datavorous/entrance-exam-dataset huggingface dataset:
Оп a new scale of temperature (which is linear) and called the $W$ scale, the freezing and boiling points of water are $39^{\circ} \mathrm{W}$ and $239^{\circ} \mathrm{W}$ respectively. What will be the temperature on the new scale, corresponding to a temperature of $39^{\circ} \mathrm{C}$ on the Celsius scale?

Options:
A $78^{\circ} \mathrm{W}$
B $117^{\circ} \mathrm{W}$
C $200^{\circ} \mathrm{W}$
D $139^{\circ} \mathrm{W}$

The correct answer is:
$117^{\circ} \mathrm{W}$

Solution:
Given,
\(\begin{aligned}
& \mathrm{T}_{\mathrm{f}}=39^{\circ} \mathrm{W} \\
& \mathrm{T}_{\mathrm{b}}=230^{\circ} \mathrm{W}
\end{aligned}\)
We know that,
\(\begin{aligned}
& \frac{\mathrm{T}_{\mathrm{b}}-\mathrm{T}_{\mathrm{f}}}{100^{\circ} \mathrm{C}-\mathrm{o}^{\circ} \mathrm{C}}=\frac{\mathrm{T}-39^{\circ} \mathrm{C}}{39^{\circ} \mathrm{C}} \\
& \frac{239-39}{100}=\frac{\mathrm{T}-39}{39} \\
& \frac{200}{100}=\frac{\mathrm{T}-39}{39} \\
& \mathrm{~T}-39=2 \times 39=78^{\circ} \mathrm{W} \\
& \mathrm{T}=39+78=117^{\circ} \mathrm{W}
\end{aligned}\)

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2008 (Screening)
A 'new' variety of rice was patented by a foreign company, though such varieties have been present in India for a long time. This is related to

Options:
A Lerma Rojo
B Sharbati Sonora
C Co-667
D Basmati

The correct answer is:
Basmati

Solution:
ln 1997, an American company got patent right on Basmati rice through the US Patent and Trademark Office. This 'new' variety of Basmati had actually been derived from Indian farmer's varieties. This 'new' variety was produced by crossing Indian Basmati with semi-dwarf varieties.
Lerma Rojo and Sharbati Sonora are high yielding varieties of wheat. Co-667 is a variety of soybean.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2018
A Newtonian fluid fills the clearance between a shaft and a sleeve. When a force of 800 N is applied to the shaft, parallel to the sleeve, the shaft attains a speed of 2 cm s-1. If a force of 2.4 kN is applied instead, the shaft would move with a speed of

Options:
A 2 cm s-1
B 15 cm s-1
C 6 cm s-1
D none of these

The correct answer is:
6 cm s-1

Solution:
Viscous force is given by
F=Î·Avd
F1F2=v1v2
8002400=2v2
v2=6 cm/s

Topic: Physics, Mechanical Properties of Fluids, NEET
A nitrogen fixing microbe associated with Azolla in rice fields is

Options:
A Spirulina
B Anabaena
C Frankia
D Tolypothrix

The correct answer is:
Anabaena

Solution:
Anabaena azollae is a free-living nitrogen fixing blue-green alga or cyanobacterium but it may also live symbiotically in the leaf cavities of Azolla, an aquatic, free floating, fresh water pteridophyte (fern). This cyanobacterium has nitrogenase enzyme, therefore can fix nitrogen. If this fern is grown in paddy (rice) fields, a remarkable $50 \%$ increase in yield can be noticed.

Topic: Biology, Mineral Nutrition, NEET, NEET 2012 (Screening)
A non stoichiometric compound Cu1.8S is formed due to incorporation of Cu2+ ions in the lattice of cuprous sulphide. What percentage of Cu2+ ion in the total copper content is present in the compound?

