datavorous/entrance-exam-dataset huggingface dataset:
For an ideal solution of two components A & B, which of the following is true?

Options:
A Δ H mixing > 0 (zero)
B A-B interaction is stronger than A-A & B-B. interactions
C A-A, B-B & A-B interactions are identical
D Δ H mixing < 0 (zero)

The correct answer is:
A-A, B-B & A-B interactions are identical

Solution:
When two components A & B form an ideal solution, the interactions A – A, B – B and A – B are similar due to which no heat is absorbed or released and also volume after mixing remains same.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2014 (19 Apr Online)
For an infinitely long solid cylinder of radius R having a uniform density, the gravitational field at a distance r 1 is E1 and at a distance r 2 is E 2 , then

Options:
A E1<E2 for r1<r2<R
B E1>E2 for R<r1<r2
C both 1 and 2
D none of the above

The correct answer is:
both 1 and 2

Solution:
Graph of gravitational field versus distance from axis for a solid cylinder is given below

Topic: Physics, Gravitation, JEE Main
For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is 45, then the probability that he is unable to solve less than two problems is

Options:
A 20151549
B 164251548
C 316254548
D 5454549

The correct answer is:
5454549

Solution:
Let, X=The number of questions, student are unable to solve.
p=P (the student answers a question correctly) =45
⇒ q=1-p=15 (probability that the student does not answer a question correctly)
Also, n=50 (Total number of question)
Hence, required probability:
P(X<2)
=PX=0+PX=1
= 50C0 4550+ 50C1454915
=545 4549

Topic: Mathematics, Probability, JEE Main, JEE Main 2019 (12 Apr Shift 2)
For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is 45, then the probability that he is unable to solve less than two problems is

Options:
A 20151549
B 164251548
C 316254548
D 5454549

The correct answer is:
5454549

Solution:
Let, X= number of questions student is unable to solve.
p=P (the student answers a question correctly) =45
⇒ q=1-p=15 (probability that the student does not answer a question correctly)
Also, n=50 (Total number of question)
Hence, required probability:
P(X<2)
=PX=0+PX=1
= 50C0 4550+ 50C1454915
=545 4549

Topic: Mathematics, Probability, JEE Main
For an intrinsic semiconductor at temperature $T=0 \mathrm{~K}$, which of the following statement is true?

Options:
A All energy states in the valence band are filled with electrons and all energy states in the conduction band are empty of electrons.
B All energy states in the valence band are empty of electrons and all energy states in the conduction band are filled with electrons.
C All energy states in the valence and conduction band are filled with holes.
D All energy states in the valence and conduction band are filled with electrons.

The correct answer is:
All energy states in the valence band are filled with electrons and all energy states in the conduction band are empty of electrons.

Solution:
Intrinsic Semiconductors:
- A pure semiconductor is called an intrinsic semiconductor. It has thermally generated current carriers.
- They have four electrons in the outermost orbit of the atom and atoms are held together by a covalent bond.
- Because of fewer charge carriers at room temperature, intrinsic semiconductors have low conductivity so they have no practical use.
Explanation:
- An ideal, perfectly pure semiconductor (with no impurities) is called an intrinsic semiconductor.
- At absolute zero temperature valence band is full of electrons and the conduction band is empty, hence there are no free electrons in the conduction band and holes in the valence band.
- The charge carrier concentration is zero. Hence intrinsic semiconductor behaves like an insulator

Topic: Physics, Semiconductors, JEE Main
For an ionic crystal of the general formula $A X$ and coordination number 6 , the value of radius ratio will be

Options:
A Greater than 0.73
B In between 0.73 and 0.41
C In between 0.41 and 0.22
D Less than 0.22

The correct answer is:
In between 0.73 and 0.41

Solution:
(b) The structural arrangement of co-ordination number ' 6 ' is octahedral and its radius ratio is $0.414-0.732$. The example of octahedral is $\mathrm{KCl}$ and $\mathrm{NaCl}$.

