datavorous/entrance-exam-dataset huggingface dataset:
In a thermodynamic process, pressure of a fixed mass of a gas is changed in such a manner that the gas molecules gives out $20 \mathrm{~J}$ of heat and $10 \mathrm{~J}$ of work is done on the gas. If the initial internal energy of the gas was $40 \mathrm{~J}$, then the final internal energy will be

Options:
A $30 \mathrm{~J}$
B $20 \mathrm{~J}$
C $60 \mathrm{~J}$
D $40 \mathrm{~J}$

The correct answer is:
$60 \mathrm{~J}$

Solution:
$\begin{gathered}\Delta Q=\Delta U+\Delta W=\left(U_f-U_i\right)+\Delta W \\ \Rightarrow 30=\left(U_f-40\right)+10 \Rightarrow U_f=60 \mathrm{~J}\end{gathered}$

Topic: Physics, Thermodynamics, JEE Main
In a thin prism of glass (refractive index 1.5), which of the following relations between the angle of minimum deviations $\delta_m$ and angle of refraction $r$ will be correct

Options:
A $\delta_m=r$
B $\mathcal{S}_m=1.5 r$
C $\mathcal{S}_m=2 r$
D $\delta_m=\frac{r}{2}$

The correct answer is:
$\delta_m=r$

Solution:
$\mathcal{S}_m=(\mu-1)(2 r)=(1.5-1) 2 r=0.5 \times 2 r=r$

Topic: Physics, Ray Optics, JEE Main
In a titration experiment, 10 mL of an FeCl2 solution consumed 25 mL of a standard K2Cr2O7 solution to reach the equivalent point. The standard K2Cr2O7 solution is prepared by dissolving 1.225 g of K2Cr2O7 in 250 mL water. The concentration of the FeCl2 solution is closest to

[Given : molecular weight of K2Cr2O7=294 g mol-1]

Options:
A 0.25 N
B 0.50 N
C 0.10 N
D 0.04 N

The correct answer is:
0.25 N

Solution:
In titration at equivalence point, number of equivalents of both reactants must be same.

∴ N1V1=N2V2

N (normality) of FeCl2=N×V of K2Cr2O7V of FeCl2

N (normality) of K2Cr2O7= Mass ×1000 Equivalent mass × Volume of solution prepared

Equivalent mass = Molar mass n factor

=2946

=49 g/equi.

(K2Cr2O7 as oxidising agent is reduced to 2Cr3+).

Oxidation state of 2 Cr atoms change from +6 to +3.

Total change is3×2=6, (∴n-factor =6)

∴ N (of K2Cr2O7) =1225×100049×250=0.1N

∴ N (of FeCl2) =0.1×2510

=0.25 N

Topic: Chemistry, Redox Reactions, JEE Main
In a tournament with five teams, each team plays against every other team exctly once. Each game is won by one of the playing teams and the winning team scores one point, while the losing team scores zero. Which of the following is NOT necessarily true?

Options:
A There are at least two teams which have at most two points each.
B There are at least two teams which have at least two points each.
C There are at most three teams which have at least threee points each
D There are at most four teams which have at most two points each

The correct answer is:
There are at most four teams which have at most two points each

Solution:
Let teams be $T_{1}, T_{2}, T_{3}, T_{4} \& T_{5}$
Now, we can have 5 teams with the scores of 2 points each matches are
(I) $\mathrm{T}_{1} \mathrm{~T}_{2}$
(II) $\mathrm{T}_{1} \mathrm{~T}_{3}$
(III) $\mathrm{T}_{1} \mathrm{~T}_{4}$
$(\mathrm{IV}) \mathrm{T}_{1} \mathrm{~T}_{5}$
(M) $\mathrm{T}_{2} \mathrm{~T}_{3}$
$(\mathrm{NI}) \mathrm{T}_{2} \mathrm{~T}_{4}$
(MI) $\mathrm{T}_{2} \mathrm{~T}_{5}$
$(\mathrm{VIII}) \mathrm{T}_{3} \mathrm{~T}_{4}$
$(\mathbb{X}) \mathrm{T}_{3} \mathrm{~T}_{5}$
$(\mathrm{X}) \mathrm{T}_{4} \mathrm{~T}_{5}$
This score board contradicts, option D $\therefore D$ is not always necssarily true.

