datavorous/entrance-exam-dataset huggingface dataset:
The number and type of bonds between two carbon atoms in calcium carbide are

Options:
A
One sigma, one pi

B
One sigma, two pi

C
Two sigma, one pi

D
Two sigma, two pi

The correct answer is:
One sigma, two pi

Solution:

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2005
The number and type of bonds between two carbon atoms in calcium carbide are :

Options:
A One sigma, one pi
B One sigma, two pi
C Two sigma, one pi
D Two sigma, two pi

The correct answer is:
One sigma, two pi

Solution:
One sigma, two pi Calcium carbide \(\left(\mathrm{CaC}_2\right)\) is an ionic compound \(\left(\mathrm{Ca}^{2+} \mathrm{C}_2^{2-}\right)\) having anion \(\mathrm{C}_2^{2-}\). The calcium atom is bonded to the two carbon atoms through ionic bonding \([: C \equiv C:]^{2-}\). It has one sigma and two pi bonds.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The number and type of bonds between two carbon atoms in calcium carbide are

Options:
A
One sigma, one pi

B
One sigma, two pi

C
Two sigma, one pi

D
Two sigma, two pi

The correct answer is:
One sigma, two pi

Solution:

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The number and type of bonds between two carbon atoms in \( \mathrm{CaC}_{2} \) are

Options:
A One sigma \( (\sigma) \) and one pi \( (\pi) \) bond
B One sigma \( (\sigma) \) and one and half pi \( (\pi) \) bond
C One sigma \( (\sigma) \) bond
D One sigma \( (\sigma) \) and two pi \( (\pi) \) bond

The correct answer is:
One sigma \( (\sigma) \) and two pi \( (\pi) \) bond

Solution:
The structure of CaC2 is :
C - ||| C - Ca + 2

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The number and type of bonds in C22- ion in CaC2 are:

Options:
A Two σ-bonds and two π-bonds
B One σ-bond and two π-bonds
C One σ-bond and one π-bond
D Two σ-bonds and one π-bond

The correct answer is:
One σ-bond and two π-bonds

Solution:
The structure of calcium carbide is:
Ca2+C≡C2-
Both C atoms are sp hybridised. The carbide anion, having 2 units of negative charge between two carbon atoms, has 1 σ and 2 π bonds.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2014 (09 Apr Online)
The number density of free electrons in copper is nearly 8×1028 m-3. A copper wire has its area of cross-section =2×10-6 m2 and is carrying a current of 3.2 A. The drift speed of the electrons is _____×10-6 m s-1.

Options:

The correct answer is:
125

Solution:
The formula to calculate the current through any conductor is given by

I=nAevd ...1

Substitute the values of the given parameters into equation (1) and solve to calculate the required drift velocity of the electrons.

3.2=8×1028×2×10-6×1.6×10-19×vd⇒vd=3.216×1.6×1000 m s-1=125×10-6 m s-1

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The number obtained after dividing the number formed by the last three digits of 17256 by 100 is

Options:

The correct answer is:
6.81

Solution:
Last three digits of 17256
=289128=-1+290128
=128C0-1128-128C1⋅290+128C2⋅2902+.......
=1-128×290+64×127×292×100+.......
=1-......120+.....800+.......
=681
Hence, 681100=6.81

Topic: Mathematics, Binomial Theorem, JEE Main
The number of 2×2 matrices A with real entries, such that A+AT=3I and AAT=5I, is equal to

Options:
A 0
B 1
C 2
D Infinite

The correct answer is:
2

Solution:
Let, A=abcd, AT=acbd
Given, A+AT=3I
2ab+cb+c2d=3003⇒a=32,d=32,b=-c
A=32b-b32
AAT=32b-b3232-bb32⇒94+b20094+b2=5005
⇒94+b2=5⇒b2=5-94
⇒b=±112

Hence, 2 matrices are possible

Topic: Mathematics, Matrices, JEE Main
The number of 2-centre-2-electron and 3-centre-2-electron bonds in B2H6, respectively, are:

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

The correct answer is:
4 and 2

Solution:
B2H6 has 2 bridge bonds 3C-2e- and 4 terminal bonds 2C-2e-.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2019 (10 Jan Shift 2)
The number of 3Ã3 matrices M with entries from {0,1,2}, such that the sum of the diagonal elements of MTM is 5, are

Options:
A 198
B 126
C 135
D 162

The correct answer is:
198

Solution:
Let, M=abcdefghi

∴ Sum of the diagonal elements, Tr(MT​ M)=a2+b2+c2+d2+e2+f2+g2+h2+i2=5, where entries are 0,1,2

Only two cases are possible.

