Science Practice Paper 2018-19 Questions and Answers

Welcome to the Science Practice Paper 2018-19 Questions and Answers. This practice paper is designed to help students strengthen their understanding of key concepts in Science, improve problem-solving skills, and prepare effectively for their exams. The questions are based on the CBSE curriculum and cover various topics in Physics, Chemistry, and Biology. Each question is followed by a clear and detailed answer to ensure that students can grasp the concepts easily and learn the correct method of solving problems.

This resource is ideal for revision, self-assessment, and practice to boost confidence and enhance academic performance. Students are encouraged to attempt the questions first and then refer to the solutions for better understanding.

Science Practice Paper 2018-19 Questions and Answers from Textbook

SECTION A

1. Name a common nutrient that is absorbed in the small intestine and reabsorbed by the kidney tubules. Ans: Glucose.

2. The presence of a particular group of bacteria in water bodies indicates contamination. Identify the group. Ans: Coliform bacteria.

SECTION B

3. How is Magnesium Chloride formed by the transfer of electrons? Why does the solution of Magnesium chloride conduct electricity?

Ans: Magnesium has an electronic configuration of 2, 8, 2, and Chlorine has 2, 8, 7. Magnesium will lose its two valence electrons to form a stable Mg2+ ion. Each chlorine atom needs one electron to achieve a stable electronic configuration. Therefore, one magnesium atom transfers one electron to each of the two chlorine atoms, forming a MgCl2​ molecule. The electrostatic attraction between the positively charged Mg2+ ion and the two negatively charged Cl− ions results in an ionic bond.

The solution of Magnesium chloride conducts electricity because it contains free-moving ions (Mg2+ and Cl−) which act as charge carriers in the solution.

4. In a flowering plant, summarize the events that take place after fertilization.

Ans: After fertilization, the following events take place:

  1. The zygote divides several times to form an embryo within the ovule.
  2. The ovule develops a tough coat and gradually converts into a seed.
  3. The ovary ripens to form the fruit.
  4. The remaining floral parts, such as the petals, sepals, and stamens, fall off.

5. A ray of light enters into benzene from air. If the refractive index of benzene is 1.50, by what percent does the speed of light reduce on entering the benzene?

Ans: The refractive index (n) is given by the formula n=speed of light in air/speed of light in medium. Given, n=1.50. Let the speed of light in air be ca​ and in benzene be cb​. 1.50=ca​/cb​⟹cb​=ca​/1.50.

The reduction in speed is ca​−cb​=ca​−(ca​/1.50)=ca​(1−1/1.50)=ca​(1−0.67)=0.33ca​.

The percentage reduction is (0.33ca​/ca​)×100%=33%. The speed of light reduces by 33% on entering the benzene.

OR

For the same angle of incidence in media A, B and C, the angles of refraction are 200, 306 and 400 respectively. In which medium will the velocity of light be maximum? Give reason in support of your answer.

Ans: The velocity of light will be maximum in medium C.

According to Snell’s law, n=sini/sinr, where ‘i’ is the angle of incidence and ‘r’ is the angle of refraction. The velocity of light is inversely proportional to the refractive index of the medium. Given that the angle of incidence is the same for all three media, the refractive index will be inversely proportional to the sine of the angle of refraction. n∝1/sinr. Since the velocity of light is inversely proportional to the refractive index, it is directly proportional to the sine of the angle of refraction. Velocity ∝sinr. The angle of refraction is highest in medium C (400), so the velocity of light will be maximum in medium C.

SECTION C

6. What happens when aqueous solutions of Sodium sulphate and Barium chloride are mixed? Give a balanced equation for the reaction with state symbols. Name and define the type of chemical reaction involved in the above change.

Ans: When aqueous solutions of sodium sulphate (Na2​SO4​) and barium chloride (BaCl2​) are mixed, a white precipitate of barium sulphate (BaSO4​) is formed. This is a precipitation reaction. Sodium chloride (NaCl) is also formed in the solution. The balanced chemical equation is:

Na2​SO4​(aq)+BaCl2​(aq)→BaSO4​(s)+2NaCl(aq).

This is a Double Displacement Reaction. In this type of reaction, the ions of two different reacting compounds are exchanged to form two new compounds.