Topic: Chemistry, Solid State, JEE Main
For an ionic crystal of the type $A B$, the value of (limiting) radius ratio is 0.40 . The value suggests that the crystal structure should be

Options:
A Octahedral
B Tetrahedral
C Square planar
D Plane triangle

The correct answer is:
Tetrahedral

Solution:
The crystals in which radius ratio value is found between $0.225-0.414$ shows tetrahedral crystal structure.

Topic: Chemistry, Solid State, JEE Main
For an isothermal expansion of a perfect gas, the value of $\frac{\Delta P}{P}$ is equal to

Options:
A $-\gamma^{1 / 2} \frac{\Delta V}{V}$
B $-\frac{\Delta V}{V}$
C $-\gamma \frac{\Delta V}{V}$
D $-\gamma^{2} \frac{\Delta V}{V}$

The correct answer is:
$-\frac{\Delta V}{V}$

Solution:
Differentiate $P V=$ constant w.r.t $V$
$\Rightarrow P \Delta V+V \Delta P=0 \quad \Rightarrow \frac{\Delta P}{P}=-\frac{\Delta V}{V}$

Topic: Physics, Thermodynamics, JEE Main
For an $L E D$ to emit light in visible region of the electromagnetic spectrum, it can have energy band gap in the range of, (Plank's constant, $h=6.6 \times 10^{-34} \mathrm{Js}$ and speed of light, $e=3 \times 10^8 \mathrm{~ms}^{-1}$ in vacuum)

Options:
A 0.1 eV to 0.4 eV
B 0.9 eV to 1.6 eV
C 1.7 eV to 3.1 eV
D 0.5 eV to 0.8 eV

The correct answer is:
1.7 eV to 3.1 eV

Solution:
Key Idea Frequency of visible rays range from $4 \times 10^{14} \mathrm{~Hz}$ to $7 \times 10^{14} \mathrm{~Hz}$.
Maximum and minimum energy of photon in visible light,
$$
\begin{aligned}
E_{\max } & =h v=\frac{6.6 \times 10^{-34} \times 7 \times 10^{14}}{1.6 \times 10^{-19}} \\
& =2.88 \mathrm{eV} \simeq 3 \mathrm{eV} \\
E_{\min } & =\frac{6.6 \times 10^{-34} \times 4 \times 10^{14}}{1.6 \times 10^{-19}} \\
& =1.65 \mathrm{eV} \simeq 1.7 \mathrm{eV}
\end{aligned}
$$
So, the correct option matches between the energy $1.7 \mathrm{eV}$ to $3 \mathrm{eV}$ is $(\mathrm{c})$,

Topic: Physics, Semiconductors, JEE Main
For an object placed at a distance 2.4 m from a lens, a sharp focused image is observed on a screen placed at a distance 12 cm from the lens. A glass plate of refractive index 1.5 and thickness 1 cm is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen?

Options:
A 0.8 m
B 3.2 m
C 1.2 m
D 5.6 m

The correct answer is:
3.2 m

Solution:
Given:
object using the lens formula:
\(\begin{aligned}& \frac{1}{f}=\frac{1}{v \prime}-\frac{1}{w^{\prime}} \\\Rightarrow & \frac{21}{240} \mathrm{~cm}=\frac{3}{35} \mathrm{~cm}-\frac{1}{u^{\prime}} \\\Rightarrow \frac{1}{u^{\prime}} & =\frac{3}{35} \mathrm{~cm}-\frac{21}{240} \mathrm{~cm} \\\Rightarrow & \frac{1}{u^{\prime}}=\frac{1}{5}\left[\frac{3}{7} \mathrm{~cm}-\frac{21}{48} \mathrm{~cm}\right] \\\Rightarrow \frac{1}{u^{\prime}} & =\frac{1}{5}\left[\frac{144-147}{336}\right] \\\Rightarrow u^{\prime} & =-560 \mathrm{~cm}=-5.6 \mathrm{~m}\end{aligned}\)
Hence, the object should be at \(5.6 \mathrm{~m}\) in front of the lens.
Hence, the oblect should be moved by 5.6-24= \(3.2 \mathrm{~cm}\).

