Students must start practicing the questions from CBSE Sample Papers for Class 10 Science with Solutions Set 2 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Science Set 2 with Solutions
Time: 3 Hours
Maximum Marks: 80
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 2 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 3 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 5 marks each. Answers 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 4 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.
Question 1.
A reactive metal (M) is treated with H2SO4 (dil). The gas is evolved and is collected over the water as shown in the figure.
The correct conclusion drawn is/are (1)
(a) the gas is hydrogen
(b) the gas is lighter than the air
(c) the gas is SO2 and is lighter than air
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)
When any reactive metal (M) reacts with the acid H2SO4 (dil.), it evolves hydrogen gas (H2), which is lighter than air.
M (s) + H2SO4 (dil.) → MSO4 + H2 (g)
Question 2.
The following reaction is an example of (1)
CaO (s) + H2O (l) → Ca(OH)2 (aq)
(a) Displacement and combination reaction
(b) Decomposition and exothermic reaction
(c) Combination and exothermic reaction
(d) Combination and endothermic reaction
Answer:
(c) Combination and exothermic reaction
The given reaction is an example of both a combination and exothermic reaction because calcium oxide and water are combined to form a single product, i.e. calcium hydroxide, and produce heat during the reaction.
Question 3.
Name the given compound. (1)
(a) Propanone
(b) Butanal
(c) Propane
(d) Propanoic acid
Answer:
(a) Propanone
Question 4.
The composition of aqua-regia is (1)
(a) dil.HCI : conc. HNO3 (3 : 1)
(b) conc. HCl : dil. HNO3 (3 : 1)
(c) conc. HCl : conc. HNO3 (3 : 1)
(d) dil. HCl : dil. HNO3 (3 : 1)
Answer:
(c) conc. HCl : conc. HNO3 (3 : 1)
conc. HCl and conc. HNO3 in 3 : 1 ratio form aqua-regia. It is a highly corrosive, fuming liquid. It can dissolve all metals even gold and platinum also.
Question 5.
Which of the following acids does not give hydrogen gas on reacting with metals (except Mn and Mg)? (1)
(a) HNO3
(b) HCl
(c) H2SO4
(d) All of the above
Answer:
(a) HNO3
Nitric acid (HNO3) reacting with metals (except Mn and Mg) does not give hydrogen gas. Because it is a strong oxidizing agent, as soon as hydrogen gas is formed in the reaction between metal and dil. HNO3, the nitric acid oxidizes this hydrogen into water.
Question 6.
Pritam added hydrochloric acid in a first test tube containing small pieces of marble and then in a second test tube containing zinc granules. He observed gas evolved from both the test tubes. The gases evolved are (1)
(a) H2 in first test tube, O2 in second test tube
(b) CO2 in the first test tube, H2 in the second test tube
(c) O2 in first test tube, Cl2 in second test tube
(d) Cl2 in the first test tube, CO2 in the second test tube
Answer:
(b) CO2 in the first test tube, H2 in the second test tube
In the first test tube, the following reaction takes place.
CO2 gas has evolved here.
In a second test tube, the following reaction takes place.
Here, H2 gas is evolved.
Question 7.
Complete the missing variables given as ‘x’ and ‘y’ in the following reaction. (1)
Pb(NO3)2 (aq) + 2KI (aq) → PbI2 (x) + 2KNO3 (y)
(a) (aq) and (aq)
(b) (s) and (s)
(d) (aq) and (s)
(d) (s) and (aq)
Answer:
(d) (s) and (aq)
When lead nitrate reacts with a potassium iodide solution, then insoluble solid precipitates of lead iodide are formed along with the potassium nitrate solution.
Question 8.
The diagram given below shows a section through an alveolus and a blood capillary.
What is the oxygen concentration in X, Y, and Z? (1)
X | Y | Z | |
(a) | High | Low | High |
(b) | High | Low | Low |
(c) | Low | High | High |
(d) | Low | High | Low |
Answer:
(c) X – Low, Y – High, Z – High
A high concentration of oxygen in the alveolar air and a low concentration in the blood create a higher diffusion gradient. Blood leaving the alveolus contains a higher concentration of oxygen.
Question 9.
The diagram given below shows a vertical section through the heart.
