datavorous/entrance-exam-dataset huggingface dataset:
Find the area of a triangle whose vertices are $(1,-1),(-4,6)$ and $(-3,-5)$.

Options:
A 8
B 24
C 16
D 12

The correct answer is:
24

Solution:
Let the given points be:
$\begin{aligned}
& A(1,-1)=\left(x_1, y_1\right) \\
& B(-4,6)=\left(x_2, y_2\right) \\
& C(-3,-5)=\left(x_3, y_3\right)
\end{aligned}$
The area of the triangle formed by the points $\left(x_1, y_1\right),\left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$ is the numerical value of the expression $1 / 2\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$.
Now, the area of triangle $A B C=(1 / 2)|1(6+5)-4(-5+1)-3(-1-6)|$
$\begin{aligned}
& =(1 / 2)|11-4(-4)-3(-7)| \\
& =(1 / 2)|11+16+21| \\
& =(1 / 2) \times 48 \\
& =24
\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main
Find the area of the triangle formed by the $X$-axis and the tangent and the normal to the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$

Options:
A $\frac{a b}{4} \sqrt{a^2+b^2}$
B $4 a b$
C $\frac{b}{4 a}\left(a^2+b^2\right)$
D $\frac{a b}{2} \sqrt{a^2+b^2}$

The correct answer is:
$\frac{b}{4 a}\left(a^2+b^2\right)$

Solution:
Ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1...(i)$
To find Equation of tangent and normal at point $\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$
We have to find $\frac{d y}{d x}$.
Slope of tangent $=\left.\frac{d y}{d x}\right|_{\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)}=m_T$
Now, differentiating Eq. (i) w.r.t. $x$
$\begin{aligned} \frac{2 x}{a^2}+\frac{2 y}{b^2} \frac{d y}{d x} & =0 \\ \frac{d y}{d x} & =-\frac{2 x}{a^2 \times \frac{2 y}{b^2}}=-\frac{b^2}{a^2}\left(\frac{x}{y}\right)\end{aligned}$
$$
\begin{aligned}
& \therefore \quad m_T=-\frac{b^2}{a^2}\left(\frac{\frac{a}{\sqrt{2}}}{\frac{b}{\sqrt{2}}}\right)=-\frac{b}{a} \\
& \text { and } m_N=-\frac{1}{m_T}=\frac{a}{b}
\end{aligned}
$$
Equation of tangent
$$
\left(y-\frac{b}{\sqrt{2}}\right)=-\frac{b}{a}\left(x-\frac{a}{\sqrt{2}}\right)...(ii)
$$
Equation of normal
$$
\left(y-\frac{b}{\sqrt{2}}\right)=\frac{a}{b}\left(x-\frac{a}{\sqrt{2}}\right)...(iii)
$$
To find Area of triangle made by tangent, normal and $X$-axis.
We have to find points on $X$-axis cut by tangent and normal, i.e $B$ and $C$
For $B$ Put $y=0$ in Eq. (ii) $\Rightarrow-\frac{b}{\sqrt{2}}=-\frac{b}{a}\left(x-\frac{a}{\sqrt{2}}\right)$
$$
\frac{a}{\sqrt{2}}=x \frac{-a}{\sqrt{2}}(x=\sqrt{2} a)
$$
$\therefore B$ is $(\sqrt{2} a, 0)$
For $C$ Put $y=0$ in Eq. (iii) $\Rightarrow-\frac{b}{\sqrt{2}}=\frac{a}{b}\left(x-\frac{a}{\sqrt{2}}\right)$
$$
\begin{array}{rlrl}
\Rightarrow & & \frac{-b^2}{\sqrt{2} a}+\frac{a}{\sqrt{2}} & =x \\
\Rightarrow & \frac{a^2-b^2}{\sqrt{2} a} & =x
\end{array}
$$
Area of $\begin{aligned} \triangle A B C & =\frac{1}{2} \times \text { Base } \times \text { Height } \\ & =\frac{1}{2} \times B C \times A M \\ & =\frac{1}{2} \times\left[\sqrt{2} a-\left(\frac{a^2-b^2}{\sqrt{2 a}}\right)\right] \times \frac{b}{\sqrt{2}} \\ & =\frac{1}{2}\left(\frac{2 a^2-a^2+b^2}{\sqrt{2} a}\right) \cdot \frac{b}{\sqrt{2}} \\ & =\frac{b}{4 a}\left(a^2+b^2\right)\end{aligned}$

