CBSE EXAMINATION PAPER-2010
NEET-ZOOLOGY
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 200 questions. All questions are compulsory.
- This question paper is divided into 1 sections.
- Section A – questions number 1 to 200 are multiple choice questions
- There is no overall choice given in the question paper. However, an internal choice has been provided in few questions.
- Use of calculator is NOT allowed.
Section A
The number of atoms in 0.1 mol of a triatomic gas is: (N_A = 6.02 × 10²³ mol⁻¹)
The number of atoms in 0.1 mol of a triatomic gas
= 0.1 × 3 × 6.023 × 10²³
= 1.806 × 10²³
Which of the following represents the correct order of increasing electron gain enthalpy with negative sign for the elements O, S, F and Cl?
O < S < F < Cl
Electron gain enthalpy of the given elements are −141, −200, −333 and −349 kJ mol⁻¹ respectively.
The correct order of increasing ionic radii among the following isoelectronic species are:
Among the isoelectronic species, size increases with the increase in negative charge. Thus S²⁻ has the highest negative charge and hence largest in size followed by Cl⁻, K⁺ and Ca²⁺.
In which one of the following species the central atom has the type of hybridization which is not the same as that present in the other three?
For SbCl₅²⁻: 5 + 5 + 2 / 2 = 6 means sp³d² hybridisation.
I₃⁻, SF₄, and PCl₅ ; all have sp³d hybridization.
In which of the following pairs of molecules/ions, the central atoms have sp² hybridization?
BF₃: 3+3/2 = 3 means sp² hybridisation.
NO₂⁻: 5+1/2 = 3 means sp² hybridisation.
Which one of the following species does not exist under normal conditions?
Bond order of Be₂ = 0, hence Be₂ cannot exist.
Standard entropies of X₂, Y₂ and XY₃ are 60, 40 and 50 JK⁻¹ mol⁻¹, respectively.
(1 / 2) X₂ + (3 / 2)Y₂ ⇌ XY₃, ΔH = −30 kJ
to be at equilibrium, the temperature should be:
ΔS for the reaction (1 / 2) X₂ + (3 / 2)Y₂ ⇌ XY₃,
ΔS = 50 − (30 + 60) = −40 J
For equilibrium,
ΔG = 0 = ΔH − TΔS
T = ΔH / ΔS = −30000 / -40 = 750 K
In which of the following equilibrium K_c and Kₚ are not equal?
2C(s) + O₂(g) ⇌ 2CO₂(g)
Δn = 2 − 1 = +1
∴ K_c and Kₚ are not equal.
Which one of the following molecular hydrides acts as a Lewis acid?
Boron in B₂H₆ is electron deficient.
If pH of a saturated solution of Ba(OH)₂ is 12, the value of its K_sp is:
.
What is [H⁺] in mol/L of a solution that is 0.20 M in CH₃COONa and 0.10 M in CH₃COOH? (For CH₃COOH, Kₐ = 1.8 × 10⁻⁵)
pH = pKₐ + log[Salt / Acid]
log[H⁺] = log Kₐ− log[Salt / Acid]
log[H⁺] = log Kₐ + log[Acid / Salt]
[H⁺] = Kₐ[Acid / Salt]
= 1.8 × 10⁻⁵ × 0.1 / 0.2 = 9 × 10⁻⁶ M
In a buffer solution containing equal concentration of B⁻ and HB, the K_b for B⁻ is 10⁻¹⁰. The pH of the buffer solution is:
K_b = 10⁻¹⁰, Kₐ = 10⁻⁴ or pKₐ = 4
For the buffer solution containing equal concentration of B⁻ and HB,
pH = pKₐ + log1
pH = pKₐ = 4
Which of the following alkaline earth metal sulphates has hydration enthalpy higher than the lattice enthalpy?
Be²⁺ is very small, hence its hydration enthalpy is greater than its lattice enthalpy. Also, hydration energy decreases down the group.
Property of the alkaline earth metals that increases with their atomic number:
The magnitude of hydration energy for the hydroxides of alkaline earth metals remains almost the same whereas lattice energy decreases appreciably down the group. Hence, solubility increases down the group.
The tendency of BF₃, BCl₃ and BBr₃ to behave as Lewis acid decreases in the sequence:
In BF₃, p–p overlap between B and F is maximum due to the identical size and energy of p-orbitals, so electron deficiency in boron of BF₃ is neutralized partially to the maximum extent by back donation. Also, the tendency to back donate decreases from F to I. So the order will be:
BF₃ > BCl₃ > BBr₃.
In the following the most stable conformation of n-butane is:
-
(A)
-
(B)
-
(C)
-
(D)
The bulky methyl groups are maximum away from each other.
Liquid hydrocarbons can be converted to a mixture of gaseous hydrocarbons by :
During cracking, higher hydrocarbons (liquid) are converted to lower gaseous hydrocarbons.
Which one is most reactive towards electrophilic reagent?
Among –OH, –CH₂OH, –NHCOCH₃ and –OCH₃, methoxy (–OCH₃) group has the highest +M effect.
AB crystallizes in a body centred cubic lattice with edge length 'a' = 387 pm. The distance between two oppositely charged ions in the lattice is:
For bcc lattice, diagonal = a√3
The distance between the two oppositely charged ions =
(a / 2)√3 = 387 × 1.732 / 2 ≃ 335pm
25.3 g of sodium carbonate, Na₂CO₃, is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ions, Na⁺, and carbonate ions, CO₃²⁻, are respectively. (Molar mass of Na₂CO₃ = 106 g mol⁻¹)
Concentration of
Na₂CO₃ = 25.3/106 × 1000/250 = 0.955 M
[Na⁺] = 2 × 0.955 = 1.91 M
[CO₃²⁻] = 0.955 M
An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase?
When the aqueous solution of one molal KI is diluted with water, concentration decreases; therefore, the vapour pressure of the resulting solution increases.
A solution of sucrose (molar mass = 342 g mol⁻¹) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be (K_f for water = 1.86 K kg mol⁻¹):
ΔT_f = K_f × (1000W₂ / M₂W₁)
= (1.86 × 1000 × 68.5) / (342 × 1000)
= 0.372°C
T_f = −0.372°C ∵ (0 − ΔT_f)
An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to:
Strong electrolytes are completely ionised at all concentrations. On increasing dilution, the number of ions remains the same but the ionic mobility increases.
For the reduction of silver ions with copper metal, the standard cell potential was found to be +0.46 V at 25°C. The value of standard Gibbs energy, ΔG°, will be: (F = 96500 C mol⁻¹)
Cu + 2Ag⁺_(aq) → Cu²⁺_(aq) + 2Ag₍ₛ₎
Here,
n = 2, E_cell° = +0.46 V
ΔG° = −nE°F = (−2 × 0.46 × 96500) / 1000 kJ = −89 kJ
For the reaction [N₂O₅(g) → 2NO₂ (g) + 1/2O₂(g)] the value rate of disappearance of N₂O₅ is given as 6.25 × 10⁻³ mol L⁻¹ s⁻¹. The rate of formation of NO₂ and O₂ is given respectively as:
N₂O₅ (g) ⟶ 2 NO₂ (g) + 1 / 2 O₂ (g)
− d / dt [N₂O₅] = + (1 / 2) (d / dt) [NO₂] = 2 (d / dt) [O₂]
d / dt [NO₂] = 1.25 × 10⁻² mol L⁻¹ s⁻¹ and
d / dt [O₂] = 3.125 × 10⁻³ mol L⁻¹ s⁻¹
During the kinetic study of the reaction, 2A + B → C + D following results were obtained:
In case of (II) and (III), keeping the concentration of [A] constant, when the concentration of [B] is doubled, the rate quadruples. Hence, it is second order with respect to B.
