datavorous/entrance-exam-dataset huggingface dataset:
Primary amines on heating with \( \mathrm{CS}_{2} \) are followed by a reaction with excess of mercuric chloride yields isothiocyanates. This reaction is called

Options:
A Hofmann mustard oil reaction.
B Perkin reaction
C Fries reaction
D Diels-Alder reaction

The correct answer is:
Hofmann mustard oil reaction.

Solution:
Primary amine undergoes nucleophilic addition reaction on carbon disulphide gas and forms a compound.

This compound on reaction with mercuric chloride forms ethyl isothiocyanate and gives the black coloured precipitate of HgS.

Isothiocyanate has a smell similar to mustard oil .

Secondary and tertiary amines do not give black coloured precipitates.

Topic: Chemistry, Amines, JEE Main
Primary coil of a transformer is connected to 220 V AC. Primary and secondary turns of the transforms are 100 and 10 respectively. Secondary coil of transformer is connected to two series resistances as shown in figure. The output voltage V0 is :

Options:
A 7 V
B 15 V
C 44 V
D 22 V

The correct answer is:
7 V

Solution:
Ratio of EMF developed will be given by,

εpεs=NpNs=10010⇒εs=22 V

Therefore, current in secondary circuit will be

I=2222×103=1 mA

Hence, potential drop across 7 kΩ will be V0=I×7 kΩ=7 V

Topic: Physics, Alternating Current, JEE Main, JEE Main 2024 (30 Jan Shift 1)
Primary roots and its branches constitute the

Options:
A Tap root system
B Adventitious root system
C Tertiary root system
D Fibrous root system

The correct answer is:
Tap root system

Solution:
The tap root system consists of a tap root, secondary roots, tertiary roots and rootlets. The first root produced from seed is called radicle which grows up directly into the main or primary root. In dicotyledonous plants this root becomes more prominent and is known as tap root and many-branched roots arise from this by forming a tap root system.

Topic: Biology, Morphology of Flowering Plants, NEET
Primary, secondary and tertiary amines can be separated using:

Options:
A Para-Toluene sulphonyl chloride
B Chloroform and KOH
C Benzene sulphonic acid
D Acetyl amide

The correct answer is:
Para-Toluene sulphonyl chloride

Solution:
Primary amines react with Para-toluene sulphonyl chloride to form a precipitate that is soluble in NaOH.

Secondary amines reacts with Para-toluene sulphonyl chloride to give a precipitate that is insoluble in NaOH.

Tertiary amines do not react with Para-toluene sulphonyl chloride.

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (17 Mar Shift 2)
\( { }^{\prime} \mathrm{X}^{\prime} \) is an alkaline earth metal. It imparts brick red colour to the Bunsen flame. Its oxide is basic in nature. The atomic
number of \( ^{\prime} \mathrm{X}^{\prime} \) is

Options:

The correct answer is:
20

Solution:
Element 'X' is calcium. Calcium imparts brick red colour to the Bunsen flame. Atomic number of calcium is 20 .

Topic: Chemistry, s Block Elements, JEE Main
Principal, azimuthal and magnetic quantum numbers are respectively related to

Options:
A Size, shape and orientation
B Shape, size and orientation
C Size, orientation and shape
D None of the above

The correct answer is:
Size, shape and orientation

Solution:
$n, l$ and $m$ are related to size, shape and orientation respectively.

Topic: Chemistry, Structure of Atom, JEE Main
Probability that A will pass the exam is 14, B will pass the exam is 25 and C will pass the exam is 23. The probability that exactly one of them will pass the exam is

Options:
A 25
B 320
C 920
D 45

The correct answer is:
920

Solution:
PA=14,PB=25,PC=23
∵A,B,C are independent events, then
the probability that exactly one of them passes
=PA∩BC∩CC+PAC∩B∩CC+PAC∩BC∩C
=PA⋅PBC⋅PCC+PAC⋅PB⋅PCC+PACPBC⋅PC
=14×35×13+34×25×13+34×35×23=920

Topic: Mathematics, Probability, JEE Main
Product A and B formed in the following set of reactions are:

Options:
A
B
C
D

The correct answer is:

Solution:
The reaction is known as hydroboration-oxidation. In this reaction, B2H6 adds to the double bond of propene, resulting in the formation of an organoborane intermediate. Subsequent treatment with H2O2 converts the organoborane intermediate to an alcohol. Hydration follows anti-Markownikoff's rule.

In this hydration reaction, 1-methylcyclohexene (an alkene) is reacted with aqueous H3O+ (formed from water and a strong acid such as H2SO4), resulting in Markovnikov addition of water across the pi bond. The product is an alcohol.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Product 'A' in the reaction is:
\(\mathrm{CH}_3 \mathrm{CO}-\mathrm{CH}_2 \mathrm{COOC}_2 \mathrm{H}_5 \xrightarrow{\mathrm{NaOH}, \mathrm{H}_2 \mathrm{O}} A+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+\mathrm{CO}_2\)

Options:
A \(\mathrm{CH}_3 \mathrm{COOH}\)
B \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\)
C \(\mathrm{CH}_3 \mathrm{COCH}_3\)
D \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{CHO}\)

The correct answer is:
\(\mathrm{CH}_3 \mathrm{COCH}_3\)

Solution:
Ketonic hydrolysis:
\(\mathrm{CH}_3 \mathrm{CO}-\mathrm{CH}_2 \mathrm{COOC}_2 \mathrm{H}_5 \xrightarrow{\mathrm{NaOH}, \mathrm{H}_2 \mathrm{O}} \mathrm{CH}_3 \mathrm{COCH}_3+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+\mathrm{CO}_2\)

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Product (A) is

Options:
A No reaction
B
C
D

The correct answer is:

Solution:
It occurs as follows

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Product 'A' of following sequence of reactions is

Options:
A
B
C
D

The correct answer is:

Solution:
In the first step ethyl benzene undergoes electrophilic substitution reaction to form 1-Bromo-4-ethylbenzene.

In the next step 1-Bromo-4-ethylbenzene undergoes free radical substitution at benzylic position followed by dehydrohalogenation(last step) in presence of alcoholic KOH to form 4-bromo styrene.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Product [B] is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Product B is:

Options:
A
B
C
D

The correct answer is:

Solution:
A must be an aldehyde

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Product 'C' is

Options:
A
B
C
D

The correct answer is:

Solution:
It undergoes intramolecular aldol condensation

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Product \(\mathrm{P}\) is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (04 Apr Shift 2)
Product obtain on treatment of ammonia with excess of ethyl chloride is

Options:
A Diethyl amine
B Ethane
C Tetraethyl ammonium chloride
D Methyl amine

The correct answer is:
Tetraethyl ammonium chloride

Solution:
If NH3 is in excess, then 1o amine will be the main product, if C2H5Cl is in excess then mixture of 1o,2o,3o and quaternary amine is obtained.

Topic: Chemistry, Amines, NEET
Product of any $r$ consecutive natural numbers is always divisible by

Options:
A $\mathrm{r}$ !
B $(r+4) !$
C $(r+1) !$
D $(r+2) !$

The correct answer is:
$\mathrm{r}$ !

Solution:
Hints : $(n+1)(n+2) \ldots \ldots .(n+r)$
$$
\begin{aligned}
& =\frac{(n+r) !}{n !} \\
& =\frac{(n+r) !}{n ! r !} r !=r !^{n+r} C_n
\end{aligned}
$$

Topic: Mathematics, Permutation Combination, JEE Main
Product of real roots of the equation $t^2 x^2+|x|+9=0$

Options:
A
is always positive

B
is always negative

C
does not exist

D
none of these

The correct answer is:
does not exist

Solution:
No Solution Available

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2002
Product of the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Product should be

Options:
A
B
C
D

The correct answer is:

Solution:
It is an example of internal Cannizzaro reaction; where -CHO is oxidized while ketonic group of the same molecule is reduced. Since keto group can't be oxidized, alcohol corresponding to -CHO group will not be formed.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Product (X and Y) of the following reactions (1 and 2) are

(1) 2NaOH(Cold and dilute)+Cl2âNaCl+X+H2O

(2) 6NaOH(Hot and conc.)+3Cl2âNaCl+Y+3H2O

Options:
A X=NaClO3 and Y=NaOCl
B X=NaOCl and Y=NaClO3
C X=NaHClO3 and Y=NaOCl
D X=NaClO3 and Y=NaHClO3

The correct answer is:
X=NaOCl and Y=NaClO3

Solution:
Cl2+2NaOH(dil.)→coldNaCl+NaOCl(X)+H2O
3Cl2+6NaOH(conc.)→hot5NaCl+NaClO3(Y)+3H2O

Topic: Chemistry, Redox Reactions, JEE Main
Product (X) is

Options:
A
B
C
D

The correct answer is:

Solution:
Here Benzene is formed as follows

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Products A and B formed in the following reactions are respectively:

Options:
A
B
C
D

The correct answer is:

Solution:
Initially compound reacts with nitrous acid to form diazonium salt, which undergo diazo coupling reaction with aniline

Topic: Chemistry, Amines, JEE Main
Products A and B formed in the following reactions are respectively:

Options:
A
B
C
D

The correct answer is:

Solution:
Initially compound reacts with nitrous acid to form diazonium salt, which undergo diazo coupling reaction with aniline

Topic: Chemistry, Amines, JEE Main, JEE Main 2018 (16 Apr Online)
Products of the reaction which is given below will be

Options:
A Vic diol
B Vic dicarbonyl compound
C and CH3–CHO
D and CH3–COOH

The correct answer is:
and CH3–CHO

Solution:
Here acetone and acetaldehyde formed as follows

Topic: Chemistry, Hydrocarbons, JEE Main
Projectiles A and B are thrown at angles of 45° and 60° with vertical respectively from top of a 400 m high tower. If their times of flight are same, the ratio of their speeds of projection vA:vB is:

Options:
A 1:3
B 2:1
C 1: 2
D 1:2

The correct answer is:
1:2

Solution:
For same time of flight, vertical component of velocity of projection must be same. Therefore,

vAsin45°=vBsin30°⇒vAvB=1212⇒vAvB=12

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Prolonged __________X_________ leads to leads to a complex disorder called diabetes mellitus, associated with loss of ___________ through urine. Select the correct option for X and Y.

Options:
A X- hyperglycemia; Y- glucose
B X- hypoglycemia; Y- glucose
C X- hyperglycemia; Y- ketone bodies
D both a and c

The correct answer is:
both a and c

Solution:
Prolonged hyperglycemia is a condition in which blood glucose remains high in a person that leads to a complex disorder called diabetes mellitus. It is associated with the loss of glucose and harmful compounds ketone bodies through urine.

Topic: Biology, Chemical Coordination and Integration, NEET
Propan-1-ol can be prepared from propene

Options:
A H 2 O/H 2 SO 4
B Hg(OAc) 2 ,  H 2 Ofollowed by NaBH 4
C B 2 H 6 followed by H 2 O 2
D CH 3 COOH,  H 2 SO 4

The correct answer is:
B 2 H 6 followed by H 2 O 2

Solution:
Propan-1-ol can be prepared from propene by using B 2 H 6 followed by H 2 O 2
6CH 3 −CH =  CH 2  +  B 2 H 6 → H 2 O 2 CH 3 − CH 2 − CH 2 OH

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Propene, $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2$ can be converted to 1-propanol by oxidation. Which set of reagents among the following is ideal to effect the conversion

Options:
A Alkaline $\mathrm{KMnO}_4$
B $\mathrm{B}_2 \mathrm{H}_6$ and alkaline $\mathrm{H}_2 \mathrm{O}_2$
C $\mathrm{O}_3 / \mathrm{Zn}$ dust
D $\mathrm{OsO}_4 / \mathrm{CH}_4, \mathrm{Cl}_2$

The correct answer is:
$\mathrm{B}_2 \mathrm{H}_6$ and alkaline $\mathrm{H}_2 \mathrm{O}_2$

Solution:
Hydroboration oxidation (Industrial preparation of alcohol)
$\begin{aligned} & 3 \mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2+\frac{1}{2} \mathrm{~B}_2 \mathrm{H}_6 \xrightarrow[\text { ether }]{\text { Dry }}\left(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3\right)_3 \mathrm{~B} \\ &\left(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3\right)_3 \mathrm{~B} \xrightarrow{\mathrm{H}_2 \mathrm{O}_2} 3 \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{OH}\end{aligned}$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Propene on reaction with hypochlorous acid gives?

Options:
A
B
C
D

The correct answer is:

Solution:
Propene on reaction with hypochlorous acid gives
according to Markovnikov's rule.

Topic: Chemistry, Hydrocarbons, JEE Main
Property of the alkaline earth metals that increases with their atomic number

Options:
A solubility of their hydroxides in water
B solubility of their sulphates in water
C ionisation energy
D electronegativity

The correct answer is:
solubility of their hydroxides in water

Solution:
Electronegativity as well as ionisation energy both usually decrease on moving downward a group with increase in atomic number. The hydroxides and sulphates of alkaline earth metals are ionic solids and the solubility of ionic solids is governed by two factors, viz, lattice energy and hydration energy. For solubility, hydration energy > lattice energy.
Hydration energy varies inversely with size, ie, decreases with increase in size. However, lattice energy in case of sulphates, remains almost same with increase in the atomic number of alkaline earth metals, due to large size of sulphate ion. Hence, hydration energy only governs the solubility in this case. Thus, solubility of alkaline earth metal sulphates decreases as the hydration energy decreases on moving downward the II A group.
On the other hand, in case of hydroxides, the lattice energies are different because of medium size of hydroxide ions, and decreases on moving from $\mathrm{Be}$ to $\mathrm{Ba}$. This tends to increase the solubility and to overcome the counter-effect produced by the decrease in hydration energy. Hence, the solubility of alkaline earth metal hydroxides increases with increase in the atomic number of alkaline earth metals.

Topic: Chemistry, s Block Elements, NEET, NEET 2010 (Screening)
Property of the colloidal solution is due to

Options:
A Nature of dispersed phase
B Nature of dispersion medium
C Physical state of dispersed phase
D Temperature of the system

The correct answer is:
Physical state of dispersed phase

Solution:
Correct option is 3. Physical state of dispersed phase
Properties of the colloidal solution depend upon the physical state of the dispersed phase and molecular weight.

Topic: Chemistry, Surface Chemistry, JEE Main
Propionic acid is subjected to reduction with hydroiodic acid in the presence of a little $P$, the product formed is

Options:
A Ethane
B Propane
C Butane
D None of these

The correct answer is:
Propane

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Propionic acid is subjected to reduction with hydroiodic acid in the presence of a little P, the product formed is

Options:
A Ethane
B Propane
C Butane
D None of these

The correct answer is:
Propane

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Propionic acid with $\mathrm{Br}_2-\mathrm{P}$ yields a dibromo product. Its structure would be

Options:
A $\mathrm{CH}_2 \mathrm{Br}-\mathrm{CHBr}-\mathrm{COOH}$
B

C $\mathrm{CH}_2 \mathrm{Br}-\mathrm{CH}_2-\mathrm{COBr}$
D

The correct answer is:

Solution:
Key Idea $\mathrm{Br}_2$ in presence of $\mathrm{P}$ attacks on $\alpha$-carbon atom of acid and $\alpha$ hydrogen is substituted by bromine.
This reaction is called Hell-Volhard Zelinsky (HVZ) reduction.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Propionic acid with $\mathrm{Br}_2 / \mathrm{P}$ yields a dibromo product, Its structure would be -

Options:
A

B $\mathrm{CH}_2 \mathrm{Br}-\mathrm{CHBr}-\mathrm{COOH}$
C

D $\mathrm{CH}_2 \mathrm{Br}-\mathrm{CH}_2-\mathrm{COBr}$

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, NEET, NEET 2009 (Mains)
Proteins are built up of

Options:
A Dicarboxylic acids
B Amino acids
C Alcohols
D Hydroxy acids

The correct answer is:
Amino acids

Solution:
Proteins are polymers of amino acids.
Amino acid $\rightarrow$ Dipeptide $\rightarrow$ Polypeptide $\rightarrow$ Protein.

Topic: Chemistry, Biomolecules, JEE Main
Proteins are hydrolysed by enzymes into

Options:
A Dicarboxylic acids
B Hydroxy acids
C Amino acids
D Aromatic acids

The correct answer is:
Amino acids

Solution:
Protein $\xrightarrow{\text { enzyme }} \underset{\text { (Acidic medimm in stomach) }}{\text { Amino acid }}$

Topic: Chemistry, Biomolecules, JEE Main
Proton, Deuteron and alpha particle of the same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particle are respectively $r_p, r_d$ and $r_\alpha$. Which one of the following relations is correct?

Options:
A
$r_\alpha=r_p=r_d$

B
$r_\alpha=r_p < r_d$
C
$r_\alpha>r_d>r_p$

D
$r_\alpha=r_d>r_p$

The correct answer is:
$r_\alpha=r_p < r_d$

Solution:
$r=\frac{\sqrt{2 m K}}{B q}$
$r \propto \frac{\sqrt{m}}{q}$
$r_\alpha=r_p < r_d$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2012 (Offline)
Proton (p) and electron (e) will have same deBroglie wavelength when the ratio of their momentum is (assume, $m_p=1849 \mathrm{~m}_{\mathrm{e}}$ )

Options:
A $1: 43$
B $43: 1$
C $1: 1849$
D $1: 1$

The correct answer is:
$1: 1$

Solution:
De Broglie wavelength is $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$
$\lambda_{\mathrm{p}}=\lambda_{\mathrm{e}} \Rightarrow \frac{\mathrm{h}}{\mathrm{m}_{\mathrm{p}} \mathrm{v}_{\mathrm{p}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{e}}}$
$\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{e}}=\mathrm{m}_{\mathrm{p}} \mathrm{v}_{\mathrm{p}} \Rightarrow \mathrm{p}_{\mathrm{e}}=\mathrm{p}_{\mathrm{p}}$
$\therefore \frac{\mathrm{p}_{\mathrm{p}}}{\mathrm{p}_{\mathrm{e}}}=\frac{1}{1}$

Topic: Physics, Dual Nature of Matter, JEE Main
Proton (P) and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume, mp=1849 me)

Options:
A 1 : 1
B 1 : 1849
C 1 : 43
D 43 : 1

The correct answer is:
1 : 1

Solution:
The De Broglie wavelength is λ=hmv

The momentum is given by p=mv

The momentum of proton is pp=mpvp

The momentum of electron is p'=meve

Since h is the Planck's constant and it is given that both wavelengths are same, i.e.

λp=λe, so pp=p'.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Proton with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of 1012 m/s2 by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is 1.6×10-27kg )

Options:
A 0.71mT
B 7.1mT
C 0.071mT
D 71mT

The correct answer is:
0.71mT

Solution:
∴K.E.=1.6×10-13=12×1.6×10-27v2
v=2×107m/s
∴Bqv=ma
B=1.6×10-27×10121.6×10-19×2×107
=0.71×10-3T
So,

B=0.71 mT

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (08 Jan Shift 1)
Protozoa are classified on the basis of:

Options:
A Mode of reproduction
B Locomotory organelles
C Mode of nutrition
D None of these

The correct answer is:
Locomotory organelles

Solution:
Protozoa include all single-celled, eukaryotic organisms without a cell wall and they also show the heterotrophic mode of nutrition, but main reason for classifying these organisms into separate phylum is due to presence of locomotory organelles. Amoeboid protozoa form cytoplasmic extensions called pseudopodia for locomotion. Trypanosoma, Trichomonas vaginalis and Leishmania donovani possess flagella for locomotion and Ciliated Protozoans (Paramoecium) moves with the help of cilia.

Topic: Biology, Biological Classification, NEET
Prove that, if an insulated, uncharged conductor is placed near a charged conductor and no other conductors are present, the uncharged body must intermediate in potential between that of the charged body and that of infinity.

Options:

None

Solution:
The main concept used is the electric field $\mathrm{E}=\frac{\mathrm{dV}}{\mathrm{dr}}$ that suggest the electric potential decreases along the direction of electric field.
Consider a charged conductor and an insulated uncharged conductor.
Now take any path from the charged conductor to the uncharged conductor along the direction of electric field. So, the electric potential decrease along this path from a charge conductor body to infinity.
Now, another path from the uncharged conductor to infinity will again continually lower the potential further. This ensures that the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

Topic: Physics, Electrostatic Potential and Capacitance
Prove that the Greatest Integer Function $f: R \rightarrow$ R given by $f(x)=[x]$, is neither one-one nor onto, where $[\mathbf{x}]$ denotes the greatest integer less than or equal to $x$.

Options:

None

Solution:
$f: R \rightarrow R$ given by $f(x)=[x]$
(a) $f(1.2)=1$, $f(1.5)=1 \Rightarrow f$ is not one-one
(b) All the images of $x \in R$ belonging to its domain have integers as the images in codomain. But no fraction
proper or improper belonging to codomain of $f$ has any pre-image in its domain.
$\Rightarrow \mathrm{f}$ is not onto.

Topic: Mathematics, Relations and Functions (Class 12)
Provide the systematic name of the compound shown

.

Options:
A 4-Butyl-2-ethyl-1-methylcycloheptane
B 1-Butyl-4-ethyl-3-methylcycloheptane
C 2-Butyl-4-ethyl-1-methylcycloheptane
D 4-Butyl-1-ethyl-2-methylcycloheptane

The correct answer is:
4-Butyl-1-ethyl-2-methylcycloheptane

Solution:
The name of the compound is 4-Butyl-1-ethyl-2-methylcycloheptane.

Topic: Chemistry, General Organic Chemistry, JEE Main
$\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_4\right] \mathrm{Cl}_2$ is

Options:
A Square planar
B Tetrahedral
C Pyramidal
D Pentagonal

The correct answer is:
Square planar

Solution:
Since hybridisation is $d s p^2$ so it is square planar,

Topic: Chemistry, Coordination Compounds, JEE Main
Pt(s)H2( g)(1bar)H+(aq)(1M)M3+(aq),M+(aq)Pt(s)

The Ecell for the given cell is 0.1115 V at 298 K

When M+(aq)M3+(aq)=10a

The value of a is ___________

Given : Eθ=M3+/M+0.2 V

2.303RTF=0.059 V

Options:

The correct answer is:
3

Solution:
Overall cell reaction :-

H2( g)+M(aq)3+⟶M(aq)++2H(aq)+

Using Nernst equation:
ECell =ECathode o-Eanode o-0.0592logM+×12M+31

⇒0.1115=0.2-0.0592logM+M+3

⇒3=logM+M+3

∴a=3

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Pure ammonia is placed in a vessel at temperature where its dissociation constant $\alpha)$ is appreciable. At equilibrium

Options:
A $K_p$ does not chan ge significantly with pressure
B $\alpha$ does not change with pressure
C Concentration of $\mathrm{NH}_3$ does not change with pressure
D Concentration of $\mathrm{H}_2$ is less than that of $\mathrm{N}_2$

The correct answer is:
$K_p$ does not chan ge significantly with pressure

Solution:
$K_p$ is constant and does not change with pressure.

Topic: Chemistry, Chemical Equilibrium, JEE Main
Pure inductance of $3.0 \mathrm{H}$ is connected as shown below. The equivalent inductance of the circuit is

Options:
A $1 \mathrm{H}$
B $2 \mathrm{H}$
C $3 \mathrm{H}$
D $9 \mathrm{H}$

The correct answer is:
$1 \mathrm{H}$

Solution:
The inductances are in parallel $\Rightarrow$
$L_{e q}=\frac{L}{3}=\frac{3}{3}=1 \mathrm{H}$

Topic: Physics, Electromagnetic Induction, JEE Main
Pure N2 can be obtained by heating

Options:
A NH4NO3
B BaN32
C NH42Cr2O7
D NH3 with CuO

The correct answer is:
BaN32

Solution:
Metal nitrides on heating give pure nitrogen
BaN32→ΔBa+3N2
Ammonium nitrate on heating gives ammonia and not nitrogen
NH4NO3→ΔN2O+2H2O
NH42Cr2O7→ΔN2impure+Cr2O3+4H2O

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Pure Si at $500 \mathrm{~K}$ has equal number of electron $\left(n_e\right)$ and hole $\left(n_h\right)$ concentrations of $1.5 \times 10^{16} \mathrm{~m}^{-3}$. Doping by indium increases $n_h$ to $4.5 \times 10^{22} \mathrm{~m}^{-3}$. The doped semiconductor is of

Options:
A $n$-type with electron concentration $n_e=5 \times 10^{22} \mathrm{~m}^{-3}$
B $p$-type with electron concentration $n_e=2.5 \times 10^{10} \mathrm{~m}^{-3}$
C n-type with electron concentration $n_e=2.5 \times 10^{23} \mathrm{~m}^{-3}$
D p-type having electron concentration $n_e=5 \times 10^9 \mathrm{~m}^{-3}$

The correct answer is:
p-type having electron concentration $n_e=5 \times 10^9 \mathrm{~m}^{-3}$

Solution:
$$
\begin{aligned}
& n_i^2=n_e n_h \\
& n_e=\frac{\left(n_i\right)^2}{n_h} \\
& n_e=\frac{\left(1.5 \times 10^{16}\right)^2}{\left(4.5 \times 10^{22}\right)} \\
& n_e=5 \times 10^9 \mathrm{~m}^{-3}
\end{aligned}
$$
So, $n_{f_1} \gg n_e$ semiconductor is $p$-type.

Topic: Physics, Rotational Motion, JEE Main