Topic: Mathematics, Application of Derivatives, JEE Main
Find the atomic number for an atom that emits Kα radiation with wavelength 4λ, if Kα wavelength emitted by an atom of atomic number Z=11 is λ .

Options:
A Z=6
B Z=4
C Z=11
D Z=44

The correct answer is:
Z=6

Solution:
1 λ ∝ z-1 2
∴ λ 1 λ 2 = z 2 - 1 2 z 1 - 1 2
1 4 = Z 2 - 1 1 1 - 1 2
Solving this, we get, Z 2 = 6

Topic: Physics, Atomic Physics, JEE Main
Find the Binding energy per nucleon for Sn50120. Mass of proton mp=1.00783 U, mass of neutron mn=1.00867 U and mass of tin nucleus msn=119.902199 U. (take 1U = 931 MeV)

Options:
A 7.5MeV
B 9.0MeV
C 8.0MeV
D 8.5MeV

The correct answer is:
8.5MeV

Solution:
Binding energy =(ΔM)C2=(AM)931

put the value of ∆M

BE=8.5MeV

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2020 (04 Sep Shift 2)
Find the centre of mass xCM of the shaded portion of the disc of radius R=24 cm.

Options:

The correct answer is:
4

Solution:
Mass of shaded portion = 3 M 4
X CM = 0 - M 4 - R 2 M - M / 4 = R 6
R = 24 cm
Therefore,
X CM = 2 4 6 = 4 cm

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Find the change in the temperature on the Fahrenheit scale and on Kelvin scale, if an iron piece is heated from 30âÂ°C to 90âÂ°C.

Options:
A 108âÂ°F,â60âK
B 100 °F, 55 K
C 100âÂ°F,â65âK
D 60 °F, 108 K

The correct answer is:
108 °F, 60 K

Solution:
ΔTC=90 °C−30 °C=60 °C
ΔTF=95ΔTC=95(60) °F=108 °F
ΔTK=ΔT⁡C⁡=60 K⁡

Topic: Physics, Thermal Properties of Matter, JEE Main
Find the closest integer wavelength value of Kα line (in picometer) in copper (Z = 29) if the wavelength of Kα = line in iron (Z = 26) is known to be equal to193 picometer.

Options:

The correct answer is:
154

Solution:
By Mosely law 1Î» â (Z â1)2
Î»2193=26-1229-12
Î»2=193Ã252282
Î»2 = 153.8 pm â 154 pm

Topic: Physics, Atomic Physics, JEE Main
Find the coordinates of the foci and the length of the latusrectum of the hyperbol(a)
$\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$

Options:
A $(0, \pm 2), \frac{32}{3}$
B $(0, \pm 5), \frac{32}{3}$
C $(\pm 5,0), \frac{32}{3}$
D $(0, \pm 5), \frac{3}{32}$

The correct answer is:
$(\pm 5,0), \frac{32}{3}$

Solution:
We have, $a=3, b=4$ and
$c=\sqrt{a^{2}+b^{2}}=\sqrt{9+16}=5$
Therefore, the coordinates of the foci are $(\pm 5,0)$.
Also, length of latusrectum $=\frac{2 b^{2}}{a}=\frac{32}{3}$

Topic: Mathematics, Hyperbola, JEE Main
Find the coordinates of the point where the line joining the points (2,-3,1) and (3,-4,-5) cuts the plane $2 x+y+z=7$

Options:
A (1,2,-7)
B (1,-2,7)
C (-1,-2,7)
D (1,2,7)

The correct answer is:
(1,-2,7)

