datavorous/entrance-exam-dataset huggingface dataset:
Two isomers A and B with Molar mass 184 g/mol and elemental compositio C, 52.2%; H, 49% and Br 42.9% gave benzoic acid and p- bromobenzoic acid, respectively on oxidation with KMnO4. Isomer 'A' is optically active and gives a pale yellow precipitate when warmed with alcoholic AgNO3. Isomer 'A' and 'B' are, respectively :

Options:
A

B

C

H3C-CHBr-C6H5 and

D

and H3C-CHBr-C6H5

The correct answer is:

H3C-CHBr-C6H5 and

Solution:
Based on the percentage composition of the elements given the molecular formula of the compound is C8H9Br

And the degree of unsaturation or double bond equivalent of C8H9Br is 4.

It is mentioned that A is optically active and gives pale yellow precipitate with AgNO3.

Hence, the structures of A and B can be deduced based on the following reactions.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Two isomers 'A' and 'B' with molecular formula C4H8 give different products on oxidation with KMnO4 in acidic medium. Isomer 'A' on reaction with KMnO4/H+ results in effervescence of a gas and gives ketone. The compound 'A' is

Options:
A But-1-ene.
B cis-But-2-ene.
C trans-But-2-ene.
D 2-methyl propene.

The correct answer is:
2-methyl propene.

Solution:
The formula C4H8 shows that it has degree of unsaturation as one. And since it react with KMnO4 in acidic medium then it should be an alkene. Only the terminal alkene gives effervescence of carbon dioxide gas. So, it should be a terminal alkene. and other product is ketone so, it should have a 3 degree carbon too. Based on these we can say that the A is

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2022 (29 Jun Shift 1)
Two large identical plates are placed in front of each other at x=d and x=2d as shown in the figure. If charges on plates are Q and -5Q , the potential versus distance graph for the region x=0 to x=3d is: (d is very small and potential at x=0 is V0 )

Options:
A
B
C
D

The correct answer is:

Solution:
Potential decrease in the direction of E .
â´ From origin potential decrease from V0 to certain value
E=dVdx ââ´ âslope of graph V-x is equal to E
E is higher in between two plate
â´ âslope between plate is maximum.

Topic: Physics, Electrostatics, JEE Main
Two large insulating plates having surface charge densities + Ï and âÏ are fixed at a distance d from each other. A small test charge q of mass m is attached to two identical springs as shown in the figure. The charge q is now released from rest with springs in natural length. Then q will (neglect gravity)

Options:
A perform SHM with angular frequency km
B perform SHM with amplitude σq2kε0
C not perform SHM but will have a periodic motion
D remain stationary

The correct answer is:
perform SHM with amplitude σq2kε0

Solution:
Electric field between plates is uniform hence electrostatic force due to this field is constant. Angular frequency does not change by constant force equilibrium position will be at the point where 2kx0=q×σ∈0;x0=σq2k∈0
x0 is also the amplitude of oscillation

Topic: Physics, Electrostatics, NEET
Two large insulating plates having surface charge densities +Ï and -Ï are fixed some distance apart in a gravity-free region and two ideal insulating springs of force constant k are connected to the plates as shown in the figure. A particle of charge q and mass m which is attached to the junction of the springs is released from rest, then the particle will cross its equilibrium position with a speed

Options:
A v=qσεokm
B v=qÏÎµ0k2m
C v=qσε02km
D v=qσ2ε0km

The correct answer is:
v=qσε0k2m

Solution:
The electric field between plates is

E=ÏÎµ0 (Constant)

The angular frequency of SHM doesn't change by a constant force, only the equilibrium position will be affected.

Ï=2km

The amplitude of SHM can be calculated by finding the distance of the equilibrium position from the rest position (starting position) of the particle.

2kA=qÏÎµ0

âA=qÏ2kÎµ0

So the particle will cross its equilibrium position with a speed

v=qÏÎµ0k2m

Topic: Physics, Oscillations, JEE Main
Two large oppositely charged insulated plates have a uniform electric field between them as shown. The distance between the plates is increased by a small amount. Which of the following statements is/are correct.

