datavorous/entrance-exam-dataset huggingface dataset:
Use of an artificial kidney during hemodialysis may result in:
a Nitrogenous waste build-up in the body
b Non-elimination of excess potassium ions
(c) Reduced absorption of calcium ions from gastro-intestinal tract
d Reduced RBC production
Which of the following option is the most appropriate?

Options:
A a and b are correct
B b and c are correct
C c and d are correct
D a and d are correct

The correct answer is:
c and d are correct

Solution:
Hemodialysis is a treatment to filter wastes and water from the blood. It leads to the elimination of nitrogenous waste that builds up in the body due to metabolism. The dialysis also removes the excess potassium from the body.

Hemodialysis reduces the calcium ions from the GI tract and also reduces the production of RBC.

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2019 (All India)
Use of anti-histamines and steroids give a quick relief from

Options:
A allergy
B nausea
C cough
D headache

The correct answer is:
allergy

Solution:
Allergy is the hypersensitiveness of a person to some foreign substances coming in contact with or entering the body. The common allergns are dust, pollen mould, spores, fabricates, bacteria, etc. During allergic reaction, there is increased release of histamine from mast cells. Use of anti-histamines and steroids give a quick relief from allergy.

Topic: Biology, Human Health and Diseases, NEET, NEET 2009 (Screening)
Use of bioresources by multinational companies and organizations without authorisation from the concerned country and its people is called

Options:
A Biodgradation
B Biopiracy
C Bio-infringement
D Bioexploitation

The correct answer is:
Biopiracy

Solution:
Biopiracy is referred to the use of bioresources by multinational companies and other organizations without proper authorization front the countries and people concerned without compensatory payment.

Bio-infringement is the commission of a prohibited act with respect to a patented invention without permission from the patent holder.

Bio-exploitation means taking advantage of the biological resources of other countries without permission.

Biodegradation is a biological breakdown of organic material by bacteria, fungi, etc.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2018
Use of bio-resources by multinational companies and other organizations without proper authorization from the countries and people concerned without compensatory payment is known as

Options:
A Genetic relatedness
B Biopiracy
C Biodegradation
D Biomagnification

The correct answer is:
Biopiracy

Solution:
Most of the industrialized nations are rich financially but poor in biodiversity and traditional knowledge. Traditional knowledge related to bioresources is used by developed countries by inadequate compensation and benefit sharing to farmers and indigenous people of a specific region/country.

Topic: Biology, Biotechnology and Its Applications, NEET
Use of Cannabis products results in

Options:
A Alteration in perception, thoughts, feelings.
B Depressed brain activity.
C Suppressed brain function and pain relief.
D Stimulation of nervous system, increase alertness and activity.

The correct answer is:
Alteration in perception, thoughts, feelings.

Solution:
Hallucinations mean perception of an image or sound in the absence of its actual stimuli. Various drugs that have a stimulatory effect on the central nervous system can result in hallucinations. The effects seen are alterations in mood, perception, thoughts etc. Many such products are obtained from Cannabis. Hence these drugs are also known as psychodelic drugs. These include bhang, ganja, charas and marijuana. These are obtained from the inflorescence of plants or flower tops or leaves etc.

Topic: Biology, Human Health and Diseases, NEET
Use the below pair of names to fill the gaps in the given statement.

"Connell's elegant field experiments showed that on rocky sea coasts of Scotland, the larger and competitively superior barnacle _______ dominates the intertidal area, and exclude smaller barnacle _______ from that zone"

Options:
A Chathamalus, Balanus
B Balanus, Balanus
C Chathamalus, chhathamalus
D Balanus, Chathamalus

The correct answer is:
Balanus, Chathamalus

Solution:
The distribution of a species is restricted to asmall geographical area due to the activity of competitively superior species, it was found to expand its distributional rangedramatically when the competing species was experimentally removed. Connell performed an elegant field experiment to show that studying Balanus and Chathamalus.