7. Identify the compound of calcium which is used for plastering of fractured bones. With the help of chemical equation show how is it prepared and what special precautions should be taken during the preparation of this compound.

Ans: The compound of calcium used for plastering fractured bones is Plaster of Paris (PoP), which is Calcium Sulphate Hemihydrate (CaSO4​⋅1/2H2​O).

It is prepared by heating Gypsum (CaSO4​⋅2H2​O) to a temperature of 373 K (100°C).

The chemical equation is: CaSO4​⋅2H2​O(s)373K​CaSO4​⋅1/2H2​O(s)+3/2H2​O(l) A special precaution to be taken during the preparation of this compound is that the temperature should be strictly controlled and not allowed to go above 373 K. If it is heated to a higher temperature, all the water of crystallization will be lost, and anhydrous calcium sulphate (CaSO4​) called “dead burnt plaster” will be formed, which does not have the plastering properties of PoP.

OR

‘Sweet tooth may lead to tooth decay’. Explain why? What is the role of tooth paste in preventing cavities? Ans: When we eat sugary food, the bacteria present in our mouth convert the sugar into acid. This acid lowers the pH of the mouth, making it acidic. This acidic environment then starts to corrode the tooth enamel, leading to tooth decay.

Toothpaste plays a crucial role in preventing cavities because it is alkaline (basic) in nature. It neutralizes the acid produced by the bacteria in the mouth, restoring the pH balance and preventing the decay of tooth enamel.

8. The electronic configuration of an element ‘X’ is 2,8,6. To which group and period of the modern periodic table does ‘X’ belong. State its valency and justify your answer in each case.

  • Period: The element ‘X’ belongs to the 3rd period. This is because it has three electron shells (K, L, and M). The number of shells determines the period number.
  • Group: The element ‘X’ belongs to Group 16. This is because it has 6 valence electrons (electrons in the outermost shell). The group number for elements with valence electrons from 3 to 8 is determined by adding 10 to the number of valence electrons.
  • Valency: The valency of element ‘X’ is 2. To achieve a stable octet configuration, it needs to gain two electrons. The valency is the number of electrons an atom needs to gain, lose, or share to become stable.

9. Pertaining to endocrine system, what will you interpret if- i) You observe swollen neck in people living in the hills

Ans: This is an indication of Goitre. This condition is caused by a deficiency of iodine in the diet. The thyroid gland in the neck swells up in an attempt to produce more thyroid hormone, which is essential for carbohydrate, protein, and fat metabolism.

ii) Over secretion of Growth Hormone takes place during childhood Over-secretion of the growth hormone during childhood leads to

Gigantism. The bones and other body parts grow excessively, and the person becomes abnormally tall.

iii) Facial hair develops in boys aged 13. The development of facial hair in boys at this age is a sign of

puberty. This is a normal process caused by the secretion of the male sex hormone,

testosterone, which is responsible for the development of secondary sexual characteristics in males.

10. A variegated leaf with green and yellow patches in used for an experiment to prove that chlorophyll is required for photosynthesis. Before the experiment the green portions (A), and the pale yellow portions (B), are observed. What will be the colour of ‘A’ just before and after the starch test? Also write the equation of photosynthesis and mark, as well as validate from which molecule the by-product is obtained.

Ans: Before the starch test, the colour of portion ‘A’ will be green due to the presence of chlorophyll. After the starch test (using iodine solution), the colour of portion ‘A’ will turn blue-black, confirming the presence of starch.

The colour of portion ‘B’ (yellow patches) will not change to blue-black, as it lacks chlorophyll and therefore cannot perform photosynthesis to produce starch. Equation for photosynthesis:

6CO2​+12H2​OSunlight, Chlorophyll​C6​H12​O6​+6O2​+6H2​O

The by-product, oxygen (O2​), is obtained from the water (H2​O) molecule. This is validated by experiments using isotopes of oxygen, which show that the oxygen gas released during photosynthesis originates from the splitting of water molecules.

11. The image of an object formed by a mirror is real, inverted and is of magnification -1. If the image is at the distance of 30 cm from the mirror, where is the object placed? Find the position of the image if the object is now moved 20 cm towards the mirror. What is the nature of the image obtained? Justify your answer with the help of ray diagram.

Ans: Since the image is real, inverted, and the magnification is -1, the mirror must be a concave mirror and the object is placed at the center of curvature (C).

