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Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Ex 2.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 2 Matrices Ex 2.3

Question 1.
Solve the following equations by the inversion method.
(i) x + 2y = 2, 2x + 3y = 3
Solution:
The given equations can be written in the matrix form as :
\(\left[\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\)
This is of the form AX = B, where

∴ A^-1 = \(\left[\begin{array}{rr}
-3 & 2 \\
2 & -1
\end{array}\right]\)
Now, premultiply AX = B by A^-1, we get,
A^-1(AX) = A^-1B
∴ (A^-1A)X = A^-1B
∴ IX = A^-1B
∴ X = \(=\left[\begin{array}{rr}
-3 & 2 \\
2 & -1
\end{array}\right]\left[\begin{array}{l}
2 \\
3
\end{array}\right]\)
∴ \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(=\left[\begin{array}{r}
-6+6 \\
4-3
\end{array}\right]\) = \(=\left[\begin{array}{l}
0 \\
1
\end{array}\right]\)
By equality of matrices,
x = 0, y = 1 is the required solution.

(ii) x + y = 4, 2x – y = 5
Solution:
x + y = 4, 2x – y = 5
The given equations can be written in the matrix form as:
\(\left[\begin{array}{cc}
1 & 1 \\
2 & -1
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
This is of the form AX = B ⇒ X ⇒ A^-1B
A = \(\left[\begin{array}{cc}
1 & 1 \\
2 & -1
\end{array}\right]\)
|A| = -1 – 2 = -3 ≠ 0

By equality of matrices.
x = 3, y = 1

(iii) 2x + 6y = 8, x + 3y = 5
Solution:
The given equations can be written in the matrix form as :
\(\left[\begin{array}{ll}
2 & 6 \\
1 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
8 \\
5
\end{array}\right]\)
This is of the form AX = B, where
A = \(\left[\begin{array}{ll}
2 & 6 \\
1 & 3
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) and B = \(\left[\begin{array}{l}
8 \\
5
\end{array}\right]\)
Let us find A^-1.
|A| = \(\left|\begin{array}{ll}
2 & 6 \\
1 & 3
\end{array}\right|\) = 6 – 6 = 0
∴ A^-1 does not exist.
Hence, x and y do not exist.

Question 2.
Solve the following equations by reduction method.
(i) 2x + y = 5, 3x + 5y = -3
Solution:
The given equations can be written in the matrix form as :

By equality of matrices,
2x + y = 5 …(1)
7y = -21 …(2)
From (2), y = -3
Substituting y = -3 in (1), we get,
2x – 3 = 5
∴ 2x = 8 ∴ x = 4
Hence, x = 4, y = -3 is the required solution.

(ii) x + 3y = 2, 3x + 5y = 4.
Solution:
The given equations can be written in the matrix form as :
\(\left[\begin{array}{ll}
1 & 3 \\
3 & 5
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
2 \\
4
\end{array}\right]\)
By R₂ – 3R₁, we get
\(\left[\begin{array}{rr}
1 & 3 \\
0 & -4
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left(\begin{array}{r}
2 \\
-2
\end{array}\right)\)
∴ \(\left[\begin{array}{l}
x+3 \\
0-4 y
\end{array}\right]\) = \(\left[\begin{array}{r}
2 \\
-2
\end{array}\right]\)
By equality of matrices,
x + 3y = 2 …(1)
-4y = -2
From (2), y = \(\frac{1}{2}\)
Substituting y = \(\frac{1}{2}\) in (1), we get,
x + \(\frac{3}{2}\) = 2
∴ x = 2 – \(\frac{3}{2}=\frac{1}{2}\)
Hence, x = \(\frac{1}{2}\), y = \(\frac{1}{2}\) is the required solution.

(iii) 3x – y = 1, 4x + y = 6
Solution:
The given equations can be written in the matrix form as :

By equality of matrices,
12x – 4y = 4 … (1)
7y = 14 … (2)
From (2), y = 2
Substituting y = 2 in (1), we get,
12x – 8 = 4
∴ 12x = 12 ∴ x = 1
Hence, x = 1, y = 2 is the required solution.

(iv) 5x + 2y = 4, 7x + 3y = 5
Solution:
5x + 2y = 4 ………..(1)
7x + 3y = 5 …………(2)
Multiplying Eq. (1) with 7 and Eq. (2) with 5

Put y = -3 into Eq. (1)
5x + 2y = 4
5x + 2(-3) = 4
5x – 6 = 4
5x = 4 + 6
5x = 10
x = \(\frac{10}{5}\)
x = 2
Hence, x = 2, y = -3 is the required solution.

Question 3.
The cost of 4 pencils, 3 pens and 2 erasers is ₹ 60. The cost of 2 pencils, 4 pens and 6 erasers is ₹ 90, whereas the cost of 6 pencils, 2 pens and 3 erasers is ₹ 70. Find the cost of each item by using matrices.
Solution:
Let the cost of 1 pencil, 1 pen and 1 eraser be ₹ x, ₹ y and ₹ z respectively.
Then, from the given conditions,
4x + 3y + 2z = 60
2x + 4y + 6z = 90, i.e., x + 2y + 3z = 45
6x + 2y + 3z = 70
These equations can be written in the matrix form as :

By equality of matrices,
x + 2y + 3z = 45 …….(1)
– 5y – 10z = – 120 …….(2)
5z = 40
From (3), z = 8
Substituting z = 8 in (2), we get,
– 5y – 80 = -120
∴ – 5y = -40 ∴ y = 8
Substituting y = 8, z = 8 in (1), we get,
x + 16 + 24 = 45
∴ x + 40 = 45 ∴ x = 5
∴ x = 5, y = 8, z = 8
Hence, the cost is ₹ 5 for a pencil, ₹ 8 for a pen and ₹ 8 for an eraser.

Question 4.
If three numbers are added, their sum is 2. If 2 times the second number is subtracted from the sum of first and third numbers, we get 8 and if three times the first number is added to the sum of second and third numbers, we get 4. Find the numbers using matrices.
Solution:
Let the three numbers be x, y and z. According to the given conditions,
x + y + z = 2
x + z – 2y = 8, i.e., x – 2y + 2 = 8
and y + z + 3x = 4, i.e., 3x + y + z = 4
Hence, the system of linear equations is
x + y + z = 2
x – 2y + z = 8
3x + y + z = 4
These equations can be written in the matrix form as :

By equality of matrices,
x + y + z = 2 ……(1)
-3y = 6 ……(2)
– 2y – 2z = -2 ……..(3)
From (2), y = -2
Substituting y = -2 in (3), we get,
-2(-2) – 2z = -2
∴ -2z = -6 ∴ z = 3
Substituting y = -2, z = 3 in (1), we get,
x – 2 + 3 = 2 ∴ x = 1
Hence, the required numbers are 1, -2 and 3.

Question 5.
The total cost of 3 T.V. sets and 2 V.C.R.s is ₹ 35000. The shop-keeper wants profit of ₹ 1000 per television and ₹ 500 per V.C.R. He can sell 2 T. V. sets and 1 V.C.R. and get the total revenue as ₹ 21,500. Find the cost price and the selling price of a T.V. sets and a V.C.R.
Solution:
Let the cost of each T.V. set be ₹ x and each V.C.R. be ₹ y. Then the total cost of 3 T.V. sets and 2 V.C.R.’s is ₹ (3x + 2y) which is given to be ₹ 35,000.
∴ 3x + 2y = 35000
The shopkeeper wants profit of ₹ 1000 per T.V. set and of ₹ 500 per V.C.R.
∴ the selling price of each T.V. set is ₹ (x + 1000) and of each V.C.R. is ₹ (y + 500).
∴ selling price of 2 T.V. set and 1 V.C.R. is
₹ [2(x + 1000) + (y + 500)] which is given to be ₹ 21,500.
∴ 2(x + 1000) + (y + 500) = 21500
∴ 2x + 2000 + y + 500 = 21500
∴ 2x + y = 19000
Hence, the system of linear equations is
3x + 2y = 35000
2x + y = 19000
These equations can be written in the matrix form as :

By equality of matrices,
2x + y = 19000 ……….(1)
-x = -3000 ……….(2)
From (2), x = 3000
Substituting x = 3000 in (1), we get,
2(3000) + y = 19000
∴ y = 13000
∴ the cost price of one T.V. set is ₹ 3000 and of one V.C.R. is ₹ 13000 and the selling price of one T.V. set is ₹ 4000 and of one V.C.R. is ₹ 13500.

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 13 Nuclear Chemistry and Radioactivity Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 13 Nuclear Chemistry and Radioactivity

1. Choose correct option.

Question A.
Identify nuclear fusion reaction

Answer:
Among the given options, reactions (i) and (ii) represent nuclear fusion reaction wherein lighter nuclei combine to form a heavy nucleus.

Question B.
The missing particle from the nuclear reaction is

Answer:
(A) \({ }_{15}^{30} \mathrm{P}\)

Question C.
\({ }_{27}^{60} \mathrm{CO}\) decays with half-life of 5.27 years to produce \({ }_{28}^{60} \mathrm{Ni}\). What is the decay constant for such radioactive disintegration ?
a. 0.132 y^-1
b. 0.138
c. 29.6 y
d. 13.8%
Answer:
a. 0.132 y^-1

Question D.
The radioactive isotope used in the treatment of Leukemia is
a. ⁶⁰Co
b. ²²⁶Ra
c. ³²P
d. ¹³¹I
Answer:
c. ³²P

Question E.
The process by which nuclei having low masses are united to form nuclei with large masses is
a. chemical reaction
b. nuclear fission
c. nuclear fusion
d. chain reaction
Answer:
c. nuclear fusion

2. Explain

Question A.
On the basis of even-odd of protons and neutrons, what type of nuclides are most stable ?
Answer:

• Nuclides with even number of protons (Z) and even number of neutrons (N) are most stable.
• These nuclides tend to form proton-proton and neutron-neutron pairs.
• This impart stability to the nucleus.

Question B.
Explain in brief, nuclear fission.
Answer:
i. Nuclear fission: It is a process which involves splitting of the heavy nucleus of an atom into two nearly equal fragments accompanied by release of the large amount of energy.
e.g. Nuclear fission of ²³⁵U

ii. When a uranium nucleus absorbs neutron, it breaks into two lighter fragments and releases energy (heat), more neutrons, and other radiation. This can be given as,

iii. Characteristics of nuclear fission reactions:

• The mass of the fission products is less than the parent nucleus. A large amount of energy corresponding to the mass loss is released in each fission.
• When one uranium 235 nucleus undergoes fission, three neutrons are emitted, which subsequently disintegrate three more uranium nuclei and thereby produce nine neutrons. Such a chain continues by itself.
• In a very short time enormous amount of energy is liberated, which can be utilized for destructive or peaceful purposes.
• Energy released per fission is approximately 200 MeV.

Note:

• Each fission may lead to different products.
• There is no unique way for fission of ²³⁵U that produces Ba and Kr. There are 400 ways for fission of ²³⁵U leading to 800 fission products.
• Many of these fission products are radioactive which undergo spontaneous disintegrations giving rise to new elements in the periodic table.

Question C.
The nuclides with odd number of both protons and neutrons are the least stable. Why ?
Answer:

• The nuclides with odd number of both protons and neutrons are the least stable because, odd number of protons and neutrons results in the presence of two unpaired nucleons.
• These unpaired nucleons result in instability. Hence, such nuclides are the least stable.

Question D.
Referring the stabilty belt of stable nuclides, which nuclides are β^– and β⁺ emitters ? Why ?
Answer:

• Beta decay occurs when an unstable nucleus emits a beta particle and energy. A beta particle is either an electron or a positron. An electron is a negatively charged particle, and a positron is a positively charged electron (or anti-electron).
• When the beta particle is an electron, the decay is called beta-minus (β^–) decay. In beta-minus decay, a neutron breaks down to a proton and an electron, and the electron is emitted from the nucleus.
• When the beta particle is a positron, the decay is called beta-plus (β⁺) decay. In beta-plus decay, a proton breaks down to a neutron and a positron, and the positron is emitted from the nucleus.
• Thus, beta-minus decay occurs when a nucleus has too many neutrons relative to protons (i.e., N/Z > 1) and beta-plus decay occurs when a nucleus has too few neutrons relative to protons (i.e., N/Z < 1).
• By referring the stability belt of stable nuclides, nuclides with N/Z > 1 are to the left of the stability zone. Such nuclides are beta-minus emitters as they become stable when a neutron converts to a proton.
• Nuclides with N/Z < 1 are to the right of the stability zone. Such nuclides are beta-plus emitters as they become stable when a proton converts to a neutron.

Question E.
Explain with an example each nuclear transmutation and artifiacial radioactivity. What is the difference between them ?
Answer:
i. Nuclear transmutation: It involves transformation of a stable nucleus into another nucleus takes place which can be either stable or unstable.
ii. Artificial (induced) radioactivity: It is nuclear transmutation where the product nucleus is radioactive. The product nucleus decays spontaneously with emission of radiation and particles.

Step-I can be considered as nuclear transmutation as it produces a new nuclide \({ }_{7}^{13} \mathrm{~N}\).
However, the new nuclide is unstable (radioactive). Hence, step-I involves artificial (induced) radioactivity. Thus, in artificial transmutation, a stable element is collided with high speed particles to form another radioactive element.

Question F.
What is binding energy per nucleon ? Explain with the help of diagram how binding energy per nucleon affects nuclear stability ?
Answer:

i. Binding energy per nucleon (\(\overline{\mathrm{B}}\)), for nucleus containing (A) nucleons with binding energy (B.E.) is given as,
\(\overline{\mathrm{B}}\) = B.E./A
ii. Mean binding energy per nucleon (\(\overline{\mathrm{B}}\)) for the most stable isotopes as a function of mass number is shown above. This plot leads to the following inferences:
a. Light nuclides: (A < 30)
The peaks with A values in multiples of 4. For example, \({ }_{2}^{4} \mathrm{He},{ }_{6}^{12} \mathrm{C},{ }_{8}^{16} \mathrm{O}\) are more stable.
b. Medium mass nuclides: (30 < A < 90)
\(\overline{\mathrm{B}}\) increases typically from 8 MeV for A = 16 to nearly 8.3 MeV for A between 28 and 32 and it remains nearly constant 8.5 MeV beyond this and shows a broad maximum. The nuclides falling on the maximum are most stable which turns possess high values. ⁵⁶Fe with \(\overline{\mathrm{B}}\) value of 8.79 MeV is the most stable.
c. Heavy nuclides (A > 90)
\(\overline{\mathrm{B}}\) decreases from maximum 8.79 MeV to 7.7 MeV for A ≅ 210, 209Bi is the stable nuclide. Beyond this, all nuclides are radioactive (α-emitters).

Question G.
Explain with example α-decay.
Answer:
i. The emission of α-particle from the nuclei of an radioelement is called α-decay.
ii. The charge on an α-particle is +2 with a mass of 4 u.
It is identical with helium nucleus and hence an α-particle is designated as \({ }_{2}^{4} \mathrm{He}\).
iii. In the α-decay process, the parent nucleus \({ }_{\mathrm{z}}^{\mathrm{A}} \mathrm{X}\) emits an α-particle and produces daughter nucleus Y. The parent nucleus thus loses two protons (charge +2) and two neutrons. The total mass lost is 4 u. The daughter nucleus will therefore, have mass 4 units less and charge 2 units less than its parent.
iv. General equation for α-decay process can be given as:

In α-decay process of radium, radon (daughter nuclei) is formed with loses of two protons (charge +2) and two neutrons. The total mass lost is 4 u.
Thus, radon has a mass of 4 units less and charge 2 units less than its parent radium.

Question H.
Energy produced in nuclear fusion is much larger than that produced in nuclear fission. Why is it difficult to use fusion to produce energy ?
Answer:

• Nuclear fusion involves the fusion of lighter nuclei to form a heavy nucleus which is accompanied by an enormous amount of energy (heat).
• Fusion reaction requires extremely high temperature typically of the order of 108 K.

