Important Questions For You!

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CBSE Class 10 Science highlights important concepts which include chemical reactions, principles, equations and a lot more. TopperLearning provides study materials for CBSE Class 10 Science which offers a systematic way to prepare for the final examination. Take our study notes, for example.

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Deleted Portion (Theory)

  • Under Unit I: Chemical Substances - Nature and Behaviour
    o   Metals and Non-metals: Basic Metallurgical processes; Corrosion and its prevention
    o   Carbon and its Compounds: Nomenclature of carbon compounds containing functional groups (halogens, alcohol, ketones, aldehydes, alkanes and alkynes), difference between saturated hydro carbons and unsaturated hydrocarbons.
    o   Chemical properties of carbon compounds (combustion, oxidation, addition and substitution reaction). Ethanol and Ethanoic acid (only properties and uses), soaps and detergents.
  •  Under Unit II: World of Living
    o   Control and co-ordination in animals and plants: Tropic movements in plants; Introduction of plant hormones; Control and co-ordination in animals: Nervous system; Voluntary, involuntary and reflex action; Chemical co-ordination: animal hormones.
    o   Heredity and Evolution: Basic concepts of evolution.
  • Under Unit III: Natural Phenomena
    o   The Human Eye and the Colourful World: Functioning of a lens in Human eye, defects of vision and their corrections, applications of spherical mirrors and lenses.
  • Under Unit IV: Effects of Current
    o   Magnetic Effects of Electric Current: Electric Generator, Direct current.
    o   Alternating current: frequency of AC. Advantage of AC over DC. Domestic electric circuits.
  • Under Unit V: Natural Resources
    o   Sources of energy: Different forms of energy, conventional and nonconventional sources of energy: Fossil fuels, solar energy; biogas; wind, water and tidal energy; Nuclear energy. Renewable versus non-renewable sources of Energy.


Chapter 1: Chemical Reactions and Equations

1. Why are decomposition reactions called the opposite of combination reactions?                                                            [1M]

2. When potassium chlorate (KClO3) is heated in the presence of manganese dioxide catalyst, it decomposes to form potassium chloride and oxygen gas. Represent this in the form of a balanced chemical equation.                                                       [1M]

3. A solution of potassium chloride when mixed with silver nitrate solution forms an insoluble white substance. Write the chemical reaction involved and mention the type of chemical reaction.                                                                                          [3M]

4. Equal lengths of Mg ribbon are taken in test tubes A and B. Hydrochloric acid is added to test tube A, while acetic acid is added to test tube B. In which case the reaction would occur more vigorously and why? Write the chemical equations for reactions in test tubes A and B.                                                                                                                                                                     [3M]

5. Describe an activity to show the decomposition reaction of ferrous sulphate in the laboratory.                                       [3M]

6. A solution of substance 'X' is used for white washing.                                                                                                 [3M]

  1.  Name substance 'X' and write its formula.
  2.  Write the reaction of substance 'X' named in (i) above with water.
  3.  Write the balanced equation for the following chemical reaction:
      Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride

7.                                                                                                                                                                                   [5M]
    a. Write chemical equations for the reactions taking place when 

  1. Zinc sulphide is heated in air.
  2. Zinc carbonate is calcined.
  3. Manganese dioxide is heated with aluminium powder.

    b. 

  1. What happens when a strip of lead metal is placed in a solution of copper chloride? Write the balanced chemical equation for  the eaction along with the colour changes observed during the reaction.
  2.  What are precipitation reactions? Give one example of a precipitation reaction.

8.                                                                                                                                                                                 [5M]

  1. A water-insoluble calcium compound (A) on reacting with dil. H2SO4 released a colourless and odourless gas (B) with brisk    effervescence. When gas (B) was passed through lime water, the lime water turned milky and again formed compound A. Identify A and B, and write the chemical equations for the reactions involved. 
  2. A brown substance 'X' on heating in air forms a compound 'Y'. When hydrogen gas is passed over 'Y', it changes to 'X' again.
    1. Name substances 'X' and 'Y'.
    2. Name the processes occurring during the two changes.
    3. Write the chemical equations involved.


Chapter 2: Acids, Bases and Salts

1. Fresh milk has a pH of 6. How do you think pH will change as it turns into curd? Explain.                                           [1M]

2. Classify the following into acidic oxides and basic oxides: Na2O, SO2, MgO, CO2.                                                       [1M]

3.                                                                                                                                                                                [3M]

  1. Write the chemical name and formula of washing soda. How is it prepared? Write the chemical equation of the reaction.
  2. Why should Plaster of Paris be stored in a moisture-proof container?
  3. If we have hydrochloric acid and acetic acid in equal concentration, which will be the stronger acid and why?

4. Define 'water of crystallisation'.                                                                                                                            [3M]
    Give two examples of substances having water of crystallisation.
    Write their chemical formulae.

5.                                                                                                                                                                               [3M]

  1. Name the compound which is obtained from baking soda and is used to remove the permanent hardness of water. 
  2. Write its chemical formula.
  3. What happens when it is recrystallised from its aqueous solution?

6.                                                                                                                                                                               [3M]

  1. What change will you observe in the colour of red litmus paper when it is dipped into a solution of sodium sulphate? Give reason to explain your observation. 
  2. A bottle filled up to the brim with concentrated sulphuric acid is left open in the atmosphere by mistake. Will there be any change in the level of liquid? Explain your answer with reasons.

7.                                                                                                                                                                               [5M]
   a. Write word equations and balanced equations for the reaction taking place when  

  1. Dilute sulphuric acid reacts with zinc granules
  2. Dilute hydrochloric acid reacts with magnesium ribbon

   b. What is meant by 'hydrated' and 'anhydrous' salts? Explain with examples.
   c. Give two important uses of washing soda.

8.                                                                                                                                                                           [5M]
  a. Write the chemical equation for the preparation of 

  1.   Bleaching powder
  2.   Plaster  of Paris
  3.   Caustic soda

  b. What is the chlor-alkali process? Give two uses of NaOH obtained from this process.

9. Give suitable reasons for the following statements:                                                                                                 [5M]

  1.  Rain water conducts electricity, whereas distilled water does not.
  2.  We feel a burning sensation in the stomach when we overeat.
  3.  On rubbing a tarnished copper vessel with lemon or tamarind juice, it begins to shine again.
  4.  A solution of sodium carbonate is not acidic.
  5.  Dry ammonia has no effect on litmus paper, whereas a solution of NH3 in water has.


Chapter 3: Metals and Non-Metals

1. What happens when iron nails are kept in an aqueous solution of CuSO4?                                                                 [1M]

2. Name a non-metal which is lustrous and a metal which is non-lustrous.                                                                    [1M]

3. What are amphoteric oxides? Give two examples. Element X on reaction with oxygen forms a dioxide XO2, which turns blue litmus red. Is element X a metal or a non-metal?                                                                                                                  [3M]

4.                                                                                                                                                                                 [3M]

  1. What are the two main allotropes of carbon? 
  2. Distinguish these two allotropes on the basis of hardness and electrical conduction.

5. What happens when                                                                                                                                              [3M]

  1. Zinc reacts with copper sulphate
  2. Magnesium reacts with HCl
  3. Sodium reacts with water

6.                                                                                                                                                                                 [5M]

  1. Write the electronic configurations of sodium and chlorine.
  2. Show the formation of sodium chloride from sodium and chlorine by the transfer of electrons.
  3. State the type of bonding formed.
  4. List two properties of the compounds formed by this bonding.


Chapter 4: Carbon and its Compounds
1. Write the name of the following compound. Name the acid and alcohol from which it might be prepared.                      [1M]

2. Name the functional group present in each of the following organic compounds:                                                        [1M]

  1. CH3COCH3
  2. C2H5COOH

3. What is meant by a homologous series of carbon compounds? Classify the following carbon compounds into two homologous series and name them. [3]
C3H4, C3H6, C4H6, C4H8, C5H8, C5H10 

4.                                                                                                                                                                                [5M]

  1. State two properties of carbon which lead to a large number of carbon compounds.
  2.  Explain isomerism. State any four characteristics of isomers. Draw the structures of possible isomers of butane, C4H10.       Also,explain why we cannot have isomers of the first three members of the alkane series.


Chapter 5: Periodic Classification of Elements

1. How were the positions of cobalt and nickel resolved in the modern periodic table?                                                    [1M]

2. What are lanthanides and actinides?                                                                                                                      [1M]

3. How does the metallic character of elements change along a period of the periodic table from left to right and why?     [2M]

4. How does the modern periodic law justify one position for isotopes?                                                                          [2M]

5. State Mendeleev’s periodic law. Which group of elements was missing from Mendeleev’s original periodic table? Besides gallium, which two other elements have since been discovered for which Mendeleev had left gaps in his periodic table?                 [3M]

6. How could the modern periodic law remove various anomalies of Mendeleev’s periodic table? Explain with examples.    [3M]

7. Atoms of eight elements—A, B, C, D, E, F, G and H—have the same number of electronic shells but different number of electrons in their outermost shell. Elements A and G combine to form an ionic compound which can also be extracted from sea water. This ionic compound is added to a small amount of almost all vegetables and dishes during cooking. Oxides of elements A and B are basic in nature, while those of elements E and F are acidic. However, the oxide of element D is almost neutral. Based on the above information, answer the following questions:                                                                                                              [5M]

  1. To which group or period of the periodic table do these elements belong?
  2. What would be the nature of the compound formed by a combination of elements B and F?
  3. Which two of these elements could definitely be metals?
  4. Which one of the eight elements is most likely to be found in the gaseous state at room temperature?
  5. If the number of electrons in the outermost shell of elements C and G are 3 and 7, respectively, write the formula of the      compound formed by the combination of C and G.

8. Explain:                                                                                                                                                              [5M]

  1. Larger the atomic size, more metallic is the element.
  2. Size of the atom changes when it loses or gains electrons.
  3. K is more reactive than Li.
  4. Electronegativity of Cl is higher than S.
  5. Group 17 elements are non-metals, while Group 1 elements are metals.

9.

Select from the table:

  1.  Which is the most electronegative?
  2.  How many valence electrons are present in G?
  3.  Write the formula of the compound between B and H.
  4.  In the compound between F and J, what type of bond will be formed?
  5.  Draw the electron dot structure for the compound formed between C and K.


Chapter 6: Life Processes

1. Why do fish die when taken out of water?                                                                                                                 [1M]

2. Why is diffusion insufficient to meet the oxygen requirements of multicellular organisms?                                             [1M]

3. How do guard cells regulate the opening and closing of stomata?                                                                                [2M]

4. What advantage over an aquatic organism does a terrestrial organism have with regard to obtaining oxygen for respiration? [2M]

5. Why is small intestine longer in herbivores as compared to carnivores?                                                                        [2M]

6. Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds?                                        [3M]

7. What are the different ways in which glucose is oxidised to provide energy in various organisms?                                   [3M]

8. How are fats digested in our body? Where does this process take place?                                                                       [3M]

9.                                                                                                                                                                               [5M]

  1. Draw a well-labelled diagram of the human digestive system. 
  2. Describe the role of the following in the process of digestion:
    (i) Bile (ii) Salivary amylase (iii) HCl

10.                                                                                                                                                                            [5M]

  1. Define excretion.
  2. Name the basic filtration unit present in the kidney.
  3. Draw the excretory system in human beings and label the organs of the excretory system which perform the following        functions:
  1.  forms urine
  2.  is a long tube which collects urine from the kidney
  3.  stores urine until it is passed out

11.                                                                                                                                                                            [5M]

  1. Mention any two components of blood. 
  2. Trace the movement of oxygenated blood in the body.
  3. Write the function of valves present in between the atria and ventricles.
  4.  Write one structural difference between the composition of arteries and veins.


Chapter 8: How do Organisms Reproduce?

1. Name one sexually transmitted disease each caused by bacterial infection and viral infection.                                        [1M]

2. Where does fertilisation take place in human females?                                                                                                [1M]

3. A couple wants to space the birth of their second child. Suggest one preventive method which could be observed?          [1M]

  1. By the husband
  2. By the wife for the same

4. Differentiate between self pollination and cross pollination.                                                                                        [2M]

5. What is the function of the Cowper's gland and prostate gland?                                                                                  [2M]

6. What is vegetative propagation? State two advantages and two disadvantages of this method.                                      [3M]

7. Name the process by which amoeba reproduces. Draw the various stages of its reproduction in a proper sequence.         [3M]

8. List any four methods of contraception used by humans. How does their use have a direct effect on the health and prosperity of a family?                                                                                                                                                                        [3M]

9. Mention the post fertilisation changes that occur in a flower.                                                                                        [3M]

10.                                                                                                                                                                                  [5M]

  1. List three differences between sexual and asexual types of reproduction.
  2. Explain why variations are observed in the offspring of sexually reproducing organisms.

11.                                                                                                                                                                                 [5M]

  1. Write the function of the following parts in the human female reproductive system:
     (i) Ovary (ii) Oviduct (iii) Uterus
  2. Describe in brief the structure and function of the placenta.

12.                                                                                                                                                                                [5M]

  1. Distinguish between cross-pollination and self-pollination. Mention the site and product of fertilisation in a flower.
  2. Draw a labelled diagram of a pistil showing the following parts: Stigma, Style, Ovary, Female Germ Cell


Chapter 9: Heredity and Evolution

1. Why is DNA copying necessary during reproduction?                                                                                                 [1M]

2. Mendel crossed a pure white recessive pea plant with a dominant pure red flowered plant. What will be the first generation F1 hybrids? [1]

3. A Mendelian experiment consisted of breeding pea plants bearing violet flowers with pea plant bearing white flowers. What will be the result in the F1 progeny?                                                                                                                                      [1M]

4. Why did Mendel choose pea plant for his experiment?                                                                                               [2M]

5. Only variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement? Why or why not?                                                                                                                                                       [3M]

6. State the three Mendel's laws of inheritance.                                                                                                           [3M]

7. Give the basic features of the mechanism of inheritance.                                                                                          [5M]

8.                                                                                                                                                                                 [5M]

  1. ‘Sex of children is determined by what they inherit from the father and not from the mother.’ Justify.
  2.  Explain the result of a monohybrid cross.


Chapter 10: Light - Reflection and Refraction

1. Which mirror is used as a rear-view mirror in vehicles?                                                                                            [1M]

2. If the radius of curvature of a spherical mirror is 60 cm, what is its focal length?                                                        [1M]

3. What is the nature and position of the image when an object is placed at the focus of a concave mirror?                      [1M]

4. If a magnification of -1 is to be obtained using a convex lens of focal length 6 cm, then the object must be placed        [1M]

  1. within 12 cm
  2. at 6 cm
  3. iat 12 cm
  4. beyond 12 cm

5. Draw a neat and labelled ray diagram when an object is placed in front of a convex mirror between infinity and the pole.  [3M]

6. What is the speed of light in a medium of refractive index 1.8 if its speed in air is 300000 km/s?                                     [3M]

7. Define refractive index. If light enters from air to glass having a refractive index 1.5, then calculate the speed of light in glass.                                                                                                                                                                                           [3M]

8. Describe the Cartesian sign convention for a spherical lens. Draw a neat and labelled diagram to illustrate the sign convention.                                                                                                                                                                                        [3M]

9. If a concave mirror has a focal length of 15 cm, find the two positions where an object can be placed to give, in each case, an image twice the height of the object.                                                                                                                             [3M]

10. Draw a diagram to represent a convex mirror. On this diagram, mark the principal axis, principal focus F and centre of curvature C if the focal length of a convex mirror is 5 cm. Also, comment on the characteristics of the image if an object is placed 15 cm from the mirror.                                                                                                                                                                  [5M]

11.                                                                                                                                                                                 [5M]

  1. Two lenses A and B have a power of +3 D and –4 D, respectively. What is the nature and focal length of each lens?
  2.  A combination of lens contains two converging lens of focal length 30 cm and 50 cm and a diverging lens of focal length 40 cm. Find the power and focal length of the combination.


Chapter 11: Human Eye and the Colourful World

1. Why does the Sun appear white at noon?                                                                                                                   [1M]

2. What is meant by dispersion of white light?                                                                                                                [1M]

3. In the formation of spectrum of white light by a prism:                                                                                               [2M]

  1. which colour is deviated least?
  2. which colour is deviated most? 

4. What is Tyndall effect? Explain with an example.                                                                                                         [3M]

5. Why does the sky appear blue during the day and red at sunset?                                                                                  [3M]

6. Draw and explain the dispersion of white light through a prism. Also, comment on the deviation, wavelength, frequency and speed of the dispersed light.                                                                                                                                                   [5M]

7. Write a short note on the formation of a rainbow with a neat and labelled diagram.                                                        [5M]


Chapter 12: Electricity

1. Define electric power and state its SI units.                                                                                                                 [1M]

2. The values of potential difference V applied across a resistor and the corresponding values of current I flowing in the resistor are given below:                                                                                                                                                                 [2M]

  1. What is the nature of the V-I graph plotted for the above values of potential difference and current?
  2. Which law is illustrated by such type of graph?

3. What are the factors on which the heating effect of electric current depend? Explain with a practical example.                    [3M]

4. State and explain Ohm’s law with a graph.                                                                                                                    [3M]

5. What is resistivity? State the factors on which it depends.                                                                                              [3M]

6. Resistors of 8 Ω and 12 Ω are connected in parallel to each other, while a resistor of 6 Ω is connected in series to a 6-V battery. Calculate:                                                                                                                                                                      [5M]

  1. Total resistance of the circuit
  2. Total current in the circuit
  3. Total potential difference across the 6-Ω resistor

Draw a neat and labelled circuit diagram for the same.

7. An electric iron draws 2.4 amperes of current from a 240-V source. Find                                                                        [3M]

  1. Resistance of the electric iron
  2.  Power consumed by the electric iron

8. Give reasons for the following:                                                                                                                                   [5M]

  1. Filament-type electric bulbs are not power efficient.
  2. Coils of heating devices are made of alloys rather than pure metals such as copper.
  3. An electric bulb is not filled with normal atmospheric air but is filled with argon or nitrogen.
  4. Metals such as copper and aluminium are used on a large scale for the transmission of electricity.
  5.  Parallel combination is preferred over series combination for connecting electric appliances in houses.

9.                                                                                                                                                                                    [5M]

  1. Derive the expression for heat produced due to current (I) flowing for a time period (t) through a resistor (R) having a        potential difference (v) across the terminals.
  2. Name the relation. List all the variables along with their SI units.
  3.  How much heat will an instrument of 14 W produce in half an hour if it is connected to a battery of 70 V? 


Chapter 13: Magnetic Effect of Electric Current

1. What is the effect on magnetic field strength produced at a point near a straight conductor if the electric current flowing through it increases?                                                                                                                                                                  [1M]

2. What were the observations made by Oersted in his experiment of current-carrying conductors?                                   [1M]

3. If a current-carrying conductor is kept in a magnetic field, it experiences a force. List the factors on which the direction of this force depends.                                                                                                                                                            [3M]

4. Explain Fleming’s right-hand rule and left-hand rule.                                                                                                   [3M]

5. What is a magnetic field? List characteristics of a magnetic field. If two magnetic field lines intersect at a point, what does that indicate?                                                                                                                                                                     [3M]

6. Define and explain the phenomenon of electromagnetic induction.                                                                               [3M]

7. A coil of insulated copper wire is connected to a galvanometer. What happens if a bar magnet is                                     [3M]

  1.  Pushed into the coil
  2.  Withdrawn from the hollow space of the coil
  3.  Held stationary inside the coil

8. Explain with a neat sketch the working of an electric motor.                                                                                         [5M]


Chapter 15: Our Environment

1. Why should biodegradable and non-biodegradable wastes be discarded in two separate dustbins?                                   [1M]

2. In the following food chain, 20,000 J of energy was available to plants. How much energy would be available to man in this chain?                                                                                                                                                                                    [1M]
   Plants → Sheep → Man

3. You being an environmentalist are interested in contributing towards the conservation of natural resources. List two activities that you can do on your own.                                                                                                                                              [1M]

4. Why is it necessary to ban the use of plastic bags?                                                                                                     [2M]

5. Give reason why a food chain cannot have more than four trophic levels.                                                                      [2M]

6. What are the adverse effects of combustion of fossil fuels on the environment? List any two steps you would suggest to minimise environmental pollution caused by burning of fossil fuels.                                                                                                [3M]

7. How is ozone formed in the upper atmosphere? Why is damage to the ozone layer a cause of concern to us? What causes this damage?                                                                                                                                                                     [3M]

8. Explain the phenomenon of ‘biological magnification’. How does it affect organisms belonging to different trophic levels, particularly the tertiary consumers?                                                                                                                                                [3M]

9. What is a food chain? Why is the flow of energy in an ecosystem unidirectional? Explain briefly.                                      [3M]

10.                                                                                                                                                                                  [3M]

  1. Why should national parks be allowed to remain in their pristine form?
  2.  Why is the reuse of materials better than recycling?