  • Object position: The image is formed at a distance of 30 cm from the mirror, so the object is also at a distance of 30 cm from the mirror. This is because when m=−1, the object and image distances are equal (v=u) and the object is at C.
  • New object position: If the object is moved 20 cm towards the mirror, its new position will be at 30−20=10 cm from the mirror. Since the focal length is half the radius of curvature, f=C/2=30/2=15 cm. The new object position (10 cm) is between the pole (P) and the principal focus (F) of the concave mirror.
  • New image position and nature: When an object is placed between the pole and the principal focus of a concave mirror, a virtual, erect, and magnified image is formed behind the mirror. *

OR

What is meant by power of a lens? You have three lenses L1​, L2​ and L3​ of powers +10D, +5D and -10D respectively. State the nature and focal length of each lens. Explain which of the three lenses will form a virtual and magnified image of an object placed at 15 cm from the lens. Draw the ray diagram in support of your answer.

  • Power of a lens: The power of a lens is a measure of its degree of convergence or divergence of light rays. It is defined as the reciprocal of its focal length in meters. The unit of power is Dioptre (D).
  • Nature and focal length of lenses:
    • L1​: Power = +10D. Since the power is positive, it is a converging lens (convex lens). Focal length f=1/P=1/10=0.1 m=10 cm.
    • L2​: Power = +5D. Since the power is positive, it is a converging lens (convex lens). Focal length f=1/P=1/5=0.2 m=20 cm.
    • L3​: Power = -10D. Since the power is negative, it is a diverging lens (concave lens). Focal length f=1/P=1/−10=−0.1 m=−10 cm.
  • Virtual and magnified image: Only a convex lens can form a virtual and magnified image. This happens when the object is placed between the principal focus (F) and the optical center (O). The object distance is 15 cm.
    • For lens L1​, the focal length is 10 cm. The object is at 15 cm, which is beyond the focal length. So, it will form a real and inverted image.
    • For lens L2​, the focal length is 20 cm. The object is at 15 cm, which is between the focus (F) and the optical center (O). Therefore, lens L2​ will form a virtual and magnified image.
    • For lens L3​ (concave), a virtual, erect, and diminished image is always formed.
  • Ray diagram: *

12. Two lamps, one rated 100 W at 220 V and the other 200 W at 220V are connected (i) in series and (ii) in parallel to electric main supply of 220V. Find the current drawn in each case.

Ans: Step 1: Find resistance of each lamp
Formula: R = V² / P

For 100 W lamp:
R1 = 220 × 220 / 100 = 484 ohm

For 200 W lamp:
R2 = 220 × 220 / 200 = 242 ohm

So, R1 = 484 ohm and R2 = 242 ohm

Case (i): Series connection
Equivalent resistance Rs = R1 + R2 = 484 + 242 = 726 ohm
Current Is = V / Rs = 220 / 726 = 0.303 ampere

Case (ii): Parallel connection
1/Rp = 1/R1 + 1/R2
1/Rp = 1/484 + 1/242 = (1 + 2)/484 = 3/484
Rp = 484 / 3 = 161.3 ohm
Current Ip = V / Rp = 220 / 161.3 = 1.364 ampere

13. The figure below shows three cylindrical copper conductors along with their face areas and lengths. Compare the resistance and the resistivity of the three conductors. Justify your answer.

  • Resistivity: The resistivity of all three conductors will be the same. This is because resistivity is an intrinsic property of the material and does not depend on the dimensions (length or area) of the conductor. Since all three conductors are made of copper, their resistivity will be identical.
  • Resistance: The resistance (R) of a conductor is given by the formula R=ρ(L/A), where ρ is resistivity, L is length, and A is the cross-sectional area. Since the resistivity (ρ) is the same for all three, the resistance depends on the ratio of L/A.
    • Conductor (a): Ra​=ρ(L/A)
    • Conductor (b): Rb​=ρ(3L/(A/3))=ρ(9L/A)=9Ra​. The resistance is 9 times that of conductor (a).
    • Conductor (c): Rc​=ρ((L/3)/3A)=ρ(L/9A)=(1/9)Ra​. The resistance is 1/9th of conductor (a). Therefore, the resistance of the three conductors in increasing order is Rc​<Ra​<Rb​.

