datavorous/entrance-exam-dataset huggingface dataset:
$A$ and $B$ are two metals with threshold frequencies $1.8 \times 10^{14} \mathrm{~Hz}$ and $2.2 \times 10^{14} \mathrm{~Hz}$. Two identical photons of energy $0.825 \mathrm{eV}$ each are incident on them. Then photoelectrons are emitted by (Take $h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ )

Options:
A $B$ alone
B $A$ alone
C neither $A$ nor $B \quad$
D both $A$ and $B$

The correct answer is:
$A$ alone

Solution:
Threshold energy of $A$ is
$$
\begin{aligned}
E_{A} &=h v_{A}=6.6 \times 10^{-34} \times 1.8 \times 10^{14} \\
&=11.88 \times 10^{-20} \mathrm{~J} \\
&=\frac{11.88 \times 10^{-20}}{1.6 \times 10^{-19}} \mathrm{eV}=0.74 \mathrm{eV}
\end{aligned}
$$
Similarly, $E_{B}=0.91 \mathrm{eV}$
Since, the incident photons have energy greater than $E_{A}$ but less than $E_{B}$. So, photoelectrons will be emitted from metal $A$ only.

Topic: Physics, Dual Nature of Matter, JEE Main
A and B are two points on a uniform metal ring whose centre is C. The angle ACB=Î¸ . A and B are maintained at two different constant temperatures. When Î¸=180o , the rate of total heat flow from A to B is 1.2 W. When Î¸=90o, this rate will be

Options:
A 0.6Â W
B 0.9Â W
C 1.6 W
D 1.8Â W

The correct answer is:
1.6 W

Solution:
Let total resistance of the ring be R
for Î¸=180o , we can consider the ring as a parallel combination of two sections of resistance R2 each.
â´ Req=R4
â´ Rate of heat flow

P1=1.2=ÎTR4

ââT=0.3R
for, Î¸=90o we can consider the ring as a parallel combination of two sections of resistance, R4 & 3R4
â´ Req=R4.3R4R4+3R4=3R16
â´ P2=ÎT3R16=163R0.3R=1.6 watt

Topic: Physics, Thermal Properties of Matter, NEET
A and B are two radioactive substances whose half-lives are 1 and 2 years respectively. Initially, 10 g of A and,1 g of B is taken. The time (approximate ) after which they will have the same quantity remaining is

Options:
A 6.62 year

B 5 year

C 3.2 year

D 7 year

The correct answer is:
6.62 year

Solution:
N⁡=N⁡012t12
⇒N⁡A⁡=1012t1 and N⁡B⁡=112t2
Given
⇒1012t=12t2 ⇒10=12-t2
⇒ 1 0 = 2 t / 2
Taking log on both sides
log 1 0 = t 2 log 1 0 2 ⇒ 1 = t 2 × 0 · 3 0 1 0
⇒ t = 6 · 6 2 years

Topic: Physics, Nuclear Physics, JEE Main
A and B are two sources generating sound waves. A listener is situated at $\mathrm{C}$. The frequency of the source at $\mathrm{A}$ is $500 \mathrm{~Hz}$. A, now, moves towards $\mathrm{C}$ with a speed $4 \mathrm{~m} / \mathrm{s}$. The number of beats heard at $\mathrm{C}$ is 6 . When $\mathrm{A}$ moves away from $C$ with speed $4 \mathrm{~m} / \mathrm{s}$, the number of beats heard at $\mathrm{C}$ is 18 . The speed of sound is $340 \mathrm{~m} / \mathrm{s}$. The frequency of the source at $\mathrm{B}$ is :

Options:
A
$500 \mathrm{~Hz}$

B
$506 \mathrm{~Hz}$

C
$512 \mathrm{~Hz}$

D
$494 \mathrm{~Hz}$

The correct answer is:
$512 \mathrm{~Hz}$

Solution:
$\mathrm{f}=500 \mathrm{~Hz}$
Case 1: When source is moving towards stationary listener
apparent frequency $\eta^{\prime}=\eta\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right)$
$$
=500\left(\frac{340}{336}\right)=506 \mathrm{~Hz}
$$
Case 2 : When source is moving away from the stationary listener
$$
\eta^{\prime \prime}=\eta\left(\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right)=500\left(\frac{340}{344}\right)=494 \mathrm{~Hz}
$$
In case 1 number of beats heard is 6 and in case 2 number of beats heard is 18 therefore frequency of the source at $\mathrm{B}=512 \mathrm{~Hz}$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2013 (22 Apr Online)
A and B are two substances undergoing radioactive decay in a container. The half life of A is 15 min and that of B is 5 min. If the initial concentration of B is 4 times that of A and they both start decaying at the same time, how much time will it take for the concentration of both of them to be same? ________min.

