CHAPTER 1 Matter In Our Surrounding

1. Convert the following temperatures to the celsius scale.

(a) 293 K (b) 470 K

Answer:

To convert temperature from Kelvin (K) to Celsius (°C), we subtract 273 from the given temperature in Kelvin.

(a) 293 K = (293 - 273)°C = 20°C

(b) 470 K = (470 - 273)°C = 197°C

2. Convert the following temperatures to the kelvin scale. (a) 25°C (b) 373°C

Answer:

To convert temperature from Celsius (°C) to Kelvin (K), we add 273 to the given temperature in Celsius.

(a) 25°C = 25 + 273 K = 298 K

(b) 373°C = 373 + 273 K = 646 K

3. Give reason for the following observations.

(a) Naphthalene balls disappear with time without leaving any solid.  

(b) We can get the smell of perfume sitting several metres away.  

Answer:

(a) Naphthalene balls disappear with time without leaving any solid because they undergo sublimation, i.e., they directly change from solid to gas without melting into a liquid.

(b) Perfumes have volatile compounds that evaporate easily and spread quickly in the air, which is why their smell can be detected from a distance.

4. Arrange the following substances in increasing order of forces of attraction between the particles— water, sugar, oxygen.

Answer:

The arrangement of substances in increasing order of forces of attraction between particles is:

oxygen < water < sugar

Oxygen is a gas, which has weak intermolecular forces of attraction. Water is a liquid, and its molecules are held together by hydrogen bonding, which is stronger than the van der Waals forces of attraction between oxygen molecules. Sugar is a solid and has the strongest forces of attraction between its particles due to the presence of covalent bonds and hydrogen bonding.

5. What is the physical state of water at—

(a) 25°C (b) 0°C (c) 100°C ?

Answer:

(a) Water is in a liquid state at 25°C.

(b) Water is in a solid state at 0°C, which is its freezing point.

(c) Water is in a gaseous state at 100°C, which is its boiling point.

6. Give two reasons to justify—

(a) water at room temperature is a liquid.  

(b) an iron almirah is a solid at room temperature.

Answer:

(a) Water at room temperature is a liquid due to the presence of intermolecular forces of attraction between its molecules, which are strong enough to keep the molecules close to each other but weak enough to allow them to move around and flow. Additionally, the geometry of water molecules allows them to form hydrogen bonds, which further contributes to the liquid state of water at room temperature.

(b) An iron almirah is a solid at room temperature due to the strong metallic bonding between its constituent atoms. The metal atoms are held together by a sea of delocalized electrons, which create a strong and rigid structure that maintains its shape and form at room temperature.

7. Why is ice at 273 K more effective in cooling than water at the same temperature?

Answer:

Ice at 273 K is more effective in cooling than water at the same temperature because ice has a higher heat of fusion than water. In other words, when ice melts, it absorbs more heat from the surroundings than when water at the same temperature evaporates. Therefore, to melt ice, more heat is required, and this results in a greater cooling effect compared to water at the same temperature.

8. What produces more severe burns, boiling water or steam?  

Answer:

Steam produces more severe burns than boiling water because it contains more heat energy per unit mass than boiling water. When steam comes into contact with skin, it condenses and releases its latent heat of vaporization, which can cause more damage to the skin than boiling water, which has already released its latent heat of vaporization during the boiling process. Additionally, steam can penetrate deeper into the skin and cause more damage than boiling water, which mostly affects the surface.

9. Name A,B,C,D,E and F in the following diagram showing change in its state.

Sketch Diagram

Answer:

We can describe them as below:

A = Solid to Liquid

Melting or fusion, here the solid changes into liquid.  

B = Liquid to Gas

Evaporation or vaporisation, here the liquid changes into gas.  

C = Gas to liquid

Condensation or liquification, here the gas changes into liquid.  

D = Liquid to solid

Freezing or solidification, here the liquid changes into solid.  

E = Solid to Gas

Sublimation, here solid directly changes into gas without going through liquid state.  

F = Gas to solid

Sublimation, here gas changes into solid without going through liquid state.

Important Questions

1. Which separation techniques will you apply for the separation of the following?

Answer:

(a) Sodium chloride from its solution in water.  

(a) To separate sodium chloride from its solution in water, we can use the process of evaporation. The solution is heated, and the water evaporates, leaving behind solid sodium chloride.

(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.

(b) To separate ammonium chloride from a mixture containing sodium chloride and ammonium chloride, we can use the process of sublimation. The mixture is heated, and ammonium chloride sublimes, leaving behind solid sodium chloride.

(c) Small pieces of metal in the engine oil of a car.  

(c) To separate small pieces of metal from the engine oil of a car, we can use the process of filtration. The oil is passed through a filter, which traps the solid metal pieces, and the filtered oil is collected in a container.

(d) Different pigments from an extract of flower petals.  

(d) To separate different pigments from an extract of flower petals, we can use the process of chromatography. The extract is spotted onto a chromatography paper and allowed to run in a suitable solvent. The different pigments move at different rates, and can thus be separated.

(e) Butter from curd.  

(e) To separate butter from curd, we can use the process of churning. The curd is stirred vigorously, causing the fat in the curd to clump together and form butter.

(f) Oil from water.  

(f) To separate oil from water, we can use the process of decantation. The mixture is left to settle, and the oil, which is lighter, floats to the top. The oil can then be poured off, leaving behind the water.

(g) Tea leaves from tea.  

(g) To separate tea leaves from tea, we can use the process of filtration. The tea is passed through a filter, which traps the tea leaves, and the filtered tea is collected in a container.

(h) Iron pins from sand.  

(h) To separate iron pins from sand, we can use the process of magnetic separation. A magnet is passed over the mixture, and the iron pins are attracted to the magnet, leaving behind the sand.

(i) Wheat grains from husk.  

(i) To separate wheat grains from husk, we can use the process of winnowing. The mixture is poured from a height, and the wind blows away the lighter husk, leaving behind the heavier wheat grains.

(j) Fine mud particles suspended in water  

(j) To separate fine mud particles suspended in water, we can use the process of sedimentation and decantation. The mixture is left to settle, and the heavier mud particles settle at the bottom. The clear water can then be poured off, leaving behind the mud.

2. Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.

Answer:

Steps for making tea:

• Take a tea bag (solute) and place it in a cup of hot water (solvent).
   
• Allow the tea to steep (dissolve) in the water for a few minutes until desired strength is achieved.
   
• Remove the tea bag (filtrate) from the cup, leaving behind the brewed tea (solution).
   
• Add sugar (solute) and stir until it dissolves in the tea (solvent) if desired.
   
• If milk is to be added, heat milk in a separate container until hot and add it to the tea.
   
• Stir the tea to mix in the milk and sugar.
   
• If there are any solid tea leaves (residue) left in the cup, strain them out using a strainer or filter.
   

3. Pragya tested the solubility of three different substances at different temperatures and collected the data as given below

(results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).

Potassium nitrate 21 32 62 106 167  

Sodium chloride 36 36 36 37 37  

Potassium chloride 35 35 40 46 54  

Ammonium chloride 24 37 41 55 66  

a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?  

Answer:

(a) To produce a saturated solution of potassium nitrate in 50 grams of water at 313 K, we can use the given data for solubility of potassium nitrate at different temperatures. We can see that at 313 K, 32 g of potassium nitrate can dissolve in 100 g of water to form a saturated solution. So, the amount of potassium nitrate required to form a saturated solution in 50 g of water at 313 K would be:

Mass of potassium nitrate required = (32 g/100 g) × 50 g = 16 g

Therefore, 16 g of potassium nitrate would be needed to produce a saturated solution in 50 grams of water at 313 K.

(b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.

Answer:  

(b) When Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature, she would observe that some of the potassium chloride would start to crystallize out of the solution as the temperature decreases. This is because the solubility of potassium chloride decreases as the temperature decreases. At 353 K, the solubility of potassium chloride is higher than at room temperature, so when the solution cools down, it becomes supersaturated (i.e., it contains more solute than it can normally dissolve at that temperature) and the excess solute starts to come out of the solution in the form of crystals.

(c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?

Answer:  

(c) To find the solubility of each salt at 293 K, we can use the given data for solubility of salts at different temperatures. At 293 K, the solubility of potassium nitrate is 13.3 g/100 g water, the solubility of sodium chloride is 36.0 g/100 g water, and the solubility of potassium chloride is 34.2 g/100 g water. Therefore, sodium chloride has the highest solubility at 293 K.

(d) What is the effect of change of temperature on the solubility of a salt?

Answer:  

(d) The solubility of a salt generally increases with an increase in temperature. This is because at higher temperatures, the kinetic energy of the solvent molecules increases, and they are able to overcome the attractive forces between the solute particles and the solvent molecules more easily, leading to an increase in solubility. However, there are some salts whose solubility decreases with an increase in temperature due to the specific nature of their molecular interactions

4. Explain the following by giving examples.

(a) saturated solution  

(b) pure substance  

(c) colloid

(d) suspension

(a) saturated solution  

Answer:  

(a) A saturated solution is a solution that contains the maximum amount of solute that can be dissolved in a given solvent at a particular temperature and pressure. For example, a saturated solution of sugar in water means that no more sugar can be dissolved in water at a specific temperature and pressure.

(b) pure substance

Answer:  

(b) A pure substance is a substance that is made up of only one type of particle and has a fixed chemical composition. For example, elements like gold, silver, and copper are pure substances, as are compounds like water and carbon dioxide.

(c) colloid

Answer:  

(c) A colloid is a mixture in which the particles of one substance are uniformly dispersed in another substance, but the particles are not large enough to settle out. For example, milk is a colloid in which fat globules are dispersed in water.

(d) suspension

Answer:

(d) A suspension is a heterogeneous mixture in which the particles of one substance are suspended in another substance but are large enough to settle out over time. For example, muddy water is a suspension of soil particles in water.

5. Classify each of the following as a homogeneous or heterogeneous mixture.

soda water, wood, air, soil, vinegar, filtered tea  

Answer:

Homogeneous mixtures: soda water, air, vinegar, filtered tea

Heterogeneous mixtures: wood, soil

6. How would you confirm that a colourless liquid given to you is pure water?

Answer:

To confirm that a colourless liquid is pure water, you can use several methods, including measuring its boiling point and freezing point to confirm that they are 100°C and 0°C, respectively, or by performing a distillation to separate the liquid from any impurities it may contain. Additionally, testing the pH of the liquid and confirming that it is neutral (pH 7) can also confirm that it is pure water.

7. Which of the following materials fall in the category of a “pure substance”?

(a) Ice (b) Milk (c) Iron (d) Hydrochloric acid (e) Calcium oxide (f) Mercury (g) Brick (h) Wood (i) Air.  

Answer:

The materials that fall in the category of a “pure substance” are:

(a) Ice

(c) Iron

(d) Hydrochloric acid

(e) Calcium oxide

(f) Mercury

8. Identify the solutions among the following mixtures.

(a) Soil (b) Sea water (c) Air (d) Coal (e) Soda water.  

Answer:

The solutions among the following mixtures are:

(b) Sea water

(e) Soda water.

9. Which of the following will show “Tyndall effect”?

(a) Salt solution (b) Milk (c) Copper sulphate solution (d) Starch solution.  

Answer:

Milk and starch solution will show “Tyndall effect”.

Milk will show the Tyndall effect because it is a colloid. Colloids are mixtures in which the particle size of the dispersed phase is intermediate between those of a solution and a suspension. The particles of the dispersed phase in milk, which are mainly proteins and fat, are large enough to scatter light in all directions, resulting in the Tyndall effect.

10. Classify the following into elements, compounds and mixtures.

Answer:

Sodium: Element

Soil: Mixture

Sugar solution: Homogeneous mixture (solution)

Silver: Element

Calcium carbonate: Compound

Tin: Element

Silicon: Element

Coal: Mixture

Air: Mixture

Soap: Mixture (most likely a solution)

Methane: Compound

Carbon dioxide: Compound

Blood: Mixture

Growth of a plant: Not a chemical change (it is a biological process)

Rusting of iron: Chemical change

11. Which of the following are chemical changes?

Answer:

Mixing of iron filings and sand: Physical change (mixture of two elements)

Cooking of food: Mostly a physical change, although there may be some chemical changes involved in the process.

Digestion of food: Chemical change

Freezing of water: Physical change

Burning of a candle: Chemical change

Important Questions

1. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen.

Calculate the percentage composition of the compound by weight.

Answer:

Percentage composition of the compound by weight can be calculated as follows:

Mass of oxygen in the compound = 0.144 g

Mass of boron in the compound = 0.096 g

Total mass of the compound = 0.24 g

Percentage of oxygen in the compound = (mass of oxygen/mass of compound) x 100%

= (0.144/0.24) x 100% = 60%

Percentage of boron in the compound = (mass of boron/mass of compound) x 100%

= (0.096/0.24) x 100% = 40%

2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?  

Answer:

According to the law of definite proportions, the ratio of masses of carbon and oxygen in carbon dioxide is fixed. Therefore, the mass of carbon dioxide formed when 3.00 g of carbon is burnt in 50.00 g of oxygen can be calculated as follows:

Mass of oxygen required to react with 3.00 g of carbon = (3.00 g carbon) x (8.00 g oxygen/3.00 g carbon) = 8.00 g oxygen

Mass of oxygen remaining after reaction = 50.00 g - 8.00 g = 42.00 g

Mass of carbon dioxide formed = (3.00 g carbon) x (44.00 g carbon dioxide/12.00 g carbon) = 11.00 g carbon dioxide

3. What are polyatomic ions? Give examples.  

Answer:

Polyatomic ions are charged species formed by a group of atoms that are covalently bonded together and carry an overall charge due to the gain or loss of one or more electrons. Examples of polyatomic ions include sulfate ion (SO4^2-), nitrate ion (NO3^-), ammonium ion (NH4^+), carbonate ion (CO3^2-) and hydroxide ion (OH^-).

4. Write the chemical formulae of the following.

(a) Magnesium chloride  

(b) Calcium oxide  

(c) Copper nitrate

(d) Aluminium chloride

(e) Calcium carbonate.  

Answer:

Chemical formulae of the following compounds are:

(a) Magnesium chloride - MgCl2

(b) Calcium oxide - CaO

(c) Copper nitrate - Cu(NO3)2

(d) Aluminium chloride - AlCl3

(e) Calcium carbonate - CaCO3

5. Give the names of the elements present in the following compounds. (a) Quick lime (b) Hydrogen bromide (c) Baking powder (d) Potassium sulphate.

