datavorous/entrance-exam-dataset huggingface dataset:
The solution of the equation $\frac{d y}{d x}=y\left(e^x+1\right)$ is

Options:
A $y+\mathrm{e}^{\left(\mathrm{e}^t+x+c\right)}=0$
B $\log y=e^x+x+c$
C $\log y+e^x=x+c$
D None of these

The correct answer is:
$\log y=e^x+x+c$

Solution:
$\frac{d y}{d x}=y\left(e^x+1\right) \Rightarrow \frac{d y}{y}=\left(e^x+1\right) d x$
Integrating both sides, we get $\log y=e^x+x+c$

Topic: Mathematics, Differential Equations, JEE Main
The solution of the equation $\frac{d y}{d x}+y \tan x=x^m \cos x$ is

Options:
A $(m+1) y=x^{m+1} \cos x+c(m+1) \cos x$
B $m y=\left(x^m+c\right) \cos x$
C $y=\left(x^{m+1}+c\right) \cos x$
D None of these

The correct answer is:
$(m+1) y=x^{m+1} \cos x+c(m+1) \cos x$

Solution:
This is the linear equation of the form $\frac{d y}{d x}+P y=Q$, where $P=\tan x$ and $Q=x^m \cos x$
Now integrating factor (I.F.) $=\mathrm{e}^{\int P d x}=\mathrm{e}^{\int \tan d x}$
$=e^{\log \sec x}=\sec x$
Thus solution is given by, ${ }^{y \cdot e^{\int p d x}}=\int Q \cdot e^{\int p d x} d x+c$
$\begin{aligned} & \Rightarrow y \cdot \sec x=\int x^m \cdot \cos x \cdot \sec x d x+c \Rightarrow y \sec x=\frac{x^{m+1}}{m+1}+c \\ & \Rightarrow(m+1) y=x^{m+1} \cos x+c(m+1) \cos x\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
The solution of the equation sin2â¡x+cos2â¡3x=1 is

Options:
A x=nπ3
B x=nπ+π4
C x=nπ4
D x=nπ+π2

The correct answer is:
x=nπ4

Solution:
sin2⁡x=1-cos2⁡3x=sin2⁡3x
sin2⁡3x-sin2⁡x=0
sin⁡4xsin⁡2x=0
sin⁡2x=0⇒2x=nπ⇒x=nπ2
or
sin⁡4x=0⇒4x=nπ⇒x=nπ4
⇒x=nπ4

Topic: Mathematics, Trigonometric Equations, JEE Main
The solution of the equation tan2⁡x-1-1=1+cos⁡2x is

Options:
A x=nπ-π2
B x=nπ±π4
C x=nπ±π3
D x=nπ

The correct answer is:
x=nπ±π3

Solution:
1tan2⁡x-1=1+cos⁡2x
⇒cos2⁡xsin2⁡x-cos2⁡x=1+cos⁡2x
⇒1+cos⁡2x2-cos⁡2x-1+cos⁡2x=0
1+cos⁡2x-12cos⁡2x-1=0
⇒1+cos⁡2x=0 or 1+12cos⁡2x=0
⇒cos⁡2x=-1 or -12
⇒2x=2nπ±π or 2nπ±2π3
⇒x=nπ±π2 or nπ±π3
But for x=nπ±π2,cos⁡x=0 and hence tan⁡x is not defined
⇒x=nπ±π3

Topic: Mathematics, Trigonometric Equations, JEE Main
The solution of the equation $x^2 \frac{d^2 y}{d x^2}=\ln x$, when $x=1, y=0$ and $\frac{d y}{d x}=-1$ is

Options:
A $\frac{1}{2}(\ln x)^2+\ln x$
B $\frac{1}{2}(\ln x)^2-\ln x$
C $-\frac{1}{2}(\ln x)^2+\ln x$
D $-\frac{1}{2}(\ln x)^2-\ln x$