Solution:
The direction ratios of the line are $3-2,-4-(-3),-5-1$ i.e. 1,-1,-6
Hence equation of the line joining the given points is $\frac{x-2}{1}=\frac{y+3}{-1}=\frac{z-1}{-6}=r($ say $)$
Coordinates of any point on this line are $(r+2,-r-3,-6 r+1)$
If this point lies on the given plane $2 x+y+z=7$ then $2(r+2)+(-r-3)+(-6 r+1)=7 \Rightarrow r=-1$
Coordinates of any point on this line are $(-1+2,-(-1)-3,-6(-1)+1)$ i.e. (1,-2,7)

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Find the correct option:

Column: A	Column: B

(A)	Parenchyma	1.	Living tissue that provides mechanical support

B	Collenchyma	2.	Food storage

C	Sclerenchyma	3.	Dead tissue that provides mechanical support

Options:
A A-1;ââB-2;ââC-3
B A-2;ââB-1;ââC- 3
C A-1Â &Â 2;ââB-2;ââC-3
D None of these

The correct answer is:
A-2;ââB-1;ââC- 3

Solution:
Parenchyma function is the storage of food, photosynthesis and secretion. Collenchyma is a living tissue whose main function is mechanical support and photosynthesis also. Sclerenchyma is a dead tissue whose main function is mechanical support.

Topic: Biology, Anatomy of Flowering Plants, NEET
Find the correct order of acid strengths of the following compounds

Options:
A F>D>E>B>C>A
B D>F>E>C>B>A
C D>E>F>B>C>A
D F>E>D>B>C>A

The correct answer is:
F>D>E>B>C>A

Solution:
Electron-withdrawing substituents can increase acidity of a nearby atom, which increases with electronegativity and decreases with increasing distance to the atom.

For the compounds given:

–NO2 (meta) withdraws electrons, stabilises the phenoxide ion, hence acidity is increased.

Lower acidity in case of meta-isomer as compared to para isomer is explained by the fact m-nitrophenoxide is stabilised by inductive effect only and no resonance effect operates with -NO2 group.

Stability order: F>D>E>B>C>A.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Find the correct order of electron gain enthalpy \( \left(\Delta_{\mathrm{eg}} \mathrm{H}\right) \) of the given elements.

Options:
A \( \mathrm{O}>\mathrm{B}>\mathrm{C}>\mathrm{N} \)
B \( \mathrm{O}>\mathrm{C}>\mathrm{N}>\mathrm{B} \)
C \( \mathrm{O}>\mathrm{C}>\mathrm{B}>\mathrm{N} \)
D \( \mathrm{O}>\mathrm{N}>\mathrm{C}>\mathrm{B} \)

The correct answer is:
\( \mathrm{O}>\mathrm{C}>\mathrm{B}>\mathrm{N} \)

Solution:
Electron gain enthalpy down the group decreases and across the period it increases. Nitrogen has the positive value of electron gain enthalpy.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Find the current I in the given circuit

Options:
A 2 A
B 4.5 A
C 6 A
D 1.5 A

The correct answer is:
4.5 A

Solution:
For the batteries connected in parallel
Îµeq=122+6212+12=9 V,req=12+12-1=1 Î©
I=Îµeqreq+1=92=4.5 A

Topic: Physics, Current Electricity, JEE Main
Find the decay constant (in s −1 ) of Co55 radionuclide if its activity is known to decrease 4% per hour. The decay product is non-radioactive.

Options:
A 1.1 × 10-5 s-1
B 1.10 × 10-2 s-1

C 1.1 × 102 s-1

D 1.09 × 10-5 s-1

The correct answer is:
1.1 × 10-5 s-1

Solution:
The initial activity of nuclei A0 = λN0
Activity of nuclei at time t, A = λN
= λN0e–λt
A = A0e–λt ...(1)
Since activity decreases at 4.0% per hour, so the activity of Co55 radionuclide at t = 1 hr,
A = A0 – ηA0 = A0(1 – η), where η = 0.04 ...(2)
Taking log of Eq. (1), we get
log A⁡ A⁡ 0 = - λ t … 3
or log A⁡ 0 1 - η A⁡ 0 = - λ t
or λ = - 1 t ln 1 - η
On substituting values, we get
λ≃1.1×10-5 s-1

