datavorous/entrance-exam-dataset huggingface dataset:
In a monohybrid cross, what will be a percentage of heterozygous offspring in F3-generation?

Options:
A 25%
B 50%
C 12.5%
D 62.5%

The correct answer is:
62.5%

Solution:
A monohybrid cross has the genotypes completely dominant or completely recessive alleles (RR or rr). When these two genotypes are crossed,(RR x rr), the F1 generation will have all heterozygous plants (RrRr). In F2 generation when RrRr is crossed, Rr, RR and rr plants are formed in 2:1:1 ratio. In F3 generation when the cross was done between the plants with genotypes RR X Rr, Rr X Rr, Rr X rr, RR X rr, we will get RR, rr and Rr plants in 3:3:10 ratio. The total number of heterozygous gametes are 10; hence, the percentage of heterozygous offspring will be 10/16*100= 62.5%.

Topic: Biology, Principles of Inheritance and Variation, NEET
In a mono-Keto compound, generally which form of tautomeric structure is more stable than other?

Options:
A Keto form is more stable
B Enol form is more stable
C Equally stable
D Stability cannot be predicted.

The correct answer is:
Keto form is more stable

Solution:
Enol formation leads to decrease in stability for a mono keto structure. Hence keto form is more stable.

Topic: Chemistry, General Organic Chemistry, NEET
In a moss the sporophyte:

Options:
A manufactures food for itself, as well as for the gametophyte
B is partially parasitic on the gametophyte
C produces gametes that give rise to the gametophyte
D arises from a spore produced from the gametophyte

The correct answer is:
is partially parasitic on the gametophyte

Solution:
In mosses, the sporophyte developing from the embryo partially depends on the gametophyte. It contains chlorophyll and hence is capable of making its own food. Sporophyte is attached to the photosynthetic gametophyte so that it can get nourishment from the gametophyte. Some cells of the sporophyte undergo meiosis to produce haploid spores. The spores germinate to produce gametophyte.
Related Theory
When a moss spore germinates it first develops a protonema (sporophyte), a filamentous to sheet-like growth form. the leaves develop on the stems, giving rise to one or more leafy-stemmed plants. The protonema are ephemeral, with the leafy-stemmed plants the persistent and dominant growth form. The leafy-stemmed part is the gametophyte. because it produces gametes. Sex organs are multicellular. Male sex organ is called antheridium and produces biflagellate antherozoids. The female sex organ is called archegonium and produces a single egg.

Topic: Biology, Plant Kingdom, NEET, NEET 2006
In a mouse, duplication of centrioles in cytoplasm is completed in

Options:
A S-Phase
B G2 Phase
C G1 Phase
D both B and C

The correct answer is:
S-Phase

Solution:
The cell cycle is divided into two phases - Interphase and Mitosis phase or the M phase. The interphase is further divide into 3 phases - Gap 1 or G1 phase, Synthesis phase or S phase and Gap 2 or G2 phase. G1 phase is the interval between mitosis and initiation of DNA replication during which the cell is metabolically active and grows but does not replicate. The DNA replication begins in the S phase and the centrioles duplicate in the cytoplasm. In the G2 phase, proteins are synthesised in preparation for mitosis while cell growth continues.

Topic: Biology, Cell Cycle and Cell Division, NEET
In a muscle fibre Ca++ is stored in ___

Options:
A endoplasm
B sarcoplasmic reticulum
C mitochondria
D Chloroplasts

The correct answer is:
sarcoplasmic reticulum

Solution:
In a muscle fibre Ca++ is stored in sarcoplasmic reticulum

Topic: Biology, Locomotion and Movement, NEET
In a myofibril, the band which contains only actin filament is also known as

Options:
A Isotropic band
B Anisotropic band
C Inotropic band
D Dark band

The correct answer is:
Isotropic band

Solution:
A myofibril has two alternate dark and light bands on it. The band which contains only actin filament is also known as I band or Isotropic band or light band. The band which contains actin as well as myosin filament is called A band or Anisotropic band or dark band.

Topic: Biology, Locomotion and Movement, NEET
In a $n$-type semiconductor, which of the following statement is true?

