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11051.

f-block elements have ______ outermost shell(s) incomplete. (A) one (B) two (C) three (D) four

Answer»

Correct answer is (C) three

11052.

______ of an element is determined by the number of valence electrons present in the outermost shell of an atom.

Answer»

Valency of an element is determined by the number of valence electrons present in the outermost shell of an atom.

11053.

______ elements have two outermost shells incompletely filled.

Answer»

Transition or d-block elements have two outermost shells incompletely filled.

11054.

Which is the incomplete period in the modern periodic table?

Answer»

Seventh period is the incomplete period in the modern periodic table.

11055.

Which of the following correctly depicts the atomic radius of a certain atom having three shells?

Answer»

Figure B correctly depicts the atomic radius of a certain atom having three shells.

11056.

Name the followingFirst element of actinide series

Answer»

Thorium (Th)

First element of actinide series is thorium (Th)

Thorium
11057.

Name the followingElement having one shell and one valence electron.

Answer»

Hydrogen

Element having one shell and one valence electron is hydrogen

11058.

Name the followingA group of elements having three incomplete outermost shells

Answer»

Inner transition elements or f-block elements

A group of elements having three incomplete outermost shells is inner transition elements or f-block elements

Three incomplete outermost shell is present in f block elements.
11059.

What problem arises in using alitame as artificial sweetener?

Answer»

Alitame is a high potency sweetener. It is difficult to control the sweetness of food while using alitame as an artificial sweetener.

11060.

Deduce the molecular shape of BrF3 on the basis of VSEPR theory.

Answer»

There are three(3) bond pairs and two(2) lone pairs. The two lone pairs will occupy the position at equatorial which are cause for the distraction of molecule hence it has a distorted ‘T’ shape structure. 

11061.

What will be the names of the following compounds: ICl, BrF?

Answer»

ICl : Iodine monochloride 

BrF : Bromine monofluoride

11062.

Which of the following chemicals can be added for sweetening of food items at cooking temperature and does not provide calories?(i) Sucrose(ii) Glucose(iii) Aspartame(iv) Sucrolose

Answer»

(iv) Sucrolose

11063.

In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (i) Assertion and reason both are correct statement but reason does not explain assertion. (ii) Assertion and reason both are correct and reason explains the assertion. (iii) Both assertion and reason are wrong statement. (iv) Assertion is correct statement reason is wrong statement. (v) Assertion is wrong statement reason is correct statement.Assertion : All chemicals added to food items are called food preservatives.Reason : All these chemicals increase the nutritive value of the food.

Answer»

(iii) Both assertion and reason are wrong statement.

11064.

Explain how animals in Vertebrata are classified into further subgroups.

Answer»

Vertebrates are classified into further subgroups on following bases: 

1. Pisces: The body is streamlined. Muscular tail is present which assists in locomotion. Body is covered with scales. Paired gills are present; which can breathe oxygen dissolved in water. They are cold-blooded animals. The heart has only two chambers. They lay eggs. 

2. Tetrapoda: Animals have four limbs for locomotion and hence the name Tetrapoda. Tetrapoda is divided into four classes, viz. amphibia, reptilia, aves and mammalia

a. Amphibia: These animals are adapted to live both in water and land. Mucus glands on skin keep the skin moist. The animals breathe through skin when in water and through lungs when on land. The heart has three chambers. These are cold-blooded animals. 

Examples: Frog, toad, salamander, etc. 

b. Reptilia: These animals show crawling movement for locomotion. Skin is hardened to form scales. Most of the reptilians have three-chambered heart but crocodile has four-chambered heart. They don’t need water to lay eggs, rather eggs are covered with hard shells and laid on land. 

Examples: snakes, lizards, crocodile, turtle, etc. 

c. Aves: The body is covered with feathers. Forelimbs are modified into wings. These are warm-blooded animals. The heart has four chambers. Bones are hollow (pneumatic); which assists in flying. All the birds belong to this class. 

d. Mammalia: The body is covered with hairs. Skin has seat glands and sebaceous glands. Mammary glands are present in females and are used for nourishing the young ones. Most of the mammalians give birth to young ones and are called viviparous. Some of the mammals lay eggs and are called oviparous. 

Examples: human, chimpanzee, lion, platypus, horse, etc.

11065.

Explain The hierarchy of classification groups.