(I) Five entries are 1 and other fourare 0

∴ 9C5×1

(II) One entry is 2, one entry is 1 and others are 0.

∴ 9C2×2

Total cases=126+72=198.

Topic: Mathematics, Matrices, JEE Main
The number of $3 \times 3$ non-singular matrices, with four entries as 1 and all other entries as 0 , is

Options:
A
5

B
6

C
at least 7

D
less than 4

The correct answer is:
at least 7

Solution:
First row with exactly one zero; total number of cases $=6$
First row 2 zeros we get more cases
Total we get more than 7 .

Topic: Mathematics, Matrices, JEE Main
The number of $3 \times 3$ non-singular matrices, with four entries as 1 and all other entries as 0 , is

Options:
A
5

B
6

C
at least 7

D
less than 4

The correct answer is:
at least 7

Solution:
First row with exactly one zero; total number of cases $=6$
First row 2 zeros we get more cases
Total we get more than 7 .

Topic: Mathematics, Matrices, JEE Main, JEE Main 2010
The number of 3-digit numbers, that are divisible by either 2 or 3 but not divisible by 7 is _____ .

Options:

The correct answer is:
514

Solution:
We know that total number of three-digit number will be 900,

Now let nA be number of three-digit number which are divisible by 2, i.e,

A=100,102,...,998

⇒nA=450

nB be number of three-digit number which are divisible by 3, i.e, B=102,105,...,999

nB=300

Numbers divisible by both 2 and 3 is

102,108,...996 i.e, 150

Numbers divisible by both 2 and 7 are 112,126,...,994 i.e., 64 numbers.

Numbers divisible by both 3 and 7 is 105,126,...987 i.e, 43 numbers.

Numbers divisible by 2,3 & 7 is 126,168,...,966 i.e, 21 numbers.

Required number

=450+150-64-43+21=514

Hence, 514 three-digit number are there which are divisible by 2 or 3 but not divisible by 7.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (01 Feb Shift 1)
The number of 3 digit numbers, that are divisible by either 3 or 4 but not divisible by 48 , is

Options:
A 472
B 432
C 507
D 400

The correct answer is:
432

Solution:
We know that,

Total number of three digit number will be 999-99=900,

Now,

Number of 3- digit number which are divisible by 3 will be 9003=300,

Number of 3- digit number which are divisible by 4 will be 9004=225,

Number of 3- digit number which are divisible by 3 & 4 both will be 90012=75,

So , number of 3- digit number which are divisible by 3 or 4 will be 300+225-75=450,

Now, number of 3- digit number which are divisible by 48 is 90048=18

So, number of 3- digit number which are divisible by 3 or 4 but not 48 will be

=450-18=432

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (29 Jan Shift 2)
The number of 3-digit numbers, formed using the digits $2,3,4,5$ and 7 , when the repetition of digits is not allowed, and which are not divisible by 3 , is equal to__________

Options:

The correct answer is:
36

Solution:
total number of three digit numbers not divisible by 3 will be formed by using the digits
$\begin{aligned}
& (4,5,7) \\
& (3,4,7) \\
& (2,5,7) \\
& (2,4,7) \\
& (2,4,5) \\
& (2,3,5) \\
& \text { number of ways }=6 \times 3 !=36
\end{aligned}$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2024 (08 Apr Shift 1)
The number of 4-digit numbers which are neither multiple of 7 nor multiple of 3 is

.