Options:
A Electrons are majority carriers and trivalent atoms are dopants
B Electrons are minority carriers and pentavalent atoms are dopants
C Holes are minority carriers and pentavalentf atoms are dopants
D Holes are majority carriers and trivalent atoms are dopants

The correct answer is:
Holes are minority carriers and pentavalentf atoms are dopants

Solution:
The r-type sem-iconductor can be produced by doping an impurity atom of valence 5 i.e., pentavalent atoms. i.e., phosphorus.

Topic: Physics, Semiconductors, NEET, NEET 2013 (All India)
In a new system of units, the unit of force is 100 N, unit of length is 10 m and unit of time is 100 s. Unit of mass in this system can be expressed as

Options:
A 10 5 kg
B 10 6 kg
C 10 2 kg
D 10 3 kg

The correct answer is:
10 5 kg

Solution:
Here, force (F) = [MLTâ2] = 100 N ...(i)
Length (L) = [L] = 10 m ...(ii)
Time (t) = [T] = 100 s ...(iii)
Substituting values of L and T from Equations. (ii) and (iii) in Equation. (i), we get
M Ã 10 Ã (100)â2 = 100
or Mâ¡ Ã 1 0 1 0 0 Ã 1 0 0 = 1 0 0
or M = 100 Ã 1000 kg
M= 10 5 kg

Topic: Physics, Units and Dimensions, NEET
In a non-ideal solution, what type of plots of vapor pressure versus mole fractions are observed for solutions that exhibit positive deviations from Raoult's law?

Options:
A Linear
B Concave upward
C Concave downward
D Exponential

The correct answer is:
Concave upward

Solution:
In non-ideal solutions with positive deviations, the plots of vapor pressure versus mole fractions are typically concave upward.

Topic: Chemistry, Solutions, NEET
In a nonpolar covalent bond, where is the electron pair situated with respect to the two identical nuclei?

Options:
A Closer to one nucleus
B Equally between both nuclei
C Far from both nuclei
D Closer to the smaller nucleus

The correct answer is:
Equally between both nuclei

Solution:
In a nonpolar covalent bond, the electron pair is situated equally between both identical nuclei

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
In a normal spinel types structure, the oxide ions are arranged in ccp whereas 1/8 tetrahedral holes are occupied by Zn2+ions and 50% of octahedral holes are occupied by Fe3+ ions. The formula of the compound is -

Options:
A Zn2Fe2O4
B ZnFe2O3
C ZnFe2O4
D ZnFe2O2

The correct answer is:
ZnFe2O4

Solution:
In one unit cell no. of O2+=4
The no. of Zn2+=18× 8=1
The no. of Fe3+=12× 4=2
∴ molecular formula of given spinel structure is ZnFe2O4

Topic: Chemistry, Solid State, NEET
In a n-p-n transistor, the collector current is 10 mA. If 90% of the electrons emitted reach the collector, then the emitter current will be

Options:
A 9 mA

B 11 mA
C 1 mA
D 0.1 mA

The correct answer is:
11 mA

Solution:
In an n-p-n transistor emitter current (ie) is the sum of base current (ib) and collector current (ic).
Given, 90100ie=ic ...(i)
Also, ic=10 mA ...(ii)
From Eqs. (i) and (ii), we get
ie=10×10090=11 mA

Topic: Physics, Semiconductors, NEET
In a nuclear reactor, the fuel is consumed at the rate of 1 mg s-1. The power generated in kilowatt is

Options:
A 9Â ÃÂ 104
B 9 ×107
C 9Â Ã108
D 9Â Ã1012

The correct answer is:
9 ×107

Solution:
From Einstein's mass energy relation the energy released is
âE = âmc2
where âm is mass and c is speed is light.
Given
Δ m = 1mg = 1 × 1 0 - 6 kg, c = 3 × 1 0 8 m / s
∴ âE = 1 × 10-6 × 3 × 1082
âE = 9×1010 J
The rate at which energy is dissipated is known as power, ie,
P=â Et =9 ×10101=9×1010W
Since, 103 W = 1k W
∴ P= 9 × 107 kW

Topic: Physics, Nuclear Physics, NEET
In a nuclear reactor, the function of the moderator is to decrease

Options:
A Number of neutrons
B Speed of neutrons
C Escape of neutrons
D Temperature of the reactor

The correct answer is:
Speed of neutrons

Solution:
In a nuclear reactor, the function of the moderator is to decrease the speed of neutrons.

