.
In production of food grains India stands third after –
(a) America and Australia
(b) America and China
(c) America and Canada
(d) China and Canada.
Answer:
(b) America and China

Question 4.
Antyodaya ration cards are meant for those who are –
(a) Below poverty line
(b) Above poverty line
(c) Poorest of the poor
(d) None of the above.
Answer:
(c) Poorest of the poor

Question 5.
Target-ted Public Distribution System was introduced in the year –
(a) 1997
(b) 1998
(b) 1999
(d) 2000.
Answer:
(a) 1997

Fill in the blank:

1. Reserve stock of food grains to meet …………… situations.
2. ……………. cards for people below poverty line.
3. ……………. cards for the poorest of the poor.
4. In India two kinds of wheat are grown……………… and mikrani wheat.
5. India is second largest producer of ………………… in the world.

Answer:

1. Emergency
2. BPL
3. Antyodaya
4. Walgair wheat
5. Rice.

MP Board Class 9th Social Science Chapter 18 Very Short Answer Type Questions

Question 1.
Name the main requirements of life?
Answer:
Food, clothing and housing.

Question 2.
How does the World Development Report 1986? Define Food security.
Answer:
According to the World. Development Report 1986, “Food security is the availability of adequate food at all time for active and healthy life for all”.

Question 3.
On what three factors does food security depend?
Answer:
Food security depends on three factors Public Distribution System, alertness of government and the measures taken in a situation of crisis.

Question 4.
Name the major cereals?
Answer:
The major Cereals are rice, wheat, mullets and maize.

Question 5.
Name some rabi crops?
Answer:
Some rabi crops are wheat, oat, gram, etc.

Question 6.
Name rice producing states of India?
Answer:
Rice producing states of India are West Bengal, Uttar Pradesh, Andhra Pradesh, Chhattisgarh, Madhya Pradesh, Bihar, Tamil Nadu, Orissa, Punjab amd Assam.

Question 7.
When was Public Distribution System amended?
Answer:
Public Distribution System was amended in January 1992.

Question 8.
What are the parts of Public Distribution System?
Answer:
Public Distribution System includes fair price shops, soft coke depot, super market and kerosene shops.

Question 9.
Name two crops used as fodder for cattle?
Answer:
Jowar and Bajra.

Question 10.
What type of economy have we?
Answer:
We have developing economy.

MP Board Class 9th Social Science Chapter 18 Short Answer Type Questions

Question 1.
What does food security generally imply? Mention the stages of food security?
Answer:
The food security generally implies that the whole population at all time, should have access to minimum quantity of cereals, but due to continuous changes in a developing country there can be following stages of food security:

• Availability of adequate food grains.
• Availability of food grains and pulses in adequate quantity.
• Availability of milk and milk products along with food grains and pulses.
• Availability of food-grains, pulses, milk and milk products, vegetables, fruits etc.

Question 2.
Describe the chief crops of India and their cultivation regions.
Answer:
The chief crops of India paddy (rice), wheat, bajra, jowar, maize, gram and pulses. Their cultivation regions

1. Paddy (rice):
West Bengal, Uttar Pradesh, Andhra Pradesh, Chhattisgarh, Madhya Pradesh, Bihar, Tamilnadu, Orissa, Punjab, Assam.

2. Wheat:
Uttar Pradesh, Punjab, Haryana, Madhya Pradesh, Bihar, Rajasthan, Maharashtra, West Bengal, Uttarakhand, Gujarat.

3. Bajra:
Rajasthan, Maharashtra, Tamilnadu, Punjab, Uttar Pradesh, Gujarat.

4. Jowar:
Madhya Pradesh, Maharashtra, Karnataka, Andhra Pradesh.

5. Maize:
Uttar Pradesh, Karnataka, Andhra Pradesh, Madhya Pradesh, Punjab, Bihar, Rajasthan.

6. Gram & Pulses:
Madhya Pradesh, Uttar Pradesh, Maharashtra, Punjab, Karnataka.

Question 3.
What do you understand by ‘Fair Price Shop’? What are its benefits?
Answer:
Sugar, cereals, kerosene etc. are distributed to the ration card holders through ration shops which are also known as ‘Fair Price Shops’. Any family with a ration card can buy stipulated amount of these items, every month from nearby ration shop. At these shops all the items are sold at a price lower than the market price. Today there are about 4.6 lakh ration shops in the country.

