datavorous/entrance-exam-dataset huggingface dataset:
Magnetism at the centre of Bar magnet is ______

Options:
A Minimum
B Zero
C Negative
D Maximum

The correct answer is:
Zero

Solution:
The magnetic field lines near the center of a bar magnet are parallel to the bar magnet and have the least density. If a plane is passed through the center of the magnet, there will be an equal number of dipoles on each side with opposite polarities, which cancel each other out and leave a net magnetic field of zero.

Topic: Physics, Magnetic Properties of Matter, JEE Main
Magnifying power of a simple microscope is (when final image is formed at $D=25 \mathrm{~cm}$ from eye)

Options:
A $\frac{D}{f}$
B $1+\frac{D}{f}$
C $1+\frac{f}{D}$
D $1-\frac{D}{f}$

The correct answer is:
$1+\frac{D}{f}$

Solution:
The correct option is 2.$1+\frac{D}{f}$
A Simple Microscope uses Convex lens for magnification. When the object is placed inside the focal point(f) of the lens, then virtual, erect and magnified image is formed at near point ( $\mathrm{D}=$ $25 \mathrm{~cm})$ from the eye.
The linear magnification $\mathrm{m}$, for the image formed at the near point $\mathrm{D}$, by a simple microscope can be obtained by using the relation : $\mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}$, where $\mathrm{v}$ is the image distance, $\mathrm{u}$ is the object distance from the lens.
Using Lens Formula, $\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}$
$m=v\left(\frac{1}{v}-\frac{1}{f}\right)=1-\frac{v}{f}$
According to Sign Convetion, $\mathrm{v}$ is negative (image is formed at the left side of the lens instead of right side) and equal to $D$
$m=1+\frac{D}{f}$
So Magnifying Power is given by $1+\frac{D}{f}$

Topic: Physics, Ray Optics, JEE Main
Magnitude of kinetic energy in an orbit is equal to

Options:
A Half of the potential energy
B Twice of the potential energy
C One fourth of the potential energy
D None of the above

The correct answer is:
Half of the potential energy

Solution:
For Kinetic and potential of electron in Bohr's orbit:

Kinetic energy of an electron moving around nucleus in its orbit,

K.E. =12(Ze)e4πεr=Ze28πεr

Here, e = charge of electron

Ze= charge of nucleus

ε=permittivity of free space

r = radius of electron's orbit

And Potential energy,

P.E.=-Ze24πεr

Now, by comparing, we get
KE=12PE

i.e., magnitude of kinetic energy is half of the potential energy.

Topic: Chemistry, Structure of Atom, JEE Main
Magnitude of magnetic field (in SI units) at the centre of a hexagonal shape coil of side 10 cm,50 turns and carrying currentI (Ampere) in units of μ0Iπ is :

Options:
A 2503
B 503
C 5003
D 53

The correct answer is:
5003

Solution:
The magnetic field due to the current-carrying finite length wire is given by

B=μ0I4πRsinθ1+sinθ2

Here for hexagonal shape coil, θ1=θ2=30°.

So, the net magnetic field at the centre will be B=50×6×μ0i4π10100cos 30°sin 30°+sin 30°

B=2×75×10μ0i3π12+12

B=15003μ0iπ =5003μ0iπ

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (03 Sep Shift 1)
Magnitude of which of the following property of alkali metals increases with the increase of atomic number

Options:
A Electronegativity
B Ionic radius
C First ionization energy
D Melting point

The correct answer is:
Ionic radius

Solution:

Topic: Chemistry, s Block Elements, JEE Main
Main axis of diatomic molecule is \( Z \). The orbitals \( P_{x} \) and \( P_{y} \) overlap to form

Options:
A II - molecular orbita
B \( \sigma \) - molecular orbital
C \( \delta \) - molecular orbital
D No bond is formed

The correct answer is:
II - molecular orbita

Solution:
If \( z \)-axis is the molecular axis, then \( \Pi \) - molecular orbitals are formed by overlap of \( p x \) and py .The orbitals perpendicular
to the axis takes part in overlapping for \( \Pi \) - molecular orbitals formation, hence, \( \Pi \) - bond is formed.

Topic: Chemistry, Chemical Bonding and Molecular Structure, KCET, KCET 2016
Main constituent of coal tar is

Options:
A Aliphatic compounds
B Aromatic compounds
C Heterocyclic compounds
D Cycloalkanes

The correct answer is:
Aromatic compounds

Solution:
Coal tar is a main source of aromatic hydrocarbons like benzene, naphthalene, anthracene, phenol etc

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
Main function of lenticel is:

Options:
A transpiration.
B guttation.
C gaseous exchange.
D bleeding.

