datavorous/entrance-exam-dataset huggingface dataset:
An element $X$ belongs to fourth period and fifteenth group of the periodic table. Which of the following statements is true?

Options:
A It has completely filled s-orbital and a partialy filled $d$ -orbital
B it has completely filled s-and $p$ -orbitals and a partially filled $d$ -orbital
C It has completely filled $s$ - and $p$ -orbitals and a hat filled $d$ -orbital
D It has a half filled $p$ -orbital and completely fled s-and $d$ -orbitals

The correct answer is:
It has a half filled $p$ -orbital and completely fled s-and $d$ -orbitals

Solution:
The element belonging to 4 th period and 15 th group is arsenic (As). The outer electronic configuration of 15 th group element is $n s^{2} n \rho^{3}$. Since arsenic belongs to 4 th period, therefore it has filled 'd' orbital too. Thus, the electronic configuration of As is $[A r] 4 s^{2} 3 d^{10} 4 \rho^{3} .$
So, it has filled's' and'd orbital and half-fliled porbital.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
An element X shows +3, oxidation state in its compounds. Out of the four compounds given below, choose the incorrect formula for the element X.

Options:
A X2O3
B X2SO43
C XPO4
D X2Cl3

The correct answer is:
X2Cl3

Solution:
Oxidation state of X is +3. The formula should be XCl3 and not X2Cl3.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2015 (11 Apr Online)
An elemental crystal has a density of 8570 kg/m3. The packing efficiency is 0.68. The closest distance of approach between neighbouring atom is 2.86 Ã. What is the mass in amu of one atom approximately?
[Report your answer by rounding it upto nearset whole number]

Options:

The correct answer is:
93

Solution:
The packing efficiency=0.68, means the given lattice is BCC.
The closest distance of approach=2r
2r=2.86 A o = 3 a 2 â 2Ã2.86 3 = 3.30 A o .
Let, at. wt. of the element= w
â´ 2Ãw6Ã1023Ã3.33Ã10â30= 8570
w=92.39 ≃ 93 amu

Topic: Chemistry, Solid State, JEE Main
An elevator in a building can carry a maximum of 10 persons, with the average mass of each person being 68 kg . The mass of the elevator itself is 920 kg and it moves with a constant speed of 3 m/s . The frictional force opposing the motion is 6000 N . If the elevator is moving up with its full capacity, the power delivered by the motor to the elevator g=10m/s2 must be at least:

Options:
A 56300W
B 62360W
C 48000W
D 66000W

The correct answer is:
66000W

Solution:
Net force on motor will be
Fm=920+6810g+6000
=22000 N
So, required power for motor
Pm=F→m.v→
=22000×3
=66000 watt

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2020 (07 Jan Shift 2)
An elevator is going upward and the variation in the velocity of the elevator is as given in the graph. What is the height to
which the elevator takes the passengers?

Options:
A \( 3.6 \mathrm{~m} \)
B \( 28.8 \mathrm{~m} \)
C \( 36.0 \mathrm{~m} \)
D \( 32.4 \mathrm{~m} \)

The correct answer is:
\( 36.0 \mathrm{~m} \)

Solution:
Area under graph will give displacement
Height = 1 2 Ã4Ã3.6+3.6Ã8=36âm

Topic: Physics, Motion In One Dimension, JEE Main
An elevator of mass m is connected to a rope which pulls the elevator up with a maximum acceleration equal to a . If the maximum bearable tension in the rope is T , then for a safe journey the minimum diameter of the rope is [ g = acceleration due to gravity]

Options:
A 2mg+aπT12
B 4mg+aπT12
C mg+aπT12
D mg+a2πT12

The correct answer is:
4mg+aπT12

Solution:
The maximum tension in the rope =m(g+a)
Stress in the rope =mg+aπr2
∴ T=mg+aπr2=mg+aπd22
∴ T=4mg+aπd2
∴ d2= 4mg+aπT
∴ d=4mg+aπT12

Topic: Physics, Laws of Motion, JEE Main
An elevator weighing $6000 \mathrm{~kg}$ is pulled upward by a cable with an acceleration of $5 \mathrm{~ms}^{-2}$. Taking $g$ to be $10 \mathrm{~ms}^{-2}$, then the tension in the cable is

Options:
A $6000 \mathrm{~N}$
B $9000 \mathrm{~N}$
C $60000 \mathrm{~N}$
D $90000 \mathrm{~N}$

The correct answer is:
$90000 \mathrm{~N}$

Solution:
$T=m(g+a)=6000(10+5)=90000 \mathrm{~N}$

Topic: Physics, Laws of Motion, JEE Main
An Ellingham diagram provides information about:

Options:
A the conditions of pH and potential under which a species is thermodynamically stable.
B the temperature dependence of the standard Gibbs energies of formation of some metal oxides.
C the pressure dependence of the standard electrode potentials of reduction reactions involved in the extraction of metals.
D the kinetics of the reduction process.