11. Explain some harmful effects of agricultural practices on the environment.                                                                  [5M]

 

ASSERTION – REASONING QUESTIONS

Two statements are given - one labelled Assertion (A) and the other labelled Reason (R).
Select the correct answer to these questions from the codes (i), (ii), (iii) and (iv) given below.

  1. Both A and R are true, and R is the correct explanation of the assertion.
  2. Both A and R are true, but R is not the correct explanation of the assertion.
  3. A is true, but R is false.
  4. A is false, but R is true.


Chapter 1: Chemical Reactions and Equations

1. Assertion: Aluminium foil is commonly used for packing food preparations.
Reason: A coating of aluminium oxide is deposited on the metal surface and protects it against corrosion.

2. Assertion: When carbon dioxide gas is passed through lime water, a white precipitate is initially formed.
Reason: White precipitate is of calcium carbonate which is formed during the reaction.

3. Assertion: In the electrolysis of water, the volume of hydrogen liberated is twice the volume of oxygen formed.
Reason: It is because water has hydrogen and oxygen in the ratio 1:2.

4. Assertion: Silver chloride turns grey in sunlight.
Reason: Silver chloride decomposes into silver and chloride by light.

5. Assertion: Copper can easily displace silver on reacting with an aqueous solution of silver nitrate.
Reason: Silver can be easily precipitated since it is insoluble in water.


Chapter 2: Acids, Bases and Salts

1. Assertion: Reaction of zinc granules with dilute sulphuric acid releases a gas with a pop sound.
Reason: When a metal reacts with an acid, it releases water vapour.

2. Assertion: An acid is a substance which contains one or more hydrogen atoms in its molecule.
Reason: An aqueous solution of acid turns blue litmus red.

3. Assertion: Acetic acid is a weak acid.
Reason: Acetic acid gets partially ionised in aqueous solution.

4. Assertion: Milk of magnesia is taken to get rid of pain in the stomach during indigestion.
Reason: Milk of magnesia is a base and it neutralises the excess acid in the stomach.

5. Assertion: Dry HCl gas does not change the colour of litmus paper.
Reason: Dry HCl gas can behave as an acid.


Chapter 3: Metals and Non-metals

1. Assertion: Metals can be drawn into wires and reshaped.
Reason: Metals are ductile and malleable.

2. Assertion: Rusting of iron does not take place in ordinary water.
Reason: Ordinary water contains dissolved oxygen.

3. Assertion: Positive ions are known as cations.
Reason: Cations are released at the anode.

4. Assertion: When zinc nitrate reacts with aluminium, aluminium nitrate is formed.
Reason: Aluminium is more reactive than zinc.

5. Assertion: Sodium chloride is not soluble in water.
Reason: Electrovalent compounds are generally soluble in water.


Chapter 4: Carbon and its Compounds

1. Assertion: Diamond is the hardest crystalline allotropic form of carbon.
Reason: Carbon atoms in diamond are tetrahedral in nature.

2. Assertion: A large number of carbon compounds exists due to the self-linking property of carbon known as catenation.
Reason: Strength of carbon to carbon bonds is very high.

3. Assertion: During fermentation, glucose is converted to ethyl alcohol with the help of the enzyme maltase.
Reason: Alcoholic fermentation is carried out by certain enzymes present in yeast.

4. Assertion: Zinc nitrate reacts with aluminium.
Reason: Aluminium is more reactive than zinc.

5. Assertion: Sodium chloride is not soluble in water.
Reason: Electrovalent compounds are generally soluble in water.


Chapter 5: Periodic Classification of Elements

1. Assertion: Chlorine is the most reactive member of the halogen family.
Reason: Size of the chlorine atom is greater than that of the fluorine atom.

2. Assertion: In the long form of periodic table, the elements are arranged in order of increasing masses
Reason: The properties of the elements are related to the atomic number.

3. Assertion: In a triad, the three elements present have the same gaps of atomic masses.
Reason: Elements in a triad have similar properties.

4. Assertion: Tendency to lose electrons will decrease across a period.
Reason: Effective nuclear charge increases across a period.

5. Assertion: Boron and silicon have the same properties.
Reason: Boron and silicon are in the same block.


Chapter 6: Life Processes

1. Assertion: Diffusion does not meet the high energy requirements of multicellular organisms.
Reason: Diffusion is a fast process but only occurs at the surface of the body.

2. Assertion: Saliva contains an enzyme called amylase.
Reason: Amylase helps to break down simple sugars like glucose into complex molecules like starch.

3. Assertion: Bile helps in the emulsification of fats.
Reason: Bile makes acidic food coming from the stomach alkaline so that pancreatic enzymes can act on it.

4. Assertion: The inner lining of the small intestine has numerous finger-like projections called villi.
Reason: Villi increase the surface area for absorption.

5. Assertion: Rings of cartilage are present in the throat.
Reason: They ensure that the air passage does not collapse.


Chapter 8: How do Organisms Reproduce?

1. Assertion: Reproduction enables the continuity of life for generations.
Reason: Reproduction is a biological process in which an organism gives rise to young ones similar to itself.

2. Assertion: Paramoecium reproduces by budding.
Reason: All unicellular organisms reproduce by asexual methods.

3. Assertion: Errors during DNA copying are a source of variations.
Reason: Variations are useful for ensuring the survival of the species.

4. Assertion: Tail of the sperm consists of an acrosome and mitochondria.
Reason: Acrosome contains enzymes which assist in fertilisation.

5. Assertion: Contraceptives can be effectively used for population control.
Reason: They can be used to avoid unwanted pregnancy in females.


Chapter 9: Heredity and Evolution

1. Assertion: In a monohybrid cross, the F1 generation indicates dominant characters.
Reason: Dominance occurs only in the heterozygous state

2. Assertion: The father is responsible for the sex of the child.
Reason: Boys inherit the Y chromosome and girls inherit the X chromosome from the father.

3. Assertion: Mendel was successful in his hybridisation experiments.
Reason: Garden pea proved ideal experimental material.

4. Assertion: Mendel chose pea plant for his experiments.
Reason: Pea plant provides diverse visible traits and has a short life span.

5. Assertion: When TT and tt pea plants were crossed, only tall plants were obtained in F1 progeny.
Reason: This was because tall allele was dominant over the dwarf allele.


Chapter 10: Light - Reflection & Refraction

1. Assertion: Concave mirrors are used in torches.
Reason: The light bulb placed at the focus of the concave reflector produces a strong beam of light.

2. Assertion: Magnification produced by a concave mirror of focal length 10 cm for the object distance 15 cm and image distance 30 cm is –2.
Reason: The image formed by a concave mirror is always real, inverted and magnified irrespective of the position of the object.

3. Assertion: When a beam of light is incident on a glass slab perpendicularly, there is no refraction.
Reason: The rays of light which fall perpendicularly on the glass slab reach, enter and leave the glass slab at the same instant of time.

4. Assertion: When the direction of the rays of light is reversed, they retrace their path.
Reason: According to the dual theory of light, light is a particle and a wave simultaneously.

5. Assertion: When we stand close to the convex mirror our image is virtual, erect and diminished.
Reason: When object lies anywhere between pole and infinity of convex mirror, the image formed is virtual, erect and diminished.


Chapter 11: Human Eye and the Colourful World

1. Assertion: The Sun appears for about two minutes even after it has set below the horizon.
Reason: The refraction of light rays coming from the Sun is caused by the Earth’s atmosphere.

2. Assertion: Red colour light after dispersion is deviated the most.
Reason: Red colour has the maximum speed in a glass prism.

3. Assertion: White light splits up into seven colours on passing through a transparent medium.
Reason: Dispersion of white light occurs because the colours of white light travel at the same speed through the glass prism.

4. Assertion: The Sun appears white when it is overhead in the sky.
Reason: Light coming from the Sun has to travel a relatively shorter distance through the atmosphere to reach us.


Chapter 12: Electricity

1. Assertion: The property of a conductor due to which it opposes the flow of current through it is called resistance.
Reason: The resistance of a conductor depends on thickness and temperature.

2. Assertion: Electrical wires are made of copper or aluminium.
Reason: Copper and aluminium have very low electrical resistance.

3. Assertion: If the length of the wire is doubled, the resistance gets doubled.
Reason: Resistance of a conductor is inversely proportional to its length.

4. Assertion: Nichrome wire is used to make the filament of an electric bulb.
Reason: Nichrome does not undergo oxidation even at a high temperature.

5. Assertion: In a parallel combination of circuits, each appliance has its separate switch and thus can be switched on/off independently.
Reason: In a parallel combination, each appliance gets the same voltage as that of the power supply.


Chapter 13: Magnetic Effect of Electric Current

1. Assertion: Motion of a charged particle under the action of the magnetic field is not always with constant speed.
Reason: Magnetic force does not have any component either along or opposite to the direction of the charged particle.

2. Assertion: Current generated by powerhouse generators reverses its direction after equal intervals of time.
Reason: Direct current can be transmitted for long distances without much loss of electric current.

3. Assertion: When a bar magnet is held stationary inside the coil, there is no deflection in the galvanometer.
Reason: If there is no relative motion between the magnet and the coil, the current will not be induced in the coil.

4. Assertion: The magnitude of the current induced in the coil can be increased by increasing the number of turns in the coil.
Reason: By reversing the direction of motion of the coil, the direction of current can be reversed.

5. Assertion: Freely suspended current carrying solenoid comes to rest in N-S direction just like a bar magnet.
Reason: On one end of current carrying straight solenoid behaves as a North pole and other end as a South pole.


Chapter 15: Our Environment

1. Assertion: Ecology is the study of the relationship between living organisms and their environment.
Reason: The biotic community and the non-living environment of an area function together to form an ecosystem.

2. Assertion: Producers are called nature’s recyclers.
Reason: They produce food by photosynthesis and make it available to other links of the ecosystem.

3. Assertion: Food chains are limited to 4-5 trophic levels.
Reason: The flow of energy within trophic levels follows the 10% law.

4. Assertion: Most air conditioning units are not eco-friendly.
Reason: They cause ozone depletion.

5. Assertion: Plastic is biodegradable.
Reason: Biodegradable substances can be broken down by biological processes.


PARAGRAPH BASED QUESTIONS

1. Read the following and answer any four questions from (i) to (v) Newlands’ Law of Octaves In 1864, Newlands arranged the known 56 elements in the order of increasing atomic masses. He observed that the properties of every eighth element are similar to the properties of the first element. Based on this observation, he proposed the Law of Octaves for the classification of elements.
Law of Octaves: When the elements are arranged in the increasing order of their atomic masses, the properties of every eighth element are similar to the first.


Many limitation were found in Newlands’ octaves. This law was found to be applicable only up to calcium. Newlands fitted all the known elements in a table of 7 X 8 that is 56 boxes. Newlands placed two elements each in some boxes to accommodate all the
known elements in the table. For example, Co and Ni, Ce and La. Moreover, he placed some elements with different properties under the same note in the octave. For example, Newlands placed the metals Co and Ni under the note ‘Do’ along with halogens, while Fe, having similarity with Co and Ni, away from them along with the nonmetals O and S under the note ‘Ti’. Also, Newlands’ octaves did not have provision to accommodate the newly discovered elements. The properties of the new elements discovered later on did not fit in the Newlands’ law of octaves.

i) Newland’s rule known as _____________ .

  1. Law of triad
  2. Law of octave
  3. Law of periodic table
  4. Periodic law

ii) Newland’s law of octaves based upon_______ .

  1. Increasing order of atomic number
  2. Increasing order of atomic mass
  3. Increasing order of electron
  4. Increasing order of atomic size

iii) Newland’s law of octaves is applicable to ___________ .

  1. Sodium
  2. Magnesium
  3. Calcium
  4. Sulphur

iv) A and B are two elements having similar properties which obey the law of octave. How many elements are there in between A and B?

  1. 6
  2. 7
  3. 8
  4. 5

v) Which of the following is true regarding Newland’s Law of Octaves?

  1. It worked well with only lighter elements.
  2. It was applicable only up to calcium.
  3. Both are correct.
  4. Both are incorrect.

 

2. Read the following and answer any four questions from (i) to (v)
M is an element in the form of a powder. M burns in oxygen and the product obtained is soluble in water. The solution is tested with litmus.

i) If M is a metal, then the litmus will turn _____.

  1. Red
  2. Blue
  3. Yellow
  4. Orange

ii) If M is a non-metal, then the litmus will turn _____.

  1. Red
  2. Blue
  3. Yellow
  4. Orange 

iii) If M is a reactive metal, then _____ will be evolved when M reacts with dilute sulphuric acid.

  1. Carbon dioxide gas
  2. Hydrogen gas
  3. Nitrogen gas
  4. Oxygen gas

iv) If M is a metal, it will form _____ oxide, which will form ______ solution with water.

  1. Basic, alkaline
  2. Acidic, acidic
  3. Basic, acidic
  4. Acidic basic

v) If M is a non-metal, it will not conduct electricity in the form of ______.

  1. Diamond
  2. Graphite
  3. Cesium
  4. Zinc

3. Read the following and answer any four questions from (i) to (v)
It is an allotrope of carbon containing clusters of 60 carbon atoms joined together to form spherical molecules. There are 60 carbon atoms in a molecule of buckminsterfullerene, so its formula is C60. The allotrope was named buckminsterfullerene after the American architect Buckminster Fuller.

(i) How many hexagons of carbon atoms are present in one molecule of Buckminster fullerene?

  1. 20
  2. 15
  3. 30
  4. 18

(ii) How many pentagons of carbon atoms are present in one molecule of Buckminster fullerene?

  1. 10
  2. 12
  3. 15
  4. 20

(iii) Which of the following molecule is called buckminsterfullerene?

  1. C90
  2. C60
  3. C70
  4. C120

(iv)Which allotrope of carbon exists as spherical molecules?

  1. Diamond
  2. Coke
  3. Graphite
  4. Fullerene

(v) Which of the following is true about C60?

  1. Each carbon bonded covalently to 3 other carbon atoms in a hexagonal ball like structure.
  2. Each carbon bonded covalently to 4 other carbon atoms in layers
  3. A giant lattice structure
  4. Pentagonal in shape


4. Read the following and answer any four questions from (i) to (v)
Metal A burns in air, on heating, to form an oxide A2O3 whereas another metal B bums in air only on strong heating to form an oxide BO. The two oxides A2O3 and BO can react with hydrochloric acid as well as sodium hydroxide solution to form the corresponding salts and water. And element E forms an oxide E2O. An aqueous solution of E2O turns red litmus paper blue.

(i) What is the nature of oxide A2O3?

  1. aAcidic oxide
  2. Basic oxide
  3. Amphoteric oxide
  4. Neutral oxide

(ii) What is the nature of oxide BO?

  1. Acidic oxide
  2. Basic oxide
  3. Amphoteric oxide
  4. Neutral oxide

(iii) Name one metal like A.

  1. Sulphur
  2. Gold
  3. Aluminium
  4. Iron

(iv)Name one metal like B.

  1. Sulphur
  2. Gold
  3. Zinc
  4. Iron

(v) State the nature of oxide element E.

  1. Acidic
  2. Basic
  3. Amphoteric
  4. Neutral

5. Read the following and answer any four questions from (i) to (v)
Sample of five metals ‘A’, ‘B’, ‘C’, ‘D’ and ‘E’ was taken and added to the following solution one by one. The results obtained have been tabulated as follows.

Use the above table to answer the following questions about the given metals.

i) Which of them is most reactive?

  1. A
  2. B
  3. D
  4. E

ii) What would you observe if ‘B’ is added to CuSO4?

  1. Reddish brown depositCBSE
  2. Grey deposit
  3. Greyish silver deposit
  4. Black deposit

iii) Arrange ‘A’, ‘B’, ‘C’, ‘D’ and ‘E’ in the increasing order of reactivity.

  1. E>A>B>C>D
  2. E>B>A>D>C
  3. E>B>A>C>D
  4. C>D>A>E>B

iv) Container of which metal can store zinc sulphate and silver nitrate solution?

  1. A
  2. B
  3. C
  4. D

v) Which of the above solution(s) can be stored in a container made of any of these metals?

  1. Aluminium sulphates
  2. Magnesium sulphates
  3. Both (a) & (b)
  4. None of these

6. Read the following paragraph and answer the questions that follow:
The food material taken in during the process of nutrition is used in cells to provide energy for various life processes. Diverse organisms do this in different ways – some use oxygen to breakdown glucose completely into carbon dioxide and water, some use other pathways that do not involve oxygen.

i) Which three carbon molecule is formed during the breakdown of glucose?

  1. Lactic acid
  2. Glyceraldehyde
  3. Acetic acid
  4. Pyruvate

ii) The process in which pyruvate may be converted into ethanol and carbon dioxide is

  1. Germination
  2. Fermentation
  3. Cellular respiration
  4. Oxidation

iii) Breakdown of pyruvate using oxygen takes place in the

  1. Cytoplasm
  2. Stroma
  3. Mitochondria
  4. Cellular matrix

iv) The breakdown of pyruvate in the absence of oxygen producing lactic acid occurs primarily in the

  1. Muscle cells
  2. Brain cells
  3. Cardiac cells
  4. Nerve cells

v) Observe the graph and interpret which of the following is true w.r.t. rate of respiration?

  1. Increases with an increase in temperature
  2. Increases first and then decreases with an increase in temperature
  3. Decreases with an increase in temperature
  4. Decreases first and then increases with an increase in temperature

7. Read the following paragraph and answer the questions that follow:
Plants have tissues to transport water, nutrients and minerals. Xylem transports water and mineral salts from the roots up to other parts of the plant, while phloem transports sucrose and amino acids between the leaves and other parts of the plant.

(i) Which of the following processes will not occur in the absence of xylem?

  1. Transport of water
  2. Conduction of food
  3. Transport of minerals
  4. Both A and C

(ii) Transport of food by the phloem is called

  1. Transpiration
  2. Translocation
  3. Guttation
  4. Adhesion

(iii) Which of the following is not a characteristic of xylem?

  1. Living cells
  2. Lack of cytoplasm
  3. Impermeable to water
  4. Presence of lignin

(iv)In phloem transport occurs between where the substances are made i.e. ________
and where they are used or stored i.e. ________.

  1. Sink, source
  2. Source, sink
  3. Origin, destination
  4. Destination, origin

(v) ____________ is the process which involves transport of water and minerals.

  1. Transpiration stream
  2. Translocation
  3. Guttation
  4. Adhesion

8. Read the following paragraph and answer the questions that follow:
Mendel experimented on a pea plant and considered 7 main contrasting traits in the plants. In monohybrid experiment, Mendel took two pea plants of opposite traits (one short and one tall) and crossed them. He found the first generation offspring were tall and called it F1 progeny. Then he crossed F1 progeny and obtained both tall and short plants in the ratio 3:1. In a dihybrid cross experiment, Mendel considered two traits, each having two alleles. He crossed wrinkled-green seed and round-yellow seeds and observed that all the first generation progeny (F1 progeny) were round-yellow. This meant that dominant traits were the round shape and yellow colour. He then self-pollinated the F1 progeny and obtained 4 different traits wrinkled-yellow, round-yellow, wrinkled-green seeds and round-green in the ratio 9:3:3:1.

i) Which of the following was not a trait considered by Mendel for his experiments?

  1. Flower colour
  2. Pod shape
  3. Stem height
  4. Seed type

ii) What is the phenotypic ratio of the F2 progeny in a dihybrid cross?

  1. 3:1
  2. 1:2:1
  3. 9:3:3:1
  4. 3:4

iii) What is the genotypic ratio of the F2 progeny in a monohybrid cross?

  1. 3:1
  2. 1:2:1
  3. 9:3:3:1
  4. 3:4

iv) A cross between tall and dwarf plants will produce which type of plants in the F1 generation?

  1. Tall
  2. Dwarf
  3. Semi-dwarf
  4. Short

v) Because of his contribution to the field of genetics, Mendel is known as the

  1. Father of genetics
  2. Father of botany
  3. Father of medicine
  4. Father of plant physiology

9. Read the following paragraph and answer the questions that follow:
The growing size of the human population is a cause of concern for all people. The rate of birth in a given population will determine its size. Reproduction is the process by which organisms increase their population. The process of sexual maturation for reproduction is gradual and takes place while general body is still going on. Some degree of sexual maturation does not necessarily mean that mind or body is ready for sexual acts or for having and bringing up children. Various contraceptive devices are being used by human beings to control the size of population.

i) Which of the following is a common sign of sexual maturation in both boys and girls?

  1. Development of breasts
  2. Growth of pubic hair
  3. Adam’s apple
  4. Broadening of hips

ii) Which of the following is an IUCD?

  1. Copper-T
  2. Diaphragm
  3. Oral pill
  4. Tubectomy 

iii) Which among the following is not a sexually transmitted disease?

  1. Gonorrhoea
  2. AIDS
  3. Syphilis
  4. Cholera

iv) A couple wants to space the birth of their second child. Which of the following preventive measure could be taken by the husband?

  1. Oral pills
  2. Diaphragms
  3. Tubectomy
  4. Condoms

v) A pregnant woman visits the doctor to determine the sex of the child? Why is she denied of this testing?

