Wednesday, October 23, 2013

CBSE Science X Sample Paper 2014[SA_II]

CBSE Science X Question paper of 2011 solution for preparation of 2014
 1. What are the various steps in a food chain called?
Solution: Trophic levels.
2. What is the important function of presence of ozone in earth’s atmosphere?
Solution: The ozone layer prevents the harmful UV rays from entering into the earth’s atmosphere.
3. Write the electron dot structure of ethane molecule, C2H6

Self
4. What makes the earth’s atmosphere a heterogeneous mixture?
Solution: Different gases like  oxygen, nitrogen, argon and CO2 with other gases make the earth’s atmosphere a heterogeneous mixture.
5. List any four characteristics of a good fuel.
Solution: Characteristics of a good fuel are :
i. low ignition temperature.
ii. High calorific value
iii. Environmental friendly.
iv. Easily available and accessible
6. What are non-renewable resources of energy? Give two examples of such resources.
Solution: Non-renewable resources of energy are those energy resources that cannot be replenished in reasonable period. Two examples of non-renewable resources are coal and petroleum.
7. (i) How do you calculate the possible valency of an element from the electronic configuration of its atoms?
(ii) Calculate the valency of element X whose atomic number is 9.
Solution: (i) The electrons present in the last shell determine the valency of a particular element.
If,
Number of valence electrons ≤ 4: valency = number of valence electrons
Number of valence electrons > 4: valency = 8 - number of valence electrons
(ii) The electronic configuration of element X with atomic number 9 is 2,7
Its valency = 8–7 =1
Hence, the valency of element with atomic number 9 is 1.
8. On the basis of electronic configuration, how will you identify the first and the last element of a period?
Solution: If the number of valence electron is 1 then the element is said to be the first element of a period, If the number of valence electrons are 8, the element is said to be the last element of a period.

9. State the two laws of reflection of light.

10. The stars appear higher from horizon than they actually are. Explain why it is so.
Solution:
Stars appear slightly higher from the horizon due to atmospheric refraction. The atmosphere bends the starlight towards the normal. This apparent positioning is not stationary but keeps changing.
11. Explain why the planets do not twinkle but the stars twinkle.
Solution:
The planets being closer to the Earth do not appear as point sources of light. The planet covers a small circular area in space, when viewed from Earth. So, even if light from one point in the circular disc is blocked, light from other points reaches our eyes. As the light from the planets are not completely blocked from our sight and they do not twinkle. Stars on the other hand are point sized objects and twinkle by atmospheric refraction of light.
12. Write any two differences between binary fission and multiple fission in a tabular form as observed in cells of organisms.
Solution: Two differences between binary fission and multiple fission are as follows:
Binary Fission
Multiple fission
The single cell divides into halves.                    
The single cell divides into many daughter cells
Binary fission is of two types; along any
plane or longitudinally.
Multiple fission occurs along only one plane.

Amoeba (along any plane), Leishmania (longitudinally)
Example: Plasmodium

13. Why is DNA copying an essential part of the process of reproduction?
Solution: DNA copying mechanism is significant in reproduction because it enables to make copies of the blueprint of the body design of organism of a particular species. This in turn ensures that the basic body design of organisms of a particular species is same and they look similar.
 14. Explain the terms: (i) Speciation (ii) Natural selection
Solution:
(i) Speciation - Speciation is an evolutionary phenomenon by which new species are formed. There are three models of speciation:
1. When the groups that evolve to be separate species are in different geographic locations and are isolated geographically from each other
2. When the groups that evolve to be separate species are geographic near and the individuals can move one species area to others.
3. When the groups that evolve to be separate species occur together  in the same geographic area.
(ii) Natural selection – Natural selection is a key mechanism in evolution.  It is a process that results in an increased survival and reproductive success of individuals that are well adjusted to the environment. The theory of natural selection was given by Charles Darwin.
15. Explain how equal genetic contribution of male and female parents is ensured in the progeny.
Solution:
The genetic material of a cell is stored in the DNA. During sexual reproduction, the genetic material from both the parents is transferred to the offspring. All human beings have 23 pairs of chromosomes out of which two
chromosomes are sex chromosomes. These sex chromosomes are responsible for determining the sex of a child.
(i) The sex chromosomes are represented by X and Y. Females have two X chromosomes (XX) and males have one X and one Y chromosome (XY)
(ii) The gametes receive half of the chromosomes so the male gametes would be 22 + X or 22 + Y and since females have X chromosomes their gametes have 22 + X chromosomes.
(iii) So the sex of the baby is determined by the type of male gamete which fuses with the female.

