datavorous/entrance-exam-dataset huggingface dataset:
The density of a liquid is 1.2 g/mL. There are 35 drops in 2 mL. The number of molecules in 1 drop is
(Molecular weight of liquid = 70)

Options:
A 1.235NA
B 1352NA
C 1.2(35)2NA
D 1.2NA

The correct answer is:
1.2(35)2NA

Solution:
Volume of one drop =235mL
Number of moles in one drop =2×1.235×70=1.2(35)2
Number of molecules in one drop =1.2(35)2×NA

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
The density of a material, in the shape of a cube, is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are 1.5 % and 1 %, respectively, the maximum error in determining the density is

Options:
A 6 %
B 2.5 %
C 3.5 %
D 4.5 %

The correct answer is:
4.5 %

Solution:
ÏâML3
ÎÏÏâÎMM+3ÎLL
ÎÏÏÃ100â1.5+3Ã1
% relative error =4.5 %

Topic: Physics, Units and Dimensions, NEET
The density of a newly discovered plant is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth $R$, the radius of the planet would be:

Options:
A $2 R$
B $4 R$
C $\frac{1}{4} R$
D $\frac{1}{2} R$

The correct answer is:
$\frac{1}{2} R$

Solution:
We know that:
$$
\begin{aligned}
g=\frac{G M}{R^2} & =\frac{G \frac{4}{3} \pi R^3 \rho}{R^2} \\
& =\frac{4}{3} \pi G \rho R
\end{aligned}
$$
New according to the question
$$
\begin{aligned}
g_{\text {planel }} & =g_{\text {earh }} \\
\Rightarrow \quad R_{\text {planel }} & =\frac{R_{\text {earth }}}{2}
\end{aligned}
$$

Topic: Physics, Gravitation, NEET, NEET 2004
The density of copper is 9Ã103 kg m-3 and its atomic mass is 63.5 u. Each copper atom provides one free electron. Estimate the number of free electrons per cubic metre in copper.

Options:
A 1019
B 1023
C 1025
D 1028

The correct answer is:
1028

Solution:
The density of copper,
Ï=9Ã103 kg m-3=9Ã106 g m-3
Avogadro number, NA=6.023Ã1023
Mass of 1 mole of the copper atom, M=63.5 g
Thus, the number of free electrons per volume is
n=NAMÏ=6.023Ã102363.5Ã9Ã106 =8.5Ã1028 m-3

Topic: Physics, Current Electricity, NEET
The density of gas A is twice that B. At the same temperature the molecular weight of gas B is twice that of A. The ratio of pressure of gas A and B will be

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

The correct answer is:
4:1

Solution:
dA=2dB;2MA=MB;
As PM=dRT
So PAPB×MAMB=dAdB×RTRT
PAPB×12=2
PAPB=41

Topic: Chemistry, Solutions, NEET
The density of KBr is 2.75 g cm-3 , length of the unit cell is 654 pm. K = 39, Br = 80, then what can be true about the predicted nature of the solid?
(Given: N A =6.023Ã 10 23 )

Options:
A Solid has face centred cubic system with co-ordination number = 6
B Solid has simple cubic system with co-ordination number = 8
C Solid has face centred cubic system with co-ordination number =12
D None of the above

The correct answer is:
Solid has face centred cubic system with co-ordination number = 6

Solution:
Given that âÏ=2.75âgâc m â3
we need to find type of unit cell

â´âââÏ= ZÃM N A a 3 âµâââââZ= ÏÃ N A Ã a 3 M âââââââââââ=â 2.75Ã6.023Ã 10 23 Ã ( 654Ã 10 â10 ) 3 119
â´âââZâ4ââSoâfccâunitâcell
In FCC 1 atm surrounded by 6 atoms so coordination number is 6.

Topic: Chemistry, Solid State, NEET
The density of material in the CGS system of units is 4 g cm-3. In a system of units in which unit of length is 10 cm and the unit of mass is 100 g, the value of density of the material will be

Options:
A 0.4
B 40
C 400
D 0.04

The correct answer is:
40

Solution:
In the CGS system,

dâ=â4 g cm-3

The unit of mass is 100 g and unit of length is 10 cm, so

density =â4(100 g100)(1010 cm)3

=â(4100)(110)3â(100 g)(10 cm)3

=â4100âÃâ(10)3â100 g(10 cm)3

=â40âunits

Topic: Physics, Units and Dimensions, NEET
The density of neon will be highest at:

Options:
A STP
B 0Â°C,2 atm
C 273Â°C,1 atm
D 0Â°C, 1.5 atm

The correct answer is:
0Â°C,2 atm

Solution:
PV=wMRT
or P=WV.mRT,d=wV
or P=dmRT
â´Pâd,dâ1T

Topic: Chemistry, States of Matter, NEET
The density of solid argon is 1.65 g/mL at -2330C . If the argon atom is assumed to be sphere of radius 1.54Ã10-8 cm , what percentage of solid argon has apparently empty space? (At. wt. of Ar =

40)

Options:
A 62%
B 72%
C 52%
D 42%

The correct answer is:
62%

Solution:
Volume of one atom of Ar =43Ïr3
Also, No. of atoms in 1.65g =1.6540Ã6.023Ã1023
Total volume of all atoms of Ar in solid state.
=43Ïr3Ã1.6540Ã 6.023Ã1023
=43Ã227Ã1.54Ã10-83Ã1.6540Ã6.023Ã1023
=0.380 cm3
Volume of solid argon =1 cm3
% Empty space =[[1-0.380]1Ã100=62%

Topic: Chemistry, Solid State, NEET
The density of the solution is $2.15 \mathrm{~g} \mathrm{~mL}^{-1}$, then mass of $2.5 \mathrm{~mL}$ solution in correct significant figures is

Options:
A $5375 \times 10^{-3} \mathrm{~g}$
B $5.4 \mathrm{~g}$
C $5.38 \mathrm{~g}$
D $53.75 \mathrm{~g}$

The correct answer is:
$5.4 \mathrm{~g}$

Solution:
$\begin{aligned}
& \text {Explanation: Density }=\frac{\text { Mass }}{\text { Volume }} \\
& \text {Density }=2.15 \mathrm{~g} / \mathrm{mL} \text {, Volume of solution }=2.5 \mathrm{~mL} \\
& \therefore 2.15=\frac{\text { Mass }}{2.5} \Rightarrow \text { Mass }=5.375 \mathrm{~g}
\end{aligned}$
The correct significant figure should have the same number of significant digits as the number with least number of significant digits.
$\therefore \text { Mass }=5.4 \mathrm{~g}$

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2022 (Phase 2)
The density of water at $20^{\circ} \mathrm{C}$ in $998 \mathrm{~kg} / \mathrm{m}^3$ and at a $40^{\circ} \mathrm{C} 992 \mathrm{~kg} / \mathrm{m}^3$. The cocfficient of volume expansion of water is:

Options:
A $3 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
B $2 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
C $6 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
D $10^{-3} \rho^{\circ} \mathrm{C}$

The correct answer is:
$3 \times 10^{-4} /{ }^{\circ} \mathrm{C}$

Solution:
According to the question
$$
\begin{aligned}
& T_1=20^{\circ} \mathrm{C}, T_2=40^{\circ} \mathrm{C} \\
& \rho_r=998 \mathrm{~kg} / \mathrm{m}^3 \rho_{40}=992 \mathrm{~kg} / \mathrm{m}^3 \\
& \text { So, } \rho_{\mathrm{T}_2}=\frac{\rho_{\mathrm{T}_1}}{(1+\mathrm{v} \Delta T)}=\frac{\rho_{\mathrm{T}_1}}{1+\mathrm{v}\left(T_2-T_1\right)}
\end{aligned}
$$
Putting the values we get
$$
\begin{aligned}
& 992=\frac{998}{1+v(40-20)} \\
& 992=\frac{998}{1+20 \mathrm{v}} \\
& \Rightarrow 992(1+20 \mathrm{v})=998 \\
& \Rightarrow \quad 1+20 \mathrm{v}=\frac{998}{992} \\
& \Rightarrow \quad 20 \mathrm{v}=\frac{6}{992} \\
& \Rightarrow \mathrm{v}=\frac{6}{992} \times \frac{1}{20}=3 \times 10^{-4} /{ }^{\circ}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2013 (Karnataka)
The density of water at the surface of ocean is Ï. If the bulk modulus of water is B, then the density of ocean water at depth, when the pressure is Î±p0 where p0 is the atmospheric pressure, is

