datavorous/entrance-exam-dataset huggingface dataset:
Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths $\lambda_1: \lambda_2$ emitted in the two cases is

Options:
A $7 / 5$
B $27 / 20$
C $27 / 5$
D $20 / 7$

The correct answer is:
$20 / 7$

Solution:
Here, for wavelength $\lambda_1$
$n_1=4$ and $n_2=3$
and for $\lambda_2, n_1=3$ and $n_2=2$
We have $\frac{h c}{\lambda}=-13.6\left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right]$
So, for $\lambda_1$
$\begin{aligned}
\Rightarrow \frac{h c}{\lambda_1} & =-13.6\left[\frac{1}{(4)^2}-\frac{1}{(3)^2}\right] \\
\frac{h c}{\lambda_1} & =13.6\left[\frac{7}{144}\right]
\end{aligned}$
Similarly, for $\lambda_2$
$\begin{aligned}
\Rightarrow \frac{h c}{\lambda_2} & =-13.6\left[\frac{1}{(3)^2}-\frac{1}{(2)^2}\right] \\
\frac{h c}{\lambda_2} & =13.6\left[\frac{5}{36}\right]
\end{aligned}$
Hence, from Eqs. (i) and (ii), we get
$\frac{\lambda_1}{\lambda_2}=\frac{20}{7}$

Topic: Physics, Atomic Physics, NEET, NEET 2012 (Screening)
Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths Î»1: Î»2 emitted in the two cases is

Options:
A 7 5
B 27 20
C 27 5
D 20 7

The correct answer is:
20 7

Solution:
According to Rydberg formula
1 λ = Râ¡ 1 n fâ¡ 2 - 1 n iâ¡ 2
In first case, nf = 3, ni = 4
∴ 1 λ 1 = Râ¡ 1 3 2 - 1 4 2 = Râ¡ 1 9 - 1 1 6 = 7 1 4 4 Râ¡ … i
In second case, nf = 2, ni = 3
∴ 1 λ 2 = Râ¡ 1 2 2 - 1 3 2 = Râ¡ 1 4 - 1 9 = 5 3 6 Râ¡ … i i
Divide (ii) by (i) we get
λ 1 λ 2 = 5 3 6 × 1 4 4 7 = 2 0 7

Topic: Physics, Atomic Physics, NEET
Electron Transport System (ETS) is located in mitochondrial ________.

Options:
A outer membrane
B inter membrane space
C inner membrane
D matrix

The correct answer is:
inner membrane

Solution:
ATP synthase is an enzyme that helps in generating ATP which is used for all cell activities. It is a globular protein having two subunits -a headpiece -F1 and integral base- F0 attached to the inner membrane of the mitochondria.
Electron transport system (ETS) or electron transport chain (ETC) is a metabolic pathway with series of coenzymes and cytochromes where electron is passed to from one carrier to another until the ultimate acceptor. This process occurs at the inner membrane of the mitochondria and the different parts of the ETS are arranged in five kinds of sequential complexes.

Topic: Biology, Respiration in Plants, NEET
Electron will be lost from which molecular orbital when N2 is converted to N2+?

Options:
A σ*2px
B σ2pz
C π2px
D π*2px

The correct answer is:
σ2pz

Solution:
This can be done by the molecular orbital theory.

First, write the molecular orbital electronic configuration of N2molecule.

The total number of electrons present in the dinitrogen molecules is=14.

All the 14 electrons distributed in the bonding and antibonding molecular orbitals given below.

