datavorous/entrance-exam-dataset huggingface dataset:
A compass needle free to turn in a horizontal plane is placed at the centre of a circular coil of 30 turns and radius 12 cm . The coil is in a vertical plane making an angle of 45 ∘ with the magnetic meridian. The needle points west to east when the current in the coil is 0.35 A. The horizontal component of the earth's magnetic field (in T) at that location is

Options:
A 3.9×10-5
B 3.9×10-6
C 5×10-5
D 2.6×10-5

The correct answer is:
3.9×10-5

Solution:
Given, number of turns in the coil n = 30
Current in the coil I = 0.35 A
Radius of circular coil = 12 cm = 0.12 m

Let N - S be the line of magnetic of 45o with the magnetic meridian. C1 and C2 be the plane of coil. The needle points west to east. The magnetic field produced due to the coil is B. The direction of B is along the axis of coil i.e., it makes an angle of 45o with east. The needle points west to east only if the direction of magnetic field B is at 45o from east.
Use the formula, magnetic field produced by a current carrying coil
B = Î¼ 0 4 Ï Â· 2 Ï I r = 1 0 Ã 2 Ã 3 . 1 4 Ã 0 . 3 5 Ã 3 0 0 . 1 2
= 1.83 x 10-6 x 30 = 5.49 x 10-5 T
The horizontal component of magnetic field
H = B sin 45o = 5.49 x 10-5 Ã 1 2
= 3.9 x 10-5 T.

Topic: Physics, Magnetic Effects of Current, NEET
A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It

Options:
A will become rigid showing no movement
B will stay in any position
C will stay in north-south direction only
D will stay in east-west direction only

The correct answer is:
will stay in any position

Solution:
It will stay in any position at geomagnetic north and south poles.

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2012 (Screening)
A competitive inhibitor of an enzyme

Options:
A Decreases the Km
B Decreases the Vmax
C Decreases both the Km & Vmax
D Increases the Km

The correct answer is:
Increases the Km

Solution:
In competitive inhibition of enzyme catalysis, binding of an inhibitor prevents binding of the target molecule of the enzyme, also known as the substrate. This is accomplished by blocking the binding site of the substrate: the active site. Competitive inhibition can be overcome by adding more substrate to the reaction, which increases the chances of the enzyme and substrate binding. As a result, competitive inhibition alters only the Km, leaving the Vmax the same.

Topic: Chemistry, Biomolecules, NEET
A competitive inhibitor of succinic dehydrogenase is

Options:
A malonate
B oxaloacetate
C o-ketoglutarate
D malate

The correct answer is:
malonate

Solution:
Succinic dehydrogenase oxidised the succinate to fumarate. In eukaryotes succinate dehydrogenase is tightly bound to the inner mitochondrial membrane, in prokaryotes to the plasma membrane. Electrons pass from succinate through the FAD and iron-sulphur centres before entering the chain of electron carriers in the mitochondrial inner membrane Malonate, an analogue of succinate is a strong competitive inhibitor of succinate dehydrogenase and therefore, blocks the activity of citric acid cycle.
In Krebs cycle the reversible hydration of fumarate to malate is catalysed by fumarase enzyme.
In the last reaction of Krebs cycle NAD-linked L-malate dehydrogenase catalyses the oxidation of L-malate to oxaloacetate. The isocitrate dehydrogenase in Krebs cycle catalyses oxidative decarboxylation of isocitrate to form $\alpha$-ketoglutarate, which in next step through oxidative decarboxylation converted to succinyl Co-A.

Topic: Biology, Biomolecules (B), NEET, NEET 2008 (Screening)
A complete set of chromosome or cell genes inherited as a unit from a parent is known as

Options:
A Genome
B Genotype
C Karyotype
D Chromosomes

The correct answer is:
Genome

Solution:
A complete set of chromosomes or a gametic numbers of chromosomes is called the genome. It is one set of a chromosome which is inherited from one parent through a gamete.

Topic: Biology, Principles of Inheritance and Variation, NEET
A completely four-chambered heart is present in

Options:
A all reptiles, all birds and all mammals
B some reptiles, all birds and all mammals
C no reptiles, all birds and all mammals
D some reptiles, some birds and all mammals

The correct answer is:
some reptiles, all birds and all mammals

Solution:
Chordates show a closed circulatory system i.e. proper system of heart and blood vessels. The heart has evolved from two chambers to four chambers. Pisces shows the presence of two chambers: one ventricle and one auricle. Amphibians show three chambers: two auricles and one ventricle due to which mixed blood flows in the heart. Reptiles like crocodiles, birds and mammals show four chambers (two auricles and two ventricles) and thus proper closed circulation is seen. Except for crocodilians, which have a four-chambered heart, all reptiles have a three-chambered heart consisting of two atria and one ventricle.

Topic: Biology, Animal Kingdom, NEET
A complex of the type [M(AA)₂X₂]ⁿ⁺ is known to be optically active. What does this indicate about the structure of the complex?

Options:
A The complex has a tetrahedral geometry, resulting in optical activity.
B The complex has a square planar geometry, leading to optical activity.
C The complex has a linear geometry, causing optical activity.
D The complex has a trigonal bipyramidal geometry, exhibiting optical activity.

The correct answer is:
The complex has a square planar geometry, leading to optical activity.

Solution:
Correct Option is : (B)
The complex has a square planar geometry, leading to optical activity.

Topic: Chemistry, Coordination Compounds, NEET
A Complex P of composition Cr(H2O)6Brn has a spin only magnetic moment of 3.83 BM. It reacts with AgNO3 and shows geometrical isomerism. The IUPAC nomenclature of P is

Options:
A Hexaaqua chromium (III) bromide
B Dibromidotetraaqua chromium (IV) bromide dihydrate
C Tetraaquadibromido chromium (IV) bromide dihydrate
D Tetraaquadibromido chromium (III) bromide dihydrate

The correct answer is:
Tetraaquadibromido chromium (III) bromide dihydrate

Solution:
This compound can be identified and named as follows
Cr(H2O)6Brn
As (μcomplex)spin= 3.8 B.M. =n(n+2)
On solving n=3
Hence the complex is [Cr(H2O)6]Br3
oxidation number of Cr =+3.
As compound also shows geometrical isomerism so it will be [Cr(H2O)4Br2]Br.2H2O
It is named as → Tetraaquadibromido chromium (III) bromide dihydrate

Topic: Chemistry, Coordination Compounds, NEET
A compound (A) of boron reacts with NMe3to give an adduct (B) which on hydrolysis gives a compound (C) and hydrogen gas. Compound (C) is an acid. Identify the compounds A, B and C. Give the reactions involved.

Options:
A A = B2H6, B = B2H6·NMe3, C = H3BO3
B A = B2H6, B = B2H6·NMe3, C = HBO2
C A = B2H6·NMe3, B = B2H6, C = HBO2
D A = B2H6·NMe3, B = B2H6, C = H3BO3

The correct answer is:
A = B2H6, B = B2H6·NMe3, C = H3BO3

Solution:
Correct Option is : (A)
A = B2H6, B = B2H6·NMe3, C = H3BO3

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
A compound A of formula C3H6Cl2 on reaction with alkali can give B of formula C3H6O or C of formula C3H4. B on oxidation gave a compound of the formula C3H6O2. C with dilute H2SO4 containing Hg2+ ion gave D of formula C3H6O, with which bromine and NaOH gave the sodium salt of acid of formula C2H4O2. Then A is

Options:
A CH3CH2CHCl2
B CH3CCl2CH3
C CH2ClCH2CH2Cl
D CH3CHClCH2Cl

The correct answer is:
CH3CH2CHCl2

Solution:
Since, B and D are different thus, B is CH3CH2CHO and so A CH3CH2CHCl2.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
A compound AB has a rock type structure with A : B = 1 : 1. The formula weight of AB is 6.023 y amu and the closest A – B distance is y1/3 nm. The density of lattice is?

Options:
A 1.5&nbspkgm-3
B 2.5&nbspkgm-3
C 5.0&nbspkgm-3
D 7.5&nbspkgm-3

The correct answer is:
5.0&nbspkgm-3

Solution:
AB has rock-salt (A : B :: 1 : 1) structure, i.e., fcc structure (n = 4) and formula weight of AB is 6.023 y g having closest distance A – B y1/3 nm. Therefore, edge length of unit cell
(a)=2(A++B-)=2×y1/3×10-9m
So density of AB =n×mol.wt.Av.no.×V=n×mol.wt.6.023×1023×a3
=4×6.023×y×10-36.023×1023×(2y1/3×10-9)3 (mol. wt. in kg)
=5.0&nbspkgm-3

Topic: Chemistry, Solid State, NEET
A compound alloy of gold and copper crystallises in cubic lattice in which gold atoms occupy the lattice corners and copper atoms occupy the centre of each of cube faces. The formula of the compound is

Options:
A AuCu3
B Au2Cu
C AuCu
D Au3Cu

The correct answer is:
AuCu3

Solution:
Au atom = =8×18=1
Cu atoms =6×12=3
So formula AuCu3

Topic: Chemistry, Solid State, NEET
A compound C5H8 reacts with ammonical AgNO3 to give a white precipitate and reacts with excess of KMnO4 solution to give CH32CH–COOH . The compound is –

Options:
A CH2=CH–CH=CH–CH3
B CH32CHC≡CH
C CH3CH22C≡CH
D CH32C=C=CH2

The correct answer is:
CH32CHC≡CH

Solution:

Topic: Chemistry, Hydrocarbons, NEET
A compound C5H10O(A) forms a phenylhydrazone and gives negative Tollen's test and a positive iodoform reaction. It also gives n-pentane on reduction. The compound (A) is

Options:
A pentanal
B 2-pentanone
C 3-pentanone
D amyl alcohol

The correct answer is:
2-pentanone

Solution:

Topic: Chemistry, Aldehydes and Ketones, NEET
A compound contains 28% N2 and 72% of a metal by weight. 3 atoms of the metal combines with 2 atoms of N. The atomic weight of metal is

Options:
A 12
B 32
C 24
D 16

The correct answer is:
24

Solution:
Here 3 atoms of the metal combines with 2 atoms of N
Hence, the formula of the compound =M3N2
Suppose the atomic weight of metal = X
28 g of nitrogen combines =3Xg metal
From question, 28 g nitrogen combines = 72 g metal
3X=72
So X=24
Atomic weight of metal = 24.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A compound contains 1.08 mol of Na, 0.539 mol of Cu and 2.16 mol of F. Its aqueous solution shows osmotic pressure which is three times that of urea having same molar concentration. The formula of the compound is :

Options:
A Na4[CuF6]
B Na[CuF4]
C Na2[CuF4]
D Na2[CuF3]

The correct answer is:
Na2[CuF4]

Solution:
Mol ratio of Na, Cu and F
= 1 . 0 8 0 . 5 3 9 : 0 . 5 3 9 0 . 5 3 9 : 2 . 1 6 0 . 5 3 9 = 2 : 1 : 4
Empirical formula = Na2CuF4 : Van't Hoff factor (i)
= 3 (given)
Hence, formula of the compound :
Na2[CuF4] â 2Na+ + [CuF4]2-

Topic: Chemistry, Coordination Compounds, NEET
A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?

Options:
A Empirical formula is CHCl and Molecular formula is C2H2Cl2
B Empirical formula is CH2Cl and Molecular formula is C2H4Cl2
C Empirical formula is C3H2Cl and Molecular formula is C6H4Cl2
D Empirical formula is CH2Cl3and Molecular formula is C2H4Cl6

The correct answer is:
Empirical formula is CH2Cl and Molecular formula is C2H4Cl2

Solution:
Correct Option is : (B)
Empirical formula is CH2Cl and Molecular formula is C2H4Cl2

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A compound contains atoms X, Y and Z. The oxidation number of X is +2, Y is +5 and Z is -2. The possible formula of the compound is

Options:
A XYZ2
B Y2(XZ3)2
C X3(YZ4)2
D X3(Y4Z)2

The correct answer is:
X3(YZ4)2

Solution:
Here compound in option C is correct choice as in it sum of all oxidation numbers is zero.
X3(YZ4)2—3×(2)+2×(+5)+8×(-2)=0
=6+10-16=0

Topic: Chemistry, Redox Reactions, NEET
A compound contains three elements A, B and C, if the oxidation of A=+2,B=+5 and C=-2, the possible formula of the compound is

Options:
A A3(B4C)2
B A3(BC4)2
C A2(BC3)2
D ABC2

The correct answer is:
A3(BC4)2

Solution:
Here compound A3(BC4)2 is possible as in it sum of oxidation numbers is zero as follows
3(+2)+2×5+8(-2)=0

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A compound CuCl has face centred cubic structure. Its density is 3.4 g cm-3 . The length of unit cell is

Options:
A 5.783âÃ
B 7.783 Å
C 6.783 Å
D 8.783 Å

The correct answer is:
5.783 Å

Solution:
Molecular weight of CuCl =63.5+35.5=99
For face centred cubic cell
z = 4
Density =  z × molecular weight a 3  ×  N A
3.4=4×99a3×6.023×1023
a=5.783× 10 −8  cm
= 5.783 Å .

Topic: Chemistry, Solid State, NEET
A compound formed by element $\mathrm{X}$ and $\mathrm{Y}$ crystallises in a cube and the $\mathrm{X}$ atoms are at the corners of a cube and the $\mathrm{Y}$ atoms are at the face centres. The formula of the compound is:

Options:
A $\mathrm{XY}_3$
B $\mathrm{X}_3 \mathrm{Y}$
C XY
D $\mathrm{XY}_2$

The correct answer is:
$\mathrm{XY}_3$

Solution:
In a unit cell, $\mathrm{X}$ atoms at the corners
$$
=\frac{1}{8} \times 8=1
$$
$Y$ atoms at the face centres $=\frac{1}{2} \times 6=3$
Ratio of $\mathrm{X}$ and $\mathrm{Y}=1: 3$. Hence formula is $\mathrm{XY}_3$.

Topic: Chemistry, Solid State, NEET, NEET 2004
A compound forms a hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?

Options:
A Total Voids = 6, Tetrahedral Voids = 2
B Total Voids = 12, Tetrahedral Voids = 6
C Total Voids = 18, Tetrahedral Voids = 12
D Total Voids = 24, Tetrahedral Voids = 8

The correct answer is:
Total Voids = 18, Tetrahedral Voids = 12

Solution:
Correct Option is : (C)
Total Voids = 18, Tetrahedral Voids = 12

Topic: Chemistry, Solid State, NEET
A compound has the formula C2HCl2Br . The number of non-identical structures that are possible is

Options:
A one
B two
C three
D four

The correct answer is:
three

Solution:
The non-identical structures are
(i)
(ii)
(iii)

Topic: Chemistry, General Organic Chemistry, NEET
A compound is formed by cation C and anion A . The anions form hexagonal close packed (hcp) lattice and the cations occupy 75% of octahedral voids. The formula of the compound is:

Options:
A C2A3
B C3A2
C C3A4
D C4A3

The correct answer is:
C3A4

Solution:
In HCP lattice, effective number of atoms present in unit cell
=6

Thus, the number of anions per unit cell =6

The number of octahedral voids is equal to the effective number of atoms in the unit cell.

According to the question, cations are occupying 75% of the octahedral voids,

Thus, the number of cations per unit cell =6×75100=92

The formula of the ionic compound represents the simplest ratio of the ions present in it. The cation is written first and then the anion.

Thus, the formula of the given compound=C92A6⇒C9A12=C3A4

Hexagonal close packed (hcp) refers to layers of spheres packed so that spheres in alternating layers overlie one another.

Topic: Chemistry, Solid State, NEET, NEET 2019 (All India)
A compound is formed by two elements A and B. The element B forms cubic close packed structure and atoms of A occupy 13 of tetrahedral voids. If the formula of the compound is AxBy, then the value of x+y is in option

Options:
A 2
B 5
C 4
D 3

The correct answer is:
5

Solution:
The number of effective atoms per unit cell in the cubic close packed arrangement is 4. Hence, there are four atoms of B per unit cell. In cubic close packed arrangement, there are two tetrahedral voids per atom. So, eight tetrahedral voids are possible. Given that 1/3 of tetrahedral voids are occupied. Hence, the possible formula is

A83B4 and the simplest formula is A8B12≡A2B3.

So, x=2 & y=3

⇒x+y=5

Topic: Chemistry, Solid State, NEET, NEET 2023 (All India)
A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound?

Options:
A MN
B M2N3
C M2N
D MN3

The correct answer is:
M2N3

Solution:
Correct Option is : (B)
M2N3

Topic: Chemistry, Solid State, NEET
A compound is formed by two elements X and Y. Atoms of the element Y (as anions) make ccp and those of the element X (as cations) occupy all the octahedral voids. What is the formula of the compound?

Options:
A XY2
B X2Y
C XY
D X2Y3

The correct answer is:
XY

Solution:
Correct Option is : (C)
XY

Topic: Chemistry, Solid State, NEET
A compound leaf is shown in the diagram:

Select the option with correct identification of the type of leaf and its example.

Options:
A Pinnately compound, Peepal
B Pinnately compound, Neem
C Palmately compound, Banana
D Palmately compound, Silk cotton

The correct answer is:
Pinnately compound, Neem

Solution:
Compound leaf: If the incisions reach the midrib and dividing the lamina into a number of leaflets. It may be:
(i) Pinnately compound leaf: Example: Neem.
(ii) Palmately compound leaf: Example: Silk cotton.

Topic: Biology, Morphology of Flowering Plants, NEET
A compound MpXq has cubic close packing (ccp) arrangement of X. Its unit cell structure is shown below. The empirical formula of the compound is

Options:
A M2X
B M5X14
C MX2
D MX

The correct answer is:
MX2

Solution:
X=ccp=8×18+6×12=4
M=4×14+1=2
M2X4 – empirical formula is MX2

Topic: Chemistry, Solid State, NEET
A compound MX2 has observed and normal molar masses 65.6 and 164 respectively. Calculate the apparent degree of ionization of MX2

Options:
A 75%
B 85%
C 65%
D 25%

The correct answer is:
75%

Solution:
i=NormalmolarmassObservedmolarmass
=16465.6=2.5
Î±=i-1n-1
n=3 (number of ions)
α= 2.5−1 3−1 = 1.5 2 =0.75
So percentage ionization of MX2 will be 75%.

Topic: Chemistry, Solutions, NEET
A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image at the least distance of distinct vision (25cm)?

Options:
A 2.5 cm
B 3.75 cm
C 4.0 cm
D 5.0 cm

The correct answer is:
2.5 cm

Solution:
Correct Option is : (A)
2.5 cm

Topic: Chemistry, Ray Optics, NEET
A compound microscope having magnifying power 35 with its eye-piece of focal length 10 cm. Assume that the final image is at least distance of distinct vision then the magnification produced by the objective is

Options:
A - 4

B 5
C 10
D - 10

The correct answer is:
- 10

Solution:
For a compound microscope, the magnifying power is given by
MP=meÃm0
When the final image is at least distance of distance vision then
Me=1+Dfe
So, MP=m01+Dfe
â -35=m01+2510
â -35=m0Ã3.5
â m0=- 10
The negative sign indicates that the image formed by objective is inverted.

Topic: Physics, Ray Optics, NEET
A compound of sulphur possesses 8% sulphur by mass. Calculate the least molecular mass of compound.

Options:
A 300
B 400
C 200
D 255

The correct answer is:
400

Solution:
∵ 8 gm sulphur is present in 100 gm of substance
∴ 32gm sulphur will present =1008×32=400

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A compound of vanadium chloride has spin only magnetic moment of 1.73 BM. Its formula is

Options:
A VCl2
B VCl5
C VCl4
D VCl3

The correct answer is:
VCl4

Solution:
Spin only Magnetic Moment
Î¼ = n n + 2 BM = 1 . 7 3 = 3
â´ n = 1 (unpaired e-)
Electronic configuration of V= 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 4s 2
to get 1 unpaired electron V has to loose 4 electron to form V 4+.
So, formula of chloride is VCl 4

Topic: Chemistry, d and f Block Elements, NEET
A compound that easily undergoes bromination is-

Options:
A Phenol
B Toluene
C Benzene
D Benzoic acid

The correct answer is:
Phenol

Solution:
A compound that undergoes bromination easily is phenol. Due to presence of â OH(+M) group the ring becomes much more active in substitution reactions. The bromination occurs due to availability of electrons on ortho and para position.
In toluene CH3 group show +HC which is less activating group than -OH & -COOH (-M) ring deactivating group.
more activating group given E + substitution (bromination) easily.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
A compound that gives a positive iodoform test is

Options:
A 1- pentanol
B 3-pentanone
C 2- pentanone
D pentanal

The correct answer is:
2- pentanone

Solution:
For iodoform test, group is required

Topic: Chemistry, Aldehydes and Ketones, NEET
A compound was found to contain nitrogen 28 g and oxygen 80 g. The formula of the compound is (N = 14, O = 16)

Options:
A NO
B N2O3
C N2O5
D N2O4

The correct answer is:
N2O5

Solution:
Moles of N=2814=2
Moles of O=8016=5
So, the formula of oxide will be N2O5

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A compound which is a strong oxidizing agent and has orange coloured crystal. It is used in the preparation of azo compounds. Identify the compound:

Options:
A Hydrogen peroxide
B Potassium permanganate
C Sodium chromate
D Potassium dichromate

The correct answer is:
Potassium dichromate

Solution:
Potassium dichromate K 2 Cr 2 O 7 is an orange-red crystalline soluble solid substance. It is a good oxidizing agent and is used in making chrome pigments,azo dyes and as a bleaching agent.

Topic: Chemistry, d and f Block Elements, NEET
A compound with a molecular formula of $\mathrm{C}_6 \mathrm{H}_{14}$ has two tertiary carbons. Its IUPAC name is :

Options:
A 2-methylpentane
B 2,3-dimethylbutane
C 2,2-dimethylbutane
D n-hexane

The correct answer is:
2,3-dimethylbutane

Solution:

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2024
A compound with molecular mass 180 is acylated with CH3COCl to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is

Options:
A 1
B 3
C 5
D 4

The correct answer is:
5

Solution:
R-NH2M+CH3COCl→RNHCOCH3M+42+HCl
If the increase in molecular mass is 42, the reactant has one NH2 group.
Actual increase in molecular mass =390-180=210
Number of NH2 groups =21042=5

Topic: Chemistry, Carboxylic Acid Derivatives, NEET
A compound 'X' on treatment with Br2/NaOH , provided C3H9N (Y) which gives positive carbylamine test. Compound 'X' is

Options:
A CH3COCH2NHCH3
B CH3CH2COCH2NH2
C CH3CH2CH2CONH2
D CH3CON(CH3)2

The correct answer is:
CH3CH2CH2CONH2

Solution:
X→Br2/NaOHC3H9NY
As Y gives +ve carbylamine test so it is a primary amine and X is an amide which on Hoffmann's bromamide degradation gives amine Y.
CH 3 CH 2 CH 2 CONH 2 = X → Br 2 +NaOH CH 3 CH 2 CH 2 NH 2
Option (2) is not possible as it is not an amide. (1,4) are not possible as they will not give 1° amine on degradation.

Topic: Chemistry, Amines, NEET
A compound ' $\mathrm{X}^{\prime}$ upon reaction with $\mathrm{H}_2 \mathrm{O}$ produced a colourless gas ' $Y$ ' with rotten fish smell. Gas ' $\mathrm{Y}$ ' is absorbed in a solution of $\mathrm{CuSO}_4$ to give $\mathrm{Cu}_3 \mathrm{P}_2$ as one of the products. Predict the compound ' $\mathrm{X}$ '

Options:
A $\mathrm{Ca}_3 \mathrm{P}_2$
B $\mathrm{NH}_4 \mathrm{Cl}$
C $\mathrm{As}_2 \mathrm{O}_3$
D $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$

The correct answer is:
$\mathrm{Ca}_3 \mathrm{P}_2$

Solution:
$$
\text { The given road map problem is }
$$
As compound ' $\mathrm{X}$ ' produces a colorless gas ' $\mathrm{Y}$ ' with rotten fish smell, it can be $\mathrm{Ca}_3 \mathrm{P}_2$ (calcium phosphide). The reaction takes palce as follows :
$$
\mathrm{Ca}_3 \mathrm{P}_2+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Ca}(\mathrm{OH})_2+\mathrm{PH}_3
$$
Colorless gas with rotten fish smell. ' $\mathrm{Y}$ '
$\mathrm{CuSO}_4+\underset{\mathrm{Y}^{\prime}}{\mathrm{PH}_3} \longrightarrow \mathrm{Cu}_3 \mathrm{P}_2+\mathrm{H}_2 \mathrm{SO}_4$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET, NEET 2019 (Odisha)
A compound Z with molecular formula C3H9N reacts with C6H5SO2Cl to give a solid, insoluble in alkali. Identify Z.

Options:
A Methylamine
B Ethylmethylamine
C Ethylamine
D Methoxyamine

The correct answer is:
Ethylmethylamine

Solution:
Correct Option is : (B)
Ethylmethylamine

Topic: Chemistry, Amines, NEET
A concave lens of focal length 20 cm product an image half in size of the real object. The distance of the real object is

Options:
A 20 cm
B 30 cm
C 10 cm
D 60 cm

The correct answer is:
20 cm

Solution:
We know that
m=ff+u
12=-20-20+u or-20+u=-40
Or u=-40+20 or u=-20 cm

Topic: Physics, Ray Optics, NEET