datavorous/entrance-exam-dataset huggingface dataset:
The value of Z, the slope of the line (regression coefficient) is _______ for mammals in the tropical forest of different continents.

Options:
A 0.6
B 0.8
C 1.15
D 1.21

The correct answer is:
1.15

Solution:
Alexander von Humboldt derived the speciesâarea relationship and observed that within a region, species richness increased with the increasing explored area, but only up to a limit.
The relationship is a straight line described by the equation
log S = log C +Z log A
where,
S = Species richness, A = Area, Z Slope of the line (regression coefficient), C = Y-intercept.
The value of Z lies in the range of 0.1 to 0.2, regardless of the taxonomic group or the region. If we analyze the speciesâarea relationships among very large areas like the entire continents, the slope of the line will be much steeper (Z values in the range of 0.6 to 1.2). For frugivorous (fruit-eating) birds and mammals in the tropical forests of different continents, the slope is found to be 1.15.

Topic: Biology, Biodiversity and its Conservation, NEET
The values of dissociation constant of some bases are given below. Which is the weakest base?

Options:
A 1.8×10-5
B 4.8×10-10
C 7.2×10-11
D 7.07×10-7

The correct answer is:
7.2×10-11

Solution:
For base BOH
BOH⇌B++OH-
Dissociation constant Kb=B+OH-[BOH]
∴ Higher the value of Kb, more will be dissociation of base and stronger will be base and vice-versa.
∴ Weakest base among given choices is having Kb=7.2×10-11.

Topic: Chemistry, Ionic Equilibrium, NEET
The values of Λeq∞ for NH4Cl , NaOH and NaCl are, respectively, 149.74, 248.1 and 126.4 ohm-1cm2eq-1 . The value of Λeq∞NH4OH is
in  ohm −1 cm −2 eq −1

Options:
A 371.44
B 271.44
C 71.44
D It cannot be calculated from the data given

The correct answer is:
271.44

Solution:
Λeq∞ for NH4Cl can be find out by the relation given below
Λeqo or Λ∞(NH4OH)=Λ∞eq(NH4Cl)+Λ∞eq(NaOH)-Λ∞eq(NaCl)
=(149.74+248.1-126.4)
=271.44  ohm −1   cm 2   eq −1

Topic: Chemistry, Electrochemistry, NEET
The values of $\Delta H$ and $\Delta S$ for the reaction, $\mathrm{C}_{\text {(graphite) }}+\mathrm{CO}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}(\mathrm{g})$ are $170 \mathrm{~kJ}$ and $170 \mathrm{JK}^{-1}$, respectively. This reaction will be spontaneous at

Options:
A $710 \mathrm{~K}$
B $910 \mathrm{~K}$
C $1110 \mathrm{~K}$
D $510 \mathrm{~K}$

The correct answer is:
$1110 \mathrm{~K}$

Solution:
Key Idea For spontaneous process, $\Delta \mathrm{G} < 0$
$$
\begin{aligned}
& \Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S} \\
& \text { Given, } \Delta \mathrm{H}=170 \mathrm{~kJ}=170 \times 10^3 \mathrm{~J} \\
& \Delta \mathrm{S}=170 \mathrm{JK}^{-1} \\
& \mathrm{~T}=\text { ? } \\
& \Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S} \\
& 0 < 170 \times 10^3-\mathrm{T} \times 170 \\
& \mathrm{~T}>1000 \\
& \therefore \quad \mathrm{T}=1110 \mathrm{~K} \\
&
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2009 (Screening)
The values of \(K_{p_1}\) and \(K_{p_2}\) for the reactions
\(X \rightleftharpoons Y+Z\) ...(i)
and \(A \rightleftharpoons 2 B\) ...(ii)
are in ratio of \(9: 1\). If degree of dissociation of \(X\) and \(A\) be equal, then total pressure at equilibrium (i) and (ii) are in the ratio

