Students must start practicing the questions from CBSE Sample Papers for Class 10 Science with Solutions Set 4 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Science Set 4 with Solutions
Time: 3 Hours
Maximum Marks: 80
General Instructions:
- This question paper consists of 39 questions in 5 sections.
- All questions are compulsory. However, an internal choice is provided in some questions. A student is expected to attempt only one of these questions.
- Section A consists of 20 objective type questions carrying 1 mark each.
- 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.
- 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.
- 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.
- 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.
Two pink coloured flowers on crossing resulted in 1 red, 2 pink and 1 white flower progeny. What can be infered about the nature of the cross?
(a) Double fertilisation
(b) Self-pollination
(c) Cross fertilisation
(d) No fertilisation
Answer:
(c) Cross fertilisation
Explanation: In the given cross, pink flowering plants have two dissimilar forms of the gene controlling flower colour. So, on selling of two pink flowers (Rr), red (RR), pink (Rr) and white (rr) flower are obtained in ratio 1 : 2 : 1.
Question 2.
A light ray enters from medium A to medium B as shown in the figure. The refractive index of medium B relative to A will be:
(a) greater than unity
(b) less than unity
(c) equal to unity
(d) zero
Answer:
(a) greater than unity
Explanation: As the light ray when travelled from medium A to medium B, then they bend towards the normal which means that medium B has higher refractive index and less speed of light with respect to medium A, So, refractive index of medium B w.r.t. medium A will be greater than unity.
Question 3.
Which of the following mirror is used by a dentist to examine a small cavity?
(a) Convex mirror
(b) Plane mirror
(c) Concave mirror
(d) Combination of convex and concave mirror
Answer:
(c) Concave mirror
Explanation: As concave mirror gives the dentist a magnified reflection of the mouth while also refracting a bit of light. It forms an image in the mirror as larger and brighten and thus it makes easier for the dentist to see the cavity.
Question 4.
In which of the following aspects does multiple fission differ from binary fission?
(i) Number of offspring produced.
(ii) Level of genetic variation in offspring.
(iii) Number of parents involved.
(iv) Multiple fission occurs in Plasmodium, whereas binary fission occurs in Leishmania.
(a) Only (i) is correct
(b) Both (i) and (iv) are correct
(c) (iii) is correct
(d) (ii) is correct
Answer:
(b) Both (i) and (iv) are correct
Explanation: Multiple fission produces many offspring whereas binary fission produces only two. Off-spring produced through multiple fission as well as binary fission are genetically identical to each other and to their parents. Both multiple fission and binary fission require only one parent. Plasmodium, the protozoan that causes malaria reproduces through multiple fission. Leishmania causes Kala-azar and it reproduces through binary fission.
Question 5.
Which of the following endocrine glands is unpaired?
(a) Adrenal
(b) Pituitary
(c) Testes
(d) Ovary
Answer:
(b) Pituitary
Explanation: The pituitary gland, also known as the hypophysis, is a pea-sized endocrine gland. It is a protrusion at the base of the brain, at the bottom of the hypothalamus. It is a gland which is not paired. Adrenal, ovary and testes are paired glands.
Question 6.
If a person has five resistors, each of value \(\frac{1}{5}\) Ω, then the maximum resistance he can obtain by connecting them is:
(a) 1 Ω
(b) 5 Ω
(c) 10 Ω
(d) 25 Ω
Answer:
(a) 1 Ω
Explanation:
Resistance of one resistor = \(\frac{1}{5}\) Ω
Number of resistors = 5
Maximum resistances can be obtained by combining the resistors in a series:
Rs = R1 + R2 + R3 + R4 + R5
Hence, a person on combining five resistors in a series gets resistance 1 Ω.
Question 7.
Which of the following is not a straight-chain hydrocarbon?
Answer:
Explanation: In straight-chain hydrocarbons, carbon atoms are connected through covalent bonds in one continuous chain with no branches. In the structure (a), (b) and (c), all the carbon atoms are connected in a continuous straight chain:
In structure (d), a – CH3 group is attached to the second carbon atom of the chain forming a branch. Hence, compound in structure (d) is a branched chain hydrocarbon.
Question 8.
A plant is grown in a sealed container with a controlled environment containing carbon dioxide, water, and sunlight. After some time, the plant starts to show growth and produces oxygen. Which of the following statements is most likely true?
