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4251.

Answer the following briefly : (i) Three functions of WHO. (ii) Three advantages of a small family. (iii) Explain the terms : (1) Population density (2) Natality.

Answer»

(i) Three functions of WHO: 

(a) To direct and co-ordinate international health projects. 

(b) To encourage and conduct scientific research. 

(c) To organise international campaigns, to control and eliminate diseases like AIDS. 

(ii) Three advantages of a small family: 

(a) Children get proper education, 

(b) Children can get proper medical facilities. 

(c) Children can get proper diet and nutrition. 

(iii) (1) Population density: The sum total of individuals in a given geographic region at a specified period is known as the population density. It is the number of individuals in an unit area, such as per square kilometre. 

(2) Natality: It is the number of live births per 1000 people of population per year.

4252.

Given below is an example of a certain structure and its special functional activity. On a similar pattern fill in the blanks with suitable functions : Example : Chloroplast and Photosynthesis : (i) Xylem and ........................(ii) Ciliary Body and ..................(iii) Seminiferous Tubule and ...................(iv) Thyroid Gland and ..............(v) Eustachian Tube an ...............

Answer»

(i) Xylem and Water transport. 

(ii) Ciliary body and Accommodation of eye lens

(iii) Seminiferous tubule and Spermatogenesis 

(iv) Thyroid gland and basal Metabolism 

(v) Eustachian tube and Body equilibrium

4253.

Name the following: (i) A membrane that disappears during late prophase. (ii) A fluid that occupies the larger cavity of the eyeball behind the lens. (iii) The ground substance present in a chloroplast. (iv) A specific part of a chromosome that determines hereditary characteristics. (v) A neurotransmitter stored at the terminal end of the axon.

Answer»

(i) Nuclear Membrane 

(ii) Vitreous humour 

(iii) Stroma.

(iv) Genes 

(v) Acetylcholine

4254.

The following paragraph is related to absorption of water from the soil. Copy and complete the following paragraph by selecting the correct word from those given in the box. You may use the term only once. Exosmosis, Hypertonic, Osmosis, Isotonic, Hypotonic, Cortical, Endosmosis Water enters the root hair from the soil by the process of ……… .This is because the solution in the soil is ……… whereas the cell sap in the root hair cell is ……… . The water then passes through the cells by cell to cell ………. and reaches the xylem of the root.

Answer»

Water enters the root hair from the soil by the process of Endosmosis. This is because the solution in the soil is hypotonic whereas the cell sap in the root hair cell is hypertonic. The water then passes through the cortical cells by cell to cell osmosis and reaches the xylem of the root.

4255.

Give one point of difference between the following on the basis of what is given in the brackets :(i) Myopia and Hypermetropia. (cause of the defect)(ii) Cerebrum and Spinal cord. (arrangement of cytons and axons of neuron)(iii) Genotype and Phenotype. (definition)(iv) Karyokinesis and Cytokinesis. (explain the term.)(v) Light reaction and Dark reaction. (site of occurrence)

Answer»

(i)

Myopia Hypermetropia
Eye ball lengthened from front to back.Eye ball shortened from front to back.


(ii)

CerebrumSpinal Cord
Cyton outside and axons inside.Axrins outside and cytons inside.


(iii) 

GenotypePhenotype
It is the genetic constitution.These are the observable characters controlled by genes.


(iv) 

KaryokinesisCytokinesis
It is the division of the nucleus.It is the division of the cytoplasm during cell division.


(v)

Light reactionDark reaction
In grana.In stroma of the chloroplast.
4256.

Given one point of difference between the following on the basis of what is given in the brackets : (i) Myopia and Hypermetropia. (cause of the defect) (ii) Cerebrum and Spinal cord. (arrangement ofcytons and axons of neuron) (iii) Genotype and Phenotype. (definition) (iv) Karyokinesis and Cytokinesis. (explain the term) (v) Light reaction and Dark reaction. (site of occurrence)

Answer»

(i)

MyopiaHypermetropia
Lengthened of the eyeball from front to back.Shortening of the eyeball from

(ii)

CerebrumSpinal cord
Axons (white matter) on the inner side and cytons (gray matter) on the outer side are present.Cytous (Gray matter) on the inner side and Axons (white matter) on the outer side.


