datavorous/entrance-exam-dataset huggingface dataset:
A disk of radius $\frac{a}{4}$ having a uniformly distributed charge $6 \mathrm{C}$ is placed in the $x-y$ plane with its centre at $\left(\frac{-a}{2}, 0,0\right)$. A rod of length a carrying a uniformly distributed charge $8 \mathrm{C}$ is placed on the $x$-axis from $x=\frac{a}{4}$ to $x=\frac{5 a}{4}$. Two point charges $-7 \mathrm{C}$ and $3 \mathrm{C}$ are placed at $\left(\frac{a}{4}, \frac{-a}{4}, 0\right)$ and $\left(\frac{-3 a}{4}, \frac{3 a}{4}, 0\right)$, respectively. Consider a cubical surface formed by six surfaces $x=\pm \frac{a}{2}, y=\pm \frac{a}{2}, z=\pm \frac{a}{2}$.
The electric flux through this cubical surface is

Options:
A
$\frac{-2 C}{\varepsilon_0}$

B
$\frac{2 C}{\varepsilon_0}$

C
$\frac{10 C}{\varepsilon_0}$

D
$\frac{12 C}{\varepsilon_0}$

The correct answer is:
$\frac{-2 C}{\varepsilon_0}$

Solution:
Total enclosed charge as already shown is
$$
q_{\text {net }}=\frac{6 C}{2}+\frac{8 C}{4}-7 C=-2 C
$$
From Gauss-theorem, net flux,
$$
\varphi_{\text {net }}=\frac{q_{\text {net }}}{\varepsilon_0}=\frac{-2 C}{\varepsilon_0}
$$

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2009 (Paper 1)
A disk of radius R with uniform positive charge density σ is placed on the xy plane with its center at the origin. The Coulomb potential along the z-axis is
Vz=σ2ϵ0R2+z2-z

A particle of positive charge q is placed initially at rest at a point on the z-axis with z=z0 and z0>0. In addition to the Coulomb force, the particle experiences a vertical force F→=-ck^ with c>0. Let β=2cε0qσ. Which of the following statement(s) is(are) correct?

Options:
A For β=14 and z0=257R, the particle reaches the origin.
B For β=14 and z0=37R, the particle reaches the origin.
C For β=14 and z0=R3, the particle returns back to z=z0
D For β>1 and z0>0, the particle always reaches the origin.

The correct answers are:
For β=14 and z0=257R, the particle reaches the origin., For β=14 and z0=R3, the particle returns back to z=z0, For β>1 and z0>0, the particle always reaches the origin.

Solution:
For the particle to reach at origin, work done by the force should be greater than the increase in the potential energy of the charged particle. Therefore,

cz≥σq2ϵ0R-R2+z2-z⇒c2ϵ0σq≥R-R2+z2-zz⇒β≥R-R2+z2-zz

For option A

R-R2+z2-zz=R-R1+62549-257R257≃0.242

Since, 0.25>0.242

Option A is correct.

For option B

R-R2+z2-zz=R-R1+949-37R37≃0.79

Since, 0.25<0.79

Option B is incorrect

For option C

R-R2+z2-zz=R-R1+13-13R13≃0.73

Since, 0.25<0.73

Therefore, the particle returns to z0

Hence, option C is correct.

For option D,

For any value of z>0

R-R2+z2-zz<1

Hence, option D is correct.

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2022 (Paper 2)
A double star system consists of two stars $A$ and $B$ which have time period $T_A$ and $T_B$. Radius $R_A$ and $R_B$ and mass $M_A$ and $M_B$. Choose the correct option.

Options:
A
If $T_A>T_B$, then $R_A>R_B$

B
If $T_A>T_B$, then $M_A>M_B$

C
$\left(\frac{T_A}{T_B}\right)^2=\left(\frac{R_A}{R_B}\right)^3$

D
$T_A=T_B$

The correct answer is:
$T_A=T_B$

Solution:
In case of binary star system angular velocity and hence the time period of both the stars are equal.

