datavorous/entrance-exam-dataset huggingface dataset:
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ defined by $\mathrm{f}(\mathrm{x})=3-4 \mathrm{x}$
(ii) $f: R \rightarrow R$ defined by $f(x)=1+x^2$

Options:

None

Solution:
(i) $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ defined by $3-4 \mathrm{x}$, $\mathrm{f}\left(\mathrm{x}_1\right)=3-4 \mathrm{x}_1, \mathrm{f}\left(\mathrm{x}_2\right)=3-4 \mathrm{x}_2$
(a) $f\left(x_1\right)=f\left(x_2\right) \Rightarrow 3-4 x_1=3-4 x_2$
$\Rightarrow x_1=x_2$. This shows that $f$ is one-one
(b) $f(x)=y=3-4 x \quad \therefore x=\frac{3-y}{4}$
For every value of y belonging to its codomain. There is a pre-image in its domain $\Rightarrow$ fis onto. Hence, $f$ is one-one onto
(ii) $f: R \rightarrow R$ given by $f(x)=1+x^2$
(a) $\mathrm{f}(1)=1+1=2, \mathrm{f}(-1)=1+1=2$
$\therefore \mathrm{f}(-1)=\mathrm{f}(1)=2$ i.e. $-1$ and 1 have the same image 2 . $\Rightarrow \mathrm{f}$ is not one-one.
(b) No negative number belonging to its codomain has its pre-image in its domain
$\Rightarrow \mathrm{f}$ is not onto. Thus $\mathrm{f}$ is neither one-one nor onto.

Topic: Mathematics, Relations and Functions (Class 12)
In each of the following pairs of ions in which I ion is more stable than II

Options:
A
B
C
D

The correct answer is:

Solution:
(A) I: Benzyl carbocation is more stable than allyl carbocation
(B) II: Alkyl carbocation is more stable than vinyl carbocation
(C) II: Allyl carbocation is more stable than alkyl carbocation
(D) II: Hetero atom present on α-carbon stabilish the carbocation

Topic: Chemistry, General Organic Chemistry, JEE Main
In electrolysis of aqueous copper sulphate, the gas at anode and cathode is

Options:
A $\mathrm{O}_2$ and $\mathrm{H}_2$
B $\mathrm{SO}_2$ and $\mathrm{H}_2$
C $\mathrm{H}_2$ and $\mathrm{O}_2$
D $\mathrm{SO}_3$ and $\mathrm{O}_2$

The correct answer is:
$\mathrm{O}_2$ and $\mathrm{H}_2$

Solution:
Cathode: $2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2+2 \mathrm{OH}^{-}$
Anode: $\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}^{+}+\frac{1}{2} \mathrm{O}_2+2 \mathrm{e}^{-}$

Topic: Chemistry, Electrochemistry, JEE Main
In electrolytic refining of metals, electrolysis of an aqueous solution of its complex salt is done with impure metal as anode and an strip of pure metal as cathode. This method cannot be used for the refining of the metal

Options:
A Silver
B Copper
C Aluminium
D Zinc

The correct answer is:
Aluminium

Solution:
Correct Option is 3 Aluminium
Aluminium can be extracted by Hall's process.
The ore is first converted into pure aluminium oxide by the Bayer Process, and this is then electrolysed in solution in molten cryolite - another aluminium compound.
The aluminium oxide has too high a melting point to electrolyse on its own. The usual aluminium ore is bauxite.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
In electrolytic refining, the impure metal is made is used to make

Options:
A Cathode
B Anode
C Electrolytic bath
D None of these

The correct answer is:
Anode

Solution:
Impure metal as anode and pure metal as cathode.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, E and B stand for dimensions of electric and magnetic fields respectively, while Ïµ0 and Î¼0 stand for dimensions of the permittivity and permeability of free space respectively. L and T are dimensions of length and time respectively. All the quantities are given in SI units.
The relation between E and B is

Options:
A E=BLT
B E=BL-1T
C E=BLT-1
D E=BL-1T-1

The correct answer is:
E=BLT-1

Solution:
We have EC=B
â´ B=EC=EL-1T1
â E=BLT-1

Topic: Physics, Units and Dimensions, JEE Main
In Engelmann's experiment, bacteria were used to

Options:
A split light into its spectral components.
B detect the sites of O2 evolution.
C absorb the spectra of chlorophyll a and b using Cladophora.
D detect the sites of CO2 evolution and monitor the algal photosynthesis.

