Category: MCQ Questions

Answer: (b) plasmodesmata

Answer: (b) flagellin
Explanation:
Bacterial flagella is made up of protein flagellin.

Answer: (d) Exosmosis

Answer: (b) Robert hooke

Answer: (a) 23 sr RNA

Answer: (a) Cytoplasm

Answer: (b) semi-permeable
Explanation:
Plasma membrane is a selective permeable membrane that allows only selective molecules to pass through it.
The permeability depends on the electric charge and polarity of the molecules.

Answer: (b) calcium pectate
Explanation:
Middle lamella is made up of calcium pectate.

Answer: (a) 1-D, 2-A, 3-B, 4-C
Explanation:
A filamentous protein structure present in the cytoplasm is called cytoskeleton.
Flagella is a hair-like outgrowths of the cell membrane.
The central part of the proximal region of the centriole is called hub.
Fimbriae are small bristle like fibres sprouting out of the cell.

Answer: (a) Mycoplasma
Explanation:
Mycoplasma does not have cell wall.
PPLO (Pleuro Pneumonia Like Organisms)

Answer: (a) Cytoplasm

Answer: (d) nerve cells
Explanation:
Nerve cells are the longest cells.

Answer: (c) SER
Explanation:
The main site for synthesis of lipid is smooth endoplasmic reticulum.

Answer: (c) protein synthesis
Explanation:
Ribosomes are called protein factories because they synthesize proteins.

Answer: (b) Lysosome

Answer: (a) both DNA and histones

Answer: (a) Nerve cell

Answer: (c) While lipids can rarely flip-flop, proteins can not

Answer: (c) Golgi apparatus

Answer: (d) Cristae
Explanation:
The inner membrane forms a number of foldings called cristae towards the matrix. These cristae increases the surface area.

Answer: (c) Cotton field
Explanation:
Rice is grown in flooded fields, a situation that depletes the soil of oxygen. Soils that are anaerobic (lacking oxygen) allow the bacteria that produce methane from decomposing organic matter to thrive. Some of this methane then bubbles to the surface, but most of it is diffused back into the atmosphere through the rice plants themselves. Hence, rice Paddy field that produces maximum methane gas into the atmosphere.

Answer: (b) Fluorides in water
Explanation:
Skeletal fluorosis is a bone disease caused by excessive accumulation of fluoride in the bones. In advanced cases, skeletal fluorosis causes painful damage to bones and joints.

Answer: (d) CO.
Explanation:
Motor vehicle emissions contribute to air pollution and are a major ingredient in the creation of smog in some large cities. Transportation contributes to more than half of the carbon monoxide, sulphur dioxide and nitrogen oxides, and almost a quarter of the hydrocarbons emitted into our air. Carbon monoxide (CO) is a colourless, odourless, toxic non-irritating gas. It is a product by incomplete combustion of fuel such as natural gas, coal or wood. Vehicular exhaust is a major source of carbon monoxide.

Answer: (d) Freons.
Explanation:
Fluorocarbons such as freon-1 (CFCl₃) and freon-12 (CF₂Cl₂) emitted as propellants in aerosol spray cans, refrigerators, fire fighting reagents etc. are stable compounds and chemically inert. They do not react with any substance with which they come in contact and thus float through the atmosphere unchanged and eventually enter the stratosphere. There they absorb UV radiations and break down liberating free atomic chlorine which causes decomposition of ozone. This results in the depletion of the ozone layer.

Answer: (c) Soil
Explanation:
Photochemical smog is produced when pollutants from the combustion of fossil fuels react with sunlight. The energy in the sunlight converts the pollutants into other toxic chemicals. For photochemical smog to form, there must be other pollutants in the air, specifically nitrous oxides and other volatile organic compounds (VOCs).

Answer: (c) Microorganisms present in the soil
Explanation:
CO is converted into CO₂ by microorganism present in soil.

Answer: (b) Biodegradable pollution
Explanation:
Domestic waste mostly constitutes biodegradable pollution.

Answer: (d) Nuclear Fallout.
Explanation:
Nuclear fallout, or simply fallout, is the residual radioactive material propelled into the upper atmosphere following a nuclear blast or a nuclear reaction conducted in an unshielded facility, so called because it “falls out” of the sky after the explosion and the shock wave have passed. While deforestation, soil erosion and increasing deserts can be controlled, nuclear fallout causes genetic mutation in humans and destruction that cannot be controlled or cured.

Answer: (c) Thermal power stations
Explanation:
Thermal power plant, agro industry and mining all are responsible for soil and water pollution. (Due to extraction of chemicals which cause pollution)

Answer: (c) Amount of BOD
Explanation:
The amount of oxygen required to break down a certain amount of organic matter is called the biological oxygen demand (BOD). The amount of BOD in water is an indicator of the level of pollution.

Answer: (c) Microorganisms present in the soil
Explanation:
CO is converted into CO₂ by microorganism present in soil.

Answer: (d) Ammonia acts as a sink for NO_x.
Explanation:
The average residence time of NO is 4 days.

Answer: (a) Minamata disease
Explanation:
Disease caused by eating fish found in water contaminated with industrial waste having mercury is minamata disease.

Answer: (d) Freons.
Explanation:
Fluorocarbons such as freon-1 (CFCl₃) and freon-12 (CF₂Cl₂) emitted as propellants in aerosol spray cans, refrigerators, fire fighting reagents etc. are stable compounds and chemically inert. They do not react with any substance with which they come in contact and thus float through the atmosphere unchanged and eventually enter the stratosphere. There they absorb UV radiations and break down liberating free atomic chlorine which causes decomposition of ozone. This results in the depletion of the ozone layer.

Answer: (c) Soil
Explanation:
Photochemical smog is produced when pollutants from the combustion of fossil fuels react with sunlight. The energy in the sunlight converts the pollutants into other toxic chemicals. For photochemical smog to form, there must be other pollutants in the air, specifically nitrous oxides and other volatile organic compounds (VOCs).

Answer: (d) Ozone layer does not permit infrared radiation from the sun to reach the earth.
Explanation:
Allotropes of Oxygen-
O₂ (Dioxygen) and O₃ (Ozone)
Infrared radiation is most important for heating up the atmosphere and the surface of the earth. These rays have high wavelength and low frequency and are not very harmful. Ozone does not hinder infrared radiation.

Answer: (b) CH₂ = CH₂

Answer: (c) O₃
Explanation:
Ultraviolet radiations cause photochemical splitting of oxygen molecules in the stratosphere. The nascent oxygen combines with molecular oxygen to form ozone. In troposphere nascent oxygen comes from nitrogen dioxide.

