datavorous/entrance-exam-dataset huggingface dataset:
The axial filament of the sperm arises from:

Options:
A Proximal centriole
B Distal centriole
C Acrosome
D Nucleus

The correct answer is:
Distal centriole

Solution:
The posterior centriole of sperm is known as the distal centriole and gives rise to the axial filament of the sperm.

Topic: Biology, Human Reproduction, NEET
The axon terminals of nerves contains

Options:
A synaptic vesicles
B dendrites
C myelin sheath
D axon hillock

The correct answer is:
synaptic vesicles

Solution:
The axon terminal release neurotransmitters from presynaptic knob which binds with the receptors of postsynaptic membrane to transmit electrical impulses across synapse.

Topic: Biology, Neural Control and Coordination, NEET
The axon terminals of nerves contains

Options:
A Vesicles filled with neurotransmitters
B Vesicles filled with calcium ions
C Nissl's granules
D Vesicles filled with neurohormones

The correct answer is:
Vesicles filled with neurotransmitters

Solution:
The axon terminal releases neurotransmitters from presynaptic knob which binds with the receptors of the postsynaptic membrane to transmit electrical impulses across a synapse.

Topic: Biology, Neural Control and Coordination, NEET
The axoneme is found in

Options:
A Cilia
B Flagella
C Microbodies
D Both (A) and (B)

The correct answer is:
Both (A) and (B)

Solution:
The electron microscopic study of a cilium or the flagellum shows that they are covered with the plasma membrane. Their core called the axoneme, possesses a number of microtubules running parallel to the long axis. The axoneme usually has nine doublets of radially arranged peripheral microtubules, and a pair of centrally located microtubules. Such an arrangement of axonemal microtubules is referred to as the 9+2 array.

Topic: Biology, Cell - The Unit of Life, NEET
The B–A–B angle in trigonal planar geometry is:

Options:
A 90
B 180
C 120
D 109.5

The correct answer is:
120

Solution:
Correct Option is : 3
120

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The B–A–B angle of H2O molecule is:

Options:
A 109.5
B 104.5
C 119.5
D 107

The correct answer is:
104.5

Solution:
Correct Option is : 2
104.5

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The B–A–B angle of SO2molecule is:

Options:
A 109.5
B 107
C 104.5
D 119.5

The correct answer is:
119.5

Solution:
Correct Option is : 4
119.5

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The back flow of faecal matter in the large intestine is prevented by the presence of

Options:
A Sphincter of Oddi
B Ileo-caecal valve
C Gastric-oesophageal sphincter
D Pyloric sphincter

The correct answer is:
Ileo-caecal valve

Solution:
The undigested, unabsorbed substances called faeces enter into the caecum of the large intestine through ileo-caecal valve, which prevents the backflow of the faecal matter. It is temporarily stored in the rectum till defaecation.

Topic: Biology, Digestion and Absorption, NEET
The back-side attack on 2-bromobutane by methoxide (CH3O-) gives the product shown below. Which Fischer projection represents 2-bromobutane used as the reactant in this reaction?

Options:
A
B
C
D

The correct answer is:

Solution:
In SN2 reaction, inversion takes place, so configuration get reversed.

Topic: Chemistry, General Organic Chemistry, NEET
The backside of a truck is open and a box of 40 kg is placed 5 m away from the rear end. The coefficient of friction of the box with the surface of the truck is 0.15. The truck starts from rest with acceleration 2 m s-2. Calculate the distance covered by the truck when the box falls off.

Options:
A 20 m

B 30 m
C 40 m
D 50 m

The correct answer is:
20 m

Solution:
The various forces acting on the block are as shown
As the truck moves in forward direction with acceleration , the box experiences a force F in backward direction,
F=ma=40×2= 80 N
in the backward direction.
Its motion will be opposed by the force of friction
f=μ N=μ mg=0.15×40×10=60 N
The acceleration of the box relative to the truck toward the rear end is
a=F-fm=80-6040=0.5 m s-2
If t be the time taken by the box to fall off the truck
s=ut+12at2
5=0+12×0.5×t2
t=20 s
During this time distance covered by the truck
x=0×t+12×2×202=20 m

Topic: Physics, Laws of Motion, NEET
The bacterial cell divides in every minute. If it takes one hour to fill up a cup, how much time will be taken to fill half the cup?