14. What is biogas? Describe the steps involved in obtaining biogas.

Biogas is a mixture of gases, primarily methane (CH4​), produced by the anaerobic decomposition of organic matter like agricultural waste, animal dung, and sewage.

The process of obtaining biogas involves the following steps:

  1. Preparation of slurry: Organic waste materials such as cow dung, water, and agricultural residues are mixed in a large tank to create a slurry.
  2. Feeding into the digester: This slurry is then fed into the digester, a sealed tank where anaerobic microorganisms break down the complex organic compounds.
  3. Anaerobic decomposition: The breakdown of the organic matter by bacteria in the absence of oxygen produces a mixture of gases, mainly methane and carbon dioxide.
  4. Gas collection: The biogas produced rises and is collected in a gas holder at the top of the digester, from where it is supplied through pipelines for use.
  5. Sludge removal: The remaining spent slurry, called sludge, is rich in nutrients and is used as an excellent fertilizer.

15. How is ozone both beneficial and damaging? How can depletion of ozone layer be prevented?

  • Beneficial role of ozone: The ozone layer in the stratosphere is a shield that absorbs harmful ultraviolet (UV) radiation from the sun. This prevents UV radiation from reaching the Earth’s surface, protecting living organisms from skin cancer, cataracts, and other health issues.
  • Damaging role of ozone: At ground level, ozone is a pollutant that is harmful to humans and other living organisms. It can damage lung tissue and cause respiratory problems.
  • Prevention of ozone layer depletion: The depletion of the ozone layer is primarily caused by man-made chemicals called Chlorofluorocarbons (CFCs), which are used in refrigerants, fire extinguishers, and aerosol sprays.
    • The depletion can be prevented by phasing out the production and use of these ozone-depleting substances.
    • International agreements like the Montreal Protocol have been successful in reducing the use of CFCs globally.
    • Using alternatives to CFCs that do not harm the ozone layer is essential.

OR

The flow of energy between various components of the environment has been extensively studied. List four findings.

  1. Unidirectional flow: The flow of energy in an ecosystem is unidirectional. Energy flows from the sun to producers, and then to consumers in a linear path. It does not flow back from consumers to producers.
  2. Ten percent law: At each trophic level, only about 10% of the energy is transferred to the next trophic level. The rest is lost to the environment as heat or is used for metabolic processes by the organisms.
  3. Decreasing energy at higher trophic levels: Due to the 10% law, the amount of energy available to organisms decreases significantly at successive trophic levels. This is why food chains typically do not have more than 3-4 trophic levels.
  4. Energy loss: The continuous loss of energy at each level means that ecosystems require a constant and continuous supply of energy from the sun to sustain the food chain.

SECTION D

16. a) How will you show experimentally that metals are good conductors of heat.

b) Describe the extraction of Mercury metal from its ore Cinnabar (HgS).

Ans: a) An experiment to show that metals are good conductors of heat:

  1. Take a metallic rod and fix some wax-coated pins on it at equal intervals.
  2. Heat one end of the metallic rod with a burner.
  3. Observe the pins. You will notice that the pins start falling off one by one, starting from the pin closest to the heated end. Conclusion: This shows that heat is transferred from the heated end to the other end of the rod, and the pins fall as the wax melts. The rate at which the pins fall indicates that heat is conducted along the metal rod, proving that metals are good conductors of heat. b) The extraction of Mercury metal from its ore Cinnabar (HgS) involves a two-step process:
  4. Roasting: The ore Cinnabar (HgS) is heated in the presence of air. This converts the sulphide ore into its oxide, and sulphur dioxide gas is released.
    • 2HgS(s)+3O2​(g)Heat​2HgO(s)+2SO2​(g)
  5. Reduction: The mercury oxide (HgO) is then heated further. Since mercury is a less reactive metal, its oxide can be reduced by heating alone. The mercury oxide decomposes to form mercury metal and oxygen gas.
    • 2HgO(s)Heat​2Hg(l)+O2​(g)

17. A compound A (C2​H4​O2​) reacts with Na metal to form a compound ‘B’ and evolves a gas which burns with a pop sound. Compound ‘A’ on treatment with an alcohol ‘C’ in presence of an acid forms a sweet smelling compound ‘D’ (C4​H8​O2​) On addition of NaOH to ‘D’ gives back B and C. Identify A, B, C and D write the reactions involved.