Topic: Mathematics, Permutation Combination, JEE Main
In a town $35.4 \%$ of the people are not literates, $27 \%$ have education up to primary school, $18.6 \%$ have education up to middle school. The people with education up to high school are twice the number of people with education up to Pre-University. Of the remaining, 660 are graduates. If the population of the town is 15000 , then what is the number of people with education up to high school?

Options:
A 3120
B 1560
C 1460
D None of these

The correct answer is:
1460

Solution:
No. of people whoare illiterates $=15000 \times \frac{34.5}{100}=5175$ No. of people who have education up to primaryschool
$=15000 \times \frac{27}{100}=4050$
Similarly, No. of peoplewho have education upto middle school $=2790$ Let the no. of people who have education upto high school $=x$ $\therefore$ According to the question. No. of people who have education upto pre-university
$=\frac{x}{2}$
So, Total no. of people who are not graduates
$=5175+4050+2790+x+\frac{x}{2}=12150+x+\frac{x}{2}$
Since, 660 are graduates, therefore
$15000-\left(12150+x+\frac{x}{2}\right)=660$
$\Rightarrow 15000-660=12150+\frac{3 x}{2}$
$\Rightarrow 2(15000-660)=24300+3 x$
$\Rightarrow 28680-24300=3 x$
$\Rightarrow \frac{4380}{3}=x \Rightarrow x=1460$
Hence, 1460 students have education upto high school.

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 2)
In a transformer, number of turns in the primary coil are 140 and that in the secondary coil are 280. If current in primary coil is $4 \mathrm{~A}$, then that in the secondary coil is

Options:
A
4 A

B
2 A

C
6 A

D
10 A

The correct answer is:
2 A

Solution:
$I_1 N_1=I_2 N_2 \Rightarrow I_2=\frac{4 \times 140}{280}=2 \mathrm{~A}$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2002
In a transformer the number of primary turns is four times that of the secondary turns. Its primary is connected to an AC source of voltage V. Then

Options:
A current through its secondary is about four times that of the current through its primary.
B voltage across its secondary is about four times that of the voltage across its primary.
C voltage across its secondary is about two times that of the voltage across the primary.
D voltage across its secondary is about 122 times of the voltage across its primary.

The correct answer is:
current through its secondary is about four times that of the current through its primary.

Solution:
In a transformer the primary and secondary currents are related by
Is=NpNsIp
And the voltage are related by
Vs=NsNpVp
Where subscripts p and s refer to the primary and secondary of the transformer.
Here, Vp=V, NpNs=4
∴Is=4Ip
And Vs=14V=V4

Topic: Physics, Alternating Current, JEE Main
In a transistor circuit shown here the base current is $35 \mu$. The value of the resistor $R_8$ is

Options:
A $123.5 \mathrm{k} \Omega$
B $257 \mathrm{k} \Omega$
C $380.05 \mathrm{k} \Omega$
D None of these

The correct answer is:
$257 \mathrm{k} \Omega$

Solution:
$V_b=i_b R_b \Rightarrow R_b=\frac{9}{35 \times 10^{-6}}=257 \mathrm{k} \Omega$.

Topic: Physics, Semiconductors, JEE Main
In a transistor, the base is made very thin and lightly doped with an impurity, because

Options:
A to save the transistor from heating effect
B to enable the emitter to emit small number of electrons and holes
C to enable the collector to collect 95% of the holes or electron coming from the emitter side
D none of the above

The correct answer is:
to enable the collector to collect 95% of the holes or electron coming from the emitter side

Solution:
Base is made thin so that very less amount of charge is exchanged with emitter and remaining charges interacts with collector.

