datavorous/entrance-exam-dataset huggingface dataset:
Carbon monoxide is carried around a closed cyclic process abc, in which bc is an isothermal process, as shown in the diagram. The gas absorbs 7000 J of heat as its temperature is increased from 300 K to 1000 K in going from a to b. The quantity of heat ejected by the gas during the process ca is

Options:
A 4200 J
B 500 J
C 9000 J
D 9800 J

The correct answer is:
9800 J

Solution:
Δ Q ab = + 7 0 0 0 = μ C v 1 0 0 0 - 3 0 0 ...(i)
For the process ca:
T a = 3 0 0 K , T c = T b = 1 0 0 0 K
Δ Q ca = μ C p 3 0 0 - 1 0 0 0 = - μ C p × 7 0 0
= - μ C v + R 7 0 0 ...(ii)
For carbon monoxide:
T a = 3 0 0 K , T c = T h = 1 0 0 0 K
Δ Q ca = μ C p 3 0 0 - 1 0 0 0 = - μ C p × 7 0 0
= - μ C v + R 7 0 0
For carbon monoxide:
γ = 7 5
C v = R γ - 1 = R 7 5 - 1 = 5 R 2 ...(iii)
Hence, from Eq. (i)
7 0 0 0 = μ 5 R 2 × 7 0 0 or μ R = 2 0 5 = 4
Δ Q ca = - 7 0 0 0 + 4 × 7 0 0 = - 9 8 0 0 J
Negative sign shows that heat is ejected.

Topic: Physics, Thermodynamics, JEE Main
Carbon, silicon and germanium atoms have four valence electrons each. Their valence and conduction bands are separated energy band gaps represented $\left(E_g\right)_C,\left(E_g\right)_{S i}$, and $\left(E_g\right)_{G e}$ respectively. Which one of the following relationship is true in their case?

Options:
A $\left.\left(\mathrm{E}_g\right)_{\mathrm{C}}>\mathrm{E}_g\right)_{\mathrm{Si}}$
B $\left.\left(\mathrm{E}_g\right)_{\mathrm{C}} < \mathrm{E}_g\right)_{\mathrm{Si}}$
C $\left.\left(\mathrm{E}_g\right)_{\mathrm{C}}=\mathrm{E}_g\right)_{\mathrm{Si}}$
D $\left(\mathrm{E}_g\right)_{\mathrm{C}}>\left(\mathrm{E}_g\right)_{\mathrm{Ge}}$

The correct answer is:
$\left.\left(\mathrm{E}_g\right)_{\mathrm{C}}>\mathrm{E}_g\right)_{\mathrm{Si}}$

Solution:
Band gap of carbon is $5.5 \mathrm{eV}$ while that of silicon is $1.1 \mathrm{eV}$
$$
\left(E_g\right)_C>\left(E_B\right)_S
$$

Topic: Physics, Semiconductors, JEE Main
Carbon, silicon and germanium have four valence electrons each. At room temperature which one of the following statements is most appropriate?

Options:
A The number of free electrons for conduction is significant only in Si and Ge but small in C
B The number of free conduction electrons is significant in $\mathrm{C}$ but small in Si and $\mathrm{Ge}$
C The number of free conduction electrons is negligibly small in all the three
D The number of free electrons for conduction is significant in all three

The correct answer is:
The number of free electrons for conduction is significant only in Si and Ge but small in C

Solution:
Carbon (C), silicon (Si) and germanium (Ge) have the same lattice structure and their valence electrons are 4. For $\mathrm{C}$, these electrons are in the second orbit, for $\mathrm{Si}$ it is third and for germanium, it is the fourth orbit. In solid-state, the higher the orbit, the greater the possibility of overlapping of energy bands. Ionization energies are also less therefore Ge has more conductivity compared to Si. Both are semiconductors. Carbon is an insulator.

Topic: Physics, Semiconductors, JEE Main
Carbon, silicon and germanium have four valence electrons each. At room temperature which one of the following statements is most appropriate?

Options:
A
The number of free conduction electrons is significant in $\mathrm{C}$ but small in $\mathrm{Si}$ and $\mathrm{Ge}$.

B
The number of free conduction electrons is negligible small in all the three.

C
The number of free electrons for conduction is significant in all the three.