Question 4.
Write a short note on Buffer Stock?
Answer:
If the production of food grains is less than to face such crises of shortage and to distribute them through Public Distribution System, the stock of food grains kept by government is known as Buffer Stock. Buffer Stock is the stock of food grains namely wheat and rice procured by government through Food Corporation of India (FCI).

The FCI purchases wheat and rice from farmers in states where there is surplus production. These food grains are stored in large granaries. It helps in resolving the problem of shortage of food grains during emergencies.

MP Board Class 9th Social Science Solutions Chapter 18 Food Security in India

Question 1.
Describe coarse crops.
Answer:
Sorghum (Jowar), bajra and maize are coarse crops.

1. Sorghum (Jowar):
Jowar is used as fodder for cattle and as food for human beings. In India it is food of the poor and in foreign countries it is used to prepare starch and glucose. About 87% of the total production of jowar in the country is produced in Madhya Pradesh, Maharashtra, Karnataka and Andhra Pradesh.

2. Bajra:
It is used as fodder for cattle. India is the largest producer of bajrai in the world. The main bajra producing states of India are Rajasthan, Gujarat, Maharashtra, Uttar Pradesh, Tamilnadu, Haryana etc. Of the total production of bajra in the country 96% is grown here.

3. Maize:
It is used as food and fodder. In India it is grown in about all states but mainly it is grown in Uttar Pradesh, Punjab, Bihar, Madhya Pradesh, Rajasthan, Gujarat and Karnataka.

Question 2.
Why has food security become necessary in India? What are the causes responsible for this?
Answer:

Food security related to the food – related needs of man. In simple words food security means availability of nutritive food to all. Also people should have purchasing power (money) for the arrangement and availability of food at reasonable prices. According to the World Development Report, 1986 “Food security is the availability of adequate food at all time for active and healthy life for all”.

Necessity of Food Security:
In the present Indian situation food security has become very important. Our economy is developing but the population is also increasing rapidly. So to meet the increasing demand, food security has become necessary. The causes responsible for this can be divided into two parts internal causes and external causes.

I. Internal causes:
Internal causes include those which are related to the internal conditions of the country. Following factors are included in these.

1. Basis of life:
India is a country of large population and the birth rate is also very high. Therefore food security is necessary.

2. Dependence on Monsoon:
Majority of crops in India are dependent on monsoon for irrigation, but the monsoon is always uncertain and irregular. The distribution of rains is uneven too. Consequently droughts and famines are common features of our economy. Therefore food security is necessary.

3. Low Productivity:
In India the productivity of food grains regarding per hectare and per labor is low. From this point of view also food security is necessary.

4. Natural Calamities:
Other than the problem of monsoon, flood, insects and pets, cold waves, soil erosion etc. also destroy the food crops in some or the – other part of the country. So the problem of shortage of food crops arises. The famine in Orissa of 1835 in Punjab and Madhya Pradesh in 1877 and in West Bengal in 1943, lacks of people died of hunger. So food security is necessary to face these natural calamities.

5. Continuously Rising Prices:
The prices of food grains are increasing continuously which results in starvation. So food security is necessary to overcome this problem.

6. Progress of the Country:
No country can progress without self-sufficiency of food, and for this food security is necessary.

II. External causes:
External causes include those causes which are related to the relation of the other countries with our country. Following are the external causes:

1. Dependence on Foreign Countries:
Food is the basic need of human beings. So when this requirement is not fulfilled it becomes the primary duty of the Government to fulfill these needs of the people. If there is a shortage of food we have to depend on foreign countries. Whenever there is inadequate supply of food-grains in our country we have to import even if food graii are costly or cheap the quality is good or bad. Thus dependence on foreign countries increases.

2. Decrease in Foreign Exchange:
Whenever we import things like food grains we have to spend our foreign exchange unnecessarily. We can meet our demand for food ourselves but we cannot. This results in shortage of foreign exchange to purchase very important commodities.

