datavorous/entrance-exam-dataset huggingface dataset:
In the light of recent classification of living organisms into three domains of life (bacteria, archaea, and eukarya), which one of the following statements is true about archaea?

Options:
A Archaea resemble eukarya in all respects
B Archaea have some novel features that are absent in other prokaryotes and eukaryotes
C Archaea completely differ from both prokaryotes and eukaryotess
D Archaea completely differ from prokaryotes

The correct answer is:
Archaea have some novel features that are absent in other prokaryotes and eukaryotes

Solution:
A domain of prokaryotic organisms containing the archaebacteria includes the methanogens which produce methane, the thermoacidophilic bacteria which live in an extremely hot and acidic environment and the halophilic bacteria which can only function at high salt concentrations and are abundant in the oceans.

Topic: Biology, Biological Classification, NEET
In the line spectra of hydrogen atom, difference between the largest and the shortest wavelengths of the Lyman series is 305 A∘. The corresponding difference for the Paschan series in A∘ is:________

Options:

The correct answer is:
10553.14

Solution:
1λmin=R1=R ; n=∞ to 1

1λmax=R1−14=3R4 ; n=2 to 1

⇒ λmax−λmin=43R−1R

304=13R ......a

Paschen : Yλ'min=R19

and Yλ'max=R19−116=7R16×9

λ'max−λ'min=16×97R−9R=817R ...b

ba=x3×304=817

⇒ x=10553.14

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2020 (04 Sep Shift 1)
In the long form of the periodic table, the valence shell electronic configuration of 5s25p4 corresponds to the element present in:

Options:
A Group 16 and period 5
B Group 17 and period 5
C Group 16 and period 6
D Group 17 and period 6

The correct answer is:
Group 16 and period 5

Solution:
As given, electronic configuration of valence shell (outermost shell) is 5s2 5p4..that means principle quantum number of valence shell is 5..As we know,principle quantum number of valence shell shows the 'period' of that element in periodic table..Hence period of the element is 5As there are total 6 electrons in valence shell or it similar to general valence shell configuration of group 16 i.e. ns2 np4,hence its group is 16..element correspond to period 5 and group 16

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2015 (10 Apr Online)
In the main group elements (i) as we proceed down the same group in the periodic table and (ii) as we proceed from left to right in the same period, the atomic radius

Options:
A (i) Increase continuously; (ii) Decreases continuously
B (i) Decreases continuously; (ii) Increases continuously
C (i) Increases continuously; (ii) Decreases upto the group IV and then increases upto the end of the period
D (i) Decreases continuously; (ii) Decreases upto the group IV and then increases upto the end of the period

The correct answer is:
(i) Increase continuously; (ii) Decreases continuously

Solution:
(i) Increase continuously; (ii) Decreases continuously

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
In the majority of angiosperms:

Options:
A Egg has a filiform apparatus
B There are numerous antipodal cells
C Reduction division occurs in the megaspore
D A small central cell is present in the embryo sac

The correct answer is:
Reduction division occurs in the megaspore

Solution:
MEGASPOROGENESIS:
Definition: The process of formation of megaspores from megaspore mother cell (MMC) is called megasporogenesis.
i) Embryosac generally develop from a single megaspore that differentiate from a single megaspore mother cell (MMC) in the micropylar region of the nucellus.
ii) MMC is a large cell containing dense cytoplasm and a prominent nucleus.
iii) The MMC undergoes meiosis and forms a linear tetrad of four haploid megaspores. Out of which one remains functional (chalazal end) and three degenerate (micropylar end).

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2016 (Phase 2)
In the mean life of a radioactive sample

Options:
A about 13 of substance disintegrates
B about 23 of the substance disintegrates
C about 90% of the substance disintegrates
D almost all the substance disintegrates

The correct answer is:
about 23 of the substance disintegrates

Solution:
Substance remained after time t,
N=N0e-λt

Mean life of radioactive sample is, tmean=1λ
∴ Substance remained after mean life
N=N0e-1
N=0.37N0
∴ Substance disintegrated
N0-N=0.63N0
i.e. 63%, nearly 23.

