datavorous/entrance-exam-dataset huggingface dataset:
At what temperature the centigrade (Celsius) and Fahrenheit, readings are the same

Options:
A $-40^{\circ}$
B $+40^{\circ}$
C $36.6^{\circ}$
D $-37^{\circ}$

The correct answer is:
$-40^{\circ}$

Solution:
The correct answer is option(1). The formulas for converting between degree Celsius and degree Fahrenheit are:
$\begin{aligned}{ }^{\circ} \mathrm{F} & =\left({ }^{\circ} \mathrm{C} \times \frac{9}{5}\right)+32 \\ { }^{\circ} \mathrm{C} & =\left({ }^{\circ} \mathrm{F}-32\right) \times \frac{5}{9}\end{aligned}$
To find the temperature when both are equal, we use an old algebra trick and just set ${ }^{\circ} \mathrm{F}={ }^{\circ} \mathrm{C}$ and solve one of the equations.
$\begin{aligned} & { }^{\circ} \mathrm{C}=\left({ }^{\circ} \mathrm{C} * \times \frac{9}{5}\right)+32 \\ & { }^{\circ} \mathrm{C}-\left({ }^{\circ} \mathrm{C} \times \frac{9}{5}\right)=32 \\ & -\frac{4}{5} \times{ }^{\circ} \mathrm{C}=32 \\ & { }^{\circ} \mathrm{C}=-32 \times \frac{5}{4} \\ & { }^{\circ} \mathrm{C}=-40 \\ & { }^{\circ} \mathrm{F}=\left({ }^{\circ} \mathrm{F} \times \frac{9}{5}\right)+32 \\ & { }^{\circ} \mathrm{F}-\left({ }^{\circ} \mathrm{F} \times \frac{9}{5}\right)=32 \\ & -\frac{4}{5} \times{ }^{\circ} \mathrm{F}=32 \\ & { }^{\circ} \mathrm{F}=-32 \times 54 \\ & { }^{\circ} \mathrm{F}=-40\end{aligned}$
So the temperature when both the Celsius and Fahrenheit scales are the same is - 40 degrees.

Topic: Physics, Thermal Properties of Matter, JEE Main
At what temperature the $\mathrm{RMS}$ velocity of $\mathrm{SO}_2$ be same as that of ${ }^{\mathrm{O}_2}$ at $303 \mathrm{~K}$

Options:
A $273 K$
B $606 \mathrm{~K}$
C $303 K$
D $403 \mathrm{~K}$

The correct answer is:
$606 \mathrm{~K}$

Solution:
$\frac{U_{\mathrm{SO}_2}}{U_{O_2}}=\sqrt{\frac{M_{O_2} T_{\mathrm{SO}_2}}{M_{\mathrm{SO}_2} T_{O_2}}}=\sqrt{\frac{32 \times T_{\mathrm{SO}_2}}{64 \times} \frac{303}{}}=1$
$1=\frac{32 \times}{64} \frac{T_{\mathrm{SO}_2}}{303} \Rightarrow T_{S_{O_2}}=606 \mathrm{k}$

Topic: Chemistry, States of Matter, JEE Main
At what temperature would N2 molecules have same average speed as CO molecules at 200 K .

Options:
A -73oC
B 200oC
C 700oC
D none

The correct answer is:
-73oC

Solution:
Average  speed =  8RT πM
Both N2 and CO have same molecular mass, So, for same average speed, temperature should be same
â´âTâ=â200âK=â 73 o C

Topic: Chemistry, States of Matter, JEE Main
At what value of $X$, will $\left|\begin{array}{ccc}x+\omega^2 & \omega & 1 \\ \omega & \omega^2 & 1+x \\ 1 & x+\omega & \omega^2\end{array}\right|=0$

