datavorous/entrance-exam-dataset huggingface dataset:
Assume that the nuclear binding energy per nucleon $(B / A)$ versus mass number $(A)$ is as shown in the figure. Use this plot to choose the correct choice (s) given below.

Options:
A
Fusion of two nuclei with mass numbers lying in the range of $1 < A < 50$ will release energy
B
Fusion of two nuclei with mass numbers lying in the range of $51 < A < 100$ will release energy
C
Fission of a nucleus lying in the mass range of $100 < A < 200$ will release energy when broken into two equal fragments
D
Fission of a nucleus lying in the mass range $200 < A < 260$ will release energy when broken into two equal fragments

The correct answers are:
Fusion of two nuclei with mass numbers lying in the range of $51 < A < 100$ will release energy,
Fission of a nucleus lying in the mass range $200 < A < 260$ will release energy when broken into two equal fragments

Solution:
In fusion two or more lighter nuclei combine to make a comparatively heavier nucleus. In fission, a heavy nucleus breaks into two or more comparatively lighter nuclei. Futher, energy will be released in a nuclear process if total binding energy increases. $\therefore$ correct options are (b) and (d).

Topic: Physics, Nuclear Physics, JEE Advanced, JEE Advanced 2008 (Paper 1)
Assuming that Hund's rule is violated, the bond order and magnetic nature of the diatomic molecule $\mathrm{B}_2$ is

Options:
A
1 and diamagnetic

B
0 and diamagnetic

C
1 and paramagnetic

D
0 and paramagnetic

The correct answer is:
1 and diamagnetic

Solution:
$\mathrm{B}_2$ : Total electrons $=10$
Configuration : $\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^1=\pi 2 p_y^1$
If Hunds rule is violated, then
$$
\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \quad \pi 2 p_x^2=\pi 2 p_y^0
$$
So, diamagnetic
$$
\text { Bond order }=\frac{6-4}{2}=1
$$
Chemical bonding
Conceptual
II

Topic: Chemistry, Structure of Atom, JEE Advanced, JEE Advanced 2010 (Paper 2)
At $400 \mathrm{~K}$, the root mean square (rms) speed of a gas $X$ (molecular weight $=40$ ) is equal to the most probable speed of gas $Y$ at $60 \mathrm{~K}$. The molecular weight of the gas $Y$ is

Options:

The correct answer is:
4

Solution:

Topic: Chemistry, States of Matter, JEE Advanced, JEE Advanced 2009 (Paper 2)
At time $t=0$, a battery of $10 \mathrm{~V}$ is connected across points $A$ and $B$ in the given circuit. If the capacitors have no charge initially, at what time (in second) does the voltage across them become $4 \mathrm{~V}$ ?
[Take $: \ln 5=1.6, \ln 3=1.1]$

Options:

The correct answer is:
2

Solution:
Voltage across the capacitors will increase from 0 to $10 \mathrm{~V}$ exponentially. The voltage at time $t$ will be given by
$$
\begin{aligned}
& V=10\left(1-e^{-t / \tau} \mathrm{C}\right) \\
& \text { Here } \tau_c=C_{\text {net }} R_{\text {net }} \\
& =\left(1 \times 10^6\right)\left(4 \times 10^{-6}\right)=4 \mathrm{~s} \\
& \therefore \quad V=10\left(1-e^{-t / 4}\right) \\
&
\end{aligned}
$$
Substituting $V=4$ volt we have,
$$
4=10\left(1-e^{-t / 4}\right)
$$
Substituting $V=4$ volt we have,
$$
4=10\left(1-e^{-t / 4}\right)
$$
or $e^{-t / 4}=0.6=\frac{3}{5}$
Taking $\log$ both sides we have,
$$
-\frac{t}{4}=\ln 3-\ln 5
$$
or $\quad t=4(\ln 5-\ln 3)=2 \mathrm{~s}$.
Hence, the answer is 2 .

