SCIENCE
SET I
Section – A
Q. 1. Write title number of covalent bonds in the molecule of ethane.
Answer.
Q. 2. Name the life process of an organism that helps in the growth of its population.
Answer. Reproduction is the process that helps in the growth of population of an organism.
Q. 3. What will be the amount of energy available to the organisms of the 2nd trophic level of a food chain, if the energy available at the first trophic level is 10,000 joules?
Answer. According to 10% law, only 10% of energy entering a particular trophic level of organisms is available for transfer to the next higher trophic level.
If energy in the 1st trophic level is 10000 J.
Then, energy in 2nd trophic level = 10% of 10000 J = 10/100 x 10000 J = 1000 Joules
Q. 4. The absolute refractive indices of glass and water are 4/3 and 3/2 respectively. If the speed of light is 2 x 108 m/s, calculate the speed of light in (i) vacuum, (ii) water.
Answer.
Q. 5. List -two main causes of the pollution of water of the river Ganga. State how pollution and contamination of river water prove harmful for the health of the people of neighbouring areas.
Answer. Two main cause of the pollution of water of the river Ganga :
(i) Disposal of industrial effluents into tire river.
(ii) Human activities like bathing, washing, immersion of ashes; etc.
Pollution and contamination of river wafer has led to various harmful effects on health of people in neighbouring areas who use river water for domestic or drinking purposes. It has led to various types of water borne diseases such as typhoid or cholera. Also human and industrial activities have lead to contamination of marine life which can be hazardous for consumption.
Q. 6. What is biodiversity? What will happen if biodiversity of an area is not preserved? Mention one effect of it.
Answer. Biodiversity is the diversity of life in an area. It is the number and range of variety of species of life forms in an area.
The loss of biodiversity of an area will lead to the loss of ecological stability of the eco- system. It can result in environment disasters such as floods, forest fires, etc.
Q. 7. List two tests for experimentally distinguishing between an alcohol and a carboxylic acid and describe how these tests are performed.
Answer. Test-1—Litmus Test. Take 2 strips of blue litmus paper. Place a drop each of alcohol and carboxylic acid on these strips separately. The blue litmus paper turns red in the case of carboxylic acid and remains unaffected in the case of alcohol.
Test-2—Sodium hydrogen carbonate Test /Sodium carbonate Test. A pinch of sodium hydrogen carbonate or sodium carbonate is added to both alcohol and a carboxylic acid separately. If brisk effervescence with the evolution of a colourless gas is observed, it ! indicates the presence of carboxylic acid whereas no effervescence is seen in case of an alcohol.
Q. 8. Draw the electron-dot structure for ethyne. A mixture of ethyne and oxygen is burnt for welding. In your opinion, why cannot we use a mixture of ethyne and air for this purpose?
Answer. In pure oxygen, ethyne undergoes complete combustion and high temperature suitable for welding is attained. Whereas air contains less percentage/amount of oxygen which results in incomplete combustion of ethyne and the temperature required for the welding is not attained. Hence we cannot use a mixture of ethyne and air for this purpose.
Q. 9. Two elements ‘P’ and ‘Q’ belong to the same period of the modern periodic table and are in Group-1 and Group-2 respectively. Compare their following characteristics in tabular form:
(a) The number of electrons in their atoms
(b) The sizes of their atoms
(c) Their metallic characters
(d) Their tendencies to lose electrons
(e) The formula of their oxides
(f) The formula of their chlorides
Answer.
| Characteristics | P | Q |
| (a) No. of electrons in their atoms | Less than Q31119 | More than P41220 (any one pair) |
| (b) Size of the atom | Bigger | Smaller |
| (c) Metallic character | More metallic | Less metallic |
| (d) Tendency to lose electrons | More | Less |
| (e) Formula of their oxides | P2O | QO |
| (f) Formula of their chlorides | PCl | QCl2 |
Q. 10. Taking the example of an element of atomic number 16, explain how the electronic configuration of the atom of an element relates to its position in the modern periodic table and how valency of an element is calculated on the basis of its atomic number.
Answer. Atomic number of the element = 16 Electronic configuration = K L M = 2, 8, 6
Since this element has 3 shells, the period number will be 3 as period number is equal to the number of shells that start filling up in it.
