Please see below Class 10 Science Sample Paper Term 2 Set E with solutions. We have provided Class 10 Science Sample Papers with solutions designed by Science teachers for Class 10 based on the latest examination pattern issued by CBSE. We have provided the following sample paper for Term 2 Class 10 Science with answers. You will be able to understand the type of questions which can come in the upcoming exams.

CBSE Sample Paper for Class 10 Science Term 2 Set E

SECTION – A

1. The table shows the electronic structures of four elements.

(a) Identify which element(s) will form covalent bonds with carbon.
(b) “Carbon reacts with an element in the above table to form several compounds.” Give suitable reason.
Ans. (a) Elements P and R will form covalent bond as both are non-metals.
(b) Carbon forms several compounds because it has valency four (tetravalency) and can form long chains (catenation).

2. The diagram below shows part of the periodic table.
(a) Which elements would react together to form covalent compounds?
(b) Between the two elements W and Z, which will have a bigger atomic radius? Why 

Ans. (a) Element Y and Z both are non-metals and react with each other to form covalent compounds.
(b) ‘W’ will have a bigger atomic radius as it has four shells which means it has larger distance between nucleus and outermost shell while element ‘Z’ will have only two shells.

3. (a) Trace the path a male gamete takes to fertilise a female gamete after being released from the penis.
(b) State the number of sets of chromosomes present in a zygote.
Ans. (a) A male gamete follows the following path to fertilise a female gamete after being released from the penis:
Sperm   →   Enter into   →  Uterus  →   Fallopian    →    Zygote
                  the vagina                          tube                  (2n)
                                           (Fertilisation takes place)
(b) A zygote is formed by the fusion of two sets of gametes i.e.,(n) from male parent and (n) from female parent. Therefore, it will have two sets of chromosomes (2n).

4. Rajesh observed a patch of greenish black powdery mass on a stale piece of bread.
(a) Name the organism responsible for this and its specific mode of asexual reproduction.
(b) Name its vegetative and reproductive parts.
Ans. (a) The greenish black powdery mass on a stale piece of bread is due to the growth of bread mould fungus (Rhizopus) which reproduce by the method of spore formation.
(b) Rhizopus possess thread-like projections called hyphae (vegetative structure) from which small spherical structure called sporangium (reproductive structure) is developed.

5. Mustard was growing in two fields – A and B. While field A produced brown coloured seeds, field B produced yellow coloured seeds.
It was observed that in field A, the offsprings showed only the parental trait for consecutive generations, whereas in field B, majority of the offsprings showed a variation in the progeny.
 What are the probable reasons for these?
Ans. In field A, the parental traits seen in consecutive generations of the offspring is due to self-pollination because self pollination maintained parental characteristics in the offspring whereas in field B, where majority of the offsprings showed a variation in progeny due to the occurrence of cross-pollination because in cross-pollination, genetic diversity is introduced within the various species of same plant.

OR

In an asexually reproducing species, if a trait X exists in 5% of a population and trait Y exists in 70% of the same population, which of the two trait is likely to have arisen earlier? Give reason.
Ans. If trait X exists in 5% (smaller fraction) and trait Y exists in 70% (larger fraction) of the population then the larger fraction i.e., trait Y must have arisen earlier because in asexual reproduction variation do not occur, as the cell produces identical copies of DNA. New trait come in the population only due to sudden mutation and further inherited. 70% of the  population with trait Y is likely to have been replicating that trait for longer period than 5% of population with trait X.

6. A simple motor is made in a school laboratory. A coil of wire is mounted on an axle between the poles of a horseshoe magnet, as illustrated.

In the example above, coil ABCD is horizontal and the battery is connected as shown.
(a) For this position, state the direction of the force on the arm AB.
(b) Why does the current in the arm BC not contribute to the turning force on the coil?
Ans. (a) From the Fleming’s left hand rule, Force on arm AB is downwards.
(b) Because BC is in the same direction as the direction of field lines. Force is zero when the direction of current in the conductor is the same as that of the magnetic field.

OR

A circuit contains a battery, a variable resistor and a solenoid. The figure below shows the magnetic field pattern produced by the current in the solenoid. 

(a) State how the magnetic field pattern indicates regions where the magnetic field is stronger.
(b) What happens to the magnetic field when the current in the circuit is reversed?
Ans. (a) Relative closeness of magnetic field lines indicates the strength of magnetic field. Since field lines are crowded around the ends of the solenoid, hence these are the regions of strong magnetic field.
(b) The direction of the magnetic field will be reversed on reversing the direction of electric current.

