datavorous/entrance-exam-dataset huggingface dataset:
A loop carrying current $I$ lies in the $x-y$ plane as shown in the figure. The unit vector $\hat{k}$ is coming out of the plane of the paper. The magnetic moment of the current loop is

Options:
A $a^{2} I \hat{k}$
B $\left(\frac{\pi}{2}+1\right) a^{2} I \hat{k}$
C $-\left(\frac{\pi}{2}+1\right) a^{2} I \hat{k}$
D $(2 \pi+1) a^{2} I \hat{k}$

The correct answer is:
$\left(\frac{\pi}{2}+1\right) a^{2} I \hat{k}$

Solution:
Magnetic moment of a current carrying loop $\vec{M}=N I \vec{A}$

Here $N=1, A=a^{2}+2 \pi\left(\frac{a}{2}\right)^{2}=a^{2}\left[1+\frac{\pi}{2}\right]$

From Screw law, direction of $m$ is outward or in +ve z-direction.

$\therefore \quad \vec{M}=I a^{2}\left[1+\frac{\pi}{2}\right] \hat{k}$

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2012 (Paper 2)
A magnetic field $\mathbf{B}=B_0 \hat{\mathbf{j}}$ exists in the region $a < x < 2 a$ and $\mathbf{B}=B_0 \hat{\mathbf{j}}$ in the region $2 a < x < 3 a$, where $B_0$ is a positive constant. A positive point charge moving with a velocity $\mathbf{v}=v_0 \hat{\mathbf{i}}$, where $v_0$ is a positive constant, enters the magnetic field at $x=a$. The trajectory of the charge in this region can be like

Options:
A

B

C

D

The correct answer is:

Solution:
$\mathbf{F}_m=q(\mathbf{v} \times \mathbf{B})$
Correct option is (a).

Topic: Physics, Magnetic Effects of Current, JEE Main
A magnetic field $\mathbf{B}=B_0 \hat{\mathbf{j}}$ exists in the region $a < x < 2 a$ and $\mathbf{B}=B_0 \hat{\mathbf{j}}$ in the region $2 a < x < 3 a$, where $B_0$ is a positive constant. A positive point charge moving with a velocity $\mathbf{v}=v_0 \hat{\mathbf{i}}$, where $v_0$ is a positive constant, enters the magnetic field at $x=a$. The trajectory of the charge in this region can be like

Options:
A

B

C

D

The correct answer is:

Solution:
$\mathbf{F}_m=q(\mathbf{v} \times \mathbf{B})$
Correct option is (a).

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2007 (Paper 2)
A man walks a distance of 3 units from the origin towards the North-East $\left(\mathrm{N} 45^{\circ} \mathrm{E}\right)$ direction. From there, he walks a distance of 4 units towards the North-West ( $\mathrm{N} 45^{\circ} \mathrm{W}$ ) direction to reach a point $P$. Then, the position of $P$ in the Argand plane is

Options:
A
$3 e^{i \pi / 4}+4 i$

B
$(3-4 i) e^{i \pi / 4}$

C
$(4+3 i) e^{i \pi / 4}$

D
$(3+4 i) e^{i \pi / 4}$

The correct answer is:
$(3+4 i) e^{i \pi / 4}$

Solution:
Let $O A=3$, so that the complex number associated with $A$ is $3 e^{i \pi / 4}$. If $z$ is the complex number associated with $P$, then
$$
\begin{array}{rlrl}
& & \frac{z-3 e^{i \pi / 4}}{0-3 e^{i \pi / 4}} & =\frac{4}{3} e^{-i \pi / 2}=-\frac{4 i}{3} \\
\Rightarrow & 3 z-9 e^{i \pi / 4} & =12 i^{i \pi / 4} \\
\Rightarrow & z & z & =(3+4 i) e^{i \pi / 4}
\end{array}
$$

Topic: Mathematics, Complex Number, JEE Advanced, JEE Advanced 2007 (Paper 1)
A massless rod $B D$ is suspended by two identical massless strings $A B$ and $C D$ of equal lengths. A block of mass ' $m$ ' is suspended point $P$ such that $B P$ is equal to ' $x$ ', if the fundamental frequency of the left wire is twice the fundamental frequency of right wire, then the value of $x$ is

