datavorous/entrance-exam-dataset huggingface dataset:
The total energy of an electron in the first excited state of hydrogen atom is about $-3.4 \mathrm{eV}$. Its kinetic energy in this state is:

Options:
A $3.4 \mathrm{eV}$
B $6.8 \mathrm{eV}$
C $-3.4 \mathrm{eV}$
D $-6.8 \mathrm{eV}$

The correct answer is:
$3.4 \mathrm{eV}$

Solution:
$$
\mathrm{K} . \mathrm{E} .=\left|\frac{1}{2} P . E\right|
$$
But P.E is negative
$$
\begin{aligned}
\therefore \text { Total energy } & =\left|\frac{1}{2} P \cdot E\right|-P \cdot E \\
& =-\frac{P \cdot E}{2}=-3.4 \mathrm{eV} \\
\mathrm{K} . \mathrm{E} & =+3.4 \mathrm{eV}
\end{aligned}
$$

Topic: Physics, Atomic Physics, NEET, NEET 2005
The total energy of an electron in the first excited state of hydrogen is about -3.4eV. Its kinetic energy in this state is

Options:
A -Â 3.4Â eV
B -Â 6.8Â eV
C 6.8Â eV
D 3.4 eV

The correct answer is:
3.4 eV

Solution:
The kinetic energy of an electron
K =Ze28ÏÎµ0r
The potential energy of an electron
U =-14ÏÎµ0Ze2r
â´ Total energy
E = K +U = Ze28ÏÎµ0r- Ze24ÏÎµ0r
Or E=-Ze28ÏÎµ0r
Or E=-K
Or K=-E= -(-3.4)
Or =3.4 eV

Topic: Physics, Atomic Physics, NEET
The total energy of an electron in the first excited state of the hydrogen atom is about -3.4 eV. The potential energy of the electron in this state is

Options:
A -6.8 eV
B -8.6 eV
C 6.8 eV
D 8.6 eV

The correct answer is:
-6.8 eV

Solution:
The kinetic energy of electron (KE) = KZe 2 2r
Potential energy of electron (PE) = - KZe 2 r ⇒ PE = - 2 KE
In this calculation, the electric potential and potential energy are zero at infinity.
Total energy = PE + KE = - 2KE + KE = - KE
PE of electron in this first excited state
= - 2 KE = -2 (3.4) = - 6.8 eV

Topic: Physics, Atomic Physics, NEET
The total energy of an electron in the nth stationary orbit of the hydrogen atom can be obtained by

Options:
A En=13.6n2eV
B En=-13.6n2eV
C En=-1.36n2eV
D En=-13.6×n2eV

The correct answer is:
En=-13.6n2eV

Solution:
Bohr’s Postulates are as follows: The electron revolves around the nucleus at stable orbits without radiating energy. The stable orbits are called stationary orbits. These orbits are unique and have different radii. The electron cannot orbit in between two stationary orbits, meaning the orbits are uniquely determined. The stationery orbits are obtained by equating the angular momentum of the electron to an integral multiple of reduced Planck’s constant, mvr=nh2π, where v is the velocity of the electron in the nth orbit and r is the radius of the nth orbit. The orbits have definite energies called the energy shells or energy levels. For hydrogen, En=-13.6n2 eV.

Topic: Physics, Atomic Physics, NEET, NEET 2020 (Phase 2)
The total energy of an electron revolving in the second orbit of a hydrogen atom is

Options:
A -13.6 eV
B -1.51eV
C -3.4 eV
D Zero

The correct answer is:
-3.4 eV

Solution:
E=-13.6n2 eV
For n=2, E=-13.622=-3.4 eV

Topic: Physics, Atomic Physics, NEET
The total energy of electron in the ground state of hydrogen atom is $-13.6 \mathrm{eV}$. The kinetic energy of an electron in the first excited state is:

Options:
A $6.8 \mathrm{eV}$
B $13.6 \mathrm{eV}$
C $1.7 \mathrm{eV}$
D $-3.4 \mathrm{eV}$.

