CBSE EXAMINATION PAPER-2014
NEET-ZOOLOGY
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 180 questions. All questions are compulsory.
- This question paper is divided into 1 sections.
- Section A – questions number 1 to 180 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
When 22.4 litres of H₂(g) is mixed with 11.2 litres of Cl₂(g), each at S.T.P., the moles of HCl(g) formed is equal to:
H₂(g) + Cl₂(g) → 2HCl(g)
22.4 L 11.2 L 0 Initial
= 1mole = 0.5 mole
1 − 0.5 0.5 − 0.5 2 × 0.5 Final
Moles of HCl formed = 2 × 0.5 = 1
1.0 g of magnesium is burnt with 0.56 g O₂ in a closed vessel. Which reactant is left in excess and how much? (At. wt. Mg = 24; O = 16)
Calculate the energy in joule corresponding to light of wavelength 45 nm: (Planck’s constant h = 6.63 × 10⁻³⁴ Js; speed of light c = 3 × 10⁸ ms⁻¹)
E=hc /λ
= 6.63 × 10⁻³⁴ × 3 × 10⁸/ 45 × 10⁻⁹
= 4.42 × 10⁻¹⁸ J
What is the maximum number of orbitals that can be identified with the following quantum numbers? n = 3, l = 1, m = 0
Given: n = 3, l = 1, m = 0 Hence orbital is 3p. For p-orbital, m values are: Hence the number of orbital identified by m = 0 can be one only.
Which of the following orders of ionic radii is correctly represented?
All answers are incorrect.
H⁻ > H > H⁺ ; O²⁻ > F⁻ < Na⁺
O²⁻ > F⁻ > Na⁺ ; N³⁻ > Mg²⁺ > Al³⁺
For isoelectronic species, the size is determined by Zeff.
Higher the Zeff, lower is the size of the ionic species.
Which of the following molecules has the maximum dipole moment?
Dipole moment of NH₃ > NF₃
Be²⁺ is isoelectronic with which of the following ions?
Be²⁺ = (4 − 2) = 2 is isoelectronic with Li⁺ (3 − 1 = 2) Since both have same number of electrons in their outermost shell.
Equal masses of H₂, O₂ and methane have been taken in a container of volume V at temperature 27°C in identical conditions. The ratio of the volumes of gases H₂ : O₂ : methane would be:
According to Avogadro's law, "At same temperature and pressure, equal volume contains equal number of moles."
ⁿH₂ = w/2
ⁿO₂ = w/32
ⁿCH) = w/16
V_H₂ : V_O₂_: V_CH₄
= ⁿH₂ : ⁿO₂ : ⁿCH₄
= w/2 : w/32 : w/16
= 16 : 1 : 2
Which one of the following species has plane triangular shape?
Hybridization of NO₃⁻
= 1/2 (5 + 0 + 1 − 0)
= 6/2 = 3
= sp²
Hence geometry is trigonal planar.
NO₂⁻ (nitrite ion) also has sp² hybridization and gives a trigonal planar geometry, but because there are only two outer atoms, the molecular geometry is bent with D120°.bond angels.
‘Which property of colloidal solution is independent of charge on the colloidal particles:-
Tyndall‘ effect is an optical property, and it is . independent of charge on colloidal particles.
For the reversible reaction, N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + heat The equilibrium shifts in forward direction:
Given reaction, is exothermxc reaction. Hence: acooldmg to Le-Chatelier's prmclplc low temperature favours the forward reaction and on increasing pressure equilibrium will shift, towards lesser number of moles i.e. forward direction.
For the reaction: X₂O₄(l) → 2XO₂(g)
ΔU = 2.1 kcal, ΔS = 20 cal K⁻¹ at 300 K Calculate ΔG is:
Given:
ΔU = 2.1 kcal
ΔS = 20 cal K⁻¹
T = 300 K
∴ ΔH = ΔU + Δn_g RT
Putting the values in the equation:
ΔH = 2.1 + 2 × (2/1000) × 300
= 2.1 + 1.2
= 3.3 kcal
Now,
ΔG = ΔH − TΔS
= 3.3 − 300 × (20/1000)
= −2.7 kcal
Which of the following salts will give highest pH in water?
Na₂CO₃ is a salt of strong base (NaOH) and weak acid (H₂CO₃). On hydrolysis this salt will produce a strongly basic solution, i.e. pH will be highest (pH > 7) for this solution.
Others are combinations of:
KCl = Strong acid + Strong base
→ Neutral solution (pH ≈ 7)
NaCl = Strong acid + Strong base
→ Neutral solution (pH ≈ 7)
CuSO₄ = Strong acid + Weak base
→ Acidic solution (pH < 7)
For a given exothermic reaction, Kₚ and K′ₚ are the equilibrium constants at temperatures T₁ and T₂ respectively. Assuming that heat of reaction is constant in temperature range between T₁ and T₂, it is readily observed that:
n exothermic reactions, the value of Kₚ decreases on increasing temperature.
∴ Kₚ > K′ₚ (Assuming T₁ < T₂)
Using the Gibbs energy change, ΔG° = +63.3 kJ, for the following reaction,
Ag₂CO₃(s) ⇌ 2Ag⁺(aq) + CO₃²⁻(aq) The Kₛₚ of Ag₂CO₃(s) in water at 25°C is: (R = 8.314 J K⁻¹ mol⁻¹)
ΔG° = −2.303 log K
Here K = [Ag⁺]² [CO₃²⁻] = Kₛₚ
63.3 × 10⁺³ = −2.303 × 8.314 × 298 log Kₛₚ
∴ log Kₛₚ = −63.3 × 10⁺³ / 5705.8 = −11.09
∴ Kₛₚ = Antilog (−11.09) = 8 × 10⁻¹²
(i) H₂O₂ + O₃ → H₂O + 2O₂
(ii) H₂O₂ + Ag₂O → 2Ag + H₂O + O₂
Role of hydrogen peroxide in the above reactions is respectively:
.
In the Kjeldahl’s method for estimation of nitrogen present in a soil sample, ammonia evolved from 0.75 g of sample neutralized 10 mL of 1 M H₂SO₄. The percentage of nitrogen in the soil is:
10 mL, 1 M H₂SO₄ = 20 mL, 1 M NH₃
∴ wt of N in one mole NH₃ = 14
∴ 20 × 10⁻³ mol NH₃→20 × 10⁻³ × 14 nitrogen
∴ 0.75 g of sample contains
(14 × 20 × 10⁻³) / 0.75 × 100
= 37.33%
Acidity of diprotic acids in aqueous solutions increases in the order:
The weakening of M–H bond with increase in size of M (where M = S, Se, Te) explains the acid character of hydrides.
Which of the following organic compounds has same hybridization as its combustion product (CO₂)?
The combustion reaction of ethylene is:
C₂H₄ + 5/2O₂ → 2CO₂ + 2H₂O
Both H–C ≡ CH and CO₂ have sp hybridization.
Magnetic moment 2.83 BM is given by which of the following ions? (At. nos. Ti = 22, Cr = 24, Mn = 25, Ni = 28):-
Magnetic moment: √n(n + 2)
‘Which of the following is not a common component of Photochemical Smog?
The oxidised hydrocarbons and ozone in presence of humidity cause photochemical smog.
Hydrocarbons + O₂, NO₂, NO,O, O₃ → Peroxides, aldehydes, peroxyacetyl nitrate (PAN), acrolein, etc.
