datavorous/entrance-exam-dataset huggingface dataset:
Consider a system of blocks $X, A$ and $B$ as shown in the figure. The blocks $A$ and $B$ have equal mass and are connected by a massless string through a massless pulley. The coefficient of friction between block $A$ and $X$ or $B$ and $X$ is 0.5 . If block $X$ moves on the horizontal frictionless surface what should be Its minimum acceleration such that blocks $A$ and $B$ remain stationary. $(g=$ acceleration due to gravity.)

Options:
A $\frac{g}{3}$
B $3 g$
C $\frac{g}{4}$
D $\frac{3 g}{4}$

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

Solution:
Given, Mass of block $A$ and $B, m_A=m_B=m$. Coefficient of friction between each block and surface $\mu=0.5$.
Block $\not X$ moves with acceleration $a$ such that block $A$ and $B$ remains stationary. Block $B$ is stationary only when, Tension force on block $B$ is equal to friction force on $B$.
i.e.,
$$
\begin{aligned}
& T=\mu R \\
& T=\mu m a \quad \text {... (i) }[\because R=m a]
\end{aligned}
$$
(i) $[\because R=m a]$
Block $A$ is stationary only when friction force on block $A$ is equal to the sum of tension force on block $A$ and applied force on block $A$.
$$
\begin{aligned}
\mu R & =T+m a \\
\mu m g & =\mu m a+m a
\end{aligned}
$$
[From Eq. (i),]
$$
\begin{aligned}
\mu g & =\mu a+a \\
\mu g & =a(\mu+1) \\
a=\frac{\mu g}{\mu+1} & =\frac{0.5 g}{0.5+1}=\frac{0.5 g}{1.5}=\frac{g}{3} \\
\therefore \quad & =\frac{g}{3}
\end{aligned}
$$
Thus, $g / 3$ should be its minimum acceleration such that blocks $A$ and $B$ remains stationary.

Topic: Physics, Laws of Motion, TS EAMCET, TS EAMCET 2019 (03 May Shift 1)
Consider a tank made of glass (refractive index 1.5 ) with a thick bottom. It is filled with a liquid of refractive index μ. A student finds that, irrespective of what the incident angle i (see figure) is for a beam of light entering the liquid, the light reflected from the liquid glass interface is never completely polarized. For this to happen, the minimum value of μ is:

Options:
A 35
B 43
C 53
D 53

The correct answer is:
35

Solution:
For reflected light not be completely polarized,
tanâ¡râ Î¼gÎ¼w=1.5Î¼
For minimum Î¼ r is maximum. r is maximum when i is maximum i.e. i=90o
â sinâ¡r=1Î¼
â tanâ¡r=1Î¼2-1
â´1Î¼2-1=1.5Î¼
â Î¼2=2.25 Î¼2-1
â 1.25 Î¼2=2.25
â Î¼=2.251.25
=95=35

Topic: Physics, Wave Optics, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Consider a thin metallic sheet perpendicular to the plane of the paper moving with speed v in a uniform magnetic field B going into the plane of the paper (see figure). If charge densities Ï1 and Ï2 are induced on the left and right surfaces respectively of the sheet, then (ignore fringe effects)

Options:
A σ1=-ϵ0vB2, σ2=ϵ0vB2
B σ1=ϵ0vB, σ2=-ϵ0vB
C σ1=ϵ0vB2, σ2=-ϵ0vB2
D σ1=σ2=ϵ0vB

The correct answer is:
σ1=ϵ0vB, σ2=-ϵ0vB

Solution:
σ1=-σ2 [due to magnetic force]
E=σε0=vB
σ=ε0vB

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2016 (10 Apr Online)
Consider a thin metallic sheet perpendicular to the plane of the paper moving with speed v in a uniform magnetic field B going into the plane of the paper (see figure). If charge densities Ï1 and Ï2 are induced on the left and right surfaces respectively of the sheet, then (ignore fringe effects)

Options:
A σ1=-ϵ0vB2, σ2=ϵ0vB2
B σ1=ϵ0vB, σ2=-ϵ0vB
C σ1=ϵ0vB2, σ2=-ϵ0vB2
D σ1=σ2=ϵ0vB

