datavorous/entrance-exam-dataset huggingface dataset:
Two balls, having linear momenta $\overrightarrow{\mathbf{p}}_1=p \hat{\mathbf{i}}$ and $\overrightarrow{\mathbf{p}}_2=-p \hat{\mathbf{i}}$, undergo a collision in free space. There is no external force acting on the balls. Let $\overrightarrow{\mathbf{p}}_1^{\prime}$ and $\overrightarrow{\mathbf{p}}_2^{\prime}$ be their final momenta. The following option(s) is/are NOT ALLOWED for any non-zero value of $p, a_1, a_2, b_1, b_2, c_1$ and $c_2$

Options:
A
$\overrightarrow{\mathbf{p}}_1^{\prime}=a_1 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}+c_1 \hat{\mathbf{k}}, \overrightarrow{\mathbf{p}}_2^{\prime}=a_2 \hat{\mathbf{i}}+b_2 \hat{\mathbf{j}}$

B
$\overrightarrow{\mathbf{p}}_1^{\prime}=c_1 \hat{\mathbf{k}}, \overrightarrow{\mathbf{p}}_2^{\prime}=c_2 \hat{\mathbf{k}}$

C
$\overrightarrow{\mathbf{p}}_1^{\prime}=a_1 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}+c_1 \hat{\mathbf{k}}$

D
$\overrightarrow{\mathbf{p}}_1^{\prime}=a_1 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}$
$\overrightarrow{\mathbf{p}}_2^{\prime}=a_2 \hat{\mathbf{i}}+b_2 \hat{\mathbf{j}}-c_1 \hat{\mathbf{k}}$
$\overrightarrow{\mathbf{p}}_2^{\prime}=a_2 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}$

The correct answers are:
$\overrightarrow{\mathbf{p}}_1^{\prime}=a_1 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}+c_1 \hat{\mathbf{k}}, \overrightarrow{\mathbf{p}}_2^{\prime}=a_2 \hat{\mathbf{i}}+b_2 \hat{\mathbf{j}}$
,
$\overrightarrow{\mathbf{p}}_1^{\prime}=a_1 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}$
$\overrightarrow{\mathbf{p}}_2^{\prime}=a_2 \hat{\mathbf{i}}+b_2 \hat{\mathbf{j}}-c_1 \hat{\mathbf{k}}$
$\overrightarrow{\mathbf{p}}_2^{\prime}=a_2 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}$

Solution:
Initial momentum of the system $\overrightarrow{\mathbf{p}_1}+\overrightarrow{\mathbf{p}_2}=0$
$\therefore$ Final momentum $\overrightarrow{\mathbf{p}_1^{\prime}}+\overrightarrow{\mathbf{p}_2^{\prime}}$ should also be zero.
Option (b) is allowed because if we putc $c_1=-c_2 \neq 0, \overrightarrow{\mathbf{p}_1^{\prime}}+\overrightarrow{\mathbf{p}_2^{\prime}}$ will be zero. Similarly, we can check other options.
$\therefore$ correct options are (a) and (d).

Topic: Physics, Center of Mass Momentum and Collision, JEE Advanced, JEE Advanced 2008 (Paper 1)
Two bars of radius ' $r$ ' and ' $2 r$ ' are kept in contact as shown. An electric current $I$ is passed through the bars. Which one of the following is correct?

Options:
A
Heat produced in bar $B C$ is 4 times the heat produced in bar $A B$

B
Electric field in both halves is equal

C
Current density across $A B$ is doubled that of across $B C$

D
Potential difference across $A B$ is 4 times that of across $B C$

The correct answer is:
Heat produced in bar $B C$ is 4 times the heat produced in bar $A B$

Solution:
Current flowing through both the bars is equal. Now, the heat produced is given by
$$
\begin{aligned}
H & =I^2 R t \text { or } H \propto R \text { or } \frac{H_{A B}}{H_{B C}}=\frac{R_{A B}}{R_{B C}} \\
& =\frac{\left(\frac{1}{2 r}\right)^2}{\left(\frac{1}{r}\right)^2} \\
& =\frac{1}{4} \text { or } H_{B C}=4 H_{A B}
\end{aligned}
$$
$$
\left(\text { as } R \propto \frac{1}{A} \propto \frac{1}{r^2}\right)
$$

Topic: Physics, Current Electricity, JEE Main
Two bars of radius ' $r$ ' and ' $2 r$ ' are kept in contact as shown. An electric current $I$ is passed through the bars. Which one of the following is correct?