Topic: Physics, Nuclear Physics, NEET
In the mean value theorem, $f(b)-f(a)=(b-a) f^{\prime}(c)$ if $a=4, b=9$ and $f(x)=\sqrt{x}$ then the value of $c$ is

Options:
A $8.00$
B $5.25$
C $4.00$
D $6.25$

The correct answer is:
$6.25$

Solution:
$f(x)=\sqrt{x}$
$\begin{aligned}& \therefore f(a)=\sqrt{4}=2, \quad f(b)=\sqrt{9}=3 ; \quad f^{\prime}(x)=\frac{1}{2 \sqrt{x}} \\
& \text { Also, } f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}=\frac{3-2}{9-4}=\frac{1}{5}\end{aligned}$
Also, $\therefore \frac{1}{2 \sqrt{c}}=\frac{1}{5} \Rightarrow c=\frac{25}{4}=6.25$

Topic: Mathematics, Application of Derivatives, JEE Main
In the measurement of a physical quantity $X=\frac{A^2 B}{C^{1 / 3} D^3}$. The percentage errors introduced in the measurements of the quantities $A, B, C$, and $D$ are $2 \%, 2 \%, 4 \%$ and $5 \%$, respectively. Then, the minimum amount of percentage error in the measurement of $X$ is contributed by

Options:
A $A$
B $B$
C $C$
D $D$

The correct answer is:
$C$

Solution:
$\begin{aligned}
& x=\frac{A^2 B}{C^{1 / 3} D^3} \\
& \% \text { error in } x, \\
& \frac{\Delta x}{x}=2 \frac{\Delta A}{A}+\frac{\Delta B}{B}+\frac{1}{3} \frac{\Delta C}{C}+\frac{3 \Delta D}{D} \\
& 2\left(\frac{\Delta A}{A} \times 100 \%\right)+\left(\frac{\Delta B}{B} \times 100 \%\right) \\
& +\frac{1}{3}\left(\frac{\Delta C}{C} \times 100 \%\right)+3\left(\frac{\Delta D}{D} \times 100 \%\right)
\end{aligned}$
Minimum percentage error
$=\frac{1}{3}\left(\frac{\Delta C}{C} \times 100 \%\right)=\frac{1}{3} \times 4=1.1 \%$

Topic: Physics, Mathematics in Physics, JEE Main
In the measurement of a physical quantity $X=\frac{A^2 B}{C^{1 / 3} D^3}$. The percentage errors introduced in the measurements of the quantities $A, B, C$, and $D$ are $2 \%, 2 \%, 4 \%$ and $5 \%$, respectively. Then, the minimum amount of percentage error in the measurement of $X$ is contributed by

Options:
A $A$
B $B$
C $C$
D $D$

The correct answer is:
$C$

Solution:
$\begin{aligned}
& x=\frac{A^2 B}{C^{1 / 3} D^3} \\
& \% \text { error in } x, \\
& \qquad \frac{\Delta x}{x}=2 \frac{\Delta A}{A}+\frac{\Delta B}{B}+\frac{1}{3} \frac{\Delta C}{C}+\frac{3 \Delta D}{D} \\
& 2\left(\frac{\Delta A}{A} \times 100 \%\right)+\left(\frac{\Delta B}{B} \times 100 \%\right) \\
& +\frac{1}{3}\left(\frac{\Delta C}{C} \times 100 \%\right)+3\left(\frac{\Delta D}{D} \times 100 \%\right)
\end{aligned}$
Minimum percentage error
$=\frac{1}{3}\left(\frac{\Delta C}{C} \times 100 \%\right)=\frac{1}{3} \times 4=1.1 \%$

Topic: Physics, Mathematics in Physics, JEE Main
In the measurement of resistance of a wire using Ohm's law, the plot between V and I is drawn as shown, then resistance of given wire will be

Options:
A 0.833Â Î©
B 0.9Â Î©
C 1 Î©
D none of these

The correct answer is:
1 Î©

Solution:
We know V∝I, so V-I curve for a ohmic devices is a straight line passing through origin. Due to some errors/carelessness during experiments, all points may not come on the same line. In this situation, we draw the most appropriate curve.

From given V-I curve,
⇒R=2.52.5=1 Ω

Topic: Physics, Experimental Physics, JEE Main
In the mechanism of action of a protein hormone, one of the secondary messengers is

Options:
A cyclic AMP.
B insulin.
C T3.
D gastrin.

The correct answer is:
cyclic AMP.