7. DDT was sprayed in a lake to regulate breeding of mosquitoes. How would it affect the trophic levels in the following food chain associated with a lake? Justify your answer. 

Ans. Pesticides like DDT, are non-biodegradable pesticides which enter the food chain from the first trophic level i.e., planktons (producers).
Non-biodegradable chemicals accumulate into the body of organisms through the food chain which go on increasing in its concentration at each trophic level by the phenomenon called biomagnification. In the given food chain, hawk will have the highest level of pesticides because hawk is present on the top of food chain and is a top carnivore.

OR

In the following food chain, vertical arrows indicate the energy lost to the environment and horizontal arrows indicate energy transferred to the next trophic level. Which one of the  three vertical arrows (A, C and E) and which one of the two horizontal arrows (B and D) will represent more energy transfer? Give reason for your answer. 

Ans. According to the given food chain, A will represent more energy transfer as compared to C and E. B will represent more energy transfer as compared to D.
When green plants are eaten by primary consumers, a great deal of energy is lost as heat to the environment, some amount goes into digestion and in doing work and the rest goes towards growth and reproduction. An average of 10% of the food eaten is made available for the next level of consumers. This loss of energy takes place at every trophic level.

SECTION – B

8. Choose an element from period 3 of modern periodic table that matches the description given below in each instance. Give reason for your choice.
(a) It has a similar structure to diamond.
(b) It has same valency as lithium.
(c) It has variable valency and is a member of the oxygen family (group 16).
Ans. (a) Silicon forms carborundum which has a crystal structure like that of diamond and is almost as hard.
(b) Sodium will have same valency as that of lithium The electronic configurations of Li (At. no. 3) and Na (At. no. 11) are 2, 1 and 2, 8, 1 respectively. 
From above electronic configuration, it can be seen that both have same valency of 1.
(c) Sulphur is the member of oxygen family and third period which shows variable valency when it  combines with different elements. It forms oxides like SO₂ and SO₃.

9. (a) How many isomers are possible for the compound with the molecular formula C4H8? Draw the electron dot structure of branched chain isomer.
(b) How will you prove that C₄H₈ and C₅H₁₀ are homologues?
Ans. (a) The possible structural isomers of compound C₄H₈ are as follows:

(b) C₄H₈ and C₅H₁₀ differ by —CH₂ group. They possess the same general formula C_nH_2n and show similar chemical properties and gradation in physical properties such as melting point, boiling point, density etc. hence they belong to same homologous series.

OR

A carbon compound ‘A’ having melting point 156 K and boiling point 351 K, with molecular formula C₂H₆O is soluble in water in all proportions.
(a) Identify ‘A’ and draw its electron dot structure.
(b) Give the molecular formula of any two homologues of ‘A’.
Ans. (a) The compound ‘A’ with melting point 156 K and boiling point 351 K is ethanol i.e. CH₃CH₂OH. 
The electron dot structure is

(b) The compounds of homologous series have similar chemical properties, but gradually changing physical properties thus the molecular formulae of two homologues of A is CH₃OH or (CH₄O) and  CH₃CH₂CH₂OH or (C₃H₈O).

10. Two pea plants – one with round yellow seeds (RRYY) and another with wrinkled green (rryy) seeds produce F₁ progeny that have round, yellow (RrYy) seeds.
When F₁ plants are self-pollinated, which new combination of characters is expected in F₂ progeny? How many seeds with these new combinations of characters will be produced when a total 160 seeds are produced in F₂ generation? Explain with reason.
Ans.

Out of 160 seeds, 30 round green and 30 wrinkled yellow seeds are new combination of characters in F₂ progenies.
The new combination of the characteristics are produced because of the independent assortment of seed shape and seed colour trait.

11. (a) It would cost a man ` 3.50 to buy 1.0 kW h of electrical energy from the main electricity board. His generator has a maximum power of 2.0 kW. The generator produces energy at this maximum power for 3 hours. Calculate how much it would cost to buy the same amount of energy from the main electricity board.
(b) 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
Ans. (a) Energy = Power × Time
So, Energy (E) = 3 × 2 = 6 kWh
Cost of buying electricity from the main electricity board = 6 × 3.50 = Rs 21.0
(b) 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.  

12. 