  • Compound A: C2​H4​O2​. Since it reacts with Na metal to produce a gas that burns with a pop sound (hydrogen gas), it must be a carboxylic acid. The compound is Ethanoic acid (CH3​COOH).
  • Compound B: When ethanoic acid reacts with sodium metal, it forms Sodium ethanoate (CH3​COONa) and hydrogen gas.
    • Reaction 1: 2CH3​COOH+2Na→2CH3​COONa+H2​
  • Compound C: Compound A (CH3​COOH) reacts with alcohol ‘C’ to form a sweet-smelling compound ‘D’. This is an esterification reaction. Since the product ‘D’ is an ester with the formula C4​H8​O2​, the alcohol ‘C’ must have two carbon atoms. Therefore, C is Ethanol (C2​H5​OH).
  • Compound D: The reaction between ethanoic acid and ethanol produces a sweet-smelling ester, Ethyl ethanoate (CH3​COOC2​H5​) and water.
    • Reaction 2: CH3​COOH+C2​H5​OHAcid​CH3​COOC2​H5​+H2​O
  • Reaction of D with NaOH: The addition of NaOH to the ester ‘D’ gives back B and C. This is a saponification reaction.
    • Reaction 3: CH3​COOC2​H5​+NaOH→CH3​COONa+C2​H5​OH
    • The products are Sodium ethanoate (B) and Ethanol (C).

OR

a) Explain why carbon forms covalent bond? Give two reasons for carbon forming a large number of compounds.

b) Explain the formation of ammonia molecule.

Ans: a) Carbon forms covalent bonds because its atomic number is 6, and its electronic configuration is 2, 4. To achieve a stable noble gas configuration, it needs to either gain four electrons (to form a C4− ion) or lose four electrons (to form a C4+ ion).

  • Gaining four electrons would be difficult due to the nucleus’s inability to hold ten electrons.
  • Losing four electrons would require a large amount of energy to remove them from the nucleus.
  • Therefore, carbon overcomes these difficulties by sharing its four valence electrons with other atoms to form covalent bonds.

Two reasons for carbon forming a large number of compounds are:

  1. Catenation: Carbon has a unique ability to form bonds with other carbon atoms, creating long chains, branched chains, and closed rings. This property, called catenation, allows it to form a vast number of compounds.
  2. Tetravalency: Carbon has a valency of four, which means it can form bonds with four other atoms of carbon or other monovalent elements like hydrogen, halogens, or oxygen. This allows for a wide variety of compounds to be formed. b) Formation of an ammonia molecule (NH3​): The atomic number of Nitrogen is 7, with an electronic configuration of 2, 5. It needs three more electrons to complete its octet. The atomic number of Hydrogen is 1, and it needs one more electron to complete its duplet. To form an ammonia molecule, one nitrogen atom shares its three unpaired electrons with three hydrogen atoms. Each hydrogen atom, in turn, shares its single electron with the nitrogen atom.

This sharing of electrons results in a covalent bond between the nitrogen atom and each of the three hydrogen atoms, forming a stable ammonia molecule.

18. a) Draw the diagram of female reproductive system and match and mark the part(s):

i) Where block is created surgically to prevent fertilization.

ii) Where Copper-T is inserted?

iii) Inside which condom can be placed.

b) Why do more and more people prefer to use condoms? What is the principle behind use of condoms?

Ans: a)

  • i) Fallopian Tube: The fallopian tubes are surgically cut or tied (tubal ligation) to create a block and prevent fertilization.
  • ii) Uterus: Copper-T is a contraceptive device that is inserted into the uterus.
  • iii) Vagina: A condom, a barrier contraceptive, can be placed inside the vagina to prevent sperm from reaching the cervix and uterus.b) People prefer to use condoms for two main reasons:
  1. Protection from STIs: They are effective in preventing the transmission of sexually transmitted infections (STIs), including HIV/AIDS.
  2. Contraception: They are a reliable method of contraception, preventing unwanted pregnancies. The principle behind the use of condoms is barrier contraception. A condom acts as a physical barrier that prevents sperm from entering the female reproductive tract, thus stopping fertilization from occurring.

19. Name the phenomenon that governs the following: – i) Green beetles living in green bushes are not eaten by the crows.