Question I.
How does N/Z ratio affect the nuclear stability ? Explain with a suitable diagram.
Answer:

• When the graph of number of neutrons (N) against protons (Z) is drawn, and all the stable isotopes are plotted on it, there is quite a clear correlation between N and Z. This graph is shown in the adjacent figure.
• A large number of elements have several stable isotopes and hence, the curve appears as a belt or zone called stability zone. All stable nuclides fall with this zone and the nuclei that are to the left or to the right of the stability zone are unstable and exhibit radioactivity. Below the belt, a straight line which represents the ratio N/Z to be nearly unity (i.e., N = Z) is shown.
• For nuclei lighter than \({ }_{20}^{40} \mathrm{Ca}\), the straight line (N = Z) passes through the belt. The lighter nuclides are therefore stable (N/Z being 1).
• The N/Z ratio for the stable nuclides heavier than calcium gives a curved appearance to the belt with gradual increase of N/Z (> 1). The heavier nuclides therefore, need more number of neutrons than protons to attain stability. The heavier nuclides with increasing number of protons render large coulombic repulsions. With increased number of neutrons, the protons within the nuclei get more separated, which renders them stable.
• Thus, nuclear stability is linked to the number of nucleons (neutrons and protons). In general, the lighter stable nuclei have equal numbers of protons and neutrons while heavier stable nuclei have increasingly more neutrons than protons.

[Note: Atoms with unstable nuclei are radioactive (exhibit radioactivity). To become more stable, the nuclei undergo radioactive decay.]

Question J.
You are given a very old sample of wood. How will you determine its age ?
Answer:
The age of the wood sample can be determined by radiocarbon dating as ¹⁴C becomes a part of a plant due to the photosynthesis reaction (i.e., absorption of [¹⁴CO₂ + ¹²CO₂]).
i. The activity (N) of given wood sample and that of fresh sample of live plant (N₀) is measured, where, N₀ denotes the activity of the given sample at the time of death.
ii. The age of the given wood sample. can be determined by applying following Formulae:

Note: The oldest rock found so far in Northern Canada is 3.96 billion years old.

3. Answer the following question

Question A.
Give example of mirror nuclei.
Answer:
Example of mirror nuclei: \({ }_{1}^{3} \mathrm{H}\) and \({ }_{2}^{3} \mathrm{He}\)

Question B.
Balance the nuclear reaction:

Answer:

Question C.
Name the most stable nuclide known. Write two factors responsible for its stability.
Answer:
The most stable nuclide known is lead (\({ }_{82}^{208} \mathrm{~Pb}\)).
Two factors responsible for its stability are as follows:

• It is a nuclide with even number of both protons (Z) and neutrons (N).
• It has two magic numbers i.e., 82 (for protons) and 126 (for neutrons).

Question D.
Write relation between decay constant of a radioelement and its half life.
Answer:
Relation between decay constant of a radioelement and its half-life is given as, λ = \(\frac{0.693}{\mathrm{t}_{1 / 2}}\)
Where, λ = Decay constant, t_1/2 = Half-life of a radioelement

Question E.
What is the difference between an α-particle and helium atom ?
Answer:

• Helium atom is composed of 2 protons and 2 neutrons (or 1 neutron) along with 2 electrons in the outer shell.
• On the other hand, α-particle constitutes 2 protons and 2 neutrons bound together to form a particle which is similar to helium (except presence of electrons).
• Helium is one of the inert gas which is stable (duplet complete) whereas α-particle is unstable and highly reactive.

Question F.
Write one point that differentiates nuclear reations from chemical reactions.
Answer:
Chemical reactions:

• Rearrangement of atoms by breaking and forming of chemical bonds.
• Different isotopes of an element have same behaviour.

Nuclear reactions:

• Elements or isotopes of one element are converted into another element in a nuclear reaction.
• Isotopes of an element behave differently.

Question G.
Write pairs of isotones and one pair of mirror nuclei from the following :

Answer:
Isotones: i. \({ }_{5}^{10} \mathrm{~B} \text { and }{ }_{6}^{11} \mathrm{C}\)
ii. \({ }_{13}^{27} \mathrm{Al} \text { and }{ }_{14}^{28} \mathrm{~S}\)
Mirror nuclei: Since there are no isobars the given set of nuclides does not contain a pair of mirror nuclei.

Question H.
Derive the relationship between half life and decay constant of a radioelement.
Answer:
Equation for the decay constant is given as,
λ = \(\frac{2.303}{t} \log _{10} \frac{\mathrm{N}_{0}}{\mathrm{~N}}\) …(i)
Where, λ = Decay constant
N = Number of nuclei (atoms) present at time t
At t = 0, N = N₀.
Hence, at t = t_1/2, N = N₀/2
Substitution of these values of N and t in equation (i) gives,

Question I.
Represent graphically log₁₀ (activity /dps) versus t/s. What is its slope ?
Answer:
Equation for a decay constant (λ) is given as,

Hence, instead if log₁₀N versus t, log₁₀ \(\left(\frac{-\mathrm{d} \mathrm{N}}{\mathrm{dt}}\right)\) which is log₁₀ (activity) is plotted.
The graph of log₁₀ (activity/dps) versus t/s gives a straight line which can be represented as follows:

Question J.
Write two units of radioactivity. How are they interrelated ?
Answer:
The unit of radioactivity is curie (Ci).
1 Ci = 3.7 × 10¹⁰ dps
ii. Other unit of radioactivity is Becquerel (Bq).
1 Bq = 1 dps
Thus, 1 Ci = 3.7 × 10¹⁰ dps = 3.7 × 10¹⁰ Bq

Question K.
Half life of ²⁴Na is 900 minutes. What is its decay constant?
Answer:

Question L.
Decay constant of ¹⁹⁷Hg is 0.017 h^-1. What is its half life ?
Answer:

Question M.
The total binding energy of ⁵⁸Ni is 508 MeV. What is its binding energy per nucleon ?
Answer:
Given: B.E. of ⁵⁸Ni = 508 MeV,
A = 58
To find: Binding energy per nucleon \(\bar{B}\)

Question N.
Atomic mass of \({ }_{16}^{32} \mathrm{~S}\) is 31.97 u. If masses of neutron and H atom are 1.0087 u and 1.0078 u respectively. What is the mass defect ?
Answer:
Given: m = 31.97 u, Z = 16, A = 32
m_n = 1.0087 u
m_H = 1.0078 u
To find: Δm
Formula: Δm = Zm_H + (A – Z)m_n – m
Calculation: Δm = Zm_H + (A – Z)m_n – m
= 16 × 1.0078 + (16 × 1.0087) – 31.97
= [16.1248 + 16.1392] – 31.97
= 0.294 u
Ans: The mass defect is 0.294 u.

Question O.
Write the fusion reactions occuring in the Sun and stars.
Answer:
Fusion reactions occurring in the Sun and stars are can be represented as,

Question P.
How many α and β – particles are emitted in the trasmutation
\({ }_{90}^{232} \mathrm{Th} \longrightarrow{ }_{82}^{208} \mathrm{~Pb}\)
Answer:
\({ }_{90}^{232} \mathrm{Th} \longrightarrow{ }_{82}^{208} \mathrm{~Pb}\)
The emission of one α-particle decreases the mass number by 4 whereas the emission of β-particles has no effect on mass number.
Net decrease in mass number = 232 – 208 = 24.
This decrease is only due to α-particles. Hence, number of α-particles emitted = \(\frac {24}{4}\) = 6
Now, the emission of one α-particle decrease the atomic number by 2 and one β-particle emission increases it by 1.
The net decrease in atomic number = 90 – 82 = 8
The emission of 6 α-particles causes decrease in atomic number by 12. However, the actual decrease is only 8. Thus, atomic number increases by 4. This increase is due to emission of 4 β-particles.
Thus, 6 α and 4 β-particles are emitted.

Question Q.
A produces B by α- emission. If B is in the group 16 of periodic table, what is the group of A ?
Answer:

When α-emission occurs, atomic number decreases by 2 and atomic mass number by 4.
Thus, if ‘B’ belongs to group 16 of periodic table, that means outermost orbit will contain 6 electrons.
Thus, ‘A’ will have 8 electrons in its valence shell and it will belong to group 18 of the periodic table.

Question R.
Find the number of α and β- particles emitted in the process
\({ }_{86}^{222} \mathrm{Rn} \longrightarrow{ }_{84}^{214} \mathrm{PO}\)
Answer:
The emission of one α-particle decreases the mass number by 4 whereas the emission of β-particles has no effect on mass number.
Net decrease in mass number = 222 – 214 = 8. This decrease is only due to α-particle. Hence, number of α-particle emitted = 8/4 = 2
Now, the emission of one α-particle decreases the atomic number by 2 and one β-particle emission increases it by 1.
The net decrease in atomic number = 86 – 84 = 2
The emission of 2 α-particles causes decrease in atomic number by 4. However, the actual decrease is only 2. It means atomic number increases by 2. This increase is due to emission of 2 β-particles.
Thus, 2 α and 2 β-particles are emitted.

[Note: The above question is modified to include the final decay product so as to determine the number of α-particles and β-particles emitted in the process. Here, the final decay product is assumed to be Po-214.]

4. Solve the problems

Question A.
Half life of ¹⁸F is 110 minutes. What fraction of ¹⁸F sample decays in 20 minutes ?
Answer:
Given: t_1/2 = 110 min
t = 20 min
To find: Fraction of ¹⁸F simple that decays


∴ Fraction of ¹⁸F sample that decays = 1 – 0.882 = 0.118
Ans: Fraction of ¹⁸F sample that decays in 20 minutes is 0.118.

Question B.
Half life of ³⁵S is 87.8 d. What percentage of ³⁵S sample remains after 180 d ?
Answer:
Given: t_1/2 = 87.8 d,
N₀ = 100,
t = 180 d
To find: % of ³⁵S that remains after 180 days

Question C.
Half life ⁶⁷Ga is 78 h. How long will it take to decay 12% of sample of Ga ?
Answer:
Given: t_1/2 = 78 h,
N₀ = 100,
N = 100 – 12 = 88
To find: t

Question D.
0.5 g Sample of ²⁰¹Tl decays to 0.0788 g in 8 days. What is its half life ?
Answer:
Given: N₀ = 0.5 g,
N = 0.0788 g,
t = 8 days
To find: t_1/2

Question E.
65% of ¹¹¹In sample decays in 4.2 d. What is its half life ?
Answer:
Given: N₀ = 100,
N = 100 – 65 = 35,
t = 4.2d
To find: t_1/2

Question F.
Calculate the binding energy per nucleon of \({ }_{36}^{84} \mathrm{Kr}\) whose atomic mass is 83.913 u. (Mass of neutron is 1.0087 u and that of H atom is 1.0078 u).
Answer:
Given: A = 84, Z = 36,
m = 83.913 u
m_n = 1.0087 u
m_H = 1.0078 u
To find: Binding energy per nucleon \((\bar{B})\)

Question G.
Calculate the energy in Mev released in the nuclear reaction
\({ }_{77}^{174} \mathrm{Ir} \longrightarrow{ }_{75}^{170} \mathrm{Re}+{ }_{2}^{4} \mathrm{He}\)
Atomic masses : Ir = 173.97 u,
Re = 169.96 u and
He = 4.0026 u
Answer:
Given: m_Ir= 173.97 u
m_Re = 169.96 u
m_He = 4.0026 u
To find: Energy released
Formulae: i. Δm = (mass of ¹⁷⁴Ir) – (mass of ¹⁷⁰Re + mass of ⁴He)
ii. E = Δm × 931.4 MeV
Calculation:i. Δm = (mass of ¹⁷⁴Ir) – (mass of ¹⁷⁰Re + mass of ⁴He)
= 173.97 – (169.96 + 4.0026)
= 7.4 × 10^-3 u
ii. E = Δm × 931.4
= 7.4 × 10^-3 × 931.4
= 6.89236 MeV ≈ 6.892 MeV
Ans: The energy released in given nuclear reaction is 6.892 MeV.

Question H.
A 3/4 of the original amount of radioisotope decays in 60 minutes. What is its half life ?
Answer:

Question I.
How many – particles are emitted by 0.1 g of ²²⁶Ra in one year?
Answer:
Given: t = 1 y,
Amount of sample = 0.1 g
To find: Number of particles emitted

Activity = \(\frac{-\mathrm{d} \mathrm{N}}{\mathrm{dt}}\) = λN
= 4.28 × 10^-4 × 2.665 × 10²⁰ atoms
= 1.141 × 10¹⁷ particles/year
Ans: Particles emitted by 0.1 g of ²²⁶Ra in one year = 1.141 × 10¹⁷ particles/year.
[Note: The half-life of radium is 1620 years. In order to apply appropriate textual concept, we have used this value in calculation.]

Question J.
A sample of ³²P initially shows activity of one Curie. After 303 days the activity falls to 1.5× 10⁴ dps. What is the half life of ³²P ?
Answer:

Question K.
Half life of radon is 3.82 d. By what time would 99.9 % of radon will be decayed.
Answer:
Given: t_1/2 = 3.82 d,
N₀ = 100
N = 100 – 99.9 = 0.1
To find: t

Question L.
It has been found that the Sun’s mass loss is 4.34 × 10⁹ kg per second. How much energy per second would be radiated into space by the Sun ?
Answer:
Given: Sun’s mass loss = 4.34 × 10⁹ kg per second
To find: Energy radiated per second into space by Sun
calculation: Δm = 4.34 × 10⁹ kg per second
Now, 1.66 × 10^-27 kg = 1u
∴ Δm = \(\frac{4.34 \times 10^{9}}{1.66 \times 10^{-27}}\) u per second
= 2.614 × 10³⁶ u per second
Now, 1 u = 931.4 MeV
2.614 × 10³⁶ u per second = 2.614 × 10³⁶ × 931.4
= 2.435 × 10³⁹ MeV/s
Now, 1 MeV = 1.6022 × 10^-19 J and 1 eV = 1 × 10^-6 MeV
1 MeV = 1.6022 × 10^-13 J
= 1.6022 × 10^-16 LJ
E = 2.435 × 10³⁹ MeV/s × 1.6022 × 10^-16 kJ/MeV
= 3.901 × 10²³ kJ/s
Ans: Energy radiated per second into space by Sun is 3.901 × 10²³ kJ/s.

Question M.
A sample of old wood shows 7.0 dps/g. If the fresh sample of tree shows 16.0 dps/g, How old is the given sample of wood ? Half life of ¹⁴C 5730 y.
Answer:

Activity :

1. Discuss five applications of radioactivity for peaceful purpose.
Answer:

• Development in earth sciences: Like to understand various geographical changes occurring on earth.
• Development in space technology: To study nuclear reactions in stars which may lead to new discoveries.
• Development in medical sciences: Diagnosis and treatment of various diseases.
• Development in industries: As a potent source of electricity or a power generator.
• Development in agriculture: To study or monitor changes in soil like uptake of nutrients from the soil etc.

[Note: Students can use above points are reference to discuss topic in class].

2. Organize a trip to Bhabha Atomic Reasearch Centre, Mumbai to learn about nuclear reactor. This will have to be organized through your college.
Answer:
Students are expected to visit the place to understand more about nuclear reactors.

11th Chemistry Digest Chapter 13 Nuclear Chemistry and Radioactivity Intext Questions and Answers

Do you know? (Textbook Page no. 190)

Question 1.
How small is the nucleus in comparison to the rest of the atom?
Answer:
The radius of nucleus is of the order of 10^-15 m whereas that of the outer sphere is of the order of 10^-10 m. The size of outer sphere, is 10⁵ times larger than the nucleus i.e., if we consider the atom of size of football stadium then its nucleus will be the size of a pea.

(Textbook Page no. 191)

Question 1.
Identify the following nuclides as: isotopes, isobars and isotones.

