datavorous/entrance-exam-dataset huggingface dataset:
The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vectors $\hat{\mathbf{a}}, \hat{\mathbf{b}}, \hat{\mathbf{c}}$ such that $\hat{\mathbf{a}} \cdot \hat{\mathbf{b}}=\hat{\mathbf{b}} \cdot \hat{\mathbf{c}}=\hat{\mathbf{c}} \cdot \hat{\mathbf{a}}=\frac{1}{2}$. Then, the volume of the parallelopiped is

Options:
A
$\frac{1}{\sqrt{2}} \mathrm{cu}$ units

B
$\frac{1}{2 \sqrt{2}}$ cu units

C
$\frac{\sqrt{3}}{2}$ cu units

D
$\frac{1}{\sqrt{3}} \mathrm{cu}$ units

The correct answer is:
$\frac{1}{\sqrt{2}} \mathrm{cu}$ units

Solution:
The volume of the parallelopiped with coterminus edges as $\hat{\mathbf{a}}, \hat{\mathbf{b}}, \hat{\mathbf{c}}$ is given by $[\hat{\mathbf{a}} \hat{\mathbf{b}} \hat{\mathbf{c}}]=\hat{\mathbf{a}} \cdot(\hat{\mathbf{b}} \times \hat{\mathbf{c}})$
$$
\begin{aligned}
& \text { Now, }[\hat{\mathbf{a}} \hat{\mathbf{b}} \hat{\mathbf{c}}]^2=\left|\begin{array}{ccc}
\hat{\mathbf{a}} \cdot \hat{\mathbf{a}} & \hat{\mathbf{a}} \cdot \hat{\mathbf{b}} & \hat{\mathbf{a}} \cdot \hat{\mathbf{c}} \\
\hat{\mathbf{b}} \cdot \hat{\mathbf{a}} & \hat{\mathbf{b}} \cdot \hat{\mathbf{b}} & \hat{\mathbf{b}} \cdot \hat{\mathbf{c}} \\
\hat{\mathbf{c}} \cdot \hat{\mathbf{a}} & \hat{\mathbf{c}} \cdot \hat{\mathbf{b}} & \hat{\mathbf{c}} \cdot \hat{\mathbf{c}}
\end{array}\right|=\left|\begin{array}{ccc}
1 & 1 / 2 & 1 / 2 \\
1 / 2 & 1 & 1 / 2 \\
1 / 2 & 1 / 2 & 1
\end{array}\right|=\frac{1}{2} \\
& \Rightarrow \quad[\hat{\mathbf{a}} \hat{\mathbf{b}} \hat{\mathbf{c}}]^2=\frac{1}{2} \\
&
\end{aligned}
$$
Thus, the required volume of the parallelopiped $=\frac{1}{\sqrt{2}} \mathrm{cu}$ units

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2008 (Paper 1)
The electrochemical cell shown below is a concentration cell. $\mathrm{M} \mid \mathrm{M}^{2+}$ (saturated solution of a sparingly soluble salt, $\left.\mathrm{MX}_{2}\right) \| \mathrm{M}^{2 \perp}\left(0.001 \mathrm{~mol} \mathrm{dm}^{-3}\right) \mid \mathrm{M}$.

The emf of the cell depends on the difference in concentrations of $\mathrm{M}^{2+}$ ions at the two electrodes. The emf of the cell at $298 \mathrm{~K}$ is $0.059 \mathrm{~V}$.

Question: The value of $\Delta \mathrm{G}\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$ for the given cell is (take 1 $\mathrm{F}=96500 \mathrm{Cmol}^{-1}$ )

Options:
A $-5.7$
B $5.7$
C $11.4$
D $-11.4$

The correct answer is:
$-11.4$

Solution:
At anode : $M(\mathrm{~s})+2 X^{-}(\mathrm{aq}) \longrightarrow M X_{2}(\mathrm{aq})+2 \mathrm{e}^{-}$

At cathode $: M^{+2}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow M(\mathrm{~s})$