Topic: Physics, Semiconductors, NEET
In a transistor, the collector current varies by 0.49 mA and emitter current varies by 0.50 A . Current gain Î² measured is

Options:
A 49
B 150
C 99
D 100

The correct answer is:
49

Solution:
Given,
ÎIC=0.49 mA=0.49Ã10-3 A
ÎIE=0.50 mA=0.50Ã103 A
Î±=ÎICÎIE
Î²=Î±1-Î±=0.490.501-0.490.50
=49501-4950
=49Ã505050-49
Î²=49

Topic: Physics, Semiconductors, JEE Main
In a transverse wave the distance between a crest and neighbouring trough at the same instant is $4.0 \mathrm{~cm}$ and the distance between a crest and trough at the same place is $1.0 \mathrm{~cm}$. The next crest appears at the same place after a time interval of $0.4 \mathrm{~s}$. The maximum speed of the vibrating particles in the medium is:

Options:
A
$\frac{3 \pi}{2} \mathrm{~cm} / \mathrm{s}$

B
$\frac{5 \pi}{2} \mathrm{~cm} / \mathrm{s}$

C
$\frac{\pi}{2} \mathrm{~cm} / \mathrm{s}$

D
$2 ~\pi \mathrm{~cm} / \mathrm{s}$

The correct answer is:
$\frac{5 \pi}{2} \mathrm{~cm} / \mathrm{s}$

Solution:
No Solution Available

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2013 (25 Apr Online)
In a travelling plane electromagnetic wave, the maximum magnetic field is $1.26 \times 10^{-4} \mathrm{~T}$. The intensity of the wave is (Assume, $\mu_0=1.26 \times 10^{-6} \mathrm{H} / \mathrm{m}$ )

Options:
A $1.56 \times 10^6 \frac{\mathrm{W}}{\mathrm{m}^2}$
B $1.89 \times 10^6 \frac{\mathrm{W}}{\mathrm{m}^2}$
C $8.92 \times 10^5 \frac{\mathrm{W}}{\mathrm{m}^2}$
D $4.62 \times 10^6 \frac{\mathrm{W}}{\mathrm{m}^2}$

The correct answer is:
$1.89 \times 10^6 \frac{\mathrm{W}}{\mathrm{m}^2}$

Solution:
Given, maximum magnetic field,
$\begin{aligned}
& B_0=1.26 \times 10^{-4} \mathrm{~T} \\
& \mu_0=1.26 \times 10^{-6} \mathrm{H} / \mathrm{m}
\end{aligned}$
$\therefore$ Intensity of EMW (Electromagentic wave) is given by
$\begin{aligned}
I & =\frac{1}{2} \frac{B_0^2 c}{\mu_0}=\frac{1}{2} \times \frac{\left(1.26 \times 10^{-4}\right)^2 \times 3 \times 10^8}{1.26 \times 10^{-6}} \\
& =\frac{1}{2} \times \frac{1.26 \times 1.26 \times 10^{-8} \times 3 \times 10^8}{1.26 \times 10^{-6}} \\
& =0.63 \times 3 \times 10^6=1.89 \times 10^6 \mathrm{~W} / \mathrm{m}^2
\end{aligned}$

Topic: Physics, Electromagnetic Waves, JEE Main
In a triangle ABC, coordinates of A are 1, 2 and the equations of the medians through B and C are respectively, x+y=5 and x=4. Then area of ΔABC (in sq. units) is :

Options:
A 12
B 4
C 9
D 5

The correct answer is:
9

Solution:
Let C4, α be on x=4.

Now, midpoint of AC is 52, α+22 which lies on x+y=5.

∴α+22=52⇒α=3

∴C4, 3

Similarly, let Bβ, 5-β be on x+y=5.

The midpoint of AB =β+12, 7-β2 which lies on x=4⇒β=7⇒B7, -2.