N2→KK0σ2s2σ*2s2π2px2=π2py2σ2pz2

N2+→KKσ2s2σ*2s2π2px2=π2py22pz1

As per the molecular orbital electronic configuration, we can understand that the last electron is removing from the σ2pz bonding molecular orbital.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Electronegative of fluorine is more than either boron or phosphorus. What conclusion can be drawn from the fact that BF3 has no dipole moment but PF3 does

Options:
A BF3 is not spherically symmetrical but PF3 is
B BF3 molecule must be linear
C The atomic radius of P is larger than the atomic radius of B
D The BF3 molecule must be planar triangular

The correct answer is:
The BF3 molecule must be planar triangular

Solution:
BF3 molecule must be planar triangular thus net dipole becomes zero while PF3 is pyramidal.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Electronegativity values for elements are useful in predicting

Options:
A Polarity of bonds in molecules
B Positions of elements in electrochemical series
C Co - ordination number of elements
D Oxidation number of elements

The correct answer is:
Polarity of bonds in molecules

Solution:
At the electronegativity differences increases electron density increase at toward the more electronegative atoms as the polarity reduces.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
Electronic configuration of some elements are given
A)  1s 2 2s 2
B)  1s 2 2s 2 2p 6
C)  1s 2 2s 2 2p 6 3s 2
D)  1s 2 2s 2 2p 3
E)  1s 2 2s 2 2p 5
The most ionic compound will be formed between

Options:
A A and D
B A and E
C C and E
D C and D

The correct answer is:
C and E

Solution:
As Δ EN increases ionic character also increases so here C and E will form most ionic compound.
Here C is Mg and E is F so compound is MgF2 .

Topic: Chemistry, Structure of Atom, NEET
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν0) and work function (W0) of the metal

Options:
A v0= 2.91 × 1014s–1and W0=4.41 × 10–19J
B v0= 3.54 × 1014s–1and W0=1.21 × 10–19J
C v0= 4.41 × 1014s–1and W0=2.91 × 10–19J
D v0= 2.35 × 1014s–1and W0=5.54 × 10–19J

The correct answer is:
v0= 4.41 × 1014s–1and W0=2.91 × 10–19J

Solution:
Correct Option is : (C)
v0= 4.41 × 1014s–1and W0=2.91 × 10–19J

Topic: Chemistry, Structure of Atom, NEET
Electrons from excited chlorophyll molecule of photosystem II are first accepted by:

Options:
A Cytochrome-b
B Cytochrome-f
C Plastoquinone
D Ferredoxin

The correct answer is:
Plastoquinone

Solution:
Electrons from excited chlorophyll molecule of photosystem II are accepted first by plastoquinone. Photosystem II is a photosynthetic pigment system along with some electron carriers that is located in the appressed part of the grana thylakoids. Photosystem II has chlorophyll a, band carotenoids. Other components of PS II are phaeophytin, plastoquinone (PQ), cytochrome complex and blue coloured copper containing plastocyanin

Topic: Biology, Photosynthesis in Higher Plants, NEET
Electrons from excited chlorophyll molecule of photosystem-II are accepted first by

Options:
A cytochrome-b
B cytochrome-f
C quinone
D ferredoxin

The correct answer is:
quinone

Solution:
Photosystem-II is a photosynthetic pigment system alongwith some electron carriers that is located in the apressed part of grana thylakoids. Photocentre has a dimer of special chlorophyll-a molecules called $\mathrm{P}_{680} . \mathrm{P}_{680}$ of photosystem-II absorbs light energy, gets excited and transfers its electrons to electron acceptor molecule phaeophytin. Phaeophytin on accepting electrons become strong reducing agent. It donates, its electrons to downstream component of ETC, $i e$, quinone, cyt- $b$ cyt- $f$ complex, plastocyanin. While passing over cytochrome complex, electron losses sufficient energy to synthesize ATP. Plastocyanin is a copper containing soluble protein, which transfers electrons to $P_{700}$ reaction centre of photosystem-I. On geting excited, $P_{700}$ hands over electrons to a special chlorophyll molecule $\mathrm{X}$ from where electrons are passed to membrane bound iron-sulphur protein for transfer to another soluble protein ferredoxin.

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2008 (Screening)
Electrons of mass, m with de-Broglie wavelength, λ fall on the target in an X-ray tube. The cutoff wavelength, λ0 of the emitted X-ray is

Options:
A λ0=2mcλ2h
B λ0=2hmc
C λ0=2m2c2λ3h2
D Î»0=Î»

The correct answer is:
λ0=2mcλ2h

Solution:
Momentum, P=hλ⇒E=P22m⇒h22mλ2=hcλ0

⇒λ0=hch22mλ2

=2mcλ2h.