Thus, here $n=2$

$\Delta G=-n F E_{\text {cell }}$

$=-2 \times 96500 \times 0.059 \times 10^{-3} \mathrm{~kJ} / \mathrm{mole}=-11.4 \mathrm{~kJ} / \mathrm{mole}$

Topic: Chemistry, Electrochemistry, JEE Advanced, JEE Advanced 2012 (Paper 2)
The electrochemical cell shown below is a concentration cell. $\mathrm{M} \mid \mathrm{M}^{2+}$ (saturated solution of a sparingly soluble salt, $\left.\mathrm{MX}_{2}\right) \| \mathrm{M}^{2 \perp}\left(0.001 \mathrm{~mol} \mathrm{dm}^{-3}\right) \mid \mathrm{M}$.

The emf of the cell depends on the difference in concentrations of $\mathrm{M}^{2+}$ ions at the two electrodes. The emf of the cell at $298 \mathrm{~K}$ is $0.059 \mathrm{~V}$.

Question: The solubility product $\left(\mathrm{K}_{s p} ; \mathrm{mol}^{3} \mathrm{dm}^{-9}\right)$ of $\mathrm{MX}_{2}$ at $298 \mathrm{~K}$ based on the information available for the given concentration cell is (take $2.303 \times \mathrm{R} \times 298 / \mathrm{F}=0.059 \mathrm{~V}$ )

Options:
A $1 \times 10^{-15}$
B $4 \times 10^{-15}$
C $1 \times 10^{-12}$
D $4 \times 10^{-12}$

The correct answer is:
$4 \times 10^{-15}$

Solution:
$M \mid M^{2+}$ (aq) $\| M^{2+}$ (aq) $\mid M$

$0.001 \mathrm{M}$

Anode : $\quad M \longrightarrow M^{2+}(\mathrm{aq})+2 e^{-}$

Cathode : $M^{2+}(\mathrm{aq})+2 e^{-} \longrightarrow M$

$M^{2+}(\mathrm{aq})_{c} \rightleftharpoons M^{2+}(\mathrm{aq})_{a}$

$\begin{array}{c}

E_{\text {cell }}=0-\frac{0.059}{2} \log \left\{\frac{M^{2+}(\mathrm{aq})_{a}}{10^{-3}}\right\} \\

\Rightarrow 0.059=-\frac{0.059}{2} \log \left\{\frac{M^{2+}(\mathrm{aq})_{a}}{10^{-3}}\right\} \\

-2=\log \left\{\frac{M^{2+}(\mathrm{aq})_{a}}{10^{-3}}\right\}

\end{array}$

$\Rightarrow \quad 10^{-2} \times 10^{-3}=\mathrm{M}^{2+}(\mathrm{aq})_{a}=$ solubility $=s$

$\quad M X_{2} \rightleftharpoons M^{2+}+2 X^{-}$

$K_{s p}=S .(2 S)^{2} \quad 2 S$

$\Rightarrow \quad K_{s p}=4 s^{3}=4 \times\left(10^{-5}\right)^{3}=4 \times 10^{-15}$

Topic: Chemistry, Electrochemistry, JEE Advanced, JEE Advanced 2012 (Paper 2)
The ellipse $E_{1}: \frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_{2}$ passing through the point $(0,4)$ circumscribes the rectangle $R$. The eccentricity of the ellipse $E_{2}$ is

Options:
A $\frac{\sqrt{2}}{2}$
B $\frac{\sqrt{3}}{2}$
C $\frac{1}{2}$
D $\frac{3}{4}$

The correct answer is:
$\frac{1}{2}$

Solution:
As rectangle $A B C D$ circumscribed the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1, \quad \therefore A=(3,2)$

Let the ellipse circumscribing the rectangle $A B C D$ is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1...(i)$

Given that ellipse (i) passes through $(a, 4)$ $\therefore b^{2}=16$

Also ellipse (i) passes through $A(3,2)$

$\therefore \quad \frac{9}{a^{2}}+\frac{4}{16}=1 \Rightarrow a^{2}=12$

$\therefore \quad e=\sqrt{1-\frac{12}{16}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$

