datavorous/entrance-exam-dataset huggingface dataset:
The number of correct statement/s from the following is _____ .

A.Larger the activation energy, smaller is the value of the rate constant.

B. The higher is the activation energy, higher is the value of the temperature coefficient.

C. At lower temperatures, increase in temperature causes more change in the value of k than at higher temperature.

D. A plot of ln k vs 1 T is a straight line with slope equal to -EaR

Options:

The correct answer is:
3

Solution:
A:k=Ae-EaRT(Arrhenius rate equation)

As Ea(Activation energy) increases k(rate constant) decreases

B :Higher activation energy, greater is the effect of given temperature rise on rate constant 'k'. Activation energy is the minimum energy needed for the reaction to occur. When temperature increases, Kinetic Energy also increases; as temperature increases, more molecules have higher Kinetic Energy, and thus the fraction of molecules that have high enough Kinetic Energy to overcome the energy barrier also increases.

Temperature coefficient =kT+10kT

Option (C) is wrong. Δk may be greater or lesser depending on temperature.

D: lnk=lnA-EaRT in this equation the slope is -EaR.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2023 (24 Jan Shift 1)
The number of correctly matched combination is/are

Ore	Element

Chalcopyrite	Cu

Limestone	Ca

Corundum	Al

Magnetite	Fe

Pyrolusite	Mn

Cassiterite	Zn

Cinnabar	Hg

Calamine	Ca

Siderite	Sn

Cerussite	Pb

Options:

The correct answer is:
7

Solution:
The correctly matched ores are

Ore	Element

Chalcopyrite	Cu

Limestone	Ca

Corundum	Al

Magnetite	Fe

Pyrolusite	Mn

Cassiterite	Zn

Cinnabar	Hg

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
The number of coulombs required for the deposition of $108 \mathrm{~g}$ of silver is

Options:
A 96500
B 48250
C 193000
D 10000

The correct answer is:
96500

Solution:
Amt. deposited $=\frac{\text { E. wt } \times \text { Q }}{96500}$;
$$
107.870=\frac{107.870}{96500} \times \mathrm{Q} ; \quad \therefore \mathrm{Q}=96500 \mathrm{C}
$$

Topic: Chemistry, Electrochemistry, JEE Main
The number of critical points of the function $f(x)=(x-2)^{2 / 3}(2 x+1)$ is

Options:
A 1
B 2
C 0
D 3

The correct answer is:
2

Solution:
$\begin{aligned} & \mathrm{f}(\mathrm{x})=(\mathrm{x}-2)^{2 / 3}(2 \mathrm{x}+1) \\ & \mathrm{f}^{\prime}(\mathrm{x})=\frac{2}{3}(\mathrm{x}-2)^{-1 / 3}(2 \mathrm{x}+1)+(\mathrm{x}-2)^{2 / 3} \\ & \mathrm{f}^{\prime}(\mathrm{x})=2 \times \frac{(2 \mathrm{x}+1)+(\mathrm{x}-2)}{3(\mathrm{x}-2)^{1 / 3}} \\ & \frac{3 \mathrm{x}-1}{(\mathrm{x}-2)^{1 / 3}}=0\end{aligned}$
Critical points $\mathrm{x}=\frac{1}{3}$ and $\mathrm{x}=2$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2024 (08 Apr Shift 1)
The number of $d$ electrons in $\mathrm{Fe}{ }^{+2}$ (atomic number of $F e=26$) is not equal to that of the

Options:
A $P$-electrons in $\mathrm{Ne}$ (At. No. $=10$)
B $s$-electrons in $Mg$ (At. No. $=12$)
C $d$-electrons in $\mathrm{Fe}$
D $P$-electrons in $\mathrm{Cl}^{-}$(At. No. of $\mathrm{Cl}=17$)

The correct answer is:
$P$-electrons in $\mathrm{Cl}^{-}$(At. No. of $\mathrm{Cl}=17$)

Solution:
${ }_{26} \mathrm{Fe}=[\mathrm{Ar}] 3 d^6, 4 s^2$
$\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 d^6, 4 s^0$
Number of $d$-electrons $=6$
${ }_{17} \mathrm{Cl}=[\mathrm{Ne}] 3 s^2, 3 p^5$
$C l^{-}=[\mathrm{Ne}] 3 s^2, 3 p^6$
Number of $p$-electrons $=6$.