What are the functions of the numbered blood vessels? (1)
Carries blood to the body | Carries blood to lungs | Carries blood from lungs | Carries blood from the body | |
(a) | 1 | 2 | 3 | 4 |
(b) | 1 | 3 | 4 | 2 |
(c) | 2 | 4 | 3 | 1 |
(d) | 3 | 1 | 4 | 2 |
Answer:
(d) 1. Pulmonary arteries – carry blood to the lungs.
2. Vena cava – carries blood from the body.
3. Aorta – carries blood from heart to body.
4. Pulmonary veins – carry blood from the lungs to the heart.
Question 10.
When a new plant is formed as a result of cross-pollination from different varieties of a plant, the newly formed plant is (1)
(a) dominant plant
(b) mutant plant
(c) Both (a) and (b)
(d) hybrid plant
Answer:
(d) hybrid plant
When a new plant is formed as a result of cross-pollination from different varieties of plants, the newly formed plant is hybrid as it contains alleles from two different plants. So, it exhibits the characteristics of both plants.
Question 11.
Which of the following options represents the function of nodes of Ranvier? (1)
(a) It is a functional unit of nerve
(b) It conducts impulses toward the nerve cell body
(c) It speeds up the impulse transmission
(d) It provides electric insulation
Answer:
(c) It speeds up the impulse transmission
The nodes of Ranvier are gaps along the myelin sheath that covers the axon of neuron cells. They speed up impulse transmission that runs along the axon.
Question 12.
Choose the mode of asexual reproduction exhibited by Bryophyllum and Plasmodium. (1)
(a) Bryophyllum, Vegetative propagation Plasmodium, Multiple fission
(b) Plasmodium, Multiple fission Bryophyllum, Vegetative propagation
(c) Planaria, Budding Plasmodium, Binary fission
(d) Hydra, Budding Rhizopus, Spore formation
Answer:
(a) Bryophyllum, Vegetative propagation Plasmodium, Multiple fission
Bryophyllum reproduces by vegetative propagation. Plasmodium reproduces by multiple fission.
Question 13.
At noon, the sun appears white as (1)
(a) light is least scattered
(b) all the colours of the white light are scattered away
(c) the blue colour is scattered the most
(d) the red colour is scattered the most
Answer:
(b) all the colours of the white light are scattered away
At noon, the sun appears white because the light from the sun is directly over H’s head and travels a relatively shorter distance. The sun appears white because a small amount of the blue and the violet colour are scattered.
Question 14.
Where is the magnetism of a bar magnet maximum? (1)
(a) At the centre of a bar magnet
(b) Near the poles of a bar magnet
(c) Far away from the poles of a bar magnet
(d) None of the above
Answer:
(b) Near the poles of a bar magnet
The magnetism of a bar magnet is maximum near the poles and minimum at the centre of a magnet.
Question 15.
Villi present on the internal wall of the intestine help in the (1)
(a) emulsification of fats
(b) breakdown of proteins
(c) absorption of digested food
(d) digestion of carbohydrates
Answer:
(c) absorption of digested food
The small finger-like projections, i.e. Villi present on the internal wall of the intestine increase the surface area for better absorption of digested food.
Question 16.
Which of the following structures is involved in gaseous exchange in the woody stem of a plant? (1)
(a) Stomata
(b) Lenticels
(c) Guard cell
(d) Epidermis
Answer:
(b) In the stems of woody plants, the exchange of respiratory gases takes place through lenticels. These are small openings in the pits of the bark.
Direction (Q. Nos. 17-20): These 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 but 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): Food cans are coated with tin and not with zinc. (1)
Reason (R): Zinc is more reactive than tin.
Answer:
(a) Both A and R are true and R is the correct explanation of A.
Question 18.
Assertion (A): The rate of breathing in aquatic organisms is much slower than that seen in terrestrial organisms. (1)
Reason (R): The amount of oxygen dissolved in water is very low compared to the amount of oxygen in air.
Answer:
(d) A is false but R is true.
Assertion is false but Reason is true and Assertion can be corrected as The rate of breathing in aquatic organisms is much faster than that seen in terrestrial animals this is due to the lesser amount of dissolved oxygen in water.
Question 19.
Assertion (A): Pyruvate is a six-carbon molecule. (1)
Reason (R): It is prepared in the cytoplasm as the first step of cellular respiration.
