datavorous/entrance-exam-dataset huggingface dataset:
If the $x$-intercept of some line $L$ is double as that of the line, $3 x+4 y=12$ and the $y$-intercept of $L$ is half as that of the same line, then the slope of $L$ is :

Options:
A
$-3$

B
$-3 / 8$

C
$-3 / 2$

D
$-3 / 16$

The correct answer is:
$-3 / 16$

Solution:
Given line $3 x+4 y=12$ can be rewritten as
$$
\begin{aligned}
& \frac{3 x}{12}+\frac{4 y}{12}=1 \Rightarrow \frac{x}{4}+\frac{y}{3}=1 \\
& \Rightarrow x \text {-intercept }=4 \text { and } y \text {-intercept }=3
\end{aligned}
$$
Let the required line be
$\mathrm{L}: \frac{x}{a}+\frac{y}{b}=1$ where
$a=x$-intercept and $b=y$-intercept
According to the question
$a=4 \times 2=8$ and $b=3 / 2$
$\therefore$ Required line is $\frac{x}{8}+\frac{2 y}{3}=1$
$\Rightarrow 3 x+16 y=24$
$\Rightarrow y=\frac{-3}{16} x+\frac{24}{16}$
Hence, required slope $=\frac{-3}{16}$.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2013 (22 Apr Online)
If the Young's modulus of the material is 3 times its modulus of rigidity, then its volume elasticity will be

Options:
A Zero
B Infinity
C $2 \times 10^{10} \mathrm{~N} / \mathrm{m}^2$
D $3 \times 10^{10} \mathrm{~N} / \mathrm{m}^2$

The correct answer is:
Infinity

Solution:
$\quad I=2 (\eta)(1+\sigma) \Rightarrow 3 \eta=2 (\eta)(1+\sigma)=\sigma=\frac{3}{2}-1=\frac{1}{2}$
Now substituting the value of $\sigma$ in the following expression.
$Y=3 K(1-2 \sigma) \Rightarrow K=\frac{Y}{3(1-2 \sigma)}=\infty$

Topic: Physics, Mechanical Properties of Solids, JEE Main
If $\alpha+\beta-\gamma=\pi$, then $\sin ^2 \alpha+\sin ^2 \beta-\sin ^2 \gamma=$

Options:
A $2 \sin \alpha \sin \beta \cos \gamma$
B $2 \cos \alpha \cos \beta \cos \gamma$
C $2 \sin \alpha \sin \beta \sin \gamma$
D None of these

The correct answer is:
$2 \sin \alpha \sin \beta \cos \gamma$

Solution:
$\begin{aligned} & \text {We have } \alpha+\beta-\gamma=\pi \text {. } \\ & \text { Now } \sin ^2 \alpha+\sin ^2 \beta-\sin ^2 \gamma \quad=\sin ^2 \alpha+\sin (\beta-\gamma) \sin (\beta+\gamma) \\ & \left.=\sin ^2 \alpha+\sin (\pi-\alpha) \sin (\beta+\gamma) \quad \alpha+\beta-\gamma=\pi\right) \\ & =\sin ^2 \alpha+\sin \alpha \sin (\beta+\gamma)=\sin \alpha\{\sin \alpha+\sin (\beta+\gamma)\} \\ & =\sin \alpha\{\sin (\pi- \beta+\gamma)+\sin (\beta+\gamma)\} \\ & =\sin \alpha\{-\sin (\gamma-\beta)+\sin (\gamma+\beta)\} \\ & =\sin \alpha\{2 \sin \beta \cos \gamma\}=2 \sin \alpha \sin \beta \cos \gamma\end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
if
then

Options:
A $\lim _{x \rightarrow 0^+} f(x)=e$
B $\lim _{x \rightarrow 0^-} f(x)=e^2$
C $f(x)$ is discontinuous at $x=0$
D None of these

The correct answer is:
$\lim _{x \rightarrow 0^-} f(x)=e^2$

Solution:
$\lim _{x \rightarrow 0^-} f(x)=\lim _{x \rightarrow 0}\left[(1+2 x)^{1 / 2 x}\right]^2=e^2$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If there are $m$ sources and $n$ destinations in a transportation matrix, the total number of basic variables in a basic feasible solution is