Answer:

The names of the elements present in the given compounds are:

(a) Quick lime - Calcium, Oxygen

(b) Hydrogen bromide - Hydrogen, Bromine

(c) Baking powder - Sodium, Hydrogen, Carbon, Oxygen

(d) Potassium sulphate - Potassium, Sulphur, Oxygen

6. Calculate the molar mass of the following substances.

(a) Ethyne, C2H2  

(b) Sulphur molecule, S8  

(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)

(d) Hydrochloric acid, Hcl

(e) Nitric acid, HNO3

Answer:

Molar mass of the following substances are:

(a) Ethyne, C2H2 - Molar mass = 2 x Atomic mass of Carbon + 2 x Atomic mass of Hydrogen = (2 x 12.01 u) + (2 x 1.01 u) = 26.04 u

(b) Sulphur molecule, S8 - Molar mass = 8 x Atomic mass of Sulphur = (8 x 32.06 u) = 256.48 u

(c) Phosphorus molecule, P4 - Molar mass = 4 x Atomic mass of Phosphorus = (4 x 31.00 u) = 124.00 u

(d) Hydrochloric acid, HCl - Molar mass = Atomic mass of Hydrogen + Atomic mass of Chlorine = (1.01 u) + (35.45 u) = 36.46 u

(e) Nitric acid, HNO3 - Molar mass = Atomic mass of Hydrogen + Atomic mass of Nitrogen + 3 x Atomic mass of Oxygen = (1.01 u) + (14.01 u) + (3 x 16.00 u) = 63.01 u

7. What is the mass of—

(a) 1 mole of nitrogen atoms?  

(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?  

(c) 10 moles of sodium sulphite (Na2SO3)?  

Answer:

(a) The atomic mass of nitrogen is 14 g/mol. Therefore, 1 mole of nitrogen atoms will have a mass of 14 grams.

(b) The atomic mass of aluminium is 27 g/mol. Therefore, 4 moles of aluminium atoms will have a mass of 4 × 27 = 108 grams.

(c) The formula mass of sodium sulphite (Na2SO3) is 2 × 23 + 32 + 3 × 16 = 126 g/mol. Therefore, 10 moles of sodium sulphite will have a mass of 10 × 126 = 1260 grams.

8. Convert into mole.

(a) 12 g of oxygen gas  

(b) 20 g of water  

(c) 22 g of carbon dioxide

Answer:

(a) The molar mass of oxygen is 32 g/mol. Therefore, 12 g of oxygen gas is equivalent to 12/32 = 0.375 moles.

(b) The molar mass of water is 18 g/mol. Therefore, 20 g of water is equivalent to 20/18 = 1.111 moles.

(c) The molar mass of carbon dioxide is 44 g/mol. Therefore, 22 g of carbon dioxide is equivalent to 22/44 = 0.5 moles.

9. What is the mass of:

(a) 0.2 mole of oxygen atoms?  

(b) 0.5 mole of water molecules?

Answer:

(a) The atomic mass of oxygen is 16 g/mol. Therefore, 0.2 mole of oxygen atoms will have a mass of 0.2 × 16 = 3.2 grams.

(b) The molar mass of water is 18 g/mol. Therefore, 0.5 mole of water molecules will have a mass of 0.5 × 18 = 9 grams.

10. Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur

Answer:

To calculate the number of molecules of sulfur present in 16 g of solid sulfur, we can use the following steps:

Find the molar mass of sulfur (S8):

Molar mass of S8 = 8 x atomic mass of S = 8 x 32 g/mol = 256 g/mol

Calculate the number of moles of sulfur:

Number of moles = mass / molar mass

Number of moles = 16 g / 256 g/mol

Number of moles = 0.0625 mol

Use Avogadro's number to convert the number of moles to the number of molecules:

Number of molecules = number of moles x Avogadro's number

Number of molecules = 0.0625 mol x 6.022 x 10^23 molecules/mol

Number of molecules = 3.76 x 10^22 molecules

Therefore, there are 3.76 x 10^22 molecules of sulfur (S8) present in 16 g of solid sulfur.

11. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.

(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)

Answer:

To calculate the number of aluminum ions present in 0.051 g of aluminum oxide, we first need to determine the number of moles of aluminum oxide.

The molar mass of aluminum oxide (Al2O3) can be calculated as follows:

Molar mass of Al2O3 = (2 x molar mass of Al) + (3 x molar mass of O)

= (2 x 27 u) + (3 x 16 u)

= 54 u + 48 u

= 102 u

Thus, the number of moles of Al2O3 present in 0.051 g can be calculated using the formula:

Number of moles = Mass / Molar mass

Number of moles of Al2O3 = 0.051 g / 102 g/mol

= 0.0005 mol

Since aluminum oxide has a 2:3 ratio of aluminum ions to oxide ions, we can calculate the number of aluminum ions present as follows:

Number of aluminum ions = 0.0005 mol x 2

= 0.001 mol

Finally, we can convert this to the number of aluminum ions using Avogadro's number:

Number of aluminum ions = 0.001 mol x 6.022 x 10^23 ions/mol

= 6.022 x 10^20 ions

Therefore, there are 6.022 x 10^20 aluminum ions present in 0.051 g of aluminum oxide.

Important Questions

1. Compare the properties of electrons, protons and neutrons.

Answer:  

Electrons, protons and neutrons are subatomic particles present in an atom.  

Electrons have a negative charge and negligible mass, protons have a positive charge and a mass of approximately 1 atomic mass unit, while neutrons have no charge and a mass similar to that of protons.

2. What are the limitations of J.J. Thomson’s model of the atom?

Answer:  

The main limitation of J.J. Thomson's model of the atom was that it could not explain the observations of the alpha particle scattering experiment conducted by Rutherford.

3. What are the limitations of Rutherford’s model of the atom?

Answer:  

The main limitation of Rutherford's model of the atom was that it could not explain the stability of an atom as per the laws of electromagnetic radiation. According to classical electromagnetic theory, an accelerated charged particle should continuously lose energy in the form of radiation and ultimately fall into the nucleus.

4. Describe Bohr’s model of the atom.

Answer:

Bohr's model of the atom proposed that electrons revolve around the nucleus in certain specific circular orbits or shells. The electrons in the innermost shell have the lowest energy, while those in the outermost shell have the highest energy. The model could explain the spectral lines observed in the hydrogen spectrum, which could not be explained by previous models.

5. Compare all the proposed models of an atom given in this chapter.

Answer:

The proposed models of an atom have evolved over time based on new experimental observations and theoretical understanding. Thomson's model proposed that an atom is a sphere of positive charge with negatively charged electrons embedded in it. Rutherford's model proposed that an atom has a small, positively charged nucleus with electrons orbiting it. Bohr's model incorporated the idea of electrons revolving in specific shells or orbits. Modern quantum mechanical models propose that electrons do not revolve in fixed orbits but occupy specific energy levels or orbitals around the nucleus.

6. Summarise the rules for writing of distribution of electrons in various shells for the first eighteen elements.

Answer:

The rules for writing the distribution of electrons in various shells for the first eighteen elements are:

The first shell can hold a maximum of 2 electrons.

The second shell can hold a maximum of 8 electrons.

The third shell can hold a maximum of 18 electrons.

Electrons are filled in shells in order of increasing energy levels.

The maximum number of electrons in any shell can be calculated using the formula 2n^2, where n is the shell number.

For example, the distribution of electrons in the first eighteen elements is:

Hydrogen: 1

Helium: 2

Lithium: 2, 1

Beryllium: 2, 2

Boron: 2, 3

Carbon: 2, 4

Nitrogen: 2, 5

Oxygen: 2, 6

Fluorine: 2, 7

Neon: 2, 8

Sodium: 2, 8, 1

Magnesium: 2, 8, 2

Aluminium: 2, 8, 3

Silicon: 2, 8, 4

Phosphorus: 2, 8, 5

Sulfur: 2, 8, 6

Chlorine: 2, 8, 7

Argon: 2, 8, 8

7. Define valency by taking examples of silicon and oxygen.

Answer:

Valency is the combining capacity of an atom or a group of atoms with other atoms. It is determined by the number of valence electrons in an atom. For example, silicon has four valence electrons, so its valency is 4. Oxygen has six valence electrons, so its valency is 2 (as it needs two more electrons to complete its outermost shell).

Valency is defined as the combining capacity of an element with other elements to form a compound. It can be determined by the number of electrons that an atom gains, loses, or shares when it combines with another atom.

8. Explain with examples

(i) Atomic number,

(ii) Mass number,

(iii) Isotopes and

(iv) Isobars. Give any two uses of isotopes.

Answer:  

Silicon has an atomic number of 14, and its electronic configuration is 2, 8, 4. It has four valence electrons in its outermost shell, so its valency is 4.

Oxygen has an atomic number of 8, and its electronic configuration is 2, 6. It has six valence electrons in its outermost shell, so its valency is 2.

8. Explain with examples 

(i) Atomic number, 

(ii) Mass number, 

(iii) Isotopes and 

(iv) Isobars. 

Give any two uses of isotopes.

Answer:

(i) Atomic number: It is the number of protons present in the nucleus of an atom. It is denoted by Z. For example, the atomic number of carbon is 6, which means that there are 6 protons in the nucleus of a carbon atom.

(ii) Mass number: It is the sum of the number of protons and neutrons present in the nucleus of an atom. It is denoted by A. For example, the mass number of carbon-12 is 12, which means that there are 6 protons and 6 neutrons in the nucleus of carbon-12.

(iii) Isotopes: Atoms of the same element having the same atomic number (Z) but different mass numbers (A) are called isotopes. For example, carbon-12 and carbon-14 are isotopes of carbon.

(iv) Isobars: Atoms of different elements having the same mass number (A) but different atomic numbers (Z) are called isobars. For example, chlorine-37 and potassium-37 are isobars.

Two uses of isotopes are:

(a) In the field of medicine, isotopes are used in diagnostic techniques like X-ray and CT scans.

(b) Isotopes are used in agriculture to study plant metabolism, photosynthesis and nitrogen fixation.

9. Na+ has completely filled K and L shells. Explain.

Answer:

Na+ has completely filled K and L shells because Na+ is a positively charged ion, which means that it has lost one electron from its outermost shell (M shell) to form a cation. As a result, the K and L shells (which are inner shells) now appear to be completely filled for Na+.

10. If bromine atom is available in the form of, say, two isotopes 79 35 Br (49.7%) and 81 35 Br (50.3%), calculate the average atomic mass of bromine atom.  

Answer:

Average atomic mass of bromine atom = (49.7/100 × 79) + (50.3/100 × 81)

= 39.263 + 40.743

= 80.006 u

Therefore, the average atomic mass of bromine is approximately 80.006 u.

11. The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes 16 8 X and 18 8 X in the sample?

Answer:

Let x be the percentage of isotope 16

8 X and y be the percentage of isotope 18

8 X.

Given, the average atomic mass of element X is 16.2 u.

The atomic mass of isotope 16

8 X is 16 u and the atomic mass of isotope 18

8 X is 18 u.

Using the formula for calculating the average atomic mass:

Average atomic mass = (x/100) * 16 + (y/100) * 18

Substituting the given values:

16.2 = (x/100) * 16 + (y/100) * 18

Simplifying:

162 = 16x + 18y

Since x and y represent percentages, they must add up to 100.

x + y = 100

Solving the two equations simultaneously:

x = 30

y = 70

Therefore, the percentages of isotopes 16

8 X and 18

8 X in the sample are 30% and 70% respectively.

12. If Z = 3, what would be the valency of the element? Also, name the element.

Answer:

The element with atomic number Z = 3 is Lithium (Li).

Valency is defined as the combining capacity of an element with other elements to form a compound. The valency of an element can be determined by the number of electrons in the outermost shell of its atom.

In the case of Lithium, the atomic number is 3 which means it has 3 electrons in total. These electrons are distributed in 2 shells -  

The first shell contains 2 electrons and the second shell contains only 1 electron. Since the outermost shell has only 1 electron, the valency of Lithium is 1.

13. Composition of the nuclei of two atomic species X and Y are given as under

Column X Column Y  

Protons = 6……….6  

Neutrons = 6……….8  

Give the mass numbers of X and Y.  

What is the relation between the two species?

Answer:

Mass number = Number of protons + Number of neutrons

Mass number of X = 6 + 6 = 12

Mass number of Y = 6 + 8 = 14

The two species X and Y have the same atomic number (number of protons) but different mass numbers. Therefore, they are isotopes of the same element.

14. For the following statements, write T for True and F for False.

Answer:

(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.

False (J.J. Thomson proposed that atoms contain negatively charged electrons embedded in a positively charged sphere)

(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.

False (A neutron is a subatomic particle that is neutral and is composed of one up quark and two down quarks.)

(c) The mass of an electron is about 1/2000 times that of proton.

True (The mass of an electron is approximately 1/2000 times that of a proton)

(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.

True (Isotope iodine-131 is used to make tincture iodine, which is used as a medicine)

Put tick (ü) against correct choice and cross (×) against wrong choice in questions 15, 16 and 17.

15. Rutherford’s alpha-particle scattering experiment was responsible for the discovery of

(a) Atomic Nucleus  

(b) Electron  

(c) Proton

(d) Neutron  

Answer:

The correct option is  (a) Atomic Nucleus

Explanation: Rutherford's alpha-particle scattering experiment was a landmark experiment in the field of atomic physics. In this experiment, a beam of alpha particles (positively charged particles) was directed at a thin sheet of gold foil. Rutherford expected the alpha particles to pass straight through the foil, as per Thomson's plum pudding model of the atom, where the positive charge was assumed to be uniformly distributed throughout the atom.

However, to Rutherford's surprise, some of the alpha particles were deflected at large angles and some even bounced back in the opposite direction. This led Rutherford to conclude that the positive charge in the atom was concentrated in a very small region at the center, which he called the atomic nucleus. This discovery laid the foundation for the modern understanding of the atomic structure, which includes the existence of protons, neutrons, and electrons.

16. Isotopes of an element have

(a) the same physical properties  

(b) different chemical properties  

(c) different number of neutrons

(d) different atomic numbers.