Topic: Physics, Alternating Current, JEE Advanced, JEE Advanced 2010 (Paper 2)
Axis of a parabola is $y=x$ and vertex and focus are at a distance $\sqrt{2}$ and $2 \sqrt{2}$ respectively, from the origin. Then, equation of the parabola is

Options:
A
$(x-y)^2=8(x+y-2)$

B
$(x+y)^2=2(x+y-2)$

C
$(x-y)^2=4(x+y-2)$

D
$(x+y)^2=2(x-y+2)$

The correct answer is:
$(x-y)^2=8(x+y-2)$

Solution:
As distance of vertex from origin is $\sqrt{2}$ and focus is $2 \sqrt{2}$. $\therefore V(1,1)$ and $F(2,2)$
[i.e. lying on $y=x$ ] where, length of latusrectum
$$
\begin{aligned}
& =4 a, \\
& =4 \sqrt{2}
\end{aligned}
$$
$$
[\because \text { where } a=\sqrt{2}]
$$
$\therefore$ By definition of parabola
$$
P M^2=(4 a)(P N)
$$
where, $P N$ is length of perpendicular upon $x+y-2=0$
[i.e. tangent at vertex]
$$
\begin{array}{ll}
\Rightarrow & \frac{(x-y)^2}{2}=4 \sqrt{2}\left(\frac{x+y-2}{\sqrt{2}}\right) \\
\therefore & (x-y)^2=8(x+y-2)
\end{array}
$$
Hence, the equation of parabola is $(x-y)^2=8(x+y-2)$.

Topic: Mathematics, Parabola, JEE Main
Axis of a parabola is $y=x$ and vertex and focus are at a distance $\sqrt{2}$ and $2 \sqrt{2}$ respectively, from the origin. Then, equation of the parabola is

Options:
A
$(x-y)^2=8(x+y-2)$

B
$(x+y)^2=2(x+y-2)$

C
$(x-y)^2=4(x+y-2)$

D
$(x+y)^2=2(x-y+2)$

The correct answer is:
$(x-y)^2=8(x+y-2)$

Solution:
As distance of vertex from origin is $\sqrt{2}$ and focus is $2 \sqrt{2}$. $\therefore V(1,1)$ and $F(2,2)$
[i.e. lying on $y=x$ ] where, length of latusrectum
$$
\begin{aligned}
& =4 a, \\
& =4 \sqrt{2}
\end{aligned}
$$
$$
[\because \text { where } a=\sqrt{2}]
$$
$\therefore$ By definition of parabola
$$
P M^2=(4 a)(P N)
$$
where, $P N$ is length of perpendicular upon $x+y-2=0$
[i.e. tangent at vertex]
$$
\begin{array}{ll}
\Rightarrow & \frac{(x-y)^2}{2}=4 \sqrt{2}\left(\frac{x+y-2}{\sqrt{2}}\right) \\
\therefore & (x-y)^2=8(x+y-2)
\end{array}
$$
Hence, the equation of parabola is $(x-y)^2=8(x+y-2)$.

Topic: Mathematics, Parabola, JEE Advanced, JEE Advanced 2006
$\mathrm{B}(\mathrm{OH})_3+\mathrm{NaOH} \rightleftharpoons \mathrm{NaBO}_2+\mathrm{Na}\left[\mathrm{B}(\mathrm{OH})_4\right]+\mathrm{H}_2 \mathrm{O}$.
How can this reaction is made to proceed in forward direction?