Options:
A $3: 1$
B $1: 9$
C $36: 1$
D $1: 1$

The correct answer is:
$36: 1$

Solution:
From equation.
\(\begin{array}{lcc}
& X & \rightleftharpoons & Y & + & Z \\
\text {Initial mole } & 10 & & 0 & & 0 \\
\text {mole at equilibrium } & (1-a) & & a & & a
\end{array}\)
$\begin{aligned}
& K_{p_1}=\frac{p_Y \times p_z}{p_X} \\
& =\frac{\left[\frac{\alpha \times p_1}{1+\alpha}\right]\left[\frac{\alpha \times p_1}{1+\alpha}\right]}{\left[\frac{1-\alpha}{1+\alpha}\right] p_1} \\
& K_{P_1}=\frac{\alpha^2 p_1}{1-\alpha^2}
\end{aligned}$
From equation
\(\begin{array}{lcc}
& A & \rightleftharpoons & 2 B \\
\text {Initial mole } & 10 & & 0 \\
\text {mole at equilibrium } & (1-a) & & 2 a
\end{array}\)
$\begin{aligned}
& K_{p_2}=\frac{\left[\frac{2 \alpha}{1+\alpha} \cdot p_2\right]^2}{\left[\frac{1-\alpha}{1+\alpha}\right] p_2}=\frac{4 \alpha^2 p_2}{1-\alpha^2}
\end{aligned}$
From Eqs (i) and (ii)
$\frac{K_{P_1}}{K_{P_2}}=\frac{p_1}{4 p_2}$
$\begin{array}{ll}
\Rightarrow & \frac{9}{1}=\frac{p_1}{4 p_2} \\
\Rightarrow & \therefore \frac{p_1}{p_2}=\frac{36}{1}
\end{array}$

Topic: Chemistry, Biomolecules, NEET, NEET 2008 (Screening)
The values of Ksp of two sparingly soluble salts \(\mathrm{Ni}(\mathrm{OH}) 2\) and \(\mathrm{AgCN}\) are \(2.0 \times 10^{-15}\) and \(6 \times 10^{-17}\) respectively. Which salt is more soluble?

Options:
A \(\mathrm{Ni}(\mathrm{OH}) 2\)
B \(\mathrm{AgCN}\)
C Both are equally soluble
D Cannot be determined from the given information

The correct answer is:
\(\mathrm{Ni}(\mathrm{OH}) 2\)

Solution:
Correct Option is : (A)
\(\mathrm{Ni}(\mathrm{OH}) 2\)

Topic: Chemistry, Ionic Equilibrium, NEET
The values of osmotic potential (Ï) and pressure potential (Ï) of cells A, B, C and D are given below.

Cell	Ï	Ï

A	-1.0	0.5

B	-0.6	0.3

C	-1.2	0.6

D	-0.8	0.4

Identify the correct sequence that shows the path of movement of water from among the following.

Options:
A D→C→A→B
B B→D→A→C
C B→C→D→A
D C→B→A→D

The correct answer is:
B→D→A→C

Solution:
The movement of water occurs from low DPD cell to high DPD cell. DPD is equal but opposite to water potential (algebraic sum of solute or osmotic potential and pressure potential).
Î¨w = Î¨s + Î¨p

Cell	Water potential
(osmotic potential + pressure potential	DPD

A	-1+0.5=-0.5	+0.5

B	-0.6+0.3=-0.3	+0.3

C	-1.2+0.6=-0.6	+0.6

D	-0.8+0.4=-0.4	+0.4

Topic: Biology, Transport in Plants, NEET
The van der Waal's constant 'b' of a gas is 4 π × 1 0 - 4 L/mol . How near can the centres of the two molecules approach each other ? [Use : NA=6×1023]

