CBSE Sample Papers for Class 10 Science Set 7 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 10 Science with Solutions Set 7 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Science Set 7 with Solutions

Time: 3 Hours
Maximum Marks: 80

General Instructions:

1. This question paper consists of 39 questions in 5 sections.
2. All questions are compulsory. However, an internal choice is provided in some questions. A student is expected to attempt only one of these questions.
3. Section A consists of 20 objective type questions carrying 1 mark each.
4. 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.
5. 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.
6. 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.
7. Section E consists of 3 source-based/case-based units of assessment of 04 marks each with sub-parts.

Section – A

Select and write the most appropriate option out of the four options given for each of the questions 1 – 20.
There is no negative mark for incorrect response.

Question 1.
Pentane has the molecular formula C₅H₁₂. It has:
(a) 5 covalent bonds
(b) 12 covalent bonds
(c) 16 covalent bonds
(d) 17 covalent bonds
Answer:
(c) 16 covalent bonds
Explanation: Structural formula of pentane is:

Number of C – C covalent bonds = 4
Number of C – H covalent bonds = 12
Hence, the total number of covalent bonds in the structure of pentane is 16.

Question 2.
Mineral acids are stronger acids than carboxylic acids because
(i) Mineral acids are completely ionized.
(ii) Carboxylic acids are completely ionized.
(iii) Mineral acids are partially ionized.
(iv) Carboxylic acids are partially ionized.
(a) (i) and (iv)
(b) (ii) and (iii)
(c) (i) and (ii)
(d) (iii) and (iv)
Answer:
(a) (i) and (iv)
Mineral acids like HCl are completely ionized in solvents like water whereas carboxylic acids such as CH₃COOH are only partially ionized.

Question 3.
A few drops of iodine solution were added to rice water. The solution turned blue-black. This indicates that rice water contains :
(a) Complex proteins
(b) Simple proteins
(c) Starch
(d) Fats
Answer:
(c) Starch
Explanation: Due to the presence of starch, when iodine solution is added to rice water, the solution becomes blue-black. Iodine forms a Starch Iodide complex when it comes in contact with the amylose structure of starch. The blue-black color comes from the starch iodide complex.

Question 4.
A coating of zinc will be observed in

(a) I and III
(b) Only I
(c) Only II
(d) III and IV
Answer:
(a) I and III
Zn, being more reactive than iron and copper, therefore displaces the iron and copper from their solutions. Hence, Fe and Cu will form a coating on zinc due to displacement reaction.

Question 5.
When a white insulating sheath is used to enclose the axon in the majority of the neurons, then it is known as
(a) Myelin sheath
(c) Nodes of Ranvier
(b) Neurolemma
(d) Synaptic clefts
Answer:
(a) Myelin sheath
Explanation: The myelin sheath is a protective covering made up of specialized cells called Schwann cells in the peripheral nervous system and oligodendrocytes in the central nervous system. It appears ‘ white due to its high lipid content.

Question 6.
Amit checked the odour of onion juice and then he added a few drops of onion juice in dil. HCl and in dil. NaOH solution in separate test tubes. Which of the following observations he would not observe?
(a) Odour of onion juice in sodium hydroxide vanished.
(b) Odour of onion juice in dilute hydrochloric acid remained the same.
(c) Onion acts as an olfactory indicator.
(d) Odour of onion juice in dilute HCl vanished.
Answer:
(d) Odour of onion juice in dilute HCl vanished.
(a), (b) and (c) observations will be observed by him. Onion is an olfactory indicator whose odour changes or vanishes in a basic medium and remains unchanged in an acidic medium.

Question 7.
A copper wire is held between the poles of a magnet:

The current in the wire can be reversed. The pole of the magnet can also be changed over. In how many of the four directions shown can the force act on the wire?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2
Explanation: By Fleming’s left-hand rule, we know that the force on the wire is perpendicular to the current in the wire and the magnetic field.

Question 8.
The sequence of anaerobic respiration in our muscle cells during heavy exercise is


Answer:

Conversion of glucose into pyruvate takes place in mitochondria. This pyruvate converts to ethanol and lactic acid in human muscle cells. This causes muscle cramps after prolonged exercise.