D
The number of free electrons for conduction is significant only in $\mathrm{Si}$ and $\mathrm{Ge}$ but small in $\mathrm{C}$.

The correct answer is:
The number of free electrons for conduction is significant only in $\mathrm{Si}$ and $\mathrm{Ge}$ but small in $\mathrm{C}$.

Solution:
No Solution Available

Topic: Physics, Semiconductors, JEE Main, JEE Main 2007
Carbon, silicon and germanium have four valence electrons each. These are characterized by valence and conduction bands separated by energy band gap respectively equal to \( \left(E_{g}\right)_{C},\left(E_{g}\right)_{S i} \) and \( \left(E_{g}\right)_{G e} . \) Which of the following statements is true?

Options:
A \( \left(E_{g}\right)_{S i} < \left(E_{g}\right)_{G e} < \left(E_{g}\right)_{C} \)
B \( \left(E_{g}\right)_{C} < \left(E_{g}\right)_{G e}>\left(E_{g}\right)_{S i} \)
C \( \left(E_{g}\right)_{C}>\left(E_{g}\right)_{S i}>\left(E_{g}\right)_{G e} \)
D \( \left(E_{g}\right)_{C}=\left(E_{g}\right)_{S i}=\left(E_{g}\right)_{G e} \)

The correct answer is:
\( \left(E_{g}\right)_{C}>\left(E_{g}\right)_{S i}>\left(E_{g}\right)_{G e} \)

Solution:
The energy band gap is maximum for carbon, less for silicon and least for germanium.

The energy band gap, Eg, is determined from the linear part of the optical absorption spectral. Therefore, we must remember the basic idea of energy band gap for common elements.
EgC>EgSi>EgGe

Topic: Physics, Semiconductors, JEE Main
Carbonic anhydrase enzyme exists

Options:
A mainly in plasma
B only in plasma
C mainly in RBC
D only in RBC

The correct answer is:
mainly in RBC

Solution:
The carbonic anhydrases form a family of enzymes that catalyze the interconversion between carbon dioxide and water and the dissociated ions of carbonic acid. The active site of most carbonic anhydrases contains a zinc ion. It is mainly found in RBC and little amount is also in plasma.

Topic: Biology, Breathing and Exchange of Gases, NEET
Carboxylic acids are easily ionised. The main reason of this statement

Options:
A Absence of α-hydrogen
B Resonance stabilisation of carboxylate ion
C Reactivity of α-hydrogen
D Hydrogen bond

The correct answer is:
Resonance stabilisation of carboxylate ion

Solution:
Carboxylic acids are easily ionized because there is resonance in carboxylate ion due to $\pi$ - electron shifting so $\mathrm{H}^{+}$get ionised very easily.

Topic: Chemistry, General Organic Chemistry, JEE Main
Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their

Options:
A more extensive association of carboxylic acid via van der Waal's force of attraction.
B formation of carboxylate ion.
C formation of intramolecular \( \mathrm{H} \)-bonding.
D formation of intermolecular \( \mathrm{H} \)-bonding.

The correct answer is:
formation of intermolecular \( \mathrm{H} \)-bonding.

Solution:
Carboxylic acid has a higher boiling point than aldehyde, ketone and even alcohols of comparable molecular mass.
This is due to their more extensive association through intermolecular H-bonding and it forms a dimer.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Carboxylic acids react with diazomethane to form

Options:
A Amine
B Alcohol
C Ester
D Amide

The correct answer is:
Ester

Solution:
Diazomethane provides the necessary $\mathrm{CH}_2$ group to form an ester (esterification using diazomethane.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Carbylamine forms from aliphatic or aromatic primary amine via which of the following intermediates?

Options:
A
Carbanion

B
Carbene

C
Carbocation

D
Carbon radical

The correct answer is:
Carbene

Solution:
Mechanism

Topic: Chemistry, Amines, JEE Main
Carbylamine forms from aliphatic or aromatic primary amine via which of the following intermediates?