Topic: Physics, Nuclear Physics, NEET
In a nuclear reactor, U235 undergoes Fission liberating 200 MeV of energy. The reactor has a 10% efficiency and produces 1000 MW power. If the reactor is to function for 10 yr, find the total mass of uranium required.

Options:
A 38470
B 38490
C 48490
D 48470

The correct answer is:
38470

Solution:
The reactor produces 1000 MW power or 109J/s. The reactor is to function for 10 yr. Therefore, total energy which the reactor will supply in 10 yr is
E = (power) (time)
= (109J/s) (10 x 365 x 24 x 3600s)
= 3. 1536 x 1017 J
But since the efficiency of the reactor is only 10%, therefore actual energy needed is 10 times of it or 3. 1536 x 1018 J. One uranium atom liberates 200 MeV of energy or 200 x 1. 6 x 10-13 J or 3. 2 x 10-11 J of energy. So, number of uranium atoms needed are
3.1536×10183.2×10-11=0.9855×1029
or number of kg - moles of uranium needed are
n=0.9855×10296.02×1026=163.7
Hence, total mass of uranium required is
m = (n)M
or = (163. 7) (235) kg
or m = 38470 kg
or m = 38470 kg

Topic: Physics, Nuclear Physics, NEET
In a nucleophilic substitution reaction of alcohols, what is the role of the leaving group?

Options:
A To accept electrons
B To donate electrons
C To stabilise the transition state
D To form a covalent bond with the nucleophile

The correct answer is:
To form a covalent bond with the nucleophile

Solution:
In a nucleophilic substitution reaction of alcohols, the leaving group forms a covalent bond with the nucleophile as the hydroxyl group is replaced.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
In a nucleophilic substitution reaction, what is the leaving group that replaces the hydroxyl group (-OH) in alcohols?

Options:
A Hydrogen ion (H+)
B Alkoxide ion (RO-)
C Halide ion (X-)
D Water molecule (H2O)

The correct answer is:
Halide ion (X-)

Solution:
In a nucleophilic substitution reaction of alcohols, the hydroxyl group (-OH) is replaced by a halide ion (X-).

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
In a nucleophilic substitution reaction, what is the product when a cyanide ion (CN-) attacks through the carbon atom?

Options:
A Alkyl cyanide
B Alkyl nitrite
C Alkyl isocyanide
D Alkyl nitroalkane

The correct answer is:
Alkyl cyanide

Solution:
When CN- attacks through the carbon atom, it forms an alkyl cyanide.

Topic: Chemistry, Amines, NEET
In a nucleophilic substitution reaction, what is the product when a nitroalkane reacts with a nucleophile?

Options:
A Alkyl cyanide
B Alkyl nitrite
C Alkyl isocyanide
D Alkyl nitroalkane

The correct answer is:
Alkyl nitroalkane

Solution:
When a nitroalkane reacts with a nucleophile, it forms an alkyl nitroalkane.

Topic: Chemistry, Amines, NEET
In a nucleophilic substitution reaction, what type of compound is formed when a hydrocarbon reacts with a nucleophile?

Options:
A Alcohol
B Ether
C Alkane
D Alkyl chloride

The correct answer is:
Alkane

Solution:
When a hydrocarbon reacts with a nucleophile, it forms an alkane.

Topic: Chemistry, Amines, NEET
In a nucleophilic substitution reaction, which carbon atom in a haloalkane is often attacked by the nucleophile?

Options:
A The carbon bonded to hydrogen
B The carbon bonded to the halogen
C The carbon bonded to oxygen
D The carbon bonded to nitrogen

The correct answer is:
The carbon bonded to the halogen

Solution:
The carbon atom bonded to the halogen in a haloalkane is often attacked by the nucleophile.

Topic: Chemistry, Amines, NEET
In a one dimensional collision of two particles, velocities are interchanged when:
(i) Collision is elastic and masses are equal
(ii) Collision is inelastic but masses are unequal
Select the correct alternative.