Options:
A
Addition of cis 1, 2 diol

B
Addition of borax

C
Addition of trans 1, 2 diol

D
Addition of $\mathrm{Na}_2 \mathrm{HPO}_4$

The correct answer is:
Addition of cis 1, 2 diol

Solution:
This reaction is reversible reaction because sodium metaborate, $\mathrm{Na}^{+}\left[\mathrm{B}(\mathrm{OH})_4\right]^{-}$formed by the reaction between $\mathrm{B}(\mathrm{OH})_3$ and $\mathrm{NaOH}$ gets hydrolysed to regenerate $\mathrm{B}(\mathrm{OH})_3$ and $\mathrm{NaOH}$
If, some quantity of polyhydroxy compound like cis-1, 3-diol, catechol, glycerol etc. is added to the reaction mixture. The $\mathrm{B}(\mathrm{OH})_3$ combines with such polyhydroxy compound to give chelated complex compound. Due to complex compound formation, stability increases and due to higher stability of complex, reaction moves in forward direction.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
$\mathrm{B}(\mathrm{OH})_3+\mathrm{NaOH} \rightleftharpoons \mathrm{NaBO}_2+\mathrm{Na}\left[\mathrm{B}(\mathrm{OH})_4\right]+\mathrm{H}_2 \mathrm{O}$.
How can this reaction is made to proceed in forward direction?

Options:
A
Addition of cis 1, 2 diol

B
Addition of borax

C
Addition of trans 1, 2 diol

D
Addition of $\mathrm{Na}_2 \mathrm{HPO}_4$

The correct answer is:
Addition of cis 1, 2 diol

Solution:
This reaction is reversible reaction because sodium metaborate, $\mathrm{Na}^{+}\left[\mathrm{B}(\mathrm{OH})_4\right]^{-}$formed by the reaction between $\mathrm{B}(\mathrm{OH})_3$ and $\mathrm{NaOH}$ gets hydrolysed to regenerate $\mathrm{B}(\mathrm{OH})_3$ and $\mathrm{NaOH}$
If, some quantity of polyhydroxy compound like cis-1, 3-diol, catechol, glycerol etc. is added to the reaction mixture. The $\mathrm{B}(\mathrm{OH})_3$ combines with such polyhydroxy compound to give chelated complex compound. Due to complex compound formation, stability increases and due to higher stability of complex, reaction moves in forward direction.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Advanced, JEE Advanced 2006
Balanoglossus has

Options:
A no coelom
B well developed coelom
C reduced coelom
D coelom only in embryonic stage

The correct answer is:
well developed coelom

Solution:
Animals are classified as non-chordates or chordates depending on the absence and presence of notochord. Non-chordates are further divided into many phyla beginning from phylum Porifera to phylum Hemichordates. Phylum Porifera includes the simplest animals while chordates include advanced animals as many features are seen evolving. One such important feature is the presence of body coelom, as the development of organs and organ systems will largely depend on this feature. Porifera, Coelenterate, Ctenophora, Platyhelminthes are acoelomates. Aschelminthes are pseudocoelomates as the coelom is not completely formed. All animals from phylum Annelids to chordates show presence of true coelom and hence are eucoelomates. Thus hemichordates are also eucoelomates and it includes Balanoglossus, Saccoglossus etc.

Topic: Biology, Animal Kingdom, NEET
Based on VSEPR theory, the number of 90 degree $\mathrm{F}-\mathrm{Br}-\mathrm{F}$ angles in $\mathrm{BrF}_5$ is

Options:

The correct answer is:
0

Solution:
In $\mathrm{BrF}_5$ number of electrons $=6$ $(1 l p+5 b p)$
So, the structure is supposed to be square pyramidal but will be distorted because of additional lp- $b p$ interaction.
Additional $\quad l p-b p$
interaction reduces the all bond angle and do not let any angle to be $90^{\circ}$.
Chemical bonding
Conceptual
IV

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Advanced, JEE Advanced 2010 (Paper 1)
Bleaching powder and bleach solution are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry.