  1. It is a complicated test.
  2. It may result in female foeticide.
  3. It is an expensive test.
  4. It is harmful test for the developing foetus.

10. Read the following paragraph and answer the questions that follow:
We eat various types of food which has to pass through the same digestive tract. Naturally the food has to be processed to generate particles which are small and of the same texture. The process of digestion of food in humans involves the alimentary canal and associated digestive glands.

(i) Which of the following helps in the breakdown of starch?

  1. Salivary amylase
  2. Trypsin
  3. Peptidase
  4. Chyme

(ii) What is the role of the mucus?

  1. Creates alkaline medium for digestion
  2. Movement of food along the digestive tract
  3. Protect the inner lining of the stomach from the action of the acid
  4. Breakdown of starch into sugar

(iii) Which of the following will have a longer small intestine?

  1. Deer
  2. Lion
  3. Tiger
  4. Fox

(iv) Which of the following processes would have been obstructed in the absence of villi?

  1. Digestion of fats
  2. Breakdown of lipids
  3. Absorption of food
  4. Removal of undigested food

(v) Which of the following is not processed in the small intestine?

  1. Pancreatic juice
  2. Bile
  3. Saliva
  4. Intestinal juice

11. Study the given diagram and answer the following.
A very thin and narrow beam of white light is made incident on the glass objects shown below. Comment on the nature and behaviour of the emergent beam in all the three cases.
(Given: sin 45° = 1/√2 ⁄, sin 30° = ½)

(i) There is a similarity between two of the emergent beams. Identify the two.

  1. (A) and (B)
  2. (A) and (C)
  3. (B) and (C)
  4. Insufficient data given to identify the similarity

(ii) When light enters from air to glass, the angles of the incidence and refraction in air
and glass are 45° and 30° respectively. The refractive index of the glass is

  1. √2
  2. 1/2
  3. 1
  4. 1/√2

(iii) The phenomena which takes place when the light rays emerge out in case (B)

  1. reflection
  2. refraction
  3. dispersion
  4. scattering

(iv)What is the unit of refractive index?

  1. °A
  2. cm
  3. degree
  4. no unit

(v) The ratio of sine of angle of incidence and sine of angle of refraction for particular pair of media is

  1. Zero
  2. Unity
  3. Constant
  4. None of these

 

12. The figure given below illustrates the ray diagram for the formation of image by a concave mirror. The position of the is beyond the centre of the curvature of the concave mirror. On basis of the given figure answer the questions given below.

i) If focal length of the concave mirror is 10 cm, the image formed will be at a distance _____________________.

  1. Between 10 cm and 15 cm
  2. Between 10 cm and 20 cm
  3. Beyond 20 cm
  4. At 20 cm

ii) In case of concave mirror, the image distance from the pole of the mirror is

  1. Always positive
  2. Always negative
  3. Negative or positive depending upon the position of the object
  4. None of these

iii) If the size of the object in the given figure is 5 cm and the magnification produced is – 0.5. The size of the image is

  1. – 2.5 cm
  2. – 0.1 cm
  3. 2.5 cm
  4. 0.1 cm 

iv) A negative sign in the magnification value indicate that the image is

  1. Real and inverted
  2. Real and erect
  3. Virtual and erect
  4. Virtual and inverted

v) If the value of magnification is greater than 1 then it indicates that the image formed is

  1. Diminished
  2. The same size as that of the object
  3. Enlarged
  4. Value of magnification cannot specify whether the image is diminished or magnified

 

13. I n the below circuit, a nichrome wire of length 'L' is connected between points X and Y and note the ammeter reading. The experiment is performed and repeated by inserting another nichrome wire of same thickness but twice the length i.e. ‘2L’.

i) What are the changes observed in the ammeter readings?

  1. Ammeter readings decreases, becomes half
  2. Ammeter readings increases, becomes two times
  3. Ammeter readings increases becomes quadrupled
  4. Ammeter reading decreases becomes one – fourth

ii) What change is occurred in ammeter reading if instead of changing the length the area of cross – section is doubled?

  1. Ammeter readings decreases, becomes half
  2. Ammeter readings increases, becomes two times
  3. Ammeter readings increases becomes quadrupled
  4. Ammeter reading decreases becomes one – fourth 

iii) If the resistors of 5 ohms and 10 ohms are connected in series in the above circuit. What is the ratio of the current passing through the two resistors?

  1. 2:1
  2. 3:1
  3. 1:2
  4. 1:1

iv) If the resistors are connected in parallel

  1. Current across each resistor is same and voltage changes
  2. Current and voltage across each resistor is same
  3. Current across each resistor varies and voltage remains same
  4. Current changes, voltage changes

v) SI unit of current is denoted as

  1. A
  2. C
  3. I
  4. J

14. Shama has a set of five substances. She has a chart stating resistivities of all the substances.
Observe the table

She has to choose an appropriate substance for performing electrical tasks. Which of the above substance according to you –

i) Can be used as an insulator

  1. A
  2. B
  3. B as well as C
  4. E

ii) Can be used for domestic wiring

  1. A
  2. B
  3. A as well as C
  4. D

iii) Can be utilised in making solar cells and transistors

  1. A
  2. B
  3. C
  4. D

iv) Is an alloy

  1. A
  2. B
  3. C
  4. E

v) Behaves as a semiconductor

  1. A
  2. D
  3. C
  4. E

15. Ruchi fixes a sheet of white cardboard sheet on a drawing board. She places a bar magnet in the centre of it. She sprinkles iron fillings uniformly around the bar magnet. Then she taps the board gently and observes that the iron fillings arrange themselves in particular pattern.

(i) What does the pattern shape look like?

  1. Straight lines
  2. Squares
  3. Closed curves
  4. Parallel lines

(ii) Why do iron fillings arrange in a pattern?

  1. Due to poles of magnet
  2. Due to force exerted by magnet within its magnetic field
  3. Due to repulsion between the poles and filings
  4. None of the above

(iii) What do the lines along which iron filings align represent?

  1. Magnetic field lines
  2. Magnetic force
  3. Magnetic induction
  4. Magnetic susceptibility

(iv) What does the crowding of iron filings at the end of the magnet indicate?

  1. Magnetic field is weak at poles and stronger at middle
  2. Magnetic field is stronger at poles and weak at middle
  3. Magnetic field strength goes on decreasing from north to south pole
  4. Magnetic field strength goes on decreasing from south to north pole

(v) How the strength of magnetic field is indicated?

  1. Magnetic field strength cannot be indicated by magnetic field lines
  2. Far the magnetic field lines more is the magnetic strength
  3. Closer the magnetic field lines less are the magnetic field strength
  4. Closer the magnetic field lines more is the magnetic strength

 

Chapter 1: Chemical Reactions and Equations

1. Why are decomposition reactions called the opposite of combination reactions?                                                      [1]

Solution: In a combination reaction, two substances combine to form one compound. In a decomposition reaction, a compound  breaks down into two or more substances. So, they are opposite to each other.

2. When potassium chlorate (KClO3) is heated in the presence of manganese dioxide catalyst, it decomposes to form potassium chloride and oxygen gas. Represent this in the form of a balanced chemical equation.                                                 [1]

Solution: The balanced chemical equation is

begin mathsize 12px style KCIO subscript 3 left parenthesis straight s right parenthesis end subscript rightwards arrow with MnO subscript 2 increment on top space KCI subscript left parenthesis straight s right parenthesis end subscript plus straight O subscript 2 left parenthesis straight g right parenthesis end subscript end style

3. A solution of potassium chloride when mixed with silver nitrate solution forms an insoluble white substance. Write the chemical reaction involved and mention the type of chemical reaction.                                                                                     [3]

Solution: When a solution of potassium chloride is mixed with silver nitrate solution, an insoluble white silver chloride solution is formed. The balanced equation is
KCl(aq) + AgNO3(aq) → AgCl(s) + KNO3(aq)
It is a double displacement and precipitation reaction.

4. Equal lengths of Mg ribbon are taken in test tubes A and B. Hydrochloric acid is added to test tube A, while acetic acid is added to test tube B. In which case the reaction would occur more vigorously and why? Write the chemical equations for reactions in test tubes A and B.                                                                                                                                                                [3]

Solution: Fizzing in the reaction is due to the evolution of hydrogen gas by the action of metal on acid.
In test tube A:
Mg(s) + 2HCl(aq) →MgCl2(aq) + H2(g)
In test tube B:
Mg + CH3COOH(aq) → (CH3COO)2Mg + H2(g)
Since hydrochloric acid is a strong acid and acetic acid is a weak acid, the evolution of H2 gas occurs more readily in case of HCl.

5. Describe an activity to show the decomposition reaction of ferrous sulphate in the laboratory.                                  [3]

Solution:

  1. Take about 2 grams of ferrous sulphate crystals (green coloured) in a dry boiling tube.
  2. Heat the boiling tube over a burner (by keeping the mouth of the boilingtube away from the face).
  3. The green colour of ferrous sulphate crystals first changes to white and then a brown solid (ferric oxide) is formed.
  4. A gas having the smell of burning sulphur comes out of the boiling tube.
    FeSO4.7 H2O → FeSO4 + 7 H2O
    2 FeSO4(S) → Fe2O3(S) + SO2(g) + SO3(g)

6. A solution of substance 'X' is used for white washing.                                                                                          [3]

  1.  Name substance 'X' and write its formula.
  2.  Write the reaction of substance 'X' named in (i) above with water.
  3.  Write the balanced equation for the following chemical reaction:
      Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride
Solution: 
  1. Substance 'X' is calcium oxide. Its chemical formula is CaO.
  2. Calcium oxide reacts vigorously with water to form calcium hydroxide.
  3. 3BaCl2+ Al2 (SO4)2  3BaSO4 2AlCl3

7.                                                                                                                                                                           [5]
    a) Write chemical equations for the reactions taking place when 

  1. Zinc sulphide is heated in air.
  2. Zinc carbonate is calcined.
  3. Manganese dioxide is heated with aluminium powder.

    b)

  1. What happens when a strip of lead metal is placed in a solution of copper chloride? Write the balanced chemical equation for  the eaction along with the colour changes observed during the reaction.
  2.  What are precipitation reactions? Give one example of a precipitation reaction.
Solution: 

a.
  begin mathsize 12px style straight i. space 2 ZnS subscript left parenthesis straight s right parenthesis end subscript plus 3 straight O subscript 2 left parenthesis straight g right parenthesis end subscript rightwards arrow with heat on top 2 ZnO subscript left parenthesis straight s right parenthesis end subscript plus 2 SO subscript 2 left parenthesis straight g right parenthesis end subscript

ii. space ZnCO subscript 3 left parenthesis straight s right parenthesis end subscript rightwards arrow with heat on top ZnO subscript left parenthesis straight s right parenthesis end subscript plus CO subscript 2 left parenthesis straight g right parenthesis end subscript

iii. space 3 MnO subscript 2 left parenthesis straight s right parenthesis end subscript plus 4 AI blank subscript left parenthesis straight s right parenthesis end subscript rightwards arrow with heat on top 3 Mn subscript left parenthesis straight l right parenthesis end subscript plus 2 Al subscript 2 space straight O subscript 3 left parenthesis straight s right parenthesis end subscript end style

b.
 i. a When a strip of lead metal is placed in a solution of copper chloride, lead chloride solution and copper metal are formed. The   green colour of copper chloride fades and the solution becomes colourless. A red brown coating of copper metal is deposited on     the lead strip. Lead is more reactive than copper; hence, it is able to displace it from its solution.
 Pb(s) + CuCl2(aq) → PbCl2(aq) + Cu(s)
 ii. b Any reaction in which an insoluble solid (precipitate) is formed and separates out from the solution is called a precipitation     reaction.
 BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaCl(aq)

8.                                                                                                                                                                        [5]

  1. A water-insoluble calcium compound (A) on reacting with dil. H2SO4 released a colourless and odourless gas (B) with brisk  effervescence. When gas (B) was passed through lime water, the lime water turned milky and again formed compound A.    Identify A and B, and write the chemical equations for the reactions involved. 
  2. A brown substance 'X' on heating in air forms a compound 'Y'. When hydrogen gas is passed over 'Y', it changes to 'X' again.
  1. Name substances 'X' and 'Y'.
  2. Name the processes occurring during the two changes.
  3. Write the chemical equations involved.

Solution: 
a.

 CaCO3 +dil. H2SO4 → CaSO4+ H2O + CO2
 (water insoluble)
     (A)                            (B)
 Ca (OH)2 + CO2 → CaCO3 + H2O
 (Lime water)               (A)
                                 Milky
 A: CaCO3 (limestone)
 B: CO2(g)

b.
  i. X: Copper (Cu)
     Y: Copper oxide (CuO)
 ii. First: Oxidation of X; Second: Reduction of Y
 iii. 2 Cu + O2 → 2 CuO
     CuO + H2 → Cu + H2O

 

Chapter 2: Acids, Bases and Salts

1. Fresh milk has a pH of 6. How do you think pH will change as it turns into curd? Explain.                                   [1]

Solution:The pH of milk decreases from 6 as it turns into curd. Curd is more acidic than milk.

2. Classify the following into acidic oxides and basic oxides: Na2O, SO2, MgO, CO2.                                               [1]

Solution: Acidic oxides: SO2, CO2

Basic oxides: Na2O, MgO

3.                                                                                                                                                                        [3]

  1. Write the chemical name and formula of washing soda. How is it prepared? Write the chemical equation of the reaction.
  2. Why should Plaster of Paris be stored in a moisture-proof container?
  3. If we have hydrochloric acid and acetic acid in equal concentration, which will be the stronger acid and why?
Solution:
  1. The chemical name of washing soda is sodium carbonate decahydrate.
    The formula is Na2CO3.10 H2O.
    It is obtained by heating baking soda, followed by recrystallisation.
    2NaHCO3 → Na2CO3 + H2O + CO2
    Na2CO3 + 10 H2O → Na2CO3.10 H2O
  2. Plaster of Paris should be kept in a moisture-proof container because it reacts readily with water (moisture) to give gypsum. This will make the Plaster of Paris useless after sometime.
    CaSO4·(0.5 H2O) + 3 H2O → 2 CaSO4·2 H2O
  3. Hydrochloric acid will be a stronger acid than acetic acid because it completely ionises in water to produce a large amount of hydrogen ions. On the other hand, acetic acid partially ionises in water to produce only a small amount of hydrogen ions.

4. Define 'water of crystallisation'.                                                                                                                    [3]
    Give two examples of substances having water of crystallisation.
    Write their chemical formulae.

Solution:  Water of crystallisation is the number of water molecules which combine chemically in a definite molecular proportion with the concerned salt in the crystalline state.
Two correct examples:
Copper sulphate, chemical formula: CuSO4.5 H2O
Washing soda, chemical formula: Na2CO3.10 H2O

5.                                                                                                                                                                      [3]

  1. Name the compound which is obtained from baking soda and is used to remove the permanent hardness of water. 
  2. Write its chemical formula.
  3. What happens when it is recrystallised from its aqueous solution?
Solution:  
  1. Sodium carbonate or washing soda is obtained from baking soda which is used for removing the permanent hardness of water.
  2. Chemical formula: Na2CO3
  3. Anhydrous sodium carbonate becomes hydrated when it is recrystallised from its aqueous solution.

6.                                                                                                                                                                     [3]

  1. What change will you observe in the colour of red litmus paper when it is dipped into a solution of sodium sulphate? Give reason to explain your observation. 
  2. A bottle filled up to the brim with concentrated sulphuric acid is left open in the atmosphere by mistake. Will there be any change in the level of liquid? Explain your answer with reasons.
Solution: 
  1. It will not undergo any colour change because the solution of Na2SO4 (sodium sulphate) in water is almost neutral.
  2. Concentrated sulphuric acid is highly hygroscopic. It absorbs moisture from the air and gets diluted. Since the volume increases, the acid starts flowing out of the bottle.

7.                                                                                                                                                                    [5]
   a. Write word equations and balanced equations for the reaction taking place when 

  1. Dilute sulphuric acid reacts with zinc granules
  2. Dilute hydrochloric acid reacts with magnesium ribbon

   b. What is meant by 'hydrated' and 'anhydrous' salts? Explain with examples.
   c. Give two important uses of washing soda.

Solution: 


  1. i. Dilute sulphuric acid reacts with zinc granules.
      begin mathsize 12px style stack straight H subscript 2 SO subscript 4 left parenthesis dil right parenthesis end subscript with Sulphuric space acid below plus stack Zn subscript left parenthesis straight s right parenthesis end subscript with space Zinc below space rightwards arrow space stack ZnSO subscript 4 left parenthesis aq right parenthesis end subscript with Zinc space sulphate below space plus stack straight H subscript 2 left parenthesis straight g right parenthesis end subscript with space Hydrogen below end style
    ii. Dilute hydrochloric acid reacts with magnesium ribbon
    begin mathsize 12px style stack HCI subscript left parenthesis dil right parenthesis end subscript with Hydrochloric space acid below plus stack Mg subscript left parenthesis straight s right parenthesis end subscript with space magnesium below space rightwards arrow space stack MgCI subscript 2 left parenthesis aq right parenthesis end subscript with magnesium space chloride below plus stack straight H subscript 2 left parenthesis straight g right parenthesis end subscript with space Hydrogen below end style 
  2. Salts which contain water of crystallisation are called hydrated salts.
    Example: Copper sulphate crystals contain 5 molecules of water of crystallisation.
    Salts which have lost their water of crystallisation are called anhydrous salts.
    Example: On strong heating, copper sulphate crystals lose all the water of crystallisation to form anhydrous copper sulphate.
  3. Uses of washing soda:
  1. For removing permanent hardness of water
  2. In the manufacture of glass, soap and paper

8.                                                                                                                                                                   [5]
  a. Write the chemical equation for the preparation of 

  1.   Bleaching powder
  2.   Plaster  of Paris
  3.   Caustic soda

  b. What is the chlor-alkali process? Give two uses of NaOH obtained from this process.

Solution:

a. Chemical equations for the preparation of
i. Bleaching powder

  begin mathsize 12px style stack Ca left parenthesis OH right parenthesis subscript 2 with Calcium space Oxide below plus stack CI subscript 2 with space chlorine below space rightwards arrow space stack CAOCI subscript 2 with Calcium space oxychloride below space plus stack straight H subscript 2 straight O with space Water below end style

ii. Plaster of Paris

  begin mathsize 12px style stack CaSO subscript 4.2 straight H subscript 2 straight O with Gypsume below space rightwards arrow with Heat.373 straight K on top space stack CaSO subscript 4.1 divided by 2 straight H subscript 2 straight O with Plaster space of space Paris below space plus 1 space 1 divided by 2 stack straight H subscript 2 straight O with space Water below end style

iii. Caustic soda

  begin mathsize 11px style stack 2 NaCl with Sodium space Chloride below plus stack 2 straight H subscript 2 straight O with space Water below space rightwards arrow with electrtcity on top space stack 2 NaOH with Caustic space Soda below space plus stack Cl subscript 2 with space Chlorine below plus stack straight H subscript 2 with Hydrogen below end style

b. When electricity is passed through an aqueous solution of sodium chloride (brine), it decomposes to form sodium hydroxide. This process is called the chlor-alkali process.

  begin mathsize 11px style stack 2 NaCl with Sodium space Chloride below plus stack 2 straight H subscript 2 straight O with space Water below space rightwards arrow with electrtcity on top space stack 2 NaOH with Caustic space Soda below space plus stack Cl subscript 2 with space Chlorine below plus stack straight H subscript 2 with Hydrogen below end style

Uses of NaOH obtained from the chlor-alkali process:

  1. For de-greasing metals, soaps and detergents
  2. For making paper and artificial fibres

9. Give suitable reasons for the following statements:                                                                                               [5]

  1.  Rain water conducts electricity, whereas distilled water does not.
  2.  We feel a burning sensation in the stomach when we overeat.
  3.  On rubbing a tarnished copper vessel with lemon or tamarind juice, it begins to shine again.
  4.  A solution of sodium carbonate is not acidic.
  5.  Dry ammonia has no effect on litmus paper, whereas a solution of NH3 in water has.

Solution: 

  1. Distilled water is a pure form of water and it does not contain any ionic species. Therefore, it does not conduct electricity.
    Rain water, being an impure form of water, contains many ionic species such as acids, and therefore, it conducts electricity.
  2. Our stomach produces gastric juice which mainly contains hydrochloric acid (pH about 1.4). Overeating or eating spicy foods gives a burning sensation in the stomach because of excess of acid secretion which is called acidity.
  3. Lemon juice contains citric acid and tamarind juice contains tartaric acid. These acids react with the basic layer of copper carbonate on the surface to form soluble salts which are easily removed and the surface shines.
  4. Sodium carbonate is a salt of the weak acid carbonic acid and the strong base sodium hydroxide. The pH of sodium carbonate is 9. So, it is alkaline in nature.
  5. Dry ammonia is neutral in nature, so it has no effect on litmus paper. A solution of ammonia is basic in nature, so it turns red litmus blue.