Options:
A $x=0$
B $x=1$
C $x=-1$
D None of these

The correct answer is:
$x=0$

Solution:
$\left|\begin{array}{ccc}x+\omega^2 & \omega & 1 \\ \omega & \omega^2 & 1+x \\ 1 & x+\omega & \omega^2\end{array}\right|=0$
Check at $x=0$, we get $\left|\begin{array}{ccc}\omega^2 & \omega & 1 \\ \omega & \omega^2 & 1 \\ 1 & \omega & \omega^2\end{array}\right|$
$\begin{aligned} & =\omega^2\left(\omega^4-\omega\right)-\omega\left(\omega^3-1\right)+1\left(\omega^2-\omega^2\right) \\ & =\omega^2(\omega-\omega)-\omega(1-1)+0=0 \text { Or }\end{aligned}$
$\Delta=\left|\begin{array}{ccc}1+\omega+\omega^2+x & \omega & 1 \\ 1+\omega+\omega^2+x & \omega^2 & 1+x \\ 1+\omega+\omega^2+x & x+\omega & \omega^2\end{array}\right|$ by $C_1 \rightarrow C_1+C_2+C_3$
$\begin{aligned} & =\left|\begin{array}{ccc}x & \omega & 1 \\ x & \omega^2 & 1+x \\ x & x+\omega & \omega^2\end{array}\right|,\left(\quad 1+\omega+\omega^2=0\right) \\ & =0, \text { if } x=0 .\end{aligned}$

Topic: Mathematics, Determinants, JEE Main
At which carbon chlorination would be yielding highest?

Options:

The correct answer is:
1

Solution:
The given reaction is a halogenation reaction through the free radical mechanism.

The major product of this reaction depends upon the stability of free radicals. So halogenation takes place at the side where the stability of free radical is higher after remove hydrogen.

At the position first, free radical is benzylic which is the most stable hence halogenation occurs at this side.

Topic: Chemistry, Hydrocarbons, JEE Main
At which of the $\mathrm{f}$ (a) For one gm mole; average kinetic energy $=\frac{3}{2} R T$. the molecules of a gas have twice the average kinetic energy they have at $20^{\circ} \mathrm{C}$

Options:
A $40^{\circ} \mathrm{C}$
B $80^{\circ} \mathrm{C}$
C $313^{\circ} \mathrm{C}$
D $586^{\circ} \mathrm{C}$

The correct answer is:
$313^{\circ} \mathrm{C}$

Solution:
Average kinetic energy $\propto$ Temperature
$\begin{aligned}& \frac{E_1}{E_2}=\frac{T_1}{T_2} \Rightarrow \frac{E}{2 E}=\frac{T_1}{T_2} \Rightarrow T_2=2 T_1 \\\Rightarrow & T_2=2(273+20)=586 \mathrm{~K}=313^{\circ} \mathrm{C}\end{aligned}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
At which point the line $\frac{x}{a}+\frac{y}{b}=1$, touches the curve $y=b e^{-x / a}$

Options:
A $(0,0)$
B $(0, a)$
C $(0, b)$
D $(b, 0)$

The correct answer is:
$(0, b)$

Solution:
Let the point be $\left(x_1, y_1\right), \therefore y_1=b e^{-x_1 / a}$
Also, curve $y=b e^{-x / a} \Rightarrow \frac{d y}{d x}=\frac{-b}{a} e^{-x / a}$
$\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=\frac{-b}{a} e^{-x_1 / a}=\frac{-y_1}{a}$..........(by (i))
Now, the equation of tangent of given curve at point $\left(x_1, y_1\right)$ is $y-y_1=\frac{-y_1}{a}\left(x-x_1\right) \Rightarrow \frac{x}{a}+\frac{y}{y_1}=\frac{x_1}{a}+1$
Comparing with $\frac{x}{a}+\frac{y}{b}=1$, we get
$y_1=b \text { and } 1+\frac{x_1}{a}=1 \Rightarrow x_1=0$
Hence, the point is $(0, b)$.

