datavorous/entrance-exam-dataset huggingface dataset:
An organism has a gene 'X'. Its dominant allele expresses character C-1 while its recessive allele expresses character C-2. This dominant allele is completely dominant over the recessive allele. It has another gene 'Y' which in its dominant form doesn't allow any expression of gene 'X'. Which of these genotypes of that organism will produce character C-1?

Options:
A xxyy or XxYy
B XXYY or XxYY
C XxYy or XXYy
D Xxyy or XXyy

The correct answer is:
Xxyy or XXyy

Solution:
This is an example of dominant epistatsis. Here, Y is the epistatic gene. In its dominant form that is YY or Yy it will not allow expression of X. That means for expression of X, the genotype should be yy. C-1 is dominant character expressed by gene X. Hence, C-1 can be expressed by genotypes XX or Xx. Thus, organisms which express character C-1 have genotype XXyy or Xxyy.

Topic: Biology, Principles of Inheritance and Variation, NEET
An organism has two unlinked genes P and Q. Its genotype is PpQq. How many different types of gametes will be formed?

Options:
A Two
B Six
C Four
D Eight

The correct answer is:
Four

Solution:
If genes are unlinked than they separate during gamete formation and a dihybrid PpQq forms 4 types of gamete. A gamete can have only one allele of a gene thus four types of gametes are formed PQ, Pq, pQ, pq.

Topic: Biology, Principles of Inheritance and Variation, NEET
An organism is in the same class but not in the same family. It may belong to same:

Options:
A Genus
B Species
C Order
D More than one option is correct

The correct answer is:
Order

Solution:
Taxonomy is the branch of biological science which includes nomenclature and classification of the organisms into groups based on their shared features and characteristics.
Classification is the process of grouping organisms based on observable features in to categories called 'taxa'. This process involves hierarchy of steps where each taxa represents a rank. All the steps/taxon collectively is referred as taxonomic hierarchy.
Species is the basic unit of classification and consist of similar individuals that can interbreed and produce fertile offspring.
Group of related species which has more characters in common in comparison to species of other genera is referred to as genus.
Order is a rank in classification that is higher than family but lower than class.

Topic: Biology, The Living World, NEET
An organism used as a biofertilizer for raising soyabean crop is

Options:
A Azospirillum
B Rhizobium
C Nostoc
D Azotobacter

The correct answer is:
Rhizobium

Solution:
Rhizobium leguminosarum is a symbiotic bacteria found in root nodules of legume. This bacterium has nitrogen fix nif gene and fixing $\mathrm{N}_2$. Soyabean is a legume. Thus, Rhizobium is used as a biofertilizer for raising soyabean crop.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2011 (Screening)
An organism used as a biofertilizer for raising soyabean crop is

Options:
A Azotobacter
B Azospirillum
C Rhizobium
D Nostoc

The correct answer is:
Rhizobium

Solution:
Rhizobium is used as a biofertilizer for raising crop. Rhizobium japonicum forms symbiotic association in the roots of the leguminous plant, soyabean.

Topic: Biology, Microbes in Human Welfare, NEET
An organism with two identical alleles for a given trait is

Options:
A homozygous.
B heterozygous.
C dominant.
D hermaphrodite.

The correct answer is:
homozygous.

Solution:
The homozygote is pure for the character and breeds true, that is, it gives rise to offspring having the same character on self breeding. e.g., TT or tt.

Topic: Biology, Principles of Inheritance and Variation, NEET
An orthotropous ovule is one, in which micropyle and chalaza are

Options:
A at right angles to funicle
B parallel to the funicle
C in straight line of funicle
D parallel along with ovule

The correct answer is:
in straight line of funicle

Solution:
Orthotropus ovule is an erect ovule. In this the micropyle, chalaza and funicle are in the same line. The micropyle is upward and the chalaza is downward. It can be represented by the following figure

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
An oscillator circuit contains an inductor 0.05 H and a capacitor of capacity 80 Î¼F. When the maximum voltage across the capacitor is 200 V, the maximum current (in amperes) in the circuit is

Options:
A 4
B 8
C 10
D 16

The correct answer is:
8

Solution:
Given L=0.05 H, C=80 μF
Vmax=200 V
∵ Voltage equation Vt=Vmsin⁡ωt
∴ Current (i) =cdvdt=cddtVmsin⁡ωt=CVmωcos⁡ωt
∵ω=1LC
∴ i=VmCLcos⁡ωt
∴ Maximum current im, VmCL=200×80μ0.05
im=8 A

