datavorous/entrance-exam-dataset huggingface dataset:
Choose the correct Statements from the following:
(A) Ethane-1, 2-diamine is a chelating ligand.
(B) Metallic aluminium is produced by electrolysis of aluminium oxide in presence of cryolite.
(C) Cyanide ion is used as ligand for leaching of silver.
(D) Phosphine act as a ligand in Wilkinson catalyst.
(E) The stability constants of Ca2+ and Mg2+ are similar with EDTA complexes.
Choose the correct answer from the options given below:

Options:
A (B), (C), (E) only
B (C), (D), (E) only
C (A), (B), (C) only
D (A), (D), (E) only

The correct answer is:
(A), (B), (C) only

Solution:
Statement A

Ethane−1,2−diamine is a chelating agent as it contains two coordinating sites that are two nitrogen atoms which donate their lone pairs of electrons and can form a 5-membered ring with the metal ion.

Statement B

Based on Hall-Heroults process-

The Hall-Heroult process is widely used in the extraction of aluminium. In Hall-Heroults process, pure Al2O3 is mixed with CaF2 or Na3AlF6. This results in lowering the melting point of the mixture and increases its ability to conduct electricity. A steel vessel with a lining of carbon and graphite rods is used.

Statement C

Cyanide ion is used as ligand for leaching of silver.

The cyanide process, which is also known as the Macarthur-forest process, is widely used in extracting gold or silver from the ores by dissolving them in a dilute solution of potassium cyanide or sodium cyanide.

Statement D

RhPPh33Cl Wilkinson's catalyst

Statement E

Ca++ ion forms more stable complex with EDTA

Hence option C is the answer.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2024 (30 Jan Shift 1)
Choose the incorrect match from the following.

Options:
A Inspiratory capacity = TV + IRV
B Expiratory capacity = TV + ERV
C Vital capacity = ERV + TV + IRV
D Residual volume = TV + VC - IRV

The correct answer is:
Residual volume = TV + VC - IRV

Solution:
Residual volume (RV) is the volume of air that remains in the lungs after maximum forceful expiration. In other words, it is the volume of air that cannot be expelled from the lungs. This volume remains unchanged regardless of the lung volume at which expiration was started.

Topic: Biology, Breathing and Exchange of Gases, NEET
Choose the incorrect match in the following.

Options:
A Inspiratory capacity =IRV+TV
B Functional residual capacity = RV + ERV
C Total lung capacity = VC – TV
D Vital capacity = ERV + TV + IRV

The correct answer is:
Total lung capacity = VC – TV

Solution:
Residual volume (RV)
It is the volume of air that remains in the lungs after maximum forceful expiration. In other words, it is the volume of air that cannot be expelled from the lungs. This volume remains unchanged regardless of the lung volume at which expiration was started.
Inspiratory Capacity
IC = TV + IRV
It is the total volume of air that a person can inspire after a normal
expiration.
Expiratory Capacity
EC = TV + ERV
It is the total volume of air that a person can expire after a normal
inspiration.
Functional Residual Capacity
FRC = ERV + RV
It refers to the volume of air that remains in the lungs after a
normal expiration.
Vital Capacity
VC = ERV + TV + IRV
It is the maximum volume of air that a person can inspire after a
forced expiration. Or
It is the maximum volume of air a person can expire after a forced
inspiration.
Total Lung Capacity
TLC = ERV+TV+IRV+RV = VC+RV
It refers to the total volume of air that can be accommodated in the
lungs at the end of a forced inspiration.

Topic: Biology, Breathing and Exchange of Gases, NEET
Choose the incorrect pair from the following with reference to the human ear.

Options:
A malleus, incus and stapes - provides the bony framework of the ear
B pinna - collects the vibrations in the air which produce sound
C Eustachian tube - helps in equalising the pressures on either sides of the eardrum
D macula and crista - responsible for the maintenance of the balance of the body and posture

The correct answer is:
malleus, incus and stapes - provides the bony framework of the ear

Solution:
The middle ear contains three ossicles called malleus, incus and stapes which are attached to one another in a chain-like fashion. The malleus is attached to the tympanic membrane and the stapes is attached to the oval window of the cochlea. The ear ossicles increase the efficiency of transmission of sound waves to the inner ear.

Topic: Biology, Neural Control and Coordination, NEET
Choose the incorrect statement:

Options:
A Callus is an unorganised mass of cells
B Growth of callus culture is faster than suspension culture
C In callus culture, if high auxin to cytokinin ratio is maintained then roots are formed
D In callus culture, if high cytokinin to auxin ratio is maintained then shoots are formed

The correct answer is:
Growth of callus culture is faster than suspension culture

Solution:
Callus is an unorganized amorphous mass of cell which develops in tissue culture medium. Growth of callus is slow in comparision to suspension culture. In callus, auxin promotes root formation and cytokinin promotes shoot formation.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
Choose the incorrect statement.

Options:
A The fringe width in a Young's double slit experiment reduces when the width between the two slits increases.
B The central fringe observed for in a single slit diffraction experiment is a bright fringe.
C The resolving power of the microscope is reciprocal to the maximum separation of two points seen as distinct.
D Polarisation phenomenon is observed only for transverse waves.

The correct answer is:
The resolving power of the microscope is reciprocal to the maximum separation of two points seen as distinct.

Solution:
(a) In Young's double slit experiment, the
fringe width is given by
$\beta=\frac{\lambda D}{d} \Rightarrow \beta \propto \frac{1}{d}$
So, if width between two slit $(d)$ increases, the fringe width reduces.
(b) In diffraction due to single slit, the path difference between rays reaching centre of screen is zero, so its intensity at centre is maximum i.e., bright fringe is observed at centre.
(c) The resolving power of microscope is
$=\frac{1}{d_{\min }}=\frac{2 n \sin \beta}{1.22 \lambda}$
So, it is the reciprocal of the minimum separation $\left(d_{\min }\right)$ of two points seen as distinct.
(d) The polarisation is possible only in those waves, that have vibrations in different direction, which is in case of transverse waves. Hence, only statement given in option (c) is incorrect.

Topic: Physics, Wave Optics, JEE Main
Choose the incorrect statement about the plasma membrane.

Options:
A The bilipid layer is made up of phosphoglycerides.
B The membrane of the erythrocyte has approximately 40 per cent protein and 52 per cent lipids.
C Peripheral proteins lie on the surface of membrane while the integral proteins are partially or totally buried in the membrane.
D The fluid nature of the membrane is important for functions like cell growth and formation of intercellular junctions.

The correct answer is:
The membrane of the erythrocyte has approximately 40 per cent protein and 52 per cent lipids.

Solution:
The lipid component of the membrane mainly consists of phosphoglycerides.
The ratio of protein and lipid varies considerably in different cell types. In human beings, the membrane of the erythrocyte has approximately 52 per cent protein and 40 per cent lipids.
Peripheral proteins lie on the surface of membrane while the integral proteins are partially or totally buried in the membrane.
The fluid mosaic model was introduced by Singer and Nicholson. The fluid nature of the membrane is important for functions like cell growth and formation of intercellular junctions.

Topic: Biology, Cell - The Unit of Life, NEET
Choose the incorrect statement about the two circles whose equations are given below:

x2+y2-10x-10y+41=0 and x2+y2-16x-10y+80=0

Options:
A Distance between two centres is the average of radii of both the circles.
B Both circles' centres lie inside region of one another.
C Both circles pass through the centre of each other.
D Circles have two intersection points.

The correct answer is:
Both circles' centres lie inside region of one another.

Solution:
r1=3,c1(5,5)

r2=3,c2(8,5)

C1C2=3,r1=3,r2=3

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (17 Mar Shift 1)
Choose the incorrect statement from the following:

Options:
A The speed of satellite in a given circular orbit remains constant
B For a planet revolving around the sun in an elliptical orbit, the total energy of the planet remains constant
C The linear speed of a planet revolving around the sun remains constant
D When a body falls towards earth, the displacement of earth towards the body is negligible

The correct answer is:
The linear speed of a planet revolving around the sun remains constant

Solution:
A planet revolves round the Sun by the action of the gravitational force between the sun and the planet. Depending up on the distance from the sun in the elliptic orbit, the speed of revolution of the planet changes.

The formula to calculate the linear speed of a planet in an elliptical orbit can be written as

v=GM2r-1a

From the above expression, it can be concluded that the linear speed of the planet is not constant during its revolution.

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (06 Apr Shift 2)
Choose the incorrect statement from the following.

Options:
A The electric field in electrostatics obey principle of superposition
B The electric field inside a perfect conductor is zero
C The electric dipole will try to orient in a direction opposite to the external electric field
D The electric flux passing through any closed surface enclosing the charges remains constant

The correct answer is:
The electric dipole will try to orient in a direction opposite to the external electric field

Solution:
Third statement is incorrect because electric dipole will always try to orient in the direction of electric field in order to achieve minimum potential energy and maximum stability.

Topic: Physics, Electrostatics, TS EAMCET, TS EAMCET 2018 (07 May Shift 1)
Choose the incorrect statement from the following.

Options:
A The electric field in electrostatics obey principle of superposition
B The electric field inside a perfect conductor is zero
C The electric dipole will try to orient in a direction opposite to the external electric field
D The electric flux passing through any closed surface enclosing the charges remains constant

The correct answer is:
The electric dipole will try to orient in a direction opposite to the external electric field

Solution:
Third statement is incorrect because electric dipole will always try to orient in the direction of electric field in order to achieve minimum potential energy and maximum stability.

Topic: Physics, Electrostatics, JEE Main
Choose the incorrect statement from the following regarding magnetic lines of field -

Options:
A The direction of magnetic field at a point is taken to be the direction in which the north Pole of a magnetic compass needle points.
B Magnetic field lines are closed curves.
C If magnetic field lines are parallel and equidistant, they represent zero field strength.
D More than one of the above

The correct answer is:
If magnetic field lines are parallel and equidistant, they represent zero field strength.

Solution:
- Magnetic field lines form closed loops and do not intersect.
- At any point, the magnetic field direction is tangent to the magnetic field line.
- If the magnetic field lines are parallel and equidistant, the magnetic field represents a uniform magnetic field rather than a zero field strength.

Topic: Physics, Magnetic Properties of Matter, JEE Main
Choose the incorrect statements.

Options:
A BeCO3 is preserved in an atmosphere of CO2 as it is thermally least stable.
B BeF2 forms a complex compound with excess NaF, in which the complex entity containing Be, is a cation.
C Beryllium dissolves in an alkali to form [Be(OH)4]2- ion.
D Beryllium exhibits no diagonal relationship with sodium.

The correct answer is:
BeF2 forms a complex compound with excess NaF, in which the complex entity containing Be, is a cation.

Solution:
BeF 2 + NaF → Na 2 BeF 4
Complex entity is an anion.

Topic: Chemistry, s Block Elements, JEE Main
Choose the polar molecule from the following :

Options:
A CCl4
B CO2
C CH2 = CH2
D CHCl3

The correct answer is:
CHCl3

Solution:
The solvents having an appreciable net dipole moment are known as polar solvents.

CHCl3 is polar due the presence of a net dipole moment. There is a difference in electronegativity of the central atom, carbon, hydrogen, and chlorine. This will create a dipole in the molecule. As a dipole has formed in chloroform molecule, due to electronegativity difference, which makes it polar in nature.

Hence, option D is the answer.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Choose the product of the following reaction :

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Choose the right one among the statements given below.

Options:
A IUDs are generally inserted by the user herself
B IUDs increase phagocytosis reaction in the uterus
C IUDs suppress gametogenesis
D IUDs once inserted need not be replaced

The correct answer is:
IUDs increase phagocytosis reaction in the uterus

Solution:
IUD's are intrauterine devices. As the word suggests these are inserted in the uterus. The main difference between barrier methods like condoms and IUD'S are that the IUD's are inserted by a medical practioner and condoms are self-used. The main function of these devices are to kill the sperms so that they do not reach the oviduct. These suppress the endometrium lining too so that implantation is also avoided. Examples are CuT, Lippes loop, Hormonal IUD's etc. They can later be removed with the help of a professional or medical expert.

Topic: Biology, Reproductive Health, NEET
Choose the statement which is not correct about pedigree analysis.

Options:
A It helps to understand whether the trait in question is dominant or recessive.
B It confirms that the trait is linked to one of the auto-some.
C It helps to trace the inheritance of a specific trait.
D It confirms that DNA is the carrier of genetic information.

The correct answer is:
It confirms that DNA is the carrier of genetic information.

Solution:
Pedigree chart is a diagram which shows appearance of phenotype of a particular gene. It is used to identify that the trait mentioned is dominant or recessive. It is also used to identify that the gene for this trait is present on autosome or sex-chromosome. It also traces inheritance of a particular trait. On this basis pedigree is mainly of following types - (i) Autosomal Dominant (ii) Autosomal Recessive (iii) Sex-linked Dominant (iv) Sex-linked Recessive.

Topic: Biology, Principles of Inheritance and Variation, NEET
Choose the wrong statement

Options:
A the radius of path of a charged particle moving in a uniform magnetic field is proportional to the momentum of the particle
B an electron beam is moving towards east, on which a perpendicular magnetic field is acting upwards. The beam will be deflected towards the north direction
C a positive charge is going straight away from the observer. The magnetic line of force produced due to it are in clockwise direction.
D while passing through a given place, the path of electron remains straight line. It can be definitely said that the magnetic field is not present at that place

The correct answer is:
while passing through a given place, the path of electron remains straight line. It can be definitely said that the magnetic field is not present at that place

Solution:
If the path of electron remains straight line then either magnetic field is not present at that place or velocity of electron is parallel to the magnetic field.

Topic: Physics, Magnetic Effects of Current, NEET
Choose the wrong statement for zero error and zero correction :

Options:
A If the zero of the vernier scale does not coincide with the zero of the main scale then the instrument is said to be having a zero error.
B Zero correction has a magnitude equal to zero error but sign is opposite to that of the zero error.
C Zero error is positive when the zero of vernier scale lies to the left of the zero of the main scale.
D Zero error is negative when the zero of vernier scale lies to the left of zero of main scale

The correct answer is:
Zero error is positive when the zero of vernier scale lies to the left of the zero of the main scale.

Solution:
if the zero line of the vernier scale is not coinciding with the main scale the zero error exists. Knowing the zero error necessary correction can be made to find correct measurement. Such a correction is called zero correction.
In positive zero error zero of vernier scale lies to right of the zero of the main scale.

Topic: Physics, Units and Dimensions, NEET
Choose the WRONG statement regarding the Hardy-Weinberg principle.

Options:
A The sum total of all the allelic frequencies in a population is 1.
B Variation due to genetic drift results in a changed frequency of genes and alleles in future generations.
C Natural selection can lead to stabilization, directional change, or disruption.
D Genetic recombination helps in maintaining the Hardy-Weinberg equilibrium.

The correct answer is:
Genetic recombination helps in maintaining the Hardy-Weinberg equilibrium.

Solution:
The Hardy-Weinberg equilibrium is a principle which states that the genetic variations (frequencies of alleles, gene versions, genotypes ) in a population from one generation to another generation will always remain constant if there are no evolutionary disturbing factors. Any variations will result in changed frequency of genes and alleles in the coming generations.

Topic: Biology, Evolution, NEET
Chord joining two distinct points Pa,4b and Qc,-16b (both are variable points) on the parabola y2=16x always passes through a fixed point α,β. Then, which of the following statements is correct?

Options:
A α+β=2
B α-β=4
C α+β=8
D α=β

The correct answer is:
α-β=4

Solution:
Let, P=a,4b=4t12,8t1 & Q=c,-16b=4t22,8t2
ât1=b2 and t2=-2b
ât1t2=-1
â Chord PQ is a focal chord that passes through the focus
âÎ±,Î²=4,0

Topic: Mathematics, Parabola, JEE Main
Chord length of ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1_{\text {with mid point }}$ $\left(1, \frac{2}{5}\right)_{\text {is equal to }}$ $\qquad$

Options:
A $\frac{\sqrt{1681}}{5}$
B $\frac{\sqrt{1691}}{5}$
C $\frac{\sqrt{691}}{5}$
D $\frac{\sqrt{961}}{5}$

The correct answer is:
$\frac{\sqrt{1691}}{5}$

Solution:
Correct option is (2) $\frac{\sqrt{1691}}{5}$
$\begin{aligned} & T=S_1 \Rightarrow \frac{x}{25}+\frac{y \cdot \frac{2}{5}}{16}=\frac{1}{25}+\frac{4}{16 \times 25} \\ & \frac{x}{25}+\frac{y}{8 \times 5}=\frac{5}{100} \\ & 4 x+\frac{5}{2} y=5 \Rightarrow 8 x+5 y-10=0 \\ & \frac{x^2}{25}+\frac{y^2}{16}=1 \Rightarrow(4 x)^2+(5 y)^2=400 \\ & \Rightarrow(4 x)^2+(8 x-10)^2=400 \\ & \Rightarrow 4 x^2+(4 x-5)^2=100 \quad \Rightarrow 20 x^2-40 x-75=0 \\ & \Rightarrow 4 x^2-8 x-15=0\end{aligned}$
$\begin{aligned} & L-\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}-\sqrt{\left(x_2-x_1\right)^2+\left\{\frac{10-8 x_2}{5}-\frac{\left(10-8 x_1\right)}{5}\right\}^2} \\ & =\sqrt{\left(x_2-x_1\right)^2+\frac{64}{25}\left(x_2-x_1\right)^2}=\sqrt{\frac{89}{4}} \sqrt{\left(x_2-x_1\right)^2} \\ & =\sqrt{\frac{89}{4}+\sqrt{\left(x_1-x_2\right)^2-4 x_1 x_2}}=\sqrt{\frac{89}{5}} \sqrt{4+4 \times \frac{15}{4}}=\frac{\sqrt{89} \times \sqrt{19}}{5} \\ & =\frac{\sqrt{1691}}{5}\end{aligned}$

Topic: Mathematics, Ellipse, JEE Main
Chords of an ellipse are drawn through the positive end of the minor axis. Then their mid point lies on

Options:
A A Circle
B A Parabola
C An Ellipse
D A Hyperbola

The correct answer is:
An Ellipse

Solution:
an ellipse Let the ellipse is,
Then, the equation of the chord drawn from mid point $M(h, k)$ is $\mathrm{T}=\mathrm{S}_1 \quad$ (using transformation)
$\frac{h x}{a^2}+\frac{k y}{b^2}-1=\frac{h^2}{a^2}+\frac{k^2}{b^2}-1$ .....(1)
Now, equation (1) passing through $(0, b)$
$\frac{k}{b}=\frac{h^2}{a^2}+\frac{k^2}{b^2}$
Hence the locus of $(h, k)$ is (replace $(h, k) \rightarrow(x, y)$ )
$\begin{aligned} & \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{y}{b} \\ & \frac{x^2}{a^2}+\frac{\left(y-\frac{b}{2}\right)^2}{b^2}=\frac{1}{4}\end{aligned}$
Which is an ellipse

Topic: Mathematics, Ellipse, JEE Main
Chromatographic technique/s based on the principle of differential adsorption is/are
A. Column chromatography B. Thin layer chromatography C. Paper chromatography

Choose the most appropriate answer from the options given below:

Options:
A B only
B A only
C A & B only
D C only

The correct answer is:
A & B only

Solution:
Adsorption chromatography is defined as a type of chromatography in which separation occurs based on adsorption. Partition chromatography is a type of chromatography in which separation is based on partition. Adsorption chromatography is a liquid-solid extraction. Partition chromatography is a liquid-liquid extraction.

Column chromatography and Thin layer chromatography are examples of adsorption chromatography. Paper chromatography is an example of partition chromatography.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (29 Jan Shift 2)
Cinnabar is an ore of

Options:
A $\mathrm{Hg}$
B $\mathrm{Cu}$
C $\mathrm{Pb}$
D $\mathrm{Zn}$

The correct answer is:
$\mathrm{Hg}$

Solution:
Cinnabar is HgS, so it is an ore of Hg.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Circle $x^2+y^2+6 y=0$ touches

Options:
A $y$-axis at the origin
B $x$-axis at the origin
C $x$-axis at the point $(3,0)$
D The line $y+3=0$

The correct answer is:
$x$-axis at the origin

Solution:
Centre is $(0,-3)$ and $R=\sqrt{0^2+9+0}=3$

Topic: Mathematics, Circle, JEE Main
Circles $x^2+y^2-2 x-4 y=0$ and $x^2+y^2-8 y-4=0$

Options:
A Touch each other internally
B Touch each other externally
C Cuts each other at two points
D None of these

The correct answer is:
Touch each other internally

Solution:
$C_1(1,2), C_2(0,4), R_1=\sqrt{5}, R_2=2 \sqrt{5}$
$C_1 C_2=\sqrt{5}$ and $C_1 C_2=\left|R_2-R_1\right|$
Hence circles touch internally.

Topic: Mathematics, Circle, JEE Main
Cl 2 g + Ba OH 2 → X aq. + BaCl 2 + H 2 O X + H 2 SO 4 → Y + BaSO 4 Y → Δ > 365 K Δ Z + H 2 O + O 2

Y and Z are respectively :

Options:
A HClO 4 ,  ClO 2
B HClO 3 ,  ClO 2
C HClO 3 ,  ClO 6
D HClO 4 ,  Cl 2 O 7

The correct answer is:
HClO 3 ,  ClO 2

Solution:
6 Cl 2 + 6 Ba OH 2 → Ba ClO 3 2 + 5 BaCl 2 + 6 H 2 O X Ba ClO 3 2 + H 2 SO 4 → 2 HClO 3 + BaSO 4 ↓ Y
2 HClO 3 → Δ > 365 K Δ 2 ClO 2 + H 2 O + 1 2 O 2

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Cl˙+CH4→A+B

A and B in the above atmospheric reaction step are

Options:
A C2H6 and Cl2
B C˙HCl2 and H2
C CH3˙ and HCl
D C2H6 and HCl

The correct answer is:
CH3˙ and HCl

Solution:
In summer season, chlorine atoms forming chlorine sinks, preventing much ozone depletion, whereas In winter, special type of clouds called polar stratospheric clouds are formed over Antarctica.

C•lg+CH4g→C•H3g+HClg

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2022 (26 Jul Shift 1)
Class Crustacea differs from Insecta in having :

Options:
A two pairs of antennae
B jointed foot
C chitinous cuticle
D none of the above

The correct answer is:
two pairs of antennae

Solution:
The main characteristics of class Crustacea and Insecta are as follows :
(1) Class Crustacea : Mostly aquatic with gills, two pairs of antennae and atleast three pairs of jaws, appendages mostly biramous and always more than four in the trunk region.
Example : prawn, crab, lobster etc.
(2) Class Insecta : Hexapod arthropods with a pair of antennae and three pairs of Jaws.
Example : cockroach, grasshopper, bees, housefly etc.

Topic: Biology, Animal Kingdom, JIPMER, JIPMER 2006