Options:

The correct answer is:
5143

Solution:
The total possible 4 digit numbers are 9×10×10×10=9000

Let, X is the 4 digit number divisible by 3

1002, 1005, ………, 9999

a=1002, d=3 & an=9999

9999=1002+n3-13

n3=3000

Let, Y is the 4 digit numbers divisible by 7

1001,1008, ………, 9996

a=1001, d=7 & an=9996

9996=1001+n7-17

n7=1286

Now, 4 digit numbers divisible by both 3 and 7

1008,1029,...........9996

a=1008, d=21, an=9996

Now, 9996=1008+n-121

n=429

Hence, 4 digit numbers neither divisible by 7 nor 3

=9000-1286-3000+429

=5143

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2021 (31 Aug Shift 2)
The number of 4f electrons in the ground state electronic configuration of Gd2+ is [Atomic number of Gd=64]

Options:

The correct answer is:
7

Solution:
Electronic configuration of Gd=[Xe]4f75 d16 s2

Due to more stable exactly half-filled 4f7 shell, electron filling occur in a little higher energy 5d-orbital

So configuration of Gd2+=[Xe]4f75d1

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The number of 4-letter words, with or without meaning, each consisting of 2 vowels and 2 consonants, which can be formed from the letters of the word UNIVERSE without repetition is _____.

Options:

The correct answer is:
432

Solution:
Given,

We have to make four letter word from UNIVERSE each consisting 2 vowels and 2 consonants,

Now in UNIVERSE, we have,

E, E, I, U, (Vowels) + N, R, S, V (Consonants)

2 different vowels + 2 consonants can be arrange in 3C2·4C2·4!=3×6×24=432 ways

{C23 is taken because of repetition of letter E }

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (06 Apr Shift 2)
The number of 4 letter words (with or without meaning) that can be formed from the eleven letters of the word EXAMINATION is

Options:

The correct answer is:
2454

Solution:
EXAMINATION
2N,2A,2I,E,X,M,T,O
Case I: All are different, so 8P4=8!4!=8.7.6.5=1680
Case II: 2 same and 2 different, so 3C1.C2.4!2!=3.21.12=756
Case III: 2 same and 2 same, so 3C2.4!2!2!=3.6=18
Total number of words=1680+756+18=2454

Topic: Mathematics, Permutation Combination, JEE Main
The number of 4 letter words (with or without meaning) that can be formed from the eleven letters of the word EXAMINATION is

Options:

The correct answer is:
2454

Solution:
EXAMINATION
2N,2A,2I,E,X,M,T,O
Case I: All are different, so 8P4=8!4!=8·7·6·5=1680
Case II: 2 same and 2 different, so 3C1C274!2!=3·21·12=756
Case III: 2 same and 2 same, so 3C24!2!2!=3·6=18
Total number of words=1680+756+18=2454

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2020 (08 Jan Shift 2)
The number of 5-digited number which are not divisible by 5 and which consists of different odd digits is

Options:
A $24$
B $32$
C $96$
D $120$

The correct answer is:
$96$

Solution:
The numbers are formed from the digits
$$
1,3,5,7,9
$$
Total no. of five digits $=5$ !
Total no. of five digits divisible by $5=4$ !
$\therefore$ Required number of ways
$$
=5 !-4 !=4.24=96
$$

Topic: Mathematics, Permutation Combination, JEE Main
The number of 6 digit number that can be formed using the digits 0, 1, 2, 5, 7 and 9 which are divisible by 11 and no digit is repeated is:

Options:
A 36
B 60
C 72
D 48

The correct answer is:
60

Solution:
Only one possibility is there, (9,1,2) or (0,5,7) as in even or odd places
∴ Total number of required 6 digit numbers

=(9,1,2 in odd and 0,5,7 in even places)+(9,1,2 in even places and 0,5,7 in odd places)
=3!×3!+3! –2×3!
=36+24
=60

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2019 (10 Apr Shift 1)
The number of 6 -digit numbers of the form ababab (in base 10$)$ each of which is a product of exactly 6 distinct primes is

