Class 10 Science Term 2 Answer key
Class 10 Science Term 2 Answer Key: The Central Board of Secondary Education is all set to conduct CBSE Class 10 Science Term 2 Exam on 10th May 2022, Tuesday. After completion of the examination, we will upload the unofficial Class 10 Science Term 2 Answer Key on this page. Class 10 Science Term 2 Answer Key is error-free and prepared by the expert faculties of Adda247. Instead of visiting different websites stay tuned with us and match your responses with the Class 10 Science Term 2 Answer Key provided on this page. Bookmark this page to get the Class 10 Term 2 Answer Key of all the subjects.
CBSE Class 10 Term 2 Science Answer Key
Here we have provided the miscellaneous information of CBSE Class 10 Term 2 Science Answer Key. The students who appeared in the term 2 science examination must check the information given in the table below:
|CBSE Class 10 Answer Key Science Term 2|
|Exam Conducting Body||Central Board of Secondary Education|
|Exam & Subject Name||CBSE Class 10 Science|
|Exam Date||10th May 2022, Tuesday|
|Unofficial Answer Key||10th May 2022, Tuesday|
|Official Answer Key||To be notified|
CBSE Class 10 Science Term 2 Answer Key & Paper Solution
CBSE Class 10th Science Question Paper for Answer keys(all sets)
CBSE Class 10 Science Term 2 Answer Key 2022: Paper Code 31/1/2
Q.1 (a)What is the ozone? how is it formed in the upper layer of the earth’s atmosphere? how does ozone affect our ecosystem?
Ozone: Ozone is a pale blue gas composed of three oxygen atoms bonded together. It occurs naturally high up in the Earth’s atmosphere, where it protects the surface from harmful ultraviolet (UV) rays unless dissipated by natural or human phenomena. It is also considered a pollutant with adverse effects for humans and other creatures when present closer to the ground.
Ozone is formed in the upper atmosphere by the reaction of ultraviolet UV radiations on oxygen O2 molecule. The damage to ozone layer is a cause of concern to us as due to its damage more ultraviolet rays reach the earth’s surface causing various health hazards. cause of this damage is the presence of large amount of chlorofluorocarbons in the atmosphere.
The biggest damage that ground-level ozone can do, aside from having implications to our own health, is that of the damage that it can do to plants, wildlife and the ecosystems on the planet.
Among the detrimental effects that ozone pollution can cause are:
- Plants experiencing a reduction in photosynthesis, a process whereby sunlight is turned into the energy that allows them to grow and live.
- Plants, trees and vegetations also suffering from a slower growth rate.
- Furthermore, all plant life will have a reduced chance of fighting off disease, insects, other pollutants and harsh weather.
Q. 1(b)(i) List two human-made ecosystems?
Garden and pond are man-made ecosystems.
Q.1(b)(ii) we do not have to clean the pond in the same manner as we do in an aquarium. Give a reason to justify this statement?
Ponds and lakes are natural ecosystems and they contain decomposers. Decomposers act as cleansing agents here, whereas an aquarium is an artificial ecosystem, it does not contain decomposers that cleanse it.
An aquarium is an artificial ecosystem that is an incomplete ecosystem when compared to a pond or lake which is a natural and complete ecosystem. Hence, an aquarium needs to be cleaned periodically.
Artificial Ecosystem: The artificial ecosystem includes dams, gardens, parks which are made by humans. The zoos, aquariums and botanical gardens are examples of artificial ecosystems which are maintained with the objective of conserving biodiversity. The plants and animals are placed in well-protected areas similar to their natural habitats.
Natural Ecosystem: Forests, lakes and deserts are the natural ecosystems. These are formed naturally without any human intervention. It refers to a community of living and non-living organisms interacting with each other through physical, biological and chemical processes.
Q. 2 (a)(i) Name and state the rule to determine the direction of force experienced by a current-carrying straight conductor placed in a uniform magnetic field which is perpendicular to it.
Fleming’s Left-Hand rule gives the direction of force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it. Fleming’s Right-Hand rule gives the direction of current induced in a coil due to its rotation in a magnetic field.
Q.2(a)(ii) An alpha particle while passing through a magnetic field gets projected towards north. In which direction will an electron project when it passes through the same magnetic field ?
By Fleming’s Left-hand rule,
Keeping the thumb, forefinger and middle finger in the mutually perpendicular directions. The middle finger represents the direction of the current (in the direction of motion of positive charge: towards the West), the thumb represents the force (towards the North), then the forefinger represents the direction of the field (upward)
The direction of the Magnetic field will be upward.