Answer: (b) Due to run-off excess fertilizers
Explanation:
Inorganic plant nutrients found in fertilizers. Run-off carries excess fertilizers into nearby water bodies, causing high levels of plant nutrients which promote the excessive growth of algae and other aquatic plants.

Answer: (b) CO
Explanation:
Important Air pollutants
Oxides of carbon (CO,CO₂)
Oxides of sulphur (SO₂ & SO₃)
Oxides of nitrogen
Chlorofluoro carbons (CFC)
Hydrocarbons Like CH₄
​Photochemical and Industrial smog.
Dust & other particulate matter from industries.

Answer: (d) Kinesiology – Fossil study

Answer: (d) Adolf Engler and Karl Prantl

Answer: (a) their failure to interbreed
Explanation:
Two taxonomic species are distinguished from each other by their failure to interbreed.

Answer: (a) Housefly
Explanation:
Musca domestica is the common name of housefly.

Answer: (a) important for ecosystem
Explanation:
Keystone species are plants or animals species that play a unique and crucial role in the way an ecosystem functions.
They are very important for ecosystem to function properly.

Answer: (d) pisciculture

Answer: (d) Lucknow
Explanation:
National Botanical Research Institute is located in Lucknow.

Answer: (d) The classification of organisms based on their evolutionary history and establishing their phylogeny on the totality of various parameters from all fields of studies

Answer: (c) Characters of flowers are conservative

Answer: (a) International code for Botanical nomenclature
Explanation:
The scientific names of plants are given by International code for Botanical nomenclature.

Answer: (c) open system

Answer: (c) Genera
Explanation:
Solanum, Panthera, Homo are examples of generic names.

Answer: (a) Musca domestica Order – Diptera Family – Muscidae Phylum – Arthropoda
Explanation:
Mangifera indica belongs to Phylum Angiospermae.
Triticum aestivum belongs to family Poaceae.
Panthera leo belongs to order Carnivora.

Answer: (c) Carolus Linnaeus
Explanation:
The system of providing scientific names with two components i.e. generic name and specific epithet is called Binomial Nomenclature.
Binomial Nomenclature was given by Carolus Linnaeus.

Answer: (d) All of these
Explanation:
Linnaeus used Systema Naturae as the title of his publication.
The scope of systematics includes identification, nomenclature and classification.
Systematics takes into account evolutionary relationship between organisms.

Answer: (b) Animalia

Answer: (d) Key
Explanation:
Taxon is a group of one or more population of organisms.
Kingdom, division and species come under taxon but key is taxonomical aid.

Answer: (b) stratosphere

Answer: (d) All of the above.
Explanation:
Taxonomic studies depends on ecological information of organisms, structure of cell and development process of organism and external and internal structure of organisms.

Answer: (a) Angiospermae
Explanation:
Wheat belongs to Angiospermae division.

Answer: (c) 1, 4-dibromobutane
Explanation:
The derivative is 1, 4 dibromobutane. This on reaction with sodium metal gives cyclobutane.

This reaction is an example of internal Wurtz reaction.

Answer: (b) Ozonolysis
Explanation:
Ozonolysis is the cleavage of an alkene or alkyne with ozone to form organic compounds in which the multiple carbon-carbon bonds have been replaced by a double bond to oxygen. The outcome of the reaction depends on the type of multiple bonds being oxidized.

Bromine water can be also used to identify the position of a double bond. In this reaction, red-brown colour of bromine gets turned into coulorless indicating that there is a double bond.

Answer: (b) 1, 3-Dinitrobenzene
Explanation:
NO₂ is an m-directing group and hence, 1, 3-dinitrobenzene is formed.

Answer: (d) -NO₂
Explanation:
The correct order of deactivation is
−NO₂ > −C ≡ N > −SO₃₂ H > −COOH

Answer: (a) 1 – Phenyl – 2 – butene
Explanation:
C₆H₅CH₂−CH = CH−CH₃ (1-phenyl-2-butene) exhibits the phenomenon of geometrical isomerism due to C_ab = C_ad structure, so its two isomers are possible which are given are follow :

Answer: (d) (iv) > (ii) > (i) > (iii)
Explanation:
Benzene having any activating group i.e., OH, R etc., undergoes electrophilic substitution very easily as compared to benzene itself. Thus toluene (C₆H₅CH₃), phenol (C₆H₅OH) undergo electrophilic substitution very readily than benzene. Chlorine with +E and +M effect deactivates the ring due to strong -I effect. So, it is difficult to carry out the substitution in chlorobenzene than in benzene, so correct order is Phenol (iv) > Toluene (ii) > Benzene (i) > Chlorobenzene (iii)

Answer: (a) Soda lime decarboxylation
Explanation:
Methane cannot be prepared by either Wurtz reaction, Kolbes electrolytic method or by reduction of alkenes with H₂. While acetic acid salt on heating with soda lime gives methane.
CH₃COONa^– → CH₄ + Na₂CO₃

Answer: (a) C ≡ C
Explanation:
−C ≡ C−is most reactive because sp-hybridization.

Answer: (d) Ammonical silver nitrate
Explanation:
CH₃−C ≡ C−H + AgNO₃ → CH₃ −C ≡ C−Ag
Propyne Ammonical Siver salt of Propyne

Answer: (c) 2, 3 – Dimethylbutane
Explanation:
CHMe₂ − Br + 2Na + Br − CHMe₂ + Br − CHMe₂ \(\underrightarrow { dryEther } \) Me₂CH − CHMe₂ + 2NaBr

Answer: (b) 1, 3-Dinitrobenzene
Explanation:
NO₂ is an m-directing group and hence, 1, 3-dinitrobenzene is formed.

Answer: (d) 9°44
Explanation:
According to Baeyers strain theory, the amount of the strain is directly proportional to the angle through which a valency bond has deviated from its normal position . i.e., Amount of deviation) in cyclobutane (d) = (109∘28 − 90∘)/(2) = 9∘44

Answer: (c) Hydrocarbon Gases
Explanation:
During fractionation or distillation of petroleum, gases light hydrocarbons in the form of gases are obtained from first fraction.

Answer: (a) 2 – Phenyl – 2 – propanol
Explanation:
Acidic hydration of 2-phenyl propene follows electrophilic reaction mechanism forming an intermediate 3° carbocation (more stable), there by forming 2-phenyl-2-propanol.