Options:
A 8
B 10
C 13
D 15

The correct answer is:
13

Solution:
$N=a b a b a b$
$\begin{array}{llll}1 < a \leq 9 & 0 < b \leq 9 & a, b \in \mathrm{l}\end{array}$
$N=10^{5} a+10^{4} b+10^{3} a+10^{2} b+10 a+b$
$=\left(10^{4}+10^{2}+1\right)(10 a+b)$
$=\left(10^{2}+10+1\right)\left(10^{2}-10+1\right)(10 a+b)$
$=3 \times 37 \times 13 \times 7(10 a+b)\quad ......\text {(1)}$
$\begin{array}{lcc}\text { then } & 10 \mathrm{a}+\mathrm{b} =\mathrm{P}_{1} \times \mathrm{P}_{2} & \mathrm{p}_{1}, \mathrm{p}_{2} \in \text { prime and } 10 \leq 10 \mathrm{a}+\mathrm{b} \leq 99 \\ \mathrm{a} & \mathrm{b} & 10 \mathrm{a}+\mathrm{b} \\ 1 & 0 & 10=2 \times 5 \\ 2 & 2 & 22=2 \times 11 \\ 3 & 4 & 34=2 \times 17 \\ 3 & 8 & 38=2 \times 19 \\ 4 & 6 & 46=2 \times 33 \\ 5 & 5 & 55=5 \times 11 \\ 5 & 8 & 58=2 \times 29 \\ 6 & 2 & 62=2 \times 31 \\ 7 & 4 & 74=2 \times 37 \\ 8 & 2 & 82=2 \times 41 \\ 8 & 5 & 85=5 \times 17 \\ 9 & 4 & 94=2 \times 47 \\ 9 & 5 & 95=5 \times 19\end{array}$

Topic: Mathematics, Permutation Combination, JEE Main
The number of 7 digit integers abcdefg, where a<b<c<d>e>f>g such that a,b,c,d,e,f,g∈1,2,3,....,9, are

Options:
A 700
B 20
C 720
D 800

The correct answer is:
720

Solution:
Case 1: d=7
6C3×3C3=20
Case 2: d=8
7C3×4C3=140
Case 3: d=9
8C3×5C3=560
Total = 20+140+560=720

Topic: Mathematics, Permutation Combination, JEE Main
The number of 9 digit numbers, that can be formed using all the digits of the number 123412341 so that the even digits occupy only even places, is ______

Options:

The correct answer is:
60

Solution:
According to the question,

1, 1, 1, 2, 2, 3, 3, 4, 4

there are four even digits as 2 and 4 are repeating two times each.
So, the even digits will occupy the even places in 4!(2! 2!) = 6 ways

3 and 1 are repeating two and three times respectively.
So, the remaining places will be occupied by odd digits in 5!(3! 2!)=10 ways

Therefore, total 9 digits that can be formed = 6×10 = 60.
Hence, the answer is 60.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (24 Jan Shift 1)
The number of acyclic structural isomers (including geometrical isomers) for pentene are

Options:

The correct answer is:
6

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (22 Jul Shift 1)
The number of aldol reaction (s) that occurs in the given transformation is

Options:
A 4
B 3
C 2
D 1

The correct answer is:
3

Solution:
In this sequence of reaction three times aldol reaction occurs first than cross Cannizaro reaction takes place as follows
CH3-CHO+3HCHOâOH-Aldol condensation 3timesCH2OH3-C-CHO
CH2OH3-C-CHOâOH-Cross cannizzaro reactionC-CH2OH4

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The number of aldol reaction(s) that occurs in the given transformation is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The number of alkenes (including stereoisomers) which can produce 2-butanol by the successive treatment of (i) B2H6 in tetrahydrofuran solvent and (ii) alkaline H2O2 solution is/are

Options:

The correct answer is:
2

Solution:
Structure of butan-2-ol is
CH3-C|HOH-CH2-CH3
Now, the simple format of HBO (hydroboration oxidant) reaction is
RâCH= CH 2 + B 2 H 6 â âTHFâ (Râ CH 2 CH 2 ) 3 B â â H 2 O 2 / O â Hâ RCH 2 CH 2 OH |
So, CH3-CH=CH-CH3 definitely an option as the mechanistic approach for the BH3 addition is as
( CH 3 CH 2 − CHCH 3 ) 3 B →   H 2 O 2 / O − H  CH 3 CH 2 − C | H OH CH 3
Now, CH3-CH=CH-CH3 can have both cis and trans-forms.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The number of alkenes (including stereoisomers) which can produce 2-butanol by the successive treatment of (i) \( \mathrm{B}_{2} \mathrm{H}_{6} \) in
tetrahydrofuran solvent and (ii) alkaline \( \mathrm{H}_{2} \mathrm{O}_{2} \) solution is/are