Options:
A $m+n$
B $m+n+1$
C $m+n-1$
D $\mathrm{m}$

The correct answer is:
$m+n-1$

Solution:
If $x_{i j} \geq 0$, is the number of units shipped from $\mathrm{i}^{\text {th }}$ source to $\mathrm{j}^{\text {th }}$ destination, then the equivalent LPP model will be
Minimize $Z=\sum_{i=1}^m \sum_{j=1}^n c_{i j} x_{i j}$
Subjected to:
$\begin{aligned}
& \sum_{i=1}^m x_{i j} \leq b_i(\text { demand }) \\
& \sum_{j=1}^n x_{i j} \leq a_i(\text{supply})
\end{aligned}$
If total supply $=$ total demand then it is a balanced transportation problem otherwise it is called an unbalanced transportation problem.
There will be $(m+n-1)$ basic independent variables out of $(m \times n)$ variables.

Topic: Mathematics, Linear Programming, JEE Main
If there are $n$ capacitors in parallel connected to $V$ volt source, then the energy stored is equal to

Options:
A $\mathrm{CV}$
B $\frac{1}{2} n C V^2$
C $C v^2$
D $\frac{1}{2 n} C V^2$

The correct answer is:
$\frac{1}{2} n C V^2$

Solution:
$U=\frac{1}{2} C_{e q} V^2=\frac{1}{2}(n C) V^2$

Topic: Physics, Capacitance, JEE Main
If there are n capacitors in parallel connected to V volt source, then the energy stored is equal to

Options:
A
$\mathrm{CV}$

B
$\frac{1}{2} \mathrm{nCV}^2$

C
$\mathrm{CV}^2$

D
$\frac{1}{2 n} CV^2$

The correct answer is:
$\frac{1}{2} \mathrm{nCV}^2$

Solution:
$E=\sum \frac{1}{2} C V^2=\frac{1}{2} n C V^2$

Topic: Physics, Capacitance, JEE Main, JEE Main 2002
If there are $n$ independent trials, $p$ and $q$ the probability of success and failure respectively, then probability of exactly $r$ successes
or
Let $p$ be the probability of happening an event and $q$ its failure, then the total chance of $r$ successes in $n$ trials is

Options:
A ${ }^n C_{n+1} \cdot p^{\prime} q^{n-1}$
B ${ }^n C_r p^{r-1} q^{r+1}$
C ${ }^n C_r q^{n-1} p^r$
D ${ }^n C_r p^{r+1} q^{r-1}$

The correct answer is:
${ }^n C_r q^{n-1} p^r$

Solution:
it is a fundamental concept

Topic: Mathematics, Probability, JEE Main
If there are n number of seats and m number of people have to be seated, then the number of possible ways to do this is (here, m<n)

Options:
A nPm
B Â nCm
C Â nCnÃ(m-1)!
D Â n-1Pm-1

The correct answer is:
nPm

Solution:
âµ Total number of seats=n
and number of people=m
1st person can be seated in n ways
2nd person can be seated in n-1 ways
â¯â¯â¯â¯â¯
â¯â¯â¯â¯â¯
â¯â¯â¯â¯
m th person can be seated in n-m+1ways
â´ Total number of ways
=nn-1n-2â¦n-m+1= nPm

Topic: Mathematics, Permutation Combination, JEE Main
If there are nine straight lines of which five are concurrent at a point and the other four are concurrent at another point and no two of these nine lines are parallel, then the number of points of intersection is equal to

Options:
A 20
B 22
C 36
D 38

The correct answer is:
22

Solution:
Nine straight lines will intersect at C29 points
Five are concurrent at one point so there will be loss of C25 points
Four other are concurrent at another point, so there is further loss of C24 points.
But we will have these two points at which the lines are concurrect.
So, total number of points =C2-C25-C24+29=36-10-6+2=22

Topic: Mathematics, Permutation Combination, JEE Main
If \( \theta \) lies in the second quadrant and \( 3 \tan \theta+4=0 \), then the value of \( 2 \cot \theta-5 \cos \theta+\sin \theta \) is \( \frac{6}{k} \) then find \( k \).