∴ Area of Δ=121217-21431

=121-2-3-27-4+121+8=1218=9 sq. units

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2018 (15 Apr)
In a triangle ABC, if |BC→|=3,|CA→|=5 and |BA→|=7, then the projection of the vector BA→ on BC→ is equal to

Options:
A 192
B 132
C 112
D 152

The correct answer is:
112

Solution:
We have,

a→=3,b→=5,c→=7

Now,

cosθ=a→2+c→2-b→22a→c→

⇒cosθ=9+49-252·3·7

⇒cosθ=1114

Projection of c→ on b→

=|c→|cosθ

=7×1114

=112

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (20 Jul Shift 2)
In a triangle ABC, if |BC→|=8,|CA→|=7,|AB→|=10, then the projection of the vector AB→ on AC→ is equal to :

Options:
A 254
B 8514
C 12720
D 11516

The correct answer is:
8514

Solution:
a→=8,b→=7,c→=10

cosθ=|b→|2+|c→|2-|a→|22|b→||c→|=1728

Projection of c→ on b→

=|c→|cosθ

=10×1728

=8514

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (18 Mar Shift 2)
In a triangle ABC, if cosA+2 cosB+cosC=2 and the lengths of the sides opposite to the angles A and C are 3 and 7 respectively, then cos A-cos C is equal to

Options:
A 97
B 107
C 57
D 37

The correct answer is:
107

Solution:
Given,

AB=7, BC=3.

⇒c=7, a=3.

Apply cosine rule in the given equation cosA+2cosB+cosC=2, we get

⇒b2+c2-a22bc+2c2+a2-b22ac+a2+b2-c22ab=2

⇒b2+49-914b+49+9-b221+9+b2-496b=2

⇒b2+4014b+58-b221+b2-406b=2

⇒b3-5b2-16b+80=0

⇒b-4b+4b-5=0

⇒b=-4,4,5.

But b≠-4.

Also, if b=4 triangle cannot be constructed.

That means b=5.

⇒cosA-cosC=b2+c2-a22bc-a2+b2-c22ab

⇒cosA-cosC=52+72-3225×7-32+52-7223×5

⇒cosA-cosC=6570--1530=195+105210=300210

⇒cosA-cosC=107

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2023 (12 Apr Shift 1)
In a triangle ABC, if sin⁡Acos⁡B=14 and 3tan⁡A=tan⁡B, then cot2⁡A is equal to

Options:
A 2
B 3
C 4
D 5

The correct answer is:
3

Solution:
3sin⁡Acos⁡B=sin⁡Bcos⁡A
⇒cos⁡Asin⁡B=34
⇒sin⁡A+B=1
⇒C=π2,B=π2-A
3tan⁡A=tan⁡π2-A=cot⁡A
⇒cot2⁡A=3

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
In a triangle ABC, right angle at vertex A, if the position vectors of A,B and C are respectively 3i^+ j^- k^, -i^+3j^+pk^ and 5i^+qj^-4k^ , then the point p,q lies on a line:

Options:
A Making an obtuse angle with the positive direction of x-axis
B Parallel to x-axis
C Parallel to y-axis
D Making an acute angle with the positive direction of x-axis

The correct answer is:
Making an acute angle with the positive direction of x-axis

Solution:
AB→= -4i^+2j^+p+1k^

AC→= 2i^+q-1j^-3k^

AB→⊥AC→

⇒ AB→ · AC→=0
-8+2q-1-3p+1=0
3p-2q+13=0
p,q lies on 3x-2y+13=0
Slope =32
∴ The required line makes an acute angle with the positive direction of x-axis.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2016 (09 Apr Online)
In a triangle $A B C$, coordianates of $A$ are $(1,2)$ and the equations of the medians through $B$ and $C$ are $x+y=5$ and $x=4$ respectively. Then area of $\triangle A B C$ (in sq. units) is

Options:
A
5

B
9

C
12

D
4

The correct answer is:
9

Solution:
Median through $C$ is $x=4$
So the $x$ coordianate of $C$ is 4 . let $C \equiv(4, y)$, then the midpoint of $A(1,2)$ and $C(4, y)$ is $D$ which lies on the median through $B$.
$$
\therefore \quad D \equiv\left(\frac{1+4}{2}, \frac{2+y}{2}\right)
$$
Now $\frac{1+4+2+y}{2}=5 \Rightarrow y=3$.
So, $C \equiv(4,3)$.
The centroid of the triangle is the intersection of the mesians. Here the medians $x=4$ and $x+4$ and $x+y=5$ intersect at $G(4,1)$.
The area of triangle $\triangle A B C=3 \times \triangle A G C$
$$
=3 \times \frac{1}{2}[1(1-3)+4(3-2)+4(2-1)]=9 .
$$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
In a triangle $A B C$, if $\cos \frac{A}{2}=\sqrt{\frac{b+c}{2 c}}$, then

