Students can access the CBSE Sample Papers for Class 10 Science with Solutions and marking scheme Set 5 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 10 Science Set 5 with Solutions
Time Allowed: 3 hours
Maximum Marks: 80
General Instructions:
(i) This question paper consists of 39 questions in 5 sections.
(ii) All questions are compulsory. However, an internal choice is provided in some questions. A student is expected to attempt only one of these questions.
(iii) Section A consists of 20 Objective-type questions carrying 1 mark each.
(iv) Section B consists of 6 Very Short questions carrying 02 marks each. Answers to these questions should be in the range of 30 to 50 words.
(v) Section C consists of 7 Short Answer type questions carrying 03 marks each. Answers to these questions should be in the range of 50 to 80 words.
(vi) Section D consists of 3 Long Answer type questions carrying 05 marks each. Answer to these questions should be in the range of 80 to 120 words.
(vii) Section E consists of 3 source-based/case-based units of assessment of 04 marks each with sub-parts.
Section – A (20 Marks)
Select and termite the most appropriate option out of the four options given for each of the questions 1 – 20. There is no negative mark for an incorrect response.
Question 1.
The graph given below depicts a neutralization reaction (acid + alkali → salt + water).
The pH of a solution changes as we add excess of acid to an alkali.
Which letter denotes the area of the graph where both acid and salt are present?
(A) A
(B) B
(C) C
(D) D
Answer:
(D) D
Explanation: When both add and salt are present, the pH of the solution becomes less than 7. Since, from the graph, the pH is less than 7 at only part D.
Question 2.
Which of the following is not observed in a homologous series? Give a reason for your choice.
(A) Change in chemical properties.
(B) Difference in -CH2, and 14u molecular mass.
(C) Gradation in physical properties.
(D) Same functional group.
Answer:
(A) Change in chemical properties.
Explanation: Change in chemical properties is not observed in a homologous series. The chemical properties of all compounds in a series remain the same.
Question 3.
Sodium hydroxide is termed an alkali while Ferric hydroxide is not because:
(A) sodium hydroxide is a strong base, while Ferric hydroxide is a weak base.
(B) sodium hydroxide is a base which is soluble in water while Ferric hydroxide is also a base but it is not soluble in water.
(C) sodium hydroxide is a strong base while Ferric hydroxide is a strong acid.
(D) sodium hydroxide and Ferric hydroxide both are strong bases but the solubility of sodium hydroxide in water is comparatively higher than that of Ferric hydroxide.
Answer:
(B) sodium hydroxide is a base which is soluble in water while Ferric hydroxide is also a base but it is not soluble in water.
Explanation: An alkali is a base that dissolves in water hence, all alkalis are bases but all bases are not alkalis.
Sodium hydroxide is termed as alkali as it is a strong base as well as highly soluble in water. Ferric hydroxide is also a base, but it is not an alkali as it is not soluble in water.
Question 4.
Metal oxides generally react with acids, but few oxides of metal also react with bases. Such metallic oxides are:
(i) MgO
(ii) ZnO
(iii) Al2O3
(iv) CaO
(A) (i) and (ii)
(B) (ii) and (iii)
(C) (iii) and (iv)
(D) (i) and (iv)
Answer:
(B) (ii) and (iii)
Explanation: Some metal oxides, such as aluminum oxide, zinc oxide, etc., show both addict as well as basic behavior. Such metal oxides can react with both adds as well as bases to produce salts and water. Metal oxides of this category are known as amphoteric oxides.
Question 5.
The name of the salt used to remove the permanent hardness of water is:
(A) Sodium hydrogen carbonate (NaHCO3)
(B) Sodium chloride (NaCl)
(C) Sodium carbonate decahydrate (Na2CO3.10H2O)
(D) Calcium sulphate hemihydrate (CaSO4.12H2O)
Answer:
(C) Sodium carbonate decahydrate (Na2CO3.1OH2O)
Explanation: Sodium carbonate decahydrate (washing soda: Na2CO3.10H2O) is used to remove the permanent hardness of water.