Options:

The correct answer is:
2

Solution:
Structure of butan-2-ol is
CH3-C|HOH-CH2-CH3
Now, the simple format of HBO (hydroboration oxidant) reaction is
RâCH= CH 2 + B 2 H 6 â âTHFâ (Râ CH 2 CH 2 ) 3 B â â H 2 O 2 / O â Hâ RCH 2 CH 2 OH |
So, CH3-CH=CH-CH3 definitely an option as the mechanistic approach for the BH3 addition is as
( CH 3 CH 2 − CHCH 3 ) 3 B →   H 2 O 2 / O − H  CH 3 CH 2 − C | H OH CH 3
Now, CH3-CH=CH-CH3 can have both cis and trans-forms.

Topic: Chemistry, Hydrocarbons, JEE Main
The number of all 2 -digit numbers $n$ such that $n$ is equal to the sum of the square of digit in its tens place and the cube of the digit in units place is

Options:
A 0
B 1
C 2
D 4

The correct answer is:
2

Solution:
$\mathrm{n}=\mathrm{ab}$
$\mathrm{ab}=\mathrm{a}^{2}+\mathrm{b}^{3}$
$\Rightarrow 10 \mathrm{a}+\mathrm{b}=\mathrm{a}^{2}+\mathrm{b}^{3}$
$\Rightarrow \mathrm{a}(10-\mathrm{a})+\mathrm{b}(1-\mathrm{b})(1+\mathrm{b})=0$
$\Rightarrow \mathrm{a}(10-\mathrm{a})=(\mathrm{b}-1)(\mathrm{b})(\mathrm{b}+1)$
If $\mathrm{b}=2 ; \mathrm{a}(10-\mathrm{a})=6 \quad \Rightarrow$ no value of $^{\prime} \mathrm{a}^{\prime}$
$\mathrm{b}=3 ; \mathrm{a}(10-\mathrm{a})=24 \quad \Rightarrow \mathrm{a} \in\{4,6\}$
{nos. are $43 \& 63\}$
$\mathrm{b}=4 ; \mathrm{a}(10-\mathrm{a})=60 \quad \Rightarrow$ no value of a
$\mathrm{b}=5 ; \mathrm{a}(10-\mathrm{a})=120 \quad \Rightarrow$ no need to check further
$\therefore$ nos. are $43 \& 63 .$

Topic: Mathematics, Binomial Theorem, JEE Main
The number of all 3×3 matrices A, with entries from the set -1,0,1 such that the sum of the diagonal elements of AAT is 3, is ___________.

Options:

The correct answer is:
672

Solution:
Let, A=aii3×3
tr(AAT)=3
a112+a122+a132+a212+.......+a332=3
Possible cases
0,0,0,0,0,0,1,1,1→10,0,0,0,0,0,-1,-1,-1→10,0,0,0,0,0,1,1,-1→30,0,0,0,0,0,-1,1,-1→3 9C6×8=84×8=672

Topic: Mathematics, Matrices, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The number of all 3 -digit numbers abc (in base10) for which $(a \times b \times c)+(a \times b) 6+(c \times a)+a+b+c=29$ is

Options:
A 6
B 10
C 14
D 18

The correct answer is:
14

Solution:
$\begin{array}{l}
(a \times b \times c)+(a \times b)+(c \times a)+(a+b+c)=29 \\
(1+a)(1+b)(1+c)=30 \\
=2 \times 3 \times 5 \rightarrow(a, b, c) \Rightarrow(1,2,3) \Rightarrow 6 \\
=1 \times 6 \times 5 \rightarrow(a, b, c) \Rightarrow(0,5,4) \Rightarrow 4 \\
=1 \times 3 \times 10 \rightarrow(a, b, 1) \Rightarrow(0,2,9) \Rightarrow 4 \\ \quad \frac{14}{} &{ }
\end{array}$

