datavorous/entrance-exam-dataset huggingface dataset:
A small mass $m$ is attached to a massless string whose other end is fixed at $P$ as shown in the figure. The mass is undergoing circular motion in the $x-y$ plane with centre at $O$ and constant angular speed $\omega$. If the angular momentum of the system, calculated about $O$ and $P$ are denoted by $\vec{L}_{O}$ and $\vec{L}_{P}$ respectively, then

Options:
A $\vec{L}_{O}$ and $\vec{L}_{P}$ do not vary with time
B $\vec{L}_{O}$ varies with time while $\vec{L}_{P}$ remains constant
C $\vec{L}_{O}$ remains constant while $\vec{L}_{P}$ varies with time
D $\vec{L}_{O}$ and $\vec{L}_{P}$ both vary with time

The correct answer is:
$\vec{L}_{O}$ remains constant while $\vec{L}_{P}$ varies with time

Solution:
For all locations of $\mathrm{m}$ the angular momentum of the mass $m$ about $O$ i.e., $L_{o}$ is $m r^{2} \omega$ and is directed toward $+z$ direction. The angular momentum of mass $m$ about $P$ i.e., $L_{p}$ is $m v l$ and is directed for the given location of $m$ as shown in the figure. For different location of $m$, the direction of $\vec{L}_{P}$ remains changing.

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2012 (Paper 1)
A small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches up to a maximum height of $\frac{3 v^2}{4 g}$ with respect to the initial$$
\text { position. The object is }
$$

Options:
A
ring

B
solid sphere

C
hollow sphere

D
disc

The correct answer is:
disc

Solution:
$$
\begin{aligned}
\frac{1}{2} m v^2+\frac{1}{2} I\left(\frac{v}{R}\right)^2 & =m g\left(\frac{3 v^2}{4 g}\right) \\
I & =\frac{1}{2} m R^2
\end{aligned}
$$
So, body is disc.
The correct option is (d).

Topic: Physics, Work Power Energy, JEE Main
A small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches up to a maximum height of $\frac{3 v^2}{4 g}$ with respect to the initial$$
\text { position. The object is }
$$

Options:
A
ring

B
solid sphere

C
hollow sphere

D
disc

The correct answer is:
disc

Solution:
$$
\begin{aligned}
\frac{1}{2} m v^2+\frac{1}{2} I\left(\frac{v}{R}\right)^2 & =m g\left(\frac{3 v^2}{4 g}\right) \\
I & =\frac{1}{2} m R^2
\end{aligned}
$$
So, body is disc.
The correct option is (d).

Topic: Physics, Work Power Energy, JEE Advanced, JEE Advanced 2007 (Paper 2)
A small roller of diameter 20 cm has an axle of diameter 10 cm (see figure below on the left). It is on a horizontal floor and a meter scale is positioned horizontally on its axle with one edge of the scale on top of the axle (see figure on the right). The scale is now pushed slowly on the axle so that it moves without slipping on the axle, and the roller starts rolling without slipping. After the roller has moved 50 cm, the position of the scale will look like (figures are schematic and not drawn to scale)

Options:
A

B

C

D

The correct answer is:

Solution:
Angular displacement θ=S010=5

Displacement of point B is,S=r.θ=15×5=75 cm

Displacement of roller =50 cm

Relative displacement =75-50=25 cm

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2020 (Paper 1)
A solid sphere is in pure rolling motion on an inclined surface having inclination $\theta$.

Options:
A
frictional force acting on sphere is $f=\mu m g \cos \theta$

B
$f$ is disspative force

C
friction will increase its angular velocity and decreases its linear velocity

D
If $\theta$ decreases, friction will decrease

The correct answers are:
friction will increase its angular velocity and decreases its linear velocity
,
If $\theta$ decreases, friction will decrease

Solution:
In case of pure rolling,
$$
\begin{aligned}
& f=\frac{m g \sin \theta}{1+\frac{m R^2}{I}} \\
\therefore & f \propto \sin \theta
\end{aligned}
$$
(upwards)
Therefore, as $\theta$ decreases force of friction will also decrease.

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2006
A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho=k r^a$, where $k$ and $a$ are constants and $r$ is the distance from its centre. If the electric field at $r=\frac{R}{2}$ is $\frac{1}{8}$ times that at $r=R$, find the value of $a$.