No. of valence electrons = 6
The group number will be 10 + 6 = 16
The valency of an element is determined by the number of valence electrons present in the outermost shell.
Valency of the element will be = 8 – valence electrons = 8-6 = 2
Q. 11. List six specific characteristics of sexual reproduction.
Answer. Characteristics of sexual reproduction are:
- In sexual reproduction, two parents are involved (male and female).
- The new organism produced is genetically different from both parents.
- During gamete formation meiosis occurs. After fertilisation all divisions are mitotic.
- Sexual reproduction helps in evolution.
- Fertilisation of gametes leads to zygote formation. This zygote grows and develops to form a new organism.
- Humans, fish, dogs, hens, cats, cows, horses, deer, rabbit, lions and tigers all reproduce by the method of sexual reproduction. Most of the flowering plants also reproduce by sexual reproduction.
Q. 12. What are chromosomes? Explain how in sexually reproducing organisms the number of chromosomes in the progeny is maintained.
Answer. Chromosomes are thread-like structures made up of DNA and proteins. They are found in the nucleus at the time of cell division.
During gamete formation meiosis occurs and the original number of chromosomes becomes half. Hence, when two gametes combine the zygote formed contains the full set of chromosomes and the original number of chromosomes gets restored in the progeny.
Q. 13. List four points of significance of reproductive health in a society. Name any two areas related to reproductive health which have improved over the past 50 years in our country.
Answer. The significance of reproductive health of society are:
- Regular medication and check-ups have led to development of reproductive health.Healthy mothers give birth to healthy children.
- Reproductive health should be maintained in order to prevent sexually transmitted diseases (STDs).
- Family planning by using various contraceptives enables a couple to decide on the number of children they want to have and when to have them. If a couple has less number of children they can provide good food, clothes and education to each child. So a small family is a happy family.
- Having fewer children also keeps the mother in good health. This will reduce the cases of maternal mortality as well as new born mortality.
Areas which have improved:
- Better family planning has led to reduction in family size and better economic stability.
- Decrease in STD cases due to more awareness and wider use of contraceptives.
Q. 14. Explain with an example for each, how the following provides evidences in favour of evolution in organisms: (a) Homologous organs (b) Analogous organs (c) Fossils
Answer.
(a) Homologous organs : Homologous organs are defined as the organs of different animals that are having a similar structure but differ in their functions. The structure of the two different animals resemble but the functions of their organs vary.
(b) Analogous organs : Analogous organs are defined as the organs of different animals that are having different structure but perform the same functions.
(c) Fossils : Fossils are the preserved remains, or traces of remains, of ancient organisms. Fossils are not the remains of the organism itself! They are rocks.
Q. 15. Explain the following:
(a) Speciation (b) Natural Selection
Answer.
(a) Speciation : Speciation is a process within evolution that leads to the formation of new, distinct species that are reproductively isolated from one another.
Anagenesis, or ‘phyletic evolution’, occurs when evolution acts to create new species, which are distinct from their ancestors, along a single lineage, through gradual changes in physical or genetic traits. In this instance, there is no split in the phylogenetic tree. Conversely, ‘speciation’ or cladogenesis arises from a splitting event, where a parent species is split into two distinct species, often as the result of geographic isolation or another driving force involving the separation of populations.
(b) Natural Selection : Natural selection can be defined as the process through which species adapt to their environment in order to survive. Natural selection occurs when traits that predispose organisms to survive in an environment are passed from parents to their offspring.
- Organisms that are unable to adapt to changes in the environment are unlikely to survive or reproduce.
- Those that can adapt are much more likely not just to survive, but also to thrive and reproduce.
- As offspring are born, they will have the advantageous genetic traits passed on from their parents.
- As a result, the species will change over time.
- Why? Because offspring will develop traits or characteristics that predispose them to do well in their natural environment.
Q. 16. If the image formed by a mirror for all positions of the object placed in front of it is always erect and diminished, what type of mirror is it? Draw a ray diagram to justify your answer. Where and why do we generally use this type of mirror?
Answer. Convex mirror always forms an erect, virtual and diminished image for all positions of the object placed in front of it.
Uses:
1. Convex mirrors are used as rear-view mirrors in vehicles to see the traffic at the rear side (or back side) because —
(i) a convex mirror always produces an erect image of the object;
(ii) the image formed in a convex mirror is highly diminished due to which a convex mirror gives a wide field of view.
