Class 9

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 5 What is inside the Atom? Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 5th Lesson What is inside the Atom?

9th Class Physical Science 5th Lesson What is inside the Atom? Textbook Questions and Answers

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Question 1.
What are the three subatomic particles? (AS 1)
Answer:
The three sub-atomic particles are electrons, protons, and neutrons.

Question 2.
Compare the subatomic particles electron, proton, and neutron. (AS 1)
Answer:

Question 3.
What are the limitations of J.J. Thomson’s model of the atom? (AS 1)
Answer:
The main limitation of J.J. Thomson’s model of atom was that he is unable to explain, how the positively charged particles are shielded from negatively charged particles without getting neutralized.

Question 4.
What were the three major observations Rutherford made in the gold foil experiment? (AS 1)
Answer:
The three major observations Rutherford made in the gold foil experiment were

1. Most of the space inside the atom is empty.
2. All the positive charge must be concentrated in a very small space within the atom called nucleus.
3. The size of the nucleus is very small as compared to the size of the atom.

Question 5.
Sketch Rutherford’s atomic model. Why is Rutherford’s model of the atom called the planetary model? (AS 5)
Answer:

Rutherford’s model is called planetary model because the motion of the electrons around the nucleus resembles the motion of the planets around the sun.

Question 6.
Put tick (✓) against correct choice and cross (✗) against wrong choice: (AS 1)
i) In Rutherford’s gold foil experiment, majority of alpha particles passed directly through the gold foil. This observation leads to which conclusions?
a) The positively charged region of the atom is very small. (✗)
b) The atom must consist of empty space. (✓)
c) The alpha particles makes a direct hit on the positively charged region of the atom. (✗)
d) The positively charged region of the atom is very dense. (✗)

ii) In Rutherford’s gold foil experiment, occasionally the alpha particles veered from a straight line path. This observation leads to which conclusion?
a) The positively charged region of the atom is very small. (✗)
b) The majority of the space in the atom is empty. (✗)
c) The alpha particle makes a direct hit on the positively charged region of the atom. (✗)
d) The positively charged region of the atom is very dense. (✓)

Question 7.
Which one of the following is the correct electronic configuration of sodium? (AS 1)
a) 2, 8
b) 8, 2, 1
c) 2, 1, 8
d) 2, 8, 1
Answer:
d) 2, 8, 1

Question 8.
Give the main postulates of Bohr’s model of an atom. (AS 1)
Answer:
The main postulates of Bohr’s model of atom are

1. Only certain special, discrete orbits of electrons are allowed inside the atom. These orbits or shells are called energy levels.
2. While revolving in these discrete orbits the electrons do not radiate energy and this helps that the electrons do not crash into the nucleus.
3. These orbits or shells are represented by the letters K, L, M, N, ……… or the numbers n = 1, 2, 3, …………

Question 9.
Compare all the proposed models of an atom given in this chapter. (AS 1)
Answer:
In this chapter, four atomic models were discussed. The main postulates of those models are

1. Dalton’s proposal :
a) Atoms are indivisible.
b) Atoms of an element are all identical to each other and different from the atoms of other elements.

2. Thomson’s proposal:
a) An atom is considered to be a sphere of uniform positive charge and electrons are embedded into it.
b) The total mass of the atom is considered to be uniformly distributed throughout the atom.

3. Rutherford’s proposal:
a) All the positively charged material in an atom formed a small dense centre, called the nucleus of the atom.
b) Negatively charged electrons revolve around the nucleus in well defined orbits.
c) Size of the nucleus is very small as compared to the size of the atom.

4. Neils Bohr’s proposal :
a) Electrons are revolving around the nucleus in special, discrete orbits called en¬ergy levels or shells.
b) While revolving in these discrete orbits the electron do not radiate energy and this helps that the electrons do not crash into the nucleus.
c) These orbits or shells are represented by the letters K, L, M, N,…. or the numbers n = 1, 2, 3,

Question 1o.
Define valency by taking examples of nitrogen and boron. (AS 1)
Answer:
Valency: The number of electrons present in outer most orbit of an atom is called its valency. Valency of Nitrogen :
a) Atomic number of nitrogen is 7.
b) The distribution of electrons is 2, 5.
c) The outer most orbit has 5 electrons.
d) Hence its valency should be 5. But it is easier to nitrogen to gain 3 electrons than to loose 5 electrons for becoming octet.
e) Hence the valency of nitrogen is ‘3’.

Valency of Boron :
a) Atomic number of boron is 5.
b) The distribution of electrons is 2, 3.
c) The outer most orbit has 3 electrons.
d) Hence the valency of boron is 3.

Question 11.
State the valencies of the following elements : magnesium and sodium. (AS 1)
Answer:
Magnesium :
a) Atomic number of magnesium is 12.
b) Distribution of electrons is 2, 8, 2.
c) Hence the valency is 2.

Sodium :
a) Atomic number of sodium is 11.
b) Distribution of electrons is 2, 8, 1.
c) Hence the valency is 1.

Question 12.
If Z = 5, what would be the valency of the element? (AS 2)
Answer:
1) If Z = 5, the distribution of electrons is 2, 3.
2) Hence the valency is ‘3’.

Question 13.
Write the atomic number and the symbol of an element which has mass number 32 and the number of neutrons 16 in the nucleus. (AS 1)
Answer:
Mass number (A) = 32 ; Number of neutrons (N) = 16
Number of protons (Z) = A- N = 32-16 = 16
∴ Atomic number =16
The element is sulphur.
The symbol of sulphur is S’.

Question 14.
Cl- has completely filled K and L shells. Explain. (AS 1)
Answer:
Atomic number of Cl is 17, but Cl^– has one more electron when compared with Cl atom. Distribution of electrons in Cl^– is

K	L	M
2	8	8

K shell can accommodate 2 electrons and L shell cab accommodate 8 electrons accord¬ing to the formula 2n2.
Hence the K and L shells are completely filled.

Question 15.
What is the main difference among the isotopes of the same element? (AS 1)
Answer:
The main difference between isotopes of the same element is
a) The number of neutrons is different.
b) Their physical properties are different but the chemical properties are similar.

Question 16.
For the following statements, write T for True and F for False. (AS 1)
a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
c) The mass of an electron is 1836 times that of proton.
Answer:
a) False
b) False
c) True

Question 17.
Fill in the missing information in the following table. (AS 4)

Answer:

Question 18.
How do you appreciate the efforts made by scientists to explain the structure of atom by developing various atomic models? (AS 6)
Answer:

• Structure of atom, till today it is mysterious and challenging the scientists.
• We have to appreciate the scientists right from Lavoisier, who proposed law of conservation of mass, Proust who proposed law of constant proportions, John Dalton for his first model of atom, Rutherford for giving planetary model of atom and Neils Bohr for his model of atom.
• Till today scientists are trying to know the existence of more and sub-atomic particles besides electrons, protons and neutrons.
• Hence the efforts of scientists are highly appreciable, for making our lives comfortable and leaving many challenges before us to unveil or discover them.

Question 19.
Geeta got a doubt, “Why does atomic nucleus contain proton and neutrons? Why can’t electrons and neutrons be in it”. Can you help to clarify her doubt? Explain. (AS 1)
Answer:
Nucleus contains protons and neutrons inside it but not electrons and neutrons. If it would have happen, then

1. the alpha particles in the Rutherford’s alpha particle scattering experiment would have not been deflected or scattered.
2. the idea of nucleus would have not been evolved because the mass of electron is negligible, it is most unstable.

Question 20.
Collect information about various experiments conducted and theories proposed by scientists starting from John Dalton to Neils Bohr. Prepare a story with a title “History of atom”. (AS 4)
Answer:
History of atom
John Dalton proposed atomic theory based on law of conservation of mass and law of constant proportion as :

1. Atoms were indivisible.
2. Atoms of an element are all identical to each other and different from the atoms of other elements.

Later on various experiments conducted by Thomson, Goldstein, etc. proved that atom is divisible and consists sub-atomic particles like electrons, protons and neutrons. Based on this J.J. Thomson proposed a model of atom in 1898.

According to Thomson,

1. An atom is considered to be a sphere of uniform positive charge and electrons are embedded into it.
2. The total mass of the atom is considered to be uniformly distributed throughout the atom.
3. The negative and the positive charges are supposed to be balance out and the atom as a whole is electrically neutral.

This model is also called as plum pudding model or watermelon model.

Thomson’s student Ernest Rutherford conducted alpha particle scattering experiment and got the results which were not in favour of Thomson’s model. Based on his experiment, Rutherford proposed a model of atom. According to him,

1. All the positively charged material in an atom formed a small dense centre, called the nucleus of the atom. The electrons were not a part of nucleus.
2. Negatively charged electrons revolve around the nucleus in well – defined orbits like planets revolve around the sun.
3. The size of nucleus is very small as compared to the size of the atom.

This model could not account for stability of atom, as revolving electron must lose energy and eventually crash into the nucleus, as a result matter would not exist in the form that we see it now.

In 1913, Neils Bohr proposed another model to overcome Rutherford’s defect. According to Bohr,

1. Only certain special, discrete orbits of electrons are allowed in side the atom. These orbits or shells are called energy levels.
2. While revolving in these discrete orbits the electrons do not radiate energy and this helps that the electrons do not crash into the nucleus.
3. These orbits or shells are represented by K, L, M, N ………… or the numbers 1, 2, 3,

This model could not predict the spectra of atoms.

Hence this journey continues

9th Class Physical Science 5th Lesson What is inside the Atom? InText Questions and Answers

9th Class Physical Science Textbook Page No. 75

Question 1.
Why are the atoms of different elements different?
Answer:
Nature and properties of elements depends on the arrangement of atoms. We know that different elements behave differently. This is due to the difference in their atoms.

Question 2.
Is there anything inside atom that make them to be same or different?
Answer:
The arrangement of sub atomic particles inside the atom is responsible to make them to be same or different.

Question 3.
Are atoms indivisible?
Answer:
No, atom is divisible. There are many sub-atomic particles inside the atom according to the recent experiments.

9th Class Physical Science Textbook Page No. 77

Question 4.
If an atom consists of sub-atomic particles like protons, neutrons and electrons, how are they arranged in the atom ?
Answer:
The arrangement of sub-atomic particles like protons, neutrons and electrons has been explained by many scientists like Rutherford, Neils Bohr, etc. According to them, atom consists a central mass called nucleus. Nucleus consists protons and neutrons. Electrons revolve around the nucleus in fixed shells.

9th Class Physical Science Textbook Page No. 80

Question 5.
Why is atom stable?
Answer:
In an atom, the number of protons in the nucleus is equal to the number of electrons out side the nucleus. Hence the positive and negative charges in an atom are equal. So, atom is electrically neutral. So, atom is stable. But the stability of atom was explained by Neils Bohr in a different way.

Question 6.
Can you suggest any other arrangement of subatomic particles in the atom which prevents the revolving electron to fall into the nucleus?
Answer:
Electrons have to revolve around the nucleus in definite orbits such that the centripetal and centrifugal forces acting on the electron must be equal in magnitude and opposite direction. Then the revolving electron do not fall into the nucleus.

9th Class Physical Science Textbook Page No. 82

Question 7.
How many electrons can be accommodated in each shell of an atom?
Answer:
The number of electrons that can be accommodated in each shell of an atom depends on the shell number. First shell (K) consists 2 electrons, second (L) shell consists 8 electrons, third (M) shell consists 18 electrons, fourth shell (N) consists 32 electrons, and so on.

Question 8.
Can a particular shell have just one electron?
Answer:
No, shell has just one electron.

Question 9.
What is the criteria to decide number of electrons in a shell?
Answer:
The number of electrons in a shell can be decided by using a formula 2n². (Where n is shell number).

9th Class Physical Science Textbook Page No. 83

Question 10.
What is the valency of oxygen that you can calculate by the method discussed above?
Answer:

• Oxygen has 8 electrons in its atom. The distribution of electrons is 2, 6.
• The outer most shell consists 6 electrons, this number is hear to 8.
• Hence the valency of oxygen is 8 – 6 = 2.

9th Class Physical Science Textbook Page No. 85

Question 11.
Should we consider the number of neutrons as a characteristic of an atom?
Answer:
The mass of an atom which is a characteristic of an atom depends on the number of neutrons and protons that its nucleus contains. Hence the number of neutrons can be considered as a characteristic of an atom.

9th Class Physical Science Textbook Page No. 76

Question 12.
An atom is electrically neutral. But the electrons present in it are negatively charged particles. If only negative charges were present, the atom would not be neutral. Then, why are atoms considered to be neutral?
Answer:

• This is the idea before Rutherford’s model.
• According to Rutherford’s model, number of protons inside the nucleus and number of electrons outside the nucleus are equal.
• Hence the net negative charge is equal to net positive charge. So, the atom is electrically neutral.

9th Class Physical Science Textbook Page No. 80

Question 13.
Compare Rutherford and Thomson’s models of the atom on the following basis :
1) Where is the positive charge placed?
2) How are the electrons placed?
3) Are they stationary inside the atom or moving?
Answer:

1. According to Thomson, the positive charge is uniformly distributed throughout the atom. Whereas according to Rutherford, the positively charged protons are inside the nucleus.
2. According to Thomson, electrons are embedded in the positively charged atom, but according to Rutherford, electrons are revolving around the nucleus in welldefined orbits.
3. According to Thomson, electrons are stable inside the atom but according to Rutherford, electrons are moving inside the atom.

9th Class Physical Science Textbook Page No. 83

Question 14.
Phosphorus and sulphur show multiple valency. See table 2. Why do some elements show multiple valency? Discuss with your Mends and teachers.
Answer:

• For sulphur, the number of electrons in outer most orbit is 6.
• Hence the valency should be (8 – 6 =) 2.
• But sulphur exists in so many forms.
• In the excited state, these 6 electrons also tend to participate in the bond formation.
• Hence sometimes it shows the valency 6. Ex : SO₂, SO₃, etc.
• Same situation happens for phosphorus. Ex . PCl₃, PCl₅, etc.

9th Class Physical Science 5th Lesson What is inside the Atom? Activities

Activity – 1

Question 1.
Sketch the structure of atom as you imagine.

We learnt about electron, proton and neutron.
a) Suppose you had to arrange them in an atom, how do you do it?
Answer:
Many arrangements are possible. Think that atom looks like a room, we can arrange the particles in alternating rows.

b) In how many ways can you arrange these sub-atomic particles in a spherical shape?
Answer:
Protons are positively charged, electrons are negatively charged and neutrons are
neutral. Hence neutrons and protons can be kept nearer and electrons can be kept farther or near the edge of the sphere. This is only an assumption. We can arrange in so many ways like this.

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 4 Atoms and Molecules Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 4th Lesson Atoms and Molecules

9th Class Physical Science 4th Lesson Atoms and Molecules Textbook Questions and Answers

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Question 1.
Draw the diagram to show the experimental setup for the law of conservation of mass. (AS 5)
(OR)
Draw the experimental arrangement used in verifying law of conservation of mass. Write the law of conservation of mass.
Answer:

Question 2.
Explain the process and precautions in verifying law of conservation of mass. (AS 5)
(OR)
Explain the procedure to prove in a chemical reaction the mass neither destroyed.

Answer:
Aim :
To verify law of conservation of mass.

Material required :
Sodium sulphate,Barium chloride, distilled water, conical flask, spring balance, small test tube, rubber cork, thread, retort stand.

Procedure:

1. Prepare a solution of sodium sulphate by dissolving approximately 2 gm of sodium sulphate in 100 ml distilled water in a 250 ml conical flask.
2. Prepare a Barium chloride solution by dissolving approximately 2 gm of potassium iodide in 100 ml water in another conical flask.
3. Take 100 ml solution of sodium sulphate in 250 ml conical flask.
4. Also take 4 ml solution of Barium chloride in test tube.
5. Hang the test tube in the flask carefully without mixing the solutions. Put a cork on the flask.
6. Weigh the flask with its contents carefully by spring balance.
7. Now tilt and swirl the flask, so that the two solutions mix.
8. Weigh the flask again by the spring balance.

Observations:

1. Weight of flask and contents before mixing = m₁ g
2. Weight of flask and contents after mixing = m₂ g

Conclusion :

1. We have observed that the two weights i.e., mj and m2 are equal.
2. This proves the law of conservation of mass.

Precautions:

1. Care should be taken while handling chemicals.
2. Glass apparatus may slip and break down. Hence make sure that they should not slip from your hands.
3. Contents of the conical, ffhsk should not mix before weighing first time.
4. Tie a thick thread to the conical flask, so that it will not slip while weighing.

Question 3.
15.9g of copper sulphate and 10.6g of sodium carbonate react together to give 14.2g of sodium sulphate and 12.3 g of copper carbonate. Which law of chemical combination is obeyed? How? (AS 1, AS 2)
Answer:
Reactants:
Mass of copper sulphate = 15.9 g ; Mass of sodium carbonate = 10.6 g
Total mass of reactants = 15.9 + 10.6 = 26.5 g

Products:
Mass of sodium sulphate = 14.2 g ; Mass of copper carbonate = 12.3 g
Total mass of products = 14.2 + 12.3 = 26.5 g
∴ Total mass of reactants is equal to total mass of products. This is the “Law of conservation of Mass”.

Question 4.
Carbon dioxide is added to 112 g of calcium oxide. The product formed is 200 g of calcium carbonate. Calculate the mass of carbon dioxide used. Which law of chemical combination will govern your answer? (AS 1, AS 2)
Answer:

1. Let x g of carbon dioxide is added to 112 g of calcium oxide.
2. The product is 200 g of calcium carbonate.
3. According to law of conservation of mass, ,
   Total mass of reactants = Total mass of products
   x+ 112 = 200 g
   x = 88 g
   ∴ 88 g of carbon dioxide is used.

Question 5.
0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.144 g of oxygen and 0.096 g of boron. Calculate the percentage composition of the compound by weight. (AS 1)
Mass of compound of oxygen and boron = 0.24 g
On analysis,
Mass of oxygen in the compound = 1.44 g
Mass of boron in the compound = 0.096 g

Question 6.
In a class, a teacher asked students to write the molecular formula of oxygen. Shamita wrote the formula as 02 and Priyanka as O. Which one is correct? State the reason. (AS 1, AS 2)
Answer:
Shamitha’s answer is correct.
Reason :

1. Oxygen is diatomic.
2. Two atoms of oxygen combine to form oxygen molecule.
3. Hence the formula of oxygen molecule will be ‘O₂‘.

Question 7.
Imagine what would happen if we do not have standard symbols for elements. (AS 2)
(OR)
Is it necessary to use symbols for elements? Write your opinion.
Answer:

• Chemistry involves a lot of reactions.
• If we do not have symbols, we have to write their names to represent the reactions.
• This is very tedious work.
• To avoid this, we need standard symbols to elements, which are universally accepted.
• In advanced studies, balancing of equations, atoms present in a compound, etc. will not be understood without symbols.
• Simply chemistry will not be developed unless symbols, formulae, etc. are not known.

Question 8.
Mohith said “H₂ differs from 2H.” Justify. (AS 1)
Answer:
H₂ is the hydrogen molecule in which two hydrogen atoms are combined to form one hydrogen molecule.

2H is the hydrogen atom. Here 2 hydrogen atoms are ready to participate in chemical reaction.

Question 9.
Lakshmi gives a statement “CO and Co both represent element”. Is it correct? State reason. (AS 1, AS 2)
Answer:
Lakshmi’s statement is incorrect.

Reason :

1. CO stands for carbon monoxide, a compound, which consists carbon and oxygen atoms.
2. This can be identified with the help of both C and O are capital (upper case) letters.
3. Co stands for cobalt, an element.
4. This can be identified with the help of ‘C’ is capital (upper case) letter and ‘o’ small (lower case) letter.

Question 10.
The formula of water molecule is H₂O. What information do you get from this formula? (AS 1)
Answer:

• Water is a combination of hydrogen and oxygen.
• Two hydrogen atoms and one oxygen atom combine to form one water molecule.
• Molecular weight of water molecule is 18. [Hydrogen 1, Oxygen 16. H₂O ⇒ 2 × 1 + 16=18]
• 18 g of water molecule contains 6.022 × 10²³ particles in it.
• Valency of hydrogen is 1 and oxygen is 2.