The correct answer is:
detect the sites of O2 evolution.

Solution:
In his experiment, T.W Engelmann (1843 â 1909) split light into its spectral components using a prism and then illuminated a green alga, Cladophora, placed in a suspension of aerobic bacteria. The bacteria were used to detect the sites of O2 evolution. He observed that the bacteria accumulated mainly in the region of blue and red light of the split spectrum. A first action spectrum of photosynthesis was thus described. It resembles roughly the absorption spectra of chlorophyll a and b.

Topic: Biology, Photosynthesis in Higher Plants, NEET
In ethane and cyclohexane which one of the following pairs of conformations are more stable

Options:
A Eclipsed and chair conformations
B Staggered and chair conformations
C Staggered and boat conformations
D Eclipsed and boat conformations

The correct answer is:
Staggered and chair conformations

Solution:
In ethane staggard form and in cyclohexane chair form is more stable.

Topic: Chemistry, General Organic Chemistry, JEE Main
In eukaryotic cell transcription, RNA splicing and RNA capping take place inside the

Options:
A ribosomes
B nucleus
C dictyosomes
D ER

The correct answer is:
nucleus

Solution:
Transcription involves transfer of genetic information from DNA to functional mRNA. In eukaryotes, the modification of newly formed primary mRNA transcripts (pre-mRNA) to produce functional $m$ RNA is called RNA processing or post transcriptional modification. It occurs in the nucleus and involves gene splicing (ie, removal of introns from primary transcript and splicing together of exons) and capping. Capping involves formation of a cap at $5^{\prime}$ end by condensation of guanylate residues. The fully processed mRNA molecule is exported from the nucleus to the ribosomes for translation in the cytoplasm.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2010 (Mains)
In $\mathrm{Fe}(\mathrm{CO})_5$, the $\mathrm{Fe}-\mathrm{C}$ bond possesses

Options:
A
$\pi$-character only

B
both $\sigma$ and $\pi$ characters

C
ionic character

D
$\sigma$-character only

The correct answer is:
both $\sigma$ and $\pi$ characters

Solution:
No Solution Available

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2006
In fifth group, $\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3$ is added to precipitate out the carbonates. We do not add $\mathrm{Na}_2 \mathrm{CO}_3$ because

Options:
A $\mathrm{CaCO}_3$ is soluble in $\mathrm{Na}_2 \mathrm{CO}_3$
B $\mathrm{Na}_2 \mathrm{CO}_3$ increases the solubility of fifth group carbonates
C $\mathrm{MgCO}_3$ will be precipitated out in fifth group
D None of these

The correct answer is:
$\mathrm{MgCO}_3$ will be precipitated out in fifth group

Solution:
As $\mathrm{Na}_2 \mathrm{CO}_3$ is a strong electrolyte that will cause the precipitation of $\mathrm{MgCO}_3$. So $\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3$ is used as it's a weaker electrolyte than $\mathrm{Na}_2 \mathrm{CO}_3$

Topic: Chemistry, Practical Chemistry, JEE Main
In figure, a particle is placed at the highest point A of a smooth sphere of radius r. It is given slight push, and it leaves the sphere at B, at a depth h vertically below A. The value of h is

Options:
A r6
B 14r
C 13r
D 12r

The correct answer is:
13r

Solution:
If v velocity acquired at B, then
v2=2gh
The particle will leave the sphere at B, when
mv 2 r = mgcos θ
2 gh r = g . r - h r
Which gives h=r3

Topic: Physics, Laws of Motion, JEE Main
In figure, a straight line is given for Freundrich Adsorption (y=3x+2.505). The value of 1n and logK are respectively.

