datavorous/entrance-exam-dataset huggingface dataset:
Paragraph:
Tangents are drawn from the point $P(3,4)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ touching the ellipse at points $A$ and $B$.Question:
The orthocentre of the triangle $P A B$ is

Options:
A
$\left(5, \frac{8}{7}\right)$

B
$\left(\frac{7}{5}, \frac{25}{8}\right)$

C
$\left(\frac{11}{5}, \frac{8}{5}\right)$

D
$\left(\frac{8}{25}, \frac{7}{5}\right)$

The correct answer is:
$\left(\frac{11}{5}, \frac{8}{5}\right)$

Solution:
$$
\text { Equation of } A B \text { is }
$$
$$
\begin{aligned}
y-0 & =\frac{\frac{8}{5}}{-\frac{9}{5}-3}(x-3)=\frac{8}{-24}(x-3) \\
\Rightarrow \quad y & =-\frac{1}{3}(x-3) \\
\Rightarrow \quad x & +3 y=3
\end{aligned}
$$
Equation of the straight line perpendicular to $A B$ through $P$ is $3 x-y=5$.
Equation of $P A$ is $x-3=0$.
The equation of straight line perpendicular to $P A$ through $B\left(\frac{-9}{5}, \frac{8}{5}\right)$. is $y=\frac{8}{5}$.
Hence, the orthocentre is $\left(\frac{11}{5}, \frac{8}{5}\right)$.

Topic: Mathematics, Straight Lines, JEE Advanced, JEE Advanced 2010 (Paper 2)
Paragraph:
The capacitor of capacitance $C$ can be charged (with the help of a resistance $R$ ) by a voltage source $V$, by closing switch $S_1$ while keeping switch $S_2$ open. The capacitor can be connected in series with an inductor $L$ by closing switch $S_2$ and opening $S_1$.
Question:
Initially, the capacitor was uncharged. Now, switch $S_1$ is closed and $S_2$ is kept open. If time constant of this circuit is $\tau$, then

Options:
A
after time interval $\tau$, charge on the capacitor is $C V / 2$

B
after time interval $2 \tau$, charge on the capacitor is $C V /\left(1-e^{-2}\right)$

C
the work done by the voltage source will be half on the heat dissipated when the capacitor is fully charged

D
after time interval $2 \tau$, charge on the capacitor is $C V\left(1-e^{-1}\right)$

The correct answer is:
after time interval $2 \tau$, charge on the capacitor is $C V /\left(1-e^{-2}\right)$

Solution:
Charge on capacitor at time $t$ is $q=q_0\left(1-e^{-t / \tau}\right)$
$\begin{array}{ll}\text { Here, } & q_0=C V \text { and } t=2 \tau \\ \therefore & q=C V\left(1-e^{-2 \tau / \tau}\right)=C V\left(1-e^{-2}\right)\end{array}$

Topic: Physics, Alternating Current, JEE Advanced, JEE Advanced 2006
Paragraph:
The capacitor of capacitance $C$ can be charged (with the help of a resistance $R$ ) by a voltage source $V$, by closing switch $S_1$ while keeping switch $S_2$ open. The capacitor can be connected in series with an inductor $L$ by closing switch $S_2$ and opening $S_1$.
Question:
After the capacitor gets fully charged, $S_1$ is opened and $S_2$ is closed so that the inductor is connected in series with the capacitor. Then,

Options:
A
at $t=0$, energy stored in the circuit is purely in the form of magnetic energy

B
at any time $t>0$, current in the circuit is in the same direction

C
at $t>0$, there is no exchange of energy between the inductor and capacitor

D
at any time $t>0$, maximum instantaneous current in the circuit may $V \sqrt{\frac{C}{L}}$

The correct answer is:
at any time $t>0$, maximum instantaneous current in the circuit may $V \sqrt{\frac{C}{L}}$

Solution:
From conservation of energy, $\frac{1}{2} L I_{\max }^2=\frac{1}{2} C V^2$
$$
\therefore \quad I_{\max }=V \sqrt{\frac{C}{L}}
$$