(a) The plant is undergoing photosynthesis, a form of autotrophic nutrition.
(b) The plant is obtaining nutrients from the surrounding soil, a form of heterotrophic nutrition.
(c) The plant is undergoing cellular respiration, a form of heterotrophic nutrition.
(d) The plant is absorbing nutrients directly from the air, a form of autotrophic nutrition.
Answer:
(a) The plant is undergoing photosynthesis, a form of autotrophic nutrition.
Explanation: Autotrophic nutrition is the process by which organisms produce their own food using simple inorganic substances, such as carbon dioxide, water, and sunlight, to synthesize organic compounds, like glucose. In the given scenario, the plant is grown in a sealed container with carbon dioxide, water, and sunlight, and it produces oxygen. This is a classic indication of photosynthesis, j where plants use sunlight energy to convert carbon dioxide and water into glucose and release oxygen as a byproduct.
Question 9.
When a new plant is formed as a result of cross-pollination from different varieties of a plant, the newly formed plant is called:
(a) Dominant plant
(b) Mutant plant
(c) Hybrid plant
(d) All of these
Answer:
(c) Hybrid plant
Explanation: The process of crossing pollen from one flower to the pistils of another flower is known as cross-pollination. When a new plant is formed as a result of cross-pollination from different varieties of a plant, the newly formed plant is called a hybrid plant.
Question 10.
The equivalent resistance of a series combination of two resistances is X ohm. If the resistances are of 10 Ω and 40 Ω respectively, the value of X will be:
(a) 10 Ω
(b) 20 Ω
(c) 50 Ω
(d) 40 Ω
Answer:
(c) 50 Ω
Explanation:
We know that
Total Resistance
R = R1 + R2
= 10 + 40
= 50 Ω
Hence, the value of X is 50 Ω.
Question 11.
The number of chromosomes in parents and offspring of a particular species remains constant due
(a) doubling of chromosomes after zygote formation.
(b) halving of chromosomes during gamete formation.
(c) doubling of chromosomes after gamete formation.
(d) halving of chromosomes after gamete formation.
Answer:
(b) halving of chromosomes during gamete formation.
Explanation: The number of chromosomes in parents and offspring of a particular species remains constant due to halving of chromosomes during gamete formation.
The gametes are a special type of cells which contain only half the amount of DNA as compared to normal cells of an organism. So, when a male gamete combines with a female gamete during sexual reproduction, then the new cell ‘zygote’ will have the normal amount of DNA.
Question 12.
If we place the magnetic compass near the north pole of the magnet, which pole of the needle will point towards it?
(a) North pole
(b) South pole
(c) Keep deflecting
(d) None of these
Answer:
(b) South pole
Explanation: As like poles repel each other and unlike poles attract each other. Therefore when the North pole of a bar magnet is brought near the compass, it gets deflected in the south direction.
Question 13.
A container of vegetable oil was left open for several days, and upon examination, an unpleasant smell and taste were noticed. The oil had turned rancid. Which of the following is the most likely cause of the rancidity observed in the vegetable oil?
(a) Exposure to high temperatures
(b) Microbial contamination
(c) Presence of light
(d) Oxidation of fats
Answer:
(d) Oxidation of fats
Explanation: When oils and fats are exposed to oxygen in the air, they undergo oxidation, resulting in the breakdown of their chemical structure. This oxidation process leads to the development of unpleasant odours, flavours, and the degradation of nutritional value in the oil.
Options (a), (b), and (c) are not the primary cause of rancidity in this case. While exposure to high temperatures, microbial contamination, and light can contribute to the deterioration of oils and fats, oxidation is the primary factor responsible for rancidity.
Question 14.
A patient was diagnosed with a condition that resulted in the obstruction of the bile duct. As a result, the patient experienced symptoms such as jaundice and fatty stools. Which of the following is the primary function of bile juice that is impaired in the patient with a blocked bile duct?
(a) Emulsification of fats
(b) Neutralization of stomach acid
(c) Activation of digestive enzymes
(d) Absorption of water and electrolytes
Answer:
(a) Emulsification of fats
Explanation: The primary function of bile juice is the emulsification of fats. Bile is produced by the liver and stored in the gallbladder. It is released into the small intestine to aid in the digestion and absorption of dietary fats.
Bile contains bile salts, which act as emulsifiers. Emulsification is the process of breaking down large fat globules into smaller droplets.