In case of (I) and (IV), keeping the concentration of [B] constant, when the concentration of [A] is increased four times, the rate also increases four times. Hence, the order with respect to A is one.
hence
Rate = k [A] [B]²
For an endothermic reaction, energy of activation is Eₐ and enthalpy of reaction is ΔH (both of these in kJ mol⁻¹). Minimum value of Eₐ will be:
Oxidation states of P in H₄P₂O₅, H₄P₂O₆, and H₄P₂O₇ are respectively:
Which one of the following compounds is a peroxide?
The correct order of increasing bond angles in the following species is:
(i) As the number of lone pair of electrons increases, bond angle decreases due to repulsion between Ip — Ip.
(ii) As the electronegativity of the central atom increases, bond angle increases. Hence, the correct order of bond angle:
Which of the following ions will exhibit colour in aqueous solutions?
La³⁺ : 54 e⁻ = [Xe]
Ti³⁺ : 19 e⁻ = [Ar] 3d¹ (Colour)
Lu³⁺ : 68 e⁻ = [Xe] 4f¹⁴
Sc³⁺ : 18 e⁻ = [Ar]
Which one of the following ions has electronic configuration [Ar] 3d⁶?
(At. Nos. Mn = 25, Fe = 26, Co = 27, Ni = 28)
Ni³⁺ : [Ar] 3d⁷
Mn³⁺ : [Ar] 3d⁴
Fe³⁺ : [Ar] 3d⁵
Co³⁺ : [Ar] 3d⁶
Which one of the following pairs has the same size?
Due to lanthanide contraction, the size of Zr and Hf (atoms and ions) become nearly similar.
The existence of two different coloured complexes with the composition [Co(NH₃)₄Cl₂]⁺ is due to:
.
Which one of the following complex ions is not expected to absorb visible light?
[Ni(CN)₄]²⁻ : Number of unpaired electrons = 0
[Cr(NH₃)₆]³⁺ : Number of unpaired electrons = 3
[Fe(H₂O)₆]²⁺ : Number of unpaired electrons = 4
[Ni(H₂O)₆]²⁺ : Number of unpaired electrons = 2
Crystal field stabilization energy for high-spin d⁴ octahedral complex is:
The reaction of toluene with Cl₂ in presence of FeCl₃ gives 'X' and reaction in presence of light gives 'Y'. Thus, 'X' and 'Y' are:
Which one is most reactive towards S_N1 reaction?
S_N1 reactions involve the formation of carbocations; hence, the higher the stability of intermediate carbocation, the more reactivity of the parent alkyl halide. The tertiary carbocation formed from (c) is stabilized by two phenyl groups and one methyl group, hence it is the most stable.
The correct order of increasing reactivity of C - X bond towards nucleophile in the following compounds is
Tertiary alkyl halide is most reactive towards nucleophilic substitution because the corresponding carbocation (3°) is most stable. Aryl halide is least reactive due to the partial double bond character of the C–Cl bond.
Presence of –NO₂ groups at ortho and para positions increases the reactivity of the C–Cl bond towards nucleophiles.
Which one of the following compounds has the most acidic nature?
-
(A)
-
(B)
-
(C)
-
(D)
Phenol is most acidic because its conjugate base is stabilised due to resonance, while the rest three compounds are alcohols, hence, their corrosponding conjugate bases do not exhibit resonance.
Which of the following reactions will not result in the formation of carbon–carbon bonds?
In a set of reactions, ethylbenzene yielded a product D.
-
(A)
-
(B)
-
(C)
-
(D)
Given are cyclohexanol (I), acetic acid (II), 2,4,6-trinitrophenol (III) and phenol (IV). In these, the order of decreasing acidic character will be:
Explanation: Presence of three –NO₂ groups in o-, p- positions to phenolic groups (in III) makes phenol strongly acidic because its corresponding phenate ion (conjugate base) is highly stabilised due to resonance.
Conjugate base of CH₃COOH (II) (i.e., CH₃COO⁻) shows resonance hybrid of two equivalent structures. The conjugate base of phenol (IV) is stabilized due to resonance (note that here all resonating structures are not equivalent). The conjugate base of cyclohexanol, I does not exhibit resonance, hence, not formed.
Among the given compounds, the most susceptible to nucleophilic attack at the carbonyl group is:
Cl⁻ is the weakest base and hence a better leaving group.
Which of the following statements about primary amines is False?
Aryl amines react with nitrous acid to produce diazonium salt and not phenol.
Acetamide is treated with the following reagents separately. Which one of these would yield methylamine?
Aniline in a set of the following reactions yielded a coloured product ‘Y’
The structure of ‘Y’ would be ;
-
(A)
-
(B)
-
(C)
-
(D)
Which one of the following does not exhibit the phenomenon of mutarotation ?
Sucrose does not have free –CHO or CO group; hence, it does not undergo mutarotation.
Which one of the following is employed as a tranquilizer drug?
Valium is used as a sedative.
Which of the following structures represents neoprene polymer?
-
(A)
-
(B)
-
(C)
-
(D)
Neoprene is a polymer of chloroprene (2-chloro-1,3-butadiene).
Single-celled eukaryotes are included in:
Single celled eukaryotes are included in protista. Protista includes all unicellular and colonial eukaryotes except green and red algae. It is also known as kingdom of unicellular eukaryotes.
Membrane-bound organelles are absent in:
Membrane-bound organelles are absent in Streptococcus. It is a bacterium that is included under kingdom Monera. Monerans are prokaryotes, which lack membrane-bound organelles like mitochondria, E.R., Golgi etc. Saccharomyces, Chlamydomonas and Plasmodium are eukaryotes that have membrane-bound organelles.
Virus envelope is known as:
Virus envelope is known as capsid. The capsid is composed of protein subunits called capsomere.
Algae have cell wall made up of:
Algae possess a definite cell wall containing cellulose, galactans and mannans.
Some hyperthermophilic organisms that grow in highly acidic (pH₂) habitats belong to the two groups:
Hyperthermophilic organisms that grow in highly acidic habitats belong to eubacteria and archaea groups.
Male and female gametophytes are independent and free-living in:
Sphagnum is a bryophyte in which male and female gametophytes are independent and free-living. In Pinus (a gymnosperm), mustard and castor (angiosperms), the main plant body is sporophytic. Gametophyte is highly reduced and is completely dependent on sporophyte.
Which one of the following kinds of animals are triploblastic?
Triploblastic condition can be seen in flatworms. Ctenophores, sponges and corals are diploblastic.
One example of animals having a single opening to the outside that serves both as mouth as well as anus is:
Fasciola, a flatworm, has a single body cavity to the outside that serves both as mouth for ingestion and anus for egestion of undigested food. This is known as blind sac plan.
Which one of the following statements about certain given animals is correct?