(iii)

GenotypePhepotype
It is the genetic makeup of an organism.It is the physical appearance organism.


(iv)

KaryokinesisCytokinesis
The division of nucleus during cell division are called karyokinesis.The division of cytoplasm after karyokinesis is called cytokinesis.

(v)

Light ReactionDark Reaction
It takes place in the grana of chloroplast.It takes place in the grana of chloroplast.

4257.

Given below is the outline of the male reproductive system:(i) Name the parts labelled 1 to 5(ii) State the function of the parts labelled 1 and 4.(iii) Name the cells of part 5 that produce testosterone.(iv) Why is the structure 5 present outside the body in the scrotal sacs?(v) What is semen ?

Answer»

1- Prostate gland 

2-Cowper's gland

3- Urethra

4- Sperm duct

5- Testis

(ii) 1- Prostate gland-It secretes an alkaline matter which neutralizes the acidity of the urethra.

4- Sperm duct -It carries sperms from epididymis to the urethra.

(iii) Cells of Leydig.

(iv) Testis is located in the scrotal sacs so that they can maintain a low temperature than the body cavity. This trow temperature is necessary for the maturation of sperms.

(v) Semen is the mixture of sperms and secretions of the seminal vesicles, prostate gland and Cowper's gland.

4258.

Given below are sets of 5 terms each. Without changing the first term, rearrange the remaining four so as to be in logical sequence as per the directions given in brackets for each. One has been done for you as an example.Example: Pathogen, active immunity, produces antibodies, lymphocytes, antigen, (defence mechanism of the body) Answer : Pathogen —> antigen —> lymphocytes —> produces antibodies —> active immunity. (i) Destarched plant, iodine added, washed in water, a leaf boiled in alcohol, placed in sunlight. (testing for presence of starch) (ii) Interphase, Anaphase, Prophase, Telophase, Metaphase (sequential stages in Karyokinesis). (iii) Seminiferous tubule, peris, urethra, epidodymis, vas deferens. (course of passage of sperms in man). (iv) Pinna, cochlea, tympanum, ear ossicles, auditory canal (route through which vibrations of sound enter the ear). (v) Soil water, xylem, cortex, enaodermis, root hair (coduction of water)

Answer»

(i) Destarched plant —> washed in water —> a leaf boiled in alcohol —> placed in sunlight —> Iodine added. 

(ii) Interphase —> Prophase —> Metaphase —> Anaphase —> Telophase. 

(iii) Seminiferous tubule —> Epididymis —> Vas deferens —> Urethra —> Penis. 

(iv) Pinna —> auditory canal tympanum —> ear ossicles —> cochlea. 

(v) Soil water Root hair —> Cortex —> Endodermis Xylem.

4259.

The diagram given below represents a certain stage of mitosis:(i) Identify the stage of cell division. (ii) Name the parts labelled A and B. (iii) What is the unique feature observed in this stage ? (iv) How many daughter cells are formed from this type of cell division?

Answer»

(i) Anaphase

(ii) 

A - Spindle fibre 

B - Centromere

(iii) Two sister chromatids of each chromosome separate and are drawn apart towards opposite poles.

(iv) Two daughter cells are formed from this type of cell division.

4260.

Given below is the outline of the male reproductive system :(i) Name the parts labelled 1 to 5. (ii) State the functions of the parts labelled 1 and 4. (iii) Name the cells of part 5 that produce testosterone. (iv) Why is the structure 5 present outside the body in the scrotal sacs ? (v) What is semen ?

Answer»

(i) 1—Seminal vesicle 

2—Prostate gland 

3— Urethra 

4— Sperm duct/vas deferens 

5—Testis. 