Statement I: The electric field strength decreases.
Statement II: The electrostatic force of attraction between the plates increases.
Statement III: The potential difference between the plates increases

Options:
A (I) only
B (II) only
C (III) only
D (I) and (III) only

The correct answer is:
(III) only

Solution:
Since charge and charge density on plate remains constant, the electric field between the plates does not change due to the increase in separation of plates. Now charge on one plate and electric field in its surroundings does not change, therefore the magnitude of the electrostatic force on that plate remains constant. Only the electrostatic potential difference between the plates increases as a result of increase in their separation.

The charge remains constant. If the distance is increased, the capacitance will decrease (C ∝ A/d) and since Q = CV, the potential difference must increase by the same factor that the distance increases.

Topic: Physics, Electrostatics, NEET
Two large parallel metal plates carry charges +Q and -Q respectively. A test charge q0 placed between them experiences a force F. If the separation between the plates is doubled, then the force on the test charge will be

Options:
A F
B 2F
C F2
D F4

The correct answer is:
F

Solution:
Initially, the force on charge q0 is
F= q 0 E
The electric field is,
E= Q ε 0 A
If the separation between the plates is doubled Ewill remain same as its independent of distance.
∴F=same

Topic: Physics, Electrostatics, NEET
Two large plane mirrors PM and PN are arrange as shown. The length of the part of large screen SS' in which two images of the object placed at P can be seen is x (in m). Find the value 3x.

Options:

The correct answer is:
2

Solution:
The normals for the mirror PM is PS' and the normal to the mirror PN is PS. From the geometry of the figure it can be seen that SS'=x=23.

So, 3x=2

Topic: Physics, Ray Optics, JEE Main
Two large spherical objects of mass M each (uniformly distributed) are fixed as shown in figure. A small point mass m is projected from point A heading towards center C2 of second sphere. The minimum velocity of point mass so that it can reach upto second object at point B is n3GM5R. Then calculate n. [Neglect other gravitational forces]

Options:

The correct answer is:
08.00

Solution:
If the ball can reach the mid point of line connecting spheres, then it will get attracted towards B. (At midpoint gravitational intensity is zero)
Using energy conservation,
12mV02-GMmR-GMm9R=-2GMm5R+0
Solving, V0=83GM5R
⇒n=8

Topic: Physics, Gravitation, JEE Main
Two lenses $A$ and $B$ having focal lengths $2.0 \mathrm{~cm}$ and $5.0 \mathrm{~cm}$, respectively are placed $14 \mathrm{~cm}$ apart. Lens $A$ is placed to the left of lens $B$. An object is placed $3 \mathrm{~cm}$ to the left of lens $A$. The distance of the image from lens $A$ will be

Options:
A $\frac{40}{3} \mathrm{~cm}$
B $\frac{82}{3} \mathrm{~cm}$
C $\frac{112}{5} \mathrm{~cm}$
D $\frac{92}{5} \mathrm{~cm}$

The correct answer is:
$\frac{82}{3} \mathrm{~cm}$

Solution:
Focal length of lens $A, f_1=2 \mathrm{~cm}$, focal length lens $B_1, f_2=5 \mathrm{~cm}$
The given situation is shown is the figure.
For lens $A$,
$$
u_1=-3 \mathrm{~cm}, v_1=?
$$
From lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$$
\Rightarrow \quad \begin{aligned}
\frac{1}{v_1}+\frac{1}{3} & =\frac{1}{2} \\
v_1 & =6 \mathrm{~cm}
\end{aligned}
$$
Image formed by lens $A$ works as a object for lens $B$.
Hence, for lens $B$
$$
u_2=-(14-6)=-8 \mathrm{~cm}
$$
From lens formula,
$$
\begin{aligned}
& \frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2} \\
& \frac{1}{v_2}+\frac{1}{8}=\frac{1}{5}
\end{aligned}
$$
$$
\begin{array}{ll}
\Rightarrow & \frac{1}{v_2}=\frac{3}{40} \\
\therefore & v_2=\frac{40}{3} \mathrm{~cm}
\end{array}
$$
$\therefore$ Distance of final image $\left(A^{\prime \prime} B^{\prime \prime}\right)$ from lens $A$,
$$
=14+v_2=14+\frac{40}{3}=\frac{82}{3} \mathrm{~cm}
$$