The correct answer is:
gaseous exchange.

Solution:
Lenticels are the areas in the bark which are filled with loosely arranged cells called complementary cells. The loss of water through lenticels is negligible (about $0.1 \%$ ) in comparison to the total water loss from the plant.
Related Theory
Both lenticular and cuticular gaseous exchange can take place continuously, both during the day and night, as there is no mechanism to stop or reduce it.

Topic: Biology, Transport in Plants, NEET, NEET 2002
Main function of roasting is

Options:
A To remove volatile substances
B Oxidation
C Reduction
D Slag formation

The correct answer is:
To remove volatile substances

Solution:
Correct Option is 1 To remove volatile substances.
$\begin{aligned} & \mathrm{S}_8+8 \mathrm{O}_2 \rightarrow 8 \mathrm{SO}_2 \uparrow \\ & \mathrm{P}_4+5 \mathrm{O}_2 \rightarrow \mathrm{P}_4 \mathrm{O}_{10} \uparrow \\ & 4 \mathrm{As}+3 \mathrm{O}_2 \rightarrow 2 \mathrm{As}_2 \mathrm{O}_3 \uparrow\end{aligned}$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Main objective of production/use of herbicide resistant GM crops is to

Options:
A eliminate weeds from the field without the use of manual labour
B eliminate weeds from the field without the use of herbicides
C encourage eco-friendly herbicides
D reduce herbicide accumulation in food particles for health safety

The correct answer is:
reduce herbicide accumulation in food particles for health safety

Solution:
The chemical substances, which are used to kill or repel pest are called pesticides. The pesticides, which are used to destroy weeds are called herbicides.
Herbicides are used in agriculture to control the weeds. They inhibit plant growth by blocking the biosynthesis of amino acids or photosynthetic activities of plants. Transgenic plants resistant to several herbicides, eg, glyphosate, gluphosinate, sulphonylireas etc, have been successfully developed. Main objective of production/use of herbicide resistant genetically modified crops is to reduce herbicide accumulation in food articles for health safety.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2008 (Screening)
Main objective of production/use of herbicide resistant GM crops is to

Options:
A Reduce herbicide accumulation in food articles for health safety
B Eliminate weeds from the field without the use of manual labour
C eliminate weeds from the field without the use of herbicides
D Encourage eco-friendly herbicides

The correct answer is:
Eliminate weeds from the field without the use of manual labour

Solution:
As herbicide resistant GM crops will remain unaffected to the herbicide used, while the weeds will get killed.

Topic: Biology, Microbes in Human Welfare, NEET, NEET 2008 (Mains)
Main Products formed during a reaction of 1-methoxy naphthalene with hydroiodic acid are:

Options:
A and CH3OH
B and CH3I
C and CH3OH
D and CH3I

The correct answer is:
and CH3I

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Main steps in the formation of Recombinant DNA are given below. Arrange these steps in a correct sequence.

A. Insertion of recombinant DNA into the host cell.

B. Cutting of DNA at specific location by restriction enzyme.

C. Isolation of desired DNA fragment.

D. Amplification of gene of interest using PCR.

Choose the correct answer from the options given below:

Options:
A B, D, A. C
B C, A, B, D
C B, C, D, A
D C, B, D, A

The correct answer is:
B, C, D, A

Solution:
The process of forming a recombinant DNA molecule involves the following steps:

Isolation of gene ot interest or a desired DNA molecule from an organism.

Restriction digestion of the desired DNA into fragments to insert into the vector. This is carried out by restriction enzymes.

Isolation of these fragments of DNA (desired fragment)

The fragments can be separated by gel electrophoresis.

Amplification of the gene by PCR thermo-cycler.

Insertion of recombinant DNA into the host cell.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2023 (All India)
Mainly which hormones control menstrual cycle in human beings?

Options:
A $\mathrm{FSH}$
B $\mathrm{LH}$
C FSH, LH, estrogen
D Progesteron

The correct answer is:
FSH, LH, estrogen

Solution:
FSH, LH, Oestrogen
Menstrual cycle in humans is controlled by several endocrine hormones. In the beginning of the cycle, FSH (follicle stimulating hormone), secreted by the pituitary, initiates the development of an ovarian follicle. A growing ovarian follicle gradually secretes increasing amounts of estrogen. This, in turn, leads to a sudden surge of LH secretion by the pituitary. As the LH (luteinizing hormone) level in blood suddenly increases, there is ovulation. After ovulation, the ruptured follicle is converted to a structure called corpus luteum, which secretes mainly progesterone. Progesterone prepares the uterus for a possible pregnancy. If there is no fertilisation progesterone level falls and there is beginning of a new cycle.