Options:
A $a^2+b^2=c^2$
B $b^2+c^2=a^2$
C $c^2+a^2=b^2$
D $b-c=c-a$

The correct answer is:
$a^2+b^2=c^2$

Solution:
$\cos \frac{A}{2}=\sqrt{\frac{b+c}{2 c}} \Rightarrow \sqrt{\frac{s(s-a)}{b c}}=\sqrt{\frac{b+c}{2 c}}$
$\Rightarrow 2 s(s-a)=b^2+b c$
$\Rightarrow(a+b+c)(b+c-a)=2 b^2+2 b c$
$\Rightarrow a^2+b^2=c^2$

Topic: Mathematics, Properties of Triangles, JEE Main
In a triangle $A B C$ if the angles $A, B, C$ are in $A P$, then which one of the following is correct?

Options:
A $c=a+b$
B $c^2=a^2+b^2-a b$
C $a^2=b^2+c^2-b c$
D $b^2=a^2+c^2-a c$

The correct answer is:
$b^2=a^2+c^2-a c$

Solution:
Given, in a triangle $A B C$ if the angles $A, B, C$ are in $A P$.
$\Rightarrow 2 B=A+C$
We know that, the sum of the angles of the triangle is $180^{\circ}$.
Given, in a triangle $A B C$ if the angles $A, B, C$ are in $A P$.
$\Rightarrow A+B+C=180^{\circ}$
$\Rightarrow 2 B+B=180^{\circ}$
$\Rightarrow 3 B=180^{\circ}$
$\Rightarrow B=60^{\circ}$
By cosine formula, we have
$\cos (B)=\frac{a^2+c^2-b^2}{2 a c}$
$\cos (B)=\frac{a^2+c^2-b^2}{2 a c}$
$\Rightarrow \cos \left(60^{\circ}\right)=\frac{a^2+c^2-b^2}{2 a c}$
$\Rightarrow \frac{1}{2}=\frac{\mathrm{a}^2+\mathrm{c}^2-\mathrm{b}^2}{2 \mathrm{ac}}$
$\Rightarrow \mathrm{a}^2+\mathrm{c}^2-\mathrm{b}^2=\mathrm{ac}$
$\Rightarrow b^2=a^2+c^2-a c$
Hence, In a triangle $\mathrm{ABC}$ if the angles $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are in $\mathrm{AP}, \mathrm{b}^2=\mathrm{a}^2+\mathrm{c}^2-\mathrm{ac}$ is the correct statement.

Topic: Mathematics, Properties of Triangles, JEE Main
In a triangle $A B C$, side $\mathrm{C}=2$, angle $\mathrm{A}=45^{\circ}$, side $\mathrm{a}=2 \sqrt{2}$, then what is angle $\mathrm{C}$ equal to?

Options:
A $30^{\circ}$
B $15^{\circ}$
C $45^{\circ}$
D None of the above

The correct answer is:
$30^{\circ}$

Solution:
In a triangle
Using sine rule in given triangle,
$A B C, c=2, A=45^{\circ}, \mathrm{a}=2 \sqrt{2}$,
$\frac{a}{\sin \mathrm{A}}=\frac{c}{\sin \mathrm{C}} \Rightarrow \frac{2 \sqrt{2}}{\sin 45}=\frac{2}{\sin \mathrm{C}} \Rightarrow \sin \mathrm{C}=\frac{\sin 45}{\sqrt{2}}=\frac{1}{\sqrt{2} \sqrt{2}} \Rightarrow \sin \mathrm{C}=\frac{1}{2} \Rightarrow C=30$
Hence, option (1) is correct.