Answer:
(d) A is false, but R is true.
Assertion can be corrected as pyruvate is a 3-carbon molecule.
Question 20.
Assertion (A): When a charged particle enters in the direction of a uniform magnetic field, then it moves on a straight path without deviation. (1)
Reason (R): Magnetic force on a charged particle is zero, when it moves in the direction of magnetic field.
Answer:
(a) Both A and R are true and R is the correct explanation of A.
When a charged particle enters in the direction of the magnetic field, then no magnetic force acts on the charged particle, hence it moves on a straight path without deviation in a uniform magnetic field.
Section B
Questions No. 21 to 26 are Very Short Answer Questions.
Question 21.
Diamond and graphite show different physical properties although they are made up of carbon. Name this relationship between diamond and graphite. Give the basis of this relationship also. (2)
Answer:
This relationship between diamond and graphite is called allotropy. The physical properties are different because the carbon-carbon bonding in both the allotropes varies. Diamond is hard because in it one carbon atom is bonded with four other carbon atoms with strong covalent bonds, while graphite is soft in which each C-atom is joined to three other C-atoms by strong covalent bonds to form flat hexagonal rings. (1)
The various layers of C-atoms in graphite are quite far apart so that covalent bonds can exist between them. The various layers of carbon atoms in graphite are held together by weak van der Waals forces. They can slide over one another. (1)
Question 22.
What is a feedback mechanism? Explain this mechanism with the help of a suitable example.
Answer:
The feedback mechanism is the mechanism of the body to maintain the levels of hormones in the body to the desired limit. (1)
For example, if the sugar level in the blood rises, cells of the pancreas detect this change and respond by producing more insulin in the blood. As the sugar level falls, insulin secretion is reduced. (1)
Question 23.
What is the first sign of pregnancy in a woman? How pregnancy can be prevented surgically? (2)
Or
Which species is likely to have comparatively better chances of survival, the one reproducing asexually or the one reproducing sexually? Give a reason to justify your answer.
Answer:
The absence of a menstrual cycle may be the first indication of pregnancy in a woman. When vas deferens in males are blocked surgically, sperm transfer is prevented. Similarly, when Fallopian tubes are blocked in the females the egg will not be able to reach the uterus thereby preventing pregnancy. (2)
Or
Sexual reproduction is considered to be superior to asexual reproduction as it leads to variations, while asexual reproduction does not induce variations among progeny individuals. The advantages of variations in individuals are
- It brings adaptation in individuals.
- It helps in the survival of species.
- It is the basis of evolution.
Hence, the species that reproduce through sexual reproduction have better chances of survival. (2)
Question 24.
The value of current (I) flowing through a given resistor of resistance (R). The corresponding values of potential difference (V) across the resistor are given below.
V (Volts) | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 4.0 | 5.0 |
I (Ampere) | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.8 | 1.0 |
Plot a graph between current (I) and potential difference (V) and determine the resistance (R) of the resistor. (2)
Answer:
From the scale, At the X-axis, 1 div (1 cm) = 0.1 A
At Y-axis, 1 div (1 cm) = 0.5 V
∴ Slope of graph resistance, (R) = \(\frac{V}{I}\)
∴ R = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{1.5-1.0}{0.3-0.2}\)
= \(\frac{0.5}{0.1}\)
= 5 Ω
Question 25.
Rishi went to a palmist to show his palm. The palmist used a special lens for this purpose. (2)
(a) Where should the palmist place/hold the lens, to have a real and magnified image of an object?
(b) If the focal length of this lens is 10 cm and the lens is held at a distance of 5 cm from the palm, use the lens formula to find the position and size of the image.
Or
A narrow beam of white light is incident on two glass objects as shown above. Comment on the nature of the behavior of the emergent beam in both cases.