Topic: Physics, Gravitation, JEE Advanced, JEE Advanced 2006
A few electric field lines for a system of two charges $Q_1$ and $Q_2$ fixed at two different points on the $x$-axis are shown in the figure. These lines suggest that

Options:
A
$\left|Q_1\right|>\left|Q_2\right|$

B
$\left|Q_1\right| < \left|Q_2\right|$
C
at a finite distance to the left of $Q_1$ the electric field is zero

D
at a finite distance to the right of $Q_2$ the electric field is zero

The correct answers are:
$\left|Q_1\right|>\left|Q_2\right|$
,
at a finite distance to the right of $Q_2$ the electric field is zero

Solution:
From the behaviour of electric lines, we can say that $Q_1$ is positive and $Q_2$ is negative. Further, $\left|Q_1\right|>\left|Q_2\right|$
At some finite distance to the right of $Q_2$, electric field will be zero. Because electric field due to $Q_1$ is towards right (away from $Q_1$ ) and due to $Q_2$ is towards left (towards $Q_2$ ). But since magnitude of $Q_1$ is more, the two fields may cancel each other because distance of that point from $Q_1$ will also be more.
$\therefore$ The correct options are (a) and (d).

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2010 (Paper 1)
A field line is shown in the figure. This field cannot represent

Options:
A
Magnetic field

B
Electrostatic field

C
Induced electric field

D
Gravitational field

The correct answers are:
Magnetic field
,
Gravitational field

Solution:
Electrostatic and gravitational field do not make closed loops.

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2006
A gas described by van der Waals' equation

Options:
A
behaves similar to an ideal gas in the limit of large molar volumes

B
behaves similar to an ideal gas in the limit of large pressures

C
is characterised by van der Waals' coefficients that are dependent on the identity of the gas but are independent of the temperature

D
has the pressure that is lower than the pressure exerted by the same gas behaving ideally

The correct answers are:
behaves similar to an ideal gas in the limit of large pressures
,
is characterised by van der Waals' coefficients that are dependent on the identity of the gas but are independent of the temperature
,
has the pressure that is lower than the pressure exerted by the same gas behaving ideally

Solution:
The gas described by van der Waals' equation in the limit of large molar volume, in the intermediate range of pressure experience balancing by two factors.
$$
\left(p+\frac{a}{V^2}\right)(V-b)=R T \Rightarrow p V=\frac{a}{V}-p b+\frac{a b}{V^2}=R T
$$
(We can neglect the product of two very small constants, $\frac{a b}{V^2}$ )
So,
$$
p V+\frac{a}{V}-p b=R T
$$
As
$$
\frac{a}{V}=p b
$$
The gas behaves ideally, $p V=R T$
- The van der Waals' constants ' $a$ ' and ' $b$ ' are characteristics of the gas. The value of these constants are determined by the critical constants of the gas
$$
a=3 p_C V_C^2=3 p_C\left(\frac{3 R T_C}{8 p_C}\right)^2=\frac{27\left(R T_C\right)^2}{64 p_C} ; b=\frac{1}{3}\left(\frac{3 R T_C}{8 p_C}\right)=\frac{R T_C}{8 p_C}
$$
Question is based on description by van der Waals' equation, that's why we have to ignore the effect of temperature on $a$ and $b$. Actually, the so-called constant varies to some extent with temperature and this shows that the van der Waals' equation is not a complete solution of the behaviour of real gas

Topic: Chemistry, States of Matter, JEE Advanced, JEE Advanced 2008 (Paper 1)
A glass tube of uniform internal radius $(r)$ has a valve separating the two identical ends. Initially, the valve is in a tightly closed position. End 1 has a hemispherical soap bubble of radius $r$. End 2 has sub-hemispherical soap bubble as shown in figure. Just after opening the valve,

Options:
A
air from end 1 flows towards end 2. No change in the volume of the soap bubbles

B
air from end 1 flows towards end 2. Volume of the soap bubble at end 1 decreases

C
no change occurs

D
air from end 2 flows towards end 1. Volume of the soap bubble at end 1 increases

The correct answer is:
air from end 1 flows towards end 2. Volume of the soap bubble at end 1 decreases

Solution:
$\Delta p_1=\frac{4 T}{r_1}$ and $\Delta p_2=\frac{4 T}{r_2}$
$$
\begin{array}{cc}
& r_1 < r_2 \\
\therefore & \Delta p_1>\Delta p_2
\end{array}
$$
Air will flow from 1 to 2 and volume of bubble at end- 1 will decrease.
Therefore, correct option is (b).