Options:
A only (i) is correct
B only (ii) is correct
C both (i) and (ii) are correct
D both (i) and (ii) are wrong

The correct answer is:
only (i) is correct

Solution:
For elastic collision,
v2-v1=u1-u2 ……(i)

Momentum conservation,
m1u1+m2u2=m1v1+m2u2
For equal masses,
u1+u2=v1+v2 ……(ii)
Solving (i) & (ii) we get,
v1=u2 and v2=u1

Topic: Physics, Center of Mass Momentum and Collision, NEET
In a p-n junction diode, change in temperature due to heating

Options:
A does not affect resistance of p-n junction
B affects only forward resistance
C affects only reverse resistance
D affects the overall V-I characteristics of p-n junction

The correct answer is:
affects the overall V-I characteristics of p-n junction

Solution:
Due to heating, the number of electron-hole pairs will increase, so overall resistance of diode will change. Due to which forward and reverse biasing both are changed. Hence, in a PN junction diode, change in temperature due to heating affects the overall V-I characteristics of PN junction

Topic: Physics, Semiconductors, NEET, NEET 2018
In a p-n junction diode, change in temperature due to heating

Options:
A does not affect resistance of p-n junction
B affects only forward resistance
C affects only reverse resistance
D affects the overall V-I characteristics of p-n junction

The correct answer is:
affects the overall V-I characteristics of p-n junction

Solution:
Due to heating, the number of electron-hole pairs will increase, so overall resistance of diode will change. Due to which forward and reverse biasing both are changed. Hence, in a PN junction diode, change in temperature due to heating affects the overall V-I characteristics of PN junction

Topic: Physics, Semiconductors, NEET
In a $p-n$ junction photocell, the value of the photo electromotive force by monochromatic light is proportional to:

Options:
A the barrier voltage at the $p-n$ junction
B the intensity of the light falling on the cell
C the frequency of the light falling on the cell
D the voltage applied at the $p-n$ junction

The correct answer is:
the intensity of the light falling on the cell

Solution:
In a p-n function photo cell, the value of the photo electromotive force produced by monochromatic light is proportional to the intensity of the light falling on the cell.

Topic: Physics, Semiconductors, NEET, NEET 2004
In a $p-n$ junction:

Options:
A high potential is at $n$-side and low potential at $p$-side
B high potential is at $p$-side and low potential at $n$-side
C $p$ and both are at same potential
D undetermined

The correct answer is:
high potential is at $n$-side and low potential at $p$-side

Solution:
In a p-n junction high potential is at $n$-side and law potential at $p$-side.

Topic: Physics, Semiconductors, NEET, NEET 2002
In a particular geographical area, if leopards and lions feed on similar prey, they may show __________.

Options:
A interspecific struggle
B intraspecific struggle
C commensalism
D mutualism

The correct answer is:
interspecific struggle

Solution:
When organisms of different species compete for same resources, such struggle is called interspecific struggle. Leopard and lions show interspecific struggle as they have similar prey. The species with a lower growth rate and reproducing ability will not survive.

Topic: Biology, Organisms and Populations, NEET
In a particular isomer of $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]^0$, the $\mathrm{Cl}-\mathrm{Co}-\mathrm{Cl}$ angle is $90^{\circ}$, the isomer is known as:

Options:
A optical isomer
B cis-isomer
C position isomer
D linkage isomer

The correct answer is:
cis-isomer

Solution:
The structure is shown below:

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2013 (Karnataka)
In a particular region, the same fruit is available in multiple varieties, colours, with different tastes and textures. The primary reason for this could be:

Options:
A Species diversity
B Ecological diversity
C Genetic diversity
D Hybridization

The correct answer is:
Genetic diversity

Solution:
Genetic diversity: A single species might show high diversity at the genetic level over its distributional range. The genetic variation shown by the medicinal plant Rauwolfia vomitoria growing in different Himalayan ranges might be in terms of the potency and concentration of the active chemical (reserpine) that the plant produces. India has more than 50,000 genetically different strains of rice, and 1,000 varieties of mango.