Hence, coloured fumes are not common component of photochemical smog.
Reason of lanthanoid contraction is:
The shape of f-orbitals is very much diffused and they have poor shielding effect.
out Of the following 0.10 M aqueous solutions, which one will exhibit the largest freezing point depression?
Colligative properties are properties of particles. Since Al₂(SO₄)₃ contains maximum number of particles, it will have the largest value of freezing point depression.
The pair of compounds that can exist together is:
Both are reducing agents.
When 0.1 mol MnO₄²⁻ is oxidised, the quantity of electricity required to completely oxidise MnO₄²⁻ to MnO₄⁻ is:
(⁺⁶) (⁺⁷)
MnO₄²⁻ → MnO₄⁻ + e⁻
0.1 mole
Quantity of electricity required = 0.1 F
= 0.1 × 96500
= 9650 C
In acidic medium, H₂O₂ changes Cr₂O₇²⁻ to CrO₅, which has two (–O–O–) bonds. Oxidation state of Cr in CrO₅ is:
The structure of CrO₅ is:
Hence, CrO₅ has two peroxy linkages.
Now suppose the oxidation no. of Cr is x then:
x +(− 1x 4) + (-2) = 0
x = + 6
The weight of silver (at. wt. = 108) displaced by a quantity of electricity which displaced 5600 mL of O₂ at STP will be:
w_O₂ = n_O₂ × 32
w_O₂ = 5600/22400 × 32 = 8 g
= 1 equivalent of O₂
1 equivalent of Ag = 108
The reaction of aqueous KMnO₄ with H₂O₂ in acidic conditions gives:
2KMnO₄ + 3H₂SO₄+5H₂O → K₂SO₄ + 2MnSO₄ + 8H₂O + 5O₂
i.e. Mn²⁺ ion and O₂
Among the following complexes the one which shows zero crystal field stabilization energy (CFSE) is:
Due to d⁵ configuration CFSE is zero.
Which of the following complexes is used as an anti-cancer agent: _________.
cis-platin
Among the sets of reactants which one produces anisole?
Phenols react with alkyl halides in alkaline medium to form ethers. Therefore:
Which of the following compounds will undergo racemisation with solution of KOH hydrolysis?
The compound in option (iv) will undergo S_N1 reaction with KOH solution. Hence, only this compound will undergo partial racemisation . Compounds given in other options are not chiral.
‘Which of the following will not be soluble in sodium hydrogen carbonate?
o-nitrophenol will not be soluble in NaHCO₃. Due to intramolecular hydrogen bonding, hydrogen on OH is strongly bound. So, it cannot behave as an acid and cannot react with sodium bicarbonate.
What products are formed when the following compounds are treated with Br₂ in the presence of FeBr₃?
-
(A)
-
(B)
-
(C)
-
(D)
Methyl group is ortho-para directing but due to steric hindrance effect, generated by two CH₃ groups, substitution will not take place on position (I). Hence, only two products are possible.
Identify Z in the sequence of reactions:
.
In the following reaction, the product (A)
-
(A)
-
(B)
-
(C)
-
(D)
.
Which one is most reactive towards Nucleophilic addition reaction?
-
(A)
-
(B)
-
(C)
-
(D)
Any substituent in the carbonyl compound that:increases the positive charge on the carbonyl -carbon will increase reactivity towards nucleophilic addition.
Which of the following will be most stable diazonium salt RN₂⁺ X⁻ ?
Arene diazonium salts are most stable among the given options because of the dispersal of +ve charge on the benzene ring due to resonance.
Which one of the following is an example of a thermoset- ting polymer?
-
(A)
-
(B)
-
(C)
-
(D)
Thermosetting polymers undergo, chemical changes
when heated and set to hard mass when cooled e.g: Bakelite
T
Which of the following compounds polymer- izes to form the polyester Dacron?
Which of the following hormones is produced under the condition of stress which stimulates glycogenolysis in the liver of human beings?
Adrenaline is a hormone produced by adrenal glands during high stress or exciting situations. This hormone is part of the human body's acute stress system, also called the fight or flight response.
Artificial sweetner which is stable under cold conditions only is:
Aspartame is stable under cool conditions.
If a is the length of the side of a cube, the distance between the body centred atom and one corner atom in the cube will be
In BCC, 2 atoms are present. One atom lies at the centre of the cube while the other lies at the corners of the cube. The distance between the body-centred atom and a corner atom is half of the body diagonal, i.e. blood group matched atom id half of the body diagonal ,i.e,. even to any other group.
Five kingdom system of classification suggested by RH.‘Whittaker is not based on:
Fivekingdom system of ‘classification was proposed by
RKWhmker(1969). Thefiwldngdnmchssfimfimisbased on the following criteria :
• Complexity of cell structure — Prokaryotes or eukaryotes.
• Complexity of organisms body — Unicellular or multicellular.
• Mode of obtaining nutrition — Autotrophic or heterotrophic.
• Phylogenetic relationships.
Archaebacteria differ from eubacteria in:
Archaebacteria differ from other bacteria in having a different cell wall structure. They lack peptidoglyan in cell wall and possess a monolayer of branched fatty acids attached to glycerol by ether bonds in their cell membranes.
The motile bacteria are able to move by:
Motile bacteria have thin filamentous extensions on their cell wall called flagella.
‘Which one of the following fungi contains hallucinogens?
Several mushrooms such as Amanitamuscaria, Psilocybe mexicana and Panaeolus spp. secrete hallucinogenic substances like psilocybin and psilocin. These substances may destroy brain cells and power of perception in human beings.
Which of the following shows coiled RNA strand and capsomeres?
TMV (Tobacco Mosaic Virus) is a rod-shaped virus. The rod has a core which contains helically coiled single stranded RNA. There is a protective covering of protein called capsid around the infective part. Capsid consists of small subunits called capsomeres and has antigenic property.
Viruses have:
Allviruses are nucleoproteins (Nucleic acid + Protein) in the structure. The nucleic acid (DNA and RNA) is the genetic material. In a particular virus either DNA or RNA is the genetic material. Both are never present in a virus. Single stranded RNA or ss RNA - Tobacco mosaic virus (TMV) Virus envelope is known as capsid. The capsid is composed of protein subunits called capsomere. ’
‘Which one of the following shows isogamy with non-flagellated gametes?
In Spirogyra, sexual reproduction occurs through conjugation. Gametes are non-flagellated, morphologically similar. But physiologically different (isogamy with physiological anisogamy).
Conjugation is the biological process by which one bacterium transfers genetic material to another bacterium through direct contact. During conjugation, one bacterium serves as the donor of the genetic material, and the other serves as the recipient.
The donor bacterium carries a DNA sequence called the fertility factor F-factor
‘Which one of the following is wrong about Chara?
Chara belongs to Chlorophyceae, is a green alga found attached to the bottoms of shallow water bodies such as ponds, pools and lakes. Male sex organ is called antheridium.Female sex organ is called oogonium. Oogonium is borne at the top of the four-celled filament.
An alga which can be employed as food for human being is:
Chlorella and Spirulina are unicellular algae, rich in proteins and are used as food supplements by space travellers.
‘Which of the following is responsible for peat formation?
Sphagnum, a moss, provides peat that has long been used as fuel.It has the capacity to retain water for long periods and, as such, is used to cover the plant roots during transportation.