Options:
A 88.88
B 11.11
C 99.8
D 89.8

The correct answer is:
11.11

Solution:
Let xCu2+ and (1.8-x)Cu+ ions are present in the compound Cu1.8S. Compound is electrically neutral.
So +2x+(1.8-x)-2=x=0.2
%Cu2+=0.21.8×100=11.11

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A non-conducting ring of radius r has charge q distributed uniformly over it. If it rotates with an angular velocity Ï, then the equivalent current is

Options:
A qω2π
B 2πqω
C qω2πr
D qrω

The correct answer is:
qω2π

Solution:
Non conducting charge ring rotates with an angular velocity Ï is equivalent to electric current
I=qT = qv2Ïr = qrÏ2Ïr=qÏ2Ï

Topic: Physics, Current Electricity, NEET
A non-ideal solution was prepared by mixing 30 mL chloroform and 50 mL acetone. The volume of mixture will be

Options:
A >80 mL
B <80 mL
C =80 mL
D ≥80 mL

The correct answer is:
<80 mL

Solution:
Chloroform and acetone form a non-ideal solution, in which A...B type interaction are more than A...A and B...B type interactions due to H-bonding. Hence, the solution shows negative deviation form Raoult's law. i.e.,
△Vmix=-ve,
△Hmix=-ve,
∴ Total volume of solution = less than (30+50)mL
or 80mL

Topic: Chemistry, Solutions, NEET
A non-medullated axon does not have

Options:
A Schwann cells.
B nodes of Ranvier.
C axolemma.
D neurofibrils.

The correct answer is:
nodes of Ranvier.

Solution:
Nodes of Ranvier are found to occur along a myelinated axon of the neuron. It is also called as myelin-sheath gaps. The axolemma or the membrane at the node is exposed to the extracellular space. These gaps are not insulated and highly enriched in ion channels. The nodal area participates in the exchange of ions which is necessary for regeneration of the action potential.
Non-medullated axons are without myelin sheath.
Schwann cells, a type of glial cells, help in keeping the peripheral nerve fibres alive. It is found in both myelinated and non-myelinated axons. Schwann cells form the myelin sheath in the myelinated axon.

Topic: Biology, Neural Control and Coordination, NEET
A nonmetallic element of group 13, used in making bulletproof vests is an extremely hard solid of black color. It can exist in many allotropic forms and has an unusually high melting point. Its trifluoride acts as a Lewis acid towards ammonia. The element exhibits a maximum covalency of four. Identify the element and write the reaction of its trifluoride with ammonia.

Options:
A Carbon
B Boron
C Silicon
D Phosphorus

The correct answer is:
Boron

Solution:
Correct Option is : (B)
Boron

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
A non-nutrient chemical which serves as an intercellular messenger and is secreted in trace amounts is called ___.

Options:
A Exocrine gland
B Endocrine gland
C Hormone
D None of the above

The correct answer is:
Hormone

Solution:
A non-nutrient chemical which serves as an intercellular messenger and is secreted in trace amounts is called hormone

Topic: Biology, Chemical Coordination and Integration, NEET
A non-proteinaceous enzyme is:

Options:
A Lysozyme
B Ribozyme
C Ligase
D Deoxyribonuclease

The correct answer is:
Ribozyme

Solution:
Aâ¯ribozymeâ¯is a ribonucleic acid (RNA) enzyme that catalyses a chemical reaction. It is also called catalytic RNA, ribozymesâ¯are found in the ribosome where they join amino acids to form protein chains. In the bacterial ribosomes, 23S rRNA of the large subunit of ribosome helps in forming the peptide bond during prokaryotic translation.

Topic: Biology, Biomolecules (B), NEET, NEET 2016 (Phase 2)
A non-stoichiometric compound Cu1.8S is formed due to the incorporation of Cu2+ ions in the lattice. What is the mole percentage of Cu2+ present in the compound?