2. Big convex mirrors are used as ‘shop security mirrors’. By installing a big convex mirror at a strategic point in the shop, the shop owner can keep an eye on the customer to look for thieves and shoplifters among them as convex mirrors always form a virtual, diminished and erect image.
Q. 17. What is meant by scattering of light? Use this phenomenon to explain why the clear sky appears blue or the sun appears reddish at sunrise.
Answer. Scattering of light is the phenomenon to throw light in various random directions. Light is scattered when it falls on various types of suspended particles in its path. The colour of the scattered light depends on the size of the scattering particles in the atmosphere.
— The larger particles of dust and water droplets present in the atmosphere scatter the light of longer wavelengths due to which the scattered light appears white.
— The extremely minute particles such as air molecules present in the atmosphere scatter mainly blue light present in the white sunlight.
Colour of the sky appears blue. The molecules of air and other fine particles in the atmosphere have a size smaller than the wavelength of visible light. So these particles scatter more effectively the light rays of shorter wavelength at the blue end than light of longer wavelength at the red end. When the scattered blue light enters our eyes, it gives us the feeling of a blue sky.
Colour of the sun appears red at sunrise and sunset. The sun at sunrise and sunset is very near to the horizon, and near the horizon most of the blue light of shorter wave-lengths is scattered away by the particles in the atmosphere. Therefore, the light that reaches our eyes is of longer wavelengths that gives rise to the reddish appearance of the sun.
Q. 18. Differentiate between biodegradable and non-biodegradable substances with the help of one example each. List two changes in habit that people must adopt to dispose non- biodegradable waste, for saving the environment.
Answer. Difference between biodegradable substances and non-biodegradable substances. See Q. 2, 2011 (II Outside Delhi).
Habits for disposing non-biodegradable waste:
- Disposal of wastes after separating them into biodegradable and non-biodegradable wastes in separate dustbins and treating wastes properly before dumping in landfills.
- Encourage the use of gunny bags, jute bags and paper bags in place of polythene/ plastic bags.
- Recycle the plastic and glass objects after use.
Q. 19. Both soap and detergent are some type of salts. What is the difference between them? Describe in brief the cleansing action of soap. Why do soaps not form lather in hard water? List two problems that arise due to the use of detergents instead of soaps
Answer.
A soap is the sodium or potassium salt of a long chain carboxylic acid.
Example : C17H35 COO– Na+ Sodium stearate
C15H31 COO– Na+ Sodium palmitate
A detergent is ammonium or sulphonate salt of a long chain carboxylic acid.
Example : CH3 – (CH2)11 – C6H4 – SO3– Na+
CH3 – (CH2)10 – CH2 – SO4– Na+
Cleansing Action of Soaps :Most of the dirt is oily in nature and oil does not dissolve in water. The molecule of soap constitutes sodium or potassium salts of long-chain carboxylic acids. In the case of soaps, the carbon chain dissolves in oil and the ionic end dissolves in water. Thus, the soap molecules form structures called micelles. In micelles, one end is towards the oil droplet and the other end which is the ionic faces outside. Therefore, it forms an emulsion in water and helps in dissolving the dirt when we wash our clothes.
Soap is a kind of molecule in which both the ends have different properties.
- Hydrophilic end
- Hydrophobic end
The first one is the hydrophilic end which dissolves water and is attracted to it whereas the second one is the hydrophobic end that is dissolved in hydrocarbons and is water repulsive in nature. If on the surface of the water, soap is present then the hydrophobic tail which is not soluble in water will align along the water surface.
Soaps don’t form lather in hard water
Soap is sodium or potassium salt of certain long chain fatty acids such as stearic acid, palmitic acid, oleic acid etc. When hard water is treated with soap solution, Ca2+and Mg2+ ions present in hard water react with the anions of fatty acids present in the soap to form a scum (curdy white precipitate).
Ca2+(aq) + 2C17H35COO–(aq) → (C17H35COO)2 Ca ↓
Mg2+(aq) + 2C17H35COO–(aq) → (C17H35COO)2 Mg ↓
Soap will not produce a lather with hard water until all the calcium and magnesium ions have been precipitated as stearates. Hard water thus wastes soap.