Answer: (c) Middle Oil
Explanation:
The second fraction of coal-tar distillation which distills at 170°−230°C is called middle-oil or olic oil. It contains mainly phenol, naphthalene etc.

Answer: (a) There are three carbon-carbon single bonds and three carbon-carbon double bonds.
Explanation:
There are three carbon-carbon single bonds and three carbon-carbon double bonds.

Answer: (b) Anhydrous Aluminium Chloride
Explanation:
It is a catalyst-based electrophilic substitution reaction. Lewis acids like anhydrous AlCl₃, anhydrous FeCl₃, BF₃ and BiCl₃ can be used as catalysts. The catalyst facilitates more effective attack by the electrophile (R⁺, RCO⁺ etc.) on the benzene ring.

Answer: (a) Alkanes
Explanation:
Alkyl halides react with dialkyl copper reagents to give Alkanes.

Answer: (c) But – 2 – ene
Explanation:
The lowest alkene which is capable of exhibiting geometrical isomerism is 2-Butene

Answer: (d) Deactivates the ring towards electrophilic substitution
Explanation:
−NO₂ group shows −M effect, so withdraws the electron density from the ring and hence deactivate the ring towards electrophilic aromatic substitution.

Answer: (c) Absolute Entropy
Explanation:
The Third Law of Thermodynamics is concerned with the limiting behavior of systems as the temperature approaches absolute zero. Most thermodynamics calculations use only entropy differences, so the zero point of the entropy scale is often not important. However, the Third Law tells us about the completeness as it describes the condition of zero entropy.

Answer: (b) Hydrogen Gas
Explanation:
The measure of randomness of a substance is called entropy. Greater the randomness of molecules of a substance greater is the entropy. Here hydrogen gas has more entropy as it shows more randomness/disorderliness due to less molar mass than all the given substances and also in the gas phase.

Answer: (a) 84 J mol^-1K^-1
Explanation:
∆S = (q_rev)/ (T)
= (\(\frac {8400}{100}\))
= 84 J mol^-1K^-1

Answer: (b) Cl₂(g)
Explanation:
This is possible only for elements, chlorine is a gas at this temperature, but bromine is a liquid, so it is possible only for chlorine.

Answer: (c) 350 kJ
Explanation:
∆E = q + w
= (600 – 250)
∆E = 350 J

Answer: (a) ∆S = 0
Explanation:
Equilibrium
Therefore, ∆ G = 0 = ∆ H – T ∆ S
Or T∆ S = ∆H

Answer: (b) 1.11 × 10⁴ kcal
Explanation:
n = (Mass)/ (Molecular weight)
= (6.4 × 10³)/ (64)
= 100
For 1 mole of CaC₂
∆ H = ∆H_f (CaC) + H_f (CO) – H_f (CaO)
= -14.2 – 26.4 + 151.6 = 111.1 kcal
For 100 moles, ∆H = 1.11 × 10⁴ Kcal

Answer: (c) -47 kJ
Explanation:
w = -1.2 (32) × 101.3
= – 3.89 KJ
= -4 (approx.)
= ∆ E = – 51.0 KJ
Therefore, E = q + w
– 51 = q – 4
Therefore, q = – 47 KJ

Answer: (b) 10 kJ
Explanation:
At constant volume w = 0
Therefore, ∆ U = q = 10 kJ

Answer: (c) 5P₁V₁
Explanation:
Work done = Area under P-V graph = (\(\frac {1}{2}\)) (5P₁) (2V₁) = 5P₁ V₁

Answer: (c) Sublimation of Naphthalene
Explanation:
The order of entropy in solid, liquid and gas is gas > liquid > solid .Hence, in sublimation of naphthalene, maximum increase in entropy is observed.

Answer: (c) 83-85
Explanation:
C–C bond 83–85 kcal/mol

It is the energy required to break the bond .It is defined as the standard enthalpy change when a bond is cleaved by homolysis, with reactants and products of the homolysis reaction at 0 K (absolute zero)

Answer: (d) Gibbs Free Energy
Explanation:
Gibbs free energy combines the effect of both enthalpy and entropy. The change in free energy (ΔG) is equal to the sum of the change of enthalpy (∆H) minus the product of the temperature and the change of entropy (∆S) of the system.
∆G = ∆H – T∆S
ΔG predicts the direction in which a chemical reaction will go under two conditions: (1) constant temperature and (2) constant pressure.
If ΔG is positive, then the reaction is not spontaneous (it requires the input of external energy to occur) and if it is negative, then it is spontaneous (occurs without the input of any external energy).

Answer: (c) -110.5 kJ
Explanation:
C(s) + O₂ à CO₂ ∆H₁ = -393.5
CO + (\(\frac {1}{2}\)) O₂ à CO₂ ∆H₂ = -283.0
C(s) + (\(\frac {1}{2}\)) O₂ à CO ∆H = ∆H₁ – ∆H₂
= -393.5 + 283 = -110.5 KJ

Answer: (a) -100 × kJ/mol
Explanation:
Normality of NaOH = Molarity × acidity
= 0.5 × 1 = 0.5 N
Total heat q produced = x kJ
Heat of neutralisation
= [(q)/ (Volume of acid or base)] ×1000× (1/normality of acid or base)
= (\(\frac {x}{20}\)) × 1000 × (\(\frac {1}{0.5}\))
= 100 x kJmol^-1
Since heat is liberated, heat of neutralisation = −100 x kJmol^-1

Answer: (d) For a cyclic process, q = -w
Explanation:
(1) ΔU = q + w. For an isochoric process, w = −PΔV = 0. Hence, ΔU = qv
(2) For an adiabatic process, q = 0. Hence, ΔU = w
(3 ) For an isothermal process, ΔU = 0 Hence, q = −w
(4) For a cyclic process , ΔU = 0. Hence, q = −w

Answer: (b) 10 kJ
Explanation:
At constant volume w = 0
Therefore, ∆ U = q = 10 kJ

Answer: (d) Adiabatic Expansion
Explanation:
In adiabatic process heat is neither added nor removed from system. So the work done by the system (expansion) in adiabatic process will result in decrease of internal energy of that system (from first law).
As internal energy is directly proportional to the change in temperature there will be temperature drop in an adiabatic process.

Answer: (a) KNO₃ (∆H = 35.4 kJ mol^-1)
Explanation:
More the heat absorbed, more will be the cooling effect. Hence, more the positive value of ∆ H, more the cooling effect.