Options:
A 0.3 and log 2.505
B 0.3 and 0.7033
C 3 and 2.505
D 3 and 0.7033

The correct answer is:
3 and 2.505

Solution:
The expression for the Freundlich isotherm can be represented by the following equation

xm=KpP1/n

Taking log on both sides

logxm=logk+1nlogP

On comparing with given equation of straight line, we get

Y=3x+2.505, 1n=3, logK=2.505

So, the answer is 3 and 2.505

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2023 (01 Feb Shift 2)
In figure, the optical fiber is l=2 m long and has a diameter of d=20 μm. If a ray of light is incident on one end of the fiber at angle θ1=40° , the number of reflections it makes before emerging from the other end is close to:
(refractive index of fiber is 1.31 , sin 40°=0.64 and sin-1⁡0.49=30° .)

Options:
A 55000
B 57000
C 45000
D 66000

The correct answer is:
57000

Solution:
Let n1 and n2 be the refractive indices of the air and optical fibre respectively. Applying Snell's law at Q, we get:

n1sinθ1=n2sinθ2

sin40∘=1.31×sinθ2

⇒sinθ2=0.641.31=0.4885≈0.5

⇒θ2=30∘

The distance covered along the length of the optical fibre after one reflection:

QN=RN cotθ2

=20×10-6×cot30∘=203×10-6 m

The reflections take place throughout the length of the optical fibre.

Length of the wire=2 m

Total number of reflections along its length=Length of the wiredistance covered in one reflection along the length

=2×106203=1000003=57735.02≈57000

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (08 Apr Shift 1)
In first order reaction, the concentration of the reactant decreases from $0.8 \mathrm{M}$ to $0.4 \mathrm{M}$ in 15 minutes. The time taken for the concentration to change from $0.1 \mathrm{M}$ to $0.025 \mathrm{M}$ is

Options:
A
30 minutes

B
60 minutes

C
$7.5$ minutes

D
15 minutes

The correct answer is:
30 minutes

Solution:
$\mathrm{t}_{1 / 2}=15$ minutes
$\therefore$ No. of half lives $s=2$
( $\therefore$ for change of $0.1$ to $0.025$ ) is 30 minutes

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2004
In first transition series, the melting point of $M n$ is low because

Options:
A Due to $d^{10}$ configuration, metallic bonds are strong
B Due to $d^7$ configuration, metallic bonds are weak
C Due to $d^5$ configuration, metallic bonds are weak
D None of these

The correct answer is:
Due to $d^5$ configuration, metallic bonds are weak

Solution:
Due to $d^5$ configuration, metallic bonds are weak. $d^5$ orbital is half filled as a result $3 d$ electrons are more tightly held by the nucleus and this reduces the de-localization of electrons resulting in weaker metallic bonding

Topic: Chemistry, d and f Block Elements, JEE Main
In five kingdom system, the main basis of classification is:

Options:
A structure of nucleus
B nutrition
C structure of cell wall
D asexual reproduction

The correct answer is:
nutrition

Solution:
The main criteria of the five kingdom classification were cell structure, body organisation, mode of nutrition and reproduction, and phylogenetic relationships.
Related Theory
The animal kingdom is classified on the basis of levels of organisation, body symmetry, germinal layers, nature of coelom, segmentation and presence of notochord.

Topic: Biology, Biological Classification, NEET, NEET 2002
In flowering plants, haustorial cell is

Options:
A The cell of suspensor present at the micropylar end and formed during embryogenesis.
B The cell of suspensor present at the chalazal end and formed during embryogenesis.
C The cell of suspensor present at the micropylar end and formed during the formation of the endosperm.
D The cell of suspensor present at the chalazal end and formed during the formation of the endosperm.

The correct answer is:
The cell of suspensor present at the micropylar end and formed during embryogenesis.