The correct answer is:
σ1=ϵ0vB, σ2=-ϵ0vB

Solution:
σ1=-σ2 [due to magnetic force]
E=σε0=vB
σ=ε0vB

Topic: Physics, Electromagnetic Induction, JEE Main
Consider a thin uniform square sheet made of a rigid material. If its side is a , mass m and moment of inertia I about one of its diagonals, then:

Options:
A I=ma224
B ma224<I<ma212
C I>ma212
D I=ma212

The correct answer is:
I=ma212

Solution:
In a uniform square plate due to symmetry moment of Inertia about all the axis passing through centre and lying in the plane of the plate is same.
∴ Idiagonal=ma212

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2015 (10 Apr Online)
Consider a toroid of circular cross-section of radius b , major radius R much greater than minor radius b , (see diagram) find the total energy stored in magnetic field of toroid -

Options:
A B2π2b2R2μ0
B B2π2b2R4μ0
C B2π2b2R8μ0
D B2π2b2Rμ0

The correct answer is:
B2π2b2Rμ0

Solution:
B=μ0Ni2R
ϕ=πb2×B×N
ϕ=Li
L=ϕi=μ0N2b22R , with b<<<R
E=12Li2=μ0N2i24Rb2

Topic: Physics, Magnetic Effects of Current, JEE Main
Consider a toroid with rectangular cross section, of inner radius $a$, outer radius $b$ and height $h$, carrying $n$ number of turns. Then the self-inductance of the toroidal coil when current $I$ passing through the toroid is

Options:
A $\frac{\mu_0 n^2 h}{2 \pi} \ln \left(\frac{b}{a}\right)$
B $\frac{\mu_0 n h}{2 \pi} \ln \left(\frac{b}{a}\right)$
C $\frac{\mu_0 n^2 h}{2 \pi} \ln \left(\frac{a}{b}\right)$
D $\frac{\mu_0 n h}{2 \pi} \ln \left(\frac{a}{b}\right)$

The correct answer is:
$\frac{\mu_0 n^2 h}{2 \pi} \ln \left(\frac{b}{a}\right)$

Solution:
Given, a toroid with a rectangular
cross-section of inner radius $a$ and outer radius $b$.
Height of the solenoid $=h$
Magnetic field inside a rectangular toroid is given by
$B=\frac{\mu_0 n I}{2 \pi r}$
using the infinitesimal cross-sectional area element,
$d x=h d r$
$\therefore$ Flux passing through the cross-section of toroid.
$\begin{aligned} \phi & =\int B \cdot d x=\int_a^b \frac{\mu_0 n I}{2 \pi r} \cdot(h d r) \\ & \phi=\frac{\mu_0 n I h}{2 \pi} \int_a^b \frac{1}{r} d r \\ \Rightarrow \quad & \phi=\frac{\mu_0 n h I}{2 \pi}\left[\log r l_a^b\right. \\ \phi & =\frac{\mu_0 n h I}{2 \pi}(\log b-\log a) \\ & \phi=\frac{\mu_0 n h I}{2 \pi} \ln \left(\frac{b}{a}\right)\end{aligned}$
Now, self inductance of rectangular toroid,
$L=\frac{n \phi}{I}$
Putting the value of $\phi$, we get
$L=\frac{\mu n^2 h}{2 \pi} \ln \left(\frac{b}{a}\right)$

Topic: Physics, Electromagnetic Induction, JEE Main
Consider a triangle $\mathrm{ABC}$ in the $\mathrm{xy}$-plane with vertices $\mathrm{A}=(0,0), \mathrm{B}=(1,1)$ and $\mathrm{C}=(9,1)$. If the line $\mathrm{x}=\mathrm{a}$ divides the triangle into two parts of equal area, then a equals

Options:
A 3
B $3.5$
C 4
D $4.5$

The correct answer is:
3

Solution:
area of $P Q C=\frac{1}{2}$ area of $A B C$
$$
\begin{array}{l}
\frac{1}{2}(9-a)\left(1-\frac{a}{9}\right)=\frac{1}{2} \times \frac{1}{2}(8 \times 1) \\
(9-a)^{2}=36 \Rightarrow a=3
\end{array}
$$

Topic: Mathematics, Straight Lines, JEE Main
Consider a triangle having vertices A(-2,3), B(1,9) and C(3, 8). If a line L passing through the circum-centre of triangle ABC, bisects line BC, and intersects y-axis at point 0,α2, then the value of real number α is _______.