Acoelomates are animals that have no body cavity or coelom. The examples are poriferans, coelenterates, ctenophores, platyhelminthes and nemertean. Pseudo-coelomates are animals that have false or pseudo coelom. Examples are aschelmin-thes.
Which one of the following statements about all the four of Spongilla, Leech, Dolphin and Penguin is correct?
Spongilla is a fresh water sponge that belongs to phylum Porifera. It has special collared cells called choanocytes. Choanocytes are not found in leech, dolphin and penguin.
In a unilocular ovary with a single ovule, the placentation is:
In basal type of placentation, the ovary is unilocular with a single ovule. In this, the placenta is at the base of the ovary. It is seen in Polygonum.
Ovary is half-inferior in the flowers of:
A flower in which floral parts arise from around the ovary is called perigynous. In this, the ovary is half inferior and half superior. It can be seen in the flowers of plum, peach etc.
The technical term used for the androecium in a flower of China rose (Hibiscus rosa-sinensis) is:
Hibiscus belongs to family Malvaceae. It possess numerous stamens that are called monadelphous.
Keel is characteristic of the flowers of:
Keel is the characteristic of the flowers of family papilionaceae. For example, bean, gram, pea etc.
The chief water conducting elements of xylem in gymnosperms are:
Tracheids are chief water conducting elements of xylem in gymnosperms. They are devoid of protoplasm and hence dead. The wall constituting the tracheids is hard, thick and lignified. These are elongated cells with tapering ends.
Which one of the following is not a lateral meristem?
Meristem is divided onthe basis of position in plant bodies into apical meristem, lateral meristem and intercalary meristem. Lateral meristem is present on the lateral sides, that is fascicular and interfascicular cambium and cork cambium (phellogen).
Heartwood differs from sapwood in:
Heartwood differs from sapwood in having dead and non-conducting elements. In old trees, the inner region that comprises dead elements with highly lignified walls is called heartwood.
The kind of epithelium which forms the inner walls of blood vessels is:
Squamous epithelium is formed of thin discoidal and polygonal cells that fit like tiles in a floor, so is also called pavement epithelium. It is found in the walls of blood vessels, in the alveoli of lung for exchange of gas, and in Bowman’s capsule of nephron for ultra filtration.
Which one of the following structures between two adjacent cells is an effective transport pathway?
Plasmodesmata are the structures present between two adjacent cells that permits the transport and communication between them. They are the fine cytoplasmic strands that connect the protoplasts of adjacent plant cells by passing through the cell walls.
Which one of the following has its own DNA?
Mitochondria has its own DNA. It is a structure within cytoplasm of eukaryotic cells that carries out aerobic respiration. It is the site of Kreb’s cycle and ETS. Therefore, it is also called as cell’s energy production site.
The main arena of various types of activities of a cell is:
The main arena of various types of activities of a cell is cytoplasm. It forms the living protoplasm of a cell excluding the nucleus. It consists of proteins, fats, carbohydrates, nucleic acids, vitamins, waste metabolites and all organelles.
The plasma membrane consists mainly of:
Plasma membrane comprises of mainly proteins embedded in a phospholipid bilayer. Protein molecules occur at places both inside and outer side of the lipid bilayer.
During mitosis, ER and nucleolus begin to disappear at:
During mitosis, endoplasmic reticulum and nucleus begin to disappear at early prophase. Also, nucleus and cell become spherical and DNA molecules condense to form shortened chromosome.
Which stages of cell division do the following figures A and B represent respectively?
-
(A)
-
(B)
-
(C)
-
(D)
The figure A represents late anaphase while figure B represents prophase stage of mitosis.
Which one of the following is not a micronutrient?
Magnesium is a macronutrient. Micronutrients are essential elements that are required by plants in small amounts. They include Fe, Cu, B, Mo, Mn, Cl, and Ni. Macronutrients, on the other hand, are those essential elements that are required by plants in larger amounts. The examples include C, H, O, N, S, P, K, Ca and Mg.
An element playing important role in nitrogen fixation is:
Molybdenum is a micronutrient that plays an important role in nitrogen fixation. It is a part of nitrate reductase enzyme that helps in nitrogen fixation.
One of the free-living, anaerobic nitrogen-fixer is:
Rhodospirillum is a free-living, anaerobic nitrogen fixer bacteria. It can synthesize its food in presence of light under anaerobic conditions.
Beijernickia and Azotobacter are free living aerobic nitrogen fixing bacteria. Rhizobium is a symbiotic nitrogen fixing bacteria.
PGA as the first CO₂ fixation product was discovered in photosynthesis of:
Phosphoglyceric acid (PGA) is the first stable product of photosynthesis. It was first discovered by Calvin, Benson and their colleagues in Chlorella, (alga).
C₄ plants are more efficient in photosynthesis than C₃ plants due to:
C₄ plants are more efficient in photosynthesis than C₃ plants due to photorespiration. Photorespiration is an inhibitory process that decreases the rate of photosynthesis. It is absent in C₄ plants and present only in C₃ plants.
The energy-releasing metabolic process in which substrate is oxidised without an external electron acceptor is called:
The energy releasing metabolic process in which substrate is oxidised without an external electron acceptor is called fermentation.
Coiling of garden pea tendrils around any support is an example of:
The coiling of garden pea tendrils around any support is an example of thigmotropism. Thigmotropism is the growth movement in response to touch. The stems and tendrils of the climbers are positively thigmotrophic in their response.
Phototropic curvature is the result of uneven distribution of:
Phototropic curvature is the result of uneven distribution of auxin. The experiments conducted by Charles Darwin concluded that the tip of coleoptile of canary grass contain auxin that causes the bending of the entire coleoptile towards the light source.
Photoperiodism was first characterised in:
Photoperiodism is the response to duration and timings of light and dark period. It was first characterised in tobacco plant.
If for some reason our goblet cells are non functional, this will adversely affect:
Goblet cells, found in the intestinal mucosal epithelium, secrete mucus. The mucus lubricates the food for an easy passage. So, if for some reason, goblet cells become non-functional, it will adversely affect the smooth movement of food down the intestine.
Listed below are four respiratory capacities (i–iv) and four jumbled respiratory volumes of a normal human adult:
Respiratory capacities | Respiratory volumes |
|---|---|
(i) Residual volume | 2500 mL |
(ii) Vital capacity | 3500 mL |
(iii) Inspiratory reserve volume | 1200 mL |
(iv) Inspiratory capacity | 4500 mL |
Which one of the following is the correct matching of two capacities and volumes?
The correct matching of respiratory capacities with their respiratory volumes are:
Which two of the following changes (a–d) usually tend to occur in the plain dwellers when they move to high altitudes (3,500 m or more)?
(i) Increase in red blood cell size
(ii) Increase in red blood cell production
(iii) Increased breathing rate
(iv) Increase in thrombocyte count
Changes occurring are:
At high altitude, the body undergoes numerous changes in order to increase oxygen delivery to cells and improve the efficiency of oxygen usage. The early changes include increased breathing rate and increased red blood cell production.
ABO blood groups in humans are controlled by the gene I. It has three alleles—Iᴬ, Iᴮ and i. Since there are three different alleles, six different genotypes are possible. How many phenotypes can occur?
The three alleles in ABO blood groups in humans can produce six different genotypes and four different phenotypes.
What is true about RBCs in humans?