Options:
A
$1 / 5$

B
$1 / 4$

C
$41 / 5$

D
$31 / 4$

The correct answer is:
$1 / 5$

Solution:
$f \propto v \propto \sqrt{T}$
Further $\Sigma \tau_p=0$
or
or
$$
f_{A B}=2 f_{C D} \Rightarrow T_{A B}=4 T_{C D}
$$
$$
\begin{aligned}
T_{A B}(x) & =T_{C D}(l-x) \\
4 x & =l-x\left(\text { as } T_{A B}-4 T_{C D}\right) \\
x & =\frac{l}{5}
\end{aligned}
$$

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2006
A metal rod $A B$ of length $10 x$ has its one end $A$ in ice at $0^{\circ} \mathrm{C}$ and the other end $B$ in water at $100^{\circ} \mathrm{C}$. If a point $P$ on the rod is maintained at $400^{\circ} \mathrm{C}$, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is $540 \mathrm{calg}^{-1}$ and latent heat of melting of ice is $80 \mathrm{calg}^{-1}$. If the point $P$ is at a distance of $\lambda x$ from the ice end $A$, find the value of $\lambda$. [Neglect any heat loss to the surrounding.]

Options:

The correct answer is:
9

Solution:
Heat will flow both sides from point $P$.
$$
\begin{aligned}
L_1 \frac{d m_1}{d t} & =\left(\frac{\text { Temperature difference }}{\text { Thermal resistance }}\right)_1 \\
& =\frac{400}{(\lambda x) / k A} \\
L_1 \frac{d m_2}{d t} & =\frac{400-100}{(100-\lambda) x / k A}
\end{aligned}
$$
In about two equations,
$$
\begin{aligned}
& \frac{d m_1}{d t}=\frac{d m_2}{d t} \quad \text { (given) } \\
& L_1=80 \mathrm{calg}^{-1} \text { and } L_2=540 \mathrm{calg}^{-1}
\end{aligned}
$$
Solving these two equations we get $\lambda=9$.

Topic: Physics, Thermal Properties of Matter, JEE Main
A metal rod $A B$ of length $10 x$ has its one end $A$ in ice at $0^{\circ} \mathrm{C}$ and the other end $B$ in water at $100^{\circ} \mathrm{C}$. If a point $P$ on the rod is maintained at $400^{\circ} \mathrm{C}$, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is $540 \mathrm{calg}^{-1}$ and latent heat of melting of ice is $80 \mathrm{calg}^{-1}$. If the point $P$ is at a distance of $\lambda x$ from the ice end $A$, find the value of $\lambda$. [Neglect any heat loss to the surrounding.]

Options:

The correct answer is:
9

Solution:
Heat will flow both sides from point $P$.
$$
\begin{aligned}
L_1 \frac{d m_1}{d t} & =\left(\frac{\text { Temperature difference }}{\text { Thermal resistance }}\right)_1 \\
& =\frac{400}{(\lambda x) / k A} \\
L_1 \frac{d m_2}{d t} & =\frac{400-100}{(100-\lambda) x / k A}
\end{aligned}
$$
In about two equations,
$$
\begin{aligned}
& \frac{d m_1}{d t}=\frac{d m_2}{d t} \quad \text { (given) } \\
& L_1=80 \mathrm{calg}^{-1} \text { and } L_2=540 \mathrm{calg}^{-1}
\end{aligned}
$$
Solving these two equations we get $\lambda=9$.

Topic: Physics, Thermal Properties of Matter, JEE Advanced, JEE Advanced 2009 (Paper 2)
A metal rod of length $L$ and mass $m$ is pivoted at one end. A thin disc of mass $M$ and radius $R( < L)$ is attached at its centre to the free end of the rod. Consider two ways the disc is attached. Case A-the disc is not free to rotate about its centre and case $B$-the disc is free to rotate about its centre. The rod-disc system performs
SHM in vertical plane after being released from the same displaced position. Which of the following statement(s) is/are true?