The correct answer is:
$-\frac{1}{2}(\ln x)^2-\ln x$

Solution:
$\frac{d^2 y}{d x^2}=\frac{\ln x}{x^2} \Rightarrow \frac{d y}{d x}=\frac{-(\ln x+1)}{x}+c$
$\begin{aligned} & \text {At } \frac{d y}{d x}=-1, \quad x=1, y=0, \quad \therefore c=0 \\ & \Rightarrow \quad y=-\int \frac{\ln x+1}{x} d x=-\frac{1}{2}(\ln x)^2-\ln x\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
The solution of the equation $\left(x-4 y^3\right) \frac{d y}{d x}-y=0,(y>0)$ is

Options:
A $x=y^3+c y$
B $x+2 y^3=c y$
C $y=x^3+c x$
D $y+2 x^3=c x$

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

Solution:
Here, the differential equation
$$
\begin{aligned}
& \left(x-4 y^3\right) \frac{d y}{d x}-y=0,(y>0) \\
& \Rightarrow \quad\left(x-4 y^3\right) \frac{d y}{d x}=y \quad \Rightarrow \quad \frac{d x}{d y}=\frac{x-4 y^3}{y} \\
& \Rightarrow \quad \frac{d x}{d y}=\frac{x}{y}-4 y^2 \quad \Rightarrow \quad \frac{d x}{d y}-\frac{x}{y}=-4 y^2 \\
& \therefore \quad P=-\frac{1}{y} \cdot Q=-4 y^2
\end{aligned}
$$
Integrating factor,
$$
\text { I.F. }=e^{\int P d y}=e^{\int-\frac{1}{y} d y}=e^{-\log y}=e^{\log y-1}=\frac{1}{y}
$$
Now, the solution is given by
$$
\begin{aligned}
& x \cdot(\text { I.F. })=\int(\text { I.F. }) \cdot Q d y+C \\
& \frac{x}{y}=\int \frac{1}{y} \times\left(-4 y^2\right) d y+C \\
& \frac{x}{y}=\int-4 y d y+C \\
& \frac{x}{y}=-4 \frac{y^2}{2}+C
\end{aligned}
$$
$$
\begin{aligned}
& x=-2 y^3+C y \\
& x+2 y^3=C y
\end{aligned}
$$
$\therefore \quad$ This is the required solution.

Topic: Mathematics, Differential Equations, JEE Main
The solution of the equation $x \frac{d y}{d x}+3 y=x$ is

Options:
A $x^3 y+\frac{x^4}{4}+c=0$
B $x^3 y=\frac{x^4}{4}+c$
C $x^3 y+\frac{x^4}{4}=0$
D None of these

The correct answer is:
$x^3 y=\frac{x^4}{4}+c$

Solution:
$x \frac{d y}{d x}+3 y=x \quad \Rightarrow \quad \frac{d y}{d x}+\frac{3 y}{x}=1$
It is in the form of $\frac{d y}{d x}+P y=Q$
So, I.F. $=e^{\int P d x}=e^{3 \int \frac{1}{x} d x}=e^{3 \log x}=x^3$
Hence required solution is
$y x^3=\int x^3 1 d x=\frac{x^4}{4}+c \Rightarrow y x^3-\frac{x^4}{4}+c$

Topic: Mathematics, Differential Equations, JEE Main
The solution of the equation z-z=1+2i is

Options:
A 32+2i
B 32-2i
C 3-2i
D None of these

The correct answer is:
32-2i

Solution:
Let z=x+iy
Given, z-z=1+2i
⇒ x2+y2-x+iy=1+2i
⇒ x2+y2-x=1, y=-2
⇒ x2+4-x=1
⇒ x2+4=1+x2
⇒ 2x=3 ⇒ x=32
∴ z=32-2i

Topic: Mathematics, Complex Number, JEE Main
The solution of the initial value problem \( (2 \ln x) \frac{d y}{d x}+\frac{y}{x}=\frac{1}{y} \cos x, y > 0, x>1 \) and \( y\left(\frac{3 \pi}{2}\right)=0 \) is given by which of the following options?