(ii) Part 1—Its secretion serves as a medium for the transportation of the sperms. 

Part 2—It carry the sperms. 

(iii) Leydig cells or interstitial cells of part 5 produces testosterone. 

(iv) The structure 5 is present outside the body in the scrotal sac because sperms are produced at 2 to 3°C lower than that of body temperature so, when too hot the skin of scrotal sac looses and testes are away from the body and when it is cold the skin contracts and testis are closer to the body for warmth. 

(v) Semen—The mixture of seminal vesicles secretios and sperms produces a milky fluid, which is called semen.

4261.

State whether the following statements are True or False. If False rewrite the correct form of the statement by only changing the last word of the statement. (i) The alpha cells dfthe pancreas secrete insulin. (ii) Duplicated chromosomes remain attached at a point termed centrosome. (iii) The number of pairs ofautosomes in man is 22. (iv) Penicillin obtained from si fungus is an example of an antibody. (v) Plants that manufacture their own food are termed heterotrophs.

Answer»

(i) The alpha cells of the pancreas secrete glucagon. 

(ii) Duplicated chromosomes remain attached at a point termed centromere. 

(iii) True. 

(iv) Penicillin obtained from a fungus is an example of an antibiotic. 

(v) Plants that manufacture their own food are termed autotrophs.

4262.

Given alongside is an experimental set up to study a particular process:(i) Name the process being studied. (ii) Explain the process named in (i) above. (iii) Why is the pot covered with a plastic sheet ? (iv) Mention one way in which this process is beneficial to the plant. (v) Suggest a suitable control for this experiment.

Answer»

(i) Transpiration. 

(ii) ‘Loss of water as water vapour from the aerial parts of the plant is called as transpiration’. 

(iii) Covering the pot with a plastic sheet would prevent the escape of water vapour from the pot.

(iv) Transpiration helps in the ascent of sap by producing a suction force acting from the top of a plant. 

(v) A similar empty plastic sheet with its mouth tied, with no potted plant kept in sunlight will show no drops of water.

4263.

Explain the following terms : (i) Antibiotics (ii) Antiseptic (iii) Hormones (iv) Diffusion (v) Destarched plant.

Answer»

(i) Antibiotics: The chemical substances which are produced by a micro¬organism, can stop growth of or kill another micro-organism. e.g., Pencillin. 

(ii) Antiseptics—These are chemical substances which destroy some bacteria and prevent the growth of others. Antiseptics are used locally in contact of body tissues because they are mild and are harmless to the tissues they work upon. Examples of antiseptics—Tincture, benzoic acid, boric acid, hydrogen peroxide, dettol and mercurochrome. 

(iii) Hormones: According to Selye (1948), “Hormones are the physiological organic compounds produced by certain cells for the sole purpose of directing theractivities to distant parts of the same origins.” A hormone is defined as “a chemical messenger secreted by an endocrine gland which reaches its destination by the blood stream and which has the power of influencing the activity of other distant target organs.” 

(iv) Diffusion: It is the process of movement of molecules of a substance (solid, liquid or gas) from the region of their higher concentration to the region of their lower concentration through the concentration gradient. 

Or 

Diffusion may be defined as the process of random movement of molecules of a substance from a region of higher concentration to a region of lower concentration. 

(v) Destarched plant: The plant which has no starch in the leaves and the leaf remains yellowish brown when the leaf is tested for photosynthesis with iodine solution

4264.

Given below is an example of a certain structure and its special functional activity: Example : (0) Ribosomes and Protein synthesis. On a similar pattern complete the following: (i) Hypothalamus and ……… . (ii) Suspensory ligaments and ……… . (iii) Semi circular canals and ……… . (iv) Mitochondria and ……… . (v) Seminiferous tubules and ……… .

Answer»

(i) Hypothalamus and control pituitary function. 

(ii) Suspensory ligaments and to held lens in position. 

(iii) Semi circular canals and balancing the body.

(iv) Mitochondria and release of energy.