3. Foreign Pressure:
Countries which supply food grains to other countries become influential and then they pursue them to follow their policies. These countries dominate those countries which import food grains from them, as a result they lose their freedom to decide their foreign policies.

In the year 1965 – 66 and 1966 – 67 due to the failure of monsoon India had to face a devastating drought and therefore wheat was imported from America. During these frequent emergencies of food grains India experienced that food security is very essential to save people from starvation, to protect self – respect that honor and sovereignty and for the development of the country.

MP Board Class 9th Social Science Solutions

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.1 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.1

MP Board Class 12th Physics Important Questions Chapter 13 Nuclei

Nuclei Important Questions 

Nuclei Objective Type Questions

Question 1.
Choose the correct answer of the following:

Half-life formula is the time required for the amount of something to fall to half its initial value.

Question 1.
The relation between half life time Tm and decay constant of a radioactive material is:

Answer:

Question 2.
The decayed part of a radioactive sample in two half lives will be :
(a) One fourth
(b) Half
(c) Three fourth
(d) Full part
Answer:
(c) Three fourth

Question 3.
The α – particle is a :
(a) Hydrogen nucleus
(b) Deuterium nucleus
(c) Helium nucleus
(d) Tritium.
Answer:
(c) Helium nucleus

Question 4.
The value of A and Z in the nuclear reaction \(_{ 238 }^{ 92 }{ U }\) → \(_{ A }^{ Z }{ Th }\) + \(_{ 4 }^{ 2}{ He }\) will be :
(a) 234,94
(b) 234,90
(c) 238,94
(d) 238,90.
Answer:
(b) 234,90

Question 5.
The source of energy in the sun is :
(a) Nuclear fission
(b) Nuclear fusion
(c) Chemical reaction
(d) Photo electric reaction.
Answer:
(b) Nuclear fusion

Question 6.
For the nuclear fusion which of the following is suitable :
(a) Heavy nucleus
(b) Lighter nucleus
(c) Atom bomb
(d) Radio active decay.
Answer:
(b) Lighter nucleus

Question 7.
In the symbol of nucleus \(_{ Z }^{ A }{ X }\) there are :
(a) Z – neutrons, (A – Z) protons
(b) Z – protons, (A – Z) neutrons
(c) Z – protons, A neutrons
(d) A protons. Z neutrons.
Answer:
(b) Z – protons, (A – Z) neutrons

Question 8.
In the gamma ray emission of a nucleus :
(a) Only proton number varies
(b) Number of neutrons and protons both varies
(c) There is no change in number of protons and neutrons
(d) Only number of neutrons varies.
Answer:
(c) There is no change in number of protons and neutrons

Question 2.
Fill in the blanks :

1. In the controlled chain reaction the fast moving neutrons are slowed down by using moderator. These neutrons are called …………..
2. The Einstein’s mass energy equivalence relation is …………..
3. The unstable nucleus obtained by artificial radioactivity is called …………..
4. The mass of neutrons in a nucleus is nearly equal to mass of …………..
5. In the process of ………….. the lighter nuclei together makes a heavy nucleus.
6. In nuclear reactor the heavy water is used as …………..
7. The SI unit of radio activity is …………..
8. In the range mass number A = 30 to A = 170 the value of ………….. is nearly constant.

Answer:

1. Thermal neutrons
2. E = me²
3. Radio active isotopes
4. Protons
5. Fusion
6. Moderator
7. Becquerel (Bq)
8. Per nucleon binding energy.

Question 3.
Match the columns :

Answer:

1. (d)
2. (e)
3. (a)
4. (c)
5. (b)

Question 4.
Write the answer in one word/sentence :

1. Write the equation showing the decay of free neutron.
2. The energy distribution of P – rays is continuous. Why?
3. In the α and β particles, which one has more ionization power?
4. The ionization power of α – particles is high. Why?
5. The penetrating power of γ(gamma) – rays is very high. Why?
6. Write an equation showing nuclear fusion.
7. Which part of the electromagnetic spectrum has highest penetrating power?
8. What is order of size of nucleus?
9. Does the density of all nuclei remain same? If yes then write its order.
10. How is nuclear radius related to atomic mass?