Options:
A 59 minutes
B 30 minutes
C 60 minutes
D 29 minutes

The correct answer is:
59 minutes

Solution:
The number of bacteria becomes double after each division.

Topic: Biology, Cell Cycle and Cell Division, NEET
The bacterial genome consists of

Options:
A DNA and Histone
B DNA or Histone
C DNA without Histone
D Neither DNA nor Histone

The correct answer is:
DNA without Histone

Solution:
Bacteria is a prokaryotic cell, thus lacks a well-defined nucleus. The genetic material is basically naked and not enveloped by nuclear membrane. The genome is single circular double-stranded DNA not associated with histone protein.

Topic: Biology, Cell - The Unit of Life, NEET
The bacterium Bacillus thuringiensis is widely used in contemporary biology as a/an

Options:
A indicator of water pollution
B insecticide
C agent for production of dairy products
D source of industrial enzyme

The correct answer is:
insecticide

Solution:
Bacitlus thuringiensis is a Gram-positive, soil dwelling bacterium, also occurs naturally in the gut of caterpillars of various types of moths and butterflies.
During sporulation, B. thuringiensis forms crystals of proteinaceous insecticidal by cry genes. It was determined that the oy genes are harbored in the plasmids of $B$. thuringiensis strains. Cry toxins have specific activities against species of the orderLepidoptera (moths and butterflies), Diptera (flies and mosquitoes) and Coleoptera (beetles). Thus, $B$. thuringiensis serves as an important reservoir of cry toxins and cry genes for the production of biological insecticides and insect resistant genetically modified crops.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2009 (Screening)
The bacterium (Clostridium botulinum) that causes botulism is:

Options:
A an obligate aerobe
B a facultative anaerobe
C an obligate anaerobe
D a facultative aerobe

The correct answer is:
an obligate anaerobe

Solution:
The bacterium Clostridium botulinum is an obligate anaerobic, endosporeforming, gram positive, rod-shaped bacterium found in soil and in many freshwater sediments.
Related Theory
Clostridium botulinum causes botulism (food poisoning) occurs when a toxin produced by the bacteria is consumed in improperly preserved foods. The disease is caused by a potent neurotoxin produced by the bacteria.

Topic: Biology, Biological Classification, NEET, NEET 2006
The ball starts falling from some height. After falling height h, it breaks into two parts m1 and m2. Finally, they reach the earth surface. Choose the correct option.

Options:
A R will depend on h
B If m1>m2, then x1<x2
C The centre of mass of the two balls (system) will shift towards the heavier part
D None of the above

The correct answer is:
If m1>m2, then x1<x2

Solution:
If there is no external force acting then linear momentum of system will be conserved.
From conservation of momentum theorem, the X component of momentum is conserved
So, m1V1 = m2V2
â R will not depend on the vertical height h.
if m1 > m2
Then, v1 < v2
So, x1 < x2
The CM of the system of two balls will lie on the same vertical line. It will come down vertically due to gravitational force but it will not move horizontally as (Fext)x = 0

Topic: Physics, Center of Mass Momentum and Collision, NEET
The balloon-shaped structures called tyloses

Options:
A Originate in the lumen of vessels
B Characterize the sapwood
C Are extensions of xylem parenchyma cells into vessels
D Are linked to the ascent of sap through xylem vessels

The correct answer is:
Are extensions of xylem parenchyma cells into vessels

Solution:
Tyloses are balloon-shaped structures in xylem vessels developed by xylem parenchyma cells.
Malpighi (1686) for the first time reported the ‘balloon-shaped sacs’ in the lumen of vessels of heartwood and he named it as tyloses based on the Greek word ‘Thyllen’ meaning ‘bag’ or ‘container’. Tyloses development occurs through the pit-pairs connecting parenchyma and tracheids or vessels. Sometimes many tyloses originate from a single parenchymatous cell. The tyloses completely fill the lumen rich in the tracheids and vessels of heartwood than sapwood.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2016 (Phase 2)
The Balmer series in the hydrogen spectrum corresponds to the transition from n = 2 to n = 3,4,... This series lies in the visible region. Calculate the wave number of line associated with the transition in Balmer series when the electron moves to n = 4 orbit. (Given RH = 109677 cm⁻¹).