Question 6.
When you add a few drops of acetic acid to a test-tube containing sodium bicarbonate powder, which one of the following is your observation?
(A) No reaction takes place.
(B) Acolourless gas with a pungent smell is released with brisk effervescence.
(C) A brown-colored gas is released with brisk effervescence.
(D) Formation of bubbles of a colorless and odorless gas.
Answer:
(D) Formation of bubbles of a colorless and odorless gas.
Explanation: When a few drops of acetic add are added toa test-tube containing sodium bicarbonate powder, it leads to the formation of sodium acetate along with the release of carbon dioxide gas, which is a colourless and odorless gas.
Question 7.
How will you protect yourself from the heat generated while diluting a concentrated acid?
(A) By adding acid to water with constant stirring.
(B) By adding water to acid with constant stirring.
(C) By adding water to acid followed by base.
(D) By adding base to acid with constant stirring.
Answer:
(A) By adding acid to water with constant stirring.
Explanation: The mixing of water with add is highly exothermic in nature. So, if water is added to an add, it produces a very large amount of heat. This can break the container and sometimes even causes burning. Hence, it is advised to add concentrated add to water with constant stirring.
Question 8.
The number of chromosomes in parents and offsprings of a particular species undergoing sexual reproduction remains constant due to:
(A) doubling of chromosomes after zygote formation.
(B) halving of chromosomes after zygote formation.
(C) doubling of chromosomes before gamete formation.
(D) halving of chromosomes at the time of gamete formation.
Answer:
(D) halving of chromosomes at the time of gamete formation.
Explanation: The number of chromosomes in parents and offspring of a particular species undergoing sexual reproduction remains constant due to the halving of chromosomes through the process of meiosis at the time of gamete formation.
Question 9.
Sensory nerve of a reflex arc carries information from the receptor cells to the:
(A) Spinal cord
(B) Brain
(C) Muscles of the effector organ
(D) Bones of the receptor organ
Answer:
(A) Spinal cord
Explanation: Sensory nerve of a reflex arc carries information from the receptor cells to the spinal cord. It is also called an afferent neuron.
Question 10.
Given below are two columns, Column I shows enzymes secreted by the glands in the alimentary canal of human beings and Column II indicates the components of food on which enzymes act. Choose the options showing correct matching.
Column I(Enzymes) | Column II(Component) |
(a) Pepsin | Starch |
(b) Trypsin | Proteins |
(c) Lipase | Proteins |
(d) Amylase | Emulsified fat |
Answer:
(B) is correct
Explanation: Trypsin breaks down proteins into smaller peptides in the duodenum of the small intestine. Pepsin is a stomach enzyme that serves to digest proteins found in ingested food. Lipase is an enzyme the body uses to break down fats in food so they can be absorbed in the intestines. Amylase helps your body break down starches.
Question 11.
In a person, the tubule part of the nephron is not functioning at all. What will its effect be on urine formation?
(A) The urine will not be formed.
(B) Quality and quantity of urine is unaffected.
(C) Urine is more concentrated.
(D) Urine is more diluted.
Answer:
(D) Urine is more diluted.
Explanation: The function of a tubule is the re¬ absorption of useful substances such as glucose and amino acids, salts and a major amount of water into the blood capillaries. Therefore, if the tubule part of the nephron is not functioning, the urine is more diluted since it contains both useful and waste substances.
Question 12.
During pollination, plants ensure that the pollen grain from a species germinates on the stigma of the same species. Which of the following ensures this?
(A) Hydrotropism
(B) Chemotropism
(C) Phototropism
(D) Geotropism
Answer:
(B) Chemotropism
Explanation: Chemotropism in plants leads to the growth of pollen tubes towards the ovules and thus helps in the fertilisation process.
Question 13.
Consider the following properties of virtual images:
(i) Cannot be obtained on the screen.