Question 11.
How would you write 2 molecules of Oxygen and 5 molecules of Nitrogen? (AS 1)
Answer:
2 molecules of oxygen → 2O₂
Reason :

1. Oxygen is diatomic element.
2. Two oxygen atoms combine to form one oxygen molecule.
3. The formula of oxygen molecule is O₂.

5 molecules of nitrogen → 5N₂
Reason :

1. Nitrogen is also diatomic element.
2. Two nitrogen atoms combine to form one nitrogen molecule.
3. Molecular formula of nitrogen is N₂.

Question 12.
The formula of a metal oxide is MO. Then write the formula of its chloride. (AS 1)
Answer:

• The valency of oxide is 2 i.e., O^-2.
• The formula of a metal oxide is given as MO.
• Hence the valency of the given metal must be 2 i.e., M⁺².
• Valency of chloride is 1 i.e., C^–.
• Therefore according to criss-cross method, the formula of given metal chloride will be MCl₂.

Question 13.
Formula of calcium hydroxide is Ca(OH)₂ and zinc phosphate is Zn₃(PO₄)₂. Then write the formula to calcium phosphate. (AS 1)
(OR)
Formula of calcium hydroxide is Ca(OH)₂ and zinc phosphate is Zn₃(PO₄)₂. Then write the valencies of calcium and phosphate and then write the formula of calcium phosphate.
Answer:

• Formula of calcium hydroxide is Ca(OH)₂.
• From criss-cross method we know that the valency of calcium is 2 i.e., Ca⁺² and hydroxide is 1 i.e., OH^–.
• Formula of zinc phosphate is Zn₃(PO₄)₂.
• Valency of Zn is 2 i.e., Zn^+2, and valency of phosphate is 3 i.e., PO₄^-3.
• Now the formula of calcium phosphate according to criss-cross method is Ca₃(PO₄)₂.

Question 14.
Find out the chemical names and formulae for the following common household substances. (AS 1)
a) Common salt
b) Baking soda
c) Washing soda
d) Vinegar
Answer:

Common household substance	Chemical name	Formula
a) Common salt	Sodium chloride	NaCl
b) Baking soda	Sodium bicarbonate	NaHCO3
c) Washing soda	Sodium carbonate	Na2CO3
d) Vinegar	Impure dilute acetic acid	CH3COOH

Question 15.
Calculate the mass of the following. (AS 1)
a) 0.5 mole of N₂ gas
b) 0.5 mole of N atoms
c) 3.011 × 10²³ number of N atoms
d) 6.022 × 10²³ number of N₂ molecules
Answer:
a) 0.5 mole of N₂ gas :
Mass of one mole of N₂ gas = 28 g. (∵ Molecular wt. of N₂ = 28)
Mass of 0.5 mole of N₂ gas = 28 × 0.5 = 14 g

b) 0.5 mole of N atoms :
Mass of one mole of N atoms = 14 g (∵ Atomic wt. of N = 14)
Mass of 0.5 mole of N atoms = 14 × 0.5 = 7 g

c) 3.011 × 10²³ number of N atoms :
Mass of 6.022 × 10²³ number of N atoms = 14 g

d) 6.022 × 10²³ number of N₂ molecules :
Mass of 6.022 × 10²³ number of N₂ molecules = 28 g

Question 16.
Calculate the number of particles in each of the following. (AS 1)
a) 46 g of Na
b) 8 g of O₂
c) 0.1 mole of hydrogen
Answer:
a) 46 g of Na :
Atomic weight of Na = 23
Number of particles in 23 g of Na atom = 6.022 × 10²³

b) 8 g of O₂ :
Molecular weight of O₂ is 32.
Number of particles in 32 g of O₂ molecule = 6.022 × 10²³

c) 0.1 mole of hydrogen :
Atomic weight of hydrogen is 1.
Number of particles in 1 mole of hydrogen = 6.022 × 10²³
Number of particles in 0.1 mole of hydrogen= \(\frac{0.1}{1}\) × 6.022 × 10²³ = 6.022 × 10²²

Question 17.
Convert into moles. (AS 1)
a) 12 g of O₂ gas
b) 20 g of water
c) 22 g of carbon dioxide
Answer:
a) 12 g of O₂ gas :
Molecular weight of O₂ is 32.
∴ Number of moles of 32 g of O₂ gas = 1

b) 20 g of water :
Molecular weight of water (H₂O) is 18.
Number of moles of 18 g of water = 1
Number of moles of 20 g of water = \(\frac{20}{18}\) x 1 =1.11

c) 22 g of carbon dioxide :
Molecular weight of carbon dioxide (CO₂) is 44.
∴ Number of moles of 44 g of CO₂ = 1

Question 18.
Write the valencies of Fe in FeCl₂ and FeCl₃. (AS 1)
Answer:

1. In FeCl₂, the valency of Fe is 2.
2. In FeCl₃, the valency of Fe is 3.

Question 19.
Calculate the molar mass of sulphuric acid (H₂SO₄) and glucose (C₆H₁₂O₆). (AS 1)
Answer:
a) Formula of sulphuric acid is H₂SO₄.
Molecular mass of H₂SO₄ = 2 × 1 + 1 × 32 + 4 × 16 = 2 + 32 + 64 = 98 u
∴ Molar mass of H₂SO₄ = 98 g

b) Formula of glucose is C₆H₁₂O₆.
Molecular mass of C₆H₁₂O₆ = (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 180 u
∴ Molar mass of C₆H₁₂O₆ = 180 g.

Question 20.
Which has more number of atoms – 100 g of sodium or 100 g of iron? Justify your answer. (Atomic mass of sodium = 23 u, atomic mass of iron = 56 u) (AS 1)
Answer:
100g of sodium has more number of atoms than 100g of iron.

Justification :
1) Atomic mass of sodium = 23 u
23 g of sodium contains 6.022 × 10²³ atoms.
100 g of sodium contains = \(\frac{100}{23}\) × 6.022 × 10²³ = 26.1826 × 10²³ atoms of sodium.

2) Atomic mass of iron = 56 u
∴ 56 g of iron contains 6.022 × 10²³ atoms.
100 g of iron contains = \(\frac{100}{56}\) × 6.022 × 10²³ = 10.7535 × 10²³ atoms of iron.

Question 21.
Complete the following table. (AS 1)

Answer:

Question 22.
Fill the following table. (AS 1)

Answer:

Question 23.
Make placards with symbols and valencies of the atoms of the elements separately. Each student should hold two placards, one with the symbol in the right hand and the other with the valency in the left hand. Keeping the symbols in place, students should criss-cross their valencies to form the formula of a compound.
Answer:
Student’s activity.

Question 24.
Take empty blister packs of medicines. Cut them into pieces having

Hence the formula of sodium carbonate will be Na₂ CO₃
Hence the forrnu A. Student’s activity.

9th Class Physical Science 4th Lesson Atoms and Molecules InText Questions and Answers

9th Class Physical Science Textbook Page No. 56

Question 1.
Does the weight of iron rod increase or decrease, on rusting?
Answer:
The weight of iron rod decreases on rusting.

Question 2.
Where does the matter charcoal go?
Answer:
The charcoal, on burning, gives off CO₂ which is mixed in atmosphere. The residue is remained as ash.

Question 3.
Wet clothes dry after some time – where does the water go?
Answer:
Water evaporates and mixed in the atmosphere.

Question 4.
What happens to magnesium on burning it in air?
Answer:
Magnesium on burning in air gives a bright light and ash is remained. The ash is magnesium oxide.

Question 5.
What happens to sulphur on burning it in air?
Answer:
Sulphur on burning, changes its state and colour.

9th Class Physical Science Textbook Page No. 57

Question 6.
Did you observe any precipitate in the reaction?
Answer:
In the flask, a reaction takes place between lead nitrate and potassium iodide.

Question 7.
Do you think that a chemical reaction has taken place in the flask? Give reason.
Answer:
Yes, the contents in the flask are changed as lead iodide and potassium nitrate.

9th Class Physical Science Textbook Page No. 58

Question 8.
Do the weights of the flask and its contents change during the activity?
Answer:
The weights of the flask and its contents do not change before and after reaction.

Question 9.
What are your conclusions?
Answer:
Mass was neither created nor destroyed.

Question 10.
What do you observe from table – 1?
Answer:
The components of a compound are mixed at same proportions in any sample.

9th Class Physical Science Textbook Page No. 59

Question 11.
What difference do you observe in percentage of copper, carbon, and oxygen in two samples?
Answer:
The percentage of copper, carbon and oxygen are same in two samples, i.e., they are mixed at same proportions.

9th Class Physical Science Textbook Page No. 60

Question 12.
Are elements also made of atoms?
Answer:
When the particles of a substance contain only one type of atoms, that substance is called an element. In elements the smallest particle that exist may be atoms or molecules.

9th Class Physical Science Textbook Page No. 62

Question 13.
How do we write the symbols for calcium, chlorine, chromium?
Answer:
We have only 26 alphabets in English, but there are over 100 known elements. We cannot write the same symbol for carbon, calcium, chromium, etc.

9th Class Physical Science Textbook Page No. 63

Question 14.
Would you be able to recognise the elements of the table – 2, have symbols of this category?
Answer:
Yes. They are iron, gold, sodium, and potassium.

9th Class Physical Science Textbook Page No. 64

Question 15.
Observe the atomicity and fill the following table.

Answer:

Name of the element	Formula	Atomicity
Argon	Ar	Monoatomic
Helium	He	Monoatomic
Sodium	Na	Monoatomic
Iron	Fe	Monoatomic
Aluminium	Al	Monoatomic
Copper	Cu	Monoatomic
Hydrogen	H2	Diatomic
Oxygen	O2	Diatomic
Nitrogen	N2	Diatomic
Chlorine	Cl2	Diatomic
Ozone	O3	Triatomic
Phosphorus	P4	Tetratomic
Sulphur	S8	Octatomic

Question 16.
What is valency?
Answer:
Every element reacts with other element according to its combining capacity, which we call as its valency.

9th Class Physical Science Textbook Page No. 66

Question 17.
Can you write the formula of carbon dioxide and carbon monoxide? Try to write formula for them as we have done in case of water molecule.
Answer:
Carbon dioxide :
The elements present are carbon and oxygen. One atom of carbon and one atom of oxygen are present in a molecule of carbon monoxide. Hence the formula of carbon monoxide is CO.

Carbon dioxide :
The elements present are carbon and oxygen. One atom of carbon and 2 atoms of oxygen are present in a molecule of carbon dioxide. Hence the formula of carbon dioxide is CO₂.

9th Class Physical Science Textbook Page No. 69

Question 18.
How many molecules are there in 18 grams of water?
Answer:
6.022 × 10²³ molecules are there in 18 grams of water.

Question 19.
How many atoms are there in 12 grams of carbon?
Answer:
6.022 × 10²³ atoms are there in 12 grams of carbon.

9th Class Physical Science Textbook Page No. 58

Question 20.
Do you get the same result if the conical flask is not closed?
Answer:

1. No, we cannot get the same result.
2. When the conical flask is not closed, some gases will leave out the flask during chemical reaction.

Question 21.
Recall the burning of the magnesium ribbon in air. Do you think mass is conserved during this reaction?
Answer:

1. Yes, but we cannot observe the conservation of mass.
2. When the experiment is conducted in a closed container where there is no scope for oxygen to escape, we can observe the conservation of mass. But in this condition this experiment is not possible.

9th Class Physical Science Textbook Page No. 59

Question 22.
100 g of mercuric oxide decompose to give 92.6 g of mercury and 7.4 g of oxygen. Let us assume that 10 g of oxygen reacts completely with 125 g of mercury to give mercuric oxide. Do these values agree with the law of constant proportions?
Answer:
Proportion of oxygen = 7.4 : 10
Proportion of mercury = 92.6 : 125
\(\Rightarrow \frac{7.4}{10}=\frac{92.6}{125} \Rightarrow 0.74=0.74\)
∴ They follow law of constant proportions.

Question 23.
Discuss with your friends if the carbon dioxide that you breathe out and the carbon dioxide they breathe out are identical. Is the composition of the carbon dioxide of different sources same? (Page – 73)
Answer:
Same.
This can be justified with the help of law of constant proportions.

9th Class Physical Science Textbook Page No. 60

Question 24.
Which postulate of Dalton’s theory is the result of the law of conservation of mass?
Answer:
First postulate of Dalton’s theory i.e. “Matter consists of indivisible particles called atoms”, is the result of law of conservation of mass.

Question 25.
Which postulate of Dalton’s theory can explain the law of constant proportions?
Answer:
Third postulate of Dalton’s theory i.e. “Atoms of a given element have identical mass and chemical properties. Atoms of different elements have different masses and chemical properties”, is the result of law of constant proportions.

9th Class Physical Science 4th Lesson Atoms and Molecules Activities

Activity – 1

Question 1.
Some elements and their possible symbols are given. Correct them and give reasons for your corrections.
Answer:

Activity – 2

Question 2.
Write the symbols for given elements.

Answer:

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 3 Is Matter Pure? Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 3rd Lesson Is Matter Pure?

9th Class Physical Science 3rd Lesson Is Matter Pure? Textbook Questions and Answers

Improve Your Learning

Question 1.
Which separation techniques will you apply for the separation of the following? (AS 1)
a) Sodium chloridfe from its solution in water.
b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.
c) Small pieces of metal in the engine oil of a car.
d) Different pigments from an extract of flower petals.
e) Butter from curd.
f) Oil from water.
g) Tea leaves from tea.
h) Iron pins from sand.
i) Wheat grains from husk.
j) Fine mud particles suspended in water.
Answer:

Mixture	Separation technique
a) Sodium chloride from its solution in water	Crystallization
b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride	Sublimation
c) Small pieces of metal in the engine oil of a car	Filtration
d) Different pigments from an extract of flower petals	Chromatography
e) Butter from curd	Centrifugation
f) Oil from water	Separation funnel
g) Tea leaves from tea	Filtration
h) Iron pins from sand	Magnetism
i) Wheat grains from husk	Winnowing
j) Fine mud particles suspended in water	Sedimentation and decantation (or) Filtration using filter paper

Question 2.
Write the steps you would use for making tea. Use the words given below and write the steps for making tea. (AS 7)

Answer:

• Take a cup of milk (solvent) in a tea kettle.
• Add one table spoon of sugar (solute), and one table spoon of tea powder (insoluble) to the solvent.
• Heat the tea kettle on the stove.
• The sugar (solute) dissolves in the milk (solvent) and the tea powder remains undissolved.
• Now filter the solution so formed.
• The filtrate is the tea (solution).
• The residue remained in the sieve is the insoluble component of tea powder.

Question 3.
Explain the following giving examples. (AS 1)
a) Saturated solution
b) Pure substance
c) Colloid
d) Suspension
Answer:
a) Saturated solution :
When no more solute can be dissolved in the solution at a certain temperature, it is said to be a saturated solution.

In a saturated solution, equilibrium with the undissolved solute at a certain temperature.

Ex :

1. Take 50 ml of water in a cup.
2. Add one spoon of sugar to the cup and stir still it dissolves.
3. Keep on adding sugar to the water in the cup and stir till no more sugar can be dissolved.
4. The solution so formed is a saturated solution.

b) Pure substance :
A substance is pure i.e., homogeneous if the com-position doesn’t change, no matter which part of the substance we take for examination.

Ex :

1. Take a small part of pure gold biscuit as a sample.
2. The composition is found to be same throughout it.

c) Colloid :
Colloids are heterogeneous mixtures in which the particle size is too small to be seen with the naked eye, but is big enough to scatter light.
Ex : Milk, butter, cheese, cream, gel, etc.

d) Suspension :
Suspension is a heterogeneous mixture in which the solute particles didn’t dissolve and the particles are visible to naked eye.
Ex : Syrups, chalk powder mixed with water, etc.

Question 4.
Classify each of the following as a homogeneous or heterogeneous mixture. Give reasons. (AS 1)

Answer:

Homogeneous mixtures	Heterogeneous mixtures
Soda water	Soil
Air	Wood
Vinegar
Filtered tea
Reason : Components in the above mixtures are uniformly distributed and we cannot see the components separately.	Reason : Components in the above mixtures are not uniformly distributed.

Question 5.
How would you confirm that a colourless liquid given to you is pure water? (AS 1)
Answer:

• Observe the smell. We should not find any smell.
• Observe with a naked eye we should not find any suspended particles or fumes or air bubbles.
• Pass a beam of light. It should not scatter.
• The temperature should be normal.
  Then the given colourless liquid is pure water.

Question 6.
Which of the following materials fall in the category of a “pure substance”? Give reasons. (AS 1)
a) Ice
b) Milk
c) Iron
d) Hydrochloric acid
e) Calcium oxide
f) Mercury
g) Brick
h) Wood
i) Air
Answer:

• Except brick and wood remaining materials given in the list can be treated as pure substances.
• Take any small part of ice, milk, iron, hydrochloric acid, calcium oxide, mercury and air and test for their components.
• We find that the composition is same throughout them.

Question 7.
Identify the solutions among the following mixtures. (AS 1)
a) Soil
b) Sea water
c) Air
d) Coal
e) Soda water
Answer:
The solutions are : sea water, air and soda water.

Question 8.
Which of the following will show “Tyndall effect”? How can you demonstrate “Tyndall effect” in them? (AS 1, AS 3)
a) Salt solution
b) Milk
c) Copper sulphate solution
d) Starch solution
Answer:
Milk shows Tyndall effect.

Demonstration :

1. Prepare the milk, copper sulphate, salt and starch solutions in different beakers.
2. Allow a beam of light through each of them.
3. The path of the light beam is clearly visible to us through milk.
4. The path of the light beam is not visible through remaining.
5. This experiment will be effective if it is performed in a dark room.

Question 9.
Classify the following into elements, compounds and mixtures. (AS 1)
a) Sodium
b) Soil
c) Sugar solution
d) Silver
e) Calcium carbonate
f) Tin
g) Silicon
h) Coal
i) Air
j) Soap
k) Methane
l) Carbondioxide
m) Blood
Answer:

Elements	Compounds	Mixtures
Sodium	Calcium carbonate	Soil
Silver	Coal	Sugar solution
Tin	Methane	Air
Silicon	Carbondioxide Soap	Blood

Question 10.
Classify the following substances in the below given table. (AS 1)

Answer:

Solution	Suspension	Colloidal dispersion
Soda water	Ink	Fog
Fruit salad	Nail polish	Aerosol sprays
Black coffee	Starch solution	Boot polish
Air Brass		Milk
		Blood
		Oil and water

Question 11.
Take a solution, a suspension, a colloidal dispersion in different beakers. Test whether each of these mixtures shows the Tyndall effect by focusing a light at the side of the container. (AS 3)
Answer:

• Take sugar solution (solution), starch solution (suspension) and milk solution (colloidal dispersion) in three different beakers.
• Focus a beam of light by torch or a laser beam at the side of each container and observe.
• We can see that the path of beam of light is clearly visible through all the solutions.
• Hence all the three solutions show ‘Tyndall effect”.

Question 12.
Draw the figures of arrangement of appatus for distillation and fractional distillation. What do you find the major difference in these apparatus? (AS 5)
Answer:

The main difference between these two apparatus is that a fractionating column is fitted in between the distillation flask and the condenser.

Question 13.
Determine the mass by mass percentage concentration of a 100 g salt solution which contains 20 g salt. (AS 1)
Answer:
Mass of salt = 20 g; Mass of salt solution = 100g

Question 14.
Calculate the concentration in terms of mass by volume percentage of the solution containing 2.5 g potassium chloride in 50 ml of potassium chloride (KCl) solution. (AS 1)
Answer:
Mass of potassium chloride = 2.5 g
Volume of potassium chloride solution = 50 ml

9th Class Physical Science 3rd Lesson Is Matter Pure? InText Questions and Answers

9th Class Physical Science Textbook Page No. 40

Question 1.
Can you give few more examples of this kind?
Answer:
Some more examples of homogeneous mixtures are sugar solution, lemon squash, fruit juices, syrups and tonics used in medicine, etc.

