datavorous/entrance-exam-dataset huggingface dataset:
A player caught a cricket ball of mass $150 \mathrm{~g}$ moving at a speed of $20 \mathrm{~m} / \mathrm{s}$. If the catching process is completed in $0.1 \mathrm{~s}$, the magnitude of force exerted by the ball on the hand of the player is:

Options:
A $3 \mathrm{~N}$
B $300 \mathrm{~N}$
C $150 \mathrm{~N}$
D $30 \mathrm{~N}$

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

Solution:
$\begin{aligned} & \mathrm{F}=\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}}=\frac{\mathrm{mv}-0}{0.1} \\ & =\frac{150 \times 10^{-3} \times 20}{0.1}=30 \mathrm{~N}\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2024 (08 Apr Shift 1)
A player kicks a football with an initial speed of 25 m s-1 at an angle of 45° from the ground. What are the maximum height and the time taken by the football to reach at the highest point during motion? (Take g=10 m s-2)

Options:
A hmax=15.625 m, T=1.77 s
B hmax=3.54 m, T=0.125 s
C hmax=10 m, T=2.5 s
D hmax=15.625 m, T=3.54 s

The correct answer is:
hmax=15.625 m, T=1.77 s

Solution:
θ=45°

H=u2sin2θ2g=(25)2×122×10=1258 m

and time t=T2

=usinθg=251210=25102=522s

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2021 (27 Aug Shift 2)
A player $X$ has a biased coin whose probability of showing heads is $p$ and a player $Y$ has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If $X$ starts the game, and the probability of winning the game by both the players is equal, then the value of ' $p$ ' is

Options:
A
$\frac{1}{3}$

B
$\frac{1}{5}$

C
$\frac{1}{4}$

D
$\frac{2}{5}$

The correct answer is:
$\frac{1}{3}$

Solution:
If the outcome is one of the following: $H, T T H, T T T T H, \ldots$, then $X$ wins As subsequent tosses are independent, so the probability that $X$ wins is
$$
p+\frac{p}{4}+\frac{p}{16}+\ldots=\frac{4 p}{3} .
$$
Similarly $Y$ wins if the outcome is one of the following: $T H, T T T H, T T T T T H, \ldots$
Therefore, the probability that $Y$ wins is
$$
\frac{1-p}{2}+\frac{1-p}{8}+\frac{1-p}{32}=\frac{2(1-p)}{3}
$$
Since, the probability of winning the game by both the players is equal then, we have
$$
\frac{4 p}{3}=\frac{2(1-p)}{3} \Rightarrow p=\frac{1}{3}
$$

Topic: Mathematics, Probability, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
A plumb bob is hanging from the ceiling of car. If the car move with acceleration $a$, the angle made by the string with the vertical is

Options:
A $\tan ^{-1}\left(\frac{a}{g}\right)$
B $\tan ^{-1}\left(\frac{g}{a}\right)$
C $\cos ^{-1}\left(\frac{g}{a}\right)$
D $\cos ^{-1}\left(\frac{a}{g}\right)$

The correct answer is:
$\tan ^{-1}\left(\frac{a}{g}\right)$

Solution:
When car moves towards right with acceleration $a$ then due to Pseudo force the bob will tilt in backward direction making an angle $\theta$ with vertical.
\
$$\theta=\tan ^{-1}\left(\frac{a}{g}\right)$$

Topic: Physics, Laws of Motion, JEE Main
A p-n junction (D) shown in the figure can act as a rectifier. An alternating current source $(V)$ is connected in the circuit.

Options:
A

B

C

D

The correct answer is:

Solution:
Given figure is half wave rectifier

Topic: Physics, Semiconductors, JEE Main, JEE Main 2009
A p-n photodiode is fabricated from a semiconductor with a band gap of $2.5 \mathrm{eV}$. It can detect a signal of wavelength :

Options:
A $496 Å$
B $6000 Å$
C $4000 \mathrm{~nm}$
D $6000 \mathrm{~nm}$

The correct answer is:
$496 Å$

Solution:
$$
\text { } \begin{aligned}
\lambda_{\max } & =\frac{\mathrm{hc}}{\mathrm{eV}} \\
& =\frac{1242 \mathrm{eV} Å}{2.5 \mathrm{eV}}=4960 Å
\end{aligned}
$$

