Please refer to Class 10 Science Sample Paper Term 2 Set A with solutions below. The following CBSE Sample Paper for Class 10 Science has been prepared as per the latest pattern and examination guidelines issued by CBSE. By practicing the Science Sample Paper for Class 10 students will be able to improve their understanding of the subject and get more marks.
CBSE Class 10 Science Sample Paper for Term 2
SECTION – A
1. The diagram below shows the electron arrangement in a compound formed between element X and fluorine.
(a) (i) What is the formula of this compound?
(ii) Is this an ionic or covalent compound? Give your reason.
(b) In which group of the periodic table can you find element X ?
Ans. (a) (i) XF4
(ii) It is a covalent compound.
The diagram shows that an atom of X shares four pairs of electrons with four atoms of fluorine to form the molecule XF4. Sharing of electrons is a characteristic of covalent compound.
(b) Element X has four valence electrons hence, it belongs to group 14 of the periodic table.
2. Consider a metal M with electronic configuration : 2, 8, 3.
(a) To which group and period of periodic table does M belong?
(b) Write the formulae of its sulphate and phosphate.
Ans. (a) As, there are 3 electrons in valence shell hence M belongs to group 13. Also, as electrons are filled in 3 shells, (K, L, M) thus ‘M’ belongs to 3rd period.
(b) As, ‘M’ has 3 electrons in its valence shell, thus it contributes 3 electrons in bonding. Formulae for sulphate and phosphate are M2(SO4)3 and MPO4 respectively.
3. Name two organisms each which reproduce by fragmentation, regeneration and budding.
Ans. Fragmentation — Spirogyra and starfish
Regeneration — Hydra and Planaria
Budding — Hydra and yeast
4. Colonies of yeast fail to multiply in water, but multiply in sugar solution. Give one reason for this.
Ans. Yeast multiplies by the process of budding. It requires energy for sustaining all its life activities. Sugar provides energy to yeast whereas water does not provide energy, so it fails to multiply in water because of inadequate energy.
5. In human beings, the statistical probability of getting either a male or female child is 50 : 50. Justify.
Ans. Human females are homogametic (44 + XX), that is they produce only one type of ova (22 + X). Human males are heterogametic. They produce two types of sperms : (22 + X) and (22 + Y) in equal proportion that is 50 : 50 ratio. The chance of getting male or female child is also 50 : 50, as there is equal chance of androsperm (22 + Y) or gynosperm (22 + X) fertilising an ovum.
OR
(a) Mendel selected garden pea plant as an experimental plant due to its advantage over other plants. Write any two selection criteria for selecting garden pea plant.
(b) What do you mean by a true breeding plant?
Ans. (a) The two selection criterias are as follows :
(i) Garden pea plant has short life cycle which makes it possible to study several generation in a short time period.
(ii) Garden pea plant shows several well defined contrasting characters such as plant height, pod and flower colour, etc.
(b) A true breeding plant is the one that when selffertilised, produces offspring with the same traits. They will be either homozygous dominant or homozygous recessive.
6. Can you think of magnetic field configuration with three poles?
Ans. Magnetic poles always exist in pairs. However one can imagine magnetic field configuration with three poles – when North poles of two magnets are glued together or South poles of two magnets are glued together to provide three field configuration.
OR
What are magnetic field lines? Justify the following statements : Magnetic field are closed curves.
Ans. Magnetic field lines are curved imaginary lines used to show the magnetic field in a given region.
It is taken by convention that the field lines emerges from North pole and merge at the South pole. Inside the magnet, the direction of field lines is from its South pole to its North pole. Thus, the magnetic field lines are closed curves.
7. The following organisms form a food chain.
Insect, Hawk, Grass, Snake, Frog
Which of these organisms will have the highest concentration of non-biodegradable chemicals? Name the phenomenon.
Ans. Highest concentration of harmful nonbiodegradable chemicals are found in top consumer hawk. The phenomenon exhibited here is biomagnification.
OR
Ecosystem is a community where all the living organisms interact with the nonliving constituents of the environment. Is an aquarium a complete ecosystem? Explain.
Ans. Yes, a balanced and large aquarium can be a complete ecosystem consisting of both biotic and abiotic components. Water, oxygen supply source, light source are abiotic factors, whereas aquatic plants, small animals and decomposers serve as biotic components.