Question 2.
Can you prove this with an experiment?
Answer:

• Take some thick milk in a test tube.
• Pass a beam of light from torch or a laser light.
• We cannot observe the path of light through the solution.

Question 3.
If the solution is diluted, can the path of light be visible?
Answer:

• Take some thick milk in a test tube.
• Dilute it by adding some water.
• Now pass a beam of light from torch or a laser light.
• We cannot observe the path of light through the solution.

Question 4.
What would happen if you add a little more solute to a solvent?
Answer:
The solution becomes concentrated.

Question 5.
How do you determine the percentage of the solute present in a solution?
Answer:

• Take 100 ml of water in a beaker.
• Take 50 g. of sugar in a plate.
• Add a spoon of sugar to water and stir it tell the sugar dissolve in water.
• Go on adding sugar till you reach a situation that the sugar cannot be dissolved in water.
• Now weigh the sugar remained in the plate.
• Subtract thin weight from 50 g. The weight so obtained is dissolved in water.
• Hence the maximum amount of solute present in 100 ml of solvent is the percentage of solute (solubility).

9th Class Physical Science Textbook Page No. 44

Question 6.
Did you ever observe this phenomenon in the cinema halls?
Answer:
In cinema halls when we observe the projector while the movie is running, we can observe the phenomenon of “Tyndall effect”. We can see the beams of light in which dust particles also observed.

9th Class Physical Science Textbook Page No. 46

Question 7.
Is the mixture heterogeneous? Give reasons.
Answer:
The mixture of ammonium chloride and salt is a heterogeneous mixture. Even though these two are white in colour their particles do not mix.

Question 8.
How do we separate the salt and ammonium chloride?
Answer:
We can separate the salt and ammonium chloride by the method of sublimation.

9th Class Physical Science Textbook Page No. 49

Question 9.
Can you give any examples where we use fractional distillation technique?
Answer:
We use this technique in separating the components of crude oil i.e., petrol, naphthalene, kerosene, greese, etc.

9th Class Physical Science Textbook Page No. 38

Question 10.
How does a laundry dryer squeeze out water from wet clothes?
Answer:

• The laundry dryer contains a cylindrical vessel with holes on its walls.
• When wet clothes are dropped in it, it is rotated with high speed with the help of an electric motor.
• Due to centrifugation, the water from the clothes reaches to the walls of the cylinder and comes out through the holes.
• Hence the clothes are dried up.

9th Class Physical Science Textbook Page No. 40

Question 11.
a) “All the solutions are mixtures, but not all mixtures are solutions”. Discuss about the validity of the statement and give reasons to support your argument.
Answer:

• You take any solution like salt solution, sugar solution, air, etc. all are homogeneous mixtures.
• Consider a mixture of sand and iron fehlings. It is not homogeneous. Hence this is not a solution.

b) Usually we think of a solution as a liquid that contains either a solid, liquid or a gas dissolved in it. But, we can have solid solutions. Can you give some examples?
Answer:
Examples of solid solutions are :

1. Steel used in constructions (a homogeneous mixture of iron and carbon).
2. Brass (a homogeneous mixture of zinc and copper).

9th Class Physical Science Textbook Page No. 43

Question 12.
1) Have you ever observed carefully the syrup that you take for cough? Why do you shake it before consuming?
2) Is it a suspension or colloidal solution?
Answer:

1. The syrup used for cough will be shook before consuming because it consists some undissolved particles settled down.
2. Hence cough syrup is a suspension.

9th Class Physical Science Textbook Page No. 45

Question 13.
Is there any difference between a true solution and colloidal solution? If you find the differences, what are those differences?
Answer:
Differences between true solutions and colloidal solutions :

Question 14.
Why do we use different separation techniques for mixtures like grain and husk as well as ammonium chloride and salt though both of them are heterogeneous mixtures? What is the basis for choosing a separation technique to separate mixtures?
Answer:
The basis for choosing a separation technique to separate mixtures is the property of a component in the mixture i.e., solubility in water, evaporation, appearance, etc.

9th Class Physical Science Textbook Page No. 47

Question 15.
Is it possible to find out adulteration of kerosene in petrol with this technique?
Answer:
The adulteration of kerosene in petrol can be found by using density meter.

9th Class Physical Science Textbook Page No. 50

Question 16.
a) Arrange the gases present in air in increas you observe?
Answer:

Gas	B.P
Helium	268.93°C
Hydrogen	252.9°C
Neon	246.08°C
Nitrogen	195.8°C
Argon	185.8°C
Oxygen	183°C
Methane	164°C
Krypton	153.22°C
Xenon	108.12°C
Carbondioxide	78°C

b) Which gas forms the liquid first as the air is cooled?
Answer:
Oxygen forms the liquid first as the air is cooled.

9th Class Physical Science 3rd Lesson Is Matter Pure? Activities

Activity – 1

Question 1.
How can we separate cream from milk?
Answer:

• Take some milk in a vessel.
• Spin it with a milk churner for some time.
• After some time you observe, separation of a paste like solid out of the milk.
• The paste like solid is called cream.

Activity – 2

Question 2.
Explain a demonstration to identify homogeneous and heterogeneous mixtures.
Answer:

• Take two test tubes.
• Now add one tea spoon of salt to both the test tubes.
• Fill one test tube with water and another with kerosene and stir them.
• In the first test tube (water), the salt dissolves completely.
• This is a homogeneous mixture.
• In the second test tube (kerosene), the salt is not dissolved.
• This is a heterogeneous mixture.

Activity – 3

Question 3.
Describe an activity to prepare saturated and unsaturated solutions.
Answer:
Preparation of saturated solution :

1. When no more solute can be dissolved in the solution at a certain temperature, it is said to be a saturated solution.
2. Take 50 ml of water in an empty cup.
3. Add one spoon of sugar to the water in the cup.
4. Stir the water until it dissolves.
5. Keep on adding sugar to the cup and stir till no more sugar can be dissolved in it.
6. Thus formed solution is called saturated solution.
7. In a saturated solution, equilibrium with the undissolved solute at a certain temperature.

Preparation of unsaturated solution :

1. If the amount of solute present in a solution is less than that in the saturated solution, is called an unsaturated solution.
2. Now take the solution prepared by you into a beaker.
3. Heat that solution slowly by 5 to 6°C above the room temperature.
4. The undissolved solute dissolves.
5. Add some more sugar to this solution.
6. You notice that more sugar dissolves in it easily when it is heated.
7. Thus we prepared an unsaturated solution.

Activity – 4

Question 4.
What are the factors affecting the rate of dissolving ? How do you prove them?
Answer:
Factors affecting the rate of dissolving are :

1. Temperature of the solvent.
2. Size of the solute particles.
3. Stirring the solution.

Proof:

1. Take three glass beakers and fill each of them with 100 ml of water.
2. Add two spoons of salt to each beaker.
3. Place the first beaker undisturbed.
4. Stir the contents of the second beaker.
5. Heat gently the third beaker.
6. In all the cases, the salt dissolves but the time taken to dissolve is different.
7. When the beaker is heated, the salt dissolved quickly.
8. When we stir the contents, the salt dissolved but slower than heating.
9. When we observe the undisturbed beaker, the salt dissolves but at the slowest rate.
10. This shows that the temperature of the solvent, size of the solute particles, stirring of contents are the factors affecting the rate of dissolving.

Activity – 5

Question 5.
Describe an experiment to identify suspensions and colloids.
Answer:

• Take some chalk powder in a test tube.
• Take a few drops of milk in another test tube.
• Add water to these samples and stir with a glass rod.
• Now do the following steps and write your observations in the table given.

Step 1 :
Direct a beam of light from a torch or a laser beam on the test tubes. Observe the path of the light through the solutions.

Step 2 :
1) Leave the mixture undisturbed for some time.
2) See whether the solute settles down after some time.

Step 3 :
Filter the mixtures and observe any residue found on the filter paper.
Now read your observations :

Observations :

1. In the chalk mixture, the particles of chalk settled at the bottom of the test tube and on filtration, we can observe a residue on the filter paper.
2. Hence the chalk mixture is a suspension.
3. In the milk mixture, the particles of milk are uniformly spread throughout the mixture and no residue is found on the filter paper.
4. Hence milk mixture is a colloidal solution.

Activity – 6

Question 6.
Describe an example for the separation of mixtures by sublimation.
Describe a method of separating ammonium chloride from the mixture of ammonium chloride and common salt.
Answer:
Aim :
To separate ammonium chlo-ride from the mixture of ammonium chloride and common salt.

Materials required :
China dish, funnel, cotton, ammonium chloride, common salt and stove.

Procedure:

1. Take one table spoon of ammo-nium chloride and one table spoon of common salt and mix them.
2. Take the mixture in a China dish.
3. Take a glass funnel.
4. Plug the mouth of the funnel with cotton.
5. Invert the funnel over the dish.
6. Heat the dish on the stove and observe the walls of the funnel.

Observations :
Initially we find vapours of ammonium chloride and then solidified ammonium chloride on the walls of the funnel.

Activity – 7

Question 7.
Describe a method to separate the dye present in ink.
(or)
Describe an example for the separation of a mixture by the process of evaporation.
Answer:
Aim :
To separate the dye present in ink by the process of evaporation.

Materials required :
Beaker, watch glass, water, ink and stove.

Procedure :

1. Take a beaker and fill it to half its volume with water.
2. Keep 3, glass on the mouth of a beaker.
3. Put few drops of ink on the watch glass.
4. Heat the beaker and observe the watch glass.

Observations:

1. We observe some fumes coming from the watch glass.
2. Continue heating till you do not observe any further change on the watch glass.
3. A small residue will be remained on the watch glass.

Inference :

1. We know that ink is a mixture of a dye in water.
2. The residue remained on the watch glass is the dye present in the ink.

Lab Activity

Question 8.
Describe paper chromatography activity to observe the colours present in a marker ink.
(OR)
How can you perform chromatography activity in your laboratory.
Answer:
Aim :
Separating the components of ink using paper chromatography.

Materials required :
Beaker, rectangular shaped filter papers, black marker (non-permanent), water, pencil and cello tape.

Procedure:

1. Draw a thick line just above the bottom of the filter paper using the marker.
2. Pour some water into the beaker.
3. Hang the paper strip with help of a pencil and tape in such a way that it should just touch the surface of water.
4. Make sure that the ink line or mark does not touch the water.
5. Allow the water to move up the paper for 5 minutes and then remove the strip from water.
6. Let it dry.
7. Repeat the process with green marker, a permanent marker, etc.

Observations :

1. When black marker is used, we observe different colours like red, green, violet, black, etc. on the filter paper after drying.
2. When green marker is used, we observe yellow, blue, green colours on the filter paper.
3. When permanent marker is used, we cannot find any change in the mark.

Activity – 8

Question 9.
How do you separate water and kerosene from the mixture of kerosene and water?
Answer:
Aim :
To separate water and kerosene from the mixture of kerosene and water.

Materials required :
Kerosene, water, separating funnel, beakers.

Procedure:

1. Pour the mixture of kerosene and water in a separating funnel.
2. Let it stand undisturbed for some time, so that the layers of oil and water are formed.
3. Open the stopcock of the separating funnel and pour out the lower layer of water carefully.
4. Close the stopcock of the separating funnel as the oil reaches the stop cock.

Principle involved :
The immiscible liquids separate out into layers depending on their densities.

Activity – 9

Question 10.
Explain the method of separation of two miscible liquids by distillation.
Answer:
Aim :
To separate two miscible liquids (water and acetone) by distillation.

Materials required :
Stand, distillation flask, thermometer, condenser, beaker, acetone and water, one holed rubber cork.

Procedure:

1. Take a mixture of acetone and water in a distillation flask.
2. Fit it with a thermometer and clamp it to stand.
3. Attach the condenser of the flask on one side.
4. On the other side of the condenser keep a beaker to collect distillate.
5. Heat the mixture slowly.
6. Keep a close watch on the thermometer.
7. The acetone vapourizes and condenses in the condenser.
8. The acetone can be collected from the condenser outlet.
9. Water remains in the distillation flask.
10. The separation technique used above is called distillation.

Activity – 10

Question 11.
How do you separate copper metal from the mixture of copper sulphate and aluminium?
Answer:

• Take a concentrated solution of copper sulphate into a beaker.
• Drop an aluminium foil in the beaker.
• After some time, we observe a layer of copper deposited on the aluminium foil.
• The solution becomes colourless.
• A chemical reaction takes place among the copper ions present in the solution with aluminium and copper metal is separated.

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 2 Laws of Motion Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 2nd Lesson Laws of Motion

9th Class Physical Science 2nd Lesson Laws of Motion Textbook Questions and Answers

Improve Your Learning

Question 1.
Explain the reasons for the following. (AS 1)
a) When a carpet is beaten with a stick, dust comes out of it.
Answer:

1. The dust particles in the carpet are at rest.
2. When the carpet is beaten with a stick, the state of rest of the dust particles is disturbed.
3. Due to inertia, the dust particles comes out.

b) Luggage kept on the roof of a bus is tied with a rope.
Answer:

1. Luggage kept on the roof of a bus is in the state of rest.
2. As the bus moves, the luggage also moves with a velocity equal to the velocity of the bus.
3. If the bus suddenly stops, the luggage resists to change its state of motion.
4. Hence due to inertia it will fall down.
5. To avoid this, the luggage is tied with a rope.

c) A pace bowler in cricket runs in from a long distance before he bowls.
Answer:

1. When he runs in from a long distance, he gains momentum of inertia.
2. Due to this larger inertia, larger force is applied in a short interval of time.
3. Hence the momentum will be more.

Question 2.
Two objects have masses 8 kg and 25 kg. Which one has more inertia? Why? (AS 1)
Answer:

1. The object with mass 25 kg has more inertia.
2. The resistance to change the state of object will be more for a body of larger mass.

Question 3.
Keep a small rectangular shaped piece of paper on the edge of a table and place an old five rupee coin on its surface vertically as shown in the figure below. Now give a quick push to the paper with your finger. How do you explain inertia with this experiment?

Answer:

1. The coin and the paper are in inertia of rest.
2. When we give a quick push to the paper, paper comes to inertia of motion and the coin remains in its original state i.e., inertia of rest.
3. As a result, the paper will come out and the coin remains on the table without changing its position.

Question 4.
If a car is travelling westwards with a.constant speed of 20 m/s, what is the resultant force acting on it? (AS 1, AS 7)
Answer:

1. A car is moving with a constant speed.
2. Hence the net force on the car is zero both in horizontal and vertical directions.

Question 5.
What is the momentum of a 6.0 kg bowling ball with a velocity of 2.2 m/s? (AS 1)
Answer:
Mass of the ball (m) = 6.0 kg
Velocity of the ball (v) = 2.2 m/s
Momentum (p) = nv = 6.0 kg × 2.2 m/s = 13.2 kg m/s (or) 13.2 N-s

Question 6.
Two people push a car for 3 sec, with a combined net force of 200 N. (AS 1)
a) Calculate the impulse provided to the car.
Answer:
F_net = 200 N ; ∆t =3 sec
Impulse ∆p = F_net. ∆t = 200 × 3 = 600 N – sec.

b) If the car has a mass of 1200 kg, what will be its change in velocity?
Answer:
Mass of the car (m) = 1200 kg.; Net force (F_net) = 200 N
Time ∆t = 3 sec. ; Change in velocity ∆v =?

Question 7.
What force is required to produce an acceleration of 3 m/sec2 in an object of mass 0.7 kg? (AS 1)
Answer:
Mass of the object (m) = 0.7 kg.; Acceleration (a) = 3 m/sec²
Force required (F) = ?
F = ma = 0.7 x 3 = 2.1 N

Question 8.
A force acts for 0.2 sec on an object having mass 1.4 kg initially at rest. The force stops to act but the object moves through 4 m in the next 2 seconds, find the magnitude of the force. (AS 1)
Answer:

Velocity after 0.2 sec v = u + at = 0 + 0.2 a = 0.2a ……….. (1)
After 0.2 s, the body moves with uniform velocity, acceleration is zero, because force is removed.
∴ Velocity v = \(\frac{s}{t}=\frac{4}{2}\) =2 m/s. ………. (2)
From (1) & (2)
v = 0.2a ⇒ 2 = 0.2a
⇒ a = \(\frac{2}{0.2}\) = 10 m/s²
∴ Force applied F = ma = 1.4 kg × 10 m/s² = 14 N.

Question 9.
An object of mass 5 kg is moving with a velocity of 10 ms^-1. A force is applied so that in 15 s, it attains a velocity of 25 ms^-1. What is the force applied on the object? (AS 1)
Answer:
Mass (m) = 5 kg ; Initial velocity (u) = 10 m/s.; Time (t) = 15 s
Final velocity (v) = 25 m/s.
Acceleration a = \(=\frac{v-u}{t}=\frac{25-10}{15}=\frac{15}{15}=1 \mathrm{~m} / \mathrm{s}^{2}\).
Force applied on the object F = ma = 5 × 1=5 N.

Question 10.
Find the acceleration of body of mass 2 kg from the figures shown. (AS 1)

Answer:
1) Force 30 N is acting downwards on weight of (2 × 10) = 20 kg.
The acceleration a = \(\frac{30-20}{2}=\frac{10}{2}\) = 5 m/s²

2) m₁ = 2 kg, m₂ = 3 kg.
m₂ pulls the body mt with a weight 3 × 10 = 30 N.
∴ Acceleration of m₁ = \(\frac{30-20}{3+2}=\frac{10}{5}\)
= 2 m/s².

Question 11.
Take some identical marbles. Make a path or a track keeping your notebooks on either side so as to make a path in which marbles can move. Now use one marble to hit the other marbles. Take two, three marbles and make them to hit the other marbles. What can you explain from your observations? (AS 5)
Answer:

1. When one marble is hit by another marble, both the marbles move with some velocity.
2. When the marble is hit by two, three marbles, all marbles move with a velocity which is more than in the previous case.
3. As we are hitting with more marbles, the mass increases. So that the net momentum also increases.

Question 12.
A man of mass 30 kg uses a rope to climb which bears only 450 N. What is the maximum acceleration with which he can climb safely? (AS 1, AS 7)
Answer:
Mass m = 30 kg. ; Force F = 450 N
Acceleration a = ?
F = ma
∴ a = \(\frac{\mathrm{F}}{\mathrm{m}}=\frac{450}{30}\) =15 m/sec²
∴ The required acceleration =15 m/sec²

Question 13.
An vehicle has a mass of 1500 kg. What must be the force between the vehicle and the road if the vehicle is to be stopped with a negative acceleration of 1.7 m/sec²? (AS 1, AS 7)
Answer:
Mass of the vehicle, m = 1500 kg.; Acceleration (-a) = 1.7 m/sec²
Force, F =?
F = m (-a) = 1500 × (-1.7) = ( – ) 2550 N
∴ The force between the vehicle and road is 2550 N, in the direction opposite to that of the vehicle.

Question 14.
If a fly collides with the windshield of a fast moving bus, is the impact force experienced, same for the fly and the bus? Why? (AS 1, AS 2)
Answer:
The impact force experienced by the fly will be more, because the mass of fly is negligible when compared to the mass of the bus.

