datavorous/entrance-exam-dataset huggingface dataset:
The molarity of solution obtained by mixing 750 mL of 0.5(M) HCl with 250 mL of 2(M) HCl will

Options:
A 0.875 M
B 0.975 M
C 1.75 M
D 1.00 M

The correct answer is:
0.875 M

Solution:
M f  =  M 1 V 1  +  M 2 V 2 V 1  +  V 2  =  (0.5)(750) + (2)250 1000  = 0.875 M

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
The mole fraction of a solute in its one molal aqueous solution is:

Options:
A 0.018
B 0.027
C 0.036
D 0.048

The correct answer is:
0.018

Solution:
w=1000 g H2O; n(solute)=1 mol
(as 1 molal aqueous solution implies 1 mol of solute in 1 kg of water)
N (water) =WM=100018=55.55
Whereas
n= number of moles of solute
N= number of moles of water H 2 O
Xsolute=nn+N=11+55.55
=0.018

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
The mole friction of the solute in one molal aqueous solution is:

Options:
A 0.009
B 0.018
C 0.027
D 0.036

The correct answer is:
0.018

Solution:
1 mole of solute present in $1 \mathrm{~kg}$ of $\mathrm{H}_2 \mathrm{O}$
1 mole of solute present in $\frac{1000}{18} \mathrm{~g}$. mole of $\mathrm{H}_2 \mathrm{O}$
$$
\begin{aligned}
\mathrm{X}_{\text {solute }} & =\frac{1}{\frac{1000}{18}+1}=\frac{18}{1018} \\
& =0.01768 \approx 0.018 .
\end{aligned}
$$
Related Theory
The molar fraction can be represented by $X$. If the solution consists of components $A$ and $B$, then the mole fraction is,
Mole fraction of solute
$$
=\frac{\text { Moles of Solute }}{\text { Moles of Solute }+ \text { Moles of Solven }}
$$
$$
=\frac{n_A}{n_A+n_B}
$$
Mole fraction of solute $=\frac{n_B}{n_A+n_B}$
Therefore, the sum of mole fraction of all the components is always equal to one.
Caution
Students misread the word 'molal' as 'molar'.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2005
The molecular formula C3H9N cannot represent

Options:
A 1°amine
B 2°amine
C 3°amine
D Quaternary salt

The correct answer is:
Quaternary salt

Solution:
C3H9N represent following structures
CH3CH2CH2NH2
Propanamine
(1°amine)
C H 3 C H 2 − N | H −C H 3
N-methyl ethanamine
(2°amine)
C H 3 − N | C H 3 −C H 3
N,N-dimethyl methanamine
(3°amine)

Topic: Chemistry, Amines, NEET
The molecular formula C3H9N cannot represent

Options:
A 1°amine
B 2°amine
C 3°amine
D Quaternary salt

The correct answer is:
Quaternary salt

Solution:
C3H9N represent following structures
CH3CH2CH2NH2
Propanamtextne
(1°amtextne)
C H 3 C H 2 − N | H −C H 3
N-methyl ethanamtextne
(2°amtextne)
C H 3 − N | C H 3 −C H 3
N,N-dimethyl methanamtextne
(3°amtextne)

Topic: Chemistry, Amines, NEET
The molecular formula of a non-stoichiometric tin oxide containing Sn (II) and Sn (IV) ions is Sn4.44O8.

Therefore, the molar ratio of Sn (II) to Sn (IV) is approximately

Options:
A 1 : 8
B 1 : 6
C 1 : 4
D 1 : 1

The correct answer is:
1 : 4

Solution:
Let in one mole of the given substance the number of moles of Sn (II) = x moles and number of moles of Sn (IV) = y moles.
∴ x+y=4.44 and 2x+4y=16
By solving x=0.88 and y=3.56
∴ mole ratio of Sn II to Sn IV is 0.247 : 1≃1 : 4

Topic: Chemistry, Solid State, NEET
The molecular formula of diphenyl methane:
How many structural isomers are possible when one of the hydrogen is replaced by a chlorine atom?

Options:
A 6
B 4
C 8
D 7

The correct answer is:
4

Solution:
The structural isomer for diphenyl methane are:

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2004
The molecular formula of diphenyl methane,

How many structural isomers are possible when one of the hydrogen is replaced by a chlorine atom?

Options:
A 6
B 4
C 8
D 7

The correct answer is:
4

Solution:
Only four structural isomers are possible for monochloro derivative of diphenyl methane.