Answer:

The correct option is  (c) "different number of neutrons"

Explanation: Isotopes of an element are atoms that have the same atomic number (same number of protons) but different mass numbers (different number of neutrons). Since the atomic number is the number of protons in the nucleus, it determines the chemical properties of an element. Therefore, isotopes of an element have the same chemical properties but different physical properties, such as density and boiling/melting points. The different number of neutrons affects the stability and nuclear properties of the atom, such as radioactive decay and nuclear binding energy.

17. Number of valence electrons in Cl– ion are:

(a) 16 (b) 8 (c) 17 (d) 18 

Answer:

The correct option is (b) 8.

Explanation: Chlorine (Cl) has 7 valence electrons in its outermost shell. When it gains one electron to form a Cl– ion, it attains a stable noble gas configuration with 8 electrons in its outermost shell (similar to that of argon). Therefore, Cl– ion has 8 valence electrons.

18. Which one of the following is a correct electronic configuration of sodium?

(a) 2,8 (b) 8,2,1 (c) 2,1,8 (d) 2,8,1.  

Answer:

The correct option is (d) 2,8,1

Explanation: The correct electronic configuration of sodium is (d) 2,8,1 because sodium has 11 electrons. The first two electrons occupy the first shell (K) and the remaining 8 electrons occupy the second shell (L). The last electron goes into the third shell (M) which can hold up to 8 electrons. Therefore, the electronic configuration of sodium is 2,8,1.

19. Complete the following table.  

At No...Mass No... No. of Neutrons...No. of Protons...No. of Electrons...At Species

9 -19- 10 -9- -9- -Fluorine-

16 32 -16- -16- -16- Sulphur

12 24 -12- 12 -12- -Magnesium-

-1- 2 -1- 1 -1- -Deuterium-

-1- 1 0 1 0 -Hydrogen (Ion)-

Important Questions

1. Make a comparison and write down ways in which plant cells are different from animal cells.  

Comparison between plant and animal cells:

Answer:

Plant Cells--------Animal Cells

Have a cell wall made up of cellulose surrounding the cell membrane-------- No cell wall, only a plasma membrane.

Large central vacuole present.-------- Small or no vacuoles present.

Chloroplasts present for photosynthesis.-------- No chloroplasts.

Rectangular or fixed shape.-------- Round or irregular shape.

2. How is a prokaryotic cell different from a eukaryotic cell?

Answer:

Differences between prokaryotic and eukaryotic cells:

Prokaryotic Cells-------- Eukaryotic Cells

Do not have a nucleus.-------- Have a well-defined nucleus.

Genetic material floats freely in the cytoplasm.-------- Genetic material is enclosed within a nuclear membrane.

Simple cell structure with no specialized organelles.-------- Complex cell structure with specialized organelles.

Smaller in size.-------- Larger in size.

Example: Bacteria------ Examples: Animal and plant cells

3. What would happen if the plasma membrane ruptures or breaks down?

Answer:

If the plasma membrane ruptures or breaks down, the cell's contents will leak out, and the cell will die.

4. What would happen to the life of a cell if there was no Golgi apparatus?

Answer:

Golgi apparatus is involved in packaging, processing, and transporting proteins and lipids. Without the Golgi apparatus, the cell would not be able to transport and process these molecules properly, and they would not be able to reach their final destination. This could result in a malfunction of the cell and ultimately lead to cell death.

5. Which organelle is known as the powerhouse of the cell? Why?  

Answer:

Mitochondria are known as the powerhouse of the cell because they generate energy in the form of ATP through cellular respiration. This energy is essential for the cell to carry out its functions and maintain its structure.

6. Where do the lipids and proteins constituting the cell membrane get synthesised?  

Answer:

The lipids and proteins that constitute the cell membrane are synthesized in the endoplasmic reticulum (ER) and the Golgi apparatus.

7. How does an Amoeba obtain its food?  

Answer:

Amoeba obtains its food through the process of phagocytosis. It forms pseudopodia around the food particle and engulfs it, and then digests it using enzymes.

8. What is osmosis?  

Answer:

Osmosis is the process of movement of water molecules from an area of high water concentration (dilute solution) to an area of low water concentration (concentrated solution) across a semi-permeable membrane until the concentration of water is the same on both sides of the membrane.

9. Carry out the following osmosis experiment: Take four peeled potato halves and scoos each one out to make potato cups. One of these potato cups should be made from a boiled potato. Put each potato cup in a trough containing water. Now,

(a) Keep cup A empty  

(b) Put one teaspoon sugar in cup B  

(c) Put one teaspoon salt in cup C

(d) Put one teaspoon sugar in the boiled potato cup D.

Keep these for two hours. Then observe the four potato cups and answer the following:

(i) Explain why water gathers in the hollowed portion of B and C.

Answer:

(i) Water gathers in the hollowed portions of B and C because of the process of osmosis. Osmosis is the movement of water molecules from a region of higher concentration to a region of lower concentration through a selectively permeable membrane. In this experiment, the potato cups act as selectively permeable membranes. Sugar and salt are solutes that are dissolved in water. In cup B, the sugar concentration is higher than that of pure water. Therefore, water molecules move from the region of higher concentration (water) in the trough to the region of lower concentration (sugar solution) inside the potato cup. Similarly, in cup C, the salt concentration is higher than that of pure water. Therefore, water molecules move from the region of higher concentration (water) in the trough to the region of lower concentration (salt solution) inside the potato cup.

(ii) Why is potato A necessary for this experiment?

Answer:

(ii) Potato A is necessary for this experiment because it acts as a control. Since it is empty, it does not have any solute concentration to influence the movement of water molecules. Therefore, it helps to compare the results of the other potato cups with a baseline.

(iii) Explain why water does not gather in the hollowed out portions of A and D.

Answer:

(iii) Water does not gather in the hollowed out portions of A and D because there is no solute concentration to influence the movement of water molecules. In the case of cup A, there is no solute, and in the case of cup D, the sugar concentration inside the potato cup is the same as that of the water in the trough. Therefore, there is no net movement of water molecules, and the water level remains the same.

10. Which type of cell division is required for growth and repair of body and which type is involved in formation of gametes?  

Answer:

The type of cell division required for growth and repair of the body is called mitosis. During mitosis, a single cell divides into two identical daughter cells, each with the same number of chromosomes as the parent cell. This type of cell division is involved in the growth and repair of tissues and organs in the body. It is also responsible for the asexual reproduction of many organisms, including some plants and animals.

On the other hand, meiosis is the type of cell division involved in the formation of gametes. During meiosis, a single cell divides into four daughter cells, each with half the number of chromosomes as the parent cell. This results in the production of gametes (sperm or egg cells) with half the number of chromosomes as the parent cell. When fertilization occurs, the gametes combine to form a zygote with a full complement of chromosomes. This type of cell division is crucial for sexual reproduction in most organisms.

Important Questions

1. Define the term “tissue”.

Answer:

Tissue is defined as a group of cells that are similar in structure and function, and work together to perform a specific task or function.

2. How many types of elements together make up the xylem tissue? Name them.

Answer:

Xylem tissue is composed of four types of cells: tracheids, vessels, xylem parenchyma and xylem fibres.

3. How are simple tissues different from complex tissues in plants?

Answer:

Simple tissues are composed of only one type of cell and have a similar structure and function. Complex tissues, on the other hand, are composed of more than one type of cell and have different structures and functions.

4. Differentiate between parenchyma, collenchyma and sclerenchyma on the basis of their cell wall.

Answer:

Parenchyma cells have thin cell walls and are involved in photosynthesis, storage, and other metabolic functions. Collenchyma cells have thickened cell walls that provide mechanical support to the plant. Sclerenchyma cells have thick, lignified cell walls that provide structural support to the plant.

5. What are the functions of the stomata?

Answer:

The stomata are small openings or pores on the surface of leaves, stems and other plant organs that allow for gas exchange. The main function of stomata is to regulate the exchange of gases such as oxygen, carbon dioxide, and water vapor between the plant and the environment. They also play a role in controlling water loss from the plant through transpiration.

6. Diagrammatically show the difference between the three types of muscle fibres.

Answer:

Sketch

Striated muscle fibers, are long and cylindrical with multiple nuclei located on the periphery of the cell. They have a striped or striated appearance under a microscope due to the arrangement of contractile proteins. These fibers are found in skeletal muscles and are responsible for voluntary movements.

Sketch

Unstriated muscle fibers, also known as smooth muscle fibers, are spindle-shaped and have a single nucleus. They do not have striations and are found in the walls of hollow organs like the stomach, uterus, and blood vessels, where they help to control involuntary movements.

Sketch

Cardiac muscle fibers, are branched and have a single nucleus. They have striations similar to skeletal muscle fibers but are connected by intercalated discs, which allow them to contract in a coordinated manner. These fibers are found only in the heart and are responsible for involuntary contractions that pump blood throughout the body.

7. What is the specific function of the cardiac muscle?

Answer:

The specific function of the cardiac muscle is to contract and pump blood throughout the body, specifically within the chambers of the heart.

8. Differentiate between striated, unstriated and cardiac muscles on the basis of their structure and site/location in the body.

Answer:

The differences between striated, unstriated, and cardiac muscles are as follows:

Sketch

Striated muscles have a striped appearance due to the arrangement of actin and myosin filaments. They are voluntary muscles found in skeletal muscles and are attached to bones.

Sketch

Unstriated muscles do not have striations, and their contractile proteins are arranged in a random pattern. They are involuntary muscles found in the walls of hollow organs like the stomach, uterus, and blood vessels.

Sketch

Cardiac muscles have striations like skeletal muscles but are involuntary and are found only in the heart. They have intercalated discs that allow them to contract in a coordinated manner.

I apologize, as an AI language model, I cannot create visual diagrams

9. Draw a labelled diagram of a neuron.  

Sketch

10. Name the following.

Answer:

(a) Tissue that forms the inner lining of our mouth.  

(a) The tissue that forms the inner lining of our mouth is called stratified squamous epithelium.

(b) Tissue that connects muscle to bone in humans.  

(b) The tissue that connects muscle to bone in humans is called tendon.

(c) Tissue that transports food in plants.

(c) The tissue that transports food in plants is called phloem.

(d) Tissue that stores fat in our body.

(d) The tissue that stores fat in our body is called adipose tissue.

(e) Connective tissue with a fluid matrix.

(e) The connective tissue with a fluid matrix is called blood.

(f) Tissue present in the brain.  

(f) The tissue present in the brain is called nervous tissue.

11. Identify the type of tissue in the following:

skin, bark of tree, bone, lining of kidney tubule, vascular bundle.

Answer:

The types of tissue in the following are:

Skin: The skin is composed of multiple tissue types, including stratified squamous epithelium, connective tissue, and nervous tissue.

Bark of tree: The bark of a tree is composed of several tissue types, including cork cambium, cork, phloem, and secondary phloem.

Bone: Bone is a type of connective tissue that is composed of osteocytes, collagen fibers, and mineral salts.

Lining of kidney tubule: The lining of the kidney tubule is composed of epithelial tissue, specifically simple cuboidal epithelium.

Vascular bundle: A vascular bundle is composed of two main types of tissue, xylem and phloem, along with other supporting tissues.

12. Name the regions in which parenchyma tissue is present.

Answer:

Parenchyma tissue is present in various regions of a plant body such as stems, leaves, roots, fruits, and seeds.

13. What is the role of epidermis in plants?

Answer:

The epidermis is the outermost layer of cells in a plant, which covers the entire surface of leaves, stems, roots, and other organs.  

Function: The role of the epidermis is to protect the plant from physical damage, water loss, and invasion by pathogens or pests. It also helps in regulating gas exchange, by allowing the diffusion of gases like oxygen, carbon dioxide, and water vapor in and out of the plant.

14. How does the cork act as a protective tissue?

Answer:

Cork is a protective tissue that develops from the cork cambium, a type of lateral meristem in woody plants.  

Cork cells have a thick, impermeable cell wall made up of suberin, which prevents water loss and protects the plant from physical damage, infection, and other environmental stresses.  

As the cork cells accumulate, they form a layer of protective tissue known as cork bark, which can be several centimeters thick in some species.  

The cork bark acts as a barrier against mechanical damage, pathogens, and pests, and can also provide insulation against extreme temperatures and other environmental stresses.

15. Complete the following chart:  

Sketch

Important Questions

1. What are the advantages of classifying organisms?  

Answer:

The advantages of classifying organisms are as follows:

It helps in understanding the diversity of living organisms on earth.

It helps in identifying and naming new species.

It helps in predicting the characteristics and behavior of an organism based on its classification.

It facilitates the study of evolutionary relationships among organisms.

It aids in the conservation and management of biodiversity.

It provides a systematic framework for the study of living organisms.

2. How would you choose between two characteristics to be used for developing a hierarchy in classification?

Answer:

One should consider the following factors, between two characteristics to be used for developing a hierarchy in classification:

Relevance: The characteristics chosen should be relevant to the group of organisms being classified and should reflect their shared evolutionary history.

Distinctiveness: The characteristics chosen should be distinctive enough to separate different groups of organisms from each other.

Stability: The characteristics chosen should be stable and not subject to frequent changes or modifications.

Ease of observation: The characteristics chosen should be easy to observe and measure accurately.

Universality: The characteristics chosen should be present in all members of the group being classified.

3. Explain the basis for grouping organisms into five kingdoms.

Answer:

Organisms are classified into five kingdoms based on their characteristics and evolutionary relationships. The five kingdoms of organisms are:

Monera: Unicellular prokaryotic organisms, such as bacteria and cyanobacteria.

Protista: Mostly unicellular eukaryotic organisms, such as protozoans and algae.

Fungi: Eukaryotic organisms that obtain nutrients by absorbing them from dead or decaying organic matter.

Plantae: Multicellular eukaryotic organisms that carry out photosynthesis and produce their food.

Animalia: Multicellular eukaryotic organisms that obtain their food by consuming other organisms.

These five kingdoms are based on the fundamental differences in cellular organization, mode of nutrition, and other characteristics among different groups of organisms.

4. What are the major divisions in the Plantae? What is the basis for these divisions?

Answer:

The major divisions in the Plantae are as follows:

Thallophyta: These are the simplest and most primitive plants, such as algae and fungi. They lack specialized tissues and organs and have a simple body structure.

Bryophyta: These are small, non-vascular plants, such as mosses, liverworts, and hornworts. They have a simple, primitive body structure and lack true roots, stems, and leaves.

Pteridophyta: These are vascular plants that have true roots, stems, and leaves. They reproduce by spores and include ferns, horsetails, and club mosses.

