datavorous/entrance-exam-dataset huggingface dataset:
A charged particle of mass $m_{1}$ and charge $q_{1}$ is revolving in a circle of radius $r .$ Another charged particle of charge $q_{2}$ and mass $m_{2}$ is situated at the centre of the circle. If the velocity and time period of the revolving particle be $v$ and $T$ respectively, then

Options:
A $v=\sqrt{\frac{q \cdot q_{2} r}{4 \pi \varepsilon_{0} m_{1}}}$
B $v=\frac{1}{m_{1}} \sqrt{\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r}}$
C $T=\sqrt{\frac{16 \pi^{3} \varepsilon_{0} m_{1}^{2} r^{3}}{q_{1} q_{2}}}$
D None of the above

The correct answers are:
None of the above

Solution:
The necessary centripetal force, to move in a circular path is provided by the Coulomb's force.
$$
\begin{array}{r}
\frac{m_{1} v^{2}}{\gamma}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}} \\
v=\sqrt{\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q_{1} q_{2}}{r m_{1}}}
\end{array}
$$
We know, $v=r\omega$
$$
\omega=\frac{2 \pi}{T}
$$
$$
\begin{array}{l}
T=\frac{2 \pi}{\omega} \\
=\frac{2 \pi \cdot r}{v}
\end{array}
$$
Substitution the value of $v$
$T=\frac{2 \pi r}{\sqrt{\frac{q_{1} \cdot q_{2}}{4 \pi \varepsilon_{0} R \cdot m_{1}}}}$
$T=\sqrt{\frac{4 \pi^{2} r^{2} \times 4 \pi \varepsilon_{0} \cdot r \cdot m_{1}}{q q_{2}}}$
$T=\sqrt{\frac{16 \pi^{3} \varepsilon_{0} r^{3} \cdot m_{1}}{q_{1} \cdot q_{2}}}$
None of the given options is correct.
Both the charges are positive or both negative. (Sino answers have the product $q A_{2}$ inside square loy Therefore, circular motion is not possible. The statement of question is wrong. Either $q_{1}$ or should be negative.

Topic: Physics, Magnetic Effects of Current, JEE Main
A charged particle of mass ‘ m ’ and charge ‘ q ’ moving under the influence of uniform electric field E →i^ and a uniform magnetic field B →k^ follows a trajectory from point P to Q as shown in figure. The velocities at P and Q are respectively, vi→ and -2vj→ . Then which of the following statements (A, B, C, D) are the correct? (Trajectory shown is schematic and not to scale)
A E=32mv2qa
B Rate of work done by the electric field at P is 32mv3a
C Rate of work done by both the fields at Q is zero
D The difference between the magnitude of angular momentum of the particle at P and Q is 2mav .

Options:
A A,C,D
B B,C,D
C A,B,C
D A,B,C,D

The correct answer is:
A,B,C

Solution:
(A) by work energy theorem

Wmag+Wele=12m(2v)2-12m(v)2

0+qE02a=32mv2
E0=34mv2qa

(B) Rate of work done at A= power of electric force

=qE0V
=34mv3a

(C) at Q,dwdt=0 for both forces

D ΔL→=(-m2 v 2ak^)-(- mvak )

|ΔL→|=3mva

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (09 Jan Shift 1)
A charged particle of mass $\mathrm{m}$ and charge $q$ travels on a circular path of radius $r$ that is perpendicular to a magnetic field $\mathrm{B}$. The time taken by the particle to complete one revolution is

Options:
A
$\frac{2 \pi \mathrm{mq}}{\mathrm{B}}$

B
$\frac{2 \pi q^2 B}{m}$

C
$\frac{2 \pi q B}{m}$

D
$\frac{2 \pi m}{q B}$

The correct answer is:
$\frac{2 \pi m}{q B}$

Solution:
$\mathrm{m} \omega^2 \mathrm{r}=\mathrm{Bq} \omega \mathrm{r}$
$\omega=\mathrm{Bq} / \mathrm{m}$
$\mathrm{T}=\frac{2 \pi \mathrm{m}}{\mathrm{qB}}$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2005
A charged particle q is placed at the centre O of a cube of length L ABCDEFGH. Another same charge q is placed at a distance L from O. If the total electric flux through the cube is 6Ï then the value of Ï is

Options:
A q4ε0
B q6ε0
C q2ε0
D q3ε0

The correct answer is:
q6ε0

Solution:
The net flux through the cube due to the external charge is zero and for the charge inside the cube, it is qÎµ0 through all the surfaces. For one surface, it is q6Îµ0.