Select the Taxon mentioned that represents both marine and fresh water species:
Members of Cteqophora, Cephalochordata and Eghnodermata are exclusively marine.
‘Which one of the following living organisms completely lacks a cell wall?
Gorgonia (sea-fan)isan animal. All animals lack cell wall.Gorgonia is a genus of soft corals, sea fans in the family Gorgoniidae.
Planaria possesses high capacity of:
Planaria is a flatworm which possesses a high capacity of regeneration.
A marine cartilaginous fish that can produce electric current is:
Torpedo is a sluggish fish. It is carnivorous, The prey is killed due to electric shock. The shock can also be harmful for human beings.
An example of edible underground stem is:
Phyllotaxy is the pattern of arrangement of leaves on the stem or branch.
• Ocimum, Guava, Calotropis —> Opposite phyllotaxy
• China rose, Suriflower, mustard —> Alternate phyllotaxy
• Alstonia, Nerium (Oleander) —> whorled phyllotaxy
‘When the margins of sepals or petals overlap one another without any particular direction, the condition is termed as:
If the margins of sepals or petals overlap one another but not in any particular direction, as in Cassia and gulmohur, the aestivation is called imbricate.
Placenta and pericarp are both edible portions in:
In tomato the edible part is pericarp and placent
An aggregate fruit is one which develops from:
Aggregate fruit (etaerio) develops from the multicarpellary apocarpous ovary. They are of following types: etaerio of follicles, etaerio of achenes, etaerio of berries, etaerio of drupes.
Tracheids differ from other tracheary elements in:
The walls of vessels (tracheary elements) are lignified and less thicker than tracheids. The lumen is wider. Vessels differ from tracheids in having cell fusions arising through the dissolution of end walls.
You are given a fairly old piece of dicot stem and dicot root. Which of the following anatomical structures. will you use to distinguish between the two?
Stem of maize has water-containing cavities in vascular bundles.
Choose the correctly matched pair:
(i) Innerlining of salivary ducts - Compound epithelum
(ii) Moist surface of buccal cavity - Compound epithelium
(iii) Tubular parts of nephorns - Cuboidal epithelium.
(iv) Inner surface of ‘bronchioles — Ciliated epithelium.
Choose the correctly matched pair:
Tendon is dense regular connective tissue. Adipose tissue is a type of loose connective tissue located mainly beneath the skin. Cartilage is a type of specialised connective tissue.
Which structures perform the function of mitochondria in bacteria ?
In some bacteria (e.g., Bacillus subtilis) the plasma membrane form certain invaginations or infoldings colled ‘mesosomes in the cytoplasm. The mesosomes have various functions, viz., respiratory, secretory etc.
The solid linear cytoskeletal elements having a diameter of 6 nm and made up of a single type of monomer are known as:
Microtubule, microfilament and intermediate fila- ‘ments along with ER form cytoskeleton. Microfilaments are non-living, solid and consist of actin protein. They are " 4-6 nm in diameter.
The osmotic expansion of a cell kept in water is chiefly regulated by:
The vacuole is bound by a single membrane called tonoplast. It also functions as semipermeable membrane. It segregates vacuolar contents from cytoplasm, allows osmotic entry or exit of water, concentration and storage of nutrients as well as wastes.
Match the following and select the correct answer:
Centrosome is an organelle usually containing two cylindrical structures called centrioles. The centrioles form the basal body of cilia or flagella. In chloroplast a number of organised flattened membranous sacs called the thylakoids are present in the stroma. Chlorophyll pigments are present in the thylakoids. Each mitochondrion is a double membrane bound structure. The inner membrane forms a number of infoldings called the cristae towards the matrix. The cristae increase the surface area.
Select the option which is not correct with respect to enzyme action:
Inhibitions of succinic dehydrogenase by malonate is an example of competitive inhibition. Thus it is a reversible reaction. On increasing the subsrate (succinate) concentration the effect of inhibitor is removed and Vmax remains same.
Which one of the following is a non-reducing carbohydrate?
Sucrose is classified under non reducing sugars because it does not have any free aldehyde or keto group.
During which phase(s) of cell cycle, amount of DNA ina cell remains at 4C level if the initial amount is denoted as 2C?
In M-phase both 4C and 2C of DNA are present in different stages.
In ‘S’ phase of the cell cycle:
DuringS or synthesis phase, replication or duplication of chromosomal DNA and synthesis of histone proteins takes place. During this time the amount of DNA per cell doubles.
The enzyme recombinase is required at which stage of meiosis:
The enzyme recombinase is required at Pachytene stage of meiosis. It catalyzes the exchange of short pieces of DNA between two long DNA strands, particularly the exchange of homologous regions between the paired maternal and paternal chromosomes.
Deficiency symptoms of nitrogen and potassium are visible first in:
The parts of the plants that show the deficiency symptoms also depend on the mobility of the element in the plant. For element that are actively mobilised within the plants and exported to young developing tissues, the deficiency symptoms tend to appear first in the older tissues. For example, nitrogen,potassium and magnesium are visible first in the senescent leaves.
Anoxygenic photosynthesis is characteristic of:
In Rhodospirillum, electron donor is organic compound during photosynthesis. It helps an organisms to trap light energy and store it as chemical energy.
In which one of the following processes CO₂ in not released?
Lactic acid fermentation occurs in lactic acid bacteria (Lactobacillus) and muscles (Human). Pyruvic acid produced in glycolysis is reduced by NADH, to form lactic acid without producing carbon dioxide.
Dr. F. Went noted that if coleoptile tips were removed and placed on agar for one hour, the agar would produce a bending when placed on one side of freshly-cut coleoptile stumps. Of what significance is this experiment?
Charles Darwin and his son Francis Darwin observed that the coleoptiles of Oat (Avena sativa) and canary grass (Phalaris canariensis) responded to unilateral illumination by growing towards the light source (phototropic curvature or phototropism). After a series of experiments, it was concluded that the tip of the coleoptile was the site of production of a substance, that caused the bending of coleoptile.
A few normal seedlings of tomato were kept in a dark room.After a few days they were found to have become white-coloured like albinos. Which of the following terms will you use to describe them?
Etiolation is depigmentation of leaf when plant is placed in dark for more than 36 hrs.
Which one of the following growth regulators is known as “stress hormone’?
Abscisic Acid (ABA) is called stress hormane which works in adverse environmental condition when there is low water content in atmosphere or in drought conditions. ABA causes the stomatal closure of leaves due to which the water loss by the plant is minimized.
The initial step in the digestion of milk in humans is carried out by
Rennin (also called chymosin) is an enzyme that occurs in gastric juice and is a constituent of rennet. It coagulates milk by converting caseinogen to casein. The initial step in the digestion of milk in humans is carried out by rennin.
Fructose is absorbed into the blood through mucosa cells of intestine by the process called:
Facilitated transport is a form of passive transport in which materials are moved across the plasma membrane by a transport protein down their concentration gradient; hence, it does not require energy.
Approximately seventy percent of carbon-dioxide absorbed by the blood will be transported to the lungs:
CO₂ from the respiratory tissues to the lungs is transported by the blood in three ways:
(i) In dissolved state (physical solution): A very small amount is physically dissolved in plasma (7%, i.e.. ≅ 0.3 mL of CO₂ per 100 mL of blood).
(ii) Bicarbonate ions: About . ≅ 70% (i.e. ≅ 2.5 mL per 100 mL of blood). CO₂ diffuses into plasma and then into RBCs where it combines with H₂O to form carbonic acid, which dissociates into hydrogen ions and bicarbonate ions.