16. Out of HCl and CH3COOH, which one is a weak acid and why? Describe an activity to support your answer.
Solution:
CH3COOH is a weak acid when compared to HCl because HCl completely dissociates into ions in solution whereas CH3COOH being weak acid is partially dissociated into ions in solution.
This can be proved by the following activity:
Two iron nails are fitted on a cork and are kept in a 100 mL beaker. The nails  are then connected to the two terminals of a 6-volt battery through a bulb and a switch. Some dilute HCl is poured in the beaker and the current is switched  on. The same experiment is then performed with CH3COOH.
17. Two elements X and Y belong to group 1 and 2 respectively in the same period of periodic table. Compare them with respect to:
(i) the number of valence electrons in their atoms
(ii) their valencies
(iii) metallic character
(iv) the sizes of their atoms
(v) the formulae of their oxides
(vi) the formulae of their chlorides
Solution:
(i) The number of valence electrons in element X is 1 while in element Y is 2.
(ii) The valency of element X is 1 and that of element Y is 2.
(iii) Element X is more metallic than element Y.
(iv) The size of atom X is more than that of atom Y.
(v) The formula of oxide of element X is X2O and that of element Y is YO.
(vi) The formula of chloride of element X is XCl and that of element Y is YCl2.
18. Draw the ray diagram and also state the position, the relative size and the nature of image formed by a concave mirror when the object is placed at the centre of curvature of the mirror.
19. (i) “The refractive index of diamond is 2.42”. What is the meaning of this statement?
(ii) Name a liquid whose mass density is less than that of water but it is optically denser than water.
Solution:
(i) “The refractive index of diamond is 2.42.” This means the ratio of sine of angle of incidence to the sine of angle of refraction is equal to 2.42.
(ii) Kerosene has refractive index of 1.44, it’s mass density is less than that of water but it is optically denser than water (kerosene: 1.44 and water: 1.33).
20. What eye defect is hypermetropia? Describe with a ray diagram how this defect of vision can be corrected by using an appropriate lens.
21. (a) List two sexually transmitted disease in each of the following cases:
(i) Bacterial infections
(ii) Viral infections
(b) How may the spread of such diseases be prevented?
Solution:
(a) Two sexually transmitted disease in each of the following cases are:
(i) Bacterial infections: Syphilis, Gonorrhoea
(ii) Viral infections: AIDS, Genital Herpes
(b) The spread of sexually transmitted diseases can be prevented by using condoms and avoiding multiple partners.
22. Explain Mendel’s law of independent inheritance. Give one example.
Solution:
Mendel’s law of independent assortment states that
(i) When two pairs of traits are combined in a hybrid, one pair of character segregates independent of the other pair of character.
(ii) In a dihybrid cross between two plants having round yellow (RRYY) and wrinkled green seeds (rryy), four types of gametes (RY, Ry, rY, ry) are produced. Each of these segregate independent of each other, each having a frequency of 25% of the total gametes produced.
23. (a) If the image formed by a lens is diminished in size and erect, for all positions of the object, what type of lens is it?
(b) Name the point on the lens through which a ray of light passes undeviated.
(c) An object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. Find (i) the position (ii) the magnification and (iii) the nature of the image formed.
Solution:
(a) If the image formed by a lens is diminished in size and erect, for all positions of the object, then the lens is a concave lens.
(b) The point on the lens through which a ray of light passes undeviated is known as Pole.
(c) 1/f = 1/v + 1/u => 1/20 = 1/v + 1/(-30)  => v = 60cm
The image is formed at a distance of 60 on the other side of the optical centre
