datavorous/entrance-exam-dataset huggingface dataset:
Under the same reaction conditions, initial concentration of $1.386 \mathrm{~mol} \mathrm{dm}^{-3}$ of a substance becomes half in 40 s and 20 s through first-order and zero-order kinetics, respectively. Ratio $\left(\frac{k_1}{k_0}\right)$ of the rate constants for first order $\left(k_1\right)$ and zero $\operatorname{order}\left(k_0\right)$ of the reaction is

Options:
A
$0.5 \mathrm{~mol}^{-1} \mathrm{dm}^3$

B
$1.0 \mathrm{~mol} \mathrm{dm}^3$

C
$1.5 \mathrm{~mol} \mathrm{dm}^3$

D
$2.0 \mathrm{~mol}^{-1} \mathrm{dm}^3$

The correct answer is:
$0.5 \mathrm{~mol}^{-1} \mathrm{dm}^3$

Solution:
First order kinetics, $k_1=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{40} \mathrm{~s}^{-1}$
Zero order kinetics, $k_1=\frac{C_0}{2 t_{1 / 2}}=\frac{1.386}{2 \times 20}$
Hence, $\frac{k_1}{k_0}=\frac{0.693}{1.386}=0.5$

Topic: Chemistry, Chemical Kinetics, JEE Advanced, JEE Advanced 2008 (Paper 1)
Using the data provided, calculate the multiple bond energy $\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$ of a $\mathrm{C} \equiv \mathrm{C}$ bond in $\mathrm{C}_{2} \mathrm{H}_{2}$. That energy is (take the bond energy of a $\mathrm{C}-\mathrm{H}$ bond as $350 \mathrm{~kJ} \mathrm{~mol}^{-1}$ ) $2 \mathrm{C}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{HC}=\mathrm{CH}(\mathrm{g}) ; \Delta \mathrm{H}=225 \mathrm{~kJ} \mathrm{~mol}^{-1}$ $2 \mathrm{C}(\mathrm{s}) \longrightarrow 2 \mathrm{C}(\mathrm{g}) ; \Delta \mathrm{H}=1410 \mathrm{~kJ} \mathrm{~mol}^{-1}$ $\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}(\mathrm{g}) ; \Delta \mathrm{H}=330 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Options:
A 1165
B 837
C 865
D 815

The correct answer is:
815

Solution:
(i) $2 \mathrm{C}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}(\mathrm{g})$

$\Delta \mathrm{H}=225 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(ii) $2 \mathrm{C}(\mathrm{s}) \longrightarrow 2 \mathrm{C}(\mathrm{g}) \Delta \mathrm{H}=1410 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(iii) $\mathrm{H}_{2}$ (g) $\longrightarrow 2 \mathrm{H}$ (g) $\Delta \mathrm{H}=330 \mathrm{~kJ} \mathrm{~mol}^{-1}$

From equation (i) :

$\begin{array}{l}

225=\left[2 \times \Delta \mathrm{H}_{\mathrm{C}(\mathrm{s}) \longrightarrow \mathrm{C}(\mathrm{g})}+1 \times \mathrm{BE}_{\mathrm{H}-\mathrm{H}}\right] \\

-\left[2 \times \mathrm{BE}_{\mathrm{C}-\mathrm{H}}+1 \times \mathrm{BE}_{\mathrm{C} \equiv \mathrm{C}}\right] \\

225=[1410+1 \times 330]-\left[2 \times 350+1 \times \mathrm{BE}_{\mathrm{C} \equiv \mathrm{C}}\right] \\

225=[1410+330]-\left[700+\mathrm{BE}_{\mathrm{C} \equiv \mathrm{C}}\right] \\

225=1740-700-\mathrm{BE}_{\mathrm{C} \equiv \mathrm{C}} \\

\mathrm{BE}_{\mathrm{C} \equiv \mathrm{C}}=815 \mathrm{~kJ} \mathrm{~mol}^{-1}

\end{array}$

Topic: Chemistry, Thermodynamics (C), JEE Advanced, JEE Advanced 2012 (Paper 2)
Water is filled up to a height $h$ in a beaker of radius $R$ as shown in the figure. The density of water is $\rho$, the surface tension of water is $T$ and the atmospheric pressure is $p_0$. Consider a vertical section $A B C D$ of the water column through a diameter of the beaker. The force on water on one side of this section by water on the other side of this section has magnitude.

