We have provided Class 10 Science Sample Paper as per the latest CBSE examination pattern for the current academic year. The following CBSE Sample Papers for Class 10 Science has been prepared based on the guess papers issued recently. Students will be able to practice these papers and get good marks in upcoming Science exams for Class 10.
CBSE Sample Papers for Class 10 Science
Class 10 Science Sample Paper Term 1 With Solutions Set A
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) P and R
(b) Carbon has a valency of four or etravalency & 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) Y and Z
(b) W has bigger atomic radius.
Reason: Down the group, number of shells increases.
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) Male gamete (sperm) travels in the female reproductive tract after being released. The path which it takes to fertilise the female gamete (egg) is vagina, uterus, fallopian tube where it fuses with the egg cell resulting in the formation of a zygote (Alternative: labelled figure of human female reproductive system indicating the passage of sperm from vagina to uterus and then to fallopian tube for fertilisation, resulting in the formation of a zygote.)
(b) Zygote has 2 sets of chromosomes.
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 bread mould (Rhizopus) which reproduces by spore formation.
(b) Hyphae or thread like structures are the vegetative part and tiny blob like structures or sporangia are the reproductive parts.
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 reason for parental trait in consecutive generations of the offsprings is self-pollination.
In field B, variation is seen because of recombination of genes as cross-pollination is taking place.
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 traits is likely to have arisen earlier? Give reason.
Ans. Trait Y, which exists in 70% (larger fraction) of the population, is likely to have arisen earlier because in asexual reproduction, identical, copies of DNA are produced and variations do not occur.
New traits come in the population due to sudden mutation and then are inherited. 70 % of the population with trait Y is likely to have been replicating that trait for a 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) downwards
(b) Because BC is in the same direction as the direction of field lines. Force is minimum when the direction of current in the conductor is the same as that of the magnetic field. BC will not contribute as the force on this part of the coil will be cancelled by the force on DA.
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 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 strongest magnetism.
(b) The direction of the field will also be reversed.
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. 1. DDT being a non-biodegradable pesticide will enter the food chain from the first trophic level i.e., Plankton.
2. Non-biodegradable pesticides accumulate progressively at each trophic level. This phenomenon is known as biological magnification.
3. Hawk will have the highest level of pesticide.
OR
In the following food chain, vertical arrows indicate the energy lost to the environment and horizontal arrows indicate the 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. 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 amount of energy is lost as heat in to the environment, some amount goes into digestion and in doing work and the rest goes towards growth and reproduction. An average of 0% of the food eaten is made available for the next level of consumers. his loss of energy takes place at every trophic level.
(Alternatively accept- In accordance with 0% law of transfer of energy in a food chain, only 0% of energy available at one trophic
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 6).
Ans. (a) Silicon
Reason: Tetrahedral structure
(b) Sodium
Reason: It has valence electron like
Lithium
(c) Sulphur
Reason: It forms oxides SO2 and SO3
9. (a) How many isomers are possible for the compound with the molecular formula C4H8 ? Draw electron dot structure of branched chain isomer.
(b) How will you prove that C4H8 and C5H10 are homologous?
Ans. (a) Five
(b) C4H8 and C5H10 are homologous as they differ in
1. “–CH2 –”
2. differ in 4u molecular mass
3. Same functional group
4. Same general formula
OR
A carbon compound ‘A’ having melting point 56K and boiling point 35K, with molecular formula
C2H6O is soluble in water in all proportions.
(a) Identify ‘A’ and draw its electron dot structure.
(b) Give the molecular formulae of any two homologous of ‘A’.
Ans. (a) Ethanol; C2H5OH
(b) CH3OH and C3H7OH are homologous of ethanol. OR
CH4O and C3H8O
10. Two pea plants-one with round yellow (RRYY) seeds and another with wrinkled green (rryy) seeds produce F progeny that have round, yellow (RrYy) seeds.
When F 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 of 60 seeds are produced in F2 generation? Explain with reason.
Ans. Round green: 30
Wrinkled yellow: 30
New combinations are produced because of the independent inheritance of seed shape and seed colour trait.
11. (a) It would cost a man `3.50 to buy .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) E = P × T
So, E = 3 × 2 = 6 kWh
Cost of buying electricity from the main electricity board = 6 x 3.50 = `2.0
(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 2A, what would be the value of R?
Ans. 5 ohm, 0 ohm and R are in parallel connection
1/Rp = 1/5 + 1/10 + 1/R1
1/Rp = (2+1)/10 + 1/R1
= 3/10 + 1/R1
1/Rp = (3R1 + 10)/10R1
Rp = 10R1/(3R1 + 10)
Now, 6 ohm, 6 ohm and Rp are in series
Thus,
Req = 12 + 10 R1/(3R1 + 10) ….(i)
V = I Req
From the circuit
Req = 30/2 = 15 Ω …(ii)
Equating (i) and (ii)
12 + 10 R1 /(3R1 + 10) = 15
10R1/(3R1 + 10) = 3
10R1 = (9R1 + 30)
Thus, R1 = 30 ohm.
OR
Calculate the total resistance of the circuit and find the total current in the circuit.
Ans.
R3 and R4 are in series, hence the equivalent resistance of those two = R5 = R3 + R4 = 0 ohm.
R5 and R2 are in parallel, Let R6 be the equivalent resistance for them. Hence, R6 = (R5R2)/(R5 + R2) = 00/20 = 5 ohm
Now R and R6 are in series and hence the final equivalent resistance of the entire circuit is R = R + R6 = 2 ohm.
By Ohm’s Law, we know that V = IR, hence I = V/R.
Hence the current in the circuit is 24/2 A = 2A
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 980’s.
(a) Identify Gas A. How is it formed at higher levels of the atmosphere?
(b) Why is it essential for all living beings? State the cause for the depletion of this gas.
Ans. (a) Gas A is Ozone. Ozone at the higher levels of the atmosphere is a product of UV radiation acting on oxygen (O2) molecule. The higher energy UV radiations split apart some molecular oxygen (O2) into free oxygen (O) atoms. These atoms then combine with molecular oxygen to form ozone.
UV
O2 → O + O
O + O2 → O3
(b) Ozone shields the surface of the earth and protects living organisms from ultraviolet (UV) radiations released by the sun. Chlorofluorocarbons (CFC’s) which are used as refrigerants in fire extinguishers lead to depletion of ozone layer
Section – 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 F generation.
(a) What set of genes will be present in the F generation?
(b) Give reason, why only tall plants are observed in F progeny.
(c) When FF 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) Tt
(b) Both the recessive and dominant trait are passed on in F progeny. Out of the two trait only one of them is able to express itself in the progeny, which is called the dominant (T), while the other one is called the recessive (t). Here, the ‘tall’ trait is the dominant one.
(c) Out of 800 plants, 600 plants will be tall and 200 plants will be small. :2: (TT:Tt:tt)
OR
When F plants were cross-pollinated with plants having tt genes, a total of 800 plants were produced. How many of these would be tall, medium, or short plants? Give the genotype of F2 generation.
Ans. In the cross between Tt X tt, 400 Tall (Tt) and 400 short (tt) plants will be produced. 1 1:1(Tt:tt)
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 fluctuating induced current in the coil due to relative motion between the magnet and the coil. Accordingly the galvanometer needle will fluctuate on either side of 0 mark. Changing the magnetic field around the coil generates induced current.
(c) Using a stronger magnet, or using a coil with more number of turns.
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 /flux associated with the coil as the magnet moves in and out.
Alternatively, the flux increases when the magnet goes in and it decreases when the magnet goes out.
