Study MaterialsImportant QuestionsLakhmir Singh Biology Class 10 Solutions Model Test Paper 4

Lakhmir Singh Biology Class 10 Solutions Model Test Paper 4

S Chand Biology Class 10 Solutions Model Test Paper 4

Question 1.
Name a reducing agent that may be used to obtain manganese from manganese dioxide.
Aluminium powder is used as the reducing agent to obtain manganese from its oxide.
2Al(s) + 3MgO → Al2O3 + 3Mg(s)

Question 2.
Why does a ray of light bend when it travels from one medium into another?
When a light ray travels from one medium to another, then due to change in the density of the medium, speed of light changes. Change in the speed of the light at the boundary of two media force the light ray to bend from its original path.

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    Question 3.
    A compound which is prepared from gypsum has the property of hardening when mixed with a proper quantity of water. Identify the compound. Write the chemical equation for its preparation. For what purpose is it used in hospitals?
    When gypsum is heated at 373 K, it gives plaster of paris,
    Lakhmir Singh Biology Class 10 Solutions Model Test Paper 4 1
    Plaster of paris becomes hard when the appropriate amount of water is available.
    Lakhmir Singh Biology Class 10 Solutions Model Test Paper 4 2
    In hospitals, it is used as a plaster to support broken or fractured bones.

    Question 4.
    What is the minimum number of rays required for locating the image formed by a concave mirror for an object. Draw a ray diagram to show the formation of a virtual image by a concave mirror.
    Only two rays are required for the image to be located.
    Image diagram to show the formation of a virtual image formed by a concave mirror.
    Lakhmir Singh Biology Class 10 Solutions Model Test Paper 4 3

    Question 5.
    Name the type of respiration in which the end products are.
    (a) ethanol, carbon dioxide and energy.
    (b) carbon dioxide, water and energy.
    (a) ethanol, carbon dioxide and energy – Anaerobic respiration
    (b) carbon dioxide, water and energy – Aerobic respiration

    Question 6.
    Name the functional group of organic compounds that can be hydrogenated. With the help of suitable example, explain the process of hydrogenation mentioning the conditions of the reaction and any one change in physical property with the formation of the product. Name any one natural source of organic compounds that are hydrogenated.
    Functional group that can be hydrogenated is alkene.
    Hydrogenation of oils(unsaturated alkene).
    Lakhmir Singh Biology Class 10 Solutions Model Test Paper 4 4
    Vegetable oil is unsaturated fatty acids having a double bond in between some of their carbon atoms. When vegetable oil is heated in the presence of nickel catalyst, then a saturated fatty acid called vegetable ghee is formed. This reaction is known as a hydogenation reaction.
    When vegetable oil is hydrogenated, it changes from liquid state to solid state.

    Vegetable ghee is generally produced by hydrogenation of vegetable oil or animal oil.

    Question 7.
    No chemical reaction takes place when granules of a solid, A, are mixed with the powder of another solid, B. However, when the mixture is heated, a reaction takes place between its components. One of the products, C, is a metal and settles down in the molten state while the other product, D, floats over it. It was observed that the reaction was highly exothermic.
    (a) Based on the given information, make an assumption about A and B and write a chemical equation for the chemical reaction indicating the conditions of reaction, physical state of reactants and products and thermal status of the reaction.
    (b) Mention any two types of reactions under which above chemical reaction can be classified.
    (a) Solid A is assumed to be aluminium (Al) and solid B to be manganese oxide. When the mixture of manganese oxide and aluminium is heated, manganese is produced in the molten state and aluminium oxide floats over it. Manganese produced in the molten state as a lot of heat energy is released in the process. Therefore, C is manganese and D is aluminium oxide.
    Al(s) + MnO2(s) → Mn(s) + Al2O3(s)
    (b) The type of reaction under which the above reaction is classified are.
    1. Redox reaction
    2. Displacement reaction