Topic: Chemistry, Structure of Atom, JEE Main
The number of d-electrons retained in $\mathrm{Fe}^{2+}$ (At.no.of $\mathrm{Fe}=26$ ) ion is

Options:
A
4

B
5

C
6

D
3

The correct answer is:
6

Solution:
$\mathrm{Fe}^{+2}=3 \mathrm{~d}^6 \cdot 4 \mathrm{~s}^0$

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2003
The number of diagonals in a polygon is 20 . The number of sides of the polygon is

Options:
A 5
B 6
C 8
D 10

The correct answer is:
8

Solution:
Hints: \({ }^{\mathrm{n}} \mathrm{C}_2-\mathrm{n}=20\)
\(\mathrm{n}=8\)

Topic: Mathematics, Permutation Combination, JEE Main
The number of diagonals of a polygon is 35 . If A, B are two distinct vertices of this polygon, then the number of all those triangles formed by joining three vertices of the polygon having $\mathrm{AB}$ as one of its sides is

Options:
A $1$
B $8$
C $10$
D $12$

The correct answer is:
$8$

Solution:
Number of diagonals $={ }^n \mathrm{C}_2-n$
$$
\begin{aligned}
& =\frac{n(n-1)}{2}-n=35 \\
& \Rightarrow n^2-n-2 n=35 \times 2 \Rightarrow n^2-3 n-70=0 \\
& \Rightarrow(n-10)(n+7)=0 ; \therefore n=10
\end{aligned}
$$
To form triangle we need 3 vertices out of which 2 are $A$ and B
$\therefore 8$ triangles can be formed by selecting 3 rd vertex as remaining 8 vertices of polygon.

Topic: Mathematics, Permutation Combination, TS EAMCET, TS EAMCET 2023 (14 May Shift 1)
The number of different 9 digit number that can be formed, from the digits of the number 445577888 by rearranging its digits, so that the odd digits occupy even positions, are

Options:
A $120$
B $60$
C $180$
D $36$

The correct answer is:
$60$

Solution:
ways
and remaing five even digits $4,4,8,8,8$ can be arranged in 5 places in $\frac{\underline{5}}{2 \underline{3}}$
Hence total no of ways $=\frac{\lfloor 4}{\lfloor 2 L 2} \times \frac{\underline{5}}{2 \underline{2} \underline{3}}=6 \times 10=60$

Topic: Mathematics, Permutation Combination, MHT CET, MHT CET 2022 (06 Aug Shift 1)
The number of different chain isomers for \(\mathrm{C}_7 \mathrm{H}_{16}\) is \(\qquad\)

Options:

The correct answer is:
9

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (04 Apr Shift 1)
The number of different seven-digit numbers that can be written using only the three digit 1 , 2 and 3 with the condition that the digit 2 occurs twice in each number is

Options:
A
${ }^7 C_2 2^5$

B
${ }^7 p_2 2^5$

C
${ }^7 C_2 5^2$

D
None of these

The correct answer is:

${ }^7 C_2 2^5$

Solution:
Other than 2, remaining five places can be filled by 1 and 3 for each place. The number of ways for five places is $2 \times 2 \times 2 \times 2 \times 2=2^5$. For 2 , selecting 2 places out of 7 is ${ }^7 C_2$. Hence, the required number of ways is ${ }^7 C_2 \times 2^5$.

Topic: Mathematics, Permutation Combination, JEE Main
The number of different terms in the expansion of 1-x2011+x+x2200 is

Options:
A 200
B 201
C 202
D 402

The correct answer is:
402

Solution:
1-x1-x3200
1-xC0 200-C1 200x3+â¦â¦â¦â¦.
So, number of required terms =2Ã201=402

Topic: Mathematics, Binomial Theorem, JEE Main
The number of different ways in which five alike dashes and eight alike dots can be arranged, using only seven of these ‘dashes’ and ‘dots’ is equal to

Options:
A 1287
B 119
C 120
D 1235520

The correct answer is:
120

Solution:
Dashes 5	Dots 8	Arrangements

5	2	C27

4	3	C37

3	4	C47

2	5	C57

1	6	C67

0	7	C77

Required number of ways =C27+C37+C47+C57+C67+C77
=27-7-1=120

Topic: Mathematics, Permutation Combination, JEE Main
The number of different ways in which the first twelve natural numbers can be divided into three groups, each having four elements, such that the numbers in each group are in arithmetic progression is

Options:

The correct answer is:
4

Solution:
No group can have a common difference ≥4 as there is no number ≥13
If the group containing 1 has the common difference =1 i.e. if one group is 1,2,3,4, there will be two ways to group others - 5,6,7,8, 9,10,11,12 and 5,7,9,11,6,8,10,12.