Topic: Physics, Electrostatics, JEE Main
A charged particle $q$ is placed at the centre $O$ of cube of length $L$ (A B C D E F G H). Another same charge $q$ is placed at a distance $L$ from $O$. Then the electric flux through $A B C D$ is

Options:
A
$q / 4 \pi \epsilon_0 L$

B
zero

C
$\mathrm{q} / 2\pi \epsilon_0 L$

D
$q / 3 \pi \epsilon_0 L$

The correct answer is:
zero

Solution:
The flux for both the charges exactly cancels the effect of each other

Topic: Physics, Electrostatics, JEE Main
A charged particle $q$ is placed at the centre $O$ of cube of length $L$ (A B C D E F G H). Another same charge $q$ is placed at a distance $L$ from $O$. Then the electric flux through $A B C D$ is

Options:
A
$q / 4 \pi \epsilon_0 L$

B
zero

C
$\mathrm{q} / 2\pi \epsilon_0 L$

D
$q / 3 \pi \epsilon_0 L$

The correct answer is:
zero

Solution:
The flux for both the charges exactly cancels the effect of each other

Topic: Physics, Electrostatics, JEE Main, JEE Main 2002
A charged particle $q$ is shot towards another charged particle $Q$ which is fixed, with a speed $v$ it approaches $Q$ upto a closest distance $r$ and then returns. If $q$ were given a speed $2 v$, the closest distances of approach would be

Options:
A
$\mathrm{r}$

B
$2 r$

C
$\mathrm{r} / 2$

D
$\mathrm{r} / 4$

The correct answer is:
$\mathrm{r} / 4$

Solution:
By principle of conservation of energy
Finally, $\frac{1}{2} m(2 v)^2=\frac{K q Q}{r^2}$
Equation (i) $\div$ (ii),
$$
\begin{aligned}
& \frac{1}{4}=\frac{r^{\prime}}{r} \\
& \Rightarrow r^{\prime}=\frac{r}{4} .
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2004
A charged particle with charge $\mathrm{q}$ enters a region of constant, uniform and mutually orthogonal fields $\mathrm{\vec{E}}$ and $\mathrm{\vec{B}}$ with a velocity $\mathrm{\vec{v}}$ perpendicular to both $\mathrm{\vec{E}}$ and $\mathrm{\vec{B}}$, and comes out without any change in magnitude or direction of $\mathrm{\vec{v}}$. Then

Options:
A
$\mathrm{\vec{v}=\vec{E} \times \vec{B} / B^2}$

B
$\mathrm{\vec{v}=\vec{B} \times \vec{E} / B^2}$

C
$\mathrm{\vec{v}=\vec{E} \times \vec{B} / E^2}$

D
$\overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{E}} / \mathrm{E}^2$

The correct answer is:
$\mathrm{\vec{v}=\vec{E} \times \vec{B} / B^2}$

Solution:
$\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{E}}$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2007
A chemical signal that has both endocrine and neural roles is:

Options:
A Melatonin
B Calcitonin
C Epinephrine
D Cortisol

The correct answer is:
Epinephrine

Solution:
Epinephrine (also known as adrenaline) is a hormone as well as a neurotransmitter.

Melatonin is involved in blood pressure and autonomic cardiovascular regulation, immune system regulation, and also various physiological functions such as retinal functions, detoxification of free radicals, and antioxidant actions. It also helps in maintaining the balance of circadian rhythms.

Calcitonin (Thyrocalcitonin) acts antagonistically to PTH to reduce calcium levels in the blood.

Cortisol is the main glucocorticoid, helps in maintaining blood glucose level balance as well as maintaining metabolism.