Options:

The correct answer is:
2

Solution:
From Gauss theorem,
$$
\begin{aligned}
& E \propto \frac{q}{r^2} \quad(q=\text { charge enclosed }) \\
& \therefore \quad \frac{E_2}{E_1}=\frac{q_2}{q_1}=\frac{r_1^2}{r_2^2} \\
& 8=\frac{\int_0^R\left(4 \pi r^2\right) k r^a d r}{\int_0^{R / 2}\left(4 \pi r^2\right) k r^a d r} \times \frac{(R / 2)^2}{(R)^2} \\
&
\end{aligned}
$$
Solving this equation we get, $a=2$

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2009 (Paper 2)
$$
\text { A solid sphere of radius } R \text { has moment of inertia } J \text { about its qeometrical axis. If it is }
$$
melted into a disc of radius $r$ and thickness $t$. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to $I$, then the value of $r$ is equal to

Options:
A
$\frac{2}{\sqrt{15}} R$

B
$\frac{2}{\sqrt{5}} R$

C
$\frac{3}{\sqrt{15}} R$

D
$\frac{\sqrt{3}}{\sqrt{15}} R$

The correct answer is:
$\frac{2}{\sqrt{15}} R$

Solution:
$\frac{2}{3} M R^2=\frac{1}{2} M r^2+M r^2$ or $\frac{2}{3} M R^2=\frac{3}{2} M r^2 \Rightarrow r=\frac{2}{\sqrt{15}} R$

Topic: Physics, Rotational Motion, JEE Main
$$
\text { A solid sphere of radius } R \text { has moment of inertia } J \text { about its qeometrical axis. If it is }
$$
melted into a disc of radius $r$ and thickness $t$. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to $I$, then the value of $r$ is equal to

Options:
A
$\frac{2}{\sqrt{15}} R$

B
$\frac{2}{\sqrt{5}} R$

C
$\frac{3}{\sqrt{15}} R$

D
$\frac{\sqrt{3}}{\sqrt{15}} R$

The correct answer is:
$\frac{2}{\sqrt{15}} R$

Solution:
$\frac{2}{3} M R^2=\frac{1}{2} M r^2+M r^2$ or $\frac{2}{3} M R^2=\frac{3}{2} M r^2 \Rightarrow r=\frac{2}{\sqrt{15}} R$

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2006
A solution of a metal ion when treated with $\mathrm{KI}$ gives a red precipitate which dissolved in excess KI to give a colourless solution. Moreover, the solution of metal ion on treatment with a solution of cobalt (II) thiocyanate gives rise to a deep blue crystalline precipitqte. The metal ion is

Options:
A
$\mathrm{Pb}^{2+}$

B
$\mathrm{Hg}^{2+}$

C
$\mathrm{Cu}^{2+}$

D
$\mathrm{Co}^2$

The correct answer is:
$\mathrm{Hg}^{2+}$

Solution:
$$
\begin{aligned}
& \mathrm{Hg}^{2+}+2 \mathrm{I}^{-} \longrightarrow \mathrm{HgI}_2, \mathrm{HgI}_2+\mathrm{H}^{-} \\
& \text {(Excess) }\left.\longrightarrow \mathrm{HgI}_4\right]^{2-} \\
& \text { Soluble }^{2-} \\
& \mathrm{Hg}^{2+}+\mathrm{Co}(\mathrm{SCN})_2 \longrightarrow \mathrm{Hg}(\mathrm{SCN})_2
\end{aligned}
$$
Blue crystalline ppt.

Topic: Chemistry, Practical Chemistry, JEE Advanced, JEE Advanced 2007 (Paper 2)
A solution of colourless salt $\mathrm{H}$ on boiling with excess $\mathrm{NaOH}$ produces a non-flammable gas. The gas evolution ceases after sometime. Upon addition of $\mathrm{Zn}$ dust to the same solution, the gas evolution restarts. The colourless salt (s) $\mathrm{H}$ is (are)

Options:
A
$\mathrm{NH}_4 \mathrm{NO}_3$

B
$\mathrm{NH}_4 \mathrm{NO}_2$

C
$\mathrm{NH}_4 \mathrm{Cl}$

D
$\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4$

The correct answers are:
$\mathrm{NH}_4 \mathrm{NO}_3$
,
$\mathrm{NH}_4 \mathrm{NO}_2$