Gymnosperms: These are vascular plants that produce seeds, but the seeds are not enclosed within fruits. They include conifers, cycads, and ginkgo trees.

Angiosperms: These are flowering plants that produce seeds enclosed within fruits. They are the most diverse and widespread group of plants, and include herbs, shrubs, and trees.

The basis for these divisions is primarily the differences in their reproductive structures, body organization, and mode of reproduction. Additionally, the development of specialized tissues and organs, as well as the ability to adapt to various environmental conditions, has also contributed to the division of plants into these categories.

5. How are the criteria for deciding divisions in plants different from the criteria for deciding the subgroups among animals?

Answer:

The criteria for deciding divisions in plants and subgroups among animals are different because of the unique characteristics and evolutionary history of each group.  

In plants, the major divisions are based on their reproductive structures, body organization, and mode of reproduction, while the criteria for deciding subgroups among animals are primarily based on anatomical, physiological, and developmental differences.

For example, in plants, the divisions are based on the presence or absence of specialized tissues and organs, such as roots, stems, and leaves, as well as the presence or absence of vascular tissue. In contrast, the subgroups among animals are based on features such as the presence or absence of a backbone, the type of digestive system, the type of symmetry, and the presence or absence of a coelom.

6. Explain how animals in Vertebrata are classified into further subgroups.

Answer:

Animals in Vertebrata are classified into further subgroups based on various criteria such as the presence or absence of jaws, fins, limbs, and other anatomical features. The classification of vertebrates is primarily based on the presence or absence of a backbone or vertebral column.  

The five major subgroups of vertebrates are as follows:

Fishes: Vertebrates that live in water and have gills for respiration. They are further classified into two groups: cartilaginous fishes, such as sharks and rays, and bony fishes, such as salmon and trout.

Amphibians: Vertebrates that live both in water and on land. They have moist, scaleless skin and reproduce in water. Examples include frogs, toads, and salamanders.

Reptiles: Vertebrates that are adapted to live on land and have dry, scaly skin. They reproduce by laying eggs on land. Examples include snakes, lizards, and crocodiles.

Birds: Vertebrates that are adapted for flight and have feathers and a beak. They reproduce by laying eggs and have a four-chambered heart. Examples include eagles, sparrows, and penguins.

Mammals: Vertebrates that have hair or fur and mammary glands for producing milk to nourish their young. They have a four-chambered heart and give birth to live young. Examples include humans, dogs, cats, and whales.

The classification of animals is an ongoing process, and new discoveries and advancements in molecular biology and genetics are leading to revisions and refinements in the classification system.

Important Questions

1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Answer:

The diameter of the circular track is given as 200 m.  

Diameter=  2r  =200 m.
Radius, r = 100 m.

In 40 s, the given athlete covers a distance of 200π.

In 1 s, the given athlete covers a distance = 200π / 40  

The athlete runs for 2 minutes 20 s = 140 s  

∴Total distance covered in 140 s = 140 x 200π / 40  

we solve as = 2200m (pie =22/7)

The distance covered in 140 sec is 2200 m  

Displacement

For each complete round the displacement is zero, because he will reach the same place he started from.

40 seconds= one round

40 seconds= one round

40 seconds= one round

20 seconds= half round

Displacement is straight line = 2  times radius = 2 x 100 m=200 m

Thus, the distance covered is 2200 m and the displacement is 200 m.

2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

Answer:

(a) Average speed from A to B:

The distance covered by Joseph from A to B is 300 m. The time taken by him to cover this distance is 2 minutes 30 seconds = 150 seconds.

Therefore, the average speed of Joseph from A to B is:

Average speed = Total distance / Total time taken = 300 m / 150 s = 2 m/s

The average velocity is the displacement per unit time. As Joseph moves from A to B in a straight line without changing direction, the displacement is simply the distance between A and B, which is 300 m.

Average velocity = Displacement / Total time taken = 300 m / 150 s = 2 m/s (in the direction from A to B)

(b) Average speed from A to C:

The distance covered by Joseph from A to C is (300 + 100) = 400 m. The time taken by him to cover this distance is (2 minutes 30 seconds + 1 minute) = 210 seconds.

Therefore, the average speed of Joseph from A to C is:

Average speed = Total distance / Total time taken = 400 m / 210 s = 1.9 m/s (approx)

The average velocity is the displacement per unit time. As Joseph moves from A to B and back to C, the displacement is AC Is 2 min 30 sec +1 min= 3 min 30 sec= 180 sec+ 30 sec=210 sec

Average velocity = Displacement / Total time taken = 300-100 / 210 s = 200/210= 0.95 m/s

3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h–1. On his return trip along the same route, there is less traffic and the average speed is 30 km h–1. What is the average speed for Abdul’s trip?

Answer:

We can use the formula for average speed:

Average speed = Total distance covered ÷ Total time taken

Let's assume the distance from Abdul's home to school is 'd'. Then, the time taken for Abdul to reach his school at 20 km/h would be:

Time taken = d ÷ 20

Similarly, the time taken for Abdul to travel the same distance back at 30 km/h would be:

Time taken = d ÷ 30

The total distance covered by Abdul would be 2d (as he travels the same distance twice, to and fro).

The total time taken for the trip would be:

Total time = d ÷ 20 + d ÷ 30

We can simplify the above equation to get the total time taken as:

Total time = (3d + 2d) ÷ 60 = 5d ÷ 60 = d ÷ 12

Now, using the formula for average speed, we get:

Average speed = Total distance covered ÷ Total time taken

Average speed = 2d ÷ (d ÷ 12) = 24 km/h

Therefore, the average speed for Abdul's trip is 24 km/h.

4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s–2 for 8.0 s.

How far does the boat travel during this time?

Answer:

It is Given that:

Initial velocity, u = 0 (since the boat is starting from rest)

Acceleration, a = 3.0 m/s^2

Time, t = 8.0 s

We need to find the distance travelled, which can be found using the equation of motion:

s = ut + (1/2)at^2

Substituting the values, we get:

s = 0 + (1/2)(3.0)(8.0)^2

s = 96.0 m

Therefore, the boat travels a distance of 96.0 m during this time.

5. A driver of a car travelling at 52 km h–1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h–1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

Answer:

Let's first convert the given speeds to m/s:

Speed of the first car = 52 km/h = 14.44 m/s

Speed of the second car = 3 km/h = 0.83 m/s

For the first car:

Initial velocity, u = 14.44 m/s (in the positive direction)

Final velocity, v = 0 m/s

Time taken, t = 5 s

Acceleration, a = (v-u)/t = (-14.44)/5 = -2.89 m/s^2 (in the negative direction)

For the second car:

Initial velocity, u = 0.83 m/s (in the positive direction)

Final velocity, v = 0 m/s

Time taken, t = 10 s

Acceleration, a = (v-u)/t = (-0.83)/10 = -0.083 m/s^2 (in the negative direction)

Plotting the speed versus time graphs for the two cars on the same graph paper:

As we can see from the graph, the area under the speed-time graph shows the distance travelled.  

The first car comes to rest in a shorter time and also has a higher initial speed, so its graph is a triangle with a larger area.  

The second car's graph is a trapezium with a smaller area. Therefore, the first car travels farther after the brakes are applied.  

6. Fig 8.11 shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions: Fig. 8.11  

Answer:

(a) Which of the three is travelling the fastest?

(a) It is known that the slope of a distance-time graph represents the speed of the object. We can see, slope B is greatest in the given distance-time graph. Hence, object B is moving fastest.

(b) Are all three ever at the same point on the road?  

(b) No, all three are never at the same point on the road because their distance-time graphs never intersect.

(c) How far has C travelled when B passes A?

(c).On the graph, it can be clearly seen that C is at 8 km when B passes A.

(d) How far has B travelled by the time it passes C?

(d).When B passes C, C has travelled a distance of 4/7 (for 7 boxes make 4 km so 1 box is 4/7 km) hence for 9 boxes it comes to be 4/7 X 9 = 5.143 km

7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the ground?  

(a)What velocity will it strike the ground?

Answer:

It is given that:

Initial velocity, u = 0 m/s

Final velocity, v = ?

Acceleration, a = 10 m/s^2

Distance, s = 20 m

We know that, v^2 = u^2 + 2as

Substituting the values, we get:

v^2 = 0 + 2 × 10 × 20

v^2 = 400

v = √400

v = 20 m/s

So, the ball will strike the ground with a velocity of 20 m/s.

(b)What velocity will it strike the ground?

Answer:

It is given that:

Acceleration a=10ms−2
Time of fall, t=?
We know that

v2−u2=2as
substituting values

v2−0=2×10×20

= 400 or v = 20ms−1
we know  

V = u+at  

we have
20=0+10×t or t=2s

8. The speed-time graph for a car is shown is Fig. 8.12. Fig. 8.12

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.

Answer:

Add the squares under the area slope in the graph

56 full squares plus 12 half squares = 62 squares for 4 seconds.

The total Area of the squares, will give the distance travelled by the car in 4 seconds.
From the graph on the time axis, 5 squares = 2 s that means 1 square=2/5 s

From the graph on the speed axis, 3 squares = 2m s- 1 that means 1 square=2/3 ms-1

calculate area for 1 square= 2/5 x 2/3=4/15m

Area of 62 Square= 4/15 x 62=248/15 =16.53 m

Therefore, car travels 16.53 m in first 4 seconds.

(b) Which part of the graph represents uniform motion of the car.

Answer:

(b) Part MN of the graph is straight line parallel to the time axis, thus this portion of graph represents uniform motion of car.  

9. State which of the following situations are possible and give an example for each of these:  

(a) an object with a constant acceleration but with zero velocity

Answer:

(a) Yes, an object can have a constant acceleration but zero velocity. For example, a ball thrown straight up in the air at its highest point has zero velocity but still has a constant acceleration due to gravity.

(b) an object moving with an acceleration but with uniform speed.

Answer:

(b) No, an object cannot move with an acceleration but with uniform speed. If an object has acceleration, its speed must be changing.

(c) an object moving in a certain direction with an acceleration in the perpendicular direction.

Answer:

(c) Yes, an object can move in a certain direction with an acceleration in the perpendicular direction. For example, a car moving along a straight road can have an acceleration perpendicular to the direction of motion due to a circular turn.

10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Answer:

We can use the formula for the speed of an object in circular motion:

v = (2πr) / T

where v is the speed, r is the radius of the circular orbit, and T is the time period of revolution.

In this case, the radius of the orbit is 42250 km, and the time period of revolution is 24 hours (or 86400 seconds). Substituting these values into the formula, we get:

v = (2π × 42250) / 86400

v = 3.07 km/s

Therefore, the speed of the artificial satellite is 3.07 km/s.

Important Questions

1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity?

If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.

Answer:

Yes, it is possible for an object to experience a net zero external unbalanced force and still be travelling with a non-zero velocity. This happens when the object is moving with a constant velocity in a straight line. The magnitude and direction of the velocity must remain constant for this to happen.

2. When a carpet is beaten with a stick, dust comes out of it. Explain.

Answer:

When a carpet is beaten with a stick, the dust particles that were trapped in the carpet get dislodged due to the force applied by the stick. This force causes the carpet to vibrate, and the dust particles, which were not tightly bound to the carpet fibers, come off due to the vibrations.

3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Answer:

It is advised to tie any luggage kept on the roof of a bus with a rope to prevent it from falling off due to the force of inertia. When the bus accelerates or decelerates suddenly, the luggage tends to remain in its state of motion due to its inertia, and may fall off the roof if not tied down securely.

4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because

(a) the batsman did not hit the ball hard enough.

(b) velocity is proportional to the force exerted on the ball.

(c) there is a force on the ball opposing the motion.

(d) there is no unbalanced force on the ball, so the ball would want to come to rest

Answer:

The correct option is (c) there is a force on the ball opposing the motion.  

Explanation: The ball slows down due to the force of friction acting between the ball and the ground, which opposes the motion of the ball. This force eventually brings the ball to a stop.

5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)

Answer:

Given, initial velocity, u = 0 (as the truck starts from rest)

Distance travelled, s = 400 m

Time taken, t = 20 s

Using the formula, s = ut + (1/2)at^2

We get, 400 = 0 + (1/2) a (20)^2

=> a = 400/200 = 2 m/s^2

The mass of the truck, m = 7 tonnes = 7000 kg

Force, F = ma

Substituting the values, we get,

F = 7000 x 2 = 14000 N

Thus, the acceleration of the truck is 2 m/s^2 and the force acting on it is 14000 N.

6. A stone of 1 kg is thrown with a velocity of 20 m s–1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

Answer:

To find the force of friction between the stone and the ice, we need to use the equation:

frictional force = mass x acceleration

Since the stone comes to rest, its final velocity is 0 m/s. We can use the following equation of motion to find the acceleration:

v^2 = u^2 + 2as

Where,

v = final velocity = 0 m/s

u = initial velocity = 20 m/s

s = distance traveled = 50 m

Substituting the given values in the equation, we get:

0^2 = 20^2 + 2a(50)

400 = 100a

a = 4 m/s^2

Now we can find the force of friction using:

frictional force = mass x acceleration

Substituting the given values, we get:

frictional force = 1 kg x 4 m/s^2

frictional force = 4 N

Therefore, the force of friction between the stone and the ice is 4 N.

7. A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (a) the net accelerating force and

(b) the acceleration of the train.

Answer:

(a) The net accelerating force on the train is given by the difference between the force applied by the engine and the friction force offered by the track:

Net force = Force applied by engine - Friction force

Net force = 40000 N - 5000 N

Net force = 35000 N

(b) To find the acceleration of the train, we need to use the formula:

Net force = mass × acceleration

where mass is the total mass of the train and wagons.

Total mass of the train and wagons = mass of engine + mass of 5 wagons

Total mass = 8000 kg + 5 × 2000 kg

Total mass = 18000 kg

Substituting the values into the formula, we get:

35000 N = 18000 kg × acceleration

acceleration = 35000 N / 18000 kg

acceleration = 1.94 m/s^2

Therefore, the acceleration of the train is 1.94 m/s^2.   

8. An auto mobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s–2?

Answer:

Using Newton's Second Law of Motion, we know that force (F) is equal to mass (m) multiplied by acceleration (a):

F = m * a

Here, the mass of the vehicle is given as 1500 kg and the acceleration with which it needs to be stopped is given as 1.7 m/s^2. Substituting these values, we get:

F = 1500 kg * (-1.7 m/s^2)

F = -2550 N

The negative sign indicates that the force must act in the opposite direction to the motion of the vehicle, i.e., it must be a braking force. Therefore, the force between the vehicle and the road must be 2550 N for the vehicle to be stopped with a negative acceleration of 1.7 m/s^2.

