Class 10 Science Sample Paper Term 2 Set C
Please see below Class 10 Science Sample Paper Term 2 Set C 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 C
SECTION – A
1. A current of 10 A flows through a conductor for two minutes.
(a) Calculate the amount of charge passed through any area of cross section of the conductor.
(b) If the charge of an electron is 1.6 × 10–19 C, then calculate the total number of electrons flowing.
Ans. Given that : I = 10 A, t = 2 min = 2 × 60 s = 120 s
(a) Amount of charge Q passed through any area of cross-section is given by I = Q / t
or Q = I × t ∴ Q = (10 × 120) A s = 1200 C
(b) Since, Q = ne
where n is the total number of electrons flowing and e is the charge on one electron
∴ 1200 = n × 1.6 × 10–19
or n = 1200 / 1.6 × 10–19 = 7.5 x 1021
2. (a) List three factors on which the resistance of a conductor depends.
(b) Write the SI unit of resistivity.
Ans. (a) Resistance of a conductor depends upon the following factors:
(i) Length of the conductor : Greater the length (l) of the conductor more will be the resistance (R).
R ∝ l
(ii) Area of cross-section of the conductor: Greater the cross-sectional area of the conductor, less will be the resistance.
R ∝ 1 / A
(iii) Nature of conductor.
(b) SI unit of resistivity is Ω m or ohm-meter.
In the given circuit calculate
(a) the total resistance of the circuit
(b) current flowing through the circuit
(c) potential difference across the lamp and resistor.
Ans. (a) Total resistance Rs = 18 + 6 = 24 W
(b) Current I = V Rs = 6 / 24 = 0.25 A
(c) Potential difference across lamp = 0.25 × 18 = 4.5 V
∴ Potential difference across resistor
= 6 – 4.5 V = 1.5 V
3. Compare the powers used in 2 W resistor in each of the following two 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.
4. What are magnetic field lines? Magnetic field are closed curves. Justify.
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.
5. Draw a diagram representing the magnetic field inside and outside a solenoid through which a current is flowing and mark with arrows, the direction of the current in the solenoid and the direction of the magnetic lines of force. Also mark the polarity at the faces of the solenoid.
Figure shows the magnetic lines of force due to current carrying solenoid. The direction of current in the solenoid at the face A is clockwise, so it will have the south (S) polarity and the face B of the solenoid will have the north (N) polarity.
A rectangular coil ABCD is placed between the pole pieces of a horse-shoe magnet as shown in figure.
(a) What is the direction of force on each arm?
(b) What is the effect of the forces on the coil?
Ans. (a) In figure, the current in the coil is in direction DCBA. By Fleming’s left hand rule, in the arm AB, the force is outward at right angle to the plane of the coil. On the arm BC no force acts. On the arm CD, the force is inwards perpendicular to the plane of the coil. On the arm DA, no force acts.
(b) The force on the arms AB and CD are equal in magnitude, but opposite in direction. They form a clockwise couple. So the coil will rotate clockwise with the arm AB coming out and the arm CD going in.
6. Explain different ways to induce current in a coil.
Ans. Current is induced in a coil in the following ways:
(i) When a magnet is moved towards or away from coil or there is a relative motion between coil and magnet, a current is induced in the coil circuit.
(ii) When a current passing through a coil changes, then a current is induced in a coil placed near it.
7. What is an ecosystem? Name two types of ecosystem.
Ans. Ecosystem is the structural and functional unit of the environment comprising of the living organisms and their non-living components that interact by means of food chains and chemical cycles resulting in energy-flow, biotic diversity and material cycling to form a stable system. Natural and artificial are the two types of ecosystem.
What do you mean by biological magnification?
Ans. Biomagnification is the process of increase in amount of some toxic, non-biodegradable substances such as DDT and heavy metals in successive trophic levels of a food chain. It results in accumulation of highest concentration of these toxins in topmost trophic level.
SECTION – B
8. Two elements A and B belong to the 3rd period of modern periodic table and are in group 2 and 13 respectively. Compare their following characteristics in tabular form.
(a) Their tendencies to lose electrons
(b) Their metallic characters
(c) The formula of their oxides and chlorides
Ans. Electronic configuration of A = 2, 8, 2 i.e., Mg
Electronic configuration of B = 2, 8, 3 i.e., Al
9. The molecular formulae of alkenes X and Y are CxH8 and C10Hy respectively. What are the relative molecular masses of X and Y?
Ans. The general formula of an alkene is CnH2n.
Thus, the molecular formula of X is C4H8, and the molecular formula of Y is C10H20.
Molecular mass of X = (4 × 12) + (8 × 1) = 56 u
Molecular mass of Y = (10 × 12) + (20 × 1) = 140 u
What is the maximum number of structural isomers possible for pentane with molecular formula C5H12?
Ans. The maximum number of isomers for a pentane is 3.
10. Sexually transmitted diseases (STDs) have increasingly become a threat to public health due to changing pattern of sexual behaviour and increasing resistance of pathogens to the antibiotics. What are sexuallytransmitted diseases? How can the transmission of such diseases prevented?