    Question 8.
    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 shells. It was found that elements A and G combine to form an ionic compound which can also be extracted from sea water. Oxides of the elements A and B are basic in nature while those of E and F are acidic.
    The oxide of element D is almost neutral. Answer the following Qs based on the information given herein.
    (a) To which group or period of the periodic table do the listed elements belong?
    (b) Which one of the eight elements is likely to be a noble gas?
    (c) Which one of the eight elements would have the largest atomic radius?
    (d) Which two elements amongst these are likely to be non-metals?
    (e) Which one of the these eight elements is likely to be a semi-metal or metalloid?
    (a) A and G form an ionic compound so A must be a metal of group 1 and G must be a non-metal of group 17. As the ionic compound of A and G is obtained from seawater so that ionic compound must be sodium chloride. Therefore, A and G belong to the third period of the periodic table.
    (b) Because all the given elements belong to the third period and H being the last element of the third period will be the noble gas.
    (c) A belongs to the first group of the periodic table, so it has the largest radius.
    (d) Oxides of E and F are acidic in nature, so they are non-metals.
    (e) The oxide of D is almost neutral, so it is a metalloid.

    Question 9.
    State, giving reasons, which of the following reactions will occur and which will not.
    (a) MgSO4 (aq) + Cu(s) → CuSO4 (aq) + Mg(s)
    (b) CuSO4 (aq) + Fe (s) → FeSO4 (aq) + Cu(s)
    (c) MgSO4 (aq) + Fe (s) → FeSO4 (aq) + Mg(s)
    (a) MgSO4 (aq) + Cu(s) → CuSO4 (aq) + Mg(s)
    Copper is less reactive than magnesium, so copper cannot displace magnesium. Hence, this reaction will not occur.
    (b) CuSO4 (aq) + Fe (s) → FeSO4 (aq) + Cu(s)
    Iron is more reactive than copper, so it will displace copper from copper sulphate and form iron sulphate.
    (c) MgSO4 (aq) + Fe (s) → FeSO4 (aq) + Mg(s)
    Iron is less reactive than magensium, so iron can not displace magnesium. Hence, this reaction will not occur.

    Question 10.
    Draw three labelled ray-diagrams to show the defect of vision called hypermetropia and how it is corrected by using a lens. Also name the lens used.
    Hypermetropia or long-sightedness is a common eye defect, in which a person is not able to nearby objects whereas the farther objects appear clear to the person.
    Hypermetropia deffect.
    Lakhmir Singh Biology Class 10 Solutions Model Test Paper 4 5
    Correction made with a convex lens for the hypermetropic eye.
    Lakhmir Singh Biology Class 10 Solutions Model Test Paper 4 6
    A convex lens is used to correct the defect of hypermetropia.

    Question 11.
    At what distance should an object be placed from a convex lens of focal length 18 cm to obtain an image at 36 cm from it on the other side? What will be the magnification produced in this case?
    The focal length of the convex lens, f = +18 cm
    Image distance, v = + 36 cm
    Using lens formula, \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
    ⇒ \(\frac{1}{36}-\frac{1}{u}=\frac{1}{18}\)
    ⇒ \(\frac{1}{u}=\frac{1}{36}-\frac{1}{18}=-\frac{1}{36}\)
    ⇒ u = -36 cm
    Magnification of the lens = m = \(\frac{v}{u}=\frac{36 \mathrm{cm}}{-36 \mathrm{cm}}\) = -1

    Question 12.
    The magnetic field associated with a current carrying straight conductor is in an anticlockwise direction. If the conductor was held along the east-west direction, what will be the direction of current through it? Name and state the rule applied to determine the direction of the current.
    The direction of the flow of current can be found out by Maxwell’s Right Hand Thumb Rule. It states that if we point our right-hand thumb in the direction of current then the fingers of our hand curl along the direction magnetic field.
    Now it is given that the magnetic field is in the anti-clockwise direction. If we look at the conductor from the right and apply the Thumb rule, the direction of current would be towards the East. However, if you look at the field from the left and apply this rule, the direction of current would be towards the west.