Topic: Mathematics, Permutation Combination, JEE Main
The number of all 8 digit odd numbers is

Options:
A $45 \times 10^6$
B $90 \times 10^6$
C $9 \times 10^8$
D $9 \times 10^6$

The correct answer is:
$45 \times 10^6$

Solution:
For this, we must ensure that the last digit (8th digit) is odd, which means it can only be $1,3,7,5$ or 9 and first digit can't be zero.
$\therefore$ The total no. of 8-digit odd numbers.
$=($ Choices for first digit $) \times($ Choices for the remaining 6 digits) $\times$ Choices for the last digit.
$=9 \times 10^6 \times 5=45 \times 10^6$

Topic: Mathematics, Permutation Combination, AP EAMCET, AP EAMCET 2023 (17 May Shift 1)
The number of all possible positive integral value of α for which the roots of the quadratic equation 6x2-11x+α=0 are rational numbers is:

Options:
A 5
B 3
C 4
D 2

The correct answer is:
3

Solution:
For ax2+bx+c=0, we have D=b2-4ac

D=121-24α
For rational roots, D must be perfect square.
∴α=3, 4, 5
∴ Number of values of α=3

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The number of all possible symmetric matrices of order 3×3 with each entry 1 or 2 and whose sum of diagonal elements is equal to 5, is

Options:

The correct answer is:
24

Solution:
Since, trace is equal to 5â diagonal elements are 1,2,2
âµ Matrix is also symmetric
Therefore, possible number of such matrices =3!2!Ã23=3Ã8=24

Topic: Mathematics, Matrices, JEE Main
The number of all possible three digit even numbers which can be formed with the condition that if 5 is one of the digits, then 7 is the next digit is equal to

Options:
A 5
B 325
C 345
D 365

The correct answer is:
365

Solution:
If 5 is at hundreds place, 7 will be at tens place and ones place will have 0,2,4,6 or 8. So, 5 numbers are possible.
If 5 is at tens place then, 7 will be at ones place but it won’t be an even number.
If 5 is at ones place, then it won’t be an even number. So, 5 can only be at hundreds place
5_7_x_→x=0,2,4,6,8→5 numbers
8_×9_×5_=360 numbers
(any number except 0,5 at hundreds place, any number except 5 at tens place and 0,2,4,6,8 at ones place)
Total numbers possible =5+360=365

Topic: Mathematics, Permutation Combination, JEE Main
The number of all possible values of $\theta$, where $0 < \theta < \pi$, for which the system of equations
$$
\begin{array}{r}
(y+z) \cos 3 \theta=(x y z) \sin 3 \theta \\
x \sin 3 \theta=\frac{2 \cos 3 \theta}{y}+\frac{2 \sin 3 \theta}{z}
\end{array}
$$
and $(x y z) \sin 3 \theta=(y+2 z) \cos 3 \theta$
$+y \sin 3 \theta$
have a solution $\left(x_0, y_0, z_0\right)$
with $\quad y_0 z_0 \neq 0$, is

Options:

The correct answer is:
3

Solution:
Given equations can be written as
$$
\begin{gathered}
x \sin 3 \theta-\frac{\cos 3 \theta}{y}-\frac{\cos 3 \theta}{z}=0 \\
x \sin 3 \theta-\frac{2 \cos 3 \theta}{y}-\frac{2 \sin 3 \theta}{z}=0 \\
\text { and } x \sin 3 \theta-\frac{2}{y} \cos 3 \theta \\
-\frac{1}{z}(\cos 3 \theta+\sin 3 \theta)=0
\end{gathered}
$$
Eqs. (ii) and (iii), implies
$$
\begin{aligned}
& 2 \sin 3 \theta=\cos 3 \theta+\sin 3 \theta \\
\Rightarrow & \sin 3 \theta=\cos 3 \theta \\
\therefore & \tan 3 \theta=1 \\
\Rightarrow & 3 \theta=\frac{\pi}{4}, \frac{5 \pi}{4}, \frac{9 \pi}{4} \text { or } \theta=\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{9 \pi}{12} .
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of \( \alpha \)-particles emitted during the radioactive decay chain from \( { }_{88}^{226} \mathrm{Ra} \) and ending at \( { }_{82}^{206} \mathrm{~Pb} \) is