Options:

The correct answer is:
15

Solution:
For a first order reaction:
⇒[A]t=[A]0e-kt

For A : Let [A]t be y and [A]0 be x
For a first order reaction:
∴k=ln2t1/2=ln215 min

⇒y=xe-kt

=xe-ln215t ....(1)

Similarly For B:
[B]t=[B]0e-kt

Let [B]t=y;[B]0=4x;k=ln2t1/2=ln25 min

⇒y=4xe-ln25t ....(2)
Equating equation(1) and (2) we get:

⇒xe-ln215t=4xe-ln25t

⇒t×ln 25-ln 215=ln 4

⇒t×ln215-115=2 ln2

⇒etln 25-ln 215=4

⇒t=15 min

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2023 (01 Feb Shift 1)
$\vec{A}$ and $\vec{B}$ are two vectors given by $\vec{A}=2 \hat{i}+3 \hat{j}$ and $\vec{B}=\hat{i}+\hat{j}$. The magnitude of the component of $\vec{A}$ along $\vec{B}$ is

Options:
A $\frac{5}{\sqrt{2}}$
B $\frac{3}{\sqrt{2}}$
C $\frac{7}{\sqrt{2}}$
D $\frac{1}{\sqrt{2}}$

The correct answer is:
$\frac{5}{\sqrt{2}}$

Solution:
Magnitude of components of $\vec{A}$ along $\vec{B}=\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}=\frac{(2 \hat{i}+3 \hat{j})(\hat{i}+\hat{j})}{\sqrt{2}}=\frac{5}{\sqrt{2}}$

Topic: Physics, Mathematics in Physics, JEE Main
$\mathrm{A}$ and $\mathrm{B}$ are two wires. The radius of $\mathrm{A}$ is twice that of $\mathrm{B}$. They are stretched by the same load. Then the stress on B is

Options:
A equal to that on $\mathrm{A}$
B four times that on A
C two times that on A
D half that on $\mathrm{A}$

The correct answer is:
four times that on A

Solution:
Stress $=\frac{\text { Force }}{\text { Area }} \therefore$ Stress $\propto \frac{1}{\pi r^{2}}$
$$
\frac{S_{B}}{S_{A}}=\left(\frac{r_{A}}{r_{B}}\right)^{2}=(2)^{2} \Rightarrow S_{B}=4 S_{A}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main
'a' and 'b' are van der Waals' constants for gases. Chlorine is more easily liquefied than ethane because

Options:
A
a and $b$ for $\mathrm{Cl}_2 < \mathrm{a}$ and $\mathrm{b}$ for $\mathrm{C}_2 \mathrm{H}_6$
B
a for $\mathrm{Cl}_2 < $ a for $\mathrm{C}_2 \mathrm{H}_6$ but b for $\mathrm{Cl}_2>$ b for $\mathrm{C}_2 \mathrm{H}_6$
C
a for $\mathrm{Cl}_2>$ a for $\mathrm{C}_2 \mathrm{H}_6$ but $b$ for $\mathrm{Cl}_2 < $ b for $\mathrm{C}_2 \mathrm{H}_6$
D
a and $b$ for $\mathrm{Cl}_2>a$ and $b$ for $\mathrm{C}_2 \mathrm{H}_6$

The correct answer is:
a for $\mathrm{Cl}_2>$ a for $\mathrm{C}_2 \mathrm{H}_6$ but $b$ for $\mathrm{Cl}_2 < $ b for $\mathrm{C}_2 \mathrm{H}_6$

Solution:
ease of liquefaction $\propto \frac{a}{b}$ for ethane $a=5.49, b=0.0638$ for $\mathrm{Cl}_2 \mathrm{a}=6.49, \mathrm{~b}=0.0562$

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2011
A and B are

Options:
A Me3C-CH2Cl,Me3C-CH2Cl
B
C
D

The correct answer is:

Solution:
The formation of B does not involve carbocation formation and rearrangement, so the product B is Me3CH2Cl .