The correct answer is:
$-3.4 \mathrm{eV}$.

Solution:
Energy of $n^{\text {th }}$ orbit of hydrogen atom is given by
$E_n=-\frac{13.6}{n^2} e V$
for ground state $n=1$
$E_1=-\frac{13.6}{12}=-13.6 \mathrm{eV}$
for excited state $n=2$
$E_2=-\frac{13.6}{22}=-3.4 \mathrm{eV}$

Topic: Physics, Atomic Physics, NEET, NEET 2007
The total energy radiated from a black body source is collected for 1 min and is used to heat a quantity of water. The temperature of water is found to increase from 20 â to 20.5 â. If the absolute temperature of the black body is doubled and the experiment is repeated with the same quantity of water at 20 â, the temperature of water will be

Options:
A 21 °C
B 22 Â°C
C 24 Â°C
D 28 Â°C

The correct answer is:
28 Â°C

Solution:
From Stefan's law of radiation,
E∝T4 ⇒ E1E2=T14T24
Given, T1=T, T2=2T
∴ E1E2=T42T4=124=116
E2=16 E1
Heat taken by water from radiation
E=mc∆θ
Where c is specific heat, ∆θ the change in temperature and mthe mass.
∴ E=m×1×(20.5-20)
E=m×0.5 ...(i)
When energy supplied is 16 times the previous one, then let the temperature rise to θ′
∴ 16E=m×1×(θ′-20) ...(ii)
Dividing Eq. (i) by (ii), we get
116=0.5θ′-20
⇒ θ′-20=16×0.5=8
⇒ θ′=20+8=28 ℃

Topic: Physics, Thermal Properties of Matter, NEET
The total flux (in S.I units) through a closed surface constructed around a positive charge of 0.5 C placed in a dielectric medium of dielectric constant 10 is

Options:
A 5.65×109
B 1.13×1011
C 9×109
D 8.85×10-12

The correct answer is:
5.65×109

Solution:
ϕE=qKε0=0.510×8.85×1012=5.65×109

Topic: Physics, Electrostatics, NEET
The total kinetic energy of a mixture of 4 g of H2 and 4 g of He at 300 K is

Options:
A 1800 R
B 1750 R
C 1950 R
D 2500 R

The correct answer is:
1950 R

Solution:
U=52×42R×300+32×44R×300
=5R+32R×300=13R2×300=1950 R

Topic: Physics, Kinetic Theory of Gases, NEET
The total lung capacity includes which of the following volumes?

Options:
A IRV + ERV+RV+TV
B ERV+TV+IRV
C IRV+TV+FRC
D IC+EC+FRC

The correct answer is:
IRV + ERV+RV+TV

Solution:
Total Lung Capacity (TLC = ERV+TV+IRV+RV = VC+RV)
It refers to the total volume of air that can be accommodated in the lungs at the end of a forced inspiration.
It is approximately 6 litres.

Topic: Biology, Breathing and Exchange of Gases, NEET
The Total Lung Capacity(TLC) is the total volume of air accomodated in the lungs at the end of a forced inspiration. This includes:

Options:
A

RV; IC (Inspiratory Capacity); EC (Expiratory Capacity); and ERV

B RV;ERV;IC and EC
C RV; ERV; VC (Vital Capacity) and FRC (Functional Residual Capacity)
D RV (Residual Volume); ERV (Expiratory Reserve Volume); TV (Tidal Volume); and IRV (Inspiratory Reserve Volume)

The correct answer is:
RV (Residual Volume); ERV (Expiratory Reserve Volume); TV (Tidal Volume); and IRV (Inspiratory Reserve Volume)

Solution:
Total Lung Capacity (TLC): Total volume of air accommodated in the lungs at the end of a forced inspiration is called TLC. Mathematically, it can be represented as:

TLC=RV+VC,

But, VC=TV+ERV+IRV,

Therefore, TLC=RV+TV+ERV+IRV

Topic: Biology, Breathing and Exchange of Gases, NEET, NEET 2020 (Phase 2)
The total molarity and normality of all the ions present in a solution containing 0.1 M of CuSO4 and 0.1 M of Al2(SO4)3 is

Options:
A 0.2 M, 0.2 N
B 0.7 M, 1.6 N
C 0.8 M, 1.6 N
D 1.6 M, 1.6 N

The correct answer is:
0.7 M, 1.6 N

Solution:
mole of Cu2+=0.1 , mole of SO42-=0.1
mole of Al3+=0.2 , mole of SO42-=0.3
So total mole = 0.1 + 0.1 + 0.2 + 0.3 = 0.7
Also, eq. of Cu2+=0.2 , eq. of SO42-=0.2
Eq. of Al3+=0.6, Eq. of SO42-=0.6
So total eq. =0.2+0.2+0.6+0.6=1.6 .

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
The total number of atomic orbitals in fourth energy level of an atom is

Options:
A 16
B 32
C 4
D 8

The correct answer is:
16

Solution:
Number of atomic orbitals in an orbit $=n^2=4^2=16$

Topic: Chemistry, Structure of Atom, NEET, NEET 2011 (Screening)
The total number of atomic orbitals in the fourth energy level of an atom is :

Options:
A 8
B 16
C 32
D 4

The correct answer is:
16

Solution:
Number of orbitals = n 2
Where n is shell number or principal quantum number
= 4 2 = 1 6

Topic: Chemistry, Structure of Atom, NEET
The total number of ATP formed in the aerobic respiration from one glucose molecule is

Options:
A 30
B 38
C 40
D 24

The correct answer is:
38

Solution:
The energetics of aerobic respiration is as follows:

Energetics of Glycolysis

Reaction Type	Reaction	Products of reaction	In terms of ATP

ATP Forming Reactions

Substrate-level phosphorylation	2 x 1,3 Biphosphoglycerate + 2ADP âBiphosphoglycerokinase 2 x 3-Phosphoglycerate + 2ATP	2 ATP	2 ATP

Substrate-level phosphorylation	2 x PEP + 2ADP âK+, Mg2+Pyruvate Kinase 2 x Pyruvate + 2 ATP	2 ATP	2 ATP

Dehydrogenation	2 x 3 Phosphoglyceraldehyde + 2 Pi + 2 NAD+ âDehydrogenase 2 x 1,3 Biphosphoglycerate + 2 NADH2	2 NADH2	6 ATP

ATP Consuming Reactions

Phosphorylation-I	Glucose + ATP âMg2+Hexokinase Glucose 6-Phosphate + ADP	-1 ATP	-1 ATP

Phosphorylation-II	Fructose 6-Phosphate + ATP âMg2+Phosphofructokinase Fructose 1,6-Biphosphate	-1 ATP	-1 ATP

Net Gain	8 ATP

Energetics of Link reaction

Reaction Type	Reaction	Products of reaction	In terms of ATP

Dehydrogenation	Pyruvate + CoA + NAD+ âMg+2, TPP, Lipoic acidPyruvate dehydrogenase Acetyl CoA + NADH2 + CO2	2 x NADH2	6 ATP

Energetics of Kreb's cycle/TCA cycle

Reaction Type	Reaction	Products of reaction	In terms of ATP

Dehydrogenation-I	Isocitrate + NAD+ âMn+2Isocitrate dehydrogenase Oxalosuccinate + NADH2	2 x NADH2	6 ATP

Oxidative decarboxylation	Î±-Ketoglutarate + CoA + NAD+ âMg+2, TPP, Lipoic acidÎ±-Ketoglutarate dehydrogenase Succinyl CoA + NADH2 + CO2	2 x NADH2	6 ATP

Dehydrogenation III	Succinate + FAD âSuccinic dehydrogenase Fumarate + FADH2	2 x FADH2	4 ATP