Topic: Physics, Ray Optics, JEE Main
Two lenses of power $-15 \mathrm{~D}$ and $+5 \mathrm{~D}$ are in contact with each other. The focal length of the combination is

Options:
A
$-20 \mathrm{~cm}$

B
$-10 \mathrm{~cm}$

C
$+20 \mathrm{~cm}$

D
$+10 \mathrm{~cm}$

The correct answer is:
$-10 \mathrm{~cm}$

Solution:
$\begin{aligned}
& P=P_1+P_2=-10 \\
& f=\frac{1}{P} \Rightarrow-0.1 \mathrm{~m} \Rightarrow-10 \mathrm{~cm}
\end{aligned}$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2007
Two light balls are suspended as shown in figure. When a stream of air passes through the space between them, the distance between the balls will

Options:
A Remain same
B Increase
C May increase or decrease, depending on speed of air
D Decrease

The correct answer is:
Decrease

Solution:
When a stream of air passes through the space between the balls, the pressure reduces between them as compared to the atmospheric pressure on either side. So, according to Bernoulli’s theorem, the pressure energy should be constant and is compensated by movement of balls toward each other. So, the distance between the balls will decrease.

Topic: Physics, Mechanical Properties of Fluids, MHT CET, MHT CET 2019 (Shift 2)
Two light beams of intensities 4I and 9I interfere on a screen. The phase difference between these beams on the screen at point A is zero and at point B is π. The difference of resultant intensities, at the point A and B, will be _____ I.

Options:

The correct answer is:
24

Solution:
The resultant intensity is given by Inet =I1+I2+2I1I2cosϕ

Now, for Imax, phase angle is ϕ=0 and for Imin, phase angle is ϕ=π.

Thus, Imax=I1+I22=9I+4I2=25I

And Imin=I1-I22=9I-4I2=I

The difference of resultant intensities is Imax-Imin=25I-I=24I

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2022 (29 Jul Shift 1)
Two light beams of intensities in the ratio of 9:4 are allowed to interfere. The ratio of the intensity of maxima and minima will be :

Options:
A 9:4
B 16:81
C 25:169
D 25:1

The correct answer is:
25:1

Solution:
Data provided in the question is, the ratio of the intensities,

I1I2=94

After taking square root of above equation on both sides, we get,

I1I2=32

Apply componendo and dividendo,
⇒I1+I2I1-I2=3+23-2=51
⇒I1+I2I1-I22=ImaxImin=251

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2022 (24 Jun Shift 2)
Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod AB of length l and mass m. The rod is pivoted at its center 'O' and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is:

Options:
A 12π3km
B 12πkm
C 12π6km
D 12π2km

The correct answer is:
12π6km

Solution:
Torque on the rod about O,
Ï=kx.l2Ã2
I Î±â=k.l2.Î¸.l2Ã2
I Î±â=kl22.Î¸
â´ Ï2=kl22Ml212=6kM
f=12Ï6kM

Topic: Physics, Oscillations, JEE Main, JEE Main 2019 (12 Jan Shift 1)
Two light waves having the same wavelength λ in vacuum are in phase initially. Then the first wave travels a path L1 through a medium of refractive index n1while the second wave travels a path of length L2 through a medium of refractive index n2. After this the phase difference between the two waves is:

Options:
A 2πλL2n1-L1n2
B 2πλL1n1-L2n2
C 2πλn1L1-n2L2
D 2πλn2L1-n1L2

The correct answer is:
2πλn1L1-n2L2

Solution:
The optical path between any two points is proportional to the time of travel.