(ii) m = -v/u = -60/-20 = 2
(iii)Image formed is inverted.
OR
(a) One-half of a convex lens is covered with a black paper. Will such a lens produce an image of the complete object? Support your answer with a ray diagram.
(b) An object 5 cm high is held 25 cm away from a converging lens of focal length 10 cm.
(i) Draw the ray diagram and
(ii) Calculate the position and size of the image formed.
(iii) What is the nature of the image?
Solution: (a) if one half of the lens is darkened, the image will form. However, the brightness of the image will be less as compared to that of the image formed without wrapping the lens.
(b) (i) Draw the ray diagram
(ii) 1/f = 1/v + 1/u => 1/10 = 1/v + 1/(-25)  => v = 16.67cm
 (iii)  m = -v/u = -hi/h  => hi =  (v xho)/u =  (16.67 x 5)/(-25) = - 3.34 cm
 Negative sign indicates that the image is real & inverted.
24. (a) Draw a diagram of the longitudinal section of a flower and label on it sepal, petal, ovary and stigma.
(b) Write the names of male and female reproductive parts of a flower.
Solution: (b) The male reproductive part of flower is called stamen that consists of filaments and anthers.
The female reproductive part of the flower is pistil that consists of stigma, style and ovary.
OR
(a) What is fragmentation in organisms? Name a multicellular organism which reproduces by this method.
(b) What is regeneration in organism? Describe regeneration in Planaria with the help of a suitable diagram.
Solution: (a) Fragmentation is a mode of asexual reproduction. It is the unintentional cutting up of the body of an organism. Each fragment develops into a new individual. Example: Spirogyra
(b) Regeneration: Regeneration involves the capacity of an organism to give rise to an entire individual from a cut portion. This type of regeneration occurs in Planaria. When Planaria gets cut unintentionally, the cut fragments give rise to a new individual. In this type, active cell division occurs before the replacement. The newly formed mass undergoes differentiation to form specialized cells.
 25. (a) In a tabular form, differentiate between ethanol and ethanoic acid under the following heads:
(i) Physical state   (ii) Taste (iii) NaHCO3 test (iv) Ester test
(b) Write a chemical reaction to show the dehydration of ethanol 
Solution: (a)
(i) Physical state  : Ethanol is a colourless liquid with pleasant odour where as Ethanoic acid is colourless, pungent smelling liquid
(ii) Taste : Ethanol is bitter to taste where as  Ethanoic acid is sour to taste
(iii) NaHCO3 test: Ethanol does not react with sodium bicarbonate where as When ethanoic acid reacts with sodium NaHCO3 with the evolution of carbon dioxide gas.
(iv) Ester test Ethanol on reaction with ethanoic acid in the presence of acid forms ester where as Ethanoic acid on reaction with ethanol in the presence of acid forms ester
(b) Ethanol undergoes dehydration to form ethane.
2CH3CH2 OH ---D-------à 2CH2 = CH2 + 2H2O
OR,
(a) What is a soap ? Why are soaps not suitable for washing clothes when the water is hard?
(b) Explain the action of soap in removing an oily spot from a piece of cloth.
Solution:
(a) A soap can be defined as a sodium or potassium salt ofhigher fatty acids such as oleic acid (C17H33COOH), stearic acid (C17H35COOH), palimitic acid (C15H31COOH), etc.
Soap does not work properly when the water is hard. A soap is a sodium or potassium salt of long chain fatty acids. Hard water contains salts of calcium and magnesium. When soap is added to hard water, calcium and magnesium ions present in water displace sodium or potassium ions from the soap molecules forming an insoluble substance called scum. A lot of soap is wasted in the process.
(b) Cleansing action of soaps:
The oily spot present on clothes is organic in nature and insoluble in water. Therefore, it cannot be removed by only washing with water. When soap is dissolved in water, its hydrophobic ends attach themselves to the oily spot and remove it from the cloth. Then, the molecules of soap arrange themselves in the form of micelle and trap the dirt at the centre of the cluster. These micelles remain suspended in the water. Hence, the oily spots are easily rinsed away by water.
Section: B
1. A student was given two permanent slides, one of binary fission in amoeba and other of budding in yeast. He was asked to identify any one difference in the nucleus of the two. One such difference, he identified correctly was
(a) Presence of one nucleus in amoeba, two in yeast cell and one in bud.
(b) Presence of two nuclei in centrally constricted amoeba, one in yeast cell and one in its bud.
(c) Presence of two distant nuclei in amoeba, one in yeast cell and two in bud.
(d) Presence of a single nucleus each in amoeba, yeast cell and its attached bud.
Solution: (b)
2. To determine the percentage of water absorbed by raisins, raisins are soaked in water for:
(a) 30 seconds
(b) 10 minutes
(c) 2 to 3 hours
(d) 24 hours
Solution: (c)
3. Raisins are wiped off gently before final weighing with help of
(a) a filter paper
(b) a cotton piece
(c) a cloth piece
(d) a polythene piece
Solution: (a)
4. The step(s) necessary for determining the percentage of water absorbed by raisins is/are:
(a) Raisins should be completely immersed in water
(b) Raisins should be soaked in water for sufficient time
(c) Gently wipe dry the soaked raisins
(d) All of the above steps.
Solution: (d)
5. Mohan obtained a sharp inverted image of a distant tree on the screen placed behind the lens. He then moved the screen and tried to look through the lens in the direction of the object. He would see:
(a) a blurred image on the wall of the laboratory.
(b) an erect image of the tree on the lens.
(c) no image as the screen has been removed
(d) an inverted image of the tree at the focus of the lens.
Solution: (a)
6. For finding the focal length of a convex lens by obtaining the image of a distant object, one should use as the object.
(a) a well lit distant tree
(b) window grill in the class room
(c) any distant tree
(d) a lighted candle kept at the other end of the table.
Solution: (a)
7. To find the focal length of a concave mirror Rahul focuses a distant object with this mirror. The chosen object should be
(a) a tree
(b) a building
(c) a window
(d) the sun
Solution: (d)
8. The colour of an aqueous solution of zinc sulphate as observed in the laboratory is:
(a) Green
(b) Yellow
(c) Blue
(d) Colourless
Solution: (d)
9. To show that zinc is a more active metal than copper, the correct procedure is to:
(a) add dilute nitric acid on strips of both the metals.
(b) observe transmission of heat through strips of zinc and copper.
(c) prepare solution of zinc sulphate and hang strip of copper into it.
(d) prepare solution of copper sulphate and hang strip of zinc into it.
Solution: (d)
10. Acetic acid smells like:
(a) a banana
(b) vinegar
(c) an orange
(d) a lemon
Solution: (b)
11. Acetic acid solution turns:
(a) blue litmus red
(b) red litmus blue
(c) blue litmus colourless
(d) red litmus colourless
Solution: (a)
12. On adding NaHCO3 to acetic acid, a gas is evolved which turns lime water milky due to the formation of:
(1) Calcium bicarbonate
(2) Calcium hydroxide
(3) Calcium carbonate
(4) Calcium acetate
Solution: (c)
13. A yeast cell in which budding occurs was seen to have:
(a) one bud cell
(b) two bud cell
(c) three bud cell
(d) a chain of bud cells
Solution: (a)
14. Rahim recorded the following sets of observations while tracing the path of a ray of light passing through a rectangular glass slab for different angles of incidence.

S. No.
<i
<r
<e
I
450
410
450
I I
400
380
380
III
450
410
400
IV
410
450
410






(a) I
(b) II
(c) III
(d) IV
Solution: (a) because while light is passing through the rectangular glass slab angle of refraction is less than angle of incidence. 
For more paper :  CBSE X Sample Paper 2014

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