Answer:
(a) The palmist will hold the lens where the palm is in between the focus and pole of the lens. (1)
(b) Given, focal length, f = 10 cm and object distance, u = -5 cm
For lens formula, \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
⇒ \(\frac{1}{v}=\frac{1}{10}+\frac{1}{-5}=\frac{1}{10}-\frac{1}{5}\)
⇒ v = -10 cm
Hence, the magnification
m = \(\frac{v}{u}=\frac{-10}{-5}\) = 2
Hence, the image is on the same side of the lens as an object (palm) and it is virtually erect and magnified. (1)
Or
(a) The incident beam of light after refraction through a prism splits into a band of seven colors which are violet, indigo, blue, green, yellow, orange, and red (VIBGYOR). These coloured rays emerge out of the prism in different directions and become distinct. Therefore, dispersion of white light takes place. (1)
(b) When the incident beam passes through the first prism, it gets splitted into a band of seven colours. But when those coloured rays are incident on an identical inverted prism. Then, the recombination of the coloured rays takes place. This emergent light is parallel to the incident beam but slightly shifted outward. (1)
Question 26.
How does the study of the food chain in an area or habitat help us? Give an example of a food chain operating in a large lake.
Answer:
The study of the food chain in an area or habitat helps in
- understanding the energy transfer through organisms.
- understanding harmful human activities and disruption of ecological balance, if any.
An example of the four steps of the food chain operating in a large lake is as follows:
Algae → Protozoan → Small fish → Big fish. (2)
Section C
Questions No. 27 to 33 are Short Answer Questions.
Question 27.
An element A reacts with water to form compound B which is used in whitewashing. The compound B on heating forms an oxide C which on treatment with water gives back B. Identify A, B, and C and give the reactions involved. (3)
Or
Why does sodium form sodium hydroxide when reacts with water whereas, aluminium forms only aluminium oxide?
Answer:
Element A is calcium (Ca). When it reacts with water, it forms calcium hydroxide. This compound B is calcium hydroxide Ca(OH)2, which is used in whitewashing. (1)
Or
The metals placed lower in the reactivity series are less reactive towards water. Sodium metal is placed above aluminium in the reactivity series. Hence, it reacts with water to form sodium oxide which further dissolves in water to give a sodium hydroxide solution. Whereas, aluminium reacts with oxygen to form aluminium oxide which does not dissolve in water to form aluminium hydroxide. (3)
Question 28.
Baking soda is used in small amounts for making bread and cake. It helps to make these soft and spongy. An aqueous solution of baking soda turns red litmus blue. It is also used in soda acid fire extinguishers. Use this information to answer the following questions: (3)
(a) How does it help in extinguishing fire?
(b) What is the reaction involved when it is heated?
(c) Is the pH value of baking soda solution lower than or higher than 7?
Answer:
(a) 2NaHCO3 (Baking Soda) (s) + H2SO4 (Acid) (aq) → Na2SO4 (s) + 2H2O (l) + 2CO2 (g)
The CO2 gas produced by the reaction of baking soda and acid in the soda-acid fire extinguisher helps in extinguishing fire. (1)
(b) When sodium bicarbonate is heated, it disintegrates into sodium carbonate, water and carbon dioxide.
2NaHCO3 \(\stackrel{\text { Heat }}{\longrightarrow}\) Na2CO3 (s) +H2O (l) + CO2 (g) ↑ (1)
(c) pH value of baking soda solution is higher than 7, i.e. it is alkaline. (1)
Question 29.
A child questioned his teacher why organisms resemble their parents more as compared to grandparents. In which way will the teacher explain to the child?
Answer:
The two parents involved in sexual reproduction produce gametes which fuse forming a zygote. It gradually develops into a young child showing certain similarities with the parents. Since a child inherits its characters from both parents the resemblance between them is very close. The grandparents and the child resemble less closely because a gap in the gene pool is created by the parents of the child. Variations of two generations mixing and the addition of new variations from parents increase the difference between them to a greater extent. Hence, a child resembles more closely its parents than the grandparents.
Question 30.
Shyam wants to experiment to show angle of incidence (i) is equal to the angle of refraction (r) through a transparent slab. (3)
(a) Specify the value of i at which angle of i incidence (i) and refraction (r) are equal in the transparent slab.
(b) What is the angle between the incident ray and the emergent ray through the glass slab?
(c) If the refractive index of the glass slab is 1.5, then find the speed of light in the glass slab.
(Take, the speed of light in air = 3 × 108 m/s)
Answer:
(a) When the incident ray falls normally on the glass slab, it will refract without deviation, i.e. along the normal in the glass slab, ∠i = ∠r = 0. (1)
(b) When the refraction through a glass slab takes place, then the emergent ray becomes parallel to the incident ray. So, the angle between the incident ray and the emergent ray will be zero. (1)
(c) Given, the refractive index of the glass slab = 1.5
We know that,
Question 31.