Options:
A
Restoring torque in case $A=$ Restoring torque in case $B$

B
Restoring torque in case $A < $ Restoring torque in case $B$
C
Angular frequency for case $A>$ Angular frequency for case $B$

D
Angular frequency for case $A < $ Angular frequency for case $B$

The correct answers are:
Restoring torque in case $A=$ Restoring torque in case $B$
,
Restoring torque in case $A < $ Restoring torque in case $B$

Solution:
$\tau_A=\tau_B=m g \frac{L}{2} \sin \theta+\operatorname{MgL} \sin \theta$
$=$ Restoring torque about point $O$.
In case $A$, moment of inertia will be more. Hence, angular acceleration $(\alpha=\tau / I)$ will be less. Therefore, angular frequency will be less.
$\therefore$ Correct options are (a) and (d).
Analysis of Question
Question is difficult from my point of view. Because this type of SHM is rarely taught in the class and questions of this type are not given in standard books.

Topic: Physics, Oscillations, JEE Advanced, JEE Advanced 2011 (Paper 1)
A meter bridge is set up as shown in figure, to determine an unknown resistance $X$ using a standard $10 \Omega$ resistor. The galvanometer shows null point when tapping key is at $52 \mathrm{~cm}$ mark. The end-corrections are $1 \mathrm{~cm}$ and $2 \mathrm{~cm}$ respectively for the ends $A$ and $B$. The determined value of $X$ is

Options:
A
$10.2 \Omega$

B
$10.6 \Omega$

C
$10.8 \Omega$

D
$11.1 \Omega$

The correct answer is:
$10.6 \Omega$

Solution:
Using the concept of balanced Wheatstone bridge, we have
$$
\begin{array}{rlrl}
& \frac{P}{Q} & =\frac{R}{S} \\
\therefore & & \frac{X}{(52+1)} & =\frac{10}{(48+2)} \\
\therefore & & X & =\frac{10 \times 53}{50}=10.6 \Omega
\end{array}
$$
$\therefore$ Correct option is (b).
Analysis of Question
Question is moderately tough, because normally end corrections are not taught in the coaching/schools in this type of problem.

Topic: Physics, Current Electricity, JEE Advanced, JEE Advanced 2011 (Paper 1)
A mixture of 2 moles of helium gas (atomic mass $=4 \mathrm{amu}$ ) and 1 mole of argon gas (atomic mass $=40 \mathrm{amu}$ ) is kept at $300 \mathrm{~K}$ in a container. The ratio of the rms speeds $\left(\frac{v_{\mathrm{rms}}(\text { helium })}{v_{\mathrm{rms}}(\operatorname{argon})}\right)$ is

Options:
A $0.32$
B $0.45$
C $2.24$
D $3.16$

The correct answer is:
$3.16$

Solution:
Using $V_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}} \Rightarrow V_{r m s} \propto \frac{1}{\sqrt{M}}$ $\frac{v_{\text {rms (helium) }}}{v_{\text {rms (argon) }}}=\sqrt{\frac{M_{\text {argon }}}{M_{\text {helium }}}}=\sqrt{\frac{40}{4}}=\sqrt{10} \approx 3.16$

Topic: Physics, Kinetic Theory of Gases, JEE Advanced, JEE Advanced 2012 (Paper 1)
A monoatomic ideal gas undergoes a process in which the ratio of $P$ to $V$ at any instant is constant and equals to 1 . What is the molar heat capacity of the gas?

Options:
A
$\frac{4 R}{2}$

B
$\frac{3 R}{2}$

C
$\frac{5 R}{2}$

D
0

The correct answer is:
$\frac{4 R}{2}$

Solution:
The molar heat capacity for any process is given by a following expression
$C=C_V+\frac{R}{1-\gamma}$ when $p V^\gamma=$ constant and $C_p / C_V=\gamma$
Here, $\frac{p}{V}=1$ i.e., $p V^{-1}=$ constant
$$
\therefore \quad C=\frac{3}{2} R+\frac{R}{1-(-1)}=\frac{3}{2} R+\frac{R}{2}=\frac{4}{2} R
$$

Topic: Chemistry, Thermodynamics (C), JEE Advanced, JEE Advanced 2006
A parallel plate capacitor $C$ with plates of unit area and separation $d$ is filled with a liquid of dielectric constant $K=2$. The level of liquid is $\frac{d}{3}$ initially. Suppose the liquid level decreases at a constant speed $v$, the time constant as a function of time $t$ is