Answer:

(Textbook Page No. 194)

Question 1.
i. What do you understand by the term rate of decay and give its mathematical expression.
ii. Why is minus sign required in the expression of decay rate?
Answer:
i. Rate of decay of a radioelement denotes the number of nuclei of its atoms which decay in unit time. It is also called activity of radioelement.
Rate of decay at any time t can be expressed as follows:
Rate of decay (activity) = \(-\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}\)
where, dN is the number of nuclei that decay within time interval dt.
ii. Minus sign in the expression indicates that the number of nuclei decreases with time. Therefore, dN is a negative quantity. But, the rate of decay is a positive quantity. The negative sign is introduced in the rate expression to make the rate positive.

Try this. (Textbook Page No. 194)

Question 1.
Prepare a chart of comparative properties of the above three types of radiations.
Answer:

Just think (Textbook Page No. 195)

Question 1.
Does half-life increase, decrease or remain constant? Explain.
Answer:
Half-life of a particular radioelement remains constant at a given instant. A radioactive half-life refers to the amount of time it takes for half of the original isotope to decay. It is related to decay constant by the expression: t_1/2 = 0.693 / λ

From the expression, it is evident that half-life of a radio isotope is dependent only on the decay constant and is independent of the initial amount of the radio isotope. Each successive half-life in which the amount of radio isotope decreases to its half value is the same.

Thus, half-life remains constant.

Try this (Textbook Page No. 198)

Question 1.
²⁴Mg and ²⁷Al, both undergo (α, n) reactions and the products are radioactive. These emit β particles having positive charge (called positrons). Write balanced nuclear reactions in both.
Answer:

Do you know? (Textbook Page No. 198)

Question 1.
What is the critical mass of ²³⁵U?
Answer:
i. The critical mass is the minimum mass of uranium-235 required to achieve a self-sustaining fission chain reaction under stated conditions.
ii. The chain reaction in fission of U-235 becomes self-sustaining when the critical mass of uranium-235 is about 50 kilograms.

Activity (Textbook Page No. 200)

Question 1.
You have learnt in Std. 9th, medical, industrial and agricultural applications of radioisotopes. Write at least two applications each.
Answer:
i. The uses of radioactive isotopes in the field of medicine:
a. Polycythaemia: The red blood cell count increases in the disease polycythaemia. Phosphorus-32 is used in its treatment.
b. Bone cancer: Strontium-89, strontium-90, samarium-153 and radium-223 are used in the treatment of bone cancer.

ii. The uses of radioactive isotopes in the industrial field:
a. Luminescent paint and radioluminescence: The radioactive substances radium, promethium, tritium with some phosphorus are used to make certain objects visible in the dark.
e.g. Hands of a clock, krypton-85 is used in HID (High Intensity Discharge) lamps.
b. Use in ceramic articles:
1. Luminous colours are used to decorate ceramic tiles, utensils, plates, etc.
2. Uranium oxide was earlier used to colour ceramics.

iii. The uses of radioactive isotopes in the agriculture field:
a. The genes and chromosomes that give seeds its properties like fast growth, higher productivity, etc., can be modified by means of radiation.
b. Onions and potatoes are irradiated with gamma rays from cobalt-60 to prevent their sprouting.

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 11 Adsorption and Colloids Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 11 Adsorption and Colloids

1. Choose the correct option.

Question A.
The size of colloidal particles lies between
a. 10^-10 m and 10^-9 m
b. 10^-9 m and 10^-6 m
c. 10^-6 m and 10^-4 m
d. 10^-5 m and 10^-2 m
Answer:
b. 10^-9 m and 10^-6 m

Question B.
Gum in water is an example of
a. true solution
b. suspension
c. lyophilic sol
d. lyophobic sol
Answer:
c. lyophilic sol

Question C.
In Haber process of production of ammonia K₂O is used as
a. catalyst
b. inhibitor
c. promotor
d. adsorbate
Answer:
c. promotor

Question D.
Fruit Jam is an example of-
a. sol
b. gel
c. emulsion
d. true solution
Answer:
b. gel

2. Answer in one sentence :

Question A.
Name type of adsorption in which van der Waals focres are present.
Answer:
Physical adsorption or physisorption.

Question B.
Name type of adsorption in which compound is formed.
Answer:
Chemical adsoiption or chemisorption.

Question C.
Write an equation for Freundlich adsorption isotherm.
Answer:
Freundlich proposed the following empirical equation for adsorption of a gas on solid.
\(\frac{x}{\mathrm{~m}}\) = k P^1/n (n > 1) ……(i)
where,
x = Mass of the gas adsorbed
m = Mass of the adsorbent
\(\frac{x}{\mathrm{~m}}\) = Mass of gas adsorbed per unit mass of adsorbent
P = Equilibrium pressure
k and n are constants which depend on the nature of adsorbate, adsorbent and temperature.

3. Answer the following questions:

Question A.
Define the terms:
a. Inhibition
b. Electrophoresis
c. Catalysis.
Answer:
a. Inhibition:
The phenomenon in which the rate of chemical reaction is reduced by an inhibitor is called inhibition.

b. Electrophoresis:
The movement of colloidal particles under an applied electric potential is called electrophoresis.

c. Catalysis:
The phenomenon of increasing the rate of a chemical reaction with the help of a catalyst is known as catalysis.

Question B.
Define adsorption. Why students can read blackboard written by chalks?
Answer:

• Adsorption is the phenomenon of accumulation of higher concentration of ‘one substance on the surface of another (in bulk) due to unbalanced/unsatisfied attractive forces on the surface.
• When we write on blackboard using chalk, the chalk particles get adsorbed on the surface of the blackboard.

Hence, students can read blackboard written by chalks.

Question C.
Write characteristics of adsorption.
Answer:
Following are the characteristics of adsorption:

• Adsorption is a surface phenomenon.
• It depends upon the surface area of the adsorbent.
• It involves physical forces (van der Waals forces) or chemical forces (chemical or covalent bonds).
• Adsorbate is always present in higher concentration on the surface of an adsorbent than in the bulk.
• Adsorption is dependent on temperature (of the surface) and pressure (of adsorbate gas).
• It takes place with the evolution of heat (with some exceptions).

Question D.
Distinguish between Lyophobic and Lyophilic sols.
Answer:
Lyophobic sols (colloids):

1. Lyophobic sols are formed only by special methods.
2. They are irreversible.
3. These are unstable and hence, require traces of stabilizers.
4. Addition of small amount of electrolytes causes precipitation or coagulation of lyophobic sols.
5. Viscosity of lyophobic sol is nearly the same as the dispersion medium.
6. Surface tension of lyophobic sol is nearly the same as the dispersion medium.

Lyophilic sols (colloids):

1. Lyophilic sols are formed easily by direct mixing.
2. They are reversible.
3. These are self-stabilized.
4. Addition of large amount of electrolytes causes precipitation or coagulation of lyophilic sols.
5. Viscosity of lyophilic sol is much higher than that of the dispersion medium.
6. Surface tension of lyophilic sol is lower than that of dispersion medium.

Question E.
Identify dispersed phase and dispersion medium in the following colloidal dispersions.
a. milk
b. blood
c. printing ink
d. fog
Answer:

Question F.
Write notes on :
a. Tyndall effect
b. Brownian motion
c. Types of emulsion
d. Hardy-Schulze rule
Answer:
a. Tyndall effect:
i. Tyndall observed that when light passes through true solution, the path of light through it cannot be detected.
ii. However, if the light passes through a colloidal dispersion, the particles scatter some light in all directions and the path of the light through colloidal dispersion becomes visible to observer standing at right angles to its path.
iii. The phenomenon of scattering of light by colloidal particles and making path of light visible through the dispersion is referred as Tyndall effect and the bright cone of the light is called Tyndall cone.
iv. Tyndall effect is observed only when the following conditions are satisfied.

• The diameter of the dispersed particles is not much smaller than the wavelength of light used.
• The refractive indices of dispersed phase and dispersion medium differ largely.

v. Significance of Tyndall effect:

• It is useful in determining number of particles in colloidal system and their particle size.
• It is used to distinguish between colloidal dispersion and true solution.

b. Brownian motion:
i. The colloidal or microscopic particles undergo ceaseless random zig-zag motion in all directions in a fluid. This motion of dispersed phase particles is called Brownian motion.
ii. Cause of Brownian motion:

• Particles of the dispersed phase constantly collide with the fast-moving molecules of dispersion medium (fluid).
• Due to this, the dispersed phase particles acquire kinetic energy from the molecules of the dispersion medium.
• This kinetic energy brings about Brownian motion.

c. Types of emulsion:
iii. There are two types of emulsions:
a. Emulsion of oil in water (o/w type): An emulsion in which dispersed phase is oil and dispersion medium is water is called emulsion of oil in water.
e.g. 1. Milk consists of particles of fat dispersed in water.
2. Other examples include vanishing cream, paint, etc.
b. Emulsion of water in oil (w/o type): An emulsion in which dispersed phase is water and dispersion medium is oil is called emulsion of water in oil.
e.g. 1. Cod liver oil consists of particles of water dispersed in oil.
2. Some other examples of this type include butter, cream, etc.

d. Hardy-Schulze rule:
i. Generally, greater the valency of the flocculating ion added, greater is its power to cause precipitation. This is known as Hardy-Schulze rule.
ii. In the coagulation of negative sol, the flocculating power follows the following order:
Al³⁺ > Ba²⁺ > Na⁺
iii. Similarly, in the coagulation of positive sol, the flocculating power is in the following order:
[Fe (CN)₆]^4- > PO₄^3- > SO₄^2- > Cl^–

Question G.
Explain Electrophoresis in brief with the help of diagram. What are its applications ?
Answer:
i. Electrophoresis: Electrophoresis set up is shown in the diagram below.

• The diagram shows U tube set up in which two platinum electrodes are dipped in a colloidal solution.
• When electric potential is applied across two electrodes, colloidal particles move towards one or other electrode.
• The movement of colloidal particles under an applied electric potential is called electrophoresis.
• Positively charged particles move towards cathode while negatively charged particles migrate towards anode and get deposited on the respective electrode.

ii. Applications of electrophoresis:

• On the basis of direction of movement of the colloidal particles under the influence of electric field, it is possible to know the sign of charge on the particles.
• It is also used to measure the rate of migration of sol particles.
• Mixture of colloidal particles can be separated by electrophoresis, since different colloidal particles in mixture migrate with different rates.

Question H.
Explain why finely divided substance is more effective as adsorbent?
Answer:

• Adsorption is a surface phenomenon and hence, the extent of adsorption depends upon surface area of the adsorbent.
• Adsorption increases with increase in surface area of the adsorbent.
• Finely divided powdered substances provide larger surface area for a given mass.

Hence, finely divided substance is more effective as adsorbent.

Question I.
What is the adsorption Isotherm?
Answer:
The relationship between the amount of a substance adsorbed per unit mass of adsorbent and the equilibrium pressure (in case of gas) or concentration (in case of solution) at a given constant temperature is called an adsorption isotherm.

Question J.
Aqueous solution of raw sugar, when passed over beds of animal charcoal, becomes colourless. Explain.
Answer:

• When aqueous solution of raw sugar is passed over beds of animal charcoal, charcoal adsorbs the coloured particles from the raw sugar.
• Thus, due to the adsorption of coloured particles, raw sugar becomes colourless when passed over beds of animal charcoal.

Question K.
What happens when a beam of light is passed through a colloidal sol?
Answer:
i. When a beam of light is passed through colloidal sol, it is observed that the colloidal particles scatter some of the incident light in all directions.
ii. Because of this scattering of light, the path of light through the colloidal dispersion becomes visible to observer standing at right angles to its path and the phenomenon is known as Tyndall effect.
iii.

Question L.
Mention factors affecting adsorption of gas on solids.
Answer:
Adsorption of gases on solids depends upon the following factors:

• Nature of adsorbate (gas)
• Nature of solid adsorbent
• Surface area of adsorbent
• Temperature of the surface
• Pressure of the gas

Question M.
Give four uses of adsorption.
Answer:
i. Catalysis (Heterogeneous catalysis):

• The solid catalysts are used in many industrial manufacturing processes.
• For example, iron is used as a catalyst in manufacturing of ammonia, platinum in manufacturing of sulphuric acid, H₂SO₄ (by contact process) while finely divided nickel is employed as a catalyst in hydrogenation of oils.

ii. Gas masks:

• It is a device which consists of activated charcoal or mixture of adsorbents.
• It is used for breathing in coal mines to avoid inhaling of the poisonous gases.

iii. Control of humidity: Silica and alumina gels are good adsorbents of moisture.
iv. Production of high vacuum:

• Lowering of temperature at a given pressure, increases the rate of adsorption of gases on charcoal powder. By using this principle, high vacuum can be attained by adsorption.
• A vessel evacuated by vacuum pump is connected to another vessel containing coconut charcoal cooled by liquid air. The charcoal adsorbs the remaining traces of air or moisture to create a high vacuum.

Question N.
Explain Bredig’s arc method.
Answer:

• Colloidal sols can be prepared by electrical disintegration using Bredig’s arc method.
• This process involves vaporization as well as condensation.
• Colloidal sols of metals such as gold, silver, platinum can be prepared by this method.
• In this method, electric arc is struck between electrodes of metal immersed in the dispersion medium.
• The intense heat produced vapourizes the metal which then condenses to form particles of colloidal sol.

Question O.
Explain the term emulsions and types of emulsions.
Answer:
i. A colloidal system in which one liquid is dispersed in another immiscible liquid is called an emulsion.
ii. There are liquid-liquid colloidal systems in which both liquids are either completely or partially immiscible.
iii. There are two types of emulsions:
a. Emulsion of oil in water (o/w type): An emulsion in which dispersed phase is oil and dispersion medium is water is called emulsion of oil in water.
e.g. 1. Milk consists of particles of fat dispersed in water.
2. Other examples include vanishing cream, paint, etc.
b. Emulsion of water in oil (w/o type): An emulsion in which dispersed phase is water and dispersion medium is oil is called emulsion of water in oil.
e.g. 1. Cod liver oil consists of particles of water dispersed in oil.
2. Some other examples of this type include butter, cream, etc.

4. Explain the following :

Question A.
A finely divided substance is more effective as adsorbent.
Answer:

• Adsorption is a surface phenomenon and hence, the extent of adsorption depends upon the surface area of the adsorbent.
• Adsorption increases with an increase in surface area of the adsorbent.
• Finely divided powdered substances provide a larger surface area for a given mass. Hence, a finely divided substance is more effective as an adsorbent.

Question B.
Freundlich adsorption isotherm, with the help of a graph.
Answer:
Graphical representation of the Freundlich adsorption isotherm:

i. Freundlich proposed the following empirical equation for adsorption of a gas on solid.
\(\frac{x}{\mathrm{~m}}\) = k P^1/n (n > 1) ………(i)
where,
x = Mass of the gas adsorbed
m = Mass of the adsorbent
\(\frac{x}{\mathrm{~m}}\) = Mass of gas adsorbed per unit mass of adsorbent
P = Equilibrium pressure
k and n are constants which depend on the nature of adsorbate, adsorbent and temperature.
ii. The graphical representation of Freundlich equation is as shown in the adjacent plot of x/m vs ‘P’.
iii. In case of solution, P in the equation (i) is replaced by the concentration (C) and thus,
\(\frac{x}{\mathrm{~m}}\) = k C^1/n ………(ii)
iv. By taking logarithm on both sides of the equation (ii),
we get
log \(\frac{x}{\mathrm{~m}}\) = log k + \(\frac{1}{n}\) log C ……..(iii)
v. On plotting a graph of log \(\frac{x}{\mathrm{~m}}\) against log C or log P, a straight line is obtained as shown in the adjacent plot. The slope of the straight line is and intercept on Y-axis is log k.
vi. The factor \(\frac{1}{n}\) ranges from 0 to 1. Equation (iii) holds good over limited range of pressures.
a. When \(\frac{1}{n}\) → 0, \(\frac{x}{\mathrm{~m}}\) → constant, the adsorption is independent of pressure.
b. When \(\frac{1}{n}\) = 1, \(\frac{x}{\mathrm{~m}}\) = k P, i.e., \(\frac{x}{\mathrm{~m}}\) ∝ P, the adsorption varies directly with pressure.
c. The experimental isotherms tend to saturate at high pressure.