Answer»

Charles Darwin put forward the idea of evolution in 1859, in his book, The Origin of Species. Ernst Haeckel, Robert Whittaker and Carl Woese have tried to classify living organisms into broad categories, called kingdoms. Robert Whittaker, in 1969 proposed ‘Five kingdoms classification’ of living organisms.

The hierarchy can be represented as Kingdom sub grouping into Phylum for animals or Division for plants, Class, Order, Family, Genus and Species. Hence, the basic unit of classification is species. Species includes all the organisms that are similar to breed and produce fertile offspring.

11066.

What are the advantages of classifying organisms?

Answer»

Classification of organisms is done to make their study simpler and easier. Organisms are classified on many bases like presence or absence of cell wall, their nature i.e. prokaryotic and eukaryotic, their habitat, etc. The major characteristics considered for classifying all organisms are:

  • Type of cellular organization: whether the organism is prokaryotic or eukaryotic 
  • Body organization: whether the organism is multicellular or unicellular 
  • Mode of obtaining food: whether the organism is autotrophic of heterotrophic 
  • Presence or absence of cell wall
11067.

How would you choose between two characteristics to be used for developing a hierarchy in classification?

Answer»

We do not list out the whole hierarchy groups it belongs to. Instead we limit ourselves to writing the name of the genus and species of that particular organism.

11068.

How are the criteria for deciding divisions in plants different from the criteria for deciding sub groups among animals?

Answer»

We can see that, even in these few questions that we have asked, a hierarchy is developing. The characteristics of body design used for classification of plants will be very different from those important for classifying animals. This is because the basic designs are different, based on the need to make their own food (plants), or acquire it (animals). Therefore these design features (having a skeleton, for example) are to be used to make sub-groups, rather than making broad groups.

11069.

What are the major divisions in the Plantae? What is the basis for these divisions?

Answer»

The major divisions of Plantae and the basis for these divisions are as follows:

  • Thallophyta: Simple body design; with no differentiation into root, stem and leaves.
  • Bryophyta: Body is differentiated into stem and leaf-like structures. Vascular system is absent.
  • Pteridophyta: Body is differentiated into root, stem and leaves. Vascular system is present. Reproductive organs are inconspicuous. Seeds are not produced.
  • Gymnosperms: Seeds are naked.
  • Angiosperms: Seeds are covered.
11070.

How would you choose between two characteristics to be used for developing a hierarchy in classification?

Answer»

Before developing a hierarchy we need to finalise the elements which will define the basis for classification.

 Based on these, the common characteristics are used for forming sub-groups.

For example, plants differ from animals in their body design, process of food formation i.e. photosynthesis, etc. But, the main classification between plants & animals is based on the property of locomotion because plants cannot move but animals can. Hence, locomotion is considered as the fundamental characteristic for distinguishing between plants and animals.

11071.

What are the major divisions in the plantae? What is the basis for these divisions?

Answer»

The first level of classification among plants depends on whether the plant body has well differentiated, distinct components. The next level of classification is based on whether the differentiated plant body has special tissues for the transport of water and other substances within it. Further classification looks at the ability’ to bear seeds and whether the seeds are enclosed within fruits.

11072.

Name two plants that belong to herbs?

Answer»

Tomato and potato.

11073.

Part of the plant which conducts water to all parts of the plant is called as ____ Leaf b. Root c. Flower d. Stem

Answer»

Part of the plant which conducts water to all parts of the plant is called as Stem.

11074.

What are the major parts of plants?

Answer»

Stem, root, leaves and flower.

11075.

Differentiate briefly the characteristics of kingdom-Plantae and Animalia.

Answer»
PlanteaAnimalia
They are autotrophic in natureThe are heterotrophic in nature
They are nonmotileThey are motile
They have rigid cell walls. They don't have cell of glycogen
Their growth is unlimitedThey have limited, controlled growth.
They store food in form of starchThey store food in form of glycogen

11076.

What is Animalia?

Answer»

These are organisms which are eukaryotic, multicellular and heterotrophic. Their cells do not have cell wall. Most animals are mobile. Animals are classified on the basis of different features like cellular or tissue level of body organization, body symmetry, type of body cavity called as coelom, presence or absence of segmentation and presence or absence of a backbone.

Classification of the kingdom Animalia includes Invertebrata and Vertebrata.

11077.