Chapter 3: Metals and Non-Metals

1. What happens when iron nails are kept in an aqueous solution of CuSO4?                                                              [1]

Solution:  When an iron nail is immersed in an aqueous solution of copper sulphate, a brown coating of copper is developed over the iron nail. The blue copper sulphate solution becomes pale green since iron is more reactive than copper.
Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s)

2. Name a non-metal which is lustrous and a metal which is non-lustrous.                                                                 [1]

Solution: 

  1. Non-metal which is lustrous: Iodine (I)
  2. Metal which is non-lustrous: Sodium (Na)

3. What are amphoteric oxides? Give two examples. Element X on reaction with oxygen forms a dioxide XO2, which turns blue litmus red. Is element X a metal or a non-metal?                                                                                                              [3]

Solution: 

  1. Amphoteric oxides are those oxides which show properties of both acids as well as bases to form salts and water.
  2. Aluminium oxide (Al2O3) and ZnO are examples of amphoteric oxides.
  3. X is a non-metal. Non-metallic oxides are acidic in nature, and hence, the turn blue litmus red.

4.                                                                                                                                                                            [3]

  1. What are the two main allotropes of carbon? 
  2. Distinguish these two allotropes on the basis of hardness and electrical conduction.
Solution: Diamond and graphite are the two allotropes of carbon.
  1. Diamond:
    i. hardest substance
    ii. electrical insulator
  2. Graphite:
  1. comparatively soft; it is slippery over layers
  2. good electrical conductor

5. What happens when                                                                                                                                         [3]

  1. Zinc reacts with copper sulphate
  2. Magnesium reacts with HCl
  3. Sodium reacts with water
Solution: 
  1. Zinc reacts with copper sulphate to give zinc sulphate and copper metal.
    CuSO4(aq) + Zn(s) → ZnSO4(aq) + Cu(s)
  2. Magnesium reacts with HCl to give magnesium chloride and hydrogen gas.
    Mg(s) + 2 HCl(aq) → MgCl2 + H2(g)
  3. Sodium reacts with water to give sodium hydroxide and hydrogen gas.
    2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)

6.                                                                                                                                                                           [5]

  1. Write the electronic configurations of sodium and chlorine.
  2. Show the formation of sodium chloride from sodium and chlorine by the transfer of electrons.
  3. State the type of bonding formed.
  4. List two properties of the compounds formed by this bonding.

Solution: 

  1. Formation of ionic compounds – Sodium Chloride
    i. Sodium has one electron in its outermost shell, and chlorine has seven electrons in its outermost shell.
    ii. Sodium donates its one electron to chlorine and develops a net positive charge (Na+).
    iii. Chlorine accepts the electron and develops a net negative charge (Cl-).
    iv. Sodium and chloride ions, being oppositely charged, attract each other and are held by strong electrostatic forces of attraction.
    v. This results in the formation of an electrovalent or ionic bond.

  2. This type of bonding is called electrovalent or ionic bonding.
  3. Properties shown by compounds formed by this bonding:
    i. Ionic compounds have high melting points and boiling points.
    ii. They are strong electrolytes and conduct electricity in the molten state and in aqueous solutions.


Chapter 4: Carbon and its Compounds
1. Write the name of the following compound. Name the acid and alcohol from which it might be prepared.                [1]

Solution: The name of the given compound is butyl acetate. The name of the acid is acetic acid and the name of the alcohol is butanol.

2. Name the functional group present in each of the following organic compounds: [1]

  1. CH3COCH3
  2. C2H5COOH
Solution: 
  1. Ketone
  2. Carboxylic acid

3. What is meant by a homologous series of carbon compounds? Classify the following carbon compounds into two homologous series and name them.                                                                                                                                                 [3]
C3H4, C3H6, C4H6, C4H8, C5H8, C5H10 

Solution:  It is a group of organic compounds with a similar structure and similar chemical properties in which the successive compounds differ by a -CH2 group.
Molecules in a homologous series:

4.                                                                                                                                                                         [5]

  1. State two properties of carbon which lead to a large number of carbon compounds.
  2.  Explain isomerism. State any four characteristics of isomers. Draw the structures of possible isomers of butane, C4H10.       Also,explain why we cannot have isomers of the first three members of the alkane series.
Solution: 

a.

  1. Catenation: The property of the carbon element due to which its atoms can join to one another to form long carbon chains is called catenation.
  2. Tetravalency: Carbon has a valency of four. So, it is capable of bonding with four other atoms of carbon or atoms of some other mono-valent element. Compounds of carbon are formed with oxygen, nitrogen, hydrogen, sulphur, chlorine and many other elements, giving rise to compounds with specific properties which depend on elements other than the carbon present in the molecule.

b. Isomers: Organic compounds having the same molecular formula but different structural arrangement of atoms in their molecules are called isomers.
Characteristics of isomers:

  1. They have the same molecular formula but different structural formulae.
  2. They show similar properties only when they contain the same functional group.
  3. Two isomers can have different boiling points. For example, in isomers of pentane, the branched chain pentane will have lower boiling point than linear pentane because the boiling point depends on the surface area which is more in case of n-pentane (linear).
  4. Isomers can have different functional groups. For example, aldehyde and ketone are two isomers, but they contain different functional groups.
    Two isomers of butane:

Isomers for the first three members of the alkane series are not possible because

  1. The parent carbon must contain the most number of carbon atoms.
  2. The branching cannot be done from either the first or the last carbon atom.


Chapter 5: Periodic Classification of Elements

1. How were the positions of cobalt and nickel resolved in the modern periodic table?                                               [1]

Solution: The modern periodic table is based on the law that the properties of elements are a periodic function of their atomic numbers. So, the problem was resolved because cobalt has a lower atomic number (27) than nickel (28).

2. What are lanthanides and actinides?                                                                                                                 [1]

Solution: The two rows of elements at the bottom of the modern periodic table are called the lanthanides (or lanthanoids) and actinides (or actinoids).
Lanthanides: Ce (Z = 58) to Lu (Z = 71)
Actinides: Th (Z = 90) to Lr (Z = 103) 

3. How does the metallic character of elements change along a period of the periodic table from left to right and why? [2]

Solution: Metallic character decreases from left to right along the period of the periodic table, because on moving from left to right, the size of the atoms decreases, and hence, the tendency to release electrons decreases. Thus, the electropositive character decreases.

4. How does the modern periodic law justify one position for isotopes?                                                                     [2]

Solution: According to the modern periodic law, elements are arranged in the modern periodic table in the increasing order of their atomic numbers. Isotopes have the same atomic number and different atomic mass. So, although they have different atomic masses, they are still given the same position in the modern periodic table.

5. State Mendeleev’s periodic law. Which group of elements was missing from Mendeleev’s original periodic table? Besides gallium, which two other elements have since been discovered for which Mendeleev had left gaps in his periodic table?            [3]

Solution:

  1. Mendeleev’s periodic law: The properties of elements are periodic functions of their atomic masses.
  2. The zero group of noble gas elements was missing from Mendeleev’s original periodic table.
  3. Scandium (Sc) and Germanium (Ge) are the other two elements for which gaps were left by Mendeleev in his periodic table.

6. How could the modern periodic law remove various anomalies of Mendeleev’s periodic table? Explain with examples.[3]

Solution: The modern periodic table could remove various anomalies of Mendeleev’s periodic table:

In the modern periodic table, the elements are arranged in the increasing order of their atomic number, removing the anomaly regarding certain pairs of elements in Mendeleev’s periodic table.

  1. Atomic number of cobalt is 27and nickel is 28. Hence, cobalt will come before nickel even though its atomic mass is greater.
  2. All isotopes of the same elements have different atomic masses but the same atomic number; therefore, they are placed in the same position in the modern periodic table

7. Atoms of eight elements—A, B, C, D, E, F, G and H—have the same number of electronic shells but different number of electrons in their outermost shell. Elements A and G combine to form an ionic compound which can also be extracted from sea water. This ionic compound is added to a small amount of almost all vegetables and dishes during cooking. Oxides of elements A and B are basic in nature, while those of elements E and F are acidic. However, the oxide of element D is almost neutral. Based on the above information, answer the following questions:                                                                                                             [5]

  1. To which group or period of the periodic table do these elements belong?
  2. What would be the nature of the compound formed by a combination of elements B and F?
  3. Which two of these elements could definitely be metals?
  4. Which one of the eight elements is most likely to be found in the gaseous state at room temperature?
  5. If the number of electrons in the outermost shell of elements C and G are 3 and 7, respectively, write the formula of the      compound formed by the combination of C and G.
Solution:.
  1. Third period; Groups 1, 2, 13, 14, 15, 16, 17, 18, respectively
  2. Nature of compound: Electrovalent/ionic
  3. Metals: A and B; Non-metals: E, F, G
  4. G/H
  5. CG3

8. Explain:                                                                                                                                                             [5]

  1. Larger the atomic size, more metallic is the element.
  2. Size of the atom changes when it loses or gains electrons.
  3. K is more reactive than Li.
  4. Electronegativity of Cl is higher than S.
  5. Group 17 elements are non-metals, while Group 1 elements are metals.
Solution:
  1. Larger the atomic size, farther is the valence electron from the nucleus and lesser is the pull exerted on it. As a result, the electron can be easily removed from the valence shell, and hence, more metallic is the element.
  2. When an atom loses or gains an electron to form an ion, the number of electrons present in the outermost shell also changes. Corresponding to that, the effective nuclear charge on the changed number of electrons changes which further changes the size of an atom as there is an inverse relation between the effective nuclear charge and the size of the atom.
  3. K and Li belong to Group 1, i.e. metals. For metals, chemical reactivity of elements increases down the group, because chemical reactivity increases as the electropositive or metallic character increases.
  4. Electronegativity of chlorine is higher than sulphur because both belong to the third group and chlorine follows sulphur. We know that within a period, electronegativity increases as we move from left to right because of a decrease in the atomic size and increase in the nuclear charge.
  5. Group 17 elements are non-metals because they have 7 electrons in their valence shell and ionise by accepting 1 electron to form an anion.
    For example, the Group 17 elements F, Cl, Br and I all have 7 electrons in their valence shell and ionise by accepting 1 electron to form F- , Cl- , Br- and I-.
    Group 1 elements are metals because they have a tendency to lose the one electron present in their valence shell to form a positive ion.
    For example, Group 1 elements Li, Na, K, Rb and Cs have a tendency to lose the one electron present in their valence shell to form the positive ions Li+, Na+, K+,Rb+ and Cs+.

9.                                                                                                                                                                          [5]

Select from the table:

  1.  Which is the most electronegative?
  2.  How many valence electrons are present in G?
  3.  Write the formula of the compound between B and H.
  4.  In the compound between F and J, what type of bond will be formed?
  5.  Draw the electron dot structure for the compound formed between C and K.

Solution:

  1. The most electronegative is J.
  2. Valence electrons present in G are 5.
  3. B contains 1 valence electron and H contains 6 valence electrons. So, the valency of B is +1 and the valency of H is -2.
  4. In the compound between F and J, the type of bond formed will be covalent.
  5. The electron dot structure for the compound formed between C and K is



Chapter 6: Life Processes

1. Why do fish die when taken out of water?                                                                                                     [1]

Solution:

  • Fish utilise the oxygen dissolved in water for their respiration.
  • When fish are taken out of water, the supply of oxygen to the fish is cut as
  • they cannot absorb and breathe using the oxygen present in the atmosphere.Hence, the fish die after some time.

2. Why is diffusion insufficient to meet the oxygen requirements of multicellular organisms?                                   [1]

Solution:

  • In multicellular organisms such as man, the volume of the human body is large, and hence, oxygen cannot diffuse into all the cells of the human body quickly.
  • Diffusion, being a slow process, takes a lot of time to circulate oxygen to all the body cells.
  • Hence, diffusion is insufficient to meet the oxygen requirements of multicellular organisms like humans.

3. How do guard cells regulate the opening and closing of stomata?                                                                     [2]

Solution:

  • The opening and closing of stomata is controlled by guard cells.
  • When water flows into the guard cells, they swell up and their curved surface causes the stomata to open.
  • When the guard cells lose water, they shrink and become flaccid and straight, thus closing the stomata.

4. What advantage over an aquatic organism does a terrestrial organism have with regard to obtaining oxygen for respiration? [2]

Solution: Adaptation of terrestrial organism over aquatic organism for efficient uptake of oxygen from air:

  • Increased respiratory surface
  • Very fine and delicate surface for easy exchange of oxyen and carbon dioxide
  • Placement of respiratory surface within the body for protection
  • Mechanism for moving the air in and out of the respiratory surface where the oxygen is absorbed.

5. Why is small intestine longer in herbivores as compared to carnivores?                                                            [2]

Solution: Herbivores have longer small intestine to allow the cellulose to be digested completely. Herbivores have longer intestine than carnivores to digest grass. The intestine would host many small bacteria that process and breakdown cellulose into glucose.

6. Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds?                            [3]

Solution:

  • Warm-blooded animals such as birds and mammals maintain a constant body temperature by cooling themselves when they are in a hotter environment and by warming their bodies when they are in a cooler environment.
  • Hence, these animals require more oxygen (O2) for cellular respiration so that they can produce more energy to maintain their body temperature.
  • Thus, it is necessary for them to separate oxygenated and deoxygenated blood so that their circulatory system is more efficient and can maintain a constant body temperature.

7. What are the different ways in which glucose is oxidised to provide energy in various organisms?                      [3]

Solution: The first step towards obtaining energy is the breakdown of six-carbon glucose into two molecules of three-carbon pyruvate. This process takes place in the cytoplasm.

Respiration in the presence of oxygen in mitochondria:

  • Breakdown of glucose in the presence of oxygen is called aerobic respiration. This process takes place in the mitochondria.
  • The process of aerobic respiration releases carbon dioxide, water and energy.

Respiration in the absence of oxygen in muscle cells:

  • Breakdown of glucose in the absence of oxygen is called anaerobic respiration.
  • During heavy physical exercise such as cycling, running or lifting heavy weights,the body is often deprived of oxygen.
  • The demand for energy is high, while the supply of oxygen to the body is limited.
  • Therefore, muscle cells perform anaerobic respiration to fulfill the increasing energy demands of the body. Here, glucose gets converted to lactic acid.

Respiration in the absence of oxygen in yeast:

  • Organisms which do not require the presence of oxygen for respiration are called anaerobes.
  • Unicellular organisms such as yeast and some bacteria are examples of anaerobes.
  • The process of anaerobic respiration in yeast results in the formation of ethanol along with the release of carbon dioxide and energy. Water is not released in this process.

8. How are fats digested in our body? Where does this process take place?                                                            [3]

Solution:

  • Fats can be digested with the help of bile released from the liver.
  • Fats are present in the small intestine in the form of large globules, which makes it difficult for the enzymes to act on them.
  • Bile salts break fats into smaller globules to increase the action of enzymes.
    This process is known as emulsification.
  • Later, steapsin, a pancreatic lipase, acts on the emulsified fats and breaks them down into fatty acids and glycerol.
  • This aids in the easy digestion of food.
  • Fat digestion takes place in the small intestine.
  • This digested fat is absorbed into the body through the villi by diffusion.

9.                                                                                                                                                                    [5]

  1. Draw a well-labelled diagram of the human digestive system. 
  2. Describe the role of the following in the process of digestion:
    (i) Bile (ii) Salivary amylase (iii) HCl
Solution:
  1. Human digestive system

  2. Role in digestion:
  1. Bile: Bile helps to emulsify fats and thus aids in their absorption in the small intestine. It also makes the acidic food coming from the stomach alkaline, so as to enable the action of pancreatic enzymes.
  2. Salivary amylase: The hydrolytic action of salivary amylase hydrolyses starch into maltose.
  3. HCl: Hydrochloric acid gets mixed with food and kills the bacteria present in food. It also destroys the structure of proteins so that the enzymes can digest them easily. HCl also creates acidic conditions necessary for the action of the enzyme pepsin. It activates pepsin to act on proteins.

10.                                                                                                                                                                  [5]

  1. Define excretion.
  2. Name the basic filtration unit present in the kidney.
  3. Draw the excretory system in human beings and label the organs of the excretory system which perform the following        functions:
  1.  forms urine
  2.  is a long tube which collects urine from the kidney
  3.  stores urine until it is passed out
Solution:
  1. Excretion is the process of removal of harmful and unwanted substances, especially nitrogenous wastes, from the body.
  2. Nephron is the basic filtration unit of the kidneys.
  1. Kidneys: Form urine
  2. Ureter: Long tube which collects urine from the kidney
  3. Urinary bladder: Stores urine until it is passed out

11.                                                                                                                                                                  [5]

  1. Mention any two components of blood. 
  2. Trace the movement of oxygenated blood in the body.
  3. Write the function of valves present in between the atria and ventricles.
  4.  Write one structural difference between the composition of arteries and veins.
Solution:
  1. Components of blood: Plasma and blood cells (corpuscles)
  2. Movement of oxygenated blood in the body:

  3. Valves present in between atria and ventricles help to restrict the backflow of the blood from the ventricle to the atrium when the ventricle contracts.

  4. Differences between artery and vein:



Chapter 8: How do Organisms Reproduce?

1. Name one sexually transmitted disease each caused by bacterial infection and viral infection.                              [1]

Solution: Diseases caused by bacterial infection: Chlamydia, Gonorrhea, Syphilis
Diseases caused by viral infection: AIDS, Hepatitis B

2. Where does fertilisation take place in human females?                                                                                      [1]

Solution: Fertilisation in human females takes place in the fallopian tube just outside the ovary.

3. A couple wants to space the birth of their second child. Suggest one preventive method which could be observed? [1]

  1. By the husband
  2. By the wife for the same
Solution:
  1. By the husband: Use of condoms
  2. By the wife: Use of oral pills or loop or Copper-T

4. Differentiate between self pollination and cross pollination.                                                                               [2]

Solution: Differences between self pollination and cross pollination:

5. What is the function of the Cowper's gland and prostate gland?                                                                          [2]

Solution:
Function of the Cowper's gland: It secretes a white, viscous and alkaline secretion which acts as a lubricant.
Function of the prostate gland: The secretions of this gland keeps the sperms motile and active.

6. What is vegetative propagation? State two advantages and two disadvantages of this method.                              [3]

Solution: Vegetative propagation is a type of reproduction in which several plants are capable of reproducing naturally through the vegetative parts of plants such as roots, stems and leaves.

Advantages of vegetative propagation:

  • Plants not capable of reproducing sexually reproduce by this method.
  • It is a fast and certain method to obtain plants with desired features.

Disadvantages of vegetative propagation:

  • There is no possibility for variation.
  • The new plant grows in the same area as the parent plant which leads to competition for resources.

7. Name the process by which amoeba reproduces. Draw the various stages of its reproduction in a proper sequence. [3]

Solution: Amoeba reproduces by the process of binary fission. In this method, a single parent cell splits and gets divided into two daughter cells.

8. List any four methods of contraception used by humans. How does their use have a direct effect on the health and prosperity of a family?                                                                                                                                                              [3]

Solution:

Methods of contraception used by humans:

  • Barrier methods: Use of condoms in males and diaphragms in females
  • Chemical methods: Use of various hormonal pills
  • IUDs: Use of IUDs like copper-T and Lippes loop
    Surgical methods: Tubectomy in females and vasectomy in males

Impact of these techniques on the health and prosperity of the family:

  • They help to prevent unwanted pregnancies and control population growth.
  • They offer protection from sexually transmitted diseases like AIDS, syphilis etc.
  • They ensure proper health of the mother and child by preventing frequent pregnancies.
  • These techniques offer sufficient gap between the offspring for proper care of children.

9. Mention the post fertilisation changes that occur in a flower.                                                                             [3]

Solution: Post fertilisation changes that occur in a flower:

  • The diploid zygote develops into an embryo that forms the future plant.
  • The endosperm cells provides the required nutrition for the developing embryo.
  • The ovule forms the seed.
  • The ovary forms the fruit.
  • The outer and inner integuments of the ovule forms the testa or the seed coat of the seed.
  • Petals and sepals fall off.

10.                                                                                                                                                                      [5]

  1. List three differences between sexual and asexual types of reproduction.
  2. Explain why variations are observed in the offspring of sexually reproducing organisms.
Solution: 1. Differences between sexual and asexual reproduction:

2.
  • In sexual reproduction, two parents are involved which are different from one another.
  • The male and female gametes are formed by meiosis which allows crossing over and recombination which generates variation in the offspring.
  • The fusion of male and female gametes combines two different DNA copies and results in new combinations of genes and increases genetic variation.
  • Hence, variations are observed in the offspring of sexually reproducing organisms.

11.                                                                                                                                                                      [5]

  1. Write the function of the following parts in the human female reproductive system:
     (i) Ovary (ii) Oviduct (iii) Uterus
  2. Describe in brief the structure and function of the placenta.
Solution:

a. unctions:

i.Ovary:
  • Produces ova or female gametes.
  • Secretes the female hormones oestrogen and progesterone which areresponsible for changes in the female body at the time of puberty.
ii.Oviduct:
  • Acts as the site for the fertilisation of male and female gametes.
  • After fertilisation, the ovum travels down to the uterus through the oviduct.
iii.Uterus:
  • Protects and nourishes the developing embryo with the help of placenta.

b. Structure and function of the placenta:
The placenta is embedded in the uterine wall and serves as a connecting link between the mother’s body and the baby.