Solution:
Hormonal action depends on whether the hormone is Lipid or water soluble. Lipid soluble hormones can cross the target cell membrane easily, get attached to the mobile receptors within the cytoplasm and reach the nucleus to perform their functions. However water soluble hormones like insulin, cannot cross the cell membrane easily. These get attached to the static receptors on the cell membrane. This attachment triggers the formation of secondary messengers that help these hormones to act on cells. These also cause amplification of hormonal action so that many targets are affected by a single hormone. This is the main difference between how a lipid-soluble or water soluble hormone act. Secondary messengers commonly seen in cells are cyclic AMP. Other examples are inositol triphosphate (IP3), calcium and cyclic GMP.

Topic: Biology, Chemical Coordination and Integration, NEET
In the metallurgical extraction of copper, following reaction is used :

FeO+SiO2→FeSiO3

FeO and FeSiO3 respectively are.

Options:
A gangue and flux.
B flux and slag
C slag and flux.
D Gauge and slag

The correct answer is:
Gauge and slag

Solution:
During extraction of copper, iron oxide present as impurity is removed by adding SiO2 and converting it into FeSiO3

FeO+SiO2→FeSiO3 Flux Slag

Hence,FeO is gangue and FeSiO3 is slag.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2022 (28 Jun Shift 2)
In the metallurgy of alumminium, cryolite is mixed in the molten state because it

Options:
A Increases the melting point of alumina
B Oxidises alumina
C Reduces alumina
D Decreases the melting point of alumina

The correct answer is:
Decreases the melting point of alumina

Solution:
$N a_3 A l F_6$
Cryolite
(1) Decreases the melting point of alumina
(2) Increases conductivity of the solution

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In the metallurgy of iron, when limestone is added to the blast furnace, the calcium ion ends up in:-

Options:
A Slag
B Gangue
C Metallic Ca
D CaCO3

The correct answer is:
Slag

Solution:
Lime stone acts as flux in this process.
CaCO3â CaO+C02
CaO+SiO2â CasiO3(slag)

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
In the middle of the depletion layer of a reverse biased p-n junction, the

Options:
A
electric field is zero

B
potential is maximum

C
electric field is maximum

D
potential is zero

The correct answer is:
electric field is maximum

Solution:
No Solution Available

Topic: Physics, Semiconductors, JEE Main, JEE Main 2003
In the middle of the depletion layer of a reverse biased PN junction, the

Options:
A Potential is zero
B Electric field is zero
C Potential is maximum
D Electric field is maximum

The correct answer is:
Electric field is maximum

Solution:
The electric field strength versus distance curve across the $P-N$ junction is as follows

Topic: Physics, Semiconductors, JEE Main
In the mixture of Na2CO3+NaHCO3 volume of HCl required is x mL with phenolphthalein indicator and y mL with methyl orange indicator in the same titration. Hence volume of HCl for complete reaction of Na2CO3 is:

Options:
A 2x
B y
C x2
D y–x

The correct answer is:
2x

Solution:
If x mL is with phenolphthalein the rxn occurring is
Na2CO3+HClâNaHCO3+NaCl
So the volume of HCI for complete reaction of Na2CO3 would be double of ' x ' because the reaction will be
Na2CO3+2HClâNa2CO3+2NaCl
So, 2x

Topic: Chemistry, Redox Reactions, JEE Main
In the modern blast furnaces, the charge consists of a mixture of

Options:
A Calcined iron oxides + lime + anthracite coal
B Calcined iron oxides + limestone + coke
C Hydrated iron oxides + dolomite + coke
D Iron pyrites + lime +bituminous coal

The correct answer is:
Calcined iron oxides + limestone + coke

Solution:
Modern blast furnaces consists of
\(\underset{1}{\mathrm{CaCO}_3} \underset{\vdots}{+} \underset{4}{\text { Coke }} \underset{\vdots}{+} \underset{8}{\text { Calcinedironoxide }}\)
and following reaction takes place
\(\mathrm{Fe}_2 \mathrm{O}_3+3 \mathrm{CO} \rightarrow 2 \mathrm{Fe}+3 \mathrm{CO}_2\)

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
In the modern periodic table, the acidic nature of oxides

Options:
A increases from left to right and decreases from top to bottom
B decreases from right to left and increases from top to bottom
C remains same across a period and increases from top to bottom
D decreases from left to right and remains the same down the group

The correct answer is:
increases from left to right and decreases from top to bottom