The correct answer is:
the temperature dependence of the standard Gibbs energies of formation of some metal oxides.

Solution:
Ellingham diagram is graph of ΔG° vs T of any metal/element oxide. since

ΔG°=ΔH°-TΔS°

for most metal oxide formation metal (s) + oxygen (g) →metal oxide (s)

ΔH°=-ve

ΔS°=-ve

so graph will be a straight line with - ve y-intracept & + ve slope.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2020 (05 Sep Shift 1)
An ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1, \mathrm{a}>\mathrm{b}$ and the parabola $\mathrm{x}^{2}=4(\mathrm{y}+\mathrm{b})$ are such that the two foci of the ellipse and the end points of the latus rectum of parabola are the vertices of a square. The eccentricity of the ellipse is

Options:
A $\frac{1}{\sqrt{13}}$
B $\frac{2}{\sqrt{13}}$
C $\frac{1}{\sqrt{11}}$
D $\frac{2}{\sqrt{11}}$

The correct answer is:
$\frac{2}{\sqrt{13}}$

Solution:
As it is square
$$
\begin{array}{l}
\mathrm{ae}=2 \\
4=\sqrt{(\mathrm{ae}-2)^{2}+(1-\mathrm{b})^{2}} \\
\mathrm{~b}=-3 \quad \quad \mathrm{~b}=5 \\
\mathrm{a}^{2}=13 \quad \mathrm{a}^{2}=29 \\
\quad\left(\text { from realation } \mathrm{b}^{2}=\mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right)\right. \\
\mathrm{ae}=2 \\
\mathrm{e}=\frac{2}{\sqrt{13}}
\end{array}
$$

Topic: Mathematics, Ellipse, JEE Main
An ellipse has eccentricity \( 1 / 2 \) and one focus at point \( \mathrm{P}(1 / 2,1) \). Its one directrix nearer to point \( \mathrm{P} \) is the common tangent, to the circle \( x^{2}+y^{2}=1 \) and the hyperbola \( x^{2}-y^{2}=1 \). The equation of the ellipse is

Options:
A \( 3 x^{2}+4 y^{2}-2 x-8 y+4=0 \)
B \( 3 x^{2}+4 y^{2}+2 x+8 y-4=0 \)
C \( 3 x^{2}+4 y^{2}-2 x+8 y-4=0 \)
D None of these

The correct answer is:
\( 3 x^{2}+4 y^{2}-2 x-8 y+4=0 \)

Solution:
Any point on the hyperbola x 2 - y 2 = 1 is (sec θ , tan θ ).
Then tangent at (sec θ , tan θ ) to x 2 - y 2 = 1 is
x sec θ - y tan θ = 1 .........(1)
this will also be a tangent to x 2 + y 2 = 1 if radius = length of perpendicular from (0,0) to the equation (1)
or 1 = 1 sec 2 θ + tan 2 θ
or sec 2 θ + tan 2 θ = 1
or 2 tan 2 θ = 0
or tan θ = 0 ∴ θ = 0 , π
∴ putting for θ in (1), the common tangents are ± x = 1 ,
x = 1 & x + 1 = 0
x = 1 is nearer to F 1 2 , 1 than x + 1 = 0
∴ the ellipse has the following focus = 1 2 , 1
corresponding directrix is x -1 = 0 and e = 1 2
∴ by focus - directirx property, the equation of the ellipse is
x - 1 2 2 + y - 1 2 = 1 2 x - 1 1 2 + 0 2
or 3x2 + 4y2 - 2x - 8y + 4 = 0

Topic: Mathematics, Hyperbola, JEE Main
An ellipse has foci 4,2, 2,2 and it passes through the point P2,4. The eccentricity of the ellipse is