Options:
A 88.88
B 89.8
C 63.5%
D 11.11

The correct answer is:
11.11

Solution:
In one mole, Let there be Cu+2 =x moles
â´ââ Cu + =1.8âx moles
∵ 2x+ 1.8−x 1=+2
x=0.2
% Cu +2 = 0.2 1.8 ×100=11.11%

Topic: Chemistry, Solid State, NEET
A non-uniform rod of length l having mass density Î»(x)=A+Bx2 is placed on -x-axis with its ends at a,0 and a+l, 0. The force it would exert on a point mass m kept at the origin is

Options:
A GmAlaa+2+Bl
B GmAlaa-l-Bl
C GmAlaa+l+Bl
D GmAlaa+l-Bl

The correct answer is:
GmAlaa+l+Bl

Solution:
F=∫aa+lGm dMx2=Gm∫aa+l(A+Bx2)dxx2
=Gm∫aa+lAx2dx+∫aa+lB  dx
=Gm[A[-1x]aa+l+Bl]
=GmA1a-1a+l+Bl

=GmAlaa+l+Bl

Topic: Physics, Gravitation, NEET
A non-uniform thin rod of length L is placed along the x-axis such that one of its ends is at the origin. The linear mass density of the rod is Î»=Î»0 x. Find the distance of the centre of mass of the rod from the origin.

Options:
A L2
B 2L3
C L4
D L5

The correct answer is:
2L3

Solution:
The mass of considered element is
dm = Î» dx = Î» 0 xdx

â´ x cm = â« 0 L xdm â« dm = â« 0 L x Î» 0 xdx â« 0 L Î» 0 xdx
= Î» 0 x 3 3 0 L Î» 0 x 2 2 0 L = Î» 0 L 3 3 Î» 0 L 2 2 = 2 3 L

Topic: Physics, Center of Mass Momentum and Collision, NEET
A non-viscous liquid is flowing through a horizontal pipe as shown in the figure. Three tubes A, B and C are connected to the pipe. The radii of the tubes A, B and C at the junction are 2 cm, 1 cm and 2 cm respectively. It can be said that the

Options:
A height of the liquid in the tube A is maximum
B height of the liquid in the tubes A and B is the same
C height of the liquid in all the three tubes is the same
D height of the liquid in the tubes A and C is the same

The correct answer is:
height of the liquid in the tubes A and C is the same

Solution:
As cross-section areas of both the tubes A and C are same and tube is horizontal.
Hence according to equation of continuity
vA=vC
Therefore according to Bernoulli's theorem
PA=PC
i.e. height of liquid is same in both the tubes A and C.

Topic: Physics, Mechanical Properties of Fluids, NEET
A non-zero external force acts on a system of particles. The velocity and acceleration of the centre of mass are found to be v0 and a0 at an instant t. It is possible that

Options:
A v0=0,a0=0
B v0=0,a0â 0
C v0â 0,a0=0
D v0≠0,a0=1

The correct answer is:
v0=0,a0≠0

Solution:
If the external force is non-zero then the acceleration of the centre of mass must be non-zero a0=FextMâ 0. However, at a particular instant of time velocity of the centre of mass may be zero or non-zero. Hence v0=0,a0â 0
From the equation of velocity is in +ve direction and acceleration is in -ve direction, then at some moment velocity will become 0.

Topic: Physics, Center of Mass Momentum and Collision, NEET
A normal girl, whose mother is haemophilic marries a male with no ancestral history of haemophilia.
What will be the possible phenotypes of the offsprings ?
(a) Haemophilic son and haemophilic daughter.
(b) Haemophilic son and carrier daughter.
(c) Normal daughter and normal son.
(d) Normal son and haemophilic daughter.
Choose the most appropriate answer from the options given below :

Options:
A (a) and (b) only
B (b) and (c) only
C (a) and (d) only
D (b) and (d) only

The correct answer is:
(b) and (c) only

Solution:
The girl will be carrier (Heterozygous) for Hemophilia while the male have no history of Haemophilia. This can be explained with the help of following cross:

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2022 (Phase 2)
A normal visioned man whose father was color blind marries a normal vision woman whose father was also color blind. They have their first child as daughter. What are the chances that this child would be color blind?