SECTION – B
8. Draw the structures of the following compounds :
(i) Acetic acid (ii) Chloroform (iii) Ethanol
Ans.
9. Two elements A and B belong to group 1 and 2 respectively in the same period. Compare them with respect to :
(a) the number of valence electrons
(b) valency
(c) metallic character
(d) size of the atoms.
Ans. (a) Number of valence electrons in A = 1 and B = 2.
(b) Valency of A is one while B is two.
(c) Element A (group 1) is more metallic as compared to B (group 2).
(d) Element B (group 2) is smaller than element A (group 1) in size.
OR
Four elements A, B, C and D along with their configurations are given below : Element A – 2, 1; Element B – 2, 8; Element C – 2, 8, 1; Element D – 2, 8, 8.
(i) Which two elements belong to the same period?
(ii) Which two elements belong to the same group?
(iii) Which element out of A and C is more reactive and why?
(iv) Which element out of A and B forms more number of compounds?
Ans. (i) Elements in a period have same number of shells. Thus, elements A and B belong to the second period while C and D belong to the third period.
(ii) Elements in a group have same number of valence electrons. Thus, elements A and C belong to the same group 1, and elements B and D belong to the same group 18.
(iii) Element C is more reactive than element A due to its bigger size and weaker force of attraction of the nucleus on the valence electrons.
(iv) Elements B, being a noble gas does not form compounds. Hence, element A forms more number of compounds than element B.
10. Mendel performed several experiments that include monohybrid, dihybrid and trihybrid crosses. On the basis of these experiments, mendel formulated laws of heredity. How do Mendel’s experiment shows that traits are inherited independently?
Ans. In a dihybrid cross given by Mendel, it was observed that when two pairs of traits or characters were considered, each trait expressed independent of the other. Thus, Mendel was able to propose the Law of Independent Assortment, which state that pair of genes separate independently of each other during gamete formation. This could be explained clearly
from the given cross:
F2 generation ratio : 9 (Round-yellow) : 3 (Round-green) : 3 (Wrinkled-yellow) : 1 (Wrinkled-green)
11. Write Joule’s law of heating.
Ans. The Joule’s law of heating implies that heat produced in a resistor is
(i) directly proportional to the square of current for a given resistance,
(ii) directly proportional to resistance for a given current, and
(iii) directly proportional to the time for which the current flows through the resistor.
i.e., H = I2Rt
12. In a house 4 bulbs of 100 watt each are lighted for 5 hours daily, 2 fans of 50 watt each are used for 10 hours daily and a refrigerator of 1.00 kW is used for half an hour daily. Calculate the total energy consumed in a month of January and its cost at the rate of ` 3.60 per kWh.
Ans. Power of each bulb = 100 watt
Total power of 4 bulbs, P1 = 4 × 100 = 400 watt
Energy consumed by bulbs in 1 day
E1 = P1 × t = 400 watt × 5 hours = 2000 Wh = 2 kWh
Power of each fan = 50 watt
Total power of 2 fans = 2 × 50 watt P2 = 100 watt
Energy consumed by fans in 1 day
E2 = P2 × t = 100 watt × 10 hours
= 1000 watt hour = 1 kWh
Energy consumed by refrigerator,
E3 = 1 kW × 1/2 h = 0.5 kWh
Total energy consumed in one day
E = E1 + E2 + E3 = (2 + 1 + 0.5) kWh = 3.5 kWh
Total energy consumed in a month of January (31 days)
= E × 31 = (3.5 × 31) kWh = 108.5 kWh
Cost of energy consumed = ` (108.5 × 3.60) = ` 390.6
OR
Compare the power used in 2 W resistor in each of the following circuits.
Ans. In circuit A, Total resistance, R = 1 + 2 = 3 Ω
Voltage across 2 Ω = V Total / R Total × 2 Ω = 6 / 3 × 2 = 4 V
∴ Power used in 2 Ω resistor, P = V2 / R = (4)2 / 2 = 8 W
In circuit B, voltage across both the resistance is same i.e. 4 V and both are connected in parallel combination.
∴ Power used in 2 Ω resistor = V2 / R = (4)2 / 2 = 8 W
∴ Power used in 2 Ω resistor in each case is same i.e., 8 W.