The distance traversed by light in a medium of refractive index μ in time t is given by d=vt .......(i)

Where v is velocity of light in the medium. The distance traversed by light in a vacuum in this time,

s=ct

=c×dv [from equation (i)]

=dcv=μd .....(ii) (Sinceμ=cv)

This distance is the equivalent distance in vacuum and is called optical path.

Here, optical path for first ray =n1L1

Optical path for second ray =n2L2

Path difference =n1L1-n2L2

Now, phase difference =2πλ×path difference

=2πλ×n1L1-n2L2

Topic: Physics, Ray Optics, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Two light waves of wavelengths 800 and 600 nm are used in Young's double slit experiment to obtain interference fringes on a screen placed 7 m away from plane of slits. If the two slits are separated by 0.35 mm, then shortest distance from the central bright maximum to the point where the bright fringes of the two wavelength coincide will be ______ mm.

Options:

The correct answer is:
48

Solution:
Fringe width for both cases can be written as.ω1=λ1Dd & ω2=λ2Dd.

Using values of wavelength, we get ω1=16 mm & ω2=12 mm.

Let y be the common distance of the bright fringes by the both wavelength, then

y=n1ω1=n2ω2.

As LCM ω1,ω2=48 mm, therefore at y=48 mm distance both bright fringes will be found.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Two line segments $\mathrm{AB}$ and $\mathrm{CD}$ are constrained to move along the $\mathrm{x}$ and $\mathrm{y}$ axes, respectively, in such a way that the points A, B, C, D are concyclic. If $A B=a$ and $C D=b$, then the locus of the centre of the circle passing through A, B, C, D in polar coordinates is

Options:
A $r^{2}=\frac{a^{2}+b^{2}}{4}$
B $r^{2} \cos 2 \theta=\frac{a^{2}-b^{2}}{4}$
C $r^{2}=4\left(a^{2}+b^{2}\right)$
D $r^{2} \cos 2 \theta=4\left(a^{2}-b^{2}\right)$

The correct answer is:
$r^{2} \cos 2 \theta=\frac{a^{2}-b^{2}}{4}$

Solution:
$$
\begin{array}{l}
2 \sqrt{g^{2}-c}=a \\
2 \sqrt{f^{2}-c}=b
\end{array}
$$
Polar coordinates of centre of circle be $(\cos \theta, r \sin \theta)$
$$
\begin{array}{ll}
g=-r \cos \theta \quad \text { and } g^{2}-f^{2}=\frac{a^{2}-b^{2}}{4} \\
f=-r \sin \theta \quad \therefore \quad r^{2} \cos 2 \theta=\frac{a^{2}-b^{2}}{4}
\end{array}
$$

Topic: Mathematics, Circle, JEE Main
Two lines $\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}$ and $\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}$

intersect at the point $R$. The reflection of $R$ in the $x y$ - plane has coordinates:

Options:
A (2,-4,-7)
B (2,4,7)
C (2,-4,7)
D (-2,4,7)

The correct answer is:
(2,-4,-7)

Solution:
Let the coordinate of $P$ with respect to line

$\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}=\lambda$

$\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}=\mu$

$L_{1}=(\lambda+3,3 \lambda-1,-\lambda+6)$

and coordinate of $P$ w.r.t.

line $L_{2}=(7 \mu-5,-6 \mu+2,4 \mu+3)$

$\therefore \quad \lambda-7 \mu=-8,3 \lambda+6 \mu=3, \lambda+4 \mu=3$

From above equation : $\lambda=-1, \mu=1$

$\therefore \quad$ Coordinate of point of intersection $R=(2,-4,7)$

Image of $R$ w.r.t. $x y$ plane $=(2,-4,-7)$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (11 Jan Shift 2)
Two lines L1 and L2 of slope 1 are tangent to y2=4x and x2+2y2=4 respectively, such that the distance d units between L1 and L2 is minimum, then the value of d is equal to

