Students can access the CBSE Sample Papers for Class 10 Science with Solutions and marking scheme Term 2 Set 2 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 10 Science Term 2 Set 2 with Solutions
Time : 2 Hours
Max. Marks : 40
General Instructions :
- All questions are compulsory.
- The question paper has three sections and 15 questions. All questions are compulsory.
- Section-A has 7 questions of 2 marks each; Section-B has 6 questions of 3 marks each; and Section-C has 2 case based questions of 4 marks each.
- Internal choices have been provided in some questions. A student has to attempt only one of the alternatives in such questions.
Section – A
Question 1.
The electronic configuration of an element is 2, 8, 4. State its: (2)
(a) Group and period in the Modem Periodic Table.
(b) Name and write its one physical property.
OR
An element belongs to third period and group 16 of Modem Periodic Table.
(a) Determine the number of valence electrons and valency of the element.
(b) Name the element X and state whether it is metallic or non-metallic.
Answer:
(a) The given element with electronic configuration 2, 8, 4 is silicon (Si)
1. It means that it belongs to 3rd period and 14th group.
2. The name of element is silicon.
(b) The name of the element is silicon, it is a metalloid, (i.e., element that has properties of both metals and non-metals).
OR
(a) Since the element belongs to group 16, therefore is electronic configuration is 2, 8, 6 and hence it has 6 valence electrons. Its valency is 8 – 6 = 2.
(b) Element X is Sulphur and it is a Non-metal.
Question 2.
Carbon has the unique property to form bonds with other carbon atoms: (2)
(a) Name the unique property of carbon.
(b) Give reason for unique property of carbon atom.
(c) Draw the structure of cyclohexane.
Answer:
(a) Catenation.
(b) Unique property of carbon atom is because of tetra valency of carbon atom.
(c) Cyclohexane.
Question 3.
Answer the following questions: (2)
(a) Name the method by which spirogyra reproduces under favourable conditions. Is this method sexual or asexual?
(b) What is gestation period? Explain Parturition.
Answer:
(a) Spirogyra reproduces under favourable conditions by fragmentation which is an asexual mode of reproduction.
(b) Gestation period is the time from fertilisation till the birth of the new born. The delivery of full term baby from the uterus of mother after the end of gestation period is called parturition.
Question 4.
(a) What are dominant and recessive traits? (2)
(b) “Is it possible that a trait is inherited but may not be expressed in the next generation? Give a suitable example to justify this statement.
OR
The sex of a new born child is a matter of chance and none of the parents may be considered responsible for it. Draw a flowchart showing determination of sex of a newborn to justify this statement.
Answer:
(a) The trait which can express its effect over contrasting trait is called dominant trait whereas the trait
which cannot express its effect over contrasting trait or which gets suppressed by the contrasting trait is called recessive trait. The inherited trait which is not expressed will be a recessive trait.
(b) In Mendel’s experiment, when pure tall pea plants were crossed with pure dwarf pea plants, only tall pea plants were obtained in F1 generation. On selfing the pea plants of F1 generation both tall and dwarf pea plants were obtained in F2 generation. Reappearance of the dwarf pea plants in F2 generation proves that the dwarf trait was inherited but not expressed in F2 generation. The recessive trait does not express itself in the presence of the dominant trait. So, it is possible that one trait may be inherited but may not be expressed in an organism.
OR
In human beings, there are two types of sex chromosome X and Y; females have XX chromosome whereas males have XY chromosome. Females produce eggs which carry only X chromosomes but males contain half of the sperms with X chromosomes and other half with Y chromosomes. During fertilisation when X carrying sperms fuse with an egg which contains X chromosome the offsprings will be a female (XX). But when Y bearing sperms fuses with an egg (X) the offspring will be male (XY). Thus the sex of a child is determined by the type of sex chromosome X or Y received by the male gamete.
Question 5.
Explain why fertilisation is possible if mating takes place during the middle of menstrual cycle? (2)
Answer:
Mostly in a healthy woman ovulation occurs on 14th day of 28 days menstrual cycle which is the middle day of the cycle hence if mating occurs in the middle of cycle there is maximum chance of fertilisation.
Question 6.
In the following food chain, grass provides 4000 J of energy to the grasshopper. (2)
Grass, grasshopper, frogs, snakes.
How much energy will be available to snakes and frogs?
Answer:
If grass provides 4000 J energy, then according to 10 per cent law, it will give 10% of its energy to next trophic level.