Question: Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is

Options:
A $\mathrm{Cl}_{2} \mathrm{O}$
B $\mathrm{Cl}_{2} \mathrm{O}_{7}$
C $\mathrm{ClO}_{2}$
D $\mathrm{Cl}_{2} \mathrm{O}_{6}$

The correct answer is:
$\mathrm{Cl}_{2} \mathrm{O}$

Solution:
$\mathrm{Ca}(\mathrm{OCl}) \mathrm{Cl} \longrightarrow \mathrm{Ca}^{2+}+-\mathrm{OCl}+\mathrm{Cl}^{-}$

-OCl (Hypochlorite ion) is anion of the acid HOCl which on dehydration gives $\mathrm{Cl}_{2} \mathrm{O}$.

$2 \mathrm{HOCl} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{Cl}_{2} \mathrm{O}$

Topic: Chemistry, Redox Reactions, JEE Advanced, JEE Advanced 2012 (Paper 2)
Bleaching powder and bleach solution are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry.

Question: $25 \mathrm{~mL}$ of household solution was mixed with $30 \mathrm{~mL}$ of $0.50$ M KI and $10 \mathrm{~mL}$ of $4 \mathrm{~N}$ acetic acid. In the titration of the liberated iodine, $48 \mathrm{~mL}$ of $0.25 \mathrm{~N~} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$ was used to reach the end point. The molarity of the household bleach solution is

Options:
A $0.48 \mathrm{M}$
B $0.96 \mathrm{M}$
C $0.24 \mathrm{M}$
D $0.024 \mathrm{M}$

The correct answer is:
$0.24 \mathrm{M}$

Solution:
Bleach $+2 \mathrm{KI} \longrightarrow \mathrm{I}_{2}+$ Products

$\mathrm{I}_{2}+2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \longrightarrow \mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}+2 \mathrm{NaI}$

Number of millimole of hypo $=0.25 \times 48$

$=2 \times$ millimole of $\mathrm{I}_{2}$

$\therefore$ Number of millimole of $\mathrm{I}_{2}=\frac{0.25 \times 48}{2}=6$

millimole of $\mathrm{I}_{2}=$ millimole of bleach

Molarity of bleaching solution

$=\frac{\text { Millimoles of bleach }}{\text { Vol.(in } \mathrm{mL} \text { ) of bleach }}=\frac{6}{25}=0.24$

Topic: Chemistry, Redox Reactions, JEE Advanced, JEE Advanced 2012 (Paper 2)
Bombardment of aluminium by $\alpha$-particle leads to its artificial disintegration in two ways, (i) and (ii) as shown. Products $X, Y$ and $Z$ respectively are

Options:
A
proton, neutron, positron

B
neutron, positron, proton

C
proton, positron, neutron

D
positron, proton, neutron

The correct answer is:
proton, neutron, positron

Solution:

Topic: Chemistry, Structure of Atom, JEE Advanced, JEE Advanced 2011 (Paper 1)
Both $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ and $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ are diamagnetic. The hybridisations of nickel in these complexes, respectively, are

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

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

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

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

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

Solution:
In $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$, the oxidation state of $\mathrm{Ni}$ is zero $(0)$.
$$
\mathrm{Ni}(28)[\mathrm{Ar}] 4 \mathrm{~s}^2, 3 d^8
$$
$$
\mathrm{CO} \text { is a strong ligand, causes coupling, thus }
$$
In $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$, the oxidation state of $\mathrm{Ni}$ is $+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 d^8, 4 s^0$
$$
\mathrm{CN}^{-} \text {is strong ligand causes coupling. }
$$

Topic: Chemistry, Coordination Compounds, JEE Main
Both $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ and $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ are diamagnetic. The hybridisations of nickel in these complexes, respectively, are

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

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

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

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

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

Solution:
In $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$, the oxidation state of $\mathrm{Ni}$ is zero $(0)$.
$$
\mathrm{Ni}(28)[\mathrm{Ar}] 4 \mathrm{~s}^2, 3 d^8
$$
$$
\mathrm{CO} \text { is a strong ligand, causes coupling, thus }
$$
In $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$, the oxidation state of $\mathrm{Ni}$ is $+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 d^8, 4 s^0$
$$
\mathrm{CN}^{-} \text {is strong ligand causes coupling. }
$$

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2008 (Paper 2)
$C_V$ and $C_p$ denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then

Options:
A
$C_p-C_V$ is larger for a diatomic ideal gas than for a monoatomic ideal gas

B
$C_p+C_V$ is larger for a diatomic ideal gas than for a monoatomic ideal gas.