Solution:
$$
\begin{aligned}
& \mathrm{NH}_4 \mathrm{NO}_3+\mathrm{NaOH} \longrightarrow \mathrm{NaNO}_3+\mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} \\
& \mathrm{NaNO}_3+8[\mathrm{H}] \mathrm{NaOH}+\mathrm{NH}_3+2 \mathrm{H}_2 \mathrm{O} \\
& \mathrm{NH}_4 \mathrm{NO}_2+\mathrm{NaOH} \longrightarrow \mathrm{NaHO}_2+\mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} \\
& \mathrm{NaNO}_2+6[\mathrm{H}] \longrightarrow \mathrm{NaOH}+\mathrm{NH}_3+\mathrm{H}_2 \mathrm{O}
\end{aligned}
$$

Topic: Chemistry, d and f Block Elements, JEE Advanced, JEE Advanced 2008 (Paper 1)
A solution when diluted with $\mathrm{H}_2 \mathrm{O}$ and boiled, it gives a white precipitate. On addition of excess $\mathrm{NH}_4 \mathrm{Cl} / \mathrm{NH}_4 \mathrm{OH}$, the volume of precipitate decreases leaving behind a white gelatinous precipitate. Identify the precipitate which dissolves in $\mathrm{NH}_4 \mathrm{OH} / \mathrm{NH}_4 \mathrm{Cl}$.

Options:
A
$\mathrm{Zn}(\mathrm{OH})_2$

B
$\mathrm{Al}(\mathrm{OH})_3$

C
$\mathrm{Mg}(\mathrm{OH})_2$

D
$\mathrm{Ca}(\mathrm{OH})_2$

The correct answer is:
$\mathrm{Zn}(\mathrm{OH})_2$

Solution:
A solution of $\mathrm{Zn}^{2+}$ ion reacts with water which on boiling give white ppt. of $\mathrm{Zn}(\mathrm{OH})_2$. This precipitates dissolve in excess of $\mathrm{NH}_4 \mathrm{OH}$ in presence of $\mathrm{NH}_4 \mathrm{Cl}$ because $\mathrm{Zn}(\mathrm{OH})_2$ on reaction with $\mathrm{NH}_4 \mathrm{OH}$ in presence of $\mathrm{NH}_4 \mathrm{Cl}$ forms tetrammine zinc (II) complex (soluble).

Topic: Chemistry, Practical Chemistry, JEE Advanced, JEE Advanced 2006
A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, $A$ is the point of contact. $B$ is the centre of the sphere and $C$ is its topmost point. Then,

Options:
A
$\overrightarrow{\mathbf{v}}_C-\overrightarrow{\mathbf{v}}_A=2\left(\overrightarrow{\mathbf{v}}_B-\overrightarrow{\mathbf{v}}_C\right)$

B
$\overrightarrow{\mathbf{v}}_C-\overrightarrow{\mathbf{v}}_B=\overrightarrow{\mathbf{v}}_B-\overrightarrow{\mathbf{v}}_A$

C
$\left|\overrightarrow{\mathbf{v}}_C-\overrightarrow{\mathbf{v}}_A\right|=2\left|\overrightarrow{\mathbf{v}}_B-\overrightarrow{\mathbf{v}}_C\right|$

D
$\left|\overrightarrow{\mathbf{v}}_C-\overrightarrow{\mathbf{v}}_A\right|=4\left|\overrightarrow{\mathbf{v}}_B\right|$

The correct answers are:
$\overrightarrow{\mathbf{v}}_C-\overrightarrow{\mathbf{v}}_B=\overrightarrow{\mathbf{v}}_B-\overrightarrow{\mathbf{v}}_A$
,
$\left|\overrightarrow{\mathbf{v}}_C-\overrightarrow{\mathbf{v}}_A\right|=2\left|\overrightarrow{\mathbf{v}}_B-\overrightarrow{\mathbf{v}}_C\right|$

Solution:
$$
v_A=0, v_B=v \text { and } v_C=2 v
$$

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2009 (Paper 2)
A spherical metal shell $A$ of radius $R_A$ and a solid metal sphere $B$ of radius $R_B < \left(R_A\right)$ are kept far apart and each is given charge $+Q$. Now, they are connected by a thin metal wire. Then

Options:
A
$E_A^{\text {inside }}=0$

B
$Q_A>Q_B$

C
$\frac{\sigma_{\mathrm{A}}}{\sigma_B}=\frac{R_B}{R_A}$

D
$E_A^{\text {on surface }} < E_B^{\text {on surface }}$

The correct answers are:
$E_A^{\text {inside }}=0$
,
$Q_A>Q_B$
,
$\frac{\sigma_{\mathrm{A}}}{\sigma_B}=\frac{R_B}{R_A}$
,
$E_A^{\text {on surface }} < E_B^{\text {on surface }}$