In the case of a blocked bile duct,, the patient experiences symptoms such as jaundice (due to the accumulation of bilirubin, a bile pigment) and fatty stools (due to the malabsorption of dietary fats).
Question 15.
In the circuit shown below, what is the direction of the current?
(a) No current flowing
(b) Anti-clock wise
(c) Clock wise
(d) Data insufficient
Answer:
(c) Clock wise
Explanation: If the current flows from North to South the compass needle will move towards the last.
Question 16.
The opening and closing of stomatal pores depends upon :
(a) Oxygen
(b) Water in guard cells
(c) Concentration of carbon dioxide in stomata
(d) Temperature
Answer:
(b) Water in guard cells
Explanation: The entry of water into guard cells aids in the opening of guard cells, the guard cell I becomes turgid because of this. Water going out from guard cells aids in the closing of guard cells, because of this the guard cells become flaccid.
Question No 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 current carrying rod is suspended between a U-shaped magnet and the rod deflects. Reason: A force is exerted on the rod due to a magnetic field.
Answer:
(a) Both A and R are true and R is the correct explanation of A
Explanation: A force is always exerted due to magnetic field in the same way electric current flowing 1 through any conductor produces a magnetic field. And in this case, Fleming’s left-hand rule is used to predict the directions of the magnetic field, current and displacement.
Question 18.
Assertion: Silver articles become black after sometime when exposed to sunlight.
Reason: It is because silver reacts with carbonates present in the air.
Answer:
(c) A is true but R is false
Explanation: Silver reacts with sulphur present in the air and forms a layer of silver sulphide, therefore, silver articles get tarnished or becomes black after sometime when exposed to sunlight.
Question 19.
Assertion: Pancreatic amylase digests starch to maltose.
Reason: Pancreatic amylase breaks the peptide bond of protein.
Answer:
(c) A is true but R is false
Explanation: Pancreatic amylase is a starch-splitting enzyme like salivary amylase, so it cann digest proteins.
Question 20.
Assertion: Electric current flowing through a metallic wire is directly proportional to the pote difference across its ends.
Reason: Ohm’s law expression V = IR, where R (resistance) of the wire is always varying.
Answer:
(c) A is true but R is false
Explanation: Ohm’s law states that the electric current flowing through a metallic wire is directly proportional to the potential difference across its two ends. The expression is written as :
V = IR
Here, R (resistance of the wire) is constant value then only the statement will be valid.
V ∝ I only if \(\frac{V}{I}\) = constant
Section – B
Question No. 21 to 26 are very short answer questions.
Question 21.
(i) What is myopia?
(ii) How can it be corrected?
OR
Draw a neat diagram to show the refraction of a light ray through a triangular glass prism. Mark the angle of incidence, angle of emergence, incident ray, refracted ray, emergent ray and the angle of deviation.
Answer:
(i) The defect of the human eye in which a person can see the objects lying at short distances clearly but cannot see the far objects distinctly is called myopia or short-sightedness.
(ii) A myopic eye can be corrected by using spectacles with concave lenses of suitable power or focal length.
OR
The labelled diagram has been shown in the figure given below, in which
PE – Incident ray ∠i – Angle of incidence
EF – Refracted ray ∠r – Angle of refraction
FS – Emergent ray ∠e – Angle of emergence
∠A – Angle of the prism ∠D – Angle of deviation
Question 22.
What is the importance of photosynthesis in the life of the following :
(i) Green plants
(ii) Non-green plants
(iii) Animals
Answer:
(i) Green plants can build up complex energy-rich molecules of carbohydrates which are further used for different metabolic activities of cells.
(ii) non-green plants such as saprophytes and parasites use the food prepared by green plants during photosynthesis as a source of their own nutrition.
(iii) Animals eat green plants or eat animals that feed on green plants.
Question 23.
How can a chemical equation be made very informative and why?
Answer:
A chemical equation can be made more informative by mentioning the physical states of the reactants and products along with their chemical formula. The gaseous, liquid, aqueous, and solid states of the reactants and products are represented by the notations, (g), (l), (aq) and (s) respectively. The word aqueous is written if the reactant or product is present as a solution in water. When we use the symbol (g) with H20 it means in the reaction water is used as steam.
Example: 3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g)2(g)
In the given chemical reaction, Fe in the solid state reacts with water in steam form and forms iron oxide in solid form and releases hydrogen gas.
