datavorous/entrance-exam-dataset huggingface dataset:
What is the approximate standard free energy change per mole of $\mathrm{Zn}$ (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) for a Daniel cell at $298 \mathrm{~K}$ ?

Options:
A -212.3
B 230.0
C 0.0
D -1.10

The correct answer is:
-212.3

Solution:
For Daniell cell we have
$$
\mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s)
$$
Gibbs free energy $=-n F E$
$n=$ number of electrons $=2$
$F=$ Faraday's constant $=96500 \mathrm{C}$
$E=\operatorname{emf}$ of cell $=1.1 \mathrm{~V}$
$\therefore$ Gibbs free energy $=-2 \times 96500 \times 1.1 \mathrm{~J}=-212.3 \mathrm{~kJ}$

Topic: Chemistry, Electrochemistry, JEE Main
What is the area between the curve
$y=\cos 3 x, 0 \leq x \leq \frac{\pi}{6}$ and the co-ordinate axes?

Options:
A 1 squale unit
B $\frac{1}{2}$ squaue unit
C $\frac{1}{3}$ square unit
D $\frac{1}{4}$ square unit

The correct answer is:
$\frac{1}{3}$ square unit

Solution:
Required Area $=\int_{0}^{\pi / 6} \cos 3 x d x$
$=\left.\frac{\sin 3 x}{3}\right|_{0} ^{\pi / 6}=\frac{\sin 3(\pi / 6)}{3}-\sin 0$
$=\frac{1}{3} \sin \pi / 2-0=\frac{1}{3}(1)=\frac{1}{3}$ sq. unit.

Topic: Mathematics, Area Under Curves, JEE Main
What is the area bounded by \(y=\tan x, y=0\) and \(x=\frac{\pi}{4} ?\)

Options:
A \(\ell n ~2~ \mathrm{sq}\). units
B \(\frac{\ell n 2}{2} ~\mathrm{sq}\). units
C \(2~(\ell n ~2)~ \mathrm{sq}\). units
D None of these

The correct answer is:
\(\frac{\ell n 2}{2} ~\mathrm{sq}\). units

Solution:
Required area \(=\int_0^{\frac{\pi}{4}} \tan x d x\)
\(=\ln |\sec 1|_0^{\frac{\pi}{4}}=\ln \sqrt{2}=\frac{\ln 2}{2}\)

Topic: Mathematics, Area Under Curves, JEE Main
What is the area of a loop of the curve $\mathrm{r}=\operatorname{asin} 3 \theta$ ?

Options:
A $\frac{\pi \mathrm{a}^{2}}{6}$
B $\frac{\pi \mathrm{a}^{2}}{8}$
C $\frac{\pi \mathrm{a}^{2}}{12}$
D $\frac{\pi \mathrm{a}^{2}}{24}$

The correct answer is:
$\frac{\pi \mathrm{a}^{2}}{24}$

Solution:
If curve $\mathrm{r}=\mathrm{a} \sin 3 \theta$
To trace the curve, we consider the following table :
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline 3 \theta= & 0 & \frac{\pi}{2} & \pi & \frac{3 \pi}{2} & 2 \pi & \frac{5 \pi}{2} & 3 \pi \\
\hline\theta= & 0 & \frac{\pi}{6} & \frac{\pi}{3} & \frac{\pi}{2} & \frac{2 \pi}{3} & \frac{5 \pi}{6} & \pi \\
\hline \mathrm{r}= & 0 & \mathrm{a} & 0 & -\mathrm{a} & 0 & \mathrm{a} & 0 \\
\hline
\end{array}
Thus there is a loop between $\theta=0 \& \theta=\frac{\pi}{3}$ as r varies from $\mathrm{r}=0$ to $\mathrm{r}=0$.