Options:
A ρBB-α-1p0
B ρBB+α-1p0
C ρBB-αp0
D ρBB+αp0

The correct answer is:
ρBB-α-1p0

Solution:
From the definition of bulk modulus
B = - dp dV V
As we move from surface to place where pressure changes to α p 0, let us assume volume changes by Δ V, then
B = - V Δ p Δ V = V α - 1 p 0 Δ V
New volume, V ′ = V - Δ V = 1 - α - 1 p 0 B V
Density at the given depth, ρ ′ = ρ V V ′, where ρ is density at surface
ρ ′ = ρ × B B - α - 1 p 0

Topic: Physics, Mechanical Properties of Solids, NEET
The density of water is 1000 kg $~{{m}^{-3}}$. The density of water vapour at 100 °C and 1 atm pressure is 0.6 kg $~{{m}^{-3}}$. The volume of a molecule multiplied by the total number gives ,what is called, molecular volume. What is the estimated volume of a water molecule using the given data?

Options:
A 3 x ${{10}^{-30}}~{{m}^{3}}$
B 3 x ${{10}^{-30}}~{{m}^{3}}$
C 3 x ${{10}^{-30}}~{{m}^{3}}$
D 3 x ${{10}^{-30}}~{{m}^{3}}$

The correct answer is:
3 x ${{10}^{-30}}~{{m}^{3}}$

Solution:
Correct Option is : (B)
3 x ${{10}^{-30}}~{{m}^{3}}$

Topic: Chemistry, Kinetic Theory of Gases, NEET
The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly

Options:
A The depression in freezing point increases because acetic acid is less polar than the other two acids, leading to weaker interactions with water molecules.
B Among H, Cl, and F, H is least electronegative while F is most electronegative. Then, F can withdraw electrons towards itself more than Cl and H. Thus, trifluoroacetic acid can easily lose H+ ions i.e., trifluoroacetic acid ionizes to the largest extent. Now, the more ions produced, the greater is the depression of the freezing point
C The depression in freezing point increases because trifluoroacetic acid has the highest molecular weight, resulting in stronger van der Waals forces with water.
D The depression in freezing point increases because acetic acid is a weaker acid, leading to more significant changes in colligative properties.

The correct answer is:
Among H, Cl, and F, H is least electronegative while F is most electronegative. Then, F can withdraw electrons towards itself more than Cl and H. Thus, trifluoroacetic acid can easily lose H+ ions i.e., trifluoroacetic acid ionizes to the largest extent. Now, the more ions produced, the greater is the depression of the freezing point

Solution:
Correct Option is : (B)
Among H, Cl, and F, H is least electronegative while F is most electronegative. Then, F can withdraw electrons towards itself more than Cl and H. Thus, trifluoroacetic acid can easily lose H+ ions i.e., trifluoroacetic acid ionizes to the largest extent. Now, the more ions produced, the greater is the depression of the freezing point

Topic: Chemistry, Solutions, NEET
The depth d at which the value of acceleration due to gravity becomes 1n times the value at the earth's surface is (R = radius of the earth )

Options:
A d=Rnn-1
B d=Rn-12n
C d=Rn-1n
D d=R2n-1n

The correct answer is:
d=Rn-1n

Solution:
At depth d we have
gâ²=g1-dR
gâ²=gn
gn=g1-dR
â´ 1n=1-dR
â´ dR=1-1n=n-1n
d=Rn-1n

Topic: Physics, Gravitation, NEET
The desirable properties for making permanent magnets are

Options:
A high retentivity and high coercive force
B high retentivity and low coercive force
C low retentivity and high coercive force
D low retentivity and low coercive force

The correct answer is:
high retentivity and high coercive force

Solution:
The desirable properties for making permanent magnets are high retentivity, high coercivity and high relative magnetic permeability.

Topic: Physics, Magnetic Properties of Matter, NEET
The destruction of the biological nature and activity of proteins by heat or chemical agent is called–

Options:
A Dehydration
B Denaturation
C Denitrogenation
D Deammination

The correct answer is:
Denaturation

Solution:
The process brings in changes in physical as we as biological activity of the proteins is called denaturation.

Topic: Chemistry, Biomolecules, NEET
The deuteron is bound by nuclear forces just as H-atom is made up of p and e bound by electrostatic forces. If we consider the force between neutron and proton in deuteron as given in the form of a Coulomb potential but with an effective charge e′:
What is the estimated value of the ratio (e'/e) if the binding energy of a deuteron is 2.2 MeV?