Options:
A 1 0 - 7 m
B 1 0 - 10 m
C 5 × 1 0 - 1 1 m
D 5 × 1 0 - 9 m

The correct answer is:
1 0 - 10 m

Solution:
b = 4 Ã 4 3 Ã Ï r 3 Ã N A ;
We know that, 1L=1000 cm3
4 Ã Ï Ã 1 0 - 4 Ã 1 0 0 0 = 4 Ã 4 3 Ã Ï Ã r 3 Ã 6 Ã 1 0 2 3
r = 5 Ã 1 0 - 9 cm
Distance of closest approach =2r
= 1 0 - 8 cm or 1 0 - 10 m

Topic: Chemistry, States of Matter, NEET
The Vander Waal's constant 'a' for the gases O2,N2,NH3 and CH4 are 1.3,1.390,4.170 and 2.253 L2atmmol-2 respectively. The gas which can be most easily liquefied is

Options:
A CH4
B O2
C NH3
D N2

The correct answer is:
NH3

Solution:
Value of constant a is greater than other for NH3 that’s why NH3 can be most easily liquefied.

Topic: Chemistry, States of Matter, NEET
. The Van’t Hoff factor for complete dissociation of K2SO4is:

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Correct Option is : 3
3

Topic: Chemistry, Solutions, NEET
The Van’t Hoff factor for complete dissociation of NaCl is:

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
Correct Option is : 2
2

Topic: Chemistry, Solutions, NEET
The van't Hoff factor, $i$ for a compound which undergoes dissociation in one solvent and association in other solvent is respectively.

Options:
A less than one and less than one
B greater than one and less than one
C greater than one and greater than one
D less than one and greater than one

The correct answer is:
greater than one and less than one

Solution:
For dissociation, $i>1$.
For association, $i < 1$

Topic: Chemistry, Solutions, NEET, NEET 2011 (Screening)
The van't Hoff factor i for a dilute aqueous solution of the strong electrolyte barium hydroxide is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
3

Solution:
BaOH2⟶Ba2++2OH-
∴ van't Hoff factor =1+2=3.

Topic: Chemistry, Solutions, NEET, NEET 2016 (Phase 2)
The vapour density of a volatile chloride of a metal is 95 and the specific heat of the metal is 0.13 Cal/g. The exact atomic weight of metal is

Options:
A 50
B 49.23
C 19.692
D 98.12

The correct answer is:
50

Solution:
Molecular weight =2ÃV.D.=2Ã95=190
Approx atomic weight =6.4 specific heat=6.40.13=49.23
Suppose the formula of metal chloride =MClX
Molecular weight = 49.23+35.5X=190
So, X = 4(Nearest whole no.)
Formula is MCl 4
mol.âwt.= at. wt. of metal exact + 35.5Ã4
190â140=at.wt.
at.âwt.=â50

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
The vapour pressure of benzene at 90°C is 1020 torr. A solution of 15 g of a solute in 58.5 g benzene has a vapour pressure of 990 torr. The molecular weight of the solute is

Options:
A 78.2
B 204.2
C 148.2
D 226.6

The correct answer is:
226.6

Solution:
Using Raoult's law for relative lowering of vapour pressure as follows
Po-PsPo=w×Mm×W
Here Po=VP or pure solvent, Ps= VP of solution
or 1020-9901020=5×78m×58.5
So m=226.6.

Topic: Chemistry, Solutions, NEET
The vapour pressure of mercury is 0.002 mm Hg at 27°C
Hg(l)⇌Hg(g)
The value of equilibrium constant KC is

Options:
A 1.068×10-7 M
B 0.002 M
C 8.12×10-5 M
D 3.9×10-5 M

The correct answer is:
1.068×10-7 M

Solution:
Hg(l)⇌Hg(g)
Δng=1
Kp=0.002760=2.63×10-6 atm
Kp=Kc(RT)Δn
Kc=KpRT=2.63×10-60.0821×300
=1.068×10-7 M

Topic: Chemistry, Chemical Equilibrium, NEET
The vapour pressure of pure benzene at 25°C is 640 mm Hg and that of the solute A in benzene is 630 mm of Hg. The molality of solution is

Options:
A 0.2 m
B 0.4 m
C 0.5 m
D 0.1 m

The correct answer is:
0.2 m

Solution:
P =  P o X A
630=640XA
XA=630640=0.984
XB=0.016
m=XB×1000XAMA=0.016×10000.984×78
=0.2 m

Topic: Chemistry, Solutions, NEET
The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass 78 g mol-1). Vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance?