Question 9.
On which of the given factors, does resistance not depend?
(a) Length of conductor
(b) Area of cross-section
(c) Temperature
(d) Density
Answer:
(d) Density
Explanation: The resistance of a wire can be expressed as:
R = ρ\(\frac{\mathrm{L}}{\mathrm{A}}\)
Where, A = Area of cross-section of the conductor
L = Length of the conductor
ρ = Resistivity
From the above relation, we can see that the resistance of a wire is directly proportional to its length and inversely proportional to the area of the cross-section. Hence, resistance does not depend on the density.

Question 10.
Which of the following substances breaks down starch?
(a) Lipase
(b) Amylase
(c) Complex sugars
(d) Trypsin
Answer:
(b) Amylase
Amylase is a carbohydrate-digesting enzyme, which is responsible for the breakdown of starch into smaller units.

Question 11.
Which is the first enzyme to mix with food in the digestive tract?
(a) Amylase
(b) Pepsin
(c) Trypsin
(d) Cellulase
Answer:
(a) Amylase
Explanation: Amylase is the first enzyme in the digestive tract to mix with food. It is secreted in the mouth and acts on starch to break it down into smaller molecules.

Question 12.
Which of the following compounds do not cause the thinning of the ozone layer?
(a) Chlorofluorocarbon
(b) Carbon dioxide
(c) Nitrogen oxide
(d) Methane
Answer:
(d) Methane
Methane is not responsible for the depletion of the ozone layer. The primary cause of ozone depletion is the presence of chlorine-containing source gases (primarily CFCs and related halocarbons), CO₂, and oxides of nitrogen.

Question 13.
Raman was absent from the class because of muscle pain which he claims to be due to excessive physical exercise he had done yesterday. This pain is due to:
(a) Formation of lactic acid
(b) Formation of acetic acid
(c) Formation of pyruvic acid
(d) Formation of Hydrochloric acid
Answer:
(a) Formation of lactic acid
Explanation: When Raman claims that his muscle pain is due to excessive physical exercise, it indicates that he experienced a condition known as muscle soreness or delayed onset muscle soreness (DOMS). DOMS is often caused by strenuous physical activity or exercise, especially when the muscles are subjected to unfamiliar or intense exertion.

During intense exercise, the production of lactic acid increases in the muscles. Lactic acid is a byproduct of the anaerobic breakdown of glucose. When lactic acid accumulates in the muscles, it can lead to muscle fatigue, discomfort, and pain, which is often felt as muscle soreness after exercise.

Question 14.
The splitting of white light into seven colours on passing through a glass prism is due to
(a) dispersion
(b) refraction
(c) scattering
(d) reflection
Answer:
(a) dispersion
The splitting of white light into its component colours on passing through glass is called dispersion.

Question 15.
Which of the following is a sink tissue in plants?
(a) Leaves
(b) Stems
(c) Developing fruits
(d) Tubers or bulbs
Answer:
(c) Developing fruits
Explanation: Developing fruits serve as sink tissues in plants because they actively accumulate and store nutrients from the plant’s vascular system. As the fruit develops, it draws nutrients from the phloem, acting as a sink for carbohydrates, minerals, and other essential substances. This nutrient i accumulation supports the growth and maturation of the fruit, ensuring its successful development and eventual dispersal of seeds.

Question 16.
Which structure among the following options forms villi towards the embryo, creating a large surface area for the exchange of nutrients, gases, etc. between the mother and the embryo?
(a) Placenta
(b) Lacteals
(c) Oviduct
(d) Cervix
Answer:
(a) Placenta
The placenta forms villi towards the embryo for the exchange of nutrients, gases, and waste materials between the mother and embryo.

Questions 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: Plants raised by vegetative propagation can bear flowers and seeds earlier than those produced from seeds.
Reason: Plants which have lost the capacity to bear viable seeds, can propagate through vegetative propagation.
Answer:
(b) Both A and R are true and R is not the correct explanation of A
Explanation: The plant which is unable to produce seeds reproduce using vegetative propagation. This process is a type of asexual reproduction which is much faster than sexual reproduction.