Topic: Physics, Mechanical Properties of Fluids, JEE Advanced, JEE Advanced 2008 (Paper 2)
A hollow pipe of length $0.8 \mathrm{~m}$ is closed at one end. At its open end a $0.5 \mathrm{~m}$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is $50 \mathrm{~N}$ and the speed of sound is $320 \mathrm{~ms}^{-1}$, the mass of the string is

Options:
A
$5 \mathrm{~g}$

B
$10 \mathrm{~g}$

C
$20 \mathrm{~g}$

D
$40 \mathrm{~g}$

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

Solution:
$2\left(\frac{v_1}{2 l_1}\right)=\frac{v_2}{4 l_2}$ $\therefore \quad \frac{\sqrt{T / \mu}}{l_1}=\frac{320}{4 l_2}$
( $\mu=$ mass per unit length of wire)
or $\frac{\sqrt{50 / \mu}}{0.5}=\frac{320}{4 \times 0.8}$
Solving we get $\mu=0.02 \frac{\mathrm{kg}}{\mathrm{m}}=20 \frac{\mathrm{g}}{\mathrm{m}}$
$\therefore$ Mass of string
$$
=\left(20 \frac{\mathrm{g}}{\mathrm{m}}\right)(0.5 \mathrm{~m})=10 \mathrm{~g}
$$
$\therefore$ The correct option is (b).

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2010 (Paper 2)
A horizontal force F is applied at the centre of mass of a cylindrical object of mass m and radius R, perpendicular to its axis as shown in the figure. The coefficient of friction between the object and the ground is μ. The centre of mass of the object has an acceleration a. The acceleration due to gravity is g. Given that the object rolls without slipping, which of the following statement(s) is (are) correct?

Options:
A For the same F, the value of a does not depend on whether the cylinder is solid or hollow.
B For a solid cylinder, the maximum possible value of a is 2μg.
C The magnitude of the frictional force on the object due to the ground is always μmg.
D For a thin-walled hollow cylinder, a=F2m.

The correct answers are:
For a solid cylinder, the maximum possible value of a is 2μg., For a thin-walled hollow cylinder, a=F2m.

Solution:
Let moment of inertia =I

For pure rolling, a=αR

and F−f=ma ...1

and fR=Iα

or fR=IaR

or f=IaR2

Therefore, F−IaR2=ma

∴ F=m+IR2a

⇒ For the same value of F, value of a depends on I.

⇒ For a solid cylinderI=mR22 (to calculate maximum a)

f=μmg

∴ μmg=mR22×aR2

Therefore, a=2μg

⇒f is dependent on applied force.

⇒ For thin-walled hollow cylinderI=mR2,

F=m+mR2R2a=2ma

∴ a=F2m

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2021 (Paper 1)
A hyperbola, having the transverse axis of length $2 \sin \theta$, is confocal with the ellipse $3 x^2+4 y^2=12$. Then, its equation is