Answer:

1. ₀n¹ → ₁H¹ + _-1B⁰ + \(\overline { ν }\)
2. Because during the β – decay anti – neutrino is emitted
3. α – particles
4. Because α – particles have large mass and large size and their velocity is low
5. Because their velocity is very high and they do not have any charge,
6. ₁H² + ₁H² → ₂He⁴ + Q
7. γ – rays
8. 10^-15 m
9. Yes, 10¹⁷ per cubic metre
10. R = R₀ A^1/3

Nuclei Very Short Answer Type Questions

Question 1.
What will be the ratio of radii of two nucleus of mass number A₁ and A₂?
Solution:
From the formula

Question 2.
What will be the energy equivalent to 10 milligram?
Solution:
E = mc²
= 10 x 10^-6 x (3 x 10⁸)²
= 9 x 10⁹ joule.

Question 3.
What are thermal neutrons?
Answer:
In the process of controlled nuclear fission, the fast moving neutrons are slowed down with the help of moderation. These slow stray neutrons are called as thermal neutrons.

Question 4.
Does the ratio of neutron and proton after emission of α – particle in a nucleus increase, decrease or remain constant?
Answer:
The ratio can increase, decrease or can remain same. It depends upon the nature of nucleus.

Question 5.
Nuclear fusion is not possible in laboratory. Why?
Answer:
For nuclear fusion very high temperature (≈ 10⁷K) and very high pressure is required which is not possible in laboratory.

Question 6.
Is a free neutron a stable particle?
Answer:
No, free neutron is decayed in a proton, an electron and an antineutrino (\(\overline { ν }\)).
₀n¹ → ₁P¹ + _-1e⁰ + \(\overline { ν }\)

Question 7.
In heavy nucleus the number of neutrons is more than number of protons. Explain why?
Answer:
In a nucleus there is a coulomb’s repulsive force between protons in addition to a strong attraction nuclear force. On the other hand in case of neutrons, within the nucleus, there is only a short range attractive nuclear force. For heavy nuclei to be stable the repulsive force must be less. This is possible only if the number of neutrons is more than number of protons.

Question 8.
If a radioactive substance that is capable to emit a, p and y rays is kept on a piece of paper, then which rays has the maximum possibility to be stoped?
Answer:
As the mass of α – particles is more and its penetrating power is least, so it will be stopped.

Question 9.
Why is heavy water used as moderator in a nuclear reactor?
Answer:
Heavy water contains protons (of mass nearly equal to the mass of neutrons). Fast moving neutrons undergo elastic collision with these slow moving neutrons and thus get slowed down. Hence heavy water is used as moderator.

Question 10.
An atom expressed by _YX^A emits n α – particles. How many proton will remain in it? What will be the new atomic number of new atom?
Answer:
In an α – particles there are 2 protons. If n α – particles are emitted then 2n proton will be reduced.
∴ No. of remaining protons = Z – 2n
Atomic number = Y – 2n.

Question 11.
Penetrating power of β – particles is more than that of α – particles but its penetrating power is less. Why?
Answer:
β – particles have small size and high velocity so its penetrating power is more than α – particles. Due to small size and high velocity the probability of collision with gas molecules decreases and hence ionization power becomes less.

Question 12.
The nuclear fusion could not be used as an experimental and controlled source of energy till now. Why?
Answer:
The nuclear fusion takes place at very high temperature. Due to such very high temperature it is not possible to control it.

Question 13.
Why is a neutron preferred as a bombarding particle in nuclear fission?
Answer:
Neutron is a neutral particle i.e., it has no charge. So it is neither repelled nor attracted by the nucleus and hence can penetrate deep into the nucleus to cause a nuclear reaction.

Question 14.
β – rays are emitted from nucleus. They are made of fast moving electron though there are no electrons in the nucleus. Why it is so?
Answer:
Nucleus has no electron in it. Actually due to decay of neutron into proton and electron, the β – rays (particle) are produced.

Question 15.
Justify that gamma rays have more penetrating power and α – particles have more ionizing power.
Answer:
Gamma rays are electromagnetic waves. Its speed is equal to the speed of light. So they can penetrate deeply into the matter i.e., their penetrating power is high, α – particles has high mass and posses greater kinetic energy. So when they collide with any atom, it transfer its energy to orbital electrons and they are ejected. Hence α – particles has high ionizing power.