(ii) Are formed by both concave and convex lens.
(iii) Are always erect.
(iv) Are always inverted.
The correct properties are:
(A) (i) and (iv)
(B) (i) and (ii)
(C) (i), (ii) and (iii)
(D) (i), (ii) and (iv)
Answer:
(C) (i), (ii) and (iii)
Explanation: A virtual image is formed when reflected rays appear to meet. Such images cannot be obtained on screen. Plane mirrors, convex mirror and concave lens always form a virtual image. They are always erect.
Question 14.
When we enter a dark room coming from outside, immediately, the things inside the room do not appear clear to our eyes. This is because:
(A) pupils do not open at all in the dark.
(B) pupils take time to adjust.
(C) light travels slower in a dark room.
(D) pupils open very quickly in the dark.
Answer:
(B) pupils take time to adjust.
Explanation: When we enter a dark room coming from outside, immediately, the things inside the room do not appear clear to our eyes. This is because pupils take time to adjust.
Question 15.
Pyramid of energy is the:
(A) total energy in an ecosystem.
(B) net energy in an ecosystem.
(C) energy consumed by various organisms.
(D) graphic representation of energy levels at each trophic level.
Answer:
(D) graphic representation of energy levels at each trophic level.
Explanation: When the energy level is represented in the form of a pyramid, it is known as a pyramid of energy.
Question 16.
Excessive exposure of humans to UV rays results in
(i) damage to immune system.
(ii) damage to lungs.
(iii) skin cancer.
(iv) peptic ulcers.
(A) (i) and (ii)
(B) (ii) and (iv)
(C) (i) and (iii)
(D) (iii) and (iv)
Answer:
(C) (i) and (iii)
Explanation: Excessive exposure of humans to ultraviolet (UV)-rays results in:
(i) Skin cancer.
(ii) Damage to the immune system of the body.
Question Nos. 17 to 20 consist of two statements – Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below:
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Question 17.
Assertion (A): The melting point and boiling point of ethanol are lower than that of sodium chloride.
Reason (R): The forces of attraction between the molecules of ionic compounds are very strong.
Answer:
(A) Both A and R are true, and R is the correct explanation of A.
Explanation: The melting point and boiling point of ethanol are indeed lower than that of sodium chloride. This is because ethanol is a covalent compound, while sodium chloride is an ionic compound. Covalent compounds have weaker intermolecular forces than ionic compounds, which means less energy is required to break the bonds holding the molecules together. Therefore, covalent compounds tend to have lower melting and boiling points than ionic compounds.
Question 18.
Assertion (A): Amoeba takes in food using finger like extensions of the cell surface.
Reason (R): In all unicellular organisms, the food is taken in by the entire cell surface.
Answer:
(C) A is true but R is false.
Explanation: Amoeba takes in food using temporary finger-like extensions of the cell surface, called pseudopodia, which extend and fuse over the food particle forming a food vacuole. Inside the food vacuole, complex substances are broken down into simpler ones which then diffuse into the cytoplasm. A unicellular organism does not need a specific organ for taking in food, because the entire surface of the organism is in contact with the environment.
Question 19.
Assertion (A): AC load line is used for long distance transmission.
Reason (R): It has very less loss of energy in long distance transmission.
Answer:
(A) Both A and R are true, and R is the correct explanation of A.
Explanation: AC load line can be easily transmitted over long distance without much loss in energy.
Question 20.
Assertion (A): Greater number of individuals is present in lower trophic levels.
Reason (R): The flow of energy is unidirectional.
Answer:
(B) Both A and R are true, and R is not the correct explanation of A.
Explanation: There are generally a greater number of individuals at the lower trophic levels of an ecosystem; the greatest number is of the producers. The flow of energy in an ecosystem is always linear or unidirectional. The energy captured from producers does not revert solar input. Also, the energy that passes to the herbivores does not come back to autotrophs.
Section – B (12 Marks)
(Q. no. 21 to 26 are very short answer questions.)