Solution:
More the electropositivity of central atom, the more basic is the oxide and more electronegative central atom, more acidic is the oxide.
As we know, the electronegativity is the property of the element to attract electrons pairs. Hence, acidic character of the oxides "decreases down the group" because the electronegativity decreases down the group. Electronegativity decreases down the group because the "distance between the nucleus and valences shell electron increases. Hence, acidic nature of oxides increases left to right and decreases from top to bottom.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, AP EAMCET, AP EAMCET 2021 (25 Aug Shift 1)
In the molecular orbital diagram for the molecular ion N2+ , the number of electrons in the σ2p molecular orbital is

Options:
A 0
B 2
C 3
D 1

The correct answer is:
1

Solution:
N2+→σ1s2, σ*1s2, σ2s2, σ*2s2, [π2px2=π2py2], σ2pz1
The number of electron in σ2pz is 1.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2018 (15 Apr)
In the molecular orbital diagram for the molecular ion, $\mathrm{N}_2^{+}$, the number of electrons in the $\sigma_{2 p}$ molecular orbital is :

Options:
A
$0$

B
$2$

C
$3$

D
$1$

The correct answer is:
$1$

Solution:
Total electrons in $\mathrm{N}_2^{+}=(7 \times 2)-1=13$
$\mathrm{N}_2^{+} \rightarrow \sigma_{1 s^2}, \sigma_{1 s^2}^*, \sigma_{2 s^2}, \sigma_{2 s^2}^*,\left[\pi_{2p_x}^2=\pi_{2 p_y}^2\right] \sigma_{2 p_z}^1$
Number of electron in $\sigma_{2 p_z}$ is $1$.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
In the motion of a projectile freely under gravity, its

Options:
A Total energy is conserved
B Momentum is conserved
C Energy and momentum both are conserved
D None is conserved

The correct answer is:
Total energy is conserved

Solution:
An external force by gravity is present throughout the motion so momentum will not be conserved.

Topic: Physics, Motion In Two Dimensions, JEE Main
In the network shown below, the charge accumulated in the capacitor in steady state will be:

Options:
A 10.3 μC
B 4.8 μC
C 12 μC
D 7.2μC

The correct answer is:
7.2μC

Solution:
In steady state, capacitor behaves as open circuit, so no current will flow.

The total resistance of the circuit is, R=6+4 Ω=10 Ω

The current flowing through the circuit is

I=3 V10 Ω

Potential difference on 6 Ω resistor

V'=310×6=1.8 V

The charge on the capacitor is,

Q=CV'=4 μF×1.8 V=7.2 μC

Topic: Physics, Capacitance, JEE Main, JEE Main 2023 (13 Apr Shift 2)
In the network shown in the Fig, each resistance is \(1 \Omega\). The effective resistance between \(A\) and \(B\) is

Options:
A \(\frac{4}{3} \Omega\)
B \(\frac{3}{2} \Omega\)
C \(7 \Omega\)
D \(\frac{8}{7} \Omega\)

The correct answer is:
\(\frac{8}{7} \Omega\)

Solution:
At A current is distributed and at B currents are collected. Between A and B, the distribution is symmetrical. It has been shown in the figure. It appears that current in \(\mathrm{AO}\) and \(\mathrm{OB}\) remains same. At \(\mathrm{O}\), current \(\mathrm{i}_4\) returns back without any change. If we detach \(\mathrm{O}\) from \(\mathrm{AB}\) there will not be any change in distribution.
Now, CO & OD will be in series hence its total resistance \(=2 \Omega\)
It is in parallel with \(\mathrm{CD}\), so, equivalent resistance
\(=\frac{2 \times 1}{2+1}=\frac{2}{3} \Omega\)
This equivalent resistance is in series with \(\mathrm{AC}\) & \(\mathrm{DB}\), so, total resistance
\(=\frac{2}{3}+1+1=\frac{8}{3} \Omega\)
Now \(\frac{8}{3} \Omega\) is parallel to \(\mathrm{AB}\), that is, \(2 \Omega\), so total resistance
\(=\frac{8 / 3 \times 2}{8 / 3+2}=\frac{16 / 3}{14 / 3}=\frac{16}{14}=\frac{8}{7} \Omega\)
[Alt:
Between \(\mathrm{C}\) & D, the equivalent resistance is given by
\(1 / \mathrm{r}=\frac{1}{\mathrm{r}_3}+\frac{1}{\left(\mathrm{r}_4+\mathrm{r}_5\right)}=1+\frac{1}{2}=\frac{3}{2}\)
Equivalent resistance along
\(\mathrm{ACDB}=1+\frac{2}{3}+1=\frac{8}{3}\)
\(\therefore\) Effective resistance between \(\mathrm{A}\) and \(\mathrm{B}\) is
\(\left.\frac{1}{\mathrm{R}}=\frac{3}{8}+\frac{1}{2}=\frac{7}{8} \text { or } \mathrm{R}=\frac{8}{7}\right]\)