Topic: Biology, Human Reproduction, NEET, NEET 2002
Maintenance of personal and public hygiene is necessary for prevention and control of many infectious diseases. Justify the statement giving suitable examples.

Options:

None

Solution:
Infectious diseases are diseases that are transmitted easily from one person to another. Prevention and control of such diseases requires maintenance of personal and public hygiene, some common preventive measures that should be taken for thus purpose are as follows
(i) Education-People should be made aware about communicable diseases so as to protect themselves from such diseases.
(ii) Isolation-Person infected should be kept in isolation to minimise the spread of infection.
(iii) Vaccination Vaccination on time is essential to avoid infection.
(iv) Sanitation-Sanitation should be improved so as to avoid infection from contaminated food, polluted water etc.
(v) Vector eradication-The places where vectors breed must be destroyed and adult vectors should be eliminated by suitable methods.
(vi) Sterilisation-The surroundings and articles of use of the patient should be completely sterilised to reduce the chances of infection.

Topic: Biology, Human Health and Diseases
Major product \(B\) of the following reaction has _______ \(\pi\)-bond.

Options:

The correct answer is:
5

Solution:
Major product B is $\rightarrow$
Total number of $\pi$ bonds in $B$ are 5

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Major product 'B' of the following reaction sequence is

Options:
A

B

C

D

The correct answer is:

Solution:
First step of the reaction involves electrophilic addition of Br2 in CH3OH to 2-methyl but-2-ene to form A which contains an ether linkage.

A, undergoes cleavage reaction with HI by SN1 mechanism to form B as the major product as shown in the reactions below

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2022 (27 Jul Shift 2)
Major product C is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main
Major product formed in the following reaction is a mixture of:

Options:
A
B
C
D

The correct answer is:

Solution:
Tertiary butyl cyclohexyl ether on treatment with HI forms tertiary butyl iodide and cyclohexanol.

SN1 mechanism is followed as the tertiary carbocation formed is stable.

The iodide nucleophile gets attached to the tertiary carbon atom to form tertiary butyl iodide and the cyclohexyl group that is connected to oxygen forms a bond with hydrogen coming from the dissociation of hydrogen iodide to form cyclohexanol.

The equation of the reaction is:

Hence, the answer is option D.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Major product obtained in the following reaction r1,r2 and r3 in respectively is :-

Options:
A
B
C
in all reaction
D
in all reaction

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Major product of the following reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:
During an electrophilic addition reactions double bonds donate lone pair electrons to an electrophile.

Markovnikov's Rule: The addition of a proton acid to the double bond of an alkene results in a product with the acid hydrogen bound to the carbon atom that already has the greater number of hydrogen attached.

In the given reaction electrophile D+ adds to the double bond, a carbocation is formed. Then Cl- is added at the electron deficient carbocation.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (31 Jan Shift 2)
Major product of the following reaction is

Options:
A Butene-1
B Butene-2
C Butane
D Butyne-1

The correct answer is:
Butene-2

Solution:
\(\mathrm{CH}_3-\stackrel{\huge \mathrm{Br} \atop |~~~~}{\mathrm{CH}}-\mathrm{CH}_2-\mathrm{CH}_3+\underset{\text { (alc) }}{\mathrm{KOH}} \rightarrow\)$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3+\mathrm{KBr}+\mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, Hydrocarbons, JEE Main
Major product of the following reaction is

Options:
A

B

C

D

The correct answer is:

Solution:
The product of reaction will be based on stability of carbocation intermediate.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2022 (25 Jul Shift 2)
Major product of the given below reaction:

Options:
A

B

C

D

The correct answer is:

Solution:
Free Radical Polymeristation

Topic: Chemistry, Polymers, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Major product of the given reaction is:

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Major product of the reaction is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Major product of this reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
Here anti elimination take place to give the product as follows

Topic: Chemistry, Hydrocarbons, JEE Main
major products formed in the reaction of t-butyl methyl ether with HI are :

Options:
A
B
C
D

The correct answer is:

Solution:
SN-1 Reaction, more stable carbocation is formed

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Major products of the following reaction are:

Options:
A CH3OH and HCO2H
B HCOOH and
C and
D CH3OH and

The correct answer is:
HCOOH and

Solution:
Given reaction is a Cross Cannizzaro Reaction, since there are no ∝-H present on the Carbonyl compound.
Mechanism

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2019 (10 Apr Shift 1)
Major products of the following reaction are:

Options:
A CH3OH and HCO2H
B HCOOH and
C and
D CH3OH and

The correct answer is:
HCOOH and

Solution:
Given reaction is a Cross Cannizzaro Reaction, since there are no ∝-H present on the Carbonyl compound.
Mechanism

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Male and female gametophytes are independent and free-living in

Options:
A Mustard
B Castor
C Pinus
D Sphagnum

The correct answer is:
Sphagnum

Solution:
Sphagnum is a Bryophyte in which male and female gametophytes are independent and free living. In pinus (a gymnosperm), mustard and castor (angiosperms), the main plant body is sporophytic. Gametophyte is highly reduced and is completely dependent on sporophyte.

