Please refer to Class 10 Science Sample Paper Term 2 With Solutions Set A provided below. The Sample Papers for Class 10 Science have been prepared based on the latest pattern issued by CBSE. Students should practice these guess papers for class 10 Science to gain more practice and get better marks in examinations. The Term 2 Sample Papers for Science Standard 10 will help you to understand the type of questions which can be asked in upcoming examinations.
Term 2 Sample Paper for Class 10 Science With Solutions Set A
SECTION – A
Question 1. The atomic number of three elements are given below:
Write the symbol of the element which belongs to (a) group 13, (b) group 15, of the periodic table. State the period of the periodic table to which these elements belong. Give reason to support your answer.
Answer : (a) A belongs to group 13 because its electronic configuration is 2, 3, i.e., it has 3 valence electrons.
(b) ‘B’ belongs to group 15 because its electronic configuration is 2, 5, i.e., it has 5 valence electrons.
They belong to the 2nd period as they both have two shells.
Question 2. An organic molecule has the following structure:
(i) To which homologous series does this molecule belong ?
(ii) What is the general formula of this homologous series?
Answer : (i) Alcohols
(ii) CnH2n+1OH
Question 3. Ovaries are primary sex organs of human female which are located in the lower part of abdominal cavity near the kidneys. Name the hormones secreted by human female ovaries. Write down their roles.
Answer : Hormones secreted by human female ovaries are:
(i) Estrogen : Responsible for development of secondary sexual characters in females.
(ii) Progesterone : Prepares the uterus for the reception of fertilized ovum.
Question 4. A student noticed that an organism by mistake was cut into two parts. After sometimes, he noticed that both the parts developed into new individuals.
(a) Name the mode of reproduction used by the organism. State an example of such organism which can multiply by this process.
(b) State the type of cells which can carry out this process.
Answer : (a) The mode of reproduction used by the organism is regeneration. Hydra or Planaria.
(b) It is carried out by the specialized cells which can proliferate and make a large number of cells by cell division.
Question 5. Different species use different strategies to determine sex of a newborn individual. It can be determined by environmental cues or can be genetically determined. Explain the statement by giving examples for each strategy.
OR
In human beings, the statistical probability of getting either a male or a female child is 50%. Give reasons.
Answer : Environmental Cue:
(i) In some animals, the temperature at which fertilized eggs are kept determines whether the developing animal in the egg is male or female.
(ii) In some animals like snail, individuals can change sex.
Genetic Cue: A child who inherits an X-chromosome from the father will be a girl and one who inherits a Y-chromosome from the father will be a boy.
OR
A child who inherits an “X” chromosome from the father would be a girl (XX) while a child who inherits a “Y”. chromosome from the father would be a boy (XY).
Since, the sex of the child is determined by what he/ she inherit from father, therefore the probability of getting either male or female child is 50%. 1+1
Question 6. Describe the flow of energy to the owl, if the tree provides 1500 calories of energy to the insects.
OR
Differentiate between biodegradable and non-biodegradable wastes.
Answer : The insects would provide 10 percent (one tenth) of 1500 calories, or 150 calories, of energy to the shrews.
The shrews would provide one tenth of 150 calories, or 15 calories, to the owl.
OR
Question 7. How can you show that the magnetic field produced by a given electric current in the wire decreases as the distance from the wire increases?
OR
A student boils water in an electric kettle for 20 minutes. Using the same mains supply he wants to reduce the boiling time of water. To do so should he increase or decrease the length of the heating element? Justify your answer.
Answer : If we take a magnetic compass slowly away from the current-carrying conductor, the deflection of the compass decreases. This shows that the magnetic field decreases as we move away from the current-carrying conductor.
OR
To reduce the boiling time using the same mains supply, the rate of heat production should be large. We know that P = V²/R.
Since V is constant, R should be decreased. Since R is directly proportional to l so length should be decreased.
SECTION – B
Question 8. Potassium, bromine and krypton are elements in period 4 of the Periodic Table.
(a) In which group of the periodic table can these elements be found?
(b) Bromine exists as a molecule. Draw a ‘dot-and-cross’ diagram to show the bonding in a molecule of bromine.
(c) Krypton does not react with either potassium or bromine. Explain the unreactive nature of krypton.
Answer : (a) Potassium is found is group 1, bromine in group 17 and krypton in group 18.
(b)
(c) Krypton has a stable electronic configuration, with 8 electrons in its valence shell. Hence, it
does not lose, gain or share electron(s) with another atom.
Question 9. Study the organic compound given in the box. Answer the following questions.
Ethane , Ethene, Ethanoic acid, Ethyne, Ethanol
(a) The compound having functional groups—OH and –COOH.
(b) Gas used in welding.
(c) Homologue of the homologous series with general formula CnH2n+2.
OR
(a) Why two carbon atoms cannot be linked by more than three covalent bonds?
(b) Give three differences between diamond and graphite.
Answer : (a)
(b) Ethyne
(c) Alkane : CnH2n+2
Ethane : C2H6
OR
(a) Since the maximum angle strain is obtained when the two carbon atoms are linked by three covalent bonds, therefore, two carbon atoms cannot be linked to each other by more than three covalent bonds.
(b) Differences between diamond and graphite :
Question 10. In one of his experiments with pea plants, Mendel observed that when a pure tall pea plant is crossed with a pure dwarf pea plant, in the first generation F1, only tall plants appear. What happens to the traits of the dwarf plants in this case? When the F1 generation plants were selffertilised, he observed that, in the plants of second generation F2, both tall plants and dwarf plants were present. Why it happened? Explain briefly.