Options:
A \( y=a \sqrt{\frac{1-\sin x}{\ln x}} \)
B \( y=a \sqrt{\frac{1+\sin x}{\ln x}} \)
C \( y=a \sqrt{\frac{1-\cos x}{\ln x}} \)
D \( y=a \sqrt{\frac{1+\cos x}{\ln x}} \)

The correct answer is:
\( y=a \sqrt{\frac{1+\sin x}{\ln x}} \)

Solution:
Given differential equation is
2ydydx+y21xlnx=cosxlnx
⇒ dzdx+zxlnx=cosxlnx
On substituting y2=z and 2ydydx=dzdx, we get

⇒ dzdx+zxlnx=cosxlnx
Which is a LDE,
∴ y2lnx=zlnx=sinx+C
Now, y3π2=0 ⇒C=1
∴ y=±1+sinxlnx
y=1+sinxlnx ∵ y>0

Required solution y=a1+sinxlnx where a is any constant.

Topic: Mathematics, Differential Equations, JEE Main
The solution of the system of equations x+cosâ¡Î±y=1 and cosâ¡Î±x+4y=2 satisfy xâ¥45 and yâ¤12, then the value of Î± can lie in the interval

Options:
A α∈π4,π3
B α∈0,π6
C α∈π6,π3
D α∈π3,π2

The correct answer is:
α∈π3,π2

Solution:
Δ=1cos⁡αcos⁡α4=4-cos2α

Δ1=1cos⁡α24=22-cos⁡α
Δ2=11cos⁡α2=2-cos⁡α
∵ x=22+cos⁡α≥45⇒5≥2cos⁡α+4⇒cos⁡α≤12

∵ y=12+cos⁡α≤12⇒2+cos⁡α≥2⇒cos⁡α≥0

⇒α∈π3,π2

Topic: Mathematics, Determinants, JEE Main
The solution of $x d x+y d y=x^2 y d y-x y^2 d x$ is

Options:
A $x^2-1=C\left(1+y^2\right)$
B $x^2+1=C\left(1-y^2\right)$
C $x^2-1=C\left(1-y^2\right)$
D $x^2+1=C\left(1-y^2\right)$

The correct answer is:
$x^2-1=C\left(1+y^2\right)$

Solution:
We have,
$\begin{aligned} x d x+y d y & =x^2 y d y-x y^2 d x \\ \Rightarrow \quad x d x+x y^2 d x & =x^2 y d y-y d y \\ \Rightarrow \quad\left(1+y^2\right) x d x & =-\left(1-x^2\right) y d y\end{aligned}$
On integrating both sides,
$\begin{aligned} \int \frac{2 x d x}{-1+x^2} & =\int \frac{2 y d y}{1+y^2} \\ \Rightarrow \quad \log \left(-1+x^2\right) & =\log \left(1+y^2\right)+\log c \\ \Rightarrow \quad x^2-1 & =C\left(1+y^2\right)\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
The solution of $\left(y-3 x^2\right) d x+x d y=0$ is

Options:
A $y(x)=\sin x+\frac{1}{x^2}+C$
B $y(x)=\cos x-\frac{1}{x^2}+C$
C $y(x)=x^2+\frac{C}{x}$
D $y(x)=\sqrt{x}+\frac{C}{x}$

The correct answer is:
$y(x)=x^2+\frac{C}{x}$

Solution:
We have,
$$
\begin{array}{rlrl}
& & \left(y-3 x^2\right) d x+x d y & =0 \\
\Rightarrow & y d x-3 x^2 d x+x d y & =0 \\
\Rightarrow & & y d x+x d y & =3 x^2 d x \\
\Rightarrow & & d x y & =3 x^2 d x
\end{array}
$$
On integrating both sides, we get
$$
x y=x^3+C \Rightarrow y=x^2+\frac{C}{x}
$$

Topic: Mathematics, Differential Equations, JEE Main
The solution of $y d x-x d y+3 x^2 y^2 e^{x^3} d x=0$ is

Options:
A $\frac{x}{y}+e^{x^3}=c$
B $\frac{x}{y}-e^{x^3}=0$
C $\frac{-x}{y}+e^{x^3}=0$
D None of these

The correct answer is:
$\frac{x}{y}+e^{x^3}=c$

Solution:
$\begin{aligned}
& y d x-x d y+3 x^2 y^2 e^{x^3} d x=0 \\
& \frac{y d x-x d y}{y^2}+3 x^2 e^{x^3} d x=0 \Rightarrow d\left(\frac{x}{y}\right)+d e^{x^3}=0
\end{aligned}$
On integrating, we get $\frac{x}{y}+e^{x^3}=c$