Match the parts of plant given in Column I with their function in Column IICOLUMN ICOLUMN II(a) Flower(i) Excretion(b) Leaf(ii) Photosynthesis(c) Stem(iii) Reproduction(d) Root(iv) Bears branches(v) Anchorage

Answer» a– (iii);

b– (ii);

c– (iv);

d– (v)
11078.

What is Animalia?

Answer»

These are multicellular and heterotophs. Cell wall is absent.

11079.

Boojho wanted to test the presence of starch in leaves. He performed the following steps.(1) He took a leaf and boiled it in water,(2) He placed the leaf in a petri dish and poured some iodine over it.He did not get the expected result. Which step did he miss? Explain.

Answer» He did not boil the leaf in spirit to remove the chlorophyll.
11080.

Which is the largest group among animals?

Answer»

Arthropoda is the largest group among animals.

11081.

Will a leaf taken from a potted plant kept in a dark room for a few days turn blue black when tested for starch? Give reasons for your answer.

Answer»

No, all the starch stored in the leaf would have been used up by the plant. No starch would be synthesised afresh due to non-availability of sunlight.

11082.

Select the incorrect statement from given statements: (a) A tonic used for cardiac arrest is obtained from Digitalis purpuria (b) Medicine used to treat Rheumatic pain is extracted from Capsicum annum (c) An anti-malarial drug is isolated from Cinchona officinalis (d) Anti-carcinogenic property is not seen in Catharanthus roseus.(e) All the above are correct

Answer»

(e) All the above are correct

11083.

Which of the following is not a consequence of deforestation? A. increase in soil erosion B. increase in earth’s temperature C. decrease in rainfall D. decrease in floods

Answer»

C. decrease in rainfall 

Deforestation leads to increase in floods. It leads to a decrease in rainfall and water level.

11084.

Does soil erosion affect the fertility of soil? How?

Answer»

Yes. Soil erosion removes the fertile top layer of the soil thereby, exposing the hard rocky lower layers which are less fertile.

11085.

How does deforestation cause soil erosion?

Answer»

Deforestation means cutting of trees. When trees are cut, the soil becomes loose. When there is heavy rainfall or flood, the soil being loose is washed away and this leads to soil erosion. This leads to flood and landsliding.

11086.

What is venation? Explain its types.

Answer»

Venation is the phenomenon of arrangement of veins in the lamina of the leaf surface. The veins are not only the conducting channels for water, minerals and organic food, they also provide firmness to the lamina and keep it expanded.

Venation is of three types : Reticulate, parallel and furcate.

1. Reticulate Venation : When the veinlets form a network the venation is termed as reticulate venation. E.g., leaves of dicot plants.

Reticulate venation has two sub-types :

(i) Pinnate or Unicostate reticulate venation: The lamina has a single principal vein or midrib extending from base to the apex. It produces lateral veins, e.g., Peepal, Shoe flower etc.

(ii) Palmate or Multicostate reticulate venation:

The lamina has more than one equally prominent or principal veins. They give rise to lateral veins connected by reticulations of veinlets. Multicostate reticulate venation has two forms.

(a) Convergent: The principal veins converge towards the apex of the lamina.

(b) Divergent : The principal veins diverge towards the margin.

2. Parallel Venation: When the veins arising from mid rib or main veins, run parallel to each other towards the margin or the apex of the lamina, venation is termed as parallel venation. Leaves of monocot plants. 

Parallel venation is of two sub-types :

(i) Pinnate or Unicostate parallel venation : The lamina has a single prominent vein or midrib running from the base to the apex of lamina. e.g., Banana, Canna, etc.

(ii) Palmate or multicostate parallel venation :
The lamina has several principal veins arising from the base and running towards the apex or margin of the lamina. Multicostate parallel venation has two forms :
(a) Convergent: The principal veins converge towards the apex, e.g., Bamboo, Grass.
(b) Divergent : The principal veins diverge towards the margin, e.g., Fan palm.

3. Furcate Venation : Here, the veins give off dichotomous branching. The finer branches do not form reticulum. Furcate venation is common in ferns, circeaster.

11087.

Name two types of nephrons found in human kidney.

Answer»

(i) Juxta medullary nephron 

(ii) Cortical nephron

11088.

The mechanism of concentration of filtrate is also known as counter current mechanism. Justify the statement. 

Answer»

(in the ascending limb) the out flow runs parallel to and in the opposite direction of the inflow in the descending limb.

11089.