  • It is a disc of specialised tissue which provides food and oxygen to the foetus.
  • It contains blood spaces on the mother’s side and small projections called villi on the foetal side. Here, the mother’s blood and foetal blood come in contact with each other.
  • It provides a large surface area for the exchange of nutrients and oxygen between the mother and the foetus.
  • The foetus gives away waste products and carbon dioxide to the mother’s blood for excretion.
  • It also functions as an endocrine gland and secretes the hormones necessary to maintain pregnancy.

12.                                                                                                                                                                     [5]

  1. Distinguish between cross-pollination and self-pollination. Mention the site and product of fertilisation in a flower.
  2. Draw a labelled diagram of a pistil showing the following parts: Stigma, Style, Ovary, Female Germ Cell
Solution:
a. Differences between self-pollination and cross-pollination:
 

The site of fertilisation in flowers is the ovule which is present in the ovary.
The product of fertilisation is a zygote which later develops into a fruit.

b. Diagram of pistil:

 


Chapter 9: Heredity and Evolution

1. Why is DNA copying necessary during reproduction?                                                                                       [1]

Solution: DNA copying is necessary during reproduction as DNA carries genetic information. Therefore, for an organism to produce similar offspring it is necessary to copy DNA.

2. Mendel crossed a pure white recessive pea plant with a dominant pure red flowered plant. What will be the first generation F1 hybrids?                                                                                                                                                     [1]

Solution:The first generation F1 hybrids would all produce red flowers.

3. A Mendelian experiment consisted of breeding pea plants bearing violet flowers with pea plant bearing white flowers. What will be the result in the F1 progeny?                                                                                                                            [1]

Solution: The F1 progeny would bear violet flowers as violet colour is a dominant character over white colour.

4. Why did Mendel choose pea plant for his experiment?                                                                                    [2]

Solution: Mendel chose pea plant for his experiment because of the following reasons:

  • They grow quickly.
  • They have a short life-cycle.
  • They are easily self- and cross-pollinated and produces many offspring in one cross.
  • They have easily observed characteristics.

5. Only variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement? Why or why not?                                                                                                                                            [3]

Solution: No. This statement is mostly true, but not always. Variations that are advantageous to an organism will help the organism to survive better in the environment. So these will survive in the population. Sometimes variations that do not confer any particular advantage or disadvantage to an organism may also survive in the population. Such non-advantageous variations may become advantageous in future when the environmental conditions change.

6. State the three Mendel's laws of inheritance.                                                                                                [3]

Solution: Mendel's Laws of inheritance can be described as follows:

  1. The Law of Dominance: The offspring always exhibits a dominant trait. From the two alleles received from parents, the only dominant allele is expressed.
  2. The Law of Segregation: The two copies of each chromosome will be separated from each other, causing the two distinct alleles located on those chromosomes to segregate from one another.
  3. The Law of Independent Assortment: The traits inherited through one gene will be inherited independently of the traits inherited through another gene because the genes reside on different chromosomes that are independently assorted into daughter cells during meiosis.

7. Give the basic features of the mechanism of inheritance.                                                                                 [5]

Solution: The process of transfer of genetic characters from parents to offspring is called inheritance.

The basic features of inheritance are as follows:

  • Characters are controlled by genes.
  • Each gene controls one character.
  • There may be two or more forms of the gene.
  • One form may be dominant over the other.
  • Genes are present on chromosomes.
  • An individual has two forms of the gene whether similar or dissimilar.
  • The two forms separate at the time of gamete formation.
  • The two forms are brought together in the zygote.

8.                                                                                                                                                                        [5]

  1. ‘Sex of children is determined by what they inherit from the father and not from the mother.’ Justify.
  2.  Explain the result of a monohybrid cross.
Solution:
a.
  • A male has one X chromosome and one Y chromosome (XY). Therefore, half of the male gametes or sperms will have X chromosomes and the other half will have Y chromosomes.
  • A female has both X chromosomes (XX). Therefore, all the female gametes or ova always have X chromosomes.
  • If a sperm carrying the X chromosome fertilises an ovum which always carries the X chromosome, then the combination of sex chromosomes will be XX, and hence, the child born will be a female (girl).
  • If a sperm carrying the Y chromosome fertilises an ovum, then the combination of sex chromosomes will be XY, and hence, the child born will be a male (boy).
  • Thus, the sex of children is determined by what they inherit from the father and not from the mother.
b.
  • Monohybrid cross involves only a single pair of contrasting characters.
  • Consider pea plants with a pair of contrasting characters—tallness and dwarfness—with respect to the height of the stem.
  • If pure tall pea plants (TT) are crossed with pure dwarf pea plants (tt), then all progeny plants obtained will be tall. This is called the first filial or F1 generation seeds.
  • The F1 generation has genetic constitution Tt. It is genotypically a hybrid and a heterozygous plant with two different alleles.
  • Phenotypically, the plant is tall because the allele or the gene T for tallness masks the effect of its corresponding recessive gene t.
  • F1 plants are self-pollinated to obtain the F2 generation.
  • The second filial generation F2 has a genotypic ratio 1 TT:2 Tt:1 tt.
  • In this case, because the allele T for tallness is dominant, the pea plants with genotype Tt will be tall.
  • The phenotypic ratio is 3 tall:1 dwarf.
  • Genotypically, it shows 3 types of plants: 1 TT (which is homozygous tall), 2 Tt (which are heterozygous tall) and 1 tt (which is homozygous dwarf). Thus, the genotypic ratio is 1 TT:2 Tt:1 tt.


Chapter 10: Light - Reflection and Refraction

1. Which mirror is used as a rear-view mirror in vehicles?                                                                                     [1]

Solution: A convex mirror is used as a rear-view mirror in vehicles.

2. If the radius of curvature of a spherical mirror is 60 cm, what is its focal length?                                                [1]

Solution: R = 60 CM

begin mathsize 12px style straight f equals straight R over 2
therefore space straight f equals 60 over 2 equals 30 space cm end style

Thus, the focal length of this spherical mirror is 30 cm. 

3. What is the nature and position of the image when an object is placed at the focus of a concave mirror?              [1]

Solution: The nature of image formed when an object is placed at focus F is real, inverted and highly magnified, and the image is formed at infinity.

4. If a magnification of -1 is to be obtained using a convex lens of focal length 6 cm, then the object must be placed [1]

  1. within 12 cm
  2. at 6 cm
  3. iat 12 cm
  4. beyond 12 cm
Solution: (iii) at 12 cm
For magnification of -1, the object must be placed at 2F’ = 2 × 6 = 12 cm

5. Draw a neat and labelled ray diagram when an object is placed in front of a convex mirror between infinity and the pole. [3]

Solution:

When an object is placed between infinity and the pole of a convex mirror, the image formed is virtual, erect and diminished.

6. What is the speed of light in a medium of refractive index 1.8 if its speed in air is 300000 km/s?                         [3]

Solution:

The refractive index of medium, αnm = 1.8
Speed of light in air = 300000 km/s = 3 × 108 m/s

.begin mathsize 12px style straight n presubscript straight alpha subscript straight m space equals straight n subscript straight m over straight n subscript straight alpha equals straight v subscript straight alpha over straight v subscript straight m
rightwards double arrow straight v subscript straight m equals fraction numerator 3 cross times 10 to the power of 8 cross times 1 over denominator 1.8 end fraction left parenthesis refractive space index space of space air space equals 1 right parenthesis end style
vm =1.66 × 108m/s
Thus, the velocity of light in the medium is 1.66 × 108 m/s.


7. Define refractive index. If light enters from air to glass having a refractive index 1.5, then calculate the speed of light in glass.[3]

Solution: The ratio of the speed of light in vacuum to the speed of light in a medium is called the refractive index of the medium.

begin mathsize 12px style straight n equals fraction numerator speed space of space light space in space air over denominator speed space of space light space in space glass end fraction
1.5 equals fraction numerator 3 cross times 10 to the power of 8 over denominator speed space of space light space in space glass end fraction
speed space of space light space in space glass equals fraction numerator 3 cross times 10 to the power of 8 over denominator 1.5 end fraction end style
Thus, the speed of light in glass is 2 × 108 m/s.

8. Describe the Cartesian sign convention for a spherical lens. Draw a neat and labelled diagram to illustrate the sign convention. [3]

Solution: Cartesian sign convention used in optics are as follows:

  1. All distances are measured from the optical centre of the lens.
  2. All distances measured in the same direction as that of incident light rays are taken as positive.
  3. The distances measured against the direction of the incident ray are taken as negative.
  4. The distances measured above and perpendicular to the principal axis are taken as positive.
  5. The distances measured below and perpendicular to the principal axis are taken as negative.

9. If a concave mirror has a focal length of 15 cm, find the two positions where an object can be placed to give, in each case, an image twice the height of the object.                                                                                                                  [3]

Solution: f = -15 cm (focal length of the concave mirror)

begin mathsize 12px style therefore straight v space equals negative 2 straight u
1 over straight v plus 1 over straight u equals 1 over straight f
fraction numerator 1 over denominator negative 2 straight u end fraction plus 1 over straight u equals fraction numerator 1 over denominator negative 15 end fraction end style

m 2 or -2 ( image must be two times the height of an object)
Solving this we get,
u = - 7.5 cm
Thus,
v = -2 (-7.5) = 15 cm
Thus, if the object is placed at –7.5 cm, it will form an image of the size twice the size of an object.
For v = 2u

begin mathsize 12px style fraction numerator 1 over denominator 2 straight u end fraction plus 1 over straight u equals fraction numerator 1 over denominator negative 15 end fraction end style
Solving this we get,
u = -22.5 cm
Thus,
v = 2 (-22.5) = -45 cm
Thus, if the object is placed at –22.5 cm, it will also form an image of size twice the size of an object.

10. Draw a diagram to represent a convex mirror. On this diagram, mark the principal axis, principal focus F and centre of curvature C if the focal length of a convex mirror is 5 cm. Also, comment on the characteristics of the image if an object is placed 15 cm from the mirror.                                                                                                                                                         [5]

Solution:

As f = 5 cm
Centre of curvature C will be at a distance of 10 cm.
If an object is placed at a distance of 15 cm from the convex mirror, then the image formed will be behind the mirror and it will be virtual, erect and diminished.
By the mirror formula,
begin mathsize 12px style 1 over straight f equals 1 over straight v plus 1 over straight u
therefore 1 fifth equals 1 over straight v plus fraction numerator 1 over denominator left parenthesis negative 15 right parenthesis end fraction end style
v =3.75 cm
The image will be formed between the pole and the focus.

11.                                                                                                                                                                      [5]

  1. Two lenses A and B have a power of +3 D and –4 D, respectively. What is the nature and focal length of each lens?
  2.  A combination of lens contains two converging lens of focal length 30 cm and 50 cm and a diverging lens of focal length 40 cm. Find the power and focal length of the combination.

Solution: Power of lens = +3 D

begin mathsize 12px style straight p equals fraction numerator 1 over denominator straight f left parenthesis in space metre right parenthesis end fraction
straight f equals fraction numerator 1 over denominator plus 3 end fraction equals 0.33 m end style


Focal length = +0.33 m
The positive sign indicates that the lens is a convex lens.
Power of lens = -4 D
begin mathsize 12px style straight p equals fraction numerator 1 over denominator straight f left parenthesis in space metre right parenthesis end fraction
straight f equals fraction numerator 1 over denominator negative 4 end fraction equals 0.25 space straight m end style

The negative sign indicates that the lens is a concave lens.

b.

f1 = +30 cm = +0.3 m
f2 = +50 cm = +0.5 m
f3 = -40 cm = -0.4 m

P=P1+P2+P3

begin mathsize 12px style 1 over straight f equals 1 over straight f subscript 1 plus 1 over straight f subscript 2 plus 1 over straight f subscript 3
1 over straight f equals fraction numerator 1 over denominator 0.3 end fraction plus fraction numerator 1 over denominator 0.5 end fraction plus fraction numerator 1 over denominator negative 0.4 end fraction
1 over straight f equals plus 2.8 straight D end style

Thus, the power of the combination of lenses is +2.8 D.
begin mathsize 12px style straight f equals 1 over straight p equals fraction numerator 1 over denominator plus 2.8 end fraction end style
And the focal length of the combination of lenses is +0.35 m.


Chapter 11: Human Eye and the Colourful World

1. Why does the Sun appear white at noon?                                                                                                       [1]

Solution: The Sun is nearly overhead at noon. The sunlight has to pass through much smaller portion of the Earth's atmosphere. The scattering is much less and thus, Sun looks white.

2. What is meant by dispersion of white light?                                                                                                      [1]

Solution: The splitting up of white light into seven colours on passing through a transparent medium like a glass prism is called dispersion of light.

3. In the formation of spectrum of white light by a prism:                                                                                     [2]

  1. which colour is deviated least?
  2. which colour is deviated most? 
Solution:
  1. Red
  2. Violet

4. What is Tyndall effect? Explain with an example.                                                                                              [3]

Solution: When a beam of light strikes fine particles of smoke, dust, water droplets etc., the beam of light becomes visible. This phenomenon of the scattering of light by colloidal particles is known as the Tyndall effect.
Example: The sky appearing blue is an example of the Tyndall effect. The molecules of air and other fine particles in the atmosphere have a size smaller than the wavelength of visible light. Thus, they are more effective in scattering light of shorter wavelengths at the blue end than light of longer wavelengths at the red end. Red light has a wavelength greater than blue light. Thus, when sunlight passes through the atmosphere, the fine particles in the air scatter blue colour (shorter wavelengths) more strongly than red. The scattered blue light enters our eyes, and hence, the sky appears blue.

5. Why does the sky appear blue during the day and red at sunset?                                                                      [3]

Solution:

a.

  1. Red light has a wavelength greater than blue light.
  2. Thus, when sunlight passes through the atmosphere, the fine particles in air scatter blue colour (shorter wavelengths) more strongly than red.
  3. The scattered blue light enters our eyes, and hence, the sky appears blue.

b.

  1. At sunrise or sunset, the Sun is at the horizon.
  2. Light from the Sun near the horizon passes through a thicker layer of air and a larger distance in the Earth’s atmosphere before reaching our eyes.
  3. Near the horizon, most of the blue light and shorter wavelengths are scattered. Thus, the Sun appears reddish in colour.

6. Draw and explain the dispersion of white light through a prism. Also, comment on the deviation, wavelength, frequency and speed of the dispersed light.                                                                                                                                       [5]

Solution:

  1. The band of seven colours formed on the screen when a beam of light passes through a glass prism is called the spectrum of light.
  2. The seven colours are Violet, Indigo, Blue, Green, Yellow, Orange and Red. It is commonly known by the acronym VIBGYOR.
  3. The splitting of white light into its seven constituent colours after passing through the transparent medium is called dispersion of light.
  4. The dispersion of white light occurs due to the different speeds of these colours while passing through the prism.
  5. Red colour has the maximum speed and deviates the least, while violet colour has the minimum speed and deviates the most.
  6. The colours in the order of increasing frequency but decreasing wavelengths are Red, Orange, Yellow, Green, Blue, Indigo and Violet.

7. Write a short note on the formation of a rainbow with a neat and labelled diagram.                                            [5]

Solution:

  1. A rainbow is a natural spectrum which appears in the sky after rainfall.
  2. It is caused by the dispersion of sunlight by water droplets in the atmosphere. It always forms in the direction opposite to the Sun.
  3. The water droplets act like tiny prisms which refract and disperse sunlight. Then, they reflect light internally and refract light again.




Chapter 12: Electricity

1. Define electric power and state its SI units.                                                                                                    [1]

Solution: Electric power is electrical work done per unit time.
The SI unit of power is Watt (W).

2. The values of potential difference V applied across a resistor and the corresponding values of current I flowing in the resistor are given below:                                                                                                                                                    [2]

  1. What is the nature of the V-I graph plotted for the above values of potential difference and current?
  2. Which law is illustrated by such type of graph?
Solution:
  1. Straight line graph
  2. Ohm’s law is illustrated when nature of V- I graph is straight line.

3. What are the factors on which the heating effect of electric current depend? Explain with a practical example.    [3]

Solution: According to the law, the heat produced in a resistor is

  • Directly proportional to the square of the current in the resistor
  • Directly proportional to the resistance of the resistor
  • Directly proportional to the time for which the current flows through the resistance

The heating effect is used in bulbs. The current passing through the filament of a bulb causes the bulb to heat up and give out light and heat. The filament is made up of an element with a high melting point. The bulb is also filled with chemically inactive nitrogen or argon gas to prolong the life of the filament.

4. State and explain Ohm’s law with a graph.                                                                                                      [3]

Solution: Ohm’s law states that ‘the potential difference V across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same’.

begin mathsize 12px style straight V proportional to straight I
straight V over straight I equals constant space equals straight R end style

V=IR
If we plot the V-I graph for a conductor, then it shows a linear nature.

5. What is resistivity? State the factors on which it depends.                                                                                 [3]

Solution: Resistivity is defined as the resistance of a material of unit cross-section area and unit length. Resistivity is a characteristic property of a substance.
Resistivity of a material depends on the nature of substance and temperature.
Note: Resistivity does not depend on length and/or thickness of the conductor.

6. Resistors of 8 Ω and 12 Ω are connected in parallel to each other, while a resistor of 6 Ω is connected in series to a 6-V battery. Calculate:                                                                                                                                                         [5]

  1. Total resistance of the circuit
  2. Total current in the circuit
  3. Total potential difference across the 6-Ω resistor

Draw a neat and labelled circuit diagram for the same.

Solution:

R1 = 10 Ω, R2 = 20 Ω, R3 = 30 Ω
V = 6 V


Total resistance of the circuit:
begin mathsize 12px style 1 over straight R subscript straight p equals 1 over straight R subscript 1 plus 1 over straight R subscript 2 equals 1 over 8 plus 1 over 12 end style
Rp=4.8 Ω
Total effective resistance R = R +R =4.8 + 6 = 10.8Ω

Total current in the circuit:
V = IR
begin mathsize 12px style straight I equals fraction numerator 6 over denominator 10.8 end fraction equals 0.55 space straight A end style
Total potential difference across a 6-Ω resistor:
Current through R3 = 0.55 A
R3 =6 Ω
Thus,
Potential difference, V = 0.55× 6 =3.3 V

7. An electric iron draws 2.4 amperes of current from a 240-V source. Find                                                            [3]

  1. Resistance of the electric iron
  2.  Power consumed by the electric iron
Solution:

         I = 2.4 A
        V = 240 V

  1. R = V/I = 240/2.4 = 100 Ω
  2. P = V × I = 2.4 × 240 = 576 W

8. Give reasons for the following:                                                                                                                       [5]

  1. Filament-type electric bulbs are not power efficient.
  2. Coils of heating devices are made of alloys rather than pure metals such as copper.
  3. An electric bulb is not filled with normal atmospheric air but is filled with argon or nitrogen.
  4. Metals such as copper and aluminium are used on a large scale for the transmission of electricity.
  5.  Parallel combination is preferred over series combination for connecting electric appliances in houses.
Solution:

i.

  • Most of the electric power consumed by the filament of the bulb appears as heat due to which the electric bulb becomes hot.
  • A small amount of electric power is converted to light.
  • Thus, filament-type bulbs are not power efficient.

ii.

  • Resistivity of an alloy is much higher than that of a pure metal like copper.
  • An alloy does not undergo oxidation easily even at high temperature, thus avoiding the damage of the electric appliance.

iii.

  • If there is atmospheric air in an electric bulb, the extremely hot tungsten filament will catch fire due to the oxygen.
  • Gases like argon and nitrogen are chemically unreactive and do not cause the ungsten filament to catch fire, hence prolonging the life of the electric bulb.

iv.

  • Metals like copper and aluminium have low resistivity. Thus, they are good conductors of electricity.
  • They are cheaply available conductors.

v.

  • We can have separate switches for each appliance.
  • Each appliance can be operated separately. Thus, the working of one appliance does not affect the working of other appliances.
  • The potential difference across each appliance is the same, i.e. 220 V, as the power supply line.
  • Thus, each appliance can draw the required amount of current.