Options:
A
Heat produced in bar $B C$ is 4 times the heat produced in bar $A B$

B
Electric field in both halves is equal

C
Current density across $A B$ is doubled that of across $B C$

D
Potential difference across $A B$ is 4 times that of across $B C$

The correct answer is:
Heat produced in bar $B C$ is 4 times the heat produced in bar $A B$

Solution:
Current flowing through both the bars is equal. Now, the heat produced is given by
$$
\begin{aligned}
H & =I^2 R t \text { or } H \propto R \text { or } \frac{H_{A B}}{H_{B C}}=\frac{R_{A B}}{R_{B C}} \\
& =\frac{\left(\frac{1}{2 r}\right)^2}{\left(\frac{1}{r}\right)^2} \\
& =\frac{1}{4} \text { or } H_{B C}=4 H_{A B}
\end{aligned}
$$
$$
\left(\text { as } R \propto \frac{1}{A} \propto \frac{1}{r^2}\right)
$$

Topic: Physics, Current Electricity, JEE Advanced, JEE Advanced 2006
Two batteries of different emfs and different internal resistances are connected as shown. The voltage across $A B$ in volt is

Options:

The correct answer is:
5

Solution:
$V_{A B}=$ Equivalent emf of two batteries in parallel.
$$
\begin{aligned}
& =\frac{E_1 / r_1+E_1 / r_2}{1 / r_1+1 / r_2} \\
& =\frac{(6 / 1)+(3 / 2)}{(1 / 1)+(1 / 2)}=5 \mathrm{~V}
\end{aligned}
$$
$\therefore$ Answer is 5 .
Analysis of Question
(i) In my opinion this is the simplest question of JEE 2011.
(ii) When no current is drawn from this equivalent battery, then
$$
V_{A B}=V=E
$$
Otherwise, $V=E \pm$ ir
(iii) Similar type of question was asked in JEE 1981 and cancelled paper of JEE 1997.

Topic: Physics, Current Electricity, JEE Advanced, JEE Advanced 2011 (Paper 2)
Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is $60^{\circ}$ ). In the position of minimum deviation, the angle of refraction will be

Options:
A
$30^{\circ}$ for both the colours

B
greater for the violet colour

C
greater for the red colour

D
equal but not $30^{\circ}$ for both the colours

The correct answer is:
$30^{\circ}$ for both the colours

Solution:
At minimum deviation $\left(\delta=\delta_m\right)$ :
$$
r_1=r_2=\frac{A}{2}=\frac{60^{\circ}}{2}=30^{\circ}
$$
(For both colours)
$\therefore$ correct answer is (a).

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2008 (Paper 1)
Two identical discs of same radius $R$ are rotating about their axes in opposite directions with the same constant angular speed $\omega$. The discs are in the same horizontal plane. At time $t=0$, the points $P$ and $Q$ are facing each other as shown in the figure. The relative speed between the two points $P$ and $Q$ is $v_{r}$. In one time period $(T)$ of rotation of the discs, $v_{r}$ as a function of time is best represented by

Options:
A
B
C
D

The correct answer is:

Solution:
At $t=0, t=\frac{T}{2}$ and $t=T$ therelative velocity will bezero. At $t=\frac{T}{4}$ and $t=\frac{3 T}{4}$, the relative velocity will be maximum in magnitude

Hence graph (a) correctly depicts $v_{r}$ versus $t$ graph.