Topic: Physics, Alternating Current, JEE Advanced, JEE Advanced 2006
Paragraph:
The capacitor of capacitance $C$ can be charged (with the help of a resistance $R$ ) by a voltage source $V$, by closing switch $S_1$ while keeping switch $S_2$ open. The capacitor can be connected in series with an inductor $L$ by closing switch $S_2$ and opening $S_1$.
Question:
Initially, the capacitor was uncharged. Now, switch $S_1$ is closed and $S_2$ is kept open. If time constant of this circuit is $\tau$, then

Options:
A
after time interval $\tau$, charge on the capacitor is $C V / 2$

B
after time interval $2 \tau$, charge on the capacitor is $C V /\left(1-e^{-2}\right)$

C
the work done by the voltage source will be half on the heat dissipated when the capacitor is fully charged

D
after time interval $2 \tau$, charge on the capacitor is $C V\left(1-e^{-1}\right)$

The correct answer is:
after time interval $2 \tau$, charge on the capacitor is $C V /\left(1-e^{-2}\right)$

Solution:
Charge on capacitor at time $t$ is $q=q_0\left(1-e^{-t / \tau}\right)$
$\begin{array}{ll}\text { Here, } & q_0=C V \text { and } t=2 \tau \\ \therefore & q=C V\left(1-e^{-2 \tau / \tau}\right)=C V\left(1-e^{-2}\right)\end{array}$

Topic: Physics, Alternating Current, JEE Main
Paragraph:
The capacitor of capacitance $C$ can be charged (with the help of a resistance $R$ ) by a voltage source $V$, by closing switch $S_1$ while keeping switch $S_2$ open. The capacitor can be connected in series with an inductor $L$ by closing switch $S_2$ and opening $S_1$.
Question:
If the total charge stored in the $L C$ circuit is $Q_0$, then for $t \geq 0$

Options:
A
the charge on the capacitor is $Q=Q_0 \cos \left(\frac{\pi}{2}+\frac{t}{\sqrt{L C}}\right)$

B
the charge on the capacitor is $Q=Q_0 \cos \left(\frac{\pi}{2}-\frac{t}{\sqrt{L C}}\right)$

C
the charge on the capacitor is $Q=-L C \frac{d^2 Q}{d t^2}$

D
the charge on the capacitor is $Q=-\frac{1}{\sqrt{L C}} \frac{d^2 Q}{d t^2}$

The correct answer is:
the charge on the capacitor is $Q=-L C \frac{d^2 Q}{d t^2}$

Solution:
Comparing the LC oscillations with normal SHM, we get
$$
\frac{d^2 Q}{d t^2}=-\omega^2 Q
$$
Here,
$$
\begin{aligned}
\omega^2 & =\frac{1}{L C} \\
Q & =-L C \frac{d^2 Q}{d t^2}
\end{aligned}
$$
$$
\therefore \quad Q=-L C \frac{d^2 Q}{d t^2}
$$

Topic: Physics, Alternating Current, JEE Advanced, JEE Advanced 2006
Paragraph:
The circle $x^2+y^2-8 x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$.Question:
Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

Options:
A
$2 x-\sqrt{5} y-20=0$

B
$2 x-\sqrt{5} y+4=0$

C
$3 x-4 y+8=0$

D
$4 x-3 y+4=0$

The correct answer is:
$2 x-\sqrt{5} y+4=0$

Solution:
Equation of tangent to hyperbola having slope $m$ is
$$
y=m x+\sqrt{9 m^2-4}
$$
Equation of tangent to circle is
$$
y=m(x-4)+\sqrt{16 m^2+16}
$$
Eqs. (i) and (ii) will be identical for $m=\frac{2}{\sqrt{5}}$ satisfy.
$\therefore$ Equation of common tangent is $2 x-\sqrt{5} y+4=0$

Topic: Mathematics, Hyperbola, JEE Advanced, JEE Advanced 2010 (Paper 1)
Paragraph:
The circle $x^2+y^2-8 x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$.Question:
Equation of the circle with $A B$ as its diameter is