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (25 Jul Shift 2)
For an RLC circuit driven with voltage of amplitude vm and frequency ω0=1LC the current exhibits resonance. The quality factor, Q is given by

Options:
A CRω0
B ω0LR
C ω0RL
D R(ω0C)

The correct answer is:
ω0LR

Solution:
The quality factor of an L.C.R. circuit is given by,

(Quality factor) Q=1RLC...(i)

Where L,R,C are the inductance, resistance, and conductance of the L.C.R. circuit. The given resonance frequency is, ω0=1LC

⇒C=1ω0L ...(ii)

From (i) and (ii)

⇒Q=1RL×ω0L⇒Q=ω0LR

Topic: Physics, Alternating Current, JEE Main, JEE Main 2018 (08 Apr)
For an RLC circuit driven with voltage of amplitude \( v_{m} \) and frequency \( \omega_{0}=\frac{1}{\sqrt{L C}} \) the current exhibits resonance. The quality factor, \( Q \) is given by

Options:
A \( \frac{C R}{\omega_{0}} \)
B \( \frac{\omega_{0} L}{R} \)
C \( \frac{\omega_{0} R}{L} \)
D \( \frac{R}{\left(\omega_{0} C\right)} \)

The correct answer is:
\( \frac{\omega_{0} L}{R} \)

Solution:
The quality factor of an L.C.R. circuit is given by,

(Quality factor) Q=1RLC...(i)

Where L,R,C are the inductance, resistance, and conductance of the L.C.R. circuit. The given resonance frequency is, ω0=1LC

⇒C=1ω0L ...(ii)

From (i) and (ii)

⇒Q=1RL×ω0L⇒Q=ω0LR

Topic: Physics, Alternating Current, JEE Main
For any θ∈π4,π2, the expression 3sinθ-cosθ4+6sinθ+cosθ2+4 sin6θ equals:

Options:
A 13-4cos2θ+6cos4θ
B 13-4cos2θ+6sin2θcos2θ
C 13-4cos6θ
D 13-4cos4θ+2sin2θcos2θ

The correct answer is:
13-4cos6θ

Solution:
3sinθ-cosθ4+6cosθ+sinθ2+4sin6θ

=3sinθ-cosθ22+6cosθ+sinθ2+4sin6θ

=3cos2θ+sin2θ-2sinθcosθ2+6cos2θ+sin2θ+2sinθcosθ+4sin6θ

=31-2sinθcosθ2+61+2sinθcosθ+4sin6θ

=31-4sinθcosθ+4sin2θcos2θ+61+2sinθcosθ+4sin6θ

=9+12sin2θcos2θ+4sin6θ

=9+121-cos2θcos2θ+41-cos2θ3

=9+12cos2θ-12cos4θ+41-3cos2θ+3cos4θ-cos6θ

=13-4cos6θ

Topic: Mathematics, Trigonometric Equations, JEE Main
For any θ∈π4,π2, the expression 3sinθ-cosθ4+6sinθ+cosθ2+4 sin6θ equals:

Options:
A 13-4cos2θ+6cos4θ
B 13-4cos2θ+6sin2θcos2θ
C 13-4cos6θ
D 13-4cos4θ+2sin2θcos2θ

The correct answer is:
13-4cos6θ

Solution:
3sinθ-cosθ4+6cosθ+sinθ2+4sin6θ

=3sinθ-cosθ22+6cosθ+sinθ2+4sin6θ

=3cos2θ+sin2θ-2sinθcosθ2+6cos2θ+sin2θ+2sinθcosθ+4sin6θ

=31-2sinθcosθ2+61+2sinθcosθ+4sin6θ

=31-4sinθcosθ+4sin2θcos2θ+61+2sinθcosθ+4sin6θ

=9+12sin2θcos2θ+4sin6θ

=9+121-cos2θcos2θ+41-cos2θ3

=9+12cos2θ-12cos4θ+41-3cos2θ+3cos4θ-cos6θ

=13-4cos6θ

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2019 (09 Jan Shift 1)
For any given series of spectral lines of atomic hydrogen, let Δv-=v-max-v-min be the difference in maximum and minimum wave number in cm-1.