Blood transports oxygen from respiratory organs to the tissue cells and also transports carbon dioxide from the tissue cells to the respiratory membrane. About 97% of oxygen is transported by RBCs in the blood while the remaining three percent of oxygen is carried in a dissolved state through plasma. Nearly 20-25% of carbon dioxide is transported by RBCs whereas 70% is carried as bicarbonate. About 7% of CO₂, is carried in dissolved state through plasma.
If due to some injury the chordae tendinae of the tricuspid valve of the human heart is partially non-functional, what will be the immediate effect?
Chordae tendinae are thread like bands of fibrous tissue which attach on one end to the edges of the tricuspid and mitral valves of the heart and on the other end to the papillary muscles, that serve to anchor the valves. Tricuspid valve is the valve in the heart between the right atrium and right ventricle. The valve opens to allow blood to flow from atrium into the ventricle. Thus if tricuspid valve is partially non-functional, then, the flow of blood into the pulmonary artery will be reduced.
Which one of the following statements in regard to the excretion by the human kidneys is correct?
Urine formation involves three main processes called, glomerular filtration, reabsorption and secretion. A comparison of the volume of the filtrate formed per day (which is 180 litres per day) with that of urine released (about 1.5 litres) suggests that nearly 99 percent of the glomerular filtrate is resorbed by the renal tubules. The descending limb of loop of Henle is permeable to water but impermeable to electrolytes.
The ascending limb is impermeable to water but allows transport of electrolytes. Reabsorption of sodium ions and water takes place in distal convoluted tubule.
Also, it is capable of reabsorption of HCO₃.
The principal nitrogenous excretory compound in humans is synthesised:
The principal nitrogenous excretory compound in humans is urea. It is synthesized in the mitochondrial matrix and cytosol of liver cells and eliminated through kidneys.
Consider the following four statements (i–iv) regarding kidney transplant and select the two correct ones out of these.
(i) Even if a kidney transplant is proper, the recipient may need to take immune suppresants for a long time.
(ii) The cell-mediated immune response is responsible for graft rejection.
(iii) The B lymphocytes are responsible for rejection of the graft.
(iv) The acceptance or rejection of a kidney transplant depends only on specific interferons.
The two correct statements are:
Tissue and blood group matching are essential before undertaking kidney transplant. Even if kidney transplant is proper, the recipient may need to take immuno suppresant all his/her life. The ability of body to differentiate self and non-self and the cell-mediated immune response is responsible for graft rejection.
Low Ca⁺⁺ in the body fluid may be the cause of:
Tetany is caused by reduction in the calcium level due to underactive parathyroid hormone.
The nerve centres which control the body temperature and the urge for eating are contained in:
Hypothalamus contains important nerve centres that controls the body temperature, thirst, hunger and eating, water balance and sexual function.
Injury to adrenal cortex is not likely to affect the secretion of which one of the following?
Adrenal gland has two parts—cortex and medulla. The medulla is stimulated by sympathetic nervous tissue to produce adrenaline and nor-adrenaline while the cortex is stimulated by pituitary hormone to release cortisol, aldosterone and estrogens. Thus injury to adrenal cortex does not likely to affect the secretion of adrenaline.
Which one of the following pairs is incorrectly matched?
Glucagon is secreted by α-cells of the islets of Langerhans in the pancreas. It increases the concentration of glucose in the blood by stimulating the breakdown of glycogen.
Toxic agents present in food which interfere with thyroxine synthesis lead to the development of:
Goitre is caused by deficiency of iodine in diet. Iodine is needed for the synthesis of thyroxine. Toxic agents present in food interfere with thyroxine synthesis and lead to goitre.
Transfer of pollen grains from the anther to the stigma of another flower of the same plant is called
Geitonogamy is the process of transfer of pollen grains from the anther to the stigma of another flower of the same plant. Example, maize.
Wind pollinated flowers are
Pollination by wind is called anemophily. Wind-pollinated flowers are small in size, producing large number of dry pollen grains. Pollens are small, dry and light in weight. Grasses are anemophilous plants.
Apomictic embryos in citrus arise from
Apomixis is an asexual type of reproduction in which the plant embryos grow from egg cells without being fertilized by pollen—the male part of the plant. In Citrus, apomictic embryos arise from maternal sporophytic tissue like nucellus and integuments in ovule. This type of embryony is called adventive embryony.
The scutellum observed in a grain of wheat or maize is comparable to which part of the seed in other monocotyledons?
Scutellum is the modified cotyledon observed in a grain of wheat or maize. It lies between embryo and the endosperm.
Sertoli cells are found in
Sertoli cells are found in the walls of seminiferous tubules of the testes. They anchor and provide nutrition to the developing germ cells especially the spermatids.
Vasa efferentia are the ductules leading from
Vasa efferentia are ductuless leading from rete testis to vas deferens. The rete testis is an anastomosing network of tubules located in the hilum of the testicles that carries sperm from the seminiferous tubules to the vasa efferentia.
Seminal plasma in human males is rich in :
Seminal plasma in human males are rich in fructose, calcium and certain enzymes. They provide a medium for transport of sperms, nourishes and activates sperms.
Which one of the following statements about human sperm is correct?
Acrosome a small pointed structure at the tip of nucleus. It breaks down just before fertilization, releasing hydrolytic enzymes that assist penetration between follicle cells that surrounds the ovum, thus facilitating fertilization.
The part of fallopian tube closest to the ovary is
The part of fallopian tube closest to the ovary is infundibulum. Infundibulum possess finger-like projections called fimbriae that help in collection of ovum after ovulation. It leads to wider part of oviduct called ampulla. The last part of oviduct is isthmus that has a narrow lumen and joins the uterus.
The second maturation division of the mammalian ovum occurs
In Oogenesis, the second maturation division occurs until after the ovum has been penetrated by a sperm. Oogenesis involves the formation of haploid female gametes, ova from diploid egg mother cells, oogonia of ovary of female organism.
In vitro fertilisation is a technique that involves transfer of which one of the following into the fallopian tube?
In vitro fertilization is a process where fertilization of egg occurs outside the mother’s womb. This method is used as a major treatment for infertility. This method involves the removal of eggs from the female ovaries. This egg is then allowed to fertilize with sperm in a fluid medium in a test tube. The zygote is allowed to develop for a week. Either zygote or early embryo up to 8 blastomeres is then transferred into the fallopian tube to complete its further development. If the embryo is with more than 8 blastomeres, it is transferred into uterus for successful pregnancy. A baby conceived by fertilization that occurs outside mother body is called test tube baby.
The signals for parturition originate from
Parturition is the process of delivery of the foetus. The signals for parturition originate from the fully developed foetus and placenta that induce uterine contraction called foetal ejection reflex. This triggers the release of oxytocin hormone from pituitary. This in turn leads to expulsion of the baby out of uterus.
The first movements of the foetus and appearance of hair on its head are usually observed during which month of pregnancy?
In human beings, after one month of pregnancy, the embryo’s heart is formed. By end of second month pregnancy, the foetus develops 1imbs and digits. By end of 12 weeks, major organ systems are formed. After fifth month, the first movement of foetus and appearance of hair on its head are observed. By the end of 24 weeks, the body is covered with hairs, eye lids separate and eyelashes are formed.
Which one of the following statements about morula in humans is correct?
Morula is a solid ball contained within the zone pellucida. It has almost equal quantity of cytoplasm as an uncleaved zygote but much more DNA.