Dehydrogenation IV	Malate + NAD+ âMalate dehydrogenase Oxaloacetate + NADH2	2 x NADH2	6 ATP

Substrate-level Phosphorylation	Succinyl CoA + GDP + Pi âSuccinyl CoA synthetase/Succinyl thiokinase Succinate + GTP + CoA	2 x 1 GTP	2 ATP

Net gain	24 ATP

Net gain in aerobic respiration (Glycolysis + Link reaction + Kreb's cycle)	8 ATP + 6 ATP + 24 ATP	38 ATP

Topic: Biology, Respiration in Plants, NEET
The total number of π -bond electrons in the following structure is:

Options:
A 4
B 8
C 12
D 16

The correct answer is:
8

Solution:
π bond electrons = No. of π bonds×2=4×2=8

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2015 (Phase 1)
The total number of π-bond electrons in the following structure is

Options:
A 8
B 16
C 4
D 12

The correct answer is:
8

Solution:
Total π bonds =4
Total π electrons =8

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The total number of bones in your right arm is

Options:
A 30
B 32
C 35
D 40

The correct answer is:
30

Solution:
Total bones of an arm are 30.

Each arm consists of the following 30 bones: 1 humerus, 1 radius, 1 ulna, 8 carpal bones, 5 metacarpal bones, 5 digits (14 phalanges). Phalangeal formula: 2, 3, 3, 3, 3.

Topic: Biology, Locomotion and Movement, NEET
The total number of contributing structures showing hyper conjugation (involving C – H bonds) for the following carbocation is

Options:
A 4
B 5
C 6
D 8

The correct answer is:
6

Solution:
Number of hyper conjugation structures ∝ number of α -hydrogen atoms
As 6αH - atoms are there so it can have 6 hyper conjugation structures.

Topic: Chemistry, General Organic Chemistry, NEET
The total number of isomers for the aromatic compound having molecular formula C7H8O is

Options:
A 2
B 3
C 4
D 5

The correct answer is:
5

Solution:
The total number of isomers for the aromatic compound having molecular formula C7H8O is 5 and compounds may be alcohol, ether or phenol.

Topic: Chemistry, General Organic Chemistry, NEET
The total number of lattice arrangements in different crystal systems is

Options:
A 3
B 7
C 8
D 14

The correct answer is:
7

Solution:
The three dimensional arrangrment of atoms inside a crystal. The seven basic crystal lattice arrangements are :- Cubic, Tetragonal, Orthorhombic, Monoclinic, Hexagonal, Rhombohedral and Triclinic.

Topic: Chemistry, Solid State, NEET
The total number of lobes and alveoli present in both the lungs of man are

Options:
A 17 and 30 million, respectively
B 5 and 300 million, respectively
C 19 and 300 million, respectively
D 18 and 300 lakh, respectively

The correct answer is:
5 and 300 million, respectively

Solution:
In man, the total number of lobe present in both the lungs is 5 of which three lobes, i. e., anterior, posterior, and azygous are present in right lung and two lobes called left anterior and left posterior in the left lung. The basic functional units of lungs are alveoli. The number of alveoli in human beings is 300 million.

Topic: Biology, Breathing and Exchange of Gases, NEET
The total number of lone − pairs of electrons and sp3 hybridized nitrogen atoms in melamine are respectively

Options:
A 6, 6
B 3, 3
C 6, 3
D 3, 6

The correct answer is:
6, 3

Solution:
In Melamine there are 6 number of lone â pairs of electrons and 3  sp 3 hybridized nitrogen atoms. All NH2 groups are having sp2 hybridized nitrogen atoms.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
The total number of lone pairs of electrons in N2O3 is

Options:
A 2
B 4
C 6
D 8

The correct answer is:
8

Solution:
The total number of lone pairs of electrons in N2O3 is 8 as per as its Lewis structure given below

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The total number of N-atoms present in veronal is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
2

Solution:
The structure of veronal is

It means veronal has 2 N-atoms.