Options:
A 170 g mol-1
B 264 g mol-1
C 56 g mol-1
D 358 g mol-1

The correct answer is:
170 g mol-1

Solution:
Correct Option is : (A)
170 g mol-1

Topic: Chemistry, Solutions, NEET
The vapour pressure of pure liquid A decreases from 0.80 atm to 0.6 atm on mixing a non-volatile solute B to liquide A . The mole fraction of B in the solution is

Options:
A 0.25
B 1
C 0.50
D 0.75

The correct answer is:
0.25

Solution:
According to Raoult’s Law
P0-PsP0=xB (Mole fraction of solute)
xB=0.8-0.60.8=0.25.

Topic: Chemistry, Solutions, NEET
The vapour pressure of pure liquid a is 70 torr at 27° C. It forms an ideal solution with another liquid B. The mole fraction of B is 0.2 and total vapour pressure of the solution is 84 torr at 27° C. The vapour pressure of pure liquid B at 27° C, is

Options:
A 14
B 140
C 156
D 70

The correct answer is:
140

Solution:
Mole fraction of B(XB)=0.2
Mole fraction of A(XA)=0.8
Total vapour pressure P=XA.PAo+XBPBo
84=0.8×70+0.2PBo
P B o  = 140 torr

Topic: Chemistry, Solutions, NEET
The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K . Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg.

Options:
A Composition of A and B in liquid phase is 0.4 and 0.6 respectively.
B Composition of A and B in liquid phase is 0.5 and 0.5 respectively.
C Composition of A and B in liquid phase is 0.2 and 0.8 respectively.
D Composition of A and B in liquid phase is 0.1 and 0.9 respectively.

The correct answer is:
Composition of A and B in liquid phase is 0.4 and 0.6 respectively.

Solution:
Correct Option is : (A)
Composition of A and B in liquid phase is 0.4 and 0.6 respectively.

Topic: Chemistry, Solutions, NEET
The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K . Total vapour pressure is 600 mm Hg.What is the composition of the two liquids in vapour phase?

Options:
A Composition of A and B in vapour phase is 0.4 and 0.6 respectively.
B Composition of A and B in vapour phase is 0.2 and 0.8 respectively.
C Composition of A and B in vapour phase is 0.3 and 0.7 respectively.
D Composition of A and B in vapour phase is 0.5 and 0.5 respectively.

The correct answer is:
Composition of A and B in vapour phase is 0.3 and 0.7 respectively.

Solution:
Correct Option is : (C)
Composition of A and B in vapour phase is 0.3 and 0.7 respectively.

Topic: Chemistry, Solutions, NEET
The vapour pressure of two liquids $P$ and $\mathrm{Q}$ are 80 and 60 torr respectively. The total vapour pressure of solution obtained by mixing 3 mole of $\mathrm{P}$ and $2 \mathrm{~mol}$ of $\mathrm{Q}$ would be:

Options:
A 72 torr
B 140 torr
C 68 torr
D 20 torr

The correct answer is:
72 torr

Solution:
By Raoult's Law $\mathrm{P}_{\mathrm{T}}=\mathrm{P}_{\mathrm{P}}^{\circ} \mathrm{X}_{\mathrm{P}}+\mathrm{P}_{\mathrm{Q}}^{\circ} \mathrm{X}_{\mathrm{Q}}$
$$
\begin{aligned}
& \mathrm{P}_{\mathrm{P}}^{\circ}=80 \text { torr } \\
& \mathrm{X}_{\mathrm{P}}=\frac{3}{5} \\
& \mathrm{P}_{\mathrm{Q}}^{\circ}=60 \text { torr }
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{X}_{\mathrm{Q}} & =\frac{2}{5} \\
\mathrm{P}_{\mathrm{T}} & =80 \times \frac{3}{5}+60 \times \frac{2}{5} \\
& =48+24 \\
\mathrm{P}_{\mathrm{T}} & =72 \text { torr. }
\end{aligned}
$$
Related Theory
Solutes which dissociate or associate in a particular solution will not obey Raoult's law. Raoult's law applies to solutions containing non-volatile solute only. Raoult's law cannot be applied to concentrated solutions.