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2012 (Paper 2)
Two identical uniform discs roll without slipping on two different surfaces AB and CD (see figure) starting at A and C with linear speeds υ1 and υ2 , respectively, and always remain in contact with the surfaces. If they reach B and D with the same linear speed and υ1=3 m/s, then υ2 in m/s is g=10 ms2

Options:

The correct answer is:
7

Solution:
K.Ed=K.ET+K.ERot
=12mv2+12Icmw2
=12mv2+12mr22vr2
=34mv2
KA+Wg=KB=34mv12+mg×30=KB
KC+Wg=KD=34mv22+mg×27=KD
KB=KD (Conversation of energy)
⇒34m32-34mv22+3mg=0
⇒v22=49
⇒v=7

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2015 (Paper 1)
Two large vertical and parallel metal plates having a separation of $1 \mathrm{~cm}$ are connected to a DC voltage source of potential difference $X$. A proton is released at rest midway between the two plates. It is found to move at $45^{\circ}$ to the vertical JUST after release. Then $X$ is nearly

Options:
A $1 \times 10^{-5} \mathrm{~V}$
B $1 \times 10^{-7} V$
C $1 \times 10^{-9} V$
D $1 \times 10^{-10} V$

The correct answer is:
$1 \times 10^{-9} V$

Solution:
According to the question, a proton is released at rest mid-way between the two plates and is found to move at $45^{\circ}$, so net force is at $45^{\circ}$ from vertical and two forces acting on the proton just after the release are as shown in the figure.

$q E=m g$

$\therefore \quad V=\frac{m g d}{q}=\frac{1.67 \times 10^{-27} \times 10 \times 10^{-2}}{1.6 \times 10^{-19}}=10^{-9} \mathrm{~V}$

Topic: Physics, Laws of Motion, JEE Advanced, JEE Advanced 2012 (Paper 1)
Two metallic rings $A$ and $B$, identical in shape and size but having different resistivities $\rho_A$ and $\rho_B$, are kept on top of two identical solenoids as shown in the figure. When current $I$ is switched on in both the solenoids in identical manner, the rings $A$ and $B$ jump to heights $h_A$ and $h_B$, respectively, with $h_A>h_B$. The possible relation(s) between their resistivities and their masses $m_A$ and $m_B$ is(are)

Options:
A
$\rho_A>\rho_B$ and $m_A=m_B$

B
$\rho_A < \rho_B$ and $m_A=m_B$
C
$\rho_A>\rho_B$ and $m_A>m_B$

D
$\rho_A < \rho_B$ and $m_A < m_B$

The correct answers are:
$\rho_A < \rho_B$ and $m_A=m_B$,
$\rho_A < \rho_B$ and $m_A < m_B$

Solution:
Induced emf $e=-\frac{d \phi}{d t}$. For identical rings induced emf will be same. But currents will be different. Given $h_A>h_B$. Hence, $v_A>v_B$ as $\left(h=\frac{v^2}{2 g}\right)$.
If $\rho_A>\rho_B$, then, $I_A < I_B$. In this case given condition can be fulfilled if $m_A < m_B$.
If $\rho_A < \rho_B$, then $I_A>I_B$. In this case given condition can be fulfilled of $m_A \leq m_B$.

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2009 (Paper 2)
Two moles of ideal helium gas are in a rubber balloon at $30^{\circ} \mathrm{C}$. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to $35^{\circ} \mathrm{C}$. The amount of heat required in raising the temperature is nearly (take $R=8.31 \mathrm{~J} / \mathrm{mol} . \mathrm{K}$ )

Options:
A $62 \mathrm{~J}$
B $104 \mathrm{~J}$
C $124 \mathrm{~J}$
D $208 \mathrm{~J}$

The correct answer is:
$208 \mathrm{~J}$

Solution:
The heat is supplied at constant pressure. i.e., the process is isobaric

$\begin{array}{l}

\therefore Q=n C_{p} \Delta t \\

=2\left[\frac{5}{2} R\right] \times \Delta t=2 \times \frac{5}{2} \times 8.31 \times 5=208 \mathrm{~J} \\

\left(\because C_{p}=\frac{5}{2} R \text { for mono-atomic gas }\right)

\end{array}$

Topic: Physics, Thermodynamics, JEE Advanced, JEE Advanced 2012 (Paper 2)
Two parallel chords of a circle of radius 2 are at a distance $\sqrt{3}+1$ apart. If the chords subtend at the centre, angles of $\frac{\pi}{k}$ and $\frac{2 \pi}{k}$, where $k>0$, then the value of $[k]$ is
[Note : $[k]$ denotes the largest integer less than or equal to $k$ ]