Options:
A
Carbanion

B
Carbene

C
Carbocation

D
Carbon radical

The correct answer is:
Carbene

Solution:
Mechanism

Topic: Chemistry, Amines, JEE Main, JEE Main 2013 (25 Apr Online)
Carbylamine reaction is given by

Options:
A $1^{\circ}$ amine
B $3^{\circ}$ amine
C $2^{\circ}$ amine
D Quarternary salts

The correct answer is:
$1^{\circ}$ amine

Solution:
Carbylamine reaction is given by aliphatic and aromatic primary amines

Topic: Chemistry, Amines, JEE Main
Carbylamine test is used in the detection of

Options:
A Aliphatic $2^{\circ}$ amine
B Aromatic $1^{\circ}$ amine
C Aliphatic $1^{\circ}$ amine
D Both aliphatic and aromatic $1^{\circ}$ amines

The correct answer is:
Both aliphatic and aromatic $1^{\circ}$ amines

Solution:
Carbylamine test is used in the detection of both aliphatic and aromatic $1^{\circ}$ amines

Topic: Chemistry, Amines, JEE Main
Carbylamine test is used to detect the presence of primary amino group in an organic compound. Which of the following compound is formed when this test is performed with aniline?

Options:
A
B
C
D

The correct answer is:

Solution:
CARBYL amine given by 1° amine

Carbylamine reaction can be used to identify primary amino group in a given organic compound. When primary amines are treated with chloroform and alcoholic KOH, they give isocyanides (carbylamines) having very unpleasant smell which can be easily detected.

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (25 Feb Shift 2)
Carnallite is a mineral of

Options:
A $\mathrm{Ca}$
B $\mathrm{Na}$
C $M g$
D $\mathrm{Zn}$

The correct answer is:
$M g$

Solution:
Carnallite is a mineral of $\mathrm{Mg}$
Carnallite $\mathrm{KCl} . \mathrm{MgCl}_2, 6 \mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Carnellite is a mineral of

Options:
A Ca
B Na
C Mg
D Zn

The correct answer is:
Mg

Solution:
Carnellite is KCl â MgCl 2 â 6H 2 O
It is a white or reddish mineral consisting of a hydrated chloride of potassium ( K ) and magnesium ( Mg ) . Therefore, Carnellite is a mineral of Mg.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Carnivorous plants exhibit nutritional adaption. Citing an example explain this fact.

Options:

None

Solution:
Carnivorous plants fulfill their nutritional requirements by feeding on small animals, like insects or protozoans. e.g. Nepenthes, Venus fly trap, Utricularia etc. Carnivorous plants grow in soil deficient in nitrogen. In pitcher plant leaves are modified into pitcher which stores the juice to lure an insect. When the insect come to suck this juice, chemicals present in nectar dissolve the skin of the prey and the plant obtains nutrients (mainly nitrogen) from its skin.

Topic: Biology, Mineral Nutrition
Carnot cycle of an engine is given below
Total work done by the gas in one cycle is

Options:
A $\mu R T_{2} \log \frac{V_{2}}{V_{1}}-\mu R T_{1} \log \frac{V_{3}}{V_{4}}$
B $\mu R T_{1} \log \frac{V_{2}}{V_{1}}-\mu R T_{2} \log \frac{V_{3}}{V_{4}}$
C $\mu R T_{1} \log \frac{V_{2}}{V_{1}}+\mu R T_{2} \log \frac{V_{3}}{V_{4}}$
D Zero

The correct answer is:
$\mu R T_{1} \log \frac{V_{2}}{V_{1}}-\mu R T_{2} \log \frac{V_{3}}{V_{4}}$

Solution:
Total work done by the gas
$$
W_{\text {Total }}=W_{A \rightarrow B}+W_{B \rightarrow C}+W_{C \rightarrow D}+W_{D \rightarrow A}
$$
Work done in isothermal process by ideal gas
$$
W_{A \rightarrow B}=\mu R T, \log \frac{V_{2}}{V_{1}}
$$
Work done in adiabatic process by ideal gas
$$
W_{B \rightarrow C}=\frac{\mu R\left(T_{1}-T_{2}\right)}{r-1}
$$
Work done in isothermal process by ideal gas,
$$
W_{C \rightarrow D}=-\mu R T_{2} \log \frac{V_{3}}{V_{4}}
$$
Work done in adiabatic process, by ideal gas
$$
W_{D \rightarrow A}=\frac{-\mu R\left(T_{1}-T_{2}\right)}{r-1}
$$
Adding Eqs. (i), (ii), (iii) and (iv) we get
$$
W_{\text {Total }}=\mu R T_{1} \log \frac{V_{2}}{V_{1}}-\mu R T_{2} \log \frac{V_{3}}{V_{4}}
$$