Options:
A
$\left|2 p_0 R h+\pi R^2 \rho g h-2 R T\right|$

B
$\left|2 p_0 R h+R \rho g h^2-2 R T\right|$

C
$\left|p_0 \pi R^2+R \rho g h^2-2 R T\right|$

D
$\left|p_0 \pi R^2+R \rho g h^2+2 R T\right|$

The correct answer is:
$\left|2 p_0 R h+R \rho g h^2-2 R T\right|$

Solution:
Force from right hand side liquid on left hand side liquid.
(i) Due to surface tension force $=2 R T$ (towards right)
(ii) Due to liquid pressure force
$$
\begin{aligned}
& =\int_{x=0}^{x=h}\left(p_0+\rho g h\right)(2 R \cdot x) d x \\
& =\left(2 p_0 R h+R \rho g h^2\right) \text { (towards left) }
\end{aligned}
$$
$\therefore$ Net force is $\left|2 p_0 R h+R \rho g h^2-2 R T\right|$ Correct option is (b).

Topic: Physics, Mechanical Properties of Fluids, JEE Advanced, JEE Advanced 2007 (Paper 2)
Water (with refractive index $=\frac{4}{3}$ ) in a tank is $18 \mathrm{~cm}$ deep. Oil of refractive index $\frac{7}{4}$ lies on water making a convex surface of radius of curvature $R=6 \mathrm{~cm}$ as shown. Consider oil to act as a thin lens. An object $S$ is placed $24 \mathrm{~cm}$ above water surface. The location of its image is at $x \mathrm{~cm}$ above the bottom of the tank. Then, $x$ is

Options:

The correct answer is:
2

Solution:
Two refractions will take place, first from spherical surface and the other from the plane surface.
So, applying
$$
\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}
$$
two times with proper sign convention. Ray of light is travelling downwards. Therefore, downward direction is taken as positive direction.
$$
\begin{gathered}
\frac{7 / 4}{v}-\frac{1.0}{-24}=\frac{7 / 4-1.0}{+6} \\
\frac{4 / 3}{(18-x)}-\frac{7 / 4}{v}=\frac{4 / 3-7 / 4}{\propto}
\end{gathered}
$$
Solving these equations, we get
$$
x=2 \mathrm{~cm}
$$
$\therefore$ Answer is 2 .
Analysis of Question
(i) Question is moderately difficult from calculation point of view, otherwise it is simple.
(ii) $\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}$ can be applied for
plane surface also with $R=\propto$

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2011 (Paper 2)
We have taken a saturated solution of $\mathrm{AgBr}, K_{\mathrm{sp}}$ is $12 \times 10^{-14}$. If $10^{-7}$ mole of $\mathrm{AgNO}_3$ are added to 1 litre of this solution, find conductivity (specific conductance) of this solution in terms of $10^{-7} \mathrm{Sm}^{-1}$ units.
$$
\text { Given, } \begin{aligned}
\lambda_{\left(\mathrm{Ag}^{+}\right)}^{\circ} & =6 \times 10^{-3} \mathrm{Sm}^2 \mathrm{~mol}^{-1}, \lambda_{\left(\mathrm{Br}^{-}\right)}^{\circ}=8 \times 10^{-3} \mathrm{Sm}^2 \mathrm{~mol}^{-1}, \\
\lambda_{\left(\mathrm{NO}_3^{-}\right)} & =7 \times 10^{-3} \mathrm{Sm}^2 \mathrm{~mol}^{-1},
\end{aligned}
$$

Options:

The correct answer is:
55

Solution:
Suppose the solubility of $\mathrm{AgBr}$ in $10^{-7} \mathrm{MAgNO}_3$ is $\mathrm{mol} / \mathrm{L}$
$$
\begin{aligned}
& \underset{s \mathrm{~mol} / \mathrm{L}}{\mathrm{AgBr}} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Br}^{-} \\
& \mathrm{AgNO}_3 \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{NO}_3^{-} \\
& 10^{-7} \mathrm{M} \quad 10^{-7} \mathrm{M} \times 10^{-7} \mathrm{M} \\
& \text { Total }\left[\mathrm{Ag}^{+}\right]=\left(s+10^{-7}\right) \mathrm{M} \\
& K_{\mathrm{sp}} \text { of } \mathrm{AgBr}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Br}^{-}\right] \\
& 12 \times 10^{-14}=\left(s+10^{-7}\right)(s)=s^2+10^{-7} s \\
& s^2+10^{-7} s-12 \times 10^{-14}=0 \\
&
\end{aligned}
$$
On solving $s=3 \times 10^{-7} \mathrm{M}$
Hence, $\left[\mathrm{Br}^{-}\right]=3 \times 10^{-7} \mathrm{M}=3 \times 10^{-7} \times 10^3 \mathrm{~m}^3=3 \times 10^{-4} \mathrm{~m}^3$
$$
\left[\mathrm{Ag}^{+}\right]=\left(3 \times 10^{-7}+10^{-7}\right) \mathrm{M}=4 \times 10^{-7} \times 10^3 \mathrm{~m}^3=4 \times 10^{-4} \mathrm{~m}^3
$$
$$
\left[\mathrm{NO}_3^{-}\right]=10^{-7} \mathrm{M}=10^{-7} \times 10^{+3}=1 \times 10^{-4} \mathrm{~m}^3
$$
$$
\because \quad \lambda=\frac{k}{C} \text { or } k=\lambda \times C
$$
$$
\begin{aligned}
\therefore \quad k_{\mathrm{Br}^{-}} & =3 \times 10^{-4} \times 8 \times 10^{-3} \mathrm{Sm}^{-1}=24 \times 10^{-7} \mathrm{Sm}^{-1} \\
k_{\mathrm{Ag}^{+}} & =4 \times 10^{-4} \times 6 \times 10^{-3} \mathrm{Sm}^{-1}=24 \times 10^{-7} \mathrm{Sm}^{-1} \\
k_{\mathrm{NO}_3^{-}} & =1 \times 10^{-4} \times 7 \times 10^{-3}=7 \times 10^{-7} \mathrm{Sm}^{-1} \\
k_{\text {Total }} & =k_{\mathrm{Br}^{-}}+k_{\mathrm{Ag}^{+}}+k_{\mathrm{NO}_3^{-}} \\
& =\left(24 \times 10^{-7}+24 \times 10^{-7}+7 \times 10^{-7} \mathrm{Sm}^{-1}\right. \text { ) } \\
& =55 \times 10^{-7} \mathrm{Sm}^{-1} \\
& \left.=\mathbf{5 5} \mathbf{S m}^{-1} \text { (in term of } 10^{-7} \mathrm{Sm}^{-1}\right)
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, JEE Advanced, JEE Advanced 2006
$$
\text { What are } N \text { and } M \text { ? }
$$

Options:
A
6,6

B
6,4

C
4,4

D
3, 3

The correct answers are:
6,4

Solution:
In these, (iii) and (iv) are enantiomers and (v) and (vi) are also enantiomers to each other. Thus total six isomers will be possible for $N$. In these only four isomers will be obtained on fractional distillation without resolution because $d$ and $l$ isomers (i.e., (iii) \& (iv) and (v) \& (vi)) cannot be separated by fractional distillation. Hence $N=6$ and $M=4$.

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2006
What are we referring to when we say 'simple organisms' or 'complex organisms'?What are we referring to when we say 'simple organisms' or 'complex organisms'?

Options:

None

Solution:
The organisms that have simple structural and functional organisation and are considered primitive are simple organisms. Whereas organisms that have higher and complex levels of structural and functional organisation are complex organisms. These are more advanced and said to have arisen from simple organisms.

Topic: Biology, Evolution
What is the advancement in ex situ conservation?

Options:
A Keeping threatened species in enclosures
B Gametes preservation through cryopreservation techniques
C Fertilization of eggs in vivo
D Propagation of plants using tissue culture methods

The correct answer is:
Gametes preservation through cryopreservation techniques

Solution:
In recent years, ex situ conservation has advanced beyond keeping threatened species in enclosures to include the preservation of gametes of threatened species through cryopreservation techniques.

Topic: Biology, Biodiversity and its Conservation, NEET
What is the limitation of treatments using metal salts like aluminium and magnesium hydroxide?

Options:
A They control symptoms and not the cause.
B They are easily administered to patients.
C They are ineffective in advanced stages.
D They prevent the development of ulcers.

The correct answer is:
They control symptoms and not the cause.

Solution:
Treatments with metal salts control only symptoms and not the cause.

Topic: Chemistry, Chemistry in Everyday Life, NEET
What is the term for cryopreservation of gametes of threatened species in viable and fertile conditions?

Options:
A Advanced Ex-situ conservation of biodiversity.
B In-situ conservation by sacred groves.
C In-situ cryo-conservation of biodiversity.
D In-situ conservation of biodiversity.

The correct answer is:
Advanced Ex-situ conservation of biodiversity.

Solution:
In recent years, Ex-situ conservation has advanced beyond keeping threatened species in enclosures. Now gametes of threatened species can be preserved in viable and fertile condition for long periods using cryo preservation techniques.