(iii) Carbaminohaemoglobin: ≅ About 23% (i.e .≅ 1 mL of CO₂ per 100 mL of blood). CO₂ combines with haemoglobin forming an unstable compound
Person with blood group AB is considered as universal recipient because he has:
Karl Landsteiner first reported ABO blood groups in human beings. A, B and O blood groups were discovered by Landsteiner in 1900 while AB blood group was found out by de Castello and Steini (1902). ABO blood groups are determined by the gene I (isoagglutinin). There are three alleles—Iᴬ, Iᴮ and Iᵒ of this gene. Proteins produced by the Iᴬ and Iᴮ alleles are called A antigen and B antigen. Individuals with AB blood group have both antigen A and antigen B on their RBCs, and no antibodies for either of the antigens in their plasma, so they can receive blood from any blood type.
How do parasympathetic neural signals affect the working of the heart?
Control of heart beat by nervous system can be done by medulla oblongata. Medulla oblongata has two regulatory centre:
(i) Accelerator centre-It functions through sympathetic nervous system (SNS) and increases heart beat by the secretion of epinephrine or adrenaline.
(ii) Depressor centre-It functions through parasympathetic nervous system (PSNS) by the secretion of acetylcholine. It decreases heart beat, speed of conduction of action potential gnd thereby the cardiac output.
‘Which of the following causes an increase in sodium reabsorption in the distal convoluted tubule?
Aldosterone hormone; released by the adrenal glands, helps the body regulate blood pressure. Aldosterone causes the tubules of the kidneys to increase the reabsorption of sodium and water into the blood. This increases the volume of fluid in the body, which also increases blood pressure.
Select the correct matching of the type of the joint with the example in human skeletal system:
A gliding joint is a common type of synovial joint formed between bones that meet at flat or nearly flat articular surfaces. Gliding joints allow the bones to glide past one another in any direction along the plane of the joint-upand . dowh, left and right, and diagonally. Many gliding joints are formed in the appendicular skeleton between the carpal bones of the Wrist; between the carpals and the nmetacarpals of the palm; between the tarsal bones of the ankle; and between the tarsals and the metatarsals of the foot.
Stimulation of a muscle fiber by a motor neuron occurs at:
The neuromuscular junction connects the nervous system to the-muscular system via synapses between efferent nerve fibres and muscle fibres.
Injury localized to the hypothalamus would most likely disrupt:
The hypothalamus is a highly complex structure in the brain that regulates many important brain chemicals. The “hypothalamis is responsible for hormone production. The ‘hormones produced by this area govern ‘body temperature, thirst, hunger, sleep, circadian thythm, ‘moods, sex drive, and the release of other hormones in the body. This area of the brain controls the pituitary gldnd and other glands in the body.
‘Which one of the following statements is not correct?
Retinal is a derivative of vitamin A. Retinal is a polyene chromophore, and bound to proteins called opsins, is the chemical basis of animal vision. Bound to proteins called type I rhodopsins, retinal allows ceftain microorganisms to convert light into metabolic energy.
Identify the hormone with its correct matching of source and function:
Oxytocin, a neurohypophysial hormone, is producedby the hypothalamus and stored and secreted by the poste-rior pituitary gland. It stimulates contractions of the smoothmuscle tissue in the wall of the uterus during childbirth.Progesterone is a female hormone produced by the ovariesduring release of a mature egg from an ovary (ovulation).Progesterone helps prepare the lining of the uterus (endo-metrium) to receive the egg if it becomes fertilized by asperm. If the egg is not fertilized, progesterone levels dropand menstrual bleeding begins. Atrial natriuretic factor isa hormonal substance produced by the right atrium of the
heart that stimulates the excretion of sodium and water by
the kidneys and helps regulate blood pressure.
Fight-or-flight reactions cause activation of:
Adrenal Medulla is called emergency gland and secretes adrenaline (Epinephrine) and nor-adrenaline (nor-cpinephrine) hormones.Adrenaline (epinephrine) is commonly called as ‘emergency hormone’ or 3F–hormone (For fear, fight & flight). It stimulates sweating, heart beat and breathing rate. It causes the dilation of coronary artery (supplying blood to the heart muscles), bronchioles (for increasing inspiratory volume) and pupil (for better vision).
Geitonogamy involves:
Geitonogamy is the transfer of pollen grains in different flowers of the same plant.
Male gametopyte with least number of cell is present in
Male gametophyte is highly reduced in angiosperm and is known as pollen grain. It is 2 or 3-celled.
Function of filiform apparatus is to:
Filiform apparatus helps in the entry of pollen tube into a synergid in ovule. Filiform apparatus is in form of finger-like projection comprising a core of micro fibrils enclosed in a sheath. It resembles transfer cells meant for short distance movement of metabolites. It is responsible for the absorption of food from the nucleus.
Non-albuminous seed is produced in:
Exalbuminous (non-endospermic) seeds usually store reserve food material in cotyledons. In these seeds, the endosperm is used up and not present in mature seeds, e.g., bean, gram and pea.
The shared terminal duct of the reproductive and urinary system in the human male is:
Urethra is a tube that connects the urinary bladder to the genitals for the removal of fluids from the body. The urethra travels through the penis, and carries semen as well as urine.
The main function of mammalian corpus luteum is to produce:
The corpus luteum is a temporary endocrine structure involved in ovulation and early pregnancy. The main secretory product of corpus luteum is progesterone, which is required for the establishment and maintenance of pregnancy.
Select the correct option describing gonadotropin activity in a normal pregnant female:
Synthesis of estrogen and progesterone due to high level of HCG is a normal gonadotropic activity in a normal pregnant female.
Tubectomy is a method of sterilization in which:
Tubectomy also referred to as tubal sterilization is a surgical procedure done on women as a permanent method of contraception. In this method fallopian tubes are blocked or cut, so the sperm are prevented from travelling to the ovary. This prevents any future pregnancies to occur after the surEical procedure.
Which of the following is a hormone releasing Intra Uterine Device (IUD)?
An intrauterine device (IUD) is a small contraceptive device, often "T"-shaped, that contains either copper or levonorgestrel, which is inserted into the uterus. They are one form of long-acting reversible contraception which is the most effective types of reversible birth control. The LNG-20 is so called because it releases 20 micrograms of levonorgestrel per 24 hours, compared with 30 micrograms of levonorgestrel in some mini-Pills.
Assisted reproductive technology, IVF involves transfer of:
IVF (In vitro Fertilization) is the process of fertilization by manually combining an egg and sperm in a laboratory dish.
The process involves monitoring and stimulating a woman's ovulatory process, removing ovum or ova (egg or eggs) from the woman's ovaries and letting sperm fertilise them in a fluid medium in a laboratory. The fertilised egg (zygote) is cultured for 2–6 days in a growth medium and is then transferred to the mother's uterus with the intention of establishing a successful pregnancy.
Fruit colour in squash in an example of
Epistasis is the phenomenon of suppression of phenotypic expression of gene by a nonallelic gene which shows its own effect. A dominant epistatic allele suppresses the expression of a nonallelic gene whether the latter is dominant or recessive. For example, fruit colour of Summer Squash (Cucurbita pepo) is governed by a gene which produces yellow colour in dominant state (Y-) and green colour in recessive state (yy).