Ans: Natural Selection. The green colour of the beetles provides camouflage, allowing them to survive and reproduce.

ii) Number of blue beetles in green bushes increases, only because the red beetles living there were trampled by a herd of elephants.

Ans: Genetic Drift. The random event of elephants trampling the red beetles leads to a change in the frequency of the blue beetle gene in the population.

iii) No ‘medium height plants’ are obtained in F1 generation, upon crossing pure tall and dwarf pea plants. Ans: Dominance. The gene for tallness is completely dominant over the gene for dwarfness, so the heterozygous F1 generation expresses only the tall phenotype.

iv) Tails of mice were surgically removed for several generations; still mice had tails in the following generations.

Ans: Inheritance of Acquired Traits is not possible. The removal of a tail is an acquired trait and does not affect the genes in the germ cells, so it is not passed on to the next generation.

v) A migrant beetle reproduces with the local population; as a result genes of migrant beetle enter the new population.

Ans: Gene Flow. This is the transfer of genes from one population to another, which can change the gene frequencies in the new population.

OR

a) What are fossils and how is age of fossils determined? b) During artificial selection, which features of wild cabbage were selected to give rise to i) Cabbage ii) Cauliflower

Ans: a) Fossils are the preserved remains or traces of organisms that lived in the past. They can be the remains of the organism itself, such as bones or shells, or traces like footprints, molds, or casts.

The age of fossils can be determined by two methods:

  1. Relative Dating: This method involves comparing the position of a fossil in different rock layers. Fossils found in deeper layers of the earth are considered older than those found in the upper layers.
  2. Dating by radioisotopes: This method involves analyzing the ratios of different isotopes of elements in the fossil. For example, by measuring the ratio of different carbon isotopes, the age of the fossil can be determined. b) During artificial selection, different features of wild cabbage were selected:
  • i) To give rise to Cabbage, the farmers selected for very short distances between leaves.
  • ii) To give rise to Cauliflower, the farmers selected for arrested flower development.

20. (a)What is meant by the term ‘power of accommodation’? Name the component of eye that is responsible for the power of accommodation.

(b) A student sitting at the back bench in a class has difficulty in reading. What could be his defect of vision? Draw ray diagrams to illustrate the image formation of the blackboard when he is seated at the

(i) back seat

(ii) front seat.

State two possible causes of this defect.

Explain the method of correcting this defect with the help of a ray diagram.

Ans: a) Power of accommodation is the ability of the eye lens to adjust its focal length to focus on objects at different distances. This is done by the ciliary muscles of the eye, which change the curvature of the eye lens.

b) The student’s defect of vision is Myopia (nearsightedness). This is a condition where a person can see nearby objects clearly but cannot see distant objects distinctly.

  • Ray diagrams for image formation:
    • (i) Back seat: When he is at the back seat, the image of the distant blackboard is formed in front of the retina.
    • (ii) Front seat: When he is at the front seat, the objects are nearby. The light rays from nearby objects are correctly focused on the retina.
  • Two possible causes of Myopia:
    1. Excessive curvature of the eye lens.
    2. Elongation of the eyeball.
  • Correction of Myopia: Myopia can be corrected by using a concave lens of appropriate power. The concave lens diverges the light rays before they enter the eye, allowing the eye lens to focus the image correctly on the retina. *

21. (i) With the help of an activity, explain the method of inducing electric current in a coil with a moving magnet. State the rule used to find the direction of electric current thus generated in the coil. (ii) Two circular coils-1 and coil-2 are kept close to each other as shown in the diagram. Coil-1 is connected to a battery and key and coil-2with a galvanometer. State your observation in the galvanometer:

(a) When key K closed;

(b) when key K is opened; Give reason for your observations.

Ans: i) Activity to induce electric current with a moving magnet:

  1. Take a coil of insulated copper wire and connect its ends to a galvanometer.
  2. Take a bar magnet and move its north pole towards the coil.
  3. Observe the galvanometer. You will see a deflection, which indicates that an electric current has been induced in the coil.
  4. Move the magnet away from the coil. The galvanometer will show a deflection in the opposite direction, indicating a reversal of the induced current.
  5. If you hold the magnet stationary inside the coil, there will be no deflection, showing that no current is induced when there is no change in the magnetic field. The rule used to find the direction of the induced electric current is Fleming’s Right-Hand Rule. This rule states that if you stretch the thumb, forefinger, and middle finger of your right hand mutually perpendicular to each other, such that the forefinger points in the direction of the magnetic field, the thumb points in the direction of motion of the conductor, then the middle finger will point in the direction of the induced current.

ii) (a) When key K is closed: The galvanometer shows a momentary deflection.