Topic: Biology, Chemical Coordination and Integration, NEET, NEET 2015 (Phase 1)
A chemical which stimulates the secretion of pepsin is

Options:
A Histamine
B Cimetidine
C Zantac
D Anti histamine

The correct answer is:
Histamine

Solution:
Histamine stimulates the secretion of pepsin and hydrochloric acid and antihistamines are used to prevent the action of histamines.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2022 (26 Jun Shift 1)
A chess match between 2 players A and B is won by whoever first wins a total of 2 games. Probability of A winning, drawing and losing any particular game is 17,27 and 47 respectively (the games are independent). If the probability that A wins the match in the fourth game is p, then 343p is equal to

Options:
A 60
B 30
C 607
D 307

The correct answer is:
607

Solution:
A win match in 4th game ⇒ A win one game in first 3 games, DD,LD,DL are possibilities for remaining 2 matches.
Hence, the required probability =C311727×27+27×47+47×27×17
=6074=602401=p

⇒343p=607

Topic: Mathematics, Probability, JEE Main
A child of mass 5 kg is going round a merry-go-round that makes 1 rotation in 3.14 s. The radius of the merry-go-round is 2 m. The centrifugal force on the child will be

Options:
A 80 N
B 40 N
C 100 N
D 50 N

The correct answer is:
40 N

Solution:
The angular speed ω of the child can be calculated as follows:

ω=2π3.14=2 rad s-1

The formula to calculate the centrifugal force on the child is given by

Fr=mω2r ...(1)

Substitute the values of the known parameters into equation (1) to calculate the required centrifugal force.

Fr=5 kg×2 rad s-12×2 m=40 N

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (06 Apr Shift 2)
A child pulls a rope attached to a stone with a force of $60 \mathrm{~N}$. The rope makes an angle of 60 degrees to the ground. What is the effective force along the horizontal direction?

Options:
A $45.96 \mathrm{~N}$
B $40.3 \mathrm{~N}$
C $28 \mathrm{~N}$
D $30 \mathrm{~N}$

The correct answer is:
$30 \mathrm{~N}$

Solution:
CALCULATION:
The horizontal component is responsible to move the stone along the ground.
$\vec{F}=F_x=F \cos \theta$
$F_X=60 \cos 60^{\circ}=30 \mathrm{~N}$
Hence, the effective value of the pull tending to move the stone on the ground is $30 \mathrm{~N}$.
CONCEPT:
Resolution of vectors into components: We have a vector $(F)$ where the magnitude of the vector is $\mathrm{F}$ and the angle with horizontal is $\theta$.
The vector has two components: 1. Vertical component and 2. Horizontal component
Vertical component $\left(F_y\right)=F \sin \theta$
Horizontal component $\left(\mathrm{F}_{\mathrm{x}}\right)=\mathrm{F} \operatorname{Cos} \theta$

Topic: Physics, Mathematics in Physics, JEE Main
A child sits stationary at one end of a long trolley moving uniformly with a speed $v$ on a smooth horizontal floor.
If the child gets up and runs about on the trolley in any manner, what is the speed of the C.M. of the system?

Options:

None

Solution:
When the child gets up and runs on the trolley, the forces acting on the system (child + trolley) are internal. Since, no external force is acting on the system, the speed of the C.M. of the system remains unchanged.

Topic: Physics, System of Particles and Rotational Motion
A child stands on the edge of the cliff 10 m above the ground and throws a stone horizontally with an initial speed of 5 m s–1 . Neglecting the air resistance, the speed with which the stone hits the ground will be _____ m s–1 (given, g = 10 m s–2 ).

Options:
A 20
B 15
C 30
D 25

The correct answer is:
15

Solution:
Given here, u=5 m s-1, h=10 m

Using vy2=uy2+2gh in vertical direction, we have the velocity of stone in vertical direction as vy=2gh=200 m s-1 (since, uy=0).

The final velocity along the horizontal direction is the same as the initial horizontal velocity as there is no acceleration along the x-axis.