Topic: Biology, Plant Kingdom, NEET
Maleic acid and fumaric acids are

Options:
A
Chain isomers

B
Functional isomers

C
Tautomers

D
Geometrical isomers

The correct answer is:
Geometrical isomers

Solution:
Maleic acid and fumaric acids are geometrical isomers.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2012 (26 May Online)
Maleic anhydride

Maleic anhydride can be prepared by :

Options:
A Heating trans-but-2-enedioic acÄ±d
B Heating cis-but-2-enedioic acid
C Treating cis-but-2-enedioic acid with alcohol and acid
D Treating trans-but-2-enedioic acid with alcohol and acid

The correct answer is:
Heating cis-but-2-enedioic acid

Solution:
Heating cis-but-2-enedioic acid give maleic anhydride as follows,

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2021 (25 Jul Shift 2)
Malic acid inhibits the activity of enzyme succinate dehydrogenase. During this inhibition

Options:
A V max increases, K m decreases
B V max remains the same, K m decreases
C V max decreases, K m increases
D V max remains the same, K m increases

The correct answer is:
V max remains the same, K m increases

Solution:
Any substance that can diminish the velocity of an enzyme is known as an inhibitor. The two main types of inhibition are competitive, non-competitive. In competitive inhibition, the enzyme and inhibitor compete to bind to the same active site. Hence if substrate concentration is increased the chances of substrate binding to the enzyme will increase. Thus if (S) increases K m (Michaelis constant) also increases. In non-competitive inhibition, the substrate and inhibitor bind at different sites hence do not compete with each other. This cannot be reversed by increasing the substrate concentration and hence K m remains constant. This velocity of reaction also decreases. An example of a competitive inhibitor is malic acid. This competes with succinate (substrate) to bind at the active site of enzyme succinate dehydrogenase.

Topic: Chemistry, Biomolecules, NEET
Malonic acid → 150 o C P 2 O 5 Fouls mell A → 200 o C B +Carbon
The total number of sigma and Pi bonds present in compound A are:

Options:

The correct answer is:
8

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Maltose contains how many oxygen atoms

Options:
A $6$
B $10$
C $11$
D $22$

The correct answer is:
$11$

Solution:
$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$

Topic: Chemistry, Biomolecules, JEE Main
Maltose on hydrolysis gives

Options:
A Mannose + glucose
B Galactose + glucose
C Glucose
D Mannose + fructose

The correct answer is:
Glucose

Solution:
C12H22O11Maltose+H2OâMaltasefrom yeast2C6H12O6Glucose

Topic: Chemistry, Biomolecules, JEE Main
Maltose on treatment with dilute HCl gives:

Options:
A D-Glucose
B D-Fructose
C D-Galactose
D D-Glucose and D-Fructose

The correct answer is:
D-Glucose

Solution:
Maltose is a disaccharide which contain D-glucose as a monosaccharide. Thus, Hydrolysis of maltose with dil HCl gives D- glucose.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2019 (08 Apr Shift 1)
Manganese is required in :

Options:
A Chlorophyll synthesis
B Nucleic acid synthesis
C Plant cell wall formation
D Photolysis of water during photosynthesis

The correct answer is:
Photolysis of water during photosynthesis

Solution:
No Solution Available

Topic: Biology, Mineral Nutrition, NEET, NEET 2009 (Mains)
Manganese VI has ability to disproportionate in acidic solution. The difference in oxidation states of two ions it forms in acidic solution is

Options:

The correct answer is:
3

Solution:
In acidic solution Mn(VI) become unstable relative to Mn(VII) and Mn(IV)

3MnO42-+4H+→2MnO4-+MnO2+2H2O
So difference in oxidation state of product ions of Mn is = 3

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2022 (24 Jun Shift 2)
Mannitol is the stored food in -

Options:
A Gracillaria
B Chara
C Porphyra
D Fucus

The correct answer is:
Fucus

Solution:
No Solution Available

Topic: Biology, Plant Kingdom, NEET, NEET 2009 (Mains)
Many diseases can be diagnosed by observing the symptoms in the patient. Which group of symptoms are indicative of pneumonia?