Topic: Physics, Semiconductors, NEET, NEET 2009 (Mains)
A p-n photodiode is made of a material with a band gap of $2.0 \mathrm{eV}$. The minimum frequency of the radiation that can be absorbed by the material is nearly

Options:
A $10 \times 10^{14} \mathrm{~Hz}$
B $5 \times 10^{14} \mathrm{~Hz}$
C $1 \times 10^{14} \mathrm{~Hz}$
D $20 \times 10^{14} \mathrm{~Hz}$

The correct answer is:
$5 \times 10^{14} \mathrm{~Hz}$

Solution:
$p-n$ photodiode is a semiconductor diode that produces a significant current when illuminated. It is reversed biased but is operated below the breakdown voltage.
Energy of radiation $=$ band gap energy ie, $h v=2.0 \mathrm{eV}$
$$
\text { or } v=\frac{2.0 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}} \approx 5 \times 10^{14} \mathrm{~Hz}
$$

Topic: Physics, Semiconductors, JEE Main
A p-n photodiode is made of a material with a band gap of $2.0 \mathrm{eV}$. The minimum frequency of the radiation that can be absorbed by the material is nearly

Options:
A $20 \times 10^{14} \mathrm{~Hz}$
B $10 \times 10^{14} \mathrm{~Hz}$
C $5 \times 10^{14} \mathrm{~Hz}$
D $1 \times 10^{14} \mathrm{~Hz}$

The correct answer is:
$5 \times 10^{14} \mathrm{~Hz}$

Solution:
$$
v=\frac{E}{h}=5 \times 10^{14} \mathrm{~Hz}
$$

Topic: Physics, Semiconductors, NEET, NEET 2008 (Mains)
A point charge 2×10–2 C is moved from P to S in a uniform electric field of 30 N C–1 directed along positive x-axis. If coordinates of P and S are (1, 2, 0) m and (0, 0, 0) m respectively, the work done by electric field will be

Options:
A 1200 mJ
B 600 mJ
C -600 mJ
D -1200 mJ

The correct answer is:
-600 mJ

Solution:
As electric field is uniform, so force acting on the charged particle will be uniform.

Therefore,

WE=F→·s→=qE→·s→

=2×10-230i^·0i^+0j^+0k^-i^+2j^+0k^

=-60×10-2 J

=-600 mJ

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (29 Jan Shift 2)
A point charge causes an electric flux of \( -200 \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-1} \) to pass through spherical Gaussian surface of \( 10 \mathrm{~cm} \) radius
centered on the charge. If the radius of the Gaussian surface is doubled, the total electric flux passing through the surface is

Options:
A \( -200 \mathrm{Nm}^{2} \mathrm{C}^{-1} \)
B \( -100 \mathrm{Nm}^{-2} \mathrm{C}^{-1} \)
C \( +200 \mathrm{Nm}^{2} \mathrm{C}^{-1} \)
D \( -50 \mathrm{Nm}^{2} \mathrm{C}^{-1} \)

The correct answer is:
\( -200 \mathrm{Nm}^{2} \mathrm{C}^{-1} \)

Solution:
According to Gauss's theorem, electric flux through a closed surface is given by, ∅=qε0,

where, q is charge enclosed, ε0 is constant.

Here even if the radius is doubled, the charge enclosed is the same, so the flux through the surface is also the same.

Hence, flux is -200 Nm2 C-1.

Topic: Physics, Electrostatics, JEE Main
A point charge of 10 µC is placed at the origin. At what location on the X-axis should a point charge of 40 µC be placed so that the net electric field is zero at x=2 cm on the X-axis ?

Options:
A x=6 cm
B x=4 cm
C x=8 cm
D x=–4 cm

The correct answer is:
x=6 cm

Solution:
Let the charge 40 μC be placed at a distance X0 from the origin as shown in the figure.

For the electric field at x=2 to be 0,

EP=0⇒k×1022=k×40 X0-22

⇒12=2X0-2

⇒X0-2=4

Therefore, the charge is located at X0=4+2=6 cm

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (25 Jan Shift 2)
A point charge of +12 μC is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in figure. The magnitude of the electric flux through the square will be _______ ×103 N m2 C-1.