Topic: Physics, Dual Nature of Matter, NEET, NEET 2016 (Phase 2)
Electrons used in an electron microscope are accelerated by a voltage of $25 \mathrm{kV}$. If the voltage is increased to $100 \mathrm{kV}$ then the de-Broglie wavelength associated with the electrons would

Options:
A decrease by 2 times
B decrease by 4 times
C increase by 4 times
D increase by 2 times

The correct answer is:
decrease by 2 times

Solution:
We have
$$
\lambda=\frac{12.27}{\sqrt{V}}
$$
$\mathrm{SO}$,
$$
\begin{aligned}
& \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{V_2}{V_1}} \\
& \lambda_2=\lambda_1 \sqrt{\frac{V_1}{V_2}} \\
& \lambda_2=\lambda_1 \sqrt{\frac{25}{100}} \\
& \lambda_2=\lambda_1 \sqrt{\frac{1}{4}}=\frac{\lambda_1}{2}
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2011 (Screening)
Electrons with de-Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength for the emitted X-rays is

Options:
A Î» 0 = 2 m Î» 2 c h
B λ0=2hmc
C λ0=2m2c2λ3h2
D λ 0 = λ

The correct answer is:
λ 0 = 2 m λ 2 c h

Solution:
The momentum of striking electrons,
p = h λ
Therefore the kinetic energy of striking electrons,
K = p 2 2 m = h 2 2 m λ 2
This is also the maximum energy of X-ray photons.
Therefore, hcλ0=h22mλ2 or λ0=2mλ2ch

Topic: Physics, Electromagnetic Waves, NEET
Electrons with de-Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is

Options:
A λ0=2mcλ2h
B λ0=2hmc
C λ0=2m2c2λ2h2
D λ0=λ

The correct answer is:
λ0=2mcλ2h

Solution:
Momentum of striking electrons p=h λ
∴ Kinetic energy of striking electrons
K=p22m=h22mλ2
This is also, maximum energy of X-ray photons.
Therefore, hc λ0=h22mλ2
Or λ0=2mλ2ch

Topic: Physics, Dual Nature of Matter, NEET
Electrophilic aromatic substitution is difficult in

Options:
A p-NH2C6H4CO2H
B p-CH3OC6H4CO2H
C p-NO2-C6H4CO2H
D p-ClC6H4CO2H

The correct answer is:
p-NO2-C6H4CO2H

Solution:
As both -COOH and -NO2 are ring deactivating and meta directing so electrophilic substitution is tough and slow.

Topic: Chemistry, Hydrocarbons, NEET
Electroporation involves

Options:
A promotion of seed germination by induced imbibitions of water with electric current
B making transient pores in cell membrane to facilitate entry of gene constructs
C purification of saline water with the help of an artificial membrane
D passage of sucrose through sieve pores by electroosmosis.

The correct answer is:
making transient pores in cell membrane to facilitate entry of gene constructs

Solution:
Electroporation is the process where electric energy is passed through living tissue surface causing appearance of pores in the membrane which then can be used to pass the biological material easily.

Topic: Biology, Biotechnology - Principles and Processes, NEET
Electroporation is a technique that can be used for genetic engineering. This technique enables gene manipulation by

Options:
A Opening of stomatal pores during night by artificial light
B Purification of saline water with the assist of a membrane system
C Making transient pores in cell membrane to introduce gene constructs
D Fast passage of food through sieve pores in phloem elements with the help of electric stimulation

The correct answer is:
Making transient pores in cell membrane to introduce gene constructs

Solution:
Electroporation is the process where electric energy is passed through living tissue surface causing appearance of pores on the membrane, which then can be used to pass the biological material easily. This way we can introduce the gene of interest.