Topic: Physics, Nuclear Physics, JEE Main
Find the dimensions of electric permittivity

Options:
A A2M-1L-3T4
B A2M-1L-3T0
C AM-1L-3T4
D A2M0L-3T4

The correct answer is:
A2M-1L-3T4

Solution:
From Coulomb's law, the force of attraction/repulsion between two point charges q and q separated by distance r is
F=14πε0q2r2
⇒ ε0=14π .q2Fr2
Where ε0 is electric permittivity.
Dimensions of ε0=AT2MLT-2L2
ε0=A2M-1L-3T4

Topic: Physics, Units and Dimensions, JEE Main
Find the domain and range of the following functions:
(i) $f(x)=-|x|$
(ii) $f(x)=\sqrt{9-x^2}$

Options:

None

Solution:
(i) We have $f(x)=-|x|$
Clearly, $f=-|x|$ is $\leq 0$, for all $x \in R$
$\therefore \quad$ Domain $=\mathrm{R}$ and Range
$=\{x \in R: x \leq 0\}=R^{-}$
(ii) We have $f(x)=\sqrt{9-x^2}$
Clearly, $f$ is not defined when $\left(9-x^2\right) < 0$ i.e. $x^2>9$ $\Rightarrow \quad$ when $x>3$ or $x < -3$
Also, for each real number $x$ lying between $-3$ and 3 . $f(x)=\sqrt{\left(9-x^2\right)}$ is unique.
So, Domain $=\{x \in R:-3 \leq x \leq 3\}$
Further, $y=\sqrt{\left(9-x^2\right)} \Rightarrow y^2=9-x^2$
$\Rightarrow \quad x=\sqrt{\left(9-y^2\right)}$
Clearly, $x$ is not defined when $\left(9-y^2\right) < 0$
But $\left(9-y^2\right) < 0$
$\begin{aligned}
&\Rightarrow \quad y^2>9 \Rightarrow y>3 \text { or } y < -3 \\
&\therefore \quad \text { Range }=\{y \in R:-3 \leq y \leq 3\}
\end{aligned}$

Topic: Mathematics, Relations and Functions (Class 11)
Find the domain and range of the function $f(x)=\frac{1}{\sqrt{x-5}}$.

Options:

None

Solution:
We have $f(x)=\frac{1}{\sqrt{x-5}}$
$f(x)$ is defined, if $x-5>0 \Rightarrow x>5$
So Domain of $f=(5, \infty)$
$\operatorname{Let} f(x)=y$
$\Rightarrow y=\frac{1}{\sqrt{x-5}} \Rightarrow \sqrt{x-5}=\frac{1}{y}$ $\Rightarrow \quad x-5=\frac{1}{y^2}$ $\therefore \quad x=\frac{1}{y^2}+5$ As $x \in(5, \infty) \Rightarrow y \in R^{+}$ Hence, range of $f=R^{+}$

Topic: Mathematics, Relations and Functions (Class 11)
Find the domain and the range of the real function $f$ defind by $f(x)=|x-1|$

Options:

None

Solution:
(i) $f(x)=|x-1|$ is defined for all real values of $x$;
$\therefore$ Domain of $f=\{x: x \in R\}=R$.
(ii) $f(x)=|x-1|$ can acquire only non-negative values.;
$\therefore \text { Range }=\{y: y \in R, y \geq 0\}$

Topic: Mathematics, Relations and Functions (Class 11)
Find the domain and the range of the real function $f$ defined by $f(x)=\sqrt{(x-1)}$

Options:

None

Solution:
(i) $f(x)=\sqrt{x-1}, f$ is not defined for $x-1 < 0$ or $x < 1$;
Domain of $f(x)=\{x: x \in R, x \geq 1\}$
(ii) Let $f(x)=y=\sqrt{x-1}$
$\therefore y$ is well defined for all values of $x \geq 1$.
Range $=[0, \infty)$