Topic: Biology, Organisms and Populations, NEET
Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes:

Options:
A 35.45 g/mol
B 37.45 g/mol
C 43.56 g/mol
D 39.94 g/mol

The correct answer is:
39.94 g/mol

Solution:
Correct Option is : (D)
39.94 g/mol

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Use the following data to calculate ∆latticeH⊖for NaBr. ∆subH⊖for sodium metal = 108.4 kJ mol–1Ionization enthalpy of sodium = 496 kJ mol–1Electron gain enthalpy of bromine = – 325 kJ mol–1Bond dissociation enthalpy of bromine = 192 kJ mol–1∆fH⊖for NaBr (s) = – 360.1 kJ mol–1

Options:
A ∆latticeH⊖(NaBr) = +700.5 kJmol-1
B ∆latticeH⊖(NaBr) = +786.5 kJmol-1
C ∆latticeH⊖(NaBr) = +735.5 kJmol-1
D ∆latticeH⊖(NaBr) = +816.5 kJmol-1

The correct answer is:
∆latticeH⊖(NaBr) = +735.5 kJmol-1

Solution:
Correct Option is : (C)
∆latticeH⊖(NaBr) = +735.5 kJmol-1

Topic: Chemistry, Thermodynamics (C), NEET
Use the information and data given below to answer the question: • Stronger intermolecular forces result in higher boiling point. • Strength of London forces increases with the number of electrons in the molecule. • Boiling point of HF, HCl, HBr and HI are 293 K, 189 K, 206 K and 238 K respectively. What type of intermolecular forces are present in HF, HCl, HBr, and HI molecules?

Options:
A Dipole-dipole forces
B London forces
C Hydrogen bonding
D All of the above

The correct answer is:
All of the above

Solution:
Correct Option is : (D)
All of the above

Topic: Chemistry, States of Matter, NEET
Use the information and data given below to answer the question: • Stronger intermolecular forces result in higher boiling point. • Strength of London forces increases with the number of electrons in the molecule. Which intermolecular force is predominant in determining the boiling points of HCl, HBr, and HI based on the given information and data?

Options:
A Dipole-dipole interaction
B Hydrogen bonding
C London dispersion forces
D Ion-dipole interaction

The correct answer is:
London dispersion forces

Solution:
Correct Option is : (C)
London dispersion forces

Topic: Chemistry, States of Matter, NEET
Use the information and data given below to answer the question: • Stronger intermolecular forces result in higher boiling point. • Strength of London forces increases with the number of electrons in the molecule. Which of the following statements best explains why hydrogen fluoride (HF) has the highest boiling point among HF, HCl, HBr, and HI?

Options:
A HF has the highest dipole moment due to the highest electronegativity of fluorine and the presence of hydrogen bonding.
B HF has the lowest number of electrons in its molecule, leading to stronger intermolecular forces.
C HCl has the lowest dipole moment due to the lowest electronegativity of chlorine.
D HI has the highest boiling point because of its larger molecular size.

The correct answer is:
HF has the highest dipole moment due to the highest electronegativity of fluorine and the presence of hydrogen bonding.

Solution:
Correct Option is : (A)
HF has the highest dipole moment due to the highest electronegativity of fluorine and the presence of hydrogen bonding.

Topic: Chemistry, States of Matter, NEET
Using anhydrous $\mathrm{AlCl}_3$ as catalyst, which one of the following reactions produces ethylbenzene $(\mathrm{PhEt})$ ?

Options:
A $\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2 \mathrm{OH}+\mathrm{C}_6 \mathrm{H}_6$
B $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2+\mathrm{C}_6 \mathrm{H}_6$
C $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2+\mathrm{C}_6 \mathrm{H}_6$
D $\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_3+\mathrm{C}_6 \mathrm{H}_6$

The correct answer is:
$\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2+\mathrm{C}_6 \mathrm{H}_6$

Solution:
In Friedel craft reaction benzene with ethylene in presence of $\mathrm{HCl}$ and anhydrous $\mathrm{AlCl}_3$ reacts to form ethyl benzene.
Related Theory
Anhydrous $\mathrm{AlCl}_3$ is used as a catalyst because it acts as a Lewis acid which can accept electron by forming intermediates and also by speeding up the reaction. It also leads in the creation of carbocation which is used in the electrophilic substitution reaction.

Topic: Chemistry, Hydrocarbons, NEET, NEET 2004
Using Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state.