Options:
A
$x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1$

B
$x^2 \sec ^2 \theta-y^2 \operatorname{cosec}^2 \theta=1$

C
$x^2 \sin ^2 \theta-y^2 \cos ^2 \theta=1$

D
$x^2 \cos ^2 \theta-y^2 \sin ^2 \theta=1$

The correct answer is:
$x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1$

Solution:
The given ellipse is
$$
\begin{aligned}
& \frac{x^2}{4}+\frac{y^2}{3}=1 \Rightarrow \alpha=2, \beta=\sqrt{3} \\
& \therefore \quad 3=4\left(1-e^2\right) \Rightarrow e=\frac{1}{2} \\
&
\end{aligned}
$$
$$
\therefore \quad a e=1
$$
Hence the eccentricity $e_1$, of the hyperbola is given by
$$
\begin{array}{rlrl}
& & 1 & =e_1 \sin \theta \\
\Rightarrow & & e_1 & =\operatorname{cosec} \theta \\
\Rightarrow & b^2 & =\sin ^2 \theta\left(\operatorname{cosec}^2 \theta-1\right)=\cos ^2 \theta
\end{array}
$$
Hence, the hyperbola is
$$
\frac{x^2}{\sin ^2 \theta}-\frac{y^2}{\cos ^2 \theta}=1 \text { or } x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1
$$

Topic: Mathematics, Ellipse, JEE Advanced, JEE Advanced 2007 (Paper 1)
A lamina is made by removing a small disc of diameter $2 R$ from a bigger disc of uniform mass density and radius $2 R$, as shown in the figure. The moment of inertia of this lamina about axes passing though $O$ and $P$ is $I_{O}$ and $I_{P}$ respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $I_{P} / I_{O}$ to the nearest integer is

Options:

The correct answer is:
3

Solution:
Let $\sigma$ be the surface mass density.

Moment of inertia of the lamina about axes passing through 'O'

$\begin{array}{l}

I_{O}=\frac{1}{2} \sigma\left[\pi(2 R)^{2}\right] \times(2 R)^{2}- \\

=\frac{13}{2} \pi \sigma R^{4} \quad\left[\frac{1}{2}\left(\sigma \pi R^{2}\right)^{2}+\sigma\left(\pi R^{2}\right) \times R^{2}\right]

\end{array}$

Moment of inertia of the lamina about axes passing through 'P'

$\begin{array}{l}

I_{P}=8 \pi \sigma \mathrm{R}^{4}+\sigma \pi(2 \mathrm{R})^{2} \times(2 \mathrm{R})^{2} \\

{\left[\frac{1}{2} \sigma\left(\pi R^{2}\right) R^{2}+\sigma\left(\pi R^{2}\right)\left(\sqrt{(2 R)^{2}+R^{2}}\right)^{2}\right]}

\end{array}$

$\begin{array}{l}

=24 \pi \sigma \mathrm{R}^{4}-5.5 \sigma \pi \mathrm{R}^{4}=18.5 \pi \sigma \mathrm{R}^{4} \\

\therefore \quad \frac{I_{P}}{I_{O}}=\frac{18.5 \pi \sigma R^{4}}{\frac{13}{2} \pi \sigma R^{4}}=\frac{37}{13} \approx 3

\end{array}$

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2012 (Paper 1)
A large glass slab $\left(\mu=\frac{5}{3}\right)$ of thickness 8 $\mathrm{cm}$ is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius $R \mathrm{~cm}$. What is the value of $R$ ?

Options:

The correct answer is:
6

Solution:
$$
\frac{R}{t}=\tan \theta_C
$$
or $\quad R=t\left(\tan \theta_C\right)$
$$
\begin{aligned}
& \text { But, } \quad \sin \theta_C=\frac{1}{\mu}=\frac{3}{5} \\
& \therefore \quad \tan \theta_C=\frac{3}{4} \\
& \therefore \quad R=\frac{3}{4} t=\frac{3}{4}(8 \mathrm{~cm})=6 \mathrm{~cm} \\
&
\end{aligned}
$$
Hence the answer is 6 .