The speed with which the stone hits the ground is

vnet=vx2+vy2=25+200=15 m s-1

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (01 Feb Shift 1)
A child swinging on a swing in sitting position, stands up, then the time period of the swing will

Options:
A
increase

B
decrease

C
remains same

D
increases if the child is tall and decreases if the child is short

The correct answer is:
decrease

Solution:
$\mathrm{T}=2 \pi \sqrt{\mathrm{l}_{\mathrm{eff}} / 8} ;
\mathrm{l}_{\mathrm{eff}}$ decreases when the child stands up.

Topic: Physics, Oscillations, JEE Main, JEE Main 2002
A child swinging on a swing in sitting position stands up, then the time period of the swing will

Options:
A increase
B remain same
C decrease
D increase if the child is long and decrease if the child is short

The correct answer is:
decrease

Solution:
The time period will decrease. When the child stands up, the centre of gravity is shifted upwards and so the length of swing decreases, hence from the expression of the time period which is given by T=2Ïlg, the time period will decrease.

Topic: Physics, Oscillations, NEET
A child with mass $m$ is standing at the edge of a playground merry-go-round (A large uniform disc which rotates in horizontal plane about a fixed vertical axis in parks) with moment of inertia Vs I, radius R, and initial angular velocity w as shown in the figure. The child jumps off the edge of the merry-go-round with a velocity $\mathrm{v}$ with respect to the ground in direction tangent to periphery of the disc as shown. The new angular veloity of the merry-go-round is:

Options:
A \(\sqrt{\frac{I \omega^2-m v^2}{I}}\)
B \(\frac{\left(\mathrm{I}+\mathrm{mR}^2\right) \omega-\mathrm{mvR}}{\mathrm{I}}\)
C \(\sqrt{\frac{\left(I+m R^2\right) \omega^2-m v^2}{I}}\)
D \(\frac{\mathrm{I} \omega-\mathrm{mvR}}{\mathrm{I}}\)

The correct answer is:
\(\frac{\left(\mathrm{I}+\mathrm{mR}^2\right) \omega-\mathrm{mvR}}{\mathrm{I}}\)

Solution:
Let the angular velocity of the merry-go-round be w'.
As there is no external torque, thus \(\mathrm{L}\) is conserved.
Initial angular momentum, \(\mathrm{L}_{\mathrm{i}}=\mathrm{Iw}+\mathrm{mR}^2 \mathrm{w}\)
Final angular momentum, \(\mathrm{L}_{\mathrm{f}}=\mathrm{I} w^{\prime}+\mathrm{mvR}\)
\(\begin{aligned}
& \mathrm{L}_{\mathrm{i}}=\mathrm{L}_{\mathrm{f}} \Rightarrow \mathrm{Iw}+\mathrm{mwR}^2=\mathrm{Iw}^{\prime}+\mathrm{mvR} \\
& \Rightarrow \mathrm{w}^{\prime}=\frac{\left(\mathrm{I}+\mathrm{mR}^2\right) \mathrm{w}-\mathrm{mvR}}{\mathrm{I}}
\end{aligned}\)

Topic: Physics, Rotational Motion, JEE Main
A chimney of 20 m height standing on the top of a building subtends an angle whose tangent is 16 at a distance of 70 m from the foot of the building, then the height of building is

Options:
A 50 m
B 25 m
C 75 m
D 100 m

The correct answer is:
50 m

Solution:
Given, tanâ¡Î±=16
tanâ¡Î±+Î²=tanâ¡Î±+tanâ¡Î²1-tanâ¡Î±tanâ¡Î²
h+2070=16+h701-16âh70
h+2070=70+6h420-h
â h=50
Hence, 50 m is correct choice.

Topic: Mathematics, Heights and Distances, JEE Main
A chloride salt solution acidified with dil. HNO3 gives a curdy white precipitate, [A], on addition of AgNO3. [A] on treatment with NH4OH gives a clear solution, B.