Problems that arise due to the use of detergents instead of soaps:
- Detergents are non-biodegradable, i.e., they cannot be decomposed organisms and hence cause water pollution in lakes and rivers.
- Detergents can also cause skin problems.
Q. 20. (a) Name the human male reproductive organ that produces sperms and also secretes a hormone. Write the functions of the secreted hormone.
(b) Name the parts of the human female reproductive system where
(i) fertilisation takes place,
(ii) implantation of the fertilised egg occurs.
Explain how the embryo gets nourishment inside the mother’s -body.
Answer. (a) Testis produces sperms and secretes male hormone—testosterone.
Function of testosterone hormone is to control the development of male sex organs and male features such as deeper voice, moustaches, body hair, etc. All these changes caused by testosterone are associated with male puberty which the boys attains at an age of 13 to 14 years. It also controls the formation of male gamete, i.e., sperm.
(b) (i) Fertilisation takes place in fallopian tube (oviduct).
(ii) Implantation of the fertilised egg occurs in the uterus.
Nourishment of embryo. After fertilisation, zygote is formed which develops into an embryo. The embryo gets nutrition from the mother’s blood with the help of a special tissue called placenta. This is a disc which is embedded in the uterine wall. It contains villi on the embryo’s side of the tissue. On the mother’s side are blood spaces, which surround the villi. This provides a large surface area for glucose and oxygen to pass from the mother to the embryo. In this way, embryo gets its nutrition.
Q. 21. How do Mendel’s experiments show that the
(a) traits may be dominant or recessive,
(b) traits are inherited independently?
Answer. Mendel’s experiments show that the
(a) Traits may be dominant or recessive. When Mendel cross-bred plants of two different traits of character, a tall pea plant (TT) and a dwarf pea plant (tt) to get a progeny ( F1 generation), all F1 plants were tall. Only the dominant trait was visible in this generation. But when plants of F1 generation were self bred then the two traits of character got separated in the plants of F2 generation. All plants obtained in the F2 generation were not tall. One-fourth of the F2 plants were short.
Appearance of tall characters in both the F1 and F2 generations shows that it is a dominant character. Whereas the absence of dwarf character in F1 generation and its reappearance in F2 generation shows dwarfness is the recessive character.
(b) Traits are inherited independently. Mendel cross-bred pea plants showing two different characteristics, rather than just one. When he cross-bred pea plants of round green seeds with wrinkled yellow seeds, he got F1 generation with all such seeds which were Gametes yellow and round. So, it was concluded that round and yellow character of seeds were dominant traits in the pea plant. On selfing of F1 progeny, different types of F2 progeny were obtained.
So traits of two different characters were inherited independent of each other and made new combination characteristics independent of their previous combinations.
Q. 22. What is meant by power of a lens? Define its S.I. unit. You have two lenses A and B of focal lengths +10 cm and 10 cm respectively. State the nature and power of each lens.Which of the two lenses will form a virtual and magnified image of an object placed 8 cm from the lens? Draw a ray diagram to justify your answer.
Answer.
Power of the lens is known as the inverse of the focal length of the lens measured in meter(m).
S.I unit of power = Dioptre (D)
Power of lense = 1/ focal length in meter.
The lense used is a convex lens because the focal length is a positive value.
Lens A: FA = +10 cm = +10/100 = 0.1 m
PA = 1/FA = 1/+0.1 = +10 D
The power of a convex lens is positive therefore, lens A is a convex lens.
Lens B: FB = -10 cm = -10/100 = 0.1 m
PB = 1/FB = 1/-0.1 = -10 D
The power of a concave lens is negative therefore, lens B is a concave lens.
When an object is placed at 8 cm (i.e., between the optical centre and principal focus) only convex lens will form the virtual and magnified image.
Therefore, lens A will form a virtual and magnified image of the object placed 8 cm from it. When the object is placed between the optical centre and the focus: (i.e., between O and F) the image formed is behind the object (on the same side), virtual, erect and magnified.
Q. 23. One half of a convex lens of focal length 10 cm is covered with a black paper. Can such a lens produce an image of a complete object placed at a distance of 30 cm from the lens? Draw a ray diagram to justify your answer.
A 4 cm tall 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 15 cm. Find nature, position and size of the image.