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min respectively. Starting from an equimolar non-reactive mixture of A and B, the time taken for the concentration of A to become 16 times that of B is ______ min. (Round off to the Nearest Integer).

Options:

The correct answer is:
108

Solution:
Given A=t1/2=54 min

B=t1/2=18 min

t=0 'x' M t=0 'x' M

⇒ To calculate : At=16×Bt time =?

⇒ For I order kinetic : At=A02n

n→ no. of Half lives

⇒ Now from the relation (1)

⇒x2n1=x2n2×16⇒2n2=2n1×24

⇒n2=n1+4⇒tt1/22=tt1/21+4

⇒t118-154=4⇒t=4×18×5436

⇒t=108 min

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2021 (16 Mar Shift 2)
A and B formed in the following reactions are:
CrO2Cl2+4NaOH→A+2NaCl+2H2O

A+2HCl+2H2O2→B+3H2O

Options:
A A=Na2CrO4, B=CrO5
B A=Na2Cr2O4, B=CrO4
C A=Na2Cr2O7, B=CrO3
D A=Na2Cr2O7, B=CrO5

The correct answer is:
A=Na2CrO4, B=CrO5

Solution:
Chromyl chloride dissolves in NaOH solution to give a yellow solution. The chromate solution changes to blue colour chromium peroxide with hydrogen peroxide.

The reaction will be as follows,

CrO2Cl2+4NaOH→Na2CrO4+2NaCl+2H2OCrO42-+2H+ +2H2O2→ CrO5+3H2O

Hence, option A is the answer.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (30 Jan Shift 2)
$\text { ' } a \text { ' and ' } b \text { ' in the following reactions are: }$

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
A and B in the following reactions are:

Options:
A
B
C
D

The correct answer is:

Solution:
At first the starting material will under go diazotization in the presence of sodium nitrite & hydrochloric acid then the formed compound again will under go in the presence of potassium cyanide to form cyano benzene which on further will under go stephen reduction to form the final product as result.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2021 (24 Feb Shift 1)
A and B react as per the reaction

Ag+2Bg→Cg

From the data given below identify the rate law.

S.No. A  M B  M Rate of reaction M  sec −1 1 0.3 0.4 2× 10 –3 2 0.6 0.8 8× 10 –3 3 0.6 0.4 4× 10 –3

Options:
A R=kA2B1
B R=kAB
C R=kA1B0
D R=kAB-1

The correct answer is:
R=kAB

Solution:
Rate R=k A m B n
From Exp. (i) R1=2Ã10â3=k0.3m0.4n
From Exp. (ii) R2=8Ã10â3=k0.6m0.8n
From Exp. (iii) R3=4Ã10â3=k[0.6]m[0.4]n
R3R1â4Ã10â32Ã10â3=k[0.6]m[0.4]nk[0.3]m[0.4]n
2â²=2mâ´m=1
R2R3â8Ã10â34Ã10â3=k[0.6]m[0.8]nk[0.6]m[0.4]n
2=2nâ´n=1
∴Rate law R =k A B

Topic: Chemistry, Chemical Kinetics, JEE Main
A and B respectively are

Options:
A
B
C
D

The correct answer is:

Solution:
NaBH4 is a weak reducing reagent. So it reduces only carbonyl group while reduction of -COOH group does not happen and it remains unaffected. LiAlH4 is strong reducing reagent; so it can reduce both carbonyl and carboxylic groups.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
A and C lie in 0,π2 and B lies in 0,2π. If tanA+3cosB+6sinC=1; 3tanA+cosB+4sinC=4; 5tanA+3cosB-8sinC=-2, then B-2 A-C=

Options:
A π6
B π3
C π4
D π2

The correct answer is:
π3

Solution:
Given tanA+3cosB+6sinC=1; 3tanA+cosB+4sinC=4 and 5tanA+3cosB-8sinC=-2.

Let tanA=x, cosB=y & sinC=z, then the given equations become

x+3y+6z=1 ...i

3x+y+4z=4 ...ii and

5x+3y-8z=-2 ...iii

Let P=13631453-8, X=xyz and Q=14-2.