Topic: Physics, Thermodynamics, JEE Main
Carrier ions like $\mathrm{Na}^{+}$facilitate the absorption of substance like

Options:
A amino acids and glucose
B glucose and fatty acids
C fatty acids and glycerol
D fructose and some amino acids

The correct answer is:
amino acids and glucose

Solution:
Active transport occurs with the help of energy, usually against concentration gradient. For this, cell membrane possesses carriers and gated channels. Active transports of one substance is often accompanied by permeation of other substances. The phenomenon is called secondary active transport. It is of two main types-Co-transport (eg, glucose and some amino acids along with inward pushing of excess $\mathrm{Na}^{+}$) and counter transport $\left(\mathrm{Ca}^{2+}\right.$ and $\mathrm{H}^{+}$movement outwardly as excess $\mathrm{Na}^{+}$passes inwardly).

Topic: Biology, Transport in Plants, NEET, NEET 2010 (Screening)
Carrying capacity of a population is determined by

Options:
A predation
B natality rate
C mortality rate
D limiting resources

The correct answer is:
limiting resources

Solution:
Group of organisms belonging to same species in a particular region or habitat is called as population. Carrying capacity is the maximum population that can be sustained is a particular area. In other words, the number organisms can survive normally in a particular area. The main limiting factor for the carrying capacity is the resources available in the area or the environmental factors. If the environmental conditions are not met then the population will keep decreasing until the resources rebound.
Natality rate is the number of newborns per 1000 in a particular area for a given time. (birth rate).
Mortality is the death rate where it is a measured as number of deaths per 1000 in a particular area for a given time. Predation is the population interaction between predator and prey.

Topic: Biology, Organisms and Populations, NEET
Cast iron is used for the manufacture of :

Options:
A wrought iron and pig iron
B pig iron, scrap iron and steel
C wrought iron, pig iron and steel
D wrought iron and steel

The correct answer is:
wrought iron, pig iron and steel

Solution:
Cast iron is made from pig iron which is used for production of wrought iron & steel.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2020 (02 Sep Shift 2)
Catalyst A reduces the activation energy for a reaction by 10 kJ mol-1 at 300 K. The ratio of rate constants, kT, CatalysedkT, Uncatalysed is ex. The value of x is____[nearest integer] [Assume that the pre-exponential factor is same in both the cases. Given R=8.31 J K-1 mol-1]

Options:

The correct answer is:
4

Solution:
Arrhenius equation for the rate constant of a reaction in terms of energy of activation is given by

K=Ae-EaRT

Kcatalyst=AeEa'RT

Kuncatalysed=Ae-EaRT

KcatalystKUncatalysed=eEa-Ea'RT=e10×10008.314×300=e4

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2022 (26 Jun Shift 2)
Catalytic dehydrogenation of a primary alcohol gives a/an

Options:
A secondary alcohol
B aldehyde
C ketone
D ester

The correct answer is:
aldehyde

Solution:
Catalytic dehydration of a primary alcohol gives aldehyde i.e.,

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Catalytic dehydrogenation of $n$-haptane in presence of $\mathrm{Cr}_2 \mathrm{O}_3 / \mathrm{Al}_2 \mathrm{O}_3$ at $750 \mathrm{~K}$ gives

Options:
A iso-heptane
B 1-heptene
C toluene
D 2, 3-dimethylpentene-1

The correct answer is:
1-heptene

Solution:
Octane number is related to the percentage of iso-octane since iso-octane is $81 \%$ hence octane number is $81 \%$.