Solution:
Inside a conducting shell, electric field is always zero. Therefore, option (a) is correct. When the two are connected, their potentials become the same.
$$
\begin{array}{ll}
\therefore & V_A=V_B \\
\text { or } & \frac{Q_A}{R_A}=\frac{Q_B}{R_B} \quad\left(\because V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}\right)
\end{array}
$$
Since, $R_A>R_B$
$$
\therefore \quad Q_A>Q_B
$$
$\therefore$ Option (b) is correct.
Potential is also equal to,
$$
\begin{array}{rlrl}
V & =\frac{\sigma R}{\varepsilon_0} \\
V_A & =V_B \\
\therefore \quad \sigma_A R_A & =\sigma_B R_B \\
\text { or } \quad \sigma_B & \frac{\sigma_A}{\sigma_B} & =\frac{R_B}{R_A}
\end{array}
$$
$\therefore$ Option (c) is correct.
Electric field on surface,
$$
E=\frac{\sigma}{E_0} \text { or } E \propto \sigma
$$
or $\quad \sigma_A < \sigma_B$
Since, $\quad \sigma_A < \sigma_B$
$\therefore \quad E_A < E_B$
$\therefore$ Option (d) is also correct.
$\therefore$ Correct options are (a), (b), (c) and (d).
Analysis of Question
(i) Question is simple.
(ii) $E=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R^2}=\frac{\sigma}{\varepsilon_0}$
(On surface)
As, $\sigma=$ surface charge density
$$
=\frac{Q}{4 \pi R^2}
$$
(iii) $V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}=\frac{\sigma R}{\varepsilon_0}$

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2011 (Paper 1)
A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is

Options:
A
zero everywhere

B
non-zero and uniform

C
non-uniform

D
zero only at its centre

The correct answer is:
non-zero and uniform

Solution:
Inside the cavity, field at any point is uniform and non-zero. Therefore, correct option is (b).

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2007 (Paper 2)
A spherically symmetric gravitational system of particles has a mass density
$$
\rho=\left\{\begin{array}{l}
\rho_0 \text { for } r \leq R \\
0 \text { for } r>R
\end{array}\right.
$$
where $\rho_0$ is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed $v$ as a function of distance $r(0 < r < \infty)$ from the centre of the system is represented by

Options:
A

B

C

D

The correct answer is:

Solution:
For $r \leq R$ :
$$
\frac{m v^2}{r}=\frac{G \cdot m m^{\prime}}{r^2}
$$
Here, $m^{\prime}=\left(\frac{4}{3} \pi r^3\right) \rho_0$
Substituting in Eq. (i) we get, $v \propto r$
i.e., $v-r$ graph is a straight line passing through origin.
For $r>R$ :
$$
\frac{m v^2}{r}=\frac{G \cdot m\left(\frac{4}{3} \pi R^3\right) \rho_0}{r^2}
$$
or $\quad V \propto \frac{1}{\sqrt{r}}$
The corresponding $v-r$ graph will be as shown in option (c).
$\therefore$ correct option is (c).

Topic: Physics, Gravitation, JEE Main
A spherically symmetric gravitational system of particles has a mass density
$$
\rho=\left\{\begin{array}{l}
\rho_0 \text { for } r \leq R \\
0 \text { for } r>R
\end{array}\right.
$$
where $\rho_0$ is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed $v$ as a function of distance $r(0 < r < \infty)$ from the centre of the system is represented by

Options:
A

B

C

D

The correct answer is:

Solution:
For $r \leq R$ :
$$
\frac{m v^2}{r}=\frac{G \cdot m m^{\prime}}{r^2}
$$
Here, $m^{\prime}=\left(\frac{4}{3} \pi r^3\right) \rho_0$
Substituting in Eq. (i) we get, $v \propto r$
i.e., $v-r$ graph is a straight line passing through origin.
For $r>R$ :
$$
\frac{m v^2}{r}=\frac{G \cdot m\left(\frac{4}{3} \pi R^3\right) \rho_0}{r^2}
$$
or $\quad V \propto \frac{1}{\sqrt{r}}$
The corresponding $v-r$ graph will be as shown in option (c).
$\therefore$ correct option is (c).