9. What is the momentum of an object of mass m, moving with a velocity v?

(a) (mv) 2  

(b) mv2  

(c) ½ mv2

(d) mv

Answer:

The momentum of an object of mass m, moving with a velocity v is given by the formula:

Momentum (p) = mv

So the correct option is (d) mv.

10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

Answer:

Since the cabinet is moving at a constant velocity, the net force acting on the cabinet must be zero. Therefore, the force of friction on the cabinet must be equal in magnitude and opposite in direction to the horizontal force applied to it. Hence, the friction force exerted on the cabinet is also 200 N.

11. Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 m s-1 before the collision during which they stick together. 

What will be the velocity of the combined object after collision?

Answer:

The total momentum of the system before the collision is:

p = m1v1 + m2v2

where m1 = m2 = 1.5 kg, v1 = 2.5 m/s (to the right), and v2 = -2.5 m/s (to the left).

p = (1.5 kg)(2.5 m/s) + (1.5 kg)(-2.5 m/s) = 0

The total momentum after the collision is:

p' = (m1 + m2)v'

where v' is the velocity of the combined object.

Since there is no external force acting on the system, the total momentum is conserved, so we have:

p' = p

Therefore:

(m1 + m2)v' = 0

v' = 0

So the velocity of the combined object after collision is zero.

12. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other.

Comment on this logic and explain why the truck does not move.

Answer:

The logic that the two opposite and equal forces cancel each other out is correct, but it does not explain why the truck does not move.

When we push on the truck, we exert a force on it in one direction. According to the third law of motion, the truck exerts an equal and opposite force on us in the opposite direction. These two forces do indeed cancel each other out, but that does not mean that the truck will not move.

To determine if the truck will move or not, we need to consider all the forces acting on it. In addition to the force we are applying, there are likely other forces acting on the truck, such as the force of friction between the truck's tires and the ground. If the force we are applying is not greater than the force of friction, then the truck will not move.

Therefore, the reason why the truck does not move is not because the two opposite and equal forces cancel each other out, but because the force we are applying is not enough to overcome the force of friction.

13. A hockey ball of mass 200 g travelling at 10 m s–1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s–1.

Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Answer:

The initial momentum of the hockey ball is given by:

p1 = m1v1 = (0.2 kg)(10 m/s) = 2 kg m/s

The final momentum of the hockey ball is given by:

p2 = m1v2 = (0.2 kg)(-5 m/s) = -1 kg m/s

The change in momentum is given by:

Δp = p2 - p1 = (-1 kg m/s) - (2 kg m/s) = -3 kg m/s

The magnitude of the change in momentum is given by the absolute value of Δp, which is:

|Δp| = 3 kg m/s

Therefore, the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick is 3 kg m/s.

14. A bullet of mass 10 g travelling horizontally with a velocity of 150 m s–1 strikes a stationary wooden block and comes to rest in 0.03 s.

Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Answer:

It is given that:

Mass of bullet, m = 10 g = 0.01 kg

Velocity of bullet, u = 150 m/s

Time taken by bullet to come to rest, t = 0.03 s

Let the distance of penetration of the bullet into the block be 'x'.

Using the equation of motion, v = u + at, where v = 0, a = acceleration of bullet, and t = time taken to come to rest.

We get, 0 = 150 + a × 0.03

a = -5000 m/s^2 (negative sign indicates retardation)

Using the equation of motion, s = ut + (1/2)at^2, where u = 150 m/s, a = -5000 m/s^2, and t = 0.03 s, we get:

x = ut + (1/2)at^2

x = 150 × 0.03 + (1/2)(-5000)(0.03)^2

x = 0.135 m

The distance of penetration of the bullet into the block is 0.135 m.

Using the formula, F = ma, where m = 0.01 kg and a = -5000 m/s^2, we get:

F = 0.01 × (-5000)

F = -50 N

The magnitude of the force exerted by the wooden block on the bullet is 50 N.

15. An object of mass 1 kg travelling in a straight line with a velocity of 10 m s–1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line.

Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

Answer:

Before the impact, the momentum of the object is:

p1 = m1v1 = 1 kg × 10 m/s = 10 kg m/s

The momentum of the wooden block is zero since it is stationary.

The total momentum just before impact is:

p1 + p2 = 10 kg m/s + 0 = 10 kg m/s

After the impact, the two objects stick together and move off with the same velocity, which we can calculate using the law of conservation of momentum:

p = (m1 + m2) v

where p is the total momentum just after the impact, m1 and m2 are the masses of the object and the wooden block respectively, and v is their common velocity.

The total mass of the system after the impact is:

m1 + m2 = 1 kg + 5 kg = 6 kg

So, the total momentum just after the impact is:

p = 10 kg m/s (conservation of momentum)

This gives us:

6 kg v = 10 kg m/s

v = 10/6 m/s

v = 1.67 m/s

Therefore, the velocity of the combined object is 1.67 m/s.

Note that the direction of the velocity is the same as the direction of the initial velocity of the object.

16. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s–1 to 8 m s–1 in 6 s.

Calculate the initial and final momentum of the object.  

Also, find the magnitude of the force exerted on the object.

Answer:

It is given that:

Mass of the object, m = 100 kg

Initial velocity, u = 5 m/s

Final velocity, v = 8 m/s

Time taken, t = 6 s

Using the formula for momentum,

Initial momentum, p1 = m * u

= 100 kg * 5 m/s

= 500 kg m/s

Final momentum, p2 = m * v

= 100 kg * 8 m/s

= 800 kg m/s

Change in momentum, Δp = p2 - p1

= 800 kg m/s - 500 kg m/s

= 300 kg m/s

Using the formula for acceleration,

a = (v - u) / t

= (8 m/s - 5 m/s) / 6 s

= 0.5 m/s^2

Using Newton's second law of motion,

F = m * a

= 100 kg * 0.5 m/s^2

= 50 N

Therefore,

Initial momentum = 500 kg m/s

Final momentum = 800 kg m/s

Change in momentum = 300 kg m/s

Thus, Force exerted on the object = 50 N.

17. Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum.

Comment on these suggestions.

Answer:

Kiran's suggestion is correct, as the change in momentum of an object is directly proportional to the magnitude of the force applied to it and the duration of time for which it acts. Since the insect had a smaller mass than the motorcar, it suffered a greater change in momentum as compared to the change in momentum of the motorcar.

Akhtar's suggestion is incorrect because the force experienced by an object is equal to the rate of change of its momentum, and not directly proportional to its velocity. Although the motorcar was moving with a larger velocity, the force experienced by the insect was greater because of its smaller mass.

Rahul's explanation is also incorrect because both the motorcar and the insect did not experience the same force. The force experienced by the motorcar was much greater than that experienced by the insect due to the motorcar's larger mass.

Therefore, the correct explanation is that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar due to its smaller mass, and the force experienced by the insect was greater than that experienced by the motorcar due to the insect's smaller mass.

18. How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s–2.

Answer:

First, we need to find the velocity of the dumb-bell just before it hits the floor. We can use the equation:

v² = u² + 2as

where

u = initial velocity (0)

a = acceleration due to gravity (10 m/s²)

s = distance fallen (0.8 m)

So,

v² = 0 + 2(10)(0.8)

v² = 16

v = √16

v = 4 m/s

The momentum of the dumb-bell just before it hits the floor can be calculated as:

p = mv

where

m = mass of the dumb-bell (10 kg)

v = velocity of the dumb-bell just before it hits the floor (4 m/s)

So,

p = 10 × 4

p = 40 kg m/s

Therefore, the dumb-bell will transfer a momentum of 40 kg m/s to the floor when it falls from a height of 80 cm.

A1. The following is the distance-time table of an object in motion:  

Time in seconds: 0 1 2 3 4 5 6 7

Distance in metres: 1 8  27 64  125  216 343  

(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?  

(b) What do you infer about the forces acting on the object?  

Answer:

(a) To determine the acceleration, we can calculate the differences in velocity over equal time intervals. From the given distance-time table, we can see that the velocity of the object is increasing with time. Hence, the acceleration is increasing.

(b) From the information given, we cannot infer anything directly about the forces acting on the object. However, since the acceleration is increasing, we can conclude that there must be a net force acting on the object, and this force must be increasing with time.  

A2. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m s-2.  

With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)  

Answer:

Let the force exerted by each person be denoted as F.

When two persons are pushing the motorcar at a uniform velocity, the net force on the car is zero. Therefore,

2F = 0

F = 0 N

When three persons are pushing the motorcar to produce an acceleration of 0.2 m/s^2, the net force on the car is given by:

Fnet = ma

Where m is the mass of the car and a is the acceleration produced.

Fnet = 1200 kg × 0.2 m/s^2

Fnet = 240 N

Since three persons are pushing the car, the force exerted by each person is:

F = Fnet/3

F = 240 N/3

F = 80 N

Therefore, each person pushes the motorcar with a force of 80 N.

A3. A hammer of mass 500 g, moving at 50 m s-1, 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?  

Answer:

It is given that:

Mass of the hammer, m = 500 g = 0.5 kg

Initial velocity of the hammer, u = 50 m/s

Final velocity of the hammer, v = 0 m/s

Time taken to stop the hammer, t = 0.01 s

We know that,

Force = (Change in momentum) / (Time taken)

Initial momentum of the hammer, p = mu = 0.5 kg * 50 m/s = 25 kg m/s

Final momentum of the hammer, p' = mv = 0.5 kg * 0 m/s = 0 kg m/s

Change in momentum, Δp = p' - p = -25 kg m/s

Therefore,

Force = (Change in momentum) / (Time taken) = (-25 kg m/s) / (0.01 s) = -2500 N

The negative sign indicates that the force is in the opposite direction to the initial motion of the hammer, i.e., it is acting on the hammer in the opposite direction.

A4. A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force.  

Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.  

Answer:

It is given that-

Mass of the motorcar, m = 1200 kg

Initial velocity, u = 90 km/h = 25 m/s

Final velocity, v = 18 km/h = 5 m/s

Time taken, t = 4 s

We know that,

Acceleration, a = (v - u) / t

= (5 - 25) / 4

= -5 m/s^2 (negative sign indicates retardation)

Change in momentum, Δp = m(v - u)

Magnitude of the force required can be calculated using the relation,

F = Δp / t

Substituting the given values, we get

Δp = 1200 × (5 - 25)

= -24000 kg m/s

F = -24000 / 4

= -6000 N

Therefore, the acceleration of the motorcar is -5 m/s^2 (retardation), the change in momentum is -24000 kg m/s, and the magnitude of the force required to slow down the motorcar is 6000 N.

1. How does the force of gravitation between two objects change when the distance between them is reduced to half?

Answer:

The force of gravitation between two objects is inversely proportional to the square of the distance between them. So, if the distance between two objects is reduced to half, the force of gravitation between them becomes four times stronger.

2. Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?

Answer:

Although gravitational force acts on all objects in proportion to their masses, a heavy object does not fall faster than a light object because of the effect of air resistance. Air resistance is a force that opposes the motion of objects through air, and it is more effective in slowing down heavy objects than light objects. In a vacuum, where there is no air resistance, all objects fall at the same rate regardless of their masses.

3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 1024 kg and radius of the earth is 6.4 × 106 m.)

Answer:

The magnitude of the gravitational force between the earth and a 1 kg object on its surface can be calculated using the formula:

F = G * (m1 * m2) / r^2 <(r raise to the power 2)>

where,

G = gravitational constant = 6.67 × 10^-11 Nm^2/kg^2

m1 = mass of the earth = 6 × 10^24 kg

m2 = mass of the object = 1 kg

r = radius of the earth = 6.4 × 10^6 m

Substituting the values in the formula, we get:

F = 6.67 × 10^-11 * (6 × 10^24 * 1) / (6.4 × 10^6)^2

= 9.81 N

So, the magnitude of the gravitational force between the earth and a 1 kg object on its surface is 9.81 N.

4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?

Answer:

According to Newton's third law of motion, for every action, there is an equal and opposite reaction. So, the force with which the earth attracts the moon is equal in magnitude to the force with which the moon attracts the earth.  

Therefore, the earth attracts the moon with the same force with which the moon attracts the earth.

5. If the moon attracts the earth, why does the earth not move towards the moon?  

Answer:

The earth does move towards the moon because of the gravitational force between them. Mass of the earth is much larger than the mass of the moon. However, the earth also has its own inertia, which means it resists changes in its motion.  

As a result, instead of moving towards the moon, the earth moves in a slightly curved path around the moon, which is called an orbit. This is because the gravitational force between the earth and the moon acts as a centripetal force, which keeps the earth in its orbit around the moon.

6. What happens to the force between two objects, if

Answer:

The force between two objects is given by the universal law of gravitation:

F = G (m1m2) / r^2

where G is the gravitational constant, m1 and m2 are the masses of the objects and r is the distance between them.

(i) the mass of one object is doubled?

(i) If the mass of one object is doubled, the force of attraction between the two objects will be doubled.

(ii) the distance between the objects is doubled and tripled?

(ii) If the distance between the objects is doubled, the force of attraction between them will decrease to one-fourth of its initial value. If the distance is tripled, the force of attraction will decrease to one-ninth of its initial value.

(iii) the masses of both objects are doubled?

(iii) If the masses of both objects are doubled, the force of attraction between them will be quadrupled increased four times.

7. What is the importance of universal law of gravitation?

Answer:

The universal law of gravitation is important because it explains the force of attraction between any two objects in the universe. This law is applicable to objects of any size and distance and can be used to predict the motion of planets, moons and other celestial bodies.

8. What is the acceleration of free fall?

Answer:

The acceleration of free fall is the acceleration experienced by a body when it falls freely under the influence of gravity. It is denoted by the symbol g and has a value of approximately 9.8 m/s^2 near the surface of the earth.

9. What do we call the gravitational force between the earth and an object?

Answer:

The gravitational force between the earth and an object is called the weight of the object. It is given by the formula:

Weight = mass x acceleration due to gravity

where mass is the mass of the object and acceleration due to gravity is the acceleration experienced by the object due to the earth's gravitational pull.

10. Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator.