Then, the system of equations can be written as PX=Q.

Hence, we have X=P-1Q.

Now P=13631453-8

⇒P=1-8-12-3-24-20+69-5

⇒P=-20+132+24=136.

The cofactors of the elements of the matrix P are

A11=-8-12=-20, A12=--24-20=44, A13=9-5=4,

A21=--24-18=42, A22=-8-30=-38, A23=-3-15=12,

A31=12-6=6, A32=-4-18=14, A33=1-9=-8.Thus, the adjoint matrix of P is

adjP=-2042644-3814412-8.

And P-1=adjPP,

⇒P-1=1136-2042644-3814412-8.

The solution of the given system is

xyz=1136-2042644-3814412-814-2

⇒xyz=1136-20×1+42×4+6×-244×1+-38×4+14×-24×1+12×4+-8×-2

⇒xyz=113613613668

⇒x=tanA=1, y=cosB=-1 and z=sinC=12.

Also, A and C lie in 0, π2 and B lies in 0, 2π.

Thus tanA=1, ⇒A=π4, cosB=-1, ⇒B=π and sinC=12, C=π6.

Now B-2 A-C=π-2×π4-π6=π3.

Topic: Mathematics, Determinants, JEE Main
$A+B=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$ and
$A-2 B=\left[\begin{array}{cc}-1 & 1 \\ 0 & -1\end{array}\right]$, then $A=$

Options:
A $\left[\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right]$
B $\left[\begin{array}{ll}2 / 3 & 1 / 3 \\ 1 / 3 & 2 / 3\end{array}\right]$
C $\left[\begin{array}{ll}1 / 3 & 1 / 3 \\ 2 / 3 & 1 / 3\end{array}\right]$
D None of these

The correct answer is:
$\left[\begin{array}{ll}1 / 3 & 1 / 3 \\ 2 / 3 & 1 / 3\end{array}\right]$

Solution:
$2 A+2 B=\left[\begin{array}{ll}2 & 0 \\ 2 & 2\end{array}\right]$,$A-2 B=\left[\begin{array}{cc}-1 & 1 \\ 0 & -1\end{array}\right]$
On adding, we get $3 A=\left[\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right]$
$\Rightarrow A=\left[\begin{array}{ll}1 / 3 & 1 / 3 \\ 2 / 3 & 1 / 3\end{array}\right]$.

Topic: Mathematics, Matrices, JEE Main
$(a-b) \times(a+b)=$

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

The correct answer is:
$2(\mathbf{a} \times \mathbf{b})$

Solution:
$\begin{aligned}
(\mathbf{a}-\mathbf{b}) \times(\mathbf{a}+\mathbf{b})=\mathbf{a} \times \mathbf{a}-\mathbf{b} \times \mathbf{a}+ & \mathbf{a} \times \mathbf{b}-\mathbf{b} \times \mathbf{b} \\
& =\mathbf{a} \times \mathbf{b}-\mathbf{b} \times \mathbf{a}=\mathbf{a} \times \mathbf{b}+\mathbf{a} \times \mathbf{b}=2(\mathbf{a} \times \mathbf{b})
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
a→, b→ and c→ are coplanar unit vectors. A unit vector d→ is perpendicular to them. If a→×b→×c→×d→=326i^-213j^+613k^ and the angle between a→ and b→ is 30∘, then c→ is equal to

Options:
A 313i^-413j^+1213k^
B 27i^-37j^+67k^
C 3i^-4j^+12k^
D 13i^-13j^+13k^

The correct answer is:
313i^-413j^+1213k^

Solution:
a→⋅d→=b→⋅d→=c→⋅d→=0 (Given)
a→=b→=c→=d→=1 a→⋅b→=32
a→×b→×c→×d→=326i^-213j^+613k^
⇒a→×b→⋅d→c→-a→×b→⋅c→d→=326i^-213j^+613k^ ...(1)
∵ a→, b→, c→ are coplanar ⇒a→×b→⋅c→=0
∵d→ is perpendicular to a→,b→⇒d→ must be parallel to a→×b→
So, the angle between d→ and a→×b→ is either 0o or 180o
From 1, we get,
±a→×b→d→c→=226i^-213j^+613k^
⇒±a→b→sin⁡30od→c→=326i^-213j^+613k^
⇒±c→2=326i^-213j^+613k^⇒c→=±3i^-4j^+12k^13