Options:
A 3-12
B 3+12
C 2-1
D 3+2

The correct answer is:
3-12

Solution:
y=x+1 is a tangent to y2=4x
y=x±4+2 is a tangent to x24+y22=1
⇒ Minimum distance will be between the parallel lines
y=x+6 and y=x+1
⇒d=6-12=3-12 units

Topic: Mathematics, Ellipse, JEE Main
Two lines $L_{1}: x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$

$\mathrm{L}_{2}: \mathrm{x}=\alpha, \frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha}$ are coplanar. Then, $\alpha \mathrm{can}$ take value (s)

Options:
A 1,4,5
B 1,2,5
C 3,4,5
D 2,4,5

The correct answer is:
1,4,5

Solution:
The equations of given lines can be written

as

$$

\begin{array}{l}

\mathrm{L}_{1}: \mathrm{x}-5=\frac{\mathrm{y}}{3-\alpha}=\frac{\mathrm{z}}{-2} \\

\mathrm{~L}_{2}: \mathrm{x}-\alpha=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha}

\end{array}

$$

Since, these lines are coplanar.

$$

\begin{array}{l}

\text { Therefore, }\left|\begin{array}{ccc}

5-\alpha & 0-0 & 0-0 \\

0 & 3-\alpha & -2 \\

0 & -1 & 2-\alpha

\end{array}\right|=0 \\

\Rightarrow \quad(5-\alpha)(3-\alpha)(2-\alpha)-2=0 \\

\Rightarrow \quad(5-\alpha)\left(6-3 \alpha-2 \alpha+\alpha^{2}-2\right]=0 \\

\Rightarrow \quad(5-\alpha)\left(\alpha^{2}-5 \alpha+4\right]=0 \\

\Rightarrow \quad(5-\alpha)(\alpha-1)(\alpha-4)=0 \\

\Rightarrow \quad \alpha=1,4,5

\end{array}

$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Two lines x-12=y-23=z-34 and x-45=y-12=z1 intersect at a point P. If the distance of P from the plane 2x-3y+6z=7 is Î» units, then the value of 49Î» is equal to

Options:

The correct answer is:
84

Solution:
Point P must lies on both the given lines
From 1st line coordinate of point P2λ+1,3λ+2,4λ+3
This must satisfy 2nd line
i.e. 2λ-35=3λ+12=4λ+31
⇒4λ-6=15λ+5⇒λ=-1
So, coordinates of point P are -1,-1,-1
Hence, the distance of P from the plane is
λ=-2+3-6-77=127 units

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Two liquids $A$ and $B$ are at $32^{\circ} \mathrm{C}$ and $24^{\circ} \mathrm{C}$. When mixed in equal masses the temperature of the mixture is found to be $28^{\circ} \mathrm{C}$. Their specific heats are in the ratio of

Options:
A $3: 2$
B $2: 3$
C $1: 1$
D $4: 3$

The correct answer is:
$1: 1$

Solution:
Temperature of mixture $\theta_{\text {mix }}=\frac{\theta_A c_A+\theta_B c_B}{c_A+c_B}$
$\begin{aligned} \Rightarrow & 28=\frac{32 \times c_A+24 \times c_B}{c_A+c_B} \\ \Rightarrow & 28 c_A+28 c_B=32 c_A+24 c_B \Rightarrow \frac{c_A}{c_B}=\frac{1}{1}\end{aligned}$

Topic: Physics, Thermal Properties of Matter, JEE Main
Two liquids of densities ρ1 and ρ2ρ2=2ρ1 are filled up behind a square wall of side 10m as shown in figure. Each liquid has a height of 5m. The ratio of the forces due to these liquids exerted on upper part MN to that at the lower part NO is (Assume that the liquids are not mixing):

Options:
A 13
B 23
C 12
D 14

The correct answer is:
14

Solution:
F1F2=14

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Two liquids which do not react chemically are placed in a bent tube a shown in the figure. The heights of the liquids above their surface of separation are

Options:
A directly proportional to their densities
B inversely proportional to their densities
C directly proportional to square of their densities
D equal

