datavorous/entrance-exam-dataset huggingface dataset:
Paragraph:
Two trains $A$ and $B$ are moving with speeds $20 \mathrm{~m} / \mathrm{s}$ and $30 \mathrm{~m} / \mathrm{s}$, respectively in the same direction on the same straight track, with $B$ ahead of $A$. The engines are at the front ends. The engine of train $A$ blows a long whistle.
Assume that the sound of the whistle is composed of components varying in frequency from $f_1=800 \mathrm{~Hz}$ to $f_2=1120 \mathrm{~Hz}$, as shown in the figure. The spread in the frequency (highest frequency - lowest frequency) is thus $320 \mathrm{~Hz}$. The speed of sound in still air is $340 \mathrm{~m} / \mathrm{s}$.Question:
The speed of sound of the whistle is

Options:
A
$340 \mathrm{~m} / \mathrm{s}$ for passengers in $A$ and $310 \mathrm{~m} / \mathrm{s}$ for passengers in $B$

B
$360 \mathrm{~m} / \mathrm{s}$ for passengers in $A$ and $310 \mathrm{~m} / \mathrm{s}$ for passengers in $B$

C
$310 \mathrm{~m} / \mathrm{s}$ for passengers in $A$ and $360 \mathrm{~m} / \mathrm{s}$ for passengers in $B$

D
$340 \mathrm{~m} / \mathrm{s}$ for passengers in both the trains

The correct answer is:
$360 \mathrm{~m} / \mathrm{s}$ for passengers in $A$ and $310 \mathrm{~m} / \mathrm{s}$ for passengers in $B$

Solution:
$v_{S A}=340+20=360 \mathrm{~m} / \mathrm{s}$ $v_{S B}=340-30=310 \mathrm{~m} / \mathrm{s}$
Therefore, correct option is (b).

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2007 (Paper 2)
Paragraph:
Two trains $A$ and $B$ are moving with speeds $20 \mathrm{~m} / \mathrm{s}$ and $30 \mathrm{~m} / \mathrm{s}$, respectively in the same direction on the same straight track, with $B$ ahead of $A$. The engines are at the front ends. The engine of train $A$ blows a long whistle.
Assume that the sound of the whistle is composed of components varying in frequency from $f_1=800 \mathrm{~Hz}$ to $f_2=1120 \mathrm{~Hz}$, as shown in the figure. The spread in the frequency (highest frequency - lowest frequency) is thus $320 \mathrm{~Hz}$. The speed of sound in still air is $340 \mathrm{~m} / \mathrm{s}$.Question:
The spread of frequency as observed by the passengers in train $B$ is

Options:
A
$310 \mathrm{~Hz}$

B
$330 \mathrm{~Hz}$

C
$350 \mathrm{~Hz}$

D
$290 \mathrm{~Hz}$

The correct answer is:
$310 \mathrm{~Hz}$

Solution:
For the passengers in train $B$, observer is receding with velocity $30 \mathrm{~m} / \mathrm{s}$ and source is approaching with velocity $20 \mathrm{~m} / \mathrm{s}$.
$$
\begin{array}{ll}
\therefore & f_1^{\prime}=800\left(\frac{340-30}{340-20}\right)=775 \mathrm{~Hz} \\
\text { and } & f_2^{\prime}=1120\left(\frac{340-30}{340-20}\right)=1085 \mathrm{~Hz}
\end{array}
$$
$\therefore$ Spread of frequency $=f_2^{\prime}-f_1^{\prime}=310 \mathrm{~Hz}$
$\therefore$ The correct option is (a).

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2007 (Paper 2)
Paragraph:
Two trains $A$ and $B$ are moving with speeds $20 \mathrm{~m} / \mathrm{s}$ and $30 \mathrm{~m} / \mathrm{s}$, respectively in the same direction on the same straight track, with $B$ ahead of $A$. The engines are at the front ends. The engine of train $A$ blows a long whistle.
Assume that the sound of the whistle is composed of components varying in frequency from $f_1=800 \mathrm{~Hz}$ to $f_2=1120 \mathrm{~Hz}$, as shown in the figure. The spread in the frequency (highest frequency - lowest frequency) is thus $320 \mathrm{~Hz}$. The speed of sound in still air is $340 \mathrm{~m} / \mathrm{s}$.Question:
The distribution of the sound intensity of the whistle as observed by the passengers in train $A$ is best represented by

Options:
A

B

C

D

The correct answer is:

Solution:
For the passengers in train $A$, there is no relative motion between source and observe, as both are moving with velocity $20 \mathrm{~m} / \mathrm{s}$. Therefore, there is no change in observed frequencies and correspondingly there is no change in their intensities. Therefore, the correct option is (a).