Options:
A
$x^2+y^2-12 x+24=0$

B
$x^2+y^2+12 x+24=0$

C
$x^2+y^2+24 x-12=0$

D
$x^2+y^2-24 x-12=0$

The correct answer is:
$x^2+y^2-12 x+24=0$

Solution:
The equation of the hyperbola is $\frac{x^2}{9}-\frac{y^2}{4}=1$ and that of circle is
$$
x^2+y^2-8 x=0
$$
For their points of intersection
$$
\frac{x^2}{9}+\frac{x^2-8 x}{4}=1
$$
$$
\begin{array}{ll}
\Rightarrow & 4 x^2+9 x^2-72 x=36 \\
\Rightarrow & 13 x^2-72 x-36=0
\end{array}
$$
$$
\begin{aligned}
& \Rightarrow \quad 13 x^2-78 x+6 x-36=0 \\
& \Rightarrow \quad 13 x(x-6)+6(x-6)=0 \\
& \Rightarrow \quad x=6, x=-\frac{13}{6} \\
& x=-\frac{13}{6} \text { not acceptable } \\
&
\end{aligned}
$$
Now, for $x=6, y=\pm 2 \sqrt{3}$
Required equation is
$$
\begin{aligned}
& (x-6)^2+(y+2 \sqrt{3})(y-2 \sqrt{3})=0 \\
& \Rightarrow \quad x^2-12 x+y^2+24=0 \\
& \Rightarrow \quad x^2+y^2-12 x+24=0
\end{aligned}
$$

Topic: Mathematics, Circle, JEE Advanced, JEE Advanced 2010 (Paper 1)
Paragraph:
The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal $\mathrm{M}$ is :
$$
M(s) \mid M^{+}(\alpha q ; 0.05 \text { molar }) \| M^{+}(a q ; 1 \text { molar }) \mid M(s)
$$
For the above electrolytic cell the magnitude of the cell potential $\left|E_{\text {cell }}\right|=70 \mathrm{mV}$.Question:
For the above cell

Options:
A
$E_{\text {cell }} < 0 ; \Delta G>0$
B
$E_{\text {cell }}>0 ; \Delta G < 0$
C
$E_{\text {cell }} < 0 ; \Delta G^{\circ}>0$
D
$E_{\text {cell }}>0 ; \Delta G^{\circ} < 0$

The correct answer is:
$E_{\text {cell }}>0 ; \Delta G < 0$

Solution:
The given cell is electrolytic concentration cell, So $E_{\text {cell }}^{\circ}=0$
Anode :
$$
\begin{gathered}
M(s) \longrightarrow M^{+}(a q)+e^{-} \\
\text {Cathode }: \quad M^{+}(a q)+e^{-} \longrightarrow M(s) \\
\hline M^{+}(a q) \text { cathode } \rightleftharpoons M^{+}(a q) \text { anode } \\
E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{1} \log \frac{\left[M^{+}\right]_{\text {anodic }}}{\left[M^{+}\right]_{\text {cathodic }}} \\
=0-\frac{0.0591}{1} \log \frac{0.05}{1} \\
E_{\text {cell }}=70 \mathrm{mV}(+\text { ve }) \\
\text { Hence, } \Delta G=-n F E_{\text {cell }}=-\text { ve }
\end{gathered}
$$

Topic: Chemistry, Electrochemistry, JEE Advanced, JEE Advanced 2010 (Paper 1)
Paragraph:
The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal $\mathrm{M}$ is :
$$
M(s) \mid M^{+}(\alpha q ; 0.05 \text { molar }) \| M^{+}(a q ; 1 \text { molar }) \mid M(s)
$$
For the above electrolytic cell the magnitude of the cell potential $\left|E_{\text {cell }}\right|=70 \mathrm{mV}$.Question:
If the $0.05$ molar solution of $M^{+}$is replaced by a $0.0025$ molar $M^{+}$ solution, then the magnitude of the cell potential would be

Options:
A
$35 \mathrm{mV}$

B
$70 \mathrm{mV}$

C
$140 \mathrm{mV}$

D
$700 \mathrm{mV}$

The correct answer is:
$140 \mathrm{mV}$

Solution:
$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{1} \log \frac{0.0025}{1}$ $E_{\text {cell }}=140 \mathrm{mV}$ Electrochemistry Straight conceptual III