Question 21.
State whether the given chemical reaction is a redox reaction or not. Justify your answer
MnO2 + 4HC1 → MnCl2 + 2H2O + Cl2
Answer:
The given reaction is an example of a redox reaction because the oxidation and reduction take place simultaneously in the reaction.
(i) MnO2 + 4H+ + 2e– → Mn2+ + 2H2O
As you can see in this reaction, Mn02 is converted to Mn2+ because the addition of electrons takes place in this reaction, hence we can say that MnO2 is the compound which is reduced.
(ii) 2Cl– → Cl2 + 2e–
In this reaction, Cl– is oxidised to Cl2 because the removal of electrons takes place. Hence, we can say that the oxidised compound is HCl.
Question 22.
In the experimental set up on CO2 is released during respiration’, if one forgets to keep the vial with KOH in the conical flask, how will the result vary? Give details.
Answer:
In the absence of KOH, CO2 released by germinating seeds is not absorbed, a partial vacuum is not created in the conical flask, air pressure in the flask is not reduced, and the water level does not rise in the delivery tube.
Detailed Answer:
The rise in the level of water indicates that CO2 is produced by germinating seeds during respiration. Actually, the germinating seeds respire and produce C02, which is absorbed by KOH solution. This creates a vacuum in the conical flask. The air present in the bent glass tube moves into the conical flask. This pulls the water in the bent tube further up.
So, if one forgets to keep the vial with KOH solution in a conical flask during an experiment, then the released CO2 will not be absorbed due to which the level of water will not rise in the tube and the process of respiration will get very slow.
Question 23.
How is an electric impulse created in the human nervous system? Identify the parts of a neuron which help the nerve impulse to travel:
(a) Towards the cell body
(b) Away from the cell body
Answer:
A nerve impulse is generated when the stimulus is strong. This stimulus triggers the electrical and chemical changes in the neuron.
- Dendrites are the part of neurons where information is acquired. It receives impulses and transmits impulses toward the cell body.
- Axon is the part of the neuron through which information travels as an electric impulse. It helps in transmitting impulses away from the cell body.
OR
With the help of an example, explain how the feedback mechanism regulates the hormone secretion.
Answer:
Feedback mechanism is the mechanism in our body to maintain the hormonal levels in the body in a desirable amount. An increase or decrease in the hormonal level in our body triggers the feedback mechanism.
For example, The increase in the blood sugar level stimulates the secretion of insulin so that the sugar level is maintained. If the blood sugar level falls below normal, then it stimulates the secretion of glucagon. Glucagon stimulates the breakdown of glycogen to glucose, and thus, the normal sugar level is maintained.
Question 24.
A lens made of material with refractive index 1.5 is immersed in a liquid with refractive index 1.5. The diagram shows two rays incident on the lens when it is immersed in the liquid.
Copy the diagram and draw the light rays after they pass through the lens. Justify your diagram.
Answer:
Since the refractive index of the liquid is equal to that of the material of the lens, the light rays do not undergo refraction as they pass from the liquid to the lens and back into the liquid.
Question 25.
The figure shows two magnets X and Y kept near each other. Their poles are not marked, but the magnetic field lines are shown in the figure.
If magnet X is moved towards magnet Y as indicated by the arrow, will the two magnets attract or repel each other? Justify your answer by describing how you interpret the field lines.
Answer:
They will repel each other.
The right end of magnet X and the left end of magnet Y are both north poles since field lines start from there.
OR
What are the three types of wires used in household circuits? Pick out the wire used as a safety measure for an electrical appliance with a metallic body.
Answer:
(a) Live wire
(b) Neutral wire
(c) Earth wire
Question 26.
(a) From the following groups of organisms, create a food chain which is the most advantageous for human being in terms of energy.
(b) State the possible disadvantage if the cereal plant is growing in soil rich in pesticides.