Options:

The correct answer is:
226

Solution:
From symmetry ϕ=16qε0

=12×10-66×8.85×10-12

=225.98×103 N m2 C-1≃226×103 N m2 C-1

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (24 Feb Shift 2)
A point charge of magnitude $+1 \mu \mathrm{C}$ is fixed at $(0$, $0,0)$. An isolated uncharged spherical conductor, is fixed with its center at $(4,0,0)$. The potential and the induced electric field at the centre of the sphere is :

Options:
A
$1.8 \times 10^5 \mathrm{~V}$ and $-5.625 \times 10^6 \mathrm{~V} / \mathrm{m}$

B
$0 \mathrm{~V}$ and $0 \mathrm{~V} / \mathrm{m}$

C
$2.25 \times 10^5 \mathrm{~V}$ and $-5.625 \times 10^6 \mathrm{~V} / \mathrm{m}$

D
$2.25 \times 10^5 \mathrm{~V}$ and $0 \mathrm{~V} / \mathrm{m}$

The correct answer is:
$2.25 \times 10^5 \mathrm{~V}$ and $-5.625 \times 10^6 \mathrm{~V} / \mathrm{m}$

Solution:
$\mathrm{q}=1 \mu \mathrm{C}=1 \times 10^{-6} \mathrm{C}$
$$
\mathrm{r}=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}
$$
Potential $\mathrm{V}=\frac{\mathrm{kq}}{\mathrm{r}}$
$$
\begin{aligned}
& =\frac{9 \times 10^9 \times 10^{-6}}{4 \times 10^{-2}} \\
& =2.25 \times 10^5 \mathrm{~V} .
\end{aligned}
$$
Induced electric field $\mathrm{E}=-\frac{\mathrm{kq}}{\mathrm{r}^2}$
$$
=\frac{9 \times 10^9 \times 1 \times 10^{-6}}{16 \times 10^{-4}}=-5.625 \times 10^6 \mathrm{~V} / \mathrm{m}
$$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2013 (22 Apr Online)
A point charge q1=4q0 is placed at origin. Another point charge q2=-q0 is placed at x=12 cm. Charge of proton is q0. The proton is placed on x-axis so that the electrostatic force on the proton in zero. In this situation, the position of the proton from the origin is ______ cm.

Options:

The correct answer is:
24

Solution:
Let at point P net electric field is zero.

As direction of electric field due to both charges at point P is opposite, equating magnitude we get

14πε04q0x2=14πε0q0x-122⇒2x-12=x⇒x=24 cm

Distance of proton from origin =24 cm.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (29 Jan Shift 1)
A point charge q is kept inside an uncharged conducting spherical shell. The radius of the shell is R and the charge is kept inside the shell at a distance x from the centre. Identify the correct statement.

Options:
A Net force acting on the shell is zero
B Net force on the shell is towards right side
C Net force on the shell is towards left side
D Nothing can be said about the net direction of the net force on the shell.

The correct answer is:
Net force on the shell is towards left side

Solution:
Charge induced near the surface closer to the charged particle will be more and hence sphere will be attracted to the charge q. Hence net force is towards left.

Topic: Physics, Electrostatics, JEE Main
A point charge q is placed at a distance of R from the centre of a conducting shell of inner radius 2R and outer radius 3R. The electric potential at the centre of the shell will be

Options:
A 14πε0q2R
B 14πε04q3R
C 14πε05q6R
D 14πε05q3R

The correct answer is:
14πε05q6R

Solution:
The induced charges will be -q and +q as shown in the figure.

The potential at the centre O will be

V0=14ÏÎµ0qR-q2R+q3R=14ÏÎµ05q6R

Topic: Physics, Electrostatics, JEE Main
A point charge q is placed at a distance of R from the centre of a conducting shell of inner radius 2R and outer radius 3R. The electric potential at the centre of the shell will be

Options:
A 14πε0q2R
B 14πε04q3R
C 14πε05q6R
D 14πε05q3R

The correct answer is:
14πε05q6R

Solution:
The induced charges will be -q and +q as shown in the figure.