Options:
A 20564.44 cm⁻¹
B 273.63 cm⁻¹
C 136.815 cm⁻¹
D 41.29 cm⁻¹

The correct answer is:
20564.44 cm⁻¹

Solution:
Correct Option is : (A)
20564.44 cm⁻¹

Topic: Chemistry, Structure of Atom, NEET
The band gaps of an insulator, conductor and semiconductor are respectively E g 1, E g 2 and E g 3. The relationship between them is given as

Options:
A E g 1 > E g 2 < E g 3
B E g 1 > E g 2 > E g 3
C E g 1 < E g 2 > E g 3
D E g 1 < E g 2 < E g 3

The correct answer is:
E g 1 > E g 2 < E g 3

Solution:
Band gap of insulator is highest, while that of conductor is least. So,
E g 1 > E g 3 > E g 2
E g 1 > E g 2
E g 3 > E g 2
E g 1 > E g 2 < E g 3

Topic: Physics, Semiconductors, NEET
The barrier potential of a p-n junction depends on:
a type of semi-conductor material
b amount of doping
c temperature
Which one of the following is correct?

Options:
A a and b only
B b only
C b and c only
D a, b and c

The correct answer is:
a, b and c

Solution:
The barrier potential in a p-n junction diode is a barrier which to be overcome by additional force in order to cross.

The barrier potential depends on various factors.

Different semiconductor material will have different barrier potential for the same amount of doping thus it depends on the type of semiconductor material.

For same semiconductor it is observed that for different amount of doping barrier potential is different.

For same semiconductor barrier potential changes with temperature thus barrier potential also depends on the temperature.

The barrier potential depends on type of semiconductor (for Si Vb=0.7 volt and for Ge Vb=0.3 volt), amount of doping and also on the temperature.

Topic: Physics, Semiconductors, NEET, NEET 2014
The basic character of MgO, BaO, Na2O and FeO follow the order

Options:
A MgO < FeO < BaO < Na2O
B FeO < MgO < Na2O < BaO
C FeO < MgO < BaO < Na2O
D Na2O < MgO < FeO < BaO

The correct answer is:
FeO < MgO < BaO < Na2O

Solution:
The basic character of metal oxides decreases from left to right in a period and increases down the group.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
The basic character of the transition metal monoxides follows the order:
(Atomic nos. $\mathrm{Ti}=22, \mathrm{~V}=23, \mathrm{Cr}=24, \mathrm{Fe}$ $=26$ )

Options:
A $\mathrm{VO}>\mathrm{CrO}>\mathrm{TiO}>\mathrm{FeO}$
B $\mathrm{CrO}>\mathrm{VO}>\mathrm{FeO}>\mathrm{TiO}$
C $\mathrm{TiO}>\mathrm{FeO}>\mathrm{VO}>\mathrm{CrO}$
D $\mathrm{TiO}>\mathrm{VO}>\mathrm{CrO}>\mathrm{FeO}$

The correct answer is:
$\mathrm{TiO}>\mathrm{VO}>\mathrm{CrO}>\mathrm{FeO}$

Solution:
As we move from left side to right side in the first transition series atomic number of elements increases and no. of electrons also increases. Size of elements on moving left to right decreases because effective nuclear charge of nucleus on the outermost shell electron increases. It means electronegativity of atoms increases from left to right, then the power of donation of electrons decreases which means the basicity of that element decreases.
Caution
Students get confused on the order of size of elements in the series, because order of size of elements is not regular in the periods and groups. In periods of increasing atomic number size decreases and on the groups with the increase of atomic number size also increases.

Topic: Chemistry, d and f Block Elements, NEET, NEET 2003
The basic components of a PCR reaction includes

Options:
A amplifier, primers, dNTP's, Taq polymerase and a buffer
B amplifier, template DNA, primers, dNTP's and a buffer
C template DNA, primers, Taq polymerase, amplifier and a buffer
D template DNA, primers, dNTP's, Taq polymerase and a buffer

The correct answer is:
template DNA, primers, dNTP's, Taq polymerase and a buffer

Solution:
The basic components of a PCR reaction are a DNA template, short RNA or DNA primers, nucleotides (dNTP's), Taq polymerase (DNA polymerase), and a buffer. The DNA template usually is the sample DNA, which needs to be amplified.