Name any five endangered specie of animals listed in the Red Data Book of India.

Answer»

Asiatic lion (Panthera leo persica) Bengal tiger (Panthera tigris tigris) Blue whale (Balaenoptera musculus) Banteng (Bos javanicus) Central Kashmir vole (Alticola montosa)

11090.

Explain how, deforestation leads to reduced rainfall.

Answer»

Plants need carbon dioxide for the process of photosynthesis. Fewer trees would mean that less carbon dioxide will be used up resulting in its increased amount in the atmosphere and increase in greenhouse effect. The increase in temperature on the earth disturbs the water cycle and may reduce rainfall. When number of trees are reduced then less amount of water is evaporated and hence the reduced evaporation leads to reduced rainfall.

11091.

Explain how, deforestation leads to frequent flooding of rivers.

Answer»

Deforestation plays several roles in the flooding equation Tree roots hold the soil in place, reducing the movement of sediment that can shrink river channels downstream. However in the absence of trees, soil is loosened, and it gets washed away easily with water or torrential rains and is deposited into river. When rivers have soil and water level more than their holding capacity, it lead to over flooding and causes floods.

11092.

Which of the following is an endangered species of animals? A. Dinosaur B. Asiatic lion C. Irish deer D. Hyena

Answer»

B. Asiatic lion 

The Asiatic lion, also known as the Indian lion and Persian lion, is a lion subspecies that lives as a single population in India’s Gujarat State. It is listed as Endangered on the IUCN Red List because of its small population size.

11093.

Explain how deforestation leads to reduced rainfall

Answer»

Plants absorb water from the soil and evaporates it to form clouds. Deforestation leads to cutting down of trees. As a result less water is absorbed from the soil and it disturbs the water cycle. The formation of clouds becomes difficult which leads to reduced rainfall. 

11094.

Explain the process of urine formation in human being.

Answer»

Urine formation includes three processes: 

(i) Ultrafiltration, (ii) Selective reabsorption, (iii) Selective Secretion. 

(i) Ultrafiltration: It is filtration which occurs under high blood pressure. The afferent arteriole supplies blood to glomerulus and efferent arteriole collects blood from glomerulus. Therefore, the blood pressure in glomerulus becomes 75 mm of Hg. The smaller molecules of glucose, water, salts, ammonia etc. diffuse from glomerulus to Bowman's capsule and this process is called ultrafiltration. About 1.5-2.5% of water and solutes are removed from the blood plasma that flows through glomerulus. The quantity of glomerular filtrate formed each minutes in all nephrons of both kidneys is called the glomerular filtration rate. In a normal person, the glomerular filtration rate is about 125ml. per minute. 

(ii) Selective reabsorption: It is the process by which useful material are absorbed by the blood from glomerular filtrate. The tubular epithelial cells in different segments of nephron perform this either by active or passive mechanisms. Substances like glucose, amino acids, Na+ , etc., in the filtrate are reabsorbed actively where as the nitrogenous wastes are absorbed by passive transport. In collecting tubules some water is absorbed. 

(iii) Selective secretion: The metabolic waste products like ammonia, hippuric acid, creatinine, H+ ions etc., present in the blood diffuses into DCT. This process is called selective secretion. It changes the complete composition of glomerular filtrate which is now called urine.

11095.

Mention the role of DCT in urine formation.

Answer»
  • Conditional reabsorption of Na+ and water takes place in this segment.
  • It is also reabsorbed HCO3
  • There is selective secretion of hydrogen and potassium ions to maintain the pH and sodium-potassium balance in the blood
11096.

Define GFR (Glomerular Filtration Rate).

Answer»

The amount of filtrate formed by the kidney per minute.

11097.

Explain the process of urine formation in humans. OR Explain the step of reabsorption and secretion in file process of formation of urine.

Answer»

Urine formation involves three processes : glomerular filtration, tubular reabsorption and tubular secretion.

(a) Glomerular Filtration: 

(i) Walls of glomerular capillaries and Bowman’s capsule are very thin and ‘ semipermeable. 

(ii) They allow water and small molecules in the blood to pass across the capillary wall through tiny spaces between the cells of the Bowman’s capsule, 

(iii) The blood pressure is high in the glomerular capillaries because the efferent arteriole, which drains them, is narrower than the , afferent arteriole, which supplies them, 

(iv) This pressure is about 75 mm. Hg. in man. The net force moving the fluid from the glomerular capillaries called the filtration pressure, 25 mm. Hg. Filtration of fluid under pressure is termed ultrafiltration.