Answer. (i) Yes. If a convex lens of focal length 10 cm is covered one half with a black paper, it can produce an image of the complete object between F2 and 2F2. The rays of light coming from the object get refracted by the upper half of the lens. The image formed will be real, inverted and diminished.
Q. 24. Write the importance of ciliary muscles in the human eye. Name the defect of vision that arises due to gradual weakening of the ciliary muscles in old age. What type of lenses are required by the persons suffering from this defect to see the objects clearly? Akshay, sitting in the last row in his class, could not see clearly the words written on the blackboard. When the teacher noticed it, he announced if any student sitting in the front row could volunteer to exchange his seat with Akshay. Salman immediately agreed to exchange his seat with Akshay. He could now see the words written on the blackboard clearly. The teacher thought it fit to send the message to Akshay’s parents advising them to get his eyesight checked.
In the context of the above event, answer the following questions:
(a) Which defect of vision is Akshay suffering from? Which type of lens is used to correct this defect?
(b) State the values displayed by the teacher and Salman.
(c) In your opinion, in what way can Akshay express his gratitude towards the teacher and Salman?
Answer. Importance of ciliary muscles. The focal length of eye lens can be changed by changing its shape by the action of ciliary muscles. Ciliary muscles modify the curvature of the eye lens to enable the eye to focus objects at varying distances.
In old age, ciliary muscles become weak and the eye lens become inflexible. As a result, the eye loses its power of accommodation. Because of this, an old person cannot see the nearby objects clearly. This leads to the defect called presbyopia.
Bifocal lenses are required by the persons suffering from presbyopia in which upper part consists of a concave lens for distant vision and the lower part consists of a convex lens used to see nearby objects.
(a) Akshay is suffering from myopia or near sightedness. Myopia is corrected by using spectacles containing concave lenses. The concave lens used for correcting myopia should be of such a focal length that it produces a virtual image of the distant object (lying at infinity) at the far point of the myopic eye.
(b) • The teacher is concerned about the health of the student.
• Salman is very helpful and caring.
(c) Akshay should express his gratitude towards his teacher as well as his friend Salman by getting his eyesight checked and by also following their values in his own life.
SECTION B
Q. 25. What do we observe on pouring acetic acid on red and blue litmus papers?
(A) Red litmus remains red and blue litmus turns red.
(B) Red litmus turns blue and blue litmus remains blue.
(C) Red litmus turns blue and blue litmus turns red.
(D) Red litmus becomes colourless and blue litmus remains blue.
Answer. (A)
Q. 26. While preparing soap a small quantity of common salt is generally added to the reaction mixture of vegetable oil and sodium hydroxide. Which one of the following may be the purpose of adding common salt?
(A) To reduce the basic nature of the soap
(B) To make the soap neutral
(C) To enhance the cleansing power of the soap
(D) To favour the precipitation of the soap
Answer. (D)
Q. 27. A student takes about 4 mL of distilled water in four test tubes marked P, Q, R and S. He then dissolves in each test tube an equal amount of one salt in one test tube, namely sodium sulphate in P, potassium sulphate in Q, calcium sulphate in R and magnesium sulphate in S. After that he adds an equal amount of soap solution in each test tube. On shaking each of these test tubes well, he observes a good amount of lather (foam) in the test tubes marked
(A) P and Q (B) Q and R (C) P, Q and S (D) P, R and S
Answer. (A)
Q. 28. A student was asked to observe and identify the various parts of an embryo of a red kidney bean seed. He identified the parts and listed them as under:
I. Tegtnen II. Testa III. Cotyledon IV. Radicle V. Plumule
The correctly identified parts among these are:
(A) I, II and III (B) II, III and IV (C) III, IV and V (D) I, III, IV and V
Answer. (C)
Q. 29. Given below is the list of vegetables available in the market. Select from these the two vegetables having homologous structures: Potato, sweet potato, ginger, radish, tomato, carrot, okra (Lady’s finger)
(A) Potato and sweet potato (B) Radish and carrot
(C) Okra and sweet potato (D) Potato and tomato
Answer. (B)
Q. 30. A student obtains a sharp image of the distant window (W) of the school laboratory on the screen (S) using the given concave mirror (M) to determine its focal length. Which of the following distances should he measure to get the focal length of the mirror?