Topic: Mathematics, Application of Derivatives, JEE Main
At which temperature the r.m.s. velocity of a hydrogen molecule equal to that of an oxygen molecule at 47°C ?

Options:
A 80 K
B -73 K
C 4 K
D 20 K

The correct answer is:
20 K

Solution:
The r.m.s velocity of the molecule of a gas of molecular weight M at Kelvin temperature T is given by, Vrms=3RTM.

Let MO and MH are molecular weight of oxygen and hydrogen and TO and TH the corresponding Kelvin temperature at which Vr.m.s is same for both,

Vr.m.s=3RTOMO=3RTHMH. Hence TOMO=THMH

TO=273°C+47°C=320 K,

MO=32,

MH=2⇒TH=232×320=20 K

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2024 (30 Jan Shift 1)
At which temperature the speed of sound in hydrogen will be same as that of speed of sound in oxygen at $100^{\circ} \mathrm{C}$

Options:
A $-148^{\circ} \mathrm{C}$
B $-212.5^{\circ} \mathrm{C}$
C $-317.5^{\circ} \mathrm{C}$
D $-249.7^{\circ} \mathrm{C}$

The correct answer is:
$-249.7^{\circ} \mathrm{C}$

Solution:
Speed of sound in gases is
$v=\sqrt{\frac{\gamma R T}{M}} \Rightarrow T \propto M$
(Because $v, y$-constant). Hence
$\frac{T_{H_2}}{T_{O_2}}=\frac{M_{H_2}}{M_{O_2}}$
$\Rightarrow \frac{T_{H_3}}{(273+100)}=\frac{2}{32} \Rightarrow T_{H_2}=23.2 \mathrm{~K}=-249.7^{\circ} \mathrm{C}$

Topic: Physics, Waves and Sound, JEE Main
At $x=1$, the function $f(x)=\left\{\begin{array}{cc}x^{3}-1, & 1 < x < \infty \\ x-1, & -\infty < x \leq 1\end{array}\right.$ is

Options:
A continuous and differentiable.
B continuous and non-differentiable.
C discontinuous and differentiable.
D discontinuous and non-differentiable.

The correct answer is:
continuous and non-differentiable.

Solution:
$f(x)=\left\{\begin{array}{cc}x^{3}-1, & 1 < x < \infty \\ x-1, & -\infty < x \leq 1\end{array}\right.$
We have to check the continuity at $x=1$. $\mathrm{RHL} \Rightarrow \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}\left(x^{3}-1\right)=1-1=0$ $\mathrm{LHL} \Rightarrow \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(x-1)=1-1=0$ $f(\mathrm{l})=1-1=0$
Thus the function is continuous at $x=1$.
$f^{\prime}(x)=\left\{\begin{array}{cc}
3 x^{2}, & 1 < x < \infty \\
1, & -\infty < x \leq 1
\end{array}\right.$
Now, check the differentiability at $x=1$.
LHD at $x=1 \Rightarrow 1$
RHD at $x=1 \Rightarrow 3(1)^{2}=3$
$A s, L H D \neq R H D$, function is not differentiable.

Topic: Mathematics, Continuity and Differentiability, JEE Main
At $x=\frac{\pi^2}{4}, \frac{d}{d x}\left(\tan ^{-1}(\cos \sqrt{x})+\sec ^{-1}\left(e^x\right)\right)=$

Options:
A $\frac{1}{\sqrt{e^{\frac{\pi^2}{2}}-1}}-\frac{1}{\pi}$
B $\frac{\pi}{4}+\frac{1}{\sqrt{\mathrm{e}^{\pi^2}+\mathrm{e}^{\pi^2 / 2}}}$
C $\frac{1}{\sqrt{\mathrm{e}^{\pi^2}+\mathrm{e}^{\pi^2 / 2}}}+\frac{2}{\pi} \cot \left(\frac{\sqrt{\pi}}{2}\right)$
D $\frac{1}{\sqrt{\mathrm{e}^\pi}}+\frac{1}{\pi}$