Question 16.
The mass number of two nucleus are 1 : 2. What is the ratio of their nuclear density?
Answer:
1 : 1 because nuclear density does not depend on mass number.

Question 17.
What is effect of temperature and pressure or radio activity?
Answer:
No, effect.

Question 18.
Due to the emission of α – particles does the ratio of neutron to proton decrease, increase or remains the same?
Answer:
The ratio \(\frac {n}{p}\) increases, because number of protons decreases by 2.

Nuclei Short Answer Type Questions

Question 1.
Which particles remains inside the nucleus? What will be the number of neutrons in the nucleus _zX^A?
Answer:
The neutrons and protons remains inside the nucleus. In the nucleus _zX^A number of neutrons be A – Z.

Question 2.
Define decay constant and mean life of a radioactive substance write their units.
Answer:
Decay Constant:
The decay constant is defined as the reciprocal of that time interval during which the number of active nuclei in a given sample of a radioactive substance reduces to times of the initial value of number of nuclei.
If T is half life time and X is decay constant then
T = \(\frac {0.6931}{λ}\)
Its unit is “per second”.

Mean life:
The reciprocal of decay constant is called mean life, its unit is ‘second’.

Question 3.
What is radioactivity? Write the names of the rays emitted from radio – active substances.
Answer:
Radioactivity is that property by virtue of which the nucleus of a heavy element disintegrates itself with the emission of radiation without being forced by any external agent to do so.

Following rays are emitted from a radioactive substance :

1. α – rays
2. β – rays
3. γ – rays.

Question 4.
A radioactive nucleus decays as given below :

If the mass number of A₂ is 176 and atomic number is x then what is the atomic number and mass number of A₁ and A? Are they isotopes or isobars?
Answer:
The mass number of A₂ = 176,
atomic number of A₂ = 71.
Due to decay of α – particle mass number decreases by 4 and atomic number decreases by 2.
∴ Mass number of A₁ = 176 + 4 = 180 and
its atomic number = 71 + 2 = 73
Similarly
The mass number of A = 180 + 1 = 181 and
atomic number of A = 73 + 0 = 73
Since, the atomic number of A and A₁ are same so they are isotopes.

Question 5.
Define decay constant of a radioactive substance. Write its relation with half life?
Answer:
The decay constant is defined as the reciprocal of that time interval during which the number of active nuclei in a given sample of a radioactive substance reduces to times of the initial value of number of nuclei.
If T is half life time and X is decay constant then
T = \(\frac {0.6931}{λ}\).

Question 6.
What is nuclear fission? Give an example.
Answer:
When a heavier nucleus is bombarded with neutrons then it get splitted into two lighter nuclei. This process is called as nuclear fission.
A huge amount of energy is released in this process.
₉₂U²³⁵ + ₀n¹ → ₅₆Ba¹⁴¹ + ₃₆kr⁹² + 3 ₀n¹ + 200 MeV.

Question 7.
What is nuclear fusion? Give an example.
Answer:
When two lighter nuclei fused together to form a heavier nucleus then a huge amount of energy is released in this process. This process is called nuclear fusion.
₁H² + ₁H² → ₂He⁴+ 24MeV.

Question 8.
A radioactive element with mass number 218 and atomic number 84 emits a β – particle. What would be the mass number and atomic number after decay?
Answer:
Since, after the emission of β – particle there is no change in the mass number of element but its atomic number is increased by 1.
Therefore the mass number will be 218 and atomic number will be 84 + 1 = 85.

Question 9.
The ratio of radii of two nuclei are 1 : 2. Find out the ratio of their mass number.
Solution:
Radius of nucleus

Question 10.
The nuclear fusion process is difficult as compared to nuclear fission. Why?
Answer:
For nuclear fusion a high temperature (≈ 10⁷ K) is required. It is very difficult to achieve this temperature, so nuclear fusion is difficult.

Question 11.
Write the important properties of nuclear forces.
Answer:

1. Nuclear forces are attraction force
2. These forces does not depend on charge
3. These forces are short range forces
4. They are strong forces
5. The nuclear forces are not central forces.