Topic: Biology, Biotechnology - Principles and Processes, NEET
Electrovalent bond-formation depends on :

Options:
A ionization energy
B lattice energy
C electron affinity
D all of these

The correct answer is:
all of these

Solution:
The electrovalent bond is favoured with a metal having low ionization energy and non-metal with high electron affinity.
The amount of energy released when one mole of an ionic compound is formed is called lattice energy. Higher the lattice energy more the stability of the ionic compound.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Elements of group 14

Options:
A exhibit oxidation state of +4 only
B exhibit oxidation state of +2 and +4
C form M2–and M4+ions
D form M2+and M4+ions

The correct answer is:
exhibit oxidation state of +2 and +4

Solution:
Correct Option is : (B)
exhibit oxidation state of +2 and +4

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
Elements of Group 16 generally show a lower value of the first ionization enthalpy compared to the corresponding periods of Group 15. Why?

Options:
A Group 16 elements have more protons in their nuclei, making it easier to remove an electron.
B Group 16 elements have smaller atomic sizes, leading to stronger electron-nucleus attractions.
C Group 16 elements have more stable electron configurations when they lose an electron.
D Group 16 elements have a higher effective nuclear charge, facilitating electron removal.

The correct answer is:
Group 16 elements have more stable electron configurations when they lose an electron.

Solution:
Correct Option is : (C)
Group 16 elements have more stable electron configurations when they lose an electron.

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
Elevation in the boiling point for 1 molal solution of glucose is 2K. The depression in the freezing point of 2 molal solutions of glucose in the same solvent is 2K. The relation between Kb and Kf is

Options:
A Kb=0.5Kf
B Kb=2Kf
C Kb=1.5Kf
D Kb=Kf

The correct answer is:
Kb=2Kf

Solution:
ÎTb=Kbm
Kb=ÎTbm=21=2
â¦(i)
ÎTf=Kfm
Kf=ÎTfm=22=1
â¦(ii)
⇒KbKf=2
⇒Kb=2Kf

Topic: Chemistry, Solutions, NEET
Eli Lilly, an American company, is famous for

Options:
A Producing vitamin ‘A’ enriched rice
B Producing pest resistant plants by RNAinterference
C Producing transgenic soybean for herbicideresistance
D Preparing two DNA sequences corresponding to A and B chains of human insulin and introduced them in the plasmids of E.coli to produce insulin

The correct answer is:
Preparing two DNA sequences corresponding to A and B chains of human insulin and introduced them in the plasmids of E.coli to produce insulin

Solution:
Eli Lilly marketed the first such insulin, Humulin, in 1982. Humulin was the first medication produced using modern genetic engineering techniques in which actual human DNA is inserted into a host cell (E. coli in this case). The host cells are then allowed to grow and reproduce normally, and due to the inserted human DNA, they produce a synthetic version of human insulin. However, the clinical preparations prepared from such insulins differ from endogenous human insulin in several important respects; an example is the absence of C-peptide which has in recent years been shown to have systemic effects itself. Genentech developed the technique Lilly used to produce Humulin, although the company never commercially marketed the product themselves.

Topic: Biology, Biotechnology and Its Applications, NEET
Eli Lilly, an American company prepared two DNA sequences corresponding to A and B, chains of human insulin, and introduced them in plasmids of E. coli to produce insulin chains. Chains A and B were produced separately, extracted and combined by creating

Options:
A Peptide bonds
B Ionic bonds
C H-bonds
D Disulphide bonds

The correct answer is:
Disulphide bonds

Solution:
Human insulin is synthesized as proinsulin and consists of three polypeptide chains A, B, and C. During conversion into functional insulin C polypeptide is removed and A and B polypeptides are joined two disulpide bonds forming functional insulin. The first genetically engineered human insulin was made by an American company Eli Lilly and was marketed by name Humulin.

Topic: Biology, Biotechnology and Its Applications, NEET
Eli Lily an American company prepared two DNA sequences corresponding to A and B, chains of human insulin and introduced them in plasmids of E. coil to produce insulin chains. These chains A and B were

Options:
A Peptide bonds
B Ionic bonds
C H- bonds
D Disulphide bonds

The correct answer is:
Disulphide bonds

Solution:
Insulin molecule contains two polypeptide chains: chain A and chain B . Both these chains are linked to each other with the help of disulphide bonds. Thus, after being produced separately, disuplhide bonds will be used for combining the two chains. Insulin is initially present as a prohormone, which has an additional C peptide. This C peptide is removed during maturation of the prohormone to insulin.