Topic: Mathematics, Properties of Triangles, JEE Main
In a triangle $A B C$, the sides $A B$ and $A C$ are represented by the vectors $3 \mathrm{i}+\mathrm{j}+\mathrm{k}$ and $\mathrm{i}+2 \mathrm{j}+\mathrm{k}$, respectively. Calculate the angle $\angle A B C$

Options:
A $\cos ^{-1} \sqrt{\frac{5}{11}}$
B $\cos ^{-1} \sqrt{\frac{6}{11}}$
C $\left(90^{\circ}-\cos ^{-1} \sqrt{\frac{5}{11}}\right)$
D $\left(180^{\circ}-\cos ^{-1} \sqrt{\frac{5}{11}}\right)$

The correct answer is:
$\cos ^{-1} \sqrt{\frac{5}{11}}$

Solution:
$A B=3 i+j+\hat{k}$ and $A C=i+2 j+i$
$8 A=-(3 i+j+k)$
$\angle A B C$ is angle between $B A$ and $B C$
$B C=A C+B A$
$=A C-A B$
$=\hat{i}+2 \hat{j}+\hat{k}-(3 \hat{i}+\hat{j}+\hat{k})=-2 \hat{i}+\hat{j}$
$B A \cdot B C=|B A \| B C| \cos \theta$
$-(3 \hat{i}+\hat{j}+\hat{k}) \cdot(-2 \hat{i}+\hat{j})$
$=\left(\sqrt{3^{2}+1^{2}+1^{2}}\right) \cdot\left(\sqrt{2^{2}+1^{2}}\right) \cos \theta$
$\Rightarrow \quad 6-1=\sqrt{11 \times 5} \cdot \cos \theta$
$\cos \theta=\frac{5}{\sqrt{55}}=\sqrt{\frac{5}{11}}=\theta=\cos ^{-1}=\sqrt{\frac{5}{11}}$

Topic: Physics, Mathematics in Physics, JEE Main
In a triangle $\mathrm{ABC}, \mathrm{BC}=7, \mathrm{AC}=8, \mathrm{AB}=\alpha \in \mathrm{N}$ and $\cos \mathrm{A}=\frac{2}{3}$. If $49 \cos (3 \mathrm{C})+42=\frac{\mathrm{m}}{\mathrm{n}}$, where $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}+\mathrm{n}$ is equal to________

Options:

The correct answer is:
39

Solution:
$\begin{aligned} & \cos \mathrm{A}=\frac{\mathrm{b}^2+\mathrm{c}^2-\mathrm{a}^2}{2 \mathrm{bc}} \\ & \frac{2}{3}=\frac{8^2+\mathrm{c}^2-7^2}{2 \times 8 \times \mathrm{c}} \\ & \mathrm{C}=9 \\ & \cos \mathrm{C}=\frac{7^2+8^2-9^2}{2 \times 7 \times 8}=\frac{2}{7} \\ & 49 \cos 3 \mathrm{C}+42 \\ & 49\left(4 \cos ^3 \mathrm{C}-3 \cos \mathrm{C}\right)+42 \\ & 49\left(4\left(\frac{2}{7}\right)^3-3\left(\frac{2}{7}\right)\right)+42 \\ & =\frac{32}{7} \\ & \mathrm{~m}+\mathrm{n}=32+7=39\end{aligned}$

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2024 (06 Apr Shift 2)
In a triangle $\mathrm{ABC}$, if $\mathrm{a}=3, \mathrm{~b}=4$ and $\sin \mathrm{A}=\frac{3}{4}$, then $\angle \mathrm{B}$ equals

Options:
A $60^{\circ}$
B $90^{\circ}$
C $45^{\circ}$
D $30^{\circ}$

The correct answer is:
$90^{\circ}$

Solution:
We have, $\frac{\sin A}{a}=\frac{\sin B}{b}$
or, $\quad \sin B=\frac{b}{a} \sin A$
Since, $a=3, b=4, \sin A=\frac{3}{4}$,
we get, $\sin B=\frac{4}{3} \times \frac{3}{4}=1 \quad \therefore \angle B=90^{\circ}$