Topic: Biology, Biotechnology - Principles and Processes, NEET
The basic structural unit of silicates is

Options:
A $\mathrm{SiO}^{-}$
B $\mathrm{SiO}_4^{4-}$
C $\mathrm{SiO}_3^{2-}$
D $\mathrm{SiO}_4^{2-}$

The correct answer is:
$\mathrm{SiO}_4^{4-}$

Solution:
The basic building unit of all silicates is the tetrahedral $\mathrm{SiO}_4^{4-}$. It is represented as

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET, NEET 2013 (All India)
The basic structural unit of silicates is

Options:
A SiO44-
B SiO32-
C SiO42-
D SiO-

The correct answer is:
SiO44-

Solution:
The basic structural unit of silicates is SiO44-

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
The basis for DNA fingerprinting is

Options:
A occurrence of polymorphism.
B phenotypic difference between individuals.
C availability of cloned DNA.
D aminoacyl tRNA binding to active ribosomal site.

The correct answer is:
occurrence of polymorphism.

Solution:
DNA fingerprinting is a technique to identify a person on the basis of his DNA which is specific for every individual. This technique for first time was developed by Alec jeffreys and his colleagues lecester University in U.K. DNA of am individual carries specific sequences of nucleotides which are repeated many times throughout the length of DNA This reveals plymorphism in DNA. These are inherited.Each individual inherits these repeats from his/her parents which are used as gentic markers in personal identity test.

Topic: Biology, Molecular Basis of Inheritance, NEET
The basis for the classification of elements in the modern periodic table is

Options:
A Atomic Number
B Atomic weight
C Atomic volume
D Equivalent weight

The correct answer is:
Atomic Number

Solution:
The properties of elements are periodic function of their atomic number

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
The below image summarises the cell cycle of a typical eukaryotic cell. Using the image as a reference, determine the activities occurring at the G1 phase of the cell cycle.

Options:
A Growth of the cell
B Replication of the DNA
C Formation of mitotic spindle
D Breakdown of nuclear membrane

The correct answer is:
Growth of the cell

Solution:
G1 phase is part of the interphase in the cell cycle process of a eukaryotic cell. G1 phase is the stage between mitosis and initiation of DNA replication (S phase). In the G1 phase, the cell is found to be metabolically active and shows continuous growth. S or synthesis phase is where DNA synthesis or replication takes place in the cell.

Topic: Biology, Cell Cycle and Cell Division, NEET
The below pathway is observed in which of the following option?

Options:
A 'Flaver Savr' tomato
B Canolla
C Golden rice
D Bt-Brinjal

The correct answer is:
Golden rice

Solution:
Golden rice a variety of Oryza sativa is produced through the genetic engineering to biosynthesize beta carotene, a precursor of pro-vitamin A in the edible parts of rice. Golden rice is vitamin A rich variety developed by rDNA technology and used in the treatment of vitamin A deficiency.

Topic: Biology, Biotechnology and Its Applications, NEET
The best candidate for species-specific, narrow spectrum insecticidal applications is

Options:
A Nucleopolyhedroviruses
B DDT
C Azospirillum
D Glomus

The correct answer is:
Nucleopolyhedroviruses

Solution:
Baculoviruses are pathogens that attack insects and other arthropods. The majority of baculoviruses used as biological control agents are in the genus Nucleopolyhedrovirus. These viruses are excellent candidates for species-specific, narrow spectrum insecticidal applications. They have been shown to have no negative impacts on plants, mammals, birds, fish, or even on non-target insects. This is especially desirable when beneficial insects are being conserved to aid in overall integrated pest management (IPM) program, or when an ecologically sensitive area is being treated.

Topic: Biology, Microbes in Human Welfare, NEET
The best example for pleiotropy is:

Options:
A Skin colour
B Phenylketonuria
C Colour blindness
D ABO Blood group

The correct answer is:
Phenylketonuria

Solution:
In pleiotropy, a single gene product may produce more than one effect or control several phenotypes depending on its position. The basis of pleiotropy is the interrelationship between the metabolic pathways that may contribute towards different phenotypes.