Ans. The infectious (communicable) diseases, which spread from an infected person to a healthy person by sexual contact, are called sexually transmitted diseases. These include bacterial infections such as gonorrhoea and syphilis, and viral infections such as warts and HIV-AIDS. Using a covering, called a condom, for the penis during sex helps to prevent transmission of many of these infections to some extent.
11. Explain the process of spore formation in fungi. How spore formation differs from budding?
Ans. Spore formation is the most common method of asexual reproduction seen in fungi and bacteria. During spore formation, a structure called sporangium develops from the fungal hypha. The nucleus divides several times within the sporangium and each nucleus with a bit of cytoplasm, develops into a spore. The spores are liberated and they develop into new hypha after reaching the ground. e.g., Rhizopus, Mucor and Penicillium.
Difference between budding and spore formation are as follows:
12. (a) Do larger organisms have more number of chromosomes/cells?
(b) Can organism with fewer chromosomes reproduce more easily than organisms with more number of chromosomes ?
(c) ‘‘More the number of chromosomes/cells, greater is the DNA content’’. Justify the statement.
Ans. (a) No, there is no relationship between size of organism and its chromosome number.
(b) No, process of reproduction follows a common pattern and is not dependent on the number of chromosomes.
(c) The major component of chromosome is DNA thus, if there are more number of chromosomes in a cell, the quantity of DNA will also be more.
A tall and a short pea plant are crossed and all tall F1 progenies are obtained. F1 tall plants are selfed and produced tall and dwarf plants in 3 : 1 ratio in F2. Find out the genotype of F1 tall plant.
Ans. As one dwarf plant has been produced in F2 generation and dwarf character is recessive, dwarf F2 plant must have gained ‘t’ gene from both its parents. Thus, F1 plant must have one ‘t’ gene. And, as F1 plants have also produced tall plants with gene T, thus, it must have gene ‘T’ in its genotype. Thus F1 plant has a ‘T’ and a ‘t’ gene and its genotype will be Tt i.e., hybrid tall.
13. (a) Producers always occupy the first trophic level in any food chain. Why?
(b) In the given food chain, how much energy will be available to the lion, if 10,000 J of energy is available to the producers?
Plants → Deer → Lion
Ans. (a) Producers or green plants have chlorophyll which can trap the solar energy. The first trophic level in a food chain is a producer, i.e., those organisms which can produce food with the help of sunlight and chlorophyll by a process called photosynthesis.
(b) As per 10% law of flow of energy in an ecosystem, only 10% of energy is received by the next trophic level. Hence, in the given food chain :
Plants → Deer → Lion
10,000 J 1000 J 100 J
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. The distance between the centre of the nucleus and the outermost shell of electrons is known as atomic radius. On moving from left to right along a period, atomic radii decrease because effective nuclear charge increases. For example, the atomic size decreases regularly from Li to F in the second period and from Na to Cl in the third period. It may, however, be noted that in any period, the noble gas has the largest size. On moving down in a group, atomic radii increase.
(a) Which element has the smallest size in the third period of the periodic table?
(b) Write the name of the element having smallest size in group 13?
(c) Write the correct order of increasing atomic radii for the following elements : O, C, F, Cl, Br. Give reason.
Ans. (a) Atomic size decreases across the period but, Cl has smaller size than Ar. Argon has larger atomic size due to van der Waals’ radius.
(b) Boron is the first element of group 13, hence it is smallest in size.
(c) Atomic size decreases from left to right in a period and increases from top to bottom in a group. Thus, the order is F < O < C < Cl < Br.
Out of Al, Si, P, Mg, which element has maximum and minimum atomic radii and why?
Ans. Mg has maximum atomic radius and P has minimum atomic radius, as, on moving from left to right along a period nuclear charge increases which tends to pull the electrons closer and reduces the size of atom, while on moving down the group new shells are being added and this increases the distance between the outermost electron and nucleus, hence atomic size increases.
15. Flower is the most colourful and beautiful part of the plant. In the early stage, the plants are just leafy and flowers are borne when they reproduce, therefore flowers are the reproductive parts of plant body.
(a) Name the essential and non-essential parts of a flower.
(b) Write the term given to the flower containing only one of the reproductive parts. Give some examples.
(c) Fertilization in flowering plants is not possible without pollination. Justify.
Ans. (a) Sepals and petals form the non-essential parts of a flower whereas stamens and carpels constitute the essential parts of a flower.
(b) Unisexual flowers bear organs of only one sex i.e., either stamen or pistil. e.g., Papaya, Watermelon.
(c) Pollination allows pollen grains to reach carpel which contains the egg. Thus, fertilization which involves fusion of male and female germ cells can occur only after pollination.
Flowers of Hibiscus plant produce fruits even after removal of one of its reproductive part. What could be the possible reason behind the situation?
Ans. Stamens and carpels are the male and female reproductive parts of a flower respectively. Hibiscus is a bisexual flower. When the stamens of its flower were removed, the flower still bears the female reproductive part, that is, carpel. Through cross pollination, fertilization could take place and flower can produce fruit after fertilization.