Options:

The correct answer is:
9

Solution:
We have,

AB=-2-12+3-92=45

AC=-2-32+3-82=50

BC=3-12+8-92=5

Here,

AC2=AB2+BC2

⇒(50)2=(45)2+(5)2

Hence, ∠B=90°, so circumcentre is the mid-point of hypotenuse i.e., AC.

Circumcenter ≡12,112

Mid-point of BC=2,172

Equation of line L is

y-112=112-17212-2x-12

⇒y=2x+92

Passing though 0,α2, so

α2=92⇒α=9

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2021 (20 Jul Shift 2)
Consider a triangular plot ABC with sides AB=7 m, BC=5 m and CA=6 m. A vertical lamp-post at the mid-point D of AC subtends an angle 30° at B. The height (in m) of the lamp-post is:

Options:
A 221
B 2321
C 3221
D 73

The correct answer is:
2321

Solution:
The length of the median BD is given by

BD=122BC2+BA2-AC2

⇒BD=12 225+49-36

⇒BD=12112

⇒BD=472=27 m.

Now, let DT be the tower of height h m and it subtends ∠TBD=30° at the vertex B.

Then, in ∆TDB we have tan∠TBD=TDBD

⇒tan30°=h27

⇒13=h27

⇒h=273

⇒h=2321 m.

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2019 (10 Jan Shift 1)
Consider a triangular plot \( A B C \) with sides \( A B=7 m, B C=5 m \) and \( C A=6 m . \) A vertical lamp-post at the mid-point \( D \) of \( A C \) subtends an angle \( 30^{\circ} \) at \( B \). The height (in \( m \) ) of the lamp-post is:

Options:
A \( 2 \sqrt{21} \)
B \( \frac{2}{3} \sqrt{21} \)
C \( \frac{3}{2} \sqrt{21} \)
D \( 7 \sqrt{3} \)

The correct answer is:
\( \frac{2}{3} \sqrt{21} \)

Solution:
The length of the median BD is given by

BD=122BC2+BA2-AC2

⇒BD=12 225+49-36

⇒BD=12112

⇒BD=472=27 m.

Now, let DT be the tower of height h m and it subtends ∠TBD=30° at the vertex B.

Then, in ∆TDB we have tan∠TBD=TDBD

⇒tan30°=h27

⇒13=h27

⇒h=273

⇒h=2321 m.

Topic: Mathematics, Heights and Distances, JEE Main
Consider a two particle system with particles having masses $m_1$ and $m_2$. If the first particle is pushed towards the centre of mass through a distance $d$, by what distance should the second particle be moved, so as to keep the centre of mass at the same position?

Options:
A
$\mathrm{d}$

B
$\frac{m_2}{m_1} \mathrm{~d}$

C
$\frac{m_1}{m_1+m_2} d$

D
$\frac{\mathrm{m}_1}{\mathrm{~m}_2} \mathrm{~d}$

The correct answer is:
$\frac{\mathrm{m}_1}{\mathrm{~m}_2} \mathrm{~d}$

Solution:
$m_1 d+m_2 x=0$
$x=\frac{m_1 d}{m_2}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2006
Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible force F at a point b above its centre of mass (see figure). If the coefficient of friction is μ=0.4 , the maximum possible value of 100×ba for a box not to topple before moving is ________

Options:

The correct answer is:
50

Solution:
For no toppling
Fa2+b≤mga2
μa2+μb≤a2
0.2a+0.4b≤0.5a
0.4b≤0.3a
b≤3a2
b≤0.75a (in limiting case)
But it is not possible as b can maximum be equal to 0.5a
∴100bamax=50.00

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2020 (07 Jan Shift 2)
Consider a uniform disc in the x-y plane free to rotate about an axis parallel to the Z-axis as shown (passing through the centre). A force of 4N applied at 4,3 along y-axis, as shown produces an angular acceleration of 2 s-2 . Locate the centre of mass of the disc if the mass of the disc is 2 kg.