Topic: Physics, Gravitation, JEE Advanced, JEE Advanced 2008 (Paper 1)
A stationary source is emitting sound at a fixed frequency $f_0$, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is $1.2 \%$ of $f_0$. What is the difference in the speeds of the cars (in km per hour) to the nearest integer ? The cars are moving at constant speeds much smaller than the speed of sound which is $330 \mathrm{~ms}^{-1}$.

Options:

The correct answer is:
7

Solution:
Firstly, car will be treated as an observer which is approaching the source. Then, it will be treated as a source, which is moving in the direction of sound.
Hence, $\quad f_1=f_0\left(\frac{v+v_1}{v-v_1}\right)$
$$
\begin{aligned}
f_2 & =f_0\left(\frac{v+v_2}{v-v_2}\right) \\
\therefore \quad f_1-f_2 & =\left(\frac{1.2}{100}\right) f_0 \\
& =f_0\left[\frac{v+v_1}{v-v_1}-\frac{v+v_2}{v-v_2}\right] \\
\text { or }\left(\frac{1.2}{100}\right) f_0 & =\frac{2 v\left(v_1-v_2\right)}{\left(v-v_1\right)\left(v-v_2\right)} f_0
\end{aligned}
$$
As $v_1$ and $v_2$ are very very less than $v$. We can write, $\left(v-v_1\right)$ or $\left(v-v_2\right) \approx v$
$$
\therefore \quad\left(\frac{1.2}{100}\right) f_0=\frac{2\left(v_1-v_2\right)}{v} f_0
$$
or $\left(v_1-v_2\right)=\frac{v \times 1.2}{200}$
$$
\begin{aligned}
& =\frac{330 \times 1.2}{200}=1.98 \mathrm{~ms}^{-1} \\
& =7.128 \mathrm{kmh}^{-1}
\end{aligned}
$$
$\therefore$ The nearest integer is 7 .
Hence, $\quad f_1=f_0\left(\frac{v+v_1}{v-v_1}\right)$
$$
\begin{aligned}
f_2 & =f_0\left(\frac{v+v_2}{v-v_2}\right) \\
\therefore \quad f_1-f_2 & =\left(\frac{1.2}{100}\right) f_0 \\
& =f_0\left[\frac{v+v_1}{v-v_1}-\frac{v+v_2}{v-v_2}\right] \\
\text { or }\left(\frac{1.2}{100}\right) f_0 & =\frac{2 v\left(v_1-v_2\right)}{\left(v-v_1\right)\left(v-v_2\right)} f_0
\end{aligned}
$$
As $v_1$ and $v_2$ are very very less than $v$. We can write, $\left(v-v_1\right)$ or $\left(v-v_2\right) \approx v$
$$
\therefore \quad\left(\frac{1.2}{100}\right) f_0=\frac{2\left(v_1-v_2\right)}{v} f_0
$$
or $\left(v_1-v_2\right)=\frac{v \times 1.2}{200}$
$$
\begin{aligned}
& =\frac{330 \times 1.2}{200}=1.98 \mathrm{~ms}^{-1} \\
& =7.128 \mathrm{kmh}^{-1}
\end{aligned}
$$
$\therefore$ The nearest integer is 7 .

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2010 (Paper 1)
A steady current $I$ goes through a wire loop $P Q R$ having shape of a right angle triangle with $P Q=3 x, P R=4 x$ and $Q R=5 x$. If the magnitude of the magnetic field at $P$ due to this loop is $k\left(\frac{\mu_0 I}{48 \pi x}\right)$, find the value of $k$.

Options:

The correct answer is:
7

Solution:
Magnetic field at point $P$ due to wires $R P$ and $R Q$ is zero. Only wire $Q R$ will produce magnetic field at $P$.
$$
r=3 x \cos 37^{\circ}=(3 x)\left(\frac{4}{5}\right)=\frac{12 x}{5}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
A steady current $I$ goes through a wire loop $P Q R$ having shape of a right angle triangle with $P Q=3 x, P R=4 x$ and $Q R=5 x$. If the magnitude of the magnetic field at $P$ due to this loop is $k\left(\frac{\mu_0 I}{48 \pi x}\right)$, find the value of $k$.