Topic: Physics, Current Electricity, VITEEE, VITEEE 2022
In the network shown in the figure, each resistance is 1 Î©. The effective resistance between A and B is

Options:
A 43 Ω
B 32 Ω
C 7 Ω
D 87 Î©

The correct answer is:
87 Î©

Solution:
At A current is distributed and at B currents are collected. Between A and B, the distribution is symmetrical, as shown in the figure. It appears that current in AO and OB remains the same. At O, current i4 returns back without any change. If we detach O from AB there will not be any change in the distribution.
Now, CO & OD will be in series hence it's total resistance = 2 Î©
It is in parallel with CD, so, equivalent resistance = 2Ã12+1=23 Î©
This equivalent resistance is in series with AC & DB, so, the total resistance =23+1+1=83 Î©
.
Now, 83 Î© is parallel to AB, that is, 2 Î©, so total resistance
=8/3Ã28/3+2=16/314/3=1614=87 Î©

Topic: Physics, Current Electricity, NEET
In the network shown in the figure, each resistance is 1 Î© . The effective resistance between A and B is

Options:
A 4 3 Î©
B 3 2 Ω
C 7 Ω
D 8 7 Î©

The correct answer is:
8 7 Î©

Solution:
At A current is distributed and at B currents are collected. Between A and B, the distribution is symmetrical, as shown in the figure. It appears that current in AO and OB remains same. At O, current i4 returns back without any change. If we detach O from AB there will not be any change in the distribution.
Now, CO & OD will be in series hence its total resistance = 2 Î©
It is in parallel with CD, so, equivalent resistance = 2 Ã 1 2 + 1 = 2 3 Î©
This equivalent resistance is in series with AC & DB, so , total resistance = 2 3 + 1 + 1 = 8 3 Î©
.
Now, 8 3 Î© is parallel to AB, that is, 2 Î©, so total resistance
= 8 / 3 Ã 2 8 / 3 + 2 = 1 6 / 3 1 4 / 3 = 1 6 1 4 = 8 7 Î©

Topic: Physics, Current Electricity, NEET
In the nuclear decay given below :
$$
{ }_{\mathrm{A}}^{\mathrm{A}} \mathrm{X} \longrightarrow \stackrel{\mathrm{A}}{\mathrm{A}+1} \mathrm{Y} \longrightarrow \mathrm{A}_{\mathrm{A}-4} \mathrm{~B}^* \longrightarrow \stackrel{\mathrm{A}-4}{\mathrm{Z}-1} \mathrm{~B}
$$
The particles emitted in the sequence are :

Options:
A $\alpha, \beta, \gamma$
B $\beta, \alpha, \gamma$
C $\gamma, \beta, \alpha$
D $\beta, \gamma, \alpha$

The correct answer is:
$\beta, \alpha, \gamma$

Solution:
$$
\begin{aligned}
& \mathrm{Z}^{\mathrm{X}^{\mathrm{A}} \stackrel{\beta}{\longrightarrow}} \mathrm{Z+1} \mathrm{Y}^{\mathrm{A}} \stackrel{\alpha}{\longrightarrow} \mathrm{Z}_{-1} \mathrm{~B}^{\mathrm{A}-4} \stackrel{\gamma}{\longrightarrow} \\
& \mathrm{Z}-1 \mathrm{~B}^{\mathrm{A}-4} \\
&
\end{aligned}
$$

Topic: Physics, Current Electricity, JEE Main
In the nuclear decay given below :
$$
{ }_{\mathrm{A}}^{\mathrm{A}} \mathrm{X} \longrightarrow \stackrel{\mathrm{A}}{\mathrm{A}+1} \mathrm{Y} \longrightarrow \mathrm{A}_{\mathrm{A}-4} \mathrm{~B}^* \longrightarrow \stackrel{\mathrm{A}-4}{\mathrm{Z}-1} \mathrm{~B}
$$
The particles emitted in the sequence are :