Options:
A (1, 3)
B (2, 3)
C (0, 3)
D (4, 3)

The correct answer is:
(2, 3)

Solution:
As τ=Iα
4R=2R22 2
⇒ R=2 m
Thus, COM is at (2, 3)

Topic: Physics, Rotational Motion, JEE Main
Consider a uniform electric field in the z-direction. The potential is a constant

Options:
A
in all space

B
for any $x$ for a given $z$

C
for any $y$ for a given $z$

D
on the $x-y$ plane for a given $z$

The correct answers are:
for any $x$ for a given $z$
,
for any $y$ for a given $z$
,
on the $x-y$ plane for a given $z$

Solution:
The direction of electric field is perpendicular to the equipotential surfaces. Here the electric field is in $+z$ direction. And the electric field is always remain in the direction in which the potential decreases. Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.
So, electric field in z-direction suggests that equipotential surfaces will be the plane perpendicular to z-axis along $x-y$ plane. Therefore the potential is a constant for any $x$ for $a$ given $z$ and for any $y$ for a given $z$ and on the $x-y$ plane for a given $z$.

Topic: Physics, Electrostatic Potential and Capacitance
Consider a uniform rod of mass M=4m and length l pivoted about its centre. A mass m moving with velocity v making angle θ=π4 to the rod’s long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is:

Options:
A 372vl
B 37vl
C 327vl
D 47vl

The correct answer is:
327vl

Solution:
Loi=Lof
mV2×L2=4mL212+mL24×ω
ω=6V72L=32V7L

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2020 (08 Jan Shift 1)
Consider a uniform square plate of side $a$ and mass $M$. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its comers is

Options:
A $\frac{5}{6} M a^2$
B $\frac{1}{12} M a^2$
C $\frac{7}{12} M a^2$
D $\frac{2}{3} M a^2$

The correct answer is:
$\frac{2}{3} M a^2$

Solution:
Step 1: Given information
A uniform square plate,
Side is "a" and mass "m".
According to the question, the diagram will be:
Step 2: Calculation
The moment of inertia of a square plate around an axis perpendicular to the plane going throughout its center of mass is determined by:
$\Rightarrow \mathrm{I}_{\mathrm{CM}}=\frac{\mathrm{ma}^2}{6}$
The moment of inertia around a parallel axis through one of its corners is determined by:
$\Rightarrow \mathrm{I}=\mathrm{I}_{\mathrm{CM}}+m\left(\frac{\mathrm{a}}{\sqrt{2}}\right)^2$
By substituting the values of $\mathrm{I}_{\mathrm{CM}}$,
$\begin{aligned}
= & \frac{\mathrm{ma}^2}{6}+\frac{\mathrm{ma}^2}{2} \\
& =\frac{\mathrm{ma}^2+3 \mathrm{ma}^2}{6} \\
& =\frac{4}{6} \mathrm{ma}^2 \\
& =\frac{2}{3} \mathrm{ma}^2
\end{aligned}$
Therefore, the moment of inertia is $\frac{2}{3} \mathrm{ma}^2$.

Topic: Physics, Rotational Motion, JEE Main
Consider a uniform square plate of side ' $a$ ' and mass ' $m$ '. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is

Options:
A
$\frac{5}{6} m a^2$

B
$\frac{1}{12} m a^2$

C
$\frac{7}{12} m a^2$

D
$\frac{2}{3} m a^2$

The correct answer is:
$\frac{2}{3} m a^2$

Solution:
$$
I=I_{c m}+m\left(\frac{a \sqrt{2}}{2}\right)^2=\frac{m a^2}{6}+\frac{m a^2}{2}=\frac{2}{3} m a^2
$$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2008
Consider a uniform wire of mass M and length L. It is bent into a semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the centre is :