Options:
A
$\frac{6 \varepsilon_0 R}{5 d+3 v t}$

B
$\frac{(15 d+9 v t) \varepsilon_0 R}{2 d^2-3 d v t-9 v^2 t^2}$

C
$\frac{6 \varepsilon_0 R}{5 d-3 v t}$

D
$\frac{(15 d-9 v t) \varepsilon_0 R}{2 d^2+3 d v t-9 v^2 t^2}$

The correct answer is:
$\frac{6 \varepsilon_0 R}{5 d+3 v t}$

Solution:
After time $t$, thickness of liquid will remain $\left(\frac{d}{3}-v t\right)$.
Now, time constant as function of time
$$
\begin{aligned}
\tau_c & =C R \\
& \left.=\frac{\varepsilon_0(1) \cdot R}{\left(d-\frac{d}{3}+v t\right)+\frac{d / 3-v t}{2}} \quad \quad \text { Applying } C=\frac{\varepsilon_0 A}{d-t+\frac{t}{k}}\right) \\
& =\frac{6 \varepsilon_0 R}{5 d+3 v t}
\end{aligned}
$$
$\therefore$ correct option is (a).

Topic: Physics, Capacitance, JEE Main
A parallel plate capacitor $C$ with plates of unit area and separation $d$ is filled with a liquid of dielectric constant $K=2$. The level of liquid is $\frac{d}{3}$ initially. Suppose the liquid level decreases at a constant speed $v$, the time constant as a function of time $t$ is

Options:
A
$\frac{6 \varepsilon_0 R}{5 d+3 v t}$

B
$\frac{(15 d+9 v t) \varepsilon_0 R}{2 d^2-3 d v t-9 v^2 t^2}$

C
$\frac{6 \varepsilon_0 R}{5 d-3 v t}$

D
$\frac{(15 d-9 v t) \varepsilon_0 R}{2 d^2+3 d v t-9 v^2 t^2}$

The correct answer is:
$\frac{6 \varepsilon_0 R}{5 d+3 v t}$

Solution:
After time $t$, thickness of liquid will remain $\left(\frac{d}{3}-v t\right)$.
Now, time constant as function of time
$$
\begin{aligned}
\tau_c & =C R \\
& \left.=\frac{\varepsilon_0(1) \cdot R}{\left(d-\frac{d}{3}+v t\right)+\frac{d / 3-v t}{2}} \quad \quad \text { Applying } C=\frac{\varepsilon_0 A}{d-t+\frac{t}{k}}\right) \\
& =\frac{6 \varepsilon_0 R}{5 d+3 v t}
\end{aligned}
$$
$\therefore$ correct option is (a).

Topic: Physics, Capacitance, JEE Advanced, JEE Advanced 2008 (Paper 2)
A particle moves in the $X-Y$ plane under the influence of a force such that its linear momentum is $\mathbf{p}(t)=A[\hat{\mathbf{i}} \cos k t-\hat{\mathbf{j}} \sin k t]$, where $A$ and $k$ are constants. The angle between the force and the momentum is

Options:
A
$0^{\circ}$

B
$30^{\circ}$

C
$45^{\circ}$

D
$90^{\circ}$

The correct answer is:
$90^{\circ}$

Solution:
$$
\begin{aligned}
& \mathbf{F}=\frac{d \mathbf{p}}{d t}=-k A \sin k t \hat{\mathbf{i}}-k A \cos k t \hat{\mathbf{j}} \\
& \mathbf{p}=A \cos k t \hat{\mathbf{i}}-A \sin k \hat{\mathbf{j}}
\end{aligned}
$$
Since,
F $\cdot \mathbf{p}=0$
$\therefore$ Angle between $\mathbf{F}$ and pshould be $90^{\circ}$.
$\therefore$ Correct option is $(\mathrm{d})$.

Topic: Physics, Mathematics in Physics, JEE Advanced, JEE Advanced 2007 (Paper 2)
A particle, of mass 10-3 kg and charge 1.0 C, is initially at rest. At time t=0, the particle comes under the influence of an electric field E→t=E0sinωt i^ where E0=1.0 N C-1 and ω=103 rad s-1. Consider the effect of only the electrical force on the particle. Then the maximum speed, in m s-1, attained by the particle at subsequent times is ____________.