Topic: Biology, Biotechnology and Its Applications, NEET
Elimination reaction of 2-Bromo-pentane to form pent-2-ene is
(a) β-Elimination reaction
(b) Follows Zaitsev rule
(c) Dehydrohalogenation reaction
(d) Dehydration reaction

Options:
A (a), (c), (d)
B (b), (c), (d)
C (a), (b), (d)
D (a), (b), (c)

The correct answer is:
(a), (b), (c)

Solution:
This reaction is an example of β -elimination.

Hydrogen is removed from β -carbon and halogen from α -carbon, hence, dehydrohalgoenation reaction.

Generally in E2 reaction Zaitsev alkene is formed as major product (more stable alkene).

Topic: Chemistry, Hydrocarbons, NEET, NEET 2020 (Phase 1)
Elimination reactions (especially β-elimination) are as common as the nucleophilic substitution reaction in case of alkyl halides. Which of the following reagents can be used in both cases?

Options:
A Nucleophilic substitution: By alcoholic KOH and Nucleophilic elimination: By aqueous KOH.
B Nucleophilic substitution: By aqueous KOH and Nucleophilic elimination: By alcoholic KOH.
C Nucleophilic substitution: By aqueous KOH and Nucleophilic elimination: By aqueous KOH.
D Nucleophilic substitution: By alcoholic KOH and Nucleophilic elimination: By alcoholic KOH.

The correct answer is:
Nucleophilic substitution: By aqueous KOH and Nucleophilic elimination: By alcoholic KOH.

Solution:
Correct Option is : (B)
Nucleophilic substitution: By aqueous KOH and Nucleophilic elimination: By alcoholic KOH.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
ELISA is a technique used for early diagnosis of pathogen and is based on

Options:
A using radio labelled second antibody
B detecting the RBCs
C complement-mediated cell lysis
D detecting the antibodies synthesized against the pathogen

The correct answer is:
detecting the antibodies synthesized against the pathogen

Solution:
Enzyme-linked immunosorbent assay is based on the principle of antigen-antibody interaction. It is used for the early detection of pathogens. Infection by the pathogen is detected by the presence of antigen or by detecting the antibodies synthesized against the pathogen. In ELISA specific enzymes are added which binds with the pathogen to form an immunologically active substance that can be easily detected.

Topic: Biology, Biotechnology and Its Applications, NEET
ELISA is based on the principle of

Options:
A Antigen-antibody interaction
B Antigen-antigen interaction
C Antibody-antibody interaction
D None of these

The correct answer is:
Antigen-antibody interaction

Solution:
ELISA is based on the principle of antigen-antibody interaction. Infection by a pathogen can be detected by the presence of antigens (proteins, glycoproteins, etc.) or by detecting the antibodies synthesized against the pathogen.

Topic: Biology, Biotechnology and Its Applications, NEET
ELISA is used to detect viruses, where:

Options:
A DNA-probes arc required
B southern blotting is done
C alkaline phosphatase is the key reagent.
D catalase is the key reagent.

The correct answer is:
alkaline phosphatase is the key reagent.

Solution:
ELISA (enzyme linked immunosorbent assay ) screening test is the initial test to diagnose AIDS. The test works by detecting antibodies/substances or protein which are produced in the blood when virus is present. Alkaline phosphatase and peroxidases are commonly I used enzymes as key reagent to perform the ELISA test.

Topic: Biology, Human Health and Diseases, NEET, NEET 2004
ELISA is used to detect viruses where the key reagent is:

Options:
A alkaline phosphatase
B catalase
C DNA probe
D RNase

The correct answer is:
alkaline phosphatase

Solution:
Alkaline phosphatase and peroxidases are commonly used enzymes as key reagent to perform the ELISA test.
Related Theory
ELISA (enzyme linked immunosorbent assay) screening test is the initial test to diagnose AIDS. The test works by detecting antibodies which are produced in the blood in response to the virus attack.