Question 15.
A truck is moving under a hopper with a constant speed of 20 m/sec. Sand falls on the truck at a rate 20 kg/s. What is the force acting on the truck due to falling of sand? (AS 1, AS 7)
Answer:
Mass of the sand falling on the truck in 1 sec = 20 kg
Constant speed of the truck = 20 m/s
Acceleration in 1 sec, a = \(\frac{\Delta v}{\Delta t}=\frac{20}{1}\) = 20 m/sec²
Force applied on the truck, F = ma = 20 kg x 20 m/sec² = 400 N

Question 16.
Two rubber bands stretched to the standard length cause an object to accelerate at 2 m/sec². Suppose another object with twice the mass is pulled by four rubber bands stretched to the standard length. What is the acceleration of the second object? (AS 1)
Answer:
First object:
Let the force applied by two rubber bands = F₁ Newton
Mass of the object = m₁ kg; Acceleration a₁ = 2 m/sec²
We know F = ma
F₁ = m₁ × 2
⇒ F₁ = 2m₁ …….(1)

Second object:
The force applied by 4 rubber bands = 2F₁ Newton
Mass of the object = 2m₁ kg ; Acceleration a₂ =?
We know F = ma
2F₁ = 2m₁. a₂

∴ Acceleration of the second object = 2 m/sec²

Question 17.
Illustrate an example of each of the three laws of motion. (AS 1)
Answer:
First law of motion :
A body continues its state of rest or of uniform motion unless a net force acts on it.
Ex:

1. When the bus which is at rest begins to move suddenly, the person standing in the bus falls backward.
2. When you are travelling in bus, the sudden stop of the bus makes you fall forward.

Second law of motion :
The rate of change of momentum of a body is directly proportional to the net force acting on it and it takes place in the direction of net force.
Ex : Place a ball on the veranda and push it gently. Then the ball accelerates from rest. Thus, we can say that force is an action which produces acceleration.

Third law of motion :
If one object exerts a force on the other object, the second object exerts a force on the first one with equal magnitude but in opposite direction.

Ex :

1. When birds fly, they push the air downwards with their wings, and the air pushes back the bird in opposite upward direction.
2. When a fish swims in water, the fish pushes the water back and the water pushes the fish with equal force but in opposite direction.
3. A rocket accelerates by expelling gas at high velocity. The reaction force of the gas on the rocket accelerates the rocket in a direction opposite to the expelled gases.

Question 18.
Two ice-skaters initially at rest, push of each other. If one skater whose mass is 60 kg has a velocity of 2 m/s. What is the velocity of other skater whose mass is 40 kg? (AS 1, AS 7)
Answer:
Mass of first skater m₁ = 60 kg.;
Velocity of first skater v₁ = 2 m/s.
Mass of second skater m₂ = 40 kg.; Velocity of second skater v₂ =?
As the two skaters push each other, the resultant momentum will become zero.
The resultant momentum m₁v₁ + m₂v₂ = 0

∴ Velocity of second skater is 3 m/s, but in the direction opposite to the first skater.

Question 19.
A passenger in moving train tosses a coin which falls behind him. It means that the motion of the train is (AS 7)
a) Accelerated
b) Uniform
c) Retarded
d) Circular motion
Answer:
a) Accelerated

Question 20.
A horse continues to apply a force in order to move a cart with a constant speed. Explain. (AS 1)
Answer:

1. The cart moves when the force (in the form of pulling) is applied by the horse.
2. As the horse and cart are moving, the net momentum will be zero at any instance of time.
3. Hence when the horse comes to rest, the cart also comes to rest.
4. To avoid this and to move the cart with a constant speed, the horse must apply force continuously.

Question 21.
A force of 5 N produces an acceleration of 8 m/sec² on a mass m, and an acceleration of 24 m/sec² on a mass m₂. What acceleration would the same force provide If both the masses are tied together? (AS 1)
Answer:
For the first mass (m₁)
Force F = 5 N
Acceleration a = 8 m/s²
We know, F = ma
5 = m, . 8
m₁= \(\frac{5}{8}\) kg

For the second mass (m₂)
Force F = 5 N
Acceleration a = 24 m/sec²
We know, F = ma
5 = m₂.24
m₂ = \(\frac{5}{24}\) kg
When both the masses are tied together and the same force is applied, then

Question 22.
A hammer of mass 400 g, moving at 30 m/s, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer? (AS 1)
Answer:
Mass of the hammer (m) = 400 g = 0.4 kg. ; Velocity of the hammer (v) = 30 m/s.
Momentum (∆p) = 30 × 0.4 N – s.
The nail stops the hammer with in a time 0.01 s.
∴ ∆t = 0.01 s.
The stopping force of the nail on the hammer.

Question 23.
System is shown in figure. Assume there is no friction. Find the acceleration of the liloc-ks-and tension in the string. Take g = 10 m/s² (AS 1)

Answer:
m₁ = 3 kg, m₂ = 3 kg.
1) Acceleration ’a’ and tension T on m₁ are shown in figure.
2) Acceleration a and tension T will be as shown in figure.

Normal force = weight 3s – T = 3a ………… (2)
∴ only tension applied
Tension T = 3 kg × a = 3a
From (1) & (2)
3g – 3a = 3a ⇒ 3g = 6a ⇒ a = \(\frac{3 g}{6}=\frac{3 \times 10}{6}\) = 5 m/s².
Tension T = 3a = 3 × 5 = 15 N.

Question 24.
Three identical blocks, each of mass 10 kg, are pulled as shown on the hoii ;ontal frictionless surface. If the tension (F) in the rope is 30 N, what is the acceleration oi each block? And what are the tensions in the other ropes? (Neglect the masses of the ropes) (AS 1)

Answer:
Three blocks, each of mass 10 kg are pulled by a rope.
∴ Total mass = 10 + 10 + 10 = 30 kg
Force applied by the rope, F = 30 N
∴ Acceleration of each block, a = \(\frac{F}{m}=\frac{30}{30}\) = 1 m/sec².

Tension in first rope (T₁)
First rope pulls only one block whose mass is 10 kg with an acceleration 1 m/sec².
∴ F = ma = 10 kg × 1 m/sec² = 10 N

Tension in second rope (T₂)
Second rope pulls two blocks, each of mass 10 kg.
∴ Total mass = 10 + 10 = 20 kg.
Acceleration, a = 1 m/sec²
Force, F = ma = 20 × 1 = 20 N

Question 25.
A ball of mass’m’ moves perpendicularly to a wall with a speed v, strikes it and rebounds with the same speed in the opposite direction. What is the direction and magnitude of the average force acting on the ball due to the wall? (AS 7)
Answer:
According to the Newton s third law of motion,
Force exerted by ball on the wall = – (Force exerts by the wall on the ball)
∴ F_B.W = – F_W.B
Force exerted by ball :
Mass of ball = m, speed = v, F_BW = ma = \(\frac{\mathrm{m} \cdot \mathrm{v}}{\mathrm{t}}\)
As the wall is at rest and exerts some force on the ball of mass m, then it moves in the other direction with the same speed.

Question 26.
Divya observed a horse pulling a cart. She thought that cart also pulls the horse with same force in opposite direction. As per third law of motion, the cart should not move forward. But her observation of moving cart raised some questions in her mind. Can you guess what questions are raised in her mind? (AS 2)
Answer:

• According to Newton’s third law, when horse pulls a cart, the cart also pulls the horse with same force but in opposite direction. So the cart has to stop. Why is the cart moving?
• What is the effect of friction of ground on the cart and horse?
• Does the cart continue to move even if the horse stops pulling it?
• What makes the cart to move continuously?
• Does it become an isolated system?
• What is the action and reaction in this system?

Question 27.
How do you appreciate Galileo’s thought of “any moving body continues in the stale only until some external force acts on it”, which is contradiction to the Aristotle’s belief of “any moving body naturally comes to rest”? (AS 6)
Answer:

• Science is dynamic.
• All theories can change time to time so that the science and technology will be developed.
• Aristotle’s belief proved to be wrong only by the experiments conducted by Galileo.
• So anybody can challenge the existing theories with proper experimentation.
• Aristotle’s and Galileo’s contradictory thought lead Newton to propose most popular laws of motion.
• Newton’s third law of motion is the basic principle in rocket launching.
• Nowadays we are enjoying the results of satellites launched by rockets.
• Hence comfortable life is the effort on experiments, theories, and calculations made by scientists with a zeal to invent new.

9th Class Physical Science 2nd Lesson Laws of Motion InText Questions and Answers

9th Class Physical Science Textbook Page No. 24

Question 1.
Do all the bodies have the same inertia?
Answer:
The inertia of all bodies is not same. It depends on the mass of the object.

Question 2.
What factors can decide the inertia of a body?
Answer:
Mass is the factor, which decides the inertia of a body.

9th Class Physical Science Textbook Page No. 26

Question 3.
Is the acceleration increased when net force is increased?
Answer:
Yes, the acceleration increased, as we increased the net force without changing mass of the object.

9th Class Physical Science Textbook Page No. 29

Question 4.
What do you notice from the readings in the spring balances?
Answer:
The two spring balances stretch up to a certain limit equally.

Question 5.
Are the readings of two spring balances the same?
Answer:
Yes, the readings of two spring balances are equal.

Question 6.
Are we able to make the spring balances to show different readings by pulling them simultaneously in opposite directions? Why not?
Answer:
When same force is applied in both the directions, we are unable to make the spring balances to show different readings because the action and reaction are same in magnitude and opposite in direction. When we use two forces with different magnitudes, then the spring balances can show different readings.

9th Class Physical Science Textbook Page No. 30

Question 7.
Does the rocket exert a force on the gas expelled from it?
Answer:
The rocket also exerts a force on the gas expelled from it.

9th Class Physical Science Textbook Page No. 32

Question 8.
Why does a pole vault jumper land on thick mats of foam?
Answer:
A thick mat of foam reduces the force of impact of the jumper, so that he doesn’t have any damage to his body.

Question 9.
Is it safe to jump on sand rather than a cement floor? Why?
Answer:

• It is safe to jump on sand rather than a cement floor.
• A soft and more cushioned landing surface provides a greater stopping distance because of the longer time taken to stop.

9th Class Physical Science Textbook Page No. 24

Question 10.
You may have seen the trick where a tablecloth is jerked from a table, leaving the dishes that were on the cloth nearly in their original positions.
a) What do you need to perform this successfully?
Answer:
We need a table, a cloth and some massive objects to perform this activity. The performer drag the cloth from the table very skillfully.

b) Which cloth should we use? Is it cloth made of thick cotton or thin silk?
Answer:
We have to use a thin silk cloth to perform this activity.

c) Should the dishes possess large mass or small mass?
Answer:
The dishes must possess large mass. We should not use lighter objects like plastic cups, etc.

d) Is it better to pull the cloth with a large force or pull it with a gentle and steady force?
Answer:
The cloth must be pulled with a gentle force, but with a sudden jerk.

Question 11.
What is the velocity of a small object that has separated from a rocket moving in free space with velocity 10 km/s?
Answer:
When a small object is separated from another object which is moving with a certain velocity. The small object also moves with a velocity equal to that of the object from which it is separated. Hence, the speed of the small object is 10 km/s.

9th Class Physical Science Textbook Page No. 27

Question 12.
Observe the following diagram.

What is the upper limit of weight that a strong man of mass 80 kg can lift as shown in figure?
Answer:
Total force mg = N + T (N = normal force, T = tension)
As the person is standing on the floor, the normal force N = 0.
80 = 0 + T
∴ T = 80
∴ The upper limit of the weight that the person in the figure can lift is 80 kg.

Question 13.
What is the momentum of a ceiling fan when it is rotating?
Answer:
Ceiling fan when it is rotating, possesses angular momentum.
Angular momentum L = mvr or mr²ω.

Question 14.
Is it possible to move in a curved path in the absence of a net force?
Answer:
A body comes into curved path, when centripetal force real force acts on it. Immediately after coming into curved path, an imaginary force which acts away from the centre i.e., centrifugal force comes into existence. These two forces are equal in magnitude and opposite in direction. Hence the net force is zero. So it is possible to move in a curved path in the absence of a net force.

Question 15.
Prove that the tension throughout the string is uniform when the mass of string is considered to be zero.
Answer:

Let a body of mass m is suspended through a string. The weight of the object mg acts downwards. Now tension in the string T = mg + m_sg when m_s is the mass of the string.
Here m_s is considered as zero.
Hence T_A= mg + 0 = mg; T_B = mg ; T_C = mg ; T_D = mg
∴ The tension throughout the string is uniform when the mass of string is considered to be zero.

9th Class Physical Science Textbook Page No. 31

Question 16.
The force exerted by the earth on the ball is 8 N. What is the force on the earth by the ball?
Answer:

• The force exerted by the earth on the ball is 8 N.
• The force exerted by the ball on the earth is – 8 N.

“According to Newton’s third law, if a body A exerted a force p on another body B, the B exerts a force -p on A, the two forces acting along the same line”.

Question 17.
A block is placed on the horizontal surface. There are two forces acting on the block. One, the downward pull of gravity and other a normal force acting on it. Are these forces equal and opposite? Do they form action – reaction pair? Discuss with your friends.
Answer:
These two forces form action-reaction pair.

Question 18.
Why is it difficult for a fire fighter to hold a hose that ejects large amount of water at high speed?
Answer:
A large amount of water with high speed ejects from the hose of a fire engine, produces a large force in forward direction. According to action-reaction, the hose moves back with the same force. But the fire fighter has to resist that reaction force. Hence it becomes very difficult for him.

9th Class Physical Science Textbook Page No. 33

Question 19.
A meteorite burns in the atmosphere before it reaches the earth’s surface. What happens to its momentum?
Answer:
The momentum of the meteorite becomes zero. It doesn’t touch the ground as it burns in the atmosphere. So no mass of the meteorite hits the ground.

Question 20.
As you throw a heavy ball upward, is there any change in the normal force on your feet?
Answer:
The normal force on the feet changes its direction and acts in upward direction. As a result we raise our foot while throwing the ball.

Question 21.
When a coconut falls from a tree and strikes the ground without bouncing. What happens to its momentum?
Answer:
Its momentum doesn’t change but its impact force will be very less because it is not bouncing.

Question 22.
Air bags are used in’the cars for safety. Why?
Answer:
When a car hits another vehicle, the air bags immediately comes in between the persons in the car and the wind shield of the car, to prevent damage to life of passengers.

9th Class Physical Science 2nd Lesson Laws of Motion Activities

Activity – 1

Question 1.
Explain the motion of a pen cap kept on a thick paper ring.
Answer:

1. Make a circular strip from a thick paper.
2. Balance the hoop on the centre of the mouth of the bottle.
3. Now balance a pen cap on the paper hoop aligning it on the centre of the bottle’s mouth.
4. Give the paper hoop a sharp push with your finger as fast as you can.
5. We observe that the pen cap suddenly falls into the bottle.
6. As we push paper hoop, we applied force on the paper hoop. So it changed its state from rest to motion.
7. Pen cap cannot change its state of rest.
8. Due to gravitational force, the pen cap falls into the bottle.

Activity – 2

Question 2.
Explain the motion of the carrom coins hit by a striker.
Answer:

1. Make a stack of carrom coins on the carrom board.
2. Give a sharp hit at the bottom of the stack with striker.
3. We can find that the bottom coin will be removed from the stack.
4. The other coins in the stack will slide down.
5. When we apply force on the bottom coin, the coin will move, due to change in the state of rest.
6. The stack of remaining coins does not fall vertically due to inertia.

Activity – 3

Question 3.
Show that the object with larger mass has greater inertia.
Answer:

1. Take two rectangular wooden blocks with different masses.
2. Place them on a straight line drawn on a floor.
3. Give the same push at the same time to both the blocks with the help of a wooden scale.
4. We observe that the block with small mass will accelerate more and goes farther.
5. The block with large mass accelerates less and moves shorter, due to high inertia.
6. This shows that the bodies of higher mass have high inertia.

Activity – 4

Question 4.
Show that the larger the net force greater the acceleration.
Answer:

1. Gently push a block of ice on a smooth surface and observe how the object speeds up, in other words how it accelerates.
2. Now increase the net force and observe change in its speed.

Observation : The acceleration increases.

Conclusion : If the net force is larger, then the accelerations greater.

Activity – 5

Question 5.
Show that the larger the mass smaller the acceleration.
Answer:

1. Apply a force on an ice block.
2. It undergoes some acceleration.
3. Now take a block of ice with greater mass.
4. Then apply almost the same force on the ice block which has greater mass.

Observations :

1. In both cases the object accelerates.
2. But we can observe in the second case, it will not speed up as quickly as before.

Conclusion : If the mass is larger, then the acceleration is smaller.

Activity – 6

Question 6.
Pulling two spring balances.
Answer:
Let’s take two spring balances of equal calibrations. Connect the two spring balances as shown in figure. Pull the spring balances in opposite directions as shown in figure.

Observation :
There is no change in the reading of spring balances. We are not able to make the spring balances to show different readings by pulling them simultaneously in opposite directions.

Conclusion :
According to third law of motion, when an object exerts a force on the other object, the second object also exerts a force on the first one which is equal in magnitude but opposite in direction.

The two opposing forces are known as action and reaction pair. Newton’s third law explains what happens when one object exerts a force on another object.

Activity – 7

Question 7.
Describe the preparation of a balloon rocket. What is the principle involved in it?
Answer:

Preparation of a balloon rocket :

1. Inflate a balloon and press its neck with fingers to prevent air escaping from it.
2. Pass a thread through a straw and tape the balloon on the straw.
3. Hold one end of the thread and ask your friend to hold the other end of the thread.
4. Now release air from balloon by removing fingers from the neck of the balloon.
5. The balloon moves like a rocket towards the other end

Principle involved in it:

1. Newton’s third law of motion is the principle.
2. As the air in the balloon moves backward, the balloon moves forward.

Lab Activity

Question 8.
Describe an activity to study the action and reaction forces acting on two different objects.
Answer:
Aim : To show the action and reaction forces acting on two different objects.

Material required : Test tube, rubber cork cap, Bunsen burner, laboratory stand and thread.

Procedure :

1. Take a test tube and put a small amount of water in it.
2. Place a rubber cork cap at its mouth to close it.
3. Now suspend the test tube horizontally to a stand with the help of two strings.
4. Heat the test tube with a bunsen burner until water vapourize and the rubber cork cap blows out.

Observations :

1. Observe the movement of test tube when cork cap blows out.
2. As the cork cap blows out in forward direction, the test tube recoils back.
3. We can observe the velocities of cork cap and recoil of test tube to be same.

Activity – 8

Question 9.
Show that the impulse will be less on a soft and cushioned surface.
Answer:

• Take two eggs.
• Drop them from a certain height, so that one egg falls on a concrete floor and the other on a cushioned pillow.
• We observe that the egg that falls on a concrete floor will break.

• The reason is large force acts on the egg for short interval of time.
  ∆p = F_net∆t
• The egg which falls on a cushioned pillow doesn’t break, because a smaller force acts on the egg for a longer time.
  ∆p= F_net ∆t
• This shows that the impulse (∆P) will be less on a soft and cushioned surface.

Note : Even if the ∆p is the same in both cases, the magnitude of the net force (F_{net</sub) acting on the egg determines whether the egg will break or not.}

AP State Syllabus AP Board 9th Class Physical Science Solutions Chapter 1 Motion Textbook Questions and Answers.

AP State Syllabus 9th Class Physical Science Solutions 1st Lesson Motion

9th Class Physical Science 1st Lesson Motion Textbook Questions and Answers

Improve Your Learning

Question 1.
As shown in following figure, a point traverses the curved path.
Draw the displacement vector from given points A to B.

Answer:

As the point traverses from A to B, the displacement is the shortest distance between A and B. Hence the displacement vector will be as follows.

Question 2.
“She moves at a constant speed in a constant direction.” Rephrase the same sentence in fewer words using concepts related to motion. (AS 1)
Answer:
“She moves with constant velocity”.

Reason :
Constant speed in a constant direction is nothing but ‘constant velocity’.

Question 3.
What is the average speed of a Cheetah that sprints 100 m in 4 sec? What if it sprints 50 m in 2 sec? (AS 1, AS7)
Answer:

Question 4.
Correct your friend who says, “The car rounded the curve at a constant velocity of 70 km/h”. (AS 1)
Answer:
“The car rounded the curve at a constant speed of 70 km/h”.

Reason :
In a circular motion, speed remains constant but velocity changes.

Question 5.
Suppose that the three balls shown in figure below start simultaneously from the top of the hills. Which one reaches the bottom first? Explain.

Answer:
Ball from first hill reaches the bottom first.
Reason :

1. In the first hill, the ball has uniform rectilinear motion.
2. So, the speed and velocity have same magnitude and direction.
3. In the second and third hills, the ball takes curved path.
4. So, the direction of velocity changes.