(Hawk, Rat, cereal plant, Goat, Snake, Human Beings)
Answer:
(a) Cereal Plant → Human beings
(b) Pesticides being non-biodegradable accumulate progressively at each trophic level/leads to biomagnification.
Section – C (21 Marks)
(Q.no. 27 to 33 are short answer questions.)
Question 27.
(a) While electrolysing water before passing the current some drops of an acid are added. Why? Name the gases liberated at cathode and anode. Write the relationship between the volume of gas collected at anode and the volume of gas collected at the cathode.
(b) What is observed when silver chloride is exposed to sunlight? Give the type of reaction involved.
Answer:
(a) Water is a non-electrolyte. So, the addition of acid will make it a good conductor of electricity. By this action, ions will be produced due to the dissociation of acidified water that acts as a charge carrier for the faster conduction of electricity.
The gas liberated at the cathode is hydrogen gas.
The gas liberated at the anode is oxygen gas. y In the electrolysis of water, the ratio of volumes of hydrogen and oxygen obtained is 2:1, 2 volumes of hydrogen and volume of oxygen combine to form water
(b) When silver chloride is exposed to light, it decomposes to form silver metal and chlorine gas.
The type of reaction is photolytic decomposition.
Question 28.
When ethanol reacts with ethanoic acid in the presence of conc. H2SO4, a substance with a fruity smell is produced.
Answer the following:
(a) State the class of compounds to which the fruity smelling compounds belong. Write the chemical equation for the reaction and write the chemical name of the product formed.
(b) State the role of conc. H2SO4.
Answer:
(a) Esters.
Product’s chemical name – Ethyl ethanoate
(b) Conc. H2SO4 acts as a dehydrating agent. (Helps in the removal of water formed in the reaction.)
C4H8, it is an unsaturated hydrocarbon due to the presence of a double bond.
OR
Two carbon compounds X and Y have the molecular formula C4H8, and C5H12, respectively. Which one of these is most likely to show an addition reaction? Justify your answer. iso, give the chemical equation to explain the process of addition reaction in this case.
Answer:
C4H8, it is an unsaturated hydrocarbon due to the presence of a double bond.
Question 29.
What is geotropism? Draw a labelled diagram of a potted plant showing positive geotropism and negative geotropism.
Answer:
The upward growth of shoots and the downward growth of roots in response to the pull of the earth’s gravity is called geotropism.
Question 30.
State briefly the changes that take place in a fertilised egg till the birth of the child in the human female reproductive system. What happens to the egg when it is not fertilised?
Answer:
Changes in the fertilised egg:
(i) Zygote/fertilised egg start dividing.
(ii) Implantation of the zygote in the inner uterine wall.
(iii) Embryo starts growing with the help of the placenta which results in the development of the child.
(iv) Birth of a child as a result of rhythmic contraction of the muscles in the uterus.
When an egg is not fertilized, the inner lining of the uterus slowly breaks and comes out through the vagina as blood and mucous (Menstruation).
Question 31.
A student observes the above phenomenon in the lab as a white light passes through a prism. Among many other colors, he observed the position of the two colors: Red and Violet.
(a) What is the phenomenon called?
(b) What is the reason for the violet light to bend more than the red light?
(c) Draw a ray diagram to show the path of light when two identical
glass prisms are arranged together in an inverted position with respect to each other and a narrow beam of white light is allowed to fall obliquely on one of the focus of the first prism.
Answer:
(a) The phenomenon is called dispersion.
(b) Speed of violet light inside the prism is slowest and that of red is highest. Hence, the deviation of violet light is maximum and that of red is minimum.
(c) Ray diagram:
Question 32.
(a) What is the heating effect of electric current?
(b) Write an expression for the amount of heat produced in a resistor when an electric current is passed through it stating the meanings of the symbols used.
(c) Name two appliances based on the heating effect of electric current,
Answer:
(a) When electricity is supplied to a purely resistive conductor, the energy of electric current gets dissipated entirely in the form of heat. As a result, the resistor gets heated. The heating of a resistor because of the dissipation of electrical energy is known as the heating effect of electric current.