Options:
A HAgCl3&AgNH32Cl
B HAgCl3 & NH4Ag(OH)2
C AgCl & AgNH32Cl
D AgCl & NH4Ag(OH)2

The correct answer is:
AgCl & AgNH32Cl

Solution:
Salt solution acidified with dilute HNO3 on addition of silver nitrate gives a curdy white precipitate soluble in ammonium hydroxide solution. This indicates the presence of Cl- ions in the salt.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2023 (25 Jan Shift 2)
A chord is drawn through the focus of the parabola y⁡ 2 = 6 x⁡ such that its distance from the vertex of this parabola is 5 2 , then its slope can be

Options:
A 5 2
B 2 3
C 3 2
D 2 5

The correct answer is:
5 2

Solution:
The focus and vertex of a parabola y2=4ax are respectively a, 0 and 0, 0.

The given parabola is y2=6x ⇒a=32

Thus, the focus and vertex are respectively S≡32, 0 and V≡0, 0.

The equation of a line passing through a point x1, y1 and having slope m is y-y1=mx-x1

If m is the slope of the chord through focus 32, 0, equation of the chord is y-0=mx-32

⇒2mx-2y-3m=0

We know that the length of perpendicular from origin to a line ax+by+c=0 is ca2+b2

Given, the distance of the focal chord 2mx-2y-3m=0 from the vertex 0, 0 is 52

⇒-3m4m2+4=52

⇒-3m2m2+1=52

⇒-3mm2+1=5

Squaring both sides, we get

⇒5m2+5=9m2

⇒4m2=5

⇒m=±52.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2014 (19 Apr Online)
A chord through the point $(1,-2)$ cuts the curve $3 x^2-y^2-2 x+4 y=0$ at $P$ and $Q$. If $P Q$ subtends an angle $\theta$ at the origin, then $\theta$ equals

Options:
A $60^{\circ}$
B $15^{\circ}$
C $75^{\circ}$
D $90^{\circ}$

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

Solution:
Given curve, $3 x^2-y^2-2 x+4 y=0$
Chord passes through $(1,-2)$ cut the given curve at $P$ and $Q$
Let slope of line $P Q$ is $m$.
Then, equation of line is
$$
\begin{aligned}
& y-y_1=m\left(x-x_1\right) \\
& \Rightarrow \quad y+2=m(x-1) \\
& y+2=m x-m \\
& \Rightarrow \quad \frac{m x-y}{2+m}=1 \\
&
\end{aligned}
$$
Use Eq. (ii) in Eq. (i) as follows,
$$
\begin{aligned}
& 3 x^2-y^2-2 x\left(\frac{m x-y}{2+m}\right)+4 y\left(\frac{m x-y}{2+m}\right)=0 \\
& \Rightarrow 3(2+m) x^2-(2+m) y^2-2 x(m x-y) \\
&+ 4 y(m x-y)=0 \\
& \Rightarrow x^2(6+3 m-2 m)-y^2(2+m+4)+x y \\
&(4 m+2)=0
\end{aligned}
$$
General equation of curve is
$$
a x^2+2 h x y+b y^2+2 g x+2 f y+c=0
$$
Compare Eqs. (iii) and (iv),
$$
a=6+m, b=-6-m
$$
This gives, $a+b=0$
$\Rightarrow$ Angle subtended by these two lines at origin is $90^{\circ}$.

Topic: Mathematics, Hyperbola, JEE Main
A chromatography column, packed with silica gel as stationary phase, was used to separate a mixture of compounds consisting of A benzanilide B aniline and C acetophenone. When the column is eluted with a mixture of solvents, hexane:ethyl acetate 20:80 , the sequence of obtained compounds is:

Options:
A B,C and A
B B,A and C
C C,A and B
D A,B and C

The correct answer is:
C,A and B

Solution:
A dipole moment arises in any system in which there is a separation of charge. They can, therefore, arise in ionic bonds as well as in covalent bonds. Dipole moments occur due to the difference in electronegativity between two chemically bonded atoms.

A bond dipole moment is a measure of the polarity of a chemical bond between two atoms in a molecule. It involves the concept of electric dipole moment, which is a measure of the separation of negative and positive charges in a system.

Hence the sequence of obtained compounds is (C), (A) and (B)

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (07 Jan Shift 2)
A chromosome with sub-terminal centromere is__________.