Options:
A tanπ10
B tanπ12
C tanπ6
D tanπ8

The correct answer is:
tanπ8

Solution:
Let, 4,2=S1 and 2,2=S2 and eccentricity of ellipse is e
then S1S2=2ae and PS1+PS2=2a
(where 2a is length of major axis)
âe=S1S2PS1+PS2=222+2
âe=12+1=2-1=tanÏ8

Topic: Mathematics, Ellipse, JEE Main
An ellipse has $\mathrm{OB}$ as semi minor axis, $\mathrm{F}$ and $\mathrm{F}^{\prime}$ its focii and the angle FBF' is a right angle. Then the eccentricity of the ellipse is

Options:
A
$\frac{1}{\sqrt{2}}$

B
$\frac{1}{2}$

C
$\frac{1}{4}$

D
$\frac{1}{\sqrt{3}}$

The correct answer is:
$\frac{1}{\sqrt{2}}$

Solution:
$$
\begin{aligned}
& \because \angle \text { FBF }^{\prime}=90^{\circ} \\
& \therefore\left(\sqrt{a^2 e^2+b^2}\right)^2+\left(\sqrt{a^2 e^2+b^2}\right)^2=(2 a \mathrm{e})^2 \\
& \Rightarrow 2\left(a^2 e^2+b^2\right)=4 a^2 e^2 \\
& \Rightarrow \mathrm{e}^2=\mathrm{b}^2 / \mathrm{a}^2 \\
& \text { Also } \mathrm{e}^2=1-\mathrm{b}^2 / \mathrm{a}^2=1-\mathrm{e}^2 \\
& \Rightarrow 2 \mathrm{e}^2=1, \mathrm{e}=\frac{1}{\sqrt{2}} .
\end{aligned}
$$

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2005
An ellipse inscribed in a semi-circle touches the circular are at two distinct points and also touches the bounding diameter. its major axis is parallel to the bounding diameter. When the ellipse has the maximum possible area, its eccentricity is

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

The correct answer is:
$\sqrt{\frac{2}{3}}$

Solution:
Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
$$
x^{2}=a^{2}-\frac{a^{2} y^{2}}{b^{2}} \ldots . \cdots(i)
$$
The equation of circle will be $x^{2}+(y-b)^{2}=1$....... (ii)
Using (i) in (ii), we get
$$
\begin{array}{l}
a^{2}-\frac{a^{2} y^{2}}{b^{2}}+(y-b)^{2}=1 \\
\left(1-\frac{a^{2}}{b^{2}}\right) y^{2}+2 b y+a^{2}+b^{2}-r^{2}=0
\end{array}
$$
As you can see from the figure the ellipse touches the circle for same value of $y$ so the roots of the eqaution are equal $\therefore \mathrm{D}=0$ $4 \mathrm{~b}^{2}-4\left(1-\frac{\mathrm{a}^{2}}{\mathrm{~b}^{2}}\right)\left(\mathrm{a}^{2}+\mathrm{b}^{2}-\mathrm{r}^{2}\right)=0$ $\mathrm{r}^{2}=\frac{\mathrm{a}^{4}}{\mathrm{a}^{2}-\mathrm{b}^{2}}$ $\Rightarrow \mathrm{b}=\mathrm{a} \sqrt{1-\frac{\mathrm{a}^{2}}{\mathrm{r}^{2}}}$ Area of ellipse $=\pi \mathrm{ab}$ $\mathrm{A}=\pi \times \mathrm{a} \times \mathrm{a} \sqrt{1-\frac{\mathrm{a}^{2}}{\mathrm{r}^{2}}}$
As area of ellipse is maximun
$$
\begin{array}{l}
\therefore \frac{\mathrm{d} \mathrm{A}}{\mathrm{d} \mathrm{a}}=0 \\
\Rightarrow \mathrm{a}^{2}=\frac{2 \mathrm{r}^{2}}{3} \\
\mathrm{a}=\sqrt{\frac{2}{3}} \mathrm{r} \\
\therefore \mathrm{b}=\mathrm{a} \sqrt{1-\frac{2}{3}}=\frac{\mathrm{a}}{\sqrt{3}} \\
\mathrm{e}^{2}=1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}} \\
\Rightarrow \mathrm{e}=\sqrt{\frac{2}{3}}
\end{array}
$$

Topic: Mathematics, Ellipse, JEE Main
An ellipse is drawn by taking a diameter of the circle $(x-1)^2+y^2=1$ as its semiminor axis and a diameter of the circle $x^2+(y-2)^2=4$ as its semi-major axis. If the centre of the ellipse is the origin and its axes are the coordinate axes, then the equation of the ellipse is

Options:
A
$4 x^2+y^2=4$

B
$x^2+4 y^2=8$

C
$4 x^2+y^2=8$

D
$x^2+4 y^2=16$

The correct answer is:
$x^2+4 y^2=16$

Solution:
Semi minor axis $b=2$
Semi major axis $a=4$
Equation of ellipse $=\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad \Rightarrow \frac{x^2}{16}+\frac{y^2}{4}=1$
$\Rightarrow x^2+4 y^2=16$.