The ratio Δv-Lyman/Δv-Balmar is

Options:
A 5:4
B 27:5
C 4:1
D 9:4

The correct answer is:
9:4

Solution:
=vmax-vminlymanvmax-vminBalmar

∵ν=1λ=Ech

=Emax-EminlymanEmax-EminBalmar

E=hν, ν=Eh

=E1→∞- E1→2E2→∞- E2→3

=13.6 11-1∞-13.6 11-1413.6 14-1∞-13.6 14-19

=1-0-1+1414-0-14+19 =1/41/9=94

Topic: Chemistry, Structure of Atom, JEE Main
For any given series of spectral lines of atomic hydrogen, let Δv-=v-max-v-min be the difference in maximum and minimum wave number in cm-1.

The ratio Δv-Lyman/Δv-Balmar is

Options:
A 5:4
B 27:5
C 4:1
D 9:4

The correct answer is:
9:4

Solution:
=vmax-vminlymanvmax-vminBalmar

∵ν=1λ=Ech

=Emax-EminlymanEmax-EminBalmar

E=hν, ν=Eh

=E1→∞- E1→2E2→∞- E2→3

=13.6 11-1∞-13.6 11-1413.6 14-1∞-13.6 14-19

=1-0-1+1414-0-14+19 =1/41/9=94

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2019 (09 Apr Shift 1)
For any integer $n \geq 1$, the remainder when the expression $n^5-5 n^3+4 n+139$ is divided by 120 is

Options:
A 9
B 19
C 29
D 39

The correct answer is:
19

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Binomial Theorem, AP EAMCET, AP EAMCET 2017 (26 Apr Shift 2)
For any integer $n \geq 1$, the sum $\sum_{k=1}^n k(k+2)$ is equal to

Options:
A $\frac{n(n+1)(n+2)}{6}$
B $\frac{n(n+1)(2n+1)}{6}$
C $\frac{n(n+1)(2n+7)}{6}$
D $\frac{n(n+1)(2n+9)}{6}$

The correct answer is:
$\frac{n(n+1)(2n+7)}{6}$

Solution:
Now, $\begin{aligned} & \sum_{k=1}^n k(k+2) \\ & =\sum_{k=1}^n\left(k^2+2 k\right)=\sum_{k=1}^n k^2+2 \sum_{k=1}^n k \\ & =\frac{n(n+1)(2 n+1)}{6}+\frac{2 \cdot n(n+1)}{2} \\ & =n(n+1)\left(\frac{2 n+1}{6}+1\right) \\ & =\frac{n(n+1)(2 n+7)}{6}\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main
For any positive integer $\mathrm{n}$, if $4^{\mathrm{n}}-3 \mathrm{n}$ is divided by 9 , then what is the remainder?

Options:
A 8
B 6
C 4
D 1

The correct answer is:
1

Solution:
Using binomial theorem. $4^{\mathrm{n}}-3 \mathrm{n}=(1+3)^{\mathrm{n}}-3 \mathrm{n}$
$=1+\mathrm{n} .3+\frac{\mathrm{n}(\mathrm{n}-1)}{2 !} 3^{2}+\ldots-3 \mathrm{n}$
$=1+\frac{\mathrm{n}(\mathrm{n}-1)}{2 !} \cdot 3^{2}+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{3 !} \cdot 3^{3}+\ldots$
$\Rightarrow 4^{\mathrm{n}}-3 \mathrm{n}=9\left\{\frac{\mathrm{n}(\mathrm{n}-1)}{2 !}+\ldots\right\}+1$
Thus, when $4^{\mathrm{n}}-3 \mathrm{n}$ is divided by 9 , the remainder is 1