Topic: Chemistry, Solutions, NEET, NEET 2005
The vapour pressure of water at 20âC is 17.5 mmHg.

If 18 g of glucose (C6H12O6) is added to 178.2 g of water at 20âC, the vapour pressure of the resulting solution will be

Options:
A 17.675 mmHg
B 15.750 mmHg
C 16.500 mmHg
D 17.325 mmHg

The correct answer is:
17.325 mmHg

Solution:
Moles of glucose =18180=0.1
Moles of H2O=178.218 9.9
According to Raoult's law
PÂ°-PsPÂ°=Xsolute
17.5-PS17.5 =0.110
so, Ps=17.325 mm Hg

Topic: Chemistry, Solutions, NEET
The vapour pressure of water at 20Â°C is 17.5 mm of Hg. If 18 g of glucose C6H12O6 is added to 178.2 g of water at 20âC, what will be the vapour pressure of the resulting solution?

Options:
A 17.675 mm of Hg
B 15.750 mm of Hg
C 16.500 mm of Hg
D 17.325 mm of Hg

The correct answer is:
17.325 mm of Hg

Solution:
Moles of C6H12O6=18180=0.1
Moles of H2O=178.218 9.9
According to Raoult's law
P°-PsP°=Xsolute
17.5-Ps17.5 =0.110
so, Ps=17.325 mm of Hg

Topic: Chemistry, Solutions, NEET
The vapour pressure of water at 20℃ is 17.54 mm Hg. When 20 g of a non-ionic, substance is dissolved in 100 g of water, the vapour pressure is lowered by 0.30 mm Hg. What is the molecular mass of the substance?

Options:
A 200.8 g/mol
B 206.88 g/mol
C 210.5 g/mol
D 215.2 g/mol

The correct answer is:
206.88 g/mol

Solution:
pÂ°-pspÂ°=Xsolute
or
pÂ°-psPs=nN
â´0.317.24=20M10018 {M=molar mass of solute}
â´M=20Ã18Ã17.240.3Ã100=206.88

Topic: Chemistry, Solutions, NEET
The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Options:
A 12.08 kPa
B 63.21 kPa
C 32.46 kPa
D 87.12 kPa

The correct answer is:
12.08 kPa

Solution:
Correct Option is : (A)
12.08 kPa

Topic: Chemistry, Solutions, NEET
The vapour pressures of pure benzene and toluene are 160 and 60mm Hg respectively. The mole fraction of benzene is vapour phase in contact with equimolar solution of benzene and toluene is

Options:
A 0.073
B 0.027
C 0.27
D 0.73

The correct answer is:
0.73

Solution:
PT=160Ã12+60Ã12=110
PB=PBoÃB=160Ã12
PB=PTXBâ80=110 YB
YB=80110=0.73

Topic: Chemistry, Solutions, NEET
The vapour pressures of pure liquids A and B are 400 and 600 mmHg, respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fractional of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fraction of components A and B in vapour phase, respectively are

Options:
A 500 mm Hg, 0.5, 0.5
B 450 mm Hg, 0.4, 0.6
C 450 mm Hg, 0.5, 0.5
D 500 mm Hg, 0.4, 0.6

The correct answer is:
500 mm Hg, 0.4, 0.6

Solution:
Let XA and XB be the mole fraction of liquid A and B in the mixture.
Given, PA°=400 mmHg,PB°=600 mmHg
PTotal=XA.PA°+XB.PB°=0.5×400+0.5×600=500 mmHg
Now, mole fraction of A in vapour,
YA=PAPtotal=0.5×400500=0.4
Hence [PA=XAPA°]
and mole fraction of B in vapour, YB=1-0.4=0.6

Topic: Chemistry, Solutions, NEET
The variant of cry genes used to control corn borers is

Options:
A cryIAc
B cryIAb
C cryIIAb
D Both (A) and (C)

The correct answer is:
cryIAb

Solution:
The proteins encoded by the genes cryIAc and cryIIAb control the cotton bollworms, that of cryIAb controls corn borer.