Options:
A 9.9
B 10.1
C 10.5
D 11

The correct answer is:
11

Solution:
Correct Option is : (D)
11

Topic: Chemistry, Nuclear Physics, NEET
The development of Funaria gametophyte always initiated from

Options:
A Antheredium
B Protonema
C Archegonia
D Capsule

The correct answer is:
Protonema

Solution:
Bryophyta includes the simplest and primitive land plants, characterized by the presence of independent gametophyte and parasitic sporophyte. Spore is the first cell of gametophytic generation and in funaria the spore germinates to form protonema.

Topic: Biology, Plant Kingdom, NEET
The development of male gametophyte in angiosperms is

Options:
A Partly in-situ in anther and partly ex-situ on stigma.
B Fully In-situ in anther.
C Fully Ex-situ on stigma.
D In-vitro only.

The correct answer is:
Partly in-situ in anther and partly ex-situ on stigma.

Solution:
The male gametophyte or pollen grain starts its germination inside microsporangia; thus, it is known as in-situ germination. In 60% angiosperms, pollen grain is liberated at the 2-celled stage and after landing on stigma, it undergoes further division to form 2 male gametes. Thus, germination is partially ex-situ on stigma.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The deviation produced by a prism is

Options:
A same for all wavelengths
B greatest for red and least for violet
C greatest for violet and least for red
D the prism produces no deviation

The correct answer is:
greatest for violet and least for red

Solution:
Deviation is given by, Î´=AÎ¼-1
Also, refractive index Î¼â1Î»
As Î»R>Î»Î½
â´Î¼R<Î¼Î½
â´Î´R<Î´Î½

Topic: Physics, Ray Optics, NEET
The device that can act as a complete electronic circuit is

Options:
A Junction diode
B Integrated circuit
C Junction transistor
D Zener diode

The correct answer is:
Integrated circuit

Solution:
Integrated circuits are miniature electronic circuit produced within a single crystal of a semiconductors such as silicon. They contain a million or so transistors and resistors or capacitors. They are widely used in memory circuits, micro computers, pocket calculators and electronic watches on account of their low cost and bulk, reliability into specific regions of the semiconductor crystals.

Topic: Physics, Semiconductors, NEET, NEET 2010 (Screening)
The diagram below shows a cell cycle.

Which of the following process occur during V?

Options:
A Replication of DNA
B Replication of centrioles
C Chromosomes condense and become shorter and thicker
D High metabolic rate and synthesis of proteins and cellular organelles

The correct answer is:
High metabolic rate and synthesis of proteins and cellular organelles

Solution:
In the given diagram, the 'V' labeled phase is the G1 phase. G1 phase marks the start of interphase. G1 phase is the metabolically active phase and during this phase, the cell grows in size. Synthesis of mRNA and protein of DNA synthesis occurs in this phase.

Topic: Biology, Cell Cycle and Cell Division, NEET
The diagram below shows a fertilised ovule and carpel.

​

Mark the correct set of numbers which represent the structures that will become:

(i) The future embryo
(ii) The future testa
(iii) The future micropyle region

Options:
A

i	ii	iii

5	3	1

B

i	ii	iii

7	8	4

C

i	ii	iii

6	3	7

D

i	ii	iii

5	2	4

The correct answer is:

i	ii	iii

5	2	4

Solution:
(i – 5), (ii – 2), (iii – 4); number 5, 2 and 4 are respectively egg, integument and micropyle of ovule. They form embryo, testa and micropyle.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The diagram given below is related to which of the following set of hormones?

Options:
A Cortisol, Estradiol, Gonadotrophins
B Testosterone, Oxytocin, Estradiol
C Corticoids, Iodothyronines, Steroids
D Catecholamines, Steroids, Cortisol

The correct answer is:
Corticoids, Iodothyronines, Steroids

Solution:
All lipid-soluble hormones such as adrenocorticoids, steroidal, and Iodothyronines hormones can easily cross the target cell membrane and enter inside the target cell. These hormones bind to their specific intracellular receptor to form the hormone–receptor complex, which ultimately binds to DNA and regulates gene expressions to give the physiological response.

Topic: Biology, Chemical Coordination and Integration, NEET
The diagram given below represents the events of fertilization and embryonic development in

Options:
A Cycas a Gymnosperm plant
B Pinus a Gymnosperm plant
C Cocus a Angiosperm plant
D Cycas a Angiosperm plant

The correct answer is:
Pinus a Gymnosperm plant

Solution:
The winged pollen grains are characteristic features of Pinus, a gymnosperm plant. The diagram represents the pollination by anaemophily, siphonization, fertilization and formation of the embryo.