C
$\frac{C_p}{C_V}$ is larger for a diatomic ideal gas than for a monoatomic ideal gas

D
$C_p \cdot C_V$ is larger for a diatomic ideal gas than for a monoatomic ideal gas

The correct answers are:
$C_p+C_V$ is larger for a diatomic ideal gas than for a monoatomic ideal gas.
,
$C_p \cdot C_V$ is larger for a diatomic ideal gas than for a monoatomic ideal gas

Solution:
For monoatomic gas,
$$
C_p=\frac{5}{2} R \text { and } C_V=\frac{3}{2} R
$$
For diatomic gas,
$$
C_p=\frac{7}{2} R \text { and } C_V=\frac{5}{2} R
$$

Topic: Physics, Kinetic Theory of Gases, JEE Advanced, JEE Advanced 2009 (Paper 1)
Cellulose upon acetylation with excess acetic anhydride $/ \mathrm{H}_2 \mathrm{SO}_4$ (catalytic) gives cellulose triacetate whose structures is

Options:
A

B

C

D

The correct answer is:

Solution:
Cellulose is a straight chain polysaccharide composed of D-glucose units, which are joined by $\beta$-glycosidic linkages between $\mathrm{C}-1$ of one glucose and C-4 of the next glucose. In one, unit only three hydroxy groups are free to form acetate, that's why called cellulose triacetate

Topic: Chemistry, Biomolecules, JEE Advanced, JEE Advanced 2008 (Paper 2)
$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2+\mathrm{NOCl} \longrightarrow \mathrm{P}$ Identify the adduct

Options:
A

B

C

D

The correct answer is:

Solution:
This reaction is an example of electrophilic addition reaction and in it addition takes place according to Markownikoff's rule.
("Negative part of additive reagent adds less hydrogenated or more substituted carbon atom of double bond of unsymmetrical alkane").
It is also represented in form of following mechanism

Topic: Chemistry, Aldehydes and Ketones, JEE Advanced, JEE Advanced 2006
$\mathrm{CH}_3 \mathrm{NH}_2+\mathrm{CHCl}_3+\mathrm{KOH} \rightarrow$ Nitrogen containing compound $+\mathrm{KCl}+\mathrm{H}_2 \mathrm{O}$. Nitrogen containing compound is

Options:
A

B

C

D

The correct answer is:

Solution:
$\mathrm{CH}_3 \mathrm{NH}_2+\mathrm{CHCl}_3+3 \mathrm{KOH} \rightarrow \mathrm{CH}_3 \mathrm{NC}+3 \mathrm{KCl}+3 \mathrm{H}_2 \mathrm{O}$ $\mathrm{CH}_3 \mathrm{NC}$ or $\mathrm{CH}_3-\mathrm{N}^{+} \equiv \overline{\mathrm{C}}$ Methyl isocyanide or Methyl carbyl amine This reaction is an example of carbyl amine reaction and its is used for the distinction of $p$-amines with $s$ - and $t$-amines or identification of $p$-amino group.

Topic: Chemistry, Amines, JEE Advanced, JEE Advanced 2006
Choose the correct reason(s) for the stability of the lyophobic colloidal particles.

Options:
A Preferential adsorption of ions on their surface from the solution.
B Preferential adsorption of solvent on their surface from the solution.
C Attraction between different particles having opposite charges on their surface.
D Potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles.

The correct answers are:
Preferential adsorption of ions on their surface from the solution., Potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles.

Solution:
(a) When two or more ions are present in the dispersion medium, preferential adsorption of the ion common to the colloidal particle takes place.