5. Distinguish between the following :

Question A.
Adsorption and absorption. Give one example.
Answer:
Adsorption:

• Adsorption is a surface phenomenon as adsorbed matter is concentrated only at the surface and does not penetrate through the surface to the bulk of adsorbent.
• Concentration of the adsorbate is high only at the surface of the adsorbent.
• It is dependent on temperature and pressure.
• It is accompanied by evolution of heat known as heat of adsorption.
• It depends on surface area.
  e.g. Adsoiption of a gas or liquid like acetic acid by activated charcoal.

Absorption:

• Absorption is a bulk phenomenon as absorbed matter is uniformly distributed inside as well as at the surface of the bulk of substance.
• Concentration of the absorbate is uniform throughout the bulk of the absorbent.
• It is independent of temperature and pressure.
• It may or may not be accompanied by any evolution or absorption of heat.
• It is independent of surface area.
  e.g. Absorption of water by cotton, absorption of ink by blotting paper.

Question B.
Physisorption and chemisorption. Give one example.
Answer:
Physisorption:

1. In physisorption, the forces operating are weak van der Waals forces.
2. It is not specific in nature as all gases adsorb on all solids. For example, all gases adsorb on charcoal.
3. The heat of adsorption is low and lies in the range 20-40 kJ mol^-1.
4. It occurs at low temperature and decreases with an increase of temperature.
5. It is reversible.
6. Physisorbed layer may be multimolecular layer of adsorbed particles under high pressure.
   e.g. At low temperature N₂ gas is physically adsorbed on iron.

Chemisorption:

1. In chemisorption, the forces operating are of chemical nature (covalent or ionic bonds).
2. It is highly specific and occurs only when chemical bond formation is possible between adsorbent and adsorbate. For example, adsorption of oxygen on tungsten, hydrogen on nickel, etc.
3. The heat of adsorption is high and lies in the range 40-200 kJ mol^-1.
4. It is favoured at high temperature, however, the extent of chemical adsorption is lowered at very high temperature due to bond breaking.
5. It is irreversible.
6. Chemisorption forms monomolecular layer of adsorbed particles.
   e.g. N₂ gas chemically adsorbed on iron at high temperature forms a layer of iron nitride, which desorbs at very high temperature.

6. Adsorption is surface phenomenon. Explain.
Answer:
Consider a surface of a liquid or a solid.

• The molecular forces at the surface of a liquid are unbalanced or in unsaturation state.
• In solids, the ions or molecules at the surface of a crystal do not have their forces satisfied by the close contact with other particles.
• Because of the unsaturation, solid and liquid surfaces tend to attract gases or dissolved substances with which they come in close contact. Thus, the substance accumulates on the surface of solid or liquid i.e., the substance gets adsorbed on the surface.

Hence, adsorption is a surface phenomenon.

7. Explain how the adsorption of gas on solid varies with
a. nature of adsorbate and adsorbent
b. surface area of adsorbent
Answer:
i. a. Nature of adsorbate:
1. All solids adsorb gases to some extent. It is observed that gases having high critical temperature liquify easily and can be readily adsorbed.
2. The gases such as SO₂, Cl₂, NH₃ which are easily liquefiable are adsorbed to a larger extent as compared to gases such as N₂, O₂, H₂, etc. which are difficult to liquify.
3. Thus, the amount of gas adsorbed by a solid depends on the nature of the adsorbate gas i.e., whether it is easily liquefiable or not.

b. Nature of adsorbent: Substances which provide large surface area for a given mass are effective as adsorbents and adsorb appreciable volumes of gases.
e.g. Silica gel and charcoal are effective adsorbents due to their porous nature.

ii. Surface area of the adsorbent:

• Adsorption is a surface phenomenon. Hence, the extent of adsorption increases with increase in surface area of the adsorbent.
• Finely divided substances, rough surfaces, colloidal substances are good adsorbents as they provide larger surface area for a given mass.

Note: Critical temperature of some gases and volume adsorbed.

8. Explain two applications of adsorption.
Answer:
i. Catalysis (Heterogeneous catalysis):

• The solid catalysts are used in many industrial manufacturing processes.
• For example, iron is used as a catalyst in manufacturing of ammonia, platinum in manufacturing of sulphuric acid, H₂SO₄ (by contact process) while finely divided nickel is employed as a catalyst in hydrogenation of oils.

ii. Gas masks:

• It is a device which consists of activated charcoal or mixture of adsorbents.
• It is used for breathing in coal mines to avoid inhaling of the poisonous gases.

9. Explain micelle formation in soap solution.
Answer:

• Soap molecule has a long hydrophobic hydrocarbon chain called tail which is attached to hydrophilic ionic carboxylate group, called head.
• In water, the soap molecules arrange themselves to form spherical particles that are called micelles.
• In each micelle, the hydrophobic tails of soap molecules point to the centre and the hydrophilic heads lie on the surface of the sphere.
• As a result of this, soap dispersion in water is stable.

10. Draw labelled diagrams of the following :
a. Tyndall effect
b. Dialysis
c. Bredig’s arc method
d. Soap micelle
Answer:
a. Tyndall effect:

b. Dialysis:

c. Bredig’s arc method:

d. Soap micelle:

Activity :
Collect the information about methods to study surface chemistry.
Answer:
Following are the few methods that are employed to study surface chemistry.
i. X-ray photoelectron spectroscopy:
It is a surface-sensitive spectroscopic technique which is used to measure elemental composition of the surface, to determine elements that are present as contaminants on the surface, etc.

ii. Auger electron spectroscopy:
It is a common analytical technique which is used to study surfaces of materials.

iii. Temperature programmed desorption (TPD):
Adsorbed molecules get desorbed when the surface temperature is increased. TPD technique is used to observe these desorbed molecules and helps in providing information about binding energy between the adsorbate and adsorbent.

iv. Scanning Electron Microscopy:
In this technique, a scanning electron microscope is used to focus electron beam over the surface of the sample to be examined. The electron beam interacts with the sample and an image is obtained. This image provides information about surface structure and composition of the sample.

[Note: Students are expected to collect additional information about surface chemistry on their own.]

11th Chemistry Digest Chapter 11 Adsorption and Colloids Intext Questions and Answers

Can you tell? (Textbook Page No. 160)

Question 1.
What is adsorption?
Answer:
Adsorption is the phenomenon of accumulation of higher concentration of one substance on the surface of another (in bulk) due to unbalanced/unsatisfied attractive forces on the surface.

Try this. (Textbook Page No. 161)

Question 1.
Dip a chalk in ink. What do you observe?
Answer:
When a chalk is dipped in ink, it is observed that the ink molecules are adsorbed at the surface of chalk and the surface becomes coloured, while the solvent of the ink goes deeper into the chalk due to absorption.

Internet my friend. (Textbook Page No. 172)

Question i.
Brownian motion
Answer:
Students can search relevant videos on YouTube to visualize Brownian motion.

Question ii.
Collect information about Brownian motion.
Answer:
i. The colloidal or microscopic particles undergo ceaseless random zig-zag motion in all directions in a fluid. This motion of dispersed phase particles is called Brownian motion.
ii. Cause of Brownian motion:

• Particles of the dispersed phase constantly collide with the fast-moving molecules of dispersion medium (fluid).
• Due to this, the dispersed phase particles acquire kinetic energy from the molecules of the dispersion medium.
• This kinetic energy brings about Brownian motion.

Internet my friend. (Textbook Page No. 172)

Question 1.
Collect information about surface chemistry.
Answer:

• Surface or interface represents the boundary which separates two bulk phases.
  e.g. Boundary between water and its vapour is a liquid-gas interface.
• Certain properties of substances, particularly of solids and liquids, depend upon the nature of the surface.
• An interface usually has a thickness of a few molecules. However, its area depends on the size of the bulk phase particles.
• Commonly considered bulk phases may be pure compounds or solutions.
• A number of important phenomena, namely, dissolution, crystallization, heterogeneous catalysis, electrode processes and corrosion take place at an interface.
• Thus, study of chemistry of surfaces is critical to many applications in industry, analytical investigations and day-to-day activities such as cleaning and softening of water.
• The branch of chemistry which deals with the nature of surfaces and changes occurring on the surfaces is called surface chemistry.
• Study of surfaces requires a rigorously clean surface. An ultra-clean metal surface can be obtained under very high vacuum, of the order of 10^-8 to 10^-9 pascal.
• Adsorption, catalysis and colloids (such as emulsions and gels) are some of the important aspects of surface chemistry.

[Note: Students are expected to collect additional information about surface chemistry on their own.]

Activity. (Textbook Page No. 172)

Question 1.
Calculate surface area to volume ratio of spherical particle. See how the ratio increases with the reduction of radius of the particle. Plot the ratio against the radius.
Answer:
The graph below shows that as the radius of the spherical particle decreases, the surface to volume ratio increases steadily.

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 10 States of Matter Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 10 States of Matter

1. Select and write the most appropriate alternatives from the given choices.

Question A.
The unit of viscosity is
a. dynes
b. newton
c. gram
d. poise
Answer:
d. poise

Question B.
Which of the following is true for 2 moles of an ideal gas ?
a. PV = nRT
b. PV = RT
c. PV = 2RT
d. PV = T
Answer:
c. PV = 2RT

Question C.
Intermolecular forces in liquid are
a. greater than gases
b. less than solids
c. both a and b
d. greater than solids
Answer:
c. both a and b

Question D.
Interactive forces are ………. in ideal gas.
a. nil
b. small
c. large
d. same as that of real gases
Answer:
a. nil

Question E.
At constant temperature the pressure of 22.4 dm³ volume of an ideal gas was increased from 10⁵ kPa to 210 kPa, New volume could be-
a. 44.8 dm³
b. 11.2 dm³
c. 22.4 dm³
d. 5.6 dm³
Answer:
b. 11.2 dm³

2. Answer in one sentence.

Question A.
Name the term used for mixing of different gases by random molecular motion and ferquent collision.
Answer:
The mixing of different gases by random molecular motion and frequent collision is called diffusion.

Question B.
The pressure that each individual gas would exert if it were alone in the container, what do we call it as ?
Answer:
The pressure that each individual gas would exert if it were alone in the container is called as partial pressure.

Question C.
When a gas is heated the particles move more quickly. What is the change in volume of a heated gas if the pressure is kept constant ?
Answer:
The volume of the gas increases on heating if pressure is kept constant.

Question D.
A bubble of methane gas rises from the bottom of the North sea. What will happen to the size of the bubble as it rises to the surface ?
Answer:
According to Boyle’s law, the size of the bubble of methane gas increases as it rises to the surface.

Question E.
Convert the following temperatures from degree celcius to kelvin.
a. -15° C
b. 25° C
c. -197° C
d. 273° C
Answer:
a. T(K) = t°C +273.15
∴ T(K) = -15 °C + 273.15 = 258.15 K
b. T(K) = t°C +273.15
∴ T(K) = 25 °C + 273.15 = 298.15 K
c. T(K) = t°C + 273.15
∴ T(K) = -197 °C + 273.15 = 76.15 K
d. T(K) = t°C + 273.15
∴ T(K) = 273 °C + 273.15 = 546.15 K

Question F.
Convert the following pressure values into Pascals.
a. 10 atmosphere
b. 1 kPa.
c. 107000 Nm^-2
d. 1 atmosphere
Answer:
a. 10 atmosphere:
1 atm = 101325 Pa
∴ 10 atm = 1013250 Pa
= 1.01325 × 10⁶ Pa

b. 1 kPa:
1 kPa = 1000 Pa

c. 107000 N m^-2:
1 N m^-2 = 1 Pa
∴ 107000 Nm^-2 = 107000 Pa
= 1.07 × 10⁵ Pa

d. 1 atmosphere:
1 atm = 101325 Pa
= 1.01325 × 10⁵ Pa

Question G.
Convert:
a. Exactly 1.5 atm to pascals
b. 89 kPa to newton per square metre (N m^-2)
c. 101.325 kPa to bar
d. -100 °C to Kelvin
e. 0.124 torr to standard atmosphere
Answer:
a. Exactly 1.5 atm to pascals:
1 atm = 101325 Pa
∴ 1.5 atm = 1.5 × 101325
= 151987.5 Pa

b. 89 kPa to newton per square metre (N m^-2):
1 Pa = 1 N m^-2 and 1 Pa = 10^-3 kPa
∴ 10^-3 kPa = 1 N m^-2
∴ 89 kPa = \(\frac{1 \times 89}{10^{-3}}\) N m^-2 = 89000 N m^-2

c. 101.325 kPa to bar:
1 bar = 1.0 × 10⁵ Pa
= 1.0 × 10² k Pa
∴ 100 kPa = 1 bar
∴ 101.325 kPa = \(\frac{1 \times 101.325}{100}\)
= 1.01325 bar

d. -100 °C to Kelvin:
T(K) = t °C + 273.15
∴ T(K) = (- 100 °C) + 273.15 = 173.15 K

e. 0.124 torr to standard atmosphere:
1 atm = 760 torr
∴ 1 torr = \(\frac {1}{760}\)atm
∴ 0.124 torr = 0.124 × \(\frac {1}{760}\)
= 1.632 × 10^-4 atm

Question H.
If density of a gas is measured at constant temperature and pressure then which of the following statement is correct ?
a. Density is directly proportional to molar mass of the gas.
b. Greater the density greater is the molar mass of the gas.
c. If density, temperature and pressure is given ideal gas equation can be used to find molar mass.
d. All the above statements are correct.
Answer:
d. All the above statements are correct.

Question I.
Observe the following conversions.

Which of the above reactions is in accordance with the priciple of stoichiometry ?
Answer:
Both the reactions are in accordance with the principle of stoichiometry.
In the first reaction, both the reactants are completely consumed to form product according to reaction stoichiometry.
1 mol hydrogen + 1 mol chlorine → 2 mol hydrogen chloride
In the second reaction, chlorine is the limiting reagent and it is completely consumed to form hydrogen chloride. Excess hydrogen remains unreacted at the end of the reaction. This reaction also follows principle of stoichiometry.
2 mol hydrogen + 1 mol chlorine → 2 mol hydrogen chloride + 1 mol hydrogen

Question J.
Hot air balloons float in air because of the low density of the air inside the balloon. Explain this with the help of an appropriate gas law.

Answer:
The working of hot air balloon can be explained with the help of Charles’ law. According to Charles’ law, at constant pressure, the volume of a fixed amount of a gas varies directly with the temperature. This means that as the temperature increases, the air inside the balloon expands and occupies more volume. Thus, hot air inside the balloon is less dense than the surrounding cold air. This causes the hot air balloon to float in air.

3. Answer the following questions.

Question A.
Identify the gas laws from the following diagrams.

Answer:
a. Boyle’s law
b. Charles’ law
c. Avogadro’s law [Note: Assuming, T constant]

Question B.
Consider a sample of a gas in a cylinder with a movable piston.

Show digramatically the changes in the position of piston, if
a. Pressure is increased from 1.0 bar to 2.0 bar at constant temperature.
b. Temperature is decreased from 300 K to 150 K at constant pressure
c. Temperature is decreased from 400 K to 300 K and pressure is decreased from 4 bar to 3 bar.
Answer:

Thus, the volume of the gas remains the same.
Hence, there will be no change in the position of the piston.