Answer:

No, the friend will not agree with the weight of gold bought. The weight of an object is given by the formula:

Weight = mass x acceleration due to gravity

Since the value of acceleration due to gravity (g) is greater at the poles than at the equator, the weight of the gold at the poles will be more than its weight at the equator, even though its mass remains the same.

11. Why will a sheet of paper fall slower than one that is crumpled into a ball?  

Answer:

Air resistance or drag force is responsible for slowing down the motion of objects in air. A sheet of paper has a larger surface area and experiences more air resistance compared to a crumpled paper ball, which has a smaller surface area. This air resistance opposes the motion of the falling object, and thus, the sheet of paper falls slower than the crumpled paper ball.

12. Gravitational force on the surface of the moon is only 1/6th  as strong as gravitational force on the earth. What is the weight in newtons of a 10 kg object on the moon and on the earth?

Answer:

Weight is the force exerted by the gravitational field on an object of mass m. On the moon, the acceleration due to gravity is 1/6 times that on the earth. Therefore, the weight of a 10 kg object on the moon is:

Weight on moon = m * g_moon = 10 kg * 1.6 m/s^2 = 16 N

The weight of the same object on the earth is:

Weight on earth = m * g_earth = 10 kg * 9.8 m/s^2 = 98 N

13. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate (i) the maximum height to which it rises, (ii) the total time it takes to return to the surface of the earth.

Answer:

It is given that : Initial velocity (u) = 49 m/s, final velocity (v) = 0 m/s, acceleration (g) = -9.8 m/s^2 (negative since it is acting in the opposite direction of the initial velocity), and we need to calculate the maximum height (h) and total time taken (t).

Using the equation v^2 = u^2 + 2gh, we can find the maximum height reached by the ball:

0^2 = 49^2 + 2*(-9.8)*h

h = 24.95 meters (approx)

To find the time taken for the ball to reach the maximum height, we can use the equation v = u + gt, where v = 0 m/s (at maximum height) and u = 49 m/s:

0 = 49 - 9.8t

t = 5 seconds (approx)

Since the time taken to reach maximum height is 5 seconds, the total time taken to return to the surface of the earth would be twice that, i.e. 10 seconds.

14. A stone is released from the top of a tower of height 19.6 m.

Calculate its final velocity just before touching the ground.

Answer:

We can use the equations of motion to solve this problem. Let's take the upward direction as positive.

Initial velocity, u = 0 (the stone is released from rest)

Distance fallen, s = 19.6 m

Acceleration due to gravity, a = 9.8 m/s^2 (assuming no air resistance)

Using the equation of motion s = ut + 1/2 at^2, we can find the time taken for the stone to fall:

s = ut + 1/2 at^2

19.6 = 0 + 1/2 × 9.8 × t^2

t^2 = 4

t = 2 seconds (taking the positive root)

Now using the equation v = u + at, we can find the final velocity of the stone just before it touches the ground:

v = u + at

v = 0 + 9.8 × 2

v = 19.6 m/s (downward)

Therefore, the final velocity of the stone just before touching the ground is 19.6 m/s (downward).

15. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2 , find the maximum height reached by the stone.

What is the net displacement and the total distance covered by the stone?  

Answer:

It is given that:

Initial velocity, u = 40 m/s

Acceleration due to gravity, g = 10 m/s²

At maximum height, final velocity, v = 0 m/s (since the stone will momentarily stop before falling back down)

Using the kinematic equation:

v² = u² + 2gh

where h is the maximum height reached.

Substituting the given values, we get:

0² = 40² - 2(10)h

0 = 1600 - 20h

20h = 1600

h = 80 m

Therefore, the maximum height reached by the stone is 80 m.

To find the net displacement and total distance covered by the stone, we need to consider its entire journey from the initial point of release to its final point of impact on the ground.

The net displacement is the difference between the final and initial positions of the stone. Since the stone is thrown vertically upwards and comes back down to the same point, its net displacement is zero.

The total distance covered by the stone is equal to the sum of the upward and downward journeys. The stone travels 80 m upwards and then falls back down, covering the same distance. So the total distance covered by the stone is 2 × 80 = 160 m.

16. Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 1024 kg and of the Sun = 2 × 1030 kg.

The average distance between the two is 1.5 × 1011 m.

Answer:

We can use the universal law of gravitation to calculate the force of gravitation between the Earth and the Sun:

F = G * (m1 * m2) / r^2

where:

G = 6.67 x 10^-11 N m^2 / kg^2 is the gravitational constant

m1 = mass of the Earth = 6 x 10^24 kg

m2 = mass of the Sun = 2 x 10^30 kg

r = average distance between the Earth and Sun = 1.5 x 10^11 m

Substituting the values, we get:

F = (6.67 x 10^-11) * (6 x 10^24) * (2 x 10^30) / (1.5 x 10^11)^2

= 3.52 x 10^22 N

Therefore, the force of gravitation between the Earth and the Sun is 3.52 x 10^22 N.  

17. A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s.

Calculate when and where the two stones will meet.  

Answer:

Let's first find out the time taken by the stone dropped from the tower to reach the ground.

We know that the distance travelled by an object falling freely under gravity is given by the formula:

s = ut + 1/2 gt^2

where,  

s is the distance travelled, u is the initial velocity (zero in this case),  

g is the acceleration due to gravity, and t is the time taken.

At the point where the stone hits the ground, s = 100 m.

Therefore, 100 = 0 + 1/2 × 9.8 × t^2

Solving for t, we get t = 4.52 s (approx.)

Now, let's find out the height reached by the stone thrown upwards in this time.

Using the formula:

v^2 = u^2 + 2gh

where v is the final velocity (zero when the stone reaches the maximum height), u is the initial velocity (25 m/s in this case), g is the acceleration due to gravity, and h is the maximum height reached, we get:

0 = (25)^2 - 2 × 9.8 × h

Solving for h, we get h = 31.88 m (approx.)

Therefore, the two stones will meet at a height of 31.88 m from the ground, after 4.52 s from the time the stone was thrown upwards.

18. A ball thrown up vertically returns to the thrower after 6 s. Find

(a) the velocity with which it was thrown up,  

(b) the maximum height it reaches, and  

(c) its position after 4 s

Answer:

The ball returns to the ground after 6 seconds.

Hence, the time taken by the ball to reach to the maximum height (h) is 3 seconds i.e t=3 s

Let the velocity by which it is thrown up be denoted by u

(a). For upward motion,                

v=u+at

∴     0=u+(−10)×3                  

or u=30  m/s  

(b). The maximum height reached by the ball 1/2 at^2

h=ut+1/2 at^2

h=30×3+1/2(−10)×3^2

h=45 m

(c). After 3 seconds, the ball will starts to fall down.

Let the distance by which it fall in 1s be denoted by d

d=0+1/2 at^2

d=0+ ½ × 10 ×1^2           where t′=1 s

D= 5 m  

Thus, Its height above the ground, h′=45−5= 40m

So, after 4 Sec , the ball will be at a height of 40 m above ground.

19. In what direction does the buoyant force on an object immersed in a liquid act?

Answer:

The buoyant force on an object immersed in a liquid acts in the upward direction, opposite to the gravitational force acting on the object. This is because the buoyant force is caused by the difference in pressure between the top and bottom surfaces of the object, and this pressure difference is greater at the bottom than at the top.  

The resulting net force is directed upwards, pushing the object towards the surface of the liquid. This is known as Archimedes' principle, which states that the buoyant force on an object is equal to the weight of the fluid displaced by the object, and acts in the opposite direction to the weight of the object.

20. Why does a block of plastic released under water come up to the surface of water?

Answer:

A block of plastic released under water comes up to the surface of the water due to the buoyant force acting on it. The buoyant force is caused by the difference in pressure between the top and bottom surfaces of the block of plastic. The pressure at the bottom of the block is greater than the pressure at the top, due to the weight of the water above it. This pressure difference creates an upward force on the block of plastic, which is known as the buoyant force.

According to Archimedes' principle, the buoyant force acting on the block of plastic is equal to the weight of the water displaced by the block. Since the density of the plastic is less than that of water, the block displaces an amount of water that weighs more than the block itself. Therefore, the buoyant force is greater than the weight of the block, and it rises to the surface of the water.

In other words, the buoyant force is strong enough to overcome the force of gravity acting on the block of plastic, and hence the block of plastic rises to the surface of the water.

21. The volume of 50 g of a substance is 20 cm3. If the density of water is 1 g cm–3, will the substance float or sink?

Answer:

To determine if the substance will float or sink in water, we need to compare the density of the substance to the density of water.

Density (ρ) is defined as mass (m) per unit volume (V), i.e.,

ρ = m/V

We are given that the volume of the substance is 20 cm^3 and the mass of the substance is 50 g. Therefore, the density of the substance is:

ρ = m/V = 50 g / 20 cm^3 = 2.5 g/cm^3

We are also given that the density of water is 1 g/cm^3.

Comparing the density of the substance to the density of water, we see that the density of the substance (2.5 g/cm^3) is greater than the density of water (1 g/cm^3).

Since the density of the substance is greater than the density of water, the substance will sink in water.

22. The volume of a 500 g sealed packet is 350 cm3 . Will the packet float or sink in water if the density of water is 1 g cm–3?

What will be the mass of the water displaced by this packet?

Answer:

The density of the packet can be calculated by dividing its mass by its volume:

Density of the packet = mass/volume = 500 g/350 cm³ ≈ 1.43 g/cm³

Since the density of the packet is greater than the density of water (1 g/cm³), the packet will sink in water.

When the packet is submerged in water, it will displace an amount of water equal to its own volume (350 cm³). The mass of water displaced by the packet can be calculated by multiplying the volume of water displaced by the density of water:

Mass of water displaced = volume of water displaced × density of water

= 350 cm³ × 1 g/cm³

= 350 g

Therefore, the packet will displace 350 g of water when submerged in water.

Important Questions

1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.

Answer:

Work is defined as the product of force and displacement in the direction of the force. Based on this definition, we can reason out whether or not work is done in the following activities:

• Suma is swimming in a pond.  

Suma is swimming in a pond: Yes, work is done by Suma against the water resistance as she swims in the pond.

• A donkey is carrying a load on its back.  

A donkey is carrying a load on its back: Yes, work is done by the donkey in lifting the load against gravity.

• A wind-mill is lifting water from a well.  

A wind-mill is lifting water from a well: Yes, work is done by the wind-mill in lifting the water against gravity.

• A green plant is carrying out photosynthesis.

A green plant is carrying out photosynthesis: Yes, work is done by the plant in converting solar energy into chemical energy during photosynthesis.

• An engine is pulling a train.  

An engine is pulling a train: Yes, work is done by the engine in pulling the train against friction and air resistance.

• Food grains are getting dried in the sun.  

Food grains are getting dried in the sun: No, work is not done as there is no force acting on the food grains.

• A sailboat is moving due to wind energy.

A sailboat is moving due to wind energy: Yes, work is done by the wind in moving the sailboat against air resistance.

2. An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?

Answer:

When an object is thrown at a certain angle to the ground and moves in a curved path and falls back to the ground, the work done by the force of gravity on the object is zero. This is because the force of gravity is acting in the vertical direction, while the displacement of the object is in the horizontal direction. As a result, the angle between the force of gravity and the displacement of the object is 90 degrees, which means that the work done by gravity is zero.

3. A battery lights a bulb. Describe the energy changes involved in the process

Answer:

When a battery lights a bulb, the chemical energy stored in the battery is converted into electrical energy, which flows through the circuit and is converted into light and heat energy by the bulb. The conversion of chemical energy to electrical energy is a result of a chemical reaction that takes place inside the battery. The electrical energy produced by the battery is transferred to the bulb through the wire, where it is converted into light and heat energy through the process of Joule heating.

Overall, the energy changes involved in the process are:

Chemical energy -> Electrical energy -> Light energy + Heat energy

4. Certain force acting on a 20 kg mass changes its velocity from 5 m s–1 to 2 m s–1.

Calculate the work done by the force.

Answer:

Given total mass (m)            =20 kg.

Given Initial velocity u        =5 ms−1

Given Final velocity v         =2 ms−1

We know that, work done to change the velocity of an object =

change in kinetic energy of the object.

=K2−K1

or =12mv2−12mu2

or =12m(v2−u2)

or =12×20(22−52)

or =10(4−25)

or =−21×10

or =−210 Joule

Hence, the work done to change the velocity = 210 Joules

5. A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer

Answer:

If the line joining points A and B is horizontal, then there is no change in the vertical height of the mass. Therefore, the work done on the object by the gravitational force is zero.

Explanation:

This is because work is defined as the product of the force acting on an object and the displacement of the object in the direction of the force. In this case, the gravitational force acts vertically downwards on the object, but the displacement of the object is horizontal. Therefore, there is no displacement of the object in the direction of the gravitational force, and hence, no work is done by the gravitational force.

The gravitational force only does work when there is a vertical displacement of the object. For example, if the mass is lifted vertically from point A to point B, then the gravitational force would do positive work on the object equal to the product of the force and the vertical displacement. Similarly, if the mass is lowered vertically from point B to point A, then the gravitational force would do negative work on the object equal to the product of the force and the vertical displacement.

6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

Answer:

The potential energy of a freely falling object decreases progressively because it is being converted into kinetic energy. This does not violate the law of conservation of energy.

The law of conservation of energy states that energy cannot be created or destroyed; it can only be transformed from one form to another. In the case of a freely falling object, the decrease in potential energy is exactly balanced by the increase in kinetic energy, so the total energy of the object remains constant. As the object falls, its potential energy decreases, but its kinetic energy increases by the same amount, such that the sum of the two energies remains constant.

Therefore, the law of conservation of energy is not violated when the potential energy of a freely falling object decreases progressively. It is a result of the conversion of potential energy into kinetic energy, which maintains the total energy of the object at a constant level.

7. What are the various energy transformations that occur when you are riding a bicycle?

Answer:

When you are riding a bicycle, there are several energy transformations that occur:

Chemical energy to kinetic energy: The energy stored in the food you eat is converted into chemical energy in your body, which is then converted into kinetic energy as you pedal the bicycle.

Kinetic energy to potential energy: When you pedal the bicycle up a hill, the kinetic energy of the bicycle and rider is converted into potential energy due to the increase in elevation.

Potential energy to kinetic energy: As you ride the bicycle down the other side of the hill, the potential energy is converted back into kinetic energy, as the bicycle and rider gain speed.

Kinetic energy to heat and sound energy: Friction between the bicycle tires and the road surface, as well as air resistance, cause some of the kinetic energy to be converted into heat and sound energy.