Hence,
So, for snakes and frogs, 4 J and 40 J energy will be available by 10 per cent law.
Question 7.
Answer the following questions: (2)
(a) Define electromagnetic induction.
(b) What is a permanent magnet? Give one use of it.
OR
(a) Define a compass
(b) State Fleming’s right hand rule.
Answer:
(a) The production of electricity from magnetism is called electromagnetic induction.
(b) A permanent magnet is a magnet made from steel such that once magnetized, it does not lose its magnetism easily.
OR
(a) A compass is a device used to show magnetic field direction at a point. It consists of a tiny pivoted magnet usually in the form of a pointer which can turn freely in the horizontal plane.
(b) It states that, “Stretch your right hand in such a way that the first finger, the central finger and the thumb are mutually perpendicular to each other. If the first finger points along the direction of magnetic field and the thumb points along the direction of motion of the conductor, then the direction of induced current is given by the direction of the central finger.” This rule is also called dynamo rule.
This rule is also called dynamo rule.
Section – B
Question 8.
(a) What is periodicity in properties of elements with reference to the Modern Periodic Table? Why do all the elements of the same group have similar properties? How does the tendency of elements to gain electrons change as we move from left to right in a period? State the reason of this change.
(b) How it can be proved that the basic structure of the Modem Periodic Table is based on the electronic configuration of atoms of different elements? (3)
OR
Explain the periodicity of following properties of elements:
(a) Atomic radius
(b) Electronegativity
Answer:
(a) The occurrence of the elements with similar properties after certain regular intervals when they are arranged in increasing order of atomic number is called periodicity. The periodic repetition of the properties is due to the recurrence of similar valence shell configuration after regular interval. The elements in a group have same valence electrons thus similar chemical properties. In a period, tendency to gain electrons increases from left to right. This tendency increases because the hold of nucleus on the outermost electrons becomes weak thus it becomes easy to eject the electron.
(b) Modem periodic law states that the physical and chemical properties of an element are the periodic function of the atomic number of that element. Electronic configuration of the elements play an important role in the placement of element in the modern periodic table. The valence shell electron of an element decides its position in a particular group or period for example: if the configuration of an element is 2,1 it means that it belongs to the 2nd period and 1st group. The element will be Li = 2,1.
OR
(a) Atomic radius:
In a period, atomic radius generally decreases from left to right. In a period there is a gradual increase in the nuclear charge. Since valence electrons are added in the same shell, they are more and more strongly attracted towards nucleus. This gradually decreases atomic radii. Atomic radii increase in a group from top to bottom. As we go down a group the number of shells increases and valence electrons are present in higher shell and the distance of valence electrons from nucleus increases. Both the factors decrease the force of attraction between nucleus and valence electron. Therefore, atomic size increases on moving down a group.
(b) Electronegativity:
Electronegativity is a measure of the ability of an atom or molecule to attract pairs of electrons in a chemical bond. As we move left to right across a period, electronegativity increases because as the nuclear charge of an atom increases, its electron loving character also increases. Fluorine is the most electronegative element. On the other hand, as we move top to bottom in a group, electronegativity decreases.
Question 9.
(a) List two properties of carbon which lead to the huge number of carbon compounds we see around us, giving reason for each.
(b) Why are carbon and its compounds used as fuels for most applications? (3)
Answer:
(a) Two properties of carbon which lead to the huge number of carbon compounds are:
1. Catenation: It is the ability of carbon to form bonds with other atoms of carbon.
2. Tetravalency: With the valency of four, carbon is capable of bonding with 4 other atoms. Thus forms huge number of carbon compounds.
(b) Carbon and its compounds are used as fuels for most applications due to following reasons:
1. They give large amount of heat on combustion due to high percentage of carbon and hydrogen.
2. Carbon compounds used as fuel have optimum ignition temperature with high calorific values and are easy to handle.
3. Their combustion can be controlled.
4. Saturated hydrocarbons burn with a clean flame and no smoke is produced.
Question 10.
Study the diagram and answer the following: (3)
(a) What does the figure represents?
(b) Give an example of organism which shows this process.
(c) Describe the process shown in the picture.
Answer:
(a) The figure represents binary fission which is an asexual mode of reproduction.
(b) Amoeba and Paramecium shows this mode of reproduction.
(c) The genetic material first duplicates through mitosis leading to duplication of nucleus through karyokinesis and a constriction appears in the cell membrane which deepens and finally a single parent cell divides into two daughter cells. The division of cytoplasm is called cytokinesis. This mode of asexual reproduction is called binary fission.