Hence, the area of the loop lying in the
$$
\text { positive quadrant }=\frac{1}{2} \int_{0}^{\frac{\pi}{3}} r^{2} \mathrm{~d} \theta
$$
$$
=\frac{1}{2} \int_{0}^{\frac{\pi}{3}} \sin ^{2} \phi \cdot \frac{1}{3} \mathrm{~d} \phi
$$
$\left[\right.$ On putting, $\left.3 \theta=\phi \Rightarrow \mathrm{d} \theta=\frac{1}{3} \mathrm{~d} \phi\right]$
$$
\begin{array}{c}
=\frac{a^{2}}{6} \int_{0}^{\frac{\pi}{2}} \sin ^{2} \phi \mathrm{d} \phi \\
=\frac{\mathrm{a}^{2}}{6} \cdot \int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2 \phi}{2} \mathrm{~d} \phi\left[\because \cos 2 \theta=1-2 \sin ^{2} \theta\right]
\end{array}
$$
$=\frac{\mathrm{a}^{2}}{12} \cdot\left[\phi+\frac{\sin 2 \phi}{2}\right]_{0}^{\frac{\pi}{2}}$
$=\frac{\mathrm{a}^{2}}{12} \cdot\left[\frac{\pi}{2}+\sin \pi\right]=\frac{\mathrm{a}^{2} \pi}{24}$

Topic: Mathematics, Area Under Curves, JEE Main
What is the area of one of the loops between the curve $y=$ $\mathrm{c} \sin \mathrm{x}$ and $\mathrm{x}$ -axis ?

Options:
A $\mathrm{c}$
B $2 \mathrm{c}$
C $3 \mathrm{c}$
D $4 \mathrm{c}$

The correct answer is:
$2 \mathrm{c}$

Solution:
Area of one of loop between $\mathrm{y}=\mathrm{c} \sin \mathrm{x}$ and $\mathrm{x}$ -axis
$=\mathrm{c}+\mathrm{c}=2 \mathrm{c}$

Topic: Mathematics, Area Under Curves, JEE Main
What is the area of the parabola $y^{2}=x$ bounded by its latus rectum?

Options:
A $\frac{1}{12}$ square unit
B $\frac{1}{6}$ square unit
C $\frac{1}{3}$ square unit
D None of the above

The correct answer is:
$\frac{1}{6}$ square unit

Solution:
Required Area $=2 \int_{0}^{\frac{1}{4}} \sqrt{\mathrm{x}} \mathrm{dx}$
$=2 \cdot \frac{2}{3}\left[\mathrm{x}^{3 / 2}\right]_{0}^{1 / 4}=\frac{4}{3}\left[\frac{1}{8}-0\right]=\frac{1}{6} \mathrm{sq} .$ unit.

Topic: Mathematics, Area Under Curves, JEE Main
What is the area of the region bounded by x-axis, the curve
$\mathrm{y}=\mathrm{f}(\mathrm{x})$ and the two ordinates $\mathrm{x}=\frac{1}{2}$ and $\mathrm{x}=1 ?$
Consider the function $\mathrm{f}(\mathrm{x})=|\mathrm{x}-1|+\mathrm{x}^{2}$, where $\mathrm{x} \in \mathbf{R}$.

Options:
A $\frac{5}{12}$ square unit
B $\frac{5}{6}$ square unit
C $\frac{7}{-}$ square units
D 2 square units

The correct answer is:
$\frac{5}{12}$ square unit

Solution:
$\quad f(x)=\left\{\begin{array}{ll}x^{2}+x-1 & x \geq 1 \quad \forall x \in R \\ x^{2}-x+1 & x < 1\end{array}\right.$
Hence area required for given region is
$\begin{aligned} A_{1} &=\int_{1 / 2}^{1} f(x) d x \\ &=\int_{1 / 2}^{1}\left(x^{2}-x+1\right) d x \\ &=\left[\frac{1}{3} x^{3}-\frac{x^{2}}{2}+x\right]_{1 / 2}^{1} \\ &=\left(\frac{1}{3} \times 1-\frac{1}{2}+1\right)-\left(\frac{1}{3} \times \frac{1}{8}-\frac{1}{2} \times \frac{1}{4}+\frac{1}{2}\right) \\ A_{1}=\frac{5}{12} & \text { square unit. } \end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
What is the area of the triangle formed by the lines joining the vertex of the parabola $x^{2}=12 y$ to the ends of the latus rectum?