Topic: Mathematics, Ellipse, JEE Main
The ellipse $E_{1}: \frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_{2}$ passing through the point $(0,4)$ circumscribes the rectangle $R$. The eccentricity of the ellipse $E_{2}$ is

Options:
A $\frac{\sqrt{2}}{2}$
B $\frac{\sqrt{3}}{2}$
C $\frac{1}{2}$
D $\frac{3}{4}$

The correct answer is:
$\frac{1}{2}$

Solution:
As rectangle $A B C D$ circumscribed the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1, \quad \therefore A=(3,2)$

Let the ellipse circumscribing the rectangle $A B C D$ is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1...(i)$

Given that ellipse (i) passes through $(a, 4)$ $\therefore b^{2}=16$

Also ellipse (i) passes through $A(3,2)$

$\therefore \quad \frac{9}{a^{2}}+\frac{4}{16}=1 \Rightarrow a^{2}=12$

$\therefore \quad e=\sqrt{1-\frac{12}{16}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$

Topic: Mathematics, Ellipse, JEE Advanced, JEE Advanced 2012 (Paper 1)
The equation of a plane passing through the line of intersection of the planes $x+2 y+3 z=2$ and $x-y+z=3$

and at a distance $\frac{2}{\sqrt{3}}$ from the point $(3,1,-1)$ is

Options:
A $5 x-11 y+z=17$
B $\sqrt{2} x+y=3 \sqrt{2}-1$
C $x+y+z=\sqrt{3}$
D $x-\sqrt{2} y=1-\sqrt{2}$

The correct answer is:
$5 x-11 y+z=17$

Solution:
Equation of the plane passing through the intersection line of given planes is

$\begin{array}{ll}

& (x+2 y+3 z-2)+\lambda(x-y+z-3)=0 \\

\text { or } & (1+\lambda) x+(2-\lambda) y+(3+\lambda) z+(-2-3 \lambda)=0

\end{array}$

$\because \quad$ Its distance from the point $(3,1,-1)$ is $\frac{2}{\sqrt{3}}$

$\begin{array}{l}

\therefore\left|\frac{3(1+\lambda)+1(2-\lambda)-1(3+\lambda)+(-2-3 \lambda)}{\sqrt{(1+\lambda)^{2}+(2-\lambda)^{2}+(3+\lambda)^{2}}}\right|=\frac{2}{\sqrt{3}} \\

\Rightarrow\left|\frac{-2 \lambda}{\sqrt{3 \lambda^{2}+4 \lambda+14}}\right|=\frac{2}{\sqrt{3}} \\

\Rightarrow 3 \lambda^{2}+4 \lambda+14=3 \lambda^{2} \Rightarrow \lambda=-\frac{7}{2}

\end{array}$

$\therefore$ Required equation of plane is

$(x+2 y+3 z-2)-\frac{7}{2}(x-y+z-3)=0$

or $5 x-11 y+z=17$

Topic: Mathematics, Three Dimensional Geometry, JEE Advanced, JEE Advanced 2012 (Paper 2)
The equilibrium $2 \mathrm{Cu}^{\mathrm{I}} \rightleftharpoons \mathrm{Cu}^0+\mathrm{Cu}^{\text {II }}$ in aqueous medium at $25^{\circ} \mathrm{C}$ shifts towards the left in the presence of

Options:
A
$\mathrm{NO}_3^{-}$

B
$\mathrm{Cl}^{-}$

C
$\mathrm{SCN}^{-}$

D
$\mathrm{CN}^{-}$

The correct answers are:
$\mathrm{Cl}^{-}$
,
$\mathrm{SCN}^{-}$
,
$\mathrm{CN}^{-}$

Solution:
$\mathrm{Cl}^{-}, \mathrm{CN}^{-}$and $\mathrm{SCN}^{-}$froms precipitate with $\mathrm{Cu}$ (I), remove $\mathrm{Cu}$ (I) ion from equilibrium and reaction shift in direction shift in backward direction according to Le-Chatelier's principle.