The correct answer is:
$\frac{1}{\sqrt{e^{\frac{\pi^2}{2}}-1}}-\frac{1}{\pi}$

Solution:
$$
\begin{aligned}
& \text { } \frac{d}{d x}\left(\tan ^{-1}(\cos \sqrt{x})+\sec ^{-1}\left(\mathrm{e}^x\right)\right) \\
& =\frac{(-\sin \sqrt{x})}{1+\cos ^2 \sqrt{x}} \cdot\left(\frac{1}{2 \sqrt{x}}\right)+\frac{1}{e^x \sqrt{e^{2 x}-1}} \cdot e^x
\end{aligned}
$$
but when $x=\frac{\pi^2}{4}$
$$
\begin{aligned}
& =\frac{-\sin \frac{\pi}{2}}{1+\cos ^2 \frac{\pi}{2}}\left(\frac{1}{2}\right)\left(\frac{2}{\pi}\right)+\frac{1}{\sqrt{e^{\pi 2 / 2}-1}} \\
& =-\frac{1}{\pi}+\frac{1}{\sqrt{e^{\pi^2 / 2}-1}}
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
Atomic number of Hydrogen like species that has a wave length difference of 59.3 nm between first line of Balmer and first line of Lyman series is (R H =109678  cm −1 )

Options:

The correct answer is:
3

Solution:
Wavelength of 1st line in Balmer series,
1 λ B  =  Z 2 R H 1 2 2 − 1 3 2  =  5 36   R H Z 2
or λ B = 36 5 R H Z 2
Wavelength of 1st line in Lyman series is, 1 λ L = Z 2 R H 1 1 2 − 1 2 2
or λ L = 4 3× R H Z 2
Difference λ B − λ L =59.3× 10 −7 = 36 5 R H Z 2 − 4 3 R H Z 2
= 1 R H Z 2 36 5 − 4 3
Z 2 = 88 59.3× 10 −7 ×109678×15 =9.0
or Z=3
Hydrogen-like species is Li 2+

Topic: Chemistry, Structure of Atom, JEE Main
Atomic radii of alkali metals follow the order:

Options:
A \( \mathrm{Li}>\mathrm{Na}>\mathrm{K}>\mathrm{Cs} \)
B \( \mathrm{K}>\mathrm{Cs}>\mathrm{Li}>\mathrm{Na} \)
C \( \mathrm{Na}>\mathrm{K}>\mathrm{Cs}>\mathrm{Li} \)
D \( \mathrm{Cs}>\mathrm{K}>\mathrm{Na}>\mathrm{Li} \)

The correct answer is:
\( \mathrm{Cs}>\mathrm{K}>\mathrm{Na}>\mathrm{Li} \)

Solution:
Atomic radii of alkali metals increases from Li to Cs due to presence of an extra shell of electrons.

Hence, order of atomic radii:

Cs>K>Na>Li.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Atomic spectra is an example of

Options:
A line spectra
B continuous spectra
C band spectra
D both line spectra and continuous spectra

The correct answer is:
line spectra

Solution:
Line emission spectrum:
- Line spectra are sharp lines of definite wavelengths.
- It is the characteristic of the emitting substance.
- It is used to identify the gas.
- Atoms Free excited atoms emit line spectrum. eg. sodium vapour lamp, mercury in mercury vapour lamp.
EXPLANATION:
- From the above, it is clear that Atomic spectra is an example of line spectra. Therefore option 1 is correct.