(A) MW (B) MS
(C) SW (D) MW-MS
Answer. (B)
Q. 31. A student used a device (X) to obtain/focus the image of a well illuminated distant building on a screen (S) as shown below in the diagram. Select the correct statement about the device (X).
(A) This device is a concave lens of focal length 8 cm.
(B) This device is a convex mirror of focal length 8 cm.
(C) This device is a convex lens of focal length 4 cm.
(D) This device is a convex lens of focal length 8 cm.
Answer. (D)
Q. 32. A student traces the path of a ray of light through a rectangular glass slab for the different values of angle of incidence. He observes all possible precautions at each step of the experiment. At the end of the experiment, on analysing the measurements, which of the following conclusions is he likely to draw?
(A) ∠z = ∠e < ∠r (B) ∠z < ∠e < ∠r (C) ∠z > ∠e > ∠r (D)∠z = ∠e > ∠r
Answer. (D)
Q. 33. A student traces the path of a ray of light through a triangular glass prism for different values of angle of incidence. On analyzing the ray diagrams, which one of the following conclusions is he likely to draw?
(A) The emergent ray is parallel to the incident ray.
(B) The emergent ray bends at an angle to the direction of the incident ray.
(C) The emergent ray and the refracted ray are at right angles to each other.
(D) The emergent ray is perpendicular to the incident ray.
Answer. (B)
Q. 34. When you add sodium hydrogen carbonate to acetic acid in a test tube, a gas liberates immediately with a brisk effervescence. Name this gas. Describe the method of testing this gas.
Answer. When sodium hydrogen carbonate is added to acetic acid in a test tube, then a brisk effervescence is observed due to the liberation of CO2 gas which is colourless and odourless.
Q. 35. Students were asked to observe the permanent slides showing different stages of budding in yeast under high power of a microscope.
(a) Which adjustment screw (coarse/fine) were you asked to move to focus the slides ?
(b) Draw three diagrams in correct sequence showing budding in yeast.
Answer. (a) A fine adjustment screw is moved to focus the slides.
Q. 36. A 4 cm tall object is placed on the principal axis of a convex lens. The distance of the object from the optical centre of the lens is 12 cm and its sharp image is formed at a distance of 24 cm from it on a screen on the other side of the lens. If the object is now moved a little away from the lens, in which way (towards the lens or away from the lens) will he have to move the screen to get a sharp image of the object on it again? How will the magnification of the image be affected?
Answer. (a) The screen should be moved towards the lens to get a sharp image of the object again. (b) Magnification of the image decreases on moving the object away from the lens.
SET II
SECTION A
Q. 1. Write the number of covalent bonds in the molecule of propane, C3H8.
Answer.
Q. 2. Where is DNA found in a cell?
Answer. DNA is found in the nucleus of the cell.
Q. 3. The first trophic level in a food chain is always a green plant. Why?
Answer. The first trophic level in a food chain is always a green plant because green plants have a mechanism for trapping solar energy with the help of their green pigment called chlorophyll and convert it into chemical energy which is stored as carbohydrates in the plant for the use of the next tropic level.
Q. 5. We often observe domestic waste decomposing in the by lanes of our homes. List four ways to make the residents aware that the improper disposal of wastes is harmful to the environment and also for their own health.
Answer. The various ways to make residents aware of the harmful effects of improper disposal of wastes are as follows:
- By putting posters and hoardings.
- By distributing pamphlets to the residents.
- By conducting meetings of residents, welfare society, etc.
- By advertisements through TV/radio/cable operators, etc.
Through all these mediums we can make the residents aware of how such waste serves as a breeding ground for mosquitoes, releases harmful gases, reduces soil fertility, etc.
Q. 6. List any two advantages associated with water stored in the ground.
Answer. Advantages of underground water:
- The water stored in the ground does not evaporate.
- The water stored in the ground does not promote breeding of mosquitoes.
Q. 7. What is meant by homologous series of carbon compounds? Classify the following carbon compounds into two homologous series and name them.
C3H4 C4H6, C4H8, C5H8, C5H10
Answer. A homologous series is a group of organic compounds having similar structure and similar chemical properties in which the successive compounds differ by CH2 group.