Options:

The correct answer is:
7

Solution:
Magnetic field at point $P$ due to wires $R P$ and $R Q$ is zero. Only wire $Q R$ will produce magnetic field at $P$.
$$
r=3 x \cos 37^{\circ}=(3 x)\left(\frac{4}{5}\right)=\frac{12 x}{5}
$$

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2009 (Paper 2)
A straight line $L$ through the point $(3,-2)$ is inclined at an angle $60^{\circ}$ to the line $\sqrt{3} x+y=1$. If $L$ also intersects the $X$-axis, then the equation of $L$ is

Options:
A
$y+\sqrt{3} x+2-3 \sqrt{3}=0$

B
$y-\sqrt{3} x+2+3 \sqrt{3}=0$

C
$\sqrt{3} y-x+3+2 \sqrt{3}=0$

D
$\sqrt{3} y+x-3+2 \sqrt{3}=0$

The correct answer is:
$y-\sqrt{3} x+2+3 \sqrt{3}=0$

Solution:
A straight line passing through $P$ and making an angle of $\alpha=60^{\circ}$, is given by $\frac{y-y_1}{x-x_1}=\tan (\theta \pm \alpha)$
where, $\quad \sqrt{3} x+y=1$
$$
\Rightarrow \quad y=-\sqrt{3} x+1
$$
Then, $\quad \tan \theta=-\sqrt{3}$
$$
\begin{array}{ll}
\Rightarrow & \frac{y+2}{x-3}=\frac{\tan \theta \pm \tan \alpha}{1 \mp \tan \theta \tan \alpha} \\
\Rightarrow & \frac{y+2}{x-3}=\frac{-\sqrt{3}+\sqrt{3}}{1-(-\sqrt{3})(\sqrt{3})} \\
\text { and } & \frac{y+2}{x-3}=\frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3})(\sqrt{3})}
\end{array}
$$
$$
\begin{array}{ll}
\Rightarrow & y+2=0 \\
\text { and } & \frac{y+2}{x-3}=\frac{-2 \sqrt{3}}{1-3}=\sqrt{3} \\
\Rightarrow & y+2=\sqrt{3} x-3 \sqrt{3}
\end{array}
$$
Neglecting, $y+2=0$ as it does not intersect $Y$-axis.

Topic: Mathematics, Straight Lines, JEE Main
A straight line $L$ through the point $(3,-2)$ is inclined at an angle $60^{\circ}$ to the line $\sqrt{3} x+y=1$. If $L$ also intersects the $X$-axis, then the equation of $L$ is

Options:
A
$y+\sqrt{3} x+2-3 \sqrt{3}=0$

B
$y-\sqrt{3} x+2+3 \sqrt{3}=0$

C
$\sqrt{3} y-x+3+2 \sqrt{3}=0$

D
$\sqrt{3} y+x-3+2 \sqrt{3}=0$

The correct answer is:
$y-\sqrt{3} x+2+3 \sqrt{3}=0$

Solution:
A straight line passing through $P$ and making an angle of $\alpha=60^{\circ}$, is given by $\frac{y-y_1}{x-x_1}=\tan (\theta \pm \alpha)$
where, $\quad \sqrt{3} x+y=1$
$$
\Rightarrow \quad y=-\sqrt{3} x+1
$$
Then, $\quad \tan \theta=-\sqrt{3}$
$$
\begin{array}{ll}
\Rightarrow & \frac{y+2}{x-3}=\frac{\tan \theta \pm \tan \alpha}{1 \mp \tan \theta \tan \alpha} \\
\Rightarrow & \frac{y+2}{x-3}=\frac{-\sqrt{3}+\sqrt{3}}{1-(-\sqrt{3})(\sqrt{3})} \\
\text { and } & \frac{y+2}{x-3}=\frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3})(\sqrt{3})}
\end{array}
$$
$$
\begin{array}{ll}
\Rightarrow & y+2=0 \\
\text { and } & \frac{y+2}{x-3}=\frac{-2 \sqrt{3}}{1-3}=\sqrt{3} \\
\Rightarrow & y+2=\sqrt{3} x-3 \sqrt{3}
\end{array}
$$
Neglecting, $y+2=0$ as it does not intersect $Y$-axis.