Topic: Physics, Atomic Physics, JEE Main
Atomic weight of boron is 10.81 and it has two isotopses ${}_{5} B^{10}$ and ${}_{5} B^{11}$ then the ratio of ${}_{5} \mathrm{B}^{10}: {}_{5} \mathrm{B}^{11}$ in nature would be

Options:
A $15: 16$
B $10: 11$
C $19 : 81$
D $81 : 19$

The correct answer is:
$19 : 81$

Solution:
Given =
Let the percentage of $B^{10}$ atoms be $x$. average atomic weight be
$\begin{aligned} & =\frac{10 x+11(100-x)}{100}=10.81 \\ & x=19 \therefore \frac{N_{B 10}}{N_{B 10}}=\frac{19}{81}\end{aligned}$

Topic: Physics, Nuclear Physics, JEE Main
Atoms of element X form hcp lattice and those of element Y occupy 23 of its tetrahedral voids. The percentage of element X in the lattice is (Nearest integer)

Options:

The correct answer is:
43

Solution:
Number of effective atoms in HCP lattice is =6

Number of atoms of X=6

Number of T.V. in h.c.p. =12 2×Number of atoms

Number of Y atoms =23×12=8 atoms.

% of X=614×100=42.86%

Round off to the nearest integer = 43

Topic: Chemistry, Solid State, JEE Main, JEE Main 2022 (24 Jun Shift 1)
ATPase enzyme needed for muscle contraction is located in:

Options:
A actinin
B troponin
C myosin
D actin.

The correct answer is:
myosin

Solution:
In all myosin, the head domain is a specialized ATPase that is able to couple the hydrolysis of ATP with motion.
Related Theory
The contractile protein of skeletal muscle involving ATPase activity is myosin. Myosin comprise a family of ATP-dependent motor proteins and are best known for their role in muscle contraction and their involvement in a wide range of other eukaryotic motility processes. They are responsible for actinbased motility.

Topic: Biology, Respiration in Plants, NEET, NEET 2004
ATPase enzyme needed for muscle contraction is located in:

Options:
A globular head of actin
B globular head of myosin
C active site of troponin
D active site of tropomyosin

The correct answer is:
globular head of myosin

Solution:
Actin and Myosin are contractile proteins of muscle while Troponin and Tropomyosin are regulatory proteins. Myosin is a polymer of Meromysoin units and this is the main protein of Dark bands. Myosin is a polymer made up of Meromyosin monomers. The molecule consists of Head and Tail that are attached by cross bridges. Head consists of Actin binding site and ATPase enzyme. This enzyme is needed for muscle contraction. This Light bands consist of Actin, Troponin and Tropomyosin. Actin protein consists of two double stranded polypetides while troponin is attached on these strands at intervals. Actin shows presence of Myosin binding site and when myosin attaches to this site the muscle shows contraction due to formation of cross bridges. However this cross bridge formation is not seen when muscle relaxes as this site is masked by Troponin molecule. At intervals on actin, trimeric molecule Tropomyosin is also seen. This includes TpC, TpT and TpI.

Topic: Biology, Locomotion and Movement, NEET
Attacking or reactive or electrophilic species in nitration of benzene is or In the nitration of benzene with concentrated $\mathrm{HNO}_3$ and $\mathrm{H}_2 \mathrm{SO}_4$ the attack on ring is made by

Options:
A $\mathrm{NO}_2^{-}$
B $\mathrm{NO}_2^{+}$
C $\mathrm{NO}_3^{-}$
D $\mathrm{NO}_2$

The correct answer is:
$\mathrm{NO}_2^{+}$

Solution:
$\mathrm{HO}-\mathrm{NO}_2+\mathrm{H}^{+} \mathrm{HSO}_4^{-} \rightarrow \begin{gathered}\mathrm{H}_2 \mathrm{O}+\mathrm{NO}_2^{+}+\mathrm{HSO}_4^{-} \\ \text {Nitroniumion } \\ \text { (attackingspecies) }\end{gathered}$

Topic: Chemistry, Hydrocarbons, JEE Main
Attacking or reactive or electrophilic species in nitration of benzene is or In the nitration of benzene with concentrated $\mathrm{HNO}_3$ and $\mathrm{H}_2 \mathrm{SO}_4$ the attack on ring is made by