Topic: Chemistry, General Organic Chemistry, NEET
The molecular formula of diphenylmethane,

, is C13H12;

How many structural isomers are possible when one of the hydrogen is replaced by a chlorine atom?

Options:
A 6
B 4
C 8
D 7

The correct answer is:
4

Solution:

Topic: Chemistry, General Organic Chemistry, NEET
The molecular formula of diphenylmethane is C 13 H 12 .

How many structural isomers are possible when one of the hydrogens is replaced by a chlorine atom?

Options:
A 6
B 4
C 8
D 7

The correct answer is:
4

Solution:
There are four different positions from where hydrogen can be replaced.

Topic: Chemistry, General Organic Chemistry, NEET
The molecular formula of tripalmitin is

Options:
A C51H98O6
B C54H100O7
C C57H102O8
D C61H104 O10

The correct answer is:
C51H98O6

Solution:
Lipids are classified as simple, conjugated, and derived lipids. Simple lipids are esters of fatty acids and alcohols. If these alcohols are glycerols then such simple lipids are neutral fats. Glycerol is trihydric alcohol i.e. it has three hydroxy groups. Thus if such alcohol reacts with fatty acids to form esters, three such fatty acids are needed. This forms triglyceride. Example: Glycerol reacts with three palmitic acids to form tripalmitin and its formula is C51H98O6.

Topic: Chemistry, Biomolecules, NEET
The molecular scissors used to cut DNA sequences from a particular site is called __.

Options:
A PCR
B Restriction enzyme
C Chitinase
D Bioreactors

The correct answer is:
Restriction enzyme

Solution:
The molecular scissors used to cut DNA sequences from a particular site is called restriction enzyme.

Topic: Biology, Biotechnology - Principles and Processes, NEET
The molecule of which gas have highest speed?

Options:
A H2 at -73°C
B CH4 at 300 K
C N2 at 1027°C
D O2 at 0°C

The correct answer is:
H2 at -73°C

Solution:
urms=3RTM .
The value of TM is maximum for H2 at -73°C and equal to 2002=100 .

Topic: Chemistry, States of Matter, NEET
The molecule which does not exhibit dipole moment is

Options:
A NH3
B CHCl3
C CCl4
D H2O

The correct answer is:
CCl4

Solution:
CCl4 does not exhibit dipole moment due to its symmetrical structure.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The molecules of a given mass of gas have an rms velocity of 200 m s-1 at 27 Â°C and pressure 1 atm. When the temperature is 127 Â°C and pressure is 2 atm, the rms velocity in m s-1 will be?

Options:
A 10023
B 1002
C 4003
D None of these

The correct answer is:
4003

Solution:
RMS velocity doesn't depend upon pressure, it depends upon temperature only,
ie., vrms∝T.
⇒ v1v2=T1T2 ⇒ 200v2=(273+27)(273+127)=300400
⇒ v2=4003 m s-1

Topic: Physics, Kinetic Theory of Gases, NEET
The molecules of a given mass of gas have RMS velocity of 200 m s-1 at 27oC and 1.0×105 N m-2 pressure. When the temperature and pressure of the gas are respectively, 127oC and 0.05×105 N m-2, the r.m.s. velocity of its molecules in m s-1 is:

Options:
A 1002
B 4003
C 10023
D 1003

The correct answer is:
4003

Solution:
RMS speed of molecules is VRMS=3RTM, where R is universal gas constant, T is absolute temperature and M is molar mass of gas.
So it depends only on temperature, Vrms∝T
⇒V1V2=T1T2⇒200V2=300400
⇒200V2=32⇒V2=4003 m sec-1

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2016 (Phase 1)
The moment of inertia of a body about a given axis is 1.2 kg m2. Initially, the body is at rest. In order to produce a rotational KE of 1500 J, the angular acceleration of 25 rad s-2 must be applied about that axis for a duration of

Options:
A 4 s
B 2 s
C 8 s
D 10 s

The correct answer is:
2 s

Solution:
I=1.2 kg m2
ωi=0; α=25 rad s-2
Rotational KE =12Iωf2=1500 J

⇒ωf=50 rad s-1
⇒t=ωf-ωiα=5025=2 s

Topic: Physics, Rotational Motion, NEET
The moment of inertia of a body depends upon

Options:
A mass only
B angular velocity only
C distribution of particles only
D mass and distribution of mass about the axis

The correct answer is:
mass and distribution of mass about the axis

Solution:
mass and distribution of mass about the axis

Topic: Physics, Rotational Motion, NEET
The moment of inertia of a body is I and its coefficient of linear expansion is α if the temperature of the body rises by a small amount ΔT. Then change in the moment of inertia about the same axis is