    Question 13.
    Explain analogous organs and homologous organs. Identify the analogous and homologous organs amongst the following.
    Wings of an insect, wings of a bat, forelimbs of frog, forelimbs of a human.
    Analogous orgAnswer: These are organs that have different origins, but perform similar functions.
    Homologous orgAnswer: These are organs that are similar in origin (or are embryologically similar), but perform different functions.
    The wings of an insect and wings of a bat are examples of analogous organs.
    The forelimbs of frog and forelimbs of humans are examples of homologous organs.

    Question 14.
    (a) Explain the terms.
    (i) implantation (ii) placenta
    (b) What is the average duration of human pregnancy?
    (a) (i) Implantation is the process wherein a fertilised egg (zygote) formed in the fallopian tube gets embedded on the walls of the uterus, inside the female body.
    (ii) Placenta is a vascular structure present in the inner lining of the uterus. It is connected to the foetus by the umbilical cord. The embryo receives nutrition and oxygen from the mother’s blood via the placenta.
    (b) The average duration of human pregnancy is nine months.

    Question 15.
    Explain the mechanism of reflex action with a suitable example.
    The automatic action or response provoked by a stimulus is known as reflex action. The sensory nerves that detect the stimulus are connected to nerves that move the muscles. Such a connection is called reflex arc. The reflex arc connections meet in a bundle in the spinal cord. The signal and the response that has taken place reach the brain but the brain does not coordinate the reflex response.
    For example, we withdraw our hands on touching any hot article.

    Question 16.
    (a) Why did Mendeleev leave gaps in his periodic table?
    (b) State any three limitations of Mendeleev’s classification.
    (c) How does electronic configurations of atoms change in a period with increase in atomic number?
    (a) When Mendeleev developed the periodic table, he realized that many elements are missing in the periodic table and yet to be discovered. So, left gaps in his table for such elements and also predicted their properties.

    (b) Limitations of Mendeleev’s classification.

    • He could not explain the position of isotopes in the periodic table.
    • He could not assign a proper position to the element hydrogen in the periodic table.
    • He failed to explain the wrong order of atomic masses of some elements.

    (c) As we move from left to right in a period, the atomic number will increase due to the increase of electrons in the shell.

    Question 17.
    What is meant by refining of a metal? Name the most widely used methods of refining impure metals produced by various reduction processes. Describe with the help of a labelled diagram how this method may be used for refining of copper.
    Refining of metals. Refining of metals is nothing but the removal of impurities from crude metals or purifying the crude metal.
    Electrolytic refining is the most widely used method for the refining of metals.
    Electrolysis of copper.
    Lakhmir Singh Biology Class 10 Solutions Model Test Paper 4 7
    Electrolysis of copper from copper sulphate solution is done using copper electrodes. The crude copper serves as the anode while the thin sheet of pure copper serves as the cathode.
    At anode: Cu → Cu+2 + 2e
    At cathode: Cu+2 + 2e →Cu
    As the anode dissolves away, the cathodes on which the pure metal is deposited grow in size.

    Question 18.
    Explain the process of digestion of food in mouth, stomach and small intestine in human body.
    The process of digestion of food in mouth, stomach and small intestine is described as follows.
    1. Mouth – Digestion of food begins in the mouth. Saliva present in mouth contains a digestive enzyme, called salivary amylase, which breaks down starch into sugar.

    2. Stomach – Stomach stores and mixes the food received from the oesophagus with gastric juices. The main components of gastric juice are hydrochloric acid, mucus and pepsinogen.
    Hydrochloric acid dissolves bits of food and creates an acidic medium. In this medium, pepsinogen is converted to pepsin, which is a protein-digesting enzyme. Mucus protects the inner lining of the stomach from the action of HCl.

    3. Small intestine – Small intestine is the site for complete digestion of carbohydrates, proteins and fats. It produces intestinal juice from the glands present in its wall. The intestinal juice helps in further digestion of food. Small intestine also obtains digestive juices from liver and pancreas that helps in mixing of food.
    The liver produces bile juice that causes emulsification of fats and the pancreas produces pancreatic juice for digesting proteins and emulsified fats.
    This digested food is finally absorbed through the intestinal walls.