Kinetic energy to rotational energy: The rotational motion of the wheels of the bicycle converts some of the kinetic energy into rotational energy.

Kinetic energy to electrical energy: If the bicycle has a dynamo, the rotation of the wheels can generate electrical energy that can be used to power lights or other devices on the bicycle.

Overall, riding a bicycle involves a complex interplay of energy transformations, as energy is converted from one form to another to power the motion of the bicycle and rider.

8. Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?

Answer:

Yes, the transfer of energy takes place when you push a huge rock with all your might, even if you fail to move it. When you apply a force to the rock, you are doing work on it, and therefore transferring energy to the rock.

However, if the rock does not move, then the energy you spend is being converted into another form, such as heat or sound energy. The energy is dissipated due to factors such as friction between the rock and the ground, or between your shoes and the ground. This frictional force opposes the force you apply to the rock and converts some of the energy into heat and sound energy, rather than into the kinetic energy of the rock.

Thus, even though you may not succeed in moving the rock, you are still transferring energy to it, which is being dissipated as heat and sound energy through friction.

10. An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.

Answer:

The potential energy of an object of mass m raised to a height h above the ground is given by the formula:

Potential energy = mgh

where g is the acceleration due to gravity, which is approximately equal to 9.8 m/s² near the surface of the Earth.

Using this formula, the potential energy of the 40 kg object raised to a height of 5 m above the ground is:

Potential energy = (40 kg) x (9.8 m/s²) x (5 m) = 1960 J

When the object is halfway down, it has fallen a distance of 2.5 m. At this point, its potential energy has decreased to:

Potential energy = (40 kg) x (9.8 m/s²) x (2.5 m) = 980 J

By the law of conservation of energy, this decrease in potential energy is equal to the increase in kinetic energy of the object. Therefore, the kinetic energy of the object when it is halfway down is:

Kinetic energy = 980 J

Note that at the instant the object is halfway down, it still has potential energy due to its height above the ground, but it also has kinetic energy due to its motion.

11. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.

Answer:

The work done by the force of gravity on a satellite moving round the Earth is zero.

This is because work is defined as the product of force and displacement, with the force and displacement being in the same direction. In the case of a satellite moving in a circular orbit around the Earth, the force of gravity is always perpendicular to the direction of motion of the satellite. Therefore, the force of gravity does not do any work on the satellite, as there is no displacement in the direction of the force.

Despite the force of gravity not doing any work on the satellite, it is still responsible for maintaining the circular motion of the satellite. This is because the gravitational force provides the necessary centripetal force required to keep the satellite moving in a circular path around the Earth.

12. Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.

No, there cannot be displacement of an object in the absence of any force acting on it.

Answer:

This is because according to Newton's first law of motion, an object at rest will remain at rest and an object in motion will continue in uniform motion in a straight line, unless acted upon by an external force.  

Therefore, if there is no force acting on an object, it will either remain at rest or continue to move in a straight line at a constant speed. In either case, there will be no displacement of the object unless a force is applied to it.

It is possible, however, for an object to have zero net force acting on it, which means that the forces acting on the object are balanced and there is no acceleration. In this case, the object will continue to move with a constant velocity, but it will not experience any displacement unless a force is applied to it.

It is important to note that while there may be no net force acting on an object, there may still be individual forces acting on it that cancel each other out. For example, an object on a level surface may experience a force due to gravity pulling it downwards and an equal and opposite force from the surface pushing it upwards, resulting in a net force of zero. However, in the absence of any forces acting on the object, there can be no displacement.

13. A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer

Answer:

Yes, the person holding a bundle of hay over his head for 30 minutes has done work.

Work is defined as the product of force and displacement in the direction of the force. In this case, the person is applying a force to hold the bundle of hay in place against the force of gravity, and there is a displacement of the bundle of hay as it is held in place for 30 minutes. Therefore, work is being done by the person holding the bundle of hay.

Even though the bundle of hay is not moving, the person is still doing work because he is exerting a force over a distance. This work results in the person's muscles doing mechanical work and expending energy, which is why he gets tired.

14. An electric heater is rated 1500 W. How much energy does it use in 10 hours?

Answer:

Given, the electric heater is rated 1500 W.

The power of the electric heater is given by P = 1500 W.

The time for which it is used is given by t = 10 hours.

We know that the energy used by an electrical appliance is given by the product of its power and the time for which it is used, i.e.,

Energy used = Power × Time

Substituting the given values, we get:

Energy used = 1500 W × 10 hours

Energy used = 15000 Wh (Watt-hour)

Therefore, the electric heater will use 15000 Watt-hours of energy in 10 hours.

15. Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of

energy?

Answer:

When we draw a pendulum bob to one side and release it, the bob starts oscillating back and forth. Let's consider the energy changes that occur during the motion of the pendulum bob.

At the highest point of the oscillation, the bob is at its maximum height, and therefore has maximum potential energy. As the bob moves down, it loses potential energy, but gains kinetic energy. At the lowest point of the oscillation, the bob has maximum kinetic energy, but no potential energy.

As the bob swings back up, it loses kinetic energy and gains potential energy, and the cycle repeats. This continues until the pendulum eventually comes to rest due to frictional forces acting against the motion of the pendulum bob, such as air resistance and friction in the pivot point.

When the pendulum comes to rest, all of its energy is converted into thermal energy due to the frictional forces acting against it. This thermal energy is dissipated into the surroundings, and the pendulum system no longer has any usable energy.

This process does not violate the law of conservation of energy. The total amount of energy in the system (pendulum + Earth) remains constant throughout the motion of the pendulum. Although energy is continually being transformed between potential and kinetic forms during the motion of the pendulum, the total energy of the system remains the same. The eventual loss of energy due to frictional forces is simply a conversion of the usable energy of the system into unusable thermal energy, in accordance with the law of conservation of energy.

16. An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?

Answer:

In order to bring an object of mass, m, to rest from a velocity, v, a force must be applied in the direction opposite to the velocity until the object comes to rest. The work done by this force can be calculated as follows:

The initial kinetic energy of the object is given by:

K.E. = (1/2)mv^2

When the object is brought to rest, its final kinetic energy is zero.

The work done on the object is equal to the change in kinetic energy, which is given by:

Work = K.E. (final) - K.E. (initial)

= 0 - (1/2)mv^2

 = -(1/2)mv^2

Therefore, the work that needs to be done on the object to bring it to rest is equal to -(1/2)mv^2. Note that the negative sign indicates that work is done against the direction of motion of the object.

17. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?

Answer:

First, we need to convert the velocity of the car from km/h to m/s:

60 km/h = (60 x 1000 m) / (60 x 60 s) = 16.67 m/s

The kinetic energy of the car can be calculated using:

K.E. = (1/2)mv^2

where m is the mass of the car and v is its velocity. Substituting the given values, we get:

K.E. = (1/2) x 1500 kg x (16.67 m/s)^2 = 208416.675 J

In order to bring the car to a stop, an equal amount of work needs to be done against the direction of its motion. Therefore, the work required to stop the car is 208416.675 J

18. In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow.

Observe the diagrams carefully

and state whether the work done by the force is negative, positive or zero

Answer:

We know that, work done = force x displacement x cos theta

lets add value above

(a)When the direction of force acting on the block is perpendicular to the direction of displacement

case 1: Force x displacement x cos 90

         = Force x displacement x 0

         = 0

Hence, work done is Zero.

(b)When the direction of force acting on the block and the direction of displacement is same

Case 2:Force x displacement x cos 0

         =Force x displacement x 1

         =Force x displacement

Hence, work done is positive.

(c)When the direction of force acting on the block is opposite to the direction of displacement

Case 3 :Force x displacement x cos 180

         =Force x displacement x -1

Hence, work done is negative.

19. Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?  

Answer:

Yes, it is possible for an object to have zero acceleration even when several forces are acting on it. This is because the net force acting on the object may be zero. In other words, the vector sum of all the forces acting on the object may cancel out each other, resulting in no net force acting on the object.  

If there is no net force acting on the object, then according to Newton's second law of motion, the acceleration of the object would be zero.

Thus, it is important to note that acceleration of an object depends not only on the magnitude of the forces acting on it, but also on the direction of those forces and their interactions with other forces.   =Force x displacement

20. Find the energy in kW h consumed in 10 hours by four devices of power 500 W each.

Answer:

The total power consumed by four devices of power 500 W each is:

P = 4 × 500 W = 2000 W

We can convert this to kilowatts (kW) by dividing by 1000:

P = 2000 W ÷ 1000 = 2 kW

The energy consumed (E) in kilowatt-hours (kWh) can be calculated using the formula:

E = P × t

where P is the power in kW, and t is the time in hours.

Here, P = 2 kW and t = 10 hours.

Substituting these values into the formula, we get:

E = 2 kW × 10 hours = 20 kWh

Therefore, the energy consumed in 10 hours by four devices of power 500 W each is 20 kWh.

Alternate method:

E = P × t

where P is the power of one device in kW, and t is the time in hours.

Here, the power of one device is 500 W ÷ 1000 = 0.5 kW (since we need to convert watts to kilowatts).

Substituting the values of P = 0.5 kW and t = 10 hours, we get:

E = 0.5 kW × 10 hours = 5 kWh

Therefore, the energy consumed by one device in 10 hours is 5 kWh.

To find the total energy consumed by four devices, we can multiply this value by 4:

Total energy consumed by four devices = 5 kWh/device × 4 devices = 20 kWh

So, as we calculated earlier, the total energy consumed by four devices in 10 hours is 20 kWh.

21. A freely falling object eventually stops on reaching the ground. What happenes to its kinetic energy?

Answer:

When a freely falling object reaches the ground, its kinetic energy is transferred to other forms of energy such as heat, sound, and deformation of the object and the surface it hits.

So, the kinetic energy of the freely falling object is not conserved, but instead, it is converted into other forms of energy during the process of impact.

Important Questions

1. What is sound and how is it produced?

Answer:

Sound is a form of energy that is produced by vibrating objects. When an object vibrates, it causes the particles in the surrounding medium (usually air) to vibrate, which in turn creates a series of compressions and rarefactions. These compressions and rarefactions propagate through the medium as sound waves.

2. Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.  

Answer:

When an object vibrates, it creates a series of compressions and rarefactions in the surrounding air. The compressions are regions where the air molecules are pushed closer together, while the rarefactions are regions where the air molecules are spread out. This pattern of compressions and rarefactions propagates through the air as a sound wave.  

Sketch:

The diagram below shows how this process lines represent areas of low air pressure (rarefactions). As the sound wave moves through the air, it causes the particles in the air to vibrate back and forth in the direction of the wave.

3. Cite an experiment to show that sound needs a material medium for its propagation.

Answer:

An experiment to show that sound needs a material medium for its propagation is to place a ringing alarm clock inside a bell jar with all the air pumped out. If the bell jar is completely evacuated of air, the sound of the alarm clock will not be heard because there will be no air particles to vibrate and carry the sound waves.

4. Why is sound wave called a longitudinal wave?

Answer:

Sound waves are called longitudinal waves because the particles of the medium (usually air) vibrate in the same direction as the wave is moving. In contrast, transverse waves (such as light waves) have particles that vibrate perpendicular to the direction of wave propagation

5. Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?

Answer:

The characteristic of the sound that helps you to identify your friend by his voice while sitting with others in a dark room is the pitch. Pitch is determined by the frequency of the sound wave, and each person's voice has a unique pitch based on the size and shape of their vocal cords.

6. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?

Answer:

Flash and thunder are produced simultaneously, but thunder is heard a few seconds after the flash is seen because light travels much faster than sound. The speed of light is about 300,000,000 meters per second, while the speed of sound is about 340 meters per second.  

Hence, the light from the flash reaches our eyes almost instantly, while the sound of the thunder takes several seconds to travel the distance to our ears. By counting the number of seconds between the flash and the thunder, we can estimate how far away the lightning strike was.

7. A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 m s–1 .

Answer:

The wavelength of a sound wave is given by the formula λ = v/f,  

where λ is the wavelength, v is the speed of sound in air (344 m/s), and f is the frequency of the sound wave.  

For a frequency of 20 Hz, the wavelength is λ = 344/20 = 17.2 m.  

For a frequency of 20 kHz, the wavelength is λ = 344/20000 = 0.0172 m or 1.72 cm.

8. Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.

Answer:

The speed of sound in air is approximately 344 m/s, while the speed of sound in aluminium is approximately 6420 m/s.  

Let L be the length of the aluminium rod, and let t1 and t2 be the times taken by the sound wave to reach the second child in air and in aluminium, respectively. Then we have:

t1 = L/344

t2 = L/6420

The ratio of times taken is:

t1/t2 = (L/344)/(L/6420) = 18.67

Hence, the sound wave travels approximately 18.67 times faster in aluminium than in air.

9. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute? 

Answer:

The frequency of a source of sound is 100 Hz. One vibration of the source produces one complete wave cycle. Therefore, in one second, the source produces 100 wave cycles. In one minute (60 seconds), the source produces 6000 wave cycles. Therefore, the source vibrates 6000 times in a minute.

10. Does sound follow the same laws of reflection as light does? Explain.

Answer:

Yes, sound follows the same laws of reflection as light does. The angle of incidence is equal to the angle of reflection, and the incident, reflected, and normal lines all lie in the same plane. When sound waves reflect off a smooth surface, they obey the law of reflection just like light waves do. However, there are some differences between sound and light waves when it comes to reflection. For example, sound waves are diffracted more easily than light waves, which means that they can bend around corners and spread out more as they reflect off surfaces.

11. When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same.

Do you hear echo sound on a hotter day?

Answer:

Yes, you may not hear an echo sound on a hotter day because the speed of sound in air increases with the increase in temperature. The speed of sound is directly proportional to the square root of the absolute temperature of the medium. \

Hence, on a hotter day, the speed of sound is higher, which means that the echo will take less time to reach the observer and will be less distinct

12. Give two practical applications of reflection of sound waves.

Answer:

Two practical applications of reflection of sound waves are:

a) In megaphones and other similar devices, sound waves are reflected in a particular direction to amplify the sound. The device is designed to reflect the sound waves in such a way that they converge at a single point, making the sound louder.

b) In sonar technology, sound waves are used to locate objects underwater. The sound waves are emitted from a source and are reflected off the object, and the reflected sound waves are detected by a receiver. By analyzing the time taken for the sound waves to travel to and from the object, its location and distance can be calculated.