Question 24.
What is parasitic mode of nutrition? Give examples of both plants and animals that are parasites.
Answer:
There are different strategies adopted by the organism for nutrition depending on how the food is available. Some organisms break down food outside the body and then absorb it. Others take in food into the body and then digest it. Some organisms get their nutrition from plants and animals without killing them. These organisms are called parasites and the organism from which they derive their food is called host. Some of the parasites are lice, ticks, mites, leeches and tapeworms among animals and orchids and Cuscuta among plants.
Question 25.
The first trophic level in a food chain is always a green plant. Why?
OR
We do not dean ponds or lakes, but an aquarium needs to be cleaned. Why?
Answer:
Green plants are the producers which prepare their own food by utilising solar energy from inorganic sources and all other living organisms depend on them for food. Herbivores and carnivores depend upon green plants either directly or indirectly for food. Hence the first trophic level in a food chain is always a green plant.
OR
Ponds or lakes are natural, self-sustaining and complete ecosystem. They have decomposers like bacteria or fungi, which break down the waste material, and remain clean. But an aquarium is a man-made, incomplete ecosystem and they do not have decomposers to clean the waste material. So, an aquarium needs to be cleaned but we do not clean ponds or lakes.
Question 26.
While diluting an acid, why is it recommended that the acid should be added to water and not water to the acid?
Answer:
The process of dissolving an acid or a base in water is a highly exothermic in nature. 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. Hence, it is recommended that the acid should be added to water and not water to the acid.
Section – C
Question No. 27 to 33 are short answer questions.
Question 27.
How do Mendel’s experiments show that traits may be dominant or recessive?
OR
(i) How is the equal genetic contribution of male and female parents ensured in the progeny?
(ii) How many pairs of chromosomes are present in human beings? Out of this, how many are sex chromosomes? How many types of sex chromosomes are found in human beings?
Answer:
When Mendel crossed pure tall (TT) pea plants with pure dwarf (tt) pea plants, in F1 generation, he found that all pea plants were tall (Tt). There were no dwarf plants produced in F1 generation. When he self-pollinated these F1 plants, in F2 generation, he obtained tall and dwarf plants in the ratio 3 : 1. Thus, as three-fourths of the plant in the F2 generation are tall and one-fourth is dwarf, so tall is a dominant trait whereas dwarf is a recessive trait [which expressed itself only in homozygous condition]. So he concluded that for a particular trait [here in this example, height of the plant, it may be dominant or recessive.
OR
(i) Human beings contain 23 pairs of chromosomes-22 pairs are autosomes and one pair is sex chromosomes. During meiosis cell division gametes are formed in sex cells where the chromosome number is halved(n). At the time of fertilisation when the male gamete fuses with female gamete the diploid number (2n) is restored back in the zygote. Thus, half of the chromosomes come from the father and other half from the mother. In this way, meiosis process ensured equal genetic contribution the of male and female parents in the progeny.
(ii) There are 23 pairs of chromosomes present in human beings. One pair is sex chromosome. They are XX and XY. So there are two types of sex chromosomes.
Question 28.
What is thermal decomposition reaction? Give examples. A student heats white lead nitrate [Pb(NO3)2] powder taken in a test tube over the flame. Upon heating, the powder colour changes to yellow along with the emission of some brown fumes. What is the expected product of this decomposition reaction?
Answer:
Thermal decomposition is a decomposition reaction carried out by heating. The decomposition of calcium carbonate into calcium oxide and carbon dioxide on heating is an example of thermal decomposition. The calcium oxide formed is also called lime or quicklime.
The expected product of the decomposition reaction of white lead nitrate [Pb(NO3)2] is lead(II) oxide (PbO), yellow in colour, along with the emission of nitrogen dioxide (NO2) gas, which appears as brown fumes.
Here in the presence of heat decomposition is taking place, so, it is known as thermal decomposition.
2Pb(NO3)2 → 2PbO + 4NO2 + O2
Question 29.
(i) Sometimes after a shower of rain we can see a beautiful rainbow. Can this be explained by any particular process of light?
(ii) Show the formation of a rainbow with the help of a diagram.