(b) Tubular Reabsorption: From the Bowman’s capsule, the glomerular filtrate passes into the tubule and flows through it to the collecting duct. During this course, its composition, osmotic pressure and pH change due to selective reabsorption of materials from it.

Proximal Convoluted Tubule : 

(i) The cells lining, the proximal convoluted tubule, 

(ii) Mitochondria power the active transport of nutrient molecules back into the blood, 

(iii) Microvilli increase the surface for reabsorption, 

(iv) The cells reabsorption: glucose, almost all amino acids, most of the inorganic ions (Na+, K+, Cl-), much of the water as well as some urea, uric acid and creatinine from the filtrate and secrete them back into the blood capillaries around the tubule.

(B) Henle’s Loop : (i) In the ascending limb of Henle’s loop the remaining 25% of K+ and some Cl- and Na+ ions are actively reabsorbed, 

(ii) The ascending limb is impermeable to water. Therefore, no water is reabsorbed.

(iii) The Na+ leaving the ascending limb passively diffuses into the descending limb because now its concentration is higher in the interstitial fluid than in the descending limb.

(c) Distal Convoluted Tubule, Collecting Tubule and Collecting Duct : These regions of the nephron actively reabsorb some Na+ from the filtrate, and in exchange excrete some K+ in the urine. Some Cl- is reabsorbed by diffusion from the DCT.

(D) Tubular Secretion: (i) Cells of the distal convoluted tubule and collecting duct of the nephron excrete additional waste from the blood-stream into the filtrate by active transport, 

(ii) This process of tubular secretion adds urea, creatinine, hippuric acid, potassium and hydrogen ions (H+), ammonia (NH3) and a little uric acid, 

(iii) Removal of hydrogen ions and ammonia from the blood helps to maintain the pH of the blood.

11098.

Explain the process of urine formation.

Answer»

During urine formation, the nitrogenous waste materials are removed as urine from the blood without losing the valuable molecules of blood in the process. It involves three main stages namely.

1. Ultra filtration (Glomerular filtration)

2. Reabsorption (tubular)

3. Tubular secretion.

1. Ultra filtration: Involves the removal or filtration of substances from blood of the glomerular lumen into the Bowman’s capsular lumen through the Ultra fine endothelia capsular membrane. Glomerular filtration is caused due to changes in blood pressure, the pressure of the Bowman’s capsular fluid and osmotic pressures excreted by contents in glomerular and Bowman’s lumen. The glomerular filtrate resembles the blood plasma in its chemical composition except for the molecules of high molecular weight.

2. Tubular reabsorption: This is the second step or stage in urine formation and involves the reabsorption of physiologically important substances into the surrounding capillaries when the primary urine moves down route in the nephron tubules. The reabsorption may be passive cr active at various regions of the convoluted tubule. During this process glucose, amino acids, inorganic salts and some water is absorbed. Some of the absorptions are under the influence of hormones.

3. Tubular secretion: (Active secretion) This is the last stage or step in the urine formation at the end of which a hypertonic or highly concentrated urine is formed for excretion. During this stage, the capillaries surrounding the convoluted tubule actively secrete certain unwanted substances which could not be the filtered in the Bowman’s capsule by the glomerulus into the filtrate of the tubular lumen.

As a result of these 3 processes or stages .in urine formation, homeostasis is blood is restored and waste products remain in the tubular fluid forming concentrated urine for excretion from urine. Two hormones come into play during the process namely ADH for reabsorption of water when the body has the deficit of water and Aldosterone helping in reabsorption of ions like K+ when the body faces the deficit of these ions.

11099.

Why does filtrate become concentrated as it passes through the descending limb of Henle’s loop?

Answer»

Descending limb is permeable to water but not to solutes, so the filtrate becomes concentrated.

11100.

Define Glomerular Filtration Rate (GFR).

Answer»

Glomerular Filtration Rate (GFR): The amount of the filtrate formed by the kidneys per minute is called glomerular filtration rate (GFR). GFR in a healthy individual is approximately 125 ml/minute, i.e., 180 litres per day. On an average, 1100-1200 ml of blood is filtered by the kidneys per minute which constitute roughly 1/5th of the blood pumped out by each ventricle of the heart in a minute.