Topic: Mathematics, Straight Lines, JEE Advanced, JEE Advanced 2011 (Paper 1)
A straight line through the vertex $P$ of a $\triangle P Q R$ intersects the side $Q R$ at the point $S$ and the circumcircle of the $\triangle P Q R$ at the point $T$. If $S$ is not the centre of the circumcircle, then

Options:
A
$\frac{1}{P S}+\frac{1}{S T} < \frac{2}{\sqrt{Q S \cdot S R}}$
B
$\frac{1}{P S}+\frac{1}{S T}>\frac{2}{\sqrt{Q S \cdot S R}}$

C
$\frac{1}{P S}+\frac{1}{S T} < \frac{4}{Q R}$
D
$\frac{1}{P S}+\frac{1}{S T}>\frac{4}{Q R}$

The correct answers are:
$\frac{1}{P S}+\frac{1}{S T}>\frac{2}{\sqrt{Q S \cdot S R}}$
,
$\frac{1}{P S}+\frac{1}{S T}>\frac{4}{Q R}$

Solution:
Let a straight line through the vertex $P$ of a given $\triangle P Q R$ intersects the side $Q R$ at the point $S$ and the circumcircle of $\triangle P Q R$ at the point $T$.
Points $P, Q, R, T$ are concyclic, hence
$$
\begin{array}{rlrl}
P S \cdot S T & =Q S \cdot S R & & \\
\Rightarrow \quad P_1 & \geq 2 \sqrt{P S \cdot S T} & \\
& \text { Now, } \frac{P S+S T}{2} & \geq \sqrt{P S \cdot S T} & {[\because \mathrm{AM} \geq \mathrm{GM}]}
\end{array}
$$
and $\quad \frac{1}{P S}+\frac{1}{S T} \geq \frac{2}{\sqrt{P S \cdot S T}}=\frac{2}{\sqrt{Q S \cdot S R}}$
Also, $\quad \frac{S Q+Q R}{2} \geq \sqrt{S Q \cdot S R}$
$\Rightarrow \quad \frac{Q R}{2} \geq \sqrt{S Q \cdot S R}$
$$
\begin{array}{ll}
\Rightarrow & \frac{1}{\sqrt{S Q \cdot S R}} \geq \frac{2}{Q R} \Rightarrow \frac{2}{\sqrt{S Q \cdot S R}} \geq \frac{4}{Q R} \\
\text { Hence, } & \frac{1}{P S}+\frac{1}{S T} \geq \frac{2}{\sqrt{Q S \cdot S R}} \geq \frac{4}{Q R} .
\end{array}
$$

Topic: Mathematics, Straight Lines, JEE Advanced, JEE Advanced 2008 (Paper 1)
A student is performing the experiment of resonance column. The diameter of the column tube is $4 \mathrm{~cm}$. The frequency of the tuning fork is $512 \mathrm{~Hz}$. The air temperature is $38^{\circ} \mathrm{C}$ in which the speed of sound is $336 \mathrm{~m} / \mathrm{s}$. The zero of the meter scale coincides with the top end of the resonance column tube. When the first resonance occurs, the reading of the water level in the column is

Options:
A $14.0 \mathrm{~cm}$
B $15.2 \mathrm{~cm}$
C $16.4 \mathrm{~cm}$
D $17.6 \mathrm{~cm}$

The correct answer is:
$15.2 \mathrm{~cm}$

Solution:
Considering the end correction $[e=0.3 D$ where $\mathrm{D}=$ diameter $]$

$f=n\left[\frac{v}{4(l+e)}\right]$ For first resonance, $n=1$

$\therefore f=\frac{v}{4(l+0.3 D)} \Rightarrow l=\frac{v}{4 f}-0.3 D$

or, $l=\left(\frac{336 \times 100}{4 \times 512}\right)-0.3 \times 4=15.2 \mathrm{~cm}$

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2012 (Paper 2)
A student performed the experiment of determination of focal length of a concave mirror by $u-v$ method using an optical bench of length $1.5 \mathrm{~m}$. The focal length of the mirror used is 24 $\mathrm{cm}$. The maximum error in the location of the image can be $0.2 \mathrm{~cm}$.
The 5 sets of $(u, v)$ values recorded by the student (in $\mathrm{cm})$ are : $(42,56),(48$, $48),(60,40),(66,33),(78,39)$. The data set(s) that cannot come from experiment and is (are) incorrectly recorded, is(are)

Options:
A
$(42,56)$

B
$(48,48)$

C
$(66,33)$

D
$(78,39)$

The correct answers are:
$(66,33)$
,
$(78,39)$

Solution:
Values of options (c) and (d) don't match with the mirror formula,
$$
\frac{1}{v}+\frac{1}{u}=\frac{1}{f}
$$

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2009 (Paper 1)