Options:
A 14ML2π2
B 25ML2π2
C ML2π2
D 12ML2π2

The correct answer is:
ML2π2

Solution:
πr=L⇒r=Lπ

I=Mr2=ML2π2

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Consider a vechicle moving with a velocity $54 \mathrm{~km} / \mathrm{h}$. At a distance of $400 \mathrm{~m}$, from the traffic light brakes are applied. The acceleration of the vehicle, after the application of brakes is $-0.3 \mathrm{~m} / \mathrm{s}^2$. The vehicle's position relative to the traffic light is

Options:
A $25 \mathrm{~m}$
B $375 \mathrm{~m}$
C $425 \mathrm{~m}$
D $30 \mathrm{~m}$

The correct answer is:
$25 \mathrm{~m}$

Solution:
Given, initial velocity,
$$
\begin{aligned}
& u=54 \mathrm{~km} / \mathrm{h}=54 \times \frac{5}{18} \mathrm{~m} / \mathrm{s} \\
& u=15 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Distance of signal from vehicle, $d=400 \mathrm{~m}$ acceleration, $a=-0.3 \mathrm{~m} / \mathrm{s}^2$
Now, distance covered by vehicle, when it stops is given by $v^2=u^2+2 a s$
$$
0=(15)^2-2 \times 0.3 \times s \Rightarrow s=\frac{225}{0.6}=375 \mathrm{~m}
$$
Relative position $=400-375=25 \mathrm{~m}$

Topic: Physics, Motion In One Dimension, JEE Main
Consider a vector \( \vec{F}=4 \hat{i}-3 \hat{j} . \) Another vector that is perpendicular to \( \vec{F} \) is

Options:
A \( 4 \hat{i}+3 \hat{j} \)
B \( 6 \hat{i} \)
C \( 7 \hat{k} \)
D \( 3 \hat{i}-4 \hat{j} \)

The correct answer is:
\( 7 \hat{k} \)

Solution:
Force F lies in the x-y plane so a vector along z-axis will be perpendicular to F.

Topic: Physics, Mathematics in Physics, JEE Main
Consider a vessel filled with a liquid upto height H. The bottom of the vessel lies in the X-Y plane passing through the origin. The density of the liquid varies with Z-axis as ρz=ρ02-zH2. If P1 and P2 are the pressures at the bottom surface and top surface of the liquid, the magnitude of P1-P2 is :

Options:
A ρ0gH
B 85ρ0gH
C 32ρ0gH
D 53ρ0gH

The correct answer is:
53ρ0gH

Solution:
Given that, PZ=ρ02-ZH2

Let a small column of liquid thickness be dz.

Pressure due to column,

dP1=ρ02-ZH2gdz+P2

P1=∫0πρ02-ZH2gdz+P2

=ρ0g∫0π2-Z2H2dz+P2'

=ρ0g2H-13H2H3+P2

P1=ρ0g2H-H3+P2

P1-P2=ρ0g5H3=53ρ0gH

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Consider a water jar of radius R that has water filled up to height H and is kept on a stand of height h (see figure). Through a hole of radius r r<<R at its bottom, the water leaks out and the stream of water coming down towards the ground has a shape like a funnel as shown in the figure. If the radius of the cross-section of water stream when it hits the ground is x. Then:

Options:
A x=rHH+h14
B x=rHH+h
C x=rHH+h2
D x=rHH+h12

The correct answer is:
x=rHH+h14

Solution:
Using equation of continuity
π r 2 2 gH = π x 2 2 g H + h
x = r H H + h 1 4

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2016 (09 Apr Online)
Consider a water tank as shown in the figure. It's cross-sectional area is 0.4 m2. The tank has an opening B near the bottom whose cross-section area is 1 cm2. A load of 24 kg is applied on the water at the top when the height of the water level is 40 cm above the bottom, the velocity of water coming out the opening B is v m s-1. The value of v, to the nearest integer, is ___ .[Take the value of g to be 10 m s-2]

Options:

The correct answer is:
3

Solution:
m=24 kg

A=0.4 m2

a=1 cm2

H=40 cm

Using Bernoulli's equation,

⇒P0+mgA+ρgH+12Pv12

=P0+0+12ρv2 ...1

⇒ Neglecting v1

⇒v=2gH+2mgAρ

⇒v=8+1.2

⇒v=3.033 m s-1

⇒v≃3 m s-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Consider a Young's double slit experiment as shown in figure. What should be the slit separation d in terms of wavelength λ such that the first minima occurs directly in front of the slit (S1) ?