If the group containing 1 has the common difference =2 i.e. one group is 1,3,5,7, there will be one way to group others 2,4,6,8,9,10,11,12.

If the group containing 1 has common difference =3 i.e. if one group is 1,4,7,10 there will be one way to group others 2,5,8,11,3,6,9,12

⇒ The number of ways =2+1+1=4

Topic: Mathematics, Permutation Combination, JEE Main
The number of different ways in which the first twelve natural numbers can be divided into three groups, each having four elements, such that the numbers in each group are in arithmetic progression is

Options:

The correct answer is:
4

Solution:
No group can have a common difference â¥4 as there is no number â¥13
If the group containing 1 has the common difference =1 i.e. if one group is 1,2,3,4, there will be two ways to group others - 5,6,7,8, 9,10,11,12 and 5,7,9,11,6,8,10,12.

If the group containing 1 has the common difference =2 i.e. one group is 1,3,5,7, there will be one way to group others 2,4,6,8,9,10,11,12.

If the group containing 1 has common difference =3 i.e. if one group is 1,4,7,10 there will be one way to group others 2,5,8,11,3,6,9,12
â The number of ways =2+1+1=4

Topic: Mathematics, Permutation Combination, JEE Main
The number of different words (eight-letter words) ending and beginning with a consonant which can be made out of the letters of the word 'EQUATION' is

Options:
A 5200
B 4320
C 3000
D 2160

The correct answer is:
4320

Solution:
EQUATION-8 letters. Consonants $-\mathrm{Q}, \mathrm{T}, \mathrm{N}-3$ letters.
first letter of 8 -letter word can be any of 3 consonants Last letter of $8-$ letter word can be remaining 2 consonants. The middle 6 - letters can be arranged in $6 !$ ways. So, number of different words $=3 \times 2 \times 6 !$ $=6 \times 720=4320 .$

Topic: Mathematics, Permutation Combination, NDA, NDA 2017 (Phase 1)
The number of discontinuity of the greatest integer function fx=x, xâ-72, 100 is equal to

Options:
A 104
B 102
C 101
D 103

The correct answer is:
103

Solution:
Given, fx=x, x∈-3.5, 100
As we know greatest integer is discontinuous on integer values.
In given interval, the integer values are
(-3, -2, -1, 0, …, 99)
∴ Total numbers of integers are 103.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The number of distinct complex number(s) z, such that z=1 and z3 is purely imaginary, is/are equal to

Options:

The correct answer is:
6

Solution:
Let, z=eiθ

⇒z3=cos⁡3θ+isin⁡3θ
Since, z3 is purely imaginary
⇒cos⁡3θ=0⇒θ=2nπ3±π6
∴ Possible value of θ=30o,150o,90o,270o,210o,330o.

Topic: Mathematics, Complex Number, JEE Main
The number of distinct possible lines passing through the point 2,3 and forming a triangle with the coordinate axes whose area is 12sq. units, is/are

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Equation of any line through 2,3 is y-3=mx-2
y=mx-2m+3
From the figure,

Area of ∆OAB=12 sq. units
i.e., 122m-3m3-2m=±12
taking positive sign, we get 2m+32=0
this gives one value of m=-32 and

by taking negative sign, we get 4m2-36m+9=0 which is a quadratic in m whose discriminant is greater than 0, so it gives 2 values of m
⇒3 lines are possible.

Topic: Mathematics, Straight Lines, JEE Main
The number of distinct prime divisors of the number $512^{3}-253^{3}-259^{3}$ is

Options:
A 4
B 5
C 6
D 7

The correct answer is:
6

Solution:
$\begin{array}{l}
(512)^{3}-253^{3}-259^{3} \\
=(512)^{3}-\left[\left(253^{3}\right)+(259)^{3}\right] \\
=(512)^{3}-(253+259)\left(253^{2}+259^{2}-(253)(259)\right. \\
=(512)^{3}-(512)\left[(253+259)^{2}-2(253)(259)-(253)(259)\right] \\
=512\left[(512)^{2}-\left\{(512)^{2}-3(253)(259)\right\}\right] \\
=(512)[3(253)(259)] \\
=2^{9} .3 .(253)(259) \\
=2^{9} .3(11)(23)(7)(37)
\end{array}$
6 prime divisors.