Topic: Chemistry, Chemical Equilibrium, JEE Advanced, JEE Advanced 2011 (Paper 2)
The figure shows a system consisting of (i) a ring of outer radius $3 R$ rolling clockwise without slipping on a horizontal surface with angular speed $\omega$ and (ii) an inner disc of radius $2 R$ rotating anti-clockwise with angular speed $\omega / 2$. The ring and disc are separated by frictionless ball bearings. The point $P$ on the inner disc is at a distance $R$ from the origin, where $O P$ makes an angle of $30^{\circ}$ with the horizontal. Then with respect to the horizontal surface,

Options:
A the point $O$ has linear velocity $3 R \omega \hat{i}$
B the point $P$ has linear velocity $\frac{11}{4} R \omega \hat{i}+\frac{\sqrt{3}}{4} R \omega \hat{k}$.
C the point $P$ has linear velocity $\frac{13}{4} R \omega \hat{i}-\frac{\sqrt{3}}{4} R \omega \hat{k}$
D the point $P$ has linear velocity

$\left(3-\frac{\sqrt{3}}{4}\right) R \omega \hat{i}+\frac{1}{4} R \omega \hat{k}$

The correct answers are:
the point $O$ has linear velocity $3 R \omega \hat{i}$, the point $P$ has linear velocity $\frac{11}{4} R \omega \hat{i}+\frac{\sqrt{3}}{4} R \omega \hat{k}$.

Solution:
Velocity at centre 'O' $\quad \therefore \quad \vec{v}_{o}=3 R \omega \hat{i}$

$\vec{V}_{P}=3 R \omega \hat{i}-\frac{R \omega}{2} \sin 30^{\circ} \hat{i}+\frac{R \omega}{2} \cos 30^{\circ} \hat{k}$

$\therefore \quad \vec{V}_{P}=\left[3 R_{\omega} \hat{i}-\frac{R_{\omega}}{4} \hat{i}\right]+\frac{\sqrt{3} R_{\omega}}{4} \hat{k}$

or, $\quad \vec{V}_{P}=\frac{11}{4} R_{\omega} \hat{i}+\frac{\sqrt{3}}{4} R_{\omega} \hat{k}$

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2012 (Paper 2)
The figure shows certain wire segments joined together to form a coplanar loop. The loop is placed in a perpendicular
magnetic field in the direction going into the plane of the figure. The magnitude of the field increases with time. $I_1$ and $I_2$ are the currents in the segments $a b$ and $c d$. Then,

Options:
A
$I_1>I_2$

B
$I_1 < I_2$
C
$I_1$ is in the direction ba and $I_2$ is in the direction $c d$

D
$I_1$ is in the direction $a b$ and $I_2$ is in the direction $d c$

The correct answer is:
$I_1$ is in the direction $a b$ and $I_2$ is in the direction $d c$

Solution:
Cross $\otimes$ magnetic field passing from the closed loop is increasing. Therefore, from Lenz's law induced current will product $\operatorname{dot} \odot$ magnetic field. Hence, induced current is anticlockwise.

Topic: Physics, Electromagnetic Induction, JEE Main
The figure shows certain wire segments joined together to form a coplanar loop. The loop is placed in a perpendicular
magnetic field in the direction going into the plane of the figure. The magnitude of the field increases with time. $I_1$ and $I_2$ are the currents in the segments $a b$ and $c d$. Then,

Options:
A
$I_1>I_2$

B
$I_1 < I_2$
C
$I_1$ is in the direction ba and $I_2$ is in the direction $c d$

D
$I_1$ is in the direction $a b$ and $I_2$ is in the direction $d c$

The correct answer is:
$I_1$ is in the direction $a b$ and $I_2$ is in the direction $d c$

Solution:
Cross $\otimes$ magnetic field passing from the closed loop is increasing. Therefore, from Lenz's law induced current will product $\operatorname{dot} \odot$ magnetic field. Hence, induced current is anticlockwise.