(b) When an electric current I passes through the conductor, the amount of heat produced in time t is:
H = Vlt
Since, V = IR
So, H = l2Rt
(c) Two appliances based on the heating effect of electric current are electric iron and electric geyser.
Question 33.
Suppose your parents have constructed a two-room house and you want in the living room there be a provision of one electric bulb, one electric fan, a refrigerator and a plug point for appliances of power up to 2 kilowatts. Draw a circuit diagram showing the electric fuse and earthing as safety devices.
Answer:
(i) Four components should be labelled.
(ii) All of them should be in parallel and there should be a fuse for safety.
(iii) Live and earth wires should be there.
Section – D (15 marks)
(Qzno. 34 to 36 are long answer questions.)
Question 34.
Write the chemical equation for the following:
(a) Combustion of methane
(b) Oxidation of ethanol
(c) Hydrogenation of ethene
(d) Esterification reaction
(e) Saponification reaction
Answer:
OR
(a) Draw two structural isomers of butane.
(b) Draw the structures of propanol and propanone
(c) Name the third homologue of (a) alcohols (b) aldehydes
(d) Name the following:
(e) Show the covalent bond formation in nitrogen molecule.
Answer:
(a) Structural formulae of isomers of Butane are: n-butane and isobutane.
(c) (i) 3rd homologue of alcohol is propanol (C3H7OH).
(ii) 3rd homologue of aldehyde is propanal (CH3CH2CHO)
(d) (i) Benzene (C6H6)
(ii) 1- Butene
(e) Covalent bond formation in Nitrogen molecule:
Question 35.
Different organisms reproduce by different methods suitable to their body designs.
(a) Justify the above statement using examples of three different organisms that reproduce by different methods of asexual reproduction.
(b) Differentiate between sexual and asexual modes of reproduction.
Answer:
(a) Amoeba: Binary fission
Plasmodium: Multiple fission
Hydra: Budding
Planaria: Regeneration (Any three + Explain)
(b) Sexual: two parents; Asexual: single parent.
Detailed Answer:
(a) (i) Binary Fission in Amoeba: In this method, the nucleus first divides mitotically into two, followed by the division of the cytoplasm. The cell finally splits into two daughter cells. So, from one Amoeba parent, two daughter Amoebae are formed.
(ii) Budding in Hydra: In budding, a small part of the body of the parents grows out as a ‘bud’, which then detaches and becomes a new organism. Hydra reproduces by budding using regenerative cells. A bud develops as an outgrowth in Hydra due to repeated cell division at one specific site. When fully mature, the bud detaches itself from the parent body and develops into new independent individuals.
(iii) Regeneration in Planaria: In this method, small cut or broken parts of an organism’s body grow or regenerate into separate individuals. Planaria can be cut into any number of pieces and each piece grows into a complete organism.
(b) Differences between sexual and asexual reproduction:
S.No | Sexual Reproduction | Asexual Reproduction |
(i) | Two parents are required | Only one parent is required |
(ii) | Offsprings are genetically dissimilar from parents | Offsprings are identical to parents |
OR
Give one example each of unisexual and bisexual flowers. Differentiate between the two types of pollination that occur in flowers. What happens when pollen lands on a suitable stigma? Write about the events that occur till the seed formation in the ovary.
Answer:
Unisexual flower: Papaya/water-melon/any other. (Any one)
Bisexual flower: Hibiscus/Rose/any other.
Self-pollination: The pollen grains are transferred from the anther to the stigma of the same flower or to the flower of the same plant.
Cross-pollination: The pollen grains are transferred from the anther to the stigma of a flower of a different plant.
After pollen lands on a suitable stigma, a pollen tube grows out of pollen grain and travels through the style to reach the ovary. The male germ cell fuses with the female germ cell to form a zygote. Zygote divides several times to form an embryo within the ovule. The ovule develops tough coat and gradually gets converted into a seed.