Topic: Biology, Biodiversity and its Conservation, NEET
When $20 \mathrm{~g}$ of naphthoic acid $\left(\mathrm{C}_{11} \mathrm{H}_8 \mathrm{O}_2\right)$ is dissolved in $50 \mathrm{~g}$ of benzene $\left(K_f=1.72 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$, a freezing point depression of $2 \mathrm{~K}$ is observed. The van't Hoff factor $(i)$ is

Options:
A
$0.5$

B
1

C
2

D
3

The correct answer is:
$0.5$

Solution:
Actual molecular weight of napthoic acid $\left(\mathrm{C}_{11} \mathrm{H}_8 \mathrm{O}_2\right)=172$
$$
\begin{aligned}
\text { Molecular mass (calculated ) } & =\frac{1000 . K_f \cdot w}{W . \Delta T_f} \\
& =\frac{1000 \times 1.72 \times 20}{50 \times 2} \\
& =344
\end{aligned}
$$

Topic: Chemistry, Solutions, JEE Advanced, JEE Advanced 2007 (Paper 1)
When benzene sulphonic acid and $p$-nitrophenol are treated with $\mathrm{NaHCO}_3$, the gases released respectively are

Options:
A
$\mathrm{SO}_2, \mathrm{NO}_2$

B
$\mathrm{SO}_2, \mathrm{NO}$

C
$\mathrm{SO}_2, \mathrm{CO}_2$

D
$\mathrm{CO}_2, \mathrm{CO}_2$

The correct answer is:
$\mathrm{CO}_2, \mathrm{CO}_2$

Solution:
$$
\text { Benzene sulphonic acid and p-nitrophenol react with } \mathrm{NaHCO}_3 \text { and evolve } \mathrm{CO}_2 \text { gas. }
$$
Because benezene sulphonic acid and $p$-nitro phenol are stronger acids with $\mathrm{H}_2 \mathrm{CO}_3$ (i.e., $\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2$ ) so, they are capable to evolve $\mathrm{CO}_2$ with $\mathrm{NaHCO}_3$.
$$
\mathrm{HCO}_3^{-}+\underset{\text { Acid }}{\mathrm{H}^{+}} \rightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow
$$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
When benzene sulphonic acid and $p$-nitrophenol are treated with $\mathrm{NaHCO}_3$, the gases released respectively are

Options:
A
$\mathrm{SO}_2, \mathrm{NO}_2$

B
$\mathrm{SO}_2, \mathrm{NO}$

C
$\mathrm{SO}_2, \mathrm{CO}_2$

D
$\mathrm{CO}_2, \mathrm{CO}_2$

The correct answer is:
$\mathrm{CO}_2, \mathrm{CO}_2$

Solution:
$$
\text { Benzene sulphonic acid and p-nitrophenol react with } \mathrm{NaHCO}_3 \text { and evolve } \mathrm{CO}_2 \text { gas. }
$$
Because benezene sulphonic acid and $p$-nitro phenol are stronger acids with $\mathrm{H}_2 \mathrm{CO}_3$ (i.e., $\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2$ ) so, they are capable to evolve $\mathrm{CO}_2$ with $\mathrm{NaHCO}_3$.
$$
\mathrm{HCO}_3^{-}+\underset{\text { Acid }}{\mathrm{H}^{+}} \rightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow
$$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2006
When the following aldohexose exists in its D-configuration, the total number of stereoisomers in its pyranose form is :

$\mathrm{CHO}-\mathrm{CH}_{2}-\mathrm{CHOH}-\mathrm{CHOH}-\mathrm{CHOH}-\mathrm{CH}_{2} \mathrm{OH}$

Options:

The correct answer is:
8

Solution:
Thus, total number of stereoisomers in pyranose form of D-configuration $=2^{3}=8$

Topic: Chemistry, Biomolecules, JEE Advanced, JEE Advanced 2012 (Paper 1)
When two identical batteries of internal resistance $1 \Omega$ each are connected in series across a resistor $R$, the rate of heat produced in $R$ is $J_1$. When the same batteries are connected in parallel across $R$, the rate is $J_2$. If $J_1=2.25 J_2$ then the value of $R$ in $\Omega$ is

Options:

The correct answer is:
4

Solution:
In series, $i=\frac{2 E}{2+R}$
$$
\therefore J_1=i^2 R=\left(\frac{2 E}{2+R}\right)^2 \cdot R
$$
In parallel, $i=\frac{E}{0.5+R}$
$$
\begin{aligned}
\therefore \quad J_2 & =i^2 R=\left(\frac{E}{0.5+R}\right)^2 \cdot R \\
\frac{J_1}{J_2} & =2.25=\frac{4(0.5+R)^2}{(2+R)^2}
\end{aligned}
$$
or $1.5=\frac{2(0.5+R)}{(2+R)}$
Solving we get, $R=4 \Omega$
$\therefore$ The answer is 4 .