Options:
A $\mathrm{NO}_2^{-}$
B $\mathrm{NO}_2^{+}$
C $\mathrm{NO}_3^{-}$
D $\mathrm{NO}_2$

The correct answer is:
$\mathrm{NO}_2^{+}$

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Aufbau principle is obeyed in which of the following electronic configurations

Options:
A $1 s^2 2 s^2 2 p^6$
B $1 s^2 3 p^3 3 s^2$
C $1 s^2 3 s^2 3 p^6$
D $1 s^2 2 s^2 3 s^2$

The correct answer is:
$1 s^2 2 s^2 2 p^6$

Solution:
According to Aufbau principle electron are filling increasing order of energy. Therefore the electronic configuration $1 s^2 2 s^2 2 p^6$ obeys Aufbau principle.

Topic: Chemistry, Structure of Atom, JEE Main
Autoreduction process is used in the extraction of

Options:
A \( \mathrm{Cu} \) and \( \mathrm{Hg} \)
B \( \mathrm{Hg} \) and \( \mathrm{Zn} \)
C \( \mathrm{Cu} \) and \( \mathrm{Al} \)
D Fe and \( \mathrm{Pb} \)

The correct answer is:
\( \mathrm{Cu} \) and \( \mathrm{Hg} \)

Solution:
Autoreduction is used for sulphide ore of less electropositive metals like Hg , Pb and Cu

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
A"V" shaped rigid body has two identical uniform arms. What must be the angle between the two arms so that when the body is hung from one end, the other arm is horizontal ?

Options:
A $\cos ^{-1}(1 / 3)$
B $\cos ^{-1}(1 / 2)$
C $\cos ^{-1}(1 / 4)$
D $\cos ^{-1}(1 / 6)$

The correct answer is:
$\cos ^{-1}(1 / 3)$

Solution:
For one arm to remain horizontal the net torque about $\mathrm{O}$ must be zero (in the position shown in the figure) for this $\mathrm{OP}=\mathrm{OQ}$
$\Rightarrow \mathrm{OQ}=\frac{\ell}{2} \cos \theta$
from figure
$\mathrm{AE}=\mathrm{AC}+\mathrm{CE}$ $\Rightarrow \mathrm{AE}=\ell \cos \theta+\mathrm{OQ}$ $\Rightarrow \frac{\ell}{2}=\ell \cos \theta+\frac{\ell}{2} \cos \theta$ $\Rightarrow \cos \theta=\frac{1}{3}$ hence $\theta=\cos ^{-1}(1 / 3)$
correct Answer is (A)

Topic: Physics, Rotational Motion, KVPY, KVPY 2016 (SA)
Average $\mathrm{C}-\mathrm{H}$ bond energy is $416 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Which of the following is correct?

Options:
A $\mathrm{CH}_4(g)+416 \mathrm{~kJ} \longrightarrow \mathrm{C}(g)+4 \mathrm{H}(g)$
B $\mathrm{CH}_4(g) \longrightarrow \mathrm{C}(g)+4 \mathrm{H}(g)+416 \mathrm{~kJ}$
C $\mathrm{CH}_4(g)+1664 \mathrm{~kJ} \longrightarrow \mathrm{C}(g)+4 \mathrm{H}(g)$
D $\mathrm{CH}_4(g) \longrightarrow \mathrm{C}(g)+4 \mathrm{H}(g)+1664 \mathrm{~kJ}$

The correct answer is:
$\mathrm{CH}_4(g)+1664 \mathrm{~kJ} \longrightarrow \mathrm{C}(g)+4 \mathrm{H}(g)$

Solution:
$\mathrm{CH}_4$ has four $\mathrm{C}-\mathrm{H}$ bonds. So, $4 \times 416=1664 \mathrm{~kJ}$ is required to break $\mathrm{CH}_4$ into $\mathrm{C}$ and $4 \mathrm{H}$.