Topic: Mathematics, Binomial Theorem, NDA, NDA 2006 (Phase 2)
For any positive n, let fn=4n+4n2-12n+1+2n-1. Then, âk=140fk100 is equal to

Options:

The correct answer is:
3.64

Solution:
Let, x=2n+1 and y=2n-1âx2+y2=4x
x2-y2=2
Also, xy=4n2-1
So,
fn=x2+y2+xyx+y=x3-y3x2-y2=x3-y32
=122n+1322n-132
Substituting n=1 to 40, we get
f1=12332-132
f2=12532-332
f3=12732-532
f40=128132-7932
âân=140=128132-132=729-12=364

Topic: Mathematics, Sequences and Series, JEE Main
For any real number x, let x denote the largest integer less than or equal to x. Let f be a real-valued function defined on the interval -10,10 by

fx=x-x,if x is odd1+x-x,if x is even

Then, the value of π210∫-1010fx cosπxdx is

Options:
A 4
B 2
C 1
D 0

The correct answer is:
4

Solution:
fx=x-1,1-x,1≤x<20≤x<1

fx is periodic with period 2, and is even.

∴I=∫-1010 fx cosπx dx

= 2∫010fx cosπx dx

=2×5∫02fx cosπx dx

=10 ∫011-x cosπx dx+∫12x-1 cosπx dx=10I1+I2

I2=∫12x-1 cosπx dx

Put x-1=t

I2=-∫01tcosπtdt

I1=∫011-x cosπx dx=-∫01xcosπx dx

∴I=10-2∫01xcosπx dx=-20xsinπxπ+cosπxπ201

=-20-1π2-1π2=40π2

∴π210I=4

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (25 Jul Shift 1)
For any square matrix $A, A A^T$ is a

Options:
A Unit matrix
B Symmetric matrix
C Skew symmetric matrix
D Diagonal matrix

The correct answer is:
Symmetric matrix

Solution:
We have, $\left(A A^T\right)=\left(A^T\right)^T A^T=A A^T$ (by reversal law)
$\therefore A A^T$ is symmetric matrix.

Topic: Mathematics, Matrices, JEE Main
For any three positive real numbers a, b and c. If 925a2+b2+25c2-3ac=15b3a+c. Then

Options:
A b, c and a are in G.P.
B b, c and a are in A.P.
C a, b and c are in A.P.
D a, b and c are in G.P.

The correct answer is:
b, c and a are in A.P.

Solution:
225a2+9b2+25c2-75ac-45ab-15bc=0
⇒ 215a2+3b2+5c2-15a5c-15a3b-3b5c=0
⇒ 215a2+23b2+25c2-215a5c-215a3b-23b5c=0
⇒15a-3b2+3b-5c2+5c-15a2=0
⇒15a-3b=3b-5c=5c-15a=0
⇒15a=3b=5c
⇒ a1=b5=c3=k
⇒ a=k, b=5k, c=3k
⇒ a, c, b are in A.P.
⇒ b, c, a are in A.P.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2017 (02 Apr)
For any two real numbers $\theta$ and $\phi$ we define $\theta$. R $\phi$ if and only if $\sec ^{2} \theta-\tan ^{2} \phi=1$. The relation $R$ is

Options:
A reflexive but not transitive
B symmetric but not reflexive
C both reflexive and symmetric but not transitive
D an equivalence relation