Topic: Mathematics, Relations and Functions (Class 11)
Find the domain of each of the following functions given by
(i) $f(x)=\frac{1}{\sqrt{1-\cos x}}$
(ii) $f(x)=\frac{1}{\sqrt{x+|x|}}$
(iii) $f(x)=x|x|$
(iv) $f(x)=\frac{x^3-x+3}{x^2-1}$
(v) $f(x)=\frac{3 x}{28-x}$

Options:

None

Solution:
(i) We have, $f(x)=\frac{1}{\sqrt{1-\cos x}}$ Here, $f(x)$ is not defined, if $1-\cos x=0$ $\Rightarrow \cos x=1 \Rightarrow x=2 n \pi \forall n \in Z$
$\therefore$ Domain of $f=R-\{2 n \pi: n \in Z\}$
(ii) Given $f(x)=\frac{1}{\sqrt{x+|x|}}$
$$
\because x+|x|=x-x=0, x < 0
$$
$$
=0+0=0, x=0=x+x=2 x, x>0
$$
Hence, $f(x)$ is defined, if $x>0$.
$\therefore \quad$ Domain of $f=\mathrm{R}^{+}$
(iii) Given $f(x)=x|x|$
Clearly, $f(x)$ is defined for any $x \in \mathrm{R}$.
$\therefore$ Domain of $f=R$
(iv) Given $f(x)=\frac{x^3-x+3}{x^2-1}$
$f(x)$ is not defined, if $x^2-1=0$
$\Rightarrow(x-1)(x+1)=0 \Rightarrow x=-1,1$
$\therefore$ Domain of $f=R-\{-1,1\}$
(v) We have, $f(x)=\frac{3 x}{28-x}$
$f(x)$ is not defined, if $28-x=0 \Rightarrow x=28$
$\therefore$ Domain of $f=R-\{28\}$

Topic: Mathematics, Relations and Functions (Class 11)
Find the domain of the function $f(x)=\frac{x^2+3 x+5}{x^2-5 x+4}$

Options:

None

Solution:
Since $x^2-5 x+4=(x-4)(x-1)$, the function $f$ is defined for all real numbers except $x=4$ and $x=1$. Hence, the domain of $f$ is $R-\{1,4\}$.

Topic: Mathematics, Relations and Functions (Class 11)
Find the domain of the function
$f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}$

Options:

None

Solution:
$f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}=\frac{(x+1)^2}{(x-2)(x-6)}$
The function is not defined at $x=2,6$
Domain of $f=\{x: x \in R$ and $x \neq 2, x \neq 6\}$
$=R-\{2,6\}$

Topic: Mathematics, Relations and Functions (Class 11)
Find the domain of the real valued function \(f(x)=\left([x]^2-[x]-2\right)^{-1 / 2}\), where \([\cdot]\) is the greatest integer function.

Options:
A \(R-(-1,3]\)
B \(R-[-1,3)\)
C \(R-(-1,3)\)
D \(R-[-1,3]\)

The correct answer is:
\(R-(-1,3]\)

Solution:
Let \(y=f(x)=\left([x]^2-[x]-2\right)^{-1 / 2}\)
\(\Rightarrow \quad y^2=\frac{1}{\sqrt{[x]^2-[x]-2}}\)
For real valued
\(\begin{aligned}
& {[x]^2-[x]-2 > 0} \\
& \Rightarrow \quad\{[x]-2\}\{[x]+1\} > 0 \\
& {[x] \in \mathbf{R}-(-1,2)} \\
\end{aligned}\)
So, \(x \in \mathbf{R}-(-1,3]\)

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2020 (17 Sep Shift 1)
Find the electric field at point P (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length L carrying a charge Q. The distance of the point P from the centre of the rod is a=32L

Options:
A Q3πε0L2
B Q23πε0L2
C 3Q4πε0L2
D Q4πε0L2

The correct answer is:
Q23πε0L2

Solution:
E=kλasinθ1+sinθ2

E=14πε0×QL×13 L2×2sinθ

tanθ=L/23 L2=13

sinθ=12

E=14πε0×2Q3 L2×2×12

E=Q23πε0 L2

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (26 Feb Shift 1)
Find the electric flux (in S.I. unit) through the rectangular plate abcd of length l=2 m, width L and whose centre is at a distance OP=x0=L2 from an infinite line of charge with linear charge density Î»=136ÏÃ10-9 C m-1. Consider that the plane of the frame is perpendicular to line OP.