Options:

The correct answer is:
2.3

Solution:
3tanθ+4=0⇒tanθ=-43

cotθ=-3/4
cosθ=-3/5
sinθ=4/5
=2cotθ-5cosθ+sinθ
=2(-3/4)-5(-3/5)+4/5=2310

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If three dice are thrown together, then the probability of getting 5 on at least one of them is

Options:
A $\frac{125}{216}$
B $\frac{215}{216}$
C $\frac{1}{216}$
D $\frac{91}{216}$

The correct answer is:
$\frac{91}{216}$

Solution:
Required probability
$={ }^3 C_1\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^2$ $+{ }^3 C_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)+{ }^3 C_3\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^0=\frac{91}{216}$

Topic: Mathematics, Probability, JEE Main
If three distinct numbers a, b, c are in G.P. and the equations ax2+2bx+c=0 and dx2+2ex+f=0 have a common root, then which one of the following statements is correct?

Options:
A da,eb,fc are in A.P.
B d, e, f are in A.P.
C d, e, f are in G.P.
D da,eb,fc are in G.P.

The correct answer is:
da,eb,fc are in A.P.

Solution:
∵ a, b, c are in G.P. ⇒b2=ac …(i)
Equation ax2+2bx+c=0
Hence, D=4b2-4ac=0 ∵b2=ac
⇒ equation has equal roots ⇒ roots are =-ba, -ba ( ∵ sum of roots α+α=-2ba )
∵ Both equations have one root in common.
∴ Common root is -ba. It will satisfy equation.
∴d-ba2+2e-ba+f=0⇒db2-2aeb+a2f=0
⇒dac-2aeb+a2f=0 ∵b2=ac
⇒dc-2aeb+af=0
⇒dcac-2ebac+afac=0 (dividing by ac )
⇒da-2ebb2+fc=-⇒da+fc=2eb
⇒da,eb,fc are in A.P.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2019 (08 Apr Shift 2)
If three distinct points $A, B, C$ are given in the 2dimensional coordinate plane such that the ratio of the distance of each one of them from the point $(1,0)$ to the distance from $(-1,0)$ is equal to $\frac{1}{2}$, then the circumcentre of the triangle $A B C$ is at the point

Options:
A
$\left(\frac{5}{3}, 0\right)$

B
$(0,0)$

C
$\left(\frac{1}{3}, 0\right)$

D
$(3,0)$

The correct answer is:
$\left(\frac{5}{3}, 0\right)$

Solution:
Let $P(1,0)$ and $Q(-1,0), A(x, y)$
$$
\begin{aligned}
& \text { Given: } \frac{A P}{A Q}=\frac{B P}{B Q}=\frac{C P}{C Q}=\frac{1}{2} \\
& \Rightarrow 2 A P=A Q \\
& \Rightarrow 4(A P)^2=A Q^2
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow 4\left[(x-1)^2+y^2\right]=(x+1)^2+y^2 \\
& \Rightarrow 4\left(x^2+1-2 x\right)+4 y^2=x^2+1+2 x+y^2 \\
& \Rightarrow 3 x^2+3 y^2-8 x-2 x+4-1=0 \\
& \Rightarrow 3 x^2+3 y^2-10 x+3=0 \\
& \Rightarrow x^2+y^2-\frac{10}{3} x+1=0
\end{aligned}
$$
$\therefore$ A lies on the circle given by (1). As $B$ and $C$ also follow the same condition.
$\therefore \quad$ Centre of circumcircle of $\triangle A B C=$ centre of circle given by (1)
$$
=\left(\frac{5}{3}, 0\right) \text {. }
$$

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2012 (19 May Online)
If three fair dice are thrown and the sum is an odd number, then the probability that all the three dice show an odd number is

Options:
A 34
B 12
C 56
D 14

The correct answer is:
14

Solution:
Total cases = sum on 3 dice is an odd number =6Ã6Ã62=108
Favourable cases = all three dice show an odd number
=3Ã3Ã3=27
Required probability =27108=14

Topic: Mathematics, Probability, JEE Main
If three geometric means be inserted between 2 and 32 , then the third geometric mean will be

Options:
A $8$
B $4$
C $16$
D $12$

The correct answer is:
$16$

Solution:
$2, g_1, g_2, g_3, 32$ where
$a=2, a r=g_1, a r^2=g_2, a r^3=g_3$ and $a r^4=32$
Now $2 \times r^4=32 \Rightarrow r^4=16=(2)^4 \Rightarrow r=2$
Then third geometric mean $=a r^3=2 \times 2^3=16$
Aliter: By formula,
$G_3=2 \cdot\left(\frac{32}{2}\right)^{3 / 4}$ $=2 \cdot 8=16$