Question 36.
Analyse the following observation table showing the variation of image-distance (v) with object-distance (u) in case of a convex lens and answer the questions that follow without doing any calculations:
S. No | Object-Distance u(cm) | Image-Distance v(cm) |
1 | -100 | +25 |
2 | -60 | +30 |
3 | -40 | +40 |
4 | -30 | +60 |
5 | -25 | +100 |
6 | -15 | +120 |
(a) What is the focal length of the convex lens? Give a reason to justify your answer.
(b) Write the serial number of the observation, which is not correct. On what basis have you arrived at this conclusion?
(c) Select an appropriate scale and draw a ray diagram for the observation at S.No. 2. Also find the approximate value of magnification.
Answer:
(a) The focal length of the convex lens can be calculated from S.No. 3 as when an object is placed at a distance from the convex lens, its image is formed on the other side of the lens at the same distance from the lens. So, the focal length is + 20 cm.
(b) S.No. 6 is incorrect as the object distance is between the focus and pole and here the real image is formed as the image distance is positive. But in such situation, virtual image should form.
(c) Approximate value of magnification for object distance – 60 cm and image distance +30 cm is -0.5.
u = -60 cm; v = + 30 cm,/ = – 20 cm m = v/u = 30/-60= 1/2 = -0.5
OR
(a) A security mirror used in a big showroom has a radius of curvature 5 m. If customer is standing at a distance of 20 m from the cash counter, find the position, nature and size of the image formed in the security mirror.
(b) Neha visited a dentist in his clinic. She observed that the dentist was holding an instrument fitted with a mirror. State the nature of this mirror and the reason for its use in the instrument used by dentist.
Answer:
(a) It is a convex mirror. So focal length should be positive.
Radius of curvature R = + 5 m
∴ Focal length f = 5/2 + 2.5 m
Object distance u =- 20 m
- Nature of image = virtual and erect image1
- Size of image: diminished image
(b) Concave Mirror – Reason: to obtain an erect and enlarged image of teeth
Section – E (12 Marks)
Question 37.
On the basis of reactivity metals are grouped into three categories:
(a) Metals of low reactivity
(b) Metal of medium reactivity
(c) Metals of high reactivity
Therefore, metals are extracted in pure form from their ores on the basis of their chemical properties. Metals of high reactivity are extracted from their ores by electrolysis of the molten ore. Metals of low reactivity are extracted from their sulphide ores, which are converted into their oxides. The oxides
of these metals are reduced to metals by simple heating.
(a) Name the process of reduction used for a metal that gives a vigorous reaction with air and water both.
(b) Carbon cannot be used as a reducing agent to obtain aluminium from its oxide? Why?
(c) Describe briefly the method to obtain mercury from cinnabar. Write the chemical equation for the reactions involved in the process.
Answer:
(a) Electrolytic reduction: The metal is likely to be sodium (Na). It has a very much affinity to oxygen. So, reducing agents like carbon and aluminium can’t be used.
(b) Because aluminium has a greater affinity for oxygen than for carbon, therefore carbon cannot reduce alumina (Al2O3) to aluminium.
(c) Cinnabar (HgS-mercury (II) sulphide or mercury sulphide) is an ore of mercury. It is heated in air to give mercuric oxide (HgO). Mercuric oxide is further heated to get mercury.
The reaction involved are: \(\begin{aligned}
& 2 \mathrm{HgS}(s)+3 \mathrm{O}_2(g) \stackrel{\Delta}{\longrightarrow} 2 \mathrm{HgO}(s)+2 \mathrm{SO}_2(g) \\
& 2 \mathrm{HgO}(s) \stackrel{\Delta}{\longrightarrow} 2 \mathrm{Hg}(l)+\mathrm{O}_2(g)
\end{aligned}\)
OR
(c) Differentiate between roasting and calcination giving a chemical equation for each.