Topic: Chemistry, Chemistry in Everyday Life, NEET
The total number of nitrogenous bases in human genome is estimated to be about:

Options:
A 35 million
B 3.1 billion
C 3.5 million
D 35 thousand

The correct answer is:
3.1 billion

Solution:
International collaborative efforts were taken for Human genome project with a 15-year plan to identify and map an originally estimated 80,000-100,000 human genome and to sequence approximately 3 billion base pairs.

Topic: Biology, Molecular Basis of Inheritance, NEET
The total number of optically inactive products obtained from the complete ozonolysis of the compound given below here is

Options:
A 2
B 4
C 1
D 0

The correct answer is:
4

Solution:
Ozonolysis : CH3CHO (inactive)
Hence all the 4 molecules formed here are optically inactive.

Topic: Chemistry, Hydrocarbons, NEET
The total number of possible enantiomeric pairs that can be obtained during mono-chlorination of 2-methylbutane is

Options:
A 3
B 4
C 1
D 2

The correct answer is:
2

Solution:
Its monochloro derivatives are as follows :
(i)
It will exist as enantiomeric pair (d and l-forms)
(ii)
no asymmetric C atom
(iii)
It will exist as enantiomeric pair (d- and l-forms)
(iv)
No asymmetric carbon atom
Hence, only two enantiomeric pairs will be obtained by the monochlorination of 2-methylbutane.

Topic: Chemistry, General Organic Chemistry, NEET
The total number of species, that are known and described range between

Options:
A 0.5-1.0 million
B 1.1-1.2 million
C 1.7-1.8 million
D 2.5-3.0 million

The correct answer is:
1.7-1.8 million

Solution:
Scientists have described the range of the total number of world's species between 1.7-1.8 million as in 2010. The total number of plants over the earth is approximately 3,21,212, the total number of vertebrates is approximately 62,305, the total number of invertebrates is approximately 1,305,250, and others (bacteria, cyanobacteria, algae) are approximately 51, 563. Thus, the number of species over earth totals to 1,740,330.

Topic: Biology, Biodiversity and its Conservation, NEET
The total number of stereoisomers in open-chain aldohexose (such as glucose) is

Options:
A 8
B 4
C 16
D 2

The correct answer is:
16

Solution:
The number of chiral carbons in open-chain aldohexose (such as glucose) is four, therefore, the number of stereoisomers =24=16 .

Topic: Chemistry, Biomolecules, NEET
The total number of structural ethers possible with the molecular formula C5H12O?

Options:
A 4
B 5
C 6
D 7

The correct answer is:
6

Solution:
molecular formula =C5H12O

Total ether isomers =6

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
The total number of teeth which comes only once in a life span___________

Options:
A 20
B 8
C 32
D 12

The correct answer is:
12

Solution:
Twenty milk teeth of a child get replaced in adulthood by 20 permanent teeth. But the human adult has 32 teeth; hence, the rest 12 can never be replaced if broken because they grow only once in a man’s life.

Topic: Biology, Digestion and Absorption, NEET
The total number of vertebrae in man is

Options:
A Cervical = 8; Thoracic = 12; Lumbar = 4; Sacrum = (1) 5; Coccyx = (1) 4
Total No. of bones 33
B Cervical = 6; Thoracic = 12; Lumbar = 5; Sacrum = (1) 4; Coccyx = (1) 4
Total No. of bones 31
C Cervical = 6; Thoracic = 12; Lumbar = 5; Sacrum = (1) 5; Coccyx = (1) 4
Total No. of bones 32
D Cervical = 7; Thoracic = 12; Lumbar = 5; Sacrum = (1) 5; Coccyx = (1) 4
Total No. of bones 33

The correct answer is:
Cervical = 7; Thoracic = 12; Lumbar = 5; Sacrum = (1) 5; Coccyx = (1) 4
Total No. of bones 33