Options:
A $\frac{\pi {{a}_{o}}}{2v}e$
B $\frac{3\pi {{a}_{o}}}{v}e$
C $\frac{\pi {{a}_{o}}}{v}e$
D $\frac{2\pi {{a}_{o}}}{v}e$

The correct answer is:
$\frac{2\pi {{a}_{o}}}{v}e$

Solution:
Correct Option is : (D)
$\frac{2\pi {{a}_{o}}}{v}e$

Topic: Chemistry, Atomic Physics, NEET
Using crystal field theory, choose the correct option regarding the electronic configuration of the central metal atom/ion, and determine the magnetic moment value in the following: [FeF6]3–, [Fe(H2O)6]2+, [Fe(CN)6]4–

Options:
A [FeF6]3–: High spin, 1 unpaired electron, magnetic moment = 1.73 BM
B [Fe(H2O)6]2+: Low spin, 4 unpaired electrons, magnetic moment = 4.90 BM
C [Fe(CN)6]4–: Low spin, 0 unpaired electrons, magnetic moment = 0 BM
D [FeF6]3–: Low spin, 1 unpaired electron, magnetic moment = 1.73 BM

The correct answer is:
[Fe(H2O)6]2+: Low spin, 4 unpaired electrons, magnetic moment = 4.90 BM

Solution:
Correct Option is : (B)
[Fe(H2O)6]2+: Low spin, 4 unpaired electrons, magnetic moment = 4.90 BM

Topic: Chemistry, Coordination Compounds, NEET
Using crystal field theory, choose the correct option regarding the electronic configuration of the central metal atom/ion, and determine the magnetic moment value in the following: [CoF6]3–, [Co(H2O)6]2+, [Co(CN)6]3–

Options:
A [CoF6]3–: Low spin, 1 unpaired electron, magnetic moment = 1.73 BM
B [Co(H2O)6]2+: High spin, 3 unpaired electrons, magnetic moment = 3.87 BM
C [Co(CN)6]3–: Low spin, 1 unpaired electron, magnetic moment = 1.73 BM
D [CoF6]3–: Low spin, 1 unpaired electron, magnetic moment = 1.73 BM

The correct answer is:
[Co(H2O)6]2+: High spin, 3 unpaired electrons, magnetic moment = 3.87 BM

Solution:
Correct Option is : (B)
[Co(H2O)6]2+: High spin, 3 unpaired electrons, magnetic moment = 3.87 BM

Topic: Chemistry, Coordination Compounds, NEET
Using electrolytic method, the cost of production of 5L of oxygen at STP , is Rs X , the cost of production of same volume of hydrogen at STP , will be

Options:
A 2X
B X2
C 8X
D X8

The correct answer is:
X2

Solution:
During electrolysis of water oxygen is produced at anode and hydrogen at cathode according to reaction, 2 H 2 O→2 H 2 + O 2 H 2 and O2 are produced in 2:1

Topic: Chemistry, Electrochemistry, NEET
Using imprints from a plate with complete medium and carrying bacterial colonies, you can select streptomycin resistant mutants and prove that such mutations do not originate as adaptation. These imprints need to be used:

Options:
A on plates with and without streptomycin
B on plates with minimal medium
C only on plates with streptomycin
D only on plates without streptomycin.

The correct answer is:
only on plates with streptomycin

Solution:
Plates having streptomycin allow to propagate only those bacteria, which are resistant to the antibiotic. Plates without streptomycin act as control.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2005
Using s, p, d, f notations, describe the orbital with the following quantum numbers n = 2, l = 1

Options:
A 2f
B 2s
C 2p
D 2d

The correct answer is:
2p

Solution:
Correct Option is : (C)
2p

Topic: Chemistry, Structure of Atom, NEET
Using s, p, d, f notations, describe the orbital with the following quantum numbers n = 3, l = 2

Options:
A 3f
B 3s
C 3p
D 3d

The correct answer is:
3d

Solution:
Correct Option is : (D)
3d

Topic: Chemistry, Structure of Atom, NEET
Using s, p, d, f notations, describe the orbital with the following quantum numbers n = 4, l = 0

Options:
A 4f
B 4s
C 4p
D 4d

The correct answer is:
4s

Solution:
Correct Option is : (B)
4s

Topic: Chemistry, Structure of Atom, NEET
Using s, p, d, f notations, describe the orbital with the following quantum numbers n = 5, l = 3

Options:
A 5f
B 5s
C 5p
D 5d

The correct answer is:
5f

Solution:
Correct Option is : (A)
5f

Topic: Chemistry, Structure of Atom, NEET
Using s, p, d notations, describe the orbital with the following quantum number: (i) n=1, l=0; and (ii) n = 3; l=1