Example: Phenylketonuria, sickle-cell anemia, etc.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2020 (Phase 2)
The best material for demonstrating streaming movements of protoplasm within living cells is

Options:
A staminal hairs of Tradescantia
B onion peelings
C pith cells of angiosperms
D root hairs of Tradescantia

The correct answer is:
staminal hairs of Tradescantia

Solution:
When the movement of protoplasm occurs around different vaculoes in different directions within cell e.g., staminal hair of Tradescantia.

Topic: Biology, Plant - Growth and Development, NEET
The best method for the separation of naphthalene and benzoic acid from their mixture is::

Options:
A distillation
B sublimation
C chromatography
D crystallisation

The correct answer is:
sublimation

Solution:
Naphthalene can sublime directly to vapour upon heating, but benzoic acid do not.
Related Theory
When benzoic acid is heated in presence of $a$ strong dehydrating agent like $\mathrm{P}_2 \mathrm{O}_5$ or $\mathrm{H}_2 \mathrm{SO}_f$ it forms benzoic anhydride.
$$
2 \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH} \underset{\mathrm{BaOl}_5}{\longrightarrow}\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CO}\right)_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O}
$$
Caution
Naphthalene is volatile and benzoic acid is non-volatile due to the formation of dimer via hydrogen bonding (intermolecular).

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2005
The best method to prepare cyclohexene from cyclohexanol is by using

Options:
A Conc.HCl+ZnCl2
B Conc. H3PO4
C HBr
D Conc.Â Â HCl

The correct answer is:
Conc. H3PO4

Solution:
In the rest of the cases, substitutions take place where OH- will be replaced by halide ion.

Topic: Chemistry, Hydrocarbons, NEET
The best reagent to convert pent-3-en-2-ol into pent-3-en-2-one is

Options:
A pyridinium chloro-chromate
B chromic anhydride in glacial acetic acid
C acidic dichromate
D acidic permanganate

The correct answer is:
pyridinium chloro-chromate

Solution:
Only suitable reagent is pyridinium chloro-chromate. Other options given will affect(C=C) bond.

Topic: Chemistry, Aldehydes and Ketones, NEET
The bezylic carbon in C 6 H 5 − CH 2 − CH 2 −COOH is

Options:
A linearly placed with respect to benzene ring and carbonyl group
B trigonaly shaped
C tetrahedral in shape
D at 90° to the plane of benzene being

The correct answer is:
tetrahedral in shape

Solution:
The bezylic carbon in C 6 H 5 − CH 2 − CH 2 −COOH is sp3 hybridised so it is trigonaly shaped.
According to VSEPR theory, the orientation of these electron pairs is tetrahedral.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The binding energies per nucleon for deuteron (H21) and helium (He42) are 1.1 MeV and 7.0 MeV respectively. The energy released when two deuterons fuse to form a helium nucleus (He42) is

Options:
A 23.6âMeV
B 20.3 MeV
C 4.4 MeV
D 28 MeV

The correct answer is:
23.6 MeV

Solution:
12H  +  12H  →  24He
The binding energy of two deuterons = 2  ×  [1.1  ×  2]  =  4.4 MeV
The binding energy of one helium nucleus = 4×7=28  MeV
∴ the total energy released when two deuterons fuse to form a helium nucleus =28−4.4=23.6  MeV

Topic: Physics, Nuclear Physics, NEET
The binding energy of an element is 64 MeV. If Binding energy/nucleon is 6.4, then the number of nucleons are

Options:
A 10
B 64
C 16
D 6

The correct answer is:
10

Solution:
Binding energy = 64 MeV
Binding energy/nucleon = 6.4
∴ Number of nucleon = 6 4 6.4 = 1 0

Topic: Chemistry, Chemical Kinetics, NEET
The binding energy of deuteron is 2.2 $\mathrm{MeV}$ and that of ${ }_2^4 \mathrm{He}$ is $28 \mathrm{MeV}$. If two deuterons are fused to form one ${ }_2^4 \mathrm{He}$ then the energy released is:

Options:
A $30.2 \mathrm{MeV}$
B $25.8 \mathrm{MeV}$
C $23.6 \mathrm{MeV}$
D $19.2 \mathrm{MeV}$