Topic: Physics, Alternating Current, NEET
An out-cross will have parents

Options:
A From different breeds
B From same breeds but with common ancestors only from one side
C From same breed but with common ancestors from both sides
D From same breed but with no common ancestors

The correct answer is:
From same breed but with no common ancestors

Solution:
Outcrossing is the process of mating of animals within the same breed, but having no common ancestors on either side of their pedigree up to 4–6 generations. Outcrossing helps to overcome inbreeding depression.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
An oval mass of white mater, present above the medulla oblongata acts as

Options:
A Pneumothoracic centre
B Pneumococcal centre
C Pneumatic centre
D Pneumotaxic centre

The correct answer is:
Pneumotaxic centre

Solution:
An oval mass of white mater, present above the medulla oblongata acts as respiratory pneumotaxic centre and they are Pons varoli.
Pons Varoli is situated in front of the cerebellum below the midbrain and above the medulla oblongata. It relays impulses between the medulla oblongata and more superior part of the brain, between the hemispheres of the cerebellum and between the cerebrum and cerebellum. It contains centre that work with those in the medulla to regulate breathing.

Topic: Biology, Breathing and Exchange of Gases, NEET
An ovule which becomes curved so that the nucellus and embryo sac lie at right angles to the funicle is:

Options:
A hemitropous
B campylotropous
C anatropous
D orthotropous.

The correct answer is:
hemitropous

Solution:
Ovules have been separated into six categories based on their shapes:
Orthotropous (Atropous): Ovules is straight so that the chalaza, the funicle and the micropyle are all aligned.
Anatropous: Ovules become completely inverted during development so that the micropyle lies close to the hilum.
Hemi-anatropous: Ovules becomes at a right angle in relation to the funicle.
Campylotropous: ovule is bent and the alignment between the chalaza and micropyle is lost.
Amphitropous: ovule is very much curved that the embryo sac and is horseshoe shaped.
Circinotropous: The funicle is especially long that it creates a nearly full circle around the ovule whose micropyle is ultimately pointing upwards.
(Ref. to Fig. 28.2)
pollen tube down through the style. This tube then enters the ovary and reaches the ovule of the plant. Once that occurs, fertilization can arise as the nucleus of the pollen grain is sent
down the tube to merge with the nucleus in the embryo sac. Note that the male alternative to the ovule is pollen, which contains the male gametophytes.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2004
An P-N-P transistor circuit is arranged as shown. It is a

Options:
A common base amplifier circuit
B common-emitter amplifier circuit
C common-collector circuit
D none

The correct answer is:
common-collector circuit

Solution:
The P-N junction which is forward biased is the emitter-base pair of the transistor. Hence the common junction is collector, so it is a common collector circuit

Topic: Physics, Semiconductors, NEET
An Î±- particle after passing through a potential difference of V volts collides with a nucleus. If the atomic number of the nucleus is Z then the distance of closest approach of Î±- particle to the nucleus will be

Options:
A 14.4ZV AÂ°
B 14.4ZVÂ m
C 14.4ZVÂ cm
D all of these

The correct answer is:
14.4ZV AÂ°

Solution:
K.E. = P.E. = qV
⇒2eV=KZe2ed=qV
∴d=9×109×Z×e×2e2eV
∴d=9×109×1.6×10–19×ZV
d=14.4×10-10ZVm

Topic: Physics, Atomic Physics, NEET
An α particle and a deuteron are moving with velocities v and 2v respectively. What will be the ratio of their de - Broglie wavelengths?

Options:
A 1 : 2
B 2 :1
C 1 :1
D 2 :1

The correct answer is:
1 :1

Solution:
De - Broglie wave length λ=hP
λαλD=PαPD=mα VαmD VD
=42×V2V=1

Topic: Physics, Atomic Physics, NEET
An Î±-particle and a proton are accelerated from rest, by a potential difference of 100 V. After this, their de-Broglie wavelengths are Î»Î± and Î»p, respectively. The ratio Î»pÎ»Î±, to the nearest integer, is

Options:
A 3
B 2
C 4
D 5

The correct answer is:
3

Solution:
1 2 mv 2 = qV
â p = mv = 2 mq V
De broglie wavelength,
Î» = h p â Î» Î± Î» p = P p P Î± = m p q p m Î± q Î±
â Î» p Î» Î± = 8
The nearest integer is 3
â´ answer is 3