(d) This is called zeta potential which helps in increasing the stability of the lyophobic sol.

Topic: Chemistry, Surface Chemistry, JEE Advanced, JEE Advanced 2012 (Paper 1)
Column I contains a list of processes involving expansion of an ideal gas. Match this with Column II describing the thermodynamic change during this process. Indicate your answer by darkening the appropriate bubbles of the $4 \times 4$ matrix given in the ORS.

Options:
A
(A) q, (B) p,q, (C) p,s, (D) q,s

B
(A) q,r, (B) p,q, (C) s, (D) q

C
(A) p, (B) q, (C) p,r, (D) q,s

D
(A) p,q, (B) q, (C) p,q, (D) q

The correct answer is:
(A) q, (B) p,q, (C) p,s, (D) q,s

Solution:
(A) : In case of free expansion under adiabatic conditions, change in internal energy $\Delta U=0$.
$\therefore$ Internal energy and temperature will remain constant.
(B) $p \propto \frac{1}{v^2}$
$$
\begin{array}{rrrr}
& \therefore & p V^2 & =\text { constant } \\
& \text { or } & \left(\frac{n R T}{V}\right) V^2=\text { constant } \\
& \therefore & T \propto \frac{1}{V}
\end{array}
$$
If volume is doubled, temperature will decrease as per Eq. (ii). Further, molar heat capacity in process $p V^x=$ constant is
$$
C=C_V+\frac{R}{1-x}
$$
From Eq. (i), $x=2$
$$
\therefore \quad C=\frac{3}{2} R+\frac{R}{1-2}=+\frac{R}{2}
$$
Since molar heat capacity is positive, according to $Q=n C \Delta T, Q$ will be negative it $\Delta T$ is negative. Or gas loses heat if temperature is decreasing.
(C) :
$$
\begin{aligned}
p & \propto \frac{1}{V^{1 / 3}} \\
P V^{4 / 3} & =\text { constant } \\
\left(\frac{n R T}{V}\right) V^{4 / 3} & =\text { constant } \\
T & \propto \frac{1}{V^{1 / 3}}
\end{aligned}
$$
Further, with increase in volume temperature will decrease.
Here,
$$
\begin{array}{lc}
\text { Here, } & x=4 / 3 \\
\therefore & C=\frac{3}{2} R+\frac{R}{1-4 / 3}=-1.5 R
\end{array}
$$
As molar heat capacity is negative, $Q$ will be positive, if $\Delta T$ is negative or gas gains heat with decrease in temperature.
(D) $T \propto p V$
In expansion from $V_1$ or $2 V_1$, product of $p V$ is increasing. Therefore, temperature will increase or $\Delta U=+$ ve.
Further, in expansion work done is also positive.
Hence, $Q=W+\Delta U=+$ ve
or gas gains heat.

Topic: Physics, Thermodynamics, JEE Main
Column I contains a list of processes involving expansion of an ideal gas. Match this with Column II describing the thermodynamic change during this process. Indicate your answer by darkening the appropriate bubbles of the $4 \times 4$ matrix given in the ORS.