Topic: Mathematics, Properties of Triangles, JEE Main
In a triangle $\mathrm{ABC}$, let $\angle \mathrm{C}=\frac{\pi}{2}$. If $\mathrm{r}$ is the inradius and $\mathrm{R}$ is the circumradius of the the triangle $A B C$, then $2(r+R)$ equals

Options:
A
$b+c$

B
$a+b$

C
$a+b+c$

D
$c + a$

The correct answer is:
$a+b$

Solution:
$$
2 r+2 R=c+\frac{2 a b}{(a+b+c)}=\frac{(a+b)^2+c(a+b)}{(a+b+c)}=a+b \quad\left(\text { since } c^2=a^2+b^2\right)
$$

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2005
In a triangle $\mathrm{ABC}$, medians $\mathrm{AD}$ and $\mathrm{BE}$ are drawn. If $\mathrm{AD}=4, \angle \mathrm{DAB}=\frac{\pi}{6}$ and $\angle \mathrm{ABE}=\frac{\pi}{3}$, then the area of the $\triangle \mathrm{ABC}$ is

Options:
A
$\frac{64}{3}$

B
$\frac{8}{3}$

C
$\frac{16}{3}$

D
None of these

The correct answer is:
None of these

Solution:
$\tan 60^{\circ}=\frac{8 / 3}{x}$ or $x=\frac{8}{3 \sqrt{3}}$
Area of $\triangle \mathrm{ABD}=\frac{1}{2} \times 4 \times \frac{8}{3 \sqrt{3}}=\frac{16}{3 \sqrt{3}}$

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2003
In a triangle PQR, the co-ordinates of the points P and Q are -2,4 and 4,-2 respectively. If the equation of the perpendicular bisector of PR is 2x-y+2=0, then the centre of the circumcircle of the ΔPQR is:

Options:
A –1, 0
B –2, –2
C 0, 2
D 1, 4

The correct answer is:
–2, –2

Solution:
Perpendicular bisector of PR is 2x-y+2=0

Midpoint of PQ is M1,1

Equation of perpendicular bisector of PQ is y=x.

Point of intersection of equations PR and PQ is circumcentre.

So, circumcentre is -2,-2

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2021 (17 Mar Shift 1)
In a triangle $\mathrm{PQR}, \angle \mathrm{R}=\frac{\pi}{2}$. If $\tan \left(\frac{\mathrm{P}}{2}\right)$ and $\tan \left(\frac{\mathrm{Q}}{2}\right)$ are the roots of $a x^2+b x+c=0, a \neq 0$ then

Options:
A
$a=b+c$

B
$c=a+b$

C
$b=c$

D
$b=a+c$

The correct answer is:
$c=a+b$

Solution:
$\tan \left(\frac{\mathrm{P}}{2}\right), \tan \left(\frac{\mathrm{Q}}{2}\right)$ are the roots of $\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0$
$$
\tan \left(\frac{\mathrm{P}}{2}\right)+\tan \left(\frac{\mathrm{Q}}{2}\right)=-\frac{\mathrm{b}}{\mathrm{a}}
$$
$\tan \left(\frac{\mathrm{P}}{2}\right) \tan \left(\frac{\mathrm{Q}}{2}\right)=\frac{\mathrm{c}}{\mathrm{a}}$
$\frac{\tan \left(\frac{\mathrm{P}}{2}\right)+\tan \left(\frac{\mathrm{Q}}{2}\right)}{1-\tan \left(\frac{\mathrm{P}}{2}\right) \tan \left(\frac{\mathrm{Q}}{2}\right)}=\tan \left(\frac{\mathrm{P}}{2}+\frac{\mathrm{Q}}{2}\right)=1$
$\Rightarrow \frac{-\frac{\mathrm{b}}{\mathrm{a}}}{1-\frac{\mathrm{c}}{\mathrm{a}}}=1 \Rightarrow-\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{a}}{\mathrm{a}}-\frac{\mathrm{c}}{\mathrm{a}} \Rightarrow-\mathrm{b}=\mathrm{a}-\mathrm{c}$
$\mathrm{c}=\mathrm{a}+\mathrm{b}$.