The potential at the centre O will be

V0=14πε0qR-q2R+q3R=14πε05q6R

Topic: Physics, Electrostatics, JEE Main
A point charge q is placed at some distance d away from the centre of a grounded conducting sphere of radius r, as shown in the figure. The charge that flows from the earth to the sphere is

Options:
A -qrd-r
B -qdd-r
C -qrd
D -qdr

The correct answer is:
-qrd

Solution:
Let us assume that the charge flowing from earth to the sphere is q1. Since the sphere is grounded, its potential will be zero. Hence,

q14πε0r+q4πε0d=0

q1=-qrd

Topic: Physics, Electrostatics, JEE Main
A point charge $+ q$ is placed at the centre of a cube of side $L$. The electric flux emerging from the cube is

Options:
A $\frac{q}{\varepsilon_0}$
B Zero
C $\frac{6 q L^2}{\varepsilon_0}$
D $\frac{q}{6 L^2 \varepsilon_0}$

The correct answer is:
$\frac{q}{\varepsilon_0}$

Solution:
Electric flux coming out through a closed surface is $\frac{q}{\varepsilon_0}$

Topic: Physics, Electrostatics, JEE Main
A point charge q is situated at a distance d from one end of a thin non-conducting rod of length L having a charge Q (uniformly distributed along its length) as shown in fig. Then the magnitude of the electric force between them is:

Options:
A 14πε0qQ2d(d+L)
B 14πε02qQd(d+L)
C 14πε0qQ3d(d+L)
D 14ÏÎµ0qQd(d+L)

The correct answer is:
14πε0qQd(d+L)

Solution:
F=∫dd+Lkqdqx2=∫dd+LkqQLdxx2=kqQL∫dd+L1x2dx
F=kqQd(d+L)=14πε0qQd(d+L)
Or
The linear charge density of rod λ=QL;

dq=λdx

Force due to small element dx to charge q

dF=1(q)(λ)dx4πϵ0x2

Total force in-between point charge and rod
Fnet=qQL∫dL+dx-2dx4πϵ0=qQ4πϵ0L-1xdL+d
=qQ4πε0L1d-1L+d
=qQ4πϵ01(d)(L+d))

Topic: Physics, Electrostatics, JEE Main
A Point dipole \( \overrightarrow{\mathrm{p}}=-\mathrm{p}_{0} \widehat{x} \) is kept at the origin. The potential and electric field due to this dipole on the \( y- \) axis at a distance \( d \) are, respectively: (Taken \( V=0 \) at infinity)

Options:
A \( \frac{|\overrightarrow{\mathrm{p}}|}{4 \pi \varepsilon_{0} \mathrm{~d}^{2}}, \frac{-\overrightarrow{\mathrm{p}}}{4 \pi \varepsilon_{0} \mathrm{~d}^{3}} \)
B \( 0, \frac{|\overrightarrow{\mathrm{p}}|}{4 \pi \varepsilon_{0} \mathrm{~d}^{3}} \)
C \( \frac{|\overrightarrow{\mathrm{p}}|}{4 \pi \varepsilon_{0} \mathrm{~d}^{3}}, \frac{\overrightarrow{\mathrm{p}}}{4 \pi \varepsilon_{0} \mathrm{~d}^{3}} \)
D \( 0, \frac{-\overrightarrow{\mathrm{p}}}{4 \pi \varepsilon_{0} \mathrm{~d}^{3}} \)

The correct answer is:
\( 0, \frac{-\overrightarrow{\mathrm{p}}}{4 \pi \varepsilon_{0} \mathrm{~d}^{3}} \)

Solution:
E→=-p→4πε0r3=-p→4πε0d3
V=0

Topic: Physics, Electrostatics, JEE Main
A Point dipole p→=-p0x^ is kept at the origin. The potential and electric field due to this dipole on the y-axis at a distance d are, respectively: (Taken V=0 at infinity)

Options:
A |p→|4πε0d2,-p→4πε0d3
B 0,|p→|4πε0d3
C |p→|4πε0d3,p→4πε0d3
D 0,-p→4πε0d3

The correct answer is:
0,-p→4πε0d3

Solution:
E→=-p→4πε0r3=-p→4πε0d3
V=0

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (12 Apr Shift 1)
A point in the domain that cannot be filled in so that the resulting function is continuous is called

Options:
A Removable Discontinuity
B Non-Removable Discontinuity
C Impossible Discontinuity
D Irrelevant Discontinuity

The correct answer is:
Non-Removable Discontinuity

Solution:
We have,
A point in the domain that can't be filled in so that resulting function is discontinuous is called non-removable discontinuity.
Hence, option (b) is correct.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 2)
A point inside the circle $x^2+y^2+3 x-3 y+2=0$ is

Options:
A $(-1,3)$
B $(-2,1)$
C $(2,1)$
D $(-3,2)$

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

Solution:
Point is inside, outside or on the circle as $S_1$ is $ \lt ,\gt,=0$. For point $(-2,1), S_1 \lt 0$.