(v) Seminiferous tubules and spermatogenesis.

4265.

Given below is an example of a particular structure and its special functional activity, e.g., Glomerulous and ultra filtration. On a similar pattern complete the following:(i) Corpus luteum and ………. . (ii) Iris of the eye and ………. . (iii) Seminal vesicle and ………. . (iv) Phloem and ………. . (v) Eustachian tube and ………. . 

Answer»

(i) Corpus luteum and secretion of progesterone. 

(ii) Iris of the eye and regulates the amount of light entering the eye. 

(iii) Seminal vesicle and semen. 

(iv) Phloem and photosynthesis. 

(v) Eustachian tube and balancing the air pressure on either side of the eardrum.

4266.

Match the items in Column I with that which is most appropriate in Column II.Column IColumn II(1) Pacemaker(a) Associated with static body balance(2) Stroma(b) Chordae tendinae(3) Afferent nerve(c) Site of light reaction(4) Prolactin(d) Motor neuron(5) Saccules(e) S A node(f) Stimulates production of milk by the mammary gland(g) Site of dark reaction(h) Transmits impulses from receptor organ to spinal ftord.(i) Secreted by unterior, tube of Pituitary gland(j) Transfers impulses from spinal cord to muscles

Answer»
Column IColumn II
 (1)Pacemaker(e) S A node
(2) Stroma (c) site of light reaction
(3) Afferent nerve(h) Transmit impulses from receptor organ to spinal cord
(4) Prolactin(f) Stimulates production of milk by the mammary gland
(5) Saccules(a) Associated with static body balance.

4267.

Given below is an example of a certain structure and its special functional activity. On a similar pattern fill in the blanks with suitable functions :Example : Chloroplast and Photosynthesis (i) Xylem and ……… . (ii) Ciliary Body and ……… . (iii) Seminiferous Tubule and ……… . (iv) Thyroid gland and ……… . (iv) Eustachian Tube and ……… .

Answer»

(i) Xylem and upward transport of water and minerals. 

(ii) Ciliary body and alter the shape of the lens of eye. 

(iii) Seminiferous tubule and production of sperms in testes. 

(iv) Thyroid gland and regulate basal metabolism. 

(v) Eustachean tube and equalize air pressure on either side of the ear drum.

4268.

Study the following diagram carefully and then answer the questions that follow:(i) Name the cell labelled 1. (ii) Identify the phenomenon occurring in A. (iii) Mention two structural differences between 1 and 2. (iv) Name the process occurring m B and C and state the importance of this process in the human body.

Answer»

(i) 1-Red blood cells. 

(ii) Diapedesis 

(iii)

12
(a)  BIconcave in shapeAmeoboid in shape
(b) Nucleus  absentnucleus present

(iv) Phagocytosis Process. By this process the WBC engulfs the diseases causing germs that enter the body and thus defend our body.

4269.

Rewrite the complete the following sentences by inserting the correct word in the space indicated: (i) ………. vaccine is given to build up immunity against polio. (ii) Phenotype is the observable characteristic which is ………. controlled. (iii) Wooden doors swell up in rainy season due to ………. . (iv) The blood vessel that begins and ends in capillaries is the ………. . (v) ………. is the phenomenon of contraction of the cytoplasm from the cell wall.

Answer»

(i) Salk’s vaccine is given to build up immunity against polio. 

(ii) Phenotype is the observable characteristic which is genetically controlled. 

(iii) Wooden doors swell up in rainy season due to imbibition. 

(iv) The blood vessel that begins and ends in capillaries is the hepatic portal system. 

(v) Plasmolysis is the phenomenon of contraction of the cytoplasm hern cell wall.

4270.

Rewrite and complete the following sentences by inserting the correct word in the space indicated : (i) The phenomenon of loss of water through a cut stem or injured part of plant is called ………(ii) ……… is the scientific, name of garden pea, which Mendel used for his experiments. (iii) A fluid that occupies the larger cavity of the eye ball behind the lens is ……… (iv) Oxygen combines with haemoglobin present in RBC and forms ……… (v) ……… causes corrosion of the marble or brick surface.