Topic: Physics, Electromagnetic Induction, JEE Advanced, JEE Advanced 2009 (Paper 1)
The figure shows the $p-V$ plot of an ideal gas taken through a cycle $A B C D A$. The part $A B C$ is a semi-circle and $C D A$ is half of an ellipse. Then,

Options:
A
the process during the path $A \rightarrow B$ is isothermal

B
heat flows out of the gas during the path $B \rightarrow C \rightarrow D$

C
work done during the path $A \rightarrow B \rightarrow C$ is zero

D
positive work is done by the gas in the cycle $A B C D A$.

The correct answers are:
heat flows out of the gas during the path $B \rightarrow C \rightarrow D$
,
positive work is done by the gas in the cycle $A B C D A$.

Solution:
(A) $p-V$ graph is not rectangular hyperbola. Therefore, process $A-B$ is not isothermal.
(B) In process $B C D$, product of $p V$ (therefore temperature and intemal energy) is decreasing. Further, volume is decreasing. Hence, work done is also negative. Hence, $Q$ will be negative or heat will flow out of the gas.
(C) $W_{A B C}=$ positive
(D) For clockwise cycle on $p-V$ diagram with $P$ on $y$-axis, net work done is positive.

Topic: Physics, Thermodynamics, JEE Advanced, JEE Advanced 2009 (Paper 2)
The focal length of a thin biconvex lens is $20 \mathrm{~cm}$. When an object is moved from a distance of $25 \mathrm{~cm}$ in front of it to $50 \mathrm{~cm}$, the magnification of its image changes from $m_{25}$ to $m_{50}$. The ratio $\frac{m_{25}}{m_{50}}$ is

Options:

The correct answer is:
6

Solution:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ or
$$
\frac{u}{v}-1=\frac{u}{f}
$$
$\begin{array}{rlrl} & \text { or } & \frac{u}{v} & =\left(\frac{u+f}{f}\right) \\ \therefore & m=\frac{v}{u} & =\left(\frac{f}{u+f}\right) \\ \frac{m_{25}}{m_{50}} & =\frac{\left(\frac{20}{-25+20}\right)}{\left(\frac{20}{-50+20}\right)}=6\end{array}$
$\therefore$ answer is 6 .

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2010 (Paper 1)
The following carbohydrate is

Options:
A
a ketohexose

B
an aldohexose

C
an $\alpha$-furanose

D
an $\alpha$-pyranose

The correct answer is:
an aldohexose

Solution:
Here, the- $\mathrm{OH}$ of hemiacetal group is equatorial therefore, it is a $\beta$-pyranose of an aldohexose.

Topic: Chemistry, Biomolecules, JEE Advanced, JEE Advanced 2011 (Paper 2)
The freezing point (in ${ }^{\circ} \mathrm{C}$ ) of solution containing $0.1 \mathrm{~g}$ of $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ (molecular weight 329$)$ in $100 \mathrm{~g}$ of water $\left(K_f=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$ is

Options:
A
$-2.3 \times 10^{-2}$

B
$-5.7 \times 10^{-2}$

C
$-5.7 \times 10^{-3}$

D
$-1.2 \times 10^{-2}$

The correct answer is:
$-2.3 \times 10^{-2}$

Solution:
vant Hoff's factor $(i)=4\left\{3 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}\right\} \quad$ Molality $=\frac{0.1}{329} \times \frac{1000}{100}=\frac{1}{329}$
$$
\Rightarrow \quad-\Delta T_f=i K_f \cdot m=4 \times 1.86 \times \frac{1}{329}=2.3 \times 10^{-2} \Rightarrow T_f=-2.3 \times 10^{-2}{ }^{\circ} \mathrm{C}
$$
(As freezing point of water is $0^{\circ} \mathrm{C}$ )

Topic: Chemistry, Solutions, JEE Advanced, JEE Advanced 2011 (Paper 2)
The function $f:[0,3] \rightarrow[1,29]$, defined by $f(x)=2 x^{3}-15 x^{2}+36 x+1$, is

Options:
A one-one and onto
B onto but not one-one
C one-one but not onto
D neither one-one nor onto