Topic: Mathematics, Differential Equations, JEE Main
The solution set of $(5+4 \cos \theta)(2 \cos \theta+1)=0$ in the interval $[0,2 \pi]$, is :

Options:
A $\left\{\frac{\pi}{3}, \frac{2 \pi}{3}\right\}$
B $\left\{\frac{\pi}{3}, \pi\right\}$
C $\left\{\frac{2 \pi}{3}, \frac{4 \pi}{3}\right\}$
D $\left\{\frac{2 \pi}{3}, \frac{5 \pi}{3}\right\}$

The correct answer is:
$\left\{\frac{2 \pi}{3}, \frac{4 \pi}{3}\right\}$

Solution:
We have,
$(5+4 \cos \theta)(2 \cos \theta+1)=0$...(i)
$\cos \theta=\frac{1-\tan ^2 \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}$
$\therefore \quad \cos \theta=\frac{1-t^2}{1+t^2} \quad\left[\right.$ put $\left.\tan \frac{\theta}{2}=t\right]$
Then, Eq. (i) becomes
$\left[5+4\left(\frac{1-t^2}{1+t^2}\right)\right]\left[2\left(\frac{1-t^2}{1+t^2}\right)+1\right]=0$
$\Rightarrow\left[5+5 t^2+4-4 t^2\right]\left[2-2 t^2+1+t^2\right]=0$
$\Rightarrow \quad\left(t^2+9\right)\left(3-t^2\right)=0$
$\therefore \quad t= \pm \sqrt{3}$
$\Rightarrow \quad \tan \frac{\theta}{2}=\sqrt{3}$ or $\tan \frac{\theta}{2}=-\sqrt{3}$
$\Rightarrow \quad \frac{\theta}{2}=\frac{\pi}{3}$ or $\frac{\theta}{2}=\frac{2 \pi}{3}$
$\therefore \quad \theta=\frac{2 \pi}{3}$ or $\frac{4 \pi}{3}$

Topic: Mathematics, Trigonometric Equations, JEE Main
The solution set of the equation 4sin⁡θcos⁡θ-2cos⁡θ-23sin⁡θ+3=0 in the interval 0,2π is

Options:
A 3π4,7π4
B π3,5π3
C 3π4,π,π3,5π3
D π6,5π6,11π6

The correct answer is:
π6,5π6,11π6

Solution:
4sin⁡θcos⁡θ-2cos⁡θ-23sin⁡θ+3=0
2cos⁡θ2sin⁡θ-1-32sin⁡θ-1=0
2cos⁡θ-32sin⁡θ-1=0
⇒sin⁡θ=12,cos⁡θ=32
θ=π6,5π6,11π6

Topic: Mathematics, Trigonometric Equations, JEE Main
The solution set of the equation $\sin ^{-1} x=2 \tan ^{-1} x$ is

Options:
A $\{1,2\}$
B $\{-1,2\}$
C $\{-1,1,0\}$
D $\{1,1 / 2,0\}$

The correct answer is:
$\{-1,1,0\}$

Solution:
$\begin{array}{ll}\sin ^{-1} x=2 \tan ^{-1} x \Rightarrow \sin ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^2} \\ \Rightarrow \frac{2 x}{1+x^2}=x \Rightarrow x^3-x=0 \\ \Rightarrow x(x+1)(x-1)=0 \Rightarrow x=\{-1,1,0\}\end{array}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The solution set of the inequality tan-1⁡x2≤tan-1⁡x+6 is

Options:
A -tan⁡2,tan⁡3
B tan⁡2,-tan⁡3
C tan⁡2,tan⁡3
D -∞,∞

The correct answer is:
-∞,∞

Solution:
Let tan −1 x=u
So, inequality becomes u 2 −u−6≤0
⇒−2≤u≤3
So, −2≤ tan −1 x≤3
As − π 2 < tan −1 x< π 2
⇒−2≤ tan −1 x≤3, ∀x∈R
Therefore, the solution set of the inequality tan-1⁡x2≤tan-1⁡x+6 is -∞,∞