Topic: Mathematics, Ellipse, JEE Main
An ellipse is drawn by taking a diameter of the circle $(x-1)^2+y^2=1$ as its semiminor axis and a diameter of the circle $x^2+(y-2)^2=4$ as its semi-major axis. If the centre of the ellipse is the origin and its axes are the coordinate axes, then the equation of the ellipse is

Options:
A
$4 x^2+y^2=4$

B
$x^2+4 y^2=8$

C
$4 x^2+y^2=8$

D
$x^2+4 y^2=16$

The correct answer is:
$x^2+4 y^2=16$

Solution:
Semi minor axis $b=2$
Semi major axis $a=4$
Equation of ellipse $=\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad \Rightarrow \frac{x^2}{16}+\frac{y^2}{4}=1$
$\Rightarrow x^2+4 y^2=16$.

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2012 (Offline)
An ellipse is drawn by taking a diameter of the circle $(x-1)^2+y^2=1$ as its semiminor axis and a diameter of the circle $x^2+(y-2)^2=4$ is semi-major axis.If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is

Options:
A $4 x^2+y^2=8$
B $4 x^2+y^2=4$
C $x^2+4 y^2=8$
D $x^2+4 y^2=16$

The correct answer is:
$x^2+4 y^2=16$

Solution:
$(x-1)^2+y^2=1$
$\begin{aligned} & \text { Radius }=1 \\ & \text { Diameter }=2\end{aligned}$
$\begin{aligned} & b=2 \\ & x^2+(y-2)^2=4\end{aligned}$
Radius $=\sqrt{4}=2$
Diameter $=4$
$a=4$
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is the equation of ellipse
$\begin{aligned}
& \frac{x^2}{4^2}+\frac{y^2}{2^2}=1 \\
& \frac{x^2}{16}+\frac{y^2}{4}=1 \\
& x^2+4 y^2=16
\end{aligned}$

Topic: Mathematics, Ellipse, JEE Main
An ellipse is drawn by taking a diameter of the circle $(x-1)^2+y^2=4$ as its semi-minor axis and a diameter of the circle $x^2+(y-2)^2=16$ is a semi-major axis. If the center of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is

Options:
A $x^2+4 y^2=64$
B $4 x^2+y^2=16$
C $x^2+8 y^2=64$
D $8 x^2+y^2=64$

The correct answer is:
$x^2+4 y^2=64$

Solution:
Concept:
The equation of the circle with center at $(h, k)$ and radius $r$ is given by:
$(x-h)^2+(y-k)^2=r^2$
Calculation:
Equation of the circle is $(x-1)^2+y^2=4$, radius $=2$, diameter $=4$ Length of the semi-minor axis is 4
Equation of the circle is $x^2+(y-2)^2=16$, radius $=4$, diameter $=8$ Length of the semi-major axis is 8 , we know the equation of the ellipse
$\begin{aligned} & \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ & \frac{x^2}{(8)^2}+\frac{y^2}{(4)^2}=1 \\ & \frac{x^2}{64}+\frac{y^2}{16}=1 \\ & x^2+4 y^2=64\end{aligned}$
Hence, the equation of the ellipse is $x^2+4 y^2=64$.