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2010 (Paper 2)
A light beam is travelling from Region I to Region IV (refer figure). The refractive index in Region I, II, III and IV are $n_0, \frac{n_0}{2}, \frac{n_0}{6}$ and $\frac{n_0}{8}$, respectively. The angle of incidence $\theta$ for which the beam just misses entering Region IV is

Options:
A
$\sin ^{-1}\left(\frac{3}{4}\right)$

B
$\sin ^{-1}\left(\frac{1}{8}\right)$

C
$\sin ^{-1}\left(\frac{1}{4}\right)$

D
$\sin ^{-1}\left(\frac{1}{3}\right)$

The correct answer is:
$\sin ^{-1}\left(\frac{1}{8}\right)$

Solution:
Critical angle from region III to region IV
$$
\sin \theta_c=\frac{n_0 / 8}{n_0 / 6}=\frac{3}{4}
$$
Now applying Snell's law in region I and region III :
$$
n_0 \sin \theta=\frac{n_0}{6} \sin \theta_c
$$
or
$$
\begin{aligned}
\sin \theta & =\frac{1}{6} \sin \theta_c=\frac{1}{6}\left(\frac{3}{4}\right)=\frac{1}{8} \\
\theta & =\sin ^{-1}\left(\frac{1}{8}\right)
\end{aligned}
$$
$\therefore$ correct option is (b).

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2008 (Paper 2)
A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses $0.36 \mathrm{~kg}$ and $0.72 \mathrm{~kg}$. Taking $g=10 \mathrm{~ms}^{-2}$, find thework done (in Joule) by string on the block of mass $0.36 \mathrm{~kg}$ during the first second after the system is released from rest.

Options:

The correct answer is:
8

Solution:
$$
\text { } \begin{aligned}
\text { Now, } B & =\frac{\mu_0}{4 \pi} \frac{I}{12 x / 5}\left[\sin 37^{\circ}+\sin 53^{\circ}\right] \\
& =7\left(\frac{\mu_0 I}{48 \pi x}\right) \\
a & =\frac{\text { Net pulling force }}{\text { Total mass }} \\
& =\frac{0.72 g-0.36 g}{0.72+0.36}=\frac{g}{3} \\
s & =\frac{1}{2} a t^2=\frac{1}{2}\left(\frac{g}{3}\right)(1)^2=\frac{g}{6}
\end{aligned}
$$
$$
\begin{aligned}
& T-0.36 g=0.36 a=0.36 \frac{g}{3} \\
& \therefore \quad T=0.48 g \\
& \text { Now, } W_T=T S \cos 0^{\circ} \text { (on } 3.6 \mathrm{~kg} \text { mass) } \\
& =(0.48 g)\left(\frac{g}{6}\right)(1)=0.08\left(g^2\right) \\
& =0.08(10)^2=8 \mathrm{~J} \\
&
\end{aligned}
$$

Topic: Physics, Work Power Energy, JEE Main
A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses $0.36 \mathrm{~kg}$ and $0.72 \mathrm{~kg}$. Taking $g=10 \mathrm{~ms}^{-2}$, find thework done (in Joule) by string on the block of mass $0.36 \mathrm{~kg}$ during the first second after the system is released from rest.

Options:

The correct answer is:
8

Solution:
$$
\text { } \begin{aligned}
\text { Now, } B & =\frac{\mu_0}{4 \pi} \frac{I}{12 x / 5}\left[\sin 37^{\circ}+\sin 53^{\circ}\right] \\
& =7\left(\frac{\mu_0 I}{48 \pi x}\right) \\
a & =\frac{\text { Net pulling force }}{\text { Total mass }} \\
& =\frac{0.72 g-0.36 g}{0.72+0.36}=\frac{g}{3} \\
s & =\frac{1}{2} a t^2=\frac{1}{2}\left(\frac{g}{3}\right)(1)^2=\frac{g}{6}
\end{aligned}
$$
$$
\begin{aligned}
& T-0.36 g=0.36 a=0.36 \frac{g}{3} \\
& \therefore \quad T=0.48 g \\
& \text { Now, } W_T=T S \cos 0^{\circ} \text { (on } 3.6 \mathrm{~kg} \text { mass) } \\
& =(0.48 g)\left(\frac{g}{6}\right)(1)=0.08\left(g^2\right) \\
& =0.08(10)^2=8 \mathrm{~J} \\
&
\end{aligned}
$$