Topic: Biology, Human Health and Diseases, NEET, NEET 2003
ELISA is used to detect viruses, where

Options:
A DNA-probes are required
B Southern bloting is done
C Alkaline phosphatase is the key reagent
D Catalase is the key reagent

The correct answer is:
Alkaline phosphatase is the key reagent

Solution:
ELISA test is a technique used to detect and quantitate extremely small amount of a protein, antibody or antigen with the help of enzyme. The commonly used enzymes are peroxidase and alkaline phosphatase. Southern blotting and DNA probes are used in molecular analysis of DNA. Catalase is not involved in ELISA.

Topic: Biology, Biotechnology and Its Applications, NEET
ELISA test is done to diagnose:

Options:
A AIDS
B Typhoid
C Malaria
D Common cold

The correct answer is:
AIDS

Solution:
A widely used diagnostic test for AIDS is enzyme linked immuno-sorbent assay (ELISA).

Topic: Biology, Human Health and Diseases, NEET
Elution is

Options:
A The process of coating of DNA with micropellets of gold and tungsten
B The process of precipitation of DNA using a suitable reagent
C The process of insertion of DNA
D The process of cutting out and extracting bands of DNA from gel.

The correct answer is:
The process of cutting out and extracting bands of DNA from gel.

Solution:
Using gel electrophoresis the DNA fragments are separated according to their size on gel block in the electric field. If the gel block is cut to remove the DNA fragments than the process is called elution.

Topic: Biology, Biotechnology - Principles and Processes, NEET
Embryo with more than 16 blastomeres formed due to in-vitro fertilization is transferred into:

Options:
A Uterus
B Fallopian tube
C Fimbriae
D Cervix

The correct answer is:
Uterus

Solution:
INTRA UTERINE TRANSFER (IUT): In this technique, after In-vitro fertilization the zygote is allowed to develop in to embryo (more than 8 blastomeres). The embryo is transformed directly into the uterus for implantation and further development. Embryos formed by In-vivo fertilization could be used for such transfer to assest those females who fail to conceive.

Topic: Biology, Reproductive Health, NEET, NEET 2016 (Phase 2)
Embryoids are the

Options:
A Embryos produced from the culture of somatic plant cells
B Plants produced after hardening
C Plants which show totipotency
D Young embryos

The correct answer is:
Embryos produced from the culture of somatic plant cells

Solution:
Somatic cells, after attaining the stationary phase in a cell suspension, each cell differentiates into an independent embryo called embryoids.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
Embryological support for evolution was disapproved by

Options:
A Alfred Wallace
B Charles Darwin
C Oparin
D Karl Ernst von Baer

The correct answer is:
Karl Ernst von Baer

Solution:
EMBRYOLOGICAL evidence: Embryological support for evolution was also proposed by Ernst Haeckel based upon the observation of certain features during the embryonic stage common to all vertebrates that are absent in adults. For example, the embryos of all vertebrates including humans develop a row of vestigial gill slits behind the head, but it is a functional organ only in fish and not found in any other adult vertebrate. According to Ernst Hackel, ontogeny (development of the embryo) is the recapitulation of phylogeny (the ancestral sequence). This view is summarized by his biogenetic law: Ontogeny recapitulates phylogeny. This proposal was disapproved of a careful study performed by Karl Ernst von Baer. He noted that embryos never pass through the adult stages of other animals.

Topic: Biology, Evolution, NEET, NEET 2020 (Phase 1)
Embryological support for evolution was proposed by:

Options:
A Ernst Heckel
B Karl Ernst von Baer
C Charles Darwin
D Alfred Wallace

The correct answer is:
Ernst Heckel

Solution:
Embryological support for evolution was also proposed by Ernst Haeckel based upon the observation of certain features during the embryonic stage common to all vertebrates that are absent in adults. For example, the embryos of all vertebrates including humans develop a row of vestigial gill slits behind the head, but it is a functional organ only in fish and not found in any other adult vertebrate.

Topic: Biology, Evolution, NEET, NEET 2020 (Phase 2)