Topic: Mathematics, Ellipse, JEE Main
An ellipse is drawn by taking the diameter of the circle $(x-1)^2+y^2=1$ as its semiminor axis and the diameter of the circle $x^2+(y-2)^2=4$ as the semi-major axis, if the center of the ellipse at the origin and its axis are the coordinates axes, then the equation of the ellipse is

Options:
A $4 x^2+y^2=4$
B $x^2+4 y^2=8$
C $4 x^2+y^2=8$
D $x^2+4 y^2=16$

The correct answer is:
$x^2+4 y^2=16$

Solution:
The correct option is 4
$x^2+4 y^2=16$
Explanation for the correct option:
Find the equation of the ellipse :
Given the circle $(x-1)^2+y^2=1$
Radius $=1$ and diameter $=2$
$\therefore$ Length of semi-minor axis $=$ diameter of the circle $(x-1)^2+y^2=1$ (given)
$\therefore b=2$
And Circle $x^2+(y-2)^2=4$
Radius $=2$ and diameter $=4$
$\therefore$ length of the semi-major axis $=$ diameter of the circle $x^2+(y-2)^2=4$
(given)
$\therefore a=4$
Therefore the equation of the ellipse
$\begin{aligned} & \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ \Rightarrow & \frac{x^2}{4^2}+\frac{y^2}{2^2}=1 \\ \Rightarrow & \frac{x^2}{16}+\frac{y^2}{4}=1 \\ \Rightarrow & x^2+4 y^2=16\end{aligned}$
Hence the correct option is 4.

Topic: Mathematics, Ellipse, JEE Main
An ellipse passes through the foci of the hyperbola, 9x2-4y2=36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively. If the product of eccentricities of the two conics is 12, then which of the following points does not lie on the ellipse?

Options:
A 392,Â 3
B 132, 32
C 132,Â 6
D 13,Â 0

The correct answer is:
132, 32

Solution:
Equation of the hyperbola
x24-y29=1
Focus of hyperbola a1e1, 0 and (-a1e1, 0)
a1=2, e1= 1+94=132
∴ Foci would be +13, 0 and -13, 0
Product of eccentricity would be
132 ⋅e2=12
∴ e2= 113 .
As the major & minor axis of the ellipse coincide with axis of hyperbola then the value of a2 for ellipse would be 13 ,
e2=1-b22a22
113=1-b2213
b22=12
∴ Equation of the ellipse would be
x213+y212=1.
Option (1) 394 ⋅(13)+312=1
Satisfies the equation, hence it lies on the ellipse.
Option (2) 134 (13)+34.12=1
does not lie on the ellipse.
Option (3) 132(13)+612=1 satisfy.
Option (4) 1313+0=1 satisfy.
So, option 132, 32 is the answer.

Topic: Mathematics, Ellipse, JEE Main
An ellipse passes through the foci of the hyperbola, 9x2-4y2=36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively. If the product of eccentricities of the two conics is 12, then which of the following points does not lie on the ellipse?

Options:
A 392,Â 3
B 132, 32
C 132,Â 6
D 13,Â 0

The correct answer is:
132, 32

Solution:
Equation of the hyperbola
x24-y29=1
Focus of hyperbola a1e1, 0 and (-a1e1, 0)
a1=2, e1= 1+94=132
∴ Foci would be +13, 0 and -13, 0
Product of eccentricity would be
132 ⋅e2=12
∴ e2= 113 .
As the major & minor axis of the ellipse coincide with axis of hyperbola then the value of a2 for ellipse would be 13 ,
e2=1-b22a22
113=1-b2213
b22=12
∴ Equation of the ellipse would be
x213+y212=1.
Option (1) 394 ⋅(13)+312=1
Satisfies the equation, hence it lies on the ellipse.
Option (2) 134 (13)+34.12=1
does not lie on the ellipse.
Option (3) 132(13)+612=1 satisfy.
Option (4) 1313+0=1 satisfy.
So, option 132, 32 is the answer.

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2015 (10 Apr Online)
An ellipse passes through the point $(-3,1)$ and its eccentricity is $\sqrt{\frac{2}{5}}$. The equation of the ellipse is

Options:
A $3 x^2+5 y^2=32$
B $3 x^2+5 y^2=25$
C $3 x^2+y^2=4$
D $3 x^2+y^2=9$

The correct answer is:
$3 x^2+5 y^2=32$

Solution:
Let the equation of ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\because$ It passes through $(-3,1)$
So, $\frac{9}{a^2}+\frac{1}{b^2}=1 \Rightarrow 9+\frac{a^2}{b^2}=a^2$ ...(i)
Given eccentricity is $\sqrt{2 / 5}$
So, $\frac{2}{5}=1-\frac{b^2}{a^2} \Rightarrow \frac{b^2}{a^2}=\frac{3}{5}$ ...(ii)
From equation (i) and (ii), $a^2=\frac{32}{3}, b^2=\frac{32}{5}$
Hence required equation of ellipse is $3 x^2+5 y^2=32$.