The correct answer is:
an equivalence relation

Solution:
Given relation is defined as $\theta R \phi$ such that $\sec ^{2} \theta-\tan ^{2} \phi=1$
For Reflexive
When $\theta R \theta$
$\sec ^{2} \theta-\tan ^{2} \theta=1$
$\Rightarrow 1=1,$ which is true.
Thus, it is reflexive.
For Symmetric When $\theta R \phi$
$\begin{array}{l}
\sec ^{2} \theta-\tan ^{2} \phi=1 \\
\Rightarrow\left(1+\tan ^{2} \theta\right)-\left(\sec ^{2} \phi-1\right)=1 \\
\Rightarrow 2+\tan ^{2} \theta-\sec ^{2} \phi=1 \\
\Rightarrow \sec ^{2} \phi-\tan ^{2} \theta=1
\end{array}$
$\Rightarrow \phi F \theta$
Thus, it is symmetric.
For Transitive
When $\theta R \phi$ and $\phi$ Fiy, then
$\sec ^{2} \theta-\tan ^{2} \phi=1$
and $\sec ^{2} \phi-\tan ^{2} \psi=1$
Now, $\theta$ R $\psi$
Then. $\sec ^{2} \theta-\tan ^{2} \psi=1$
$\Rightarrow \sec ^{2} \theta-\tan ^{2} \psi+1=1+1$
$\Rightarrow \sec ^{2} \theta-\tan ^{2} \psi+\sec ^{2} \phi-\tan ^{2} \phi=1+1$
$\Rightarrow \theta R \phi$ and $\phi R \psi$
Thus, it is transitive. Hence, it is an equivalence relation.

Topic: Mathematics, Sets and Relations, JEE Main
For any two sets $\mathrm{A}$ and $\mathrm{B}, \mathrm{A}-(\mathrm{A}-\mathrm{B})$ equals

Options:
A $\mathrm{B}$
B $\mathrm{A}-\mathrm{B}$
C $\mathrm{A} \cap \mathrm{B}$
D $\mathrm{A}^{\mathrm{C}} \cap \mathrm{B}^{\mathrm{C}}$

The correct answer is:
$\mathrm{A} \cap \mathrm{B}$

Solution:
Hints: $A-(A-B)=A-\left(A \cap B^c\right)=A \cap\left(A \cap B^c\right)^c=A \cap\left(A^c \cup B\right)=\left(A \cap A^c\right) \cup(A \cap B)=A \cap B$

Topic: Mathematics, Sets and Relations, JEE Main
For any two sets A and B, the values of A-B∪BC is equal to

Options:
A AC∩BC
B A∪B
C A-B
D B-A

The correct answer is:
AC∩BC

Solution:
( A−B )∪B=( A∩ B C )∪B                           =( A∪B )∩( B C ∪B )                           =( A∪B )∩( U )                           =A∪B
Hence,
[ ( A−B )∪B ] C = ( A∪B ) C = A C ∩ B C

Topic: Mathematics, Sets and Relations, JEE Main
For any two statement p and q, the negative of the expression p∨~p∧q is

Options:
A ~p∨~q
B p∧q
C ~p∧~q
D

p↔q

The correct answer is:
~p∧~q

Solution:
~p∨~p∧q

Using ~a∨b=~a∧~b, ~a∧b=~a∨~b & ~~a=a, we get

=~p∧~~p∧q

=~p∧~~p∨q

=∼p∧p∨∼q

Using distributive law, i.e. a∧b∨c=a∧b∨a∧c, we get

=~p∧p∨~p∧~q

Using ~a∧a=C, where C represents contradiction, we get

=C∨~p∧~q

Using C∨a=a, we get

=~p∧~q.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2019 (09 Apr Shift 1)
For any two vectors $\mathbf{a}$ and $\mathbf{b},(\mathbf{a} \times \mathbf{b})^2$ is equal to

Options:
A $a^2-b^2$
B $a^2+b^2$
C $a^2 b^2-(\mathbf{a} \cdot \mathbf{b})^2$
D None of these

The correct answer is:
$a^2 b^2-(\mathbf{a} \cdot \mathbf{b})^2$

Solution:
$\begin{aligned}
&(\mathbf{a} \times \mathbf{b})^2=(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{a} \times \mathbf{b})=(a b \sin \theta \hat{\mathbf{n}})(a b \sin \theta \hat{\mathbf{n}}) \\
&=a^2 b^2 \sin ^2 \theta=a^2 b^2(1\left.-\cos ^2 \theta\right) \\
&= a^2 b^2-a^2 b^2 \cos ^2 \theta=a^2 b^2-(\mathbf{a} \cdot \mathbf{b})^2 .
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
For any vectora a the value of $(\vec{a} \times \hat{i})^2+(\vec{a} \times \hat{j})^2+(\vec{a} \times \hat{k})^2$ is equal to