Options:
A 10 %
B zero
C 50 %
D 100 %

The correct answer is:
zero

Solution:
Normal Visioned man genotype: XY. Normal visioned but carrier woman genotype: XXc (sincer her father is colour blind). Their daughter can have two genotypes XX or XXc. In both these cases the daughter will not be colourblind.

Topic: Biology, Principles of Inheritance and Variation, NEET
A normal woman whose father had haemophilia married a normal man. What is the chance of occurrence of disease in the progeny?

Options:
A 100%
B 50%
C 25%
D 75%

The correct answer is:
25%

Solution:
Haemophilia is a sex-linked recessive disease. Normally, blood clotting occurs as a result of cascade of proteins. In this disease, one of the proteins participating in this cascade is affected. As a result, the clotting of blood is also affected. A small cut may bleed for a long time. Male child may become haemophilic when his mother is at least a carrier of haemophilia.

Topic: Biology, Principles of Inheritance and Variation, NEET
A normal woman, whose father was colour-blind is married to a normal man. The sons would be:

Options:
A 75% colour-blind
B $50 \%$ colour-blind
C all normal
D all colour-blind.

The correct answer is:
$50 \%$ colour-blind

Solution:
The genotype of normal woman with colourblind father $=\mathrm{XX}^{\mathrm{h}}$
The genotype of normal $\operatorname{man}=X Y$
$\therefore 50 \%$ of the sons would be colourblind.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2004
A normal-visioned man whose father was colour blind, marries a woman whose father was also colourblind. They have their first child as a daughter. What are the chances that this child would be colourblind?

Options:
A $100 \%$
B $0 \%$
C $25 \%$
D $50 \%$

The correct answer is:
$0 \%$

Solution:
Colour blindness is an X-linked disease. So, woman whose father was colourblind will be carrier for the disease.
So, possibility of a colourblind daughter (i.e., $X^{\mathrm{c}} \mathrm{X}^{\mathrm{c}}$ in $\mathrm{F}_1$ generation is $0 \%$.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2012 (Screening)
A normal-visioned man, whose father was colourblind, marries a woman whose father was also colourblind. They have their first child as a daughter. What is the chance that this child would be colourblind?

Options:
A 100%
B Zero percent
C 25%
D 50%

The correct answer is:
Zero percent

Solution:
If a normal visioned man marries a woman whose father was also colour blind. Then his wife would be carried of this disease if her mother was normal. This trait passed in to children but daughters produce by this couple are carrier not the colour blind. 50% of sons would be colour blind.

Topic: Biology, Principles of Inheritance and Variation, NEET
A NPN transistor is connected in common emitter configuration in a given amplifier. A load resistance of 800 Ω is connected in the collector circuit and the voltage drop across it is 0.8 V. If the current amplification factor is 0.96 and the input resistance of the circuit is 192 Ω , the voltage gain and the power gain of the amplifier will respectively be:

Options:
A 4, 3.84
B 3.69, 3.84
C 4, 4
D 4, 3.96

The correct answer is:
4, 3.84

Solution:
Given,

The value of the load resistance is 800 Ω.

Voltage drop across the load resistance is, 0.8 V

Voltage gain =β.RCRB
V=0.9680192
V=96×8192=4
And power gain of the amplifier is given by,
βac.Av
=0.96×4
=3.84

Topic: Physics, Semiconductors, NEET, NEET 2016 (Phase 1)
A nuclear reaction given by ${ }_z \mathrm{X}^{\mathrm{A}} \rightarrow{ }_{z+1} \mathrm{Y}^{\mathrm{A}}+-1^{e^0}+\bar{v}$ represents:

Options:
A $\beta$-decay
B $\gamma$-decay
C fusion
D fission

The correct answer is:
$\beta$-decay

Solution:
Emission of electron \(\left(\mathrm{e}^{-}\right)+\)antineutrino \(\left(\gamma^{-}\right)\)
\(\Rightarrow \beta\)-decay.