Topic: Mathematics, Functions, JEE Main
The solution set of x∈-π,π for the inequality sin⁡2x+1≤cos⁡x+2sin⁡x is

Options:
A x∈0,π6
B x∈π6,5π6∪0
C x∈-π6,5π6
D x∈-π2,π2

The correct answer is:
x∈π6,5π6∪0

Solution:
2sin⁡xcos⁡x+1-cos⁡x-2sin⁡x≤0
2sin⁡x-1cos⁡x-1≤0
Case I: cos⁡x=1⇒x=0
Case II: otherwise cos⁡x<1
⇒2sin⁡x-1≥0⇒sin⁡x≥12
⇒x∈π6,5π6
Hence, from case I and case II,
x∈π6,5π6∪0

Topic: Mathematics, Trigonometric Equations, JEE Main
The solution set of \( x \in(-\pi, \pi) \) for the inequality \( \sin 2 x+1 \leq \cos x+2 \sin x \) is

Options:
A \( x \in\left[0, \frac{\pi}{6}\right] \)
B \( x \in\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right] \cup\{0\} \)
C \( x \in\left[-\frac{\pi}{6}, \frac{5 \pi}{6}\right] \)
D \( x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)

The correct answer is:
\( x \in\left[\frac{\pi}{6}, \frac{5 \pi}{6}\right] \cup\{0\} \)

Solution:
2sin⁡xcos⁡x+1-cos⁡x-2sin⁡x≤0
2sin⁡x-1cos⁡x-1≤0
Case I: cos⁡x=1⇒x=0
Case II: otherwise cos⁡x<1
⇒2sin⁡x-1≥0⇒sin⁡x≥12
⇒x∈π6,5π6
Hence, from case I and case II,
x∈π6,5π6∪0

Topic: Mathematics, Trigonometric Equations, JEE Main
The solutions of the equation 1+sin2xsin2xsin2xcos2x1+cos2xcos2x4sin2x4sin2x1+4sin2x=0,0<x<π, are

Options:
A π12,π6
B π6,5π6
C 5π12,7π12
D 7π12,11π12

The correct answer is:
7π12,11π12

Solution:
1+sin2xsin2xsin2xcos2x1+cos2xcos2x4sin2x4sin2x1+4sin2x=0

use R1→R1+R2+R3

⇒2+4sin2x111cos2x1+cos2xcos2x4sin2x4sin2x1+4sin2x=0

use C1→C1-C3 & C2→C2-C3

⇒2+4sin2x00101cos2x-1-11+4sin2x=0

⇒2+4sin2x=0

⇒sin2x=-12

⇒2x=π+π6,2π-π6

x=π2+π12,π-π12

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The solutions of $(x+y+1) d y=d x$ are

Options:
A $x+y+2=C e^{y}$
B $x+y+4=C \log y$
C $\quad \log (x+y+2)=C y$
D $\log (x+y+2)=C-y$

The correct answer is:
$x+y+2=C e^{y}$

Solution:
Putting $x+y+1=u,$ we have $d u=d x+d y$ and the given equation reduces to $u(d u-d x)=$ $d x$

$\Rightarrow \quad \frac{u d u}{u+1}=d x \quad \Rightarrow \quad u-\log (u+1)=x$

$\Rightarrow \log (x+y+2)=y+$ constant

$\Rightarrow x+y+2=C e^{y}$

Topic: Mathematics, Differential Equations, JEE Main
The solutions set of inequation \(\cos ^{-1} x < \sin ^{-1} x\) is

Options:
A \([-1,1]\)
B \(\left[\frac{1}{\sqrt{2}}, 1\right]\)
C \([0,1]\)
D \(\left(\frac{1}{\sqrt{2}}, 1\right)\)

The correct answer is:
\(\left(\frac{1}{\sqrt{2}}, 1\right)\)

Solution:
Hints: \(\cos ^{-1} x < \sin ^{-1} x\)
\(x \in\left(\frac{1}{\sqrt{2}}, 1\right), \quad \cos ^{-1} x < \sin ^{-1} x\)

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The source of energy in a cellular reaction is

Options:
A Light energy
B Solar radiation
C Chemical energy
D Heat energy

The correct answer is:
Chemical energy

Solution:
Cellular reaction mainly refers to metabolism which is essentially bond making and bond breaking and chemical energy is the source that fuels such reactions.