Options:
A λ5-2
B λ25-2
C λ25-2
D λ5-2

The correct answer is:
λ25-2

Solution:
Path diff. in front of a slit on screen
∆=5d-2d=2n-1λ2
(for minima)
For 1st minima n=1
d5-2=λ2
d=λ25-2

Topic: Physics, Wave Optics, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Consider an electrical circuit containing a two way switch S. Initially S is open and then T1 is connected to T2. As the current in R=6 Ω attains a maximum value of steady-state level, T1 is disconnected from T2 and immediately connected to T3. Potential drop across r=3 Ω resistor immediately after T1 is connected to T3 is ______V. (Round off to the Nearest Integer)

Options:

The correct answer is:
3

Solution:
When T1 and T2 are connected, then the steady state current in the inductor I=66=1 A

When T1 and T3 are connected then current through inductor remains same. So potential difference across 3 Ω

V=Ir=1×3=3 V

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Consider an electrochemical cell: A s A n+ aq, 2 M B 2n+ aq, 1 M B s . The value of ÎH o for the cell reaction is twice that of ÎG o at 300 âK. If the emf of the cell is zero, the â S â (in J K -1 m o l -1) of the cell reaction per mole of B formed at 300âK is ____
(Given: lnâ¡2=0.7, R =â8.3âJâ K â1 â mol â1 . H,âS and G are enthalpy, entropy and Gibbs energy, respectively.)

Options:

The correct answer is:
-11.60

Solution:
A(s)â|â A +n (aq.â2âM)â||â B +2n (aq.â1âM)â|âB(s)
ÎH o â=â 2ÎG o E cell â=â0
CellâRxââââAâ A +n + ne â ]âÃâ2
B +2n +2â ne â âB(s)
2âA(s)+ B +2n 1âM (aq)â 2â A +n 2âM (aq)+B(s)
ÎGâ=â ÎG o â+âRTâlnâ A +n 2 B +2n
ÎG o â=ââRTâln A +n 2 B +2n â=ââRTâÃâln 2 2 1 â=ââRTln4
ÎG o â=â ÎH o â TÎS o
ÎG o â=â 2ÎG o â TÎS o
ÎS o â=â ÎG o T â=ââ RTâln4 T
=ââ8.3âÃâ2âÃâ0.7â=ââ11.6âJ/Kâ mol â1

Topic: Chemistry, Electrochemistry, JEE Main
Consider an electromagnetic wave propagating in vacuum. Choose the correct statement:

Options:
A For an electromagnetic wave propagating in +y direction the electric field is E→=12Eyzx,tz^ and the magnetic field is B→=12Bzx,ty^
B For an electromagnetic wave propagating in + y direction the electric field is Eâ=12Eyzx,ty^ and the magnetic field is Bâ=12Byzx,tz^
C For an electromagnetic wave propagating in + x direction the electric field is E→=12Eyzy, z, t y^+z^ and the magnetic field is B→=12Eyzy, z, t y^+z^
D For an electromagnetic wave propagating in + x direction the electric field is E→=12Eyzx,t y^-z^ and the magnetic field is B→=12 Byzx, ty^+z^

The correct answer is:
For an electromagnetic wave propagating in + x direction the electric field is E→=12Eyzx,t y^-z^ and the magnetic field is B→=12 Byzx, ty^+z^

Solution:
Wave in x - direction means E→ and B→ should be function of x and t
y^-z^⊥y^+z^
E→ ×B→|| x^

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2016 (10 Apr Online)
Consider an electron in a hydrogen atom, revolving in its second excited state (having radius 4.65 Å ). The de-Broglie wavelength of this electron is:

Options:
A 12.9 Å
B 6.6 Å
C 9.7 Å
D 3.5 Å

The correct answer is:
9.7 Å

Solution:
According to Bohr’s atomic model
mvr=nh2π
And the de-Broglie wavelength
λ=hmv
From the above two equations we can write, λ=2πrn
In the 2nd excited state
λ=2πr3
∴ λ=23π×4.65 Å=9.7 Å

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2019 (12 Apr Shift 2)
Consider an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ .
What is the area of the greatest rectangle that can be inscribed in the ellipse?