Options:

The correct answer is:
3

Solution:
Let $\theta=\frac{\pi}{2 k}$
$$
\begin{aligned}
& \cos \theta=\frac{x}{2} \\
& \Rightarrow \quad \cos 2 \theta=\frac{\sqrt{3}+1-x}{2} \\
& \Rightarrow 2 \cos ^2 \theta-1=\frac{\sqrt{3}+1-x}{2} \\
& \Rightarrow 2\left(\frac{x^2}{4}\right)-1=\frac{\sqrt{3}+1-x}{2} \\
& \Rightarrow \quad x^2+x-3-\sqrt{3}=0 \\
& \Rightarrow \quad x=\frac{-1 \pm \sqrt{1+12+4 \sqrt{3}}}{2} \\
& =\frac{-1 \pm \sqrt{13+4 \sqrt{3}}}{2} \\
& =\frac{-1+2 \sqrt{3}+1}{2}=\sqrt{3} \\
& \therefore \quad \cos \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{6} \\
& \therefore \quad \text { Required angle }=\frac{\pi}{k}=2 \theta=\frac{\pi}{3} \\
& \Rightarrow \quad k=3 \\
&
\end{aligned}
$$

Topic: Mathematics, Circle, JEE Main
Two parallel chords of a circle of radius 2 are at a distance $\sqrt{3}+1$ apart. If the chords subtend at the centre, angles of $\frac{\pi}{k}$ and $\frac{2 \pi}{k}$, where $k>0$, then the value of $[k]$ is
[Note : $[k]$ denotes the largest integer less than or equal to $k$ ]

Options:

The correct answer is:
3

Solution:
Let $\theta=\frac{\pi}{2 k}$
$$
\begin{aligned}
& \cos \theta=\frac{x}{2} \\
& \Rightarrow \quad \cos 2 \theta=\frac{\sqrt{3}+1-x}{2} \\
& \Rightarrow 2 \cos ^2 \theta-1=\frac{\sqrt{3}+1-x}{2} \\
& \Rightarrow 2\left(\frac{x^2}{4}\right)-1=\frac{\sqrt{3}+1-x}{2} \\
& \Rightarrow \quad x^2+x-3-\sqrt{3}=0 \\
& \Rightarrow \quad x=\frac{-1 \pm \sqrt{1+12+4 \sqrt{3}}}{2} \\
& =\frac{-1 \pm \sqrt{13+4 \sqrt{3}}}{2} \\
& =\frac{-1+2 \sqrt{3}+1}{2}=\sqrt{3} \\
& \therefore \quad \cos \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{6} \\
& \therefore \quad \text { Required angle }=\frac{\pi}{k}=2 \theta=\frac{\pi}{3} \\
& \Rightarrow \quad k=3 \\
&
\end{aligned}
$$

Topic: Mathematics, Circle, JEE Advanced, JEE Advanced 2010 (Paper 2)
Two particles of mass $m$ each are tied at the ends of a light string of length $2 a$. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance $a$ from the centre $P$ (as shown in the figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force $F$. As a result, the particles move towards each other on the surface.
The magnitude of acceleration, when the separation between them becomes $2 x$, is

Options:
A
$\frac{F}{2 m} \frac{a}{\sqrt{a^2-x^2}}$

B
$\frac{F}{2 m} \frac{x}{\sqrt{a^2-x^2}}$

C
$\frac{F}{2 m} \frac{x}{a}$

D
$\frac{F}{2 m} \frac{\sqrt{a^2-x^2}}{x}$

The correct answer is:
$\frac{F}{2 m} \frac{x}{\sqrt{a^2-x^2}}$

Solution:
6.
$$
\begin{aligned}
& 2 T \cos \theta=F \\
& \therefore T=\frac{F}{2} \sec \theta \\
& \text { Acceleration of particle }=\frac{T \sin \theta}{m}=\frac{F \tan \theta}{2 m}=\frac{F}{2 m} \frac{x}{\sqrt{a^2-x^2}}
\end{aligned}
$$
$$
\therefore \text { Option (b) is correct. }
$$

Topic: Physics, Laws of Motion, JEE Advanced, JEE Advanced 2007 (Paper 1)
Two small particles of equal masses start moving in opposite directions from a point $A$ in a horizontal circular orbit. Their tangential velocities are $v$ and $2 v$ respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at $A$, these two particles will again reach the point $A$ ?