Answer: (c) 37.56
Explanation:
For gaseous reactants and products, we have a relation between standard enthalpy of vaporization (ΔH_vap) and standard internal energy (ΔE) as-
ΔH_vap = ΔE+ Δn_g​ RT whereas,
Δn_g = n₂​ − n₁, i.e., difference between no. of moles of reactant and product.
For vaporization of water,
H₂O (l) → H₂O(g)
​Therefore, Δn_g = 1 − 0 = 1
Therefore, ΔH_vap = ΔE+ Δn_g RT
⇒ ΔE = ΔH_vap − Δn_g
​RT = 40.63 − (1×8.314×10^-3 × 373) = 37.53 KJ/mol
Hence the value ΔE for this process will be 37.53KJ/mol.

Answer: (c) 0.4 atm
Explanation:
Vapour pressure depends on T only and it does not depend on container volume.

Answer: (c) Alcohol, Petrol, Water
Explanation:
The vapour pressure of the liquid is inversely proportional to the magnitude of the intermolecular forces of attraction present. Based on this, the liquid with higher vapour pressure in the different pairs is: (a) Alcohol, (b) Petrol, (c) Water.

Answer: (a) 64.0
Explanation:
Let rate of diffusion of gas x, r₁ = a
Therefore, rate of diffusion of methane, r₂ = 2 a
According to Grahams Law of Diffusion
(\(\frac {r_1}{r_2}\)) = (\(\sqrt{\frac {M_2}{M_1}}\))
M₁ = Molecular mass of gas x
M₂ = Molecular mass of Methane = 16 g
Therefore, (\(\frac {a}{2a}\)) = (\(\sqrt{\frac {16}{M_2}}\))
Squaring both the sides, (\(\frac {1}{4}\)) = (\(\frac {16}{M_2}\))
or, M₂ = 16 × 4 = 64 g

Answer: (c) Gas
Explanation:
Of the three states of matter, the gaseous state is the simplest and shows greatest uniformity in behaviour. Gases show almost similar behaviour irrespective of their chemical nature. This state is characterized by:
Gases maintain neither the volume nor the shape. They completely fill the container in which they are placed.

They expand appreciably on heating. Gases are highly compressible. The volume of the gas decreases when the pressure increases. They diffuse rapidly into space. Gases exert equal pressure in all directions.
All gases are colourless except a few e.g. chlorine (greenish yellow) bromine (reddish brown), nitrogen dioxide (reddish brown)

The behaviour of gases can be described by certain quantitative relationships called gas laws. They give the relationship between mass, pressure, volume and temperature.

Answer: (a) Attract one another
Explanation:
The basic concept of the kinetic-molecular theory give us the information why real gases deviate from ideal behavior. The molecules of an ideal gas are assumed to occupy no space and have no attractions for one another. Real molecules, however, do have finite volumes, and they do attract one another. So, a gas deviates from ideal behavior at a high pressure because its molecules attract one another.

Answer: (d) Units of Volume,Temperature and Pressure
Explanation:
The value of the gas constant R depends on the units used for pressure, volume and temperature.

Answer: (d) D
Explanation:
V_c = 3b = 3 × 4N × (4/3) πr³

Answer: (d) Mass of the gas cannot be determined by weighing a container in which it is contained
Explanation:
Mass of the gas = mass of the cylinder including gas – mass of empty cylinder. So mass of a gas can be determined by weighing the container in which it is enclosed. Thus, the statement (d) is wrong for gases.

Answer: (c) Volume occupied by the molecules
Explanation:
In van der Waals equation of state of the gas law, the constant b is a measure of the volume occupied by the molecules. It gives the effective size of the gas molecules. The greater value of b indicates a larger size of the molecules and smaller compressible volume.

Answer: (a) 30 L
Explanation:
According to the ideal gas equation, we have
PV = nRT
​PV = (\(\frac {w}{M}\)) RT
​V = ​ (\(\frac {w}{M}\)) (\(\frac {RT}{P}\))
Given values are:
w = 2.8 g
M = Molar mass of CO = 28 g mol^-1
T = 27°C = (273 + 27) = 300 K
P = 0.821 atm
R = 0.0821 L atm mol^-1 K^-1
Putting the values in the formula we get :
V = (2.8 g /28 g mol^-1) × (0.0821 L atm mol^-1 K^-1) × (300 K)/(0.821 atm)
= 3 L

Answer: (a) 20 × 150
Explanation:
At constant T, P₁V₁ = P₂V₂
1 × 20 = P₂ × 50;
P₂ = (\(\frac {20}{ 50}\)) × 1

Answer: (b) 11

Answer: (b) Liquid
Explanation:
In a liquid, particles will flow or glide over one another, but stay toward the bottom of the container. The attractive forces between particles are strong enough to hold a specific volume but not strong enough to keep the molecules sliding over each other.

Answer: (b) Capillary Action
Explanation:
Capillarity, rise or depression of a liquid in a small passage such as a tube of small cross-sectional area, like the spaces between the fibres of a towel or the openings in a porous material. Capillarity is not limited to the vertical direction. Water is drawn into the fibres of a towel, no matter how the towel is oriented.

Liquids that rise in small-bore tubes inserted into the liquid are said to wet the tube, whereas liquids that are depressed within thin tubes below the surface of the surrounding liquid do not wet the tube. Water is a liquid that wets glass capillary tubes; mercury is one that does not. When wetting does not occur, capillarity does not occur.

Capillarity is the result of surface, or interfacial, forces. The rise of water in a thin tube inserted in water is caused by forces of attraction between the molecules of water and the glass walls and among the molecules of water themselves. These attractive forces just balance the force of gravity of the column of water that has risen to a characteristic height. The narrower the bore of the capillary tube, the higher the water rises. Mercury, conversely, is depressed to a greater degree, the narrower the bore.

Answer: (d) Kinetic Molecular Theory
Explanation:
The kinetic molecular theory (KMT) is a simple microscopic model that effectively explains the gas laws described in previous modules of this chapter. This theory is based on the following five postulates described here. (Note: The term “molecule” will be used to refer to the individual chemical species that compose the gas, although some gases are composed of atomic species, for example, the noble gases.)

Gases are composed of molecules that are in continuous motion, travelling in straight lines and changing direction only when they collide with other molecules or with the walls of a container.
The molecules composing the gas are negligibly small compared to the distances between them.
The pressure exerted by a gas in a container results from collisions between the gas molecules and the container walls.