Options:

The correct answer is:
2

Solution:
m=10-3 kg, q=1 C, t=0
Force on particle will be
F=qE=qE0sinωt
At vmax, F=0 ⇒qE0sin⁡ω t o =0
⇒ωto=π
F=qE0sin⁡ωt
dvdt=qE0msinωt
∫0vdv=∫0π∕ωqE0msinωt dt
v-0=qE0mω-cosωt0π∕ω
v-0=qE0mω-cosπ--cos0
v=1×1mω×2=2 m s-1

Topic: Physics, Laws of Motion, JEE Advanced, JEE Advanced 2018 (Paper 2)
A particle of mass $m$ and charge $q$, moving with velocity $v$ enters Region II normal to the boundary as shown in the figure. Region II has a uniform magnetic field $B$ perpendicular to the plane of the paper. The length of the Region II is $l$. Choose the correct choice(s)

Options:
A
The particle enters Region III only if its velocity $v>\frac{q l B}{m}$

B
The particle enters Region III only if its velocity $v < \frac{q I B}{m}$
C
Path length of the particle in Region II is maximum when velocity $v=\frac{q I B}{m}$

D
Time spent in Region II is same for any velocity $v$ as long as the particle returns to Region I

The correct answers are:
The particle enters Region III only if its velocity $v>\frac{q l B}{m}$
,
Path length of the particle in Region II is maximum when velocity $v=\frac{q I B}{m}$
,
Time spent in Region II is same for any velocity $v$ as long as the particle returns to Region I

Solution:
$\overrightarrow{\mathbf{V}} \perp \overrightarrow{\mathbf{B}}$ in region II. Therefore, path of particle is circle in region II. Particle enters in region III if, radius of circular path, $r>1$ or $\frac{m v}{B q}>1$ or $v>\frac{B q l}{m}$
If $v=\frac{B q l}{m}, r=\frac{m v}{B q}=l$, particle will turn back and path length will be maximum as shown in figure in region II. If particle returns to region I, time spent in region II will be $t=\frac{T}{2}=\frac{\pi m}{B q}$, which is independent of $v$.
$\therefore$ correct options are (a), (c) and (d).

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2008 (Paper 1)
A particle of unit mass is moving along the x-axis under the influence of a force and its total energy is conserved. Four possible forms of the potential energy of the particle are given in column I (a and U0 are constants). Match the potential energies in column I to the corresponding statement(s) in column II.

Column I	Column II

A.	U1x=U021-xa22	P.	The force acting on the particle is zero at x = a

B.	U2x=U02xa2	Q.	The force acting on the particle is zero at x = 0 .

C.	U3x=U02xa2exp⁡-xa2	R.	The force acting on the particle is zero at x= -a

D.	U4x=U02xa-13xa3	S.	The particle experiences an attractive force towards x = 0 in the region x<a

T.	The particle with total energy U04 can oscillate about the point x= -a

Options:
A a-s,t;b-s;c-s,t;d-t;
B a-s;b-r,s;c-r,s,t;d-s,t;
C a-r,s;b-q;c-q,r;d-t;
D a-p,q,r,t;b-q,s;c-p,q,r,s;d-p,r,t;

The correct answer is:
a-p,q,r,t;b-q,s;c-p,q,r,s;d-p,r,t;

Solution:
U1x=U021-x2a22
F=-U0221-x2a2-2xa2
=2U0a4a2-x2x
F=2U0a4xa-xa+x
F=0 , at x=0, a, -a
x=-a, U=0,x=0, U=U02â Oscillate when total ME is less then U02
AâP,Q,R,T
B- U0x=U02xa2
F=-U0xa2
F=0, x=0
BâQ,S
C- U3x=U02xa2expâ¡-x2a2
F=-U022xa2expâ¡-x2a2+x2a2expâ¡-x2a2-2xa2
=-U022xa4expâ¡-x2a2a2-x2
=U0a4ex2a2 xx+ax-a
x=0, x=-a, x=+a, F=0
x=0, U=0 even function hence minima
CâP,Q,R,S
D- U4x=U02xa-x33a3
F=-U0a1a-x2a3=U02a3x2-a2
x=+a, x=-a, F=0
x=-a, U=-U03, x=+a, U=+U03
Hence oscillate about x=-a if T.M.E<U03
DâP,R,T