Topic: Biology, Biotechnology and Its Applications, NEET
The variation in potency and concentration of reserpine of Rauwolfia is an example of

Options:
A Genetic diversity
B Species diversity
C Ecological diversity
D Population diversity

The correct answer is:
Genetic diversity

Solution:
Genetic diversity is the number and types of genes as well as chromosomes present in different species and the variations in the genes and their alleles in the same species. Hence, the variation in potency and concentration of reserpine of Rauwolfia is a result of genetic diversity. The production of reserpine is genetically regulated and with variation, in alleles, different concentrations of reserpine will be produced.

Topic: Biology, Biodiversity and its Conservation, NEET
The variation of electrostatic potential with radial distance r from the centre of a positively charged metallic thin shell of radius R is given by the graph

Options:
A
B
C
D

The correct answer is:

Solution:
The electric potential inside the surface and inside the surface of a sphere in terms of charge and the radius of the sphere, Vin=VS=KQR and Vout = KQ r(r>R), so the graphical representation for this,

Topic: Physics, Electrostatics, NEET, NEET 2020 (Phase 2)
The variation of EMF with time for four types of generators are shown in the figures. Which amongst them can be called AC?
(a)
(b)
(c)
(d)

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

The correct answer is:
(a), (b), (c) and (d)

Solution:
The emf generated due to a rotating conductor in a generator is given by
$$
E=-\frac{d \phi_{\mathrm{B}}}{\mathrm{dt}}
$$
where, $\phi_B=$ magnetic flux linked with conductor From the above equation we can conclude that the emf generated is a varying function of time with opposite polarity. So, all the graph are correct for the given variations.

Topic: Physics, Electromagnetic Induction, NEET, NEET 2019 (Odisha)
The variation of extent of adsorption with pressure at a given constant temperature is given in following figure

Which of the following relation between temperature of isotherms is correct?

Options:
A T1=T2=T3
B T1<T2<T3
C T3<T2<T1
D T1<T2>T3

The correct answer is:
T3<T2<T1

Solution:
At a particular pressure, the degree of adsorption decreases with increase in temperature so T3<T2<T1.

Topic: Chemistry, Surface Chemistry, NEET
The variation of induced emf (e) with time (t) in a coil if a short bar magnet is moved along its axis with a constant velocity, is best represented as

Options:
A
B
C
D

The correct answer is:

Solution:
Polarity of emf will be opposite in the two cases while entering and while leaving the coil. Only in this option polarity is changing. Hence, the correct option is

Topic: Physics, Electromagnetic Induction, NEET
The variation of photocurrent with collector potential for different frequencies of incident radiation ν1,ν2,ν3 is as shown in the graph, then

Options:
A ν1=ν2=ν3
B ν1>ν2>ν3
C ν1<ν2<ν3
D ν3=ν1+ν22

The correct answer is:
ν1<ν2<ν3

Solution:
The smaller the incident frequency, the smaller the stopping voltage required.
ν3>ν2>ν1

Topic: Physics, Dual Nature of Matter, NEET
The variation of susceptibility $(\chi)$ with absolute temperature $(T)$ for a paramagnetic material is represented as:

Options:
A

B

C

D

The correct answer is:

Solution:
Curie law $\chi_m \propto \frac{1}{T}$ (Paramagnetic)
Susceptibility $\chi$ varies inversely with absolute temperature.