(b) When key K is opened: The galvanometer shows a momentary deflection in the opposite direction.

Reason: A current starts flowing through Coil-1 when the key K is closed, creating a magnetic field around it. This change in the magnetic field lines around Coil-2 induces an electric current, which is indicated by the galvanometer. When the key is opened, the current in Coil-1 stops, and the magnetic field collapses. This change in the magnetic field induces a momentary current in Coil-2 in the opposite direction. There is no continuous deflection because the current is only induced when there is a change in the magnetic field.

OR

Name a device which converts mechanical energy into electrical energy. Explain the underlying principle and working of this device with the help of a labelled diagram.

Ans: The device that converts mechanical energy into electrical energy is an electric generator.

Principle: An electric generator works on the principle of electromagnetic induction. This principle states that when a coil is rotated in a magnetic field, a current is induced in the coil.

Working:

  • An electric generator consists of a rectangular coil (ABCD) placed in the magnetic field of a strong magnet.
  • The two ends of the coil are connected to two carbon brushes (B1​ and B2​) which are pressed against two rings (R1​ and R2​).
  • When the axle attached to the rings is rotated, the coil also rotates in the magnetic field.
  • As the coil rotates, the magnetic flux linked with it changes, and a current is induced in the coil.
  • The direction of the induced current changes every half rotation of the coil, so an alternating current (AC) is produced.

SECTION E

22. When few drops of phenolphthalein are added to a dilute solution of sodium hydroxide a pink colour is produced. What will be the colour of the final mixture when excess of HCl is added to it? (justify your answer)

Ans: When a few drops of phenolphthalein are added to a dilute solution of sodium hydroxide (NaOH), a pink colour is produced because sodium hydroxide is a strong base, and phenolphthalein is pink in basic medium.

When excess hydrochloric acid (HCl) is added to this solution, the colour of the final mixture will become colourless. The hydrochloric acid (a strong acid) will neutralize the sodium hydroxide, and the solution will become acidic. Phenolphthalein is colourless in acidic medium, so the pink colour will disappear.

OR

Arrange the metals iron, magnesium, zinc and copper in the increasing order of their reactivity. What will be the two observations made by the student when iron filings are added to copper sulphate solution?

  • Increasing order of reactivity: Copper < Iron < Zinc < Magnesium.
  • When iron filings are added to a blue copper sulphate solution, the following two observations will be made:
    1. The blue colour of the copper sulphate solution will slowly fade and turn pale green. This is due to the formation of iron sulphate (FeSO4​).
    2. A brownish-red coating of copper metal will be deposited on the iron filings. This is a single displacement reaction because iron is more reactive than copper and displaces it from its salt solution.
    • Fe(s)+CuSO4​(aq)→FeSO4​(aq)+Cu(s)

23. From an experiment to study the properties of acetic acid. Answer the following questions: a) Name the substances which on addition to acetic acid produce carbon dioxide gas. Give relevant chemical equation for the above? b) How is CO2​ gas tested in the laboratory?

a) When sodium carbonate (Na2​CO3​) or sodium hydrogen carbonate (NaHCO3​) is added to acetic acid, carbon dioxide gas is produced.

  • Reaction with sodium carbonate:
    • 2CH3​COOH(aq)+Na2​CO3​(s)→2CH3​COONa(aq)+H2​O(l)+CO2​(g)
  • Reaction with sodium hydrogen carbonate:
    • CH3​COOH(aq)+NaHCO3​(s)→CH3​COONa(aq)+H2​O(l)+CO2​(g) b) Carbon dioxide gas is tested in the laboratory by passing it through lime water (Ca(OH)2​). Carbon dioxide reacts with lime water to form a white precipitate of calcium carbonate, which turns the lime water milky.
  • Ca(OH)2​(aq)+CO2​(g)→CaCO3​(s)+H2​O(l)

24. When observed under high power of the microscope, ‘chain of buds’ is visible in the microscopic view. In which organism can it be observed? Explain the process.

The “chain of buds” is observed in yeast.