9.                                                                                                                                                                        [5]

  1. Derive the expression for heat produced due to current (I) flowing for a time period (t) through a resistor (R) having a        potential difference (v) across the terminals.
  2. Name the relation. List all the variables along with their SI units.
  3.  How much heat will an instrument of 14 W produce in half an hour if it is connected to a battery of 70 V? 
Solution:
  1. All materials offer resistance to the flow of current through them. So, some external energy is required to make the current flow. This energy is provided by the battery. Some of this energy gets dissipated as heat energy, so the resistor becomes hot. Work done in carrying a charge Q through a potential difference V is given as
    Q = It
    Using Ohm's law,
    V = I R
    W = I2 Rt
    This work done in carrying the charge through the wire is heat energy produced.
    Thus, H = V I t = I2 Rt
  2. This relation is known as Joule’s law of heating.
    H - Heat energy produced. Its SI unit is Joule (J).
    I - Current flowing in the wire. The SI unit is Ampere (A).
    R – Resistance. Its SI unit is Ohm (Ω).
    t – Time for which the current flows in the wire. Its SI unit is seconds (s).
  3. How much heat will an instrument of 14 W produce in half an hour if it is
    connected to a battery of 70 V?
    P = 14 W
    t = 0.5 h = 30 min × 60 = 1800 s
    V = 70 V
    We know
    H = I2 R t = V I t
    We know
    P = V × I
    I = 14/70 = 0.2 A
    H = 0.2 × 1800 = 360 Joule


Chapter 13: Magnetic Effect of Electric Current

1. What is the effect on magnetic field strength produced at a point near a straight conductor if the electric current flowing through it increases?                                                                                                                                                        [1]

Solution: If the electric current flowing through a wire is increased, the strength of the magnetic field produced at a point near the straight wire also increases.

2. What were the observations made by Oersted in his experiment of current-carrying conductors?                         [1]

Solution: Oersted’s experiment shows that a current-carrying wire produces a magnetic field around it. When the direction of current flowing in the wire is reversed, the direction of the magnetic field is also reversed.

3. If a current-carrying conductor is kept in a magnetic field, it experiences a force. List the factors on which the direction of this force depends.                                                                                                                                                  [3]

Solution:

  1. A current-carrying conductor when placed in a magnetic field experiences a force.
  2. When the direction of this current is reversed, the direction of the force also gets reversed.
  3. Similarly, when the direction of the magnetic field is reversed, the direction of the force is also reversed.
  4. The maximum force on a conductor acts when the current and the field are at right angles to each other.

4. Explain Fleming’s right-hand rule and left-hand rule.                                                                                       [3]

Solution:
Fleming’s right-hand rule:

Fleming’s right-hand rule is used to find the direction of induced current.

 

Stretch the thumb, forefinger and middle finger of the right hand such that they are mutually perpendicular to each other. If the first finger points in the direction of the field and the thumb in the direction of motion of the conductor, then the middle finger gives the direction of induced current in the conductor.
Fleming’s left-hand rule:
Fleming’s left-hand rule gives the direction of the magnetic force acting on a
conductor.

 

Stretch the thumb, forefinger and middle finger of the left hand such that they are mutually perpendicular to each other. If the first finger points in the direction of the field and the middle finger in the direction of the current, then the thumb gives the direction of the motion or the force on the conductor.

5. What is a magnetic field? List characteristics of a magnetic field. If two magnetic field lines intersect at a point, what does that indicate?                                                                                                                                                           [3]

Solution: The space surrounding a magnet in which a magnetic force is exerted is called a magnetic field.

Characteristics of a magnetic field:

  1. Magnetic field lines originate from the North Pole and end at the South Pole of the magnet.
  2. Magnetic field lines come close to one another near the poles of a magnet and are widely separated at other places.
  3. Magnetic field lines do not intersect each other.

Magnetic field lines do not intersect each other at any point because the resultant force on the North Pole at any point can only be in one direction. But if two magnetic field lines intersect, it will indicate that the resultant force on the North Pole at the point of intersection is in two directions. Such a case is not possible. Thus, two magnetic field lines never appear to intersect each other.

6. Define and explain the phenomenon of electromagnetic induction.                                                                    [3]

Solution: The phenomenon in which an electric current is induced in a conductor because of a changing magnetic field is called electromagnetic induction. When the North Pole of a bar magnet is brought near a coil, the needle of the galvanometer shows a deflection in one direction. When the magnet is moved away, the deflection is reversed.

 

When the coil is moved towards the magnet, a similar effect is observed. Similarly, when the experiment is repeated with the South Pole facing the coil, the galvanometer shows deflection opposite to the case of the North Pole. The magnetic field may also change due to the relative motion between the coils placed near a current-carrying conductor. In such a case, the magnetic field may change either due to the change in the current through the conductor or due to the relative motion between the coil and the conductor. The change in the magnetic field in the coil results in the current being induced in it. This phenomenon in which the changing magnetic field in a conductor induces a current in another conductor is known as electromagnetic induction.

7. A coil of insulated copper wire is connected to a galvanometer. What happens if a bar magnet is                           [3]

  1.  Pushed into the coil
  2.  Withdrawn from the hollow space of the coil
  3.  Held stationary inside the coil
Solution:
  1. When a bar magnet is pushed into the coil or withdrawn from the coil, there will be a deflection in the galvanometer.
  2. This is because of the induced current in the coil which is generated due to the relative motion between the coil and the magnet.
  3. When held stationary near the coil, there is no relative motion and so no current is induced in the coil.

8. Explain with a neat sketch the working of an electric motor.                                                                             [5]

Solution:

  1. An electric motor is a device which converts electrical energy to mechanical energy, and it works on the principle of force experienced by a currentcarrying conductor in a magnetic field.
  2. An electric motor consists of a rectangular coil of insulated copper wire. The coil is placed between the two poles of a magnet.
  3. The current in the coil enters from the source battery through the conducting brush X and flows back to the battery through brush Y. Current in the arms
    AB and CD is in opposite directions.
  4. On applying Fleming’s left-hand rule, we find that the force acting on the arm AB pushes it downwards, while the force acting on the arm CD pushes it upwards. Thus, the coil and the axle O rotate in the anti-clockwise direction.
  5. After half rotation, the current in the coil gets reversed with the help of a commutator. This reverses the direction of the force acting on the two arms AB and CD. However, these arms have reversed positions after that half rotation.
  6. Thus, the coil and the axle rotate in the same direction, i.e. anti-clockwise.


Chapter 15: Our Environment

1. Why should biodegradable and non-biodegradable wastes be discarded in two separate dustbins?                       [1]

Solution:

  • Biodegradable and non-biodegradable wastes should be discarded in two different dustbins so that they can be collected and disposed of separately.
  • Biodegradable waste can be decomposed in a natural manner by the process of composting.
  • Non-biodegradable wastes can be sent for recycling.
  • If the two wastes are collected in a single bin, they would mix and may form toxic compounds which can cause pollution.

2. In the following food chain, 20,000 J of energy was available to plants. How much energy would be available to man in this chain?                                                                                                                                                                        [1]
   Plants → Sheep → Man

Solution:

According to the 10% law of energy transfer,
Energy available from Sun = 20000 J
Energy converted by plants = 1% of 20000
Energy available with plants = 200 J
Energy transferred to deer = 10% of 200
Energy available with deer = 20 J

3. You being an environmentalist are interested in contributing towards the conservation of natural resources. List two activities that you can do on your own.                                                                                                                                  [1]

Solution: Activities which can be done as an environmentalist to conserve natural resources:

  • Use public transport for commuting instead of a personal vehicle to save fuel.
  • Avoid using clothes, accessories or articles made of animal skin to prevent the hunting of animals.
  • Using energy-efficient electrical appliances to save electricity.
  • Ensuring no leakage of water taps and pipes at home to save water resources.

4. Why is it necessary to ban the use of plastic bags?                                                                                        [2]

Solution: Plastic bags are non-biodegradable products which pose a serious threat to the environment. Non-biodegradable products do not decompose easily and thus keep on accumulating in the environment.

There are various other reasons for which plastic bags should be banned:

  • Plastic bags, on getting into drains and sewer system, choke them up resulting in spilling of dirty water on the roads
  • Burning of plastic bags produces harmful gases, which can cause health problems
  • Street animals may consume these plastic bags, which may even result in their deaths.

5. Give reason why a food chain cannot have more than four trophic levels.                                                          [2]

Solution: Food chains are limited to 4-5 trophic levels because energy losses between trophic levels restrict the length of food chains and the biomass of higher trophic levels. This loss of energy explains why there are rarely more than four trophic levels in a food chain or web. The organisms in a trophic level are not usually entirely consumed by organisms in the next trophic level.

6. What are the adverse effects of combustion of fossil fuels on the environment? List any two steps you would suggest to minimise environmental pollution caused by burning of fossil fuels.                                                                                    [3]

Solution: Combustion of fossil fuels is responsible for several environmental issues such as accumulation of greenhouse gases, acidification, air pollution, water pollution, damage to land surface and ground-level ozone. Steps to minimise environmental pollution caused by burning of fossil fuels:

  • Increase the use of solar, wind and hydro power
  • Use smokeless appliances
  • Promote afforestation

7. How is ozone formed in the upper atmosphere? Why is damage to the ozone layer a cause of concern to us? What causes this damage?                                                                                                                                                         [3]

Solution: 

Formation of the ozone layer:

  • The ozone layer is naturally found in the upper part of the atmosphere called the stratosphere.
  • It is created when ultraviolet radiation (sunlight) strikes the stratosphere, dissociating or splitting oxygen molecules (O2) to atomic oxygen (O).
  • Atomic oxygen quickly combines with further oxygen molecules to form ozone.
 
 Damage to the ozone layer:
  • The ozone layer functions as a shield against strong UV radiations and protects the Earth from these harmful radiations.
  • If the ozone layer is damaged, then the harmful UV rays would penetrate the Earth’s surface and can cause skin cancer and damage vegetation, animal life and human beings.

Causes of ozone layer depletion:

  • Damage to the ozone layer is caused by gases such as methyl bromide, nitrogen oxides released from freezers, air conditioners, aerosol products and industrial solvents.
  • Compounds such as CFCs break down into chlorine atoms in the atmosphere.
  • These chlorine atoms break down O3 into oxygen (O2) and nascent oxygen (O) which degrade the ozone layer.

8. Explain the phenomenon of ‘biological magnification’. How does it affect organisms belonging to different trophic levels, particularly the tertiary consumers?                                                                                                                                     [3]

Solution:

The intensity of accumulation of toxic substances such as DDT increases as we move from a lower trophic level to a higher trophic level in a food chain. This phenomenon is called biomagnification.

  • Let’s take the example of DDT which has been applied on farm lands to eradicate pests.
  • DDT gets washed off from the farm land and reaches the water body.
  • In water, small phytoplankton (producers) would accumulate certain amounts of DDT which would be passed on to the next trophic level, the zooplankton
  • (primary consumers), and then to fish (secondary consumers) which feed on these zooplankton and finally to tertiary consumers.
  • Concentration of DDT would increase with each trophic level, with producers having the minimum concentration, while the tertiary consumers having the highest concentration of DDT accumulation.
  • This is because tertiary consumers have consumed many secondary consumers which in turn had consumed many primary consumers. In this way, tertiary consumers accumulate the maximum concentration of toxic chemicals.
       

9. What is a food chain? Why is the flow of energy in an ecosystem unidirectional? Explain briefly.                          [3]

Solution: The sequence of living organisms in a community in which one organism consumes another organism to transfer food energy is called a food chain.

  • Energy enters plants from the Sun during the process of photosynthesis.
  • This energy is then passed on from one organism to another in a food chain.
  • Solar energy converted by autotrophs to food energy cannot be reconverted to solar energy, and the energy which passes from the herbivores to the carnivores can never go back to herbivores.
  • The energy lost as heat cannot be returned to plants and reused during photosynthesis.
  • Therefore, the flow of energy in an ecosystem is said to be unidirectional.

10.                                                                                                                                                                  [3]

  1. Why should national parks be allowed to remain in their pristine form?
  2.  Why is the reuse of materials better than recycling?
Solution:
  1. National parks should be allowed to remain in their pristine form so that the natural habitat of wild animals and birds is preserved.
  2. Recycling is the process which involves conversion of used materials to new materials which are ready for use again. It helps in the conservation of raw materials. However, the process involves a lot of expenditure and energy consumption. Recycling units release a lot of toxic chemicals as waste materials and pollute the environment.
    Reuse involves using the same materials again and again for different purposes. This does not utilise any money or energy and does not release toxic wastes. Hence, the reuse of materials is better than recycling.

11. Explain some harmful effects of agricultural practices on the environment.                                                      [5]

Solution:

Some harmful effects of agricultural practices on the environment:

  • Soil degradation: Extensive cropping causes loss of soil fertility. Also, over time, it can lead to soil erosion and finally to desertification.
  • Pollution: Use of synthetic fertilisers and pesticides leads to soil, water and air pollution.
  • Water shortage: Excess use of groundwater for agriculture lowers the water level. This results in acute water shortage at many places.
  • Biomagnification: Chemical pesticides, being non-biodegradable, accumulate in organisms in increasing amounts at each trophic level.
  • Deforestation: Indiscriminate cutting down of trees for agriculture has resulted in the loss of habitat for wildlife. Thus, it also causes damage to the natural ecosystem.


ASSERTION – REASONING QUESTIONS

Two statements are given - one labelled Assertion (A) and the other labelled Reason (R).
Select the correct answer to these questions from the codes (i), (ii), (iii) and (iv) given below.

  1. Both A and R are true, and R is the correct explanation of the assertion.
  2. Both A and R are true, but R is not the correct explanation of the assertion.
  3. A is true, but R is false.
  4. A is false, but R is true.


Chapter 1: Chemical Reactions and Equations

1. Assertion: Aluminium foil is commonly used for packing food preparations.
Reason: A coating of aluminium oxide is deposited on the metal surface and protects it against corrosion.

Solution: Correct Answer: (i)
Both A and R are true, and the reason is the correct explanation of the assertion. Aluminium from the foil slowly combines with oxygen present in air to form Al2O3. The oxide coating of the metal is not affected by water, acids or bases and prevents food from rancidity.

2. Assertion: When carbon dioxide gas is passed through lime water, a white precipitate is initially formed.
Reason: White precipitate is of calcium carbonate which is formed during the reaction.

Solution: Correct Answer: (i)
Both A and R are true, and the reason is the correct explanation of the assertion. Lime water contains traces of calcium hydroxide dissolved in it. It reacts with carbon dioxide gas to form a white precipitate of calcium carbonate.
Ca(OH)2 +CO2 (g) → CaCO3 (s) H2 O(g)
(inlimewater)            (whiteppt.)

3. Assertion: In the electrolysis of water, the volume of hydrogen liberated is twice the volume of oxygen formed.
Reason: It is because water has hydrogen and oxygen in the ratio 1:2.

Solution: Correct Answer: (iii)
A is true, but R is false. Hydrogen and oxygen in H2O are in the ratio 2:1.

4. Assertion: Silver chloride turns grey in sunlight.
Reason: Silver chloride decomposes into silver and chloride by light.

Solution: Correct Answer: (i)
Both A and R are true, and the reason is the correct explanation of the assertion. Silver chloride turns grey in sunlight. This is due to the decomposition of silver chloride into silver and chlorine by light.
begin mathsize 12px style 2 AgCl left parenthesis straight s right parenthesis space rightwards arrow with sunlight on top 2 Ag left parenthesis straight s right parenthesis plus space Cl subscript 2 space left parenthesis straight g right parenthesis end style

5. Assertion: Copper can easily displace silver on reacting with an aqueous solution of silver nitrate.

Reason: Silver can be easily precipitated since it is insoluble in water.

Solution: Correct Answer: (iii)
A is true, but R is false. Copper can displace silver from silver nitrate solution because it is a more reactive element than silver.


Chapter 2: Acids, Bases and Salts

1. Assertion: Reaction of zinc granules with dilute sulphuric acid releases a gas with a pop sound.
Reason: When a metal reacts with an acid, it releases water vapour.

Solution: Correct Answer: (iii)
A is true, but R is false. When a metal reacts with an acid, it releases hydrogen gas which produces a pop sound.

2. Assertion: An acid is a substance which contains one or more hydrogen atoms in its molecule.
Reason: An aqueous solution of acid turns blue litmus red.

Solution: Correct Answer: (iv)
A is false, but R is true. An acid is a substance which can release one or more H+ ions in aqueous solution, and an acid turns blue litmus paper red.

3. Assertion: Acetic acid is a weak acid.
Reason: Acetic acid gets partially ionised in aqueous solution.

Solution: Correct Answer: (i)
Both A and R are true, and the reason is the correct explanation of the assertion. Strength of an acid depends on the number of hydrogen ions they release in aqueous solution. Since acetic acid releases only a small amount of hydrogen ions in aqueous solution, it is considered a weak acid.

4. Assertion: Milk of magnesia is taken to get rid of pain in the stomach during indigestion.
Reason: Milk of magnesia is a base and it neutralises the excess acid in the stomach.

Solution: Correct Answer: (i)
Both A and R are true, and the reason is the correct explanation of the assertion. Milk of magnesia (magnesium hydroxide) is most commonly found as a liquid. The magnesium in the liquid can neutralise excess stomach acid and is hence used to treat indigestion.

5. Assertion: Dry HCl gas does not change the colour of litmus paper.
Reason: Dry HCl gas can behave as an acid.

Solution: Correct Answer: (iii)
A is true, but R is false. Dry HCl does not change the colour of litmus paper because it cannot behave an acid, so it cannot dissociate in the dry state to release H+ ions.


Chapter 3: Metals and Non-metals

1. Assertion: Metals can be drawn into wires and reshaped.
Reason: Metals are ductile and malleable.

Solution: Correct Answer: (i)
Both A and R are true, and the reason is the correct explanation of the assertion. The ability of metals to be drawn into wires is called ductility. Because of malleability and ductility, metals can be given different shapes according to our needs.

2. Assertion: Rusting of iron does not take place in ordinary water.
Reason: Ordinary water contains dissolved oxygen.

Solution: Correct Answer: (iv)
A is false, but R is true. Rusting of iron can take place in ordinary water. However, it is not possible in distilled water.

3. Assertion: Positive ions are known as cations.
Reason: Cations are released at the anode.

Solution: Correct Answer: (iii)
A is true, but R is false. Cations are positive ions which migrate towards the cathode or negative electrode in an electrolytic cell. These ions are released as neutral species by losing their charge.

4. Assertion: When zinc nitrate reacts with aluminium, aluminium nitrate is formed.
Reason: Aluminium is more reactive than zinc.

Solution: Correct Answer: (i)
Both A and R are true, and the reason is the correct explanation of the assertion. Zinc nitrate reacts with aluminium, and aluminum replaces zinc because it is more reactive than zinc.

5. Assertion: Sodium chloride is not soluble in water.
Reason: Electrovalent compounds are generally soluble in water.

Solution: Correct Answer: (iv)
A is false, but R is true. Sodium chloride is soluble in water because electrovalent compounds are generally soluble in water.


Chapter 4: Carbon and its Compounds

1. Assertion: Diamond is the hardest crystalline allotropic form of carbon.
Reason: Carbon atoms in diamond are tetrahedral in nature.

Solution: Correct Answer: (i)
Both A and R are true, and the reason is the correct explanation of the assertion. Carbon atoms in diamond are tetrahedral as well as closely packed in space.

2. Assertion: A large number of carbon compounds exists due to the self-linking property of carbon known as catenation.
Reason: Strength of carbon to carbon bonds is very high.

Solution: Correct Answer: (i)
Both A and R are true, and the reason is the correct explanation of the assertion. For example, C-C bond strength (355 KJ mol-1) is very high, and it makes stable carbon to carbon bonds.

3. Assertion: During fermentation, glucose is converted to ethyl alcohol with the help of the enzyme maltase.
Reason: Alcoholic fermentation is carried out by certain enzymes present in yeast.

Solution: Correct Answer: (iv)
A is false, but R is true. During fermentation, glucose is converted to ethyl alcohol with the help of the enzyme zymase present in yeast.

4. Assertion: Zinc nitrate reacts with aluminium.
Reason: Aluminium is more reactive than zinc.

Solution: Correct Answer: (i)
Both A and R are true, and the reason is the correct explanation of the assertion. Zinc nitrate reacts with aluminium. Aluminum replaces zinc because it is more reactive than zinc.

5. Assertion: Sodium chloride is not soluble in water.
Reason: Electrovalent compounds are generally soluble in water.

Solution: Correct Answer: (iv)
A is false, but R is true. Sodium chloride is soluble in water because electrovalent compounds are generally soluble in water.


Chapter 5: Periodic Classification of Elements

1. Assertion: Chlorine is the most reactive member of the halogen family.
Reason: Size of the chlorine atom is greater than that of the fluorine atom.

Solution: Correct Answer: (iv)
A is false, but R is true. Fluorine is the most reactive element belonging to the halogen family and not chlorine. Fluorine is placed above chlorine in the periodic table.

2. Assertion: In the long form of periodic table, the elements are arranged in order of increasing masses
Reason: The properties of the elements are related to the atomic number.

Solution: Correct Answer: (iv)
A is false, bur R is true. In the long form of the periodic table, the elements are arranged in order of increasing numbers.

3. Assertion: In a triad, the three elements present have the same gaps of atomic masses.
Reason: Elements in a triad have similar properties.

Solution: Correct Answer: (iv)
A is false, but R is true. In a triad, the atomic mass of the middle element is the mean of the atomic masses of the first and third elements.

4. Assertion: Tendency to lose electrons will decrease across a period.
Reason: Effective nuclear charge increases across a period.