Topic: Physics, Current Electricity, JEE Advanced, JEE Advanced 2010 (Paper 1)
When two progressive waves $y_1=4 \sin (2 x-6 t) \quad$ and $y_2=3 \sin \left(2 x-6 t-\frac{\pi}{2}\right) \quad$ are superimposed, the amplitude of the resultant wave is

Options:

The correct answer is:
5

Solution:
Phase difference between the two waves is $90^{\circ}$. Amplitudes are added by vector method.
$\therefore$ Answer is 5 .

Topic: Physics, Oscillations, JEE Advanced, JEE Advanced 2010 (Paper 1)
Which is correct statement if $\mathrm{N}_2$ is added at equilibrium condition?

Options:
A
The equilibrium will shift to forward direction because according to IInd law of thermodynamics the entropy must increases in the direction of spontaneous reaction.

B
The condition for equilibrium is $G_{\mathrm{N}_2}+3 G_{\mathrm{H}_2}=2 G_{\mathrm{NH}_3}$ where $G$ is Gibbs free energy per mole of the gaseous species measured at the partial pressure. The condition of equilibrium is unaffected by the use of catalyst, which increases the rate of both the forward and backward reactions to the same extent.

C
The catalyst will increase the rate of forward reaction by $\alpha$ and that of backward reaction by $\beta$.

D
Catalyst will not alter the rate of either of the reaction.

The correct answer is:
The condition for equilibrium is $G_{\mathrm{N}_2}+3 G_{\mathrm{H}_2}=2 G_{\mathrm{NH}_3}$ where $G$ is Gibbs free energy per mole of the gaseous species measured at the partial pressure. The condition of equilibrium is unaffected by the use of catalyst, which increases the rate of both the forward and backward reactions to the same extent.

Solution:
$K_p$ or $K_c$ remains constant for above reaction at constant temperature.
Although in presence of catalyst rate of forward and backward reaction is increased in same order (extent) due to decreasing the activation energy of both reaction. So the equilibrium is established in short time i.e., catalyst is able to increase the concentration of reactants and products in same order, so the value of $K_C$ or $K_p$ remains constant. Hence equilibrium state is not effected in presence of catalyst. Hence, at equilibrium $\Delta G=0$
$$
\begin{aligned}
& \because \quad \Delta G=2 G \text { of } \mathrm{NH}_3-\left(\mathrm{G} \text { of } \mathrm{H}_2+3 \times G \text { of } \mathrm{NH}_3\right) \\
& \therefore G_{\mathrm{N}_2}+3 G_{\mathrm{H}_2}=2 G_{\mathrm{NH}_3}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), JEE Advanced, JEE Advanced 2006
Which of the field patterns given in the figure is valid for electric field as well as for magnetic field?

Options:
A

B

C

D

The correct answer is:

Solution:
Correct answer is (c), because induced electric field lines (produced by change in magnetic field) and magnetic field lines form closed loops.
Analysis of Question
Question is simple, provided you have an idea of induced electric field lines.

Topic: Physics, Electromagnetic Induction, JEE Advanced, JEE Advanced 2011 (Paper 2)
Which of the following complexe is used in cencer treatment?

Options:
A Na2CoCl4
B cis- [PtCl2(NH3)2]
C mer- [Co(NH3)3Cl3]
D cis- K2PtCl2Br2

The correct answer is:
cis- [PtCl2(NH3)2]

Solution:
Cis-platin is used as an anticancer unit in treatment of advanced bladder cancer, metastatic ovarian cancer, andmetastatic testicular cancer. Testicular, ovarian, bladder, head and neck, esophageal, small and non-small cell lung, breast, cervical, stomach and prostate cancers.

Topic: Chemistry, Chemistry in Everyday Life, NEET
Which of the following halides react(s) with $\mathrm{AgNO}_{3}(\mathrm{aq})$ to give a precipitate that dissolves in $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(\mathrm{aq})$ ?