Question 6.
In the figure given below distance vs time graphs showing motion of two cars A and B are given. Which car moves fast? (AS 1)

Answer:
Car A moves fast.

Reason :

1. If we draw perpendiculars to X and Y axes from A and B respectively, we can observe that A covers large distance (S₁) within a short time (t₁).
2. Find the slopes of the lines OA and OB at any instant. Slope of OA is high. Hence car A moves faster.

Question 7.
Draw the distance vs time graph when the speed of a body increases uniformly. (AS 5)
Answer:
Let us consider a car moves as shown in the table.

Time (t) sec	Distance in meters
0 sec	0 meters
1 sec	3 meters
2 sec	6 meters
3 sec	9 meters
4 sec	12 meters
5 sec	15 meters

Now draw a s-t graph.

Question 8.
Draw the distance-time graph when its speed decreases uniformly. (AS 5)
Answer:
Let us consider the movement of a car after applying brakes.

Time (t) sec	Distance in meters
0 sec	20 meters
1 sec	18 meters
2 sec	16 meters
3 sec	14 meters
4 sec	12 meters
5 sec	10 meters

Now draw distance-time graph.

Question 9.
A car travels at a speed of 80 km/h during the first half of its running time and at 40 km/h during the other half. Find the average speed of the car. (AS 1, AS 7)
Answer:
Let the total running time = x hrs

Question 10.
A car covers half the distance at a speed of 50 km/h and the other half at 40 km/h. Find the average speed of the car. (AS 1, AS 7)
Answer:
Let the total distance = x km.
First half is covered with a speed of 50 km/h.

Question 11.
Derive the equation for uniform accelerated motion for the displacement covered in its n^th second of its motion. (s_n = u + a ( n – \(\frac{1}{2}\)) (AS 1)
Answer:
We know that distance travelled by an object in t seconds Is s = ut + \(\frac{1}{2}\) at²
∴ Distance travelled in ‘n’ seconds, s_{(n sec)} = un + \(\frac{1}{2}\)an² ………. (1)
Distance travelled in (n – 1) seconds, s_{(n – 1)} = u(n – 1) + \(\frac{1}{2}\) a(n – 1)² …….. (2)

Question 12.
A particle covers 10m in first 5 sec and 10m in next 3 sec. Assuming constant acceleration. Find initial speed, acceleration and distance covered in next 2 sec. (AS 1, AS 7)
Answer:
Distance covered in first 5 sec = 10

To find the distance covered in next 2 sec, we have to find the initial speed after 8 sec
i.e., the final velocity after 8 sec.

Question 13.
A car starts from rest and travels with uniform acceleration ‘α’, for some time and then with uniform retardation ‘β’ and comes to rest. The time of motion is “t”. Find the maximum velocity attained by it. (αβt/(α+β)) (AS 1, AS 7)
Answer:
Acceleration a = a m/sec²
Initial speed u = 0 m/sec
Let the time be t₁ sec.
From equation v = u + at
⇒ v = 0 + αt₁
\(\therefore \mathrm{t}_{1}=\frac{\mathrm{v}}{\alpha} \mathrm{sec}\)
Retardation a = – β m/sec²
Initial speed ‘u’ is equal to the final
velocity with acceleration ‘α’
= u = αt₁ m/sec
Final velocity v = 0 m/sec
Let the time be t₂ sec

Question 14.
A man is 48m behind a bus which is at rest. The bus starts accelerating at the rate of 1 m/s², at the same time the man starts running with uniform velocity of 10 m/s. What is the minimum time in which the man catches the bus? (AS 1, AS 7)
Answer:
Bus is at rest.
∴ u = 0; a = 1 m/sec²
Let the bus cover the distance ‘s’ in ‘n’ seconds.

A man running with uniform velocity, v = 10 m/sec.
Distance covered by man in n seconds = 10 nm.
But after ‘n’ seconds the man catches the bus.

∴ The minimum time in which the man catches the bus is 8 sec.

Question 15.
A body leaving a certain point “O” moves with a constant acceleration. At the end of the fifth second, its velocity is 1.5 m/s. At the end of the sixth second, the body stops and then begins to move backwards. Find the distance traversed by the body before it stops. Determine the velocity with which the body returns to point “0”. (AS 1)
Answer:
Velocity in 5^th sec = 1.5 m/sec ; The body comes to rest in 6^th sec.
∴ Final velocity in 6^th sec, v = 0
∴ Acceleration in 6^th sec is v = u + at ⇒ 0 = 1.5 + a. 1 ⇒ a = -1.5 m/sec²
[The velocity in 5^th sec becomes the initial velocity for 6^th sec]

After 6 sec, the body comes to rest.
∴ v = 0, a = -1.5 m/sec², u = ?, t = 6 sec.
v = u + at ⇒ 0 = u – 1.5 × 6 ⇒ u = 9 m/sec.
∴ Distance traversed by the body in 6 sec. i.e., before it stops.
s = ut + \(\frac{1}{2}\) at² = 9 × 6 + \(\frac{1}{2}\) × – 1.5 × 6² = 54 – 27 = 27m.
For backward journey,
u = 0 m/sec, t = 6 sec, a = -1.5 m/sec²
v = u + at ⇒ v = 0 – 1.5 × 6 ⇒ v = – 9
∴ Velocity for backward journey is – 9 m/sec.

Question 16.
Distinguish between speed and velocity.(AS 1)
Answer:

Speed	Velocity
1. The distance covered in unit time is called average speed.	1. The displacement of an object per unit time is called average velocity.
2. Speed = \(\frac{\text { Distance }}{\text { Time }}\)	2. Velocity =\(\frac{\text { Displacement }}{\text { Time }}\)
3. Speed is scalar.	3. Velocity is vector.
4. Speed gives the idea of how fast the body moves.	4. Velocity gives the idea of how fast the body moves in specified direction.

Question 17.
What do you mean by constant acceleration? ((AS 1)
Answer:

• Acceleration is the rate of change of velocity.
  2
• It gives an idea how quickly velocity of a body is changing.
• Acceleration is uniform, when in equal intervals of time, equal changes of velocity occurs.
• For example, while driving a car, if we steadily increase the velocity from 30 km/h to 35 km/h in 1 sec and 35 km/h to 40 km/h in the next second and so on. In this case the acceleration is 5km/h, is said to be constant acceleration.
  

Question 18.
When the velocity is constant, can the average velocity over any time interval differ from instantaneous velocity at any instant ? If so, give an example; if not, explain why. (AS 2, AS 1)
Answer:
No. Here velocity is constant.
∴ Average velocity over any time interval is same and the instantaneous velocity at any instant is same.

Ex : Let us consider a car moves on a straight road with constant velocity say 10 m/s.
1) Now let the distance covered (AB) by the car in 1 s = 10 m.

2) Distance covered in 2s (AC) = 20 m.
Average velocity from A to C is \(\frac{20 m}{2 s}\) = 10 m/s.

3) Instantaneous velocity at A or B or C at any point = 10 m/s.

Question 19.
Can the direction of velocity of an object reverse when its acceleration is constant? If so give an example; if not, explain why? (AS 2, AS 1)
Answer:
Yes. In case of a vertically projected body, while the body is moving up, the direction of velocity is upward, whereas while it is falling down, the direction of velocity is downward. Acceleration in both the cases is constant (numerically).

Question 20.
A point mass starts moving in a straight line with constant acceleration V’. At a time t after the beginning of motion, the acceleration changes sign, without change in magnitude. Determine the time t0 from the beginning of the motion in which the point mass returns to the initial position. (AS 1)
Answer:

Question 21.
Consider a train which can accelerate with an acceleration of 20 cm/s² and slow down with deceleration of 100 cm/s². Find the minimum time for the train to travel between the stations 2.7 km apart. (AS 1)
Answer:
Let the Acceleration of the train a = 20 cm/s²
Deceleration of the train β = 100 cm/s²
Distance between the two stations s = 2.7 km = 27 × 10⁴ cm
Let the minimum time for the train to travel between the two stations is t sec.

Question 22.
You may have heard the story of the race between the rabbit and tortoise. They started from same point simultaneously with constant speeds. During the journey, rabbit took rest somewhere along the way for a while. But the tortoise moved steadily with lesser speed and reached the finishing point before rabbit. Rabbit woke up and ran, but rabbit realized that the tortoise had won the race. Draw distance vs time graph for this story. (AS 5)
Answer:

1. OX – movement of tortoise.
2. OABC – movement of rabbit
3. Rabbit and tortoise start at O’.
4. After time tj rabbit is at A and tortoise is at P.
5. Rabbit takes rest up to time t₂.
6. After time t₂, tortoise is at Q, but rabbit has no displacement.
7. After time t₃, the tortoise reaches the destination ‘X’.
8. But rabbit reaches the destination after time t₄.

Question 23.
A train of length 50 m is moving with a constant speed of 10 m/s. Calculate the time taken by the train to cross an electric pole and a bridge of length 250 m. (AS 1)
Answer:
Length of the train 50 m.; Speed of the train v = 10 m/s.
Distance travelled while crossing an electric pole = Length of the train = s = 50 m.
∴ Time taken to cross the electric pole ‘t’ = \(\frac{s}{v} \Rightarrow t=\frac{50}{10}\) = 5 s.
Length of the bridge = 250 m.
Distance travelled while crossing the bridge = Length of train + Length of bridge
= 50 + 250 = 300 m.
∴ Time taken to cross the bridge = \(\frac{300 \mathrm{~m}}{10 \mathrm{~m} / \mathrm{s}}\) = 30 sec.

Question 24.
Two trains each of having a speed of 30 km/h are headed at each other in opposite direction on the same track. A bird flies off one train to another with a constant speed of 60 km/h when they are 60 km apart till before they crash. Find the distance covered by the bird and how many trips the bird can make from one train to the other before they crash. (AS 1)
Answer:
Speed of each train = 30 km/hr
Speed of the bird = 60 km/hr
Distance between the two trains = 60 km
These two trains crash in one hour.
The bird flies a distance of 60 km till before the two trains crash.
The bird can make number of trips (infinity) before they crash.

Question 25.
A Stone dropped from top of a well reaches the surface of water in 2 seconds, find the velocity of stone while it touches the surface of water and what is the depth of the water surface from top of well (g=10m/s²) (Using V = U + at, S = Ut + 1/2 at²)
Answer:
Given that
t = 2s
u = 0 m/s [∵ free fall body]
v = ?
Depth s = ?
a = g = 10 m/s²

i) v = u + at
v = 0 + 10 × 2 = 20 m/s

ii) s = ut + – \(\frac{1}{2}\)at²
= 0 + \(\frac{1}{2}\) × 10 × 2²
= \(\frac{1}{2}\) × 10 × 4
= 20 m
Hence, velocity of stone while it touches the surface of water = 20 m/s
Depth of the water surface from the top of well = 20 m.

Question 26.
An object moving with 6m per second execute an acceleration 2 m/s² in next 3 seconds. How much distance it covered? (s = ut + 1/2 at²)
Answer:
u = 6 m/s; t = 3 sec; a = 2 m/s²
s = ut + \(\frac{1}{2}\) at²
= 6 × 3 + \(\frac{1}{2}\) × 2 × 3² = 18 + 9 = 27 m
The object covers 27 m in 3 sec.

Question 27.
A car stopped after travelling distance 8 m due to applying brakes at the speed of 40 m/s. Find acceleration and retardation of car in that period, (v² – u² = 2as)
Answer:
Here u = 40 m/s; v = 0 (vehical stopped); s = 8 m; a =?
v² – u² = 2as 0 – 40² = 2 × a × 8
a = \(\frac{-(40)(40)}{2 \times 8}\) =-100m/s
Acceleration = 100 m/s² with retordation on (-sign).

9th Class Physical Science 1st Lesson Motion InText Questions and Answers

9th Class Physical Science Textbook Page No. 1

Question 1.
If earth is in motion, why don’t we directly perceive the motion of the earth?
Answer:
Earth is in motion. We, the people on the earth also move with a speed equal to that of the earth. We cannot directly perceive the motion of the earth, because of this.

Question 2.
Are the walls of your classroom at rest or in motion? Why?
Answer:
The walls are at rest in view of our observation. When we discuss this in view of the motion of the earth, the walls are also in motion.

Question 3.
Have you ever experienced that the train in which you sit appears to move when it is at rest? Why?
Answer:
This happens when we sit in a stationary train and, the train on another track starts moving.

9th Class Physical Science Textbook Page No. 2

Question 4.
Why do we observe these changes?

Answer:
These changes are due to the point of observation. We know that earth is a sphere, the upward direction of the vertical position on its surface decisively depends upon the place on the earth’s surface, where the vertical is drawn.

Question 5.
Are the terms relative or not?
Answer:
The terms “longer”, “shorter”; “up” and “down”, etc. are relative to each other.

9th Class Physical Science Textbook Page No. 4

Question 6.
What answer may the passenger give to the driver?
Answer:
The car is in motion with respect to the observer on the road, but at rest with respect to the passenger. Because motion is a combined property of the observer and the body which is being observed.

Question 7.
How do we understand motion?
Answer:
A body is said to be in motion when its position is changing continuously with time relative to an observer.

9th Class Physical Science Textbook Page No. 6

Question 8.
Can you measure the average speed and average velocity?
Answer:
Yes, we can measure the average speed and average velocity.

Question 9.
How can you differentiate speed and average velocity?
Answer:

1. Speed gives the idea of how fast the body moves.
2. Velocity is the speed of an object in a specified direction.

9th Class Physical Science Textbook Page No. 7

Question 10.
Can you find the speed of the car at a particular instant of time?
Answer:
Yes, we can find the speed of the car at any instant of time by looking at its speedometer

Question 11.
What is the speed of the car at the instant of time ‘t₃‘ for given motion?
Answer:

The instantaneous speed is represented by the slope of the curve at a given instant of time. We can find the slope of the curve at any point on it by drawing a tangent to the curve at that point. The slope of the curve gives speed of the car at that instant.

9th Class Physical Science Textbook Page No. 8

Question 12.
In what direction does an object move? Distance vs time graph
Answer:
The object moves in the direction tangential to the direction of the motion of the string.

Question 13.
Which motion is called uniform? Why?
Answer:
The motion of the body is said to be in uniform when its velocity is constant.

9th Class Physical Science Textbook Page No. 9

Question 14.
What is the shape of the graph?
Answer:

The shape of the graph for a body which is in uniform motion is a straight line as shown in the figure.

9th Class Physical Science Textbook Page No. 10

Question 15.
a) What is the shape of the graph?
Answer:
It is a curve.

b) Is it a straight line or not? Why?
Answer:
The graph is not a straight line because the speed is changing irregularly.

Question 16.
Draw velocity vectors in the given figure at times t = 0, 1s, 2s, 3s.
Answer:

From the graph, we conclude that when the ball moves down the inclined plane its speed increases gradually but its direction remains constant.

9th Class Physical Science Textbook Page No. 11

Question 17.
Draw velocity vectors at times t = 1s, 2s, 3s in the given figure.
Answer:

From the graph, we conclude that when the ball moves up the inclined plane its speed decreases gradually, but the direction of motion remains constant.

Question 18.
Can you give few examples for motion of an object where its speed remains constant but velocity changes?
Answer:
For the bodies which are in uniform in circular motion the speed remain constant but velocity change. Ex : Rotation of earth, revolution of moon around the earth, etc.

Question 19.
Is the direction of motion constant? How?
Answer:
No, the direction of motion also changes continuously.

Question 20.
Can you give some more examples where speed and direction simultaneously change?
Answer:
Motion of a rocket, horizontally projected body, kicked football, a cricket ball bowled by a bowler, etc.

9th Class Physical Science Textbook Page No. 12

Question 21.
What is acceleration? How can we know that a body is accelerating?
Answer:

1. Acceleration gives an idea how quickly velocity of a body changing.
2. It is equal to the rate of change in velocity.
3. While travelling in a bus or car, when the driver presses the accelerator, the passen¬gers sitting in the bus experience acceleration. Their bodies press against the seats due to acceleration.

Question 22.
At which point is the speed maximum?
Answer:
At B, the speed will be maximum.

Question 23.
Does the object in motion possess acceleration or not?
Answer:
Any object which is in motion possesses acceleration.

9th Class Physical Science Textbook Page No. 5

Question 24.
What is the displacement of the body if it returns to the same point from where it started? Give one example from daily life.
Answer:
When a body returns to the same point where it is started, then the displacement is zero.
Ex: A man starts from his home, goes to a market and returns home. Then his displacement is zero.

Question 25.
When do the distance and magnitude of displacement become equal?
Answer:
The distance and the magnitude of displacement become equal when the body moves along a straight line in one direction.

9th Class Physical Science Textbook Page No. 6

Question 26.
What is the average speed of the car if it covers 200 km in 5 h?
Answer:
Distance = 200 km ; Time = 5 h

Question 27.
When does the average velocity become zero?
Answer:
The average velocity of a body becomes zero when its displacement is zero.

Question 28.
A man used his car. The initial and final odometer readings are 4849 and 5549 respectively. The journey time is 25h. What is his average speed during the journey?
Answer:
Distance covered = 5549 – 4849 = 700 km.
Time = 25h.

9th Class Physical Science Textbook Page No. 9

Question 29.
Very often you must have seen traffic police stopping motorists or scooter drivers who drive fast and fine them. Does fine for speeding depend on average speed or instantaneous speed? Explain.
Answer:
Instantaneous speed.

Question 30.
One airplane travels due north at 300 km/h and another airplane travels due south at 300 km/h. Are their speeds the same? Are their velocities the same? Explain.
Answer:

1. Speed is same.
2. Velocities are same in magnitude but differs in direction in the observer’s point of view.

Question 31.
The speedometer of the car indicates a constant reading. Is the car in uniform motion? Explain.
Answer:
Yes.

1. The indicator in speedometer changes its position even for a small change in speed.
2. As it indicates a constant reading, the car moves equal distances at equal intervals of time.
3. Hence the motion is uniform.

9th Class Physical Science Textbook Page No. 11

Question 32.
An ant is moving on the surface of a ball. Does it’s velocity change or not? Explain.
Answer:
Velocity changes.
As the ant is moving on the surface of a ball, it has to go in circular motion.

Question 33.
Give an example of motion where there is a change only in speed but no change in direction of motion.
Answer:
Motion of a bus on the road.

9th Class Physical Science Textbook Page No. 13

Question 34.
What is the acceleration of a race car that moves at constant velocity of 300 km/h?
Answer:
Velocity = 300 km/h = \(300 \times \frac{5}{18}=\frac{500}{6}\) = 83.33 m/sec
As the velocity is constant, the acceleration is also constant.
∴ Acceleration = 83.33 m/sec².

Question 35.
Which has the greater acceleration, an airplane, that goes from 1000 km/h to 1005 km/h in 10s or a skateboard that goes from zero to 5km/h in 1 second?
Answer:

Question 36.
What is the deceleration of a vehicle moving in a straight line that changes its velocity from 100 km/h to a dead stop in 10 sec?
Answer:

Question 37.
Correct your Mend who says “Acceleration gives an idea of how fast the position changes.”
Answer:
“Acceleration gives the idea of how fast the position changes in a given direction.”

9th Class Physical Science 1st Lesson Heat Activities

Activity 1 Distance and Displacement

Question 1.
Draw a graph showing the difference between distance and displacement.
Answer:

• Take a ball and throw it into the air with some angle to the horizontal.
• Observe its path and draw it on paper.
• The figure shows the path taken by the ball.
• The distance ASB gives the distance travelled by the ball.
• The length of \(\overrightarrow{\mathrm{AB}}\) gives the displacement of the ball.

Displacement:
Displacement is the shortest distance between initial and final points in a specified direction represented by a vector.

Distance :
Distance is the length of the path traversed by an object in a given time interval.

Activitie – 2 Drawing displacement vectors

Question 2.
Draw displacement vectors from A to B in the following situations.