Gas molecules exert no attractive or repulsive forces on each other or the container walls; therefore, their collisions are elastic (do not involve a loss of energy).
The average kinetic energy of the gas molecules is proportional to the kelvin temperature of the gas.

Answer: (b) Surface Tension
Explanation:
Raindrops take up the spherical shape due to the surface tension of water which is caused due to the tendency of water molecules to stick together. The spherical shape is having the least possible surface area due to which it can resist any of the external force in the atmosphere.

Answer: (b) Grahams Law
Explanation:
Grahams law states that the rate of diffusion of a gas is inversely proportional to the square root of its molecular weight. In the same conditions of temperature and pressure, the molar mass is proportional to the mass density. Therefore the rate of diffusion of different gases is inversely proportional to the square root of their mass densities.
r α (\(\sqrt{\frac {1}{d}}\))
and r α (\(\sqrt{\frac {1}{M}}\))

Answer: (a) Boyles law
Explanation:
In 1662 Robert Boyle studied the relationship between volume and pressure of a gas of fixed amount at constant temperature. He observed that volume of a given mass of a gas is inversely proportional to its pressure at a constant temperature. Boyles law, published in 1662, states that, at constant temperature, the product of the pressure and volume of a given mass of an ideal gas in a closed system is always constant. It can be verified experimentally using a pressure gauge and a variable volume container. It can also be derived from the kinetic theory of gases: if a container, with a fixed number of molecules inside, is reduced in volume, more molecules will strike a given area of the sides of the container per unit time, causing a greater pressure.
A statement of Boyles law is as follows:
The volume of a given mass of a gas is inversely related to pressure when the temperature is constant.​
V ∝ (\(\frac {1}{P}\)) meaning “Volume is inversely proportional to Pressure”, or
P ∝ (\(\frac {1}{V}\)) meaning “Pressure is inversely proportional to Volume”, or
where P is the pressure, and V is the volume of a gas, and k₁ is the constant in this equation.

Answer: (a) 2 : 1
Explanation:
According to Grahams law
(\(\frac {r_1}{r_2}\)) = (\(\sqrt{\frac {M_1}{M_2}}\))
(r_He/r_CH4) = (\(\sqrt{\frac {16}{4}}\))
= (\(\frac {1}{2}\))

Answer: (a) The particles move faster.
Explanation:
There is a great deal of empty space between particles, which have a lot of kinetic energy. The particles move very fast and collide into one another when the gas is heated up, causing them to diffuse, or spread out, until they are evenly distributed throughout the volume of the container.

Answer: (a) Strongly basic
Explanation:
Caesium hydroxide or cesium hydroxide (CsOH) is a chemical compound consisting of caesium ions and hydroxide ions. It is a strong base (p_kb = -1.76), much like the other alkali metal hydroxides such as sodium hydroxide and potassium hydroxide.

Answer: (b) (Na₂CO₃. 10H₂O)
Explanation:
The Solvay process or ammonia soda process is used for the manufacture of sodium carbonate
(Na₂CO₃. 10H₂O).

Answer: (a) Deep blue solution
Explanation:
When an alkali metal is dissolved in liquid ammonia, it results in the formation of a deep blue coloured solution. The ammoniated electrons absorb energy corresponding to a red region of visible light. Therefore, the transmitted light is blue in colour.

Answer: (a) Acid containing Oxygen
Explanation:
An oxyacid, oxoacid, or ternary acid is an acid that contains oxygen. Specifically, it is a compound that contains hydrogen, oxygen, and at least one other element, with at least one hydrogen atom bond to oxygen that can dissociate to produce the H⁺ cation and the anion of the acid. e.g., carbonic acid, H₂CO₃ (OC(OH)₂; sulphuric acid, H₂SO₄ (O₂S(OH)₂).

Answer: (c) Developing interlocking needle like crystals of hydrated silicates
Explanation:
Water develops interlocking needle ­like crystals of hydrated silicates. The reactions involved are the hydration of calcium aluminates and calcium silicates which change into their colloidal gels.

At the same time, some calcium hydroxide and aluminium hydroxides are formed as precipitates due to hydrolysis. Calcium hydroxide binds the particles of calcium silicate together while aluminium hydroxide fills the interstices rendering the mass impervious

Answer: (c) Calcined gypsum
Explanation:
The composition of gypsum is CaSO₄ ∙2H₂O. It does not have CaCO₃

Answer: (c) Mg
Explanation:
Carnallite is an evaporite mineral a hydrated potassium magnesium chloride. It is variably coloured yellow to white, reddish or blue. It occurs with a sequence of potassium and magnesium evaporite. It is an uncommon double chloride mineral that forms under specific conditions. It is an important source of potash.

Answer: (a) Mg
Explanation:
Magnesium metal is used for the preparation of the wire of flash bulb.

Answer: (d) KOH.
Explanation:
The basic strength increases down the group and decreases along a period.

Answer: (b) MgCl₂ × 5MgO × (xH₂O)
Explanation:
Mixture of MgCl₂ and MgO is called Sorels cement. It is MgCl₂ × 5MgO × (xH₂O)

Answer: (d) Cs.
Explanation:
Atomic size increases as we move down the alkali group. As a result, the binding energies of their atoms in the crystal lattice decrease. Also, the strength of metallic bonds decreases on moving down a group in the periodic table. This causes a decrease in the melting point. Among the given metals, Cs is the largest and has the least melting point.

Answer: (d) KOH.
Explanation:
The basic strength increases down the group and decreases along a period.

Answer: (b) Na⁺
Explanation:
Hydration energy depends on charge of ion and ionic radius. Higher the charge, greater the hydration energy. On the other hand, smaller the size, greater the hydration energy. Charge is considered first for comparison. Hence, Mg²⁺ has higher hydration energy than Na⁺.

Answer: (a) Ca(OCI)₂
Explanation:
Milk of lime reacts with chlorine to form bleaching powder.
2Ca(OH)₂ + 2Cl₂ → CaCl₂ + Ca(OCl)₂ + 2H₂O

Answer: (d) Both (2) and (3) are correct
Explanation:
Calcium carbonate is strongly heated until it undergoes thermal decomposition to form calcium oxide and carbon dioxide. The calcium oxide (unslaked lime) is dissolved in water to form calcium hydroxide (limewater).
CaCO₃ → CaO + CO₂

Answer: (c) Ca₃(PO₄)₂
Explanation:
Ca₃(PO₄)₂. Both Ca and P are needed by human beings. Also they prevent moisture absorbing power of other components such as MgCl₂, CaCl₂, CaSO₄ and MgSO₄ present in commercial sodium chloride.