Question D.
List the characteristic physical properties of the gases.
Answer:
Characteristic physical properties of the gases:

• Gases are lighter than solids and liquids (i.e., possess lower density).
• Gases do not possess a fixed volume and shape. They occupy entire space available and take the shape of the container.
• Gas molecules are widely separated and are in continuous, random motion. Therefore, gases exert pressure equally in all directions due to collision of gas molecules, on the walls of the container.
• In case of gases, intermolecular forces are weakest.
• Gases possess the property of diffusion, which is a spontaneous homogeneous inter mixing of two or more gases.
• Gases are highly compressible.

Question E.
Define the terms:
a. Polarizability
b. Hydrogen bond
c. Aqueous tension
d. Dipole moment
Answer:
a. Polarizability is defined as the ability of an atom or a molecule to form momentary dipoles, that means, the ability of the atom or molecule to become polar by redistributing its electrons.

b. The electrostatic force of attraction between positively polarised hydrogen atom of one molecule and a highly electronegative atom (which may be negatively charged) of other molecule is called as hydrogen bond.

c. The pressure exerted by saturated water vapour is called aqueous tension.

d. Dipole moment (p) is the product of the magnitude of the charge (Q) and the distance between the centres of positive and negative charge (r). It is designated by a Greek Letter (p) and its unit is Debye (D).

Question F.
Would it be easier to drink water with a straw on the top of the Mount Everest or at the base ? Explain.
Answer:
When you drink through a straw, the pressure inside the straw reduces (as the air is withdraw by mouth) and the liquid is pushed up to your mouth by atmospheric pressure. Thus, drinking with a straw makes use of pressure difference to force the liquid into your mouth. So, if the pressure difference is less it will be difficult to drink through a straw. On the top of the Mount Everest, atmospheric pressure is very low. Hence, it will be difficult to drink water with a straw on the top of Mount Everest as compared to at the base.

Question G.
Identify type of the intermolecular forces in the following compounds.
a. CH₃ – OH
b. CH₂ = CH₂
c. CHCl₃
d. CH₂Cl₂
Answer:
a. Hydrogen bonding (dipole-dipole attraction) and London dispersion forces
b. London dispersion forces
c. Dipole-dipole interactions and London dispersion forces
d. Dipole-dipole interactions and London dispersion forces

Question H.
Name the types of intermolecular forces present in Ar, Cl₂, CCl₄ and HNO₃.
Answer:
a. Ar: London dispersion forces
b. Cl₂: London dispersion forces
c. CCl₄: London dispersion forces
d. HNO₃: Flydrogen bonding (dipole-dipole attraction) and London dispersion forces.

Question I.
Match the pairs of the following :

Answer:
a – ii,
b – iii

Question J.
Write the statement for :
(a) Boyle’s law
(b) Charles’ law
Answer:
a. Statement of Boyle’s law: For a fixed mass (number of moles ‘n’) of a gas at constant temperature, the pressure (P) of the gas is inversely proportional to the volume (V) of the gas.
OR
At constant temperature, the pressure of fixed amount (number of moles) of a gas varies inversely with its volume.

b. Statement for Charles’ law:
‘At constant pressure, the volume of a fixed mass of a gas is directly proportional to its temperature in Kelvin.

Question K.
Differentiate between Real gas and Ideal gas.
Answer:
Ideal gas:

1. Strictly obeys Boyle’s and Charles’ law.
   \(\frac{\mathrm{PV}}{\mathrm{nRT}}\) = 1
2. Molecules are perfectly elastic.
3. No attraction or repulsion between the gas molecules i.e. collision without loss of kinetic energy (K.E.)
4. Actual volume of the gas molecules is negligible as compared to total volume of the gas.
5. Ideal gases cannot be liquified even at low temperature but continues to obey Charles’ law and finally occupies zero volume at 0 K.
6. Practically, ideal gas does not exist.

Real gas:

1. Shows deviation from Boyle’s and Charles’ law at high pressure and temperature, i.e. obeys Boyle’s law and Charles’ law at low pressure and high temperature. \(\frac{\mathrm{PV}}{\mathrm{nRT}}\) ≠ 1
2. Molecules are not perfectly elastic.
3. Intermolecular attraction is present, hence collision takes place with loss of kinetic energy.
4. Actual volume of individual gas molecule is significant at high pressure and low- temperature.
5. Real gases undergo liquefaction at low’ temperature when cooled and compressed.
6. Gases that exist in nature like H₂, O₂, CO₂, N₂, He, etc. are real gases.

4. Answer the following questions

Question A.
State and write mathematical expression for Dalton’s law of partial pressure and explain it with suitable example.
Answer:
i. Statement: The total pressure of a mixture of two or more non-reactive gases is the sum of the partial pressures of the individual gases in the mixture.
ii. Explanation:
Dalton’s law can be mathematically expressed as:
P_Total = P₁ + P₂ + P₃ …(at constant T and V)
where, P_Total is the total pressure of the mixture and P₁, P₂, P₃, … are the partial pressures of individual gases 1, 2, 3, … in the mixture.
For example, consider two non-reactive gases A and B. On mixing the two gases, pressure exerted by individual gas A in the mixture of both the gases is called partial pressure of gas A (say P₁). Likewise, partial pressure of gas B is P₂. According to Dalton’s law, total pressure of the mixture of gas A and B at constant T and V will be given as:
P_Total = P₁ + P₂

iii. Schematic illustration of Dalton’s law of partial pressures:

Question B.
Derive an Ideal gas equation. Mention the terms involved in it. Also write how it is utilised to obtain combined gas law.
Answer:
According to Boyle’s law,
V ∝ \(\frac{1}{\mathrm{P}}\) (at constant T and n) ……….(1)
According to Charles’ law,
V ∝ T (at constant P and n) ……(2)
According to Avogadro’s law,
V ∝ n (at constant P and T) ……(3)
Combining relations (1), (2) and (3), we get
V ∝ \(\frac{\mathrm{nT}}{\mathrm{P}}\)
Converting this proportionality into an equation by introducing a constant of proportionality (‘R’ known as gas constant), we get
∴ V = \(\frac{\mathrm{nRT}}{\mathrm{P}}\)
On rearranging the above equation, we get
PV = nRT
where,
P = Pressure of gas,
V = Volume of gas,
n = number of moles of gas,
R = Gas constant,
T = Absolute temperature of gas.
This is the ideal gas equation or equation of state.
[Note: In the ideal gas equation, R is called gas constant or universal gas constant, whose value is same for all the gases. In this equation, if three variables are known, fourth can be calculated. The equation describes the state of an ideal gas. Hence, it is also called as equation of state.]

The ideal gas equation is written as PV = nRT …(1)
On rearranging equation (1), we get,

The ideal gas equation used in this form is called combined gas law.

Question C.
With the help of graph answer the following –

At constant temperature,
a. Graph shows relation between pressure and volume. Represent the relation mathematically.
b. Identify the law.
c. Write the statement of law.
Answer:
a. P ∝ \(\frac{1}{\mathrm{~V}}\)
b. The graph represents Boyle’s law as it gives relation between pressure and volume at constant temperature.
c. Statement of Boyle’s law: For a fixed mass (number of moles ‘n’) of a gas at constant temperature, the pressure (P) of the gas is inversely proportional to the volume (V) of the gas.
OR
At constant temperature, the pressure of fixed amount (number of moles) of a gas varies inversely with its volume.

Question D.
Write Postulates of kinetic theory of gases.
Answer:
Postulates of kinetic theory of gases:

• Gases consist of tiny particles (molecules or atoms).
• On an average, gas molecules remain far apart from each other. Therefore, the actual volume of the gas molecules is negligible as compared to the volume of the container. Hence, gases are highly compressible.
• The attractive forces between the gas molecules are negligible at ordinary temperature and pressure. As a result, gas expands to occupy entire volume of the container.
• Gas molecules are in constant random motion and move in all possible directions in straight lines. They collide with each other and with the walls of the container.
• Pressure of the gas is due to the collision of gas molecules with the walls of the container.
• The collisions of the gas molecules are perfectly elastic in nature, which means that the total energy of the gaseous particle remains unchanged after collision.
• The different gas molecules move with different velocities at any instant and hence have different kinetic energies. However, the average kinetic energy of the gas molecules is directly proportional to the absolute temperature.

Question E.
Write a short note on
a. Vapour pressure.
b. Surface tension
c. Viscosity.
Answer:
a. Vapour pressure:

• Molecules of liquid have tendency to escape from its surface to form vapour above it. This called evaporation.
• When a liquid is placed in a closed container, the liquid undergoes evaporation and vapours formed undergo condensation.
• At equilibrium, the rate of evaporation and rate of condensation are equal.
• The pressure exerted by the vapour in equilibrium with the liquid is known as saturated vapour pressure or simply vapour pressure.
• Vapour pressure is measured by means of a manometer.
• The most common unit for vapour pressure is torr. 1 torr = 1 mm Hg.

[Note: i. The vapour pressure of water is also called aqueous tension.
ii. Water has a vapour pressure of approximately 20 torr at room temperature.]

b. Surface tension:

• The particles in the bulk of liquid are uniformly attracted in all directions and the net force acting on the molecules present inside the bulk is zero.
• But the molecules at the surface experience a net attractive force towards the interior of the liquid, or the forces acting on the molecules on the surface are imbalanced.
• Therefore, liquids have tendency to minimize their surface area and the surface acts as a stretched membrane.
• The force acting per unit length perpendicular to the line drawn on the surface of liquid is called surface tension.
• Unit: Surface tension is measured in SI unit, N m^-1 and is denoted by Greek letter ‘γ’

c. Viscosity:
i. Liquids (fluids) have tendency to flow.
ii. Viscosity measures the magnitude of internal friction in a liquid or fluid to flow as measured by the force per unit area resisting uniform flow.
iii. Different layers of a liquid flow with different velocity. This called laminar flow. Here, the layers of molecules in the immediate contact of the fixed surface remains stationary. The subsequent layers slip over one another. Strong intermolecular forces obstruct the layers from slipping over one another, resulting in a friction between the layers.
iv. Viscosity is defined as the force of friction between the successive layers of a flowing liquid. It is also the resistance to the flow of a liquid.
v. When a liquid flow through a tube, the central layer has the highest velocity, whereas the layer along the inner wall in the tube remains stationary. This is a result of the viscosity of a liquid. Hence, a velocity gradient exists across the cross-section of the tube.

vi. Viscosity is expressed in terms of coefficient of viscosity, ‘η’ (Eta). The SI unit of viscosity coefficient is N s m^-2 (newton second per square meter). In CGS system, the unit (η) is measured in poise.
1 poise = 1 g cm^-1 s^-1 = 10^-1 kg m^-1 s^-1

5. Solve the following

Question A.
A balloon is inflated with helium gas at room temperature of 25 °C and at 1 bar pressure when its initial volume is 2.27L and allowed to rise in air. As it rises in the air external pressure decreases and the volume of the gas increases till finally it bursts when external pressure is 0.3bar. What is the limit at which volume of the balloon can stay inflated ?
Answer:
Given: P₁ = Initial pressure = 1 bar
V₁ = Initial volume = 2.27 L
P₂ = Final pressure = 0.3 bar
To find: V₂ = Final volume
Formula: P₁V₁ = P₂V₂ (at constant n and T)
Calculation: According to Boyle’s law,
P₁V₁ = P₂V₂ (at constant n and T)
∴ V₂ = \(\frac{P_{1} V_{1}}{P_{2}}=\frac{1 \times 2.27}{0.3}\) = 7.566667 L ≈ 7.567 L
Ans: The balloon can stay inflated below the volume of 7.567 L.

Question B.
A syringe has a volume of 10.0 cm³ at pressure 1 atm. If you plug the end so that no gas can escape and push the plunger down, what must be the final volume to change the pressure to 3.5 atm?

Answer:
Given: P₁ = Initial pressure = 1 atm
V₁ = Initial volume = 10.0 cm³
P₂ = Final pressure = 3.5 atm
To find: V₂ = Final volume
Formula: P₁V₁ = P₂V₂ (at constant n and T)
Calculation: According to Boyle’s law,
P₁V₁ = P₂V₂ (at constant n and T)
∴ V₂ = \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{P}_{2}}=\frac{1 \times 10.0}{3.5}\)
= 2.857 L
Ans: The final volume of the gas in the syringe is 2.857 L.

Question C.
The volume of a given mass of a gas at 0°C is 2 dm³. Calculate the new volume of the gas at constant pressure when
a. The temperature is increased by 10°C.
b.The temperature is decreased by 10°C.
Answer:
Given: T₁ = Initial temperature = 0 °C = 0 + 273.15 = 273.15 K,
V₁ = Initial volume = 2 dm³
a. T₂ = Final temperature = 273.15 K + 10 = 283.15 K
b. T₂ = Final temperature = 273.15 K – 10 = 263.15 K
To find: V₂ = Final volume in both the cases
Formula: \(\frac{\mathrm{V}_{\mathrm{l}}}{\mathrm{T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\) (at constant n and P)
Calculation: According to Charles’ law,

Ans: The new volume of a given mass of gas is:
a. 2.073 dm³
b. 1.927 dm³

Question D.
A hot air balloon has a volume of 2800 m³ at 99 °C. What is the volume if the air cools to 80 °C?

Answer:
Given: V₁ = Initial volume = 2800 m³, T₁ = Initial temperature = 99 °C = 99 + 273.15 = 372.15 K,
T₂ = Final temperature = 80 °C = 80 + 273.15 K = 353.15 K
To find: V₂ = Final volume
Formula: = \(\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\) (at constant n and P)
Calculation: According to Charles’ law,
\(\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\) (at constant n and P)
∴ \(\mathrm{V}_{2}=\frac{\mathrm{V}_{1} \mathrm{~T}_{2}}{\mathrm{~T}_{1}}=\frac{2800 \times 353.15}{372.15}=\mathbf{2 6 5 7 \mathrm { m } ^ { 3 }}\)
Ans: The volume of the balloon when the air cools to 80 °C is 2657 m³.

Question E.
At 0 °C, a gas occupies 22.4 liters. How nuch hot must be the gas in celsius and in kelvin to reach volume of 25.0 literes?
Answer:
V₁ = Initial volume of the gas = 22.4 L,
T₁ = Initial temperature = 0 + 273.15 = 273.15 K,
V₂ = Final volume = 25.0 L
To find: T₂ = Final temperature in Celsius and in Kelvin

Ans: The temperature of the gas must be 31.7 °C or 304.9 K.

Question F.
A 20 L container holds 0.650 mol of He gas at 37 °C at a pressure of 628.3 bar. What will be new pressure inside the container if the volume is reduced to 12 L. The temperature is increased to 177 °C and 1.25 mol of additional He gas was added to it?
Answer:
Given: V₁ = Initial volume = 20 L, n₁ = Initial number of moles = 0.650 mol
P₁ = Initial pressure = 628.3 bar
T₁ = Initial temperature = 37 °C = 37 + 273.15 K = 310.15 K
n₂ = Final number of moles = 0.650 + 1.25 = 1.90 mol, V₂ = Final volume = 12 L
T₂ = Final temperature = 177 °C = 177 + 273.15 K = 450.15 K, R = 0.0821 L atm K^-1 mol^-1
To find: P₂ = Final pressure
Formula: PV = nRT
Calculation: According to ideal gas equation,
P₂V₂ = n₂RT₂.
∴ \(\mathrm{P}_{2}=\frac{\mathrm{n}_{2} \mathrm{RT}_{2}}{\mathrm{~V}_{2}}=\frac{1.90 \times 0.0821 \times 450.15}{12}=\mathbf{5 . 8 5 2} \mathrm{atm}\)
Ans: The final pressure of the gas is 5.852 atm.
[Note: In the above numerical, converting the pressure value to different units, we get: 5.852 atm = 4447.52 torr = 5.928 bar]

Question G.
Nitrogen gas is filled in a container of volume 2.32 L at 32 °C and 4.7 atm pressure. Calculate the number of moles of the gas.
Answer:
Given: V = 2.32 L, P = 4.7 atm, T = 32 °C = 32 + 273.15 K = 305.15 K
R = 0.0821 L atm K^-1 mol^-1
To find: n = number of moles of gas
Formula: PV = nRT
Calculation: According to ideal gas equation,
PV = nRT
∴ \(\mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{4.7 \times 2.32}{0.0821 \times 305.15}=\mathbf{0 . 4 3 5} \mathrm{moles}\)
Ans: Number of moles of N₂ gas in the given volume is 0.435 moles.