Topic: Chemistry, Environmental Chemistry, NEET
The source of time varying magnetic field may be

(A) a permanent magnet

(B) an electric field changing linearly with time

(C) direct current

(D) a decelerating charge particle

(E) an antenna fed with a digital signal

Choose the correct answer from the options given below.

Options:
A (C) and (E) only
B (D) only
C (A) only
D (B) and (D) only

The correct answer is:
(D) only

Solution:
An electric field varying linearly with time will not produce variable magnetic field as current will be constant.

A decelerating charged particle can result in a time varying magnetic field by emitting radiation.

Similarly a permanent magnet has a constant magnetic field and a direct current does not change its direction and cannot produce a magnetic field.

An antenna can modulate the digital signals .

So, the right option is the statement D.

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2023 (13 Apr Shift 1)
The source that illuminates the double $-$ slit in 'double - slit interference experiment' emits two distinct monochromatic waves of wavelength $500\mathrm{~nm}$ and $600 \mathrm{~nm}$, each of them producing its own pattern on the screen. At the central point of the pattern when path difference is zero, maxima of both the patterns coincide and the resulting interference pattern is most distinct at the region of zero path difference. But as one moves out of this central region, the two fringe systems are gradually out of step such that maximum due to on wavelength coincides with the minimum due to the other and the combined fringe system becomes completely indistinct. This may happen when path difference in $\mathrm{nm}$ is:

Options:
A
$2000$

B
$3000$

C
$1000$

D
$1500$

The correct answer is:
$1500$

Solution:
No Solution Available

Topic: Physics, Wave Optics, JEE Main, JEE Main 2013 (25 Apr Online)
The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation:
$$
\mathrm{K}(\mathrm{x})=\mathrm{K}_{\mathrm{o}}+\lambda \mathrm{x}(\lambda=\mathrm{a} \text { constant })
$$
The capacitance $\mathrm{C}$, of the capacitor, would be related to its vacuum capacitance $\mathrm{C}_{\mathrm{o}}$ for the relation :

Options:
A
$\left.\mathrm{C}=\frac{\lambda \mathrm{d}}{\ln \left(1+\mathrm{K}_{\mathrm{o}} \lambda \mathrm{d}\right.} \mathrm{C}_{\mathrm{o}}\right)$

B
$\left.\mathrm{C}=\frac{\lambda}{\mathrm{d} \cdot \ln \left(1+\mathrm{K}_{\mathrm{o}} \lambda \mathrm{d}\right.} \mathrm{C}_{\mathrm{o}}\right)$

C
$\left.\mathrm{C}=\frac{\lambda \mathrm{d}}{\ln \left(1+\lambda \mathrm{d} / \mathrm{K}_{\mathrm{o}}\right.} \mathrm{C}_{\mathrm{o}}\right)$

D
$\left.\mathrm{C}=\frac{\lambda}{\mathrm{d} . \ln \left(1+\mathrm{K}_{\mathrm{o}} / \lambda \mathrm{d}\right.} \mathrm{C}_{\mathrm{o}}\right)$

The correct answer is:
$\left.\mathrm{C}=\frac{\lambda \mathrm{d}}{\ln \left(1+\lambda \mathrm{d} / \mathrm{K}_{\mathrm{o}}\right.} \mathrm{C}_{\mathrm{o}}\right)$