Options:
A acrocentric
B acentric
C metacentric
D telocentric

The correct answer is:
acrocentric

Solution:
A chromosome consists of two chromatids that remain attached at a point called the centromere. According to the position of the centromere, the chromosome is of four types. In the acrocentric chromosome the centromere is present towards the end of the chromosome or the centromere is sub-terminal in position.

Topic: Biology, Principles of Inheritance and Variation, NEET
A circle C1 has radius 2 units and a circle C2 has radius 3 units. The distance between the centres of C1 and C2 is 7 units. If two lines, one tangent to both circles and the other passing through the center of both circles, intersect at point P which lies between the centers of C1 and C2, then the distance between P and the centre of C2 is

Options:
A 94 units
B 73 units
C 215 units
D 145 units

The correct answer is:
215 units

Solution:
Since △O1PT1~△O2PT2,

∴23=x7-x
3x=14-2x
x=145

⇒PO2=7-145=215 units

Topic: Mathematics, Circle, JEE Main
A circle C touches the line x=2y at the point 2,1 and intersects the circle C1:x2+y2+2y-5=0 at two points P and Q such that PQ is a diameter of C1. Then the diameter of C is :

Options:
A 415
B 285
C 15
D 75

The correct answer is:
75

Solution:
Family of circle touching line 2y=x at point 2,1
(x-2)2+(y-1)2+λ(x-2y)=0 ...(1)
common chord PQ is
(x-2)2+(y-1)2+λ(x-2y)-x2-y2-2y+5=0
(diameter of C1 passes through (0,-1))
4+4+2λ-1+2+5=0

λ=-7 put in equation 1

(x-2)2+(y-1)2-7(x-2y)=0

x2+y2-11x+12y+5=0

r=1214+36-5=121+1244=2454=2452⇒ diameter =245=75

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (26 Aug Shift 2)
A circle cuts a chord of length 4 a on the $x$ -axis and passes through a point on the $y$ -axis, distant $2 \mathrm{~b}$ from the origin. Then the locus of the centre of this circle, is:

Options:
A a hyperbola
B an ellipse
C a straight line
D a parabola

The correct answer is:
a parabola

Solution:
Let centre be $C(h, k)$

$C Q=C P=r$

$\Rightarrow C Q^{2}=C P^{2}$

$(h-0)^{2}+(k \pm 0)^{2}=C M^{2}+M P^{2}$

$h^{2}+(k \pm 2 \mathrm{~b})^{2}=k^{2}+4 a^{2}$

$h^{2}+k^{2}+4 b^{2} \pm 4 b k=k^{2}+4 a^{2}$

Then, the locus of centre $C(h, k)$

$x^{2}+4 b^{2} \pm 4 b y=4 a^{2}$

Hence, the above locus of the centre of circle is a

parabola.

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (11 Jan Shift 2)
A circle is drawn whose centre is on the xâaxis and it touches the y-axis. If no part of the circle lies outside the parabola y2=8x, then the maximum possible radius of the circle is

Options:

The correct answer is:
4

Solution:
Let equation of circle is x-r2+y2=r2
Solving it with y2=8x, we get,
x2-2rx+r2+8x=r2âx=0 or x=2r-8
ârmaximum=4

Topic: Mathematics, Circle, JEE Main
A circle is drawn with the two foci of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
at the end of the diameter. What is the equation of the circle?

Options:
A $x^{2}+y^{2}=a^{2}+b^{2}$
B $x^{2}+y^{2}=a^{2}-b^{2}$ (20) 10-I]
C $x^{2}+y^{2}=2\left(a^{2}+b^{2}\right)$
D $x^{2}+y^{2}=2\left(a^{2}-b^{2}\right)$

The correct answer is:
$x^{2}+y^{2}=a^{2}-b^{2}$ (20) 10-I]

Solution:
Foci of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ are given as $(a e, 0)$ and $(-a, 0)$.
Since, two foci are at the end of the diameter $\therefore$ Equation of circle, is
$(x-a e)(x+a e)+(y-0)(y-0)=0$
$\Rightarrow x^{2}-a^{2} e^{2}+y^{2}=0$
$\Rightarrow x^{2}+y^{2}-a^{2}\left(1-\frac{b^{2}}{a^{2}}\right)=0\left(\because e=\sqrt{1-\frac{b^{2}}{a^{2}}}\right)$
$\Rightarrow x^{2}+y^{2}-a^{2}+b^{2}=0$
$\Rightarrow x^{2}+y^{2}=a^{2}-b^{2}$