Question 12.
Write properties of nuclear forces. Prove that the density of nucleus is independent of mass number.
Answer:
Properties of nuclear forces :

1. Nuclear forces are attraction force
2. These forces does not depend on charge
3. These forces are short range forces
4. They are strong forces
5. The nuclear forces are not central forces.

Let the mass number of a nucleus is A and radius is r then volume of nucleus be

Thus, it is clear that material density of a nucleus is independent of its mass number.

Question 13.
Write difference between nuclear fusion and nuclear fission.
Answer:
Difference between nuclear fusion and nuclear fission :
Nuclear fission:

• In this prosses a heavy nucleus splitted into two lighter nuclei.
• This process is possible at normal temperature.
• Per fission released energy is very high (200 MeV)

Nuclear fusion:

• In this process two lighter nuclei fused together to form a heavy nucleus.
• This process is possible at very high temperature.
• Per fusion released energy is comparatively less (24 MeV).

Nuclei Long Answer Type Questions

Question 1.
Define ‘electron volt’ and ‘Atomic mass unit”. Find out the energy equivalent to mass of a proton in joule.
Answer:
Electron volt:
The energy acquired by an electron when it is being accelerated by a potential difference of 1 volt is called ‘electron volt’.
1 electron volt = 1.6 x 10^-19J.

Atomic mass unit:
An atomic mass unit is defined as (\(\frac {1}{12}\)) of the mass of 1 atom of carbon – 12.
1 atomic mass unit = 1.66 x 10^-27kg
(1 a.m.u. or only 1u)
The mass of proton = 1.673 x 10^-27 x (3 x 10⁸ )²
= 15.05 x 10^-11J
= 1.505 x 10^-10J
∴ The energy equivalent mass of 1 proton is 1.505 x 10^-10J

Question 2.
Draw the graph showing the variation of binding energy per nucleon with the mass number. Write important conclusions drawn from it. (CBSE 1994)
Or
Draw the graph showing the variation of binding energy per nucleon with mass number. Highlight the region where the nuclei are more stable. (CBSE 1996)
Or
Draw the graph showing the variation of binding energy per nucleon with the mass number. Highlight the region where the nuclear fusion takes place. (CBSE 1996)
Answer:
From the binding energy curve, it follows that:

1. The average binding energy/ nucleon for nearly all elements is 8 MeV

2. The maximum value of binding energy per nucleon is 8.79 MeV and it is for Fe. This explains the large abundance of Fe in nature

3. Since for nuclei of intermediate mass numbers, the value of the binding energy per nucleon is around the maximum value, these nuclei are the most stable

4. The binding energy/ nucleon decreases for nuclei of mass number more than 56 and its least value is 7.6 MeV.

5. Below mass number 28, there are peaks in the curve corresponding to those nuclei whose mass numbers are multiples of four. These nuclei contain an equal number of protons and neutrons. Though peaks occur corresponding to \(_{ 4 }^{ 2 }{ He}\), \(_{ 8 }^{ 4 }{ Be }\), \(_{ 12 }^{ 6 }{ C }\), \(_{ 16 }^{ 8 }{ O }\) and \(_{ 20}^{ 10 }{ Ne }\), we have shown only the first peak corresponding to \(_{ 4 }^{ 2 }{ He }\) Obviously, these nuclei have more binding energy per nucleon than their neighbours. This explains the stability of an α – particle (\(_{ 4 }^{ 2 }{ U }\))

6. The smallest value of the binding energy per nucleon is in the case of a deuteron (\(_{ 2 }^{ 1 }{ U }\)) and its value (2.2 MeV)/2 = 1.1 MeV. This is confirmed by the fact that a photon whose energy is 2.2 MeV or more, can split a deuteron into a free neutron and a free proton. This phenomenon is called the photodisintegration of a deuteron.