Options:

The correct answer is:
0.5

Solution:
If we arrange three more rectangular plates (similar to the one which is given) around the wire, then we get a rectangular box open at both ends. We bring two more plates and use them to close this rectangular box. We can see from the figure that we have a symmetric system for which the total flux is

ϕ=λlε0

Hence the flux through one plate is

ϕ=λl4ε0=0.5

Topic: Physics, Electrostatics, JEE Main
Find the equation of chord of contact to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ at the point $(1,3)$.

Options:
A $\frac{y}{16}+\frac{x}{3}=1$
B $\frac{y}{16}-\frac{x}{3}=1$
C $\frac{x}{16}+\frac{y}{3}=1$
D $\frac{x}{16}-\frac{y}{3}=1$

The correct answer is:
$\frac{x}{16}+\frac{y}{3}=1$

Solution:
The equation of chord of contact for the ellipse $\frac{x^2}{19}+\frac{y^2}{9}=1$ is given by $\frac{\mathrm{xx}_1}{16}+\frac{\mathrm{yy}_1}{9}=1$ where $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is the point of contact. Substituting $\left(x_1, y_1\right)=(1,3)$ gives us \(\frac{x .1}{16}+\frac{y .3}{9}=1\)

Topic: Mathematics, Ellipse, JEE Main
Find the equation of that diameter which bisects the chords $7 x+y-20=0$ of the hyperbola $\frac{x^2}{3}-\frac{y^2}{7}=1$.

Options:
A $y=\frac{-x}{3}$
B $x=\frac{-y}{3}$
C $x=\frac{16 y}{9}$
D $x=\frac{32 y}{9}$

The correct answer is:
$y=\frac{-x}{3}$

Solution:
Given Hyperbola: $\frac{x^2}{3}-\frac{y^2}{7}=1$
Given chord : $7 \mathrm{x}+\mathrm{y}=20$
Diameter - It is the locus of mid points of parallel chords and it passes through centre of hyperbola.
It's general equation is $y=\frac{+b^2 x}{a^2 m}$
Let the mid point of chord be $\left(\mathrm{x}_1, \mathrm{y}_1\right)$.
Equation of chord in mid point form is $S_1=T \Rightarrow \frac{\mathrm{xx}_1}{\mathrm{a}^2}-\frac{\mathrm{yy}_1}{\mathrm{~b}^2}-1=\frac{\mathrm{x}_1^2}{\mathrm{a}^2}-\frac{\mathrm{y}_1^2}{\mathrm{~b}^2}-1$ $\Rightarrow \frac{\mathrm{xx}_1}{3}-\frac{\mathrm{yy}_1}{7}=\frac{\mathrm{x}_1^2}{3}-\frac{\mathrm{y}_1^2}{7}$ This equation is identical to the equation of chord given. So, getting values of $\mathrm{x}_1, \mathrm{y}_1$ by comparing co-efficients.
$\frac{x_1}{3}=1 \& \frac{y_1}{7}=\frac{-1}{7} \Rightarrow x_1=3, y=-1$
This mid point lies on diameter $\mathrm{y}=\frac{+\mathrm{b}^2 \mathrm{x}}{\mathrm{a}^2 \mathrm{~m}}=\frac{7 \mathrm{x}}{3 \mathrm{~m}}$
So, by putting the values of $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ and getting $\mathrm{m} \Rightarrow-1=\frac{7(3)}{3(\mathrm{~m})} \Rightarrow \mathrm{m}=-7$
Putting $m=-7$ in diameter's equation,
$\begin{aligned}
& y=\frac{7 x}{3(-7)} \\
& \Rightarrow y=\frac{-x}{3}
\end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main
Find the equation of the plane passing through \( A(-1,2,3) \) and \( B(3,-5,6) \) and parallel to the line \( \frac{x-4}{2}=\frac{3-y}{-4}=\frac{z-2}{5} \) from the following.