Topic: Mathematics, Ellipse, JEE Main
An ellipse, with foci at 0,2 and 0,-2 and minor axis of length 4 , passes through which of the following points?

Options:
A 1, 22
B 2, 2
C 2, 2
D 2, 22

The correct answer is:
2, 2

Solution:
Given that be =2 and a=2
(here a<b )
∵ a2=b21-e2
∴b2=8
∴ Equation of ellipse x24+y28=1
Clearly 2, 2 lies on the ellipse.

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2019 (12 Apr Shift 2)
An ellipse with its minor and major axis parallel to the coordinate axes passes through $(0,0),(1,0)$ and $(0,2)$. One of its foci lies on the y-axis. The eccentricity of the ellipse is

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

The correct answer is:
$\sqrt{2}-1$

Solution:
Centre is
$$
\frac{\left(x-\frac{\alpha}{2}\right)^{2}}{a^{2}}+\frac{(y-\beta)^{2}}{b^{2}}=1
$$
pass through $(0,0)(0,2) \&(1,0)$
Distance between $\begin{aligned} F_{1} \& F_{2} \Rightarrow \alpha &=2 \mathrm{ae} \\ \alpha^{2} &=4 \mathrm{a}^{2} \mathrm{e}^{2} \end{aligned}$
Pass through $(0,0)$
$$
\begin{array}{l}
\Rightarrow \frac{\alpha^{2}}{4 \mathrm{a}^{2}}+\frac{\beta^{2}}{\mathrm{~b}^{2}}=1 \\
\frac{\alpha^{2}}{4 \mathrm{a}^{2}}=\mathrm{e}^{2}
\end{array}
$$
Pass through $(0,2)$
$$
\frac{\alpha^{2}}{4 \mathrm{a}^{2}}+\frac{(2-\beta)^{2}}{\mathrm{~b}^{2}}=1
$$
from these two $\Rightarrow \beta=1$
Use properties $\mathrm{PF}_{1}+\mathrm{PF}_{2}=2 \mathrm{a}$
$$
\begin{array}{l}
\mathrm{F}_{1}(0,1) \quad \mathrm{F}_{2}(\alpha, 1) \quad \mathrm{P}(0,2) \\
1+\sqrt{\alpha^{2}+1}=2 \mathrm{a} \\
\mathrm{F}_{1}(0,1) \quad \mathrm{F}_{2}(\alpha, 1) \quad \mathrm{P}(1,0) \\
\sqrt{1+1}+\sqrt{(\alpha-1)^{2}+1}=2 \mathrm{a}
\end{array}
$$
From these two
$$
\alpha=-2 \alpha+2 \sqrt{2} \mathrm{a}
$$
put $\alpha$ in any of above two equations
$$
\begin{array}{l}
a=\frac{\sqrt{2}+1}{2} \Rightarrow 2 a=\sqrt{2}+1 \\
\alpha=1 \\
\alpha=2 \mathrm{ae} \\
\alpha=1,2 \mathrm{a}=\sqrt{2}+1 \\
\text { find } \mathrm{e}=\sqrt{2}
\end{array}
$$

Topic: Mathematics, Ellipse, JEE Main
An elliptical loop having resistance R, of semi major axis a , and semi minor axis b is placed in a magnetic field as shown in the figure. If the loop is rotated about the x-axis with angular frequency ω, the average power loss in the loop due to Joule heating is :

Options:
A π2a2b2B2ω22R
B zero
C πbBωR
D π2a2b2B2ω2R

The correct answer is:
π2a2b2B2ω22R

Solution:
The expression for induced EMF in the rotating loop in uniform magnetic field is given by

ε=NABω cosωt

The average power loss

P=<ε2R>=<A2B2ω2cos2ωtR>

<cos2θ>=12

<P>=A2B2ω2R12

⇒ <P>=π2a2b2B22Rω2

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (03 Sep Shift 1)
An EM wave from air enters a medium. The electric fields are E→1=E01x^cos⁡2πvzc-t in air and E→2=E02x^cos⁡k2z-ctin medium, where the wave number k and frequency v refer to their values in the air. The medium is non-magnetic. If ϵr1 and ϵr2 refer to relative permittivities of air and medium respectively, which of the following options, is correct?