Topic: Chemistry, Thermodynamics (C), JEE Main
Average density of the earth

Options:
A
does not depend on g

B
is a complex function of g

C
is directly proportional to g

D
is inversely proportional to g

The correct answer is:
is directly proportional to g

Solution:
$\mathrm{g}=\frac{\mathrm{G} 4 \pi}{3} \rho_{\mathrm{av}} \mathrm{R}$

Topic: Physics, Gravitation, JEE Main, JEE Main 2005
Average distance of the Earth from the Sun is $L_{1}$ : If one year of the Earth $=D$ days, one year of another planet whose average distance from the Sun is $L_{2}$ will be

Options:
A $D\left(\frac{L_{2}}{L_{1}}\right)^{\frac{1}{2}}$ days
B $D\left(\frac{L_{2}}{L_{1}}\right)^{\frac{3}{2}}$ days
C $D\left(\frac{L_{2}}{L_{1}}\right)^{\frac{2}{3}}$ days
D $D\left(\frac{L_{2}}{L_{1}}\right)$ days

The correct answer is:
$D\left(\frac{L_{2}}{L_{1}}\right)^{\frac{3}{2}}$ days

Solution:
By Keplar's law,
$$
\begin{array}{l}
T^{2} \propto R^{3} \\
T=D\left(\frac{L_{2}}{L_{1}}\right)^{3 / 2}
\end{array}
$$

Topic: Physics, Gravitation, JEE Main
Average energy stored in a pure inductance $L$ when a current $i$ flows through it, is

Options:
A $L i^2$
B $2 L i^2$
C $\frac{L i^2}{4}$
D $\frac{L i^2}{2}$

The correct answer is:
$\frac{L i^2}{2}$

Solution:
As we know $e=-\frac{d \phi}{d t}=-L \frac{d i}{d t}$
Work done against back e.m.f. e in time dt and current i is
$d W=-e i d t=$ $L \frac{d i}{d t} i d t=L i d i \Rightarrow W=L \int_0^i i d i=\frac{1}{2} L i^2$

Topic: Physics, Electromagnetic Induction, JEE Main
AX is a covalent diatomic molecule where A and X are second row elements of periodic table. Based on Molecular orbital theory, the bond order of AX is 2.5. The total number of electrons in AX is_____. (Round off to the Nearest Integer).

Options:

The correct answer is:
15

Solution:
Bond order of the covalent molecule, AX is given as 2.5, As AX is a covalent diatomic molecule of second row elements of periodic table we can consider CO, NO but bond order of CO having 14 electrons which is isoelectronic to nitrogen should be equal to 3 and NO having 15 electrons will have a bond order of 2.5.

Molecular orbital configuration and bond order calculation of NO is

ABMO= antibonding molecular orbital

Bond order =Number of bonding electrons - number of anti bonding electrons.

Note: Total number of electrons equal to 13 will also have the 2.5 bond order. But in this case neutral diatomic molecule will not be possible

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Axillary bud and terminal bud are derived from the activity of:

Options:
A lateral meristem
B intercalary meristem
C apical meristem
D parenchyma

The correct answer is:
apical meristem

Solution:
The apical meristem is responsible for axillary and terminal bud formation. The apical meristem is the growth in a plant which seems at the root point of young shoots and leaves.
Related Theory
The apical meristem is also recognized as a growing tip. The main aspect this is to initiate the operation of new cell formation in new seedlings at the hints of roots and shoots which ultimately contributes to forming the buds. It help in raising the height and length of the plant.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2002
Axis of a parabola lies along x-axis. If its vertex and focus are at distances 2 and 4 respectively from the origin, on the positive x-axis then which of following points does not lie on it?