Topic: Mathematics, Vector Algebra, JEE Main
A,B and C are three biomolecules. The results of the tests performed on them are given below:

Molisch's test	Barfoed Test	Biuret Test

A	Positive	Negative	Negative

B	Positive	Positive	Negative

C	Negative	Negative	Positive

A,B and C are respectively:

Options:
A A= Glucose, B= Fructose C= Albumin
B A= Lactose, B= Glucose, C= Albumin
C A= Lactose, B= Glucose, C= Alanine
D A= Lactose, B= Fructose, C= Alanine

The correct answer is:
A= Lactose, B= Glucose, C= Albumin

Solution:
Alanine does not show Biuret test because
Biuret test is used for deduction of peptide
linkage & alanine is amino acid.
Albumine is protein so have paptide linkage
so it gives positive Biuret test.
Positive Barfoed test is shown by
monosaccharide but not disaccharide. Positive
Molisch's test is shown by glucose.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2020 (09 Jan Shift 2)
A, B and C in the given figure are respectively:

Options:
A Cell Membrane, Cell wall, RNA
B Protoplast, Cell Wall, DNA
C Cell wall, Cell Membrane, RNA
D Cell wall, Cell Membrane, DNA

The correct answer is:
Cell wall, Cell Membrane, DNA

Solution:
Bacteria reproduce mainly by fission. Here A is Cell wall, B is Cell Membrane and C is DNA.

Topic: Biology, Biological Classification, NEET
A, B, C , and D are four different physical quantities having different dimensions. None of them is dimensionless. But we know that the equation AD=C ln( BD ) holds true. Then which of the combination is not a meaningful quantity?

Options:
A CBD-A2D2C
B A2-B2C2
C AB-C
D A 2 −AC D

The correct answer is:
A 2 −AC D

Solution:
The dimension of A is not equal to the dimension of C
Hence it is not possible to subtract AC from A2

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2016 (10 Apr Online)
A, B, C and D are four different physical quantities having different dimensions. None of them is dimensionless. But we know that the equation AD=C ln( BD ) holds true. Then which of the combination is not a meaningful quantity ?

Options:
A CBD-A2D2C
B A2-B2C2
C AB-C
D A 2 −AC D

The correct answer is:
A 2 −AC D

Solution:
The dimension of A is not equal to the dimension of C
Hence it is not possible to subtract AC from A2

Topic: Physics, Units and Dimensions, JEE Main
A, B, C and D are four particles with each of mass M lying on the vertices of a square of side a. They always move along a common circle with velocity v under mutual gravitational force. Find v so that they always remain on the vertices of the square

Options:
A GM22+122a
B GM2+12a
C GM22+1a
D none

The correct answer is:
GM22+122a

Solution:
From the above figure, it is clear that
Fnet=2F1+F3=MV2R F1=F2
2 GM2a2+GM2a22=Mv2a/2
v=GM1+2222a

Topic: Physics, Gravitation, JEE Main
$\vec{a}, \vec{b}, \vec{c}$ are 3 vectors, such that $\vec{a}+\vec{b}+\vec{c}=0,|\vec{a}|=1,|\vec{b}|=2 \mid \vec{a}$ then $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ is equal to

Options:
A
1

B
0

C
$-7$

D
7

The correct answer is:
$-7$

Solution:
$\vec{a}+\vec{b}+\vec{c}=0 \Rightarrow(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})=0$
$|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$
$\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=\frac{-1-4-9}{2}=-7$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2003
$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three vectors such that $|\mathbf{a}|=3$, $|\mathbf{b}|=5,|\mathbf{c}|=7$. If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are perpendicular to the vectors $\mathbf{b}+\mathbf{c}, \mathbf{c}+\mathbf{a}, \mathbf{a}+\mathbf{b}$ respectively, then $\sqrt{(\mathbf{a}+\mathbf{b}+\mathbf{c})^2-2}=$