Question 11.
Answer the following questions: (3)
(a) Define electric power.
(b) What is a super conductor? Give two examples.
(c) What does an electric circuit mean?
Answer:
(a) The electrical work done per unit time is called electric power.
(b) A superconductor is a substance of zero resistance at very low temperatures. Example: Mercury below 4.2 K, Lead below 7.25 K.
(c)A continuous conducting path consisting of wires and other electrical components (like resistance or electric bulb, switch etc.) between the two terminals of a cell or battery, along which an electric current flows, is called an electric circuit.
Question 12.
How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of: (2)
(a) 4 Ω
(b) 1 Ω?
OR
(a) How many 176 Ω resistors in parallel are required to carry 5 A on a 220 V line?
(b) In the circuit diagram shown below, calculate:
(i) Total resistance
(ii) Current shown by the ammeter A,
Answer:
(a) The three resistors of resistances 2 Ω, 3Ω and 6 Ω have to be combined as shown in the figure to obtain 4 Ω resistance.
Equivalent resistance of 3 Ω and 6 Ω connected in parallel is,
\(\frac{1}{R_{P}}=\frac{1}{3}+\frac{1}{6}\)
= \(\frac{2 + 1}{6}=\frac{3}{6}\)
\(R_{P}=\frac{6}{3}=2 \Omega\)
or
Now, RP and 2 Ω are joined in series and the equivalent resistance is,
R = RP + 2
= (2 + 2) Ω
= 4 Ω
(b) In order to obtain 1 Ω resistance, the resistors 2 Ω, 3 Ω and 6 Ω have to be combined as shown in figure. 2 Ω,3 Ω and 6 Ω resistances are connected in parallel,
\(\frac{1}{R}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)
= \(\frac {3+2+1}{6}\)
= \(\frac { 6 }{ 6 }\)
R = 1 Ω
OR
Here, Potential difference, V = 220 V, Current, I = 5 A
∴Resistance, R = \(\frac { V }{ I }\)
= \(\frac { 220 }{ 5 }\) = 44 Ω
Let the number of 176 Ω resistors to be connected in parallel to give an equivalent resistance of 44 Ω be x. Equivalent resistance of ‘x’ 176 Ω resistance connected in parallel is \(\frac { 176 }{ x }\) Ω
But,
\(\frac { 176 }{ x }\) = 44
∴x = \(\frac { 176 }{ 44 }\) = 4
Thus, 4 resistors of 176 Q each should be connected in parallel.
(b)
(i) The two resistors of resistance 4 Q each are connected in parallel.
\(\frac{1}{R}=\frac{1}{4}+\frac{1}{4}\)
= \(\frac { 1+1 }{ 4 }\)
= \(\frac { 2 }{ 4 }\)
= \(\frac { 1 }{ 2 }\)
∴R = 2 Ω
Now, Potential difference, V = 2 V
Total resistance, R = 2 Ω
(ii) Using Ohm’s law, Current,
I = \(\frac { V }{ R }\)
= \(\frac { 2 }{ 2 }\) A
= 1 A
Question 13.
Answer the following questions: (3)
(a) What is meant by non-biodegradable waste? Identify non-biodegradable wastes from the following: Empty packet of chips, empty plastic bottle of mineral water, empty paper box of sweets, empty tin of cold drink.
(b) What is the role of decomposers in the ecosystem?
(c) The depletion of ozone layer is a cause of concern. Why?
Answer:
(a) The substances that cannot be degraded naturally by the action of microbes and persist in environment for longer period of time are called non-biodegradable substances or wastes.
From the given list of wastes, the non-biodegradable wastes are empty packet of chips, empty plastic bottle of mineral water, and empty tin of cold drink.
(b) Decomposers act upon dead and decay organisms and 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. They maintain the balance in the ecosystem and provide space for the new life in the ecosystem.
(c) Ozone layer is found in stratosphere which prevents the harmful UV rays to reach the earth’s surface. But the depletion of ozone layer allows more UV rays to pass through it and causes various harmful effects on human beings like cancer, genetic defects, cataract, etc.
Section – C
This section has 02 case-based questions (14 and 15). Each case is followed by 03 sub-questions (a, b and c). Parts a and b are compulsory. However, an internal choice has been provided in part c.
Question 14.