Solution: Correct Answer: (i)
Both A and R are true, and the reason is the correct explanation of the assertion. As the effective nuclear charge acting on the valence shell electrons increases across a period, the tendency to lose electrons will decrease.

5. Assertion: Boron and silicon have the same properties.
Reason: Boron and silicon are in the same block.

Solution: Correct Answer: (ii)
Both A and R are true, but R is not the correct explanation of the assertion. Boron and silicon have the same properties because both of these elements are metalloids.


Chapter 6: Life Processes

1. Assertion: Diffusion does not meet the high energy requirements of multicellular organisms.
Reason: Diffusion is a fast process but only occurs at the surface of the body.

Solution: Correct Answer: (iii)
Solution: A is true, but R is false. In multicellular organisms, the volume of the body is large, and hence, oxygen cannot diffuse into all the cells of the body quickly. Diffusion, being a slow process, takes a lot of time to circulate oxygen to all the body cells. Hence, diffusion is insufficient to meet the oxygen requirements of multicellular organisms like humans.

2. Assertion: Saliva contains an enzyme called amylase.
Reason: Amylase helps to break down simple sugars like glucose into complex molecules like starch.

Solution: Correct Answer: (iii)
A is true, but R is false. Saliva contains an enzyme called salivary amylase which breaks down starch, a complex molecule, to give simple sugars such as glucose.

3. Assertion: Bile helps in the emulsification of fats.
Reason: Bile makes acidic food coming from the stomach alkaline so that pancreatic enzymes can act on it.

Solution: Correct Answer: (ii)
Both A and R are true, but R is not the correct explanation of the assertion. Bile is a greenish yellow liquid secreted by the liver in the small intestine. It contains bile pigments, bile salts, cholesterol and phospholipids. It breaks down large fats into smaller globules by a process called emulsification. They also help in incorporating fatty acids and glycerol into small, spherical, water-soluble molecules called micelles.

4. Assertion: The inner lining of the small intestine has numerous finger-like projections called villi.
Reason: Villi increase the surface area for absorption.

Solution: Correct Answer: (i)
Both A and R are true, and R is the correct explanation of the assertion. The inner lining of the small intestine has numerous finger-like projections called villi which increase the surface area for absorption. Villi are richly supplied with blood vessels which take the nutrients of absorbed food to each and every cell of the body.

5. Assertion: Rings of cartilage are present in the throat.
Reason: They ensure that the air passage does not collapse.

Solution: Correct Answer: (i)
Both A and R are true, and R is the correct explanation of the assertion. Rings of cartilage are present in the throat as they provide flexibility and support to the trachea and allow it to move and bend when breathing. They also support the throat during air blockage and help in smooth air flow.


Chapter 8: How do Organisms Reproduce?

1. Assertion: Reproduction enables the continuity of life for generations.
Reason: Reproduction is a biological process in which an organism gives rise to young ones similar to itself.

Solution: Correct Answer: (i)
Both A and R are true, and R is the correct explanation of the assertion. Reproduction is a process which involves the production of new organisms similar to the parent. Producing the same kind of individual helps to perpetuate the species and enables the continuity of life for generations.

2. Assertion: Paramoecium reproduces by budding.
Reason: All unicellular organisms reproduce by asexual methods.

Solution: Correct Answer: (iv)
A is false, but R is true. All unicellular organisms, due to simplicity of their structure, reproduce by asexual methods of reproduction. Paramoecium is a unicellular organism. However, it reproduces by binary fission and not by budding.

3. Assertion: Errors during DNA copying are a source of variations.
Reason: Variations are useful for ensuring the survival of the species.

Solution: Correct Answer: (i)
Both A and R are true, and R is the correct explanation of the assertion. The DNA copying mechanism is not absolutely accurate, and the resultant errors are a source of variations in a population of an organism. Every individual organism cannot be protected by variations, but in a population, variations are useful for ensuring the survival of the species.

4. Assertion: Tail of the sperm consists of an acrosome and mitochondria.
Reason: Acrosome contains enzymes which assist in fertilisation.

Solution: Correct Answer: (iv)
A is false, but R is true. The head of the sperm consists of an acrosome which contains enzymes which help the sperm to penetrate the egg and assist in fertilisation. Mitochondria are present in the middle portion of the sperm.

5. Assertion: Contraceptives can be effectively used for population control.
Reason: They can be used to avoid unwanted pregnancy in females.

Solution: Correct Answer: (i)
Both A and R are true, and R is the correct explanation of the assertion. Contraceptives help to avoid the occurrence of unwanted pregnancy in females. This helps to prevent unwanted births and thereby helps in population control.


Chapter 9: Heredity and Evolution

1. Assertion: In a monohybrid cross, the F1 generation indicates dominant characters.
Reason: Dominance occurs only in the heterozygous state.

Solution: Correct Answer: (iii)
A is true, but R is false. A monohybrid cross involves a single pair of alleles or factors of a character. Crossing of parents produces the F1 generation or offspring which are heterozygous dominant. A dominant character is one of a pair of alleles which can express itself whether present in either homozygous or heterozygous state.

2. Assertion: The father is responsible for the sex of the child.
Reason: Boys inherit the Y chromosome and girls inherit the X chromosome from the father.

Solution: Correct Answer: (i)
Both A and R are true, and R is the correct explanation of the assertion. All children (boys or girls) always inherit an X chromosome from their mother. Thus, the sex of children is determined by what they inherit from their father. A child who inherits an X chromosome from the father will be a girl, and one who inherits a Y chromosome from the father will be a boy.

3. Assertion: Mendel was successful in his hybridisation experiments.
Reason: Garden pea proved ideal experimental material.

Solution: Correct Answer: (ii)
Both A and R are true, but R is not the correct explanation of the assertion. Mendel chose garden pea for his experiments as it had well-defined characters, bisexual flowers and assisted in easy hybridisation. Besides these features, garden pea, being able to self-fertilise, had pure lines for several generations. Therefore, any variety used was pure for the characters it carried. Mendel’s success was mainly based on the fact that he considered a single character at one time. Characters of the garden pea plant proved to be an added advantage for Mendel’s experiments.

4. Assertion: Mendel chose pea plant for his experiments.
Reason: Pea plant provides diverse visible traits and has a short life span.

Solution: Correct Answer: (i)
Both A and R are true, and R is the correct explanation of the assertion. Pea plants have a short life span and provide diverse visible traits which helped Mendel to carry out a larger number of crosses using different traits.

5. Assertion: When TT and tt pea plants were crossed, only tall plants were obtained in F1 progeny.
Reason: This was because tall allele was dominant over the dwarf allele.

Solution:Correct Answer: (i)
Both A and R are true, and R is the correct explanation of the assertion. With respect to height of the plant, tall (TT) plants are dominant over dwarf (tt) plants. Tallness is a dominant trait and dwarfness is a recessive trait. According to the law of dominance, only dominant characters are expressed while recessive characters are suppressed in the F1 generation.


Chapter 10: Light - Reflection & Refraction

1. Assertion: Concave mirrors are used in torches.
Reason: The light bulb placed at the focus of the concave reflector produces a strong beam of light.

Solution: Correct Answer: (i)
Both A and R are true, and R is the correct explanation of the assertion. Concave mirrors are used in torches because the light bulb placed at its focus converges light rays and produces a strong beam of light.

2. Assertion: Magnification produced by a concave mirror of focal length 10 cm for the object distance 15 cm and image distance 30 cm is –2.
Reason: The image formed by a concave mirror is always real, inverted and magnified irrespective of the position of the object.

Solution: Correct Answer: (iii)
A is true, but R is false. Magnification produced by a concave mirror of focal length 10 cm for the object distance 15 cm and image distance 30 cm is –2 because linear magnification produced by a mirror is equal to the ratio of the image distance to the object distance with a negative sign. A concave mirror does not always form a real, inverted and magnified image as it can also form a virtual, erect and magnified image when the object is placed between the pole and the focus of the concave mirror.

3. Assertion: When a beam of light is incident on a glass slab perpendicularly, there is no refraction.
Reason: The rays of light which fall perpendicularly on the glass slab reach, enter and leave the glass slab at the same instant of time.

Solution: Correct Answer: (i)
Both A and R are true, and R is the correct explanation of the assertion. When a beam of light is incident on the glass slab perpendicularly, there is no refraction because the rays of light which fall perpendicularly on the glass slab reach, enter and leave the glass slab at the same instant.

4. Assertion: When the direction of the rays of light is reversed, they retrace their path.
Reason: According to the dual theory of light, light is a particle and a wave simultaneously.

Solution: Correct Answer: (ii)
Both A and R are true, but R is not the correct explanation of the assertion. According to the principle of reversibility, when the direction of the rays of light is reversed, the rays retrace their path.

5. Assertion: When we stand close to the convex mirror our image is virtual, erect and diminished.
Reason: When object lies anywhere between pole and infinity of convex mirror, the image formed is virtual, erect and diminished.

Solution:Correct Answer: (i)
Both A and R are true, and R is the correct explanation of the assertion. A When object is placed anywhere between the pole and infinity of convex mirror the image formed is virtual, erect and diminished.


Chapter 11: Human Eye and the Colourful World

1. Assertion: The Sun appears for about two minutes even after it has set below the horizon.
Reason: The refraction of light rays coming from the Sun is caused by the Earth’s atmosphere.

Solution: Correct Answer: (i)
Both A and R are true, and R is the correct explanation of the assertion. The Sun appears for about two minutes even after it has set below the horizon due to atmospheric refraction.

2. Assertion: Red colour light after dispersion is deviated the most.
Reason: Red colour has the maximum speed in a glass prism.

Solution: Correct Answer: (iv)
A is false, but R is true. The red colour light has the maximum speed in a glass prism, and thus, it deviates the least.

3. Assertion: White light splits up into seven colours on passing through a transparent medium.
Reason: Dispersion of white light occurs because the colours of white light travel at the same speed through the glass prism.

Solution: Correct Answer: (iii)
A is true, but R is false. White light splits up into seven colours on passing through a transparent medium. The dispersion of white light occurs because the colours of white light travel at different speeds through the glass prism.

4. Assertion: The Sun appears white when it is overhead in the sky.
Reason: Light coming from the Sun has to travel a relatively shorter distance through the atmosphere to reach us.

Solution: Correct Answer: (i)
Both A and R are true, and R is the correct explanation of the assertion. When light coming from the Sun has to travel a relatively shorter distance through the atmosphere to reach us, the Sun appears white when it is overhead in the sky.


Chapter 12: Electricity

1. Assertion: The property of a conductor due to which it opposes the flow of current through it is called resistance.
Reason: The resistance of a conductor depends on thickness and temperature.

Solution: Correct Answer: (ii)
Both A and R are true, but R is not the correct explanation of the assertion. The property of a conductor due to which it opposes the flow of current through it is called resistance.The resistance of a conductor depends on thickness, length, nature of material and temperature.

2. Assertion: Electrical wires are made of copper or aluminium.
Reason: Copper and aluminium have very low electrical resistance.

Solution: Correct Answer: (i)
Both A and R are true, and R is the correct explanation of the assertion. Electrical wires are made of copper or aluminium because they have very low electrical resistance.

3. Assertion: If the length of the wire is doubled, the resistance gets doubled.
Reason: Resistance of a conductor is inversely proportional to its length.

Solution: Correct Answer: (iii)
A is true, but R is false. If the length of the wire is doubled, then the resistance gets doubled, because the resistance of a conductor is directly proportional to the length of the conductor.

4. Assertion: Nichrome wire is used to make the filament of an electric bulb.
Reason: Nichrome does not undergo oxidation even at a high temperature.

Solution: Correct Answer: (iv)
A is false, but R is true. Nichrome wire is used to make the heating element of electrical appliances because it has more resistivity than pure metal and does not undergo oxidation at high temperatures. Tungsten filament is used in an electric bulb.

5. Assertion: In a parallel combination of circuits, each appliance has its separate switch and thus can be switched on/off independently.
Reason: In a parallel combination, each appliance gets the same voltage as that of the power supply.

Solution: Correct Answer: (ii)
Both A and R are true, but R is not the correct explanation of the assertion. In a parallel combination of circuits, each appliance has a separate switch which can be switched on/off independently. In a parallel combination, each appliance gets the same voltage as that of the power supply.


Chapter 13: Magnetic Effect of Electric Current

1. Assertion: Motion of a charged particle under the action of the magnetic field is not always with constant speed.
Reason: Magnetic force does not have any component either along or opposite to the direction of the charged particle.

Solution: Correct Answer: (iv)
A is false, but R is true. The energy of a charged particle moving in the magnetic field alone does not change as it experiences a force perpendicular to the direction of motion. Thus, the motion of a charged particle under the action of the magnetic field is always with constant speed. The magnetic force does not have any component either along or opposite to the direction of the charged particle.

2. Assertion: Current generated by powerhouse generators reverses its direction after equal intervals of time.
Reason: Direct current can be transmitted for long distances without much loss of electric current.

Solution: Correct Answer: (iii)
A is true, but R is false. Current generated by powerhouse generators is alternating current. This current reverse its direction after equal intervals of time. Alternating current can be transmitted for long distances without much loss of electric current.

3. Assertion: When a bar magnet is held stationary inside the coil, there is no deflection in the galvanometer.
Reason: If there is no relative motion between the magnet and the coil, the current will not be induced in the coil.

Solution: Correct Answer: (i)
Both A and R are true, and R is the correct explanation of the assertion. When the bar magnet is held stationary inside the coil, there is no deflection in the galvanometer. This shows that there is no current produced in the coil as there is no relative motion between the coil of the wire and the magnet.

4. Assertion: The magnitude of the current induced in the coil can be increased by increasing the number of turns in the coil.
Reason: By reversing the direction of motion of the coil, the direction of current can be reversed.

Solution: Correct Answer: (ii)
Both A and R are true, but R is not the correct explanation of the assertion. The magnitude of the current induced in the coil can be increased by increasing the number of turns in the coil. By reversing the direction of motion of the coil, the direction of current can be reversed.

5. Assertion: Freely suspended current carrying solenoid comes to rest in N-S direction just like a bar magnet.
Reason: On one end of current carrying straight solenoid behaves as a North pole and other end as a South pole.

Solution: Correct Answer: (ii)
Both A and R are true and R is not correct explanation of A. Assertion statement tells the relation between Earth's geographical N-S pole and magnetic N-S pole whereas reason statement gives the property of solenoid.


Chapter 15: Our Environment

1. Assertion: Ecology is the study of the relationship between living organisms and their environment.
Reason: The biotic community and the non-living environment of an area function together to form an ecosystem.

Solution: Correct Answer: (i)
Both A and R are true, and R is the correct explanation of the assertion. An ecosystem is a community of living organisms such as plants and animals in association with the non-living components such as air, water and soil of the environment interacting together as a system. These biotic and non-biotic components are linked together through nutrient cycles and energy flow.

2. Assertion: Producers are called nature’s recyclers.
Reason: They produce food by photosynthesis and make it available to other links of the ecosystem.

Solution: Correct Answer: (iv)
A is false, but R is true. Producers synthesise food through photosynthesis. Organisms depend on the producers either directly or indirectly for their sustenance. Decomposers are called nature’s recyclers as they breakdown the complex organic substances into simple inorganic substances that enter the soil and are used up once more by plants.

3. Assertion: Food chains are limited to 4-5 trophic levels.
Reason: The flow of energy within trophic levels follows the 10% law.

Solution: Correct Answer: (ii)
Both A and R are true, but R is not the correct explanation of the assertion. Food chains are limited to 4-5 trophic levels because energy losses between trophic levels restrict the length of food chains and the biomass of higher trophic levels. As we pass from one trophic level to the next, only 10% of energy is transferred from the first trophic level to the next. This is because a lot of energy is lost to the surroundings and the rest is utilised by the organism.

4. Assertion: Most air conditioning units are not eco-friendly.
Reason: They cause ozone depletion.

Solution: Correct Answer: (i)
Both A and R are true, and R is the correct explanation of the assertion. Most air conditioning units emit greenhouse gases. These gases trap heat and lead to the depletion of the ozone layer. As a result, the harmful UV rays of the Sun are able to penetrate the Earth’s surface and cause skin cancer in human beings.

5. Assertion: Plastic is biodegradable.
Reason: Biodegradable substances can be broken down by biological processes.

Solution: Correct Answer: (iv)
A is false, but R is true. Biodegradable substances can be broken down into simpler substances by biological processes. Plastic is a non-biodegradable substance. It is inorganic and does not decompose. Plastics accumulate and cause tremendous harm to the environment and living organisms.


PARAGRAPH BASED QUESTIONS

1. Read the following and answer any four questions from (i) to (v) Newlands’ Law of Octaves In 1864, Newlands arranged the known 56 elements in the order of increasing atomic masses. He observed that the properties of every eighth element are similar to the properties of the first element. Based on this observation, he proposed the Law of Octaves for the classification of elements.
Law of Octaves: When the elements are arranged in the increasing order of their atomic masses, the properties of every eighth element are similar to the first.


Many limitation were found in Newlands’ octaves. This law was found to be applicable only up to calcium. Newlands fitted all the known elements in a table of 7 X 8 that is 56 boxes. Newlands placed two elements each in some boxes to accommodate all the
known elements in the table. For example, Co and Ni, Ce and La. Moreover, he placed some elements with different properties under the same note in the octave. For example, Newlands placed the metals Co and Ni under the note ‘Do’ along with halogens, while Fe, having similarity with Co and Ni, away from them along with the nonmetals O and S under the note ‘Ti’. Also, Newlands’ octaves did not have provision to accommodate the newly discovered elements. The properties of the new elements discovered later on did not fit in the Newlands’ law of octaves.

i) Newland’s rule known as _____________ .

  1. Law of triad
  2. Law of octave
  3. Law of periodic table
  4. Periodic law

ii) Newland’s law of octaves based upon_______ .

  1. Increasing order of atomic number
  2. Increasing order of atomic mass
  3. Increasing order of electron
  4. Increasing order of atomic size

iii) Newland’s law of octaves is applicable to ___________ .

  1. Sodium
  2. Magnesium
  3. Calcium
  4. Sulphur

iv) A and B are two elements having similar properties which obey the law of octave. How many elements are there in between A and B?

  1. 6
  2. 7
  3. 8
  4. 5

v) Which of the following is true regarding Newland’s Law of Octaves?

  1. It worked well with only lighter elements.
  2. It was applicable only up to calcium.
  3. Both are correct.
  4. Both are incorrect.

 Solution:

  1. (b) Newland’s rule known as ‘law of octave’.
  2. (b) Newland’s law of octaves based upon increasing order of atomic mass.
  3. (c) Newland’s law of octaves is applicable to calcium.
  4. (a) 6 elements are there in between A and B.
  5. (c) Both the statements are correct. Hence, option C is correct.

2. Read the following and answer any four questions from (i) to (v)M is an element in the form of a powder. M burns in oxygen and the product obtained is soluble in water. The solution is tested with litmus.

i) If M is a metal, then the litmus will turn _____.

  1. Red
  2. Blue
  3. Yellow
  4. Orange

ii) If M is a non-metal, then the litmus will turn _____.

  1. Red
  2. Blue
  3. Yellow
  4. Orange 

iii) If M is a reactive metal, then _____ will be evolved when M reacts with dilute sulphuric acid.

  1. Carbon dioxide gas
  2. Hydrogen gas
  3. Nitrogen gas
  4. Oxygen gas

iv) If M is a metal, it will form _____ oxide, which will form ______ solution with water.

  1. Basic, alkaline
  2. Acidic, acidic
  3. Basic, acidic
  4. Acidic basic

v) If M is a non-metal, it will not conduct electricity in the form of ______.

  1. Diamond
  2. Graphite
  3. Cesium
  4. Zinc
Solution:
  1. (b) If M is a metal, then the litmus will turn blue.
  2. (a) If M is a non-metal, then the litmus will turn red.
  3. (b) If M is a reactive metal, then hydrogen gas will be evolved when M reacts with dilute sulphuric acid.
  4. (a) If M is a metal, it will form basic oxide, which will form alkaline solution with water.
  5. (b) If M is a non-metal, it will not conduct electricity in the form of graphite.

3. Read the following and answer any four questions from (i) to (v)
It is an allotrope of carbon containing clusters of 60 carbon atoms joined together to form spherical molecules. There are 60 carbon atoms in a molecule of buckminsterfullerene, so its formula is C60. The allotrope was named buckminsterfullerene after the American architect Buckminster Fuller.

(i) How many hexagons of carbon atoms are present in one molecule of Buckminster fullerene?

  1. 20
  2. 15
  3. 30
  4. 18

(ii) How many pentagons of carbon atoms are present in one molecule of Buckminster fullerene?

  1. 10
  2. 12
  3. 15
  4. 20

(iii) Which of the following molecule is called buckminsterfullerene?

  1. C90
  2. C60
  3. C70
  4. C120

(iv)Which allotrope of carbon exists as spherical molecules?

  1. Diamond
  2. Coke
  3. Graphite
  4. Fullerene

(v) Which of the following is true about C60?