Options:
A $a b$
B $2 a b$
C $a b / 2$
D $\sqrt{a b}$

The correct answer is:
$2 a b$

Solution:
Given equation of ellipse, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Let $\mathrm{A}(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)$ be any point on ellipse(1st quadrant)
Coordinate of $\mathrm{B}=[\mathrm{a} \cos (\pi-\theta), \mathrm{b} \sin (\pi-\theta)]$
$=(-a \cos \theta, b \sin \theta)$(2nd quadrant)
Coordinate of $C=[a \cos (\pi+\theta), b \sin (\pi+\theta)]$(3rd quadrant)
Coordinate of $\mathrm{D}=[\mathrm{a} \cos (2 \pi-\theta), \mathrm{b} \sin (2 \pi-\theta)]$
$=(a \cos \theta,-b \sin \theta) \quad$ (4th quadrant)
Area of the rectangle $\mathrm{ABCD}$ $=(a \cos \theta+a \cos \theta)(b \sin \theta+b \sin \theta)$
$=2 a \cos \theta \times 2 b \sin \theta=2 a b \sin 2 \theta$
$=2 a b \times 1=2 a b$

Topic: Mathematics, Ellipse, JEE Main
Consider an ellipse, whose center is at the origin and its major axis is along the x-axis. If its eccentricity is 35 and the distance between its foci is 6, then the area (in sq. units) of the quadrilateral inscribed in the ellipse, with the vertices as the vertices of the ellipse, is:

Options:
A 32
B 80
C 40
D 8

The correct answer is:
40

Solution:
Given eccentricity of the ellipse e = 3 5 .
Distance between foci is 2ae=6
⇒ a = 5.
Also, b 2 = a 2 (1− e 2 )
⇒ b2=251-925
⇒ b=4
Area =412ab=2ab=40

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2017 (08 Apr Online)
Consider an elliptically shaped rail PQ in the vertical plane with OP = 3 m and OQ = 4 m . A block of mass 1 kg is pulled along the rail from P to Q with a force of 18 N, which is always parallel to line PQ (see the figure given). Assuming no frictional losses, the kinetic energy of the block when it reaches Q is (nÃ10) joules. The value of n is (take acceleration due to gravity = 10 m sâ2 )

Options:

The correct answer is:
5

Solution:
Using work-energy theorem

Wmg+WF=ΔKE

-mgh+Fd=ΔKE

-1×10×4+185=ΔKE

ΔKE=50

∴ n=5

Topic: Physics, Work Power Energy, JEE Main
Consider an endothermic reaction $X \rightarrow Y$ with the activation energies $E_b$ and $E_f$ for the backward and forward reactions, respectively, in general

Options:
A $E_b \lt E_f$
B $E_b \gt E_f$
C $E_b=E_f$
D There is no definite relation between $E_b$ and

The correct answer is:
$E_b \lt E_f$

Solution:
For endothermic reaction $\Delta H=+v e$
Then from equation $\Delta H=E_{a_{F, R .}}-E_{a_{B, R}} ; E_{B . R .} \lt E_{F . R}$.

Topic: Chemistry, Chemical Kinetics, JEE Main
Consider an endothermic reaction, $X \longrightarrow Y$ with the activation energies $E_b$ and $E_f$ for the backward and forward reactions, respectively. In general

Options:
A
$\mathrm{E}_{\mathrm{b}} < \mathrm{E}_{\mathrm{f}}$
B
$E_b>E_f$

C
$E_b=E_f$

D
There is no definite relation between $E_b$ and $E_f$

The correct answer is:
$\mathrm{E}_{\mathrm{b}} < \mathrm{E}_{\mathrm{f}}$

Solution:
$\Delta \mathrm{H}=\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{b}}$
For $\Delta H=$ Positive, $E_b < E_f$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2005