Mrs. Joshi, an eight months pregnant lady was suggested by her doctor to get an ultrasound done. She went to a radiologist with her husband and got the ultrasound done. When the ultrasound was done, her husband asked doctor about the sex of the baby in the womb. (4)
(a) Is it ethical to determine the sex of the foetus? Why?
(b) What is the chance of giving birth to a girl child in human beings?
(c) What has government done to stop female foeticides?
OR
A 25-year-old young man with his partner of 3 years decides not to have babies and undergoes a surgical procedure to prevent pregnancy. This led to permanent sterilisation of young man. Name the surgery performed.
Answer:
(a) No, it is not ethical to determine the sex of the foetus because sex determination may lead to female foeticides which results in death of a girl child and thus the sex ratio becomes unbalanced in the society.
(b) The chance of giving birth to a girl child in human beings is 50%.
(c) Government has imposed a ban on sex determination techniques to stop female foeticides.
OR
Vasectomy is a surgical procedure for male sterilisation or permanent contraception.
Question 15.
Answer the following questions: (4)
(a) State Ohm’s law.
(b) Why does a conductor get heated up when an electric current flows through it? List the factors on which the heat produced in a conductor depends. State Joule’s law of heating. How will the heat produced in an electric circuit be affected if the resistance in the circuit is doubled for the same current?
(c) The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:
I (ampere) | 0.5 | 1.0 | 2.0 | 3.0 | 4.0 |
V(Volt) | 1.6 | 3.4 | 6.7 | 10.2 | 13.2 |
Plot a graph between V and I and calculate the resistance of that resistor.
OR
An electric lamp of resistance 20 Ω and a conductor of resistance 4 Ω are connected to a 6 V battery as shown in the circuit. Calculate:
(i) the total resistance of the circuit,
(ii) the current through the circuit,
(iii) the potential difference across the (1) electric lamp and (2) conductor.
(iv) power of the lamp.
Answer:
(a) According to Ohm’s law, the current flowing in a conductor is directly proportional to the potential difference applied across its ends, provided the temperature and other physical conditions of the conductor remains constant.
(b) When an electric current is passed through a conductor, it gets heated. This is called the heating effect of current. Heating effect of current is due to the transformation of electrical energy into heat energy. A battery of a cell is the source of electrical energy. The chemical reaction within the cell generates the potential difference between its two terminals that sets the electrons in motion to make the current flow through a conductor. The source has to keep expending its energy. A part of the source energy in maintaining the current is consumed into useful work and the rest of it is expended as heat.
The heat produced in a wire is directly proportional to:
(i) The square of the current (I2),
(ii) Resistance of wire (R), and
(iii) Time (t) for which the current passes.
Joule’s law of heating states that when a current of I amperes flows in a wire of resistance of R ohms for time t seconds, the heat produced in the conductor is equal to the product of the square of the current, resistance of the wire and time for which the current is passed.
By Joule’s law of heating,
H = I2Rt
Since H a R, therefore, if the resistance in the circuit is doubled, heat produced will also get doubled.
(c) The graph between V and I is given below:
Let us consider two points A and B on the slope. Draw two lines, one from points B along X-axis and another from point A along Y-axis, which meet at points C.
The slope of the graph will give the value of resistance, thus
Slope = R= \(\frac{\mathrm{AC}}{\mathrm{BC}}\)
BC =
BC = 3 – 1 = 2 A
AC = 10.2—3.4=6.8V
Slope = \(\frac{6.8}{2}\)
= 3.4 Ω
Thus, resistance (R) = 3.4 Ω
OR
(i) Given, R1 =20 Ω, R2 = 4 Ω
Since, in Series R = R1 + R2
∴Total resistance of circuit:
R = 20+4
= 24 Ω
(ii) Current through circuit
V = 6V, R=24 Ω
According to Ohm’s law
V=IR
I = \(\frac{V}{R}\)
I = \(\frac{6}{24}\)
= \(\frac{1}{4}\)
= 0.25 ampere
(iii) (1) Potential difference across electric lamp:
I = \(\frac{1}{4}\) A,R1=20
V1 = IR1
V1 = \(\frac{1}{4}\) × 20
= 5 V
(2) Potential difference across conductor
V2 = IR2
= \(\frac{1}{4}\) × 4
V2 = 1 V
(iv) Power of lamp: P = I2R
= \(\left(\frac{1}{4}\right)^{2} \times 20\)
= \(\frac{1}{4} \times \frac{1}{4} \times 20=\frac{5}{4}\)
or
P = 1.25 W.