(a) The potential difference between two points in an electric circuit is 1V. What does it mean? Name a device that helps to measure the potential difference across a conductor.
(b) Why does the connecting cord of an electric heater not glow while the heating element does?
Answer:
(a) If the potential difference between two points is 1 V, it means that if a charge of 1 C is moved from one point to the other, then 1 J of work is done. The potential difference across a conductor is measured using an instrument called the voltmeter. (1)
(b) The electric power P is given by P = I2R
The resistance of the heating element is very high. A large amount of heat is generated in the heating element and it glows. The resistance of the connecting cord is very low. Thus, negligible heat is generated in the connecting cord and it does not glow. (2)
Question 32.
A student performs an activity in which a current is passed through a solenoid. Then, he observes magnetic field patterns inside and outside the solenoid.
(a) Explain the pole formation at the ends of the solenoid.
(b) Draw magnetic field lines due to current carrying solenoid.
Answer:
(a) Pole formation at the ends of the solenoid can be explained by looking at one face of the coil. If the direction of current through the coil is clockwise, then that face has South polarity and if the direction of current is anti-clockwise, then that face has North polarity. (1)
(b) Magnetic field lines force due to a current-carrying solenoid are as shown in the figure.
Question 33.
What are decomposers? What will be the consequence of their absence in an ecosystem?
Answer:
Organisms that break down the complex organic compounds present in dead and decaying matter into simpler inorganic materials are called decomposers, e.g. certain bacteria and fungi. (1)
Decomposers act as cleaning agents of the environment by decomposing dead bodies of plants and animals. The consequence of their absence in an ecosystem can be disastrous. The dead bodies would persist for a long, leading to their accumulation and thus, polluting the environment. The biogenetic nutrients associated with remains will not be returned to the environment. As a result, all the nutrients present in soil, air, and water would soon be exhausted and the whole life cycle of organisms will be disrupted. (2)
Section D
Questions No. 34 to 36 are Long Answer Questions.
Question 34.
A compound C (molecular formula, C2H4O2) reacts with Na metal to form a compound R and evolves a gas that burns with a pop sound. Compound C on treatment with alcohol A in the presence of an acid forms a sweet-smelling compound S (molecular formula = C3H6O2). In addition to NaOH to C, it also gives R and water. S on treatment with NaOH solution gives back R and A. Identify C, R, A, S and write down the reactions involved. (5)
Or
(a) Name the type of carbon compounds that can be hydrogenated. With the help of a suitable example explain the process of hydrogenation.
(b) Explain any three methods to prevent rancidity.
Answer:
Since, compound C (Molecular formula C2H4O2) contains two oxygen atoms, therefore most probably it may be a carboxylic acid, i.e. ethanoic acid (CH3COOH).
Ethanoic acid reacts with a metal (Na), to evolve a gas that burns with a pop sound alongwith the formation of compound R, therefore, R must be salt, i.e. sodium ethanoate and the gas that burns with a pop sound must be H2 gas. (1)
Compound R is sodium ethanoate. This is also supported by the observation that when ethanoic acid reacts with NaOH, it gives R, (sodium ethanoate) and water.
Since compound C on treatment with alcohol A in the presence of acid forms a sweet-smelling compound S (Molecular formula C3H6O2), therefore, S is methyl ethanoate (ester). Since ester S has three carbon atoms and the acid C has two carbon atoms, alcohol A must contain one C atom, i.e. A is methanol.
Thus, C = CH3COOH (ethanoic acid)
R = \(\mathrm{CH}_3 \mathrm{COO}^{-} \stackrel{+}{\mathrm{Na}}\) (sodium ethanoate)
A = CH3OH (methanol)
S = CH3COOCH3 (methyl ethanoate) (2)
Or
(a) Only unsaturated hydrocarbons, i.e. alkenes, and alkynes can be hydrogenated.
e.g. In the presence of a catalyst Ni/Pd, ethyne is hydrogenated into ethare.
(b) Methods by which rancidity can be prevented are as follows.