Cu ions released from copper-releasing Intra Uterine Devices (IUDs):
Cu ions released by copper-releasing intra uterine devices suppresses sperm motility. Intra-uterine devices are inserted by doctors in the uterus through vagina. They are available as the non-medicated IUDs, copper releasing IUDs and hormone releasing IUDs.
The permissible use of the technique amniocentesis is for:
Amniocentesis involves prenatal diagnosis of metabolic error and other genetic abnormalities.
Which one of the following cannot be explained on the basis of Mendel's Law of Dominance?
According to Mendel’s law of dominance, out of two contrasting allelomorphic factors only one expresses itself in an individual. The factor that expresses itself is called dominant while the other which has not shown its effect in the heterozygous individual is termed as recessive. The option (Alleles do not show any blending and both the characters recover as such in F₂ generation) in the given question cannot be explained on the basis of law of dominance. It can only be explained on the basis of Mendel’s law of independent assortment, according to which in a dihybrid cross, the two alleles of each character assort independently of the alleles of other character and separate at the time of gamete formation.
The genotype of a plant showing the dominant phenotype can be determined by:
Test cross is the cross of an individual having dominant phenotype with an individual having recessive phenotype. It is used to determine the genotype of a plant showing the dominant phenotype, that means to determine whether the individual exhibiting dominating characters are homozygous or heterozygous.
Select the correct statement from the ones given below with respect to dihybrid cross.
Linkage is the inheritance of genes of same chromosome together and capacity of these genes to retain their parental combination in subsequent generation. The strength of linkage between two genes is inversely proportional to the distance between the two. This means, that greater the distance, lower is the linkage strength. Two linked genes show higher frequency of recombination if the distance between them is higher and lower frequency if the distance is smaller.
Which one of the following symbols and its representation, used in human pedigree analysis, is correct?
-
(A)
-
(B)
-
(C)
-
(D)
Pedigree analysis is a record of the occurrence of a trait in several generations of a human family. In this, male members are shown by squares and female by circles. Sibs are represented horizontally on a line in order of birth. It helps us in giving information about genotype of an individual for trait under investigation.
Infectious proteins are present in:
Infectious proteins are rich in prions. Prions are made of proteins without nucleic acid. It is the causal agent of scrapie disease of sheep.
Which one of the following does not follow the central dogma of molecular biology?
HIV viruses does not follow central dogma. Central dogma is a one way flow of information from DNA to mRNA and then to protein.
The one aspect which is not a salient feature of genetic code, is its being:
Genetic code is non-ambiguous. There is no ambiguity for a particular codon. A particular codon will always code for the same amino acid, where ever it is found.
Select the two correct statements out of the four (a–d) given below about lac operon.
(i) Glucose or galactose may bind with the repressor and inactivate it.
(ii) In the absence of lactose the repressor binds with the operator region.
(iii) The Z-gene codes for permease.
(iv) This was elucidated by Francois Jacob and Jacque Monod.
The correct statements are:
Jacob and Monod proposed the lac operon of E. coli. The lac operon contains a promoter, an operator, and three structural genes called z, y, and a, coding for the enzyme β-galactosidase, permease and transacetylase respectively. The lac regulator gene, designated as i gene, codes for repressor. In the absence of the inducer, the repressor binds to the lac operator, preventing RNA polymerase from binding to the promoter and thus transcribing the structural gene.
Satellite DNA is useful tool in:
Satellite DNA is useful in forensic science. The polymorphism of minisatellite, microsatellite and minivariant repeats is analysed for DNA fingerprinting. DNA profiling helps in the resolution of crimes, legal disputes etc.
Darwin’s finches are a good example of:
Adaptive radiation
Widal test is used for the diagnosis of:
Widal test is used for the diagnosis of typhoid. It is an agglutination test for the presence of antibodies against the Salmonella organisms, which cause typhoid fever.
Ringworm in humans is caused by:
Ring worm is a fungal disease caused by the dermatophyte fungal species of Microsporum, Trichophyton and Epider-mophyton.
Which one of the following statements is correct with respect to AIDS?
AIDS (Acquired Immune Deficiency Syndrome) is caused by HIV retrovirus. The virus destroys the helper T lymphocytes, thus reducing their numbers.
Select the correct statement from the ones given below?
Morphine is often given to persons who have undergone surgery as a pain killer
Breeding of crops with high levels of minerals, vitamins and proteins is called
Biofortification.
An improved variety of transgenic basmati rice
An improved variety of transgenic basmati rice gives high yield and is rich in vitamin A.
A common biocontrol agent for the control of plant diseases is
Trichoderma.
The common nitrogen fixer in paddy fields is
The common nitrogen fixer in paddy field is Azospirillum. It is an anaerobic bacteria that forms loose association with roots of paddy crops.
Which one of the following is not used in organic farming
Snail
Select the correct statement from the following
Activated sludge is a process for treating sewage and industrial wastewaters using air and a biological floc composed of bacteria and protozoans. During the process, the primary effluent is taken to aeration tank that contain large number of aerobic heterotrophic microbes. They form flocs that digest a lot of organic matter. As the biological oxygen demand of waste water is reduced, it is passed into settling tank to undergo sedimentation. The sediment of the settling tank is called activated sludge that is a rich source of aerobic bacteria. Hence, the statement (Activated sludge-sediment in settlement tanks of sewage treatment plant is a rich source of aerobic bacteria) is correct.
Biogas is produced by anaerobic breakdown of biomass with the help of methanogenic bacteria. It is made up of methane, carbon dioxide with traces of nitrogen, hydrogen sulphide and hydrogen.
Which one of the following is used as vector for cloning genes info higher organisms?
Retrovirus
DNA or RNA segment tagged with a radioactive molecule is called:
DNA or RNA segment tagged with a radioactive molecule is called probe. They are used to detect the presence of complementary sequences in nucleic acid samples. Probes are used for identification and isolation of DNA and RNA,
Restriction endonucleases are enzymes which:
Make cuts at specific positions within the DNA molecule.
Stirred tank bioreactors have been designed for:
A stirred-tank bioreactor is a cylindrical or curved base to facilitate the mixing of the reaction content. It has been designed for availability of oxygen throughout the process.
Which one of the following palindromic base sequences in DNA can be easily cut at about the middle by some particular restriction enzyme?
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(A)
-
(B)
-
(C)
-
(D)
The genetically-modified (GM) brinjal in India has been developed for:
The genetically-modified (GM) brinjal in India has been developed for insect-resistance .
Some of the characteristics of Bt cotton are:
Bt toxin genes were isolated from Bacillus thuringiensis which incorporated into cotton plant to form a genetically modified crop called Bt cotton. Bt cotton has high yield and resistance to bollworms.
Genetic engineering has been successfully used for producing:
Genetic engineering has been successfully used for producing transgenic mice. Many transgenic mice are designed to increase our understanding of how genes contribute to development of diseases.
For example transgenic mice are being used to test the safety of polio vaccine.
The figure given below is a diagrammatic representation of response of organisms to abiotic factors. What do a, b and c represent respectively?
-
(A)
-
(B)
-
(C)
-
(D)
Which one of the following is one of the characteristics of a biological community?
Stratification is one of the characteristics of biological community. Natality, mortality, age, structure and sex ratio are basic characteristics of a population.