Topic: Physics, Atomic Physics, NEET
An Î±-particle and a proton are both simultaneously projected into a region of a constant magnetic field, perpendicular to the direction of the field but in opposite directions. After 10-8 s, it is found that the velocity of Î±-particle has changed in the direction by 45Â°. The angle between the velocity vectors of the Î±-particle and the proton is

Options:
A 90Â°
B 45°
C 45Â°+90Â°
D 45Â°+90Â°2

The correct answer is:
45°+90°

Solution:
ωα=2e⁡B⁡4m⁡=12e⁡B⁡m⁡
ωp⁡=e⁡B⁡m⁡
θα=45°
θp⁡=90° from their original directions
So the total angle between the two particles is 45°+90°.

Topic: Physics, Magnetic Effects of Current, NEET
An α particle is bombarded on N14. As a result, a O17 nucleus is formed and a particle is emitted. This particle is a

Options:
A neutron
B proton
C electron
D positron

The correct answer is:
proton

Solution:
714N+ 24 α→ 817O+ ZAX
From conservation of charge
7+2=8+Z
Z=1
From conservation of mass number
14+4=17+A
A=1
∴ 11X is proton.

Topic: Physics, Nuclear Physics, NEET
An $\alpha$-particle moves in a circular path of radius $0.83 \mathrm{~cm}$ in the presence of a magnetic field of $0.25 \mathrm{~Wb} / \mathrm{m}^2$. The de-Broglie wavelength associated with the particle will be

Options:
A $1 Å$
B $0.1 Å$
C $10 Å$
D $0.01 Å$

The correct answer is:
$0.01 Å$

Solution:
We knows
$R=\frac{m v}{q B}$
$\begin{aligned}
\lambda & =\frac{h}{m v} \\
\lambda & =\frac{h}{q B R} \\
& =\frac{6.6 \times 10^{-34}}{1.6 \times 10^{-19} \times 0.83 \times 10^{-2} \times 0.25} \\
& =0.01 Å
\end{aligned}$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2012 (Screening)
An Î± -particle of energy 5 MeV is scattered through 180Â° by a stationary uranium nucleus. The distance of closest approach is of the order

Options:
A 1 Å
B 10-10Â cm
C 10-12 cm
D 10-15Â cm

The correct answer is:
10-12 cm

Solution:
According to work energy theorem
KE=PE
KE=14ÏÎµ0q1q2r
r=14ÏÎµ092eâ2eE
r=9Ã109Ã92Ã21.6Ã10-1925Ã106Ã1.6Ã10-19
r=5.3×10-14m⁡≈10-12 cm.

Topic: Physics, Atomic Physics, NEET
An Î± - particle of energy 5 MeV is scattered through 180Â° by a fixed Uranium nucleus. The distance of the closest approach is (Atomic number of uranium = 92)

Options:
A 5.3×10-12 m
B 5.3×10-13 m
C 5.3×10-14 m
D 5.3×10-15 m

The correct answer is:
5.3×10-14 m

Solution:
According to the law of conservation of energy,
The kinetic energy of α - particle = Potential energy of alpha α - particle at a distance of closest approach.
⇒ 12mv2=14πε0q1q2r
⇒ 5 MeV=9×1092e92er
r=5.3×10-14 m

Topic: Physics, Nuclear Physics, NEET
An Î± - particle of energy 5 MeV is scattered through 180Â° by a fixed Uranium nucleus. The distance of the closest approach is (Atomic number of uranium = 92)

Options:
A 5.3 × 10-12
B 5.3 × 10-13
C 5.3 × 10-14
D 5.3 × 10-15

The correct answer is:
5.3 × 10-14

Solution:
According to the law of conservation of energy,
The kinetic energy of Î± - particle = Potential energy of alpha Î± - particle at a distance of closest approach.
â 12mv2=14ÏÎµ0q1q2r
â 5 MeV=9Ã1092e92er
r=5.3Ã10-14 m

Topic: Physics, Nuclear Physics, NEET
An Î±-particle with a specific charge of 2.5Ã107 C kg-1 moves with a speed of 2Ã105 m s-1 in a perpendicular magnetic field of 0.05 T. Then, the radius of the circular path described by it is

Options:
A 8 cm
B 4 cm
C 16 cm
D 2 cm

The correct answer is:
16 cm

Solution:
r=mvBq
â r=vBqm=2Ã1050.05Ã2.5Ã107
=2Ã10712.5Ã107=20012.5 cm=16 cm

Topic: Physics, Magnetic Effects of Current, NEET
An Rh-positive male gets married to an Rh-negative female. During their second pregnancy, which of the following events would be observed?