Solution:
Cervical = 7
Thoracic = 12
Lumbar = 5
Sacrum = (1) 5
Coccyx = (1) 4
Total No. of bones 33

Topic: Biology, Locomotion and Movement, NEET
The total power dissipated in watt in the circuit shown here is:

Options:
A 40
B 54
C 4
D 16

The correct answer is:
54

Solution:
Total power dissipated $=\frac{V^2}{R}$
Power $=\frac{18 \times 18}{6}=54 \mathrm{~W}$

Topic: Physics, Current Electricity, NEET, NEET 2007
The total radiant energy per unit area, normal to the direction of incidence, received at a distance $\mathrm{R}$ from the centre of a star of radius $r$, whose outer surface radiates as a black body at a temperature $T \mathrm{~K}$ is given by
(where $\sigma$ is Stefan's constant)

Options:
A $\sigma r^2 \mathrm{~T}^4 / \mathrm{R}^2$
B $\sigma r^2 T^4 / 4 \pi r^2$
C $\sigma r^4 T^4 / r^4$
D $4 \pi \sigma r^2 T^4 / R^2$

The correct answer is:
$\sigma r^2 \mathrm{~T}^4 / \mathrm{R}^2$

Solution:
If $\mathrm{r}$ is the radius of the star and $T$ its temperature, then the energy emitted by the star per second through radiation in accordance with Stefan's law will be given by $A \sigma \mathrm{T}^4=4 \pi \mathrm{r}^2 \sigma \mathrm{T}^4$
In reaching a distance $R$ this energy will spread over a sphere of radius $R$; so the intensity of radiation will be given by
$$
S=\frac{P}{4 \pi R^2}=\frac{4 \pi r^2 \sigma T^4}{4 \pi \mathrm{R}^2}
$$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2010 (Screening)
The total volume of air accommodated in the lungs at the end of a forced inspiration is called-

Options:
A Total Lung Capacity (TLC)
B Vital Capacity (VC)
C Functional Residual Capacity (FRC)
D Expiratory Capacity (EC)

The correct answer is:
Total Lung Capacity (TLC)

Solution:
Total volume of air accomodated in the lungs at the end of a forced inspiration is called Total Lung Capacity (TLC).
This includes RV, ERV, TV and IRV.
TLC = IRV+TV+ERV+RV

Topic: Biology, Breathing and Exchange of Gases, NEET
The total volume of air accommodated in the lungs at the end of a forced inspiration is called

Options:
A Total Lung Capacity (TLC)
B Vital Capacity (VC)
C Functional Residual Capacity (FRC)
D Expiratory Capacity (EC)

The correct answer is:
Total Lung Capacity (TLC)

Solution:
Total volume of air accomodated in the lungs at the end of a forced inspiration is called Total Lung Capacity (TLC).
This includes RV, ERV, TV and IRV.
TLC=IRV+TV+ERV+RV

Topic: Biology, Breathing and Exchange of Gases, NEET
The total volume of dry gaseous products at STP , when 3 moles of electrons are transferred from anode to cathode in the electrolysis of water is:
(Volume of gas a STP = 22.4 L)

Options:
A 67.2 L
B 50.4 L
C 44.8 L
D 56.0 L

The correct answer is:
50.4 L

Solution:
3OHÎâ34O2+32H2O+3e-(Anode)
3H++3e-â32H2(Cathode)
34 mole O2 (Anode) +32 mole H2 (cathode)
mole of gas =34+32=94
mole of gas =94Ã22.4
=50.4 L

Topic: Chemistry, Electrochemistry, NEET
The tough protective layer of brain is __(i)__ and inner highly vascularised protective layer is __(ii)__.