Topic: Chemistry, Hydrocarbons, JEE Main
Catalytic hydrogenation of benzene gives

Options:
A Xylene
B Cyclohexane
C Benzoic acid
D Toluene

The correct answer is:
Cyclohexane

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Cathode rays and canal rays produced in a certain discharge tube are deflected in the same direction if

Options:
A A magnetic field is applied normally
B A magnetic field is applied tangentially
C An electric field is applied normally
D An electric field is applied tangentially

The correct answer is:
A magnetic field is applied normally

Solution:
Correct option is (1). A magnetic field is applied normally
In discharge tube cathode rays (a beam of negative particles) and canal rays (positive rays) moves opposite to each other. They will experience a magnetic force in the same direction, if a normal magnetic field is switched on

Topic: Physics, Dual Nature of Matter, JEE Main
Cathode rays are made up of

Options:
A None of these
B Negatively charged particles
C Neutral particles
D Positively charged particles

The correct answer is:
Negatively charged particles

Solution:
Cathode rays are made up of negatively charged particles called electrons. These rays produce ionization of gases trough which they travel

Topic: Physics, Dual Nature of Matter, JEE Main
Cathode rays are similar to visible light rays in that

Options:
A they both can be deflected by electric and magnetic fields.
B they both have a definite magnitude of wavelength.
C they both can ionise a gas through which they pass.
D they both can expose a photographic plate.

The correct answer is:
they both can expose a photographic plate.

Solution:
Light consists of photons and cathode rays consist of electrons. So, both rays can act like particles. However, both affect the photographic plate.

The photographic plates can detect the radiation produced by radioactive substances.

Topic: Physics, Dual Nature of Matter, JEE Main
Cathode rays consist of

Options:
A Photons
B Electrons
C Protons
D $\alpha_{\text {-particles }}$

The correct answer is:
Electrons

Solution:
Let's try to analyze the effect of Electric field and magnetic field on cathode rays:
Effect of electric field: When an electric field is applied to a stream of cathode rays, they get deflected towards the positive plate. It shows that cathode rays are themselves negatively charged.

Effect of magnetic field: When a magnetic field is applied perpendicular to the path of cathode rays, they get deflected in the direction expected for negative particles. This further confirms that cathode rays are negatively charged. \(\Rightarrow\) Therefore, we can conclude that cathode rays consist of electrons.

Topic: Physics, Atomic Physics, JEE Main
Cathode rays of velocity 106 m s-1 describe an approximate circular path of the radius 1 m in an electric field 300 Vcm-1. If the velocity of cathode rays are doubled. The value of electric field so that the rays describe the same circular path, will be

Options:
A 2400 V cm-1
B 600 V cm-1
C 1200 V cm-1
D 12000 V cm-1

The correct answer is:
1200 V cm-1

Solution:
Cathode rays are composed of electrons, when they move in electric field a force
F=eE ...(i)
Acts on them, this provides the necessary centripetal force to the particles
F=mv2r ...(ii)
From Eqs. (i) and (ii), we get
eE=mv2r
â r=mv2eE=m1062e(300) ...(iii)
When velocity is doubled same circular path is followed, hence radius is same
r=m2Ã1062eE ...(iv)
Equating Eqs. (iii) and (iv), we get
mÃ1062300e=mÃ2Ã1062eE
â E=300Ã4=1200 V cm-1

Topic: Physics, Electromagnetic Waves, JEE Main
Cathode rays of velocity 106 m s-1 describe an approximately circular path of radius 1 m in an electric field 300 V m-1. If the velocity of cathode rays are doubled. The value of electric field so that the rays describe the same circular path, will be

Options:
A 2400 V m-1
B 600 V m-1
C 1200 V m-1
D 12000 V m-1

The correct answer is:
1200 V m-1

Solution:
Cathode rays are composed of electrons, when they move in electric field a force
F=eE ...(i)
Acts on them, this provides the necessary centripetal force to the particles
F=mv2r ...(ii)
From Eqs. (i) and (ii), we get
eE=mv2r
â r=mv2eE=m1062e(300) ...(iii)
When velocity is doubled same circular path is followed, hence radius is same
r=m2Ã1062eE ...(iv)
Equating Eqs. (iii) and (iv), we get
mÃ1062300e=mÃ2Ã1062eE
â E=300Ã4=1200 V cm-1

Topic: Physics, Electromagnetic Waves, JEE Main
Cathode rays ______

Options:
A Are the atoms moving toward the cathode
B Are electrons emitted by cathode and travelling towards anode
C Are electromagnetic waves
D Are negative ions travelling from cathode to anode

The correct answer is:
Are electrons emitted by cathode and travelling towards anode

Solution:
Cathode rays are a beam of electrons emitted from the cathode of a high vacuum tube.
Cathode rays are negatively charged and hence they move towards positively charged ions hence they move towards anode.
Therefore the correct option is (2).