Options:
A Rh incompatibility disease
B Erythroblastosis foetalis
C Muscular dystrophy
D Both A and B

The correct answer is:
Both A and B

Solution:
Erythroblastosis fetalis classically results from Rh incompatibility, which may develop when a woman with Rh-negative blood is impregnated by a man with Rh-positive blood and conceives a fetus with Rh-positive blood, sometimes resulting in hemolysis.

Topic: Biology, Body Fluids and Circulation, NEET
An round disc of moment of inertia $I_2$ about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia $\mathrm{I}_1$ rotating with an angular velocity $\omega$ about the same axis. The final angular velocity of the combination of discs is:

Options:
A $\frac{I_2 \omega}{I_2+I_2}$
B $\omega$
C $\frac{I_1 \omega}{I_1+I_2}$
D $\frac{\left(I_1+I_2\right) \omega}{I_1}$

The correct answer is:
$\frac{I_1 \omega}{I_1+I_2}$

Solution:
From law of conservation of angular momentum we have:
$$
\begin{aligned}
& I_1 \omega_1=\left(I_1+I_2\right) \omega_2 \\
& \therefore \quad \omega_2=\frac{I_1 \omega_1}{\left(I_1+I_2\right)}=\frac{I_1 \omega}{I_1+I_2} \\
&
\end{aligned}
$$

Topic: Physics, Rotational Motion, NEET, NEET 2004
An undeformed spring of spring constant k is connected to a bead of mass m which can move along a frictionless rod as shown in the figure. If the particle strikes the bead at an angle of 45Â° with the horizontal and sticks to it, then the maximum elongation of the spring after the collision is

Options:
A v4m2k
B v2mk
C v2m2k
D v4mk

The correct answer is:
v2mk

Solution:
Conserving the linear momentum of the system along the rod

mv cos 45∘=2mv1

v 1 = v 2 2

Applying work-energy theorem
- 1 2 kx 2 = - 1 2 2 m v 2 2 2
⇒ x=v2mk

Topic: Physics, Center of Mass Momentum and Collision, NEET
An undifferentiated layer, mesoglea, is present in

Options:
A Fasciola
B Ancylostoma
C Hirudinaria
D Adamsia

The correct answer is:
Adamsia

Solution:
Animals in which the cells are arranged in two embryonic layers, external ectoderm, and internal endoderm, are called diploblastic animals, e.g., coelenterates. An undifferentiated layer, mesoglea, is present in between the ectoderm and the endoderm.

Topic: Biology, Animal Kingdom, NEET
An unknown compound A dissociates at 500°C to give products as follows –
A(g)⇌B(g)+C(g)+D(g)
Vapour density of the equilibrium mixture is 50 when it dissociates to the extent to 10% . What will be the molecular weight of Compound A–

Options:
A 120
B 130
C 134
D 140

The correct answer is:
120

Solution:
A g âB g +C g +D g
Î±= Iintial vapour densityâVapour density at equilibrium moles of gaseous productÃvapour density at equilbrium
1 10 = Vapour densityâ50 3â1 50
Vapour density=60
Molar mass of compound A=2ÃVD=2Ã60=120 gm/mol

Topic: Chemistry, Chemical Equilibrium, NEET
An unknown element forms an oxide. What will be the equivalent weight of the element if the oxygen content is 20% by weight

Options:
A 14
B 32
C 2
D 54

The correct answer is:
32

Solution:
Let the oxide of unknown element be MO.
Given that oxygen content is 20% by weight.
Then, Eq. weight of unknown element (M) =8020×8 g
Equivalent weight of unknown element (M) =32 g .