Answer:

Activitie – 3 Measuring the average speed

Question 3.
How can the average speed be measured?
Answer:

1. Select two positions (say A and B) 50 meters apart in the ground.
2. Ask some students to stand at point A.
3. Ask another group of students with stop watches to stand at B.
4. When you clap your hand, the students at A start running towards the point B in any direction or path.
5. At the same time the students at B start their stop watches.
6. Observe that for each runner there is a student at B to measure the time taken for completing the race.
7. Note the time taken by each student to cover the distance between the points A and B in the table given below.
   

   Student	Time taken to reach B (Sec.)	Average speed (50 ft) m/s
   A1	t1	………
   A2	t2	………
   A3	t3	………

8. The student who took the least time to reach B (from A) is said to be the fastest runner.
9. The student who is fastest runner has the greatest average speed.
10. Thus we measure the average speed of any moving body.

Activity – 4 Observing the direction of motion of a body

Question 4.
Show that the direction of velocity is tangent to the path at a point of interest when a body is in uniform circular motion.
Answer:

• Carefully whirl a small object on the end of the string in the horizontal plane.
• Release the object while it is whirling on the string.
• We observe that the body along the tangent at the point where we released the body.
• Try to release the object at different points on the circle and observe the direction of motion of object after it has been released form the string.
• We will notice that the object moves on a straight line along the tangent to the circle at the point where we released it.

Activity – 5 Understanding uniform motion

Question 5.
Describe uniform motion.
Answer:
1) Consider a cyclist moving on a straight road.
2) The distance covered by him with respect to time is given in the following table.

Time (t in seconds)	Distance (s in meters)
0	0
1	4
2	8
3	12
4	16
–	–

3) Draw distance vs time graph for the given values in the table.

4) The graph will be as shown in the figure.
5) The straight line graph shows that the cyclist covers equal distances in equal inter¬vals of time.
6) If the direction of motion of the cyclist is assumed as constant, then we conclude that velocity is constant.
7) The motion of the body is said to be uniform when its velocity is constant.

Activity – 6 Observing the motion of a ball on an inclined plane

Question 6.
Describe an activity to explain the situation that “the speed changes but the direction of motion remains constant”.
Answer:
1) Set up an inclined plane as shown in the figure.
2) Take a ball and release it from the top of the inclined plane.
3) The positions of the ball at various times are shown in the figure given below.

4) On close observation we find that when the ball moves down on the inclined plane, its speed increases gradually, and the direction of motion remains constant on inclined plane.
5) Now push the ball till it acquires certain speed and release it with that speed from the bottom of the inclined plane.

6) We observe that the ball moves upward to a certain distance and comes back to the bottom.
7) From this we conclude that the speed changes but the direction of motion remains constant.

Activity – 7 Observing uniform circular motion

Question 7.
Explain with an example where “speed remains constant, but its velocity changes”.
Answer:

• Whirl continuously a stone which is tied to the end of the string.
• Draw its path of motion and velocity vectors at different positions as shown in figure.
• Assume that the speed of stone is constant.
• We observe that the path of the stone is a circle, and the direction of velocity changes at every instant of time, but the speed is constant.
• In this activity, we observe that though speed remains constant, its velocity changes.

Activity – 8 Observing the motion of an object thrown into air

Question 8.
Explain an activity to observe where speed and direction of motion change continuously.
Answer:

• Throw a stone into the air by making some angle with the horizontal.
• The path of the stone and velocity vectors are as shown in the figure.
• Here we observe that the speed of stone is not uniform as it traverses different distances at different intervals of time and finally comes to rest.
• The direction of motion is also not constant, as shown by the velocity vectors.
• In this activity, we noticed that the speed and direc¬tion of motion both change continuously.

Lab Activity

Question 9.
Describe an activity to find the acceleration and velocity of an object moving on inclined track.
Answer:
Aim : To find the acceleration and velocity of an object moving on an inclined track. Materials required: Glass marbles, book, digital clock, long plastic tubes and steel plate.
Procedure:

• Take a long plastic U type flat electrical wire cassing channel of length nearly 200 cm. Use this channel as track.
• Mark the readings in cm, along the track.
• Place one end of the track on a book and the other end on the floor.
• Keep a steel plate on the floor at the bottom of the track.
• Consider the reading at the bottom of the track as zero.
• Take a marble having enough size to travel in the track freely.
• Now release the marble freely from a certain distance say 40 cm.
• Start the digital clock when the marble is released.
• It moves down on the track and strikes the steel plate.
• Stop the digital clock when a sound is produced.
• Repeat the same experiment for the same distance 2 to 3 times and note the values of times in the table.

• Repeat the same experiment or various distances.
• Draw s -1 graph for above values.
• Do the above experiment by various slopes of the track and find acceleration in each case.

Conclusions :

1. As the slope increases, acceleration increases.
2. When iron block is used, we obtain the same conclusion as above. (The numerical values are less than the numerical values when marble is used)

AP State Syllabus AP Board 9th Class English Textbook Solutions Chapter 6C An Icon of Civil Rights Textbook Questions and Answers.

AP State Syllabus 9th Class English Solutions Chapter 6C An Icon of Civil Rights

9th Class English Chapter 6C An Icon of Civil Rights Textbook Questions and Answers

Answer the following questions.

Question 1.
The speaker talks about “creative battle” in the beginning of his speech. What does he mean by this phrase?
Answer:
Martin Luther King (Jr) describes the Civil Rights Movement of the blacks in the USA as a ‘Creative battle’. This is to show that the battle is going to ‘create’ a new world. The battle uses ‘good’ will and ‘good’ intention as weapons. It follows ‘good’ methods like ‘truth’, ‘non-violence’ and ‘love’. Its aim is to promote universal brotherhood. Hence it is a creative battle.

Question 2.
What is Martin Luther King’s speech about? List the issues he is talking about.
Answer:
Martin Luther King’s speech is about justice and equality. It is about universal brotherhood. It is about food to every body, education to every mind and dignity and respect for every spirit. It is about truth, love and peace. It is delivered as Nobel Prize acceptance speech.

Question 3.
Do you think that this is an emotive speech ? If yes, pick out the expressions that show it is an emotive speech.
Answer:
Yes, it is an emotive speech. Every part, in fact, is an example to prove the point. Yet, here are some striking expressions : 22 million Negroes are engaged in a creative battle ; our children, crying for brotherhood, were answered with fire hoses beleaguered and unrelenting struggle ………

Question 4.
What sort of future does the speaker visualize for the Americans and the mankind in general?
Answer:
Martin Luther King (Jr) is full of hope. He visualises a bright future for Americans and humanity. He dreams of a widening and lengthening super highway. Blacks and whites will travel along it in a cooperative and brotherly mood. That will lead them to an ideal land. There everyone gets food, education, justice, equality and dignity. Love, truth and peace will rule supreme.

Vocabulary

I. Given below are the words taken from the reading passage listed as key words. Match the word with the meaning as used in the text.

Key word	Choice words
afflict	affect, touch, cause pain
beleaguered	experienced criticism, shattered, humiliated
retaliation	violation, reformation, revenge
tortuous	complicated, unclear, straight
prostrate	lie flat, roll on, unmoved
turmoil	certainty, great confusion, trouble
curator	representative, person in charge, physician

Answer:

afflict (v)	cause pain
beleaguered (adj)	experienced criticism
retaliation (n)	revenge
tortuous (adj)	complicated
prostrate (adj)	lie flat
turmoil (n)	great confusion
curator(n)	person in charge

II. Read the following expressions taken from the reading passage.
1. blazing light of truth
2. wounded justice
3. majestic scorn

Do they have any specific meaning?
Why does the speaker use such expressions?

The above phrases are figurative expressions. They mean a word or a phrase used in a different way from its usual meaning in order to create a particular mental image or effect to add interest to a speech or a writing. Here the two words that convey opposite meaning are combined together to get a positive meaning.

Now read the passage once again and pick out the figurative expressions.

Find out the meanings of all the expressions including the ones given above.
Answer:

1. blazing light of truth	bright light generated by pure truth
2. wounded justice	justice destroyed ; injustice pervading everywhere
3. majestic scorn	contempt or hatred of highest order
4. sterile passivity	unproductive inactiveness
5. creative psalm of brotherhood	the feeling of being brothers is like a song praising God
6. super highway of justice	a smooth path of justice

Writing

I. You have listened to the speech delivered by Subhash Chandra Bose and read the speech by Martin Luther King Jr.
Let’s analyze their speeches.

Discuss the following questions in groups.
1) How do they begin their speeches?
2) Do you find any logical sequence of ideas in their speeches?
3) What sort of language do they use? (Persuasive, argumentative, emotive)
4) Do you notice any quotations, examples? (to support their argument)
5) Do they use any linkers for cohesion?
6) Do they maintain unity of ideas/thoughts for coherence?
7) What expressions do they use to conclude their speeches?
Answer:

1. They begin their speeches with one’s obligations to family and country and to the Civil Rights Movement in the USA.
2. Yes, there is absolute logical sequence of ideas in their speeches.
3. They use argumentative and emotive language.
4. Yes, I do notice quite a good number of examples.
5. Yes, they use linkers liberally for the purpose of cohesion.
6. Yes, they do maintain unity of ideas for coherence.
7. They conclude their speeches with expressions of hope, freedom, justice, and equality for all!

II. Prepare a speech on the following occasion in your school.
You can use some of the quotations given in the box.

Independence Day

Answer:
Respected teachers, elders and guests and my dear friends! A very brilliant morning and warm greetings of our greatest day of days to you all.

We breathe this cool morning the inspiring air of freedom. Today we stand in the protective shelter of our tricolour flag as it flies high in the sky and proclaims to the world the great achievement of Indians! But the flag also reminds us of the selfless sacrifice thousands of Indians made to gift us this invaluable independence. The flag, at the same time, advises us to be mindful of our responsibilities.

As we celebrate this great event in the lap of goddess of learning, shouldn’t we pledge to follow the path our national leaders have so brilliantly illuminated for us? As the present students and future citizens, shall we plan for our future and our mother India’s future too?

As you all know well, planning for future demands analysis of present. Are we, then, rellay free in its comprehensive sense? Are we, for example, free from fear? For, as Aristotle has well said, “He who has overcome his fears will truly be free”. Are we really aware of and following Eleanor Roosevelt when he says, “Freedom makes a huge requirement of every human being. With freedom comes responsibility. For the person who is unwilling to grow up, the person who does not want to carry his own weight, this is a frightening prospect?” Are we really respecting freedom of expression? Do we follow

S.G. Talientyre’s mantra in this regard? The mantra is : “I disapprove of what you say, but I will defend to the death your right to say it.” Are we free from hatred and jealousy? Shouldn’t we put into practice Martin Luther King’s (Jr) wise advice – Let us not seek to satisfy our thirst for freedom by drinking from the cup of bitterness and hatred – even after independence? Are we allowing our innate nature to move ahead in its own way? Do we prove Virginia Woolf right in the saying – Lock up your libraries if you like, but there is no gate, no lock, no bolt that you can set upon the freedom of my mind?

The present picture is full of challenges. Illiteracy, poverty, corruption, pollution, fast depleting natural resources – the list is too long to complete here. It haunts us day and night. Let us all work together to solve these burning problems. Let us empower ourselves first to be able to work for Mother India! Let us hope to see Mother India full of smiles! Thank you all for your encouraging attention. Jai Hind I

Project Work

Collect information about the great leaders who fought for the freedom of our country. Arrange the information in the table given below:

Discuss in groups and write down the questions you will need to get the information.

On the basis of the information collected in the table below, write a brief biographical sketch of any one of them and present it before others in class.


Answer:
A BRIEF BIOGRAPHICAL SKETCH OF SAROJINI NAIDU

Sarojini Devi was born on 13th February 1879 in Hyderabad to Aghoranath Chattopadhyaya and Varada Sundari Devi.

She passed matriculation when she was just 12. Then she went to England for higher studies in King’s college in London and later at Cambridge.

She married Mutyala Govindarajulu Naidu in 1898. Kandukuri Veeresalingam acted as the priest on the occasion.

Gopala Krishna Gokhale influenced and guided Sarojini Naidu into Freedom Struggle. She met Gandhiji in 1915 and Nehru in 1916 for the first time. She took a very active part in Congress activities and Independence movements. She inspired many men with her poetic recitals which won for her the title ‘Nightingale of India’, Recognizing her important role, Congress leaders made her the party president in 1925. But the British government imprisoned herforthe same role.

After India became independent, she was made the governor of United Provinces (now Uttar Pradesh), the first Indian woman to become so. People remember her as a patriot, poet, philospher and lover of peace. She remains a source of inspiration to many persons-particularly women.

An Icon of Civil Rights Summary in English

Martin Luther King (Junior) is one of the most influential and inspiring Afro-American leader. His Nobel Prize acceptance speech is noted for its excellent content and effective expression. He accepts the Nobel Prize on behalf of the Civil Rights Movement in America. He remembers the suffering of blacks in the USA. He praises the Indians for showing the path of non-violence. He believes in the power of truth and love. He is hopeful of bright future for humanity. He has faith in man’s ability to lay a super highway of justice on which all people will cooperatively go ahead and create a world where every body gets food, mind finds education and spirit receives dignity, equality and brotherhood. He says faith gives us courage and confidence to complete the mission of establishing universal brotherhood that has roots in love, truth and peace.

An Icon of Civil Rights Glossary

icon (n) : a famous person ; a symbol of an idea

majesty, highness, excellencies (nouns) : terms showing high respect to people of top positions

accept(v) : take; receive

determination (n) : firmness in continuing to do something despite difficulties

scorn (n) : contempt; hatred

reign (n) : rule (Note that the letter ‘g’ is silent.)

mindful (adj) : aware ; remembering

sanctuary (n) : a place that is safe

segregation (n) : separation ; isolation

debilitating (v+ing) : weakening

grinding (v+ing) (adj) : never ending ; never improving

afflicts (v) : causes suffering

rung (n) : step

beleaguered (adj) : experiencing a lot of criticism

unrelenting (v+ing : adj) : not stopping and not becoming less severe

contemplation (n) : serious thinking

antithetical (adj) : opposite

sterile (adj) : not capable of producing

elegy (n) : a poem or song about sad feelings

psalm (n) : a song praising god (Note that the letters ‘p’ and T are silent.)

revenge(n) : causing suffering to others because the caused suffering

aggression (n) : attack

retaliation (n) : action against someone because they harmed

tortuous (adj) : complicated, long, full of bends

alliances (n-plural) : groups working together to achieve what they want

audacious (adj) : daring ; willing to take risks

despair (n) : the feeling of losing all hope

flotsam and jetsam (phrase-noun) : useless things, persons with no job, no home

cynical (adj) : not believing that something good will happen

triumphant (adj) : successful; victorious

prostrate (adj) : lying on the ground facing downwards

altars (n-plural) : places of worships

redemptive (adj) : that saves from evil

dreary (adj) : that which makes one dull, bored

soared (v-past tense) : rose high ; went up

righteousness (n) : morally good behaviour

heirloom (n) : a valuable object from forefathers

AP State Syllabus AP Board 9th Class English Textbook Solutions Chapter 6B Freedom Textbook Questions and Answers.

AP State Syllabus 9th Class English Solutions Chapter 6B Freedom

9th Class English Chapter 6B Freedom Textbook Questions and Answers

Answer the following questions.

Question 1.
What sort of freedom does the poet wish to have?
Answer:
The poet wants freedom from fear and physical, intellectual, psychological, emotional, social, and financial weaknesses.

Question 2.
The poet talks about fear in the opening lines of the poem. What kind of fear is he talking about?
Answer:
The poet refers to fear of country’s future, development, social and economic equality, etc.

Question 3.
What does the expression “truth’s adventurous paths” mean?
Answer:
“Truth’s adventurous paths” means the road taken by persons speaking only truth is full of challenges and problems.

Question 4.
What does the poet mean by ‘figures’ in the poem? What sort of figures are they? What does the poet want them to be?
Answer:
‘Figures’ here are people. They are obedient and patient. The poet wants them to have their own individuality.

Question 5.
“Where figures wait with patience and obedience for the master of show.” What does the poet mean by this?
Answer:
Persons here wait for their leader. They just follow their master without any original thinking.

Question 6.
What does the ‘shackles of slumber’ mean? How does it arrest the progression of life?
Answer:
‘Shackles of slumber’ means chains of laziness, inaction and indifference. When one is a prisoner of laziness, life does not progress.

Question 7.
Do you think we are all free from fear? What kind of fears are haunting our motherland now?
Answer:
No, we are not at all free from fears. The fears of over population, over pollution, lawlessness, insecurity to women and the weak are some of the striking ones haunting our mother land.

Question 8.
What does freedom mean to you? Is it freedom from hunger? Is it freedom from physical attack? Is it freedom from illiteracy? is it freedom from social oppression? What else?
Answer:
Freedom from desire, selfishness, illiteracy, dirt and disease, hatred, poverty, etc. is what freedom means to me.

Freedom Summary in English

The poem ‘Freedom’ by Rabindranath Tagore is both philosophical and psychological. He wants a special kind of freedom for the motherland. It is freedom from fear. It is freedom from old age and related weaknesses. It is freedom from lack of vision and slumber. It is freedom from mistrust. It is freedom from uncertain destiny, cruel power and lack of individuality. It is freedom from meaningless movements, senseless statistics and heartless artificiality.

Freedom Glossary

claim (v) : demand ; achieve

beckoning (v+ing) : showing signs to call

shackles (n-plural) : chains

slumber (n) : sleep

mistrusting (v+ing) : having no faith

anarchy(n) : a situation without order, control or government

helm (n) : steering wheel or handle

puppet (n) : a person whose actions are controlled by others

mimicry (n) : copying; imitation

AP State Syllabus AP Board 9th Class English Textbook Solutions Chapter 6A A Long Walk to Freedom Textbook Questions and Answers.

AP State Syllabus 9th Class English Solutions Chapter 6A A Long Walk to Freedom

9th Class English Chapter 6A A Long Walk to Freedom Textbook Questions and Answers

Look at the following pictures and answer the questions that follow.

Question 1.
What do you know about the great persons in these pictures?
Answer:
The pictures are of Mahatma Gandhi and Nelson Mandela. Both of them are world class leaders. They sacrificed their lives for their countries. They exhibited excellent human qualities in their movements. They remain forever as models for leadership and humanity.

Question 2.
What similarities do you find in their lives?
Answer:
Mahatma Gandhi and Nelson Mandela have a lot of similarities in their lives. In fact, Mandela derived inspiration from Gandhi. Both are coloured. Both suffered at the hands of the whites. Both of them fought for their countries. They sacrificed their personal lives for their movement. Both of them spent a lot of time in jails. People regard both of them as their greatest leaders.

Comprehension

Answer the following questions.

Question 1.
Why is it difficult to fulfil the ‘twin obligations’ in a country like South Africa?
Answer:
In South Africa, coloured persons were not allowed to fulfil their twin obligations. If anyone tried to do so, he was punished and isolated. Blacks were not permitted to live like human beings.

Question 2.
What sort of freedom did Mandela enjoy as a boy? Was it real? Give your opinion.
Answer:
Mandela was born free. He could run around in their fields. He could swim in the streams nearby. He had the freedom to roast maize under the stars. He enjoyed rides on the backs of bulls. But that freedom was very limited and purely private. Later, it turned out to be an illusion.

Question 3.
How did Mandela’s understanding of freedom change with age and experience?
Answer:
As a boy, Mandela thought he was free. As a student, he knew he did not have freedom. Later, he realized that all blacks were deprived of their freedom, dignity, and self-respect. Finally, he understood that he could not enjoy his limited freedom as long as his people were not free. He felt that freedom was indivisible.

Question 4.
What does the line ‘the oppressed and the oppressor alike are robbed of their humanity’ suggest?
Answer:
Mandela is a wise leader. He has a lot of insight. He says ’the oppressed and the oppressor alike are robbed of humanity1. For an ordinary person, only the oppressed appear to be the sufferer. But the oppressor also suffers from hatred. Hence all are losers. Freedom to everyone alone is the solution.

Question 5.
What relevance does Nelson Mandela’s life have to the present society?
Answer:
Mandela’s life is relevant to any place and at any time. His life basically deals with human values. And humanity remains the same everywhere and at any time.