Topic: Chemistry, Electrochemistry, JEE Advanced, JEE Advanced 2010 (Paper 1)
Paragraph:
The coordination number of $\mathrm{Ni}^{2+}$ is 4 .
$\mathrm{NiCl}_2+\mathrm{KCN}$ (excess) $\rightarrow A$ (cyano complex)
$\mathrm{NiCl}_2+$ conc. $\mathrm{HCl}$ (excess) $\rightarrow B$ (chloro complex)Question:
Predict the magnetic nature of $A$ and $B$ are

Options:
A
both are diamagnetic

B
$A$ is diamagnetic and $B$ is paramagnetic with one unpaired electron

C
$A$ is diamagnetic and $B$ is paramagnetic with two unpaired electrons

D
both are paramagnetic

The correct answer is:
$A$ is diamagnetic and $B$ is paramagnetic with two unpaired electrons

Solution:
In it unpaired orbital is not present, so it is diamagnetic in character (square planar shape) In chloro complex $\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{Cl})_4\right]$, complex ion is $\left[\mathrm{Ni}^2\left(\mathrm{Cl}_4\right)\right]^{2-}$ and $\mathrm{Ni}$ is present $\mathrm{Ni}^{2+}$ or $\mathrm{Ni}(\mathrm{II})$, so
$$
\mathrm{Ni}^2+=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^8, 4 s^0
$$
In $\left.\left[\mathrm{Ni}_{\left(\mathrm{Cl}_4\right)}\right)\right]^2$ ion, $\mathrm{Ni}^{2+}$ is present as follows due to weaker ligand character is $\mathrm{Cl}^{-}$ion. $\left(\mathrm{Cl}^{-}\right.$is weak field ligand). So, it is unable to pair up the electron and $\mathrm{Ni}^{2+}$ needs four empty orbitals to accommodate four $\mathrm{Cl}^{-}$ligand. Thus, $\left(\mathrm{NiCl}_4\right)^{2-}$ shows $s p^3$-hybridisation (Tetrahedral shapes)
Hence, due to presence of unpaired orbitals, it is paramagnetic in character.

Topic: Chemistry, Coordination Compounds, JEE Main
Paragraph:
The coordination number of $\mathrm{Ni}^{2+}$ is 4 .
$\mathrm{NiCl}_2+\mathrm{KCN}$ (excess) $\rightarrow A$ (cyano complex)
$\mathrm{NiCl}_2+$ conc. $\mathrm{HCl}$ (excess) $\rightarrow B$ (chloro complex)Question:
Predict the magnetic nature of $A$ and $B$ are

Options:
A
both are diamagnetic

B
$A$ is diamagnetic and $B$ is paramagnetic with one unpaired electron

C
$A$ is diamagnetic and $B$ is paramagnetic with two unpaired electrons

D
both are paramagnetic

The correct answer is:
$A$ is diamagnetic and $B$ is paramagnetic with two unpaired electrons

Solution:
In it unpaired orbital is not present, so it is diamagnetic in character (square planar shape) In chloro complex $\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{Cl})_4\right]$, complex ion is $\left[\mathrm{Ni}^2\left(\mathrm{Cl}_4\right)\right]^{2-}$ and $\mathrm{Ni}$ is present $\mathrm{Ni}^{2+}$ or $\mathrm{Ni}(\mathrm{II})$, so
$$
\mathrm{Ni}^2+=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^8, 4 s^0
$$
In $\left.\left[\mathrm{Ni}_{\left(\mathrm{Cl}_4\right)}\right)\right]^2$ ion, $\mathrm{Ni}^{2+}$ is present as follows due to weaker ligand character is $\mathrm{Cl}^{-}$ion. $\left(\mathrm{Cl}^{-}\right.$is weak field ligand). So, it is unable to pair up the electron and $\mathrm{Ni}^{2+}$ needs four empty orbitals to accommodate four $\mathrm{Cl}^{-}$ligand. Thus, $\left(\mathrm{NiCl}_4\right)^{2-}$ shows $s p^3$-hybridisation (Tetrahedral shapes)
Hence, due to presence of unpaired orbitals, it is paramagnetic in character.