Topic: Mathematics, Sequences and Series, JEE Main
If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is _____$\times 10^{-2} \mathrm{MeV}$. (Given $1 \mathrm{u}=931 \mathrm{MeV} / \mathrm{c}^2$, atomic mass of helium $=4.002603 \mathrm{u}$ )

Options:

The correct answer is:
727

Solution:
Reaction :
$\begin{aligned}
& 3{ }_2^4 \mathrm{He} \longrightarrow{ }_6^{12} \mathrm{C}+\gamma \text { rays } \\
& \text { Mass defect }=\Delta \mathrm{m}=\left(3 \mathrm{~m}_{\mathrm{He}}-\mathrm{m}_{\mathrm{C}}\right) \\
& =(3 \times 4.002603-12)=0.007809 \mathrm{u}
\end{aligned}$
Energy released
$=931 \Delta \mathrm{m} \mathrm{MeV}$
$=7.27 \mathrm{MeV}=727 \times 10^{-2} \mathrm{MeV}$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2024 (05 Apr Shift 1)
If three moles of monoatomic gas γ=53 is mixed with two moles of a diatomic gas γ=75, the value of adiabatic exponent γ for the mixture is:

Options:
A 1.75
B 1.40
C 1.52
D 1.35

The correct answer is:
1.52

Solution:
Given:

f1=3,f2=5n1=3,n2=2

Now, for mixture

fmixture =n1f1+n2f2n1+n2=9+105=195

Therefore, γmixture =1+2fmixture=1+2×519=2919=1.52

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2024 (30 Jan Shift 2)
If three normals are drawn from the point c,0 to the parabola y2=4x and two of which are perpendicular, then the value of c is equal to

Options:
A 3
B 4
C 5
D 6

The correct answer is:
3

Solution:
Equation of the normal in slope form is y=mx-2m-m3 which passes through c,0
â0=mc-2m-m3
âm3+2-cm=0
âm=0 or m2+2-c=0
âm1m2=2-c1=-1
âc=3

Topic: Mathematics, Parabola, JEE Main
If three numbers be in G.P., then their logarithms will be in

Options:
A A.P.
B G.P.
C H.P.
D None of these

The correct answer is:
A.P.

Solution:
Let three number $a, b$ and $C_{\text {in G.P., then }} b^2=a c$
$\Rightarrow 2 \log _e b=\log _e a+\log _e c$ or $\log _e b=\frac{\log _e a+\log _e c}{2}$
Thus their logarithms are in A.P.

Topic: Mathematics, Sequences and Series, JEE Main
If three object $A, B$ and $C$ of mass $m, 2 m$ and $3 m$ are placed along the $x$-axis at distance $3 x, 2 x$ and $x$ from its origin. In this case, the distance of the centre of mass from the origin is

Options:
A $x_{c m}=\frac{5}{2} x$
B $x_{c m}=\frac{1}{2} x$
C $x_{c m}=\frac{5}{3} x$
D $x_{c m}=4 x$