Options:
A i: Piamater and ii: duramater
B i: duramater and ii: Piamater
C i: duramater and ii: arachnoid
D i: piamater and ii: sub-dural space

The correct answer is:
i: duramater and ii: Piamater

Solution:
The dura mater is the outermost layer, lying directly underneath the bones of the skull. It is thick, tough and inextensible.
The arachnoid mater is the middle layer of the meninges, lying directly underneath the dura mater. It consists of layers of connective tissue and is avascular.
The pia mater is located underneath the sub-arachnoid space. It is very thin, and tightly adhered to the surface of the brain and spinal cord.

Topic: Biology, Neural Control and Coordination, NEET
The toxin encoded by cry genes and produced by the bacteria- Bacillus thurengeneis is called ___.

Options:
A Cry gene
B laugh gene
C Cry proteins
D happy protein

The correct answer is:
Cry proteins

Solution:
The toxin encoded by cry genes and produced by the bacteria- Bacillus thurengeneis is called cry proteins

Topic: Biology, Biotechnology and Its Applications, NEET
The toxin produced by Bacillus thuringiensis gets activated in

Options:
A acidic pH of the stomach.
B high temperature.
C alkaline pH of the gut.
D mechanical action in the insect gut.

The correct answer is:
alkaline pH of the gut.

Solution:
Bacillus thuringiensis forms protein crystal during a particular phase of their growth. These crystals contain a toxic insecticidal protein which exist as inactive protoxin but once an insect injest the inactive toxin, it is converted into an active form of toxin due to alkaline pH of the gut which solublises the crystals.

Topic: Biology, Biotechnology and Its Applications, NEET
The tracts of forests that are restored in the places of worship are called ___.

Options:
A Garden
B Sacred groves
C Biodiversity
D Conservation grooves

The correct answer is:
Sacred groves

Solution:
The tracts of forests that are restored in the places of worship are called sacred groves

Topic: Biology, Biodiversity and its Conservation, NEET
The traditional inclusion and multiplication of genes along with the desired genes involves

Options:
A the hybridization procedures in plant and animal breeding.
B the hybridization procedures applied only in bacteria.
C the hybridization procedures only in plant breeding.
D the hybridization procedures in animal breeding.

The correct answer is:
the hybridization procedures in plant and animal breeding.

Solution:
Traditional hybridization procedures used in plant and animal breeding, very often lead to the inclusion and multiplication of undesirable genes along with the desired genes.

Topic: Biology, Biotechnology - Principles and Processes, NEET
The trajectory of a projectile in a vertical plane is y=axâbx2, where a and b are constants and x and y are respectively horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal is

Options:
A b 2 2 a , tan -1 b
B a 2 b , tan -1 2 a
C a 2 4 b , tan -1 a
D 2 a 2 b , tan -1 a

The correct answer is:
a 2 4 b , tan -1 a

Solution:
y=axâbx2
For height (or y) to be maximum
dâ¡ y dâ¡ x = 0 or a - 2 b x = 0
or x = a 2 b â¦ 1
â´ y max = a a 2 b - b a 2 b 2 = a 2 4 b â¦ 2
dâ¡ y dâ¡ x x = 0 = a = tan Î¸ 0
where Î¸0 is the angle of projection
â´ Î¸ 0 = tan -1 a

Topic: Physics, Motion In Two Dimensions, NEET
The trans-alkenes are formed by the reduction of alkynes with

Options:
A Sn-HCl
B H2-Pd/C,BaSO4
C NaBH4
D Na/liq.NH3

The correct answer is:
Na/liq.NH3

Solution:
Na/liq.NH3 reduces alkynes into trans alkene (trans addition) as its mode is anti addition while rest other are syn addition reagents.

Topic: Chemistry, Hydrocarbons, NEET
The transformation ratio in the step-up transformer is

Options:
A 1
B Greater than one
C Less than one
D The ratio greater or less than one depends on the other factors

The correct answer is:
Greater than one

Solution:
Transformation ratio,
k=NsNp=VsVp
For step-up transformer,
Ns>Np, ie.,Vs>Vp hence, k>1.

Topic: Physics, Alternating Current, NEET