Topic: Physics, Nuclear Physics, NEET, NEET 2003
A nuclear transformation is given by Yn, Î±â 3Li7. The nucleus of the element Y is

Options:
A 5Be11
B 5B10
C 5B9
D 6C12

The correct answer is:
5B10

Solution:
Yn, α means the nucleus splits into α -particle and neutrons
i.e., ZYA+ 0n1→ 3Li7+ 2He4
So, A+1=7+4
⇒ A=10
and Z+0=3+2
Z=5
Hence, the nucleus of elements Y is boron
i.e., 5Y10= 5B10

Topic: Physics, Nuclear Physics, NEET
A nucleoside differs from a nucleotide. It lacks the:

Options:
A base
B sugar
C phosphate group
D hydroxyl group

The correct answer is:
sugar

Solution:
A nucleoside only has nitogenous base and a phosphate group.

Topic: Biology, Molecular Basis of Inheritance, NEET
A nucleosome is a portion of the chromonema containing

Options:
A Only histones
B Both DNA and histones
C Only DNA
D Both DNA and RNA

The correct answer is:
Both DNA and histones

Solution:
The chromatin is formed of a series of repeating units called nucleosomes. Each nucleosome consists of a chain of DNA twist around a histone octamer. The core of the nucleosome consists of four histones namely H2A, H2B, H, and H4. Another histone namely H1 is associated with the linker region.

Topic: Biology, Molecular Basis of Inheritance, NEET
A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2:1. The ratio of their nuclear size will be

Options:
A 213 :1
B 1 :312
C 312 :1
D 1 :213

The correct answer is:
1 :213

Solution:
According to the law of conservation of momentum, we get
m1v1=m2v2
Or m1m2=v2v1
But m=43πr3 ρ or m ∝r3
⇒ m1m2=r12r23
∴ r13r23=v2v1=12
⇒ r1r2=1213
∴ r1 :r2=1 :213

Topic: Physics, Nuclear Physics, NEET
A nucleus is bombarded with a high-speed neutron so that resulting nucleus is a radioactive one. This phenomenon is called

Options:
A Artificial radioactivity
B Fusion
C Fission
D Radioactivity

The correct answer is:
Fission

Solution:
When atoms of an element are bombarded by neutrons, the atomic nuclei are (artificially) disintegrated and emit lighter particle (eg. Î±-particle, Î²-particle,proton etc.).
Sometimes a neutron is observed by the nucleus which is converted into its heavier isotope and energy is emitted in the form of Î³-photons.
This process in which heavy nucleus is broken into two nearly equal fragments is called nuclear fission.

Topic: Physics, Nuclear Physics, NEET
A nucleus ${ }_n^m X$ emits one $\alpha$-particle and two $\beta^{-}$particles. The resulting nucleus is

Options:
A ${ }_n^{m-6} Z$
B ${ }_n^{m-4} X$
C ${ }_{n-2}^{m-4} Y$
D ${ }_{n-4}^{m-6} Z$

The correct answer is:
${ }_n^{m-4} X$

Solution:
${ }_n^m X \stackrel{\alpha}{\rightarrow}{ }_{n-2} X^{m-4} \stackrel{2 \beta}{\rightarrow}{ }_n X^{m-4}$

Topic: Physics, Nuclear Physics, NEET, NEET 2011 (Screening)
A nucleus of 84Po210 originally at rest emits an Î±-particle with speed v. What will be recoil speed of the daughter nucleus?