The biomass available for consumption by the herbivores and the decomposers is called:
The biomass available for consumption by the herbivores and the decomposers is called net primary productivity.
Study the four statements (i–iv) given below and select the two correct ones out of them:
(i) A lion eating a deer and a sparrow feeding on grain are ecologically similar in being consumers.
(ii) Predator star fish Pisaster helps in maintaining species diversity of some invertebrates.
(iii) Predators ultimately lead to the extinction of prey species.
(iv) Production of chemicals such as nicotine, strychnine by the plants are metabolic disorders.
The two correct statements are:
(i) and (ii)
Which one of the following is an example of Exsitu conservation?
Seed bank
dB is a standard abbreviation used for the quantitative expression of:
dB is a standard abbreviation for the quantitative expression of noise. Unwanted sound is called noise. Generally sound above 80 dB is noise.
A renewable exhaustible natural resource is:
A renewable exhaustible natural resource is forest.
The two gases making highest relative contribution to the greenhouse gases are:
The gases that makes highest relative contribution to the green house gases are carbon dioxide (CO₂) and methane ( CH₄).
Carrier ions like Na⁺ facilitate the absorption of substances like:
Amino acids and glucose
The dimension of (1 / 2)ε₀E², where ε₀ is permittivity of free space and E is electric field, is:
(1 / 2)ε₀E² represents energy density i.e., energy per unit volume.
= [(1 / 2)ε₀E²] = [ML²T⁻²] / [L³] = [ML⁻¹T⁻²]
A particle has initial velocity (3î + 4ĵ) and has acceleration (0.4î + 0.3ĵ). It's speed after 10 s is:
Given,
𝐮⃗ = 3î + 4ĵ and 𝐚⃗ = 0.4î + 0.3ĵ
⇒ uₓ = 3 units, u_y = 4 units
aₓ = 0.4 units, a_y = 0.3 units
Along x-axis,
∴ vₓ = uₓ + aₓ × 10 = 3 + 4 = 7 units
Along y-axis,
v_y = 4 + 0.3 × 10 = 4 + 3 = 7 units
Net final velocity
∴ v = √(vₓ² + v_y²) = 7√2 units
𝐯⃗ᵢ = 3î + 4ĵ, 𝐚⃗ = 0.4î + 0.3ĵ
Time, t = 10 sec
Final velocity 𝐯⃗_f after t = 10 sec,
𝐯⃗_f = 𝐯⃗ᵢ + 𝐚⃗t
𝐯⃗_f = (3î + 4ĵ) + (0.4î + 0.3ĵ)(10) = 7î + 7ĵ
A particle moves a distance x in time t according to the equation x = (t+5)⁻¹. The acceleration of the particle is proportional to:
distance x = 1 / (t+5)
∴ velocity v = dx / dt = (-1) / (t+5)²
∴ acceleration a = d²x / dt² = 2 / (t+5)³ = 2x³
Therefore, v³ᐟ² = -(t+5)⁻³
So, a ∝ v³ᐟ²
A ball is dropped from a high-rise platform at t = 0 starting from rest. After 6 seconds, another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v? (Take g = 10 m/s²)
Clearly, distance moved by the 1ˢᵗ ball in 18 s = distance moved by the 2ⁿᵈ ball in 12 s......(i)
Now, distance moved in 18 s by the 1ˢᵗ ball, = 1 / 2 × 10 × 18²
= 90 × 18 = 1620 m
Distance moved in 12 s by the 2ⁿᵈ ball,
= ut + (1 / 2)gt² ⇒ 12v + 1 / 2 × 10 × (12)²
⇒ 12v + 720
From equation (i),
∴ 1620 = 12v + 5 × 144
⇒ v = 135 − 60 = 75 ms⁻¹
A man of 50 kg mass is standing in a gravity-free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed of 2 m/s. When the stone reaches the floor, the distance of the man above the floor will be:
No external force is acting, therefore, momentum is conserved.
By momentum conservation,
50u + 0.5 × 2 = 0
where u is the velocity of man.
u = (−1 / 50) ms⁻¹
Negative sign of u shows that the man moves upward.
Time taken by the stone to reach the ground
= 10 / 2 = 5 sec
Distance moved by the man,
= 5 × 1 / 50 = 0.1 m
∴ When the stone reaches the floor, the distance of the man above the floor is 10.1 m.
Six vectors, 𝐚⃗ through 𝐟⃗ have the magnitudes and directions indicated in the figure. Which of the following statements is true?
Using the law of vector addition, (𝐝⃗ + 𝐞⃗ ) is as shown in the fig.
∴ 𝐝⃗ + 𝐞⃗ = 𝐟⃗
A block of mass m is in contact with the cart C as shown in the Figure.
The coefficient of static friction between the block and the cart is μ. The acceleration α of the cart that will prevent the block from falling satisfies:
Forces acting on the block are as shown in the fig. Normal reaction N is provided by the force ma due to acceleration α
∴ N = ma
For the block not to fall, the frictional force,
F_f ≥ mg
⇒ μN ≥ mg
⇒ μma ≥ mg
⇒ a ≥ g / μ
An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of 2 m/s. The mass per unit length of water in the pipe is 100 kg/m. What is the power of the engine?
Amount of water flowing per second from the pipe = m / time == (m / ℓ) · (ℓ / t) = (m / ℓ) v= (m / ℓ)v
Power = K.E. of water flowing per second
= 1 / 2(m / ℓ)v.v²
= 1 / 2(m / ℓ)v³
= 1 / 2 × 100 × 8 = 400 W
A ball moving with velocity 2 m/s collides head-on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in m/s) after collision will be:
Clearly v₁ = 2 ms⁻¹, v₂ = 0
m₁ = m (say), m₂ = 2m
v₁′ = ?, v₂′ = ?
e = (v₁′ − v₂′) / (v₂ − v₁).........(i)
By conservation of momentum,
2m = mv₁′ + 2mv₂′ ...(ii)
From (i),
0.5 = (v₂′ − v₁′) / 2
∴ v₂′ = 1 + v₁′
From (ii),
2 = v₁′ + 2 + 2v₁′
⇒ v₁′ = 0 and v₂′ = 1 ms⁻¹
Two particles which are initially at rest, move towards each other under the action of their internal attraction. If their speeds are v and 2v at any instant, then the speed of centre of mass of the system will be:
If no external force action a system of particles, the centre of mass remains at rest. So, speed of centre of mass is zero.
A circular disk of moment of inertia Iₜ is rotating in a horizontal plane, its symmetry axis, with a constant angular speed ωᵢ. Another disk of moment of inertia I_b is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed ω_f. The energy lost by the initially rotating disk to friction is:
By conservation of angular momentum,
Iₜωᵢ = (Iₜ + I_b)ω_f
where ω_f is the final angular velocity of disks.
∴ ω_f = (Iₜ / (Iₜ + I_b)) ωᵢ
Loss in K.E., ΔK = Initial K.E. − Final K.E.