Answer:
(i) A rainbow is a natural spectrum appearing in the sky after a rain shower. It is caused by dispersion of sunlight by tiny droplets present in the atmosphere. A rainbow is always formed in a direction opposite to that of the Sun. The water droplets act like small prisms. They refract and disperse the incident sunlight, then reflect it internally and finally refract it again when it comes out of the rain¬drop. The dispersion of light and internal reflection make different colours visible to the observer’s eye. A rainbow is also seen on a sunny day when we look at the sky through a waterfall or through a water fountain with the sun behind us.
(ii) The diagram below explains how a rainbow is formed:
Question 30.
Write the molecular formula of the following compound and draw their electron dot structure,
(i) Ethane
(ii) Ethene
(iii) Ethyne
Answer:
(i) Ethane Molecular formula: C2H6
(ii) Ethene: Molecular formula: C2H4
(iii) Ethyne: Molecular formula C2H2
Quetsion 31.
What is ozone? Show the reactions of the formation of the ozone from oxygen in the atmosphere? Name the pollutant and its role in the depletion of ozone layer.
Answer:
Ozone is a triatomic molecule made up of three oxygen atoms. It is present as a layer in our strato-sphere. It prevents the harmful UV rays of the Sun from reaching the surface of the Earth. Ozone is formed by the absorption of UV rays coming from the Sun.
O2 \(\rightleftharpoons\) O + O
O + O2 → O3
UV radiation splits oxygen molecules in to oxygen atoms and the oxygen atoms, combine with oxygen molecule to form ozone.
CFCs Chlorofluorocarbons are mainly responsible for ozone layer depletion. CFCs release chlorine atoms which breaks ozone to oxygen, more amounts of CFCs thus, released will cause the depletion of the ozone layer.
Question 32.
(i) Write two points of difference between electrical energy and electric power.
(ii) Out of 60 W and 40 W lamps, which one has a higher electrical resistance when used?
(iii) What is the commercial unit of electrical energy? Convert it into joules.
Answer:
(i)
S. No. | Electric energy | Electric power |
1. | Electrical energy consumed by an electrical appliance is the product of its power rating and the time it is used. | Electric power is the rate at which electrical energy is consumed. |
2. | It is measured in kWh. | It is measured in watt or kilowatt. |
(ii) We know, Power (P) = \(\frac{\mathrm{V}^2}{\mathrm{R}}\)
Therefore, P is inversely proportional to R as voltage remains the same.
40 W lamp has a higher resistance.
(iii) The commercial unit of electrical energy is kWh.
⇒ 1 kWh = 1000 W × 1 hr = 1000 W × 3600s
⇒ = 36 × 105 J
⇒ = 3.6 × 106 J
Question 33.
Answer the following questions:
(i) What is the importance of pH in everyday life?
(ii) Why does tooth decay start when the pH of mouth is the lower than 5.5?
(iii) What is the change in pH value of milk when it changes into curd? Explain.
Answer:
(i) Our stomach produces hydrochloric acid which helps in digestion of food. It has pH around 1. excess acid is produced, it causes pain and irritation. It can be controlled by taking antacids controls the pH in the stomach.
(ii) Tooth enamel is the hardest substance in our body. It gets corroded slowly when the pH in mouth is below 5.5. When the pH in the mouth falls below 5.5, tooth decay starts. Bacteria in the mouth produce acid by degradation of sugar and food particles that remain in the mouth after eating. The acid produced in the mouth attacks the enamel thereby, creating tooth decay.
(iii) When milk changes into curd, the pH values will change, the pH value of milk is 6 since it is acidic in nature. When the milk is converted into curd due to the action of bacteria, lactic acid is formed which is more acidic in nature. Therefore, the pH value of the milk is reduced to the pH range of 4.5 – 5.5 as it turns to curd.
Section – D
Question No. 34 to 36 are long answer questions.
Question 34.
(i) Describe two methods for the concentration of ores.
(ii) How is copper extracted from its sulphide ore? Explain the various steps supported by chemical equations. Draw a labelled diagram for electrolytic refining of copper.
OR
(i) Explain any two physical properties of ionic compounds giving reasons.
(ii) List any two metals found a free state in earth’s crust.
(pi) Metals towards the top of the activity series cannot be obtained from their compounds by reducing with carbon. Why?
(iv) What will you observe when:
(a) Some zinc pieces are put in the copper sulphate solution.
(b) Some silver pieces are put into green-coloured ferrous sulphate solution.