Topic: Mathematics, Circle, JEE Main
A point is chosen at random inside a rectangle measuring 6 inches by 5 inches. What is the probability that the randomly selected point is at least one inch from the edge of the rectangle?

Options:
A $\frac{2}{3}$
B $\frac{1}{3}$
C $\frac{1}{4}$
D $\frac{2}{5}$

The correct answer is:
$\frac{2}{5}$

Solution:
Probability that the randomly selected point is at least one inch from the edge of the rectangle $=\frac{4 \times 3}{6 \times 5}=\frac{12}{30}=\frac{2}{5}$

Topic: Mathematics, Probability, JEE Main
A point like object is placed at distance of 1m in front of a convex lens of focal length 0.5m. A plane mirror is placed at a distance of 2m behind the lens. The position and nature of the image formed by the system is

Options:
A 2.6m from the mirror, real
B 1m from the mirror, virtual
C 1m from the mirror, real
D 2.6m from the mirror, virtual

The correct answer is:
2.6m from the mirror, virtual

Solution:
For I1 applying lens formula

1v1-1-1=10.5

v1=1

For I2 applying condition of plane mirror

for final image (I3) again applying lens formula

1v3-1-3=10.5

v2=35

The distance of final image from mirror

D=35+3

D=2.6

The image will be real (but it is given virtual in answer)

Topic: Physics, Ray Optics, JEE Main, JEE Main 2020 (06 Sep Shift 1)
A point mass ' $m$ ' is moved in a vertical circle of radius ' $r$ ' with the help of a string. The velocity of the mass is $\sqrt{7 g r}$ at the lowest point. The tension in the string at the lowest point will be: $\quad$

Options:
A $6 \mathrm{mg}$
B $7 \mathrm{mg}$
C $8 \mathrm{mg}$
D $1 \mathrm{mg}$

The correct answer is:
$8 \mathrm{mg}$

Solution:
\(\begin{gathered}\mathrm{T}-\mathrm{mg}=\frac{\mathrm{mv}}{\mathrm{r}} \\ \mathrm{T}-\mathrm{mg}=\frac{\mathrm{m}(7 \mathrm{gr})}{\mathrm{r}} \\ \mathrm{T}=8 \mathrm{mg}\end{gathered}\)

Topic: Physics, Laws of Motion, JEE Main
A point mass oscillates along the $x$-axis according to the law $\mathrm{x=x_0 \cos (\omega t-\pi / 4)}$. If the acceleration of the particle is written as
$\mathrm{a=A \cos (\omega t+\delta)}$

Options:
A
$\mathrm{A}=\mathrm{x}_0, \delta=-\pi / 4$

B
$\mathrm{A}=\mathrm{x}_0 \omega^2, \delta=-\pi / 4$

C
$\mathrm{A}=\mathrm{x}_0 \omega^2, \delta=-\pi / 4$

D
$\mathrm{A=x_0 \omega^2, \delta=3 \pi / 4}$

The correct answer is:
$\mathrm{A=x_0 \omega^2, \delta=3 \pi / 4}$

Solution:
$\mathrm{v=-x_0 \omega \sin (\omega t-\pi / 4)}$
$\mathrm{a=-x_0 \omega^2 \cos \left(\omega t+\pi-\frac{\pi}{4}\right)}$
$\mathrm{a=A \cos (\omega t+\delta)}$
$\mathrm{A=x_0 \omega^2 ; \quad \delta=\frac{3 \pi}{4}}$

Topic: Physics, Oscillations, JEE Main, JEE Main 2007
A point moves in such a way that the difference of its distance from two points (8,0) and (-8,0) always remains $4 .$ Then, the locus of the point is

Options:
A a circle
B a parabola
C an ellipse
D a hyperbola

The correct answer is:
a hyperbola

Solution:
We know by the definition of hyperbola, "A hyperbola is the locus of a point which moves in such a way that the difference between two fixed points remains constant, ie., $|P S-P S|=2 a$