Q.2(b) (i) What is a solenoid?
A solenoid is a device comprised of a coil of wire, the housing and a moveable plunger (armature). When an electrical current is introduced, a magnetic field forms around the coil which draws the plunger in. More simply, a solenoid converts electrical energy into mechanical work.
- The coil is made of many turns of tightly wound copper wire. When an electrical current flows through this wire, a strong magnetic field/flux is created.
- The housing, usually made of iron or steel, surrounds the coil concentrating the magnetic field generated by the coil.
- The plunger is attracted to the stop through the concentration of the magnetic field providing the mechanical force to do work.
Q.2(b)(ii) Draw the pattern of magnetic field lines of the magnetic field produced by a solenoid through which a steady current flows.
A solenoid is a long coil that contains a large number of close turns of insulated copper wire. Its ends are connected to a battery B through a switch X.
The magnetic field lines inside a solenoid are in the form of parallel straight lines. This pattern of field lines inside the solenoid indicates that the strength of the magnetic field is uniform (same at all points).
Q.3(a) What is variation? List two main reasons that may lead to variation in a population.
variation, in biology, any difference between cells, individual organisms, or groups of organisms of any species caused either by genetic differences (genotypic variation) or by the effect of environmental factors on the expression of the genetic potentials (phenotypic variation).
Major causes of variation include mutations, gene flow, and sexual reproduction.
- DNA mutation causes genetic variation by altering the genes of individuals in a population.
- Gene flow leads to genetic variation as new individuals with different gene combinations migrate into a population.
Q.3(b)(i) In a cross between violet flowered plants and white flowered plants, state the characteristics of the plants obtained in the F1 progeny.
A Mendelian’s experiment consists of breeding a pea plant bearing violet flowers with pea plant that bear white flowers.
The intermediate colours do not get a chance to appear in the offspring of cross-pollinated pea plants.
The result in f1 progeny will be violet flower because it is dominant over the white flower.
Q.3(b)(ii) If the plants of F1 progeny are self-pollinated, then what would be observed in the plants of F2 progeny?
In the F2 progeny, one-quarter of the plants are short and the rest are tall. It has 1:2:1 ratio of TT, Tt And tt trait combinations.
- F1 is the first generation of hybrid. This is also known as the first filial generation.
- This generation resembles exactly one of the parental types.
- The F2 progeny is obtained by making self-pollination between the F1 progeny.
- Two F1 progenies selected are self-pollinated and the F2 progeny is obtained and it can’t be cross-pollinated.
Q.3(b)(iii) If 100 plants are produced in F2 progeny, then how many plants will show the recessive trait?
75% Violet (recessive)
Q.4 (a) Which of the following flowers will have a higher possibility of self-pollination?
Mustard, Papaya, Watermelon, Hibiscus
Hibiscus: hibiscus, (genus Hibiscus), genus of numerous species of herbs, shrubs, and trees in the mallow family (Malvaceae) that are native to warm temperate and tropical regions. Several are cultivated as ornamentals for their showy flowers, and a number are useful as fibre plants.
Q.4(b)List the two reproductive parts of a bisexual flower.
Two Reproductive Parts of Bisexual of Flower: They are stamens and carpels.
- Stamens: Male reproductive part. In produces pollen grains in anther.
- Carpel: Female Reproductive part. It has three parts – stigma, style, ovary.
- Stigma: It receives pollen grains.
- Style: Through it pollen tube passes into ovule.
- Ovary: It contains ovules that contain egg cell.
Q.5 (a) Name the process shown below and define it
Regeneration: The process by which some organisms replace or restore lost or amputated body parts.
5.(b) Name the types of cells present in the organisms which exhibit this process.
Regeneration occurs in organisms like hydra, flatworms, tapeworms. They have highly adaptive regenerative capabilities. When an organism is wounded, its cells are activated and the damaged tissues and organs are remodelled back to the original state.
Q.6: What is a Döbereiner triad ? Why can’t the following three elements form such a triad?
N (14); P (31); As (75)
(Atomic masses are given in parenthesis)
Döbereiner triad: Dobereiner’s triads were groups of elements with similar properties that were identified by the German chemist Johann Wolfgang Dobereiner. He observed that groups of three elements (triads) could be formed in which all the elements shared similar physical and chemical properties.
Dobereiner stated in his law of triads that the arithmetic mean of the atomic masses of the first and third element in a triad would be approximately equal to the atomic mass of the second element in that triad. He also suggested that this law could be extended for other quantifiable properties of elements, such as density.