Question 6.
“It was this desire … that animated my life”, which desire is the narrator referring to?
Answer:
The desire Mandela refers to is the greater hunger for the freedom of his people.

Vocabulary

I. Tick (✓) appropriate meaning for each of the following underlined words.

1. I was born free.
a) able to act at will
b) having personal rights
c) not subjected to constraints
d) costing nothing
Answer:
c) not subjected to constraints

2. I was prevented from fulfilling my obligations.
a) not able to perform
b) stopped from doing
c) conditioned to do
d) forced to do
Answer:
b) stopped from doing

3. My freedom was curtailed.
a) enhanced
b) lost
c) reduced
d) blocked
Answer:
c) reduced

4. I was not a virtuous leader.
a) dignified
b) law-abiding
c) well behaved
d) honest
Answer:
d) honest

II. Read the following paragraph carefully. Fill in the blanks with the most appropriate forms of the words in brackets.

Nelson Mandela was an outstanding black ___(1)___ (lead) of South Africa, who spent his life time ___(2)___ (fight) against racial ___(3)___ (segregate). He had to spend 30 years of imprisonment to achieve ___(4)___ (free) of the coloured. Finally he ___(5)___ (creation) history when he became the first black man as the President of ___(6)___ (independence) Republic of South Africa. This great leader who has been a source of ___(7)___ (inspire)for millions of freedom lovers in the world was influenced by Mahathma Gandhi, the father of our nation!
Answer:
1) leader
2) fighting
3) segregation
4) freedom
5) created
6) independent
7) inspiration

Grammar

Defining Relative Clause :

Read the following sentences and notice the underlined parts.

1. The man who takes away another man’s freedom is a prisoner.
2. The people followed the principles that Mandela advocated.

• As you perhaps know, the clauses underlined above are called adjectival clauses because they qualify the noun in the preceding clause.
• In (1) we are able to identify who the man is with the help of the information contained in the clause, who takes another man’s freedom. Similarly, the identity of the principles is revealed by the clause, that Mandela advocated. Without these clauses, the listeners will not be able to know who the man is and which principles they are. Who in (1) refers to the man and that in (2) refers to the principles. These words in these sentences are

Relative pronouns.

• The Adjectival clauses are also called Defining Relative Clauses because they help to define the person or the object referred to.
• The whole expression containing the noun phrase and the Relative Clause now acts like a single noun phrase.

Pick out the Defining Relative Clauses and the noun phrases they define from the text. Fill in the table given below. One is done for you.

Writing

I. Read the story once again and analyze the text in the light of the following questions.

• What incidents do you find in the first paragraph?
• How does the writer reflect on (feel about) these incidents?
• What is the writer’s point of view on the incidents?

You may have witnessed several instances of discrimination in the world around you. Write an essay about one such incident. You may use the following clues:

• When and where did it take place?
• With whom did the incident occur?
• What were you doing at that time?
• How did you feel about these incidents?

Answer:
Pleasant are the ways of villagers. Sometimes the ways are mysterious too, particularly to a town boy as young as 12. Even after clear explanation by village elders, I find it difficult to understand the way the villagers behaved that day.

It was summer. As in every summer, I had been to my grandparents’ village. Born and growing in a small house in a crowded town, I always love to go to villages. And my love for my grandfather’s village is all the more as it is on a river bank with a range of hillocks on another side. I also cherish village games and countryside walks.

That particular day filled my heart with rather sad feelings. My grandfather and I were walking along a path beside the river. A man of about 25 was riding a bicycle towards the village. A young boy of below 10 years was sitting on the back seat of the cycle. He appeared to have been enjoying that ride as if it were for the first time he was riding a cycle !

Suddenly a group of 5 or 6 men stopped the cycle. They started scolding the man for allowing the boy to sit on his cycle. And as to the boy, they almost beat him. The boy’s eyes were filled with tears of fear. The trembling boy’s tender face left a deep impression on my mind. And the memory has been haunting me whenever I think of the village. Since then, though I have been going there, the feelings about the village have not been as pleasant as they used to be before that incident.

As I was at my wit’s end as to why the boy shouldn’t ride the bicycle, I asked my grandmother the reason for their behaviour. Grandmother said in a hushed voice that the boy belonged to a lower caste and the man was from a higher caste !

My mind failed to understand that and till today I see no point in that attitude ! May God bless us with the understanding that all men – nay, all living beings – are equal !

Study Skills

Read the following biographical account of a great patriot of India, which describes events in his life. After reading the text, complete the chronological table.

Subhash Chandra Bose was born on 23rd January in Cuttack in 1897. He was born in a rich family. When he was five he was admitted into a big European school. At the age of twelve, he was shifted to another school, where his headmaster, Beni Madhav Das, kindled the spirit of patriotism in him. When he was fifteen, he came under the profound influence of an outstanding spiritual leader, Swami Vivekananda.

After his graduation, Subhash left for Cambridge in 1919 to appear for the Indian Civil Service Examination (ICS). But he had made up his mind to dedicate his life to the service of his country; he resigned from the Indian Civil Service and returned to India in 1921. He took part in freedom struggle, Independence Movement and fought against the British rule in India. Bose was arrested and sent to a prison in Burma. With the cooperation of some prisoners and freedom lovers, Bose formed the Indian National Army (INA) in 1941 in Singapore. He inspired the troops to fight against the British to liberate their motherland. On 21st October 1943, the Azad Hind Government was set up in foreign soil.

Subhash is called ‘Nethaji’ (Neta- a leader) because he was a true and passionate leader of the Indian struggle for freedom.

Complete the following table based on your reading of the passage.

Year	Incident that took place/Significance
1. 1897	Subhash Chandra Bose was born on 23 January
2. 1902	Joined a big European school
3. 1909	Shifted to another school/headmaster Beni Madhav Das kindled the spirit of patriotism
4. 1912	Came under the influence of Swami Vivekananda
5. 1919	Left for Cambridge to appear for Indian Civil Service Examination
6. 1921	Resigned from Indian Civil Service and returned to India
7. 1941	Formed Indian National Army (INA) in Singapore
8. 1943	Set up Azad Hind Government – on 21st October in foreign soil.

Listening

Practise listening carefully. Then you will be bale to speak.
Listen to the ‘speech of Subhash Chandra Bose’ and answer the questions that follow.

Speech of Subhash Chandra Bose
Brave soldiers! Today you have taken an oath that you will give fight to the enemy till the last breath of your life, under the national tricolour. From today you are the soldiers of the Indian National Army of free India. You have volunteered to shoulder the responsibility of forty crores of Indians. From today your mind, might and money belong to the Indian Nation. Friends, you have the honour to be the pioneer soldiers of Azad Hind Fauj. Your names will be written in the history of Free India. Every soldier who is martyred in this holy war will have a monument, Free India. The coming generations will shower flowers on those monuments. You are very fortunate that you have got this valuable opportunity to serve your motherland. Although we are performing this ceremony in a foreign land, our heads and hearts are in our country. You should remember that your military and political responsibilities are increasing day by day and you must be ready to shoulder them competently. The drum of Indian Independence has been sounded. We have to prepare for the battle ahead. We should prepare ourselves as early as possible so that we can perform the duties we have shouldered. I assure you that the time is not far off when you will have to put to use the military skill which you possess.

Today we are taking the vow of independence under the National Flag. A time will come when you will salute this flag in the Red Fort. But remember that you will have to pay the price of freedom. It has to be got by force. Its price is blood. We will not beg freedom from any foreign country. We shall achieve freedom by paying its price. It doesn’t matter how much price we have to pay for it. I assure you that I will lead the army when we march to India together. The news of the ceremony that we are performing here has reached India. It will encourage the patriots at home, who are fighting empty-handed against the British. Throughout my life, it had been my ambition to equip an army that would capture freedom from the enemy. Today I congratulate you because the honour of such an army belongs to you. With this, I close my speech. May God be with you and give you the strength to the pledge which you have taken voluntarily today.
Inquilab Zindabad!

I. Based on the ‘speech of Subhash Chandra Bose’ answer the following question orally.

1. What is the thrust of Subhash Chandra Bose’s speech?
Answer:
The thrust (main point) of Bose’s speech is motivating the soldiers to fight for mother India’s Freedom with a sense of sacrifice and responsibility.

II. Say whether the following statements are True or False.

1. Subhash asked the soldiers to sacrifice everything for their motherland.
Answer:
True

2. Subhash dreamt that every soldier of INA would have a monument in Free India.
Answer:
True

III. Have you ever heard of &ny national leader’s speech? Talk about him/her.
Answer:
Yes. I heard of Atal Biharr Vajpayee’s speech. He uses a very simple language. Humour is clear and dignified in his talk. He quotes from poetry and himself composes poems and recites. His commitment to the cause of the country makes his words appealing. His selfless service adds force to his flow of sounds.

Oral Activity

Debating
Work in pairs. Organize a debate in dass on the following proposition. Women should work in kitchens and men in offices.
One member of the pair speaks in favour of the proposition, while the other speaks against it.
Remember

• Organize your ideas as main points and sub-points;
• Put your ideas in a proper order ( sequence);
• Give suitable examples, quotes;
• Use polite expressions;

You may use some of the following words/phrases to express your views

• In my opinion ………………
• personally feel ………………
• It’s my feeling ……………….
• think ……………..

To agree with your opponents

• I agree with my worthy opponents ……………..
• I am in favour of the ……………..
• I think they / you are right ……………..
• I support the idea ……………..

To disagree with your opponents

• I’m sorry to differ with you ……………..
• I disagree with you ……………..
• It may be your opinion but I’m not happy with this ……………..
• That’s purely your idea but the reality is different ……………..

To establish your point of view/stand

• Since I have evidence I strongly believe this ……………..
• I’m fully confident with my point as ……………..
• I’ve no doubt about this since it is a ……………..
• Therefore I conclude that ……………..

A : It is definitely very good if women remain at home and men work in, offices. It was practised in good olden days and everyone lived happily and peacefully.

B : I love to agree with your example. But that belongs to past. We live in the present. The times have changed a lot. Change, we must along with the time !

C : Women confining to homes is a good proposition. That is the place where a woman’s abilities are most useful. Work at offices men can take care of!

D : I’m really sorry to disagree with you! Women, no doubt, are best home makers. But that doesn’t mean they cannot do other things. Don’t we have examples in thousands that prove women’s abilities! What about Sakuntala Devi, Sunitha Williams ; Kalpana Chawla ; Indra Nobyi; Indir a Gandhi, and many more?

E : Hard working husband and school going students need personalised service at home. So it would always be good for women to stay at home. A hgppy father and satisfied children make a home a pleasant place to live in!

F : That is a fine idea. But what about the feelings of a woman. Particulary when she has an urge for expression of her talents! You see from one angle. Let’s try to be comprehensive.

G : Women have tremendous patience. She can attend to work at home and yet be a office goer. Allowing women to work outside is advantageous in many ways. Inconveniences, if any, can be sorted out somehow. Father and children should learn to enjoy sharing work at home. Then everyone will be happy.

H : It is a balanced opinion. What I would like to put in is that every woman has her own choice. Some women may prefer to stay at home on their own. Others may love to take up a job outside. Hence what I feel to be the best option is to give freedom to women to be a home maker or a job seeker!

A Long Walk to Freedom Summary in English

Nelson Mandela’s narration ‘A Long Walk to Freedom’ is at once soul-stirring and thought-provoking. Every man has obligations to his family and his country. Fulfilling them depends on one’s ability and nature. But in South Africa coloured people were not allowed to fulfil any of those obligations. But Mandela tried and succeeded to fulfil his obligations to the country. But he failed in the case of his family. He was born free. But soon he discovered that his freedom was not true. He realised that no black in South Africa was free. They didn’t have self-respect too. His hunger for personal freedom became the hunger for everyone’s freedom. For that, he changed himself. He became animated. He became bold. He sacrificed his personal ambitions and life. He worked for the freedom of the oppressor too. He believed that freedom was indivisible. He attempted for freedom to everyone and he achieved it too.

A Long Walk to Freedom Glossary

twin (adj) : two

obligations (n-plural) : duties ; responsibilities

humane (adj) : kind ; concerned

inevitably (adv) : as is certain to happen

inclinations (n-plural) : feelings /tendencies that make one do something

isolated (v-past tense) : separated from others

ripped (v-past tense) : torn ; separated

twilight (adj) : mysterious ; uncertain ; not clearly defined

rebellion (n) : opposition to authority

mealies (n-plural) : maize

abided by (ph.v) : accepted the law / worked according to rules

illusion (n) : something that appears to be present but actually not present

transitory (adj) : temporary

yearned (v-past tense) : wanted very much ; longed

potential (n) : quality that exists and can be developed

keep (n) : food, clothes and other basic things that one needs to live

obstructed (v-past tense) : blocked ; stopped

curtailed (v-past tense) : limited

animated (v-past tense) : made full of energy

monk (n) : a person living without family, personal possessions etc.

virtuous (adj) : behaving in a very good and moral way ; honest

oppress (v) : treat others cruelly and unfairly

oppressor (n) : a person oppressing others

the oppressed (v-past participle – used as a noun along with ‘the’) (plural) : people who are oppressed

prejudice (n) : an unreasonable dislike based on race etc.

AP State Syllabus AP Board 9th Class English Textbook Solutions Chapter 5C The Ham Radio Textbook Questions and Answers.

AP State Syllabus 9th Class English Solutions Chapter 5C The Ham Radio

9th Class English Chapter 5C The Ham Radio Textbook Questions and Answers

Answer the following questions.

Question 1.
What are the places in which disaster management becomes imperative?
Answer:
Disaster management becomes imperative in public places, business centres and road junctions etc. Large number of people move around in such areas.

Question 2.
How can we empower the disaster management crew?
Answer:
We can empower disaster management crew by providing them with best devices.

Question 3.
In what way does Amateur Radio (Ham Radio) become inevitable in times of natural calamities?
Answer:
Ham radio is inevitable in times of calamities. This is because conventional communication network fails during disasters.

Question 4.
Cite the disasters in which Amateur Radio Operators commenced disaster relief when other systems failed.
Answer:
Disasters during which Ham radio helped the crew are : 2001 Gujarath Earthquake, 2001 WTC attack in the USA, 2004 tsunami in India and many more.

Question 5.
Explore other incidents in which Ham Radio Operators can take up rescue operations.
Answer:
Ham radio operators can take up rescue operations during road and fire accidents ; war and mega social, cultural or religious gatherings.

Question 6.
What is the central theme of the essay?
Answer:
The theme of the essay is the multiple uses of ham radio.

Question 7.
List the criteria to qualify for becoming a Radio Amateur.
Answer:
To become a ham, one should qualify in
1) Morse code
2) communication procedure and
3) basic electronics.

Project Work

Get into groups and collect information relating to various disasters from Newspapers, magazines and books etc. Each group should work on one disaster. Fill in the boxes in the table given in the next page with necessary information. One is done for you.

The Ham Radio Summary in English

Disasters and terrorism may strike thickly populated areas at anytime. Communication is most important in such situations. And conventional communications like cell phones fail exactly in those emergencies. It is in those times Ham radio comes handy. Ham radio uses high frequency and automatic batteries. And the Ham volunteers are skilful and ever ready to work. Failure of the system is very less likely. The only problem is the availability of information about Ham operators. During 2001 Gujarath Earthquake, 2001 attacks on WTC in the USA, 2004 Tsunami in India, Ham radio served disaster management personnel wonderfully. Ham radio is useful in day to day problems like medical emergencies too. Anyone above 12 with no specific educational qualifications can appear for Ham radio licence test and on passing it can obtain the licence.

The Ham Radio Glossary

ham (n) : a person who operates radio signals as a hobby, not as a job

strike (v) : attack

efficient (adj) : doing thoroughly

imperative (adj) : extremely important or urgent

calamities (n-plural) : disasters

adverse (adj) : unfavourable/harsh

crew (n) : persons working as a team

amateur (adj) : working for enjoyment, not as a job

innovation (n) : new, improved creation

adept (adj) : skilful

terrestrial (adj) : related to /on land

choke points : blocking or obstructing points

improvising (v+ing) : managing with available things when the right things are not availble

wiped out (phrasal verb) : destroyed totally

AP SSC 9th Class English Textbook Solutions

AP State Syllabus AP Board 9th Class English Textbook Solutions Chapter 5B Grabbing Everything on the Land Textbook Questions and Answers.

AP State Syllabus 9th Class English Solutions Chapter 5B Grabbing Everything on the Land

9th Class English Chapter 5B Grabbing Everything on the Land Textbook Questions and Answers

Answer the following questions.

Question 1.
What is the central theme of the poem?
Answer:
The theme of the poem is the vast destructive power of tsunami and the widespread death and damage it causes.

Question 2.
What does the ‘hand’ refer to? Where was the hand born?
Answer:
The hand refers to the tide of tsunami. It is born under the surface of the sea.

Question 3.
Identify the most striking line in the poem.
Answer:
The most striking line the poem is the concluding line.

Question 4.
Describe the damage caused to mankind due to ‘Tsunami’. Locate the words or expressions which tell the fury of ‘Tsunami’.
Answer:
Tsunami killed men and animals. It uprooted trees and destroyed houses. In fact, it smashed everything that was in its way. The expressions that show the fury of tsunami are : dreadful might; anger unleashed ; white horses galloping ; dissolved under the heat; crashing; crunching; tearing

Question 5.
Why do you think only a few could understand the meaning of ‘Tsunami’?
Answer:
Majority of the people of tusunami affected areas were killed. Only a few survived. The destruction caused by tsunami is known only to those living after it.

Grabbing Everything on the Land Summary in English

Lily Usher presents in detail the terrible destructive power of ocean tides in her poem “Grabbing Everything on the Land”. The wave came like a giant hand felling trees and crushing homes. It destroyed everything in sight and no one could escape from it. The hand was born in the ocean and got its strength from the rocks underneath. Everyone was surprised to see the waves coming towards them like fast running white horses, smashing everything on their way. Crashing, crushing, crunching, and cutting was everywhere. Death danced all around. The few who managed to escape understood what ’tsunami1 meant.

Grabbing Everything on the Land Glossary

grabbing (v+ing) : seizing ; taking something suddenly and with force

foam (n) : mass of bubbles on water surface

uprooting (v+ing) : felling; pulling roots from the ground

smashing (v+ing) : crushing ; destroying

fist (n) : closed fingers ; symbol of attack

survive (v) : able to live despite difficulties

dreadful (adj) : terrbile ; evoking fear

might (n) : power; energy

tide (n) : rising surface of water

belly (n) : stomach

plates (n – plural) : (here) rocks in the earth’s crust

fury (n) : anger

unleashed (v – pt) : let loose

shriek (n) : a cry of fear

galloping (v + ing) : running very fast

dissolved (v – past tense) : melted ; (here) died

crunching (n + ing) : crushing noisily

seeping (v + ing) : flowing slowly

brutally (adv) : cruelly; mercilessly

hound (n) : a hunting dog

punch (n) : a hit; shot

tsunami (n) : an extremely large wave in the sea ; a tidal wave
(from Janapese)

AP SSC 9th Class English Textbook Solutions

AP State Syllabus AP Board 9th Class English Textbook Solutions Chapter 5A A Havoc of Flood Textbook Questions and Answers.

AP State Syllabus 9th Class English Solutions Chapter 5A A Havoc of Flood

9th Class English Chapter 5A A Havoc of Flood Textbook Questions and Answers

Look at the picture and answer the questions that follow.

Question 1.
What made the people stand on the roof-tops?
Answer:
Floods of severe intensity made people stand on the roof-tops.

Question 2.
What are the people in the helicopter trying to do?
Answer:
People in the helicopter are trying to airlift the people standing on the roof-tops.

Question 3.
Have you ever seen a situation like this in your life? If yes, when and where did you see it?
Answer:
No, I haven’t seen a situation like that. But in cinemas and on TVs I have seen such situations.