= (1 / 2) Iₜωᵢ² − 1 / 2 (Iₜ + I_b)ω_f²
= (1 / 2) Iₜωᵢ² − 1 / 2 (Iₜ + I_b) [Iₜ² / (Iₜ + I_b)²] ωᵢ²
= (1 / 2) ωᵢ² [Iₜ / (Iₜ + I_b)] (Iₜ + I_b − Iₜ)
= (1 / 2) ωᵢ² (IₜI_b) / (Iₜ + I_b)
A particle of mass M is situated at the centre of a spherical shell of same mass and radius a. The gravitational potential at a point situated at a / 2 distance from the centre, will be:
Potential at the given point = Potential at the point due to the shell + Potential due to the particle
= - GM / a - 2GM / a = -3GM / a
The radii of circular orbits of two satellites A and B of the Earth are 4R and R, respectively. If the speed of satellite A is 3 V, the speed of satellite B will be:
Orbital velocity of a satellite in a circular orbit of radius a is given by
v = √(GM / a)
⇒ v ∝ √(1 / a)
⇒ v₂ / v₁ = √(a₁ / a₂)
∴ v₂ - v₁√(4R / R) = 2v₁ = 6V
A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs, in time t?
The rate of heat flow is given by
Q / t = K.A. ΔT / ℓ
Area of Original rod A = πR²;
Area of new rod A' = πR² / 4
A' = A / 4
Volume of original rod will be equal to the volume of new rod.
So, A'ℓ'=Aℓ
or,
∴ πR²ℓ = π(R / 2)²ℓ'
⇒ ℓ' / ℓ = R² / (R² / 4) = 4
∴ Q' / Q = (A' / A).(ℓ / ℓ') = (1 / 4).(1 / 4) = 1 / 16
∴ Q' = Q / 16
The total radiant energy per unit area, normal to the direction of incidence, received at a distance R from the centre of a star of radius r, whose surface radiates as a black body at a temperature T K is given by:
(where σ is Stefan's constant)
E = (S / S₀)σT⁴ = (4πr² / 4πR²)σT⁴
= σ(r² / R²)T⁴
If ΔU and ΔW represent the increase in internal energy and work done by the system respectively in a thermodynamical process, which of the following is true?
By first law of thermodynamics,
ΔQ = ΔU + ΔW
In adiabatic process, ΔQ = 0
∴ ΔU = - ΔW
In isothermal process, ΔU = 0
∴ ΔQ = ΔW
The displacement of a particle along the x-axis is given by x = a sin²(ωt). The motion of the particle corresponds to:
x = a sin²(ωt) = a / 2(1-cos2ωt)
dx / dt = (a / 2) 2ω sin 2 ωt
d²x / dt² = (4ω²a / 2).cos2ωt
This represents an S. H. M. of frequency = ω / π
The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be:
T = 2π√(M/K)
∴ T₁/T₂ = √(M₁/M₂)
∴ T₂ = T₁√(M₂/M₁) = T₁√(2M/M)
T₂ = T₁√2 = √2T (where T₁ = T)
A transverse wave is represented by y = Asin(ωt − kx). For what value of the wavelength is the wave velocity equal to the maximum particle velocity?
y = Asin(ωt − kx)
Particle velocity,
vₚ = dy/dt = A ω cos(ωt - kx)
∴ vₚ ₘₐₓ = Aω
wave velocity = ω/k
∴ Aω = ω/k
i.e., A = 1/k But k = 2π/λ
∴ λ = 2πA
A tuning fork of frequency 512 Hz makes 4 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was:
The frequency of the piano string = 512 ± 4 = 516 or 508. When the tension is increased, beat frequency decreases to 2, it means that frequency of the string is 508 as frequency of string increases with tension.
Two positive ions, each carrying a charge q, are separated by a distance d. If F the force of repulsion between the ions, the number of electrons missing from each ion will be (e being the charge of an electron):
Let n be the number of electrons missing.
F = (1 / 4πε₀). (q² / d²)
⇒ q = √(4πε₀d²F) = ne
∴ n = √(4πε₀Fd² / e²)
A square surface of side L meter in the plane of the paper is placed in a uniform electric field E (volt/m) acting along the same plane at an angle θ with the horizontal side of the square as shown in the figure. The electric flux linked to the surface, in units of volt·m, is:
Electric flux, Φ = EA cos θ, where θ = angle between E and normal to the surface.
Here θ = π/2
Φ = 0
A series combination of n₁ capacitors, each of value C₁, is charged by a source of potential difference 4 V. When another parallel combination of n₂ capacitors, each of value C₂, is charged by a source of potential difference V, it has the same (total) energy stored in it, as the first combination has. The value of C₂, in terms of C₁, is then
In series, C_eff = C₁ / n₁
∴ Energy stored,
E_S = (1 / 2)C_eff V_S²
= (1 / 2) (C₁ / n₁)16V²
= 8V² (C₁ / n₁)
In parallel,
C_eff = n₂C₂
∴ Energy stored,
Eₚ = (1 / 2) n₂C₂V²
According to question,
Eₛ = Eₚ
∴ 8V²C₁ / n₁= (1 / 2) n₂C₂V²
⇒ C₂ = 16C₁ / n₁n₂
In producing chlorine by electrolysis 100 kW power at 125 V is being consumed. How much chlorine per minute is liberated? (E.C.E. of chlorine is 0.367×10⁻⁶ kg/C)
I = P / V = ((100 × 10³) / 125) A = (10⁵ / 60) A
E.C.E. = 0.367×10⁻⁶ kg C⁻¹
Charge per minute = (I ×60)C
= ((10⁵ × 60) / 125) C= ((6 × 10⁶) / 125)C
∴ Mass liberated = (6 ×10⁶) / 125 × 0.367 × 10⁻⁶
= (6 ×1000 × 0.367 × 10⁻³) / 125
= 17.61 × 10⁻³ kg
Consider the following two statements:
(A) Kirchhoff's junction law follows from the conservation of charge.
(B) Kirchhoff's loop law follows from the conservation of energy.
Which of the following is correct?
Junction law follows from conservation of charge and loop law is the conservation of energy
A potentiometer circuit is set up as shown. The potential gradient across the potentiometer wire is k volt/cm and the ammeter, present in the circuit, reads 1.0 A when two way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3 is plugged in, are found to be at lengths l₁ cm and l₂ cm, respectively.
The magnitudes of the resistors R and X, in ohms, are then equal, respectively, to
(i) When key between the terminals 1 and 2 is plugged in,
P.D. across R = IR = kl₁
⇒ R = kl₁ as I = 1A
(ii) When key between terminals 1 and 3 is plugged in, P.D. across
(X + R) = I(X + R) = k l₂
⇒ X + R = kl₂
∴ X = k(l₂ - l₁)
∴ R = kl₁ and X = k(l₂ - l₁)
Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency f Hz. The magnitude of magnetic induction at the centre of the ring is
When the ring rotates about its axis with a uniform frequency f Hz, the current flowing in the ring is
I = q / T = qf
Magnetic field at the centre of the ring is
B = μ₀I / 2R = μ₀qf / 2R
A circular coil ABCD carrying a current i is placed in a uniform magnetic field. If the magnetic force on the segment AB is 𝐅⃗, the force on the remaining segment BCDA is:
Here, 𝐅⃗_ AB + 𝐅⃗_BCDA = 0⃗
𝐅⃗_BCDA = -𝐅⃗_AB = -𝐅⃗
(∵ F_AB = 𝐅⃗)
A galvanometer has a coil of resistance 100 Ω and gives a full-scale deflection for 30 mA current. It is to work as a voltmeter of 30 volt range, the resistance required to be added will be:
Let the resistance to be added be R, then
30 = I_g(r+R)
R = 30 / I_g - r = (30 / 30 × 10⁻³) - 100
= 1000 - 100 = 900 Ω
Electromagnets are made of soft iron because soft iron has:
Soft iron is a soft magnetic material soft iron has low retentivity and low coercive force or coercivity suitable for electromagnets.