13. How the present environment is negatively effected due to modern lifestyle ?
Ans. With the advancement in technology over time, there has been improvement in lifestyle of people. Such changes have also changed their attitudes. When people have more resources at their end they tend to overuse and misuse it thereby generating huge amounts of waste materials. For example, the affluent lifestyle has forced people to start using more of disposable articles, e.g., plastic cups, bags, etc., which keep on accumulating in the environment and lie undecomposed, thereby negatively affecting the environment.
Similarly, excessive use of refrigerators and air conditioners, plastic foams, etc., also release high quantities of CFCs which are responsible for ozone depletion.
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. Asexual reproduction is a way by which organisms can give rise to new organisms. It is a mode of reproduction that involves a single parent and take place in unicellular as well as in multiculluar organisms.
(a) Name the mode of reproduction of the following organisms and state the important feature of each mode :
(i) Planaria
(ii) Hydra
(iii) Rhizopus
(b) We can develop new plants from the leaves of Bryophyllum. Comment.
(c) List two advantages of vegetative propagation over other modes of reproduction.
Ans. (a) (i) Planaria – Regeneration
1. Regeneration of organism from its cut body parts occurs by the process of growth and development.
2. Regeneration is an asexual mode of reproduction common in lower plants and animals.
(ii) Hydra – Budding
1. In budding, a small part of the body of the parent organism grows out as a bud which on detaching forms a new organism.
2. Budding occurs in yeast, some protozoans and certain lower animals.
(iii) Rhizopus – Spores
1. Spores are usually produced in sporangia.
2. Spore formation is a common method of an asexual reproduction in bacteria and most of the fungi.
(b) The leaves of a Bryophyllum have special type of buds in their margins. These buds may get detached from the leaves, fall to ground and then grow to produce new Bryophyllum plants. The buds can also drop to the ground together with the leaf and then grow to produce new plants.
(c) Advantages of vegetative propagation are :
1. It is a quick method of propagation.
2. The new plants produced by artificial vegetative propagation are exactly like the parent plants.
3. Many plants can be grown from one plant by vegetative propagation.
OR
Differentiate binary fission from multiple fission and give one example of each.
Ans. In binary fission the parent organism splits into two new organisms and the nucleus of the parent body divides only once to produce two nuclei, whereas in multiple fission the parent organism splits into many new organisms and the nucleus of the parent body divides repeatedly to produce many nuclei. Example of binary fission – Leishmania Example of multiple fission – Plasmodium.
15. When current is flowing through the solenoid, the magnetic field line pattern resembles exactly with those of a bar magnet with the fixed polarity, i.e., North and South pole at its ends and it acquires the repulsive and attractive properties similar to bar magnet. Hence, the current carrying solenoid behaves as a bar magnet.
(a) What are the uses of solenoid?
(b) Draw magnetic filed lines for the current carrying solenoid as shown in figure and explain that out of three points A, B and C, at which point, magnetic field strength is maximum.
(c) How does a solenoid behave like a magnet? Can you determine the North and South poles of a currentcarrying solenoid with the help of a bar magnet? Explain.
Ans. (a) Solenoid are used to form temporary magnet called electromagnet. The electromagnet has more advantages over permanent bar magnet as electromagnets can be turned on and off making them effective as controls and switches.
(b) Outside the solenoid magnetic field is minimum. At the ends of solenoid, magnetic field strength is half to that inside. So maximum field strength is at point B.
(c) When current is passed through a solenoid, it acts as a bar magnet. Suspend a bar magnet from a support. Bring a current-carrying solenoid near the North pole of the bar magnet. Observe if it gets attracted or repelled.
Now, perform the same action for south pole of the bar magnet.
Following the rule that opposite poles attract each other, one can identify the poles of the solenoid.
OR
Explain the phenomenon of electromagnetic induction briefly.
Ans.
To explain electromagnetic induction take two different coils of copper wire having large number of turns (say 100 and 50 turns respectively). Insert them over a non-conducting cylindrical roll, as shown in figure. Connect the coil-1, having large number of turns, in series with a battery and a plug key. Also connect the other coil-2 with a galvanometer as shown.
Plug in the key. Observe the galvanometer. You will observe that the needle of the galvanometer instantly jumps to one side and just as quickly returns to zero, indicating a momentary current in coil-2. Hence, the change in magnetic field lines associated with the coil-2 is the cause of induced electric current in it.