Comprehension

I. Answer the following questions.

Question 1.
What mood is highlighted throughout the description? Pick out the words that suggest the mood.
Answer:
The mood highlighted in the description is that of sorrow, shock and horror. Some of the words that suggest the mood are : terrible, damp, cold, cloudy, gloomy, roaring, shattering, frantically, trauma, heart-rending, pathetic plight, catastrophe, aghast

Question 2.
Which of the scenes in the text has moved you the most? Why?
Answer:
A 40 year old widow Rajeswari staying at a roadside shelter with her four young children without food and water moved me. A mother watching silently the suffering of tender children is unbearable and unimaginable.

Question 3.
Who played the key role in the rescue operation of the floods? What steps were taken by the government?
Answer:
Government agencies, the army, the navy and personnel from Disaster Management Response Force played a key role in rescue operations. Government opened relief camps and distributed food, clothing, medicines etc., among the victims. Government also extended all possible help.

Question 4.
“The flood victims were looking on with wide open eyes for some help to meet their needs.” Can you guess what their needs may have been?
Answer:
The immediate needs of the flood victims are food, shelter, clothing and medicines.

Question 5.
“Help always pays gratitude.” In what way is this statement true in the light of the context?
Answer:
All the needs of the victims were met by government agencies, philanthropists and NGOs. The victims were thankful for the help. Hence the saying ‘Help always pays gratitude1’ is true.

Question 6.
How did the roaring floods disturb the lives of the people of Kurnool?
Answer:
The roaring floods destroyed houses, disturbed normal life, killed men and animals, shattered their hopes in Kurnool.

Question 7.
What relief measures would you suggest for the flood victims?
Answer:
Apart from basic needs like food and clothing, I would suggest emotional support to victims and permanent solution to floods.

Question 8.
What inspiration can you draw from the last two paragraphs of the lesson?
Answer:
Rajeswari’s courage, determination, hope and concern for children teach us very valuable lessons. And her sense of thankfulness to those who helped her is highly admirable.

Vocabulary

Read the following sentence and notice the meaning of the underlined word.
They reached their dwelling places hoping a bright morning the next day.
In this sentence ‘bright’ means ‘full of light’ or ‘shining strongly’ or ‘happy.’

A single word in English (with the same spelling and the same pronunciation) may have many meanings.

The correct meaning (or the suitable meaning) of that word is understood with the help of the context.

Words with multiple meanings are called HOMONYMS. (It is very interesting to improve one’s word power with the help of HOMONYMS.)

Some commonly used Homonyms are : kind, type, plant, articles, cricket, bat, bowl, sight, light, kite …..

I. Identify the meaning of ‘bright’ in each of the foliowing sentences and write your own sentences using ’bright1 in different meanings.

1. I like bright colours.
bright = strong, thick, easy to see
My uncle bought a bright blue dress for me.

2. Tejaswini gave me a bright smile, bright = cheerful and lively
On knowing the results, their eyes turned bright.

3. Sindu is a bright student.
bright = brilliant, quick to learn
Many teachers prefer to talk to bright students.

4. Yamuna has bright ideas.
bright = helpful
The manager always seeks bright ideas.

5. This young player has a bright future.
bright = likely to be successful
The students of this school can hope for a bright future.

6. We took rest in a bright room.
bright = full of light
As she kept all the windows open, the room was bright when we entered it.

II. Several people and things are involved in rescue operations. There is a description of a flood rescue operation. Complete the concept map given below with the suitable information from the text:

Answer:

Grammar

Read the following sentences from the text and notice the underlined words/ expressions.

1. No sooner had the relief team arrived there than their joy knew no bounds.
2. They had scarcely arrived at their destinations when the rain poured down.
In the above sentences the expressions “No sooner… than” and “scarcely… when” are used to suggest that one thing happened very soon after another. The expression ‘hardly … when’ is also used to express the same.

• Pairs of words like ‘No sooner… than’ ; ‘scarcely… when’ and ‘hardly…. when’ are called ‘correlative conjunctions’.
• They connect two expressions/clauses /actions. Hence they are conjunctions.
• They show the relationship between the two actions in terms of time. Hence they are correlatives.
• Special Note

1. ‘No sooner’ always goes with ‘than’ and ‘scarcely/hardly’ always go with ‘when’. They are inseparable.
2. ‘When ‘scarcely/hardly’ are used at the beginning of the clause, the helping verb is used before the subject, not after the subject. (Subject – helping verb inversion) If they are used after the subject, the word order is not changed. In the case of ‘No sooner’, there is inversion always.

Combine the following sentences using the expressions “No sooner… than, scarcely… when, hardly… when.”
One has been done for you.
1. I put the phone down. It rang again.
A : No sooner had I put the phone down than it rang again.
or
B : I had scarcely put the phone down when it rang again.
or
C : I had hardly put the phone down when it rang again.

2. I arrived at the station. Then the bus came in.
A : No sooner had I arrived at the station than the bus came in.
B : I had scarcely arrived at the station when the bus came in
C : I had hardly arrived at the station when the bus came in.

3. I closed the door. Somebody knocked it again.
A : No sooner had I closed the door than somebody knocked it again.
B : Scarcely had I closed the door when somebody knocked it again.
C : Hardly had I closed the door when somebody knocked it again.

4. She finished the meal. She started feeling hungry again.
A : No sooner had she finished her meal than she started feeling hungry again.
B : She had scarcely finished her meal when she started feeling hungry again.
C : She had hardly finished her meal when she started feeling hungry again.

5. Madhavi opened the door. The dog entered the room.
A : No sooner had Madhavi opened the door than the dog entered the room.
B : Scarcely had Madhavi opened the door when the dog entered the room.
C : Madhavi had hardly opened the door when the dog entered the room.

Writing

I. In the reading passage, one of the victims of the flood, Rajeswari shared her sufferings with the flood relief team. On the basis of this, develop an interview by a news reporter.

News Reporter : Rajeswari, could you tell me what had happened?
Rajeswari : ……………………………………………
News Reporter : ……………………………………………
Rajeswari : ……………………………………………
News Reporter : ……………………………………………
Rajeswari : ……………………………………………
News Reporter : ……………………………………………
Rajeswari : ……………………………………………
News Reporter : ……………………………………………
Answer:
News Reporter : Rajeswari, could you tell me what had happened?

Rajeswari : Oh! It was dreadful. The night of 27! Horrible! Downpour! Nonstop! We couldn’t believe our eyes! Even multi-storeyed buildings were shaking. What about our mud shack?

News Reporter : Has someone come to help you?

Rajeswari : Not immediately. As waters entered our small shaky house, I took my four children and ran for safety.

News Reporter : Could you easily find out a safe place?

Rajeswari : Yes, very close to my house, the road has a high side wall like structure which then stood above flood waters. So we all stayed there!

News Reporter : How long did you stay there?

Rajeswari : It was for two long days in open I Without food and water. Only flood water and our tears!

News Reporter : Hasn’t help reached you even then?

Rajeswari : Help in the form of rescue team in a fibre boat came to us. They took us to the nearby relief camp, gave us clothes and food. They promised to give money to repair my damaged house. They supported us well. May God be kind to them!

News Reporter : Even in your testing times you are so thankful and you think of their welfare! Great of you! May God bless you!

II. Assume that you happened to read some of the articles/news reports on the flood rescues. You were deeply moved by the sorrows of the victims of the floods. You decided to raise funds for the cause. You wanted to share this idea with your classmates and seek their assistance in this matter.

Prepare a speech/a talk that you would like to make to convince the donors about the need for raising funds for the cause.

You may use the ideas given below.

• Floods in Kurnool
• Damage occurred
• Human suffering and deaths
• Loss of property
• Death of animals
• Loss of crops
• Shortage of basic needs like food, clothing and shelter
• Importance of relief operations
• Moral responsibility

Answer:
My dear lovers of humanity!

Times call us to rise to the occasion. Here is a situation watching which none can remain unmoved. I know your concern for our fellow beings’ suffering : Recently Kurnool witnessed floods of grave intensity. Flood water reached house roof level. Hundreds of people lost their lives. Animals in thousands were covered in water and were dead. Loss of property runs into crores of rupees. Crops were damaged when they were about to be reaped. People now are crying for basic needs like food, shelter, water, clothes, medicines. They need moral support. They need assurance from us. They lost everything except hope. It is our duty to ensure their hope continues to live. It is our moral responsibility to extend to them whatever help we can ! You know ‘Service to man is service to God! And service to man in such dire need is the best kind of worship. I very humbly appeal to you all with folded hands. Let us join our hands. Let us collect men, money, materials etc. Let us support our brothern. Let us share their sorrow and suffering. Let us prove to the world that we love humanity!

III. Assume that you were one of the victims of the flood and received some help from a donor. Write a letter thanking him and expressing your gratitude.

Bus stand centre, Kurnool. 01 October 2009. Respected Dharma Rao garu, Our gratitude makes us write thus. Pray that you and your family members are fine. Thank you so much for your generous help at a very apt time. Floods carried away everything from us except our lives and hope. When we were cursing our fate, help from you came like God-sent gift. It was really like a boat to a person drifting in waters with no shore. As we were thinking that our end was not far away, your timely help infused life and and hope into us. How much I may say, our gratitude still remains incompletely expressed. Persons like you prove to the world what humanity is, nay – what divinity is! May God bless you with what we call eternal bliss. I continue to pray to God to be kind to you and all your near and dear. I look forward to a chance to be of any use to you. I will be blessed if I get that opportunity. Once again our pranamams to you. Thankfully yours, Rakshita. To Sree Dharma Rao garu H.No. 12-85 Near Masjid KADAPA.

IV. On the basis of the diagram given below write a paragraph stating the sequential series of actions/methods (preparation before, during and after) that can be taken for disaster management.

Answer:
Disaster Management includes different activities. First disaster-prone areas are to be assessed with the help of past occurrence and present events. And probable risk to life and property is to be estimated. Then protective structures and relief camps are to be planned and got constructed. After these preparatory arrangements, advance warning systems and evacuation machinery are to be put in place. Later community at large and teachers in particular are to be trained in rescue operations. And finally the outside world is to be informed about the disaster along with the particulars of loss and quantity of aid required. Effective Disaster Management depends on the execution of these steps in sequence.

Study Skills

Note-Making
We take notes when we read a book in order to record information for future reference. Such notes help us to revise lessons easily before examinations.

Let’s know the process of note-making.

• Read the passage once quickly
• Underline the key terms during the second reading s Note only the most important information
• Condense the information
• Omit examples and illustrations
• Organize the condensed information in a suitable format
• Keep a suitable title

Some strategies for condensing information.

• Use numbers instead of words
  e.g. sixty eight written as 68
• Use short substitutes for long words
  e.g. maths for mathematics
• Use reduced verb forms
  e.g. Killed instead of was killed
• Use the ‘to-infinitive’ to indicate future time
  e.g. relief teams to help the people
• Use abbreviations and acronyms
  e.g. IAF, AIR
• Use condensed spelling of words
  e.g. Dept, for Department, Dr. for Doctor

Here is a model answer for paragraphs 2 – 5 from your Reading Passage – A.
Floods in Kurnool

28-9-2009
a)No tourists
b)No buses

27-9-2009
a) heavy rainfall
b) damp, cold and cloudy climate
c) pouring rain

28-9-2009
a) river waters entered the streets
b) flood inundating houses
c) 20 lakh cusecs inflow to the Srisailam Dam
d) threat of flood on the banks of the Krishna
e) heart-rending scenes
f) people seeking help to rescue them

Now make notes from the following passage and then summarise it.

The word “disaster” is derived from Middle French “desastre” and from old Latin “disastro”. A disaster can be defined as any tragic event that can cause damage to life, property and destroy the economic, social and cultural life.

The natural disaster is a consequence when a natural hazard affects humans or built environment. Human vulnerability and lack of appropriate emergency management lead to financial, environmental and human loss. The resulting loss depends on the capacity of the population to support or resist the disaster. Disasters occur when hazards meet vulnerability.

A natural hazard will never result in a natural disaster in areas without vulnerability. Various phenomena like earthquakes, landslides, volcanic eruptions, floods and cyclones are all natural hazards that kill thousands of people and destroy a lot of money and property each year.

Natural hazards can strike in unpopulated areas and never develop into disasters. The rapid growth of the world’s population and its increased concentration often in hazardous environments has escalated both the frequency and severity of natural disasters. Tropical climate, unstable land forms, deforestation and non-engineered constructions m^kethe disaster-prone areas more vulnerable. Developing countries suffer more or less chronically by natural disasters.
Answer:
A) Notes :
Para 1 :
disaster rooted in French, Latin
damage to life and property
destroys economic, social and cultural life

Para 2 :
natural hazards affecting populated areas
vulnerability adds to loss
hazard + vulnerability = disaster

Para 3 :
earthquakes, floods etc – hazards

Para 4 :
hazards in unpopulated areas – not disasters
increasing population, tropical climate,
unstable land forms, deforestation,
non – engineered constructions increase vulnerability
developing countries – chronic – victims.

B) The word ‘disaster’ – rooted in French and Latin – means a damage-causing event that destroys economic, social and cultural life. Hazards affecting vulnerable populations result in disasters. Lack of emergency management mechanism leads to huge loss of life and property. Hazards like earthquakes and floods destroy life and property every year. Increasing populations, tropical climate, unstable land forms, deforestation and non-engineered constructions add to the vulnerability. Developing countries are the chronic victims.

Listening

Practise listening carefully. Then you will be able to speak.
Listen to the ’News Bulletin’ and answer the following questions.

News Bulletin

This is All India Radio. The news…. read by Latika Ratnam.

The headlines first…
An unprecedented flood caused a havoc in Krishna, Guntur, Kurnool and Mahabubnagar districts of Andhra Pradesh.

The Prime Minister visits the flood-affected areas tomorrow.

India defeated Pakistan in the triangular cricket series held at Brisbane, Australia.

Now the news in detail. Massive floods wreaked a havoc in Krishna, Guntur, Kurnool and Mahabubnagar districts of Andhra Pradesh. Water has been released from the dams of Srisailam, Nagarjuna Sagar and the Prakasam Barrage. About 400 villages have been inundated with floodwaters, 4 lakhs of people have become homeless. The death toll has reached 532 so far.

However, the rescue operations are in full swing. The Chief Minister of Andhra Pradesh has made an aerial survey of the flood-hit areas. An immediate aid of two lakhs of rupees has been announced to the members of the deceased. A relief of ten thousand rupees has been announced to the homeless. The C.M. has conducted an emergency meeting with the respective collectors of the flood-hit areas and asked them to set up rehabilitation camps immediately on war-footing.

1. What was the havoc caused by the flood?
Answer:
About 400 villages were inundated. Around 4 lakh people became homeless. 532 persons died till then.

2. What relief measures were taken by the Government of A.P.?
Answer:
The Chief Minister made an aerial survey. An immediate aid of rupees two lakhs was announced to the members of the deceased. A relief of rupees ten thousand was declared to the homeless. Rehabilitation camps were set up on war footing.

3. What are the other highlights of the news bulletin?
Answer:
1) The Prime Minister visits the flood-affected areas tomorrow.
2) India defeated Pakistan in the triangular cricket series held at Brisbane, Australia.

Oral Activity

Read the lesson “A Havoc of Flood” once again. On the basis of the ideas in it, prepare a mock interview for a TV/newspaper.

Work in groups and collect the information about the havoc caused, relief operations
etc., from the following.
1. Victims
2. Officials
3. Doctors
4. N.G.Os (Non-Governmental Organisations/Voluntary Organisations)

In each group one member will be the reporter and the others will play the other roles. Afterwards each group will make the presentation.

1. Interview with Victims :
TV Reporter : Hello, the floods seem to be very severe. Have you been receiving relief from anyone?

Victims : Yes, the floods have been very severe. Never before in our lives have we seen floods of this seriousness. Yes, as to relief, Government officials and voluntary organisations have been extending maximum help. But because of the large number of victims, damaged roads , dead telephones, disrupted power supply, relief is not in sufficient quantity and at right time. But expecting more than what we get would be a sin. We hope and pray for good days ahead.

TV Reporter : Thank you for your detailed inputs. Your positive attitude in these testing times is highly admirable. I firmly believe God will bless people of your kind.

Victims : Thank you.

2. Interview with Officials :
Reporter : Good afternoon sir. How have you been coping with the disaster?

Officials :
Good afternoon. We have been putting in our best efforts. All available men and materials have been put to use. Forecast of floods well in advance has helped us in planning. Warning of floods also helped us minimise the loss of lives and property.

Reporter : Are any other departments working with you?

Officials :
Yes, almost all the government departments are working with us. Some departments are directly on the field. Other departments support us from behind. Even non-government organisations and individuals join us in extending relief.

Reporter : Thank your sir.

3. Interview with Doctors :
Reporter : Good evening doctor. How is the health scenario in the flood affected areas?

Doctor : Good evening. Floods bring in a flood of diseases too. Water is contaminated. Cleanliness is totally missing. People don’t even think of hygiene. Doctors work with their focus on present illness and possible epidemics. Cooperation from every corner eases our job to some extent.

Reporter : Thank you doctor.

4. Interview with NGOs :
Reporter : Good morning gentlemen ! As government agencies are also actively involved in relief programmes, what do you think is the importance of your role?

NGO : Good morning. Yes, you have a valid point there ! But the fact is our job is to extend to the needy whatever help they require. Government Agencies are no doubt active. But they work in a certain frame work. For us, there are no such limitations. We help those who don’t fit into government schemes for one or the other reason. Any how, every one plays a significant role in times of trouble !

Reporter : Thank you. Your analysis is really eye opening !

NGO : You’re welcome.

A Havoc of Flood Summary in English

Water, water everywhere! 10 feet deep even in houses! For three or four days! Dead bodies of cattle and people, damaged houses, breached bunds, deafening cries for help, rescue, and relief teams that was the scene in Kurnool district on 28 September 2009. Continuous downpour during the night brought floods into the city and other villages. Never before in the known history was there such a heavy downpour. Floods from the Thungabhadra, the Handri Niva rivers added to the gravity. Lakhs of people lost their shelters. Loss of lives and property was at shocking levels. Government stepped in at once. The services of the Army, Navy and Disaster Management Agencies were sought. Helicopters, inflatable boats, fibre glass boats and other equipment were pressed into use. Voluntary organisations, institutions, individuals and philanthropists joined the relief operations. Camps for homeless were organised. Food, water, blankets, milk and medicines were distributed among victims. One Ms. Rajeswari 40 exhibited rare courage in rescuing herself and her four children. She expressed her gratitude to all those who helped her during their intense suffering.

A Havoc of Flood Glossary

havoc(n) : a situation in which there is a lot of damage and destruction

basin (n) : an area of land along a river with streams running down into it

namesake (n) : the same name

destination (n) : a place where one wants to reach

gloomy (adj) : sad ;unhappy

inundating (v+ing) : submerging ; filling with water

shattering (v+ing) : destroying

forecast(v) : tell in advance what is going to happen

bore the brunt (idiom) : received or suffered the major part of something bad

fury (n) : anger, severity

submerged (v-past tense) : covered in water

heart-rending (adj) : heart-breaking ; causing a lot of sadness

frantically (adv) : in a hectic way ; quickly ; with uncontrollable emotions

rescue (v) : save from a danger

trauma (n) : an upsettingly unpleasant condition

tributaries (n-plural) : rivers or streams that flow into a larger river

requisitioned (v-past tense) : demanded officially the use of

breached (v-past tense) : damaged, broken

inflatable (adj) : that which can be filled with air

reluctant (adj) : unwilling

groping (v+ing) : trying and finding something that cannot be seen

pathetic (adj) : causing pity

plight (n) : a difficult and sad situation

catastrophe (n) : a disaster

fastened (v-past tense) : tied or joined, tethered (Note : The letter’t’ is silent in ‘fasten’.)

aghast (adj) : horrified ; surprised

reaped (v-past tense) : cut crops ; harvested

rejuvenated (v-past tense) : made to look more lively

shack(n) : a small building made of wood or metal

flee (v) : run away from danger

despair (n) : hopelessness, sorrow

beaming (adj) : cheerful; pleased

AP SSC 9th Class English Textbook Solutions