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2007 (Paper 2)
Paragraph:
When a metal rod $M$ is dipped into an aqueous colourless concentrated solution of compound $\mathrm{N}$, the solution turns light blue. Addition of aqueous $\mathrm{NaCl}$ to the blue solution gives a white precipitate $\mathrm{O}$. Addition of aqueous $\mathrm{NH}_3$ dissolves $\mathrm{O}$ and gives an intense blue solution.Question:
The final solution contains

Options:
A
$\left[\mathrm{Pb}\left(\mathrm{NH}_3\right)_4\right]^{2+}$ and $\left[\mathrm{CoCl}_4\right]^{2-}$

B
$\left[\mathrm{Al}\left(\mathrm{NH}_3\right)_4\right]^{3+}$ and $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}$

C
$\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$and $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+}$

D
$\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$and $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}$

The correct answer is:
$\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$and $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+}$

Solution:
Since in the beginning, concentrated $\mathrm{AgNO}_3$ solution was taken, some $\mathrm{AgNO}_3$ remain in solution which gives white precipitate of $\mathrm{AgCl}$ on addition of $\mathrm{NaCl}$. The precipitate finally dissolve in ammonia and $\mathrm{Cu}^{2+}$ present in solution also forms complex with ammonia as

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2011 (Paper 1)
Paragraph:
When a metal rod $M$ is dipped into an aqueous colourless concentrated solution of compound $\mathrm{N}$, the solution turns light blue. Addition of aqueous $\mathrm{NaCl}$ to the blue solution gives a white precipitate $\mathrm{O}$. Addition of aqueous $\mathrm{NH}_3$ dissolves $\mathrm{O}$ and gives an intense blue solution.Question:
The metal $\operatorname{rod} M$ is

Options:
A
$\mathrm{Fe}$

B
$\mathrm{Cu}$

C
$\mathrm{Ni}$

D
$\mathrm{Co}$

The correct answer is:
$\mathrm{Cu}$

Solution:
Appearance of blue colour on addition of the metal rod $M$ to the solution of $N$ is an indication that metal may be $\mathrm{Cu}$. Later formation of white precipitate on addition of $\mathrm{NaCl}$ and dissolution of this precipitate in aqueous ammonia confirm that while preipitate is of $\mathrm{AgCl}$. This implies that the solution $N$ is of $\mathrm{AgNO}_3$. This confirm that the metal $M$ is only $\mathrm{Cu}$.

Topic: Chemistry, d and f Block Elements, JEE Advanced, JEE Advanced 2011 (Paper 1)
Paragraph:
When a metal rod $M$ is dipped into an aqueous colourless concentrated solution of compound $\mathrm{N}$, the solution turns light blue. Addition of aqueous $\mathrm{NaCl}$ to the blue solution gives a white precipitate $\mathrm{O}$. Addition of aqueous $\mathrm{NH}_3$ dissolves $\mathrm{O}$ and gives an intense blue solution.Question:
The final solution contains

Options:
A
$\left[\mathrm{Pb}\left(\mathrm{NH}_3\right)_4\right]^{2+}$ and $\left[\mathrm{CoCl}_4\right]^{2-}$

B
$\left[\mathrm{Al}\left(\mathrm{NH}_3\right)_4\right]^{3+}$ and $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}$

C
$\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$and $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+}$

D
$\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$and $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}$

The correct answer is:
$\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$and $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+}$

Solution:
Since in the beginning, concentrated $\mathrm{AgNO}_3$ solution was taken, some $\mathrm{AgNO}_3$ remain in solution which gives white precipitate of $\mathrm{AgCl}$ on addition of $\mathrm{NaCl}$. The precipitate finally dissolve in ammonia and $\mathrm{Cu}^{2+}$ present in solution also forms complex with ammonia as