Q. 9. The elements 4Be, 12Mg and 20Ca, each having two valence electrons in their valence shells, are in periods 2, 3 and 4 respectively of the modern periodic table. Answer the following questions associated with these elements, giving reason in each case:
(a) In which group should they be?
(b) Which one of them is least reactive?
(c) Which one of them has the largest atomic size?
Answer.
(a) They all belong to 2nd group of the periodic table as all the elements have two valence electrons in their valence shell.
(b) Be is least reactive as it contains the least number of shells (i.e., 2). The tendency of an atom to lose electrons on moving down in a group increases as valence electrons move more and more away from the nucleus and nuclear charge on valence electrons decreases.
(c) Ca has the largest atomic size as it has the maximum number of shells.
Q. 11. List three distinguishing features between sexual and asexual types of reproduction, in tabular form.
Answer. Difference between Sexual and Asexual types of reproductions
| Sexual Reproduction | Asexual Reproduction |
| Two parents are involved | One parent is involved |
| Characteristics of both parents are inherited | Characteristics of only one parent are inherited |
| It requires the formation of gametes | It doesn’t require the formation of gametes |
| Special organs for reproduction are required. | Special organs for reproduction are not required |
| Examples – Mangoes, coconut, Hibiscus etc. | Examples – Potato, Jasmine, Rose, Yeast, Bryophyllum etc |
Q. 14. A pea plant with blue colour flower denoted by BB is crossbred with a pea plant with white flower denoted by ww.
(a) What is the expected colour of the flowers in their F4 progeny?
(b) What will be the percentage of plants bearing white flower in F2 generation, when the flowers of F4 plants were selfed?
(c) State the expected ratio of the genotype BB and Bw in the F2 progeny.
Answer.
Q. 18. What is an ecosystem? List its two main components. We do not clean natural ponds or lakes but an aquarium needs to be cleaned regularly. Why is it so? Explain.
Answer. An ecosystem is a self-contained unit of living things (plants, animals and decomposers) and their non-living environment (soil, air and water). An ecosystem needs only the input of sunlight energy for its functioning.
Ecosystem is made up of two main components:
(i) Abiotic component meaning non-living component.
(ii) Biotic component meaning living component.
An aquarium is an artificial ecosystem. It needs to be cleaned regularly because it has – stagnant water and no decomposers. Therefore the dead parts and excrete of the organisms of the aquarium are not decomposed and would never be returned to their original pool. The accumulation of these wastes can be dangerous for the living organisms in the aquarium.
Q. 19. What are fossils? How are they formed? Described in brief two methods of determining the age of fossils. State any one role of fossils in the study of the process of evolution.
Answer.
Fossils : fossils are the remains of organisms found in the earth’s strata (rock layers). These organisms have, in some way, been protected from the bacterial action that degrades carbon-based organisms.
Fossils are created through sudden disruptions, like earthquakes. They can also be formed through steadier processes, like exposure to the (extreme) elements.
In most cases, the environment plays a direct role, as a preservative, in whether an organism will become a fossil.
The ideal environment is silty, with moving sediment that can cover the organism. The floors of oceans, lakes, and deep ponds are prime spots for fossilization.
Imagine a sick dinosaur pulling itself to a lake for a drink. It dies. The body falls in and sinks to the bottom. The lake’s motion deposits mud and sand on the animal.
Over time, the soft tissues dissolve, and the mineral-rich sediment fills the gaps. Years pass and the minerals harden around the bones, creating a petrified structure we’ve come to know as fossils.
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The age of the fossils can be determined by the following way:
1. Radiometric dating :In this method, the age of the fossil can be determined by tracing the radioactive elements present in the rocks and examining it chemically.
2. Relative dating :The fossils are found in the sedimentary rocks in the form of layers accumulated over large span of time. So, using this fact, the geologist find the age of fossils in the way that the fossil found at the bottom of those layers are found to be older than that found above those.
3. Fossils have a significant role in evolution as they help in creating evolutionary relationships with the earlier organisms and the present day organisms.
SET III
Q. 1. Write the number of covalent bonds in the molecule of butane, C4H10.
Answer.
Q. 2. Name two simple organisms having the ability of regeneration.
Answer. Two simple organisms having the ability of regeneration are Planaria and Hydra.