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
An unknown gas having density 4 k gm-3 and pressure 1.2×105 Nm-2 . The ratio of root mean square and most probable velocity is

Options:
A 1
B 1.22
C 1.12
D Data are not sufficient

The correct answer is:
1.22

Solution:
Ratio of vrms and vmp is approx 1.22 at any pressure and given density
vrms=3Pd=300 m sec-1
v mp = 2P d =245 m sec −1
vrmsvmp=1.22

Topic: Chemistry, States of Matter, NEET
An unknown resistance R1 is connected in series with a resistance of 10 Î©. This combination is connected to one gap of the meter bridge while a resistance R2 is connected in the other gap. The balance point is at 50 cm, Now, when the 10 Î© resistance is removed the balance point shifts to 40 cm. The value of R1(in ohm) is

Options:
A 20  Ω
B 10  Ω
C 60  Ω
D 40  Ω

The correct answer is:
20  Ω

Solution:
The balance condition of a meter bridge experiment
RS=l1100-l1
Here, R=R1, S=R2
â´ R1R2=l1100-l1
case-I
R1+10R2=5050 ...(i)
â R1+10=R2
case-II
R1R2=4060
â R2=6040R1 ...(ii)
So, Eqs. (i) and (ii) give
R1+10=6040R1
â R 1 =20ââÎ©

Topic: Physics, Current Electricity, NEET
An unknown tuning fork when sounded together with a tuning fork of frequency 256 Hz emits two beats. On loading the tuning fork of frequency 256 Hz, the number of beats heard is 1 per second. The frequency (in Hz) of the unknown tuning fork is

Options:
A 257
B 258
C 256
D 254

The correct answer is:
254

Solution:
nA= Known frequency =256 Hz, nB=?
x=2 bps, which is decreasing after loading i.e. x↓
Known tuning fork is loaded so nA↓
Hence nA↓-nB=x↓ .....(i) Correct
nB-nA↓=x↓ .....(ii) Wrong
⇒ nB=nA-x=256-2=254 Hz

Topic: Physics, Waves and Sound, NEET
An unpolarised beam of intensity I0 is incident on a pair of nicol prisms making an angle of 60o with each other. The intensity of light emerging from the pair is

Options:
A I0
B I02
C I04
D I08

The correct answer is:
I08

Solution:
I02cos2⁡60o=I08

Topic: Physics, Wave Optics, NEET
An unpolarised beam of intensity I0 is incident on a pair of nicol prisms making an angle of 60°with each other. The intensity of light emerging from the pair is -

Options:
A I0
B I02
C I04
D I08

The correct answer is:
I08

Solution:
The intensity of light emerging from the first prism =I02
and intensity of light emerging from the second prism is =I02cos2⁡60o=I08 .

Topic: Physics, Wave Optics, NEET
An unpolarised light beam strikes a glass surface at Brewster's angle. Then

Options:
A The refracted light will be completely polarised.
B Both the reflected and refracted light will be completely polarised.
C The reflected light will be completely polarised but the refracted light will be partially polarised.
D The reflected light will be partially polarised.

The correct answer is:
The reflected light will be completely polarised but the refracted light will be partially polarised.

Solution:
According to Brewster's law, reflected rays are completely polarized and refracted rays are partially polarized.

Topic: Physics, Wave Optics, NEET, NEET 2024
An unsaturated hydrocarbon ‘A’ adds two molecules of H2and on reductive ozonolysis gives butane-1,4-dial, ethanal and propanone. Give the IUPAC name of ‘A’.

Options:
A 2-methylocta-2, 6-diene
B 2-methylocta-2,4-diene
C 1-methylocta-2, 6-diene
D 3-methylocta-2, 6-diene

The correct answer is:
2-methylocta-2, 6-diene

Solution:
Correct Option is : (A)
2-methylocta-2, 6-diene

Topic: Chemistry, Hydrocarbons, NEET
An unsaturated hydrocarbon X absorbs two hydrogen molecules on catalytic hydrogenation, and also gives following reaction
X → O 3 +Zn/ H 2 O A → [Ag ( NH 3 ) 2 ] + B (3-oxo-hexanedicarboxylic acid). X will be

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, NEET
An X-ray tube produces a continuous spectrum of radiation with its shortest wavelength of45Ã10-2Ã. The maximum energy of a photon in the radiation in eV is h=6.62Ã10-34J s,c=3Ã108m s-1

Options:
A 27,500
B 22,500
C 17,500
D 12,500

The correct answer is:
27,500

Solution:
E = hc λ
= 6.62 × 1 0 - 3 4 × 3 × 1 0 8 4 5 × 1 0 - 1 2
= 0.44 × 1 0 - 1 4 1.6 × 1 0 - 1 9
=0.275×105 eV = 27500 eV

Topic: Physics, Atomic Physics, NEET
An X-ray tube produces a continuous spectrum of radiation with its shortest wavelength of45Ã10-2 Ã. The maximum energy of a photon in the radiation in eV is h=6.62Ã10-34J s, c=3Ã108m s-1