Answer»

(i) The phenomenon of loss of water through a cut stem or injured part of plant is called bleeding. 

(ii) Pisum sativum is the scientific name of garden pea, which Mendel used for his experiments. 

(iii) A fluid that occupies the larger cavity of the eye ball behind the lens is vitreous humour. 

(iv) Oxygen combines with- haemoglobin present in RBC and forms, Oxyhaemoglobin. 

(v) Acid rain causes corrosion of the marble or brick surface.

4271.

Explain term Destarched plant.

Answer»

Destarched plant - It is the plant in which starch has been removed from the leaves.

4272.

Rewrite and complete the following sentences by inserting the correct word in the space indicated : (i) The phenomenon of loss of water through a cut stem or injured part of plant is called ..........(ii) ........... is the scientific name of garden pea, which Mendel used for his experiments. (iii) A fluid that occupies the larger cavity of the eye ball behind the lens is ............(iv) Oxygen combines with haemoglobin present in RBC and forms .............(v) .............. causes corrosion of the marble or brick surface.

Answer»

(i) Bleeding

(ii) Pisum sativum

(iii) Vitreous humour

(iv) Oxy-haemoglobin

(v) Acid rain

4273.

Explain how the rate of transpiration is affected on :1. a windy day    2. a foggy day

Answer»

(1) on a windy day rate of transpiration increases because of the water vapour will not be allowed to accumulate in the vicinity of the leaves and an increase in the vapour pressure gradient through the stomata will lead to increased rate of transpiration.

(2) A foggy day means more moisture in the air, so the rate of transpiration decreises with the rise in vapour pressure in the atmospheric air.

4274.

The figure given below represents an experimental set up with a weighing machine to demonstrate a particular process in plants. The experimental set up was placed in bright sunlight. Study the diagram and answer the following questions :(i) Name the process intended for study. (ii) Define the above mentioned process. (iii) When the weight of the test tube (A & B) is taken before and after the experiment, what is observed ? Give reasons to justify your observation in A & B. (iv) What is the purpose of keeping the test tube B in the experimental set up ?

Answer»

(i) Transpiration 

(ii) It is the lost of water in the form of water vapours from the aerial parts of the plant 

(iii) Weight of test tube A will decrease after the experiment because water will be lost from it through the leaves by transpiration. Weight of test tube B will remain same after the experiment because water will not be lost by transpiration as there is no plant in it and nor by evaporation as oil is spread over it, which will not allow evaporation.

(iv) It is a control experiment where the purpose of using test tube B is to compare the level of water in both test tubes.

4275.

Give biological reasons for the following :(1) Pituitary gland is also known as the master gland.(2) Gametes have a haploid number of chromosomes.

Answer»

(1) Because it controls the secretion of the hormones of other endocrine glands.

(2) Because after sexual reproduction, the diploid number of chromosome is maintained.

4276.

Give the biological / technical terms for the structure that carries visual stimuli from retina to the brain.

Answer»

Optic nerve the structure that carries visual stimuli from retina to the brain.

4277.

Given below is an experimental set up to demonstrate a particular process. Study the same and answer the questions that follow : (i) Name the physiological process being studied. (ii) Explain the process mentioned above. (iii) What is the aim of the above experiment ? (iv) What would you observe in the experimental set-up after an hour ? Give a reason to support your answer. (v) Mention any three adaptations found in plants to overcome the physiological process mentioned in (i) above.

Answer»

(i) Transpiration 

(ii) The two cobalt chloride papers are attached, one on the dorsal and the other on the ventral surface of a dorsiventral leaf with the help of glass slide and clips. The set up is left in the open for few hours. 

(iii) To show that more transpiration occurs from the under surface of the leaf. 

(iv) Cobalt chloride paper in the lower surface will turn pink taster because of presence of more stomata present on the under surface. 