Solution:
Zygote of an angiospermic embryo divides first to form two cells. One towards the micropyle is called basal cell and the other towards the chalazal end is called embryonal cell. The basal cell divides by transverse divisions to form a 6–10 celled long filament-like structure called suspensor. The main function of suspensor is to push the developing embryo into food laden endosperm to provide nutrition. The micropylar cell of the suspensor swells up and is known as haustorial-cell.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In following pairs, the one in which both transition metal ions are colourless is

Options:
A Sc3+,Zn2+
B Ti4+,Cu2+
C V2+,Ti3+
D Zn2+,Mn2+

The correct answer is:
Sc3+,Zn2+

Solution:
d-block elements form coloured ions due to presence of unpaired electrons in them. since there are no unpaired electrons in Sc3+ and Zn2+ ions, they are colourless.

No of unpaired e-

Ti3+-[Ar]3 d1,Cu2+-[Ar]3 d9

Sc3+-[Ar]3 d0,Zn2+-[Ar]3 d10

Ti4+-[Ar]3 d0 V2+-[Ar]3 d3 ,

No d-d transitions in ions with d0 & d10 configuration. Therefore they are colourless.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (29 Jul Shift 1)
In following reaction
$\mathrm{yMnO}_4^{-}+x \mathrm{H}^{+}+\mathrm{C}_2 \mathrm{O}_4^{-} \rightarrow y \mathrm{Mn}^{++}+2 \mathrm{CO}_2+\frac{x}{2} \mathrm{H}_2 \mathrm{O}$ $x$ and $y$ are

Options:
A 2 and 16
B 16 and 2
C 8 and 16
D 5 and 2

The correct answer is:
16 and 2

Solution:
$2 \mathrm{MnO}_4^{-}+16 \mathrm{H}^{+}+\mathrm{C}_2 \mathrm{O}_4^{--} \rightarrow 2 \mathrm{Mn}^{+2}+2 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, d and f Block Elements, JEE Main
In forced oscillation of a particle the amplitude is maximum for a frequency $\omega_1$ of the force, while the energy is maximum for a frequency $\omega_2$ of the force, then

Options:
A
$\omega_1=\omega_2$

B
$\omega_1>\omega_2$

C
$\omega_1 < \omega_2$ when damping is small and $\omega_1>\omega_2$ when damping is large
D
$\omega_1 < \omega_2$

The correct answer is:
$\omega_1=\omega_2$

Solution:
Both amplitude and energy get maximised when the frequency is equal to the natural frequency. This is the condition of resonance.
$$
\omega_1=\omega_2
$$

Topic: Physics, Oscillations, JEE Main, JEE Main 2004
In Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at 10.2 V. The wavelength of light emitted by hydrogen atom when excited to the first excitation level is _________ $\mathrm{nm}$. (Given $\mathrm{hc}=1245 \mathrm{eV} \mathrm{nm}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ ).

Options:

The correct answer is:
122

Solution:
$\begin{aligned} & 10.2 \mathrm{eV}=\frac{\mathrm{hc}}{\lambda} \\ & \lambda=\frac{1245 \mathrm{eV}-\mathrm{nm}}{10.2 \mathrm{eV}}=122.06 \mathrm{~nm}\end{aligned}$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2024 (06 Apr Shift 2)
In Fraunhofer diffraction experiment, L is the distance between screen and the obstacle, b is the size of obstacle and λ is wavelength of incident light. The general condition for the applicability of Fraunhofer diffraction is

Options:
A b 2 Lλ ≫1
B b 2 Lλ =1
C b 2 Lλ ≪1
D b 2 Lλ ≠1

The correct answer is:
b 2 Lλ ≪1

Solution:
The general condition for Fraunhofer diffraction isb2LÎ»&Lt;1.

Topic: Physics, Wave Optics, JEE Main
In free space, a particle A of charge 1 μC is held fixed at point P . Another particle B of the same charge and mass 4 μg is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P is:
[Take 14πϵ0=9×109 N m2 C-2 ]

Options:
A 1.0 m s-1
B 1.5×102 m s-1
C 2.0×103 m s-1
D 3.0×104 m s-1

The correct answer is:
2.0×103 m s-1

Solution:
From conservation in mechanical energy:
-ΔP.E.=ΔK.E.
14πε01×1062110-3-19×10-3=12×4×10-6×V2
9×109×10-1289×10-3=2×10-6×V2
V2=4×106
V=2×103 m s-1

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (10 Apr Shift 2)
In frequency modulated wave

Options:
A Frequency varies with time
B Both frequency and amplitude vary with time
C Amplitude varies with time
D Both frequency and amplitude are constant

The correct answer is:
Frequency varies with time

Solution:
In frequency modulated wave, only frequency varies with time, while the amplitude remains constant. This wave can be shown as

It is used in FM broadcasting, telemetry and RADAR etc.