Options:
A (i)1p and (ii)3p
B (i)1s and (ii)3p
C (i)1s and (ii)3d
D (i)1p and (ii)3d

The correct answer is:
(i)1s and (ii)3p

Solution:
Correct Option is : (B)
(i)1s and (ii)3p

Topic: Chemistry, Structure of Atom, NEET
Using s, p, d notations, describe the orbital with the following quantum numbers: (i) n = 4; l =2; (ii) n=4; l=3

Options:
A (i)4f and (ii)4p
B (i)4p and (ii)4f
C (i)4d and (ii)4f
D (i)4f and (ii)4d

The correct answer is:
(i)4d and (ii)4f

Solution:
Correct Option is : (C)
(i)4d and (ii)4f

Topic: Chemistry, Structure of Atom, NEET
Using Stock notation, represent the following compounds: HAuCl₄, Tl₂O, FeO, Fe₂O₃, CuI, CuO, MnO, and MnO₂ What is the oxidation number of manganese (Mn) in MnO₂?

Options:
A +1
B +2
C +3
D +4

The correct answer is:
+4

Solution:
Correct Option is : (D)
+4

Topic: Chemistry, Redox Reactions, NEET
Using the data provided, calculate the multiple bond energy (kJmol-1) of a Câ¡C bond in C2H2. That energy is (take the bond energy of a C-H bond as 350 kJ mol-1).
2Cs+H2gâC2H2gÎH=225 kJmol-1
2C(s)â2C(g)ÎH=1410 kJmol-1
H2gâ2HgÎH=330 kJmol-1

Options:
A 837
B 1165
C 815
D 865

The correct answer is:
815

Solution:
Let x be the bond energy of C≡C , it can be find out as follows
2C(s)+H2(g)→C2H2(g)
ΔH=ΣB.E.reactant-B.E.product
1410+330-x+2×350=225
x=815 kJmol-1

Topic: Chemistry, Thermodynamics (C), NEET
Using the data provided, calculate the multiple bond energy kJ mol-1 of a Câ¡C bond in C2H2 . That energy is (take the bond energy of a C-H bond as 350 kJ mol-1 )

2Csâ2Cg ; âH=1410 kJ mol-1

H2gâ2H(g) : âH=330 kJ mol-1

2C s + H 2 q â C 2 H 2 âââ;ââÎH=225â kJmol â1

Options:
A 1165
B 837
C 865
D 815

The correct answer is:
815

Solution:
(i) 2Cs+H2gâH-Câ¡C-H(g)
âH=225 kJ mol-1
(ii) 2Csâ2C(g)
âH1=1410 KJ mol-1
iii H2(g)â2H(g)
âH2=330 KJ mol-1
From equation (i) :
225=âH1+âH2-[ 2âHC-H+âHCâ¡C]
225= 1410+330+2Ã350+âHCâ¡C ]
225=1040-âHCâ¡C
âHCâ¡C=1040-225=815 KJ mol-1

Topic: Chemistry, Thermodynamics (C), NEET
Using the equation of state pV=nRT, What is the relation between the temperature density of a gas and gas pressure(p)?

Options:
A ρ ∝ P
B ρ ∝ P3
C ρ ∝ P1/2
D ρ ∝ P1/5

The correct answer is:
ρ ∝ P

Solution:
Correct Option is : (A)
ρ ∝ P

Topic: Chemistry, States of Matter, NEET
Using the following data
Mass hydrogen atom = 1.00783 u
Mass of neutron = 1.00867 u
Mass of nitrogen atom ( 7N14)= 14.00307 u
The calculated value of the binding energy of the nucleus of the nitrogen atom ( 7N14) is close to

Options:
A 56 MeV

B 98 MeV

C 104 MeV

D 112 MeV

The correct answer is:
104 MeV

Solution:
The binding energy of nucleus may be defined as the energy equivalent to the mass defect of the nucleus.
If ∆m is mass defect then according to Einstein's mass-energy relation.
Binding Energy
=∆ mc2 = Zmp + A-Z mn -M c2
=7 × 1.00783 + 7 × 1.00867-14.00307c2
or BE=0.1124 × 931.5 MeV
or BE = 104.6 MeV

Topic: Physics, Nuclear Physics, NEET
Using the formula $v=\sqrt{\frac{\gamma P}{\rho }}$, which of the following option is correct