Options:
A
4

B
3

C
2

D
1

The correct answer is:
2

Solution:
At first collision one particle having speed $2 v$ will rotate $240^{\circ}\left(\right.$ or $\left.\frac{4 \pi}{3}\right)$ while other particle having speed $v$ will rotate $120^{\circ}$ (or $\left.\frac{2 \pi}{3}\right)$. At first collision they will exchange their velocities. Now as shown in figure, after two collisions they will again reach at point $A$.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two small particles of equal masses start moving in opposite directions from a point $A$ in a horizontal circular orbit. Their tangential velocities are $v$ and $2 v$ respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at $A$, these two particles will again reach the point $A$ ?

Options:
A
4

B
3

C
2

D
1

The correct answer is:
2

Solution:
At first collision one particle having speed $2 v$ will rotate $240^{\circ}\left(\right.$ or $\left.\frac{4 \pi}{3}\right)$ while other particle having speed $v$ will rotate $120^{\circ}$ (or $\left.\frac{2 \pi}{3}\right)$. At first collision they will exchange their velocities. Now as shown in figure, after two collisions they will again reach at point $A$.

Topic: Physics, Center of Mass Momentum and Collision, JEE Advanced, JEE Advanced 2009 (Paper 1)
Two soap bubbles $A$ and $B$ are kept in a closed chamber where the air is maintained at pressure $8 \mathrm{Nm}^{-2}$. The radii of bubbles $A$ and $B$ are $2 \mathrm{~cm}$, respectively. Surface tension of the soap-water used to make bubbles is $0.04$ $\mathrm{Nm}^{-1}$. Find the ratio $\frac{n_B}{n_A}$, where $n_A$ and $n_B$ are the number of moles of air in bubbles $A$ and $B$, respectively. [Neglect the effect of gravity]

Options:

The correct answer is:
6

Solution:
Although not given in the question but we will have to assume that temperatures of $A$ and $B$ are same.
$$
\begin{aligned}
& \frac{n_B}{n_A}=\frac{p_B V_B / R T}{p_A V_A / R T}=\frac{p_B V_B}{p_A V_A} \\
&= \frac{p+4 s / r_A \times 4 / 3 \pi\left(r_A\right)^3}{\left(p+4 s / r_B\right) \times 4 / 3 \pi\left(r_B\right)^3} \\
&(s=\text { surface tension) }
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Advanced, JEE Advanced 2009 (Paper 2)
Two solid cylinders $P$ and $Q$ of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder $P$ has most of its mass concentrated near its surface, while $Q$ has most of its mass concentrated near the axis. Which statement(s) is(are) correct?

Options:
A Both cylinders $P$ and $Q$ reach the ground at the same time.
B Cylinders $P$ has larger linear acceleration than cylinder $Q$.
C Both cylinders reach the ground with same translational kinetic energy.
D Cylinder $Q$ reaches the ground with larger angular speed.

The correct answers are:
Cylinder $Q$ reaches the ground with larger angular speed.

Solution:
As we know, acceleration of the center of mass of cylinder rolling down an inclined plane

$a_{c}=\frac{g \sin \theta}{1+\frac{I}{M R^{2}}}$

In case of $P$ the mass is concentrated away from the axis, So $I_{P}>I_{Q}$

$\therefore \quad a_{P} < a_{Q} \Rightarrow v_{\mathrm{P}} < v_{\mathrm{Q}} \Rightarrow \omega_{\mathrm{P}} < \omega_{\mathrm{Q}}$

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2012 (Paper 2)
Two solid spheres $A$ and $B$ of equal volumes but of different densities $d_A$ and $d_B$ are connected by a string. They are fully immersed in a fluid of density $d_F$. They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if