Answer:
Two methods used for the separation of ores are:
(i) Froth Flotation Method: It is generally used to remove gangue from sulphide ores. First the ore is powdered and a suspension in water is formed. To this ore Collectors and froth stabilisers were added. The collectors generally used are pine oils, fatty acids etc. The function of collectors is to increase the non-wettability of the metal part of the ore and allows it to form a froth. Froth Stabilizers (cresols, aniline etc.) sustain the froth. The oil wets the metal and the water wets the gangue. Paddles and air constantly stir up the suspension to create the froth. This frothy metal is skimmed off the top and dried to recover the metal.
(ii) Magnetic ore Separation: This method is used in those cases where either ore or the impurities are of magnetic nature. In this method, the powdered impure ore in the form of a thin layer is allowed to fall on a rubber belt which moves horizontally over two rollers, one of which has an electromagnet attached. As the ore particles roll over the belt, the magnetic component in the ore gets attracted towards the magnet. It gets collected in a heap while the non magnetic component forms a separate heap.
(ii) Copper is extracted from suiphide ore by roasting. It is done in the presence of air:
OR
(i) The two physical properties of ionic compounds are:
1. Ionic compounds are usually crystalline solids because their oppositely charged ions attract one another strongly and forms a regular crystal structure.
2. Ionic compounds have high melting and boiling points because ionic compounds are composed of oppositely charged positive and negative ions held together by a strong electrostatic force of attraction. Therefore, a large amount of energy is required to overcome these forces.
(ii) Gold and Platinum are the two metals that are found in a free state in the earth’s crust. These metals are located at the bottom of the activity series.
(iii) Metals such as sodium, magnesium, calcium, and aluminium high up in the reactivity series are very reactive and cannot be obtained from their compounds by heating with carbon. This is because these metals have more affinity for oxygen than carbon.
(iv) (a) The blue solution will become colourless, and reddish-brown copper metal will be deposited.
(b) When some silver pieces are put into the green coloured ferrous sulphate solution, there will be no reaction because Ag is less reactive than the iron:
Ag(s) + FeSO4(aq) → No reaction
Question 35.
What is “dispersion of white light”? Draw a labelled diagram to illustrate the recombination of the spectrum of white light. Why should the two prisms used for this purpose identical and placed in an inverted position with respect to each other?
OR
If the image formed by a mirror for all positions of the object placed in front of it is always diminished, erect and virtual, state the type of the mirror and also draw a ray diagram to justify your answer. Write one use of such mirrors are put to and why?
Answer:
The phenomenon of splitting of white light into its constituent colours on passing through a prism is known as the dispersion of white light. This splitting of the light rays occurs because of the different angles of bending for each colour and this different angles of bending occur because of different component of light faces different refractive indices when passing through the glass prism.
It is essential that the two prisms used for this purpose should be identical and placed in an inverted position with respect to each other so that the second prism completely nullifies the dispersion caused by the first prism and we get the pure white light.
OR
The type of the mirror is convex mirror. The ray diagram is shown below:
Use of concex mirror: A convex mirror always produces a smaller, virtual and erect image of an object. In convex mirror, the length of the image is shorter than that of the object. Hence, it is used as a side view mirror in vehicles. The convex mirror forms images of vehicles that are spread over a relatively larger area.
Question 36.
Describe the significance of spore formation in Rhizopus.
OR
What is vegetative propagation in plants? Describe the process of vegetative propagation through leaves in plants?
Answer:
Rhizopus is a fungus that commonly grows on bread, pickle and jam when conditions are favourable for its growth. It is a simple multicellular organism but shows specific reproductive parts. The thread like structure that can be seen on moist left-over bread pieces are the hyphae of the bread mould. They are not the reproductive parts. The tiny-round headed structures on a thin stalk are the . reproductive parts. The round blobs are the sporangia, inside which are a large number of tiny cells or spores that help in giving rise to the new Rhizopus individuals. These spores have thick walls that protect them till they find a moist surface to grow. Hence, these spores are a means of asexual reproduction in Rhizopus.
OR
Vegetative propagation is a method of asexual reproduction in plants where new plants are produced from the vegetative parts of plants like root, stem and leaves. While animals cannot use this method of reproduction, plants can. This method is used to produce new plants by layering or grafting as in rose, jasmine, sugarcane and grapes for agriculture purposes. It has both benefits and some drawbacks. Some examples of vegeta¬tive propagation are shown below:
(i) Propagation by buds on leaf margins in Bryophyllum leaf. The buds that develop along the leaf margin fall on the soil and each of them develops into a new plant.