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2006
Paragraph:
The coordination number of $\mathrm{Ni}^{2+}$ is 4 .
$\mathrm{NiCl}_2+\mathrm{KCN}$ (excess) $\rightarrow A$ (cyano complex)
$\mathrm{NiCl}_2+$ conc. $\mathrm{HCl}$ (excess) $\rightarrow B$ (chloro complex)Question:
The hybridization of $A$ and $B$ are

Options:
A
$d s p^2, s p^3$

B
$s p^3, s p^3$

C
$d s p^2, d s p^2$

D
$s p^3 d^2, d^2 s p^3$

The correct answer is:
$d s p^2, s p^3$

Solution:
Hybridisation of $\mathrm{Ni}$ in $\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]$ is $d s p^2$ and in $\mathrm{K}_2\left[\mathrm{Ni}^2\left(\mathrm{Cl}_4\right)\right]$ is $s p^3$ (Detail see in question no. 25).

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2006
Paragraph:
The coordination number of $\mathrm{Ni}^{2+}$ is 4 .
$\mathrm{NiCl}_2+\mathrm{KCN}$ (excess) $\rightarrow A$ (cyano complex)
$\mathrm{NiCl}_2+$ conc. $\mathrm{HCl}$ (excess) $\rightarrow B$ (chloro complex)Question:
The IUPAC name of $A$ and $B$ are

Options:
A
Potassium tetracyanonickelate (II), potassium tetrachloronickelate (II)

B
Tetracyanopotassiumnickelate (II), tetrorachloropotassiumnickelate (II)

C
Tetracyanonickel (II), tetrachloronickel (II)

D
Potassium tetracyanonickel (II), potassium tetrachloronickel (II)

The correct answer is:
Potassium tetracyanonickelate (II), potassium tetrachloronickelate (II)

Solution:

Topic: Chemistry, Coordination Compounds, JEE Main
Paragraph:
The coordination number of $\mathrm{Ni}^{2+}$ is 4 .
$\mathrm{NiCl}_2+\mathrm{KCN}$ (excess) $\rightarrow A$ (cyano complex)
$\mathrm{NiCl}_2+$ conc. $\mathrm{HCl}$ (excess) $\rightarrow B$ (chloro complex)Question:
The IUPAC name of $A$ and $B$ are

Options:
A
Potassium tetracyanonickelate (II), potassium tetrachloronickelate (II)

B
Tetracyanopotassiumnickelate (II), tetrorachloropotassiumnickelate (II)

C
Tetracyanonickel (II), tetrachloronickel (II)

D
Potassium tetracyanonickel (II), potassium tetrachloronickel (II)

The correct answer is:
Potassium tetracyanonickelate (II), potassium tetrachloronickelate (II)

Solution:

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2006
Paragraph:
The figure shows a surface $X Y$ separating two transparent media, medium-1 and medium-2. The lines $a b$ and $c d$ represent wavefronts of a light wave travelling in medium-1 and incident on $X Y$. The lines ef and $g h$ represent wavefronts of the light wave in medium-2 after refraction.
Question:
Light travels as a

Options:
A
parallel beam in each medium

B
convergent beam in each medium

C
divergent beam in each medium

D
divergent beam in one medium and convergent beam in the other medium

The correct answer is:
parallel beam in each medium

Solution:
Wavefronts are parallel in both media. Therefore, light which is perpendicular to wavefront travels as a parallel beam in each medium. Hence, the correct option is (a).

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2007 (Paper 2)
Paragraph:
The figure shows a surface $X Y$ separating two transparent media, medium-1 and medium-2. The lines $a b$ and $c d$ represent wavefronts of a light wave travelling in medium-1 and incident on $X Y$. The lines ef and $g h$ represent wavefronts of the light wave in medium-2 after refraction.
Question:
The phases of the light wave at $c, d, e$ and $f$ are $\phi_c, \phi_d, \phi_e$ and $\phi_f$, respectively. It is given that $\phi_c \neq \phi_f$.

Options:
A
$\phi_c$ cannot be equal to $\phi_d$

B
$\phi_d$ can be equal to $\phi_e$

C
$\left(\phi_d-\phi_f\right)$ is equal to $\left(\phi_c-\phi_e\right)$

D
$\left(\phi_d-\phi_c\right)$ is not equal to $\left(\phi_f-\phi_e\right)$

The correct answer is:
$\left(\phi_d-\phi_f\right)$ is equal to $\left(\phi_c-\phi_e\right)$

Solution:
All points on a wavefront are at the same phase.
$$
\begin{array}{rlrl}
\therefore & \phi_d & =\phi_c \text { and } \phi_f=\phi_e \\
& \therefore & \phi_d-\phi_f & =\phi_c-\phi_e
\end{array}
$$
Hence, the correct option is (c).