Topic: Physics, Dual Nature of Matter, JEE Main
Cattle ranches are known to causes acute green house effect. This is due to

Options:
A mechanized milking practices
B methanogenic bacteria in rumen
C decomposition of left over fodder
D decomposition of organic remains in faeces.

The correct answer is:
methanogenic bacteria in rumen

Solution:
Cattle ranching refers to practice of raising grazing livestock such as cattle or sheep for meat and wool, etc. The area of landscape meant primarily for cattle ranching is called a ranch. The practice has lead to accelerated deforestation and contributed to increased methane gas concentration in atmosphere. Whatever cattle eat; is subjected to the action of methanogens (bacteria) harbouring the stomach (rumen) of cattle. The action of these bacteria on food material produces methane gas which is released through cattle fart into the earth's atmosphere where it absorbs heat just like carbon dioxide, contributing to global warming.

Topic: Biology, Environmental Issues, AIIMS, AIIMS 2012
C-C bond length in benzene is

Options:
A $1.39 Å$
B $1.54 Å$
C $1.34 Å$
D Different in different bonds

The correct answer is:
$1.39 Å$

Solution:
$C-C$ bond length in benzene is $1.39 Å$ which is in between $C-C \quad(1.54 \hat{A})$ and $C=C(1.34 \hat{A})$ because of resonance.

Topic: Chemistry, General Organic Chemistry, JEE Main
Cell membrane in animals is composed mainly of the molecule of :

Options:
A lipids
B proteins
C lipids and proteins
D carbohydrates

The correct answer is:
lipids and proteins

Solution:
Chemically, the plasma membrane or cell membrane is made up of $60 \%$ proteins and $40 \%$ lipids. The percentage of carbohydrates ranges from 1-10 which are in the form of glycoproteins or glycolipids. Proteins may be extrinsic (peripheral) and intrinsic (integral) while lipids may be phospholipids, glycolipids and sterols.

Topic: Biology, Cell - The Unit of Life, JIPMER, JIPMER 2006
Cells in G0 phase:

Options:
A Exit the cell cycle
B Enter the cell cycle
C Suspend the cell cycle
D Terminate the cell cycle

The correct answer is:
Exit the cell cycle

Solution:
The cells remain in an inert state in the G0 phase of the cell cycle. Hence, G0 can be either viewed as the distinct inert stage which occurs outside the cell cycle or as an extension to the G1 phase, where cell neither divides nor prepares for it.

Sometimes, G0 is also referred to as “post-mitotic” state where cells in G0 are in non-dividing phase in the outside of cell cycle.

Some types of cells, such as nerve and heart muscle cells, become post-mitotic when they reach maturity (i.e., when they are terminally differentiated) but continue to perform their main functions for the rest of the organism's life.

Multinucleated muscle cells which don’t undergo cytokinesis are also thought to be in the G0 stage. Hence, it suspends the cell cycle.

G0 stage represents actually indicates the arrest of cell cycle and onset of differentiation.

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2019 (All India)
Cells which are not dividing are likely to be at

Options:
A
$\mathrm{G}_1$

B
$\mathrm{G}_2$

C
$\mathrm{G}_0$

D
S-phase

The correct answer is:
$\mathrm{G}_0$

Solution:
In $\mathrm{G}_0$ stage, no growth takes place. It is also called the quiescent stage. Some cells of the body like heart cells, neuron which do not divide, exit at $\mathrm{G}_1$ stage and enter $\mathrm{G}_0$ of the cell cycle an inactive stage.
The cells in this stage remain metabolically active but no longer proliferate unless called on to do so depending up on the requirement of the organism.
$G_1$ phase is the first growth phase in post mitotic gap phase that lasts from the end of mitotic cycle to the initiation of DNA replication.
$\mathrm{G}_2$ phase is the second growth phase or premitotic gap phase in which cell prepares itself to enter cell division, or mitosis.
S-phase is the synthetic phase in which chromosomes replicate.

Topic: Biology, Cell Cycle and Cell Division