Options:
A 9 square units
B 12 square units
C 14 square units
D 18 square units

The correct answer is:
18 square units

Solution:
Given parabola is $\mathrm{x}^{2}=12 y$ which is of the form $\mathrm{x}^{2}=4 a y$.
$\Rightarrow 4 a=12 \Rightarrow a=3$
Now, $L M$ is the latus rectum whose length $=4 a=4 \times 3$
=12
So, area of $\Delta L M V=\frac{1}{2} \times L M \times V F$.
$=\left(\frac{1}{2} \times 12 \times 3\right)$ sq. unit
$=18$ square unit

Topic: Mathematics, Parabola, JEE Main
What is the area of the triangle formed by the lines $y-x=0, y+x=0, x=c ?$

Options:
A $c / 2$
B $c^{2}$
C $2 c^{2}$
D $c^{2} / 2$

The correct answer is:
$c^{2}$

Solution:
The shaded portion shows the required area. Since the area is symmetric about the $x$ -axis therefore Required area $=2$ area $(\Delta A O D)$
$$
\begin{array}{l}
=2 \times \frac{1}{2} \times \text { base } \times \text { height } \\
=2 \times \frac{1}{2} \times O D \times A D \\
=c \times c \\
=c^{2}
\end{array}
$$

Topic: Mathematics, Straight Lines, JEE Main
What is the area under the curve $\mathrm{y}=|\mathrm{x}|+|\mathrm{x}-1|$ between $\mathrm{x}=0$ and $\mathrm{x}=1 ?$

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

The correct answer is:
1

Solution:
$|\mathrm{x}|$ for $\mathrm{x} \geq 0$
$=\mathrm{x}$ and $|\mathrm{x}-1|$ for $\mathrm{x} \leq 1$
$=-(\mathrm{x}-1)$
so, $\int_{0}^{1}(|\mathrm{x}|+|\mathrm{x}-| 1 \mid)=$ required area
$\mathrm{a}=\int_{0}^{1} \mathrm{x} \mathrm{dx}-\int_{0}^{1}(\mathrm{x}-1) \mathrm{d} \mathrm{x}$
$=\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{1}-\left[\frac{\mathrm{x}^{2}}{2}-\mathrm{x}\right]_{0}^{1}=\frac{1}{2}-\left(\frac{1}{2}-1\right)=1 \mathrm{sq}$ units

Topic: Mathematics, Area Under Curves, JEE Main
What is the basis of classification of algae?

Options:
A on the basis of chlorophyll
B on the basis of size of plant
C on the basis of nature of plant
D none of the above

The correct answer is:
on the basis of chlorophyll

Solution:
The algae are classified on the basis of chlorophyll present in algae. They are classified into three main classes: Chlorophyceae, Phaeophyceae and Rhodophyceae.

Topic: Biology, Plant Kingdom, NEET
What is the behavior of Z for all gases at high pressures?

Options:
A Z becomes zero.
B Z remains constant at 1.
C Z increases with increasing pressure.
D Z decreases with increasing pressure.

The correct answer is:
Z increases with increasing pressure.

Solution:
At high pressures, all gases have Z values greater than 1, indicating that they are more difficult to compress.

Topic: Chemistry, States of Matter, NEET
What is the change in mass of a body, when taken 64 km below the surface of the earth? [Take radius of the earth as 6400 km]

Options:
A Increases by 2%
B Remain constant
C Increases by 1%
D Decrease by 1%

The correct answer is:
Remain constant

Solution:
We know that mass is a universal constant at every place in the above problem when we change the position of the body in upward directon, then the value of acceleration due to gravity will be changed so, the weight of the body should change but, the mass will not be changed.

Topic: Physics, Work Power Energy, TS EAMCET, TS EAMCET 2021 (04 Aug Shift 1)
What is the change in the number of carbon atoms when an amide is subjected to the Hoffmann bromamide degradation reaction?

Options:
A The number of carbon atoms remains the same.
B One carbon atom is gained.
C One carbon atom is lost.
D Two carbon atoms are lost.

The correct answer is:
One carbon atom is lost.

Solution:
In the Hoffmann bromamide degradation reaction, the amine formed contains one carbon less than that present in the amide, so one carbon atom is lost in the process.

Topic: Chemistry, Amines, NEET
What is the characteristic behavior of the pV vs. p plot for dihydrogen and helium as pressure increases?

Options:
A pV decreases
B pV remains constant
C pV first decreases, then increases
D pV increases

The correct answer is:
pV increases

Solution:
For dihydrogen and helium, as the pressure increases, the value of pV also increases.

Topic: Chemistry, States of Matter, NEET
What is the chemical nature of lecithin?