Question H.
At 25 °C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at the height where temperature is 10 °C and volume of the gas 640 mL ?
Answer:
Given: V₁ = Initial volume = 600 mL, V₂ = Final volume = 640 mL
P₁ = Initial pressure = 760 mm Hg
T₁ = Initial temperature = 25 °C = 25 + 273.15 K = 298.15 K
T₂ = Final temperature = 10 °C = 10 + 273.15 K = 283.15 K
P₂ = Final pressure
Formula: \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
Calculation: According to combined gas law.

Ans: The final pressure of a gas is 676.654 mm Hg.

Question I.
A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5g neon. If pressure of the mixture of the gases in the cylinder is 25 bar. What is the partial pressure of dioxygen and neon in the mixture?
Answer:
Given: m_O2 = 70.6 g, m_Ne = 167.5 g,
P_Total = 25 bar
To find: Partial pressure of each gas
Formula: P₁ = x₁ × P_Total
Calculation: Determine number of moles (n) of each gas using formula: n = \(\frac{\mathrm{m}}{\mathrm{M}}\)

Ans: The partial pressure of dioxygen and neon are 5.2 bar and 19.8 bar respectively.

Question J.
Calculate the pressure in atm of 1.0 mole of helium in a 2.0 dm³ container at 20.0 °C.
Answer:
Given: n = number of moles = 1.0 mol, V = volume = 2.0 dm³
T = Temperature = 20.0 °C = 20.0 + 273.15 K = 293.15 K
R = 0.0821 L atm K^-1 mol^-1
To find: Pressure (P)
Formula: PV = nRT
Calculation: According to ideal gas equation,

Ans: The pressure of the given helium gas is 12.03 atm.

Question K.
Calculate the volume of 1 mole of a gas at exactly 20 °C at a pressure of 101.35 kPa.
Answer:
Given: n = number of moles = 1 mol, P = pressure = 101.35 kPa = 1.00025 atm ≈ 1 atm
T = Temperature = 20 °C = 20 + 273.15 K = 293.15 K
R = 0.0821 dm³ atm K^-1 mol^-1
To find: Volume (V)
Formula: PV = nRT
Calculation: According to ideal gas equation,
PV = nRT

Ans: The volume of the given gas is 24.07 dm³.

Question L.
Calculate the number of molecules of methane in 0.50 m³ of the gas at a pressure of 2.0 × 10² kPa and a temperature of exactly 300 K.
Answer:
V = 0.5 m³, P = 2.0 × 10² kPa = 2.0 × 10⁵ Pa
T = 300 K, R = 8.314 J K^-1 mol^-1
To find: Number of molecules of methane gas
Formula: PV = nRT
Calculation: According to ideal gas equation,
n = \(\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{2.0 \times 10^{5} \times 0.5}{8.314 \times 300}=40 \mathrm{~mol}\)
Number of molecules = n × NA = 40 × 6.022 × 10²³ = 2.4088 × 10²³ ≈ 2.409 × 10²⁵
Ans: The number of molecules of methane gas present is 2.409 × 10²⁵ molecules.

11th Chemistry Digest Chapter 10 States of Matter Intext Questions and Answers

Do you know? (Textbook Page No. 140)

Question 1.
Consider three compounds: H₂S, H₂Se and H₂O. Identify which has the highest boiling point. Justify.
Answer:
Among the three compounds H₂O, H₂S and H₂Se, the first one, H₂O has the smallest molecular mass. But it has the highest B.P. of 100 °C. B.P. of H₂S is -60 °C and of H₂Se is -41.25 °C. The extraordinary high B.P. of H₂O is due to very strong hydrogen bonding even though it has the lowest molecular mass.

Can you tell? (Textbook Page No. 140)

Question i.
What are the various components present in the atmosphere?
Answer:
Various components present in the atmosphere are as follows:
a. Nitrogen (78%)
b. Oxygen (21%)
c. Carbon dioxide and other gases (0.03%)
d. Inert gases (mainly argon) (0.97%)
e. Traces of water vapour

Question ii.
Name five elements and five compounds those exist as gases at room temperature.
Answer:
Five elements and five compounds that exist as gases at room temperature are as follows:

Just think. (Textbook Page No. 140)

Question 1.
What is air?
Answer:

• Air is a mixture of various gases.
• One cannot see air but can feel the cool breeze.
• The composition of air by volume is around 78 percent N₂, 21 percent O₂ and 1 percent other gases including CO₂.

Use your brainpower. (Textbook Page No. 141)

Question 1.
Find the unit in which car-tyre pressure is measured.
Answer:
Car-tyre pressure is measured in the units of pounds per square inch (psi) or Newton per metre square (N m^-2).

Do you know? (Textbook Page No. 142)

Question 1.
How does a bicycle pump work?
Answer:
A bicycle pump works on Boyle’s law. Pushing a bicycle pump squashes the same number of particles into a smaller volume. This squashing means particles hit the walls of the pump more often, increasing the pressure. The increased pressure of a gas can be felt on palm by pushing in the piston of a bicycle pump.

Internet my friend (Textbook Page No. 143)

Question 1.
i. Watch Boyle’s law experiment.
ii. Find applications of Boyle’s law.
iii. Try to study how Boyle’s law helps in ‘scuba-diving’ i.e., importance of Boyle’s law in scuba diving an exhilarating sport.
Answer:
i. Students can refer to ‘Boyle’s law experiment’ on YouTube channel of ‘Socratica’.
ii. a. Syringes: When the plunger of a syringe is pulled back out, it causes the volume of the gas inside it to increase due to the reduction of pressure. This creates a vacuum in the syringe, which is constantly trying to adjust the pressure back to normal. However, since the only substance available, such as the blood or medication, is on the other side of the needle, this liquid is sucked into the vacuum, increasing the pressure and decreasing the volume of the gas. When we push the plunger back down, the pressure again increases, lowering the volume inside the syringe, and pushing the fluid out.

b. Respiration: Boyle’s law is essential for the human breathing process. When person breathes in, his/her lung volume increases and the pressure within decreases. Since air always moves from areas of high pressure to areas of low pressure, air is drawn into the lungs. The opposite happens when person exhales. Since the lung volume decreases, the pressure within increases, forcing the air out of the lungs

c. Storage of Gases: Many industries store gases under high pressure. This allows the gas to be stored at a low volume, saving plenty of storage space.
[Note: Students are expected to search more on the internet about various other applications of Boyle’s law on their own.]

iii. Importance of Boyle’s law in scuba diving:
a. Boyle’s law affects scuba diving in many ways.
b. It explains the role of pressure in the aquatic environment.
c. As divers descend, the water pressure surrounding them increases, causing air in their body and equipment to have a smaller volume. As the divers ascend, water pressure decreases, causing their body and equipment to expand to acquire a greater volume.
d. Furthermore, it is crucial that scuba divers never attempt to hold their breath when immersed in water.
e. According to Boyle’s law, if divers attempt this when they ascend to a body of water of less pressure, then the air that is trapped in their lungs will over-expand and rupture. This is known as Pulmonary Barotrauma. Thus, it is important for scuba divers to exhale as they ascend because the external pressure increases.
f. Also, if a diver returns to the surface too quickly, dissolved gases in the blood expand and form bubbles, which can get stuck in capillaries and organs (causing the ‘bends’).
[Note: Students are expected to collect additional information their own.]

Just think. (Textbook Page No. 144)

Question i.
Why does bicycle tyre burst during summer?
Answer:

• According to Charles’ law, at constant pressure, the volume of a fixed amount of a gas varies directly with the temperature. This means that as the temperature increases, the volume also increases.
• During summer, the temperature of the surrounding air is high. Due to the high temperature, the air inside the tyre gets heated. This will increase the volume of the tyres and it will burst.

Question ii.
Why do the hot air balloons fly high?
Answer:

• According to Charles’ law, at constant pressure, gases expand on heating and become less dense. Thus, hot air is less dense than cold air.
• In a hot air balloon, the air inside it is heated by a burner. Upon heating, the air inside the balloon expands and becomes lighter (less dense) than the cooler air on the outside. This causes the hot air balloon to fly high in air.

Just think. (Textbook Page No. 145)

Question 1.
i. List out various real-life examples of Charles’ law.
ii. Refer and watch Charles’ law experiments.
Answer:
i. Few real-life examples of Charles’ law:
a. Helium balloon: If we fill a helium balloon in a warm or hot room, and then take it into a cold room, it shrinks up and will look like it has lost some of the air inside it. This shows that gases expand on heating and contract on cooling.
b. A bottle of deodorant: If we expose a bottle of deodorant to sunlight and high temperatures, the air molecules inside the bottle will expand which can lead to the bursting of the deodorant bottle. This is another example of Charles’ law.
c. Basketball: You may have noticed that a basketball has less responsive bounce during winter than in summer. This yet another example of Charles’ law. When a basketball is inflated, the air pressure inside it is set to a fixed value. As the temperature falls, the volume of the gas inside the ball also decreases proportionally.
[Note: Students are expected to collect additional real-life examples on their own,]

ii. pi [Note: Students can scan the adjacent QR code to visualize Charles’ law with the aid of a relevant video.]

Use your brainpower. (Textbook Page No. 146)

Question 1.
Why does the pressure in the automobile tyres change during hot summer or winter season?
Answer:

• According to Gay-Lussac’s law, at constant volume, pressure of a fixed amount of a gas is directly proportional to its absolute temperature.
• During hot summer, the temperature of automobile tyre increases faster. Consequently, the air inside the tyre gets heated and the gas molecules starts moving faster.
• As the volume of the tyre remains constant, the pressure inside it increases.
• During winter, the temperature of automobile tyre decreases. Consequently, the air inside the tyre gets cooled and the gas molecules starts moving much slower and the pressure inside the tyre decreases.

Just think. (Textbook Page No. 149)

Question 1.
Do all pure gases and mixtures of gases obey the gas laws?
Answer:
Yes, the gas laws are also applicable to the mixtures of gases. The measurable properties of mixture of the gases such as pressure, temperature, volume and amount of gaseous mixture are all related by an ideal gas law.

Just think. (Textbook Page No. 150)

Question 1.
Where is Dalton’s law applicable?
Answer:
Air is gaseous mixture of different gases. Dalton’s law is useful for the study of various phenomena in air, for example, air pollution.

Just think. (Textbook Page No. 155)

Question 1.
What makes the oil rise through the wick in an oil lamp?
Answer:
In an oil lamp, oil rises through the wick due to the capillary action. Such a capillary rise of oil is due to surface tension of oil. The wick acts as a capillary tube. When the wick is placed in oil, the attractive forces between the oil and the inner wall of capillary (wick) pull the oil up through the wick.

Balbharti Maharashtra State Board Hindi Yuvakbharati 11th Digest रचना वृत्तांत लेखन Notes, Questions and Answers.

Maharashtra State Board 11th Hindi रचना वृत्तांत लेखन

वृत्तांत लेखन : किसी भी सभा, बैठक, कार्यक्रम आदि को लिखित रूप में प्रस्तुत करना ही वृत्तांत लेखन है।

• वृत्तांत संक्षिप्त होना चाहिए और क्रमबद्धता का विशेष ध्यान रखना चाहिए।
• वृत्तांत लेखन में उत्तम पुरुष वाचक सर्वनाम (मैं, हम) का प्रयोग नहीं करना चाहिए।
• वृत्तांत में घटना, समय, स्थान आदि का स्पष्ट उल्लेख होना चाहिए।
• यह सत्य घटना पर आधारित लेखन होता है।

1. हिंदी दिवस का वृत्तांत लिखिए।

एक शानदार हिंदी दिवम

15 सितंबर मुंबई : स्वामी विवेकानंद ज्युनियर कॉलेज आफॅ आर्ट्स एंड कॉमर्स के सभागार में दोपहर तीन बजे प्रधानाचार्य महोदय की अध्यक्षता में एक शानदार कार्यक्रम संपन्न हुआ। हिंदी विभागाध्यक्ष डॉ. किशोर सिंह ने राष्ट्रभाषा के रूप में हिंदी तथा बारहवीं कक्षा की छात्रा शिवानी और रुही में हिंदी की आज की स्थिति पर अपने विचार रखे। ग्यारहवी कक्षा के छात्रों ने राष्ट्रभक्ति पर गीत प्रस्तुत किए।

इस अवसर पर दोहों की प्रतियोगिता रखी गई जिसकी वजह से कार्यक्रम में जान आ गई थी। रमा राणे इस प्रतियोगिता में विजयी हुई जिसे पाँच सौ रुपए का पुरस्कार और सर्टीफिकेट प्रदान किया गया।

प्रधानाचार्य ने इस कार्यक्रम में घोषणा की, कि इस वर्ष हिंदी में सर्वाधिक अंक प्राप्त करने वाले ग्यारहवीं तथा बारहवीं के छात्र को एक हजार रूपए का नकद पुरस्कार दिया जाएगा। सभी उपस्थित लोगों को अध्यक्ष महोदय ने धन्यवाद दिया और सभा विसर्जित हुई।

(कार्यालय प्रतिनिधि द्वारा)

2. महाविद्यालय में आयोजित वृक्षारोपण समारोह का वृत्तांत लिखीए।

अध्यापक और छात्रों द्वारा वृक्षारोपण

17 जुलाई, दिल्ली: आज 16 जुलाई को विकास महाविद्यालय हरिनगर, दिल्ली के महाविद्यालय परिसर में वृक्षारोपण समारोह अत्यंत हर्षोल्लास एवं उत्साहपूर्वक मनाया गया। इस कार्यक्रम में शिक्षा निदेशक मुख्य अतिथि के रूप में उपस्थित थे। उन्हें पुष्पगुच्छ एवं एक पौधा देकर स्वागत किया गया।

इसके पश्चात छात्र-छात्राओं ने एक स्वर में, ‘नंगी धरती करे पुकार, वृक्ष लगाकर करो शृंगार’ गीत का समूहगान प्रस्तुत किया। अतिथि महोदय ने अपने भाषण में वृक्षों के महत्त्व पर प्रकाश डाला। उन्होंने स्वयं एक पौधा लगाकर कार्यक्रम का शुभारंभ किया।

फिर प्रधानाचार्य, उपप्रधानाचार्य तथा अध्यापकों ने वृक्षारोपण किया। छात्रों ने भी वृक्षारोपण करते हुए उनके देखभाल की जिम्मेदारी ली। पौधों के चारों ओर जाली लगाकर इनकी सिंचाई का प्रबंध किया गया ताकि पौधे फलें- फूलें और वृक्ष बन सकें। अंत में छात्रों में मिठाई वितरण करते हुए इस समारोह का समापन किया गया।

(कार्यालय प्रतिनिधि द्वारा)

3. समता विद्यालय में पुरस्कार वितरण समारोह संपन्न

(कार्यालय प्रतिनिधि द्वारा)

पुणे, 12 फरवरी: कल 11 फरवरी, को समता विद्यालय में वार्षिक पुरस्कार वितरण समारोह संपन्न हुआ। समारोह की अध्यक्षता मशहूर अभिनेता शेखर सेन ने की थी।

ईश – स्तवन और गणेश वंदना से कार्यक्रम आरंभ हुआ। पाँचवी कक्षा के छात्रों ने स्वागत गीत प्रस्तुत किया। तत्पश्चात विद्यालय के निरीक्षक श्री. अशोक कर्वे ने अध्यक्ष महोदय का परिचय और विद्यालय की विभिन्न गतिविधियों का विवरण प्रस्तुत किया।

