Science
Section – A
Q.1. Name two industries based on forest produce.
Answer: Timber industries and paper manufacturing industries are based on forest produce.
Q.2. Why are the heating elements of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer: The resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise (bum) readily at higher temperatures. Therefore, conductors of electric heating devices, such as toasters and electric irons, are made up of an alloy rather than pure metal.
Section – B
Q.3. Write the molecular formula of ethene and draw its electron dot structure. Answer: The molecular formula of ethene is C2H4
Electron dot structure of ethene.
Q.4. Given reasons:
(a) Platinum, gold and silver are used to make jewellery.
(b) Metals like sodium and potassium are stored under oil.
OR
Q.4. Silver articles become black when kept in open for some time, whereas copper vessels lose their shiny brown surfaces and gain a green coat when kept in open. Name the substances present in air with which these metals react and write the name of the products formed.
Answer: (a) Platinum, gold and silver are used to make jewellery because of its bright and shiny surface. This property is called metallic lustre.
(b) Metals like sodium and potassium are stored under oil because they are very reactive in nature, they react with oxygen present in air. Thus to prevent their oxidation they are kept in the oil.
OR
Answer: Silver articles become black when kept in open for some time, whereas copper vessel lose their shiny brown surfaces and gain a green coat when kept in open because silver articles reacts with sulphur compounds such as hydrogen sulphide present in the air to form silver sulphide (Ag2S) whereas copper reacts slowly with CO2 and water present in the air to form green coating of mixture of copper carbonate and copper hydroxide.
Q.5. The absolute refractive index of Ruby is 1.7. Find the speed of light in Ruby. The speed of light in a vacuum is 3 x 108 m/s.
Answer: We know that,
Refractive index of ruby
Section – C
Q.6. On heating blue coloured powder of copper (II) nitrate in a boiling tube, black copper oxide, O2 and a brown gas X is formed.
(a) Identify the type of reaction and gas X
(b) Write the balanced chemical equation of the reaction.
(C) Write the pH range of aqueous solution of the gas X.
Answer:
(a) Decomposition reaction
The gas X is Nitrogen dioxide (NO2)
(c) Oxides of non-metals are acidic. Therefore the aqueous solution of this gas would be acidic.
The pH would be less than 7.
Q.7. (a) While diluting an acid, why is it recommended that the acid should be added to water and not water to the acid?
(b) Dry hydrogen chloride gas does not change the colour of dry litmus paper why?
OR
Q.7. How is sodium hydroxide manufactured in industries? Name the process. In this process, a gas X is formed as a byproduct. This gas reacts with lime water to give a compound Y, which is used as a bleaching agent in the chemical industry. Identify X and Y and write the chemical equation of the reaction involved.
Answer: (a) The process of mixing concentrated acid with water is highly exothermic. So, when a concentrated acid is added to water then heat is easily absorbed by a large amount of water. Thus it is recommended to add acid to water and not water to the acid.
(b) Dry hydrogen chloride does not contain any hydrogen ions in it, so it does not show acidic behaviour and thus does not change the colour of dry litmus paper.
OR
Answer: Sodium hydroxide is manufactured by the electrolysis of a concentrated aqueous solution of sodium chloride
The process of manufacture of sodium hydroxide by electrolysis process is called the chloralkali process. Gas X is chlorine gas and compound Y is calcium oxychloride (Bleaching powder)
Q.8. What are amphoteric oxides? Give an example.
Write balanced chemical equations to justify your answer.
Answer: Those oxides which behave both acidic and basic oxides are called amphoteric oxides. Example: Al2O3 (Alumina)
Q.9. What is a homologous series of carbon compounds ? Give an example and list its three characteristics.
Answer: A homologous series is a group of organic compounds having similar structures and similar chemical properties in which the successive compounds differ by CH2 group. Example: Alkanes with general formula CnH2n+2 Characteristics:
• All the members of a homologous series can be represented by the same general formula.
• Any two adjacent homologues differ by 1 carbon atom and 2 hydrogen atoms in their molecular formulae.
• The difference in the molecular masses of any two adjacent homologues is 14 u.
Q.10. List in tabular form three distinguishing features between autotrophic nutrition and heterotrophic nutrition.
Answer:
S.No. | Autotrophic Nutrition | Heterophic Nutrition |
1. | In this mode of nutition, organisms make their own food. | In this mode of nutition, the organisms do not make their own food. |
2. | CO2 and H2O are required for the preparation of food as row materials. | They depend on autotrophs for their food either directly or indirectly. |
3. | Chlorophyll and sunlight are essentially needed. | Chlorophyll and sunlight are not needed. |
Q.11. What is transpiration? List its two functions.
OR
Q.11. (a) What is translocation? Why is it essential for plants?