Options:
A \( 47 x+14 y-30 z+109=0 \)
B \( 47 x+14 y-30 z=109 \)
C \( 47 x+14 y+30 z-109=0 \)
D None of these

The correct answer is:
\( 47 x+14 y-30 z+109=0 \)

Solution:
D.R. of AB = 4, -7, 3
D.R. of line 2 + 45
∴ D.R. Normal to the line would be
=ijk4-732+45= -47i-14j+30k
∴ Equation of the plane would =-47x-14y+30z=1
Put (-1, 2, 3) in the above equation we will get
= +109
∴ equation is 47x+14y-30z+109=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Find the equivalent capacitance between \( \mathrm{A} \) \& \( \mathrm{B} \) if each capacitor has capacitance \( \mathrm{C} \) ?

Options:
A \( \frac{4 \mathrm{C}}{3} \)
B \( \frac{8 \mathrm{C}}{3} \)
C \( 12 \mathrm{C} \)
D \( \frac{5 \mathrm{C}}{12} \)

The correct answer is:
\( \frac{8 \mathrm{C}}{3} \)

Solution:
Take equipotential points and redraw circuit and identify series and parallel
Ceq.=4 × 2 C× C2 C+C=4× 2 C3=8 C3

Topic: Physics, Capacitance, JEE Main
Find the equivalent resistance about any branch of the base of the square pyramid shown. Assume resistance of each branch is R.

Options:
A 7R15
B 8R15
C R2
D none of these

The correct answer is:
8R15

Solution:
Looking at the circuit from above, we can see that between any two adjacent points (let say A and B), the circuit is symmetric so the junction P can be disconnected.

Topic: Physics, Current Electricity, JEE Main
Find the equivalent resistance between point A and B

Options:
A 65 Ω
B 20 Ω
C 5 Ω
D 2 Ω

The correct answer is:
5 Ω

Solution:
Therefore, RAB=RAS+5×10RAS+5+10=5+5×105+5+10=5 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (26 Jun Shift 2)
Find the final product 'E'

CH 2 OH−CH =  CH 2 → (ii)  NaSO 3 H (i)  OsO 4 (A) → KHSO 4 (B) → Zn-Hg / conc. HCl (C) → CCl 4 NBS/hv (D) → (CH 3 ) 2 CuLi (E)

Options:
A But-2-ene
B But-1-ene
C Isopentene
D Pen-2-ene

The correct answer is:
But-1-ene

Solution:
Here the final product E is Butene and the reaction sequence is as follows
CH 2 OH−CH =  CH 2 → (ii)  NaSO 3 H (i)  OsO 4 CH 2 OH−CHOH− CH 2 OH (A) → KHSO 4 CH 2  = CH−CHO (B) → Zn-Hg / conc. HCl
CH 2  = CH− CH 3 (C) → CCl 4 NBS/hv CH 2  = CH− CH 2 Br (D) → (CH 3 ) 2 CuLi CH 2  = CH− CH 2 − CH 3 (E)

Topic: Chemistry, Hydrocarbons, JEE Main
Find the final product for the reaction

Options:
A

B

C

D

The correct answer is:

Solution:
The final product from this reaction is
So the correct option is (a).

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Find the gravitational force between two stones, each of mass $2 \mathrm{~kg}$ and separated by a distance $1 \mathrm{~m}$ in vacuum.