Topic: Mathematics, Hyperbola, JEE Main
A point moves in such a way that the difference of its distance from two points (8,0) and (-8,0) always remains $4 .$ Then, the locus of the point is

Options:
A a circle
B a parabola
C an ellipse
D a hyperbola

The correct answer is:
a hyperbola

Solution:
We know by the definition of hyperbola, "A hyperbola is the locus of a point which moves in such a way that the difference between two fixed points remains constant, ie., $|P S-P S|=2 a$

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2012
A point moves in such a way that the sum of its distance from $x y$-plane and $y z$-plane remains equal to its distance from $z x$-plane. The locus of the point is

Options:
A $x-y+z=2$
B $x+y-z=0$
C $x-y+z=0$
D $x-y-z=2$

The correct answer is:
$x-y+z=0$

Solution:
We know that,
Distance from \(x y\)-plane to the point \(P(x, y, z)=|z|\)
Distance from \(z x\)-plane to the point \(P(x, y, z)=|y|\)
Distance from \(y z\)-plane to the point \(P(x, y, z)=|x|\)
It is given that,
the sum of its distance from \(x y\)-plane and \(y z\)-plane remains equal to its distance from zx-plane
\(\begin{aligned}
& \Rightarrow|x|+|z|=|y| \\
& \Rightarrow x+z-y=0
\end{aligned}\)

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A point moves in the x-y plane according to x=AsinÏt and y=A1âcosÏt where A and Ï are positive constants. The distance traversed by the point during the time Ï is

Options:
A Aω2τ
B Aω2τ
C Aωτ
D Aωτ

The correct answer is:
Aωτ

Solution:
Distance travelled by point in time t is:
S=â«0ÏVâdt
âµVâ=Vx2+Vy2
Where, Vx=dxdt=AÏcosâ¡Ït and Vy=dydt=AÏsinâ¡Ït
S=â«0ÏAÏdt=AÏÏ

Topic: Physics, Motion In One Dimension, JEE Main
A point moves so that square of its distance from the point $(3,-2)$ is numerically equal to its distance from the line $5 x-12 y=13$. The equation of the locus of the point is

Options:
A $13 x^2+13 y^2-83 x+64 y+182=0$
B $x^2+y^2-11 x+16 y+26=0$
C $x^2+y^2-11 x+16 y=0$
D $13 x^2+13 y^2+83 x-64 y-182=0$

The correct answer is:
$13 x^2+13 y^2-83 x+64 y+182=0$

Solution:
$(h-3)^2+(k+2)^2=\left|\frac{5 h-12 k-13}{\sqrt{25+144}}\right|$
Replace ${ }^{(h, k)}$ by $(x, y)$, we get
$13 x^2+13 y^2-83 x+64 y+182=0$, which is the required equation of the locus of the point.

Topic: Mathematics, Straight Lines, JEE Main
A point moves, so that the sum of squares of its distance from the points (1,2) and (-2,1) is always $6 .$ Then, its locus is

Options:
A the straight line $y-\frac{3}{2}=-3\left(x+\frac{1}{2}\right)$
B a circle with centre $\left(-\frac{1}{2}, \frac{3}{2}\right)$ and radius $\frac{1}{\sqrt{2}}$
C a parabola with focus (1,2) and directrix passing through (-2,1)
D an ellipse with foci (1,2) and (-2,1)

The correct answer is:
a circle with centre $\left(-\frac{1}{2}, \frac{3}{2}\right)$ and radius $\frac{1}{\sqrt{2}}$

Solution:
Let $P$ be any point, whose coordinate is $(h, k)$
Given.

P moves, so that the sum of squares of its distances from the points $A(1,2)$ and $B(-2,1)$ is $6 .$
1.e., $(P A)^{2}+(P B)^{2}=6$
$\Rightarrow \quad(h-1)^{2}+(k-2)^{2}+(h+2)^{2}+(k-1)^{2}=6$
$\Rightarrow \quad h^{2}+1-2 h+k^{2}+4-4 k+h^{2}+4+4 h$
$$
+k^{2}+1-2 k=6
$$
$\Rightarrow \quad 2 h^{2}+2 k^{2}+2 h-6 k+4=0$
$\Rightarrow \quad h^{2}+k^{2}+h-3 k+2=0$
$\therefore$ Required locus is
$$
x^{2}+y^{2}+x-3 y+2=0
$$
Which represent a circle. Whose centre is $\left(\frac{-1}{2}, \frac{3}{2}\right)$ and radius $=\sqrt{\frac{1}{4}+\frac{9}{4}-2}=\sqrt{\frac{5}{2}-2}=\frac{1}{\sqrt{2}}$