- Keeping the food materials in air-tight containers.
- Refrigeration of cooked food at low temperature.
- Packing of food like wafers, and potato chips in packets containing nitrogen gas instead of air. (3)
Question 35.
Given below are certain situations. Analyse and describe its possible impact. (5)
(a) If we cut a part of Planaria.
(b) Stigma is removed from a flower.
(c) Style is plugged from a flower.
(d) Spermicide is applied without using a condom or diaphragm it.
(e) Fallopian tube and vasa deferens are plugged.
Or
What are the major parts of the brain? Mention the functions of different parts.
Answer:
(a) Planaria shows the property of regeneration. So, if we cut a portion of Planaria, it will develop into, a new organism from just a broken or cut part of 1 parent organism.
(b) Stigma is the terminal part of the carpel. It helps in receiving the pollen during pollination. So, if we remove the stigma then pollination will not occur.
(c) If a style is plugged from a flower then the stigma receiving the pollen grain and taking it to the ovary for fertilization will not occur as the style helps in the attachment of the stigma to the ovary.
(d) Spermicides are applied in combination with condoms to kill the sperm. If they are not applied with condoms or diaphragm then the chances of their failure will increase.
(e) If the fallopian tube is plugged then the egg will not be able to reach the uterus and thus fertilization will not take place, if the vasa deferens is plugged then the sperm transfer will be prevented.
Or
The brain is the most important coordinating centre in the body. It has three major parts or regions namely the forebrain, midbrain, and hindbrain.
Parts | Functions |
Forebrain | |
Cerebrum | The main thinking part of the brain. |
Cerebral hemispheres | Intelligence and voluntary actions. |
Olfactory lobes | Centres of smell. |
Hypothalamus | Has centres of hunger, thirst, etc. |
Midbrain | Controls reflex movements of the neck, head, and trunk in response to visual and auditory stimuli. |
Also controls the reflex movements of the eye muscles, changes in pupil size, and shape of the eye lens. | |
Hindbrain Pons | Regulates respiration. Relays information between the cerebellum and the cerebrum. |
Cerebellum | Maintains posture and balance of the body. Enables us to make precise and accurate movements. |
Medulla oblongata | Controls involuntary actions such as breathing, etc. Controlling centre for reflexes such as swallowing, coughing, vomiting, etc. |
Question 36.
(a) Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit, another set of 3 bulbs of the same wattage are connected in parallel to the same source. Will the 2 bulbs in the circuits glow with the same brightness? Justify your answer.
(b) If R1 and R2 are the resistances of the filament of 40W and 60W respectively, operating at 220V. Out of R1 and R2, which one is greater?
(c) How much energy does a 100W bulb transfer in 1 minute? (5)
Or
(a) A letter ‘A’ consists of a uniform wire of resistance 1 ohm per cm. The side of the letter is 20 cm long and the cross piece in the middle is 2 cm long while the apex angle is 60°. Find the resistance of the letter between the two ends of the legs.
(b) Calculate the amount of charge flowing in 2 min in a wire of resistance 10Ω when a potential difference of 20V is applied between the ends.
(c) In the circuit, determine the current flowing through 10Ω resistance.
Answer:
(a) In series circuit (a)
Equivalent resistance in series combination,
Rs = R + R + R = 3R, Voltage = V
Let current through each bulb in series combination be I.
(∵ Current remains same in series combination)
By Ohm’s law,
V = I × 3R ⇒ I = \(\frac{V}{3 R}\)
∴ Power consumption of each bulb in series combination,
P1 = I2 (3R)
P1 = \(\left(\frac{V}{3 R}\right)^2\) × 3R
P1 = \(\frac{V^2}{3 R}\) …….(i)
In parallel circuit (b)
The resistance of each bulb = R
The voltage across each bulb = V
(∵ voltage remains same in parallel combination)
∴ Power consumption of each bulb in parallel combination,
P2 = \(\frac{V^2}{R}\) …….(ii)
Hence, from Eqs. (i) and (ii), we get
P2/P1 = \(\frac{V^2}{R} \times \frac{3 R}{V^2}\)
P2 = 3P1
Therefore, each bulb in a parallel combination glows 3 times brighter to that of each bulb in a series combination. (3)
(b) As, power, P = \(\frac{V^2}{R}\) or R = \(\frac{V^2}{P}\)
For same voltage, R ∝ \(\frac{1}{P}\)
i.e., More the power, the lesser will be the resistance.