(b) Where do the substances in plants reach as a result of translocation?
Answer: The evaporation of water from the leaves of a plant is called transpiration. Functions of transpiration
1. It helps in the upward movement of water and minerals from the root to the leaves through the stem.
2. Helps in cooling the plant surface.
OR
Answer: (a) The transport of food from leaves to other parts of the plant is called translocation. Translocation is essential for plants because without it food prepared by the leaves cannot reach other parts of the plant for their growth and development.
(b) The substances in plants reach other tissues in plants from the leaves as a result of translocation.
Q.12. What is carpel? Write the function of its various parts.
Answer: The flask-shaped organ in the centre of a flower is called carpel. It is also called a female reproductive organ of the plant. It is made up of three parts:
1. Stigma
2. Style
3. Ovary
1. Stigma is the top part of carpel and is sticky. So, it receives the pollen from the anther of stamen.
2. Style connects stigma to ovary.
3. Ovary contains female gametes of the plant and helps in reproduction.
Q.13. A student holding a mirror in his hand directed the reflecting surface of the mirror towards the Sun. He then directed the reflected light on to a sheet of paper held close to the mirror.
(a) What should he do to bum the paper?
(b) Which type of mirror does he have?
(c) Will he be able to determine the approximate value of the focal length of this mirror from this activity ? Give reason and draw ray diagram to justify your answer in this case.
OR
Q.13. A 10 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 12 cm. The distance of the object from the lens is 18 cm. Find the nature, position and size of the image formed.
Answer: (a) He should place the sheet of paper at the focus of the mirror to burn the paper.
(b) He has a concave mirror.
(c) Yes, the sheet of paper will start burning at the focus of the mirror which will give an approximate value of focal length, i.e., the distance between the mirror and the point where the sheet of paper starts burning.
OR
Answer: A concave mirror forms a real image of the sun,
Given: Height of object, h1 = +10 cm. Focal length, f = +12 cm.
Object distance, u = -18 cm. From the lens formula,
The position of image formed is at distance of 36 cm from convex lens. Since the value of the magnification is more than 1 (it is 2), the image formed is larger than object. The minus sign of magnification shows that image is formed below the principal axis. Hence, the image formed is real and inverted,
Q.14. What are solar cells? Explain the structure of solar panel. List two principal advantages associated with solar cells.
Answer: Solar cells are the devices which convert solar energy into electricity. A simple solar cell is made up of a sandwich of a silicon-boron layer and a silicon-arsenic layer. Boron and arsenic are present in a very small amount. A piece of wire is soldered into the top of the upper layer of cell and another piece of wire is soldered at the bottom of the lower layer to pass on the current. The solar cell is covered with a glass cover for protection. Advantages:
• Solar cells have no moving parts.
• It requires no maintenance.
Q.15. Write the essential function performed by ozone at the higher levels of the Earth’s atmosphere? How is it produced ? Name the synthetic chemicals mainly responsible for the drop of amount of ozone in the atmosphere. How can the use of these chemicals be reduced?
Answer: Ozone layer absorbs most of the harmful ultraviolet radiations from the sun to the earth. It is formed high up in the atmosphere by the action of ultraviolet radiation on oxygen gas. Chlorofluorocarbons are the synthetic chemicals responsible for the drop of amount of ozone in the atmosphere.
The use of these chemicals can be reduced by:
• Replacement of chlorofluorocarbons with hydrochlorofluorocarbons because it breaks down more quickly.
• Safe disposal of old appliances such as refrigerators and freezers.
Section – D
Q.16. (a) List any three observations which posed a challenge to Mendeleev’s Periodic law.
(b) How does the metallic character of elements vary on moving (i)from left to right in a period, (ii) from top to bottom in a group of the Modem Periodic Table? Give a reason for your answer.
OR
Q.16. The electrons in the atoms of four elements A, B, C and D are distributed in three shells having 1,3,5 and 7, electrons respectively in their outermost shells. Write the group numbers in which these elements are placed in the Modem Periodic Table. Write the electronic configuration of the atoms of B and D, and the molecular formula of the compound formed when B and D combine.
Answer: (a) Three observations which posed a challenge to Mendeleev’s Periodic law are:
• The position of isotopes could not be explained.
• Wrong order of atomic masses of some elements could not be explained.
• A correct position could not be assigned to hydrogen in the periodic table.
(b) (i) On moving from left to right in a , period, the metallic character of elements decreases because electropositive character decreases.
(ii) Ongoing down in a group of the periodic table, the metallic character of elements increases because of electropostive character of elements increases.