Topic: Mathematics, Ellipse, JEE Main
A circle is given by $x^2+y^2-6 x+8 y-11=0$ and there are two points $(0,0)$ and $(1,8)$. These points lie

Options:
A Both inside the circle
B One outside and one inside the circle
C Both outside the circle
D One on and other inside the circle

The correct answer is:
One outside and one inside the circle

Solution:
From conditions that points lie inside, outside or on the circle, we see that one lies outside and one inside the circle.

Topic: Mathematics, Circle, JEE Main
A circle is inscribed in an equilateral triangle of side of length 12 . If the area and perimeter of any square inscribed in this circle are $m$ and $n$, respectively, then $m+n^2$ is equal to

Options:
A 408
B 414
C 396
D 312

The correct answer is:
408

Solution:
$\because \mathrm{r}=\frac{\Delta}{\mathrm{s}}=\frac{\sqrt{3} \mathrm{a}^2}{4 \cdot \frac{3 \mathrm{a}}{2}}=\frac{\mathrm{a}}{2 \sqrt{3}}=\frac{12}{2 \sqrt{3}}=2 \sqrt{3}$
$\begin{aligned} & \therefore A=\mathrm{r} \sqrt{2}=2 \sqrt{6} \\ & \text { Area }=\mathrm{m}=\mathrm{A}^2=24 \\ & \text { Perimeter }=\mathrm{n}=4 \mathrm{~A}=8 \sqrt{6} \\ & \therefore \mathrm{m}+\mathrm{n}^2=24+384 \\ & =408\end{aligned}$

Topic: Mathematics, Circle, JEE Main, JEE Main 2024 (06 Apr Shift 1)
A circle of radius 2 unit passes through the vertex and the focus of the parabola y2=2x and touches the parabola y=x-142+α, where α>0. Then 4α-82 is equal to ______.

Options:

The correct answer is:
63

Solution:
The vertex and focus of parabola y2=2x are V0,0 and S12,0 respectively

Let the equation of circle passing through vertex and focus of the parabola be x-h2+y-k2=4

∵ Circle passes through 0,0

⇒h2+k2=4 ⋯1

∵ Circle passes through 12,0

12-h2+k2=4

⇒h2+k2=h+154 ⋯2

On solving 1 and 2

4=h+154

⇒h=14

⇒k=634

here k=-634 is rejected as circle with centre 14,-634 can't touch the parabola y2=2x.

So the equation of circle will be x-142+k-6342=4

Since this circle touches the parabola y=x-142+α

i.e. α=2+634=8+634

4α-8=63

4α-82=63

Topic: Mathematics, Parabola, JEE Main, JEE Main 2022 (27 Jun Shift 1)
A circle of radius 2 units is touching both the axes and a circle with centre at (6, 5). The distance between their centers is

Options:
A 8 units
B 5 units
C 7 units
D None of these

The correct answer is:
5 units

Solution:
Radius of smaller circle is r=2 units (given)
Since, it is touching both the axes, so its centre is (2, 2)
The distance between their centers
=r+R= 5-22+6-22
= 32+42= 25=5 units

Topic: Mathematics, Circle, JEE Main
A circle of radius a has charge density given by Î»=Î»0cos2â¡Î¸ on its circumference, where Î»0 is a positive constant and Î¸ is the angular position of a point on the circle with respect to some reference line. The potential at the centre of the circle is

Options:
A Î»04Îµ0
B zero
C λ02ε0
D λ0ε0

The correct answer is:
λ04ε0

Solution:
dq=λadθ=λ0cos2⁡θa dθ
q= ∫θ=0θ=2πλ0cos2⁡θa dθ=λ0πa

V=q4πε0a=λ04ε0

Topic: Physics, Electrostatics, JEE Main