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2007 (Paper 2)
Paragraph:
The figure shows a surface $X Y$ separating two transparent media, medium-1 and medium-2. The lines $a b$ and $c d$ represent wavefronts of a light wave travelling in medium-1 and incident on $X Y$. The lines ef and $g h$ represent wavefronts of the light wave in medium-2 after refraction.
Question:
Speed of light is

Options:
A
the same in medium-1 and medium-2

B
larger in medium-1 than in medium-2

C
larger in medium-2 than in medium-1

D
different at $b$ and $d$

The correct answer is:
larger in medium-1 than in medium-2

Solution:
In medium-2 wavefront bends away from the normal after refraction. Therefore, ray of light which is perpendicular to wavefront bends towards the normal in medium-2 during refraction. So, medium-2 is denser or its speed in medium-1 is more. Hence, option (b) is correct.

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2007 (Paper 2)
Paragraph:
The hydrogen-like species $\mathrm{Li}^{2+}$ is in a spherically symmetric state $S_1$ with one radial node. Upon absorbing light the ion undergoes transition to a state $S_2$. The state $S_2$ has one radial node and its energy is equal to the ground state energy of the hydrogen atom.Question:
The state $S_1$ is

Options:
A
$1 s$

B
$2 \mathrm{~s}$

C
$2 p$

D
$3 \mathrm{~s}$

The correct answer is:
$2 \mathrm{~s}$

Solution:
Given : Hydrogen like species, $\mathrm{Li}^{2+}$
Spherically symmetric state, $S_1 \Rightarrow s$-orbital
$S_1$ has radial node $1 \Rightarrow 2 s \quad$ (No. of radial node $=1=n-l-1$, so $n=2$ ) Species undergoes state $S_2$
$S_2$ has energy equal to the ground state energy of $\mathrm{H}$-atom $=-13.6 \mathrm{eV}$
So, $\quad-13.6=-13.6\left(\frac{3^2}{n^2}\right)$
$n=3$
Again radial node $=1=n-1-1$
or $1=3-l-1 \quad$ so $\quad l=1$
It suggests $S_2$ is $3 p$

Topic: Chemistry, Structure of Atom, JEE Advanced, JEE Advanced 2010 (Paper 2)
Paragraph:
The hydrogen-like species $\mathrm{Li}^{2+}$ is in a spherically symmetric state $S_1$ with one radial node. Upon absorbing light the ion undergoes transition to a state $S_2$. The state $S_2$ has one radial node and its energy is equal to the ground state energy of the hydrogen atom.Question:
The orbtial angular momentum quantum number of the state $S_2$ is

Options:
A
0

B
1

C
2

D
3

The correct answer is:
1

Solution:
For $p$-orbital, orbital angular momentum quantum number $=1$ Atomic structure Conceptual data analysis IV

Topic: Chemistry, Structure of Atom, JEE Advanced, JEE Advanced 2010 (Paper 2)
Paragraph:
The hydrogen-like species $\mathrm{Li}^{2+}$ is in a spherically symmetric state $S_1$ with one radial node. Upon absorbing light the ion undergoes transition to a state $S_2$. The state $S_2$ has one radial node and its energy is equal to the ground state energy of the hydrogen atom.Question:
Energy of the state $S_1$ in units of the hydrogen atom ground state energy is

Options:
A
$0.75$

B
$1.50$

C
$2.25$

D
$4.50$

The correct answer is:
$2.25$

Solution:
Energy of state, $S_1=-13.6\left(\frac{3^2}{2^2}\right) \mathrm{eV} /$ atom $=\frac{9}{4}$ of energy of $\mathrm{H}$-atom in ground state $=2.25$ of energy of $\mathrm{H}$-atom in ground state