Options:
A Lypolipid
B Phospholipid
C Glycolipid
D Derivative of protein

The correct answer is:
Phospholipid

Solution:
Lecithin is a fat that is essential in the cells of the body. It can be found in many foods, including soybeans and egg yolks. Lecithin is taken as a medicine and is also used in the manufacturing of medicines. Lecithin is used for treating memory disorders such as dementia and Alzheimer's disease. In contrast to fats, which function as fuel molecules, lecithin serves a structural role in cell membranes. It is found in all cells. Without lecithin and other membrane phospholipids, cells would be unable to maintain their structure and probably would dissolve back into their surroundings.

Topic: Chemistry, Biomolecules, NEET
What is the chief constituent of pyrex glass?

Options:
A B2O3
B SiO2
C Al2O3
D N a 2 O

The correct answer is:
SiO2

Solution:
Pyrex glass is obtained by fusing together 60 to 80% SiO2, 10 to 25% B2O3 and remaining amount of Al2O3

Topic: Chemistry, p Block Elements (Group 13 & 14), MHT CET, MHT CET 2017
What is the chief constituent of pyrex glass?

Options:
A B2O3
B SiO2
C Al2O3
D N a 2 O

The correct answer is:
SiO2

Solution:
Pyrex glass is obtained by fusing together 60 to 80% SiO2, 10 to 25% B2O3 and remaining amount of Al2O3

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
What is the chief product obtained when $n$-butane is treated with bromine in the presence of light at $130^{\circ} \mathrm{C}$

Options:
A

B

C

D $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Br}$

The correct answer is:

Solution:
$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3+\mathrm{Br}_2 \xrightarrow[130^{\circ} \mathrm{C}]{\text { Light }}$
2-Bromobutane is the main product because $2^{\circ}$ carbonium ion is more stable than $1^{\circ}$.

Topic: Chemistry, Hydrocarbons, JEE Main
What is the coefficient of $\mathrm{x}^{3}$ in $\frac{(3-2 \mathrm{x})}{(1+3 \mathrm{x})^{3}}$ ?

Options:
A $-272$
B $-540$
C $-870$
D $-918$

The correct answer is:
$-918$

Solution:
$\frac{(3-2 \mathrm{x})}{(1+3 \mathrm{x})^{3}}=(3-2 \mathrm{x})(1+3 \mathrm{x})^{-3}$
$=(3-2 \mathrm{x})\left(1-9 \mathrm{x}+\frac{(-3)(-4)}{2 !} \cdot 9 \mathrm{x}^{2}\right.$
$\left.\quad+\frac{(-3)(-4)(-5)}{3 !} \cdot 27 \mathrm{x}^{3}+\ldots .\right)$
$\left[\operatorname{Expanding}(1+3 \mathrm{x})^{-3}\right]$
$=(3-2 \mathrm{x})\left(1-9 \mathrm{x}+54 \mathrm{x}^{2}-270 \mathrm{x}^{3}+\ldots \ldots \ldots\right)$
$\therefore$ Coefficient of $\mathrm{x}^{3}=-270 \times 3-2 \times 54$
$=-810-108=-918$

Topic: Mathematics, Binomial Theorem, JEE Main
What is the coefficient of $x^{17}$ in the expansion of $\left(3 x-\frac{x^{3}}{6}\right)^{9} ?$

Options:
A $\frac{189}{8}$
B $\frac{567}{2}$
C $\frac{21}{16}$
D None of these

The correct answer is:
$\frac{189}{8}$

Solution:
Given expansion is
$\left(3 x-\frac{x^{3}}{6}\right)^{9}$ where $a=3 x, \mathrm{~b}=\frac{-x^{3}}{6}, \mathrm{n}=9$
Now, General Term $=\mathrm{T}_{r+1}={ }^{n} C_{r}(\mathrm{a})^{\mathrm{n}-\mathrm{r}} \cdot b^{2}$
$={ }^{9} \mathrm{C}_{r}(3 x)^{9-\mathrm{r}}\left(\frac{-x^{3}}{6}\right)^{r}={ }^{9} \mathrm{C}_{r} \cdot 3^{9-r} \mathrm{x}^{9-r} \cdot \frac{(-1)^{r} x^{3 r}}{6^{r}}$
$={ }^{9} C_{r} 3^{9-r}(-1)^{r} \frac{x^{9+2 r}}{6^{r}}$
We can get coeff of $x^{17}$ when $9+2 r=17$
$\Rightarrow \quad 2 r=17-9$
$\Rightarrow \quad r=\frac{8}{2}=4$
Hence, required coefficient
$={ }^{9} \mathrm{C}_{4} \frac{3^{5}}{6^{4}}=\frac{126 \times 3}{16}=\frac{189}{8}$

Topic: Mathematics, Binomial Theorem, JEE Main
What is the commercial name for calcium oxide?