Question 18.
Assertion (A): A network of food chains existing together in an ecosystem is known as a food web.
Reason (R): An animal like a kite cannot be a part of a food web.
Answer:
(c) A is true, but R is false.
In the food web, different food chains are interconnected. Each food chain consists of different trophic levels, i.e. producers, consumers, and detrivores. So, kites can also be a part of the food web.

Question 19.
Assertion: Colour of copper sulphate solution changes when an iron nail is kept immersed in it.
Reason: The colour of copper sulphate solution changes when an iron nail is kept immersed in it due to the decomposition reaction taking place between iron and copper leading to the formation of iron sulphate.
Answer:
(c) A is true but R is false
Explanation: The colour of copper sulphate solution changes when an iron nail is kept immersed in it due to the displacement reaction iron displaces copper from copper sulphate solution leading to the formation of iron sulphate.

Question 20.
Assertion (A): A fuse is connected in series with the circuit.
Reason (R): Fans, lamps, etc, are connected in parallel in household wiring.
Answer:
(b) Both A and R are true but R is not the correct explanation of A.
A fuse is connected in series with the circuit to protect the circuit. Fans, lamps, etc. are connected in parallel in household wiring so that if any of them fuses, then it does not affect the working of other appliances.

Section – B

Questions No. 21 to 26 are very short answer questions.

Question 21.
(i) State two reasons for the following facts:
(a) Sulphur is a non-metal.
(b) Magnesium is a metal
(ii) Compound X and aluminium are used to join railway tracks.
(a) Identify the compound X
(b) Name the reaction
(c) Write down its reaction.
OR
(i) What type of oxides are formed when non-metals react with oxygen? Explain with an example.
(ii) What type of oxides are formed when metals combine with oxygen? Explain with the help of an example.
Answer:
(i) (a) Sulphur is a non-metal because:

1. It is a poor conductor of electricity.
2. Sulphur is neither malleable nor ductile.

(b) Magnesium is a metal because:

1. It is a good conductor of electricity.
2. Magnesium is malleable, not ductile.

(ii) (a) Compound X = Fe₂O₃
(b) Thermite reaction.
(c) Fe₂O₃ + 2Al → 2Fe + Al₂O₃.
OR
(i) When non-metals react with oxygen, they form acidic oxides or neutral oxides.
Example: Carbon reacts with oxygen to form an acidic oxide called carbon dioxide.

(ii) When metals combine with oxygen, they form basic oxides.
Example: Sodium reacts with oxygen to form a basic oxide called sodium oxide.

Question 22.
What could be the reasons for adopting contraceptive methods?
Answer:
The reasons for adopting contraceptive methods are as follows:

• Protection from STDs.
• Restricting the number of children (population control).
• Maintaining a sufficient gap between successive births.
• Enjoying a good reproductive health. (\(\frac{1}{2}\) × 4)

Question 23.
Write the chemical equation of the reaction in which the following changes have taken place with an example of each:
(i) Change in colour
(ii) Change in temperature
Answer:
(i) Change in colour
Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag
The solution will become blue in colour and shiny silver metal will be deposited.

(ii) Change in temperature
NaOH + HCl → NaCl + H₂O + Heat
The temperature will increase due to evolution of heat.

Question 24.
The refractive indices of the three media are given below.

Medium	Refractive Index
A	1.6
B	1.8
C	1.5

A ray of light is travelling from A and B and another ray is travelling from B to C.
(a) In which of the two cases the refracted ray bends towards the normal?
(b) In which case does the speed of light increase in the second medium?
Give a reason for your answer.
Answer:
(a) When light travels from an optically rarer medium to an optically denser medium it moves towards the normal. Since nB > nA, the light ray will bend towards the normal on passing from medium A to B. (1)
(b) The speed of the light will increase when the light travels from B to C, since n_C < n_B and v = (c/n), the speed of the light ray will increase in the second medium. (1)

Question 25.
How many X chromosomes can be found in the cells of the body of (i) a boy, and (ii) a girl?
OR
In higher animals, what does ammonia react with to produce urea?
Answer:
(i) Males have one X chromosome.
(ii) Females have two X chromosomes.
OR
Ammonia reacts with carbon dioxide in the liver to form a less toxic substance, urea which is periodically flushed out of the system.