Options:
A αIΔT
B 2αIΔT
C 4αIΔT
D αIΔT2

The correct answer is:
2αIΔT

Solution:
Let I=mr2
and âr=rÎ±âT
Î II=2Îrr=2 Î± ÎT
orâÎI =2Î±I ÎT

Topic: Physics, Thermal Properties of Matter, NEET
The moment of inertia of a disc about an axis passing through its centre and normal to its plane is I . The disc is now folded along a diameter such that the two halves are mutually perpendicular. Its moment of inertia about this diameter will now be

Options:
A I
B I2
C I2
D I4

The correct answer is:
I2

Solution:
moment of inertia of a disc about an axis passing through its centre and normal to its plane
I=12mR2
The disc is now folded along a diameter such that the two halves are mutually perpendicular
In this moment of inertia about this diameter I=Id+Id
I=2Id
Id=12I
For a diameter; Id=I2

Topic: Physics, Rotational Motion, NEET
The moment of inertia of a ring about an axis passing through the centre and perpendicular to its plane is I. It is rotating with angular velocity Ï. Another identical ring is gently placed on it so that their centres coincide. If both the rings are rotating about the same axis then loss in kinetic energy is

Options:
A Iω22
B Iω24
C Iω26
D Iω28

The correct answer is:
Iω24

Solution:
From conservation of angular momentum
I 1 ω 1 = I 2 ω 2
Iω=2I ω 2
ω 2 = ω 2
New KE=12.(2I)(ω2)2=Iω24
Loss in KE=12Iω2−Iω24=Iω24

Topic: Physics, Rotational Motion, NEET
The moment of inertia of a rod about an axis through its centre and perpendicular to it is 112ML2 (where, M is the mass and L is the length of the rod). The rod is bent in the middle so that the two halves make an angle of 60Â°. The moment of inertia of the bent rod about the same axis would be

Options:
A ML248
B ML212
C ML224
D ML283

The correct answer is:
ML212

Solution:
The total length of the rod is L. Since, the rod is bent at the middle, so each part of it will have same length L2 and mass M2
Moment of inertia of each part about an axis passing through its one end
=13M2L22
Hence, net moment of inertia about an axis passing through its middle point is
I=13M2L22+13M2L22
=13ML28+ML28=ML212

Topic: Physics, Rotational Motion, NEET
The moment of inertia of a solid cylinder about its axis is given by 12MR2. If this cylinder rolls without shipping the ratio of its rotational kinetic energy to its translational kinetic energy is-

Options:
A 1:1

B 2:2
C 1:2
D 2:3

The correct answer is:
1:2

Solution:
KR=12IÏ2 â KR=12Ã12mR2v2R2âKR=12.12mv2
KR=12KT â KRKT=12

Topic: Physics, Rotational Motion, NEET
The moment of inertia of a solid sphere of radius R about its diameter is same as that of a disc of radius 2R about its diameter. The ratio of their masses is

Options:
A 5:2
B 5:8
C 4:1
D 2:1

The correct answer is:
5:2

Solution:
The moment of inertia of a solid sphere of radius R about its diameter
Is=25MsR2
The moment of inertia of a disc of radius 2R about its diameter
Id=14Md(2R⁡)2
Given, Is=Id
25MsR2=14Md(2R⁡)2
MsMd=52

Topic: Physics, Rotational Motion, NEET
The moment of inertia of a solid sphere of radius R about its diameter is same as the that of disc of radius 2R about its diameter. The ratio of their masses is

Options:
A 5 : 2
B 5 : 8
C 4 : 1
D 2 : 1

The correct answer is:
5 : 2

Solution:
25MsR2=Md2R24
or MsMd=52

Topic: Physics, Rotational Motion, NEET
The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is $2400 \mathrm{~g} \mathrm{~cm}^2$. The length of the 400 g rod is nearly:

Options:
A 17.5 cm
B 20.7 cm
C 72.0 cm
D 8.5 cm

The correct answer is:
8.5 cm

Solution:
Moment of inertia of rod $=I=\frac{m \ell^2}{12}$
$\begin{aligned} & \Rightarrow \quad 2400=400 \frac{\ell^2}{12} \\ & \Rightarrow \quad 72=\ell^2 \\ & \Rightarrow \quad \ell=\sqrt{72}=8.48 \mathrm{~cm} \simeq 8.5 \mathrm{~cm}\end{aligned}$