अध्यक्ष महोदय के करकमलों से आदर्श विद्यार्थी, सर्वश्रेष्ठ खिलाड़ी, अभिनेता, नर्तक, गायक आदि पुरस्कार दिए गए। शालांत परीक्षा में विशेष योग्यता दिखलाने वाले मेधावी छात्रों को सम्मानित किया गया। अध्यक्ष महोदय ने अपने मार्गदर्शन पर भाषण में विद्यार्थियों को अनुशासन का पाठ पढ़ाया और उज्ज्वल भविष्य की कामना की। तत्पश्चात विद्यालय की ज्येष्ठ शिक्षिका श्रीमती चौधरी जी ने धन्यवाद यापन किया और राष्ट्रगीत के साथ समारोह का समापन हुआ।

नासिक, 7 फरवरी : विवेक महाविद्यालय के प्रांगण में बारहवीं कक्षा के छात्रों का बिदाई समारोह 6 फरवरी को संपन्न हुआ। अपने महाविद्यालय से विदा लेते समय विद्यार्थियों की आँखें छलक आईं। समारोह की अध्यक्षता महाविद्यालय के प्रधानाचार्य ने की।

ग्यारहवीं कक्षा के छात्रों ने बिदाई समारोह के आयोजन में अहं भूमिका निभाई। ठीक तीन बजे छात्र और शिक्षक प्रांगण में उपस्थित हुए थे। महाविद्यालय को रंगीन कागजों से सजाया गया था। एक छोटा सा वृक्ष का चित्र दीवार पर बना था और छात्र उस पर अपनी भावनाओं सरस्वती वंदना से कार्यक्रम की शुरुआत हुई। बारहवीं कक्षा के छात्रों ने अपने महाविद्यालय की यादें बताते हुए अपने शिक्षिकों का आभार प्रकट किया।

प्रधानाचार्य और वर्गशिक्षकों तथा ज्येष्ठ शिक्षकों ने विद्यार्थियों को मार्गदर्शन पर दो शब्द कहे। ग्यारहवीं कक्षा के छात्रों ने नृत्य एवं नाटिका प्रस्तुत कर सबका मनोरंजन किया। प्रधानाचार्य के करकमलों द्वारा आदर्श छात्र एवं छात्रा को पुरस्कृत किया गया। उसके बाद अल्पाहार दिया गया। छात्र अपने प्रिय शिक्षकों के साथ तस्वीरें लेते हुए देखे गए। बड़ा ही भावपूर्ण प्रसंग था यह, जो शाम सात बजे खत्म हुआ।

Balbharti Maharashtra State Board Hindi Yuvakbharati 11th Digest रचना संभाषण लेखन Notes, Questions and Answers.

Maharashtra State Board 11th Hindi रचना संभाषण लेखन

रोजमर्रा के जीवन में हम जो बातचीत या वार्तालाप करते हैं उसके लिखित रूप को संवाद-लेखन कहते हैं। वार्तालाप जितना चतुराई से किया गया है, उतना ही वह प्रभावशाली होता है। संवाद-लेखन करते समय ध्यान रहे –

• संवाद की भाषा सरल एवं प्रभावशाली हो।
• संवाद संक्षिप्त होने चाहिए।
• संवाद विषय और पात्र के अनुकूल होने चाहिए।
• संवाद लेखन में उचित विराम चिह्नों का प्रयोग किया जाना चाहिए।

प्रश्न 1.
निम्नलिखित जानकारी के आधार पर संवाद-लेखन कीजिए :

उत्तर :
अनन्या : हे रक्षिता! आज कितने दिनों बाद दिखाई दे रही हो; कहाँ थी?
रक्षिता : माँ के साथ दुर्घटना हो गई थी, इसलिए इधर आना नहीं हुआ।

अनन्या : ओह! क्या हुआ था?
रक्षिता : किसी बदमाश ने राह चलते उनके गले का मंगलसूत्र खींच लिया था। उनके गले पर जख्म हो गया था।

अनन्या : ओह! यह तो बहुत बुरा हुआ। आजकल दिनदहाड़े ऐसी वारदातें होने लगी हैं।
रक्षिता : काश! पुलिस अपनी जिम्मेदारियाँ ठीक से निभा पाते, तो बदमाशों की ऐसी हिम्मत नहीं होती।

अनन्या : (क्रोधित स्वर में) और हमारी पब्लिक तमाश-बिन की तरह केवल भीड़ इकट्ठा करती है।
रक्षिता : (सहमति प्रकट करते हुए) और नहीं तो क्या! हम लोग कब जिम्मेदार नागरिक बनेंगे?

अनन्या : क्या, बदमाश पकडा गया?
रक्षिता : नहीं तो! वह तो बाइक पर था, झपट्टा मारा और भाग गया।

अनन्या : खैर, तुम अपनी माँ का ख्याल रखो। मुझसे कोई सहायता चाहिए तो बिना हिचकिचाए बताना।
रक्षिता : अवश्य! चलती हूँ। अलविदा!

अनन्या : अलविदा !

प्रश्न 2.
निम्नलिखित जानकारी के आधार पर संवाद-लेखन कीजिए :

उत्तर :
पिताजी : बेटा, यहाँ आओ; बैठो। पढ़ाई के क्या हाल है? परीक्षा कब से है?
पुत्र : (घबराते हुए) परीक्षा नजदीक आई है और पढ़ाई अभी चल रही है, पूरी नहीं हुई।

पिताजी : पूरी होगी कैसे? मैं यह नहीं कहता कि दूरदर्शन मत देखो, लेकिन अपने अध्ययन के प्रति लापरवाही उचित नहीं। पूरा दिन दूरदर्शन के सामने बैठोगे तो पढ़ाई होगी ही नहीं।
पुत्र : लेकिन मैं अकेला थोड़े ही देखता हूँ, पापा? घर में सभी दूरदर्शन देखते हैं और आप मुझे ही डाँटते हो।

पिताजी : मैं तुम्हें समझा रहा हूँ। तुम्हारी दीदी को देखो, कक्षा में अव्वल आती है और तुम मात्र 50% अंक ला पाए हो।
पुत्र : (अँगूठे से जमीन कुरेदते हुए) मुझे दीदी अपने साथ पढ़ने को मना करती है।

पिताजी : मैंने दीदी को समझा दिया है, जाओ अब उसके साथ बैठकर पढ़ाई करो।
पुत्र : जी, पापा। अब मैं भी दीदी जैसे अंक लाकर दिखाऊँगा।

पिताजी : बहुत अच्छा! देर आए दुरुस्त आए। मुझे तुम पर विश्वास है। तुम जरूर अच्छे अंक ला पाओगे। भगवान तुम्हें सद्बुद्धि दें।

प्रश्न 3.
निम्नलिखित जानकारी के आधार पर संवाद-लेखन कीजिए :

उत्तर :
राम : नमस्ते श्याम! कैसे हो?
श्याम : नमस्ते! मैं ठीक हूँ। तुम कैसे हो? क्या कर रहे हो आजकल?

राम : मैं ठीक हूँ। आजकल मैं डाक टिकट इकट्ठा कर रहा हूँ।
श्याम : बहुत खूब! क्या तुम इन्हें अलबम में चिपकाओगे?

राम : हाँ! मैंने एक अलबम बना लिया है और टिकट चिपका भी दिए हैं।
श्याम : वाह! क्या, तुम्हारे पास सभी देशों के टिकट हैं?

राम : हाँ! ज्यादातर सभी देशों के टिकट मेरे पास हैं।
श्याम : (जिज्ञासा से) तो क्या इनमें, महँगी टिकटें भी हैं?

राम : मेरे पास बहुत सारी टिकटें है जिनमें कुछ टिकटें महँगी भी है।
श्याम : मेरे दोस्त, यह तो बता इस संग्रह से तुझे क्या लाभ होता है?

राम : यह मेरा शौक है जो मुझे बेहद सुख प्राप्त कराता है और हाँ, मुझे भूगोल पढ़ने में इनकी मदद मिलती है।
श्याम : बहुत अच्छा।

राम : तुम्हारा भी कोई शौक है?
श्याम : है न! मुझे जंगली फूल जमा करने का शौक है।

राम : तुम उनसे क्या करते हो?
श्याम : मैं उन्हें कागज पर चिपकाता हूँ और फिर उनका नाम लिखता हूँ।

राम : इस शौक से तुम्हें क्या लाभ होता है?
श्याम : मेरा शौक मेरा वनस्पति विज्ञान का ज्ञान बढ़ाता है।

राम : तुमसे मिलकर आज बहुत अच्छा लगा। फिर मिलेंगे, अलविदा!
श्याम : अच्छा! अलविदा!

प्रश्न 4.
निम्नलिखित जानकारी के आधार पर संवाद-लेखन कीजिए।

उत्तर :
साक्षात्कार कर्ता : जय हिंद, मेरे भाई, मेरे दोस्त!
सैनिक : जय हिंद! कहो कैसे आना हुआ?

साक्षात्कार कर्ता : आप हाल ही में सीमा पर दुश्मनों को लोहे के चने चबवाकर आए हो इसलिए मैं आपसे मिलना चाहता हूँ। आपके बारे में जानना चाहता हूँ।
सैनिक : भाई, मैंने तो सिर्फ अपना कर्तव्य निभाया है।

साक्षात्कार कर्ता : सीमा पर अपने परिवार से दूर आप कैसे रह लेते हो? क्या आपका उनके प्रति कर्तव्य नहीं है?
सैनिक : ऐसा तो नहीं। परिवार अपनी जगह है, देश अपनी जगह और देश की खातिर जो कर्तव्य है उसके आगे हमें निजी सुख-दुख बहुत छोटे नजर आते हैं।

साक्षात्कार कर्ता : धन्य हैं आप! सुना है आपका बेटा छह वर्ष का है!
सैनिक : सही सुना है आपने। मेरा छह वर्ष का बेटा है जो अभी से सैनिक बनने का सपना देख रहा है और हाँ मेरी तीन साल की बेटी भी, शत्रु के साथ युद्ध करने की बातें करती है।

साक्षात्कार कर्ता : बहुत अच्छा लगा सुनकर। आपका पूरापरिवार ही राष्ट्रभक्ति से ओत-प्रोत है। हम देशवासियों को और देश के बालकों को (जोर देते हुए) कुछ संदेश देना चाहेंगे?
सैनिक : जरूर! देश के बालकों आप देश का भविष्य हो। हर काम को सच्चाई, ईमानदारी और खुशी से करो। अपने सपनों को हकीकत में बदलो। ईश्वर तुम्हें सदबुद्धि दे। वंदे मातरम्! जय हिंद!

साक्षात्कार कर्ता : वंदे मातरम्! जय हिंद!

Balbharti Maharashtra State Board Hindi Yuvakbharati 11th Digest रचना गद्य आकलन Notes, Questions and Answers.

Maharashtra State Board 11th Hindi रचना गद्य आकलन

अपठित अर्थात जो पहले से पढ़ा / पढ़ाया न गया हो ऐसा परिच्छेद परीक्षा में दिया जाता है। इसे पढ़कर इसका आशय समझना होता है। कोई शब्द परिचित न हो और अर्थ समझ में नहीं आ रहा हो तो उसके अर्थ को वाक्य के प्रसंगानुसार ग्रहण करना चाहिए और सब कुछ समझ में आ जाने पर प्रश्न बनाना आसान हो जाएगा।

महत्त्वपूर्ण : छात्रों से अपठित गद्यांश पर आकलन हेतु मात्र प्रश्न निर्माण अपेक्षित है और प्रश्न भी ऐसे बनाने हैं जिनके उत्तर एक वाक्य में हों। हो सके उतना गद्यांश के लिए शीर्षक के बारे में प्रश्न न पूछे। आगे कुछ उदाहरण दिए हैं –

प्रश्न 1.
निम्नलिखित गद्यांश पढ़कर पाँच ऐसे प्रश्न तैयार कीजिए जिनके उत्तर एक-एक वाक्य में हों।

पंडित जवाहरलाल नेहरूजी की अंतिम इच्छा यह थी कि मैं जब मरूँ तब मैं चाहूँगा कि मेरा दाह-संस्कार हो। अगर मैं विदेश में मरूँ तो मुझे वहीं जलाया जाए पर मेरी अस्थियाँ इलाहाबाद लाई जाएँ। मुठ्ठीभर भस्म इलाहबाद की गंगा में प्रवाहित करने की मेरी इच्छा है, किंतु उसके पीछे कुछ धार्मिक भावना नहीं है, क्योंकि गंगा हमारी सदियों से पुरानी सभ्यता और संस्कृति की प्रतीक रही है।

वह मुझे हिमालय के हिमाच्छादित शिखरों और नदियों की याद दिलाती है, जिनमें मेरा लगाव और प्यार बहुत ज्यादा रहा है। गंगा मुझे शस्य-श्यामल फैले हुए मैदानों की याद दिलाती है, यहाँ मेरी जिंदगी और काम ढले हैं। गंगा में कहीं समुद्र जैसी विनाश की भी शक्ति मुझे लगती है और उसकी यह शक्ति मेरे लिए अतीत की प्रतीक व स्मृति है, जो वर्तमान में प्रवाहित है और भविष्य के महासमुद्र में आगे बढ़ते रहने की है।
उत्तर:

1. मुठ्ठीभर भस्म का विसर्जन लेखक ने कहाँ करने के लिए कहा है?
2. गंगा की कौन-सी शक्ति लेखक के लिए अतीत की प्रतीक व स्मृति है?
3. लेखक की जिंदगी और काम कहाँ ढले हैं?
4. पंडित जवाहरलाल नेहरू जी को गंगा नदी किसकी याद दिलाती है?
5. विदेश में मरने पर पंडित जवाहरलाल नेहरू जी क्या चाहते हैं?

प्रश्न 2.
निम्नलिखित गद्यांश पढ़कर पाँच ऐसे प्रश्न तैयार कीजिए जिनके उत्तर एक-एक वाक्य में हों।
वर्तमान शासन प्रणालियों में जनतंत्र से बढ़कर उत्तम कोई प्रणाली नहीं हैं, क्योंकि उसमें जनता को स्वयं यह अधिकार प्राप्त रहता है कि वह अपने प्रतिनिधियों को चुनकर विधान सभाओं और संसद में भेजें। ऐसे प्रत्यक्ष चुनाव में प्राय: वही व्यक्ति चुना जाता है, जिसका सार्वजनिक जीवन अच्छा हो और जो जनता की सेवा करता हो। इस प्रणाली में जनता को यह अधिकार है कि यदि वह किसी दल या किसी व्यक्ति के कार्यों से संतुष्ट नहीं है तो दूसरी बार उस दल या व्यक्ति को अपना मत न दें।

निर्वाचन में विरोधी दलों के भी कुछ व्यक्ति चुने जाते हैं, जो अपनी आलोचना से शासक दल के स्वेच्छाचार पर अंकुश रखते हैं। इस प्रकार देश की शासन प्रणाली में विरोधी दलों का भी महत्त्वपूर्ण स्थान होता है।
उत्तर:

1. जनता अपने प्रतिनिधियों को चुनकर कहाँ भेजती है?
2. चुनाव में कैसा व्यक्ति चुना जाता है?
3. विरोधी दल अपनी आलोचना से क्या कर सकता है?
4. जनतंत्र में जनता को किस बात का अधिकार होता है?
5. सबसे उत्तम शासन प्रणाली कौन-सी है?