Topic: Physics, Work Power Energy, JEE Advanced, JEE Advanced 2009 (Paper 2)
A light ray travelling in glass medium is incident on glass-air interface at an angle of incidence $\theta$. The reflected $(R)$ and transmitted $(T)$ intensities, both as function of $\theta$, are plotted. The correct sketch is

Options:
A

B

C

D

The correct answer is:

Solution:
After critical angle reflection will be $100 \%$ and transmission is $0 \%$. Options (b) and (c) satisfy this condition. But option (c) is the correct option. Because in option (b) transmission is given $100 \%$ at $\theta=0^{\circ}$, which is not true.
Analysis of Question
From examination point of view question is moderately difficult. Because it takes little time to understand the given graphs clearly.

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2011 (Paper 2)
A line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2 x+y+z=9$ at point $Q$. The length of the line segment $P Q$ equals

Options:
A
1

B
$\sqrt{2}$

C
$\sqrt{3}$

D
2

The correct answer is:
$\sqrt{3}$

Solution:
$$
\text { Since, } l=m=n=\frac{1}{\sqrt{3}}
$$
$\therefore$ Equation of line are
$$
\begin{gathered}
\frac{x-2}{1 / \sqrt{3}}=\frac{y+1}{1 / \sqrt{3}}=\frac{z-2}{1 / \sqrt{3}} \\
\Rightarrow \quad x-2=y+1=z-2=r \text { (say) }
\end{gathered}
$$
$\therefore$ Any point on the line is
$$
Q \equiv(r+2 r-1, r+2)
$$
$\because Q$ lies on the plane $2 x+y+z=9$
$$
\begin{aligned}
& \therefore \quad 2(r+2)+(r-1)+(r+2)=9 \\
& \Rightarrow 4 r+5=9 \Rightarrow r=1 \Rightarrow Q(3,0,3) \\
& \therefore P Q=\sqrt{(3-2)^2+(0+1)^2+(3-2)^2} \\
& =\sqrt{3}
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Advanced, JEE Advanced 2009 (Paper 2)
A long circular tube of length $10 \mathrm{~m}$ and radius $0.3 \mathrm{~m}$ carries a current I along its curved surface as shown. A wire loop of resistance $0.005 \Omega$ and of radius $0.1 \mathrm{~m}$ is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as $I=I_0 \cos 300 t$, where $I_0$ is constant. If the magnetic moment of the loop is $N \mu_0 I_0 \sin 300 t$, then $N$ is

Options:

The correct answer is:
6

Solution:
Take the circular tube as a long solenoid. The wires are closely wound. Magnetic field inside the solenoid is
$$
B=\mu_0 n i
$$
Here, $n=$ number of turns per unit length
$\therefore n i=$ current per unit length
In the given problem,
$$
\begin{aligned}
& n i=\frac{I}{L} \\
\therefore \quad B & =\frac{\mu_0 I}{L}
\end{aligned}
$$
Flux passing through the circular coil is
$$
\phi=B S=\left(\frac{\mu_0 I}{L}\right)\left(\pi r^2\right)
$$
Induced emf, $e=-\frac{d \phi}{d t}=-\left(\frac{\mu_0 \pi r^2}{L R}\right) \cdot \frac{d I}{d t}$
Induced current,
$$
i=\frac{e}{R}=-\left(\frac{\mu_0 \pi r^2}{L R}\right) \frac{d I}{d t}
$$
Magnetic moment $i A=i \pi r^2$
or $\quad M=-\left(\frac{\mu_0 \pi^2 r^4}{L R}\right) \cdot \frac{d I}{d t}$
Given, $\quad I=I_0 \cos 300 t$
$$
\therefore \quad \frac{d I}{d t}=-300 I_0 \sin (300 t)
$$
Substituting in Eq. (i), we get
$$
\begin{aligned}
M & =\left(\frac{300 \pi^2 r^4}{L R}\right) \mu_0 I_0 \sin 300 t \\
\therefore \quad N & =\frac{300 \pi^2 r^4}{L R}
\end{aligned}
$$
Substituting the values, we get
$$
\begin{aligned}
N & =\frac{300(22 / 7)^2(0.1)^4}{(10)(0.005)} \\
& =5.926
\end{aligned}
$$
or $\quad N \simeq 6$
Analysis of Question
Question is difficult to understand inside the examination hall. But 5 to $10 \%$ question in IIT JEE are always difficult. Students should not panic. Because topper of IIT JEE scores approximately $80-90 \%$.