OR
Answer: A – 1st group
B – 13th group
C- 15th group
D – 17th group
Electronic configuration B – Atomic number = 13. KLM 283 D – Atomic number = 17
KLM 287 The molecular formula of the compound when B and D combine is BD
Q.17. (a) Why is the use of iodised salt advisable? Name the disease caused due to deficiency of iodine in our diet and state its one symptom.
(b) How do nerve impulses travel in the body? Explain.
OR
Q.17. What is hydrotropism? Design an experiment to demonstrate this phenomenon.
Answer: (a) lodised salt is advisable because iodine is necessary for the formation of thyroxine hormone by the thyroid gland Goitre is the disease caused due to its deficiency. Symptom: The neck of the person appears to be swollen due to the enlargement of thyroid gland.
(b) Two neurons are not joined to one another completely. There is a small gap between a pair of neuron. This gap is called synapse. The nerve impulse are carried out to this gap by the help of neurotransmitter (chemical substance). The conduction of nerve impulse through the synapse takes place in the form of electrical nerve impulse. When a stimulus acts on the receptor an electrical impulse is produced with the help of chemical reaction. This electrical impulse passes through the synapse and then to the other neuron. Thus, in this way nerve impulses travel in the body.
OR
Answer: The movement of root of plants towards water is called hydrotropism. Take two glass troughs A and B fill each one of them two-thirds with soil. In trough A plant a tiny seedling figure
(a). In trough B plant a similar seedling and also place a small clay pot inside the soil figure
(b). Water the soil in trough A daily and uniformly. Do not water the soil in trough B but put some water in the clay pot buried in the soil. Leave both the troughs for a few days.
Now, dig up the seedlings carefully from both the trough without damaging their roots. We will find that the root of seedling in through A is straight. On the other hand, the root of seedling in trough B is found to be bent to the right side (towards the clay pot containing water) figure (b). This can be explained as follows.
In through A, the root of seedling gets water from both sides (because the soil is watered uniformly) in trough B, the roots gets water oozing out from the clay pot which is kept on the right side. So, the root of seedling in trough B grows and bends towards the source of water to the right side. The experiment shows that the root of a plant grows towards water. In other words, the root of a plant is positively hydrotropic.
Q.18. (a) What are homologous structures ? Give an example. (b) “The sex of a newborn child is a matter of chance and none of the parents may be considered responsible for it.” justify this statement with the help of a flow chart showing sex-determination in human beings.
Answer: (a) The structures which have the same basic design but different functions are called homologous structures or homologous organs. Example: Forelimbs of a man, a lizard, a frog they have same basic design of bones but perform different functions. (b) The sex of a newborn depends on what happens at the time of fertilization.
1. If a sperm carrying X chromosome fertilizes the ovum carrying X chromosome, then the girl child will be born and the child will have XX combination of sex chromosomes.
2. If a sperm carrying Y chromosome fertilizes the ovum carrying X chromosome, then the child born will be.
The above presentation clearly shows that it is matter of chance whether the newborn will be boy or girl and none of the parents may be considered responsible for it.
Q.19. When do we consider a person to be myopic or hypermetropic ? List two causes of -hypermetropia. Explain using ray diagrams how the defect associated with hypermetropic eye can be corrected.
Answer: Myopia is the defect in vision in which a person cannot see the distant objects clearly whereas in hypermetropia is the defect in which a person cannot see nearby objects clearly. Hypermetropia is caused due to:
1. Decrease in converging power of eye-lens.
2. Too short eye ball.
In a hypermetropic eye, the image of near by object lying at normal near point N (at 25 cm) is formed behind the retina.
Hypermetropic eye can be corrected using convex lenses. When a convex lens of suitable power is placed in front of hypermetropic eye, then the diverging rays of light from the object are converged first by the convex lens used. This form a virtual image of the object at another near point N’. Now, the rays can be easily focused by the eye lens to form an image on retina.
Q.20. (a) How will you infer with the help of an experiment that the same current flows through every part of a circuit containing three resistors in series connected to a battery?
(b) Consider the given circuit and find the current flowing in the circuit and potential difference across the 150 resistor when the circuit is closed.
OR
Q.20. (a) Three resistors R1, R2 and Rg are connected in parallel and the combination is connected to a battery, ammeter, voltmeter and key.
Draw suitable circuit diagram and obtain an expression for the equivalent resistance of the combination of the resistors.
(b) Calculate the equivalent resistance of the following network.
Answer: (a) Let three resistors R1, R2 and R3 are connected in series which are also connected with a battery, an ammeter and a key as shown in figure
When key is closed, the current starts flowing through the circuit. Take the reading of ammeter. Now change the position of ammeter to anywhere in between the resistors and take its reading. We will observe that in both the cases reading of ammeter will be same showing same current flows through every part of the circuit above.