Options:
A 15
B 9
C 22
D 25

The correct answer is:
9

Solution:
We have,
$$
|\mathbf{a}|=3,|\mathbf{b}|=5,|\mathbf{c}|=7
$$
$\mathbf{a}$ is perpendicular to $(\mathbf{b}+\mathbf{c})$
and $\mathbf{b}$ is perpendicular to $\mathbf{c}+\mathbf{a}$ So,
$\mathbf{b} \cdot(\mathbf{c}+\mathbf{a})=0$
and $\mathbf{c}$ is perpendicular to $(\mathbf{a}+\mathbf{b})$
Adding Eqs. (i), (ii) and (iii), we get
Now, $|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2$
$$
\begin{aligned}
& =|\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) \\
& =(3)^2+(5)^2+(7)^2+0=83
\end{aligned}
$$
Now, $\sqrt{(\mathbf{a}+\mathbf{b}+\mathbf{c})^2-2}=\sqrt{83-2}=\sqrt{81}=9$

Topic: Mathematics, Vector Algebra, JEE Main
$\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|=$

Options:
A $3 a b c+a^3+b^3+c^3$
B $3 a b c-a^3-b^3-c^3$
C $a b c-a^3+b^3+c^3$
D $a b c+a^3-b^3-c^3$

The correct answer is:
$3 a b c-a^3-b^3-c^3$

Solution:
$\left|\begin{array}{ccc}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|=\left|\begin{array}{ccc}a+b+c & a+b+c & a+b+c \\ b & c & a \\ c & a & b\end{array}\right|$, $\left(R_1 \rightarrow R_1+R_2+R_3\right)$
$=(a+b+c)\left|\begin{array}{ccc}1 & 1 & 1 \\ b & c & a \\ c & a & b\end{array}\right|=(a+b+c)\left|\begin{array}{ccc}1 & 0 & 0 \\ b & b-c & c-a \\ c & c-a & a-b\end{array}\right|$
$=3 a b c-a^3-b^3-c^3$, (After simplification)

Topic: Mathematics, Determinants, JEE Main
$\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}$ are coplanar vectors, then $(\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{d})$ is equal to

Options:
A $0$
B $1$
C a
D b

The correct answer is:
$0$

Solution:
We have, $\mathbf{a}, \mathbf{b}, \mathbf{c}$ and $\mathbf{d}$ are coplanar vectors, then $\mathbf{a} \times \mathbf{b}$ and $\mathbf{c} \times \mathbf{d}$ are perpendicular on the same plane, therefore, $\mathbf{a} \times \mathbf{b}$ and $\mathbf{c} \times \mathbf{d}$ both are parallel to each other, then $(\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{d})=0$

Topic: Mathematics, Vector Algebra, JEE Main
A,B,C,D,E,F are conducting plates each of area A and any two consecutive plates separated by a distance d. The net energy stored in the system after the switch is closed is

Options:
A 3ε0A2dV2
B 5ε0A12dV2
C ε0A2dV2
D ε0AdV2

The correct answer is:
ε0A2dV2

Solution:
Ceff=Îµ0Ad since effective field between plates A and E is zero
â´U = 12 CV2 = Îµ0A2dV2

Topic: Physics, Electrostatics, JEE Main
ABCD is a trapezium such that AB and CD are parallel and BC⊥CD. If ∠ADB=θ, BC=p and CD=q, then AB is equal to

Options:
A p2+q2p2cosθ+q2sinθ
B p2+q2sinθpcosθ+qsinθ2
C p2+q2sinθpcosθ+qsinθ
D p2+q2cosθpcosθ+qsinθ

The correct answer is:
p2+q2sinθpcosθ+qsinθ

Solution:
Let ∠BDC=α, then
tanα=BCCD=pq
AE=BC=p
tanα+θ=AEED
⇒ED=ptanα+θ
=p1-tanαtanθtanα+tanθ
=p1-pqtanθpq+tanθ
Now, AB=q-ED
=q-pq-ptanθpq+qtanθ
=pq+q2tanθ-pq+p2tanθp+qtanθ
=p2+q2tanθp+qtanθ

=p2+q2sinθpcosθ+qsinθ

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2013 (07 Apr)
A

B

C

D

The correct statement about (A), (B), (C) and (D) is:

Options:
A (B) and (C) are tranquillizers
B (B), (C) and (D) are tranquillizers
C (A) and (D) are tranquillizers
D (A), (B) and (C) are narcotic analgesics

The correct answer is:
(B) and (C) are tranquillizers

Solution:
Valium and Serotonin are tranquilizers.