Question 26.
If you have to go to a village during an AIDS prevention program, what points will you explain to the villagers?
Answer:
A few points for the prevention of AIDS are as follows.

• Use condoms while having sex.
• Pregnant women should be checked for pre-existing HIV.
• Use a disposable syringe.
• Blood transfusion should be done carefully. (\(\frac{1}{2}\) × 4)

Section – C

Questions No. 27 to 33 are short answer questions.

Question 27.
(i) A person needs a lens of power – 5.5 dioptres for correcting his distinct vision. For correcting his near vision, he needs a lens + 1.5 dioptres. What is the focal length of the lens required for correcting, the following defects?
(a) distinct vision (b) near vision?
(ii) What do you mean by Presbyopia?
Answer:
(i) (a) Power of lens needed for the correction of distant vision of the person (P) = – 5.5 D
Focal length of the lens required for correcting the distant vision is
(f) = \(\frac{1}{P}=\frac{1}{-5.5}\) m = 0.18m = 18cm.

(b) For correcting near vision, the power of lens required (P) = + 1.5 D
Focal length of lens required for correcting near vision
(f) = \(\frac{1}{P}=\frac{1}{1.5}\) m = 0.67 m = 66.7 cm.

(ii) Presbyopia is the defect of the human eye in which a person is unable to see the nearby as well as far off objects clearly.

Question 28.
Study the given diagram and answer the following questions.

(a) Write the chemical reaction involved in the process.
(b) Mention the colour of copper powder and the substance formed after heating it.
(c) How can we reverse the above reaction? Write the equation for the reverse reaction and state the substance that undergoes oxidation and the substance that undergoes reduction.
Answer:
(a) Copper powder upon heating with oxygen (present in air) to form cupric oxide or copper (II) oxide.
2Cu (s) + O₂ (g) → 2CuO (s) (Cupric oxide) (1)
(b) (i) Copper powder is brownish.
(ii) The substance formed after heating is cupric oxide. It has black colour. (1)
(c) The reaction can be reversed by heating cupric oxide in the presence of hydrogen.
CuO (s) + H₂ (g) → Cu (s) + H₂O (g)
In this reaction, H2 undergoes oxidation to H₂O, and CuO is reduced to Cu. (1)

Question 29.
Draw the structures of the following organic compounds :
(i) Propanoic acid
(ii) 2-propanol
(iii) 1-iodopropane
Answer:

Question 30.
Manoj has taken a thin converging lens to form a real magnified image of an object in front of it.
(a) Write the position of the object.
(b) Draw a ray diagram to show the image formation.
(c) If the power of the lens is +2D, then what will be the focal length of the lens?
Answer:
(a) Object is placed between F and 2F.
(b) The ray diagram is as follows:

(c) Given, P = +2D [P = Power of lens]
Since power is positive, the lens is a convex lens.
∴ Focal length, f = \(\frac{1}{P}=\frac{1}{2}\) = 0.5 m = 50 cm (1)

Question 31.
(i) Give reasons for the following observations:
(a) Ionic compounds in general have high melting and boiling points.
(b) Highly reactive metals cannot be obtained from their oxides by heating them with carbon.
(c) Copper vessels get a green coat when left exposed to air in the rainy season.
(ii) Choose the metal (from the list given below) which can displace zinc from the zinc sulphate solution, copper, magnesium, or silver. Write the equation of the chemical reaction involved.
Answer:
(i) (a) Ionic compounds in general have high melting and boiling points as their ions are held together by a stronger electrostatic force of attraction.

(b) Highly reactive metals cannot be obtained from their oxides by heating them with carbon as they have more affinity for oxygen than as compared to carbon.

(c) Copper vessels get a green coat when left exposed to air in the rainy season as it reacts with CO₂ and H₂O and forms a green coating of basic copper carbonate.