    Question 19.
    (a) Define ‘Photosynthesis’. Apart from sunlight, what other conditions are necessary for the process of photosynthesis.
    (b) Describe an experiment to show that “sunlight is essential for photosynthesis”.
    (a) Photosynthesis is the process by which plant cells prepare food (carbohydrates) from inorganic raw materials such as carbon dioxide and water in presence of sunlight and chlorophyll.
    Lakhmir Singh Biology Class 10 Solutions Model Test Paper 4 8
    The factors other than sunlight affecting the rate of photosynthesis are.

    • Carbon dioxide − Carbon dioxide acts as the substrate for photosynthesis. During this process, carbon dioxide and water are converted into carbohydrates.
    • Water − During photosynthesis, water is photolysed into oxygen and hydrogen and it provides the energy for photosynthesis.
    • Chlorophyll − Chlorophyll is the green pigment present in the leaves which absorbs sunlight.

    (b) Experiment to demonstrate that light is necessary for photosynthesis.

    • Take two potted plants. Place one plant in a dark room and the other in a room receiving sufficient sunlight.
    • After 3-4 days, take one leaf each from both the plants. Bathe the two leaves first in hot water and then in alcohol.
    • Add some drops of iodine solution on the leaves.

    It can be observed that the leaf taken from the plant kept in sunlight develops blue-black patches whereas the leaf taken from the plant kept in the dark room does not show any patches.

    Reason for the observation.
    Plants which receive sunlight prepare their own food through the process of photosynthesis. It is the presence of starch that turns the colour of iodine solution blue-black. Therefore, blue-black patches are observed on the leaf taken from the plant kept in sunlight. In contrast, the leaf of the plant kept in the dark room does not produce any starch. Therefore, no patches are observed on it.

    Question 20.
    (a) What is a magnetic field? How can the direction of magnetic field lines at a place be determined?
    (b) State the rule for the direction of the magnetic field produced around a current-carrying conductor. Draw a sketch of the pattern of field lines due to a current flowing through a straight conductor.
    (a) The magnetic field is defined as the space surrounding a magnet in which magnetic force is exerted. The direction of magnetic field lines at a place can be determined by finding the deflection in the needle of a magnetic compass.
    (b) The rule for determining the direction of the magnetic field of a current carrying wire is the right-hand thumb rule.

    Right-hand thumb rule. The right-hand thumb rule indicates the direction of the magnetic field for a know direction of the current. If the thumb of the right-hand points along the direction of current then the curled fingers of that hand gives the direction of the magnetic field due to the current.
    Lakhmir Singh Biology Class 10 Solutions Model Test Paper 4 9

    Question 21.
    Explain with the help of a labelled circuit diagram how will you find the resistance of a combination of three resistors, of resistance R1, R2 and R3 joined in parallel. Also mention how you will connect the ammeter and the voltmeter in the circuit when measuring the current in the circuit and the potential difference across one of the three resistors of the combination.
    Lakhmir Singh Biology Class 10 Solutions Model Test Paper 4 10
    Applied voltage = V
    Let the current in the circuit = I
    If total resistance of the circuit is Rnet, then
    I = \(\frac{V}{R_{\mathrm{net}}}\)
    As all the resistances are connected in parallel, so all of the resistors will get an equal voltage across their ends.
    The total current in the circuit is simply the sum of the current passing through each resistance.
    Lakhmir Singh Biology Class 10 Solutions Model Test Paper 4 11
    An ammeter is low impedance device which is always connected in the series with the circuit so that maximum of the current can pass through it and it can give a more accurate value of the current flowing in the circuit.
    A voltmeter is a high impedance device which is always connected in parallel in the circuit so that it does not draw any current and can give the actual voltage drop across the component.

    Question 22.
    Draw the given diagram in your answer book and complete it for the path of a ray of light beyond the lens.
    Lakhmir Singh Biology Class 10 Solutions Model Test Paper 4 12
    (a) What type of lens is shown in this diagram?
    (b) Is it a converging lens or a diverging lens?
    Lakhmir Singh Biology Class 10 Solutions Model Test Paper 4 13
    (a) A convex lens is shown in the diagram.
    (b) It is converging in nature.