Options:
A
$3 \vec{a}^2$

B
$\vec{a}^2$

C
$2 \vec{a}^2$

D
$4 \vec{a}^2$

The correct answer is:
$2 \vec{a}^2$

Solution:
Let $\vec{a}=x \hat{i}+y \hat{j}+z \hat{k}$
$$
\begin{aligned}
& \vec{a} \times \hat{\mathbf{i}}=z \hat{\mathbf{j}}-y \hat{\mathbf{k}} \\
& \Rightarrow(\vec{a} \times \hat{\mathbf{i}})^2=y^2+z^2
\end{aligned}
$$
similarly $(\vec{a} \times \hat{j})^2=x^2+z^2$
and $(\vec{a} \times \hat{k})^2=x^2+y^2 \Rightarrow(\vec{a} \times \hat{i})^2=y^2+z^2$
similarly $(\vec{a} \times \hat{j})^2=x^2+z^2$
and $(\vec{a} \times \hat{k})^2=x^2+y^2$
$$
\Rightarrow(\vec{a} \times \hat{i})^2+(\vec{a} \times \hat{j})^2+(\vec{a} \times \hat{k})^2=2\left(x^2+y^2+z^2\right)=2 \vec{a}^2 .
$$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2005
For being functional, a protein has to acquire at least

Options:
A Primary structure
B Secondary structure
C Tertiary structure
D Quaternary structure.

The correct answer is:
Tertiary structure

Solution:
Most of the proteins are highly functional in their tertiary structure. The long chain of peptides is folded upon itself to form a hollow ball-like structure. The disulfide linkages that are present in the proteins give the domain structure and that enables their functional activity.

Topic: Chemistry, Biomolecules, NEET
For $\alpha$ belonging to an interval of length $\beta$, suppose $(\alpha,-\alpha)$ is an interior point of the ellipse $4 x^2+5 y^2=1$. Then,$(6 \beta-4)^{201}+201=$

Options:
A $202$
B $0$
C $402$
D $201$

The correct answer is:
$201$

Solution:
$\because(\alpha,-\alpha)$ lies inside $4 x^2+5 y^2-1=0$
$\begin{aligned} & \Rightarrow \quad 4 \alpha^2+5 \alpha^2-1 < 0 \\ & \Rightarrow \quad 9 \alpha^2-1 < 0 \Rightarrow 9 \alpha^2 < 1 \\ & \Rightarrow \quad \alpha^2 < \frac{1}{9} \Rightarrow-\frac{1}{3} < \alpha < \frac{1}{3} \\ & \Rightarrow \quad \alpha \in\left(-\frac{1}{3}, \frac{1}{3}\right)\end{aligned}$
Thus, $\beta=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$
Then, $(6 \beta-4)^{201}+201$
$=\left(6 \times \frac{2}{3}-4\right)^{201}+201=201$

Topic: Mathematics, Ellipse, JEE Main
For $\mathrm{BF}_3$ molecule which of the following is true ?

Options:
A B-atom is $\mathrm{sp}^2$ hybridised.
B There is a $p \pi-p \pi$ back bonding in this molecule.
C Observed $B-F$ bond length is found to be less than the expected bond length.
D All of these

The correct answer is:
All of these

Solution:
Decrease in $B-F$ bond length is due to delocalised $p \pi-p \pi$ bonding between filled p-orbital of $F$ atom and vacant p-orbital of $B$ atom.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
For cell reaction, $\mathrm{Zn}+\mathrm{Cu}^{2+} \rightarrow \mathrm{Zn}^{2+}+\mathrm{Cu}$, cell representation is