Topic: Physics, Work Power Energy, JEE Advanced, JEE Advanced 2015 (Paper 1)
A particle $P$ starts from the point $z_0=1+2 i$, where $i=\sqrt{-1}$. It moves first horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point $z_1$. From $z_1$ the particle moves $\sqrt{2}$ units in the direction of the vector $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and then it moves through an angle $\frac{\pi}{2}$ in anti-clockwise direction on a circle with centre at origin, to reach a point $z_2$. The point $z_2$ is given by

Options:
A
$6+7 i$

B
$-7+6 i$

C
$7+6 i$

D
$-6+7 i$

The correct answer is:
$-6+7 i$

Solution:
$$
z_2^{\prime}=\left(6+\sqrt{2} \times \cos 45^{\circ}, 5+\sqrt{2} \sin 45^{\circ}\right)=(7,6)=7+6 i
$$
By rotation about $(0,0)$,
$$
\begin{aligned}
& \frac{z_2}{z_2^{\prime}}=e^{i \frac{\pi}{2}} \\
& \Rightarrow \quad z_2=z_2^{\prime}\left(e^{i \frac{\pi}{2}}\right) \\
& \Rightarrow \quad z_2=(7+6 i)\left(\cos \frac{\pi}{-}+i \sin \frac{\pi}{-}\right)=(7+6 i)(i)=-6+7 i
\end{aligned}
$$

Topic: Mathematics, Complex Number, JEE Advanced, JEE Advanced 2008 (Paper 2)
A person blows into open-end of a long pipe. As a result, a high pressure pulse of air travels down the pipe.

When this pulse reaches the other end of the pipe,

Options:
A a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is open.
B a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is open.
C a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed.
D a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed.

The correct answers are:
a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is open., a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed.

Solution:
When a sound pulse is reflected from open end of a pipe, phase changes by $180^{\circ}$. A high pressure pulse i.e., compression is reflected as a low pressure pulse i.e., rarefaction. When sound pulse is reflected through a rigid boundary (closed end of a pipe), no phase change occurs so a high pressure pulse is reflected as a high pressure pulse.

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2012 (Paper 1)
A piece of ice (heat capacity $=2100 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ and latent heat $=3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$ ) of mass $\mathrm{m}$ gram is at $-5^{\circ} \mathrm{C}$ at atmospheric pressure. It is given $420 \mathrm{~J}$ of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that $1 \mathrm{~g}$ of ice has melted. Assuming there is no other heat exchange in the process, the value of $m$ is

Options:

The correct answer is:
8

Solution:
Language of question is slightly wrong. As heat capacity and specific heat are two different physical quantities. Unit $\mathrm{J}-\mathrm{kg}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}$. The heat capacity given in the question is really the specific heat. Now applying the heat exchange equation.
$$
\begin{array}{r}
420=\left(m \times 10^{-3}\right)(2100)(5)+\left(1 \times 10^{-3}\right) \\
\left(3.36 \times 10^5\right)
\end{array}
$$
Solving this equation we get,
$$
m=8 \mathrm{~g}
$$
$\therefore$ The correct answer is 8 .

Topic: Physics, Thermal Properties of Matter, JEE Advanced, JEE Advanced 2010 (Paper 1)
A piece of wire is bent in the shape of a parabola $y=k x^2$ (y-axis vertical) with a bead of mass $m$ on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the $x$-axis with the constant acceleration $a$. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the $y$-axis is

Options:
A
$\frac{a}{g k}$

B
$\frac{a}{2 g k}$

C
$\frac{2 a}{g k}$

D
$\frac{a}{4 g k}$

The correct answer is:
$\frac{a}{2 g k}$

Solution:
$N \sin \theta=m g$
$$
\begin{aligned}
N \cos \theta & =m a \\
\tan \theta & =\frac{g}{a} \\
\cot \theta & =\frac{a}{g}=\tan \left(90^{\circ}-\theta\right)=\frac{d y}{d x}=2 k x
\end{aligned}
$$
$\therefore \quad x=\frac{a}{2 k g}$

Topic: Physics, Laws of Motion, JEE Advanced, JEE Advanced 2009 (Paper 2)
A plane passes through $(1,-2,1)$ and is perpendicular to two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, then the distance of the plane from the point $(1,2,2)$ is