Topic: Biology, Plant Kingdom, NEET
The diagram given below represents the sectional view of

Options:
A Amphitropous ovule
B Campylotropous ovule
C Anatropous ovule
D Orthotropous ovule

The correct answer is:
Orthotropous ovule

Solution:
When the micropyle, body of the ovule and funicle lie in one vertical plane, the ovule is called orthotropous. The body of ovule lies straight and upright over the funicle (atropous/erect). Hilum, chalaza and micropyle occur on the same line with hilum and chalaza nearby e.g, Piperaceae, Urticaceae, Polygonaceae.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The diagram given below represents

Options:
A inspiration
B expiration
C systole
D diastole

The correct answer is:
inspiration

Solution:
The figure shows the expansion of the lung cavity in both the planes i.e. dorsoventrally and anteroposteriorly. Expansion meaning volume increases and as the flow of gases follows Boyle's law, if volume increases pressure decreases so that air can flow from high atmospheric pressure to low atmospheric pressure i.e. inhalation can take place.

Topic: Biology, Breathing and Exchange of Gases, NEET
The diagram of a logic circuit is given below. The output A of the circuit is represented by

Options:
A P+R·Q
B P⋅(R⋅Q)
C P·R+Q
D P+(R+Q)

The correct answer is:
P+R·Q

Solution:
Output of upper OR gate =P+R
Output of lower OR gate =P+Q
â´ Net output A=P+R.P+Q
=P.P+P.Q+R.P+R.Q
=P.1+Q+R.P+R.Q
âµ 1+Q=1 and P.P=P
â´ A=P+Râ P+R.Q
=P1+R+RâQ=P+RâQ

Topic: Physics, Semiconductors, NEET
The diagram provided below is used to remove which type of pollution?

Options:
A Particulate matter
B Gaseous pollution
C SO2 pollution
D Sound pollution

The correct answer is:
Particulate matter

Solution:
It is an electrostatic precipitator which can remove 99.9% of particulate matter present in the exhaust from a thermal power plant. Electrostatic precipitator (ESP) separates the dust particle by charging them in presence of electric flield generating a corona of electrons.

Topic: Biology, Environmental Issues, NEET
The diagram represents ethanol fermentation, Identity the labeling 1, 2 and 3

Options:
A 1. The glucose breaks down into two pyruvates.
2. The pyruvate is broken down into acetaldehyde by releasing CO2.
3. The acetaldehyde is then converted to ethanol.

B 1. The glucose breaks down into two pyruvates.
2. The acetaldehyde is then converted to ethanol.
3. The pyruvate is broken down into acetaldehyde by releasing CO2.

C 1. The glucose breaks down into two acetaldehydes
2. The acetaldehyde is then converted to ethanol.
3. The pyruvate is broken down into acetaldehyde by releasing CO2.
D 1. The glucose breaks down into two pyruvates.
2. The pyruvate is broken down into acetaldehyde by releasing NH3.
3. The acetaldehyde is then converted to ethanol.

The correct answer is:
1. The glucose breaks down into two pyruvates.
2. The pyruvate is broken down into acetaldehyde by releasing CO2.
3. The acetaldehyde is then converted to ethanol.

Solution:
During ethanol fermentation, one glucose molecule breaks down into two pyruvic acids the energy released from this exothermic reaction is used to bind the inorganic phosphates to ADP and convert NAD+ to NADH. The pyruvic acids are broken down into acetaldehydes and give off CO2. The two acetaldehydes are then converted to two ethanol by using the H+ ions from NADH, converting NADH back into NAD+.

Topic: Biology, Microbes in Human Welfare, NEET
The diagram represents Miller's experiment. Choose the correct combination of labelling.

Options:
A A-Electrodes
B â NH3 + H2 + H2O + CH4
C- Cold water
D- Vacuum
E- U-trap
B A-Electrodes
B − NH4 + H2 + CO2 + CH3
C- Hot water
D- Vacuum
E- U-trap
C A-Electrodes
B − NH3 + H2O
C- Steam
D- U-trap
E- Vacuum
D A-Electrodes
B − NH3 + H2 + H2O + CH4
C- Steam
D- Vacuum
E- U-trap

The correct answer is:
A-Electrodes
B − NH3 + H2 + H2O + CH4
C- Cold water
D- Vacuum
E- U-trap

Solution:
The correct combination of labellings are:
A - Electrodes
B â NH3 + H2 + H2O + CH4
C - Cold water
D - Vacuum
E - U-trap
The MillerâUrey experiment was an experiment that simulated hypothetical conditions present on the early Earth to test what kind of environment would be needed to allow life to begin. The experiment is considered to be the classic experiment on the origin of life. It was conducted in 1953 by Stanley L. Miller and Harold C. Urey at the University of Chicago.