Topic: Mathematics, Binomial Theorem, JEE Main
The number of distinct real roots of sinxcosxcosxcosxsinxcosxcosxcosxsinx=0 in the interval -π4≤x≤π4 is:

Options:
A 4
B 1
C 2
D 3

The correct answer is:
1

Solution:
We have, sinxcosxcosxcosxsinxcosxcosxcosxsinx=0,-π4≤x≤π4

Apply: R1→R1-R2 and R2→R2-R3

sinx-cosxcosx-sinx00sinx-cosxcosx-sinxcosxcosxsinx=0

⇒sinx-cosx21-1001-1cosxcosxsinx=0

⇒sinx-cosx2sinx+2cosx=0

⇒sinx=cosx or sinx=-2cosx

∴ x=π4

Since, -π4≤x≤π4, we get -1≤tanx≤1

⇒tanx=1⇒x=π4

Hence, number of roots is one.

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2021 (25 Jul Shift 2)
The number of distinct real roots of $\left|\begin{array}{lll}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$ in the interval $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$ is

Options:
A 0
B 2
C 1
D >2

The correct answer is:
1

Solution:
Given, $\left|\begin{array}{ccc}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$
$\sin x\left(\sin ^{2} x-\cos ^{2} x\right)-\cos x\left(\sin x \cos x-\cos ^{2} x \right ) + \cos x \left(\cos ^{2} x-\sin x \cos x\right)=0$
$\Rightarrow \sin x(\sin x-\cos x)(\sin x+\cos x)$ $-\cos ^{2} x(\sin x-\cos x)-\cos ^{2} x$ $(\sin x-\cos x)=0$
$\Rightarrow(\sin x-\cos x)\left(\sin ^{2} x+\sin x \cos x-2 \cos ^{2} x\right)=0$
$\Rightarrow(\sin x-\cos x)\left(\sin ^{2} x+2 \sin x\right.\left.\cos x-\sin x \cos x-2 \cos ^{2} x\right)=0$
$\Rightarrow(\sin x-\cos x)^{2}(\sin x+2 \cos x)=0$
$\Rightarrow \sin x-\cos x=0$ or $\sin x+2 \cos x=0$
$\Rightarrow \quad \tan x=1$ or $\tan x=-2$
But $\quad x \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$
$\Rightarrow$ Only one solution i.e. $x=\frac{\pi}{4}$

Topic: Mathematics, Determinants, JEE Main
The number of distinct real roots of the equation 3x4+4x3-12x2+4=0 is _________.

Options:

The correct answer is:
4

Solution:
3x4+4x3-12x2+4=0

So,

let fx=3x4+4x3-12x2+4

∴ f'x=12xx2+x-2

=12xx+2x-1

f0=4

f1=-1

Ans. 4

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (27 Aug Shift 1)
The number of distinct real roots of the equation, cosxsinxsinxsinxcosxsinxsinxsinxcosx=0 in the interval -π4,π4 is :

Options:
A 1
B 4
C 2
D 3

The correct answer is:
2

Solution:
Given equation cosxsinxsinxsinxcosxsinxsinxsinxcosx=0

Performing row and column operations

R1→R1-R2

R2→R2-R3

⇒cosx-sinxsinx-cosx00cosx-sinxsinx-cosxsinxsinxcosx=0

C2→C2+C3

⇒cosx-sinxsinx-cosx000sinx-cosxsinxsinx+cosxcosx=0

Expanding using second row, we get

sinx+cosxcosx-sinx-sinxsinx-cosx=0

⇒cosx-sinxsinx+cosx+sinx=0

⇒cosx-sinx2sinx+cosx=0

⇒cosx-sinx=0

sinx=cosx⇒π4

Or

2sinx+cosx=0

tanx=-12

x=tan-1-12

Hence, 2 solutions are there.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2016 (09 Apr Online)
The number of distinct real roots of the equation $|x||x+2|-5|x+1|-1=0$ is_______

Options:

The correct answer is:
3

Solution:
Case-1
$\begin{aligned}
& x \geq 0 \\
& x^2+2 x-5 x-5-1=0 \\
& x^2-3 x-6=0 \\
& x=\frac{3 \pm \sqrt{9+24}}{2}=\frac{3 \pm \sqrt{33}}{2}
\end{aligned}$
One positive root
Case-2
$\begin{aligned}
& -1 \leq x < 0 \\
& -x^2-2 x-5 x-5-1=0 \\
& x^2+7 x+6=0 \\
& (x+6)(x+1)=0 \\
& x=-1
\end{aligned}$
one root in range
Case-3
$\begin{aligned}
& -2 \leq x < -1 \\
& x^2-2 x+5 x+5-1=0 \\
& x^2-3 x-4=0 \\
& (x-4)(x+1)=0
\end{aligned}$
No root in range
Case-4
$\begin{aligned}
& x < -2 \\
& x^2+7 x+4=0 \\
& x=\frac{-7 \pm \sqrt{49-16}}{2}=\frac{7 \pm \sqrt{33}}{2}
\end{aligned}$
one root in range
Total number of distinct roots are 3

Topic: Mathematics, Functions, JEE Main, JEE Main 2024 (05 Apr Shift 1)
The number of distinct real roots of x4-4x+1=0 is

Options:
A 0
B 1
C 2
D 4

The correct answer is:
2

Solution:
Let fx=x4-4x+1

f'x=4x3-4 and f'x=0⇒x=1

So the extrema of the function is at x=1

Plotting the graph we get

f''x=12x2 ∴ f''1>0

And at x=1, f1=-2<0

So, graph of fx will cut x-axis at two points,

∴ Number of real roots of fx=0 is equal to 2

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (27 Jun Shift 1)
The number of distinct real values of λ, for which the lines x-11=y-22=z+3λ2 and x-31=y-2λ2=z-12 , are coplanar is

Options:
A 2
B 4
C 3
D 1

The correct answer is:
3

Solution:
∵ Lines are coplanar
∴ 3-12-21--312λ21λ22=0
⇒ 20412λ21λ22=0
⇒ 24-λ4+4λ2-2=0
⇒ 4-λ4+2λ2-4=0
⇒ λ2λ2-2=0
⇒ λ=0, 2, -2

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2016 (10 Apr Online)
The number of distinct solutions of the equation, log12⁡sin⁡x=2-log12⁡cos⁡x in the interval 0,2π, is ________

Options:

The correct answer is:
8

Solution:
log1/2⁡sin⁡x=2-log1/2⁡cos⁡x
log1/2sin⁡xcos⁡x=2
sin⁡xcos⁡x=14
sin⁡2x=±12

Number of solution =8 .

Topic: Mathematics, Basic of Mathematics, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The number of distinct terms in the expansion of \( (x+2 y-3 z+5 w-7 u)^{n} \) is

Options:
A \( n+1 \)
B \( { }^{n+4} C_{4} \)
C \( { }^{n+4} C_{n+2} \)
D \( { }^{n} C_{4} \)

The correct answer is:
\( { }^{n+4} C_{4} \)

Solution:
We know that the number of distinct terms in the multinomial expansion of x1+x2+…+xnm is given by Cn-1n+m-1.

Therefore, number of distinct terms in the expansion of

(x+2y-3z+5w-7u)n is

=C5-1n+5-1

=C4n+4

Topic: Mathematics, Binomial Theorem, JEE Main
The number of donar sites in mercapto, glycinato, diethylene triamine and (EDTA)4- are: -

Options:
A 1 ,2, 3, 4
B 1, 2, 3, 6
C 1, 2, 4, 6
D 1, 2, 2, 6

The correct answer is:
1, 2, 3, 6

Solution:
Mercapto→HS–
Glycinate →N¨H2-CH2-COO-

Topic: Chemistry, Coordination Compounds, JEE Main
The number of eight-digit integers, with the sum of digits equal to 12 and formed by using the digits 1, 2 and 3 only are

Options:
A 255
B 277
C 288
D 266

The correct answer is:
266

Solution:
The desired number is the coefficient of x12 in x+x2+x38
= coefficient of x12 in x81+x+x28
= coefficient of x4 in 1+x31-x8
= coefficient of x4 in 1+x381-x-8
= coefficient of x4 in 1-8x3+....1-x-8
= coefficient of x4 in 1-x-8-8× coefficient of x in 1-x-8
=4+8-1C8-1-8×1+8-1C8-1
=11C7-8×8C7=11×10×93×84×3×2-8×8
=330-64=266

Topic: Mathematics, Permutation Combination, JEE Main
The number of electrons flowing per second in the filament of a $110 \mathrm{~W}$ bulb operating at $220 \mathrm{~V}$ is : $\left(\right.$ Given $\left.\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\right)$