Topic: Biology, Biodiversity and its Conservation, NEET
In a particular system, the unit of length, mass and time are chosen to be 10 cm, 10 g and 0.1 s respectively. The unit of force in this system will be equivalent to -

Options:
A 110 N
B 1 N
C 10 N
D 100 N

The correct answer is:
110 N

Solution:
Dimensionally
F=MLT-2
In C.G.S. system
1 dyne=1g 1 cm 1s-2
In the new system
1x=10 g10 cm0.1 s-2
1 dyne1 x=1 g10 gÃ1 cm10 cm10 s1 s-2
1 dyne=110,000Ã1x
104 dyne=1x
10x=105 dyne=1 N
x=110 N

Topic: Physics, Units and Dimensions, NEET
In a pea plant, the genotype Rr will be expressed as

Options:
A round seed with large starch grain
B round seed with intermediate starch grain
C round seed with small starch grain
D wrinkled seed with small starch grain

The correct answer is:
round seed with intermediate starch grain

Solution:
The gene controlling starch synthesis in pea shows a small degree of pleiotropy. Starch synthesis in pea seeds is controlled by one gene. It has two alleles B and b. Starch is synthesized effectively by BB homozygotes and therefore, large starch grains are produced. In contrast, bb homozygotes have lesser efficiency in starch synthesis and produce small starch grains. After maturation of the seeds, BB seeds are round and the bb seeds are wrinkled. Heterozygotes Bb form round seeds but the starch grains are of intermediates size. It starch grain size is considered, Bb seed shows incomplete dominance but if seed shape is considered, B allele is dominant and b is recessive.

Topic: Biology, Principles of Inheritance and Variation, NEET
In a peptide bond, which atoms are directly involved in the bond formation?

Options:
A Hydrogen and oxygen
B Carbon and nitrogen
C Carbon and oxygen
D Nitrogen and hydrogen

The correct answer is:
Carbon and nitrogen

Solution:
The peptide bond forms between the carbon atom of the carboxyl group and the nitrogen atom of the amino group in amino acids.

Topic: Chemistry, Biomolecules, NEET
In a period, atom with smallest radius generally is:

Options:
A Chalcogen
B Halogen
C Aerogen
D Pnicogen

The correct answer is:
Halogen

Solution:
In a period on moving left to right Zeffective increases, so atomic size decreases.
For inert gases, Van der waal radius is considered, so it is bigger in size as compared to halogens.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
In a photoelectric effect experiment, the slope of the graph between the stopping potential and the incident frequency will be

Options:
A 10-14
B 5 × 10-14
C 4 × 10-15
D 4 × 10-34

The correct answer is:
4 × 10-15

Solution:
When light waves of differing frequencies are made incident on the same surface, the stopping potential (Vs) differ as is evidenced from Einstein's photoelectric equation given by
hv-hv0=eVs
where h is Planck's constant, v,v0 the frequencies and e the charge.
Hence, Vs=hve-hve ...(i)
Comparing Eq. (i) with straight line equation
y=mx+c
Slope m=he
Where h=6.6Ã10-34 Js, e=1.6Ã10-19 C.
â´ m=6.6Ã10-341.6Ã10-19=4.125Ã10-15

Topic: Physics, Dual Nature of Matter, NEET
In a photoelectric experiment, the stopping potential Vs is plotted against the frequency v of the incident light. The resulting curve is a straight line which makes an angle θ with the v-axis. Then tanθ will be equal to
(Here E 0 = work function of the surface)

Options:
A he
B eh
C -ϕe
D ehϕ

The correct answer is:
he

Solution:
EK=hv-ϕ
eVs=hv-ϕ
Vs=hve-ϕe …(i)
y=mx+C …(ii)
Comparison of equation (i) and (ii) shows that the slope i.e., tanθ will be equal to he .
m(Slope)=tanθ=he .

Topic: Chemistry, Structure of Atom, NEET
In a photoelectric experiment under conditions of saturation current, we consider two different cross-section A and B as shown. Here, vA and vB represent the average speed of electrons at the two cross-sections.

Options:
A iA>iB
B vA=vB
C iA=iB
D vA>vB

The correct answer is:
iA=iB

Solution:
Current at any cross-section is equal to the number of electrons coming out of cathode and this is also equal to the number of electrons entering each area per unit time.

Also vdA=vdB, i.e, drift speeds are equal.