Topic: Physics, Rotational Motion, NEET, NEET 2024
The moment of inertia of a thin uniform rod of length L and mass M about an axis passing through a point at a distance of L3 from one of its ends and perpendicular to the rod is

Options:
A ML212
B ML29
C 7 ML248
D ML248

The correct answer is:
ML29

Solution:
ICM=ML212 (about middle point)
Using parallel axis theorem.
∴ I=ICM+Mx2
=ML212+ML62
I=ML29

Topic: Physics, Rotational Motion, NEET
The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its mid-point and perpendicular to its length is $I_0$. Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is

Options:
A $I_0+M L^2 / 4$
B $I_0+2 M L^2$
C $I_0+M L^2$
D $I_0+M L^2 / 2$

The correct answer is:
$I_0+M L^2 / 4$

Solution:
The theorem of parallel axis for moment of inertia.
$$
\begin{aligned}
& I=I_{\mathrm{CM}}+M h^2 \\
& I=I_0+M\left(\frac{L}{2}\right)^2
\end{aligned}
$$
$$
I=I_0+\frac{M L^2}{4}
$$

Topic: Physics, Rotational Motion, NEET, NEET 2011 (Screening)
The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis passing from the edge of the disc and normal to the disc is:

Options:
A $M R^2$
B $\frac{1}{2} M R^2$
C $\frac{3}{2} M R^2$
D $\frac{7}{2} M R^2$

The correct answer is:
$\frac{3}{2} M R^2$

Solution:
M.I. of disc about its normal $=\frac{1}{2} M R^2$
M.I. about its one edge $=M R^2+$ $\frac{M R^2}{2}$
$$
\therefore \quad \text { M.I. }=\frac{3}{2} M R^2
$$

Topic: Physics, Rotational Motion, NEET, NEET 2005
The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis touching the disc at its diameter and normal to the disc:

Options:
A $\frac{1}{2} M R^2$
B $M R^2$
C $\frac{2}{5} M R^2$
D $\frac{3}{2} M R^2$

The correct answer is:
$\frac{3}{2} M R^2$

Solution:
M. I. of uniform circular disc about an axis and perpendicular to the plane is:
$$
I_C=\frac{1}{2} M R^2
$$
Using the theorem of parallel axis.
M. I. of uniform circular disc about an axis touching the disc at it diameter is:
$$
\begin{aligned}
I & =I_C+M R^2 \\
& =\frac{1}{2} M R^2+M R^2 \\
& =\frac{3}{2} M R^2 .
\end{aligned}
$$

Topic: Physics, Rotational Motion, NEET, NEET 2006
The moment of inertia of a uniform semi-circular disc about an axis passing through its centre of mass and perpendicular to its plane is (Mass of this disc is M and radius is R).

Options:
A MR 2 2 - M 2 R π 2
B MR 2 2 - M 4 R 3 Ï 2
C MR 2 2 + M 4 R 3 π 2
D MR 2 2 + M 2 R π 2

The correct answer is:
MR 2 2 - M 4 R 3 π 2

Solution:
I 0 = MR 2 2
I 0 = I cm + Md 2
MR 2 2 = I cm + M 4 R 3 Ï 2
â I cm = MR 2 2 - M 4 R 3 Ï 2
I cm = MR 2 2 - M 4 R 3 Ï 2

Topic: Physics, Rotational Motion, NEET
The moment of the force, F→=4i^+5j^−6k^ at ( 2,0,−3 ), about the point ( 2,−2,−2 ), is given by

Options:
A \(-7 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}-4 \widehat{\mathrm{k}}\)
B \(-4 \hat{\mathrm{i}}-\hat{\mathrm{j}}-8 \widehat{\mathrm{k}}\)
C \(-8 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-7 \hat{\mathrm{k}}\)
D \(-7 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-8 \widehat{\mathrm{k}}\)

The correct answer is:
\(-7 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-8 \widehat{\mathrm{k}}\)

Solution:
Fâ=4i^+5j^-6k^
râ=2-2i^+0--2j^+-3--2k^=2j^-1k^
râÃFâ=2j^-1k^Ã4k^+5j^-6k^
= i^j^k^02-145-6=-7i^-4j^-8k^

Topic: Physics, Mathematics in Physics, NEET, NEET 2018
The momentum of a photon of energy 1 MeV in kg m s-1, will be