In a population of 1000 individuals 360 belong to genotype AA, 480 to Aa. and the remaining 160 to aa. Based on this data, the frequency of allele A in the population is:
According to Hardy Weinberg principle,
p² + 2pq + q² = 1 ; (p + q)² = 1
(AA) p² = 360 out of 1000 individuals
or p² = 36 out of 100
q² = 160 out of 1000 or q² = 16 out of 100
so q = √16 = 4, As p + q = 1
so p is 0.6.
A man whose father was colour blind marries a womanwhose father was colour blind and her normal father. Whatpercentage of male children of this couple will be colour blind?
According to Hardy Weinberg principle,
p² + 2pq + q² = 1 ; (p + q)² = 1
(AA) p² = 360 out of 1000 individuals
or p² = 36 out of 100
q² = 160 out of 1000 or q² = 16 out of 100
so q = √16 = 4, As p + q = 1
so p is 0.6.
A human female with Turner’s syndrome:
Turner’s syndrome is a chromosomal condition that affects development in females. A human female with Turner’s syndrome has 45 chromosomes with XO. The most common feature of Turner syndrome is short stature, which becomes evident by about age 5.
Transformation was discovered by:
Frederick Griffith (in 1928), a British Medical officer described the phenomenon of bacterial transformation. He carried out experiment with Streptococcus, pneumoniae (bacterium causing pneumonia) which is used to infect mice. By using S Strain (heat killed) and R strain- (live) it was concluded that R strain has been transformed by some material of S strain which makes R strain virulent and enable to synthesize smooth polysachharide.
Which one of the following is wrongly matched?
Process of copying genetic information from DNA to RNA is called transcription. Ata time only one DNA strand is being transcribed into RNA. The strand of DNA with polarity 3′→5’ actas template strand and the DNA strand with polarity 5′→3’ act as coding strand.
Select the correct option:
Synthesis of RNA exhibits several features that are synonymous with DNA replication. RNA synthesis requires accurate and officient initiation, elongation proceeds in the 5’′→ 3’ direction (i.e. the polymerase moves along B the template strand of DNA in the 3′→5’ direction), and RNA synthesis requires distinct and accurate termination.
Forelimbs of cat, lizard used in walking; forelimbs of whale used in swimming and forelimbs of bats used in flying are an example of:
Organs which have a common fundamental anatomical plan and similar émbryonic origin whatever varied functions they may perform are regarded as hm;mlogous organs. For examples the flippers of a whale, a bats wing, fore-limb of a horse, a bird’s wing and forelimbs of human are structurally as well as functionally different.
Which one of the following are analogous structures?
Analogous organs are those that perform similar Functions but have entirely different embryonic origins.
At which stage of HIV infection does one usually show symptoms of AIDS:-
HIV infection does usually show symptoms of AIDS. When HIV damages a large number of helper T-Lymphocytes (CD4 cells), AIDS is the stage of HIV infection that occurs when one's immune system is badly damaged and one becomes vulnerable to infections and infection-related cancers called opportunistic infections. When the number of one's CD4 cells falls below 200 cells per cubic millimetre of blood (200 cells/mm³), one is considered to have progressed to AIDS.
‘Which is the particular type of drug that is obtained from the plant whose one flowering branch is shown below?
The plant illustrated in the diagram is Datura, which has hallucinogenic properties. A hallucinogen is a substance that produces psychological effects normally associated only with dreams, schizophrenia, or religious visions. It produces changes in perception (ranging from distortions in what is sensed to perceptions of objects where there are none), thought, and feeling.
To obtain virus-fiee healthy plants from a diseased one by tissue culture technique, which part/parts of the diseased plant will be taken:
To obtain virus-free healthy plants from a diseased one by tissue culture technique, both apical and axillary meristems of the diseased plant are taken. Plant tissue culture is used to maintain or grow plant cells, tissues, or organs under sterile conditions on a nutrient culture medium of known composition
Plant tissue culture is widely used to produce clones of a plant in a method known as micro propagation.
What gases are produced in anaerobic sludge digesters?
Anaerobic digestion is a series of processes in ‘which microorganisms break down biodegradable material in the
absence of oxygen, used for industrial or domestic purposes
to manage waste and/or to release encrgy. The process of anaerobic digestion produces a biogas, consisting of methane (it will burn), carbon dioxide (it does not burn) and traces of other contaminant gases.
An analysis of chromosomal DNA using the Southern hybridization technique does not use:
PCR is a technique for enzymatically replicating DNA without using a living organism such as E. coli or Yeast. It is commonly used in medical and biological research labs for a variety of tasks like detection of hereditary diseases, identification of genetic fingerprints etc.
Blotting is the transfer of the fragmented DNA sequence to the nitrocellulose membrane or nylon membrane. The process is done by either electroblotting or capillary blotting. The DNA molecule is saturated using a NaCl solution and permanently fixed using either UV radiation or drying.
Which vector can clone only a small fragment of DNA?
Plasmids are small extranuclear circular DNAS which carry extrachromosomal genes in bacteria and some fungi.
They replicate independently. The best known vectors which
are also available commerciallly are pBR322 and pUC-18.
In vitro clonal propagation in plants is characterized
Nowa days PCR and RAPD technique are used for thie characterisation of i vifro clonal propagation in plants.
Random amplified polymorphic DNA (RAPD) is a PCR based technique for identifying genetic variation. It involves the use of a single arbitrary primer in a PCR reaction, resulting in the amplification of many discrete DNA products.
Which of the following statements is correct for the spontaneous adsorption of a gas?
For adsorption, ΔS < 0 and for a spontaneous change ΔG = −ve.
hence ΔH should be highly negative which is clear from the equation
ΔG = ΔH − TΔS
= −ΔH − T(−ΔS)
= −ΔH + TΔS
So if ΔH is highly negative, ΔG will also be (−ve).
‘Which one of the following statements is correct?
Generally, monocotyledonous seeds (e.g. grasses) are endospermic but some, as in orchids, are non-endospermic. Mango is a drupe fruit that develops from a multicarpellary syncarpous, superior ovary having one or many seeds. A sterile stamen is called a staminode. In monocots, outer covering of endosperm separates the embryo by a proteinaceous layer called aleurone layer.
Commonly used vectors for human genome sequencing are:
Human genome sequencing is a process that determines the complete DNA sequence of an organism’s genome at a single time. This requires sequencing all of an organism’s chromosomal DNA as well as DNA contained in the mitochondria and, for plants, in the chloroplast. Commonly used vectors for human genome sequencing are BAC (Bacterial-artificial chromosome) and YAC (Yeast artificial chromosome;
A P1-derived artificial chromosome (PAC) is a DNA construct that was derived from the DNA of P1 bacteriophage.
Pollen tablets are available in the market for:
Pollen grains are rich in nutrients therefore used as food supplements. Athletes and race horses use these tablets to enhance performance.
Pollen contains 22.7% of protein on average, including 10.4% of essential amino acids such as methionine, lysine, threonine, histidine, leucine, isoleucine, valine, phenylalanine, and tryptophan. Along with this, it also contains sugars, lipids and sterols.
The first human hormone produced by recombinant DNA technology is:
Mammahan hormones were among the first products prepared in bacteria by r-DNA technology. Human insulin and human growth hormone are earliest examples.
