datavorous/entrance-exam-dataset huggingface dataset:
The angular acceleration of a body, moving along the circumference of a circle, is:

Options:
A along the axis of rotation
B along the radius, away from centre
C along the radius towards the centre
D along the tangent to its position

The correct answer is:
along the axis of rotation

Solution:
In terms of linear acceleration it is given as, a→=α→×r→. Clearly direction of linear acceleration is perpendicular to both angular acceleration and radius vector.

Angular acceleration of a body, moving along the circumference of a circle is along the axis of rotation.

Topic: Physics, Laws of Motion, NEET, NEET 2023 (All India)
The angular momentum of an electron in a given orbit is J. Its kinetic energy will be

Options:
A 12J2mr2
B Jvr
C J22m
D J22n

The correct answer is:
12J2mr2

Solution:
mvr=J
mv=Jr
(mv)2=J2r2
mv2=J2mr2
K.E.=12mv2=12J2mr2

Topic: Chemistry, Structure of Atom, NEET
The angular momentum of an electron in a hydrogen atom is proportional to

Options:
A 1r
B 1r
C r
D r2

The correct answer is:
r

Solution:
L=mvr=nh2π
But r=0.529n2Z
L∝n∝r

Topic: Physics, Atomic Physics, NEET
The angular momentum of an electron in 'd' orbital is equal to
ℏ=h2π

Options:
A 2 ℏ
B 2 3 ℏ
C 0
D 6 ℏ

The correct answer is:
6 ℏ

Solution:
Orbital angular momentum = l l+1 â
for d-orbital l=2
So orbital angular momentum = 2 2+1 ℏ= 6 ℏ

Topic: Chemistry, Structure of Atom, NEET
The angular momentum of an electron in hydrogen atom is hπ. The kinetic energy of the electron is

Options:
A 13.6 eV
B 3.4 eV
C 1.51Â eVÂ
D 0.85Â eV

The correct answer is:
3.4 eV

Solution:
The angular momentum of an electron in hydrogen atom =hÏ .
As we know, from Bohr's second postulate,
mvr=hÏâv=hÏmr ...(i)
Now, kinetic energy of electron
=12mv2=12mÃhÏmr2 [from Equation (i)]
=h22Ï2mr2
=6.63Ã10-3422Ã3.142Ã9.1Ã10-31Ã5.3Ã10-112Ã1.6Ã10-19
=3.4 eV

Topic: Physics, Atomic Physics, NEET
The angular momentum of an electron moving in an orbit of hydrogen atom is $1.5\left(\frac{\mathrm{h}}{\pi}\right)$. The energy in the same orbit is nearly

Options:
A $-1.3 \mathrm{eV}$
B $-1.4 \mathrm{eV}$
C $-1.5 \mathrm{eV}$
D $-1.6 \mathrm{eV}$

The correct answer is:
$-1.5 \mathrm{eV}$

Solution:
Angular momentum of electron moving in $n^{\text {th }}$ orbit of hydrogen atom
Hence, $n=3$
Now, energy of electron in $3^{\text {rd }}$ orbit
$\begin{aligned} E & =\frac{-13.6 \times z^2}{n^2} \\ & =\frac{-13.6 \times 1^2}{3^2} \\ & =-1.51 \mathrm{eV} \\ & \approx-1.5 \mathrm{eV}\end{aligned}$

Topic: Physics, Atomic Physics, NEET, NEET 2023 (Manipur)
The angular momentum of electron in d-orbital is equal to

Options:
A 6 ℏ
B 2Â â
C 23Â â
D 0Â â

The correct answer is:
6 ℏ

Solution:
For d-electrons, l=2.
Angular orbital momentum=l(l+1)h2π
=6h2π=6 ℏ
∴h2π=ℏ

Topic: Chemistry, Structure of Atom, NEET, NEET 2015 (Phase 1)
The angular position of a line of a disc of radius r=6 cm is given by Î¸=10-5t+4t2 rad, the average angular velocity between 1 s and 3 s is

Options:
A π rad s-1
B 11 rad s-1
C 22 rad s-1
D 5.5 rad s-1

The correct answer is:
11 rad s-1

Solution:
ωav=ΔθΔt
ωav=10-53+432-10-51+4113-1
ωav=11 rad s-1

Topic: Physics, Laws of Motion, NEET
The angular resolution of a $10 \mathrm{~cm}$ diameter telescope at a wavelength of $5000 Å$ is of the order of:

Options:
A $10^6 \mathrm{rad}$
B $10^{-2} \mathrm{rad}$
C $10^{-4} \mathrm{rad}$
D $10^{-6} \mathrm{rad}$

The correct answer is:
$10^{-6} \mathrm{rad}$

Solution:
Here $R \cdot P .=\frac{1}{\Delta \theta}$
$$
\begin{aligned}
& \therefore \text { Angular resolution } \Delta \theta=\frac{1.22 \lambda}{\mathrm{D}} \\
&=\frac{1.22 \times 5000 \times 10^{-10}}{0.1} \\
&=6.10 \times 10^{-6} \\
& \cong 10^{-6}
\end{aligned}
$$

Topic: Physics, Ray Optics, NEET, NEET 2005
The angular shape of ozone molecule $\left(\mathrm{O}_3\right)$ consists of

Options:
A 1 sigma and 2 pi bonds
B 2 sigma and 2 pi bonds
C 1 sigma and 1 pi bonds
D 2 sigma and 1 pi bonds

The correct answer is:
2 sigma and 1 pi bonds

Solution:
Key Idea :
Single bond contains $=1 \sigma$ bond
Double bond contains $=1 \sigma$ bond $+1 \pi$ bond
Triple bond contains $=1 \sigma$ bond $+2 \pi$ bond
The molecule of $\mathrm{O}_3$ is bent with an angle of $116.8^{\circ}$ and equal $\mathrm{O}-\mathrm{O}$ distance of $128 \mathrm{pm}$. The ozone molecule consists of 2 sigma and 1 pi bonds. The angular shape of ozone molecule is represented as :

Topic: Chemistry, Electrochemistry, NEET, NEET 2008 (Screening)
The angular speed of a fly wheel moving with uniform angular acceleration changes from 1200 rpm to 3120 rpm in 16 seconds. The angular acceleration in rad s-2 is:

Options:
A 4π
B 12π
C 104π
D 2π

The correct answer is:
4π

Solution:
For uniform angular acceleration we can write,

ω=ω0+αt⇒α=ω-ω0t

Given, ω=3120 rpm=3120×2π60=104π rad s-1

and ω0=1200 rpm=1200×2π60=40π rad s-1

Therefore, α=104π rad s-1-40π rad s-116=4π rad s-2

Topic: Physics, Laws of Motion, NEET, NEET 2022 (Phase 1)
The angular speed of seconds hand of the clock is

Options:
A Ï1800 rad s-1
B Ï30 rad s-1
C Ï90 radÂ s-1
D Ï60 radÂ s-1

The correct answer is:
Ï30 rad s-1

Solution:
Seconds hand of the clock complete one revolution in 60 seconds, i.e. when t=60 s
θ=2π radian
∴ Angular speed
ω=Angular displacement θTime t
=2π60
=π30 rad s-1

Topic: Physics, Laws of Motion, NEET
The angular speed of the wheel of a vehicle is increased from, 360 rpm to 1200 rpm in 14 s. Its angular acceleration is,

Options:
A 2π rad s-2
B 28π rad s-2
C 120π rad s-2
D 1 rad s-2

The correct answer is:
2π rad s-2

Solution:
Given, t=14 s, ω0=36060 rps=12π rad s-1, and ω=120060 rps=40πrad s-1. Now use the first equation of motion in rotation motion, in terms of angular acceleration and angular velocities,

ω=ω0+αt⇒α=ω-ω0t=28π14=2π rad s-2.

Topic: Physics, Rotational Motion, NEET, NEET 2020 (Phase 2)
The angular velocity of a body is ω→=2i^+3j^+4k^ and a torque τ→=i^+2j^+3k^ acts on it. The rotational power will be

Options:
A 20 W
B 15 W
C 17 W
D 14 W

The correct answer is:
20 W

Solution:
We know that
P=Ïâ.Ïâ=i^+2j^+3k^.2i^+3j^+4k^
P=2+6+12=20 W

Topic: Physics, Rotational Motion, NEET
The angular velocity of second's hand of a watch will be

Options:
A Ï60 rad s-1
B π30 rad s-1
C 60π rad s-1
D 30π rad s-1

The correct answer is:
π30 rad s-1

Solution:
ω=2πrad60 s=π30rads-1

Topic: Physics, Laws of Motion, NEET
The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on 60Â° latitude becomes zero is ( Radius of the earth= 6400 km. At the poles g=10 m s-2 )

Options:
A 2.5×10-3 rad s-1
B 5.0×10-1 rad s-1
C 100 rad s-1
D 7.8×10-2 rad s-1

The correct answer is:
2.5×10-3 rad s-1

Solution:
g′=g-ω2Rcos2⁡λ 0=g-ω2Rcos2⁡60°
0=g-ω2R4 ⇒ ω=2gR=1400 rad s-1=2.5×10-3 rad s-1

Topic: Physics, Gravitation, NEET
The angular width of the central maximum in a single slit diffraction pattern is 60Â° . The width of the slit is 1 Î¼m. The slit is illuminated by monochromatic plane waves. If another slit of the same width is made near it, Young's fringes can be observed on a screen placed at a distance 50âcm from the slits. If the observed fringe width is 1âcm, what is slit separation distance? (i.e. the distance between the centres of each slit.)

Options:
A 100 Î¼m
B 25 Î¼m
C 50 Î¼m
D 75 μm

The correct answer is:
25 Î¼m

Solution:
In diffraction
dsin30°=λ
λ=d2
Young’s fringe width [ d'-the separation between two slits ]
β=λ×Dd'
10-2=d2×50×10-2d'
10-2=10-6×50×10-22×d'
d'=25 μm

Topic: Physics, Wave Optics, NEET
The angular width of the central maximum in the Fraunhofer's diffraction pattern is measured. The slit is illuminated by the light of wavelength 6000 Aâ . If the slit is illuminated by the light of another wavelength, angular width decreases by 30% . The wavelength of light used is

Options:
A 3500 Å
B 4200 Å
C 4700 Å
D 6000 Å

The correct answer is:
4200 Å

Solution:
For first diffraction minimum, dsinâ¡Î¸=Î»
And if angle is small, sinâ¡Î¸=Î¸
dÎ¸=Î»
i.e., Half angular width, Î¸=Î»d
Full angular width w=2Î¸=2Î»d
Also w′=2Î»′d
â´Î»′Î»=w′w or Î»′=Î»w′w
=6000Ã0.7=4200Â â«

Topic: Physics, Wave Optics, NEET
The animal group, where the adults are degenerated but larvae are well developed, is

Options:
A Agnatha
B Tunicates
C Amphibians
D Cephalochordates

The correct answer is:
Tunicates

Solution:
Tunicates shows retrogressive metamorphosis which results in the degeneration in adult.

Topic: Biology, Animal Kingdom, NEET
The animals with bilateral symmetry in young stage, and radial pentamerous symmetry in the adult stage, belong to the phylum:

Options:
A Annelida
B Mollusca
C Cnidaria
D Echinodermata.

The correct answer is:
Echinodermata.

Solution:
In Echinoderms, larval forms possess bilateral symmetry while adults have radial symmetry. Mollusc possess bilateral symmetry. Cnidarian have bilaterally and radial symmetry. Annelid have bilateral symmetry.
Caution
Class Gastropods (Mollusc) are characterised by their asymmetric body plan, which develops through process called torsion (rotation of visceral organs in anticlockwise direction through an angle of $180^{\circ}$ on the rest of the body during larval development).

Topic: Biology, Animal Kingdom, NEET, NEET 2004
The anion of acetylacetone (acac) forms Co $(\mathrm{acac})^{3+}$ chelate with $\mathrm{Co}^{3+}$. The rings of the chelate are:

Options:
A Five membered
B Four membered
C Six membered
D Three membered

The correct answer is:
Six membered

Solution:
Chelating rings form a stable 6 -membered ring that has conjugated double bonds.

Topic: Chemistry, Coordination Compounds, NEET, NEET 2013 (Karnataka)
The annular and spirally thickened conducting elements generally develop in the protoxylem when the root or stem is

Options:
A maturing
B elongating
C widening
D differentiating

The correct answer is:
widening

Solution:
Vessels or tracheae are made up of a row of cells, placed one above the other, with their intervening walls absent or variously pored. The walls of vessels are lignified and hard, but not very thick. The cell cavity or the lumen is wide. The thickening may be annular, spiral, scalariform, reticulate and pitted.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2009 (Screening)
The anode voltage of a photocell is kept fixed. The wavelength λ of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows

Options:
A
B
C
D

The correct answer is:

Solution:
The graph depicts the variation of λ with I.

Topic: Physics, Semiconductors, NEET
The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO4 electrolyzed (in g) during the process is (Molar mass of PbSO4=303 gmol-1 )

Options:
A 22.8
B 15.2
C 7.6
D 11.4

The correct answer is:
7.6

Solution:
A PbSO 4 s 0.05/2 +2 OH − → PbO 2 + H 2 SO 4 + 2 e − 0.05F
C PbSO 4 s 0.05/2 + 2 e − 0.05F +2 H + →Pb s + H 2 SO 4
n PbSO 4 = 0.05 2  mole
w PbSO 4 = 0.05 2 ×303=7.6 gm

Topic: Chemistry, Electrochemistry, NEET
The anomeric carbon in D(+) glucose is

Options:
A C-1 carbon
B C-2 carbon
C C-5 carbon
D C-6 carbon

The correct answer is:
C-1 carbon

Solution:
The C-1 carbon of D (+) glucose is called anomeric carbon or glycosidic carbon and the pairs of stereoisomers differ in configuration around C-1 are called anomers.

Topic: Chemistry, Biomolecules, NEET
The anterior pituitary hormone that does not stimulate other endocrine gland is

Options:
A Somatotropin
B Thyrotropin
C Gonadotropin
D Adrenocorticotropin

The correct answer is:
Somatotropin

Solution:
Growth hormone (GH or HGH), also known as somatotropin or somatropin, is a peptide hormone that stimulates growth, cell reproduction, and regeneration in humans and other animals. It is a type of mitogen which is specific only to certain kinds of cells. The growth hormone is a 191-amino acid, single-chain polypeptide that is synthesized, stored, and secreted by somatotropic cells within the lateral wings of the anterior pituitary gland. It does not stimulate other endocrine glands.

Topic: Biology, Chemical Coordination and Integration, NEET
The antibodies present in the plasma of a person with blood group A will be

Options:
A anti-A
B anti-B.
C both (A) and (B).
D none of these.

The correct answer is:
anti-B.

Solution:
Blood Group	Antigens on RBCs	Antibodies in plasma

A	A	anti-B

B	B	anti-A

AB	A, B	nil

O	nil	anti-A, anti-B

Topic: Biology, Body Fluids and Circulation, NEET
The antibodies that can bind the largest number of antigens is ________ while the antibody that is smallest in size is ________.

Choose the option that correctly fills the blanks

Options:
A IgA, IgM
B IgM, IgA
C IgG, IgD
D IgM, IgG

The correct answer is:
IgM, IgG

Solution:
Antibodies are Y-shaped proteins that form part of the immune system. They circulate in the blood, identify foreign invaders such as bacteria, viruses, fungi or parasites and then eliminate them. There are five variations of antibodies called IgM, IgA, IgD and IgE . Each of these variations plays different roles in the immune system basis of their shape. IgM is the first antibody that is produced when the body is exposed to an antigen. They bind to a large number of invaders. IgG is the most common and smallest antibody.

Topic: Biology, Human Health and Diseases, NEET
The anticoagulant hirudin is obtained form

Options:
A Ti plasmid of Agrobacterium
B Bt toxin produced by cry gene
C Seeds of Brassica napus
D None of the above

The correct answer is:
Seeds of Brassica napus

Solution:
Hirudin is an anticoagulant protein found in the leech (Hirudinaria). It is now produced through genetic engineering from the seeds of Brassica napus. The gene encoding hirudin was transferred into Brassica napus where hirudin accumulates in seeds that are purified and used medicinally.

Topic: Biology, Biotechnology and Its Applications, NEET
The anticodon of tRNA coding for alanine was found to be CGU. What will be the triplet sequence on the anti-sense strand of DNA?

Options:
A CAU
B GTA
C CGT
D GUA

The correct answer is:
CGT

Solution:
Anticodon is the triplet found on tRNA which is complimentary to mRNA and the mRNA is complimentary to the template(anti-sense) DNA strand used in transcription. Thus the sequence of tRNA will be same as the template DNA only interchanging T and U.
So, tRNA anti-codon - CGU will have DNA triplet as CGT.

Topic: Biology, Molecular Basis of Inheritance, NEET
The anticodon transfer RNA for the messenger RNA codon G-C-A is

Options:
A G –U-T
B T-G-A
C C-G-U
D A –G-T

The correct answer is:
C-G-U

Solution:
The bases of messenger RNA are complementary to those of the DNA strand.
Opposite each Adenine (A) of DNA, there appears on RNA a Uracil (U); opposite Guanine (G) is Cytosine (C); opposite cytosine (C) is Guanine(G) and opposite Thymine (T) is Adenine (A). Thus, the anticodon of G – C – A would be C – G – U.

Topic: Chemistry, Biomolecules, NEET
The aperture present in pollen grain is called germ pore where

Options:
A Exine is present
B Both exine and intine are absent
C Only exine is absent
D Only intine is absent

The correct answer is:
Only exine is absent

Solution:
Pollen grain is surrounded by two layered wall, outer wall is exine and inner wall is intine. Exine is made up of sporopollenin and is discontinuous, as it is absent at germ pore. Intine is continuous and is made up of pectocellulose.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The apical meristem of the root is present:

Options:
A only in radicals
B only in tap roots
C only in adventitious roots
D in all the roots

The correct answer is:
in all the roots

Solution:
Apical meristem is terminal in position and responsible for terminal growth of the plant. Apical meristem is present at all root tips and shoot tips.
Related Theory
If the apical meristem is removed by removing the end of the stem, the lateral buds will be released from apical dominance, and greater branching results. Eventually, one of the lateral meristems will grow more than the others and will impose apical dominance, becoming the new apical meristem.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2003
The apparent coefficient of expansion of a liquid when heated in a copper vessel is C and when heated in a silver vessel is S. If A is the linear coefficient of expansion of copper, then the linear coefficient of expansion of silver is

Options:
A C+S-3A3
B C+3A-S3
C S+3A-C3
D C+S+3A3

The correct answer is:
C+3A-S3

Solution:
Î³r=Î³a+Î³v
where Î³r = coefficient of real expansion,
Î³a = coefficient of apparent expansion and
Î³v = coefficient of expansion of vessel.
For copper Î³r=C+3Î±Cu=C+3A ....(1)
For silver Î³r=S+3Î±Ag ....(2)

from eq. (1) and (2)
â C+3A-S+3Î±Ag=0
â Î±Ag=C-S+3A3

Topic: Physics, Thermal Properties of Matter, NEET
The apparent coefficients of volume expansion of a liquid filled in vessel A and B of identical volumes are found to be Î³ 1 and Î³ 2, respectively. If Î± 1 be the coefficient of linear expansion of A then the coefficient of linear expansion of B will be

Options:
A γ 1 - γ 2 3 - α 1
B Î³ 2 - Î³ 1 3 + Î± 1
C γ 2 - γ 1 3 - α 1
D γ 1 - γ 2 3 + α 1

The correct answer is:
γ 2 - γ 1 3 + α 1

Solution:
Let V be the volume of the liquid and Δ T the rise in temperature. Since apparent expansion = true expansion - expansion of vessel
∴ Vγ1ΔT=VγΔT-V3α1ΔT
or, γ 1 = γ + 3 α 1 ...(i)
(for vessel A)
and γ 2 = γ + 3 α 2 ...(ii)
(for vessel B)
where γ is the coefficient of real expansion of the liquid.
Subtracting Eq. (ii) from Eq. (i),
γ 1 - γ 2 = 3 α 1 - α 2
or, α 1 - α 2 = γ 1 - γ 2 / 3
or, α2= γ2-γ1/3+α1

Topic: Physics, Thermal Properties of Matter, NEET
The apparent depth of an object O from AB is

Options:
A 4.29Â cm
B 5.43Â cm
C 6.19 cm
D 5.99Â cm

The correct answer is:
6.19 cm

Solution:
Apparent depth d is given by
⇒d=t1μ1+t2μ2
⇒d=41.4+51.5=2.857+3.333
⇒d=6.190 cm

Topic: Physics, Ray Optics, NEET
The appearance of antibiotic-resistant bacteria is an example of:

Options:
A adaptive radiation
B transduction
C pre-existing variation in the population
D divergent evolution

The correct answer is:
pre-existing variation in the population

Solution:
Resistant organisms/cells are appearing in a time scale of months or years and not centuries. They are examples of evolution by anthropogenic action.

Topic: Biology, Evolution, NEET
The appearance of brown spots surrounded by chlorotic veins is the prominent symptom of:

Options:
A Copper deficiency
B Molybdenum deficiency
C Magnesium toxicity
D Manganese toxicity

The correct answer is:
Manganese toxicity

Solution:
The prominent symptoms of manganese toxicity is the appearance of brown spots surrounded by chlorotic veins. Manganese competes with iron and magnesium for uptake and with magnesium for bingding with enzymes. It also inhibits calcium translocation to shoot apex.

Topic: Biology, Mineral Nutrition, NEET
The appearance of colour in solid alkali metal halides is generally due to:

Options:
A interstitial positions
B F-centres
C Schottky defect
D Frenkel defect

The correct answer is:
F-centres

Solution:
The appearance of colour in solid alkali metal halides is generally due to F-centres.
Related Theory
The anion sites occupied by electrons are called F-centre. An F-centre is a type of crystallographic defect in which an anionic vacancy in a crystal lattice is occupied by one or more unpaired electrons.

Topic: Chemistry, Solid State, NEET, NEET 2006
The appearance of recombination nodules on homologous chromosomes during meiosis characterizes:

Options:
A Bivalent
B Sites at which crossing over occurs
C Terminalization
D Synaptonemal complex

The correct answer is:
Sites at which crossing over occurs

Solution:
Prophase I in meiosis-I is divided into 5 subphases:

In Pachytene there is recombination of chromosomes. This is accomplished by the formation of recombination nodule.

There is chiasmata formation in Diplotene which depicts the past recombination event.

Terminalisation takes place of chiasmata.

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2022 (Phase 1)
The application of biotechnology includes all, except

Options:
A Biofortified crops
B Gene therapy
C Molecular diagnostics
D Conventional hybridisation

The correct answer is:
Conventional hybridisation

Solution:
Conventional breeding uses hybridization to create new combinations of genes from parent varieties.

Topic: Biology, Biotechnology and Its Applications, NEET
The application of statistical and computational methodologies to the molecular biology stream is ____

Options:
A Transcription
B Polymorphism
C Translation
D Bioinformatics

The correct answer is:
Bioinformatics

Solution:
The application of statistical and computational methodologies to the molecular biology stream is Bioinformatics.

Topic: Biology, Molecular Basis of Inheritance, NEET
The application of transform microbial metabolism to obtain valuable products is called

Options:
A gel electrophoresis.
B biotechnology.
C downstream processing.
D molecular cleaving.

The correct answer is:
biotechnology.

Solution:
Biotechnology deals with techniques of using live organisms or enzymes from organisms to produce products and processes useful to humans. In this sense, making curd, bread or wine, which are all microbe-mediated processes, could also be thought as a form of biotechnology.
The definition given by EFB is as follows: ‘The integration of natural science and organisms, cells, parts thereof, and molecular analogues for products and services’.

Topic: Biology, Biotechnology - Principles and Processes, NEET
The applications of Biotechnology include,

A Therapeutics B Diagnostics C GM crops for agriculture D Processed food

E Bioremediation F Waste treatment G Energy production

Options:
A A, B, C, E only
B C only
C B, C, D, E only
D All of these

The correct answer is:
All of these

Solution:
Biotechnology essentially deals with industrial scale production of biopharmaceuticals and biological using genetically modified microbes, plants and animals.

Topic: Biology, Biotechnology and Its Applications, NEET
The approximate depth of an ocean is 2700 m . The compressibility of water is 45.4×10-11 Pa-1 and density of water is 103 kg/m 3. What fractional compression of water will be obtained at the bottom of the ocean ?

Options:
A 0.8×10-2
B 1.0×10-2
C 1.2×10-2
D 1.4×10-2

The correct answer is:
1.2×10-2

Solution:
Pressure at the bottom of the ocean
P=ρgd
=103×10×2700
=27×106Pa
∴ Fractional compression = compressibility × pressure
=45.4×10-11Pa-1×27×106Pa
=1.2×10-2

Topic: Physics, Mechanical Properties of Solids, NEET, NEET 2015 (Phase 1)
The aqueous solution containing one mole of CoCl3.5NH3 consumed 2 mol of silver nitrate solution for precipitation of free chloride ions. The formula of the compound should be

Options:
A [Co(NH3)5Cl]Cl2
B [Co(NH3)6Cl2]Cl
C [Co(NH3)4Cl3]NH3
D [Co(NH3)5Cl3]

The correct answer is:
[Co(NH3)5Cl]Cl2

Solution:
Consumption of 2 mol of AgNO3 indicates presence of 2 ionisable chloride ions per molecule. Hence the formula is [Co(NH3)5Cl]Cl2 .

Topic: Chemistry, Coordination Compounds, NEET