  1. Each carbon bonded covalently to 3 other carbon atoms in a hexagonal ball like structure.
  2. Each carbon bonded covalently to 4 other carbon atoms in layers
  3. A giant lattice structure
  4. Pentagonal in shape
Solution:
  1. (a) There are 20 hexagons of carbon atoms are present in one molecule of Buckminster fullerene.
  2. (b) There are 12 pentagons of carbon atoms are present in one molecule of Buckminster fullerene.
  3. (b) C60 molecule is called buckminsterfullerene.
  4. iv) (d) Fullerene exists as spherical molecules.
  5. (a) Each carbon bonded covalently to 3 other carbon atoms in a hexagonal ball like structure.

4. Read the following and answer any four questions from (i) to (v)
Metal A burns in air, on heating, to form an oxide A2O3 whereas another metal B bums in air only on strong heating to form an oxide BO. The two oxides A2O3 and BO can react with hydrochloric acid as well as sodium hydroxide solution to form the corresponding salts and water. And element E forms an oxide E2O. An aqueous solution of E2O turns red litmus paper blue.

(i) What is the nature of oxide A2O3?

  1. aAcidic oxide
  2. Basic oxide
  3. Amphoteric oxide
  4. Neutral oxide

(ii) What is the nature of oxide BO?

  1. Acidic oxide
  2. Basic oxide
  3. Amphoteric oxide
  4. Neutral oxide

(iii) Name one metal like A.

  1. Sulphur
  2. Gold
  3. Aluminium
  4. Iron

(iv)Name one metal like B.

  1. Sulphur
  2. Gold
  3. Zinc
  4. Iron

(v) State the nature of oxide element E.

  1. Acidic
  2. Basic
  3. Amphoteric
  4. Neutral
Solution:
  1. (c) The nature of oxide A2O3 is amphoteric.
  2. (c) The nature of oxide BO is amphoteric.
  3. (c) Matal A is aluminium.
  4. (c) Metal B is zinc.
  5. (b) The nature of E2O is basic.

5. Read the following and answer any four questions from (i) to (v)
Sample of five metals ‘A’, ‘B’, ‘C’, ‘D’ and ‘E’ was taken and added to the following solution one by one. The results obtained have been tabulated as follows.

Use the above table to answer the following questions about the given metals.

i) Which of them is most reactive?

  1. A
  2. B
  3. D
  4. E

ii) What would you observe if ‘B’ is added to CuSO4?

  1. Reddish brown depositCBSE
  2. Grey deposit
  3. Greyish silver deposit
  4. Black deposit

iii) Arrange ‘A’, ‘B’, ‘C’, ‘D’ and ‘E’ in the increasing order of reactivity.

  1. E>A>B>C>D
  2. E>B>A>D>C
  3. E>B>A>C>D
  4. C>D>A>E>B

iv) Container of which metal can store zinc sulphate and silver nitrate solution?

  1. A
  2. B
  3. C
  4. D

v) Which of the above solution(s) can be stored in a container made of any of these metals?

  1. Aluminium sulphates
  2. Magnesium sulphates
  3. Both (a) & (b)
  4. None of these
Solution:
  1. (d) E is most reactive. Because it react with most substances.
  2. (a) Reddish brown deposit of copper will be formed since displacement has taken place.
  3. (c) E>B>A>C>D reason: The more it reacts the more reactive it is. Just count a number of displacements a metal will give.
  4. (d) Container of metal D can be used for this purpose as it does not react with any of them.
  5. (c) Aluminium and magnesium sulphates can be used to store in any container because the react with none.

6. Read the following paragraph and answer the questions that follow:
The food material taken in during the process of nutrition is used in cells to provide energy for various life processes. Diverse organisms do this in different ways – some use oxygen to breakdown glucose completely into carbon dioxide and water, some use other pathways that do not involve oxygen.

i) Which three carbon molecule is formed during the breakdown of glucose?

  1. Lactic acid
  2. Glyceraldehyde
  3. Acetic acid
  4. Pyruvate

ii) The process in which pyruvate may be converted into ethanol and carbon dioxide is

  1. Germination
  2. Fermentation
  3. Cellular respiration
  4. Oxidation

iii) Breakdown of pyruvate using oxygen takes place in the

  1. Cytoplasm
  2. Stroma
  3. Mitochondria
  4. Cellular matrix

iv) The breakdown of pyruvate in the absence of oxygen producing lactic acid occurs primarily in the

  1. Muscle cells
  2. Brain cells
  3. Cardiac cells
  4. Nerve cells

v) Observe the graph and interpret which of the following is true w.r.t. rate of respiration?

  1. Increases with an increase in temperature
  2. Increases first and then decreases with an increase in temperature
  3. Decreases with an increase in temperature
  4. Decreases first and then increases with an increase in temperature
Solution:
  1. (d) During respiration, the first step is the break-down of glucose, a six-carbon molecule, into a three-carbon molecule called pyruvate.
  2. (b) Pyruvate may be converted into ethanol and carbon dioxide. This process takes place in yeast during fermentation.
  3. (c) Breakdown of pyruvate using oxygen takes place in the mitochondria.
  4. (a) Breakdown of pyruvate in the absence of oxygen producing lactic acid occurs primarily in the muscle cells during heavy physical exercise.
  5. (b) The rate of respiration increases first with an increase in temperature and then decreases as the temperature continues to increase even further.

7. Read the following paragraph and answer the questions that follow:
Plants have tissues to transport water, nutrients and minerals. Xylem transports water and mineral salts from the roots up to other parts of the plant, while phloem transports sucrose and amino acids between the leaves and other parts of the plant.

(i) Which of the following processes will not occur in the absence of xylem?

  1. Transport of water
  2. Conduction of food
  3. Transport of minerals
  4. Both A and C

(ii) Transport of food by the phloem is called

  1. Transpiration
  2. Translocation
  3. Guttation
  4. Adhesion

(iii) Which of the following is not a characteristic of xylem?

  1. Living cells
  2. Lack of cytoplasm
  3. Impermeable to water
  4. Presence of lignin

(iv)In phloem transport occurs between where the substances are made i.e. ________
and where they are used or stored i.e. ________.

  1. Sink, source
  2. Source, sink
  3. Origin, destination
  4. Destination, origin

(v) ____________ is the process which involves transport of water and minerals.

  1. Transpiration stream
  2. Translocation
  3. Guttation
  4. Adhesion
Solution:
  1. (d) Xylem helps in the transport of water and minerals in the plant.
    (i) ((b) Transport of food by the phloem is called translocation.
  2. (a) Xylem tissue consists of dead cells.
  3. (b) From the left ventricle the blood is pumped through the aorta to different parts of the body.
  4. (b) In phloem transport occurs between where the substances are made i.e. source and where they are used or stored i.e. sink.
  5. (a) Transpiration stream is the process which involves transport of water and minerals.

8. Read the following paragraph and answer the questions that follow:
Mendel experimented on a pea plant and considered 7 main contrasting traits in the plants. In monohybrid experiment, Mendel took two pea plants of opposite traits (one short and one tall) and crossed them. He found the first generation offspring were tall and called it F1 progeny. Then he crossed F1 progeny and obtained both tall and short plants in the ratio 3:1. In a dihybrid cross experiment, Mendel considered two traits, each having two alleles. He crossed wrinkled-green seed and round-yellow seeds and observed that all the first generation progeny (F1 progeny) were round-yellow. This meant that dominant traits were the round shape and yellow colour. He then self-pollinated the F1 progeny and obtained 4 different traits wrinkled-yellow, round-yellow, wrinkled-green seeds and round-green in the ratio 9:3:3:1.

i) Which of the following was not a trait considered by Mendel for his experiments?

  1. Flower colour
  2. Pod shape
  3. Stem height
  4. Seed type

ii) What is the phenotypic ratio of the F2 progeny in a dihybrid cross?

  1. 3:1
  2. 1:2:1
  3. 9:3:3:1
  4. 3:4

iii) What is the genotypic ratio of the F2 progeny in a monohybrid cross?

  1. 3:1
  2. 1:2:1
  3. 9:3:3:1
  4. 3:4

iv) A cross between tall and dwarf plants will produce which type of plants in the F1 generation?

  1. Tall
  2. Dwarf
  3. Semi-dwarf
  4. Short

v) Because of his contribution to the field of genetics, Mendel is known as the

  1. Father of genetics
  2. Father of botany
  3. Father of medicine
  4. Father of plant physiology
Solution:
  1. (d) Seed type was not studied by Mendel.
  2. (c) The phenotypic ratio of the F2 progeny in a dihybrid cross is 9:3:3:1.
  3. (c) The genotypic ratio of the F2 progeny in a monohybrid cross is 1:2:1.
  4. (a) A cross between tall and dwarf plants will produce all tall plants in the F1 generation.
  5. (a) Mendel is known as the Father of Genetics.

9. Read the following paragraph and answer the questions that follow:
The growing size of the human population is a cause of concern for all people. The rate of birth in a given population will determine its size. Reproduction is the process by which organisms increase their population. The process of sexual maturation for reproduction is gradual and takes place while general body is still going on. Some degree of sexual maturation does not necessarily mean that mind or body is ready for sexual acts or for having and bringing up children. Various contraceptive devices are being used by human beings to control the size of population.

i) Which of the following is a common sign of sexual maturation in both boys and girls?

  1. Development of breasts
  2. Growth of pubic hair
  3. Adam’s apple
  4. Broadening of hips

ii) Which of the following is an IUCD?

  1. Copper-T
  2. Diaphragm
  3. Oral pill
  4. Tubectomy 

iii) Which among the following is not a sexually transmitted disease?

  1. Gonorrhoea
  2. AIDS
  3. Syphilis
  4. Cholera

iv) A couple wants to space the birth of their second child. Which of the following preventive measure could be taken by the husband?

  1. Oral pills
  2. Diaphragms
  3. Tubectomy
  4. Condoms

v) A pregnant woman visits the doctor to determine the sex of the child? Why is she denied of this testing?

  1. It is a complicated test.
  2. It may result in female foeticide.
  3. It is an expensive test.
  4. It is harmful test for the developing foetus.
Solution:
  1. (b) Growth of pubic hair is observed in both boys and girls during sexual maturation.
  2. (a) Copper-T is an IUCD.
  3. (d) Cholera spreads through contaminated water.
  4. (d) Husband can make us of condoms to prevent the sperms from coming in contact with the egg.
  5. (b) If the couple comes to know about the sex of the child, it may increase the chances of female foeticide as well.

10. Read the following paragraph and answer the questions that follow:
We eat various types of food which has to pass through the same digestive tract. Naturally the food has to be processed to generate particles which are small and of the same texture. The process of digestion of food in humans involves the alimentary canal and associated digestive glands.

(i) Which of the following helps in the breakdown of starch?

  1. Salivary amylase
  2. Trypsin
  3. Peptidase
  4. Chyme

(ii) What is the role of the mucus?

  1. Creates alkaline medium for digestion
  2. Movement of food along the digestive tract
  3. Protect the inner lining of the stomach from the action of the acid
  4. Breakdown of starch into sugar

(iii) Which of the following will have a longer small intestine?

  1. Deer
  2. Lion
  3. Tiger
  4. Fox

(iv) Which of the following processes would have been obstructed in the absence of villi?

  1. Digestion of fats
  2. Breakdown of lipids
  3. Absorption of food
  4. Removal of undigested food

(v) Which of the following is not processed in the small intestine?

  1. Pancreatic juice
  2. Bile
  3. Saliva
  4. Intestinal juice
Solution:
  1. (a) The saliva contains an enzyme called salivary amylase that breaks down starch which is a complex molecule to give simple sugar.
  2. (c) The mucus protects the inner lining of the stomach from the action of the acid under normal conditions.
  3. (a) Herbivores eating grass need a longer small intestine to allow the cellulose to be digested.
  4. (c) The villi are richly supplied with blood vessels which take the absorbed food to each and every cell of the body, where it is utilised for obtaining energy, building up new tissues and the repair of old tissues.
  5. (c) Saliva is processed in the mouth.

11. Study the given diagram and answer the following.
A very thin and narrow beam of white light is made incident on the glass objects shown below. Comment on the nature and behaviour of the emergent beam in all the three cases.
(Given: sin 45° = 1/√2 ⁄, sin 30° = ½)

(i) There is a similarity between two of the emergent beams. Identify the two.

  1. (A) and (B)
  2. (A) and (C)
  3. (B) and (C)
  4. Insufficient data given to identify the similarity

(ii) When light enters from air to glass, the angles of the incidence and refraction in air
and glass are 45° and 30° respectively. The refractive index of the glass is

  1. √2
  2. 1/2
  3. 1
  4. 1/√2

(iii) The phenomena which takes place when the light rays emerge out in case (B)

  1. reflection
  2. refraction
  3. dispersion
  4. scattering

(iv)What is the unit of refractive index?

  1. °A
  2. cm
  3. degree
  4. no unit

(v) The ratio of sine of angle of incidence and sine of angle of refraction for particular pair of media is

  1. Zero
  2. Unity
  3. Constant
  4. None of these

 Solution:

  1. b) (A) and (C)
    In (A) th emergent beam is white and laterally displaced. In case of (B) emergent beam is a spectrum of seven colours bent at different angles. In case (C), emergent beam from the second prism is white only.
    Thus, (A) and (B) gives similar emergent rays which are white in colour.
  2. (a) √2

    begin mathsize 12px style straight mu presuperscript straight a subscript straight g equals sini over sinr equals fraction numerator sin 45 degree over denominator sin 30 degree end fraction equals fraction numerator begin display style bevelled fraction numerator 1 over denominator square root of 2 end fraction end style over denominator begin display style bevelled 1 half end style end fraction equals square root of 2 end style
  3. (c) dispersion
    The splitting of white light into a spectrum of seven colours on passing through prism is known as dispersion.
  4. (d) no unit
    As the refractive index is the ratio of two same quantities it do not have unit.
  5. (c) constant
    The ratio of sine of angle of incidence and sine of angle of refraction for particular pair of media is constant. This constant is refractive index. This law is called Snell’s law.

12. The figure given below illustrates the ray diagram for the formation of image by a concave mirror. The position of the is beyond the centre of the curvature of the concave mirror. On basis of the given figure answer the questions given below.

i) If focal length of the concave mirror is 10 cm, the image formed will be at a distance _____________________.

  1. Between 10 cm and 15 cm
  2. Between 10 cm and 20 cm
  3. Beyond 20 cm
  4. At 20 cm

ii) In case of concave mirror, the image distance from the pole of the mirror is

  1. Always positive
  2. Always negative
  3. Negative or positive depending upon the position of the object
  4. None of these

iii) If the size of the object in the given figure is 5 cm and the magnification produced is – 0.5. The size of the image is

  1. – 2.5 cm
  2. – 0.1 cm
  3. 2.5 cm
  4. 0.1 cm 

iv) A negative sign in the magnification value indicate that the image is

  1. Real and inverted
  2. Real and erect
  3. Virtual and erect
  4. Virtual and inverted

v) If the value of magnification is greater than 1 then it indicates that the image formed is

  1. Diminished
  2. The same size as that of the object
  3. Enlarged
  4. Value of magnification cannot specify whether the image is diminished or magnified

 Solution:

  1. (b) between 10 cm and 20 cm The focal length of the mirror is 10 cm. This mean, the radius of curvature is 20 cm. From figure, it is clear that the image is formed between the focus and centre of curvature. Thus, the image is formed between 10 cm and 20 cm.
  2. (c) Positive or negative depending upon the position of object When the object is placed between the pole and focus of concave mirror, the image is formed behind the concave mirror in this case the image distance is positive. When the object is placed beyond or at the focus, the image is formed in front of the mirror and in this case the image distance is negative.
  3. (a) -2.5 cm
    m = h2/ h1
    h2 = (– 0.5 × 5)/ 10
    h2 = – 2.5 cm
  4. (a) real and inverted A negative sign in magnification value indicate that the image formed is real ad inverted.
  5. (c) enlarged When the value of magnification of the image is greater than 1 then it indicates that the image formed is enlarged.

13. I n the below circuit, a nichrome wire of length 'L' is connected between points X and Y and note the ammeter reading. The experiment is performed and repeated by inserting another nichrome wire of same thickness but twice the length i.e. ‘2L’.

i) What are the changes observed in the ammeter readings?

  1. Ammeter readings decreases, becomes half
  2. Ammeter readings increases, becomes two times
  3. Ammeter readings increases becomes quadrupled
  4. Ammeter reading decreases becomes one – fourth

ii) What change is occurred in ammeter reading if instead of changing the length the area of cross – section is doubled?

  1. Ammeter readings decreases, becomes half
  2. Ammeter readings increases, becomes two times
  3. Ammeter readings increases becomes quadrupled
  4. Ammeter reading decreases becomes one – fourth 

iii) If the resistors of 5 ohms and 10 ohms are connected in series in the above circuit. What is the ratio of the current passing through the two resistors?

  1. 2:1
  2. 3:1
  3. 1:2
  4. 1:1

iv) If the resistors are connected in parallel

  1. Current across each resistor is same and voltage changes
  2. Current and voltage across each resistor is same
  3. Current across each resistor varies and voltage remains same
  4. Current changes, voltage changes

v) SI unit of current is denoted as

  1. A
  2. C
  3. I
  4. J
Solution:
  1. (a) ammeter reading decreases, becomes half The increase in length of wire increases the resistance in the circuit and thus, the current decreases. Hence the ammeter readings are reduced and becomes half the initial readings.
  2. (b) ammeter reading increases and becomes two times When the area of wire increases, the resistance decreases thus the current in the circuit will increase. Thus, the ammeter readings will increase and becomes twice the initial readings.
  3. (d) 1:1 When resistors are connected in series the current flowing through all resistors is same. Thus, the ratio of the currents for these two resistors is 1:1
  4. (c) current across each resistor varies and voltage remains same When the resistors are connected in parallel to each other the voltage across each resistor is same while the current changes.
  5. (a) A SI unit of current is denoted as A (Ampere).

14. Shama has a set of five substances. She has a chart stating resistivities of all the substances.
Observe the table

She has to choose an appropriate substance for performing electrical tasks. Which of the above substance according to you –

i) Can be used as an insulator

  1. A
  2. B
  3. B as well as C
  4. E

ii) Can be used for domestic wiring

  1. A
  2. B
  3. A as well as C
  4. D

iii) Can be utilised in making solar cells and transistors

  1. A
  2. B
  3. C
  4. D

iv) Is an alloy

  1. A
  2. B
  3. C
  4. E

v) Behaves as a semiconductor

  1. A
  2. D
  3. C
  4. E
Solution:
  1. (d) E
    Substance E can be used as an insulator.
  2. (c) A as well as C
    Substances A and C can be used for the purpose of domestic wiring.
  3. (d) D
    Substance D can be used to make solar cells.
  4. (b) B
    An alloy has resistivity higher than a pure metal but lesser than a semiconductor. Thus, substance B is an alloy.
  5. (b) D
    Substance D is semiconductor.

15. Ruchi fixes a sheet of white cardboard sheet on a drawing board. She places a bar magnet in the centre of it. She sprinkles iron fillings uniformly around the bar magnet. Then she taps the board gently and observes that the iron fillings arrange themselves in particular pattern.

(i) What does the pattern shape look like?

  1. Straight lines
  2. Squares
  3. Closed curves
  4. Parallel lines

(ii) Why do iron fillings arrange in a pattern?

  1. Due to poles of magnet
  2. Due to force exerted by magnet within its magnetic field
  3. Due to repulsion between the poles and filings
  4. None of the above

(iii) What do the lines along which iron filings align represent?

  1. Magnetic field lines
  2. Magnetic force
  3. Magnetic induction
  4. Magnetic susceptibility

(iv) What does the crowding of iron filings at the end of the magnet indicate?

  1. Magnetic field is weak at poles and stronger at middle
  2. Magnetic field is stronger at poles and weak at middle
  3. Magnetic field strength goes on decreasing from north to south pole
  4. Magnetic field strength goes on decreasing from south to north pole

(v) How the strength of magnetic field is indicated?

  1. Magnetic field strength cannot be indicated by magnetic field lines
  2. Far the magnetic field lines more is the magnetic strength
  3. Closer the magnetic field lines less are the magnetic field strength
  4. Closer the magnetic field lines more is the magnetic strength

Solution:

  1. (c) closed curves
    The iron filings arrange themselves in the pattern of closed curves
  2. (b) Due to force exerted by magnet within its magnetic field Iron filings arrange themselves in particular pattern due to the force exerted by the magnet within its magnetic field.
  3. (a) Magnetic field lines
    The magnetic field lines represent magnetic field lines.
  4. (b) magnetic field is stronger at poles and weak at middle Crowding of the iron filings at the end of the magnet indicates that the magnetic field is strongest near the poles of the magnet and less at the middle of the magnet.
  5. (d) Closer the magnetic field lines more is the magnetic strength The strength of magnetic field is indicated by the closeness of the magnetic field lines. Closer the lines more is the strength of magnetic field and farther the lines indicate less is the magnetic field strength at that particular point.


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