Options:
A
(A) q, (B) p,q, (C) p,s, (D) q,s

B
(A) q,r, (B) p,q, (C) s, (D) q

C
(A) p, (B) q, (C) p,r, (D) q,s

D
(A) p,q, (B) q, (C) p,q, (D) q

The correct answer is:
(A) q, (B) p,q, (C) p,s, (D) q,s

Solution:
(A) : In case of free expansion under adiabatic conditions, change in internal energy $\Delta U=0$.
$\therefore$ Internal energy and temperature will remain constant.
(B) $p \propto \frac{1}{v^2}$
$$
\begin{array}{rrrr}
& \therefore & p V^2 & =\text { constant } \\
& \text { or } & \left(\frac{n R T}{V}\right) V^2=\text { constant } \\
& \therefore & T \propto \frac{1}{V}
\end{array}
$$
If volume is doubled, temperature will decrease as per Eq. (ii). Further, molar heat capacity in process $p V^x=$ constant is
$$
C=C_V+\frac{R}{1-x}
$$
From Eq. (i), $x=2$
$$
\therefore \quad C=\frac{3}{2} R+\frac{R}{1-2}=+\frac{R}{2}
$$
Since molar heat capacity is positive, according to $Q=n C \Delta T, Q$ will be negative it $\Delta T$ is negative. Or gas loses heat if temperature is decreasing.
(C) :
$$
\begin{aligned}
p & \propto \frac{1}{V^{1 / 3}} \\
P V^{4 / 3} & =\text { constant } \\
\left(\frac{n R T}{V}\right) V^{4 / 3} & =\text { constant } \\
T & \propto \frac{1}{V^{1 / 3}}
\end{aligned}
$$
Further, with increase in volume temperature will decrease.
Here,
$$
\begin{array}{lc}
\text { Here, } & x=4 / 3 \\
\therefore & C=\frac{3}{2} R+\frac{R}{1-4 / 3}=-1.5 R
\end{array}
$$
As molar heat capacity is negative, $Q$ will be positive, if $\Delta T$ is negative or gas gains heat with decrease in temperature.
(D) $T \propto p V$
In expansion from $V_1$ or $2 V_1$, product of $p V$ is increasing. Therefore, temperature will increase or $\Delta U=+$ ve.
Further, in expansion work done is also positive.
Hence, $Q=W+\Delta U=+$ ve
or gas gains heat.

Topic: Physics, Thermodynamics, JEE Advanced, JEE Advanced 2008 (Paper 2)
Column I describes some situations in which a small object moves. Column II describes some characteristics of these motions. Match the situations in Column I with the characteristics in Column II and indicate your answer by darkening appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

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

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

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

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

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

Solution:
No Solution Available

Topic: Physics, Oscillations, JEE Advanced, JEE Advanced 2007 (Paper 2)
Column I gives a list of possible set of parameters measured in some experiments. The variations of the parameters in the form of graphs are shown in Column II. Match the set of parameters given in Column I with the graphs given in Column II. Indicate your answer by darkening the appropriate bubble of the $4 \times 4$ matrix given in the ORS.

Options:
A
(A) q, (B) p,q, (C) p,s, (D) q,s

B
(A) q,r, (B) p,q, (C) s, (D) q

C
(A) p, (B) q,s, (C) s, (D) q

D
(A) p,q, (B) q, (C) p,q, (D) q

The correct answer is:
(A) p, (B) q,s, (C) s, (D) q

Solution:
(A) Potential energy is minimum at mean position.
(B) For $a=0, s=v t \longrightarrow$ option $(q)$
or $\quad s=s_0+v t \longrightarrow$ option $(r)$
For $a=$ constant,
$$
s=u t+\frac{1}{2} a t^2 \longrightarrow \text { option }(s)
$$
(C) $R=\frac{v^2 \sin 2 \theta}{g} \therefore R \propto v^2 \longrightarrow$ option (s)
(D) $T=2 \pi \sqrt{\frac{l}{g}} \therefore T^2 \propto l \longrightarrow$ option $(q)$

Topic: Physics, Experimental Physics, JEE Advanced, JEE Advanced 2008 (Paper 2)
Column I gives certain situations in which a straight metallic wire of resistance $R$ is used and Column II gives some resulting effects. Match the statements in Column I with the statements in Column II and indicate your answer by darkening appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

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

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

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

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

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

Solution:
No Solution Available

Topic: Physics, Electrostatics, JEE Main
Column I gives certain situations in which a straight metallic wire of resistance $R$ is used and Column II gives some resulting effects. Match the statements in Column I with the statements in Column II and indicate your answer by darkening appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

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

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

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

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

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

Solution:
No Solution Available

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2007 (Paper 1)