Options:
A $\mathrm{Zn}\left|\mathrm{Zn}^{2+}\right|\left|C u^{2+}\right| \mathrm{Cu}$
B $\mathrm{Cu}\left|\mathrm{Cu}^{2+} \| \mathrm{Zn}^{2+}\right| \mathrm{Zn}$
C $\mathrm{Cu}\left|\mathrm{Zn}^{2+} \| \mathrm{Zn}\right| \mathrm{Cu}^{2+}$
D $\mathrm{Cu}^{2+}\left|\mathrm{Zn} \| \mathrm{Zn}^{2+}\right| \mathrm{Cu}$

The correct answer is:
$\mathrm{Zn}\left|\mathrm{Zn}^{2+}\right|\left|C u^{2+}\right| \mathrm{Cu}$

Solution:
Electrode on which oxidation occurs is written on L.H.S and the other on the R.H.S as represented by $Z n\left|Z_n^{2+} \| C u^{2+}\right| C u$

Topic: Chemistry, Electrochemistry, JEE Main
For certain metal, incident frequency Î½ is five times threshold frequency Î½0 and the maximum velocity of the photoelectrons is 8 Ã 106 m s-1. If incident photon frequency is 2Î½0, the maximum velocity of photoelectrons will be

Options:
A 4Ã106 m s-1
B 6Ã106 m s-1
C 8Ã106 m s-1
D 1Ã106 m s-1

The correct answer is:
4Ã106 m s-1

Solution:
According to Einstein's photoelectric equation
hν=hν0+12mvmax2
or12mvmax2=h⁡ν⁡-h⁡ν⁡0
According to the given problem
12m⁡8×1062=h⁡5ν⁡0-ν⁡0…i
12m⁡vmax2=h⁡2ν⁡0-ν⁡0…ii
Dividing equation (i) by (ii)
8×1062vmax2=4ν⁡0ν⁡0
vmax2=8×10624
vmax=8×1062=4×106 m s-1

Topic: Physics, Dual Nature of Matter, JEE Main
For changing the capacitance of a given paralle plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is 34d, where d is the separation between the plates of parallel plate capacitor. The new capacitance C' in terms of original capacitance C0 is given by the following relation :

Options:
A C'=3+K4KC0
B C'=4+K3C0
C C'=4K K+3C0
D C'=43+KC0

The correct answer is:
C'=4K K+3C0

Solution:
C0=ϵ0A d

C'=C1 and C2 in series.

i.e. 1C'=1C1+1C2

1C'=3d/4ϵ0KA+d/4ϵ0 A

1C'=d4ϵ0A3+KK

C'=4KC03+K

Topic: Physics, Capacitance, JEE Main, JEE Main 2021 (16 Mar Shift 1)
For coagulation of 50 mL of a sol, 10 mL of 0.5M Cl- ion solution is required. What is the coagulating value of Cl- ion solution (Nearest integer)

NOTE: NTA question has been changed because it had errors.}

Options:

The correct answer is:
1

Solution:
Coagulation value=Number of millimoles of electrolyte required Volume of colloidal solution in L =10×0.5×100050=100

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2021 (20 Jul Shift 2)
For coagulation of arsenious sulphide sol, which of the following salt solutions will be most effective?

Options:
A Na3PO4
B NaCl
C AlCl3
D BaCl2

The correct answer is:
AlCl3

Solution:
According to Hardy's -schulz rule, greater the valency of the active ion or flocculating ion greater will be its coagulating power. S-2 is -ve charged colloid so cation having more +ve charge, more will be coagulating power.
Hence Al+3 having maximum coagulating power

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2019 (09 Jan Shift 2)
For coagulation of Arsenious sulphide sol, which one of the following salt solution will be most effective

Options:
A AlCl3
B NaCl
C BaCl2
D Na3PO4

The correct answer is:
AlCl3

Solution:
Arsenious sulphide is -ve charged colloid. Hence, cation having maximum positive charge will be most effective for coagulation. (Hardy Schulz rule)
Al3+>Ba2+>Na+ (coagulating power)

Topic: Chemistry, Surface Chemistry, JEE Main