Answer : The dwarf traits of the plants is not expressed in the presence of the dominant tall trait. In the F2 generation, both the tall and dwarf traits are present in the ratio of 3:1, respectively. This showed that the traits for tallness and dwarfness are present in the F1 generation, but the dwarfness, being the recessive trait does not express itself in the presence of tallness i.e., the dominant trait.
Question 11. Two coils of resistance R1 = 3Ω and R2 = 9Ω are connected in series across a battery of potential difference 14 V.
(a) Draw the circuit diagram.
(b) Find the electrical energy consumed in 1 min in each resistance.
Answer : (a) The circuit diagram:
(b) Given: R1 = 3 Ω, R2 = 9 Ω
In series connections, RS = R1 + R2 = 9 + 3 = 12 Ω
Now, I = V/R = 14/12 = 1.167 A
Electric energy consumed in R1
H1 = I2R1t = (1.167)2 × 3 × 60 = 245.14J
Electric energy consumed in R2
H2 = I2R2t = (1.167)2 × 9 × 60 = 735.42J
Question 12. Find the equivalent resistance of the given network between points A and B.
OR
In the circuit given below,
(a) Would any bulb glow when plug key is in open position?
(b) Write the order of brightness of the bulb when key is closed. Give reason.
Answer : In each segment of the combination 3 Ω and 2 Ω resistances are connected in series separately.
OR
(a) No bulb will glow when plug key is in open position as no current would flow through the circuit.
(b) Power of bulb, P = I2 R
For the same current, P ∝ R
but for the same voltage, P ∝ l/R or R ∝ l/P
So, resistance order of all bulb is, R25 > R40 > R60
According to Joule’s law of heating, H ∝ R (for the same current and time)
Hence, order of heating produced is H25 > H40 > H60
which is order of brightness of the bulb when key is closed.
Question 13. While teaching the chapter “Our Environment” the teacher stressed upon the harmful effects of burning of fossil fuels, plastic paper, etc. The students noticed the extensive use of plastic and polythene in daily life can be avoided and the surroundings can be kept clean. They decided to make their school “Plastic and Polythene” free and motivated each other for its minimum use.
(a) Why should the use of polythene and plastic be reduced in daily life?
(b) In what way the students would have avoided the use of plastic and polythene in their school?
Answer : (a) Because they are non-biodegradable. Plastic and polythene cannot be degraded by the activity of microorganisms.
(b) (i) Carrying tiffin and water in steel containers.
(ii) Encourage the use of ink pens instead plastic pens. (ball pens are made up of plastic)
SECTION – C
Question 14. An insulated copper wire wound on a cylindrical cardboard tube such that its length is greater than its diameter is called a solenoid. When an electric current is passed through the solenoid, it produces a magnetic field around it. The magnetic field produced by a current-carrying solenoid is similar to the magnetic field produced by a bar magnet. The field lines inside the solenoid are in the form of parallel straight lines. The strong magnetic field produced inside a currentcarrying solenoid can be used to magnetise a piece of magnetic material like soft iron, when placed inside the solenoid. The strength of magnetic field produced by a current carrying solenoid is directly proportional to the number of turns and strength of current in the solenoid.
(a) What is the strength of magnetic field inside a long current-carrying straight solenoid ?
(b) A long solenoid carrying a current produces a magnetic field B along its axis. If the current is double and the number of turns per cm is halved then what will be new value of magnetic field ?
(c) A soft iron bar is enclosed by a coil of insulated copper wire as shown in figure.
When the plug of the key is closed, which face of iron bar marked as N-pole ?
OR
Two long wires P and Q carrying current I1 and I2 are arranged as shown in figure.
Wire P carrying current I1 along x-axis. Wire Q carrying current I2 along a line parallel to y-axis given by x = 0 and z = d. Find the force exerted by wire P on wire Q.
Answer : (a) Magnetic field inside infinite solenoid is uniform. Hence it is same at all points.
(b) For a long solenoid, magnetic field B ∝ In; where I is the flowing current and n is number of turns per unit length in the solenoid. Therefore, in the given case magnetic field will remain unchanged.
(c) When the plug of the key is closed, the face B of the iron bar marked as N-pole
OR
According to right hand thumb rule, the magnetic field at Q due to wire P is along negative y – axis.
∴ Magnetic field at Q is antiparallel to current I2.
Hence, there is no force on wire Q due to wire P.
Question 15. Seema crossed pure breed of pea plants having round-yellow seeds with wrinkled green seeds and found that only A – B type of seeds were produced in the F1 generation. When in F1 generation, pea plants having A – B type of seeds were crossbred by self pollination, then in addition to the original round yellow and wrinkled green seeds, two new varieties A – D and C – B types of seeds were also obtained.
(a) What are A – B type of seeds?
(b) State whether A and B are dominant traits or recessive traits.
(c) What are A – D and C – B type of seeds?
OR
Which one will be produced in minimum and maximum number in the F2 generation?
(a) A – B
(b) C – D
Answer : (a) A – B type of seeds – Round in shape and yellow in colour 1
(b) A – B type is dominant traits. 1
(c) A – D = Round-green
C – B = Wrinkled- yellow 1+1
OR
C – D in minimum number
A – B in maximum number 1+1
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