Answer: (a) Strongly basic
Explanation:
Caesium hydroxide or cesium hydroxide (CsOH) is a chemical compound consisting of caesium ions and hydroxide ions. It is a strong base (p_kb = -1.76), much like the other alkali metal hydroxides such as sodium hydroxide and potassium hydroxide.

Answer: (d) Fe
Explanation:
Method of extraction of a metal depends on the reactivity of the metal. Iron (Fe) is not manufacture by electrolysis. Moderately reactive metals like zinc and iron are extracted by reduction of their oxides using carbon.

Answer: (a) Mg
Explanation:
Magnesium metal is used for the preparation of the wire of flash bulb.

Answer: (d) eutectic mixtue of CaCl/H₂O freezes at -55°C while that of NaCl/H₂O freezes at -18°C
Explanation:
Eutectic mixture of CaCl₂/H₂O freezes at -55°C while that of NaCl/H₂O freezes at -18°C.

When added to ice. NaCl and CaCl₂ ionise into respective ions (Na⁺, Ca²⁺ & Cl^–) this leads to increase in the number of particles (ions). As CaCl₂ has greater number of ions, Increasing the number of particles /ions increases the vant hoff factor. Vant hoff factor is directly proportional to the depression in freezing point thus lowers the freezing point of ice causing preventing the formation of ice.

Answer: (d) n, α, p, e
Explanation:
(i) (e/m) for (i) neutron = (\(\frac{0}{1}\)) = 0
(ii) α− particle = (\(\frac{2}{4}\)) = 0.5
(iii) Proton = (\(\frac{1}{1}\)) = 1
(iv) electron = (\(\frac{1}{1837}\)) = 1837.

Answer: (d) 9.84 × 10⁵ J mol^-1
Explanation:
Energy required when an electron makes transition from n = 1 to n = 2
E₂=−(1.312 × 10⁶ × (1)²)/(2²)
= −3.28 × 10⁵ J mol^-1
E₁ = −1.312 × 10⁶ J mol^-1
ΔE = E₂ − E₁
=−3.28 × 10⁵−(−13.2 × 10⁶)
ΔE = 9.84×10⁵ J mol^-1

Answer: (a) s < p < d < f
Explanation:
Order of energy is:
s < p < d < f

Answer: (d) 743 nm
Explanation:
From Law of Conservation of energy, energy of absorbed photon must be equal to combined energy of two emitted photons.
ET = E₁ + E₂ ….. (1)
Where E₁ is Energy of first emitted photon emitted and E₂is Energy of second emitted photon.
Energy E and wavelength λ of a photon are related by the equation
E= (hc)/ (λ)….. (2)
Where Plancks constant = h, c is velocity of light.
Substituting the values from (2) in (1) we get
(hc/λ_T) = (hc)/ (λ₁) + (hc)/ (λ₂)
Or (\(\frac{1}{λ_T}\)) = (\(\frac{1}{λ_1}\)) + (\(\frac{1}{λ_2}\)) …… (3)
Substituting given values in (3) we get
(\(\frac{1}{355}\)) = (\(\frac{1}{680}\)) + (\(\frac{1}{λ_2}\))
Or \(\frac{(1)}{(λ_T)}\) = (\(\frac{1}{355}\)) − (\(\frac{1}{680}\))
⇒ (\(\frac{1}{λ_2}\)) = (680 − 355)/ (355 × 680)
⇒ λ₂ = 742.77nm

Answer: (d) 3s, 3p and 3d orbitals all have the same energy
Explanation:
A hydrogen atom has 1st configuration and these its, 3p and 3d orbitals will have same energy wrt 1s orbital.

Answer: (c) Orientation of orbitals
Explanation:
The magnetic quantum number specifies orientation of orbitals.

Answer: (c) [Kr]4d¹⁰5s¹
Explanation:
The electronic configuration of Ag in ground state is [Kr]4d¹⁰5s¹

Answer: (a) K
Explanation:
K = 19 = 1s²2s²2p⁶3s²3p⁶s¹
3s²3p⁶ = m-shell
= k has only 8 electrons in M shell

Answer: (a) K⁺, Ca²⁺, Sc³⁺, Cl^–
Explanation:
Isoelectronic species are those which have same number of electrons.
K⁺ = 19 – 1 = 18; Ca²⁺ = 20 – 2 = 18; Sc³⁺ = 21 – 3 = 18; Cl^– = 17 + 1 = 18
Thus all these ions have 18 electrons in them.

Answer: (b) Chromium
Explanation:
M shell means it is third shell ⇒ n = 3
Number of electrons in M shell = 13
⇒ 3s²3p⁶3d⁵
The electronic configuration is: (1s²) (2s² 2p⁶) (3s² 3p⁶ 3d⁵) (4s¹)
The element is chromium is Cr.

Answer: (c) Spin quantum number
Explanation:
Electrons occupying the same orbital are distinguished by Spin quantum number.
For spin Quantum number it has two values +1/2 or -1/2,
Hence the value of n, l , m are same for the two electrons occupying in the same orbitals, but only the is different, which is
Therefore, Spin quantum number explains the direction through which the electron spins in an orbital. so obviously there are only 2 possible directions. Which is either clockwise or anticlockwise.
So the electron which are available in the same orbitals, must have opposite spins. Hence spin quantum number distinguished b/w the two electrons.

Answer: (b) 12 and 5
Explanation:
₂₄Cr → 1s² 2s²2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
As we know for p, l = 1 and d, l = 2
For l = 1, total number of electrons = 12 [2p⁶ and 3p⁶]
For l = 2, total number of electrons = 5 [3d⁵]

Answer: (b) 6.63 × 10^-31 m
Explanation:
m = 10 mg
= 10 × 10^-6 kg
v = 100 ms^-1
λ = \(\frac {(h)}{(mv)}\)
= (6.63×10^-34)/ (10 × 10^-6 × 100)
= 6.63 × 10^-31 m

Answer: (d) 9.84 × 10⁵ J mol^-1
Explanation:
Energy required when an electron makes transition from n = 1 to n = 2
E₂ = −(1.312 × 10⁶ × (1)²)/(2²)
= −3.28 × 10⁵ J mol^-1
E₁ = −1.312 × 10⁶ J mol^-1
ΔE = E₂ − E₁
= −3.28 × 10⁵−(−13.2 × 10⁶)
ΔE = 9.84 × 10⁵ J mol^-1

Answer: (a) 3.4 eV
Explanation:
For hydrogen atom,
The kinetic energy is equal to the negative of the total energy.
And the potential energy is equal to the twice of the total energy.
The first excited state energy of orbital = -3.4 eV
and The kinetic energy of same orbital = -(-3.4 eV) = 3.4 eV
Therefore, the kinetic energy of same orbit of hydrogen atom is 3.4 eV.