The correct answer is:
inversely proportional to their densities

Solution:
The pressure at the interface must be same, calculated via either tube. Since, both tube are open to the atmosphere, we must have
h1ρ1g=h2ρ2g
or h1ρ1=h2ρ2
or hρ=constant
or h∝1ρ

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Two liquids $X$ and $Y$ form an ideal solution. At $300 \mathrm{~K}$, vapour pressure of the solution containing 1 mol of $X$ and $3 \mathrm{~mol}$ of $Y$ is $550 \mathrm{~mm} \mathrm{Hg}$. At the same temperature, if $1 \mathrm{~mol}$ of $Y$ is further added to this solution, vapour pressure of the solution increases by $10 \mathrm{~mm} \mathrm{Hg}$. Vapour pressure (in $\mathrm{mmHg}$ ) of $X$ and $Y$ in their pure states will be, respectively :

Options:
A
200 and 300

B
300 and 400

C
400 and 600

D
500 and 600

The correct answer is:
400 and 600

Solution:
$$
\begin{aligned}
& P_T=P_X^0 x_X+P_Y^0 x_Y \\
& x_X=\text { mol fraction of } X \\
& x_Y=\text { mol fraction of } Y \\
& \therefore 550=P_x^{\circ}\left(\frac{1}{1+3}\right)+P_Y^{\circ}\left(\frac{3}{1+3}\right) \\
& =\frac{P_X^0}{4}+\frac{3 P_Y^0}{4} \\
& \therefore 550(4)=P_X^0+3 P_Y^0
\end{aligned}
$$
Further $1 \mathrm{~mol}$ of $Y$ is added and total pressure increases by $10 \mathrm{~mm} \mathrm{Hg}$.
$$
\begin{aligned}
& \therefore 550+10=\mathrm{P}_X^{\circ}\left(\frac{1}{1+4}\right)+\mathrm{P}_Y^{\circ}\left(\frac{4}{1+4}\right) \\
& \therefore 560(5)=\mathrm{P}_{\mathrm{X}}^{\circ}+4 \mathrm{P}_Y^{\circ} \ldots \ldots \ldots \ldots(2)
\end{aligned}
$$
By solving (1) and (2)
We get, $\mathrm{P}_{\mathrm{x}}^{\circ}=400 \mathrm{~mm} \mathrm{Hg}$
$$
P_Y^{\circ}=600 \mathrm{~mm} \mathrm{Hg}
$$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2009
Two long coaxial and conducting cylinders of radius a and $\mathrm{b}$ are separated by a material of conductivity $\sigma$ and a constant potential difference $\mathrm{V}$ is maintained between them, by a battery. Then the current, per unit length of the cylinder flowing from one cylinder to the other is -

Options:
A $\frac{4 \pi \sigma}{\ln (b / a)} V$
B $\frac{4 \pi \sigma}{(b+a)} V$
C $\frac{2 \pi \sigma}{\ln (b / a)} V$
D $\frac{2 \pi \sigma}{(b+a)} V$

The correct answer is:
$\frac{2 \pi \sigma}{\ln (b / a)} V$

Solution:
$\mathrm{V}=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{E} \cdot \mathrm{d} \ell=\frac{\lambda}{2 \pi \varepsilon_{0} \mathrm{r}} \ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right)$

Now, $I=\sigma \int \vec{E} \cdot \overrightarrow{d A}=\sigma \int \frac{\lambda}{2 \pi \varepsilon_{0} r} \cdot 2 \pi d r=\frac{\sigma \lambda}{\varepsilon_{0}}$

From (1)$: I=\frac{2 \sigma \pi \varepsilon_{0}}{\varepsilon_{0} \ln (b / a)}=\frac{2 \pi \sigma}{\ln (b / a)} v$