The process is called budding, which is a form of asexual reproduction. In this process, a small outgrowth or ‘bud’ forms on the parent cell. The nucleus of the parent cell divides, and a part of it moves into the bud. The bud then grows and either detaches from the parent cell to form a new individual or remains attached, forming a chain of buds.

OR

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.

Ans: The experimental setup aims to show that carbon dioxide is released during respiration. A vial containing Potassium Hydroxide (KOH) is kept in the conical flask to absorb any carbon dioxide present in the air inside the flask at the start of the experiment.

If one forgets to keep the vial with KOH in the conical flask, the result will be inconclusive or incorrect. The carbon dioxide present in the air at the beginning of the experiment will also react with the lime water, turning it milky, and a student might wrongly conclude that all the carbon dioxide was produced during respiration. The KOH is crucial to ensure that any change in the lime water is solely due to the carbon dioxide released by the respiring seeds.

25. You soak seeds of bean and observe them after 2-3 days. List four observations?

Ans: When bean seeds are soaked and observed after 2-3 days, the following four observations can be made:

  1. The seeds will have swollen due to the absorption of water.
  2. The seed coat will have split or become soft, revealing the inner parts of the seed.
  3. A small, white structure, which is the radicle (embryonic root), will have emerged from the seed.
  4. The plumule (embryonic shoot) will also be visible, indicating the start of germination.

26. The current flowing through a resistor connected in an electrical circuit and the potential difference developed across its ends are shown in the given ammeter and voltmeter. Find the least count of the voltmeter and ammeter. What is the voltage and the current across the given resistor?

  • Ammeter: The range is from 0 to 1 A. There are 10 divisions between 0 and 0.2 A.
    • Least count = 0.2 A/10=0.02 A.
    • The needle is at the 12th division from 0.
    • Current = 12×0.02 A=0.24 A.
  • Voltmeter: The range is from 0 to 3 V. There are 10 divisions between 0 and 1 V.
    • Least count = 1 V/10=0.1 V.
    • The needle is at the 15th division from 0 (past the 1 V mark).
    • Voltage = 15×0.1 V=1.5 V. The current across the resistor is 0.24 A, and the voltage is 1.5 V.

27. Consider the path of a ray of light passing through a rectangular glass slab for different angles of incidence.

(i) Which one is greater: angle of incidence or angle of refraction?

(ii) What happens to the emergent angle on increasing the incident angle at air-glass interface?

(iii) State the conditions when no bending occurs.

Ans: (i) When a ray of light passes from air (a rarer medium) into a glass slab (a denser medium), it bends towards the normal. Therefore, the angle of incidence is greater than the angle of refraction.

(ii) The emergent angle is equal to the angle of incidence. So, on increasing the incident angle at the air-glass interface, the emergent angle also increases.

(iii) No bending of light occurs when the light ray is incident on the interface normally (perpendicular to the surface).

In this case, the angle of incidence is 0°, and the angle of refraction is also 0°.

OR

Sunita takes a mirror which is depressed at the centre and mounts it on a mirror stand. An erect and enlarged image of her face is formed. She places the mirror on a stand along a meter scale at 15 cm mark. In front of this mirror, she mounts a white screen and moves it back and forth along the meter scale till a sharp, well-defined inverted image of a distant tree is formed on the screen at 35 cm mark.

(i) Name the mirror and find its focal length.

(ii) Why does Sunita get sharp image of the distant building at 35 cm mark?

Ans: (i) A mirror that is depressed at the center and forms an erect and enlarged image is a concave mirror.

To find the focal length, the sharp, inverted image of a distant object (the tree) is formed at the principal focus of the mirror.

Position of the mirror = 15 cm mark

Position of the image = 35 cm mark

Focal length = Position of image – Position of mirror = 35 cm−15 cm=20 cm.

The focal length of the concave mirror is 20 cm.

(ii) Sunita gets a sharp image of the distant tree at the 35 cm mark because the rays of light coming from a distant object are considered to be parallel to the principal axis.

After reflection from a concave mirror, these parallel rays converge and meet at the principal focus. Therefore, the sharp, inverted image is formed at the focal point, which is at the 35 cm mark.

For the official Class 10 Mathematics Solutions, you can visit:

  1. NCERT Textbooks (for Class 10):
ncert resources

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