Options:
A Quick lime
B Milk of lime
C Slaked lime
D Limestone

The correct answer is:
Quick lime

Solution:
Quick lime is commercial name of CaO.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2016 (10 Apr Online)
What is the composition of methylated spirit

Options:
A Methanol + ethanol
B Methanol
C Methanoic acid
D Methanamide

The correct answer is:
Methanol + ethanol

Solution:
5-10% methyl and remaining ethanol is called methylated spirit. It is also known as denatured alcohol because it is unfit for drinking.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
What is the composition of the mixture like when a chemical reaction reaches a state of equilibrium?

Options:
A It is unpredictable.
B It remains constant.
C It varies with temperature.
D It depends on the color of the reactants.

The correct answer is:
It remains constant.

Solution:
When a chemical reaction reaches a state of equilibrium, the composition of the mixture remains constant.

Topic: Chemistry, Chemical Equilibrium, NEET
What is the composition of the sooty material formed during the process of making fullerenes?

Options:
A Mainly C70 with traces of C60
B Mainly C60 with smaller quantity of C70 and traces of fullerenes consisting of even number of carbon atoms up to 350 or above
C Mainly C60 with smaller quantity of C70
D Mainly C60 with traces of carbon black

The correct answer is:
Mainly C60 with smaller quantity of C70 and traces of fullerenes consisting of even number of carbon atoms up to 350 or above

Solution:
The sooty material formed consists mainly of C60 with smaller quantities of C70 and traces of fullerenes consisting of even numbers of carbon atoms up to 350 or above.

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
What is the concentration of $\mathrm{H}_3 \mathrm{O}^{+}$ions in a solution having $\mathrm{POH}$ value 9 ?

Options:
A $1.0 \times 10^{-5} \mathrm{M}$
B $2.0 \times 10^{-5} \mathrm{M}$
C $1.5 \times 10^{-5} \mathrm{M}$
D $2.5 \times 10^{-5} \mathrm{M}$

The correct answer is:
$1.0 \times 10^{-5} \mathrm{M}$

Solution:
$\begin{aligned} & \mathrm{pH}=14-\mathrm{poH} \\ & =14-9=5 \\ & \therefore\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}}\end{aligned}$
or $10^{-5}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
What is the concentration of nitrate ions if equal volumes of $0.1 \mathrm{M} \mathrm{AgNO}_3$ and $0.1 \mathrm{M} \mathrm{NaCl}$ are mixed together

Options:
A $0.1 \mathrm{~N}$
B $0.2 \mathrm{M}$
C $0.05 \mathrm{M}$
D $0.25 \mathrm{M}$

The correct answer is:
$0.05 \mathrm{M}$

Solution:
$0.1 \mathrm{M} \mathrm{AgNO}_3$ will react with $0.1 \mathrm{MNaCl}$ to form $0.1 \mathrm{M} \mathrm{NaNO}_3$. But as the volume is doubled,conc. of $\mathrm{NO}_3^{-}=\frac{0.1}{2}=0.05 \mathrm{M}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
What is the concentration of nitrate ions if equal volumes of $0.1 \mathrm{MAgNO}_3$ and $0.1 \mathrm{M} \mathrm{NaCl}$ are mixed together

Options:
A $0.1 \mathrm{M}$
B $0.2 \mathrm{M}$
C $0.05 \mathrm{M}$
D $0.25 \mathrm{M}$

The correct answer is:
$0.05 \mathrm{M}$

Solution:
$0.1 \mathrm{M} \mathrm{AgNO}_3$ will react with $0.1 \mathrm{M} \mathrm{NaCl}$ to form $0.1 \mathrm{M} \mathrm{NaNO}_3$. But as the volume is doubled, conc. of
$\mathrm{NO}_3^{-}=\frac{0.1}{2}=0.05 \mathrm{M}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
What is the condition under which a substance always remains in one phase?