Morphine and Codeine are the narcotic analgesics.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2021 (27 Aug Shift 1)
\( A, B, C \) in the diagram represent rays of light incident upon a face of a right-angled prism. Before emerging from the prism, which ray (or rays) will experience total internal reflection?

Options:
A Ray \( A \)
B Ray \( B \)
C Ray \( C \)
D Ray \( A \) and \( C \)

The correct answer is:
Ray \( B \)

Solution:
For ray B, angle of incidence is 0°, so angle of refraction r1 at first face will also be zero.
And we know for a prism, r1+r2=A, where r2and A represents angle of incidence at second surface and angle of prism respectively.
⇒r2=A-r1
We can observe from above relation, that r2 will be maximum for minimum r1, and it will be maximum for ray B.
Also, to give TIR effect r2 should be greater than critical angle, and is only possible for ray B.

Topic: Physics, Ray Optics, JEE Main
ABC is a plane lamina of the shape of an equilateral triangle. D, E are mid-points of AB, AC and G is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through G and perpendicular to the plane ABC is I0. If part ADE is removed, the moment of inertia of the remaining part about the same axis is NI016 where N is an integer. Value of N is:

Options:

The correct answer is:
11

Solution:
Let mass of lamina =m, and length of side =l, then moment of inertila of lamina about an axis passing through G and perpendicular to the plane.

I0∝ml2

I0=kml2

Let moment of inertila of DEF=I1 about G

then, I1=I016

then,IADE=IBDF=IEFC=I2

3I2+I1=I0

3I2+I016=I0

⇒ I2=5I016

So, Moment of inertila of DECB=2I2+I125I016+I016

=11I016

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2020 (04 Sep Shift 1)
ABC is a right triangle, right-angled at the vertex A. A circle is drawn to touch the sides AB and AC at points P and Q respectively such that the other end points of the diameters passing through P and Q lie on the side BC. If AB=6 units, then the area (in sq. units) of the circular sector which lies inside the triangle is

Options:
A 3π+2
B π+3
C π-2
D π+2

The correct answer is:
3π+2

Solution:
AB=3r=6⇒r=2
Shaded area =14π22-122.2=π-2
Required area =4π-(π-2)
=3π+2 sq. units

Topic: Mathematics, Circle, JEE Main
\(A B C\) is a right-angled triangle in which \(\max \{A B, B C, A C\}=B C\). If the position vectors of \(B\) and \(C\) are respectively \(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(5 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}\), then
\[
\mathbf{A B} \cdot \mathbf{A C}+\mathbf{B A} \cdot \mathbf{B C}+\mathbf{C A} \cdot \mathbf{C B}=
\]

Options:
A 28
B 29
C 27
D 25

The correct answer is:
29

Solution:
Given
\(\begin{aligned}
\mathbf{B} & =3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\
\mathbf{C} & =5 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\
\mathbf{B C} & =2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}
\end{aligned}\)
\(\max \{A B, B C, A C\}=B C\)
\(\therefore \mathbf{B C}\) is hypotenuse of \(\triangle A B C\)
\(\begin{aligned} & \angle \mathbf{A}=90^{\circ} \\ & \therefore \quad \mathbf{A B} \cdot \mathbf{A C}=0 \\ & \mathbf{B A} \cdot \mathbf{B C}=|\mathbf{B A} \| \mathbf{B C}| \cos B \\ & \mathbf{C A} \cdot \mathbf{C B}=|\mathbf{C A} \| \mathbf{C B}| \cos C \\ & \therefore \mathbf{A B}-\mathbf{A C}+\mathbf{B A}-\mathbf{B C}+\mathbf{C A}-\mathbf{C B} \\ &=0+|\mathbf{B C}|(|\mathbf{B A}| \cos B+|\mathbf{C B}| \cos C) \\ &=0+|\mathbf{B C} \| \mathbf{B C}| \quad[\because \text { By projection formula }] \\ &=|\mathbf{B C}|^2-\left(\sqrt{(2)^2+3^2+4^2}\right)^2=4+9+16=29\end{aligned}\)

Topic: Mathematics, Vector Algebra, JEE Main