The first of Dobereiner’s triads was identified in the year 1817 and was constituted by the alkaline earth metals calcium, strontium and barium. Three more triads were identified by the year 1829. These triads are tabulated below.
Triad can be only when the middle element is having mean atomic weight of other two elements which N (14); P (31); As (75) does not have such triad
Q.7 The molecular formulae of two alkynes, A and B are CxH2 and C3Hy respectively. (a) Find the values of x and y.
(b) Write the names of A and B.
a)i) C2H2(value of x is 2)
(ii)C3H4(value of y is 4)
b)The name of C2H2 is Ethyne.
The name of C3H4 is Cyclopropene
Q.8 (a) Name the group of organisms which form in the first trophic level of all food chains. Why are they called so?
The first and lowest level contains the producers, green plants. The plants or their products are consumed by the second-level organisms—the herbivores, or plant eaters
(b) Why are the human beings most adversely affected by bio-magnification?
Biomagnification makes humans more prone to cancer, kidney problems, liver failure, birth defects, respiratory disorders, and heart diseases.
(c) State one ill-effect of the absence of decomposers from a natural ecosystem.
Bacteria and fungi play the role of decomposers. Decomposers break down the complex organic substances of garbage, dead animals, and plants into simpler inorganic substances that go into the soil and are used up again by the plants. In the absence of decomposers recycling of material in the biosphere will not take place which would lead to the accumulation of dead plants and animals in the environment. Additionally, the environment would be finally sapped of all its resources which are needed to maintain and sustain life.
Q:9 (a) (i) Define Electric Power and write its SI unit.
Electric power is the rate at which electric energy is transferred by an electric circuit. The SI unit of power is the watt, one joule per second. It is the rate of doing work, measured in watts, and represented by the letter . The electric power in watts produced by an electric current consisting of a charge of coulombs every seconds passing through an electric potential (voltage) difference of is
work done per unit time
is electric charge in coulombs
is time in seconds
is electric current in amperes
is electric potential or voltage in volts.
Q:9(ii) Two bulbs rated 100 W; 220 V and 60 W; 220 V are connected in parallel to an electric main of 220 V. Find the current drawn by the bulbs from the mains.
Potential difference (V) = 220 V
Power of lamp 1 (P1) = 100 W
Power of lamp 2 (P2) = 60 W
Both the bulbs are connected in parallel. Therefore, the potential difference across each of them will be 220 V, because no division of voltage occurs in a parallel circuit.
We know that
Power = Voltage x Current
P = V x I
I = P/V
I1 = P1/V
I1 = 100/220
I1 = 5/11 A
I2 = P2/V
I2 = 60/220
I2 = 3/11 A
Net current (I) = I1 + I2
I = 5/11 + 3/11
I = 8/11
I = 0.7273 A
Hence, current drawn from the line is 0.727 A
Q.9:(b) (i) State Joule’s law of heating. Express it mathematically when an appliance of resistance R is connected to a source of voltage V and the current I flows through the appliance for a time t.
State Joule’s law of heating: Joule’s law of heating states that, when a current ‘i ‘ passes through a conductor of resistance ‘r’ for time ‘t’ then the heat developed in the conductor is equal to the product of the square of the current, the resistance and time.
Hint : To answer this question, we have to use the formula for the power developed in a conductor when a current is passed through it. Then using the formula of energy in terms of power, we will get the required expression of Joule’s law of heating.
Complete step by step answer
We know that when a voltage is applied across a conductor, current flows through it. This current is constituted by the free electrons flowing within the conductor. While flowing they collide with other electrons, and also with the atoms of the conductor. This collision results in the generation of heat. We all have experienced this heating in our day to day life. Whenever an electrical appliance is used for a long time, then it gradually gets heated up.
Joule found that the power of this heating, also called the rate of heating of the electrical conductor, is equal to the square of the current and also to the resistance, that is,
Now, we know that the energy is related to the power by
Since the power is the power of heating, the energy obtained is the heat energy. So we get the mathematical expression for Joule’s law of heating as
The Joule heating is also known as Ohmic heating, or the resistive heating. This law finds its application in many of the electrical devices used. In electric fuses, this law is applied to find the current required for melting the fuse in order to break the circuit in emergency. An incandescent bulb glows when its filament is heated. This heating is governed by Joule heating.
Q.9(b)(ii) A 5Ω resistor is connected across a battery of 6 volts. Calculate the energy that dissipates as heat in 10 s.