Options:
A At the boiling point.
B At the melting point.
C At the critical temperature.
D At standard temperature and pressure.

The correct answer is:
At the critical temperature.

Solution:
At the critical temperature, a substance always remains in one phase.

Topic: Chemistry, States of Matter, NEET
What is the configuration of the carbon atom under attack in the transition state of an SN2 reaction?

Options:
A Remains unchanged
B Becomes linear
C Inverts
D Forms a double bond

The correct answer is:
Inverts

Solution:
In the transition state of an SN2 reaction, the configuration of the carbon atom under attack inverts.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
What is the conjugate base of $\mathrm{OH}^{-} ?$

Options:
A
$\mathrm{O}_2$

B
$\mathrm{H}_2 \mathrm{O}$

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

D
$\mathrm{O}^{-2}$

The correct answer is:
$\mathrm{O}^{-2}$

Solution:
$\mathrm{OH}^{-} \longrightarrow \mathrm{O}^{-2}+\mathrm{H}^{+}$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2005
What is the consequence of passing an aqueous solution of raw sugar over beds of animal charcoal?

Options:
A It becomes darker
B It turns colorless
C It solidifies
D It remains unchanged

The correct answer is:
It turns colorless

Solution:
The paragraph states that the aqueous solution of raw sugar becomes colorless as the colored substances are adsorbed by the charcoal.

Topic: Chemistry, Surface Chemistry, NEET
What is the coordination number of Cs+ in CsCl?

Options:

The correct answer is:
8

Solution:
Coordination number is defined as the no. of nearest equidistant neighbours.
The coordination number of CsCl is 8..8

Topic: Chemistry, Solid State, JEE Main
What is the coordination number of sodium in $\mathrm{Na}_2 \mathrm{O}$

Options:
A $6$
B $4$
C $8$
D $2$

The correct answer is:
$4$

Solution:
In $\mathrm{Na}_2 \mathrm{O}$, each oxide ions $\left(\mathrm{O}^{2-}\right)$ is cep-ordinated to $8 \mathrm{Na}^{+}$ ions and each $\mathrm{Na}^{+}$ ion to 4 oxide ions. Hence it has $4: 8$ co-ordination.

Topic: Chemistry, Solid State, JEE Main
What is the correct fate of ammonium ions in plants?

Options:
A It undergoes transamination with alpha ketoglutaric acid to form glumate acid which further undergoes reductive amination to form other amino acids
B It undergoes reductive amination with alpha ketoglutaric acid to glumatine which further undergoes transamination to form other amino acids
C It undergoes transamination with alpha ketoglutaric acid to glumatine which further undergoes reductive amination to form other amino acids
D It undergoes reductive amination with alpha ketoglutaric acid to glumate which further undergoes transamination to form other amino acids

The correct answer is:
It undergoes reductive amination with alpha ketoglutaric acid to glumate which further undergoes transamination to form other amino acids

Solution:
At physiological pH, the ammonia is protonated to form (ammonium) ion.
(i) Reductive amination : In these processes, ammonium ion reacts with α-ketoglutaric acid and forms glutamic acid. (ii) Transamination : It involves the transfer of amino group from one amino acid to the keto group of a keto acid. Glutamic acid is the main amino acid from which the transfer of NH2, the amino group takes place and other amino acids are formed through transamination. The enzyme transaminase catalyses all such reactions.

Topic: Biology, Mineral Nutrition, NEET
What is the correct order of spin only magnetic moment (in BM) of $\mathrm{Mn}^{2+}, \mathrm{Cr}^{2+}$ and $\mathrm{V}^{2+}$ ?