(iii) Magnesium can displace zinc from zinc sulphate solution.
Mg(s) + ZnSO₄(aq) → MgSO₄(aq) + Zn(s)

Question 32.
A spherical mirror produces an image of magnification -1 on a screen placed at a distance of 50 cm from the mirror.
(a) Write the type of mirror.
(b) Find the distance of the image from the object.
Answer:
(a) The mirror is a concave.
(b) Given, m = -1, u = -50 cm
∴ Magnification, m = \(\frac{-v}{u}\)
⇒ -1 = \(\frac{-v}{-50}\)
⇒ v = -50 cm (1)
Therefore, the image is real and inverted and of the same size as that of the object.
Here, the object is placed at the center of curvature, so the object distance is equal to the image distance, since, the screen is placed at a distance of 50 cm from the mirror, therefore object distance = image distance = 50 cm.
The distance of the image from an object is zero, i.e. the object is formed at 2f. (2)

Question 33.
Answer the following:
(i) What are decomposers? What will be the consequence of their absence in an ecosystem?
(ii) Describe how decomposers facilitate the recycling of matter to maintain the balance in the ecosystem.
Answer:
(i) Decomposers are the microbes that feed on dead and decaying organisms. Dead plants and animals will accumulate in the ecosystem as there will be no decomposers to decompose them. The decomposers will act upon dead and decayed organisms into simpler forms and get mixed in the soil which is used by producers again. But in the absence of decomposers, this whole process would not occur and the dead organisms would accumulate in the ecosystem.

(ii) Decomposers act upon dead and decay organisms to convert them into simpler forms. These simple substances get mixed up in the soil and are used as nutrients by the producers. From producers, it goes to consumers, and so on. Thus, there is the recycling of matter which is done by decomposers that maintain the balance in the ecosystem.

Section – D

Questions No. 34 to 36 are long answer questions.

Question 34.
Manish took different metals and observed that all metals reacted differently with hot water, cold water, and steam. Explain the reactions of sodium, magnesium, and iron metals with cold water, hot water, and steam.
Or
(a) An element P reacts with water to form a compound Q, which is used in whitewashing—the compound Q on heating forms an oxide R which on treatment with water gives back B. Identify P, Q, and R and give the reactions involved.
(b) How photochemical reactions have played an important role in photography?
Answer:
Reactions of Na, Mg, and Fe metals with different types of water are shown below.
With cold water: Sodium reacts violently with water. The reaction is

With hot water: Magnesium reacts with hot water to form magnesium hydroxide and hydrogen.

With steam: Iron does not react either with cold or hot water. It reacts with steam. The reactions are as follows:

Or
(a) Element P is calcium (Ca), and when it reacts with water, it forms calcium hydroxide, Thus, compound Q is calcium hydroxide Ca(OH)₂, which is used in whitewashing. (1)

(b) A photographic film used in black and white photography is a celluloid film coated with silver chloride. Its work is based on the decomposition of silver chloride in the presence of sunlight.

Question 35.
Answer the following questions:
(i) Explain the process of preparation of soap in the laboratory.
(ii) Why is common salt (sodium chloride) added during the preparation of soap?
(iii) Why is soap not suitable for washing clothes when the water is hard?
(iv) What are detergents chemically? List the merits and demerits of using detergents for cleansing.
OR
Answer the following questions:
(i) Why is it wrong to treat a bee sting with vinegar?
(ii) Why is it wrong to treat a wasp sting with baking soda solution?
(iii) How is pH paper used to find the pH of a solution?
(iv) The pH value of water is 7. What will be the pH value of:
(a) Aqueous solution of sodium hydroxide.
(b) Dil. HCl.
Answer:
(i) Soap can be prepared in the laboratory as follows:

1. Take about 20 ml of castor oil (cottonseed oil, linseed oil or soya bean oil) in a beaker.
2. Add 30 ml of 20% sodium hydroxide solution to it.
3. Heat the mixture with constant stirring till a paste of soap is formed.
4. Then add 5 to 10 grams of common salt (sodium chloride).
5. Stir the mixture well and allow it to cool. On cooling the solution, solid soap separates out.
6. When the soap sets, it can be cut into pieces called ‘soap bars’.