Topic: Chemistry, Coordination Compounds, JEE Main
Paragraph:
When a metal rod $M$ is dipped into an aqueous colourless concentrated solution of compound $\mathrm{N}$, the solution turns light blue. Addition of aqueous $\mathrm{NaCl}$ to the blue solution gives a white precipitate $\mathrm{O}$. Addition of aqueous $\mathrm{NH}_3$ dissolves $\mathrm{O}$ and gives an intense blue solution.Question:
The compound $N$ is

Options:
A
$\mathrm{AgNO}_3$

B
$\mathrm{Zn}\left(\mathrm{NO}_3\right)_2$

C
$\mathrm{Al}\left(\mathrm{NO}_3\right)_3$

D
$\mathrm{Pb}\left(\mathrm{NO}_3\right)_2$

The correct answer is:
$\mathrm{AgNO}_3$

Solution:
The compound $N$ is $\mathrm{AgNO}_3$, explained above.

Topic: Chemistry, d and f Block Elements, JEE Advanced, JEE Advanced 2011 (Paper 1)
Paragraph:
When a particle is restricted to move along $x$-axis between $x=0$ and $x=a$, where $a$ is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends $x=0$ and $x=a$. The wavelength of this standing wave is related to the liner momentum $p$ of the particle according to the de Broglie relation. The energy of the particle of mass $m$ is related to its linear momentum as $E=\frac{p^2}{2 m}$. Thus, the energy of the particle can be denoted by a quantum number $n$ taking values $1,2,3, \ldots(n=1$, called the ground state) corresponding to the number of loops in the standing wave.
Use the model described above to answer the following three questions for a particle moving in the line $x=0$ to $x=a$ [Take $h=6.6 \times 10^{-34} \mathrm{Js}$ and $e=1.6 \times 10^{-19} \mathrm{C}$ ]Question:
The speed of the particle that can take discrete values is proportional to

Options:
A
$n^{-3 / 2}$

B
$n^{-1}$

C
$n^{1 / 2}$

D
$n$

The correct answer is:
$n$

Solution:
From Eq. (i)

Topic: Physics, Atomic Physics, JEE Advanced, JEE Advanced 2009 (Paper 1)
Paragraph:
When a particle is restricted to move along $x$-axis between $x=0$ and $x=a$, where $a$ is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends $x=0$ and $x=a$. The wavelength of this standing wave is related to the liner momentum $p$ of the particle according to the de Broglie relation. The energy of the particle of mass $m$ is related to its linear momentum as $E=\frac{p^2}{2 m}$. Thus, the energy of the particle can be denoted by a quantum number $n$ taking values $1,2,3, \ldots(n=1$, called the ground state) corresponding to the number of loops in the standing wave.
Use the model described above to answer the following three questions for a particle moving in the line $x=0$ to $x=a$ [Take $h=6.6 \times 10^{-34} \mathrm{Js}$ and $e=1.6 \times 10^{-19} \mathrm{C}$ ]Question:
The allowed energy for the particle for a particular value of $n$ is proportional to

Options:
A
$a^{-2}$

B
$a^{-3 / 2}$

C
$a^{-1}$

D
$a^2$

The correct answer is:
$a^{-2}$

Solution:
$$
\begin{array}{rlrl}
& a & =\frac{n \lambda}{2} \\
\therefore & \lambda & =\frac{2 a}{n}=\frac{h}{p}=\frac{h}{\sqrt{2 E m}} \\
\text { or } & \sqrt{E} \propto \frac{1}{a} \Rightarrow E \propto \frac{1}{a^2}
\end{array}
$$

Topic: Physics, Atomic Physics, JEE Advanced, JEE Advanced 2009 (Paper 1)
Paragraph:
When a particle is restricted to move along $x$-axis between $x=0$ and $x=a$, where $a$ is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends $x=0$ and $x=a$. The wavelength of this standing wave is related to the liner momentum $p$ of the particle according to the de Broglie relation. The energy of the particle of mass $m$ is related to its linear momentum as $E=\frac{p^2}{2 m}$. Thus, the energy of the particle can be denoted by a quantum number $n$ taking values $1,2,3, \ldots(n=1$, called the ground state) corresponding to the number of loops in the standing wave.
Use the model described above to answer the following three questions for a particle moving in the line $x=0$ to $x=a$ [Take $h=6.6 \times 10^{-34} \mathrm{Js}$ and $e=1.6 \times 10^{-19} \mathrm{C}$ ]Question:
If the mass of the particle is $m=1.0 \times 10^{-30} \mathrm{~kg}$ and $a=6.6 \mathrm{~nm}$, the energy of the particle in its ground state is closest to