According to formula
V= R I
where v voltage, i current ,, R resistance..
I = 6/5
For energy dissipated , Use R= I*V
WHERE I= ∆q/∆t …….. 1.2
= ∆q/∆t…….. ∆q = 1.2×10.
= 12 ….. So,
R= ∆q*V/∆t.. 12×6 = 72
Q.10 (a) Calculate the resistance of a metal wire of length 2 m and area of cross-section 1.55 x 10-6 m2. (Resistivity of the metal is 2.8 x 10-8 m)
Q.10(b) Why are alloys preferred over pure metals to make the heating elements of electrical heating devices?
Alloys generally have high melting point as compared to the individual elements. The electrical heating devices must have high melting point so that they do not melt even at high temperature when an electrical current passes through the device. Thus alloys are used in electrical heating device.
Q.12(b)(i) Write molecular formula of benzene and draw its structure.
The structural representation of benzene is as shown in the figure below. The chemical formula for benzene is C6H6, i.e it has 6 hydrogen- H atoms and six-carbon atoms and has an average mass of about 78.112. The structure has a six-carbon ring which is represented by a hexagon and it includes 3-double bonds. The carbon atoms are represented by a corner that is bonded to other atoms.
Q.12(b)(ii) Write the number of single and double covalent bonds present in a molecule of benzene.
Six single bonds between Carbons and Hydrogens, three single bonds between six Carbons and Hydrogens ( alternatively) . Three double bonds present between carbons ( alternatively in hexagonal ring). So a total of nine single bonds, three double bonds are present in benzene ring.
Q.(12)(iii) Which compounds are called alkynes?
Alkynes are defined as the hydrocarbons in which a triple bond is present between carbon and carbon atoms. The general formula is written as where n = 1, 2, 3, 4…
For a given chemical compound of formula
n is 4, so the compound verifies the chemical formula for alkanes.
Q.13(a) List two advantages of adopting the atomic number of an element as the basis of classification of elements in the Modern Periodic Table.
Properties of elements depend upon number of valence electrons which depend upon electronic configuration. Atomic number is needed to write electronic configuration of an element. It shows that atomic number is more important to determine chemical properties of elements than atomic mass.
Q.13(b) Write the electronic configuration of the elements X(atomic number 13) and Y(atomic number 20)
The atomic number of X=20
The electronic configuration of X=2,8,8,2
Since the valence shell of X contains two electrons in the valence shell. Therefore, it loses two-electron to attain noble gas configuration and forms a dispositive ion.
The atomic number of Y=17
The electronic configuration of Y=2,8,7
Since the valence shell of Y contains seven electrons in the valence shell. Therefore, it gains one electron to attain noble gas configuration and form an anion.
Hence, one atom of X combines with two atoms of Y to form a compound XY2.
Since the bond is formed by the complete transfer of electrons. So, the bond is ionic in nature.
Electron – dot structure :
CBSE Class 10 Science Term 2 Answer Key 2022: Exam Pattern
The exam pattern given here is based on the CBSE Class 10 Science Term 2 Sample Paper released by the board on its official website. The students can calculate their marks by CBSE Class 10 Science Term 2 Answer Key 2022 given here. As per the pattern, the CBSE Class 10 science question paper consists of three sections namely Section A, Section B, and Section C. The details of all three sections are listed below:
Section A: This section contains 7 questions and each question carries 2 marks.
Section B: This section contains 6 questions and each question carries 3 marks.
Section C: This section contains 2 case-based questions. Each question carries 4 marks.
CBSE Class 10 Term 2 Science Exam Analysis 2022
The CBSE Class 10 Term 2 Science exam will be held between 10:30 and 12:30. The students can check CBSE Class 10 Term 2 Science Exam Analysis 2022 after completion of the examination. In CBSE Class 10 Term 2 Science Exam Analysis 2022, we will assess the question paper based on the errors in questions, out of syllabus questions, and difficulty level of the question paper.
CBSE Class 10 Science Term 2 Answer Key 2022: FAQs
Q. Where can I get the Class 10 Term 2 Science Answer Key 2022?
On this page, you will get the Class 10 Term 2 Science Answer Key 2022, match your responses with the answer key, and calculate your marks.
Q. Does the CBSE publish an official Answer Key for Class 10 Term 2 Science?
The CBSE Board did not inform yet about the official answer key of Class 10 Term 2 Science. When the board will publish the official answer key we will update the same here.
Q. What is the subject code of CBSE Class 10 Science?
The subject code of CBSE Class 10 Science is 086.