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2023 (Manipur)
The variation of the boiling points of the hydrogen halides is in the order HF>HI>HBr>HCl . Which of the following option explains the higher boiling point of hydrogen fluoride?

Options:
A The electronegativity of fluorine is much higher than for other elements in the group.
B There is strong hydrogen bonding between HF molecules.
C The bond energy of HF molecules is greater than in other hydrogen halides.
D The effect of nuclear shielding is much reduced in fluorine which polarizes the HF molecule.

The correct answer is:
There is strong hydrogen bonding between HF molecules.

Solution:
Due to strong H-bonding in HF molecule, boiling point is highest for HF .
HF>HI>HBr>HCl

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET, NEET 2015 (Phase 2)
The variation of the potential energy of harmonic oscillator is as shown in the figure. The spring constant is

Options:
A 1×102 N m-1
B 150 N m-1
C 66.7 N m-1
D 3×102 N m-1

The correct answer is:
150 N m-1

Solution:
Total potential energy =0.04 J
Resting potential energy =0.01 J
Maximum kinetic energy =0.04-0.01
=0.03 J
â´ 0.03 J=12 m Ï2 A2=12 kA2
â´ 0.03 =12ÃkÃ2010002
â´ k=0.06Ã2500 N m-1=150 N m-1

Topic: Physics, Oscillations, NEET
The vas deferens receives duct from the seminal vesicle and opens into urethra as

Options:
A Epididymis
B Ejaculatory duct
C Efferent ductule
D Ureter

The correct answer is:
Ejaculatory duct

Solution:
The vas deferens receives a duct from the seminal vesicle and opens into the urethra as an ejaculatory duct.

Topic: Biology, Human Reproduction, NEET
The vector for T-DNA is

Options:
A is present in the organism which grows in extreme conditions
B is present in the organism which grows in human intestine
C is present in the organism which grows in soil and is a phytopathogen
D is present in the organism which grows in the roots of plants and induces root nodules

The correct answer is:
is present in the organism which grows in soil and is a phytopathogen

Solution:
The vector is a plasmid or virus DNA used to introduce genes into a host cell, where the genes may be amplified, gene cloning or otherwise manipulated. Agrobacterium tumefaciens is the soil bacterium and a phytopathogen that naturally infects plant at the site of the wound and causes crown gall disease with delivering T-DNA from bacterial cells into host plant cells.

Topic: Biology, Biotechnology and Its Applications, NEET
The vector must also have one unique recognition site to enable foreign DNA to be inserted into the vector during the generation of an rDNA molecule. Most of the commonly used vectors contain unique recognition sites for several restriction enzymes in a small region of DNA which is referred to as a polylinker or multiple cloning site (MCS). An MCS provides:

Options:
A Ability to separate DNA fragments
B Flexibility in the choice of restriction enzyme
C Flexibility in selectable marker
D Ability of DNA to mutate itself

The correct answer is:
Flexibility in the choice of restriction enzyme

Solution:
Multiple cloning site (MCS), also called a polylinker, is a short segment of DNA which contains many (up to ~20) restriction sites. Multiple cloning site allows considerable flexibility in the choice of restriction enzymes for cloning.

Topic: Biology, Biotechnology - Principles and Processes, NEET
The vector requires a ______, which helps in identifying selectively permitting the growth of transformants.

Options:
A selectable marker
B cloning sites
C antibiotic resistance gene
D bacteriophage

The correct answer is:
selectable marker

Solution:
Vectors in biotechnology process is a carrier for the genetic material to transfer to the desired organism's genome. Vectors are also used for making copies or clones. Vector should have the following features to facilitate cloning and transfer process.
Origin of replication to enable replication of the vector and vector supporting higher copy number is preferred. Vector having a selectable marker is preferred as it will help in identifying the transformant from non-transformants. Vector with few or single restriction enzyme sites preferred to avoid fragmentation of the DNA sequence and reduce complication of the biotechnological process.

Topic: Biology, Biotechnology - Principles and Processes, NEET