The correct answer is:
onto but not one-one

Solution:
Given : $f(x)=2 x^{3}-15 x^{2}+36 x+1$

$\Rightarrow f^{\prime}(x)=6 x^{2}-30 x+36$

$=6\left(x^{2}-5 x+6\right)=6(x-2)(x-3)$

$\because f^{\prime}(x)>0 \forall x \in[0,2)$ and $f^{\prime}(x) < 0 \forall x \in(2,3)$

$\therefore f(x)$ is increasing on $[0,2)$ and decreasing

on $(2,3)$

$\therefore \quad f(x)$ is many one on $[0,3]$

Also $f(0)=1, f(2)=29, f(3)=28$

$\therefore \quad$ Absolute min $=1$ and Absolute max $=29$

$\therefore \quad$ Range of $f=[1,29]=$ codomain

Hence $f$ is onto.

Topic: Mathematics, Functions, JEE Advanced, JEE Advanced 2012 (Paper 1)
The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass.

These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless, stick, as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed $\omega$, the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the $z$-axis and (ii) a rotation

of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points $P$ and $Q$ ). Both these motions have the same angular speed $\omega$ in this case

Now consider two similar systems as shown in the figure: Case

(a) the disc with its face vertical and parallel to $x-z$ plane; Case

(b) the disc with its face making an angle of

$45^{\circ}$ with $x-y$ plane and its horizontal diameter parallel to $x$-axis. In both the cascs, the disc is welded at point $\mathrm{P}$, and the systems are rotated with constant angular speed $\omega$ about the $z$ axis.
Question: Which of the following statements about the instantaneous axis (passing through the centre of mass) is correct?

Options:
A It is vertical for both the cases (a) and (b)
B It is vertical for case (a); and is at $45^{\circ}$ to the $x-z$ plane and lies in the plane of the disc for case (b).
C It is horizontal for case (a); and is at $45^{\circ}$ to the $x$ - $z$ plane and is normal to the plane of the disc for case (b).
D It is vertical for case (a); and is $45^{\circ}$ to the $x-z$ plane and is normal to the plane of the disc for case (b).

The correct answer is:
It is vertical for both the cases (a) and (b)

Solution:
Axis of rotation is parallel to z-axis. Hence for both the cases, in stantaneous axis passing through is vertical.

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2012 (Paper 2)
The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass.

These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless, stick, as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed $\omega$, the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the $z$-axis and (ii) a rotation

of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points $P$ and $Q$ ). Both these motions have the same angular speed $\omega$ in this case

Now consider two similar systems as shown in the figure: Case

(a) the disc with its face vertical and parallel to $x-z$ plane; Case

(b) the disc with its face making an angle of

$45^{\circ}$ with $x-y$ plane and its horizontal diameter parallel to $x$-axis. In both the cascs, the disc is welded at point $\mathrm{P}$, and the systems are rotated with constant angular speed $\omega$ about the $z$ axis.
Question: Which of the following statements regarding the angular speed about the instantaneous axis (passing through the centre of mass) is correct?

Options:
A It is $\sqrt{2} \omega$ for both the cases
B It is $\omega$ for case (a); and $\omega / \sqrt{2}$ for case (b)
C It is $\omega$ for case (a); and $\sqrt{2 \omega}$ for case (b)
D It is $\omega$ for both the cases.

The correct answer is:
It is $\omega$ for both the cases.

Solution:
For rigid body $\omega$ is same for any point of the body.

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2012 (Paper 2)
The given graph represent the variations of compressibility factor $(Z)=\frac{P V}{n R T}$ versus $P$, for three real gases $A, B$ and $C$. Identify the only incorrect statement

Options:
A
For the gas $A, a=0$ and its dependence on $P$ is linear at all pressure.

B
For the gas $B, b=0$ and its dependence on $P$ is linear at all pressure.

C
For the gas $C$, which is typical real gas for which neither $a$ nor $b=0$. By knowing the minima and point of the intersection, with $Z=1, a$ and $b$ can be calculated.

D
At high pressure, the slope is positive for all real gases.

The correct answers are:
For the gas $B, b=0$ and its dependence on $P$ is linear at all pressure.