Topic: Physics, Communication System, MHT CET, MHT CET 2019 (Shift 2)
In frequency modulation

Options:
A The amplitude of the modulated wave varies as the frequency of the carrier wave.
B The frequency of the modulated wave varies as the amplitude of the modulating wave.
C The amplitude of the modulated wave varies as the amplitude of the carrier wave.
D The frequency of the modulated wave varies as the frequency of the modulating wave.

The correct answer is:
The frequency of the modulated wave varies as the frequency of the modulating wave.

Solution:
When the frequency of the carrier wave (modulated wave) is changed for the traveling of the signal wave. This is due to the low frequency of the signal or modulating wave. Hence the frequency of the modulated wave varies as the frequency of the modulating wave.
Hence the correct option is 4.

Topic: Physics, Communication System, JEE Main
In Freundlich adsorption, isotherm adsorption is proportional to pressure $P$ as

Options:
A $P^0$
B $P$
C $P^n$
D $P^{1 / n}$

The correct answer is:
$P^{1 / n}$

Solution:
$\log x / m=\log k+\frac{1}{n} \log p$; this is Freundlich isotherm.
Thus $\mathrm{p} \propto \frac{1}{n}$.

Topic: Chemistry, Surface Chemistry, JEE Main
In Friedel-Crafts alkylation of aniline, one gets:

Options:
A alkylated product with ortho and para substitution.
B secondary amine after acidic treatment.
C an amide product.
D positively charged nitrogen at benzene ring.

The correct answer is:
positively charged nitrogen at benzene ring.

Solution:
Aniline does not undergo Friedel-crafts reaction due to salt formation with aluminium chloride, the Lewis acid, which is used as a catalyst, whereas aniline is a strong base. So, an acid-base reaction will take place. Thus, aniline reacts with AlCl3 to form a salt.
Due to the positive charge on the nitrogen atom (shows -I effect and it is meta directing), electrophilic substitution in the benzene ring is deactivated. Hence, the Friedel-Crafts reaction will not take place in aniline.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2022 (29 Jun Shift 2)
In Gallilean telescope, the final image formed is

Options:
A Real, erect and enlarged
B Virtual, erect and enlarged
C Real, inverted and enlarged
D Virtual, inverted and enlarged

The correct answer is:
Virtual, erect and enlarged

Solution:
In a Galilean telescope a biconvex lens is used as a objective and a biconcave lens is used as a eye lens. The final image formed is a virtual, erect and enlarged.

Topic: Physics, Ray Optics, JEE Main
In galvanic cell, the salt bridge is used to

Options:
A Complete the circuit
B Reduce the electric resistance in the cell
C Separate cathode from anode
D Carry salts for the chemical reaction

The correct answer is:
Complete the circuit

Solution:
In galvanic cell, the salt bridge used to complete the circuit

Topic: Chemistry, Electrochemistry, JEE Main
In gamma ray emission from a nucleus

Options:
A
both the neutron number and the proton number change

B
there is no change in the proton number and the neutron number.

C
only the neutron number changes

D
only the proton number changes

The correct answer is:
there is no change in the proton number and the neutron number.

Solution:
No Solution Available

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2007
In gaseous triethyl amine the "-C-N-C-" bond angle is ____degree.