(ii) Common salt is added to the mixture to make the soap come out of the solution. Though most of the soap separates out on its own, but some of it remains in solution. Common salt is added to precipitate out all the soap from the aqueous solution.

(iii) When soap is used for washing clothes with hard water, a large amount of soap in the water reacts with the calcium and magnesium ions of hard water to form an insoluble precipitate called scum, before it can be used for the real purpose of washing.

(iv) Detergents are chemically sodium salts of sulphonic acid or ammonium salts with chlorides or bromide ions, etc.

Merits of detergents for cleansing

1. Detergents can work well with both hard water and soft water.
2. Detergents are more effective than soaps.
3. Detergents contain synthetic chemicals so they tend to provide more cleaning power.
4. Detergents are made with chemical substances so they can be modified for specific purposes such as laundry detergents etc.
5. Detergents are more easily soluble in water.

Demerits of detergents for cleansing:

1. They are expensive.
2. They can create water pollution.
3. Detergents are formed with synthetic chemicals with few natural sources therefore they are usually non-biodegradable.
4. They can cause soil pollution.

OR
(a) Bee injects acid into the skin. Vinegar is acetic acid so it can’t be used to treat bee sting.

(b) Since baking soda is basic it can’t be used to treat wasp sting because wasp injects alkaline liquid into the skin.

(c) A few drops of solution are taken on the strip of pH paper. Then the colour obtained on the paper is compared with colour and corresponding pH value given on the chart of the pH paper.

(d) (a) Aqueous solution of sodium hydroxide is basic. Therefore, its pH will be greater than 7.
(b) Dil. HCl is acidic, so its pH will be less than 7.

Question 36.
The diagram below shows a circuit in which 4Ω, 10Ω, and 20Ω resistors are connected in parallel combination. A voltmeter, an ammeter, and a 20V battery are also connected to the circuit. Calculate

(a) The total effective resistance of the circuit.
(b) The current through each resistor.
(c) Write the Joule’s law of heating.
Or
Study the following circuit and answer the questions that follow.

(a) State the type of combination of two resistors in the circuit.
(b) How much current is flowing through (i) 10Ω and (ii) 15Ω resistor?
(c) What is the ammeter reading?
Answer:
Given, R₁ = 4Ω, R₂ = 10Ω, and R₃ = 20Ω, V = 20V
(a) Total effective resistance of the circuit,
\(\frac{1}{R_{\text {eff }}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)
(Here, resistances are connected in parallel combination)

(b) Current through each resistor,

(c) According to Joule’s law of heating, the amount of heat produced in a resistor is
(i) directly proportional to the square of the current flowing through the resistor, i.e. H ∝ I²
(ii) directly proportional to the resistance of the resistor, i.e. H ∝ R
(iii) directly proportional to the time for which the current flows through the resistor.
i.e. H ∝ t ⇒ H = I²Rt (2)
Or
(a) Two resistors are in parallel combination. (1)
(b) Current through
(i) 10Ω resistor, I₁ = \(\frac{V}{R}=\frac{3}{10}\) = 0.3A
(ii) 15Ω resistor, I₂ = \(\frac{V}{R}=\frac{3}{15}\) = 0.2A (2)
(c) Ammeter reading = Total current flowing through the circuit
= 0.3 + 0.2
= 0.5A (2)

Section – E

Questions No. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in one of these sub-parts.

Question 37.
The process of dissolving an acid or a base in water is a highly exothermic one. Care must be taken while mixing concentrated nitric acid or sulphuric acid with water. The acid must always be added slowly to water with constant stirring. If water is added to a concentrated acid, the heat generated may cause the mixture to splash out and cause burns. The glass container may also break due to excessive local heating. Look out for the warning sign on the can of concentrated sulphuric acid and on the bottle of sodium hydroxide pellets. Mixing an acid or base with water results in a decrease in the concentration of ions (H₃O⁺/OH^–) per unit volume. Such a process is called dilution and the acid or the base is said to be diluted.