Options:
A
$0.8 \mathrm{meV}$

B
$8 \mathrm{meV}$

C
$80 \mathrm{meV}$

D
$800 \mathrm{meV}$

The correct answer is:
$8 \mathrm{meV}$

Solution:
From Eq. (i)
$$
E=\frac{n^2 h^2}{8 a^2 m}
$$
In ground state $n=1$
$$
\therefore \quad E_1=\frac{h^2}{8 m a^2}
$$
Substituting the values, we get
$$
E_1=8 \mathrm{meV}
$$

Topic: Physics, Atomic Physics, JEE Advanced, JEE Advanced 2009 (Paper 1)
Paragraph:
When a particle of mass $m$ moves on the $x$-axis in a potential of the form $V(x)=k x^2$, it performs simple harmonic motion.
The corresponding time period is proportional to $\sqrt{\frac{m}{k}}$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of $x=0$ in a way different from $k x^2$ and its total energy is such that the particle does not escape to infinity. Consider a particle of mass $\mathrm{m}$ moving on the $x$-axis. Its potential energy is $V(x)=\alpha x^4(\alpha>0)$ for $|x|$ near the origin and becomes a constant equal to $V_0$ for $|x| \geq X_0$ (see figure).Question:
For periodic motion of small amplitude $A$, the time period $T$ of this particle is proportional to

Options:
A
$A \sqrt{\frac{m}{\alpha}}$

B
$\frac{1}{A} \sqrt{\frac{m}{\alpha}}$

C
$A \sqrt{\frac{\alpha}{m}}$

D
$\frac{1}{A} \sqrt{\frac{\alpha}{m}}$

The correct answer is:
$\frac{1}{A} \sqrt{\frac{m}{\alpha}}$

Solution:
$[\alpha]=\left[\frac{\mathrm{PE}}{x^4}\right]=\left[\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^4}\right]=\left[\mathrm{ML}^{-2} \mathrm{~T}^{-2}\right]$
$$
\begin{array}{ll}
\therefore & {\left[\frac{m}{\alpha}\right]=\left[\mathrm{L}^2 \mathrm{~T}^2\right]} \\
\therefore & {\left[\frac{1}{A} \sqrt{\frac{m}{\alpha}}\right]=[\mathrm{T}]}
\end{array}
$$
As dimensions of amplitude $A$ is [L].
Hence, the correct option is (b).

Topic: Physics, Oscillations, JEE Advanced, JEE Advanced 2010 (Paper 1)
Paragraph:
When a particle of mass $m$ moves on the $x$-axis in a potential of the form $V(x)=k x^2$, it performs simple harmonic motion.
The corresponding time period is proportional to $\sqrt{\frac{m}{k}}$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of $x=0$ in a way different from $k x^2$ and its total energy is such that the particle does not escape to infinity. Consider a particle of mass $\mathrm{m}$ moving on the $x$-axis. Its potential energy is $V(x)=\alpha x^4(\alpha>0)$ for $|x|$ near the origin and becomes a constant equal to $V_0$ for $|x| \geq X_0$ (see figure).Question:
The acceleration of this particle for $|x|>X_0$ is

Options:
A
proportional to $V_0$

B
proportional to $\frac{V_0}{m X_0}$

C
proportional to $\sqrt{\frac{V_0}{m X_0}}$

D
Zero

The correct answer is:
Zero

Solution:
For $|x|>X_0$, potential energy is constant. Hence, kinetic energy, speed or velocity will also remain constant.
$\therefore$ Acceleration will be zero.
Hence, the correct option is (d).