∴ R2 < R1 (1)
(c) Given, P = 100 W and t = 1 min = 60 s
P = \(\frac{\text { Work done }}{\text { Time }}=\frac{\text { Energy spent }}{t}\)
∴ Energy spent = P × t = 100 × 60 = 6000 J (1)
Or
(a) Clearly, AB = BC = CD = DE = BD = \(\frac{20}{2}\) = 10 cm
and R1 = R2 = R3 = R4 = R5 = 10Ω and R5 = 2Ω
As R2 and R3 are in series, their combined resistance = 10 + 10 = 20Ω
The combination is in parallel with R5 = 2Ω
Hence, Resistance between points B and D
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 table given below shows the hints given by the quiz master in a quiz. (4)
Hints |
(i) A metal ‘A’ is used in thermite reduction. |
(ii) When metal A’ is heated with oxygen gives ‘S’, which is amphoteric. |
(iii) Metal A’ acts as a reducing agent. |
Based on the above hints answer the following questions.
(a) Identify A and B.
(b) Write down the reactions of oxide B with HCl and NaOH.
Or
Explain the process of thermite welding with reaction.
Answer:
(a) Metal A is aluminum (Al) which is used in thermite reaction. Al reacts with oxygen to form aluminium oxide, Al2O3 (B), which is amphoteric. (2)
Or
The reaction of iron (III) oxide (Fe2O3)with aluminium to produce iron is used to join railway tracks or cracked machine parts. This process is called thermite welding. It is a highly exothermic reaction. The reaction involved is as follows.
Question 38.
Mohan in his kitchen garden crossed pure-breed tall pea plants with pure-breed dwarf pea plants and obtained pea plants of F1-generation. Then he performed two types of experiments. In the first, he self-crossed the plants of F1-generation (experiment A) and in the second, he crossed the plants of F1-generation with the pure-breed dwarf parent plants (experiment B). (4)
(a) What would be the phenotypic ratio of plants in the Frgeneration?
(b) How would the genotypic ratio differ in experiment ‘B’?
(c) What would be the phenotypic and genotypic ratio of F2-generation in experiment ‘A’?
Or
How do we describe the phenotypic character that is expressed in the F1 generation? What is the term given to the contrasting character?
Answer:
(a) The phenotype of all the plants in the F1 generation would be tall.
(b) When crossed with homozygous recessive parent the genotypic ratio would be Tt : tt; 1 : 1.
(c) In experiment ‘A’, the phenotypic ratio of tall and dwarf plants would be Tall : Dwarf :: 3 : 1, whereas the genotypic ratio would be, TT : Tt : tt : 1 : 2 : 1
Or
The phenotypic character that is capable of expressing in the F1 generation is described as ‘dominant’. The contrasting character, i.e. dwarfness is the recessive character. (2)
Question 39.
The following three image formation by three concave mirrors are shown in the figure. The point O and I denote object and image, respectively. The object distance and focal length in each case are given in the figure. (4)
(a) How would you define the radius of curvature?
(b) No matter how far you stand from a mirror, your image appears erect and diminished. Identify whether this mirror is concave or convex.
(c) How does the observation of images formed help in identifying the type of mirror?
Or
Based on the text and data given in the above paragraph, out of three in which case the mirror will form the image having the same size as an object?
Answer:
(a) Radius of curvature of a mirror is defined as the radius of the sphere from which the spherical mirror was cut. (1)
(b) The mirror is a convex mirror because it always forms an erect and diminished image irrespective of the position of the object. (1)
(c) By observing the images produced by the mirror for different positions of the object, its nature can be identified as follows:
If the image formed by the mirror is of the same size as that of the object for different positions of the object, then the mirror is a plane mirror.
If the image formed by the mirror is diminished for all positions of an object, then the mirror is a convex mirror.
If the image formed behind the mirror is longer than the object, then the mirror is concave. (2)
Or
The third concave mirror will form an image of the same size as that of an object because in the third concave mirror, f = 0.1 m, so the radius of curvature R = 2f = 0.2 m and a same size image is formed when the object is at the centre of curvature. (2)