Options:
A \( 5 \)
B \( 4 \)
C \( 6 \)
D \( 3 \)

The correct answer is:
\( 5 \)

Solution:
The mass number of ending nuclei reduced to 20 units,

One α24-particle has 2+ve units of charge and 4 units of mass, therefore if w subtract the mass number of product nuclei from reactant nuclei we get 20 units of mass, which can only be compensated by 5α-particle.

Ra88226→X78206+5α1020→X78206+4β-→Pb82206

Thus, 5 α-particles emitted during the radioactive decay.

Topic: Physics, Nuclear Physics, JEE Main
The number of amino acids and number of peptide bonds in a linear tetrapeptide (made of different amino acids) are respectively

Options:
A 4 and 4
B 5 and 5
C 5 and 4
D 4 and 3

The correct answer is:
4 and 3

Solution:
Combination of four amino acid units gives linear tetrapeptide, in which three peptide bonds are present.

Topic: Chemistry, Biomolecules, JEE Main
The number of amplitude modulated broadcast stations that can be accommodated in a $300 \mathrm{kHz}$ band width for the highest modulating frequency $15 \mathrm{kHz}$ will be:

Options:
A
20

B
10

C
8

D
15

The correct answer is:
10

Solution:
Given, modulating frequency $f_m$ $=15 \mathrm{KHz}$
$\therefore \quad$ Bandwidth of one channel $=2 f_m=30 \mathrm{kHz}$
$\therefore \quad$ No of channels accommodate
$$
=\frac{300 \mathrm{kHz}}{30 \mathrm{kHz}}=10
$$

Topic: Physics, Communication System, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The number of amplitude-modulated broadcast stations that can be accommodated in a 300 kHz bandwidth for the highest modulating frequency 15 kHz will be:

Options:
A 15
B 20
C 8
D 10

The correct answer is:
10

Solution:
If modulating frequency is 15 kHz then the bandwidth of one channel =30 kHz. A number of channels accommodate. =300 kHz30 kHz=10.

Topic: Physics, Communication System, JEE Main, JEE Main 2018 (15 Apr)
The number of $\sigma$ and $\pi$ bonds in a molecule of benzene is

Options:
A $6 \sigma$ and $9 \pi$
B $9 \sigma$ and $3 \pi$
C $12 \sigma$ and $3 \pi$
D $6 \sigma$ and $6 \pi$

The correct answer is:
$12 \sigma$ and $3 \pi$

Solution:
$\begin{aligned} & C-C \pi \text { bonds }=3 \\ & C-C \sigma \text { bonds }=6 \\ & C-H \sigma \text { bonds }=6\end{aligned}$
Hence ${ }^{12 \sigma}$ and ${ }^{3 \pi}$ bonds

Topic: Chemistry, Hydrocarbons, JEE Main
The number of $\sigma$ and $\pi$-bonds in pent-2-yne is

Options:
A 12 and 2
B 11 and 2
C 13 and 2
D 12 and 3

The correct answer is:
12 and 2

Solution:
Pent-2-yne is

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The number of $\alpha$ and $\beta$-particles emitted during the transformation of ${ }_{90} \mathrm{Th}^{232}$ to ${ }_{82} \mathrm{~Pb}^{208}$ are respectively

Options:
A $4,2$
B $2,2$
C $8,6$
D $6,4$

The correct answer is:
$6,4$

Solution:
${ }_{90}^{232} \mathrm{Th} \longrightarrow{ }_{82}^{208} \mathrm{~Pb}+n \alpha+m \beta$
Number of $\alpha$-particles, $n=\frac{232-208}{4}=6$
Now, for charge balanced
$$
\begin{aligned}
90 &=82+2 n-m \\
m &=4
\end{aligned}
$$
Number of $\beta$-particles, $m=4$

Topic: Chemistry, Chemical Kinetics, JEE Main