Options:
A 0
B $6.675 \times 10^{-5} \mathrm{~N}$
C $6.675 \times 10^{-11} \mathrm{~N}$
D $2.67 \times 10^{-10} \mathrm{~N}$

The correct answer is:
$2.67 \times 10^{-10} \mathrm{~N}$

Solution:
Given, masses of two stones, $m=2 \mathrm{~kg}$
Separation, $d=1 \mathrm{~m}$
Since, gravitational force, $F=\frac{G m^2}{d^2}$
where, $G=6.67 \times 10^{11} \mathrm{~N}-\mathrm{m}^2 \mathrm{~kg}^{-2}$
$\therefore \quad F=\frac{6.67 \times 10^{-11} \times 2^2}{1^2}=2.67 \times 10^{-10} \mathrm{~N}$

Topic: Physics, Gravitation, JEE Main
Find the gravitational force of attraction between the ring and sphere as shown in the diagram, where the plane of the ring is perpendicular to the line joining the centres. If 8R is the distance between the centres of a ring (of mass m) and a sphere (mass M) where both have equal radius R

Options:
A 223GMmR2
B 89GmMR
C 138GMmR2
D 827GmMR2

The correct answer is:
827GmMR2

Solution:
Gravitational field of ring

=-GmxR2+x23/2

Force between sphere & ring

=GmM8RR2+8R23/2

=GmMR2×827

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (26 Feb Shift 1)
Find the greatest distance of the point \( P(10,7) \) from the circle \( x^{2}+y^{2}-4 x-2 y-20=0 \)

Options:

The correct answer is:
15.00

Solution:
Given circle x2+y2-4x-2y-20=0 and point P10, 7

S11= 100+49-40-14-20>0

Point is outside the circle.

C2,1; r=4+1+20=5

Greatest distance =QP=PC+r

=10-22+7-12+5

=10+5

=15

Topic: Mathematics, Circle, JEE Main
Find the incorrect pair.

Options:
A Humans - Ureotelic
B Birds - Uricotelic
C Lizards - Uricotelic
D Whale - Ammonotelic

The correct answer is:
Whale - Ammonotelic

Solution:
Animals that excrete their nitrogenous waste mainly in the form of urea are known as ureotelic and the phenomenon is known as ureotelism.
Urea can be stored in the body for considerable periods of time and is least toxic. It is eliminated in the form of urine. Ureotelism is exhibited by semi-terrestrial animals, e.g. some earthworms, adult amphibians, elasmobranch (cartilaginous fishes), and mammals. The frog like other amphibians is ammonotelic in tadpole state and ureotelic in a mature state. The earthworm is similarly ammonotelic when sufficient water is available and ureotelic when water availability is reduced.

Topic: Biology, Excretory Products and their Elimination, NEET
Find the INCORRECT pairing.

Options:
A SAN - Natural pacemaker
B Heart rate - Pulse rate
C Cardiac output - Heart rate x Stroke volume
D Heart failure - Heart attack

The correct answer is:
Heart failure - Heart attack

Solution:
The state of the heart when it is not able to pump blood effectively to meet the requirements of the body is known as heart failure. Another term used to describe heart failure is congestive heart failure as congestion of the lungs is one of the main symptom of this disease. Heart failure is not the same as heart attack or cardiac arrest. Cardiac arrest is the term used when the heart stops beating where as heart attack denotes damage to heart muscles due to inadequate blood supply.

Topic: Biology, Body Fluids and Circulation, NEET
Find the inverse of the statement, "If \( \Delta A B C \) is equilateral, then it is isosceles'

Options:
A If \( \Delta A B C \) is isosceles, then it is equilateral
B If \( \Delta A B C \) is not equilateral, then it is isosceles
C If \( \Delta A B C \) is not equilateral, then it is not isosceles
D If \( \Delta A B C \) is not isosceles, then it is not equilateral

The correct answer is:
If \( \Delta A B C \) is not equilateral, then it is not isosceles

Solution:
Here the statements are p:△ABC is equilateral; q:△ABC is isosceles.

Inverse of p⇒q is ∼p⇒ ∼q

Inverse statement is: If △ABC is not equilateral then it is not isosceles.

Topic: Mathematics, Mathematical Reasoning, JEE Main
Find the last product [D] in reaction sequence

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main