Question 3.
Calculate binding energy per nucleon of ₂₆Fe⁵⁶.
Given : m(\(_{ 26 }^{ 56 }{ Fe }\)) = 55.934939 a.m.u., m (proton ) = 1.007825 a.m.u., m (neutron) = 1.008665 a.m.u. (CBSE 1993,95, 2000 Supp.)
Solution:
Number of protons in = \(_{ 26 }^{ 56 }{ He }\) is 26 and number of neutrons is 56 – 26 = 30.
∴ Mass of 26 protons = 1 0077825 x 26
= 26.202345 a.m.u.
Mass of 30 neutrons = 1.008665 x 30
= 30.25995 a.m.u.
Total mass of nucleons = 26.202345 + 30.25995
= 56.4623 a.m.u.
Mass of ₂₆Fe⁵⁶ = 55.934939.a.m.u.,
∴ Mass defect ∆m = 56.4623 – 55.934939
= 0.527361 a. m. u.
∴ Binding energy of ₂₆Fe⁵⁶ = 0.527361 x 931
= 490.973091 MeV
Binding energy per nucleon = \(\frac {490.973091}{56}\)
= 8.7674 MeV.

Question 4.
Define the term half-life period and decay constant of a radioactive sample. Derive a relation between these terms.
Answer:
The half – life of a radioactive substance is the time it takes half of a given number of radioactive nuclei to decay. (CBSE 1995, 2001)
The SI unit of Tm is second (s).
Example:
Half life of radium is 1600 years. It means that if 1 gm of radium is taken then after 1600 years \(\frac {1}{2}\) gm of radium will decay.
Let us assume that at time t = 0, the number of radioactive nuclei present is N₀ and after time t, the number of radioactive nucleus remaining is N, then according to law of radioactive decay,
N = N₀e^-λt
Where, λ is decay constant.
If the half – life of radioactive sample is T_1/2, then at t = T_1/2
N = \(\frac { { N }_{ 0 } }{ 2 }\)
Substituting in equation (1), we get ,

From equation (2), it is clear that half life of a radioactive sample is inversely proportional to decay constant.

Question 5.
State law of radioactive decay and derive the following formula :
N = N₀ e^-λt, where symbols have their usual meaning.
Answer:
According to the radioactive decay law :
The rate at which a particular decay process occurs in a radioactive sample is proportional to the number of radioactive nuclei present (that is, those nuclei that have not yet decayed).
Let N₀ = Total number of radioactive nuclei present originally at time t = 0,
(t = 0 refers to the time when the radioactive element is freshly separated from its by products), and N = Total number of radioactive nuclei present at any time t.

According to the radioactive decay law,
– \(\frac {dN}{dt}\) ∝ N
(- ve sign indicated that the number of radioactive nuclei decreases with time).
Thus, – \(\frac {dN}{dt}\) = λN
Where A is a constant of proportionality and is called the disintegration constant or decay constant.
Equation (1) can be written as
\(\frac {dN}{dt}\) = – λdt
When t = 0, N = N₀ and when t = t,N = N.
Integrating equation (2) within proper limits,

From equation (3) it is clear that (i) the radioactive decay is an exponential process i.e., the decay is fast initially and later on it becomes slow.

Nuclei Numerical Questions

Question 1.
An imaginary fission reaction is given below :
\(_{ 92 }^{ 236 }{ X }\) → \(_{ a}^{ 141 }{ Y }\) + \(_{ 36 }^{ b }{ Z }\) + 3\(_{ 0 }^{ 1 }{ n }\)
Find out value of a and b.
Solution:
a + 36 + 0 = 92
a = 56.
and 141 + b + 3 = 236
b = 92.

Question 2.
The half life of a radioactive substance is 30 days. What will be the number of radioactive atoms after 90 days?
Solution:
Given : T = 30 days, t = 90 days
n = \(\frac {t}{T}\) = \(\frac {90}{30}\) = 3
Formula : N = N₀
= N₀(\(\frac {1}{2}\))ⁿ
= N₀(\(\frac {1}{2}\))³
= \(\frac { { N }_{ 0 } }{ 8 }\)
Thus, the number of radioactive atoms be \(\frac {1}{8}\) times of their initial value.

Question 3.
How many fission per second takes place if its half life for α – decay is 1.42 x 10¹⁷sec?
Solution:

MP Board Class 12th Physics Important Questions

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 1 Relations and Functions Ex 1.2 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 1 Relations and Functions Ex 1.2