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2011 (Paper 1)
A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral.

Options:
A
a potential difference appears between the two cylinders when a charge density is given to the inner cylinder

B
a potential difference appears between the two cylinders when a charge density is given to the outer cylinder

C
no potential difference appears between the two cylinders when a uniform line charge is kept along the axis of the cylinders

D
no potential difference appears between the two cylinders when same charge density is given to both the cylinders

The correct answer is:
a potential difference appears between the two cylinders when a charge density is given to the inner cylinder

Solution:
There will be an electric field between two cylinders (using Gauss theorem. This electric field will produce a potential difference.
$\therefore$ Answer is (a).

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2007 (Paper 1)
A long insulated copper wire is closely wound as a spiral of $N$ turns. The spiral has inner radius $a$ and outer radius $b$. The spiral lies in the XY-plane and a steady current $I$ flows through the wire. The Z-component of the magnetic field at the centre of the spiral is

Options:
A

B

C

D

The correct answer is:

Solution:
If we take a small strip of $d r$ at distancer from centre, then number of turns in this strip would be
$$
d N=\frac{N}{b-a} d r
$$
Magnetic field due to this element at the centre of the coil will be
$$
\begin{aligned}
& d B=\frac{\mu_0(d N) I}{2 r}=\frac{\mu_0 N I}{2(b-a)} \frac{d r}{r} \\
\therefore \quad & B=\int_{r=a}^{r=b} d B=\frac{\mu_0 N I}{2(b-a)} \ln \frac{b}{a}
\end{aligned}
$$
$\therefore$ Correct answer is (a).
Analysis of Question
(i) If we see this problem independently, then I will rate this question moderately difficult. But the idea of this question is taken from question number $3.245$ of IE Irodov.
(ii) Interestingly the same question was asked in IIT-JEE 2001 also.

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2011 (Paper 2)
A long straight wire carries a current, I=2 ampere. A semi-circular conducting rod is placed beside it on two conducting parallel rails of negligible resistance. Both the rails are parallel to the wire. The wire, the rod and the rails lie in the same horizontal plane, as shown in the figure. Two ends of the semi-circular rod are at distances 1 cm and 4 cm from the wire. At time t=0, the rod starts moving on the rails with a speed v=3.0 m s-1 (see the figure).

A resistor R=1.4 Ω and a capacitor C0=5.0 μF are connected in series between the rails. At time t=0, C0 is uncharged. Which of the following statement(s) is (are) correct? [μ0=4π×10−7 SI units. Take ln2=0.7]

Options:
A Maximum current through R is 1.2×10−6 ampere
B Maximum current through R is 3.8×10−6 ampere.
C Maximum charge on capacitor C0 is 8.4×10−12 coulomb.
D Maximum charge on capacitor C0 is 2.4×10−12 coulomb.

The correct answers are:
Maximum current through R is 1.2×10−6 ampere, Maximum charge on capacitor C0 is 8.4×10−12 coulomb.

Solution:
Magnetic field due the long wire, B=μ0I2πx

Motional emf in small element,

dε=Bvdx=3μ022πxdx

⇒ε=3μ0π∫141xdx

∴ ε=3μ0πln4=16.8×10−7

Maximum charge on capacitor,

qmax=εC

=16.8×10−75×10−6

=84×10−13 C=8.4×10−12 C

Maximum current in circuit,

Imax=εR=16.8×10−71.4

=1.2×10-6 A

Topic: Physics, Electromagnetic Induction, JEE Advanced, JEE Advanced 2021 (Paper 1)