The correct answer is:
$23.6 \mathrm{MeV}$

Solution:
Energy released $=28-2 \times 2.2 =28-4.1 =23.6 \mathrm{MeV}$

Topic: Physics, Nuclear Physics, NEET, NEET 2006
The binding energy of deuteron is 2.2âMeV and that of 2 4 He is 28âMeV. If two deuterons are fused to form one 2 4 He , then the energy released is

Options:
A 30.2 MeV
B 25.8 MeV
C 23.6 MeV
D 19.2 MeV

The correct answer is:
23.6 MeV

Solution:
1 2 H + 1 2 H ⟶ 2 4 He + energy
Energy released =B.E. of 2 4 He−2(B.E. of 1 2 H)
=28−2(2.2)=28−4.4=23.6 MeV.

Topic: Physics, Nuclear Physics, NEET
The binding energy of the atom of elements A and B are Ea and Eb respectively. Three atoms of the element B fuse to give one atom of element A. This fusion process is accompanied by release of energy E. Then Ea and Eb are related to each other as

Options:
A Ea+E=3E0
B Ea=3Eb
C Ea=3Eb+E
D Ea+3Eb=E

The correct answer is:
Ea=3Eb+E

Solution:
3B→A
E=Ea-3Eb

Topic: Physics, Atomic Physics, NEET
The binding energy of the electron in the lowest orbit of the hydrogen atom is 13.6 eV. The magnitudes energies from three lowest orbits of the hydrogen are

Options:
A 13.6, 6.8, 8.4 eV
B 13.6, 10.2, 3.4 eV
C 13.6, 27.2, 40.8 eV
D 13.6, 3.4, 1.5 eV

The correct answer is:
13.6, 3.4, 1.5 eV

Solution:
En=-13.6Z2n2
I orbit for (H)
Z=1,n=1
En=-13.6eV
II orbit for (H)
Z=1,n=2
En=-13.64=-3.4eV
III orbit for (H)
En=-13.69=-1.51eV

Topic: Chemistry, Structure of Atom, NEET
The binding energy per nucleon for the parent nucleus is E1 and that for the daughter nuclei is E2. Then

Options:
A E2=2E1
B E1>E2
C E2>E1
D E1=2E2

The correct answer is:
E2>E1

Solution:
After decay, the daughter nuclei will be more stable hence, binding energy per nucleon will be more than that of their parent nucleus.

Topic: Physics, Nuclear Physics, NEET
The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus, the energy released in the fusion is

Options:
A 23.6 MeV

B 2.2 MeV

C 28.0 MeV

D 30.2 MeV

The correct answer is:
23.6 MeV

Solution:
H21 + H21 → He42 + ΔE
The binding energy per nucleon of a deuteron
= 1.1 MeV
∴ Total binding energy = 2 × 1.1 = 2.2 MeV
The binding energy per nucleon of a helium nuclei
= 7 MeV
∴ Total binding energy = 4 × 7 = 28 MeV
Hence, the energy released
ΔE = (28 – 2 × 2.2) = 23.6 MeV

Topic: Physics, Nuclear Physics, NEET
The binding energy per nucleon of 16O is 7.97 MeV and that of 17O is 7.75 MeV. The energy in MeV required to remove a neutron from 17O is

Options:
A 3.52
B 3.64
C 4.23
D 7.86
E 1.68

The correct answer is:
4.23

Solution:
BE per nucleon O 16=7.97 MeV
BE per nucleon O 17=7.75 MeV
O17→n10+O16
The energy required to remove a neutron
=17×7.75-16×7.97
=4.23 MeV

Topic: Physics, Nuclear Physics, NEET
The binding energy per nucleon of deuteron 1 2 H and helium nucleus 2 4 He is 1.1 Me V and 7 Me V respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

Options:
A 13.9 Me V
B 26.9 Me V
C 23.6 Me V
D 19.2 Me V

The correct answer is:
23.6 Me V

Solution:
Total binding energy for (each deuteron)
= 2 × 1.1 = 2.2 Me V
Total binding energy for helium = 4 × 7 = 28 Me V
∴ Energy released = 28 - 2 × 2.2
= 28 - 4.4 = 23.6 Me V .

Topic: Physics, Atomic Physics, NEET