Retentivity is the residual magnetism i.e., the magnetism remains in a materials even on the removal of the magnetising field.
Magnetic hard substance like steel has high coercivity and magnetic soft substance-soft iron has low coercivity.
A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 sec in Earth's horizontal magnetic field of 24 microtesla. When a horizontal field of 18 microtesla is produced opposite to the earth's field by placing a current-carrying wire, the new time period of magnet will be:
Time period of a vibration magnetometer,
T ∝ 1 / √B
T₁ / T₂ = √(B₂ / B₁)
T₂ = T₁√(B₁ / B₂) = 2√(24×10⁻⁶) / (6×10⁻⁶) = 4s
A conducting circular loop is placed in a uniform magnetic field, B = 0.025 T, with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of 1 mm s⁻¹. The induced e.m.f. when the radius is 2 cm, is
Magnetic flux linked with the loop is ϕ = Bπr²
|e| = dϕ / dt = Bπ . 2r (dr / dt)
When r = 2 cm, dr / dt = 1 mm s⁻¹
e = 0.025 × π × 2 × 2 × 10⁻² × 10⁻³
= 0.100 × π × 10⁻⁵ = π × 10⁻⁶ V = πμV
In the given circuit, the reading of voltmeter V₁ and V₂ are 300 volts each. The reading of the voltmeter V₃ and ammeter A are respectively:
As V_L = V_C = 300 V, resonance will take place
∴ V_R = 220 V
Current, I = 220/100 = 2.2 A
∴ reading of V₃, =220 V and reading of A = 2.2 A
A 220 volt input is supplied to a transformer. The output circuit draws a current of 2.0 ampere at 440 volts. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is:
V₂ / V₁ = 0.8(I₁ / I₂)
⇒ V₂I₂ / V₁I₁ = 0.8
V₁ = 220 V, I₂ = 2.0 A, V₂ = 440 V
I₁ = V₂I₂ / V₁ × 10 / 8 = (440×2×10) / (220×8) = 5 A
Which one of the following statements is false for the properties of electromagnetic wave?
Electromagnetic waves are the combination of mutually perpendicular electric and magnetic fields. So, option (Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation of wave.) is false.
A ray of light travelling in a transparent medium of refractive index μ, falls on a surface separating the medium from air at an angle of incidence of 45°. For which of the following values of μ the ray can undergo total internal reflection?
For total internal reflection,
μ ≥ 1/sin C ≥ √2 ≥ 1.414
⇒ μ = 1.50
A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter d/2 in the central region of the lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively:
By covering aperture, focal length does not change.
But intensity is reduced by 1 / 4 times, as aperture diameter d / 2 is covered.
∴ I' = I - I / 4 = 3I / 4
∴ New focal length = f and intensity = 3I / 4.
A source S₁ is producing 10¹⁵ photons per second of wavelength 5000 Å. Another source S₂ is producing 1.02 × 10¹⁵ photons per second of wavelength 5100 Å. Then, (power of S₂) (power of S₁) is equal to:
Energy emitted/sec by S₁, P₁ = n₁(hc / λ₁)
Energy emitted/sec by S₂, P₂ = n₂.(hc / λ₂)
∴ P₂ / P₁ = (n₂ / n₁).(λ₁ / λ₂)
= (1.02×10¹⁵) / 10¹⁵ . 5000 / 5100 = 1.0
The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function 5.01 eV, when ultraviolet light of 200 nm falls on it, must be:
Kₘₐₓ = hc / λ - W = hc / λ - 5.01
= 12375 / λ(in Å) - 5.01
= 12375 / 2000 - 5.01 = 6.1875 - 5.01 = 1.17775
≈ 1.2 V
The potential difference that must be applied to stop photoelectrons = -1.2V
An alpha nucleus of energy (1 / 2)mv² bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to:
Kinetic energy of alpha nucleus is equal to electrostatic potential energy of the system of the alpha particle and the heavy nucleus. That is,
(1 / 2)mv² = (1 / 4πε₀)(q_αZe / r₀)
where r₀ is the distance of closest approach
r₀ = (2 / 4πε₀)(q_αZe) / mv²
r₀ ∝ Ze ∝ q_α ∝ 1 / m ∝ 1 / v²
The energy of hydrogen atom in the ground state is −13.6 eV. The energy of a He⁺ ion in the first excited state will be:
Energy of a H-like atom in its nᵗʰ state is given by
Eₙ = -Z² × (13.6 / n²) eV
For first excited state of He⁺
n = 2, Z = 2
∴ E_He⁺ = - 4 / 2² × 13.6 = - 13.6 eV
The mass of a ⁷₃Li nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of the ⁷₃Li nucleus is nearly:
B.E. = 0.042 × 931 ≈ 42 MeV
Number of nucleons in ⁷₃Li is 7.
∴ B.E./nucleon = 42/7= 6 MeV ≈ 5.6 MeV
The activity of a radioactive sample is measured as N₀ counts per minute at t = 0 and N₀/e counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is:
.
Which one of the following statement is FALSE?
Majority carriers in an n-type semiconductor are electrons.
Which one of the following bonds produces a solid that reflects light in the visible region and whose electrical conductivity decreases with temperature and has high melting point?
For a metal, conductivity decreases with increase in temperature.
Also, metal has high melting point.
A common emitter amplifier has a voltage gain of 50, an input impedance of 100 Ω and an output impedance of 200 Ω. The power gain of the amplifier is:
Power gain = voltage gain × current gain
= V_G.I_G = V₀/Vᵢ . I₀/Iᵢ
= V₀²/Vᵢ² . Rᵢ/R₀ = 50 × 50 × 100/200
= 2500/2 = 1250
The device that can act as a complete electronic circuit is:
Integrated circuit can act as a complete electronic circuit.
A beam of cathode rays is subjected to crossed electric (E) and magnetic field (B). The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by:
(where V is the potential difference between cathode and anode.)
As the electron beam is not deflected, then
Fₘ = Fₑ
or Bev = Ee or v = E / B ......(i)
As the electron moves from cathode to anode, its potential energy at the cathode appears as its kinetic energy at the anode. If V is the potential difference between the anode and cathode, then potential energy of the electron at cathode is eV. Also, kinetic energy of the electron at anode = (1 / 2)mv². According to the law of conservation of energy
(1 / 2) mv² = eV or v = √(2eV / m) ...(ii)
From Eqs. (i) and (ii), we have
√(2eV / m) = E / B or e / m = E² / 2VB²
To get output 1 for the following circuit, the correct choice for the input is:
The Boolen expression for the given combination is output Y=(A+B) C
A gramophone record is revolving with an angular velocity ω. A coin is placed at a distance r from the centre of the record. The static coefficient of friction is μ. The coin will revolve with the record if
When the disc spins, the frictional force between the gramophone record and the coin is μ mg.
The coin will revolve with the record if:
F_frictional ≥ F_centripetal
μ mg ≥ mω²r
μg / r ≥ ω²
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