Answer:
Calcination is defined as the process of converting ore into an oxide by heating it strongly. The ore is heated below its melting point either in the absence of air or in limited supply.
e.g., ZnCO3 → ZnO + CO2
Roasting is a process of metallurgy where ore is converted into its oxide by heating it below its melting point in the presence of excess air. An example of roasting is when Zinc sulphide is converted into zinc oxide.
2ZnS + 3O2 → 2ZnO + 2SO2
Question 38.
Mendel blended his knowledge of science and mathematics to keep the count of the individuals exhibiting a particular trait in each generation. He observed a number of contrasting visible characters controlled in pea plants in field. He conducted many experiments to arrive at the laws of inheritance.
(a) What do the F1 progeny of tall plants with round seeds and short plants with wrinkled seeds look like?
(b) Name the recessive traits in the above case.
(c) Mention the type of the new combinations of plants obtained in F2 progeny along with their ratio, if F1 progeny was allowed to self pollinate.
Answer:
(a) F1 progeny of tall plants with round seeds and short plants with wrinkled seeds will be a heterozygous tall plant with round seeds (TtRr) as tall and round is the dominant traits.
(b) The recessive traits are short plants and wrinkled seeds.
(c) The different types of combinations obtained in F2 progeny are:
Tall plants with round seeds = 9 Short plants with round seeds = 3 Tall plants with wrinkled seeds = 3 Short plants with wrinkled seeds = 1 Phenotypic ratio = Tall round: short round: tall wrinkled: short wrinkled: 9:3:3:1.
OR
(c) If 1600 plants were obtained in F2 progeny, write the number of plants having traits:
(i) Tall with round seeds.
(ii) Short with wrinkled seeds.
Write the conclusion of the above experiment.
Answer:
If 1600 plants were obtained in F2 progeny, the number of plants having traits will be:
(i) Tall plants with round seeds = 9/16 × 1600 = 900
(ii) Short plants with wrinkled seeds = 1/16 × 1600 = 100
The conclusion of the above experiment states the “Law of independent assortment”. This law states that the alleles of two (or more) different genes get sorted into gametes independently of one another.
Question 39.
A student was asked to perform an experiment to study the force on a current- carrying conductor in a magnetic field. He took a small aluminium rod AB, a strong horse shoe magnet, some connecting wires, a battery and a switch and connected them as shown. He observed that on passing the current, the rod gets displaced. On reversing the direction of the current, the direction of displacement also gets reversed. On the basis of your understanding of this phenomenon, answer the following questions:
(a) Why does the rod get displaced on passing a current through it?
(b) State the rule that determines the direction of the force on the conductor AB.
(c) (i) If the U-shaped magnet is held vertically and the aluminium rod is
suspended horizontally with its end B towards due north, then on
passing current through the rod from B to A as shown, in which direction will the rod be displaced?
(ii) Name any two devices that use current carrying conductors and magnetic field.
Answer:
(a) When a current-carrying conductor is placed in a magnetic field, it experiences a force, due to which the rod gets displaced.
(b) The rule that determines the direction of the force on the conductor AB is Fleming’s Left-Hand rule. According to this rule if the direction of electric current is perpendicular to the magnetic field, the direction of force is also perpendicular to both of them. Stretch the thumb, fore finger and middle finger of your left hand such that they are mutually perpendicular. If fore finger points in the direction of the magnetic field, the middle finger in the direction of current, then the thumb will point in the direction of motion of force.
(c) (i) According to Fleming’s Left-Hand rule, the rod will get displaced upwards.
(ii) Devices that use current-carrying conductors and magnetic fields are electric motor, electric generator, loudspeaker, microphones, etc.
OR
(c) Draw the pattern of magnetic field lines produced around a current-carrying straight conductor held vertically on a horizontal cardboard. Indicate the direction of the field lines as well as the direction of the current flowing through the conductor.
Answer:
(c) The magnetic field lines around a current carrying conductor can be represented by concentric circles.
The direction of the magnetic field can be determined by using Right-Hand Thumb rule.