The correct answer is:
$x_{c m}=\frac{5}{3} x$

Solution:
CONCEPT:
A group of particles that undergoes a particular type of motion together is called a system of particles.
A particular point where the whole of the mass of the system of particles appeared to be concentrated is called the center of mass.
Let us consider two-particle system having masses $m_1$ and $m_2$ located on $X$-axis at $x_1$ and $x_2$ respectively.
The Centre of mass of the two particles is at point $\mathrm{c}$ which is given by:
$X=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}$
$\begin{aligned} & \text { If } \mathrm{m}_1=\mathrm{m}_2=\mathrm{m} \\ & X=\frac{m x_1+m x_2}{m+m}=\frac{x_1+x_2}{2}\end{aligned}$
If we have $n$ particles having masses $m_1, m_2, m_3$, .$m_n$ on a straight line then Centre of mass is given by:
$X=\frac{m_1 x_1+m_2 x_2+m_3 x_3+\ldots \ldots m_n x_n}{m_1+m_2+m_3+\ldots \ldots \ldots . m_n}=\frac{\sum_{1-1}^n m_1 x_1}{\sum_{1-1}^n m_1}$
Now suppose we have three particles of masses $m_1, m_2$ and $m_3$ not on a straight line but in space at $\left(x_1, y_1, z_1\right),\left(x_2, y_2, z_2\right)$ and $\left(x_3, y_3, z_3\right)$. Then the Centre of the mass of the system is located at $(X, Y, Z)$ which is given by:
$\begin{aligned} & X-\text { coordinate }(X)=\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3} \\ & Y-\text { coordinate }(Y)=\frac{m_1 y_1+m_2 y_2+m_3 y_3}{m_1+m_2+m_3} \\ & Z-\text { coordinate }(Z)=\frac{m_1 z_1+m_2 z_2+m_3 z_3}{m_1+m_2+m_3}\end{aligned}$
Calculation:
Given that,
Mass of particle $A, m_1=m$
Mass of particle $A, m_2=2 m$
Mass of particle $A, m_3=3 \mathrm{~m}$
the distance of particle $\mathrm{A}$ from its origin, $\mathrm{x}_1=3 \mathrm{x}$
the distance of particle $B$ from its origin, $x_2=2 x$
the distance of particle $\mathrm{C}$ from its origin, $\mathrm{x}_3=\mathrm{x}$
Also, from the above-given explanation, we can see that centre of mass for the two-particle system can be expressed as
$X=\frac{m_1 x_1+m_2 x_2+m_3 x_3+\ldots \ldots m_n x_n}{m_1+m_2+m_3+\ldots \ldots \ldots m_n}=\frac{\sum_{l-1}^n m_1 x_4}{\sum_{1-1}^n m_4}$
Now in this case for the three-particle system, it can be modified as
C.M along $X$-coordinate $\left(x_{c m}\right)=\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3}$
$\therefore x_{c m}=\frac{m \times 3 x+2 m \times 2 x+3 m \times x}{m+2 m+3 m}=\frac{10 m x}{6 m}=\frac{5}{3} x$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
If three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is:

Options:
A 110
B 310
C 15
D 320

The correct answer is:
110

Solution:
Total number of ways of selecting three points from vertices of a regular hexagon
= C3 6=6.5.41.2.3=20
Favorable selection (forming equilateral triangle)
=2 ACE, BDF
⇒ Required probability P=Number of favourable selectionsTotal selections
=220=110

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2019 (12 Apr Shift 1)
If three positive numbers a, b and c are in A.P. such that abc=8, then the minimum possible value of b is:

Options:
A 423
B 2
C 413
D 4

The correct answer is:
2

Solution:
As a+c=2b,
Using A.M≥G.M
⇒a+b+c3≥abc13

⇒3b3≥813
b≥2

∴ bmin=2

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2017 (09 Apr Online)
If three unreactive gases having partial pressures $P_A, P_B$ and ${ }^{P_C}$ and their moles are 1,2 and 3 respectively then their total pressure will be

Options:
A $P=P_A+P_B+P_C$
B $P=\frac{P_A+P_B+P_C}{6}$
C $P=\frac{\sqrt{P_A+P_B+P_C}}{3}$
D None

The correct answer is:
$P=P_A+P_B+P_C$

Solution:
According to Dalton's law of partial pressure,
Total pressure = sum of partial pressures of individual gases
$\mathrm{P}_{\text {teftalu }}=\mathrm{P}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}+\mathrm{Pc}_C$

Topic: Chemistry, States of Matter, JEE Main
If three vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ satisfy $\mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}$ and $|\mathbf{a}|=3,|\mathbf{b}|=5,|\mathbf{c}|=7$, then the angle between $\mathbf{a}$ and $\mathbf{b}$ is

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

The correct answer is:
$60^{\circ}$

Solution:
Given, $\mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0} \Rightarrow \mathbf{a}+\mathbf{b}=-\mathbf{c}$
Squaring on both sides,
$\begin{aligned}
& \Rightarrow|\mathbf{a}|^2+|\mathbf{b}|^2+2|\mathbf{a}||\mathbf{b}| \cos \theta=|-\mathbf{c}|^2 \\
& \Rightarrow 9+25+30 \cos \theta=49 \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=60^{\circ} .
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
If time t, velocity v, and angular momentum l are taken as the fundamental units. Then the dimension of mass (m) in terms of t, v and l is:

Options:
A t-1v1l-2
B t1v2l-1
C t-2v-1l1
D t-1v-2l1

The correct answer is:
t-1v-2l1

Solution:
m∝tavbℓc

m∝[T]aLT-1bML2 T-1c

M1 L0 T0=McLb+2cTa-b-c

Comparing powers,

c=1, b=-2, a=-1

m∝t-1v-2ℓ1

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2021 (20 Jul Shift 2)
If transverse and conjugate axes of a hyperbola are equal, then its eccentricity is

Options:
A $\sqrt{3}$
B $\sqrt{2}$
C $1 / \sqrt{2}$
D $2$

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

Solution:
Hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. Here, transverse and conjugate axis of a hyperbola is equal.
i.e., $a=b \quad \therefore x^2-y^2=a^2$; which is a rectangular hyperbola. Hence, eccentricity
$e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{2}$

Topic: Mathematics, Hyperbola, JEE Main
If tube length of astronomical telescope is $105 \mathrm{~cm}$ and magnifying power is 20 for normal setting, calculate the focal length of objective

Options:
A $100 \mathrm{~cm}$
B $10 \mathrm{~cm}$
C $20 \mathrm{~cm}$
D $25 \mathrm{~cm}$

The correct answer is:
$100 \mathrm{~cm}$

Solution:
$|m|=\frac{f_o}{f_e}=20$ and $L=f_o+f_e=105 \Rightarrow f_o=100 \mathrm{~cm}$

Topic: Physics, Ray Optics, JEE Main
If two +5 D, lenses are mounted at some distance apart, the equivalent power will always be negative, if the distance is

Options:
A Greater than 40 cm

B Equal to 10 cm

C Greater than 80 cm
D Less than 10 cm

The correct answer is:
Greater than 40 cm

Solution:
When two lenses are separated by some distance x, then equivalent power
P=P₁+P₂-xP₁P₂
∴P=5+5-x×5×5
or P=10-25x
Power P will be negative if 10-25x will be negative
ie, 25x>10
or x>1025
or x>1025×100 cm
or x>40 cm

Topic: Physics, Ray Optics, JEE Main
If two different numbers are taken from the set 0, 1, 2, 3,....., 10; then the probability that their sum as well as absolute difference are both multiple of 4, is:

Options:
A 655
B 1255
C 1445
D 755

The correct answer is:
655

Solution:
Required favourable cases are 0, 4,0, 8,2, 6,2, 10,4, 8,6, 10

So, favourable cases =6

Total cases =11C2=11×102=55

Hence, the required probability =655

Topic: Mathematics, Probability, JEE Main, JEE Main 2017 (02 Apr)
If two distinct numbers a and b are selected from the set 51, 52, 53.......59, then the probability that logab is an integer is

Options:
A 718
B 59
C 736
D 316

The correct answer is:
718

Solution:
Total cases =C29
For favorable cases,
let a=5i, b=5j
logab=ji (must be an integer)
if i=1⇒j=2,3,4,5,6,7,8,9
if i=2⇒j=4,6,8
if i=3⇒j=6,9
if i=4⇒j=8
for i=5,6,7,8,9⇒ no such value of j is possible
Hence, the required probability =149C2=1436=718

Topic: Mathematics, Probability, JEE Main
If two distinct point Q, R lie on the line of intersection of the planes -x+2y-z=0 and 3x-5y+2z=0 and PQ=PR=18 where the point P is 1,-2,3, then the area of the triangle PQR is equal to