Options:
A 0.33×106
B 7Ã10-24
C 10-22
D 5Ã10-22

The correct answer is:
5×10-22

Solution:
Energy of photon is given by
E=hcÎ» ...(i)
Where h is the Planck's constant, c the velocity of light and Î» its wavelength.
de-Broglie wavelength is given by
Î»=hp
Where p is being momentum of photon.
From Eqs. (i) and (ii), we get
p= E c
Or E=pÃc
Given, E = 1 MeV=1Ã106Ã1.6Ã10-19 J.
c=3Ã108 m s-1
Hence, after putting numerical values, we obtain
p=1Ã106Ã1.6Ã10-193Ã108
= 5Ã10-22 kg m s-1

Topic: Physics, Dual Nature of Matter, NEET
The momentum of a photon of energy 1 $\mathrm{MeV}$ in $\mathrm{kg} \mathrm{m} / \mathrm{s}$ will be:

Options:
A $5 \times 10^{-22}$
B $0.33 \times 10^6$
C $7 \times 10^{-24}$
D $10^{-22}$

The correct answer is:
$5 \times 10^{-22}$

Solution:
Energy of photon $E=1 \mathrm{MeV}$
Momentum of photon $P=\frac{E}{C}$
$$
\begin{aligned}
P & =\frac{E}{C}=\frac{1 \times 10^6 \times 1.6 \times 10^{-19}}{3 \times 10^8} \\
& =5 \times 10^{-22} \mathrm{~kg} \mathrm{~m} / \mathrm{s}
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2006
The monarch butterfly is highly distasteful to its predator because

Options:
A a special chemical is present in its body
B cardiac glycosides is stored in its body
C it feeds on poisonous weed
D both a and b

The correct answer is:
both a and b

Solution:
Monarch butterfly is highly distasteful to its predator bird because is a special chemical present in its body. The butterfly acquires this chemical during its caterpillar stage by feeding on poisonous weed.

Topic: Biology, Organisms and Populations, NEET
The monomer for Teflon polymer is

Options:
A Monofluoroethene
B Difluoroethene
C Trifluoroethene
D Tetrafluoroethene

The correct answer is:
Tetrafluoroethene

Solution:
Tetrafluoroethene (CF2=CF2).

Topic: Chemistry, Polymers, NEET
The monomer of the polymer

Options:
A

B $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_3$
C $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2$
D $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_3\right)_2$

The correct answer is:

Solution:
The repeating unit in the given polymer is 2-methyl propene.
Related Theory
The given polymer is an example of cationic polymerisation.

Topic: Chemistry, Polymers, NEET, NEET 2005
The Montreal Protocol has been introduced to

Options:
A Check depletion of the ozone layer
B Mitigate the climatic change
C Check soil erosion
D Preserver water resources

The correct answer is:
Check depletion of the ozone layer

Solution:
Recognising the deleterious effects of ozone depletion, an international treaty known as the Montreal Protocol was signed at Montreal (Canada) in 1987 (effective in 1989) to control the emission of ozone-depleting substances.

Topic: Biology, Environmental Issues, NEET
The Montreal Protocol is related to:

Options:
A Persistant organic pollutants
B Global warming & climate change
C Substances that deplete the ozone layer
D Biosafety of genetically modified organisms

The correct answer is:
Substances that deplete the ozone layer

Solution:
The deleterious affects of ozone depletion, an international treaty, known as the Montreal Protocol, was signed at Montreal (Canada) in 1987 (effective in 1989) to control the emission of ozone-depleting substances.

Topic: Biology, Environmental Issues, NEET
The Montreal Protocol refers to

Options:
A Persistent organic pollutants
B Global warming and climate change
C Substances that deplete the ozone layer
D Biosafety of genetically modified organisms

The correct answer is:
Substances that deplete the ozone layer

Solution:
The Montreal Protocol on substances that deplete the ozone layer was signed in Montreal, Canada, by over 150 countries at a convention in 1987 to cut the use of CFCs (chlorofluorocarbons). The aim of the Protocol was to protect the ozone layer in the stratosphere by decreasing and eventually eliminating the use of ozone-depleting substances like CFCs. It is regarded as one of the most successful international treaties in modern history.

Topic: Biology, Environmental Issues, NEET
The moon's radius is 1 4 th that of the earth & its mass is 180 times that of earth, if g represents the acceleration due to gravity on the surface of earth, that on the surface of moon is -

Options:
A g 4
B g 5
C g 6
D g 8

The correct answer is:
g 5

Solution:
We know gravitational acceleration
g= GM R 2
for moon R m = 1 4 R & M m = M 80
so g â² = G( M 80 ) ( R 4 ) 2 = GM 5 R 2 = g 5

Topic: Physics, Gravitation, NEET
The most accepted line of descent in human evolution is:

Options:
A Australopithecus → Ramapithecus → Homo sapiens → Homo habilis
B Homo erectus → Homo habilis → Homo sapiens
C Ramapithecus → Homo habilis → Homo erectus → Homo sapiens
D Australopithecus → Ramapithecus → Homo erectus → Homo habilis → Homo sapiens.