Topic: Physics, Electrostatics, JEE Main
Two long coaxial cylindrical metal tubes (inner radius a, outer radius b ) stand vertically in a tank of dielectric oil (of mass density ρ, dielectric constant K ). The inner one is maintained at potential V and the outer one is grounded. To what equilibrium height (h) does the oil rise in the space between the tubes? [Assume this height ( h ) as an equilibrium height]

Options:
A ε02V2K-1gρb2-a2 lnba
B ε0V2K-1ρb2-a2g lnba
C 4ε0V2K-1gρb2-a2lnba
D 6ε0V2K-1ρb2-a2g lnba

The correct answer is:
ε0V2K-1ρb2-a2g lnba

Solution:
Gravitational force=mg
= ρπb2- a2gh
Net upward force F =dUdh
=12V2 dCdh
where h is the height of liquid.
Now we calculate C as a function of h
Ceq = CAir + CK
=2πε0L-hlnba+2πε0Khlnba
C = 2πε0L+K-1hlnba
F = 12V2×dCdh =12V2×2πε0K-1lnba
∴ F = mg
12V2×2πε0K-1lnba= ρπb2-a2 gh
h =πε0V2K-1ρπb2-a2g lnba
=ε0V2K-1ρb2-a2g lnba

Topic: Physics, Electrostatics, JEE Main
Two long coaxial solenoids of radius R and 2R have equal number of turns per unit length. They carry time-varying currents i1=2kt and i2=kt respectively, in the same direction. A point charge released between the solenoids at a distance r, is seen to move along a circular path. Then the value of r is

Options:
A r=R2
B r=3R2
C r=R3
D It can be any value between R and 2R

The correct answer is:
r=R2

Solution:
The magnetic field due to the inner solenoid (within its volume) is

B1=μ0n2kt

The magnetic field due to the outer solenoid (within its volume) is

B2=μ0nkt

The flux of magnetic field through the circular area of radius r is

ϕ=B1πR2+B2πr2

⇒ϕ=πμ0nkt2R2+r2

If E→ is the induced electric field at r, then

∮E→.dr→=-dϕdt

⇒E×2πr=πμ0nk2R2+r2

⇒E=μ0nkR2r+r2

The radius of the circle traversed by the charged particle is

r=mvqB2

v=Eqtm=μ0nkR2r+r2qtm

⇒μ0nkR2r+r2qtm=qrB2m

⇒μ0nktR2r+r2=rμ0nkt

⇒R2+r22=r2

⇒r=R2

Topic: Physics, Electromagnetic Induction, JEE Main
Two long concentric cylindrical conductors of radii a and b b < a are maintained at a potential difference V and carry equal and opposite currents I. An electron with a particular velocity u parallel to the axis will travel undeviated in the evacuated region between the conductors. Then u=

Options:
A 4πVμ0I lnba
B 2ÏVÎ¼0I lnab
C 2πVμ0I lnba
D 8πVμ0I lnab

The correct answer is:
2ÏVÎ¼0I lnab

Solution:
E=Î»2ÏÎµ0r
p.d. V=Î»2ÏÎµ0lnab
E=Vrlnab
B varies with r
B=Î¼0I2Ïr
B and E are perpendicular to each other
â´ quB=qE
(If particle travel undeviated)
u=EB=2ÏVÎ¼0I lnab

Topic: Physics, Electrostatics, NEET
Two long concentric cylindrical conductors of radii a and b b < a are maintained at a potential difference V and carry equal and opposite currents I. An electron with a particular velocity u parallel to the axis will travel undeviated in the evacuated region between the conductors. Then u=

Options:
A 4πVμ0I lnba
B 2πVμ0I lnab
C 2πVμ0I lnba
D 8πVμ0I lnab

The correct answer is:
2πVμ0I lnab

Solution:
E=Î»2ÏÎµ0r
p.d. V=Î»2ÏÎµ0lnab
E=Vrlnab
B varies with r
B=Î¼0I2Ïr
B and E are perpendicular to each other
â´ q u B=qE
(If particle travel undeviated)
u=EB=2ÏVÎ¼0I lnab

Topic: Physics, Electrostatics, JEE Main