Options:
A
0

B
1

C
$\sqrt{2}$

D
$2 \sqrt{2}$

The correct answer is:
$2 \sqrt{2}$

Solution:
Let the equation of plane be,
$$
a(x-1)+b(y+2)+c(z-1)=0
$$
which is perpendicular to $2 x-2 y+z=0$ and $x-y+2 z=4$
$$
\begin{array}{llrl}
\Rightarrow & 2 a-2 b+c & =0 \text { and } a-b+2 c=0 \\
\text { or } & \frac{a}{-2} 1 & =\frac{b}{2}=\frac{c}{-2} \\
\Rightarrow & -12 & 1 & -1 \\
\Rightarrow & \frac{a}{-3} & =\frac{b}{-3}=\frac{c}{0} \\
& \frac{a}{1}=\frac{b}{1}=\frac{c}{0}
\end{array}
$$
So, the equation of plane is,
$x-1+y+2=0$ or $x+y+1=0$, its distance from the point $(1,2,2)$ is $\frac{|1+2+1|}{\sqrt{2}}=2 \sqrt{2}$.

Topic: Mathematics, Three Dimensional Geometry, JEE Advanced, JEE Advanced 2006
A point charge q of mass m is suspended vertically by a string of length l. A point dipole of dipole moment p→ is now brought towards q from infinity so that the charge moves away. The final equilibrium position of the system including the direction of the dipole, the angles and distances is shown in the figure below. If the work done in bringing the dipole to this position is N×(mgh), where g is the acceleration due to gravity, then the value of N is ____________ (Note that for three coplanar forces keeping a point mass in equilibrium, Fsinθ is the same for all forces, where F is any one of the forces and θ is the angle between the other two forces)

Options:

The correct answer is:
2

Solution:
W=qV+mgh

=qkpx2+mgh ...1

For equilibrium

qEsinβ=mgsin90-α2

qEsin(180-α)=mgcosα2; qEsinα=mgcosα2

9E2sinα2cosα2=mgcosα2; qE=2mgsinα2

qkpx2=mgxsin(α2)=mgsinα2·hsinα2

qkpx2=mgh ...2

From (1) and (2)

W=2mgh

x=2

Topic: Physics, Work Power Energy, JEE Advanced, JEE Advanced 2020 (Paper 2)
A point mass is subjected to two simultaneous sinusoidal displacements in $x$-direction, $x_1(t)=A \sin \omega t$ and $x_2(t)=A \sin \left(\omega t+\frac{2 \pi}{3}\right)$. Adding a third sinusoidal displacement $x_3(t)=B \sin (\omega t+\phi)$ bring the mass to a complete rest. The values of $B$ and $\phi$ are

Options:
A
$\sqrt{2 A}, \frac{3 \pi}{4}$

B
$A, \frac{4 \pi}{3}$

C
$\sqrt{3} A, \frac{5 \pi}{6}$

D
$A, \frac{\pi}{3}$

The correct answer is:
$A, \frac{4 \pi}{3}$

Solution:
Resultant amplitude of $x_1$ and $x_2$ is $A$ at angle $\left(\frac{\pi}{3}\right)$ from $A_1$. To make resultant of $x_1, x_2$ and $x_3$ to be zero. $A_3$ should be equal to $A$ at angle $\phi=\frac{4 \pi}{3}$ as shown in figure.
$\therefore$ Correct answer is (b).
Alternate Solution
If we substitute,
$$
x_1+x_2+x_3=0
$$
or $A \sin \omega t+A \sin \left(\omega t+\frac{2 \pi}{3}\right)$
$$
+B \sin (\omega t+\phi)=0
$$
Then, by applying simple mathematics, we can prove that
$$
\begin{array}{rlrl}
B & =A \\
\text { and } & \phi & =\frac{4 \pi}{3}
\end{array}
$$
Analysis of Question
(i) Question is simple.
(ii) Question can be solved by applying mathematics also.
(iii) Amplitudes of two or more sine or cosine functions of same frequency $\omega$ can be added by vector method.

Topic: Physics, Oscillations, JEE Advanced, JEE Advanced 2011 (Paper 2)