The correct answer is:
Ramapithecus → Homo habilis → Homo erectus → Homo sapiens

Solution:
Ramapithecus → Homo habilis → Homo erectus → Homo sapiens is the consecutive line of descent in human evolution.

Topic: Biology, Evolution, NEET
The most active phagocytic white blood cells are

Options:
A neutrophils and eosinophils
B lymphocytes and macrophages
C eosinophils and lymphocytes
D neutrophils and monocytes

The correct answer is:
neutrophils and monocytes

Solution:
Neutrophils and monocytes are phagocytic white blood cells. Monocytes are largest of all leucocytes and generally change into macrophages after entering tissue spaces. Neutrophils are most numerous of all leucocytes, and have many lobed nucleus.
Eosinophils are granular white blood cells. Their number increases in people with allergic conditions such as asthma or hay fever. They are non-phagocytic and seem to play part in immune system
Lymphocytes are non-motile and non-phagocytic in nature. They produce antibodies to destroy microbes. These are found as B and T-lymphocytes. T-lymphocytes either directly attack the antigens or stimulate B-lymphocytes to produce antibodies.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2008 (Screening)
The most active phagocytic white blood cells are:

Options:
A Neutrophils and eosinophils
B Lymphocytes and macrophages
C Eosinophils and lymphocytes
D Neutrophils and monocytes

The correct answer is:
Neutrophils and monocytes

Solution:
The most active phagocytic white blood cells are neutrophils and monocytes. Neutrophil is a type of WBC (leukocyte) that has a lobed nucleus and granular cytoplasm. Neutrophils engulf bacteria and release various substance such as lysozyme and oxidizing agents. Monocyte is the largest form of WBC in vertebrates. It has a kidney shaped nucleus and is actively phagocytic, ingesting bacteria and cells debris.

Topic: Biology, Body Fluids and Circulation, NEET
The most apparent change during the evolutionary history of Homo sapiens is traced in

Options:
A remarkable increase in the brain size.
B bipedal locomotion.
C shortening of jaws.
D All of these

The correct answer is:
All of these

Solution:
The human evolution is the gradual process leading to the emergence of the modern humans. It is characterized by number of changes in the morphology, physiology, behaviour which has occurred since the split between the last common ancestor of human and chimpanzees. The most significant change observed in history of Homosapiens is the increase in cranium capacity. Other changes include the bipedalism, lengthened ontogeny.

Topic: Biology, Evolution, NEET
. The most appropriate regent for the given reaction can be –

Options:
A Conc. H2SO4/Δ
B Al2O3/Δ
C ThO2/Δ
D All of them

The correct answer is:
Al2O3/Δ

Solution:
Product (more stable alkene)

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
The most basic hydroxide from the following is

Options:
A Pr(OH)3(Z=59)
B Sm(OH)3(Z=62)
C Ho(OH)3(Z=67)
D La(OH)3(Z=57)

The correct answer is:
La(OH)3(Z=57)

Solution:
The basic nature of hydroxides decreases from La to Lu along the lanthanide series due to lanthanoid concentration. Thus, La(OH)3 is the most basic hydroxide.

Topic: Chemistry, d and f Block Elements, NEET
The most common substrate used in distilleries for the production of ethanol is

Options:
A soya meal
B ground gram
C molasses
D corn meal

The correct answer is:
molasses

Solution:
Molasses is a viscous by-product of the processing of sugarcane, grapes or sugar beets in sugar. It is the most common substrate used in distilleries for the production of ethanol. It can be used as the base material for fermentation into rum. In Australia, molasses is fermented to produce ethanol for use as an alternative fuel in motor vehicles.

Topic: Biology, Strategies for Enhancement in Food Production, NEET, NEET 2011 (Screening)
The most common type of ovule found in angiospermic plants is

Options:
A Anatropous
B Hemitropous
C Campylotropous
D Circinotropous

The correct answer is:
Anatropous

Solution:
Most common type of angiospermic ovule is anatropous ovule. it is sloa known as resupinate ovule or inverted ovule in which ovule is curved down as a results micropyle comes to lie close to funicle.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET