datavorous/entrance-exam-dataset huggingface dataset:
The number of ways of selecting 15 teams from 15 men and 15 women, such that each team consists of a man and a woman, is

Options:
A 1960
B 15!
C 15!2
D 14!

The correct answer is:
15!

Solution:
First team can be selected in 15C1.15C1 way second in 14C114C1 way's and so on
So number of ways of selections 15 teams =15!215!=15!

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways of selecting two distinct numbers from the first 15 natural numbers such that their sum is a multiple of 5, is equal to

Options:
A 20
B 36
C 21
D 16

The correct answer is:
21

Solution:
x+y=5,10,15,20,25,30
Where x≠y,x,y=y,x
The number of solutions of x+y=5 is 42!
The number of solutions of x+y=10 is 9-12!
The number of solutions of x+y=15 is 142!
The number of solutions of x+y=20 is 11-12!
∵x=5,6,…,15
The number of solutions of x+y=25 is 62!
∵x=10,11,…,15
The number of solutions of x+y=30 is 0
∵x=y=15
So, the required number of solutions
=2+4+7+5+3+0=21

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways three different natural numbers can be drawn from the set 1,2,3,4,…,10, if minimum of the chosen numbers is smaller than 4, is

Options:

The correct answer is:
85

Solution:
The selection must have atleast one element of the set 1,2,3
i.e. the required number of ways =C13⋅C27+C23⋅C17+C33⋅C07
=63+21+1=85

Topic: Mathematics, Permutation Combination, JEE Main
The number of ways to fill each of the four cells of the table with a distinct natural number, such that the sum of the
numbers is \( 10 \) and the sum of the numbers places diagonally are equal is

Options:
A \( 2 ! \times 2 ! \times 2 ! \)
B \( 4 ! \)
C \( 2(4 !) \)
D \( 2 ! \)

The correct answer is:
\( 2 ! \times 2 ! \times 2 ! \)

Solution:
The natural numbers are 1,2,3 and 4 (any others would lead to the sum exceeding 10)

Clearly, in one diagonal we have to place 1,4 And in the other 2,3

So we first select a diagonal, and then we arrange the numbers in its boxes

Number of ways of selection of the diagonal =2!

The number of arrangements of numbers=2!×2!×2!=8.

Topic: Mathematics, Permutation Combination, JEE Main
The number of white coloured salts among the following is .......

A SrSO4 B MgNH4PO4 C BaCrO4

D Mn(OH)2 E PbSO4 F PbCrO4

G AgBr H PbI2 I CaC2O4

J Fe(OH)2CH3COO

Options:

The correct answer is:
5

Solution:
The colour of the salts are as follows,

SrSO4- white
MgNH4PO4 - white
BaCrO4 - yellow
Mn(OH)2 - white
PbSO4 - white
PbCrO4 - yellow
AgBr - pale yellow
PbI2 - yellow
CaC2O4 - white

Fe(OH)2CH3COO – Brown Red

Hence the number of salts which are white in colour is 5.

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2024 (01 Feb Shift 1)
The number of words of 4 letters which can be formed with the letters of the word MATHEMATICS is α, then α300 is

Options:

The correct answer is:
8.18

Solution:
M,A,Tâ2
H,E,I,C,Sâ1

Type
Combination
Permutation

all different
8C4
8P4=1680

2 same

2 different
3C1Ã7C1
3C1Ã7C1Ã4!2!=756

2 same

2 same
3C2
3C2Ã4!2!=18

Total number of words =2454
2454300=8.18

Topic: Mathematics, Permutation Combination, JEE Main
The number of words that can be formed by using the letters of the word CALCULATE such that each word starts and ends with a consonant, are

Options:
A $5 \times 7 !$
B $\frac{9 !}{8}$
C $\frac{5 \times 7 !}{2}$
D $20 \times 7 !$

The correct answer is:
$\frac{5 \times 7 !}{2}$

Solution:
Word CALCULATE has 9 letters.
Out of which ' $C$ ' repeats 2 times,
'A' repeats 2 times,
' $L$ ' repeats 2 times,
'E', 'U' and ' $T$ ' repeats once.
$\therefore \quad$ There are 5 consonants and 4 vowels.
Two consonants out of 5 can take start and end position of the word in ${ }^5 \mathrm{P}_2$ ways.
And remaining 7 letters can take remaining 7 positions in 7 ! ways.
Also, 'C', 'A' and ' $L$ ' repeats twice each.
$\therefore \quad$ The required number of words that can be formed $=\frac{5 p_2 \times 7 !}{2 ! \times 2 ! \times 2 !}=\frac{5 \times 4 \times 3 ! \times 7 !}{3 ! \times 2 \times 2 \times 2}=\frac{5 \times 7 !}{2}$

Topic: Mathematics, Permutation Combination, MHT CET, MHT CET 2023 (12 May Shift 2)
The number of words which can be formed from the letters of the word MAXIMUM, if two consonants cannot occur together, is

Options:
A $4$ !
B $3!\times 4!$
C $7$ !
D None of these

The correct answer is:
$4$ !

Solution:
- $A \cdot 1 \cdot U \cdot$
The pointed places to be filled by $M \times M M$.
Hence required number of ways $3!\times \frac{4!}{3!}=4$ !
\{Since three vowels can be arranged in 3 ! ways also\}.

Topic: Mathematics, Permutation Combination, JEE Main
The number of words which can be made out of the letters of the word 'MOBILE' when consonants occupy odd places is

Options:
A $20$
B $36$
C $30$
D $720$

The correct answer is:
$36$

Solution:
There are 3 vowels and 3 consonants
$\therefore$ Number of ways to fill and arrange odd places
with consonants $={ }^3 C_3 \cdot 3$ !
and number of ways to fill remaining 3 places $=3$ !
$\therefore$ The number of words $={ }^3 C_3 \cdot 3 ! \cdot 3 !=36$

Topic: Mathematics, Permutation Combination, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 2)
The number of words, with or without meaning, that can be formed by taking 4 letters at a time from the letters of the word 'SYLLABUS' such that two letters are distinct and two letters are alike, is

Options:

The correct answer is:
240

Solution:
From the given word, S-2,L-2,A,B,Y,U.

Number of alike letter pairs =2

Number of ways to select one pair=C12

Number of distinct letters=5

Number of ways to select two distinct letters=C25

Number of ways to arrange four letters if two are alike=4!2!

Required number of ways =C12C25·4!2!=2·10·242=240

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2020 (05 Sep Shift 1)
The number of words (with or without meaning) that can be formed from all the letters of the word"LETTER" in which vowels never come together is.....

Options:

The correct answer is:
120

Solution:
Consonants are L, T, T, R

Vowels are E, E,

Total number of words (without meaning) from letters of word 'LETTER'=6!2!2!=180

Total number of words (with or without meaning) from letters of word 'LETTER' if vowels are together =5!2!=60

∴ Required =180-60=120

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The number of words, with or without meaning, that can be formed using all the letters of the word ASSASSINATION so that the vowels occur together, is _____ .

Options:

The correct answer is:
50400

Solution:
In the word ASSASSINATION, we will treat all the vowels that is AAAIIO as one letter only.

Now, we have AAAIIOSSSSNNT

This has 8=(7 + 1) letters of which S occurs 4 times, and N occur 2 times,

Thus, the number of ways of arranging these letters =8!4!2!=840.

Now, 6 vowels in which A occurs 3 times, and I occur two times, they can be arranged in 6!3!2!=60 ways.

Therefore, required number of ways= (840 × 60) = 50400.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (01 Feb Shift 1)
The number of x ∈0, 2π for which 2sin4x+18cos2x- 2cos4x+18sin2x=1 is:

Options:
A 2
B 6
C 4
D 8

The correct answer is:
8

Solution:
Given
2sin4⁡x+18cos2⁡x- 2cos4⁡x+18sin2⁡x=1
⇒2sin4⁡x+18cos2⁡x- 2cos4⁡x+18sin2⁡x= ±1
⇒2sin4⁡x+18cos2⁡x= ±1+ 2cos4⁡x+18sin2⁡x
By squaring both the sides, we get
2sin4⁡x+18cos2⁡x=1+2cos4⁡x+18sin2⁡x±22cos4⁡x+18sin2⁡x
⇒ 2sin4⁡x-cos4⁡x+18cos2⁡x-sin2⁡x=1±22cos4⁡x+18sin2⁡x
⇒ 2sin2⁡x-cos2⁡x+18cos2⁡x-sin2⁡x=1±22cos4⁡x+18sin2⁡x
⇒ 16cos2⁡x-sin2⁡x=1±22cos4⁡x+18sin2⁡x
⇒ 16cos⁡2x-1=±221+cos⁡2x22+91-cos⁡2x
Squaring both sides again, we get
256cos2⁡2x+1-32cos⁡2x=41+2cos⁡2x+cos2⁡2x2+91-cos⁡2x
256cos2⁡2x+1-32cos⁡2x=219-16cos⁡2x+cos2⁡2x
⇒ 254cos2⁡2x=37⇒ cos2⁡2x=37254⇒ cos⁡2x=±37254∈-1, 1
Since, the solutions lie in all four quadrants, there are 8 solutions which satisfy the equation.

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of x ∈0, 2π for which 2sin4x+18cos2x- 2cos4x+18sin2x=1 is:

Options:
A 2
B 6
C 4
D 8

The correct answer is:
8

Solution:
Given
2sin4⁡x+18cos2⁡x- 2cos4⁡x+18sin2⁡x=1
⇒2sin4⁡x+18cos2⁡x- 2cos4⁡x+18sin2⁡x= ±1
⇒2sin4⁡x+18cos2⁡x= ±1+ 2cos4⁡x+18sin2⁡x
By squaring both the sides, we get
2sin4⁡x+18cos2⁡x=1+2cos4⁡x+18sin2⁡x±22cos4⁡x+18sin2⁡x
⇒ 2sin4⁡x-cos4⁡x+18cos2⁡x-sin2⁡x=1±22cos4⁡x+18sin2⁡x
⇒ 2sin2⁡x-cos2⁡x+18cos2⁡x-sin2⁡x=1±22cos4⁡x+18sin2⁡x
⇒ 16cos2⁡x-sin2⁡x=1±22cos4⁡x+18sin2⁡x
⇒ 16cos⁡2x-1=±221+cos⁡2x22+91-cos⁡2x
Squaring both sides again, we get
256cos2⁡2x+1-32cos⁡2x=41+2cos⁡2x+cos2⁡2x2+91-cos⁡2x
256cos2⁡2x+1-32cos⁡2x=219-16cos⁡2x+cos2⁡2x
⇒ 254cos2⁡2x=37⇒ cos2⁡2x=37254⇒ cos⁡2x=±37254∈-1, 1
Since, the solutions lie in all four quadrants, there are 8 solutions which satisfy the equation.

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2016 (09 Apr Online)
The number of zeros at the end of 100! is

Options:
A 21
B 22
C 23
D 24

The correct answer is:
24

Solution:
$\mathrm{E}_5(100 !)=\left[\frac{100}{5}\right]+\left[\frac{100}{25}\right]+\left[\frac{100}{125}\right]=20+4+0=24$

Topic: Mathematics, Permutation Combination, JEE Main
The number which should be added to the numbers $2,14,62$ so that the resulting numbers may be in G.P., is

Options:
A $1$
B $2$
C $3$
D $4$

The correct answer is:
$2$

Solution:
Suppose that the added number be $x$
then $x+2, x+14, x+62$ be in G.P.
Therefore $(x+14)^2=(x+2)(x+62)$
$\Rightarrow \quad x^2+196+28 x=x^2+64 x+124$
$\Rightarrow 36 x=72 \Rightarrow x=2$
Trick : (a) Let 1 is added, then the numbers will be $3,15,63$ which are obviously not in G.P.
(b) Let 2 is added, then the numbers will be $4,16,64$ which are
obviously in G.P.

Topic: Mathematics, Sequences and Series, JEE Main
The numbers $a_n=6^n-5 n$ for $n=1,2,3, \ldots$ when divided by 25 leave the remainder

Options:
A 9
B 7
C 3
D 1

The correct answer is:
1

Solution:
Given, $a_n=6^n-5 n, n=1,2,3, \ldots$
We take; $6^n=(1+5)^n$
Expand with binomial expansion
$6^n={ }^n C_0+{ }^n C_1 5+{ }^n C_2 5^2+{ }^n C_3 5^3+\ldots$
$6^n=1+n \cdot 5+{ }^n C_2 25+{ }^n C_3 5^3+\ldots$
$\left(6^n-5 n\right)=1+25\left\{{ }^n C_2+{ }^n C_3 \cdot 5+\ldots\right\}$
$\left(6^n-5 n\right)=1+25 \cdot k$
where $k=$ positive integer.
Hence, $a_n=6^n-5 n$ divided by 25 and leave the remainder $=1$

Topic: Mathematics, Binomial Theorem, AP EAMCET, AP EAMCET 2010
The numbers $a_n=6^n-5 n$ for $n=1,2,3, \ldots$ when divided by 25 leave the remainder

Options:
A 9
B 7
C 3
D 1

The correct answer is:
1

Solution:
Given, $a_n=6^n-5 n, n=1,2,3, \ldots$
We take; $6^n=(1+5)^n$
Expand with binomial expansion
$6^n={ }^n C_0+{ }^n C_1 5+{ }^n C_2 5^2+{ }^n C_3 5^3+\ldots$
$6^n=1+n \cdot 5+{ }^n C_2 25+{ }^n C_3 5^3+\ldots$
$\left(6^n-5 n\right)=1+25\left\{{ }^n C_2+{ }^n C_3 \cdot 5+\ldots\right\}$
$\left(6^n-5 n\right)=1+25 \cdot k$
where $k=$ positive integer.
Hence, $a_n=6^n-5 n$ divided by 25 and leave the remainder $=1$

Topic: Mathematics, Binomial Theorem, TS EAMCET, TS EAMCET 2010
The numbers of arrangements of the letters of the word SALOON, if the two O's do not come together, is

Options:
A $360$
B $720$
C $240$
D $120$

The correct answer is:
$240$

Solution:
Total number of arrangements are $\frac{6!}{2!}=360$.
The number of ways in which os come together $=5!=120$.
Hence required number of ways $=360-120=240$.

Topic: Mathematics, Permutation Combination, JEE Main
The numbers of integral values of k for which the line, 3x+4y=k intersects the circle, x2+y2-2x-4y+4=0 at two distinct points is....

Options:

The correct answer is:
9

Solution:
If distance from center to line is less than radius then line cut circle at two distinct point.

Center of circle (1,2), r=1

3+8-k5<1

11-k<5

-5<11-k<5⇒ k∈(6,16)

k=7,8,9,10,11,12,13,14,15

Number of values of k=9.

Topic: Mathematics, Circle, JEE Main, JEE Main 2020 (02 Sep Shift 1)
The numbers of lone pairs and bond pairs in hydrazine are, respectively

Options:
A 2 and 4
B 2 and 6
C 2 and 5
D 1 and 5

The correct answer is:
2 and 5

Solution:
$\quad$ HYDRAZINE $\mathrm{N}_{2} \mathrm{H}_{4}$
$\mathrm{LP}=2$
$\mathrm{BP}=5$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The numbers of protons, electrons and neutrons in a molecule of heavy water are respectively :

Options:
A
$8,10,11$

B
$10,10,10$

C
$10,11,10$

D
$11,10,10$

The correct answer is:
$10,10,10$

Solution:
Heavy water is $\mathrm{D}_2 \mathrm{O}$ hence
number of electrons $=2+8=10$
number of protons $=10$
Atomic mass of $\mathrm{D}_2 \mathrm{O}=4+16=20$
hence number of neutron
$=$ Atomic mass $-$ number of protons
$$
=20-10=10
$$

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2013 (23 Apr Online)
The numerical ratio of displacement to the distance covered is always

Options:
A Less than one
B Equal to one
C Equal to or less than one
D Equal to or greater than one

The correct answer is:
Equal to or less than one

Solution:
Since displacement is always less than or equal to distance, but never greater than distance. Hence numerical ratio of displacement to the distance covered is always equal to or less than one.

Topic: Physics, Motion In One Dimension, JEE Main
The objective function $z=4 x+3 y$ can be maximized subjected to the constraints $3 x+4 y \leq 24,8 x+6 y \leq 48, x \leq$ $5, y \leq 6 ; x, y \geq 0$

Options:
A At only one point
B At two points only
C At an infinite number of points
D None of these

The correct answer is:
At an infinite number of points

Solution:
Now, to find where the function Z has maximized, let us substitute all these points in the objective function Z.
\(\begin{array}{|l|l|}
\hline Z \text { at } O(0,0) & Z=4(0)+3(0)=0+0=0 \\
\hline Z \text { at } A(0,6) & Z=4(0)+3(6)=0+18=18 \\
\hline Z \text { at } B\left(\frac{24}{7}, \frac{24}{7}\right) & Z=4\left(\frac{24}{7}\right)+3\left(\frac{24}{7}\right)=\frac{96+72}{7}=\frac{168}{7}=24 \\
\hline Z \text { at } C\left(5, \frac{4}{3}\right) & Z=4(5)+3\left(\frac{4}{3}\right)=20+4=24 \\
\hline Z \text { at } D(5,0) & Z=4(5)+3(0)=20+0=20 \\
\hline
\end{array}\)
Here, we can clearly see that, the function Z is maximized at two points B \& C giving the value 24 .
There will be infinite/multiple optimal solutions for a LPP if it has more than one set of optimal solutions that can maximize/ minimize a problem.
This will clear the fact that, the function Z will maximize at infinite number of points.

Topic: Mathematics, Linear Programming, JEE Main
The observed magnetic moment of the complex [Mn(NCS)6)]x- is 6.06 BM. The numerical value of x is

Options:

The correct answer is:
4

Solution:
The complex [Mn(NCS)6)]x- has a magnetic moment of 6.06 BM. This indicates that the complex has unpaired electrons, which contribute to its magnetic moment.

Number of unpaired electrons =5(Mn2+ : 3d5).
So Mn must be in +2 oxidation state Mn2+.

⇒ 2 + (–6) = –x ⇒ –4 = –x ⇒ x = 4
The numerical value of x is 4, so the complex is [Mn(NCS)6)]4-.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The observed osmotic pressure for a 0.10 M solution of Fe(NH4)2(SO4)2 at 25°C is 10.8 atm. The expected and experimental (observed) values of Van't Hoff factor 'i' will be respectively; (R = 0.082 L atm k- mol-1)

Options:
A 4 and 4.00
B 3 and 5.42
C 5 and 4.42
D 5 and 3.42

The correct answer is:
5 and 4.42

Solution:
π = i C R T.
1 0 . 8 = i × 0 . 1 0 × 0 . 0 8 2 × 2 9 8
i = 1 0 . 8 0 . 1 0 × 0 . 0 8 2 × 2 9 8 = 1 0 . 8 2 . 4 4 3 6 = 1 0 . 8 2 . 4 4 3 8
= 4.42 (experimental value)
FeSO 4 . NH 4 2 0 · 1 M SO 4 ⟶ Fe aq 2 + 0 1 M + 2 SO 4 aq 2 - 0 · 2 M + 2 NH 4 aq + 0 · 2 M
i (expected) = 0 · 5 M 0 · 1 M = 5

Topic: Chemistry, Solutions, JEE Main, JEE Main 2014 (19 Apr Online)
The obtuse angle between the lines $y=-2$ and $y=x+2$ is

Options:
A $120^{\circ}$
B $135^{\circ}$
C $150^{\circ}$
D $160^{\circ}$

The correct answer is:
$135^{\circ}$

Solution:
The gradient of the line $y=x+2$ is 1 . Therefore, it makes an angle of $45^{\circ}$ with $x$-axis. The second line is parallel to $x$-axis. Hence the obtuse angle between the lines is $135^{\circ}$.

Topic: Mathematics, Straight Lines, JEE Main
The octahedral complex \( \mathrm{CoSO}_{4} \mathrm{Cl} .5 \mathrm{NH}_{3} \) exists in two isomeric forms, \( \mathrm{X} \) and \( \mathrm{Y} \). Isomer \( \mathrm{X} \) reacts with \( \mathrm{AgNO}_{3} \) to give a white precipitate, but does not react with \( \mathrm{BaCl}_{2} . \) Isomer \( \mathrm{Y} \) gives a white precipitate with \( \mathrm{BaCl}_{2} \), but does not react with \( \mathrm{AgNO}_{3} . \mathrm{X} \) and \( \mathrm{Y} \) are:

Options:
A Ionization isomers
B Linkage isomers
C Coordination isomers
D Solvent isomers

The correct answer is:
Ionization isomers

Solution:
The reaction of the given octahedral complex with AgNO3 and BaCl2 produces a white precipitate individually, which indicates that the given octahedral complex has different ions. Therefore, they form different ions. So, the given complex represents ionization isomerism and the isomers are known as ionization isomers of each other.

Topic: Chemistry, Coordination Compounds, JEE Main
The octahedral complex of a metal ion M3+ with four monodentate ligands L1, L2, L3 and L4 absorb wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is

Options:
A L4 < L3 < L2 < L1
B L1 < L3 < L2 < L4
C L3 < L2 < L4 < L1
D L1 < L2 < L4 < L3

The correct answer is:
L1 < L3 < L2 < L4

Solution:
The colors of wavelengths are absorbed in decreasing order of wavelengths.
λ red > λ yellow > λ green > λ blue
as ΔE=hcλ
Δ E Red < Δ E Yellow < Δ E Green < Δ E Blue
L1 < L3 < L2 < L4
As the order of ligand field strength provides the order of relative crystal field splitting energies.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2014 (06 Apr)
The octahedral complex/complex ion which shows both facial and meridional isomers is

Options:
A triglycinato cobalt (III)
B tris (ethylene diamine) cobalt (III)
C dichloro diglycinato cobalt (III)
D trioxalate cobaltate (III)

The correct answer is:
dichloro diglycinato cobalt (III)

Solution:
Complexes of the type Maa2a2 show geometrical or fac-mer isomerism. (where, aa = bidentate ligand and a is monodentate ligand.) Among the given, dichlorodiglycinato cobalt (III), I.e., CoCl2gly2 is such salt, so it will show fac and mer isomerism.

Topic: Chemistry, Coordination Compounds, JEE Main
The octahedral diamagnetic low spin complex among the following is

Options:
A CoNH363+
B CoF63-
C CoCl63-
D NiCl42

The correct answer is:
CoNH363+

Solution:
In case of strong field ligand containing complexes, electrons get pair up against the Hund's rule to get required number of hybrid orbitals. Among the given complexes' ammonia is a strong-filed ligand and other ligands are weak filed ligands. CoNH363+ has d 2sp3 (inner orbital complex) with zero unpaired electrons. Hence, it is octahedral with diamagnetic character and low spin complex. CoF63– and CoCl63– are octahedral but having unpaired electrons.NiCl42- is not octahedral.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (10 Apr Shift 1)
The odd natural number a, such that the area of the region bounded by y=1,y=3,x=0,x=ya is 3643, equal to:

Options:
A 3
B 5
C 7
D 9

The correct answer is:
5

Solution:
Plotting the graph of the given functions, we get the shaded region as the area bounded by the curves.

Given, area of shaded region =3643

⇒∫13yady=3643

⇒ya+1a+113=3643

⇒3a+1-1a+1=3643=729-16=36-16

⇒a=5

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2022 (26 Jul Shift 1)
The OH- concentration in a mixture of 5.0 mL of 0.0504 M NH4Cl and 2 mL of 0.0210 M NH3 solution is x×10-6 M. The value of x is (Nearest integer) [Given Kw=1×10-14 and Kb=1.8×10-5

Options:

The correct answer is:
3

Solution:
NH4Cl=NH4+=0.0504M V=5.0 mLNH3=0.0210M V=2.0 mL

Kb=NH4+OH-NH3

OH-=Kb×NH3NH4+=1.8×10-5×0.0210×20.0504×5=3×10-6M

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The $\mathrm{OH}$ group of an alcohol or the$\mathrm{COOH}$ group of a carboxylic acid can be replaced by $-\mathrm{Cl}$ using:

Options:
A phosphorus pentachloride
B hypochlorous acid
C chlorine
D hydrochloride acid

The correct answer is:
phosphorus pentachloride

Solution:
$\mathrm{PCl}_5$ on reaction with alcohol and carboxylic acid, replace $-\mathrm{OH}$ group with $-\mathrm{Cl}$.
$$
\begin{aligned}
\mathrm{ROH}+\mathrm{PCl}_5 \longrightarrow \mathrm{RCl} & +\mathrm{POCl}_3+\mathrm{HCl} \\
\mathrm{RCOOH}+\mathrm{PCl}_5 \longrightarrow & \mathrm{RCOCl}^2 \\
+ & \mathrm{POCl}_3+\mathrm{HCl}
\end{aligned}
$$
Related Theory
$\mathrm{PCl}_5$ is used as a chlorinating agent and dehydrating agent for the synthesis of pesticides, phosphite antioxidants, plasticizers and stabilizers for plastic and elastomers etc

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The -OH group of methyl alcohol cannot be replaced by chlorine by the action of

Options:
A Chlorine
B Hydrogen chloride
C Phosphorus trichloride
D Phosphorus pentachloride

The correct answer is:
Chlorine

Solution:
CH 3 OH+Cl 2 →No reaction
CH3OH+HClâZnCl2CH3Cl+H2O
3  CH 3 OH+PCl 3 →3  CH 3 Cl+H 3 PO 3
CH3OH+PCl5âCH3Cl+POCl3+HCl

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The olefin, which on ozonolysis give CH 3 CH 2 CHO and CH 3 CHO is

Options:
A But-1-ene
B But-2-ene
C Pent-1-ene
D Pent-2-ene

The correct answer is:
Pent-2-ene

Solution:
CH 3 CH 2 CH = CH− CH 3 → Zn,  H 2 O O 3 CH 3 − CH 2 CHO +  CH 3 −CHO

Topic: Chemistry, Hydrocarbons, JEE Main
The one division of main scale of vernier callipers reads 1 mm and 10 divisions of Vernier scale is equal to the 9 divisions on main scale. When the two jaws of the instrument touch each other the zero of the Vernier lies to the right of zero of the main scale and its fourth division coincides with a main scale division. When a spherical bob is tightly placed between the two jaws, the zero of the Vernier scale lies in between 4.1 cm and 4.2 cm and 6th Vernier division coincides with a main scale division. The diameter of the bob will be _____ 10-2 cm.

Options:

The correct answer is:
412

Solution:
Given,

1 MSD=1 mm

10 VSD=9 MSD

Then, 1 VSD=910MSD=0.9 MSD

Least count of Vernier calliper is L.C. =0.1 mm=0.01 cm

Positive zero error=0.4 mm =0.04 cm

Observed reading of diameter of spherical bob=4.1 cm+6×0.01=4.16 cm

Real reading of diameter of bob=4.16-0.04=4.12 cm =412×10-2 cm

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2022 (27 Jul Shift 1)
The one giving maximum number of isomeric alkenes on dehydrohalogenation reaction is (excluding rearrangement)

Options:
A 1-Bromo-2-methylbutane
B 2-Bromopropane
C 2-Bromopentane
D 2-Bromo-3, 3-dimethylpentane

The correct answer is:
2-Bromopentane

Solution:
In dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms. Thus, 2-bromopentane gives pent-2-ene as the major product.

Rest all bromides are having only one type of beta hydrogen, hence, only one possible alkene is formed.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2023 (29 Jan Shift 2)
The one that can exhibit highest paramagnetic behaviour among the following is : gly= glycinato; bpy=2,2'-bipyridine

Options:
A Pd(gly)2
B Fe(en)(bpy)NH322*
C Co(OX)2(OH)2-Δ0>P
D TiNH363*

The correct answer is:
Co(OX)2(OH)2-Δ0>P

Solution:
Complex	EC	Unpaired electrons

(1) Fe(en)(bpy)NH322*	Fe2+=3d6=t2g2,2,2 ,eg0.0	0

(2) Pd(gly)2	Pd2+=4d8	0

(3) TiNH363+	Ti33=3d1	1

(4) Co(OX)2(OH)2-Δ0>P	Co5+⇒3d4⇒t2g2,1,1 ,e90,0	2

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (04 Sep Shift 2)