Options:
A
$d_A < d_F$
B
$d_B>d_F$

C
$d_A>d_F$

D
$d_A+d_B=2 d_F$

The correct answers are:
$d_A < d_F$,
$d_B>d_F$
,
$d_A+d_B=2 d_F$

Solution:
Equilibrium of $A$
$$
\begin{aligned}
V d_F g & =T+W_A \\
& =T+V d_A g
\end{aligned}
$$
Equilibrium of $B$,
$$
T+V d_F g=V d_B g
$$
Adding Eqs. (i) and (ii), we get $2 d_f=d_A+d_B$
$\therefore$ Option (d) is correct. From Eq. (i), we can see that
$$
d_F>d_A \quad \text { [as } T>0 \text { ] }
$$
$\therefore$ Option (a) is correct.
From Eq. (ii) we can see that,
$$
d_B>d_F
$$
$\therefore$ Option (a) is correct.
$\therefore$ Correct options are (a), (b) and (d).
Analysis of Question
Question is moderately difficult but conceptwise it is good.

Topic: Physics, Mechanical Properties of Fluids, JEE Advanced, JEE Advanced 2011 (Paper 2)
Two spherical bodies $A$ (radius $6 \mathrm{~cm}$ ) and $B$ (radius $18 \mathrm{~cm}$ ) are at temperatures $T_1$ and $T_2$, respectively. The maximum intensity in the emission spectrum of $A$ is at $500 \mathrm{~nm}$ and in that of $B$ is at $1500 \mathrm{~nm}$. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by $A$ to that of $B$ ?

Options:

The correct answer is:
9

Solution:
$$
\begin{aligned}
& \text { } \lambda_m \propto \frac{1}{T} \\
& \therefore \frac{\lambda_A}{\lambda_B}=\frac{T_B}{T_A}=\frac{500}{1500}=\frac{1}{3} \\
& E \propto T^4 A \text { (where } A=\text { surface area } \\
& =4 \pi R^2 \text { ) } \\
& \therefore E \propto T^4 R^2 \\
& \frac{E_A}{E_B}=\left(\frac{T_A}{T_B}\right)^4\left(\frac{R_A}{R_B}\right)^2 \\
& =(3)^4\left(\frac{6}{18}\right)^2=9
\end{aligned}
$$
$\therefore$ Answer is 9 .

Topic: Physics, Thermal Properties of Matter, JEE Advanced, JEE Advanced 2010 (Paper 1)
Two spherical planets $P$ and $Q$ have the same uniform density $\rho$, masses $M_{P}$ and $M_{Q}$ and surface areas $A$ and 4A respectively. A spherical planet $R$ also has uniform density $\rho$ and its mass is $\left(M_{P}+M_{Q}\right)$. The escape velocities from the planets $P, Q$ and $R$ are $V_{P}, V_{Q}$ and $V_{R}$, respectively. Then

Options:
A $V_{Q}>V_{R}>V_{P}$
B $V_{R}>V_{Q}>V_{P}$
C $V_{R} / V_{P}=3$
D $V_{P} / V_{Q}=\frac{1}{2}$

The correct answers are:
$V_{R}>V_{Q}>V_{P}$, $V_{P} / V_{Q}=\frac{1}{2}$

Solution:
Here Planets $P$ and $Q$ have the same uniform density ' $\rho$ ' and surface areas $A$ and $4 A$ respectively. Let the mass of $P, M_{P}$ be $m$.

Then $m=\rho \times \frac{4}{3} \pi r^{3}=\rho \times \frac{4}{3} \pi\left[\frac{A}{4 \pi}\right]^{3 / 2}$

The mass of $M_{Q}=\rho \times \frac{4}{3} \pi\left[\frac{4 A}{4 \pi}\right]^{3 / 2}=8 \mathrm{~m}$