Answer: (c) n = 3, l = 2, m = 1, s = + \(\frac {1}{2}\)
Explanation:
n = 3, l = 0 represents 3s orbital n = 3, l = 1 represents 3p orbital n = 3, l = 2 represents 3d orbital n = 4, l = 0 represents 4s orbital The order of increasing energy of the orbitals is 3s < 3p < 4s < 3d.

Answer: (c) -1
Explanation:
As we know, in Bohr model KE of an electron in an orbit = + (\(\frac {1}{2}\)) (\(\frac {e^2}{r_n}\))
Total energy of electron in an orbit = -(\(\frac {e^2}{2r_n}\))
Therefore, \(\frac {(KE)}{(TE)}\) = -1

Answer: (d) The position and velocity of the electrons in the orbit cannot be determined simultaneously
Explanation:
The position and velocity of electrons cannot be determined simultaneously does not fit in with the Bohrs model of H atom. It is a part of Heisenbergs uncertainty principle

Answer: (d) 3s, 3p and 3d orbitals all have the same energy
Explanation:
A hydrogen atom has 1s1 configuration and these its, 3p and 3d orbitals will have same energy wrt 1s orbital.

Answer: (c) 10 electrons
Explanation:
n = 6, ℓ = 2 means 6d → will have 5 orbitals.
Therefore max 10 electrons can be accommodated as each orbital can have maximum of 2 electrons.

Answer: (b) 143/11760
Hint:
Total number of cards = 52
Two cards are lost.
So remaining cards = 50
Now one card is drawn.
Probability that it is a diamond card = 13/50
Now probability that both lost cards are heart = 13/50 ×(¹¹C₂ / ⁴⁹C₂)
= 13/50 ×[{(11×10)/2}/{(49×48/2)}]
= 13/50 ×{(11×10)/(49×48)}
= {(13×11×10)/(50×49×48)}
= {(13×11)/(5×49×48)}
= 143/11760
So probability that both lost card are heart = 143/11760

Answer: (c) 16/625
Hint:
The last digit of the four whole number can be
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
The chance that any of the four numbers is divisible by 2 or 5 = 6/10 = 3/5
Hence, the chance that any of the four numbers is not divisible by 2 or 5 = 1 – 3/5 = 2/5
So, the chance that all of the four numbers are divisible by 2 or 5 = (2/5)×(2/5)×(2/5)×(2/5)
= 16/625
This is the chance that the last digit in the product will not be 0, 2, 4, 5, 6, 8 and this is also the chance that the last digit in the product is 1, 3, 7 or 9

Answer: (b) 1/36
Hint:
Total number of cases = 6³ = 216
The same number can appear on each of the dice in the following ways:
(1, 1, 1), (2, 2, 2), ………….(3, 3, 3)
So, favourable number of cases = 6
Hence, required probability = 6/216 = 1/36

Answer: (b) 1/6
Hint:
First, we choose 1 machine out of given 4.
The probability that it is fault = 2/4 = 1/2
Now, we have to pick the second fault machine.
The probability that it is fault = 1/3
So, required probability = (1/2)×(1/3) = 1/6

Answer: (d) 7/9
Hint:
When two dice are throw, then Total outcome = 36
A doublet: {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
Favourable outcome = 6
Sum is 10: {(4, 6), (5, 5), (6, 4)}
Favourable outcome = 3
Again, A doublet and sum is 10: (5, 5)
Favourable outcome = 1
Now, P(either dublet or a sum of 10 appears) = P(A dublet appear) + P(sum is 10) – P(A dublet appear and sum is 10)
⇒ P(either dublet or a sum of 10 appears) = 6/36 + 3/36 – 1/36
= (6 + 3 – 1)/36
= 8/36
= 2/9
So, P(neither dublet nor a sum of 10 appears) = 1 – 2/9 = 7/9

Answer: (d) 21
Hint:
When two dice are thrown, then total outcome = 6×6 = 36
A: Getting an odd number on the first die.
A = {(1, 1), (1, 2), (1, 3), (1, 4),(1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4),(3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4),(5, 5), (5, 6)}
Total outcome = 18
B: Getting a total of 7 on the two dice.
B = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
Total outcome = 6
C: Getting a total of greater than or equal to 8 on the two dice.
C = {(2, 6), (3, 5), (3, 6), (4, 4),(4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6),(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Total outcome = 15
Now n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
⇒ n(A ∪ B) = 18 + 6 – 3
⇒ n(A ∪ B) = 21

Answer: (a) 4/5
Hint:
Total number of ways of choosing two numbers out of six = ⁶C₂ = (6×5)/2 = 3×5 = 15
If smaller number is chosen as 3 then greater has choice are 4, 5, 6
So, total choices = 3
If smaller number is chosen as 2 then greater has choice are 3, 4, 5, 6
So, total choices = 4
If smaller number is chosen as 1 then greater has choice are 2, 3, 4, 5, 6
So, total choices = 5
Total favourable case = 3 + 4 + 5 = 12
Now, required probability = 12/15 = 4/5

Answer: (c) 9/1547
Hint:
Total number of cards = 52
Number of king card = 4
Now, 7 cards are drawn from 52 cards.
P (3 cards are king) = {⁴C₃ × ⁴⁸C₄}/⁵²C₇
= {4×(48×47×46×45)/(4×3×2×1)}/{(52×51×50×49×48×47×46)/(7×6×5×4×3×2×1)}
= {4×(48×47×46×45)×(7×6×5×4×3×2×1)}/{(4×3×2×1)×{(52×51×50×49×48×47×46)}
= (7×6×5×4×45)/(52×51×50×49)
= (6×5×4×45)/(52×51×50×7)
= (6×4×45)/(7×52×51×10)
= (6×45)/(7×13×51×10)
= (6×3)/(7×13×17×2)
= (3×3)/(7×13×17)
= 9/1547

Answer: (c) e^-1/2/8
Hint:
This question is based on Poisson distribution.
Now, λ = np = 500×(1/1000) = 500/1000 = 1/2
Now, P(x = 2) = {e^-1/2 × (1/2)²}/2! = e^-1/2/(4×2) = e^-1/2/8