Solution:
The value of dielectric constant is given as,
$$
\begin{aligned}
&\mathrm{K}=\mathrm{K}_0+\lambda \mathrm{x} \\
&\text { And, } \mathrm{V}=\int_0^{\mathrm{d}} \mathrm{Edr} \\
&\mathrm{V}=\int_0^{\mathrm{d}} \frac{\sigma}{\mathrm{K}} \mathrm{dx}
\end{aligned}
$$
$$
\begin{aligned}
&=\sigma \int_0^{\mathrm{d}} \frac{1}{\left(\mathrm{~K}_0+\lambda \mathrm{x}\right.} \mathrm{dx} \\
&=\frac{\sigma}{\lambda}\left[\ln \left(\mathrm{K}_0+\lambda \mathrm{d}-\ln \mathrm{K}_0\right]\right. \\
&=\frac{\sigma}{\lambda} \ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)
\end{aligned}
$$
Now it is given that capacitance of vacuum $=\mathrm{C}_0$.
Thus, $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}$
$$
\begin{aligned}
&=\frac{\sigma . \mathrm{s}}{\mathrm{v}}(\text { Let surface area of plates }=\mathrm{s}) \\
&=\frac{\sigma . \mathrm{s}}{\frac{\sigma}{\lambda} \ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)}
\end{aligned}
$$
$=\mathrm{s} \lambda \cdot \frac{\mathrm{d}}{\mathrm{d}} \frac{1}{\ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)}\left(\because\right.$ in vacuum $\left.\varepsilon_0=1\right)$
$\mathrm{c}=\frac{\lambda \mathrm{d}}{\ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)} \cdot \mathrm{C}_0 \quad\left(\right.$ here, $\left.\mathrm{C}_0=\frac{\mathrm{s}}{\mathrm{d}}\right)$

Topic: Physics, Capacitance, JEE Main
The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation:
$$
\mathrm{K}(\mathrm{x})=\mathrm{K}_{\mathrm{o}}+\lambda \mathrm{x}(\lambda=\mathrm{a} \text { constant })
$$
The capacitance $\mathrm{C}$, of the capacitor, would be related to its vacuum capacitance $\mathrm{C}_{\mathrm{o}}$ for the relation :

Options:
A
$\left.\mathrm{C}=\frac{\lambda \mathrm{d}}{\ln \left(1+\mathrm{K}_{\mathrm{o}} \lambda \mathrm{d}\right.} \mathrm{C}_{\mathrm{o}}\right)$

B
$\left.\mathrm{C}=\frac{\lambda}{\mathrm{d} \cdot \ln \left(1+\mathrm{K}_{\mathrm{o}} \lambda \mathrm{d}\right.} \mathrm{C}_{\mathrm{o}}\right)$

C
$\left.\mathrm{C}=\frac{\lambda \mathrm{d}}{\ln \left(1+\lambda \mathrm{d} / \mathrm{K}_{\mathrm{o}}\right.} \mathrm{C}_{\mathrm{o}}\right)$

D
$\left.\mathrm{C}=\frac{\lambda}{\mathrm{d} . \ln \left(1+\mathrm{K}_{\mathrm{o}} / \lambda \mathrm{d}\right.} \mathrm{C}_{\mathrm{o}}\right)$

The correct answer is:
$\left.\mathrm{C}=\frac{\lambda \mathrm{d}}{\ln \left(1+\lambda \mathrm{d} / \mathrm{K}_{\mathrm{o}}\right.} \mathrm{C}_{\mathrm{o}}\right)$

Solution:
The value of dielectric constant is given as,
$$
\begin{aligned}
&\mathrm{K}=\mathrm{K}_0+\lambda \mathrm{x} \\
&\text { And, } \mathrm{V}=\int_0^{\mathrm{d}} \mathrm{Edr} \\
&\mathrm{V}=\int_0^{\mathrm{d}} \frac{\sigma}{\mathrm{K}} \mathrm{dx}
\end{aligned}
$$
$$
\begin{aligned}
&=\sigma \int_0^{\mathrm{d}} \frac{1}{\left(\mathrm{~K}_0+\lambda \mathrm{x}\right.} \mathrm{dx} \\
&=\frac{\sigma}{\lambda}\left[\ln \left(\mathrm{K}_0+\lambda \mathrm{d}-\ln \mathrm{K}_0\right]\right. \\
&=\frac{\sigma}{\lambda} \ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)
\end{aligned}
$$
Now it is given that capacitance of vacuum $=\mathrm{C}_0$.
Thus, $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}$
$$
\begin{aligned}
&=\frac{\sigma . \mathrm{s}}{\mathrm{v}}(\text { Let surface area of plates }=\mathrm{s}) \\
&=\frac{\sigma . \mathrm{s}}{\frac{\sigma}{\lambda} \ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)}
\end{aligned}
$$
$=\mathrm{s} \lambda \cdot \frac{\mathrm{d}}{\mathrm{d}} \frac{1}{\ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)}\left(\because\right.$ in vacuum $\left.\varepsilon_0=1\right)$
$\mathrm{c}=\frac{\lambda \mathrm{d}}{\ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)} \cdot \mathrm{C}_0 \quad\left(\right.$ here, $\left.\mathrm{C}_0=\frac{\mathrm{s}}{\mathrm{d}}\right)$