Topic: Physics, Dual Nature of Matter, NEET
In a photoemissive cell with executing wavelength Î», the fastest electron has speed v. If the exciting wavelength is changed to 3Î»4 , the speed of fastest emitted electron will be

Options:
A Greater than v4312
B v3412
C 4312
D Less than v4312

The correct answer is:
Greater than v4312

Solution:
According to photoelectric equation,
hv-Ï0=12Â mvmax2
hcÎ»-hcÎ»0=12Â Â mvmax2
hcÎ»0-Î»Â Î»⋅Î»0=12mvmax2
vmax=2hcmÂ Î»0-Î»Â Î»⋅Î»0
When wavelength is Î» and velocity v, then
v=2hcmÂ Î»0-Î»Â Î»⋅Î»0 ..... (i)
When wavelength is 3Î»4 and velocity is v′ then
v′=2hcmÎ»0-3Î»43Î»4ÃÎ»0 ...... (ii
Divide equation (ii) by equation (i), we get
v′v=Î»0-3Î»43Î»Î»04ÃÎ»Î»0Î»0-Î»
v′=v4312Î»0-3Î»4Î»Î»0=v′>v4312

Topic: Physics, Dual Nature of Matter, NEET
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.5Ã1010 Hz and amplitude 480 V m-1. The amplitude of the oscillating magnetic field will be

Options:
A 1.52×10-8 Wb m-2
B 1.52×10-7 Wb m-2
C 1.6×10-6 Wb m-2
D 1.6×10-7 Wb m-2

The correct answer is:
1.6×10-6 Wb m-2

Solution:
If c is the speed of light or electromagnetic wave then the amplitude of electric and magnetic fields are related as,
B0=E0c
B0=4803Ã108=1.6Ã10-6 Wb m-2

Topic: Physics, Electromagnetic Waves, NEET
In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of 2.0×1010 Hz and amplitude 48 V m-1. Then the amplitude of oscillating magnetic field is: (Speed of light in free space = 3×108 m s-1)

Options:
A 1.6×10-6 T
B 1.6×10-9 T
C 1.6×10-8 T
D 1.6×10-7 T

The correct answer is:
1.6×10-7 T

Solution:
Frequency of oscillation =2×1010 Hz
Given c=3×108 m s-1,
Electric field amplitude E0=48 V m-1
Using the formula, c=E0B0
The amplitude of the oscillating magnetic field
B⁡0=E⁡0c=483×108=1.6×10-7 T.

Topic: Physics, Electromagnetic Waves, NEET, NEET 2023 (All India)
In a plant, black seed color (BB/Bb) is dominant over white seed color (bb). In order to find out the genotype of the black seed plant, with which of the following genotype will you cross it?

Options:
A bb
B Bb
C $\mathrm{BB} / \mathrm{Bb}$
D $B B$

The correct answer is:
bb

Solution:
To determine the genotype of a black seed colour at $F_2$, the black seed from $F_2$ is crossed with the white seed colour. This is called a test cross.
$\therefore$ To determine the genotype of $(\mathrm{BB} / \mathrm{Bb})$ black seed we need to cross them with white seed i.e. bb.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2024
In a plant, red fruit R dominant over yellow fruit r and tallness T is dominant over shortness t. If a plant with RRTT genotype is crossed with a plant that is rrtt. Then

Options:
A 25% will be tall with red fruit
B 50% will be tall with red fruit
C 75% will be tall with red fruit
D All of the offspring will be tall with red fruits

The correct answer is:
All of the offspring will be tall with red fruits

Solution:
A plant with RRTT genotype is crossed with a plant that is rrtt. The F1 generation will have all offspring tall with red fruits.
As

Topic: Biology, Principles of Inheritance and Variation, NEET
In a plant, red fruit $(R)$ is dominant over yellow fruit $(r)$ and tallness $(T)$ is dominant over shortness $(t)$. If a plant with RRTt genotype is crossed with a plant that is rrtt,

Options:
A $25 \%$ will be tall with red fruit
B $50 \%$ will be tall with red fruit
C $75 \%$ will be tall with red fruit
D all the offspring will be tall with red fruit.

The correct answer is:
$50 \%$ will be tall with red fruit

Solution:
Conclusion:
1. All plants are red.
2. 50 are red tall.
3. $50 \%$ are red dwarf.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2004