Options:
A 6,42
B 5,26
C 8,6
D 4,-4

The correct answer is:
8,6

Solution:
Clearly vertex is 2,0 and focus is S4,0⇒a=2 .

Hence, the equation of parabola is y2=4×2x-2

⇒y2=8x-2

⇒Point 8,6 does not lie on the parabola.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Axis of the parabola $x^2-4 x-3 y+10=0$ is

Options:
A $y+2=0$
B $x+2=0$
C $y-2=0$
D $x-2=0$

The correct answer is:
$x-2=0$

Solution:
The parabola is $(x-2)^2=(3 y-6)$. Hence axis is $x-2=0$.

Topic: Mathematics, Parabola, JEE Main
Azimuthal quantum number defines

Options:
A e/m ratio of electron
B Spin of electron
C Angular momentum of electron
D Magnetic momentum of electron

The correct answer is:
Angular momentum of electron

Solution:
Generally azimuthal quantum number defines angular momentum.

Topic: Chemistry, Structure of Atom, JEE Main
$[\mathbf{b} \times \mathbf{c}, \mathbf{c} \times \mathbf{a}, \mathbf{a} \times \mathbf{b}]$ is equal to

Options:
A $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})$
B $2[\mathbf{a} \mathbf{b} \mathbf{c}]$
C $[\mathbf{a} \mathbf{b} \mathbf{c}]^2$
D $[\mathbf{a b} \mathbf{c}]$

The correct answer is:
$[\mathbf{a} \mathbf{b} \mathbf{c}]^2$

Solution:
$\begin{aligned}
& {[\mathbf{b} \times \mathbf{c} \mathbf{c} \times \mathbf{a} \mathbf{a} \times \mathbf{b}]=(\mathbf{b} \times \mathbf{c}) .[(\mathbf{c} \times \mathbf{a}) \times(\mathbf{a} \times \mathbf{b})]} \\
& \text { Let } \mathbf{a} \times \mathbf{b}=\mathbf{d} \\
& \text { so, }(\mathbf{b} \times \mathbf{c})[(\mathbf{c} \times \mathbf{a}) \times \mathbf{d}]=(\mathbf{b} \times \mathbf{c})[(\mathbf{d} . \mathbf{a}) \mathbf{c}-(\mathbf{d} . \mathbf{c}) \cdot \mathbf{a}] \\
& =(\mathbf{b} \times \mathbf{c})[\mathbf{a} \cdot(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}-(\mathbf{a} \times \mathbf{b}) \mathbf{c} \cdot \mathbf{a}] \\
& =(\mathbf{b} \times \mathbf{c})[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{a}=\mathbf{a} \cdot[\mathbf{b} \times \mathbf{c}] \cdot[\mathbf{a} \mathbf{b} \mathbf{c}] \\
& =[\mathbf{a} \mathbf{b} \mathbf{c}][\mathbf{a} \mathbf{b} \mathbf{c}]=[\mathbf{a} \mathbf{b} \mathbf{c}]^2
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
B has a smaller first ionization enthalpy than Be. Consider the following statement:
(I) it is easier to remove 2p electron than2s electron
(II) 2p electron of B is more shielded from the nucleus by the inner core of electrons than the 2s electrons of Be
(III) 2s electron has more penetration power than 2p electron
(IV) atomic radius of B is more than Be
(atomic number B:5, Be=4 )
The correct statements are.

Options:
A (I), (II) and (IV)
B (II), (III) and (IV)
C (I), (II) and (III)
D (I), (III) and (IV)

The correct answer is:
(I), (II) and (III)

Solution:
Since, Penetration of s orbital is the maximum because of the closeness to the nucleus than the p, d and f orbitals
it is easier to remove 2p electron than 2s electron
Z eff ∝ 1 shielding effect
Z eff B > Z eff Be So, shielding effect is more on 2p electron of B than 2s electron of Be
So, BeI.E1>BI.E1

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2020 (09 Jan Shift 1)