Topic: Physics, Capacitance, JEE Main, JEE Main 2014 (12 Apr Online)
The space between the two large parallel plates is filled with a material of uniform charge density $\rho$. Assume that one of the plate is kept at $x=0$. The potential at any point $x$ between these plates is given by ( $A$ and $B$ are constants).

Options:
A $-\frac{\rho x^3}{2 \epsilon_0}$
B $-\left(\frac{\rho x^2}{2 \epsilon_0}+A x\right)$
C $-\left(\frac{\rho x^2}{2 \epsilon_0}+A x+B\right)$
D $-\left(\frac{\rho x^3}{4 \epsilon_0}+A x^2+B x\right)$

The correct answer is:
$-\left(\frac{\rho x^2}{2 \epsilon_0}+A x+B\right)$

Solution:
According to the question, the space between the two large parallel plates is filled with a material of uniform charge density pi.e., It is a homogeneous medium, therefore the potential at any point between these plates is calculated by Poisson's equation as given below,
$$
\nabla^2 V=-\frac{\rho}{\varepsilon_0}
$$
Since, we have to calculate the potential at any point along $x$-direction, hence the derivative of the potential along $y$ and $z$ directions is zero.
Hence, from Eq. (i), we get
$$
\frac{d^2 V}{d x^2}=-\frac{\rho}{\varepsilon_0}
$$
Integrate on both the sides, we get
$$
\Rightarrow \int \frac{d^2 V}{d x^2} d x=-\int \frac{\rho}{\varepsilon_0} d x \Rightarrow \frac{d V}{d x}=\frac{-\rho x}{\varepsilon_0}-A
$$
Integrate on both sides again, we get
$$
\begin{array}{rlrl}
\Rightarrow & \int \frac{d V}{d x} d x & =\int \frac{-\rho x}{\varepsilon_0} d x-\int A d x \\
V & =\frac{-\rho x^2}{2 \varepsilon_0}-A x-B \\
\therefore & V & =-\left(\frac{+\rho x^2}{2 \varepsilon_0}+A x+B\right)
\end{array}
$$

Topic: Physics, Electrostatics, JEE Main
The space inside a straight current carrying solenoid is filled with a magnetic material having magnetic susceptibility equal to 1.2×10-5. What is fractional increase in the magnetic field inside solenoid with respect to air as medium inside the solenoid?

Options:
A 1.2×10-5
B 1.2×10-3
C 1.8×10-3
D 2.4×10-5

The correct answer is:
1.2×10-5

Solution:
Given here, χ=1.2×10-5

Magnetic field inside the material in solenoid is B'=μ01+χni, here, i is current.

Now, magnetic field with air as medium inside solenoid is B=μ0ni.

So, fractional increase is B'-BB=χ=1.2×10-5

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2022 (28 Jun Shift 2)
The space wave propagation is utilized in

Options:
A television communication
B radar communication
C microwave communication
D all of the above

The correct answer is:
all of the above

Solution:
The radio waves having high frequencies are called space wave. These waves have the ability to travel through atmosphere, this is called space wave mode of propagation. At such high frequencies other method of wave propagation can't work, therefore space wave propagation is used. As television broadcast, microwave and satellite communication all use the high frequency space waves. Hence, all these use space wave mode of propagation.

Topic: Physics, Communication System, JEE Main
The space wave propagation is utilized in

Options:
A Can be reflected by any layer of earth's atmosphere
B Can be reflected by ionosphere
C Cannot be reflected by mesosphere
D Cannot be reflected by any layer of earth's atmosphere

The correct answer is:
Cannot be reflected by any layer of earth's atmosphere

Solution:
The space wave communication is utilized in all the three communication ie, television, radar and microwave communication

Topic: Physics, Communication System, JEE Main