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2005
In a triangle, the sum of lengths of two sides is $\mathrm{x}$ and the product of the lengths of the same two sides is $y$. if $x^{2}-c^{2}=y,$ where $c$ is the length of the third side of the triangle, then the circumradius of the triangle is

Options:
A $\frac{3}{2} y$
B $\frac{c}{\sqrt{3}}$
C $\frac{c}{3}$
D $\frac{y}{\sqrt{3}}$

The correct answer is:
$\frac{c}{\sqrt{3}}$

Solution:
Let two sides of triangle are $a$ and $b$.

$$

\begin{aligned}

& a+b=x \\

& a b=y \\

& x^{2}-c^{2}=y \Rightarrow(a+b)^{2}-c^{2}=a b \\

\Rightarrow &(a+b-c)(a+b+c)=a b \\

\Rightarrow & 2(s-c)(2 s)=a b \\

\Rightarrow & 4 s(s-c)=a b \\

\Rightarrow & \frac{s(s-c)}{a b}=\frac{1}{4} \\

\Rightarrow & \cos ^{2} \frac{c}{2}=\frac{1}{4} \\

\Rightarrow & \cos c=-\frac{1}{2} \Rightarrow c=120^{\circ}

\end{aligned}

$$

$\therefore$ Area of triangle is,

$\Delta=\frac{1}{2} a b\left(\sin 120^{\circ}\right)=\frac{\sqrt{3}}{4} a b$

$\because \quad R=\frac{a b c}{4 \Delta}$

$\because \quad \mathrm{R}=\frac{a b c}{\sqrt{3} a b}=\frac{c}{\sqrt{3}}$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2019 (11 Jan Shift 1)
In a triangle with sides $a, b, c, r_1>r_2>r_3$ (which are the ex-radii) then

Options:
A
a > b > c

B
a < b < c
C
a > b and b < c
D
a < b and b > c

The correct answer is:
a > b > c

Solution:
$r_1>r_2>r_3 \Rightarrow \frac{\Delta}{s-a}>\frac{\Delta}{s-b}>\frac{\Delta}{s-c} \Rightarrow s-a < s-b < s-c \Rightarrow-a < -b < -c \Rightarrow a>b>c$

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2002
In a typical combustion engine the workdone by a gas molecule is given by W=α2βe-βx2k T, where x is the displacement, k is the Boltzmann constant and T is the temperature. If α and β are constants, dimensions of α will be:

Options:
A MLT-2
B M2LT-2
C MLT-1
D M0LT0

The correct answer is:
M0LT0

Solution:
kT has dimension of energy

βx2kT is dimensionless

βL2=ML2T-2

β=MT-2

α2β has dimensions of work

α2MT-2=ML2T-2

α=M0LT0

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2021 (26 Feb Shift 1)
In a typical Wheatstone network, the resistances in cyclic order are $A=10 \Omega, B=5 \Omega, C=4 \Omega$ and $D=4$ $\Omega$ for the bridge to be balanced

Options:
A $10 \Omega$ should be connected in parallel with $A$
B $10 \Omega$ should be connected in series with $A$
C $5 \Omega$ should be connected in series with $B$
D $5 \Omega$ should be connected in parallel with $B$

The correct answer is:
$10 \Omega$ should be connected in parallel with $A$

Solution:
For a balance Wheatstone bridge.
$\begin{aligned}
& \frac{A}{B}=\frac{D}{C} \Rightarrow \frac{10}{5} \neq \frac{4}{4} \text { (Unbalanced) } \\
& \frac{A^{\prime}}{B}=\frac{D}{C} \Rightarrow \frac{A^{\prime}}{5}=\frac{4}{4} \Rightarrow A^{\prime}=5 \Omega
\end{aligned}$
$A^{\prime}(5 \Omega)$ is obtained by connecting a $10 \Omega$ resistance in parallel with $A$.

Topic: Physics, Current Electricity, JEE Main