Options:
A 4V/206
B 4V/214
C V/206
D V/214

The correct answer is:
4V/206

Solution:
Applying principle of conservation of linear momentum,
Negative sign for recoil speed
−( 206 ) V R +4V=0
V R = 4V 206

Topic: Physics, Nuclear Physics, NEET
A nucleus of mass M emits an X-ray photon of frequency n. Energy lost by the nucleus is given as

Options:
A hυ
B h2υ22MC2
C hυ1-hυ2MC2
D hυ1+hυ2MC2

The correct answer is:
hυ1+hυ2MC2

Solution:
Loss = kinetic energy of Nucleus + Energy of photon
Loss =p22M+hυ
Loss =hυC22M+hυ
Loss =hυ1+hυ2MC2

Topic: Physics, Atomic Physics, NEET
A nucleus of mass number 189 splits into two nuclei having mass number 125 and 64. The ratio of radius of two daughter nuclei respectively is:

Options:
A 4:5
B 5:4
C 25:16
D 1:1

The correct answer is:
5:4

Solution:
The relation between radius R of nuclei and mass number A is given by, R=R0A13, where R0 is fermi constant.

Thus, R1R2=A1A213=1256413=54.

Topic: Physics, Nuclear Physics, NEET, NEET 2022 (Phase 1)
A nucleus of uranium decays at rest into nuclei of thorium and helium. Then:

Options:
A The helium nucleus has less momentum then the thorium nucleus
B The helium nucleus has more momentum than the thorium nucleus
C The helium nucleus has less kinetic energy than the thorium nucleus
D The helium nucleus has more kinetic energy than the thorium nucleus

The correct answer is:
The helium nucleus has more kinetic energy than the thorium nucleus

Solution:
U→Th+α
KETh=P22mTh, KEα=P22mα
Since mα is less so KEα will be mone.

Topic: Physics, Nuclear Physics, NEET, NEET 2015 (Phase 2)
A nucleus represented by the symbol ${ }_{\mathrm{A}}^{\mathrm{A}} \mathrm{X}$ has:

Options:
A $\mathrm{Z}$ neutrons and $\mathrm{A}-\mathrm{Z}$ protons
B $\mathrm{Z}$ protons and $\mathrm{A}-\mathrm{Z}$ neutrons
C $\mathrm{Z}$ protons and $\mathrm{A}$ neutrons
D $\mathrm{Z}$ protons and $\mathrm{Z}-\mathrm{A}$ neutrons

The correct answer is:
$\mathrm{Z}$ protons and $\mathrm{A}-\mathrm{Z}$ neutrons

Solution:
$2=$ Number of protons
$A=$ Total number of protons and number
$\therefore$ Number of neutrons $\mathrm{A}=\mathrm{A}-\mathrm{Z}$

Topic: Physics, Nuclear Physics, NEET, NEET 2004
A nucleus with mass number 220 initially at rest emits an Î±-particle. If the Q value of the reaction is 5.5 MeV, then calculate the kinetic energy of the Î±-particle.

Options:
A 4.4 MeV
B 5.4 MeV
C 5.6 MeV
D 6.5 MeV

The correct answer is:
5.4 MeV

Solution:
Given that K1+K2=5.5 MeV ... (i)
From conservation of linear momentum,
p1=p2
2K1(216m)=2K2(4m) (As p=2Km)
K2=54K1 ... (ii)
Solving Eqs. (i) and (ii), we get
K2=K.E. of Î±-particle 5.4 MeV.

Topic: Physics, Nuclear Physics, NEET
A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the binding energy in the process is :

Options:
A 0.9MeV
B 9.4MeV
C 216 MeV
D 804 MeV

The correct answer is:
216 MeV

Solution:
Given binding energy per nucleon of X, Y & Z are, 7.6 MeV, 8.5 MeV & 8.5 MeV respectively.

⇒B.E. =(120×8.5)+(120×8.5)-240×7.6

⇒B.E.=2040-1824=216 MeV

Topic: Physics, Nuclear Physics, NEET, NEET 2021
A nucleus ${ }^A_Z X$ has mass represented by $M(A, Z)$. If $M_p$ and $M_n$ denote the mass of proton and neutron respectively and B.E. the binding energy in $\mathrm{MeV}$, then

Options:
A B. E. $=\left[Z M_p+(A-Z) M_n-M(A, Z)\right] c^2$
B B. E. $=\left[Z M_p+A M_n-M(A, Z)\right] c^2$
C B. E. $=M(A, Z)-Z M_p-(A-Z) M_n$
D B. E. $=\left[M(A, Z)-Z M_p-(A-Z) M_n\right] c^2$.