Options:
A $\mathrm{Mn}^{2+}>\mathrm{V}^{2+}>\mathrm{Cr}^{2+}$
B $\mathrm{V}^{2+}>\mathrm{Cr}^{2+}>\mathrm{Mn}^{2+}$
C $\mathrm{Mn}^{2+}>\mathrm{Cr}^{2+}>\mathrm{V}^{2+}$
D $\mathrm{Cr}^{2+}>\mathrm{V}^{2+}>\mathrm{Mn}^{2+}$

The correct answer is:
$\mathrm{Mn}^{2+}>\mathrm{Cr}^{2+}>\mathrm{V}^{2+}$

Solution:
Spin only magnetic moments depend upon the number of unpaired electrons, more the number of unpaired electron, greater will be the spin only magnetic moment.
${ }_{25} \mathrm{Mn}=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^5, 4 s^2$
$\mathrm{Mn}^{2+}=1 s^2, 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5, 4 s^0$
Number of unpaired electrons $=5$
${ }_{24} \mathrm{Cr}=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^5, 4 s^1$
$\mathrm{Cr}^{2+}=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^4, 4 s^0$
Number of unpaired electrons $=4$
${ }_{23} \mathrm{~V}=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^3, 4 s^2$
$\mathrm{V}^{2+}=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^3, 4 s^0$
Number of unpaired electrons $=3$
So, the correct order of spin only magnetic moment is
$\mathrm{Mn}^{2+}>\mathrm{Cr}^{2+}>\mathrm{V}^{2+}$

Topic: Chemistry, d and f Block Elements, JEE Main
What is the correct order of the following elements with respect to their density?

Options:
A Zn<Cu<Co<Fe<Cr
B Cr<Zn<Co<Cu<Fe
C Zn<Cr<Fe<Co<Cu
D Cr<Fe<Co<Cu<Zn

The correct answer is:
Zn<Cr<Fe<Co<Cu

Solution:
Among the transition series, the trend in density will be reverse of atomic radii, i.e. density increase remains almost the same and then decreases along the period.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2021 (24 Feb Shift 2)
What is the correct sequence of reagents used for converting nitrobenzene into m-dibromobenzene?

Options:
A →Sn/HCl/→KBr/→Br2/→H+
B →Br2/Fe/→Sn/HCl/→NaNO2/HCl/→CuBr/HBr
C →NaNO2/→HCl/→KBr/→H+
D →Sn/HCl/→Br2/→NaNO2/→NaBr

The correct answer is:
→Br2/Fe/→Sn/HCl/→NaNO2/HCl/→CuBr/HBr

Solution:
The process of conversion of primary aromatic amines into its diazonium salt is called diazotization.

Diazonium salts are important synthetic intermediates that can undergo coupling reactions to form azo dyes and electrophilic substitution reactions to introduce functional groups.

The correct sequence of reagents used for converting nitrobenzene into m- dibromobenzene is found in option (b).

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (24 Feb Shift 2)
What is the correct sequence of the formation of female gametophyte in angiosperms?

Options:
A Nucellus, megapore tetrad, megaspore mother cell, megaspore, female gametophyte
B Megaspore tetrad, nucellus, megaspore mother cell, megaspore, female gametophyte
C Nucellus, megaspore mother cell, megaspore tetrad, megaspore, female gametophyte
D Megaspore mother cell, megaspore tetrad, megaspore, nucellus, female gametophyte

The correct answer is:
Nucellus, megaspore mother cell, megaspore tetrad, megaspore, female gametophyte

Solution:
In angiosperms, the body of the ovule consists of a mass of parenchymatous cells called nucellus. In the hypodermal region of nucellus towards the micropylar end develops a primary archesporial cell, which by further divisions forms a diploid megaspore mother cell. The latter undergoes meiosis to form megaspore tetrad. Usually the chalazal megaspore remains functional while the other three degenerate. The functional megaspore is the first cell of female gametophyte (embryo sac).

Topic: Biology, Plant Kingdom, AIIMS, AIIMS 2015
What is the correct sequence of the increasing order of freezing points at one atmosphere of the following 1.0 M aqueous solution?
1. Urea, 2. Sodium chloride, 3. Sodium sulphate, 4. Sodium phosphate.
Select the correct answer using the codes given below

Options:
A 4, 3, 1, 2
B 3, 4, 2, 1
C 3, 4, 1, 2
D 4, 3, 2, 1

The correct answer is:
4, 3, 2, 1

Solution:
vant Hoff factors for urea, NaCl, Na2SO4 and Na3PO4 are 1, 2, 3 and 4 respectively.
As ΔTf∝im , so ΔTf for urea, NaCl, Na2SO4 and Na3PO4 are proportional to 1, 2, 3 and 4 respectively. Hence, freezing point order is Na3PO4<Na2SO4<NaCl< Urea

Topic: Chemistry, Solutions, JEE Main