(ii) A small cutting of a money plant when kept in water in a glass container or in a pot with soil grows into a new plant.
(iii) Buds in potatoes and ginger can grow into new plants under suitable conditions.
(iv) In sweet potatoes, the roots bear adventitious buds that can grow into new plants under favourable conditions.
Section – E
Question 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 reaction between an acid and a base to give a salt and water is known as a neutralization reaction. In general, a neutralization reaction can be written as –
Base + Acid → Salt + Water
1. Take about 2 ml of dilute NaOH solution in a test tube.
2. Add two drops of phenolphthalein indicator solution.
3. It turns to red or pink colour.
4. Add dilute HCl solution to the above solution drop by drop.
5. Pink colour disappears due to the reaction of NaOH (base) with HCl (acid).
6. Now add one or two drops of NaOH to the above mixture.
In the above Activity, we have observed that the effect of the base is nullified by an acid and vice-versa.
(a) Write a chemical equation for the above reaction.
(b) What conclusion can be drawn from the observation that the phenolphthalein indicator turns NaOH solution to pink?
OR
What happens when NaOH is again added to mixture?
Answer:
(a) NaOH + HCl → NaCl + H2O
Base + Acid → Salt + Water
(b) It shows that NaOH is basic in nature.
OR
Pink colour reappears on adding NaOH again to the mixture.
Question 38.
There are many plants in which parts like the root, stem and leaves develop into new plants under appropriate conditions. Unlike most animals, plants can indeed use such a mode for reproduction. This property of vegetative propagation is used in methods such as layering or grafting to grow many plants like sugarcane, roses, or grapes for agricultural purposes. Plants raised by vegetative propagation can bear flowers and fruits earlier than those produced from seeds. Such methods also make possible the propagation of plants such as banana, orange, rose and jasmine that has lost the capacity to produce seeds. Another advantage of vegetative propagation is that all plants produced are genetically similar enough to the parent plant to have all its Characteristics.
1. Take a potato and observe its surface. Can notches be seen?
2. Cut the potato into small pieces such that some contain a notch or bud and some do not.
3. Spread some cotton on a tray and wet it. Place the potato pieces on this cotton. Note where the pieces with the buds are placed.
4. Observe changes in these potato pieces over the next few days. Make sure that the cotton is kept moistened.
(a) Which parts of plants are used for vegetative propagation?
(b) Give any one advantage of vegetative propagation.
(c) Give an example of a plant which shows vegetative propagation.
OR
What observation can be seen in the above activity?
Answer:
(a) There are many plants in which parts like the root, stem and leaves are used for vegeta propagation.
(b) The advantage of vegetative propagation is that all plants produced are genetically similar em to the parent plant to have all its characteristics.
(c) Rose shows vegetative propagation.
OR
The potato pieces having buds gradually grow and develop. But there is no growth and development in potato pieces without bud.
Question 39.
Complete an electric circuit consisting of a cell, an ammeter and a nichrome wire of length 1 [say, marked (1)] and a plug key, as shown
1. Now, plug the key. Note the current in the ammeter.
2. Replace the nichrome wire with another nichrome wire of the same thickness but twice the length, that is 21 [marked (2) in the Figure].
3. Note the ammeter reading.
4. Now replace the wire with a thicker nichrome wire, of the same length 1 [marked (3)]. A thicker wire has a larger cross-sectional area. Again note down the current through the circuit.
5. Instead of taking a nichrome wire, connect a copper wire [marked (4) in Figure] in the circuit.
6. Let the wire be of the same length and area of cross-section as the first nichrome wire [marked (1)]. Note the value of the current.
(a) What happened to the ammeter reading when the length of the wire is doubled?
(b) How ammeter reading can be increased?
(c) If a wire of different material of same length and the same area of cross-section is used then how does ammeter reading affected?
OR
(d) On what factors resistance of a conductor depends?
Answer:
(a) It is observed that the ammeter reading decreases to one-half when the length of the wire is doubled.
(b) The ammeter reading is increased when a thicker wire of the same material and of the same length is used in the circuit.
(c) A change in ammeter reading is observed when a wire of different material of the same length and the same area of cross-section is used.
OR
Resistance of a conductor depends on (i) on its length, (ii) on its area of cross-section, and (iii) on the nature of its material.