Q. 3. Which of the following are always at the second trophic level of food chains? Carnivores, Auto trophs, Herbivores
Answer. Herbivores are always at the second trophic level of food chains.
Q. 4. Draw a ray diagram to show the path of the reflected ray corresponding to an incident ray of light parallel to the principal axis of a convex mirror and show the angle of incidence and angle of reflection on it.
Answer.
Q. 5. Why is sustainable management of natural resources necessary? Out of the two reuse and recycle which, in your opinion, is better to practise? Give reason.
Answer. Sustainable management is management that encourages the judicious use of natural resources to meet the current basic human needs while preserving the resources for the needs of future generations.
Reuse is a better practise than recycle as recyling requires a large amount of energy and money. In this strategy, we are encouraged to use things again and again instead of throwing them away with no expense of energy. Reuse also does not cause pollution. Example, plastic bottles of jams and pickles can be reused for storing things in the kitchen.
Q. 7. Write the name and general formula of a chain of hydrocarbons in which an addition reaction with hydrogen is possible. State the essential condition for an addition reaction. Stating this condition, write a chemical equation giving the name of the reactant and the product of the reaction.
Answer. Addition of hydrogen is possible in Alkenes CnH2n and Alkynes CnH2n-2
- Addition reaction of unsaturated hydrocarbons occurs in the presence of catalysts such as nickel (Ni) or palladium (Pd) or platinum (Pt) to form saturated hydrocarbons.
- Example, Ethene reacts with hydrogen when heated in the presence of nickel catalyst to form ethane.
Q. 9. Given below are some elements of the modern periodic table. Atomic number of the element is given in the parentheses: A(4), B(9), C(14), D(19), E(20)
(a) Select the element that has one electron in the outermost shell. Also write the electronic configuration of this element.
(b) Which two elements amongst these belong to the same group? Give reason for your answer.
(c) Which two elements amongst these belong to the same period? Which one of the two has bigger atomic radius?
Answer.
(a) D has 1 electron in the-outermost shell. Atomic no. = 19, Elec. Configuration = 2, 8, 8,1
(b) A and E belong to the same group, i.e., 2nd group of the periodic table as both have same number of valence electrons (i.e., 2) in their outermost shell.
(c) A & B and D &’E belong to the same period as they have the same no. of valence shells. A and B have 2 shells thus belong to 2nd period of the periodic table.
D and E have 4 shells thus belong to 4th period of the periodic table.
A has the bigger atomic radius among A and B and D has the bigger atomic radius among D and E as atomic size decreases on moving from left to right in a period (or atomic size decreases as the atomic number increases in a period).
Q. 12. List four categories of contraceptive methods. State in brief two advantages of adopting such preventive methods.
Answer. The contraceptive methods can be broadly classified into the following categories:
1. Natural method: It involves avoiding the chances of meeting of sperms and ovum. In this method, the sexual act is avoided from day 10th to 17th of the menstrual cycle because, during this period, ovulation is expected and therefore, the chances of fertilization are very high.
2. Barrier method: In this method, the fertilization of ovum and sperm is prevented with the help of barriers. Barriers are available for both males and females. Condoms are barriers made of thin rubber that are used to cover penis in males and vagina in females.
3. Oral contraceptives: In this method, tablets or drugs are taken orally. These contain small doses of hormones that prevent the release of eggs and thus fertilization cannot occur.
4. Implants and surgical methods: Contraceptive devices such as the loop or Copper-T are placed in the uterus to prevent pregnancy. Some surgical methods can also be used to block the gamete transfer. It includes the blocking of vas deferens to prevent the transfer of sperms known as vasectomy. Similarly, fallopian tubes of the female can be blocked so that the egg will not reach the uterus known as tubectomy.
Advantages of adopting such preventive methods:
- Contraceptive methods are used to prevent the spread of sexually transmitted diseases like syphilis, AIDS, etc.
- Contraceptives prevent unwanted and frequent pregnancies. If a couple has less number of children, they can provide good clothes, good food and a good education to each child. This will make the parents as well as children happy.
Q. 17. An object of height 5 cm is placed perpendicular to the principal axis of a concave lens of focal length 10 cm. If the distance of the object from the optical centre is 20 cm, determine the position, nature and size of the image formed using the lens formula.
Answer.