Options:
A 2338
B 4338
C 8338
D 1523

The correct answer is:
4338

Solution:
Given equations of plane are -x+2y-z=0 and 3x-5y+2z=0

Now a vector perpendicular to both the planes is given by n→=i^j^k^-12-13-52

=i^-1-j^1+k^-1

n→=-i^-j^-k^

For a point on the line of intersection of both the planes

let z=0,

so x=2y & 8x=5y

i.e. a point on the line will be 0,0,0

So equation of line of intersection of both these planes will be x1=y1=z1

The DR's of QR are 1, 1, 1

Let the point Tα,α,α be the foot of perpendicular from P to the line

⇒α-1×1+α+2×1+α-3×1=0

i.e. α=23

or T≡23,23,23

PT2=19+649+499

PT2=1149

PT=1143

If θ is ∠TPQ, then in right anged triangle TPQ,

cosθ=1143×32=579=1933

Now cos2θ=2cos2θ-1=2×1927-1=1127

and sin2θ=1-11272=42738

So required area of triangle PQR =12×1818×42738

=182×42738=4338

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (28 Jun Shift 1)
If two glass plates have water between them and are separated by very small distance (see figure), it is very difficult to pull them apart. It is because the water in between forms cylindrical surface on the side that gives rise to lower pressure in the water in comparison to atmosphere. If the radius of the cylindrical surface is R and surface tension of water is T then the pressure in water between the plates is lower by:

Options:
A TR
B T4R
C 4TR
D 2TR

The correct answer is:
2TR

Solution:
d=2R cosθ
∴ Pressure difference across a liquid surface is
∆P=T 1R1+1R2
∵ R1=R and R2=R
∆P=T1R+1R
∆P= 2TR
∴ Pressure is more in the concave side hence pressure in water between the plates is lower by 2TR

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2015 (10 Apr Online)
If two lines $L_1$ and $L_2$ in space, are defined by
$$
\begin{aligned}
& L_1=\{x=\sqrt{\lambda} y+(\sqrt{\lambda}-1), \\
& \quad z=(\sqrt{\lambda}-1) y+\sqrt{\lambda}\} \text { and } \\
& L_2=\{x=\sqrt{\mu} y+(1-\sqrt{\mu}),
\end{aligned}
$$
$$
z=(1-\sqrt{\mu}) y+\sqrt{\mu}\}
$$
then $L_1$ is perpendicular to $L_2$, for all nonnegative reals $\lambda$ and $\mu$, such that :

Options:
A
$\sqrt{\lambda}+\sqrt{\mu}=1$

B
$\lambda \neq \mu$

C
$\lambda+\mu=0$

D
$\lambda=\mu$

The correct answer is:
$\lambda=\mu$

Solution:
For $\mathrm{L}_1$,
$$
\begin{aligned}
& x=\sqrt{\lambda} y+(\sqrt{\lambda}-1) \Rightarrow y=\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}} \\
& z=(\sqrt{\lambda}-1) y+\sqrt{\lambda} \Rightarrow y=\frac{z-\sqrt{\lambda}}{\sqrt{\lambda}-1}
\end{aligned}
$$
From (i) and (ii)
$$
\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}}=\frac{y-0}{1}=\frac{z-\sqrt{\lambda}}{\sqrt{\lambda}-1}
$$
The equation (A) is the equation of line $\mathrm{L}_1$. Similarly equation of line $\mathrm{L}_2$ is
$$
\frac{x-(1-\sqrt{\mu})}{\sqrt{\mu}}=\frac{y-0}{1}=\frac{z-\sqrt{\mu}}{1-\sqrt{\mu}}
$$
Since $\mathrm{L}_1 \perp \mathrm{L}_2$, therefore
$$
\begin{aligned}
& \sqrt{\lambda} \sqrt{\mu}+1 \times 1+(\sqrt{\lambda}-1)(1-\sqrt{\mu})=0 \\
& \Rightarrow \sqrt{\lambda}+\sqrt{\mu}=0 \Rightarrow \sqrt{\lambda}=-\sqrt{\mu} \\
& \Rightarrow \lambda=\mu
\end{aligned}
$$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2013 (23 Apr Online)
If two mirrors are kept at $60^{\circ}$ to each other, then the number of images formed by them is

Options:
A
5

B
6

C
7

D
8

The correct answer is:
5

Solution:
No. of images, $n=(360 / \theta)-1$. As $\theta=60^{\circ}$ so $n=5$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2002
If two moles of an ideal gas at $546 \mathrm{~K}$ occupy a volume of 44.8 litrers, the pressure must be

Options:
A 2 atm
B 3 atm
C 4 atm
D 1 atm

The correct answer is:
2 atm

Solution:
$P=\frac{n R T}{V}=\frac{2 \times 0.0821 \times 546}{44.8 l}=$ 2 atm

Topic: Chemistry, States of Matter, JEE Main