Answer: (d) 0.0545
Hint:
Given word: INSTITUTION
Total letters = 11
The word contains 3 I, 2 N, 1 S, 3 T, 1 U and 1 O
Total number of arrangement = 11!/(3!×2!×3!) = 554400
Now, taken 3 T are together.
So total latter = 9
The number of favorable cases = 9!/(3!×2!) = 30240
Now, P(3 T are together) = 30240/554400 = 0.0545

Answer: (b) 1/9
Hint:
One person can select one house out of 3 = ³C₁ = 3
So, three persons can select one house out of three = 3×3×3 = 27
Thus, probability that all the three can apply for the same house = 3/27 = 1/9

Answer: (d) 4/9
Hint:
Total number of shocks = 5 + 4 = 9
Two shocks are pulled.
Now, P(Both are same color) = (⁵C₂ + ⁴C₂)/⁹C₂
= {(5×4)/(2×1) + (4×3)/(2×1)}/{(9×8)/(2×1)}
= {(5×4) + (4×3)/}/{(9×8)
= (5 + 3)/(9×2)
= 8/18
= 4/9

Answer: (c) 247/256
Hint:
Let x be number a discrete random variable which denotes the number of heads obtained in n (in this question n = 8)
The general form for probability of random variable x is
P(X = x) = ⁿC_x × p^x × q^n-x
Now, in the question, we want at least two heads
Now, p = q = 1/2
So, P(X ≥ 2) = ⁸C₂ × (1/2)² × (1/2)^8-2
⇒ P(X ≥ 2) = ⁸C₂ × (1/2)² × (1/2)⁶
⇒ 1 – P(X < 2) = ⁸C₀ × (1/2)⁰ × (1/2)⁸ + ⁸C₁ × (1/2)¹ × (1/2)^8-1
⇒ 1 – P(X < 2) = (1/2)⁸ + 8 × (1/2)¹ × (1/2)⁷
⇒ 1 – P(X < 2) = 1/256 + 8 × (1/2)⁸
⇒ 1 – P(X < 2) = 1/256 + 8/256
⇒ 1 – P(X < 2) = 9/256
⇒ P(X < 2) = 1 – 9/256
⇒ P(X < 2) = (256 – 9)/256
⇒ P(X < 2) = 247/256

Answer: (c) 1/2
Hint:
Given, a couple has two children.
Let A denotes both children are females i.e. {FF}
Now, P(A) = (1/2)×(1/2) = 1/4
and B denotes elder children is a female i.e. {FF, FM}
P(B) = 1/4 + 1/4 = 1/2
Now, P(A ∩ B) = 1/4
Now, P(Both the children are female if elder child is female)
P(A/B) = P(A ∩ B)/P(B)
⇒ P(A/B) = (1/4)/(1/2)
⇒ P(A/B) = 1/2

Answer: (c) e^-1/2/8
Hint:
This question is based on Poisson distribution.
Now, λ = np = 500×(1/1000) = 500/1000 = 1/2
Now, P(x = 2) = {e^-1/2 × (1/2)²}/2! = e^-1/2 /(4×2) = e^-1/2/8

Answer: (a) 1 – 3/e²
Hint:
Here m = 2
Now, P(X > 1.5) = ∑_r {(e^-2 × 2^r)/r!} {2 ≤ r ≤ ∞}
= e^-2 {2²/2! + 2³/3! + 2⁴/4! + …}
= e^-2 {(1 + 2 /1! + 2²/2! + 2³/3! + …) – 1 – 2}
= e^-2 (e² – 3)
= 1 – 3e^-2
= 1 – 3/e²

Answer: (d) 0.9
Hint:
Given, A and B are two mutually exclusive events.
So, P(A ∩ B) = 0
Again given P(A) = 0.5 and P(B ̅) = 0.6
P(B) = 1 – P(B ̅) = 1 – 0.6 = 0.4
Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ P(A ∪ B) = P(A) + P(B)
⇒ P(A ∪ B) = 0.5 + 0.4 = 0.9

Answer: (b) 2/7
Hint:
In a leap year, the total number of days = 366 days.
In 366 days, there are 52 weeks and 2 days.
Now two days may be
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday
Now there are total 7 possibilities, So total outcomes = 7
In 7 possibilities, Sunday came two times.
So, favorable case = 2
Hence, the probabilities of getting 53 Sundays in a leap year = 2/7

Answer: (a) 1 – (5/6)⁶
Hint:
Let A is the event that 6 does not occur at all.
Now, the probability of at least one 6 occur = 1 – P(A)
= 1 – (5/6)⁶

Answer: (d) 1/12
Hint:
Total cities are 4 i.e. A, B, C, D
Given, Rahul visit four cities, So, n(S) = 4! = 24
Now, sample space IS:
S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BDAC, BDCA, BCAD, BCDA, CABD, CADB, CBDA, CDAD, CDAB,CDBA, DABC, DACB, DBCA, DBAC, DCAB, DCBA}
Let G = Rahul visits A firsta and B last
⇒ G = {ACDB, ADCB}
⇒ n(G) = 2
So, P(G) = n(G)/n(S) = 2/24 = 1/12

Answer: (c) One

Answer: (c) 80 cal/cm²s

Answer: (a) 30°C

Answer: (b) Radiation

Answer: (b) t1 < t2 < t3

Answer: (c) Good-emitter

Answer: (d) 4 : 3

Answer: (b) More heat and more cold

Answer: (a) 50 sec

Answer: (d) Hollow enclosure blackened inside and having a small hole

Answer: (d) 16

Answer: (b) Convection

Answer: (b) W/ m² K²

Answer: (a) (T14 – T24)

Answer: (a) 5801 K

Answer: (c) 4p

Answer: (a) 1.36 times

Answer: (a) 1

Answer: (c) Good-emitter

Answer: (a) a + r + 1 = 1

Answer: (b) 2 s

Answer: (a) 3 km/h

Answer: (d) 6 s

Answer: (b) zero

Answer: (a) 1 m/s²

Answer: (c) m/s

Answer: (a) 4 s

Answer: (b) LT^-1

Answer: (b) [LT^-1]

Answer: (b) 5 s

Answer: (b) zero

Answer: (a) 9 m

Answer: (b) 6 m/s

Answer: (b) 5 s

Answer: (c) 2 m/s²

Answer: (a) ∏/2

Answer: (b) [LT?2]

Answer: (c) 3 m/s²

Answer: (c) m/s

Answer: (b) 2 s