Options:
A Ïµr1Ïµr2Â =12
B Ïµr1Ïµr2Â =4
C Ïµr1Ïµr2Â =2
D Ïµr1Ïµr2 =14

The correct answer is:
Ïµr1Ïµr2 =14

Solution:
∨air∨med=cc/2=2=μ0ϵ0μr2ϵr2μ0ϵ0μr1ϵr1 ϵr1ϵr2=14

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2018 (08 Apr)
An EM wave is propagating in a medium with a velocity V→=V i^. The instantaneous oscillating electric field of this em wave is along +y axis. Then the direction of oscillating magnetic field of the EM wave will be along

Options:
A -y direction
B +z direction
C -z direction
D -x direction

The correct answer is:
+z direction

Solution:
V^=E^×B^ , i^=j^×k^
∴ B^=k^ , +z direction

Topic: Physics, Electromagnetic Waves, JEE Main
An EM wave radiates outwards from a dipole antenna, with $\mathrm{E}_0$ as the amplitude of its electric field vector. The electric field $\mathrm{E}_0$ which transports significant energy from the source falls off as

Options:
A
$\frac{1}{\mathrm{r}^3}$

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

C
$\frac{1}{r}$

D
remains constant

The correct answer is:
$\frac{1}{r}$

Solution:
As we know that, the electric field is inversly proportional to $\mathrm{r}$, so $\left(\mathrm{E}_0 \propto \frac{1}{\mathrm{r}}\right)$
From a diode antenna, an electromagnetic waves are radiated outwards from dipole antenna with the amplitude of electric field vector $\left(\mathrm{E}_0\right)$ which transports significant energy from the source falls off intensity inversely as the distance (r) from the antenna, i.e.,
$$
\text { radiated energy }\left(E_0 \propto \frac{1}{r}\right)
$$

Topic: Physics, Electromagnetic Waves
An emf of 0.08 V is induced in a metal rod of length 10 cm held normal to a uniform magnetic field of 0.4 T, when move with a velocity of:

Options:
A 0.5 m s-1
B 20 m s-1
C 3.2 m s-1
D 2 m s-1

The correct answer is:
2 m s-1

Solution:
The formula to calculate the motional emf induced in a moving conductor in a magnetic field is given by

ε=Blv ...1

Substitute the values of the known parameters into equation (1) and solve to calculate the required velocity of the rod.

0.08 V=0.4 T×10 cm×1 m100 cm×v⇒v=0.08 V0.4 T×10 cm×1 m100 cm= 2 m s-1

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2023 (08 Apr Shift 2)
An e.m.f. of 5 volt is produced by a self inductance, when the current changes at a steady rate from $3 A$ to $2 A$ in 1 millisecond. The value of self inductance is

Options:
A Zero
B $5 \mathrm{H}$
C $5000 \mathrm{H}$
D $5 \mathrm{mH}$

The correct answer is:
$5 \mathrm{mH}$

Solution:
$L=\frac{e}{d i / d t}=$ $\frac{5}{3-2) / 10^{-3}}=\frac{5}{1} \times 10^{-3}=5$ milli henry

Topic: Physics, Electromagnetic Induction, JEE Main
An emf of 20 V is applied at time t=0 to a circuit containing in series 10 mH inductor and 5 Ω resistor. The ratio of the currents at time t=∞ and at t=40 s is close to: (Take e2=7.389 )

Options:
A 1.06
B 1.15
C 1.46
D 0.84

The correct answer is:
1.06

Solution:
i=i01-e-tL/R
=2051-e-t0.01/5
=41-e-500t

now at t=∞
i∞=4

at t=40 s
i40=41-e-500×40
=41-1e210000=41-17.2910000
i∞i40≈1 Slightly greater then one.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2020 (07 Jan Shift 2)
An empty LPG cylinder weighs 14.8 kg. When full, it weighs 29.0 kg and shows a pressure of 3.47 atm. In the course of use at ambient temperature, the mass of the cylinder is reduced to 23.0 kg. The final pressure inside the cylinder is _____ atm. (Nearest integer)

(Assume LPG to be an ideal gas)

Options:

The correct answer is:
2

Solution:
Initial mass of gas =29-14.8=14.2Kg

mass of gas used : =29-23=6Kg

gas left =14.2-6=8.2Kg

(1) 3.47×V=14.2×103M×R×T

(2) p×V=8.2×103M×R×T

Divide : (1)/(2)⇒ 3.47P=14.28.2

P=2.003

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2021 (01 Sep Shift 2)