Options:
A 27,500
B 22,500
C 17,500
D 12,500

The correct answer is:
27,500

Solution:
E = hc λ
= 6.62 × 1 0 - 3 4 × 3 × 1 0 8 4 5 × 1 0 - 1 2
= 0.44 × 1 0 - 1 4 1.6 × 1 0 - 1 9
=0.275×105 eV = 27500 eV

Topic: Physics, Atomic Physics, NEET
Anabolism is a/an

Options:
A Endergonic process
B Exergonic process
C Bidirectional process
D Destructive process

The correct answer is:
Endergonic process

Solution:
Anabolism or biosynthesis are endergonic reactions in which the product has more energy than the reactants. Thus, they require an input of energy.

Topic: Chemistry, Biomolecules, NEET
Anaerobic respiration of 1 mol of glucose produces

Options:
A 4CO2+4H2O
B 2C2H5OH+2CO2
C CH3CH2OH+CH3OH+CO2
D CH3OH+CO2+H2O

The correct answer is:
2C2H5OH+2CO2

Solution:
Anaerobic respiration of 1 mol of glucose produces 2 mol of C2H5OH and 2 mol of CO2 ; while aerobic respiration produces 4 mol of CO2 and 4 mol of H2O .

Topic: Chemistry, Biomolecules, NEET
Analogous organs arise due to:

Options:
A divergent evolution
B artificial selection
C genetic drift
D convergent evolution

The correct answer is:
convergent evolution

Solution:
Analogous structures are a result of convergent evolution - different structures evolving for the same function and hence having similarity.

Topic: Biology, Evolution, NEET
Analogous structures are a result of:

Options:
A Divergent evolution
B Convergent evolution
C Shared ancestry
D Stabilizing selection

The correct answer is:
Convergent evolution

Solution:
Analogous structures are a result of convergent evolution. Convergent evolution is the process whereby organisms not closely related (not monophyletic), independently evolve similar traits as a result of having to adapt to similar environments or ecological niches.

Topic: Biology, Evolution, NEET, NEET 2016 (Phase 1)
Analysis shows that nickel oxide consists of nickel ion with 96% ions having d8 configuration and 4% having d7 configuration. Which amongst the following best represents the formula of the oxide?

Options:
A Ni1.02O1.00
B Ni0.96O1.00
C Ni0.98O0.98
D Ni0.98O1.00

The correct answer is:
Ni0.98O1.00

Solution:
Ni=3d84s2 (Atomic number = 28)
So Ni2+=3d8 and Ni3+=3d7
Hence, 96% ion of Ni2+ and 4% ions of Ni3+ are present.
Let number of O2- ion present in the crystal = x
Using electroneutrality rule,
Total positive charge = Total negative charge
96×2(+vecharge)+4×3(+vecharge)
=x×2(-vecharge)
So 96×2+4×3-2x=0
So x = 1.02
Hence formula of the crystal is Ni1.00O1.02 or Ni0.98O1.00 .

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Analysis shows that nickel oxide has the formula Ni₀.₉₈O₁.₀₀. What fractions of nickel exist as Ni²⁺ and Ni³⁺ ions?

Options:
A Ni²⁺ 80%, Ni³⁺ 20%
B Ni²⁺ 70%, Ni³⁺ 30%
C Ni²⁺ 60%, Ni³⁺ 40%
D Ni²⁺ 50%, Ni³⁺ 50%

The correct answer is:
Ni²⁺ 60%, Ni³⁺ 40%

Solution:
Correct Option is : (C)
Ni²⁺ 60%, Ni³⁺ 40%

Topic: Chemistry, Solid State, NEET
Analyze the action of restriction endonuclease in the given below diagram represented as P and Q.

Options:
A

P	Q

Sticky ends	Blunt ends

B

P	Q

Blunt ends	Sticky ends

C

P	Q

Blunt ends	Staggered ends

D

P	Q

Sticky ends	Cohesive ends

The correct answer is:

P	Q

Sticky ends	Blunt ends

Solution:
Restriction endonuclease is enzymes isolated from bacteria, are used to cut DNA at a specific location. They identify the specific palindromic regions in DNA at cuts at specific locations within the palindrome known as restriction sites by producing either sticky or blunt ends.

Topic: Biology, Biotechnology - Principles and Processes, NEET