Options:
A Difficulty in respiration, fever, chills, cough, headache
B Constipation, abdominal pain, cramps, blood clots
C Nasal congestion and discharge, cough, sore throat, headache
D High fever, weakness, stomach pain, loss of appetite and constipation.

The correct answer is:
Difficulty in respiration, fever, chills, cough, headache

Solution:
Pneumonia is an inflammatory disorder of lungs. It causes inflammation of Alveoli. It is bacterial or viral in origin. It mainly affects respiratory system. Thus some symptoms can give us an idea about this disease. (Symptoms are conditions described by the patient and Signs are conditions noticed by the Doctors). Difficulty is respiration and cough gives us an idea that respiratory system is affected. Fever, chills, headache explain that body has an infection may be an inflammation. Thus these symptoms can help to diagnose the pneumonia condition.

Topic: Biology, Human Health and Diseases, NEET
Many exoplanets have been discovered by the transit method, where in one monitors, a dip in the intensity of the parent star as the exoplanet moves in front of it. The exoplanet has a radius R and the parent star has radius 100 R. If I0is the intensity observed on earth due to the parent star, then as the exoplanet transits

Options:
A the minimum observed intensity of the parent star is 0.9I0
B the minimum observed intensity of the parent star is 0.99 I0
C the minimum observed intensity of the parent star is 0.999 I0
D the minimum observed intensity of the parent star is 0.9999 I0

The correct answer is:
the minimum observed intensity of the parent star is 0.9999 I0

Solution:
Intensity of radiation (mainly visible light) emitted from surface of a star is proportional to its area.

So, I∝A or I=k A

where, k= constant.
Now, if I0= intensity of parent star.

Then, I0=kπ (100R)2=kπR2×10000

When exoplanet is in front of star, observed intensity will be minimum. Let intensity minimum is Imin, then

Imin=k π(100R)2-πR2

⇒ Imin=k π R2 (10000-1)

=kπ R2×9999

So, IminI0=kπR2×9999kπR2×10000

⇒ Imin=I0×0.9999

Topic: Physics, Thermal Properties of Matter, KVPY, KVPY 2018 (SA)
Many freshwater animals cannot live for long in sea water and vice versa mainly because of the

Options:
A Change in nitrogen levels
B Change in the levels of thermal tolerance
C Variations in light intensity
D Osmotic problems

The correct answer is:
Osmotic problems

Solution:
Organisms found in fresh water have a problem of excess internal water because of endosmosis. Organisms found in ocean or salt water have a problem of low internal water content due to exomosis. For aquatic organisms the quality (chemical composition, pH) of water becomes important. Many freshwater animals cannot live for long in sea water and vice versa because of the osmotic problems they would face.

Topic: Biology, Excretory Products and their Elimination, NEET
Mapping $f: R \rightarrow R$ which is defined as $f(x)=\cos x, x \in R$ will be

Options:
A Neither one-one nor onto
B One-one
C Onto
D One-one onto

The correct answer is:
Neither one-one nor onto

Solution:
Let $x_1, x_2 \in R$, then $f\left(x_1\right)=\cos x_1, f\left(x_2\right)=\cos x_2$, so $f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow \cos x_1=\cos x_2 \Rightarrow x_1=2 n \pi \pm x_2$
$\Rightarrow x_1 \neq x_2$, so it is not one-one.
Again the value of $f$-image of $x$ lies in between -1 to 1
$\Rightarrow f[R]=\{f(x):-1 \leq f(x) \leq 1)\}$
So other numbers of co-domain (besides -1 and 1 ) is not $f$-image. $f[R] \in R$, so it is also not onto. So this mapping is neither one-one nor onto.

Topic: Mathematics, Functions, JEE Main
Mark a cell which does not exhibit phagocytotic activity among the following.

Options:
A Monocytes
B Neutrophil
C Basophil
D Macrophage

The correct answer is:
Basophil

Solution:
Cellular components of Blood are Erythrocytes, Leukocytes and Thrombocytes. Leukocytes can be Granulocytes or Agranulocytes. Basophils are Granulocytes. These are 0-1% of the total WBC's. These secrete substances like Histamine, Serotonin and heparin. Histamine is a Vasodilator. Serotonin mainly acts as Vasoconstrictor while Heparin is an anticoagulant. Thus basophils are non-phagocytic. Phagocytic cells are Neutrophils and Monocytes. Monocytes circulate in blood for around 8 hrs and then enter the tissues as enlarged cells. These are Macrophages and hence these are too phagocytic in function.

Topic: Biology, Body Fluids and Circulation, NEET