$\therefore \quad$ The mass of Planet $R=8 \mathrm{~m}+\mathrm{m}=9 \mathrm{~m}$

If the radius of $P=r$

Then the radius of $Q=2 r$

$\left[\because r_{Q}=\left(\frac{4 A}{4 \pi}\right)^{3 / 2}=2\left(\frac{A}{4 \pi}\right)^{3 / 2}\right]$

and radius of $R=9^{1 / 3} r$

$\left[\begin{array}{l}

\because M_{R}=M_{P}+M_{Q} \\

r_{R}^{3}=r^{3}+(2 r)^{3}=9 r^{3}

\end{array}\right]$

As we know, escape velocity from the planet

$\begin{aligned} V_{e}=\sqrt{\frac{2 G M}{R}} & \therefore v_{P}=\sqrt{\frac{2 G M_{P}}{R_{p}}}=\sqrt{\frac{2 G m}{r}} \\ v_{Q} &=\sqrt{\frac{2 G M_{Q}}{R_{Q}}}=\sqrt{\frac{2 G(8 \mathrm{~m})}{2 r}}=2 v_{P} \\ v_{R} &=\sqrt{\frac{2 G(9 \mathrm{~m})}{9^{1 / 3} r}}=9^{1 / 3} v_{P} \end{aligned}$

Topic: Physics, Gravitation, JEE Advanced, JEE Advanced 2012 (Paper 2)
Two transparent media of refractive indices $\mu_1$ and $\mu_3$ have a solid lens shaped transparent material of refractive index $\mu_2$ between them as shown in figures in Column II. A ray traversing these media is also shown in the figures. In Column I different relationships between $\mu_1, \mu_2$ and $\mu_3$ are given. Match them to the ray diagram shown in Column II.

Options:
A
(A) P,R, (B) Q,S,T, (C) P,R,T, (D) Q,S

B
(A) P, (B) Q,S, (C) P,R,T, (D) Q,R,S

C
(A) P,R, (B) Q,S, (C) P,T, (D) Q,R,S

D
(A) P, (B) Q,S,T, (C) P,T, (D) Q,S

The correct answer is:
(A) P,R, (B) Q,S,T, (C) P,R,T, (D) Q,S

Solution:
(A) $\rightarrow$ since $\mu_1 < \mu_2$, the ray of light will bend towards normal after first refraction.
(B) $\rightarrow \mu_1>\mu_2$, the ray of light will bend away from the normal after first refraction.
(C) $\rightarrow$ since $\mu_2=\mu_3$ means in second refraction there will be no change in the path of ray of light.
(D) $\rightarrow$ Since $\mu_2>\mu_3$, ray of light will bend away from the normal after second refraction.
Therefore the correct options are as under.
(A) $\rightarrow$ P, R
(B) $\rightarrow$ Q, S, T
(C) $\rightarrow$ P, R, T
(D) $\rightarrow$ Q, $S$

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2010 (Paper 2)
Two wires each carrying a steady current $I$ are shown in four configurations in Column I. Some of the resulting effects are described in Column II. Match the statements in Column I with the statements in Column II and indicate your answer by darkening appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

Options:
A
A-p; B-p, s; C-p; D-q

B
A-p, q; B-r, s; C-r, s; D-r, s

C
A-q; B-s; C-s; D-p, q

D
A-q, r; B-p; C-q, r; D-q, s

The correct answer is:
A-q, r; B-p; C-q, r; D-q, s

Solution:
No Solution Available

Topic: Physics, Magnetic Effects of Current, JEE Main
Two wires each carrying a steady current $I$ are shown in four configurations in Column I. Some of the resulting effects are described in Column II. Match the statements in Column I with the statements in Column II and indicate your answer by darkening appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

Options:
A
A-p; B-p, s; C-p; D-q

B
A-p, q; B-r, s; C-r, s; D-r, s

C
A-q; B-s; C-s; D-p, q

D
A-q, r; B-p; C-q, r; D-q, s

The correct answer is:
A-q, r; B-p; C-q, r; D-q, s

Solution:
No Solution Available

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2007 (Paper 2)
Under the influence of the coulomb field of charge $+Q$, a charge $-q$ is moving around it in an elliptical orbit. Find out the correct statement(s).

Options:
A
The angular momentum of the charge $-q$ is constant

B
The linear momentum of the charge $-q$ is constant

C
The angular velocity of the charge $-q$ is constant

D
The linear speed of the charge $-q$ is constant

The correct answers are:
The angular momentum of the charge $-q$ is constant

Solution:
Net torque on $(-q$ ) about a point (say $P$ ) lying over $+Q$ is zero. Therefore, angular momentum of $(-q)$ about point $P$ should remain constant.

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2009 (Paper 2)