Topic: Physics, Oscillations, JEE Advanced, JEE Advanced 2010 (Paper 1)
Paragraph:
When a particle of mass $m$ moves on the $x$-axis in a potential of the form $V(x)=k x^2$, it performs simple harmonic motion.
The corresponding time period is proportional to $\sqrt{\frac{m}{k}}$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of $x=0$ in a way different from $k x^2$ and its total energy is such that the particle does not escape to infinity. Consider a particle of mass $\mathrm{m}$ moving on the $x$-axis. Its potential energy is $V(x)=\alpha x^4(\alpha>0)$ for $|x|$ near the origin and becomes a constant equal to $V_0$ for $|x| \geq X_0$ (see figure).Question:
If the total energy of the particle is $E$, it will perform periodic motion only if

Options:
A
$E < 0$
B
$E>0$

C
$V_0>E>0$

D
$E>V_0$

The correct answer is:
$V_0>E>0$

Solution:
If $E>V_0$, particle will escape. But simultaneously for oscillations, $E>0$. Hence, the correct answer is $V_0>E>0$ or the correct option is (c).

Topic: Physics, Oscillations, JEE Advanced, JEE Advanced 2010 (Paper 1)
Paragraph:
When liquid medicine of density $\rho$ is to be put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop.
We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension $T$ when the radius of the drop is $R$. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.Question:
If $r=5 \times 10^{-4} \mathrm{~m}, \rho=10^3 \mathrm{~kg} \mathrm{~m}^{-3}$, $g=10 \mathrm{~ms}^{-2}, T=0.11 \mathrm{Nm}^{-1}$, the radius of the drop when it detaches from the dropper is approximately

Options:
A
$1.4 \times 10^{-3} \mathrm{~m}$

B
$3.3 \times 10^{-3} \mathrm{~m}$

C
$2.0 \times 10^{-3} \mathrm{~m}$

D
$4.1 \times 10^{-3} \mathrm{~m}$

The correct answer is:
$1.4 \times 10^{-3} \mathrm{~m}$

Solution:
$$
\begin{aligned}
& \frac{2 \pi r^2 T}{R}=m g=\frac{4}{3} \pi R^3 \cdot \rho \cdot g \\
& \begin{aligned}
\therefore \quad R^4 & =\frac{3 r^2 T}{2 \rho g}=\frac{3 \times\left(5 \times 10^{-4}\right)^2(0.11)}{2 \times 10^3 \times 10} \\
& =4.125 \times 10^{-12} \mathrm{~m}^4
\end{aligned} \\
& \therefore R=1.425 \times 10^{-3} \mathrm{~m} \approx 1.4 \times 10^{-3} \mathrm{~m} \\
& \therefore \text { The correct option is (a). }
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Advanced, JEE Advanced 2010 (Paper 2)
Paragraph:
When liquid medicine of density $\rho$ is to be put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop.
We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension $T$ when the radius of the drop is $R$. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.Question:
If the radius of the opening of the dropper is $r$, the vertical force due to the surface tension on the drop of radius $R$ (assuming $r< < R$ ) is

Options:
A
$2 \pi r T$

B
$2 \pi R T$

C
$\frac{2 \pi r^2 T}{R}$

D
$\frac{2 \pi R^2 T}{r}$

The correct answer is:
$\frac{2 \pi r^2 T}{R}$

Solution:
Vertical force due to surface tension,
$$
\begin{aligned}
F_v & =F \sin \theta=(T \cdot 2 \pi r)(r / R) \\
& =\frac{2 \pi r^2 T}{R}
\end{aligned}
$$
$\therefore$ The correct option is (c).

Topic: Physics, Mechanical Properties of Fluids, JEE Advanced, JEE Advanced 2010 (Paper 2)
Paragraph:
When liquid medicine of density $\rho$ is to be put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop.
We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension $T$ when the radius of the drop is $R$. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.Question:
After the drop detaches, its surface energy is

Options:
A
$1.4 \times 10^{-6} \mathrm{~J}$

B
$2.7 \times 10^{-6} \mathrm{~J}$

C
$5.4 \times 10^{-6} \mathrm{~J}$

D
$8.1 \times 10^{-9} \mathrm{~J}$

The correct answer is:
$2.7 \times 10^{-6} \mathrm{~J}$

Solution:
Surface energy,
$$
\begin{aligned}
E & =\left(4 \pi R^2\right) T \\
& =(4 \pi)\left(1.4 \times 10^{-3}\right)^2(0.11) \\
& =2.7 \times 10^{-6} \mathrm{~J}
\end{aligned}
$$
$\therefore$ The correct option is (b).