In the above circuit, if the current reading in the ammeter A is 2 A, what would be the value of R₁?
Ans. As shown in diagram, 
5 ohm, 10 ohm and R1 are in parallel
 1 / R_p  = 1 / 5 + 1 / 10 + 1 / R₁
 1 / R_p  = 3 / 10 + 1 / R₁
 R_p  =  10R₁ / 3R₁ + 10 
Now, 6 ohm, 6 ohm and R_p  are in series Thus,
 R_eq =  10R₁ / 3R₁ + 10 + 12   ….(i)
From Ohm’s law
V = I R_eq 
From the circuit
R_eq  = 30 /2 = 15 Ω                ….(ii)
From equation (i) and (ii)
12 + 10R₁ / 3R₁ + 10 = 15  
10 R₁ = (9 R₁ + 30)
R₁ = 30 Ω

OR

Calculate the total resistance of the circuit and find the total current in the circuit.
Ans. Since R₃ and R₄ are connected in series,

So, total resistance of the circuit (R_eq) = 12 Ω
total current in the circuit (I) = 2 A

13. Gas A, found in the upper layers of the atmosphere, is a deadly poison but is essential for all living beings. The amount of this gas started declining sharply in the 1980s.
(a) Identify gas A. How is it formed at higher levels of the atmosphere?
(b) Why is it essential for all living begins? State the cause for the depletion of this gas.
Ans. (a) Gas A is ozone which is found in stratosphere layer of the atmosphere. Ozone in the earth’s atmosphere is formed by ultraviolet light striking oxygen molecules (O₂), creating two single oxygen atoms (O). The atomic oxygen (O) then combines with a molecule of O₂ to create ozone, O₃.

(b) Ozone layer is a protective shield around earth which absorbs most of the UV- radiations of sun protecting the living beings of the earth from health hazards.
Chloroflurocarbons (CFCs) used as referigerants and fire extinguishers lead to depletion of the ozone layer.

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.

14. Sahil performed an experiment to study the inheritance pattern of genes. He crossed tall pea plants (TT) with short pea plants (tt) and obtained all tall plants in F₁ generation.
(a) What will be set of genes present in the F₁ generation?
(b) Give reason why only tall plants are observed in F₁ progeny.
(c) When F₁ plants were self-pollinated, a total of 800 plants were produced.
How many of these would be tall, medium height or short plants? Give the genotype of F₂ generation.
Ans. (a)

(b) Tall (T) is a dominant trait whereas short (t) is a recessive trait and the gene corresponding to the recessive trait fails to express itself in presence of the gene representing dominant trait (TT or Tt). Therefore, all the plants appear tall in the F₁ generation.

Out of 800 plants, 200 tall plants, 400 medium height plants and 200 short plants are produced.

OR

When F₁ plants were cross-pollinated with plants having tt genes, a total of 800 plants were produced. How many of these would be tall, medium height or short plants? Give the genotype of F₂ generation.
Ans. In a cross between Tt × tt, 400 tall (TT) and 400 short (tt) plants will be produced.

Genotypic and phenotypic ratio = 2 : 2
Calculation :
2 / 4 × 800 : 2 / 4 × 800
= 400 : 400
Total – 400 tall and 400 short plants

15. Ansari Sir was demonstrating an experiment in his class with the setup as shown in the figure below.

A magnet is attached to a spring. The magnet can go in and out of the stationary coil. He lifted the magnet and released it to make it oscillate through the coil. Based on your understanding of the phenomenon, answer the following questions.
(a) What is the principle which Ansari Sir is trying to demonstrate?
(b) What will be observed when the magnet starts oscillating through the coil. Explain the reason behind this observation.
(c) Consider the situation where the magnet goes in and out of the coil. State two changes which could be made to increase the deflection in the galvanometer.
Ans. (a) Sir is trying to demonstrate the principle of Electromagnetic induction. 
(b) There will be induced current in the coil due to relative motion between the magnet and the coil. Changing the magnetic field around the coil generates induced current.
(c) Using a stronger magnet, using a coil with more number of turns would lead to increase the deflection in the galvanometer.

OR

Is there any difference in the observations in the galvanometer when the magnet swings in and then out of the stationary coil? Justify your answer. 
Ans. When the magnet moves into the coil, the galvanometer shows a momentary deflection towards one side say left. When the magnet moves out of the coil, the galvanometer shows a momentary deflection now towards right. This is due to changing magnetic field associated with the coil as the magnet moves in and out.
The flux increases when the magnet goes in and it decreases when the magnet goes out. Direction of induced current can be found by using Fleming’s right hand rule.