Topic: Chemistry, Structure of Atom, JEE Advanced, JEE Advanced 2010 (Paper 2)
Paragraph:
The key feature of Bohr's theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.Question:
In a CO molecule, the distance between $\mathrm{C}$ (mass $=12 \mathrm{amu}$ ) and $\mathrm{O}$ (mass = $16 \mathrm{amu}$ ), where $1 \mathrm{amu}=\frac{5}{3} \times 10^{-27} \mathrm{~kg}$, is close to

Options:
A
$2.4 \times 10^{-10} \mathrm{~m}$

B
$1.9 \times 10^{-10} \mathrm{~m}$

C
$1.3 \times 10^{-10} \mathrm{~m}$

D
$4.4 \times 10^{-11} \mathrm{~m}$

The correct answer is:
$1.3 \times 10^{-10} \mathrm{~m}$

Solution:
$I=\mu r^2$ ( where $\mu=$ reduced mass) $\mu=\frac{m_1 m_2}{m_1+m_2}=\frac{48}{7} \mathrm{amu}$ $=11.43 \times 10^{-27} \mathrm{~kg}$
Substituting in $I=\mu r^2$ we get,
$$
\begin{aligned}
r & =\sqrt{\frac{I}{\mu}}=\sqrt{\frac{1.87 \times 10^{-46}}{11.43 \times 10^{-27}}} \\
& =1.28 \times 10^{-10} \mathrm{~m}
\end{aligned}
$$
$\therefore$ The correct answer is (c).

Topic: Physics, Atomic Physics, JEE Advanced, JEE Advanced 2010 (Paper 2)
Paragraph:
The key feature of Bohr's theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.Question:
It is found that the excitation frequency from ground to the first excited state of rotation for the $\mathrm{CO}$ molecule is close to $\frac{4}{\pi} \times 10^{11} \mathrm{~Hz}$. Then the moment of inertia of CO molecule about its centre of mass is close to (Take $h=2 \pi \times 10^{-34} J-s$ )

Options:
A
$2.76 \times 10^{-46} \mathrm{~kg}-\mathrm{m}^2$

B
$1.87 \times 10^{-46} \mathrm{~kg}-\mathrm{m}^2$

C
$4.67 \times 10^{-47} \mathrm{~kg}-\mathrm{m}^2$

D
$1.17 \times 10^{-47} \mathrm{~kg}-\mathrm{m}^2$

The correct answer is:
$1.87 \times 10^{-46} \mathrm{~kg}-\mathrm{m}^2$

Solution:
$$
\begin{gathered}
h v=K_2-K_1=\frac{3 h^2}{8 \pi^2 I} \\
\therefore I=\frac{3 h}{8 \pi^2 f}=\frac{3 \times 2 \pi \times 10^{-34} \times \pi}{8 \times \pi^2 \times 4 \times 10^{11}}
\end{gathered}
$$
$$
=1.87 \times 10^{-46} \mathrm{~kg}-\mathrm{m}^2
$$
$\therefore$ The correct answer is (b).

Topic: Physics, Atomic Physics, JEE Advanced, JEE Advanced 2010 (Paper 2)
Paragraph:
The key feature of Bohr's theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.Question:
A diatomic molecule has moment of inertia I. By Bohr's quantization condition its rotational energy in the $n$th level ( $n=0$ is not allowed) is

Options:
A
$\frac{1}{n^2}\left(\frac{h^2}{8 \pi^2 I}\right)$

B
$\frac{1}{n}\left(\frac{h^2}{8 \pi^2 I}\right)$

C
$n\left(\frac{h^2}{8 \pi^2 I}\right)$

D
$n^2\left(\frac{h^2}{8 \pi^2 I}\right)$

The correct answer is:
$n^2\left(\frac{h^2}{8 \pi^2 I}\right)$

Solution:
$$
\begin{aligned}
L=I \omega & =\frac{n h}{2 \pi} \therefore \omega=\frac{n h}{2 \pi I} \\
K & =\frac{1}{2} I \omega^2=\frac{1}{2} I\left(\frac{n h}{2 \pi I}\right)^2 \\
& =\frac{n^2 h^2}{8 \pi^2 I}
\end{aligned}
$$
$\therefore$ The correct answer is (d).

Topic: Physics, Atomic Physics, JEE Advanced, JEE Advanced 2010 (Paper 2)