Options:
A the speed of sound in air is independent of humidity
B the speed of sound in air increases with humidity
C the speed of sound in air decreases with temperature
D the speed of sound in air is independent of pressure

The correct answer is:
the speed of sound in air is independent of pressure

Solution:
Correct Option is : (D)
the speed of sound in air is independent of pressure

Topic: Chemistry, Waves and Sound, NEET
Using the Gibbs change, ∆Go=+63.3 kJ, for the following reaction, Ag2CO3g ⇌2Ag+aq+CO32-aq the Ksp of Ag2CO3s in water at 25oC is R=8.314 JK-1mol-1

Options:
A 3.2×10-26
B 8.0×10-12
C 2.9×10-3
D 7.9×10-2

The correct answer is:
8.0×10-12

Solution:
∆Go=-2.303 RTlog⁡Ksp

63.3×103=-2.303×8.31×298log⁡Ksp

-11.09=log⁡Ksp

8×10-12=Ksp

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2014
Using the Periodic Table, predict the formula of compounds which might be formed by silicon and bromine.

Options:
A Si2Br3
B SiBr
C SiBr4
D SiBr6

The correct answer is:
SiBr4

Solution:
Correct Option is : (C)
SiBr4

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
Using the Periodic Table, predict the formulas of compounds which might be formed by aluminium and sulphur.

Options:
A Al3S5
B AlS8
C AlS
D Al2S3

The correct answer is:
Al2S3

Solution:
Correct Option is : (D)
Al2S3

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
Using the tissue sample found at the crime scene, the forensic investigator ran a DNA analysis with four suspects and the results obtained are shown in the below image. Determine the suspect having the highest probability of been present at the crime scene.

Options:
A Suspect A
B Suspect B
C Suspect C
D Suspect D

The correct answer is:
Suspect C

Solution:
DNA fingerprinting is a technique where counts of repetitive DNA sequences are used for profiling an individual as it is unique to an individual. From the image, crime scene sample and suspect C have similar banding pattern. Whereas A, B and D banding pattern is very different from the crime scene section.

Topic: Biology, Biotechnology and Its Applications, NEET
Using valence bond theory, explain the following in relation to the complexes given below: [Mn(CN)6]3–, [Co(NH3)6]3+, [Cr(H2O)6]3+, [FeCl6]4–Choose the correct option regarding inner or outer orbital complex.

Options:
A [Mn(CN)6]3–: Outer orbital complex
B [Co(NH3)6]3+: Outer orbital complex
C [Cr(H2O)6]3+: Outer orbital complex
D [FeCl6]4–: Outer orbital complex

The correct answer is:
[FeCl6]4–: Outer orbital complex

Solution:
Correct Option is : (D)
[FeCl6]4–: Outer orbital complex

Topic: Chemistry, Coordination Compounds, NEET
Using valence bond theory, explain the following in relation to the complexes given below: [Mn(CN)6]3–, [Co(NH3)6]3+, [Cr(H2O)6]3+, [FeCl6]4–Choose the correct option regarding the type of hybridization.

Options:
A [Mn(CN)6]3–: sp3d2hybridization
B [Co(NH3)6]3+: sp3d2hybridization
C [Cr(H2O)6]3+: sp3d2hybridization
D [FeCl6]4–: sp3d2hybridization

The correct answer is:
[Mn(CN)6]3–: sp3d2hybridization

Solution:
Correct Option is : (A)
[Mn(CN)6]3–: sp3d2hybridization

Topic: Chemistry, Coordination Compounds, NEET
Using valence bond theory, explain the following in relation to the complexes given below: [Mn(CN)6]3–, [Co(NH3)6]3+, [Cr(H2O)6]3+, [FeCl6]4–Choose the correct option regarding magnetic behavior.

Options:
A [Mn(CN)6]3–: High spin (paramagnetic)
B [Co(NH3)6]3+: High spin (paramagnetic)
C [Cr(H2O)6]3+: Low spin (diamagnetic)
D [FeCl6]4–: High spin (paramagnetic)

The correct answer is:
[Mn(CN)6]3–: High spin (paramagnetic)

Solution:
Correct Option is : (A)
[Mn(CN)6]3–: High spin (paramagnetic)

Topic: Chemistry, Coordination Compounds, NEET
Using valence bond theory, explain the following in relation to the complexes given below: [Mn(CN)6]3–, [Co(NH3)6]3+, [Cr(H2O)6]3+, [FeCl6]4–Choose the correct option regarding spin-only magnetic moment value.