(v) 

(1) Sunken stomata, 

(2) Presence of thick cuticle

(3) Fewer stomata.

4278.

Briefly explain the following terms : (i) Destarched plant (ii) Phenotype (iii) Death rate (iv) Power of accommodation of the eye. (v) Natural immunity

Answer»

(i) Destarched Plant: A plant from the leaves of which starch has been removed. This can be done by placing the plant in the dark for 24 to 48 hours. 

(ii) Phenotype: The expressed character whieh Is genetically controlled. 

Example : Tall pea plants (genotype TT or Tt), dwarf pea plant (genotype tt). So this expressed tallness or dwarfness is called as phenotype. 

(iii) Death Rate: This is also called as mortality It is the number of death per 1000 of population per year. 

(iv) Power of Accommodation of the eye: The process of focussing the eye at different distances is called as power of accommodation of the eye. 

(v) Natural Immunity: This is also called as native or innate immunity. This immunity is by virtue of genetic constitutional make up Tt is there in the body without any external stimulation of a previous infection.

4279.

Give the biological / technical terms for the suppressed allele of a gene.

Answer»

Recessive allele

4280.

Explain term Diffusion.

Answer»

Diffusion - It is the movement of molecules of any type, from their high to their low concentration.

4281.

Give the biological / technical terms for the a membrane which allows the passage of molecules selectively.

Answer»

Semi-permeable membrane

4282.

Explain term Hormones.

Answer»

Hormones - These are the chemicals produced by ductless glands into the blood and which act on target organs.

4283.

Give the technical / biological term for the following: (i) Onset of menstruation in a young girl around the age of 13 years. (ii) Eye defect occurring in old people whereby they are unable to see near objects. (iii) The mucous membrane lining the uterus (iv) The process of conversion of ADP to ATP during the first phase of photosynthesis. (v) The point of contact between two neurons. (vi) Protective membranes covering the human brain and spinal cord. (vii) Respiratory openings found OP the stem of woody plants. (viii) The process by which white blood cells engulf harmful microbes. (ix) The process of mixing of two different substances molecules. (x) Exudation of sap firm injured parts of a plant. 

Answer»

(i) Onset of menstruation in a young girl around the age of 13 years ‘Menarche’. 

(ii) Eye defect occurring in old people whereby they are unable to see near objects Presbyopia. 

(iii) The mucous membrane lining the uterus Endometrium 

(iv) The process of conversion of ADP to ATP during the first, phase of photosynthesis Photophosphorylation. 

(v) The point of contact between two neurons ‘Synapse’. 

(vi) Protective membranes covering the human brain and spinal cord ‘Meninges’. 

(vii) Respiratory openings found on the stem of woody plants ‘Lenticels’. 

(viii) Harmful microbes ‘Phagocytosis’. 

(ix) The process of mixing of two different substances/molecules ‘Diffusion’. 

(x) Exudation of sap from injured part of a plant ‘Bleeding’.

4284.

Given below is an example of a certain structure and its special functional activity : Example : (0) Ribosomes and Protein synthesis. On a similar pattern complete the following :(i) Hypothalamus and.........(ii) Suspensory ligaments and -------------.(iii) Semi circular canals and --------------.(iv) Mitochondria and ---------------.(v) Seminiferous tubules and ---------------.

Answer»

(i) Controlling pituitary gland.

(ii) Holding eye lens in position.

(iii) Dynamic balance.

(iv) Cellular respiration.

(v) Spermatogenesis.

4285.

Give the biological / technical terms for the a solution in which the relative concentration of water molecules and the solute on either side of the cell membrane is the same.

Answer»

Isotonic a solution in which the relative concentration of water molecules and the solute on either side of the cell membrane is the same.

4286.

Identify the gas evolved in the following reactions when : (i) Sodium propionate is heated with soda lime. (ii) Potassium sulphite is treated with dilute hydrochloric acid. (iii) Sulphur is treated with concentrated nitric acid. (iv) A few crystals of KNO3 are heated in a hard glass test tube. (v) concentrated hydrochloric acid is made to react with manganese dioxide.