प्रश्न 3.
निम्नलिखित गद्यांश पढ़कर पाँच ऐसे प्रश्न तैयार कीजिए जिनके उत्तर एक-एक वाक्य में हों।
दान देने की परिपाटी प्राचीन काल से चली आ रही है। अन्नदान, गोदान, वस्त्रदान, स्वर्णदान, भूमिदान करना भारतीय अपना परम धर्म मानते हैं। धर्म से स्वर्ग की प्राप्ति होती है, ऐसा माना जाता है। प्राचीन काल में विद्यादान को सर्वश्रेष्ठ दान माना जाता था। वर्तमान काल में कुछ नए प्रकार के दान प्रचलित हुए है – नेत्रदान, रक्तदान, किडनीदान । रक्त तो हर व्यक्ति के लिए जरूरी है। पचास वर्ष तक के निरोगी स्त्री-पुरुष रक्त दान कर सकते हैं। दुर्घटनाओं से परिपूर्ण वैज्ञानिक युग में रक्तदान, सर्वश्रेष्ठ दान माना जा रहा है। नेत्रदान करने से घबराना नहीं चाहिए क्योंकि मरणोपरांत ही आँखें निकालकर, अंधों को दी जाती हैं और वे देखने लगते हैं। बीमार के प्राण बचाने के लिए हम अपनी किडनी दान दे सकते हैं।
उत्तर :

1. प्राचीन काल से लेकर अब तक कौन-कौन से दान प्रचलित हैं?
2. किन लोगों को रक्तदान करना चाहिए?
3. वैज्ञानिक युग में कौन-सा दान श्रेष्ठ है?
4. दान करना भारतीय अपना परम धर्म क्यों मानते हैं?
5. नेत्रदान करने से घबराने की जरूरत क्यों नहीं?

प्रश्न 4.
निम्नलिखित गद्यांश पढ़कर पाँच ऐसे प्रश्न तैयार कीजिए जिनके उत्तर एक-एक वाक्य में हों।

भारत में प्राचीन काल से दहेज प्रथा चली आ रही है। कन्या के माता-पिता अपनी क्षमता के अनुसार शादी के समय दहेज देते चले आए हैं। वर एवं कन्या के परिवारवालों में आपसी प्रेम था इसलिए वरवाले कन्यावालों से किसी प्रकार की माँग करने में संकोच करते थे।

परंतु पिछले 50 वर्षों से विवाह एक व्यापार बन गया है। इससे समाज दुखी है। लड़कीवाला लड़के की योग्यता के स्थान पर धन को ही सर्वस्व मानता है और वह बड़े अमीर परिवार में अपनी लड़की को देना चाहता है। लड़का लड़कियों को देखता है।

जिस लड़की के पास धन अधिक होता है, उसे चुन लेता है। उसकी योग्यता को नहीं देखता। आज लड़की के विवाह का मूल आधार धन बन गया है। जिस दिन लड़की का जन्म होता है, उसी दिन से माता-पिता को उसके विवाह की चिंता लग जाती है।

इस बुराई को दूर करने के लिए हमें मिलकर इस प्रथा का विरोध करना चाहिए। जो दहेज लेता है, उसके लिए ऐसा कानून बनना चाहिए कि दहेज लेनेवाले को चोरी, जुआ एवं हत्या आदि अपराध करनेवालों के समान देखा जाए और सामाजिक मंच पर उसे बेइज्जत किया जाए।

इस विषय पर मात्र बोलने एवं लिखने से अब काम नहीं चलेगा। हमें एक होकर इस प्रकार के विरोध में कदम बढ़ाने होंगे।

प्रश्न :
(1) भारत में प्राचीन काल में दहेज प्रथा का स्वरूप कैसा था?
(2) माता-पिता को लड़की के विवाह की चिंता कब से लग जाती है?
(3) आज लड़की के विवाह का मूल आधार क्या बन गया है?
(4) दहेज लेने वाले के साथ कैसा व्यवहार किया जाना चाहिए?
(5) लड़की के विवाह का मूल आधार क्या बन गया है?

प्रश्न 5.
निम्नलिखित गद्यांश पढ़कर पाँच ऐसे प्रश्न तैयार कीजिए जिनके उत्तर एक-एक वाक्य में हों।
पवन पुत्र हनुमान, भीष्म पितामह, स्वामी दयानंद सरस्वती, स्वामी विवेकानंद जैसे बाल ब्रह्मचारियों ने भारत-भूमि को पावन किया है। संसार के प्राचीन ग्रंथ वेदों में लिखा है: “ब्रह्मचारी मृत्यु को जीत लेते हैं।” ‘ब्रह्म’ शब्द के अर्थ हैं ‘परमेश्वर, विद्या और शरीर-रक्षण। ब्रह्मचर्य के पालन से शरीर स्वस्थ होता है।

जिसका शरीर स्वस्थ उसीका मन स्वस्थ, जिसका मन स्वस्थ उसकी स्मरण-शक्ति बहुत होती है। स्मरण-शक्ति से आकलन शक्ति बढ़ती है । विद्यार्थी जीवन में आकलनशक्ति का अपना विशेष महत्त्व है। ब्रह्मचर्य विद्यार्थी जीवन की कमियाँ पूरी करता है।

प्राचीन भारतीय साहित्य में ब्रह्मचर्य की महिमा लिखी है। इसका पालन करनेवाला विद्यार्थी निरोगी, बुद्धिमान, संपत्तिशाली, महान बनता है। ब्रह्मचर्य की महिमा अपार है।

प्रश्न:
(1) कौन-कौन बाल ब्रह्मचारी थे?
(2) मृत्यु को कौन जीत सकते हैं?
(3) ‘ब्रह्म’ शब्द के कितने और कौन-कौन से अर्थ हैं?
(4) विद्यार्थी जीवन में किसका विशेष महत्त्व है?
(5) ब्रह्मचर्य पालन करने वाला विद्यार्थी कैसा होता है?

स्वाध्याय

निम्नलिखित गद्यांश पढ़कर पाँच ऐसे प्रश्न तैयार कीजिए जिनके उत्तर एक वाक्य में हों –

(1) हँसने का एक सामाजिक पक्ष भी होता है। हँसकर हम लोगों को अपने निकट ला सकते हैं और व्यंग्य उन्हें दूरस्थ बना देते हैं। जिसको भगाना हो उसकी थोड़ी देर हँसी खिल्ली उड़ाइए, वह तुरंत बोरिया-बिस्तर गोल कर पलायन करेगा। जितनी मुक्त हँसी होगी, उतना समीप व्यक्ति खींचेगा इसीलिए तो श्रोताओं की सहानुभूति अपनी ओर खींचने के लिए चतुर वक्ता अपना भाषण किसी रोचक कहानी या घटना से आरंभ करते हैं।

जनता यदि हँसी तो चंगुल में फँसी। सामाजिक मूल्यों और नियमों को मान्यता दिलाने और रूढ़ियों को निष्कासित करने में पुलिस या कानून सहायता नहीं करता, किन्तु वहाँ हास्य का चाबुक अचूक बैठता है। हास्य के कोड़े, उपहास-डंक और व्यंग्य-बाण मारकर आदमी को रास्ते पर लाया जा सकता है। इस प्रकार गुमराह बने समाज की रक्षा की जा सकती है।

(2) किसी भी देश या काल के लिए जवान तथा शिक्षक दोनों ही महत्त्वपूर्ण हैं। किसी एक के बिना समाज सुरक्षित नहीं रह सकता। दोनों ही समाज के रक्षक हैं, किन्तु कार्यों में भिन्नता दिखाई पड़ती है। एक शत्रु से रक्षा करता है तो दूसरा उसे (देश को) समृद्ध बनाता है।

फिर भी शिक्षक का उत्तरदायित्व जवान से कहीं बढ़कर है। भावी नागरिक निर्माण करने की जिम्मेदारी शिक्षक के ऊपर है। वह उसके शारीरिक, मानसिक तथा नैतिक विकास का जनक है, जिस पर व्यक्ति, समाज तथा राष्ट्र निर्भर है। शिक्षक के ही द्वारा कोई योग्य सैनिक बन सकता है।

आज शिक्षक ने सैनिक धाराओं में क्रांति पैदा कर दी है। हमारे अहिंसक आंदोलन ने दुनिया को दिखा दिया है कि शिक्षक सैनिक से श्रेष्ठ है। इसे बनावटी-शस्त्रों की जरूरत नहीं है। इसका आत्मिक बल सब शस्त्रों से बड़ा है।

अगर आनेवाली दुनिया इसका अनुसरण करे तो शस्त्रीकरण का नामोनिशान भू-पृष्ठ से उठ जाएगा। नैतिक शक्ति का बोल-बाला होगा, सारी दुनिया में एकात्मता की लहर फैलेगी और तब ज्ञान-विज्ञान का निर्माण विकास के लिए होगा, न कि विनाश के लिए।

(3) समाजसुधार आंदोलन को निर्भीक संन्यासी स्वामी श्रद्धानंद से नई दिशा मिली। हरिजन समस्या के समाधान में कई स्थानों पर संघर्ष का भी सामना करना पड़ा। गुरुकुल काँगड़ी के छात्रवासों और भोजनालयों में बिना किसी भेदभाव के हर जाति के विद्यार्थी रहते और खाते-पीते थे।

स्वामी जी का कहना था – मनों से छुआछूत की भावना मिटाने में आवासीय शिक्षण संस्थाओं का अच्छा योगदान हो सकता है। चौबीसों घंटे एक साथ मिलकर जब रहेंगे, खेलेंगे, कूदेंगे और पढ़ेंगे, लिखेंगे तो कहाँ तक छूत-अछूत की दीवार खड़ी रह पाएगी।

आजादी के बाद भी यदि इसी रास्ते को पकड़ा गया होता तो मंजिल बहुत पहले तय हो जाती। आवासीय पद्धति पर आश्रित ऐसे गुरुकुल उन्होंने हरियाणा में झज्जर, इंद्रप्रस्थ और कुरूक्षेत्र, गुजरात में सोनगढ़ और सूपा में भी खोले। देहरादून का कन्या गुरुकुल भी उसी श्रृंखला की कड़ी है।

(4) यश और कीर्ति पैतृक संपत्ति नहीं है। जिसका सुख-भोग संतान कर सके। वास्तविक सम्मान और यश धन के द्वारा भी प्राप्त नहीं हो सकता। ये वे पदार्थ है जो घोर परिश्रम और स्वावलंबन द्वारा ही प्राप्त हो सकते हैं। ईश्वर का वरद हस्त भी उसी के शीश पर है जो स्वत: अपनी सहायता करता है।

यदि तुम अपना जीवन धन्य बनाना चाहते हो तो खड़े हो जाओ अपने पैरों पर और संसार में एक बार शक्ति से अपने कार्यो से सुख और शांति की धारा प्रवाहित कर दो। भाग्य की भाषा पढ़ने के फेरे में जो भी डूबा वह कभी भी ऊपर नहीं आ सकता। अत: यह निश्चित है कि तुम ही अपने भाग्य विधाता हो और जीवन निर्माण करने का संपूर्ण अधिकार भी तुम ही को है।

(5) संसार में कुछ भी असाध्य नहीं है। कुछ भी असम्भव नहीं है। असम्भव, असाध्य, कठिन आदि शब्द कायरों के लिए हैं।

नेपोलियन के लिए ये शब्द उसके कोष में नहीं थे। साहसी पतले बापू ने विश्व को चकित कर दिया। क्या बापू शरीर से शक्तिशाली थे? नहीं। वह तो पतली-सी एक लंगोटी पहने लकड़ी के सहारे चलते थे, परंतु विचार सशक्त थे, भावनाएँ शक्तिशाली थीं, उनके साहस को देखकर करोड़ो भारतीय उनके पीछे थे। ब्रिटिश साम्राज्य उनसे काँप गया। अहिंसा के सहारे बिना रक्त-पात के उन्होंने भारत को स्वतंत्र कराया। यह विश्व का एक अद्वितीय उदाहरण है।

जब महात्मा गांधीजी ने अहिंसा का नारा लगाया तो लोग हँसते थे, कहते थे अहिंसा से कहीं ब्रिटिश साम्राज्य से टक्कर ली जा सकती है? परंतु वे डटे रहे, साहस नहीं छोड़ा, अंत में अहिंसा की ही विजय हुई। कहते हैं, अकेला चना क्या भाड़ फोड़ सकता है? हाँ, यदि उसमें साहस हो तो! साहसहीन के लिए सब कुछ असम्भव है। उससे अगर कहा जाए कि भाई जरा वह काम कर देना; तो वह तुरंत कहेगा, अरे! इतनी दूर!

पैदल, एक दिन में! नहीं भाई, मुझसे नहीं हो सकेगा, किसी और से करा लो। भला वह इस काम को कैसे करेगा? करने वाला दूरी और पैदल नहीं देखता! उसके मार्ग में चाहे पर्वत आकर खड़े हो जाएँ, आँधी आए या तूफान, उसको उनसे क्या वास्ता? उसको तो अपने लक्ष्य तक पहुँचना है, उसे कोई नहीं रोक सकता है। वह अपने लक्ष्य तक अवश्य पहुँच जाएगा। साहसी पुरुष दिन-रात नहीं देखा करते, आँधी तूफान, नदी-नाले, पहाड़-समुद्र नहीं देखा करते; वे तो केवल एक ही चीज देखा करते हैं कि उन्हें अपने लक्ष्य तक पहुँचना है।

Balbharti Maharashtra State Board Hindi Yuvakbharati 11th Digest रचना अनुवाद लेखन Notes, Questions and Answers.

Maharashtra State Board 11th Hindi रचना अनुवाद लेखन

अनुवाद लेखन : किसी भाषा में कही या लिखी गई बात का किसी दूसरी भाषा में सार्थक परिवर्तन अनुवाद कहलाता है। अनुवाद एक कला है। अनुवाद करते समय शब्दों का ही केवल अनुवाद नहीं करना है वाक्य में जो भाव है उसके अनुसार शब्दों का चयन और क्रम रखकर मौलिक भाव को प्रस्तुत करना होता है। आपको अंग्रेजी से हिंदी में अनुवाद करने के लिए पूछा जाएगा।

उदा.
His dreams became true
अनुवाद – उसके सपने सच हुए।

निम्नलिखित वाक्यों का हिंदी में अनुवाद कीजिए

प्रश्न 1.
Mistakes are always forgivable, if one has the courage to admit them.
उत्तर:
गलतियाँ हमेशा क्षमा की जा सकती हैं, यदि आपके पास उन्हें स्वीकारने का साहस हो।

प्रश्न 2.
As you think, so shall you become.
उत्तर:
जैसा आप सोचते हैं, वैसा आप बन जाएँगे।

प्रश्न 3.
A quick temper will make a fool of you soon enough
उत्तर:
जल्दी गुस्सा करना जल्द ही आपको मूर्ख साबित कर देगा।

प्रश्न 4.
A man is great by deeds, not by birth.
उत्तर:
व्यक्ति जन्म से नहीं कर्म से महान होता है।

प्रश्न 5.
Success and failure are both part of life and both are not permanent.
उत्तर:
सफलता और असफलता दोनों जीवन के हिस्से हैं और दोनों स्थायी नहीं होते।

प्रश्न 6.
A person who never made a mistake never tried anything new.
उत्तर:
जिस व्यक्ति ने कभी गलती नहीं की उसने कभी कुछ नया करने की कोशिश नहीं की।

प्रश्न 7.
Life should be great rather than long.
उत्तर:
जीवन लंबा होने की बजाय महान होना चाहिए।

प्रश्न 8.
Failure comes only when we forget our ideals and objectives and principles.
उत्तर:
असफलता तभी आती है जब हम अपने आदर्श, उद्देश्य और सिद्धांत भूल जाते हैं।

प्रश्न 9.
Health is the greatest gift, contentment the greatest wealth, faithfulness the best relationship.
उत्तर.
स्वास्थ्य सबसे बड़ा उपहार है, संतोष सबसे बड़ा धन है, वफादारी सबसे बड़ा संबंध है।

प्रश्न 10.
Never stop believing in hope because miracles happen everyday.
उत्तर:
उम्मीद पर विश्वास करना न छोड़ें क्योंकि चमत्कार हर दिन होते हैं।