Options:
A $6.25 \times 10^{17}$
B $1.25 \times 10^{19}$
C $6.25 \times 10^{18}$
D $31.25 \times 10^{17}$

The correct answer is:
$31.25 \times 10^{17}$

Solution:
Power $(\mathrm{P})=\mathrm{V} . \mathrm{I}$
$\begin{aligned}
& \Rightarrow 110=(220)(\mathrm{I}) \\
& \Rightarrow \mathrm{I}=0.5 \mathrm{~A}
\end{aligned}$
Now, $I=\frac{n \cdot e}{t}$
$\begin{aligned}
& \Rightarrow 0.5=\left(\frac{\mathrm{n}}{\mathrm{t}}\right)\left(1.6 \times 10^{-19}\right) \\
& \Rightarrow \frac{\mathrm{n}}{\mathrm{t}}=\frac{0.5}{1.6 \times 10^{-19}} \\
& \Rightarrow \frac{\mathrm{n}}{\mathrm{t}}=31.25 \times 10^{17}
\end{aligned}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (06 Apr Shift 2)
The number of electrons in the nucleus of $\mathrm{C}^{12}$ is

Options:
A 6
B 12
C 0
D 3

The correct answer is:
0

Solution:
In the nucleus of an atom only proton and neutrons are present.

Topic: Chemistry, Structure of Atom, JEE Main
The number of electrons involved in the reduction of permanganate to manganese dioxide in acidic medium is _____ .

Options:

The correct answer is:
3

Solution:
Permanganate is reduced to Manganese dioxide by the following reaction:
MnO4-+7+4H++3e-⟶Mn+4O2+2H2O
Here, oxidation no. of manganese was reduced by 3 from reactant to product side. Thus, three electrons are involved.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2023 (30 Jan Shift 1)
The number of electrons present in all the completely filled subshells having n=4 and s=+12 is ____ (Where n= principal quantum number and s= spin quantum number)

Options:

The correct answer is:
16

Solution:
(a) Number of sub-shells associated with n = 4:

A given value of n, l can have values from zero to (n–1).i.e.,0,1,2,3

The four sub-shells are associated with n=4, are s, p, d, and f.

Where,

The number of electrons in the sub-shells having an ms value of +1/2 for n =4: can have,

=1+3+5+7=16

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2024 (27 Jan Shift 1)
The number of electrons which can be accommodated in an orbital is

Options:
A One
B Two
C Three
D Four

The correct answer is:
Three

Solution:
The maximum number of electrons that can be accommodated in an orbital is 2.

Topic: Chemistry, Structure of Atom, JEE Main
The number of element from the following that do not belong to lanthanoids is \(\mathrm{Eu}, \mathrm{Cm}, \mathrm{Er}, \mathrm{Tb}, \mathrm{Yb}\) and \(\mathrm{Lu}\)

Options:
A 3
B 4
C 1
D 5

The correct answer is:
1

Solution:
$\mathrm{Cm}$ is Actinide

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (06 Apr Shift 1)
The number of elements in the set a,b:a2+b2=50; a, b∈Z, where Z is the set of all integers, is

Options:

The correct answer is:
12

Solution:
a2+b2=50
a=1⇒b2=49
⇒b=±7
a=2,3,4,6 do not give integer value of b .
a=5⇒b2=25⇒b=±5
a=7⇒b2=1
⇒b=±1
Similarly,
a=-1⇒b=±7
a=-5⇒b=±5
a=-7→b=±1
⇒ The required number of elements (a,b)=12

Topic: Mathematics, Sets and Relations, JEE Main
The number of elements in the set {n∈ℕ:10≤n≤100 and 3n-3 is a multiple of 7} is _______.

Options:

The correct answer is:
15

Solution:
Given that n∈10,100 and n∈N.

Let us take the values of n=1,2,3,... and check whether 3n-3 is divisible by 7 or not.

⇒31-3=0 is divisible by 7

⇒32-3=3, 3 is not divisible by 7

⇒36=7k+1 is not divisible by 7

⇒37=7α+3,

⇒37-3=7α is divisible by 7

Similarly ⇒313=7k+17α+3

=7β+3

⇒313-3=7β is divisible by 7

If we observe the pattern for n=1,7,13, 3n-3 is divisible by 7.

That means the series forms an AP with d=6.