Answer»

(i) Ethane 

(ii) Sulphur dioxide 

(iii) Nitrogen dioxide 

(iv) Oxygen 

(v) Chlorine

4287.

Give a chemical test to distinguish between the pairs of compounds:Hydrogen chloride gas and hydrogen sulphide gas.

Answer»
TestHydrogen Chloride GasHydrogen Sulphide Gas
A rod dipped in ammonium hydroxide is brought near the gasDense white fumes of ammonium chloride are formed.No reaction takes place.

4288.

Give a chemical test to distinguish between the pairs of compounds:Ethene gas and ethane gas.

Answer»
TestEtheneEthane
On shaking few drops of bromine solution in Carbon tetrachloride with the Hydrocarbon.The reddish brown bromine solution gets decolourised.No change is observed.

4289.

Give reasons for the following:(i)  A tuning fork has two prongs.(ii)  Soldiers often march ‘out-of-step” while crossing bridges.

Answer»

(i)  The two prongs of a tuning fork set each other in resonant vibrations which help to maintain the vibrations for a longer time.

(ii)  While crossing a hanging (or rope) bridge, soldiers are asked to break steps because if they are up in steps, all the separate forces being in the same phase, will produce strong energetic vibrations of a particular frequency. If the natural frequency of the bridge becomes equal to the frequency of the steps, the bridge will vibrate will larger amplitude due to resonance and hence the bridge might overturn or crumble down.

4290.

Mention any three purpose for which the plants need water.

Answer»

1) Photosynthesis 

2) Transpiration

3) Transportation

Photosynthesis:  Photosynthesis is the process by which plants make energy to grow. Photosynthesis requires sunlight, water and carbon dioxide. We demonstrated what happens when a plant cannot photosynthesise when we made our cress caterpillar.

Transpiration:  Transpiration is the process by which water moves up the stem of a plant from root to leaf when water is lost from the plant due to evaporation occurring at the leaves. This continual flow of water and nutrients  keeps the plant’s cells firm, if the cells become short of water they lose firmness and the plant starts to wilt.

We can demonstrate transpiration by placing white flowers in coloured water, the water travels up the stem to the petals which become coloured like the water.

Transportation: Transportation is a vital process in plants. Trees transport all the nutrients and water it needs for survival from its roots to the tips of the leaves.

In the case of transportation in plants, the biggest constraint is water as it ends up being a limiting factor in growth.  To overcome this problem, trees and other plants have the perfect system for the absorption and translocation of water.

4291.

An element Z has atomic number 16. Answer the following questions an Z : (i) State the period and group to which Z belongs. (ii) Is Z a metal or a non-metal ? (iii) State the formula between Z and Hydrogen. (iv) What kind of a compound is this ?

Answer»

(i) 3rd period, 16th group. 

(ii) Z is a non-metal. 

(iii) H2

(iv) Covalent compound.

4292.

Which electrode : anode or cathode is the oxidising electrode ? Why?

Answer»

Anode is the oxidising electrode because anions lose electrons at anode.

4293.

Which electrode: anode or cathode is the oxidising electrode? Why?

Answer»

Anode. Because anode is the oxidising electrode, there is loss of electrons.

4294.

Name the kind of particles present in:1. Sodium Hydroxide solution.2. Carbonic acid.3. Sugar solution.

Answer»

1. Ions i.e., Na+ and OH

2. Ions i.e., H+ and CO32– 

3. Molecules C12H22O11

4295.

An element Z has atomic number 16. Answer the following questions on Z: 1. State the period and group to which Z belongs.2. Is Z a metal or a non-metal?3. State the formula between Z and Hydrogen. 4. What kind of a compound is this?

Answer»

Z = 16 = 2, 8, 6.

1. Period No. = 3

Group No. = VI A/16

2. Non metal. 

3. Z’s valency = – 2

H = +1 So formula H2Z

4. Polar Covalent compound.

4296.