Match the following and select the correct option:
Column-I | Column-II | |
|---|---|---|
(A) Earthworm | ___ | (i) Pioneer species |
(B) Succession | ___ | (ii) Detritivore |
(C) Ecosystem service | ___ | (iii) Natality |
(D) Population growth | ___ | (iv) Pollination |
Detrivores, (e.g. earthworm) break down detritus into smaller particles. The species that invade a bare area in succession is called pioneer species. The products of ecosystem processes are termed as ecosystem services, e.g., healthy forest ecosystems purify air and water, mitigate droughts and floods etc. Natality refers to number of births during a given period in the population.
Given below is a simplified model of phosphorus cycling in a terrestrial ecosystem with four blanks (A-D). Identify the blanks:-
Phosphorus is an important element for living beings.Consumers obfain phosphorus directly or indirectly from plants. Phosphorus is also present in phosphatic rocks. It released during the decomposition of plant and animal re-‘mains. The released phosphorus may reach the deeper layers of soil and gets deposited as phosphate rocks. Al plants and animals eventually die and in due time, their organic remains or debris decay through the action of micro-organism and the phosphates are released into the wstcr for recycling.
Odum is the father of modern ecology. Clements proposed the theory of plant succession. The term ecosystem was proposed by Tansley. Elton proposed the concept of ecological pyramids. Hence, it is also called Eltonian pyramids.
A species facing extremely high risk of extinction in the immediate future is called:
Critically Endangered Themxunfacmgvcryhlghnsk - of extinction in the wild can become extinct any moment the immediate future, e.g. Sus salvinus (Pigmy Hog), Berbers nilghiriensis.
The organization which publishes the Red List of species is:
IUCN or WCN maintains a red data book which is a catalogue of threatened plants and animals facing risk of extinction. The IUCN red list (2004) document the extinction of 784 species (including 338 vertebrate, 359 invertebrates and 87 plants) in the last 500 years
Just as a person moving from Delhi to Shimla to escape the heat for the duration of hot summer, thosand of‘migratory birds from Siberia and other-extremely cold northern regions move to:
In Ex-situ conservation the threatened animals and plants are taken out of their natural habitat and are protected in special parks or areas like Zoological parks, Wild life safari parks and Botanical gardens etc. The ex-situ conservation also includes cryopreservation, fertilization of eggs in-vitro and propagation of plants through "tissue culture methods", preservation of seeds through Seed banks.
A location with luxuriant growth of lichens on the trees indicates that the:
The Keoladeo National Park is a famous avifauna sanctuary in India that plays host to thousands of birds especially during the summer season. It is formerly known as the Bharatpur Bird Sanctuary in Bharatpur, Rajasthan. October to February is the best time, for the migratory birds there as well as the residents. In August local birds start their nest building and rear their young for the next few months. October/November is when the migrants arrive. Most stay till March including the Siberian Crane.
A scrubber in the exhaust of a chemical industrial Temoves:-
Lichens are very good pollution indicators, they do not grow in polluted areas. They are sensitive to sulphur dioxide pollution.
An example of ex situ conservation is:
Scrubbers are air pollution control devices that use liquid to remove particulate matter or gases from an industrial exhaust or flue gas stream. This atomized liquid (typically water) entrains particles and pollutant gases in order to effectively wash them out of the gas flow
Scrubbers remove or scrubs SO₂ emissions from the exhaust of coal-fired power plants.
The zone of atmosphere in which the ozone layer is pressent is called:
The ozone (O₃) found in upper part of the atmosphere, i.e., Stratosphere, is Good ozone, since it acts as a shield for absorbing UV-radiations from sun.
If 20J of energy is trapped at producer level, then how much energy will be available to peacock as food in the following chain?
Plant → Mice →Snake → Peacock
According to 10% law of energy flow by Raymond Lindman. The total amount of energy that can be transferred to the next trophic level is the 0% hence, peacock will receive 0.02 J of energy as top consumer. Energy received by other organisms are
Plant → 20 J
Mice → 20 × 10% = 2 J
Snake → 2 × 10% = 0.2 J
If force (F), velocity (V) and time (T) are taken as fundamental units, then the dimensions of mass are:
Force = Mass × Acceleration
⇒ Mass = Force / Acceleration
⇒ Mass = Force / (Velocity / Time)
= [FV⁻¹T]
A particle is moving such that its position coordinate (x, y) are
(2 m, 3 m) at time t = 0
(6 m, 7 m) at time t = 2 s and
(13 m, 14 m) at time t = 5 s.
Average velocity vector (v̄ₐᵥ) from t = 0 to
t = 5 s is:
v̄ₐᵥ = Δr vector(displacement) / Δt( time taken )
= ( 13-2)î + ( 14-3)ĵ /5-0= 11/5 (î + ĵ)
A projectile is fired from the surface of the earth with avelocity of 5 ms⁻¹ and angle θ with the horizontal.Another projectile fired from another planet with a velocity of 3 ms⁻¹ at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in ms⁻²) given g = 9.8 m/s²?
Trajectory is identical for both, Horizontal range,
= u² sin 2θ / g
So that,
g_planet / g_earth = (u_planet / u_earth)²
Therefore,
g_planet = (3/5)² × (9.8 m s⁻²)
= 3.5 m s⁻²
The force ‘F’ acting on a particle of mass ‘m’ is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from zero to 8s is:
Change in momentum,
Δp = ∫F dt
= Area of F–t graph
= ar of △ − ar of ▭ + ar of ▭
= 1/2 × 2 × 6 − 3 × 2 + 4 × 3
= 12 N-s
A system consists of three masses m₁, m₂ and m₃ connected by a string passing over a pulley P. The massm₁ hangs freely and m₂ and m₃ are on a rough horizontal table (coefficient of friction = μ). The pulley is frictionless and of negligible mass. The downward acceleration of mass m₁ is: (Assume m₁ = m₂ = m₃ = m)
Acceleration =
Net force in the direction of motion / Total mass of system
= [m₁g − μ(m₂ + m₃)g] / (m₁ + m₂ + m₃)
= g/3 (1 − 2μ)
(∵ m₁ = m₂ = m₃ = m given)
A mass m is attached to thin wire and whirled in a vertical circle. The wire is most likely to break when:
T − mg = mv²/r [centripetal force] = mv²/r
⇒ T = mg + mv²/r
As the velocity is maximum at lowest point so tension is maximum at the lowest position of mass, so the chance of breaking is maximum.
A bodyof mass (4m) is lying in x-y planeat rest. It suddenly explodes into three pieces. Two pieces, each of mass (m) move perpendicular to each other with equal speeds (v). The total kinetic energy generated due to explosion is :
By conservation of linear momentum Magnitude of the momentum of heavier piece of mass (2m) =Magnitude of the vector sum of momentum of each piece of mass (m)
(2m)v₁ = √[(mv)² + (mv)²]
⇒ 2mv₁ = √2 mv
⇒ v₁ = v/√2
As two masses of each mass m move perpendicular to
each other.