Options:
A $\mathrm{HCl}$
B $\mathrm{HF}$
C $\mathrm{HBr}$
D $\mathrm{HI}$

The correct answers are:
$\mathrm{HCl}$, $\mathrm{HBr}$, $\mathrm{HI}$

Solution:
$\begin{array}{l}

\mathrm{HX}+\mathrm{AgNO}_{3} \longrightarrow \mathrm{AgX} \downarrow+\mathrm{HNO}_{3}(\mathrm{X}=\mathrm{Cl}, \mathrm{Br}, \mathrm{I}) \\

\mathrm{AgX}+2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \longrightarrow \mathrm{Na}_{3}\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right]+\mathrm{NaX}

\end{array}$

Topic: Chemistry, Practical Chemistry, JEE Advanced, JEE Advanced 2012 (Paper 1)
Which of the following molecules, in pure form, is (are) unstable at room temperature ?

Options:
A
B
C
D

The correct answers are:
,

Solution:
(b) and (c) being antiaromatic, are unstable at room temperature.

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2012 (Paper 1)
Which of the following pairs is not correctly matched?

Options:
A Vitamin B3-Pellagra
B Vitamin B12-Pernicious anemia
C Vitamin C-Scurvy
D Vitamin B6-Beriberi

The correct answer is:
Vitamin B6-Beriberi

Solution:
Vitamin B1 is also called thiamine. The deficiency causes beriberi. Symptoms of this disease of the nervous system include weight loss, emotional disturbances, Wernicke's encephalopathy (impaired sensory perception), weakness and pain in the limbs, periods of irregular heartbeat, and edema (swelling of bodily tissues).
Vitamin B3 is also called niacin. The deficiency, along with a deficiency of tryptophan causes pellagra. Symptoms include aggression, dermatitis, insomnia, weakness, mental confusion, and diarrhea. In advanced cases, pellagra may lead to dementia and death.
Vitamin C also called as ascorbic acid. The deficiency leads to a condition called scurvy and it is characterised by general weakness, anaemia, gingivitis (gum disease), and skin hemorrhages.
Vitamin B12 is also known as cobalamin. The deficiency results in macrocytic anemia, elevated homocysteine, peripheral neuropathy, memory loss, and other cognitive deficits. It is most likely to occur among elderly people, as absorption through the gut declines with age; the autoimmune disease pernicious anemia is another common cause. It can also cause symptoms of mania and psychosis. In rare extreme cases, paralysis can result.

Topic: Biology, Digestion and Absorption, NEET
Which of the following reactants on reaction with conc. $\mathrm{NaOH}$ followed by acidification gives following lactone as the

Options:
A

B

C

D

The correct answers are:

Solution:
This compound on acidification to give a compound which have $-\mathrm{CH}_2 \mathrm{OH}$ and $-\mathrm{COOH}$ group, such groups on dehydration to give lactone.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2006
$$
\text { Which of the following statement is correct in Infinitely long wire kept perpendicular }
$$
to the paper carrying current inwards the given figure?

Options:
A
net force on the loop is zero

B
net torque on the loop is zero

C
loop will rotate clockwise about axis $\mathrm{OO}^{\prime}$ when seen from $O$

D
loop will rotate anti-clockwise about $\mathrm{OO}^{\prime}$ when seen from $O$

The correct answers are:
net force on the loop is zero
,
loop will rotate clockwise about axis $\mathrm{OO}^{\prime}$ when seen from $O$

Solution:
$\mathbf{F}_{B A}=0$, because magnetic lines are parallel to this wire.
$\mathbf{F}_{C D}=0$, because magnetic lines are antiparallel to this wire.
$\mathbf{F}_{C B}$ is perpendicular to paper inwards and $\mathbf{F}_{A D}$ is perpendicular to paper outwards. These two forces (although calculated by integration) cancel each other but produce a torque which tend to rotate the loop in clockwise direction about an axis $O O^{\prime}$.

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2006
Which of the following statements is not true?

Options:
A The biodiversity decreases with increasing latitude.
B The biodiversity decreases with increasing altitude.
C The fishes show greatest biodiversity
among vertebrates
D The biodiversity of bryophytes is greater
than that of angiosperms.

The correct answer is:
The biodiversity of bryophytes is greater
than that of angiosperms.

Solution:
The biodiversity of bryophytes is less than that of angiosperms. Bryophytes are plants that can live in soil but are dependent on water for sexual reproduction.
Angiosperms are much advanced plant species which consist of male sex organ (stamen) and female sex organ (pistil) borne in a flower.

Topic: Biology, Biodiversity and its Conservation, AIIMS, AIIMS 2014