    Question 23.
    A student connects a coil of insulated wire to a galvanometer. What would be seen if a bar magnet with its north pole towards one face of the coil is.
    (a) moved quickly towards it?
    (b) moved quickly away from the coil?
    (c) placed near the face of the coil?
    Name the phenomenon involved.
    (a) When the bar magnet is moved quickly towards the coil, a deflection is observed in the galvanometer. This deflection indicates that some current is produced in the coil.

    (b) When we move the magnet quickly away from the coil, this time the deflection is indicated in the opposite direction. It indicates the current produced in the coil is in the opposite direction.

    (c) When we simply move the magnet near the face of the coil then no deflection is found in the galvanometer, because of the absence of relative motion between the coil and the magnet.

    The phenomenon involved is known as Electromagnetic induction.

    Question 24.
    A student was working in a chemistry laboratory. He found that a bottle contained a white powdery substance but half of the label on the bottle was torn. The part of label still sticking to the bottle had NaH written on it. The student took this powdery substance in a test-tube and added dil. HCl to it. A gas was evolved which extinguished a burning match-stick.
    (a) Name the gas evolved.
    (b) Write the formula of white powdery substance.
    When dilute hydrochloric acid is mixed with that salt a gas is produced which extinguished a burning match-stick that means the gas produced is carbon dioxide.
    So, the salt must be sodium bicarbonate.
    HCl + NaHCO3 → NaCl + H2O + CO2
    (a) Hence, the gas produced is carbon dioxide.
    (b) White powdery salt is sodium bicarbonate(NaHCO3).

    Question 25.
    Ravi placed a zinc plate in a glass container having a blue coloured solution of a metal sulphate MSO4 for a considerable time. He found that the blue colour of solution got lighter and lighter, and finally the solution became colourless. A red-brown deposit was also formed on the zinc plate.
    (a) Which metal, zinc or M, occurs lower in the reactivity series?
    (b) Name the substance which forms colourless solution.
    When Ravi places zinc plate in a blue colored solution of metal sulphate, he found that colour of the solution is fading that means zinc is more reactive than that metal and displacing it from the solution. As the solution becomes colourless after some time it shows that metal M will be copper.
    (a) Zinc is more reactive than metal M.
    (b) Due to the formation of zinc sulphate solution becomes colourless.
    Lakhmir Singh Biology Class 10 Solutions Model Test Paper 4 14

    Question 26.
    A student is performing an experiment to show that sunlight is necessary for photosynthesis. In the first stage, he keeps a potted plant with green leaves in a dark place for about 3 days. Towards the end of the experiment, he boils one leaf in alcohol before using iodine for testing starch.
    (a) Why is the plant kept in a dark place for about 3 days?
    (b) Why is the leaf boiled in alcohol before testing for starch with iodine?
    (a) The plant is kept in dark so as to destarch it, so that in the beginning of the experiment itself, the leaves do not have any starch.
    (b) After the plant is destarched by keeping it in dark, it is again placed in sunlight so that photosynthesis can occur. This would result in the synthesis of starch. The leaves are again destarched by boiling them in alcohol before testing them with iodine.

    Question 27.
    Pure-bred pea plants A are crossed with pure-bred pea plants B. It is found that the plants which look like A do not appear in F1 generation but re-emerge in F2 generation.
    (a) Which of the plants A and B are. (i) tall, and (ii) dwarf?
    (b) Give reason for your answer.
    (a) (i) Tall – Plant B
    (ii) Dwarf – Plant A

    (b) In pea plants, tallness is a dominant trait whereas dwarfness is a recessive trait. Plants like A do not appear in the F1 generation but reappear in the F2 generation which means that plant A is dwarf whereas plant B is tall.
    Lakhmir Singh Biology Class 10 Solutions Model Test Paper 4 15
    We can see from the cross that, tall plants appear in the F1 generation but no dwarf plants are seen.
    The dwarf plants are seen in F2 generation.

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