Total KE generated
= ½mv² + ½mv² + ½(2m)v₁²
= mv² + mv²/2
= 3/2 mv²
A solid cylinder of mass 50 kg and radius 0.5 mis free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and the other hanging freely. Tension in the string required to produce an angular acceleration of 2 revolutions s⁻² is:
Here α=2 revolutions/s²= 4π rad/s² (given)
I_cylinder =1/2 MR² =1/2 (50) (0.5)² =25/4 Kg-m²
As τ=Iα so TR= Iα
⇒ T=Iα/R=(25/4)(4π)/(0.5) N = 50πN=157N
The ratio of the accelerations for a solid sphere (mass ‘m’ and radius ‘R’) rolling down an incline of: angle ‘0’ without slipping and slipping down the incline without rolling is :
For solid sphere rolling without slipping on inclined plane, acceleration
a₁=g sin θ/ (1+K²/R²)
For solid sphere slipping on inclined plane without rolling, acceleration
a₂= g sin θ
Therefore, the required ratio is
a₁/ a₂=1/(1+K²/R²) = 1/(1+ 2/5)=5/7
Dependence of intensity of gravitational field (E) of earth with distance (r) from centre of earth is correctly represented by:
-
(A)
-
(B)
-
(C)
-
(D)
First when (r < R) E ∝ r and then when r ≥ R, E ∝ 1/r².
Hence graph (b) correctly depicts.
A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would our earth (mass = 5.98 × 10²⁴ kg)
have to be compressed to be a black hole?
From question,
Escape velocity
√ 2GM /R =c= (speed of light)
⇒R=2GM/c² =2×6.6×10⁻¹¹×5.98×10²⁴/(3×10⁸)²m
10m⁻²
Copper of fixed volume ‘V; is drawn into wire of length */’. When this wire is subjected to a constant force ‘F’, the extension produced in the wi re is ‘Al Which of the
following graphsis a straight line?
Y=F/A /Δl / l
⇒Δl =Fl/ AY
But, V=Al
A=V/l
Therefore, Δl = Fl² /VY α l²
Hence, the graph of Δl versus l² will give a straight line.
A certain number of spherical drops of a liquid of radius 'r coalesce to form a single drop of radius 'R' and volume 'V'. If 'T' is the surface tension of the liquid, then:
Volume same= n(4/3πr³) =4/3πR³
⇒n= R³ / r³
Change in energy =TΔA
=T[ 4πR²- n 4πr²]
= 4πT [R² - R³/ r³. r²]
= 3 (4/3πR³) T [ 1/R - 1/r ]
=3VT [ 1/R - 1/r ] (R>r)
=−3VT [ 1/r - 1/R ] is released
Steam at 100°C is passed into 20 g of water at 10°C. When
water acquires a temperature of 80°C, the mass of water
present will be:
[Take specific heat of water = 1 cal g⁻¹ °C⁻¹ and latent heat
of steam = 540 cal g⁻¹]
According to the principle of calorimetry, Heat lost = Heat gained
mLᵥ + ms_wΔθ = m_w s_wΔθ
⇒ m × 540 + m × 1 × (100 − 80)
= 20 × 1 × (80 − 10)
⇒ m = 2.5 g
Therefore total mass of water at 80°C
= (20 + 2.5) g
= 22.5 g
Certain quantity of water cools from 70°C to 60°C in the first 5 minutes and to 54°C in the next 5 minutes. The temperature of the surroundings is:
Let the temperature of surroundings be θₒ By Newton' law of cooling
A monoatomic gas at a pressure P, having a volume V expands isothermally to a volume 2V and then adiabatically to a volume 16V. The final pressure of the gas is :
(take {5}{3}]
For isothermal process,
P₁V₁=P₂V₂
⇒PV=P₂(2V)
⇒P₂⇒P/2
For adiabatic process,
P₂V₂⌃γ=P₃V₃⌃γ
⇒ (p/2) (2v)⌃γ =P₃ 16v⌃γ
⇒ P₃ =3/2 (1/8) ⁵ᐟ³ =P/64
A thermodynamic system undergoes cyclic process ABCDA as shown in fig. The work done by the system in the cycle is :
Work done by the system in the cycle
= Area under P-V curve and V-axis
= 1/2 (2P₀ - P₀)(2V₀ - V₀) + [-(1/2) (3P₀ - 2P₀)(2V₀ - V₀)]
= P₀V₀/2 - P₀V₀/2 = 0
The mean free path of molecules of a gas, (radius 'r') is inversely proportional to
Mean free path λₘ = 1/√2Πd²n
where d = diameter of molecule and d =2r
λₘ α 1/r²
The oscillation of a body on a smooth horizontal surface is
represented by the equation,
X = A cos(ωt)
where X = displacement at time t
ω = frequency of oscillation
Which one of the following graphs shows correctly the
variation of 'a' with 't' ?-
(A)
-
(B)
-
(C)
-
(D)
Displacement,
x=A cos(ωt) ( given)
Velocity,
v=dx/ dt=−Aωsin(ωt)
Acceleration,
a= dv /dt=−A ω²cos(ωt)
Hence, graph (c) correctly depicts the variation of a with t
If n₁, n₂ and n₃ are the fundamental frequencies of threesegments into which a string is divided, then the original fundamental frequency n of the string is given by:
n=1/ 2l √ T/m
or, n ∝1/ l
or,
n l = constant, K
Therefore,
n₁ ,l₁=K,
,n₂ l₂ =K,
n₃l₃=K
Also,
l=l₁+l₂+l₃
or,
K/n=K/n₁+K/n₂ +K/n₃
or, 1/n = 1/n₁+1/n₂ +1/n₃
The number of possible natural oscillation of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are: (velocity of sound = 340 ms⁻¹)
In case of closed organ pipe, frequency,
fₙ =(2n + 1) v/4l
For n = 0, f₀ = 100 Hz
n = 1, f₁ = 300 Hz
n = 2, f₂ = 500 Hz
n = 3, f₃ = 700 Hz
n = 4, f₄ = 900 Hz
n = 5, f₅ = 1100 Hz
n = 6, f₆ = 1300 Hz
Hence possible natural oscillation whose frequencies
< 1250 Hz = 6 (n = 0, 1, 2, 3, 4, 5)
A speeding motorcyclist sees traffic jam ahead of him. He slows down to 36 km/hour. He finds that traffic has ceased and a car moving ahead of him at 18 km/hour is honking at a frequency of 1392 Hz. If the speed of sound is 343 m/s, the frequency of the honk as heard by him will be:
According to Doppler's effect,
Apparent frequency,
n′=n(v+vₒ/ v+ vₛ)= 1392(343+10/343+5)=1412 Hz
In a region, the potential is represented byV(x, y, z) = 6x − 8xy − 8y + 6yz, where V is in volts and x, y, z are in metres.
The electric force experienced by a charge of 2 coulomb situated at point (1, 1, 1) is:
A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of the sphere respectively are:
Due to conducting sphere,
At centre, electric field E = 0
but electric potential,
V= Q/4πε₀R
Two thin dielectric slabs of dielectric constant K₁ and K₂ (K₁ < K₂) are inserted between the plates of a parallel plate capacitor, as shown in the figure. The variation of electric field 'E' between the plates with distance 'd' as measured from plate P is correctly shown by:
-
(A)
-
(B)
-
(C)
-
(D)
Electric field,
E ∝1/K
As K₁ < K₂, so E₁ > E₂.
Hence graph correctly depicts the variation of electric
field E with distance d.
Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 volt and the average resistance per km is 0.5 Ω. The power loss in the wires is:
Total resistance R = (0.5 Ω/km) × (150 km)
= 75 Ω
Total voltage drop = (8 V/km) × (150 km) = 1200 V
Power loss = (ΔV)² /R = (1200)² /75 W
= 19200 W = 19.2 kW
The resistances in the two arms of the meter bridge are 5Ω and RΩ, respectively. When the resistance R is shunted
with an equal resistance, the new balance point is at 1.6l.₁ The resistance 'R' is:
A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery used across the potentiometer wire, has an emf0of2.0 V and a negligible internal resistance. The potentiometer wire itself is 4m long, When the resistace R, connected across the given cell, has values of (i) infinity (i) 9.5Ω,
The balancing lengths’, on the potentiometer wire are found to be 3 m and 2.85 m, respectively. The value of internal resistance of the cell is
Internal resistance of the cell,
(E − V) (ℓ₁ − ℓ₂)
r = (E − V / V ) R = ((ℓ₁ − ℓ₂)/ℓ₂ ) R
= ((3 − 2.85) / 2.85) × (9.5) Ω = 0.5 Ω
Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other such that 'O' is their common point for the two. The wires carry I₁ and I₂ currents respectively. Point 'P' is lying at distance 'd' from 'O' along a direction perpendicular to the plane containing the wires. The magnetic field at the point 'P' will be:
μₒ/2Π d( I₁²+ I₂²) ¹ᐟ²) Net magnetic field B =√B₁²+ B₂²
where, B₁ =μₒI₁/2Π d , B₂ = μₒI₂ /2Π
Therefore,
μₒ/2Π d( I₁²+ I₂²) ¹ᐟ²
In an ammeter 0.2% of main current passes through galvanometer. If resistance of galvanometer is G, the resistance of ammeter will be:
As 0.2% of the main current passes through the galvanometer, hence 998/ 1000I current passes through the shunt.
(2I/1000)G=(998I/1000)S
S= G/499
Total resistance of the ammeter,
R=SG/ S+G= (G/499) G/(G/499) +G
=G/500
Following figures show the arrangement of bar magnets in different configurations. Each magnet has magnetic dipole moment Which configuration has highest net magnetic dipole moment ?
-
(A)
-
(B)
-
(C)
-
(D)
Net magnetic dipole moment = 2 M cos (θ/2)
As value of cos (θ/2) is maximum in case (c) hence net magnetic dipole moment is maximum .
A thin semicircular conducting ring (PQR) of radius ‘r’ is falling with its plane vertical ina horizontal magnetic field B, as shown in figure. The potential difference developed across the ring when its speed is v, is
Rate of decreasing of area of semicircular ring
dA/ dt = (2r)V
dt
From Faraday's law of electromagnetic induction
e= -dθ /dt = B dA/dt = -B(2r)V
As induced current in ring produces magnetic field in
upward direction hence R is at higher potential.
A transformer having efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6 A, the voltage across the secondary coil and the current in the primary coil respectively are:
Efficiency
η=VₛIₛ/ VₚIₚ
⇒0.9 = Vₛ(6)/ 3×10³
⇒Vₛ= 450 V
As, VₚIₚ=3000 so
Iₚ=3000/Vₚ=3000/200 A=15A
The angle of a prism is 'A'. One of its refracting surfaces is silvered. Light rays falling at an angle of incidence 2A on the first surface returns back through the same path after suffering reflection at the silvered surface. The refractive index, μ, of the prism is
Light with an energy flux of 25 × 10⁴ W m⁻² falls on a perfectly reflecting surface at normal incidence. If the surface area is 15 cm², the average force exerted on the surface is:
Average force,
F_av=Δp/Δt=2IA/c
(∵Power=F⋅V)
= 2×25 ×10⁴ ×15×10/3×10⁻⁴=2.50× 10⁻⁶ N
If the focal length of objective lens is increased then magnifying power of:
Magnifying power of microscope
=LD / fₒ fₑ ∝ 1/fₒ
Hence, with increase in fₒ the magnifying power of the microscope decreases.
Magnifying power of telescope fₒ/fₑ ∝ fₒ
Hence, with increase in fₒ the magnifying power of the telescope increases.
In the Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is λ is K, (λ being the wave length of light used). The intensity at a point where the path difference is λ/4, will be:
For path difference λ, phase difference
0=2π rad.
For path difference λ /4,phase difference =π/2 rad.
As
K=4Iₒ,so intensity at the given point where path difference is λ /4
K′=4Iₒcos²(π/4)(cosπ/4=cos45°)=2Iₒ=K/2.
A beam of light of λ = 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between first dark fringes on either side of the central bright fringe is:
Given:
D=2 m ;d=1 mm=1×10⁻³
λ=600nm=600×10⁻⁶ m
Width of central bright fringe (=2β) =2λD/d
2×600×10⁻⁶ × 2 / 1 ×10⁻³ m =2.4×10⁻³ m =2.24 mm
If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is
As we know, λ=h/p = h/ √2mK (∵P=√2mKE)
or λ₁/λ₂ =√K₂/K₁=√16K/K =4/1
Therefore, the percentage change in de-Broglie wavelength is
=1−4/ 4 ×100=−75%
When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is:
According to Einstein's photoelectric equation,
hν=ϕₒ+K_max
We have
hν=ϕₒ+0.5...(i)
and
1.2hν=ϕₒ+0.8...(ii)
Therefore, from the above two equations,
ϕₒ=1.0 eV
Hydrogen atom in ground state is excited by a monochromatic radiation of λ = 975 Å. Number of spectral lines in the resulting spectrum emitted will be:
For the
λ=975 Å.
1=R(1/n₁²−1/n₂²)
where R i s the Rydberg constant.
Solving, we get
n₂=n=4
(∵n₁=1, ground state)
Therefore, the number of spectral lines is
=n(n−1)/2=4(4−1)=6.
The binding energy per nucleon of ⁷₃Li and ⁴₂He nuclei are 5.60 MeV and 7.06 MeV, respectively.
In the nuclear reaction, ⁷₃Li + ¹₁H →⁴₂He + Q the value of energy Q released is:
BE of ⁴₂He =4ˣ 7.06= 28.24 MeV
BE of ⁷₃Li = =7x5.60=39.20 MeV
⁷₃Li + ¹₁H →₂He⁴ + Q
39.20 28.24 x 2 (2(=56.48 MeV)
A radio isotope "X" with a half life 1.4 × 10⁹ years decays to "Y", which is stable. A sample of the rock from a cave was found to contain "X" and "Y" in the ratio 1:7. The age of the rock is:
As
Nₓ/N_y = 1/7 (Given)
⇒ Nₓ/(Nₓ + N_y) = 1/8 = (1/2)³
Therefore, age of the rock
t = 3T₁/₂
= 3 × 1.4 × 10⁹ years
= 4.2 × 10⁹ years.
The given graph represents V–I characteristic for a semiconductor device.
Which of the following statement is correct?
The given graph represents V–I characteristics of solar cell.
The barrier potential of a p–n junction depends on:
(A) type of semiconductor material
(B) amount of doping
(C) temperature
Which one of the following is correct?
The barrier potential of a p–n junction depends on amount of doping, type of semiconductor material and temperature.
D(+) glucose reacts with hydroxylamine and yields an oxime. The structure of the oxime would be :
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(A)
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(B)
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(C)
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(D)
Given below is the representation of the extent of global diversity of invertebrates. What groups the four portion (A-D) represent respectively?
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(A)
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(B)
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(C)
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(D)
Arthropoda is the largest phylum of Animalia which includes insects. Over two-thirds of all named species in earth are arthropods. They have organ system level of organisation. Molluscass the second largest anima] phylum.
They are terrestrial or aquatic (marine or fresh),
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