The correct answer is:
B. E. $=\left[Z M_p+(A-Z) M_n-M(A, Z)\right] c^2$

Solution:
In the case of formation of a nucleus the evolution of energy equal to the binding energy of the nucleus takes place due to disappearance of a fraction of the total mass. If the quantity of mass disappearing is \(\Delta \mathrm{M}\). then the binding energy is
\(\mathrm{BE}=\Delta \mathrm{Mc}^2\)
From the above discussion, it is clear that the mass of the nucleus must be less than the sum of the masses of the constituent neutrons and protons. We can then write
\(\Delta \mathrm{M}=\mathrm{ZM}_{\mathrm{p}}+\mathrm{NM}_{\mathrm{p}}-\mathrm{M}(\mathrm{A}, \mathrm{Z})\)
Where \(M(A, Z)\) is the mass of the atom of mass number \(A\) and atomic number \(Z\). Hence, the binding energy of the nucleus is
\(\begin{aligned}
& \mathrm{BE}=\left[\mathrm{ZM}_{\mathrm{p}}+\mathrm{NM}_{\mathrm{n}}-\mathrm{M}(\mathrm{A}, \mathrm{Z})\right] \mathrm{c}^2 \\
& \mathrm{BE}=\left[\mathrm{ZM}_{\mathrm{p}}+(\mathrm{A}-\mathrm{Z}) \mathrm{M}_{\mathrm{n}}-\mathrm{M}(\mathrm{A}, \mathrm{Z})\right] \mathrm{c}^2
\end{aligned}\)
Where \(N=A-Z=\) Number of neutrons.

Topic: Physics, Nuclear Physics, NEET, NEET 2007
A nuclide of an alkaline earth metal under goes radioactive decay by emission of the $\alpha$-particles in succession. The group of the periodic table to which the resulting daughter element would belong is:

Options:
A Gr. 4
B Gr. 6
C Gr. 14
D Gr. 16

The correct answer is:
Gr. 14

Solution:
After decay of radioactive element, the last element (stable) $\mathrm{Pb}$ is obtained which belongs to group 14 of periodic table.

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2005
A number of bivalents are 8 in prophase I. What will be the number of chromosomes during anaphase II?

Options:
A 8
B 4
C 16
D 32

The correct answer is:
8

Solution:
At prophase I the number of bivalents is 8, i.e., 2n = 16. Thus, after meiosis I, the daughter cell would have 16/2 = 8 chromosomes. Therefore, the number of the chromosome at anaphase II will be 8.

Topic: Biology, Cell Cycle and Cell Division, NEET
A number of very thin, irregular-walled and vascularised bag-like structures are called as

Options:
A Cardiac notches
B Alveoli
C Nephron
D Neurons

The correct answer is:
Alveoli

Solution:
Each lung, performing gaseous exchange, consists of a small balloon-like unit called an alveolus. The alveolus is lined by simple squamous epithelium. There are around 480 millions of very thin, irregular-walled, vascularised bag-like structures called alveoli in the human lungs.

Topic: Biology, Breathing and Exchange of Gases, NEET
A nutritionally wild type organism, which does not required any additional growth supplement is known as:

Options:
A phenotype
B holotype
C auxotroph
D prototroph.

The correct answer is:
prototroph.

Solution:
Wild-type bacteria are prototrophic i.e., they can produce colonies on minimal medium (a solution containing only inorganic salts, a source of energy such as a sugar, and water). From these simple substances, the bacteria synthesize all the molecules of life. Auxotrophs are organisms that are unable to produce a particular organic compound required for their growth.
Related Theory
Auxotrophic mutants are a special case of conditional lethal mutants, in which the condition is presence or absence of a specific nutrient in the media.

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