Options:
A [Mn(CN)6]3–: 3.87 BM
B [Co(NH3)6]3+: 3.87 BM
C [Cr(H2O)6]3+: 0 BM
D [FeCl6]4–: 1.73 BM

The correct answer is:
[Co(NH3)6]3+: 3.87 BM

Solution:
Correct Option is : (B)
[Co(NH3)6]3+: 3.87 BM

Topic: Chemistry, Coordination Compounds, NEET
UTR’s are untranslated region present on

Options:
A rRNA
B tRNA
C mRNA
D hnRNA

The correct answer is:
mRNA

Solution:
A translation unit in mRNA consists of some additional sequences that are not translated and are referred to as an untranslated regions. The UTRs are present at both 5’ ends before the start codon and at 3’ end after stop codons. They are required for an efficient translation process.

Topic: Biology, Molecular Basis of Inheritance, NEET
V2O5 is red or orange in colour. it is a/an ................ oxide.

Options:
A acidic
B basic
C amphoteric
D neutral

The correct answer is:
amphoteric

Solution:
V2O5 is an amphoteric oxide.
V2O5+NaOH→ Various vanadates
V2O5+6HCl→2VOCl2+3H2O+Cl2
V2O5+10HCl→2VCl5+5H2O

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
V â I graphs for parallel and series combination of two metallic resistors are shown in figure. Which graph represents parallel combination?

Options:
A A
B B
C Both A & B
D Neither A nor B

The correct answer is:
A

Solution:
Slope of V-I graph gives the resistance and since graph A has smaller slope, hence it will represent smaller resistance. parallel combination of two resistor will always have smaller resistance compared to series combination.

Topic: Physics, Current Electricity, NEET
VSEPR theory suggests that bond length and bond angle depend upon the presence or absence of lone pairs. If the bond angle of H-O-H in H2O is 105°(approx.), then the bond angle of F-O-F in F2O will be

Options:
A greater than 105°
B lesser than 105°
C equal to 105°
D equal to 180°

The correct answer is:
lesser than 105°

Solution:
In F2O, F is more electronegative which attracts the bond pair electrons away from central oxygen atom. Which decreases the repulsion between bond pair of electrons, hence bond angle decreases.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
V-T graph for a given mass of an ideal gas is shown in the figure. Then the corresponding P-V graph would be

( BC and DA are vertical lines, CD is horizontal line)

Options:
A
B
C
D

The correct answer is:

Solution:
For AâB, Vâ T, so P=constant
For BâC, T= constant, so PV is hyperbola
For CâD, T decrease at constant V, so P decreases
For DâA, T= constant, so PV is hyperbola

Topic: Physics, Thermodynamics, NEET
V versus T curves at constant pressure P1 and P2 for an ideal gas are shown in figure. Here,

Options:
A P1>P2
B P1<P2
C P1=P2
D P1≥P2

The correct answer is:
P1>P2

Solution:
PV=μRT or P=μRTV or P=μRVT
or P=μRtan⁡θ , where tan⁡θ is the slope of the graph.
So, more the slope, lesser will be the pressure.

Topic: Physics, Kinetic Theory of Gases, NEET
V vs T curves at constant pressure P1 and P2 for an ideal gas are shown below.

Which is correct ?

Options:
A P1>P2
B P1<P2
C P1=P2
D All of the above

The correct answer is:
P1<P2

Solution:
Product PV is constant for a given temperature. As for pressure P1 , volume is more than that for P2 at a given temperature. Hence, P1 should be smaller than P2 . For such V−T graphs at constant pressure, higher the pressure lesser will be the slope.

Topic: Chemistry, States of Matter, NEET
Vaccination helps in the development of

Options:
A toxins
B lymph
C antibodies
D plasma

The correct answer is:
antibodies

Solution:
The principle of immunization or vaccination is based on the property of âmemoryâ of the immune system. In vaccination, a preparation of antigenic proteins of pathogen or inactivated/weakened pathogen (vaccine) is introduced into the body. The antibodies are produced in the body against these antigens, and they neutralize the pathogenic agents during actual infection. The vaccines also generate memory â B and T-cells that recognize the pathogen quickly on subsequent exposure and overwhelm the invaders with a massive production of antibodies.

Topic: Biology, Human Health and Diseases, NEET