Solution:
From the graph it is clear that the value of $Z$ decreases with increase of pressure. We can explain as follows on the basis of van der Waals' equation.
At high pressure, when ' $P$ ' is large, $V$ will be small and one can not ignore ' $b$ ' in comparison to $V$.
However, the term $\frac{a}{V^2}$ may be considered negligible in comparison to ' $p$ ' in van der Waals' equation
or
$$
\begin{aligned}
\left(p+\frac{a}{V^2}\right)(V-b) & =n R T \\
P(V-b) & =n R T \\
p V-p b & =n R T \\
\frac{p V}{n R T} & =1+\frac{p b}{n R T} \\
Z & =1+\frac{p b}{n R T}
\end{aligned}
$$
or
Here, $Z$ is greater than one and it increases linearly with pressure. Hence, statement (b) is false.

Topic: Chemistry, States of Matter, JEE Advanced, JEE Advanced 2006
The given graphs/data I, II, III and IV represent general trends observed for different physisorption and chemisorption processes under mild conditions of temperature and presure. Which of the following choice(s) about I, II, III and IV is (are) correct?

Options:
A I is physisorption and II is chemisorption
B $\mathrm{I}$ is physisorption and $\mathrm{III}$ is chemisorption
C IV is chemisorption and II is chemisorption
D IV is chemisorption and III is chemisorption

The correct answers are:
I is physisorption and II is chemisorption, IV is chemisorption and II is chemisorption

Solution:
Graph (I) and (III) represent physisorption because, in physisorption, the amount of adsorption decreases with the increase of temperature and increases with the increase of pressure.

Graph (II) represent chemisorption, because in chemisorption amount of adsorption increases with the increase of temperature. Graph (IV) is showing the formation of a chemical bond, hence chemisorption.

Topic: Chemistry, Surface Chemistry, JEE Advanced, JEE Advanced 2012 (Paper 2)
The graph between $1 / \lambda$ and stopping potential $(V)$ of three metals having work function $\phi_1, \phi_2$ and $\phi_3$ in an experiment of photoelectric effect is plotted as shown in the figure. Which of the following statement(s) is/are correct? [Here $\lambda$ is the wavelength of the incident ray].

Options:
A
Ratio of work functions $\phi_1: \phi_2: \phi_3=1: 2: 4$.

B
Ratio of work functions $\phi_1: \phi_2: \phi_3=4: 2: 1$.

C
$\tan \theta$ is directly proportional to $h c / e$, where $h$ is Planck's constant and $c$ is the speed of light.

D
The violet colour light can eject photoelectrons from metals 2 and 3.

The correct answers are:
Ratio of work functions $\phi_1: \phi_2: \phi_3=1: 2: 4$.
,
$\tan \theta$ is directly proportional to $h c / e$, where $h$ is Planck's constant and $c$ is the speed of light.

Solution:
From the relation, $e V=\frac{h c}{\lambda}-\phi$ or $V=\left(\frac{h c}{e}\right)\left(\frac{1}{\lambda}\right)-\frac{\phi}{e}$ This is equation of straight line.
Slope is $\tan \theta=\frac{h c}{e}$
$$
\begin{aligned}
\phi_1: \phi_2: \phi_3 & =\frac{h c}{\lambda_{01}}: \frac{h c}{\lambda_{02}}: \frac{h c}{\lambda_{03}}=\frac{1}{\lambda_{01}}: \frac{1}{\lambda_{02}}: \frac{1}{\lambda_{03}}=1: 2: 4 \\
\frac{1}{\lambda_{01}} & =0.001 \mathrm{~nm}^{-1} \text { or } \lambda_{01}=10000 Å \\
\frac{1}{\lambda_{02}} & =0.002 \mathrm{~nm}^{-1} \text { or } \lambda_{02}=5000 Å \\
\frac{1}{\lambda_{03}} & =0.004 \mathrm{~nm}^{-1} \text { or } \lambda_{03}=25000 Å
\end{aligned}
$$
Violet colour has wavelength $4000 Å$.
So, violet colour can eject photoelectrons from metal-1 and metal-2.

Topic: Physics, Dual Nature of Matter, JEE Advanced, JEE Advanced 2006