From the list given below, select the word(s) required to correctly complete blanks (1) to (5) in the following passage. The words from the list are to be used only once. Write the answers as (a) (1), (2), (3) and so on. Do not copy the passage. [ammonia, ammonium, carbonate, carbon dioxide, hydrogen, hydronium, hydroxide, precipitate, salt water]: A solution M turns blue litmus red, so it must contain (1) …………. ions; another solution O turns red litmus blue and hence, must contain (2) …………. ions. When solutions M and O are mixed together, the products will be (3) ………… and (4) ………… If a piece of magnesium was put into a solution M, (5) ………. gas would be evolved.

Answer»

1. Hydronium 

2. Hydroxide 

3. Salt 

4. Water 

5. Hydrogen

4297.

M is a metal above hydrogen in the activity series and its oxide has the formula M2O. This oxide when dissolved in water forms the corresponding hydroxide which is a good conductor of electricity. In the above context answer the following:1. What kind of combination exists between M and O? 2. How many electrons are there in the outermost shell of M? 3. Name the group to which M belongs. 4. State the reaction taking place at the cathode.5. Name the product at the anode.

Answer»

1. Electrovalent/Ionic compound is formed.

2. 1 

3. M belong to alkali metal group i.e., Group-1. 

4. M- + e → M — Reduction. 

5. Oxygen gas.

4298.

M is a metal above hydrogen in the activity series and its oxide has the formula M2O. This oxide when dissolved in water forms,the corresponding hydroxide which is a good conductor of electricity. In the above context answer the following: (i) What kind of combination exists between M and O ? (ii) How many electrons are there in the outermost shell of M ? (iii) Name the group to which M belongs. (iv) State the reaction taking place at the cathode. (v) Name the product at the anode.

Answer»

(i) Electrovalent bond exists between M and O. 

(ii) One electron is there in the outermost shell. 

(iii) M belongs to First group. 

(iv) M+ + e- -----> M (at cathode). 

(v) Oxygen gas is liberated at anode.

4299.

From the list given below, select the word(s) required to correctly complete blanks (i) to (v) in the following passage. The words from the list are to be used only once.Write the answers as (a) (i), (ii), (iii) and so on. Do not copy the passage. [ ammonia, ammonium, carbonate, carbon dioxide, hydrogen, hydronium, hydroxide, precipitate, salt, water ] (i) A solution M turns blue litmus red, so it must contain (i) .......... ions; another solution O turns red litmus blue and hence, must contain (ii) ....... ions. (ii) When solutions M and O are mixed together, the products will be (iii) ......... and (iv) ............ (iii) If a piece of magnesium was put into a solution M, (v) ......... gas would be evolved.

Answer»

(i) Hydronium 

(ii) Hydroxide 

(iii) Salt 

(iv) Water 

(v) Hydrogen

4300.

Calculate the length of PQ

Answer»

We have AP = 9 cm, CQ = 6 cm and AB = 12 cm.

For triangles △BAP & △BCQ,

∠BAP = ∠BCQ = 90°(\(\because\) Angle between radius and tangent at point of contact in 90°)

∠ABP = ∠CBQ (Vertically opposite angles)

∠APB = ∠BQC (sum of angles in a triangle in 180°)

\(\therefore\) △ BAP \(\sim\) △BCQ (By AAA similarity criteria)

\(\therefore\) AP/CQ = AB/BC

⇒ BC = \(\frac{AB\times CQ}{AP}\) = \(\frac{12\times6}9\)

 = 4 x 2 = 8 cm

Now, in △ ABP,

BP2 = AP2 + AB2

 = 92 + 122

= 81 + 144

= 225

= 152

\(\therefore\) BP = 15 cm

Now, in △BCQ,

BQ2 = BC2 + QC2

 = 82 + 62

= 64 + 36 = 100 = 102

\(\therefore\)  BP = 10 cm

Now, PQ = BP + BQ = 15 + 10

= 25 cm