(b) Given R1 = 50, R2 = 100, R3 = 150, V = 30 V
Total resistance, R = R1 + R2 + R3 [∵ 5, 10 and 15 are connected in series]
= 5 + 10 + 15
= 30
Potential difference, V = 30 V
Current in the circuit, I = ?
From Ohm’s law.
IR = 30 = 1 A
Current flowing in the circut = 1A
Potential difference across 15 resistors = 1R3 = 1 x 15 = 15 V .
OR
Answer: (a) Let R1, R2 and R3 are three resistance connected in parallel to one another and R is the equivalent resistance of the circuit. A battery of V volts has been applied across the ends of this combination. When the switch of the key is closed, current I flows in the circuit such that
10Ω and 10Ω are connected in series. Equivalent resistance of the circuit = 10Ω + 10Ω = 20Ω
Q.21. Draw the paitern of magnetic field lines produced around a current-carrying straight conductor passing perpendicularly through horizontal cardboard. State and apply the right-hand thumb rule to mark the direction of the field lines. How will the strength of the magnetic field change when the point where magnetic field is to be determined is moved away from the straight conductor ? Give reason to justify your answer.
Answer: Maxwell’s Right Hand Thumb rule states that if the current-carrying wire is imagined to be held in the right hand so that thumb points in the direction of current, then the direction in which fingers encircle the wire will give the direction of magnetic field lines around the wire. If we hold the current-carrying straight wire so that thumbs Magnetic field pattern due to a straight current-carrying wire in upward direction points the direction of current, the direction of magnetic field lines will be anticlockwise. The strength of the magnetic field is inversely proportional to the distance of the point of observation from the wire So, as we move away from the wire the strength of magnetic decreases Current (upwards).
Section – E
Q.22. A teacher provided acetic acid, water, lemon juice, aqueous solution of sodium hydrogen carbonate and sodium hydroxide to students in the school laboratory to determine the pH values of these substances using pH papers. One of the students reported the pH values of the given substances as 3, 12, 4, 8 and 14 respectively. Which one of these values is not correct? Write its correct value stating the reason.
OR
Q.22. What would a student report nearly after 30 minutes of placing duly cleaned strips of aluminium, copper, iron and zinc in freshly prepared iron sulphate solution taken in four beakers?
Answer: The value of pH for water is not correct. The correct value of pH of water is 7 because it has almost equal concentration of Hand OH, due to which it is neutral.
OR
Answer: Aluminium displaces the iron from iron sulphate and the colour of two solution changes from green to brown. No change takes place when copper strip is dipped in iron sulphate solution. No cfiange will be observed when iron strips are dipped in iron sulphate solution. The colour of the solution changes from green to colourless when zinc is added to iron sulphate solution.
Q.23. What is observed when a pinch of sodium hydrogen carbonate is added to 2 mL of acetic acid taken in a test tube? Write chemical equation for the reaction involved in this case.
Answer: CO2 gas is evolved with brisk effervescence when sodium hydrogen carbonate is added to acetic acid.
Q.24. List in proper sequence four steps of obtaining germinating dicot seeds. OR After examining a prepared slide under the high power of a compound microscope, a student concludes that the given slide shows the various stages of binary fission in a unicellular organism. Write two observations on the basis of which such a conclusion may be drawn.
Answer: 1. The root is formed when radicle of seed grows.
2. The root grows downward into the soil and absorbs water and minerals from the soil.
3. The shoot is formed from the upward growth of plumule.
4. The green leaves are developed when the shoot comes above the ground.
OR
Answer: 1. A single parent divides to form two daughter cells.
2. The nucleus of mature cell seems elongated and a grove is formed in cell which divides the nucleus.
Q.25. List four precautions which a student should observe while preparing a temporary mount of a leaf peel to show stomata in his school laboratory.
Answer: 1. Freshly plucked leaf should be taken for an epidermal peel.
2. Hold the slide by its edges.
3. Peel should be cut to a proper size.
4. The peel should be allowed to dry.
Q.26. Draw the path of a ray of light when it enters one of the faces of a glass slab at an angle of nearly 45°. Label
(i) angle of refraction
(ii) angle of emergence and
(ii) lateral displacement.
OR
Q.26. A student traces the path of a ray of light through a glass prism as shown in the diagram, but leaves it incomplete and unlabelled. Redraw and complete the diagram. Also label on it zi, ze, zr, and ZD.
Answer:
Q.27. The current flowing through a resistor connected in a circuit and the potential difference developed across its ends are as shown in the diagram by milliammeter and voltmeter readings respectively:
(a) What are the least counts of these meters?
(b) What is the resistance of the resistor?
Answer: (a) 10 mA and 0.1 V
(b) V = 2.4 volt, 1 = 250 mA = 0.25 A
From Ohm’s law.
R=v/I = 2.4/0.25 = 9.6Ω