Also n∈10,100

The required progression is 13,19,......an

⇒an=13+n-16

But 13+n-16<100

⇒n<15.5

⇒n=15∈N

Therefore, the required answer is 15.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The number of elements in the set n∈ℤ:n2-10n+19<6 is _______ .

Options:

The correct answer is:
6

Solution:
Given,

n2-10n+19<6

Now we know that,

x<a⇒-a<x<a

Now using above formula we get,

⇒-6<n2-10n+19<6

⇒n2-10n+25>0 and n2-10n+13<0

⇒n-52>0 and n-5-23n-5+23<0

⇒n∈ℤ-5 and n∈2, 3, 4, 5, 6, 7, 8

Hence, n=2, 3, 4, 6, 7, 8

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2023 (10 Apr Shift 1)
The number of elements in the set S=θ∈[0,2π]:3cos4θ-5cos2θ-2sin6θ+2=0 is

Options:
A 10
B 8
C 12
D 9

The correct answer is:
9

Solution:
Given,

3cos4θ-5cos2θ-2sin6θ+2=0

⇒cos2θ3cos2θ-5-2sin6θ+2=0

⇒1-sin2θ3-3sin2θ-5-2sin6θ+2=0

⇒sin2θ-13sin2θ+2-2sin6θ+2=0

Let sin2θ=t
⇒(t-1)(3t+2)-2t3+2=0

⇒(t-1)(3t+2)-2t3-1=0

We know that a3-b3=a-ba2+ab+b2

⇒(t-1)(3t+2)-2t-1t2+t+1=0

⇒(t-1)3t+2-2t2+t+1=0

(t-1)2t2-t=0

⇒t=0,1,12

For sin2θ=0, θ=0,π,2π

sin2θ=1, θ=π2,3π2

sin2θ=12, θ=π6,5π6,7π6,11π6

∴ Total solution =9

Hence this is the required option.

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2023 (11 Apr Shift 1)
The number of elements in the set S=θ∈-4π,4π: 3cos22θ+6cos2θ-10cos2θ+5=0 is ______.

Options:

The correct answer is:
32

Solution:
3cos22θ+6cos2θ-10cos2θ+5=0

3cos22θ+6cos2θ-51+cos2θ+5=0

cos2θ3cos2θ+1=0

cos2θ=0 or cos2θ=-13

Given θ∈-4π,4π⇒2θ∈-8π,8π

So for 2θ=2n+1π2

θ=±π4,±3π4,………±15π4 gives 16 solution

Similarly cos2θ=-13 gives 16 solution

hence, total elements in set S will be 32

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2022 (29 Jun Shift 1)
The number of elements in the set

S=x∈ℝ:2cosx2+x6=4x+4-x is

Options:
A 1
B 3
C 0
D infinite

The correct answer is:
1

Solution:
Given,

2cosx2+x6=4x+4-x

Now we know that maximum value of cos function is 1, so 2cosx2+x6≤2 and using A.M≥G.M we get 4x+4-x≥2

So, L.H.S.≤2. & R.H.s ≥2

Hence L.H.S.=2 & R.H.S.=2

Now equating both to 2 we get,

2cosx2+x6=2 and 4x+4-x=2

Now solving 2cosx2+x6=2

⇒cosx2+x6=1

⇒x2+x6=0

⇒x2+x=0

⇒x=0 or -1 ...........1

Now solving 4x+4-x=2 we get x=0 .....2

Now from equation 1 & 2 we can see common solution is x=0

Hence, possible solution is only one.

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2022 (29 Jul Shift 2)
The number of elements in the set x∈R:|x|-3x+4=6 is equal to

Options:
A 3
B 2
C 4
D 1

The correct answer is:
2

Solution:
x≠-4

x-3x+4=6

x-3=6x+4

No. of solutions =2

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (16 Mar Shift 1)
The number of elements in the set {z=a+ib∈ℂ:a,b∈ℤ and 1<z-3+2i<4} is _____.

Options:

The correct answer is:
40

Solution:
We know that z-3-2i=r represents a circle with centre 3,-2 and radius r.

1<z-3+2i<4

1<a-32+b+22<16

The ordered pairs of a,b satisfying the above inequality are 0,±2,±2,0,±1,±2,±2,±1

±2,±3,3±,±2,±1,±1,2±,±2

±3,0,0,±3,±3±1,±1,±3

i.e. total 40 elements are present in the given set

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2022 (28 Jun Shift 1)