Options:

The correct answer is:
108

Solution:
In gaseous triethyl amine the "-C-N-C-"" bond angle is 108 degree.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (27 Jul Shift 1)
In Gattermann reaction, a diazonium group is replaced by $\underline{X}$ using $\underline{Y}$. $\underline{X}$ and $\underline{Y}$ are
$\underline{X} \quad \underline{Y}$

Options:
A $\mathrm{Cl}^{\ominus} \quad \mathrm{Cu} / \mathrm{HCl}$
B $\mathrm{Cl}^{\oplus} \quad \mathrm{CuCl}_2 / \mathrm{HCl}$
C $\mathrm{Cl}^{\ominus} \quad \mathrm{CuCl}_2 / \mathrm{HCl}$
D $\mathrm{Cl}_2 \quad \mathrm{Cu}_2 \mathrm{O} / \mathrm{HCl}$

The correct answer is:
$\mathrm{Cl}^{\ominus} \quad \mathrm{Cu} / \mathrm{HCl}$

Solution:

Topic: Chemistry, Amines, JEE Main
In general in an alternating current circuit

Options:
A The average value of current is zero
B The average value of square of the current is zero
C Average power dissipation is zero
D The phase difference between voltage and current is zero

The correct answer is:
The average value of current is zero

Solution:
In general, in an alternating current circuit, the average value of the current is zero.

Topic: Physics, Alternating Current, JEE Main
In general, species diversity ............... as we move away from the equator towards the poles and with very few exceptions, tropics harbor â¦......â¦ species than temperate or polar areas.

Options:
A a-increases, b-less
B a-decreases, b-more
C a-increases, b-more
D a-decreases, b-less

The correct answer is:
a-decreases, b-more

Solution:
Biodiversity includes all the flora and fauna on Earth. All the varieties of organisms come under biodiversity.
The region near the middle of the Earth between the Tropic of Cancer and Tropic of Capricorn is considered as the tropic region. This region receives the most amount of solar energy which leads to better productivity and higher species diversity. Also, the seasonal variations in such regions are low and environment mainly remains the same supporting high species richness and promoting niche specialization. Species-area relationship show a rectangular hyperbolic function. With very few exceptions, tropics harbour more species than temperate or polar areas. A forest in a tropical region like Equador has up to ten as many species of amphibians as a forest of equal area in a temperate region like the midwest of the USA.
Temperate regions include the subtropics and polar circles. These regions faced glaciation during the ice age which was not observed in the tropic regions.
When the diversity in organisms at the species level is termed as species diversity. Species diversity decreases as we move towards the poles can be for various reasons as - temperature decreases and conditions become harsh, the amount of solar radiation decreases including intensity, reduction in resources and vegetation for survival.

Topic: Biology, Organisms and Populations, NEET
In general, the properties that decrease and increase down a group in the periodic table, respectively, are :

Options:
A Atomic radius and electronegativity
B Electronegativity and atomic radius
C Electronegativity and electron gain enthalpy
D Electron gain enthalpy and electronegativity

The correct answer is:
Electronegativity and atomic radius

Solution:
(i) Electronegativity: Top to bottom decrease.
(ii) Atomic Radius: generally Increases top to Bottom.
(iii) Electron affinity: generally top to bottom decreases.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2019 (09 Jan Shift 1)
In general, the property (magnitudes only) that shows an opposite trend in comparison to other properties across a period is :

Options:
A lonization enthalpy
B Electronegativity
C Electron gain enthalpy
D Atomic radius

The correct answer is:
Atomic radius

Solution:
On moving Left to Right along a period. Atomic Radius ⇒ decreases. Electronegativity ⇒Increases. Electron gainenthalpy ⇒ Increases. Ionisation Enthalpy ⇒ Increases.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2020 (02 Sep Shift 1)
In genetic engineering, a DNA segment (gene) of interest, is transferred to the host cell through a vector. Consider the following four agents (A-D) in this regard and select the correct OPTION about which one or more of these can be used as a vector/vectors.
A. A bacterium
B. Plasmid
C. Plasmodium
D. Bacteriophage

Options:
A A, B and D only
B A only
C A and C only
D B and D only

The correct answer is:
B and D only

Solution:
Plasmids are the extrachromosomal DNA found in bacterial cells. These are commonly used as vectors in genetic engineering programmes because these have the capacity of binding with eukaryotic DNA.
Bacteriophage is a virus that is parasitic within a bacterium. These also, are the important tools in genetic engineering as cloning vectors.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2010 (Mains)