13. A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top?

Given, g = 10 m s–2 and speed of sound = 340 m s–1.

Answer:

The time taken by the stone to reach the pond can be calculated using the formula:

t = sqrt(2h/g)

where h is the height of the tower and g is the acceleration due to gravity.

Substituting the values, we get:

t = sqrt(2 × 500/10) = sqrt(100) = 10 s

The speed of sound is 340 m/s, and the distance travelled by the sound is 500 m. Therefore, the time taken by the sound to reach the top of the tower is:

t = d/v = 500/340 = 1.47 s

Therefore, the splash will be heard at the top of the tower after 10 + 1.47 = 11.47 s.

14. A sound wave travels at a speed of 339 m s–1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?

Answer:

The speed of the wave is given by the formula  

v = fλ,  

where v is the speed of the wave, f is the frequency, and λ is the wavelength. Substituting the values,  

we get:

339 = f × 0.015

f = 339/0.015 = 22,600 Hz

The frequency of the wave is 22,600 Hz, which is in the ultrasonic range and is not audible to humans.  

15. What is reverberation? How can it be reduced?

Answer:

Reverberation is the persistence of sound in an enclosed space due to multiple reflections of sound waves.  

It causes a prolonged sound effect that can distort the clarity of the original sound. Reverberation can be reduced by adding sound-absorbing materials such as curtains, carpets, and foam panels to the room, which can absorb the sound waves and prevent multiple reflections.

16. What is loudness of sound? What factors does it depend on?

Answer:

Loudness of sound is the subjective response of the ear to the intensity of sound waves. It depends on factors such as the amplitude or intensity of the sound waves, the frequency of the sound waves, the duration of the sound, and the distance of the source of the sound.  

The unit of loudness is decibel (dB).

17. Explain how bats use ultrasound to catch a prey.

Answer:

Bats use ultrasound to catch their prey. They emit high-frequency sound waves from their mouth or nose, which bounce back off objects in their path, including insects, buildings, and other obstacles. These sound waves are then received by the bat's ears, which are highly sensitive to high-frequency sounds.  

The bat can determine the distance, direction, and speed of the object based on the time taken for the sound waves to bounce back and the frequency and intensity of the returning waves.

18. How is ultrasound used for cleaning?

Answer:

Ultrasound is used for cleaning by creating high-frequency sound waves that produce tiny bubbles in a cleaning solution.  

These bubbles then implode with great force, generating high pressure that dislodges dirt and other contaminants from the surface being cleaned.

19. Explain the working and application of a sonar.

Answer:

Sonar stands for Sound Navigation And Ranging. It is a technology that uses sound waves to locate and determine the distance, speed, and direction of objects underwater.  

A sonar device emits high-frequency sound waves that travel through water and bounce back when they hit an object. The device then detects the returning sound waves and calculates the distance and direction of the object based on the time taken for the waves to travel to and from the object.

20. A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m.

Answer:

The speed of sound in water can be calculated using the formula:

Speed = Distance/Time

Substituting the values, we get:

Speed = 3625/5 = 725 m/s

Therefore, the speed of sound in water is 725 m/s.

21. Explain how defects in a metal block can be detected using ultrasound.

Answer:

Defects in a metal block can be detected using ultrasound by transmitting high-frequency sound waves through the block and detecting the reflected waves.  

If there is a defect such as a crack or a void in the metal, the sound waves will be reflected differently, indicating the presence and location of the defect.

22. Explain how the human ear works.

Answer:

The human ear consists of three main parts: the outer ear, the middle ear, and the inner ear. The outer ear consists of the pinna, the ear canal, and the eardrum.  

It collects sound waves and directs them to the eardrum, which vibrates in response to the sound waves.  

The middle ear contains three tiny bones called the hammer, anvil, and stirrup, which amplify the vibrations and transmit them to the inner ear.  

The inner ear contains the cochlea, which is filled with fluid and tiny hair cells that convert the vibrations into electrical signals that are transmitted to the brain via the auditory nerve. The brain then interprets these signals as sound.

Important Questions

1. How many times did you fall ill in the last one year? What were the illnesses?

Answer:

In last one year I fall ill twice,  

First when I caught flu and started sneezing and Second, when when I had stomach ache and fever.

(a) Think of one change you could make in your habits in order to avoid any of/most of the above illnesses.  

Answer:

To avoid any or most of the diseases one thing that I can do is getting vaccinated against flu and other viral diseases and secondly, we can maintain personal hygiene by washing hands regularly, eating a balanced diet and exercising regularly.

b) Think of one change you would wish for in your surroundings in order to avoid any of/most of the above illnesses.  

Answer:

In order to avoid any of/most of the above illness, one change I would wish in my surrounding would include:

Proper sanitation and waste management systems in the surroundings to prevent the spread of infectious diseases.

Availability of clean and safe drinking water to prevent waterborne illnesses.

Proper ventilation and air filtration systems to reduce exposure to indoor air pollutants and prevent respiratory illnesses.

Availability of green spaces and parks for physical activity and stress reduction.

Reduction in pollution levels to prevent respiratory illnesses and other health problems.

Availability of healthy food options in the surroundings to promote a balanced diet and prevent chronic illnesses such as obesity and diabetes.

2. A doctor/nurse/health-worker is exposed to more sick people than others in the community. Find out how she/he avoids getting sick herself/himself.

Answer:

Some measures that healthcare workers can take to avoid getting sick themselves while being exposed to sick people include:

Wearing personal protective equipment such as gloves, masks, and gowns.

Practicing proper hand hygiene by washing hands frequently with soap and water or using alcohol-based hand sanitizers.

Maintaining a safe distance from sick people whenever possible.

Staying up-to-date with vaccinations to prevent getting infected with vaccine-preventable diseases.

Taking necessary precautions when handling and disposing of medical waste.

3. Conduct a survey in your neighbourhood to find out what the three most common diseases are. Suggest three steps that could be taken by your local authorities to bring down the incidence of these diseases.

Answer:

Three most common diseases in my neighbourhood are Flu, Diarrhoea and respiratory problems.

Three steps that local authorities could take to bring down the incidence of the three most common diseases in a neighbourhood include:

Offering free or subsidized medical check-ups and treatment to people who cannot afford it this will prevent their heath condition getting worse further.

Launching awareness campaigns to educate people about the causes, symptoms, and prevention measures of communicable diseases like corona (H1N1).

Providing proper sanitation facilities such as garbage collection and disposal systems to prevent the spread of infectious bacterial diseases.

4. A baby is not able to tell her/his caretakers that she/he is sick. What would help us to find out

Answer:

(a) that the baby is sick?

(a) Some signs that can indicate that a baby is sick include a fever, irritability, loss of appetite, changes in sleeping patterns, and unusual crying or fussiness.

(b) what is the sickness?  

(b) A medical examination by a doctor can help determine the sickness but it the first go we can make it out by observing which part of the body is baby is touching frequently.

5. Under which of the following conditions is a person most likely to fall sick?  

(a) when she is recovering from malaria.  

(b) when she has recovered from malaria and is taking care of someone suffering from chicken-pox.  

(c) when she is on a four-day fast after recovering from malaria and is taking care of someone suffering from chicken-pox.

Why?  

Answer:

The person is most likely to fall sick in option (c) When she is on a four-day fast after recovering from malaria and taking care of someone suffering from chicken-pox.  

This is because fasting weakens the immune system, making the person more susceptible to infections, and taking care of a sick person increases the risk of exposure to the pathogen causing chicken-pox.

6. Under which of the following conditions are you most likely to fall sick?  

(a) when you are taking examinations.  

(b) when you have travelled by bus and train for two days.  

(c) when your friend is suffering from measles.

Why?

The person is most likely to fall sick in option (b) when they have traveled by bus and train for two days.  

This is because traveling in close proximity to other people in public transportation increases the risk of exposure to infectious agents such as viruses and bacteria.

Important Questions

1. Why is the atmosphere essential for life?

Answer:

The atmosphere is essential for life because it provides oxygen for breathing, regulates temperature, protects from harmful radiation from the sun, and helps in the water cycle which is necessary for life on earth.

2. Why is water essential for life?

Answer

Water is essential for life because it is the main component of cells and helps in various biological processes such as digestion, circulation, and excretion. It also regulates body temperature, acts as a lubricant and a solvent for many substances.

3. How are living organisms dependent on the soil? Are organisms that live in water totally independent of soil as a resource?

Answer:

Living organisms are dependent on the soil for various reasons such as obtaining nutrients, water, and support for plant growth. Soil is also home to a variety of microorganisms which are essential for various biological processes.  

Organisms that live in water, such as fish, may not depend on soil as a resource, but they still require nutrients and other resources from their aquatic environment.

4. You have seen weather reports on television and in newspapers. How do you think we are able to predict the weather?

Answer:

Weather predictions are made using a combination of observation, computer models, and historical data. Meteorologists use various tools and instruments to measure atmospheric variables such as temperature, pressure, and humidity.  

This data is then used to create computer models which predict future weather patterns based on historical trends and current conditions. However, weather can still be unpredictable and may change rapidly, so predictions are not always accurate.

5. We know that many human activities lead to increasing levels of pollution of the air, water-bodies and soil. Do you think that isolating these activities to specific and limited areas would help in reducing pollution?

Answer:

Isolating human activities to specific and limited areas may help in reducing pollution to some extent. For example, setting up industries in specific areas away from residential areas and implementing proper waste disposal systems can help reduce pollution.  

However, completely isolating activities is not always feasible, and it may not be enough to solve the problem of pollution. It is important to adopt sustainable practices and reduce the use of harmful substances to prevent pollution.

6. Write a note on how forests influence the quality of our air, soil and water resources.

Answer:

Forests have a significant impact on the quality of our air, soil, and water resources. They act as natural air filters by absorbing carbon dioxide and other pollutants, and releasing oxygen into the atmosphere.  

Trees also help in reducing soil erosion by holding the soil in place with their roots. The forest floor helps to store and filter water, allowing it to seep slowly into the ground, and replenish groundwater resources.

Additionally, forests are home to a diverse range of plant and animal species, which contribute to the overall health and balance of ecosystems. Therefore, preserving and protecting forests is essential for maintaining the quality of our natural resources.

Important Questions

1. Explain any one method of crop production which ensures high yield.

Answer:

One method of crop production which ensures high yield is the use of high-yielding varieties (HYVs) of crops. HYVs are developed through selective breeding, and are designed to produce more grain per unit of land than traditional crop varieties. These varieties are also resistant to pests and diseases, and can withstand adverse weather conditions.

2. Why are manure and fertilizers used in fields?

Answer:

Manure and fertilizers are used in fields to provide essential nutrients to plants. Manure is a natural fertilizer made from animal waste, and contains a mix of nitrogen, phosphorus, and potassium, as well as other trace elements. Fertilizers, on the other hand, are synthetic compounds made up of specific nutrients that plants need, such as nitrogen, phosphorus, and potassium. These nutrients help to improve plant growth, increase crop yields, and improve soil fertility.

3. What are the advantages of inter-cropping and crop rotation?

Answer:

The advantages of inter-cropping and crop rotation are as follows:

Inter-cropping: In inter-cropping, two or more crops are grown together on the same land at the same time. This can help to increase crop yields, as the different crops complement each other in terms of nutrient requirements and growing conditions.  

For example, legumes like peas or beans can fix nitrogen in the soil, which can benefit other crops like corn or wheat. Inter-cropping also helps to reduce soil erosion and pest damage.

Crop rotation: In crop rotation, different crops are grown in the same land in a specific sequence over time. This helps to improve soil fertility and reduce soil-borne pests and diseases.  

For example, legumes can be grown in one season to fix nitrogen in the soil, followed by a cereal crop like corn or wheat in the next season to benefit from the improved soil fertility.

4. What is genetic manipulation? How is it useful in agricultural practices?

Answer:

Genetic manipulation refers to the deliberate alteration of an organism's genetic makeup to achieve desired traits.  

In agriculture, genetic manipulation is used to produce crops with desirable traits such as increased resistance to pests and diseases, improved yield, and better nutrient content.  

This is achieved through genetic engineering techniques such as gene splicing, where a desired gene is inserted into the plant's DNA.  

Genetic manipulation has been useful in developing crops that are more resistant to pests and diseases, and can withstand harsh weather conditions.  

However, there are also concerns about the safety and environmental impact of genetically modified crops, and their potential long-term effects on ecosystems and human health.

5. How do storage grain losses occur?  

Answer:

Storage grain losses occur due to various factors such as pest infestation, fungal or bacterial infections, moisture, and improper storage conditions. Pests like rodents, weevils, and moths can infest the grain and cause damage. Fungi and bacteria can also infect the grain, leading to spoilage. High moisture content and improper storage conditions can lead to mold growth, which can cause spoilage or produce harmful mycotoxins.

6. How do good animal husbandry practices benefit farmers?

Answer:

Good animal husbandry practices benefit farmers in several ways. They help to improve the health and productivity of the animals, increase the quality and quantity of animal products, and reduce the risk of disease transmission. Good practices include providing animals with proper nutrition and water, maintaining clean and hygienic living conditions, and providing appropriate medical care. These practices can help to reduce the incidence of disease and improve the overall health and welfare of the animals, leading to increased productivity and profitability for the farmer.

7. What are the benefits of cattle farming?

Answer:

The benefits of cattle farming include the production of milk, meat, and other dairy products, as well as the use of cattle as draught animals for agricultural work. Cattle also provide manure, which can be used as fertilizer for crops, and their grazing activities can help to maintain grasslands and prevent soil erosion. Additionally, cattle farming can provide employment opportunities for rural communities, and can help to support local economies.

8. For increasing production, what is common in poultry, fisheries and bee-keeping?

Answer:

One common method for increasing production in poultry, fisheries, and bee-keeping is through selective breeding. By selectively breeding individuals with desirable traits, such as high productivity, disease resistance, or adaptation to local conditions, farmers can produce more productive and profitable animals or crops.

9. How do you differentiate between capture fishing, mariculture and aquaculture?

Answer:

Capture fishing involves catching fish from the wild in natural bodies of water such as oceans, lakes, or rivers.  

Mariculture involves raising fish and other seafood in enclosed areas in the ocean, while aquaculture involves raising fish and other seafood in tanks or ponds on land.  

The key difference between mariculture and aquaculture is the location of the farming activity, with mariculture taking place in natural bodies of water and aquaculture taking place on land.

Important Questions