1. Take 10 mL water in a beaker.
2. Add a few drops of concentrated H₂SO₄ to it and swirl the beaker slowly.
3. Touch the base of the beaker.
(a) What happens to the concentration of ions when a solution is diluted?
(b) Is the process of mixing acid or base in water exothermic or endothermic?
OR
Are all bases soluble in water?
Answer:
(a) Mixing an acid or base with water results in decrease in a the concentration of ions (H₃O⁺/OH^–) per unit volume.
(b) The process of dissolving an acid or a base in water is highly exothermic.
OR
All bases do not dissolve in water. An alkali is a base that dissolves in water.

Question 38.
Priyanka has green eyes. On the other hand, her parents and brother have black eyes. Deepak (Priyanka’s husband) has black eyes. Deepak’s mother has green eyes and their father has black eyes.
(a) based on the above-given information, which eye colour is recessive and why?
(b) What is the genotype of the eye colour of Priyanka’s brother?
(c) What is the probability of Priyanka’s and Deepak’s child having green eyes?
Or
With the help of a cross, Priyanka’s brother can have 50% of the offspring with green eyes.
Answer:
(a) Green eye color is recessive based on the above-given information. As it is expressed only in homozygous conditions. (1)
(b) The genotype of Priyanka’s brother’s eye colour is CaBB or Bb. (1)
(c) Since Deepak’s mother has green eyes (bb) and father has black eyes (BB), so genotype of Deepak’s eyes (black) will be Bb. The Genotype of Priyanka’s eyes will be bb.

The probability that the offspring of Priyanka and Deepak will have green eyes is 50%. (2)
Or
Let the person marrying Priyanka’s brother be

When Priyanka’s brother marries a green eye coloured girl, there will be a 50% chance of the offspring having green eyes. (2)

Question 39.
We have studied that when a current-carrying conductor is placed in a magnetic field such that the direction of the current is perpendicular to the magnetic field, it experiences a force. This force causes the conductor to move. Now let us imagine a situation in which a conductor is moving inside a magnetic field or a magnetic field is changing around a fixed conductor. What will happen? This was first studied by English physicist Michael Faraday. In 1831, Faraday made an important breakthrough by discovering how a moving magnet can be used to generate electric currents.

To observe this effect, let us perform the following activity.
1. Take a coil of wire AB having a large number of turns. Connect the ends of the coil to a galvanometer as shown. Take a strong bar magnet and move its north pole towards end B of the coil. Do you find any change in the galvanometer needle?
2. There is a momentary deflection in the needle of the galvanometer, say to the right. This indicates the presence of a current in coil AB. The deflection becomes zero, the moment the motion of the magnet stops.
3. Now withdraw the north pole of the magnet away from the coil. Now the galvanometer is deflected toward the left, showing that the current is now set up in the direction opposite to the first.
4. Place the magnet stationary at a point near the coil, keeping its north pole towards the end B of the coil. We see that the galvanometer needle deflects toward the right when the coil is moved toward the north pole of the magnet. Similarly, the needle moves toward the left when the coil is moved away.

(a) What happens when a current-carrying conductor is placed in a magnetic field such that the direction of the current is perpendicular to the magnetic field?
(b) What is a galvanometer?
(c) When the coil is kept stationary concerning the magnet, the deflection of the galvanometer drops to zero. What do you conclude from this activity?
OR
(d) Which important discovery was made by Faraday in 1831?
Answer:
(a) When a current-carrying conductor is placed in a magnetic field such that the direction of the current is perpendicular to the magnetic field, it experiences a force. This force causes the conductor to move.

(b) A galvanometer is an instrument that can detect the presence of a current in a circuit. The pointer remains at zero (the center of the scale) for zero current flowing through it. It can deflect either to the left or to the right of the zero-mark depending on the direction of the current.

(c) This activity shows that the motion of a magnet concerning the coil produces an induced potential difference which sets up an induced electric current in the circuit.
OR
In 1831, Faraday made an important breakthrough by discovering how a moving magnet can be used to generate electric currents.