datavorous/entrance-exam-dataset huggingface dataset:
The dissolution of Al(OH)3 by a solution of NaOH results in the formation of:

Options:
A [Al(H2O)4(OH)2]+
B [Al(H2O)3(OH)3]
C [Al(H2O)2(OH)4]-
D [Al(H2O)6(OH)3]

The correct answer is:
[Al(H2O)2(OH)4]-

Solution:
Al(OH)3+NaOH(solution)→[Al(H2O)2(OH)4]Θ or NaAlO2 .

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
The dissolution of Al(OH)3 by a solution of NaOH results in the formation of

Options:
A [Al(H2O)4(OH)2]+
B [Al(H2O)3(OH)3]
C [Al(H2O)2(OH)4]-
D [Al(H2O)6(OH)3]

The correct answer is:
[Al(H2O)2(OH)4]-

Solution:
Al(OH)3 dissolves in NaOH solution to give Al(OH)4- ion which is supposed to have the octahedral complex species [Al(OH)4(H2O)2]- aqueous solution.
Al(OH)3+NaOH(aq)→[Al(OH)4(H2O)2]-(aq)+Na+(aq)

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
The dissolution of synaptonemal complex occurs during

Options:
A Diakinesis
B Leptotene
C Pachytene
D Diplotene

The correct answer is:
Diplotene

Solution:
The dissolution of synaptonemal complex occurs during diplotene stage.

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2023 (Manipur)
The distance between an object and the screen is 100 cm. A lens produces an image on the screen when placed at either of two positions 40 cm apart. The power of the lens is approximately

Options:
A 3 D
B 5 D
C 7Â D
D 9Â D

The correct answer is:
5 D

Solution:
Given, D=100 cm, x=40 cm
∴f= D 2 − x 2 4  D
If a be his maximum upward acceleration, then 70 kgg=60 kgg+60 kga
∴ a=70 kgg-60kgg60 kg
=g6=9.8 m s-26=1.63 m s-2
f=1002-4024×100
=100+40 100-40400
=140×60400=21 cm
We know that P=100fcm=10021=5 D

Topic: Physics, Ray Optics, NEET
The distance between the centres of carbon and oxygen atoms in the carbon monoxide molecule is 1.130 Ã. Locate the centre of mass of the molecule relative to the carbon atom.

Options:
A 5.428 Ã
B 1.130 Ã
C 0.6457 Ã
D 0.3260 Ã

The correct answer is:
0.6457 Ã

Solution:
X CM = m 1 x 1 + m 2 x 2 m 1 + m 2
â´ X CM = 1 2 Ã 0 + 1 6 Ã 1 . 1 3 1 2 + 1 6 = 0.6457 Ã

Topic: Physics, Center of Mass Momentum and Collision, NEET
The distance between the genes and percentage recombination shows:

Options:
A A direct relationship
B An inverse relationship
C A parallel relationship
D No relationship

The correct answer is:
A direct relationship

Solution:
If two genes are far apart on a chromosome, we would expect that the proportion of meiocytes in which a crossover occurs between the genes would be greater than in cases in which two linked loci are very close together. Thus, by determining the frequency of recombinants, we should be able to obtain a measure of the map distance between the genes.

Topic: Biology, Principles of Inheritance and Variation, NEET
The distance between the object and its real image from the convex lens is 60 cm and the height of image is two times the height of object. The focal length of the lens is

Options:
A 203Â cm
B 20Â cm
C 403 cm
D 40Â cm

The correct answer is:
403 cm

Solution:
D-U-U=-2
D-U=2U
D=3U
U=20
m=FF-20
-2=FF-20
-2F + 40=F
F=403 cm

Topic: Physics, Ray Optics, NEET
The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If $\mathrm{C}$ is its initial capacitance, its final capacitance is equal to:

Options:
A $2 \mathrm{C}$
B $C / 2$
C $4 \mathrm{C}$
D $\mathrm{C} / 4$

The correct answer is:
$\mathrm{C} / 4$

Solution:
The capacitance of parallel plate capacitor,
$\mathrm{C}=\frac{\varepsilon_0 \mathrm{~A}}{d}$
(where, the symbols have their usual meanings.)
Given, $\quad d^{\prime}=2 \mathrm{~d} \& \mathrm{~A}^{\prime}=\frac{\mathrm{A}}{2}$
Then, $\mathrm{C}^{\prime}=\frac{\varepsilon_0\left(\frac{\mathrm{A}}{2}\right)}{2 d}=\frac{\varepsilon_0 \mathrm{~A}}{4 d}=\frac{\mathrm{C}}{4}$

Topic: Physics, Capacitance, NEET, NEET 2022 (Phase 2)
The distance between two coherent sources is 1 mm. The screen is placed at a distance of 1 m from the sources. If the distance of the third bright fringe is 1.2 mm from the central fringe, the wavelength of light used is

Options:
A 4000 â«
B 5000 â«
C 6000 â«
D 7200 â«

The correct answer is:
4000 â«

Solution:
The position of maxima
yn= nDÎ»d
â Î»= yndnD
Here, n=3, yn=1.2 mm=1.2Ã10-3 m,
D=1 m, d=1 mm=1Ã10-3 m
â´ Î»= 1.2 Ã 10-3 Ã 1 Ã 10-33 Ã 1
=0.4Ã10-6 m=4Ã 10-7 m
=4000 â«

Topic: Physics, Wave Optics, NEET
The distance covered by a body of mass $5 \mathrm{~g}$ having linear momentum $0.3 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$ in $5 \mathrm{~s}$ is:

Options:
A $300 \mathrm{~m}$
B $30 \mathrm{~m}$
C $3 \mathrm{~m}$
D $0.3 \mathrm{~m}$

The correct answer is:
$300 \mathrm{~m}$

Solution:
Given,
$\begin{aligned}
m=5 \mathrm{~g} & =0.005 \mathrm{~kg}, \\
\mathrm{P}=m v & =0.3 \mathrm{~kg} \mathrm{~m} / \mathrm{s}, \text { and } \\
t & =5 \mathrm{~s},
\end{aligned}$
(where, symbols have their usual meanings.)
$\mathrm{P}=m v=0.3$
Hence, $\mathrm{V}=\frac{0.3}{m}=\frac{0.3}{0.005}=60 \mathrm{~m} / \mathrm{s}$
Hence, the distance covered $=$ speed $\times$ time
$=60 \times 5=300 \mathrm{~m}$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2022 (Phase 2)
The distance covered by a complete turn of the DNA double helix is

Options:
A 3.4Â Ã
B 34 Ã
C 3.6Â Ã
D 36Â Ã

The correct answer is:
34 Ã

Solution:
The DNA double-helix can form somewhat different three-dimensional conformations depending on environmental conditions and, to some extent, on the base sequence. Inside the cell where the DNA is fully hydrated, a highly regular structure known as the B-form is adopted. The diameter of the B-DNA is 20 angstroms, and the distance between base pairs is 3.4 angstroms. Since each helix has 10 bp the length is approximately 34 angstroms.

Topic: Biology, Molecular Basis of Inheritance, NEET
The distance covered by a particle undergoing SHM in one time period is ( amplitude $=\mathrm{A}$ )

Options:
A zero
B $\mathrm{A}$
C $2 \mathrm{~A}$
D $4 \mathrm{~A}$

The correct answer is:
$4 \mathrm{~A}$

Solution:
In a simple harmonic motion (SHM) the particle oscillates about its mean position on a straight line.
The particle moves from its mean position $(O)$ to an extreme position $(\mathrm{P})$ and then return to its mean position covering same distance of A.
Then by the conservative force, it is moved in opposite direction to a point $Q$ by distance A and then back to mean position covering a distance of A. This comprises of one time period as shown below
Hence, in one time period it covers a distance of
$$
\mathrm{x}=\mathrm{OP}+\mathrm{PO}+\mathrm{OQ}+\mathrm{QO}=\mathrm{A}+\mathrm{A}+\mathrm{A}+\mathrm{A}=4 \mathrm{~A}
$$

Topic: Physics, Oscillations, NEET, NEET 2019 (Odisha)
The distance of a geostationary satellite from the centre of earth (radius R = 6400 km) is nearest to

Options:
A 18 R
B 10 R
C 7 R
D 5 R

The correct answer is:
7 R

Solution:
Distance of geostationary satellite from surface of the earth is approx 36000 km
so from center of earth
distance = 36000+64006400≈7R

Topic: Physics, Gravitation, NEET
The distance of centre of mass of a uniform semicircular disc of radius R , from the geometric center on the axis of symmetry is

Options:
A 4R3π
B 2R3π
C 4R9π
D 2Rπ

The correct answer is:
4R3π

Solution:
Centre of mass of a uniform semicircular disc of radius R lies at a distance of hâ¡ = 4 R 3 Ï from the centre on the axis of symmetry as shown in figure.

Topic: Physics, Center of Mass Momentum and Collision, NEET
The distance of closest approach of an Î± particle fired towards a nucleus with momentum p is r. If the momentum of the Î± particle is 2p, the corresponding distance of closest approach is

Options:
A 4r

B 2r

C r2
D r4

The correct answer is:
r4

Solution:
At the distance of closest approach, r
K=14πε0.2eZer
r=14πε0.2Ze2K
∵K=P22m
∴r=44πε0.2Ze2p22m
=14πε0.4mZe2p2
Thus, r∝1p2
∴r1r2=p22p12=(2p)2p2
r1r2=41⇒∴r2=r14=r4

Topic: Physics, Atomic Physics, NEET
The distance of closest approach of an α -particle fired towards a nucleus with momentum p, is r. What will the distance of closest approach when the momentum of α -particle is 2p?

Options:
A 2r
B 4r
C r2
D r4

The correct answer is:
r4

Solution:
KE=PE
Or p22m=14ÏÎµ0.q1q2râr â1p2
Hence, at given condition distance of closest approach is r4 .

Topic: Physics, Atomic Physics, NEET
The distance of Neptune and Saturn from the Sun are nearly 1013 m and 1012 m respectively. Ratio of their time period will be

Options:
A 10
B 100
C 1010
D 1000

The correct answer is:
1010

Solution:
Using Kepler's Third law of gravitation, we get
Tn2Ts2=Rn3Rs3
Tn Ts =10 1310 123/2
=103=1010

Topic: Physics, Gravitation, NEET
The distance of the centre of mass of a hemispherical shell of radius R from its centre is

Options:
A R 2
B R 3
C 2R 2
D 2R 3

The correct answer is:
R 2

Solution:
Centre of mass of a hemispherical shell of radius R lies at a distance of h = R 2 from its centre on the axis of symmetry as shown in figure.

Topic: Physics, Center of Mass Momentum and Collision, NEET
The distance traveled by a particle is proportional to the square of time, then the particle travels with

Options:
A uniform acceleration
B uniform velocity
C increasing acceleration
D decreasing velocity

The correct answer is:
uniform acceleration

Solution:
The distance travelled by a particle is proportional to the squares of time,
∴ x∝t2
∴ v∝t
∴ a∝t0

Topic: Physics, Motion In One Dimension, NEET
The distance travelled by a particle starting from rest and moving with an acceleration $\frac{4}{3} \mathrm{~ms}^{-2}$, in the third second is

Options:
A $6 \mathrm{~m}$
B $4 \mathrm{~m}$
C $\frac{10}{3} \mathrm{~m}$
D $\frac{19}{3} \mathrm{~m}$

The correct answer is:
$\frac{10}{3} \mathrm{~m}$

Solution:
Distance travelled by the particle in $n$th second is
$S_{n t h}=u+\frac{1}{2} a(2 n-1)$
where $u$ is initial speed and $a$ is acceleration of the particle.
Here, $n=3, u=0, a=\frac{4}{3} \mathrm{~m} / \mathrm{s}^2$
$\begin{aligned}
\therefore S_{3 \text { rd }} & =0+\frac{1}{2} \times \frac{4}{3} \times(2 \times 3-1) \\
& =\frac{4}{6} \times 5 \\
& =\frac{10}{3} \mathrm{~m}
\end{aligned}$
Alternative : Distance travelled in the 3rd second $=$ distance travelled in $3 \mathrm{~s}$ - distance travelled in $2 \mathrm{~s}$
As, $u=0$,
$S_{(3 \mathrm{rd} s)}=\frac{1}{2} a \cdot 3^2-\frac{1}{2} a \cdot 2^2=\frac{1}{2} \cdot a \cdot 5$
Given $a=\frac{4}{3} \mathrm{~ms}^{-2}$
$\therefore S_{(3 \mathrm{rd} s)}=\frac{1}{2} \times \frac{4}{3} \times 5=\frac{10}{3} \mathrm{~m}$

Topic: Physics, Motion In One Dimension, NEET, NEET 2008 (Screening)
The distillation technique most suited for separating glycerol from spent-lye in the soap industry is

Options:
A Fractional distillation
B Simple distillation
C Distillation under reduced pressure
D Steam distillation

The correct answer is:
Distillation under reduced pressure

Solution:
Glycerol (B.P. = 290°C) is separated from spent lye in the soap industry by distillation under reduced pressure, as for simple distillation very high temperature is required which might decompose the component.

Topic: Chemistry, General Organic Chemistry, NEET
The division of centromere is absent in

Options:
A Anaphase
B Anaphase-I
C Anaphase-II
D All of these

The correct answer is:
Anaphase-I

Solution:
During anaphase I, chromosomal pairs separate and chromosomes start moving towards opposite poles. This separation of chromosomal pair is called disjunction, and it results in reduction in the number of chromosomes to half.

Topic: Biology, Cell Cycle and Cell Division, NEET
The DNA finger printing relies on

Options:
A Difference in patterns of genes between individuals
B Difference in order of genes between individuals
C Difference in junk DNA patterns between individuals
D All of these

The correct answer is:
Difference in junk DNA patterns between individuals

Solution:
DNA of human is almost the same for all individuals but very small amount that differs from person to person and that is used by the forensic scientists analyze to identify people. These differences are called polymorphisms (many forms) and are the key of DNA typing. It consist of variation in the length of DNA at specific loci called as restricted fragment. It is most important segment for DNA test made up of short repetitive nucleotide sequences, these are called VNTRs (variable number of tandem repeat). These sequences do not code for any protein and they are junk DNA.

Topic: Biology, Biotechnology - Principles and Processes, NEET
The DNA fragments separated on an agarose gel can be visualised after staining with:

Options:
A Bromophenol blue
B Acetocarmine
C Aniline blue
D Ethidium bromide

The correct answer is:
Ethidium bromide

Solution:
The DNA fragments separated on an agarose gel can be visualised after staining with ethidium bromide. It is an intercalating agent and a fluorescent agent. The stained DNA fragments are seen as bright orange-coloured bands under UV light. Intercalation is the insertion of molecules between the planar bases of DNA. This process is used as a method for analysing DNA.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2017
The DNA present in chloroplast is:

Options:
A Circular, double stranded
B Linear, single stranded
C Circular, single stranded
D Linear, double stranded

The correct answer is:
Circular, double stranded

Solution:
The DNA present in chloroplast is circular double stranded.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2024
The dominant mechanisms for motion of charge carriers in forward and reverse biased silicon p-n junctions are

Options:
A Drift in forward bias, diffusion in reverse bias
B Diffusion in forward bias, drift in reverse bias
C Diffusion in both forward and reverse bias
D Drift in both forward and reverse bias

The correct answer is:
Diffusion in forward bias, drift in reverse bias

Solution:
For forward biased, majority charge carrier constitute current by diffusing to other side of junction. For reversed biased, minority charge carrier constitute small amount of leakage current by drifting due toexternally applied electric field.

Topic: Physics, Semiconductors, NEET
The donor sites of (EDTA)4- are

Options:
A O-atoms only
B N-atoms only
C Two N-atoms and four O-atoms
D Three N-atoms and three O-atoms

The correct answer is:
Two N-atoms and four O-atoms

Solution:
(EDTA)4- is a hexadentate ligand and it has four O-atoms and two N-atoms as donor sites.

Topic: Chemistry, Coordination Compounds, NEET
The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20 °C and one bar will be released when 0.15g of aluminium reacts?

Options:
A 0.318 L
B 0.458 L
C 0.203 L
D 0.157 L

The correct answer is:
0.203 L

Solution:
Correct Option is : (C)
0.203 L

Topic: Chemistry, States of Matter, NEET
The drift velocity of the electrons in a copper wire of length 2 m under the application of a potential difference of 220 V is 0.5 m s-1. Their mobility (in m2 V-1 s-1)

Options:
A 2.5×10-3
B 2.5×10-2
C 5×102
D 4.5×10-3

The correct answer is:
4.5×10-3

Solution:
Drift velocity=mobility× intensity of the electric field
or vd=μE or μ=vdE
or μ=vdV/l ∵E=Vl
∴μ=0.5×2220=4.5×10−3 m2 V−1 s−1

Topic: Physics, Current Electricity, NEET
The driver of a car travelling with speed $30 \mathrm{~ms}^{-1}$ towards a hill sounds a horn of frequency $600 \mathrm{~Hz}$. If the velocity of sound in air is $330 \mathrm{~ms}^{-1}$, the frequency of reflected sound as heard by driver is

Options:
A $550 \mathrm{~Hz}$
B $555.5 \mathrm{~Hz}$
C $720 \mathrm{~Hz}$
D $500 \mathrm{~Hz}$

The correct answer is:
$720 \mathrm{~Hz}$

Solution:
Key Idea Apply Doppler's effect. Whenever there is a relative motion between a source of sound and the observer (listner), the frequency of sound heard by the observer is different from the actual frequency of sound emitted by source.
[for case I] $\quad \mathbf{n}^{\prime}=\frac{\mathrm{v}}{\mathrm{v}-30} \mathbf{n}$
[for case II] $n^{\prime \prime}=\frac{v+30}{v} n^{\prime}$
From Eqs. (i) and (ii), we get
$$
\begin{aligned}
\mathrm{n}^{\prime \prime}=\frac{\mathrm{v}+30}{\mathrm{v}-30} \mathrm{n} & =\frac{360}{300} \times 600 \\
& =720 \mathrm{~Hz}
\end{aligned}
$$

Topic: Physics, Waves and Sound, NEET, NEET 2009 (Screening)
The driver of a train moving at 72 km h-1 sights another train moving at 4 m s-1 on the same track and in the same direction. He instantly applies brakes to produce a retardation of 1 m s-2. The minimum distance between the trains so that no collision occurs is

Options:
A 32 m
B 64 m
C 128 m
D 256 m

The correct answer is:
128 m

Solution:
Relative initial velocity =72Ã518-4=16 m s-1
Now, 02-16Ã16=2-1S
Or S=128 m

Topic: Physics, Motion In One Dimension, NEET
The drug used as post operative analgesic in medicine is

Options:
A L-Dopa
B Amoxycilin
C Sulphapyridine
D Morphine

The correct answer is:
Morphine

Solution:
Analgesic are medicines used for getting relief from pain. These are of two types:
(i) Narcotics are drug which produce sleep and unconsciousness, e.g. opium alkaloids like morphine, codine, heroine.
(ii) Non-narcotics analgesics are butazolidine (phenylbutazone), brufea, etc.

Topic: Chemistry, Chemistry in Everyday Life, NEET
The dwarfness in plants of F2 generation is

Options:
A Recessive
B Dominant
C Co-dominant
D In-complete dominance

The correct answer is:
Recessive

Solution:
The factor of an allelic pair which is unable to express its effect in the presence of its contrasting factor in a heterozygote is called a recessive factor.

Topic: Biology, Principles of Inheritance and Variation, NEET
The Ea of reaction in the presence of catalyst is 4.15 KJ/mol and in the absence of catalyst is 8.3 KJ mol-1 . What is the slope of the plot of lnk vs 1T in the absence of catalyst.

Options:
A +1
B -1
C +1000
D -1000

The correct answer is:
-1000

Solution:
ln K =lnA-EaR×1T
slope =-EaR=-8.3×1038.3=-1000

Topic: Chemistry, Chemical Kinetics, NEET
The Ea of reaction in the presence of catalyst is 5.25 kJ/mol and in the absence of catalyst is 8.314 kJ mol-1 . What is the slope of the plot of lnk vs 1T in the absence of catalyst. (R = 8.314 Jâ k â1 mol â1 )

Options:
A 100
B -100
C -1000
D +1000

The correct answer is:
-1000

Solution:
ln k =lnA-EaR×1T
slope =-EaR=-8.3×1038.3=-1000

Topic: Chemistry, Chemical Kinetics, NEET
The EM3+/M2+o values for Cr, Mn, Fe and Co are -0.41, +1.57, +0.77 and +1.97 V respectively. For which one of following metal the change in oxidation state from +2 to +3 easiest?

Options:
A Co
B Mn
C Fe
D Cr

The correct answer is:
Cr

Solution:
Here ERPo of Cr = - 0.41 so its EOPo is 0.41.
As here EOPo for Cr2+/Cr3+ is maximum +0.41 V so Cr2+ will be easily oxidised to Cr3+ .
For rest of the metals EOPo will be negative so they have less tendency to be oxidized then Cr here.

Topic: Chemistry, Electrochemistry, NEET
The EM3+/M2+ values for Cr, Mn, Fe and Co are –0.41, +1.57, +0.77 and +1.97 V respectively. For which one of these metals the change in oxidation state form +2 to +3 is easiest?

Options:
A Cr
B Mn
C Fe
D Co

The correct answer is:
Cr

Solution:
ECr3+/Cr2+o=-0.41 V
EMn3+/Mn2+o=+1.57 V
EFe3+/Fe2+o=+0.77 V
ECo3+/Co2+o=+1.97 V
Reducing nature or tendency to get oxidized ∝-Eredo
As more negative value of Eredo indicates better reducing agent thus easily oxidized so oxidation of Cr2+ to Cr3+ is the easiest.
Reducing nature for these metals is as follows Cr > Fe > Mn > Co.

Topic: Chemistry, Electrochemistry, NEET
The $\mathrm{E}^{\circ}$ value for the $\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}$ couple is more positive than that of $\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}$ or $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ due to change of

Options:
A $\mathrm{d}^5$ to $\mathrm{d}^2$ configuration
B $\mathrm{d}^4$ to $\mathrm{d}^5$ configuration
C $d^3$ to $d^5$ configuration
D $d^5$ to $d^4$ configuration

The correct answer is:
$\mathrm{d}^4$ to $\mathrm{d}^5$ configuration

Solution:
$\mathrm{E}_{\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}}^{\circ}\gt\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\circ} \text { or } \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}$
Electronic configuration of $\mathrm{Mn}^{3+}=[\mathrm{Ar}] 3 d^4$
Electronic configuration of $\mathrm{Mn}^{2+}=[\mathrm{Ar}] 3 d^5$
Electronic configuration of $\mathrm{Cr}^{3+}=[\mathrm{Ar}] 3 d^3$
Electronic configuration of $\mathrm{Cr}^{2+}=[\mathrm{Ar}] 3 d^4$
As $\mathrm{Mn}^{3+}$ from $d^4$ configuration goes to more stable $d^5$ configuration (Half filled), due to more exchange energy in $d^5$ configuration.

Topic: Chemistry, d and f Block Elements, NEET, NEET 2024
The $E^{\ominus}$ values for
$\begin{aligned} & \mathrm{Al}^{+} / \mathrm{Al}=+0.55 \mathrm{~V} \text { and } \mathrm{Tl}^{+} / \mathrm{TI}=-0.34 \mathrm{~V} \\ & \mathrm{Al}^{3+} / \mathrm{Al}=-1.66 \mathrm{~V} \text { and } \mathrm{Tl}^{3+} / \mathrm{TI}=+1.26 \mathrm{~V}\end{aligned}$
Identify the incorrect statement.

Options:
A $\mathrm{Al}^{+}$is unstable in solution.
B $\mathrm{TI}$ can be easily oxidised to $\mathrm{Tl}^{+}$than $\mathrm{Tl}^{3+}$.
C $\mathrm{Al}$ is more electropositive than $\mathrm{TI}$.
D $\mathrm{Tl}^{3+}$ is a good reducing agent than $\mathrm{Tl}^{1+}$.

The correct answer is:
$\mathrm{Tl}^{3+}$ is a good reducing agent than $\mathrm{Tl}^{1+}$.

Solution:
(1) $\mathrm{Al}^{+}$is unstable in solution because $\mathrm{SRP}$ of $\mathrm{Al}^{+} / \mathrm{Al}$ is positive hence, it will get reduced in solution.
(2)
$\begin{array}{ll}\mathrm{Tl} / \mathrm{Tl}^{+}=+0.34 & (\mathrm{SOP}) \\ \mathrm{Tl} / \mathrm{Tl}^{3+}=-1.26 & (\mathrm{SOP})\end{array}$
Since, SOP TI/TI $>$ TI/T1 ${ }^{+}$
Hence, $\mathrm{TI}$ will get oxidised to $\mathrm{Tl}^{+}$easily than $\mathrm{Tl}^{3+}$
(3) Since $\mathrm{Al}^2 / \mathrm{Al}^{3+}$ has more SOP than $\mathrm{Tl} / \mathrm{Tl}^{+}$it will be more electropositive
(4)
Since $\begin{aligned} \mathrm{Tl}^{3+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Tl}^{+} ; \mathrm{E}=\frac{3 \times 1.26+0.34}{2} \\ =+2.04\end{aligned}$
Hence, $\mathrm{Tl}^{3+}$ is better oxidising agent not reducing agent

Topic: Chemistry, Electrochemistry, NEET, NEET 2023 (Manipur)
The EAN of Fe in K4FeCN6 is

Options:
A 33
B 35
C 36
D 26

The correct answer is:
36

Solution:
EAN is known as effective atomic number. It is given by the relation.
EAN=Atomic number-Oxidation state+2Ãnumber of Ligands
=26-2+2(6)=36.

Topic: Chemistry, Coordination Compounds, NEET
The EAN value of iron in potassium ferricyanide is

Options:
A 18
B 54
C 35
D 23

The correct answer is:
35

Solution:
K 3 Fe(CN) 6
Oxidation No. of Fe = +3
EAN = 26−(+3)
+ 2×6
=35
EAN of a central metal ion = (atomic no. of central atom) – oxidation state + no. of ligands × 2 × denticity.

Topic: Chemistry, Coordination Compounds, NEET
The early stage human embryo distinctly possesses

Options:
A Gills
B Gill slits
C External ear (Pinna)
D Eye brows

The correct answer is:
Gill slits

Solution:
Possession of pharyngeal gill slits and gill pouches is one of the three diagnostic characters of chordates. So, these structures do appear in the embryonic stages of all vertebrates.

Topic: Biology, Evolution, NEET
The Earth is assumed to be a sphere of radius $R$. A platform is arranged at a height $R$ from the surface of the Earth. The escape velocity of a body from this platform is $f v$, where $v$ is its escape velocity from the surface of the Earth. The value of $f$ is:

Options:
A $\frac{1}{2}$
B $\sqrt{2}$
C $\frac{1}{\sqrt{2}}$
D $\frac{1}{3}$

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

Solution:
We know that Escape velocity
$V_e=\sqrt{2 g R}$
Escape velocity of the body from platform
$\begin{aligned}
& \text {P. } \mathrm{E}+\mathrm{K} \cdot \mathrm{E}=0 \\
& -\frac{G M m}{2 \mathrm{R}}+\frac{1}{2} m V^2=0 \\
& \Rightarrow \quad V=\sqrt{\frac{G M}{R^2} \cdot R}=\sqrt{g R}
\end{aligned}$
This confirms that:
$\begin{aligned}
& fv =\frac{V}{\sqrt{2}} \\
\therefore f & =\frac{1}{\sqrt{2}} .
\end{aligned}$

Topic: Physics, Gravitation, NEET, NEET 2006