Topic: Mathematics, Circle, JEE Main
A point moves such that its distance from the point $(4,0)$ is half that of its distance from the line $x=16$ The locus of this point is

Options:
A $3 x^2+4 y^2=192$
B $4 x^2+3 y^2=192$
C $x^2+y^2=192$
D None of these

The correct answer is:
$3 x^2+4 y^2=192$

Solution:
$\sqrt{(h-4)^2+k^2}=\frac{1}{2}\left(\frac{h-16}{\sqrt{1^2+0}}\right)$
Replace $(h, k)$ by $(x, y)$, we get $3 x^2+4 y^2=192$.

Topic: Mathematics, Straight Lines, JEE Main
A point moves with uniform acceleration and its initial speed and final speed are \( 1 \mathrm{~m} \mathrm{~s}^{-1} \) and \( 2 \mathrm{~m} \mathrm{~s}^{-1} \) respectively then, the space average of velocity over the distance moved is. (in \( \mathrm{m} \mathrm{s}^{-1} \) ) :-

Options:
A \( \frac{14}{9} \mathrm{~m} \mathrm{~s}^{-1} \)
B \( \frac{3}{2} \mathrm{~ms}^{-1} \)
C \( \frac{5}{2} \mathrm{~ms}^{-1} \)
D None of these

The correct answer is:
\( \frac{14}{9} \mathrm{~m} \mathrm{~s}^{-1} \)

Solution:
<v>space=∫vdx∫dx
But a=vdvdx
So vdx=v2adv
and dx=vadv
So <v>space=∫v2adv∫vadv=∫12v2dv∫12vdv=v3312v2212=149 m s-1

Topic: Physics, Motion In One Dimension, JEE Main
A point object in air is in front of the curved surface of a plano-convex lens. The radius of curvature of the curved surface is 30 cm and the refractive index of the lens material is 1.5, then the focal length of the lens (in cm) is ___________.

Options:

The correct answer is:
60

Solution:
1f=Î¼-11R1-1R2
R1=∞
R2=â-30cm
1f=1.5-11â-1-30
1f=0.530
f = 60cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2020 (08 Jan Shift 1)
A point object is kept just beyond the first focus of a convex lens. If the lens starts moving towards the right with a constant velocity, the image will

Options:
A always move towards right
B always move towards left
C first move towards right and then towards left
D first move towards left and then towards right

The correct answer is:
first move towards left and then towards right

Solution:
dv dt = v u 2 × du dt
V image/lens = v u 2 × V object/lens
till u become 2f, m > -1 so Vimagelens> Vobjectlens and the image will move towards left and when
u become > 2f, Vimage/lens < Vobject/lens and the image will move towards the right.

Topic: Physics, Ray Optics, JEE Main
A point object is moving with a speed v in front of an arrangement of two mirrors as shown in figure. If the velocity of image in mirror M 1 with respect to image in mirror M 2 is n(v sinâÎ¸), find n

Options:

The correct answer is:
2

Solution:
Angle between vâ¡ I 1 & vâ¡ I 2 is 2 Î¸ . Their magnitudes is v.

â´ v I 1 v I 2 =| v â I 1 â v â I 2 |= v 2 + v 2 â2vââvcos2Î¸
= 2 vâ¡ sin Î¸

Topic: Physics, Ray Optics, JEE Main
A point object is moving with a speed v in front of an arrangement of two mirrors as shown in the figure. The magnitude of velocity of the image in mirror M1 with respect to the image in mirror M2 is

Options:
A 2vsinÎ¸
B vsinθ
C 3vsinθ
D 4vsinθ

The correct answer is:
2vsinθ

Solution:
Angle between vI1 and vI2 is 2Î¸. The magnitudes of these vectors is v.
â´|vâI1âvâI2|=v2+v2â2vÂ·vÂ·cos2Î¸
=2vsinÎ¸

Topic: Physics, Ray Optics, JEE Main