Topic: Physics, Mechanical Properties of Fluids, JEE Advanced, JEE Advanced 2010 (Paper 2)
Passing $\mathrm{H}_2 \mathrm{~S}$ gas into a mixture of $\mathrm{Mn}^{2+}, \mathrm{Ni}^{2+}, \mathrm{Cu}^{2+}$ and $\mathrm{Hg}^{2+}$ ions in an acidified aqueous solution precipitates

Options:
A
$\mathrm{CuS}$ and $\mathrm{HgS}$

B
$\mathrm{MnS}$ and $\mathrm{CuS}$

C
$\mathrm{MnS}$ and $\mathrm{NiS}$

D
$\mathrm{NiS}$ and $\mathrm{HgS}$

The correct answer is:
$\mathrm{CuS}$ and $\mathrm{HgS}$

Solution:
In acidic medium, $\mathrm{H}_2 \mathrm{~S}$ is very feebly ionized giving very small concentration of sulphide ion for precipitation. Therefore, the most insoluble salts $\mathrm{CuS}$ and $\mathrm{HgS}$ are precipitated only.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Advanced, JEE Advanced 2011 (Paper 2)
Photoelectric effect experiments are performed using three different metal plates $p, \quad q$ and $r$ having work functions $\varphi_p=2.0 \mathrm{eV}, \varphi_q=2.5 \mathrm{eV}$ and $\varphi_r=3.0 \mathrm{eV}$, respectively. A light beam containing wavelengths of $550 \mathrm{~nm}$, $450 \mathrm{~nm}$ and $350 \mathrm{~nm}$ with equal intensities illuminates each of the plates. The correct $I-V$ graph for the experiment is

Options:
A

B

C

D
None of these

The correct answer is:
None of these

Solution:
$$
\begin{gathered}
\text { 1. } K_p=E_p-\phi_p=\frac{1240}{550}-20=0.2545 \mathrm{eV} \\
K_q=E_q-\phi_q=\frac{1240}{450}-25=0.255 \mathrm{eV} \\
K_r=E_r-\phi_r=\frac{1240}{350}-3.0=0.543 \mathrm{eV}
\end{gathered}
$$
In the above equation $K$ represents maximum kinetic energy of photoelectrons and $E$, the energy of incident right. From the above values we can see that stopping potential, $\left|V_r\right|>\left|V_q\right|>\left|V_p\right|$
Further, their intensities are equal, but energy of individual photon of $r$ is maximum. Hence, number of photons incident (per unit area per unit time) of $r$ can be assumed to be least. Hence, saturation current of $r$ should be minimum.
Keeping these points in mind no option seems to be correct. The correct graph is shown below,
$$
\therefore \text { No choice is correct. }
$$

Topic: Physics, Dual Nature of Matter, JEE Main
Photoelectric effect experiments are performed using three different metal plates $p, \quad q$ and $r$ having work functions $\varphi_p=2.0 \mathrm{eV}, \varphi_q=2.5 \mathrm{eV}$ and $\varphi_r=3.0 \mathrm{eV}$, respectively. A light beam containing wavelengths of $550 \mathrm{~nm}$, $450 \mathrm{~nm}$ and $350 \mathrm{~nm}$ with equal intensities illuminates each of the plates. The correct $I-V$ graph for the experiment is

Options:
A

B

C

D
None of these

The correct answer is:
None of these

Solution:
$$
\begin{gathered}
\text { 1. } K_p=E_p-\phi_p=\frac{1240}{550}-20=0.2545 \mathrm{eV} \\
K_q=E_q-\phi_q=\frac{1240}{450}-25=0.255 \mathrm{eV} \\
K_r=E_r-\phi_r=\frac{1240}{350}-3.0=0.543 \mathrm{eV}
\end{gathered}
$$
In the above equation $K$ represents maximum kinetic energy of photoelectrons and $E$, the energy of incident right. From the above values we can see that stopping potential, $\left|V_r\right|>\left|V_q\right|>\left|V_p\right|$
Further, their intensities are equal, but energy of individual photon of $r$ is maximum. Hence, number of photons incident (per unit area per unit time) of $r$ can be assumed to be least. Hence, saturation current of $r$ should be minimum.
Keeping these points in mind no option seems to be correct. The correct graph is shown below,
$$
\therefore \text { No choice is correct. }
$$

Topic: Physics, Dual Nature of Matter, JEE Advanced, JEE Advanced 2009 (Paper 2)