Topic: Biology, Evolution, NEET
The diagram shows a hemispherical shell of mass m and radius R is hinged at point O placed on a horizontal surface.A ball of mass m strikes the shell at point A (as shown in the figure) moving with velocity u inclined at an angle Î¸ = tan -1 1 2 and then it stops. For the given shell to reach horizontal surface OP what minimum speed u is required?

Options:
A 2gR3
B 2 g⁡ R⁡ 3
C g⁡ R⁡ 5
D Not possible

The correct answer is:
Not possible

Solution:
Torque about O is zero as well angular momentum hence Ï = 0
â´ Not possible

Topic: Physics, Rotational Motion, NEET
The diagram shows a square loop of side 0.5 m and resistance 10 Î©. The magnetic field has a magnitude B = 1.0 T. The work done in pulling the loop out of the field slowly and uniformly in 2.0 s is

Options:
A 3.125 × 10-3 J

B 6.25 × 10- 4 J

C 1.25 × 10-2 J

D 5.0 × 10-4 J

The correct answer is:
3.125 × 10-3 J

Solution:
Speed of the loop should be
v=lengthtime =0.52=0.25 m s-1
Induced emf e=Bvl⁡=1.00.250.5=0.125 V
∴ Current in the loop i=eR=0.12510=1.25×10-2 A
The magnetic force on the left arm due to the magnetic field is
Fm=ilB=1.25×10-20.51.0
=6.25×10-3 N
To pull loop uniformly an external force of 6.25×10-3 N towards right must be applied.
∴ W = (6.25 × 10-3 N) (0.5 m)
= 3.125 × 10-3 J

Topic: Physics, Electromagnetic Induction, NEET
The diagram shows an important concept in the genetic implication of DNA. Fill in the blanks $A$ to $C$

Options:
A A-transcription, B-replication, C-James Watson
B A-translation, B-transcription, C-Erevin Chargaff
C A-transcription, B-translation, C-Francis Crick
D A-translation, B-extension, C-Rosalind Franklin

The correct answer is:
A-transcription, B-translation, C-Francis Crick

Solution:
Central dogma
$$
\mathrm{DNA} \stackrel{\text { Transcription }}{\underset{A}{\longrightarrow}} m \mathrm{RNA} \stackrel{\text { Translation }}{\underset{B}{\longrightarrow}} \text { protein } \stackrel{\text { Francis Crick }}{\longrightarrow}
$$

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2013 (All India)
The diagram shows the adiabatic curve for n moles of an ideal gas. The bulk modulus for the gas corresponding to the point P will be

Options:
A 5 nRT03V0
B nR2+T0V0
C nR1+T0V0
D 2 nRT0V0

The correct answer is:
2 nRT0V0

Solution:
For adiabatic process :
Bulk modulus : B = Î³ P
For point P : P = nRT V = nR 3 T 0 3 V 0 = nRT 0 V 0
â B = Î³ nRT 0 V 0 ...(i)
Now, TV Î³ - 1 = constant
â Î³ - 1 TdV + VdT = 0
â dV dT = - V Î³ - 1 T
For point P â - 3 V 0 3 T 0 = - 3 V 0 Î³ - 1 3 T 0
â Î³ = 2
So, from Eq. (i), B = 2 nRT 0 V 0

Topic: Physics, Thermodynamics, NEET
The diagram shows the cell cycle. During which phase do chromosomes condense and become visible?

Options:
A A
B B
C C
D D

The correct answer is:
D

Solution:
Interphase, therefore, consists of G1, S, and G2. During G1, cells make the RNA, enzymes, and other proteins needed for growth. At the end of G1, the cell becomes committed to dividing or not dividing. The nuclear division follows interphase, referred to as the M phase (M for mitosis). Growth stops temporarily during mitosis. Prophase which is the first stage of karyokinesis of mitosis follows the S and G2 phases of interphase. Prophase is marked by the initiation of condensation of chromosomal material. The chromosomal material becomes untangled during the process of chromatin condensation.

Topic: Biology, Cell Cycle and Cell Division, NEET