Options:
A $\alpha, \beta, \gamma$
B $\beta, \alpha, \gamma$
C $\gamma, \beta, \alpha$
D $\beta, \gamma, \alpha$

The correct answer is:
$\beta, \alpha, \gamma$

Solution:
$$
\begin{aligned}
& \mathrm{Z}^{\mathrm{X}^{\mathrm{A}} \stackrel{\beta}{\longrightarrow}} \mathrm{Z+1} \mathrm{Y}^{\mathrm{A}} \stackrel{\alpha}{\longrightarrow} \mathrm{Z}_{-1} \mathrm{~B}^{\mathrm{A}-4} \stackrel{\gamma}{\longrightarrow} \\
& \mathrm{Z}-1 \mathrm{~B}^{\mathrm{A}-4} \\
&
\end{aligned}
$$

Topic: Physics, Nuclear Physics, NEET, NEET 2009 (Mains)
In the nuclear fusion reaction ${ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+\mathrm{n}$ given that the repulsive potential energy between the two nuclei is $\sim 7.7 \times 10^{-14} \mathrm{~J}$, the temperature at which the gases must be heated to initiate the reaction is nearly [Boltzmann's Constant $\mathrm{k}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$ ]

Options:
A
$10^7 \mathrm{~K}$

B
$10^5 \mathrm{~K}$

C
$10^3 \mathrm{~K}$

D
$10^9 \mathrm{~K}$

The correct answer is:
$10^9 \mathrm{~K}$

Solution:
No Solution Available

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2003
In the nuclear fusion reaction ${ }_1 \mathrm{H}^2+{ }_1 \mathrm{H}^3 \rightarrow{ }_2 \mathrm{He}^4+\mathrm{n}$, if the repulsive potential energy between the two nuclei is $2.07 \times 10^{-14} \mathrm{~J}$, then the temperature at which the gases must be heated to initiate the reaction is $\left(\right.$ Boltzmann constant $\left.=1.38 \times 10^{-23} \mathrm{JK}^{-1}\right)$

Options:
A $10^9 \mathrm{~K}$
B $10^7 \mathrm{~K}$
C $10^5 \mathrm{~K}$
D $10^{12} \mathrm{~K}$

The correct answer is:
$10^9 \mathrm{~K}$

Solution:
No solution. Refer to answer key.

Topic: Physics, Nuclear Physics, JEE Main
In the nuclear reaction, \({ }_6^{11} \mathrm{C} \rightarrow{ }_5^{11} \mathrm{~B}+\beta+X, X\) stands for

Options:
A a neutron
B an electron
C a neutrino
D an anti-neutrino

The correct answer is:
a neutrino

Solution:
Nuclear reaction is given as
\({ }_6^{11} \mathrm{C} \rightarrow{ }_5^{11} \mathrm{~B}+\beta+X\) ...(i)
When positron ( \(\beta\) ) is emitted from a nuclei, then atomic number decreases by one unit, while the mass number remains the same.
Hence, Eq. (i) is written as
\({ }_6^{11} \mathrm{C} \rightarrow{ }_5^{11} \mathrm{~B}+{ }_1^0 \beta+v \text { (neutrino) }\)
Hence, \(X\) is a neutrino v.

Topic: Physics, Nuclear Physics, AP EAMCET, AP EAMCET 2020 (17 Sep Shift 1)
In the nuclear reaction \({ }_7^{14} \mathrm{~N}+\mathrm{X} \rightarrow{ }_6^{14} \mathrm{C}+{ }_1^1 \mathrm{H}\) the \(\mathrm{X}\) will be

Options:
A \({ }_{-1}^0 \mathrm{e}\)
B \({ }_1^1 \mathrm{H}\)
C \({ }_1^2 \mathrm{H}\)
D \({ }_0^1 \mathrm{n}\)

The correct answer is:
\({ }_0^1 \mathrm{n}\)

Solution:
\(\text {Hints: } \mathrm{X} \rightarrow{ }_0^1 \mathrm{n}\)

Topic: Physics, Nuclear Physics, JEE Main
In the nucleophilic substitution reactions S N 2â or S N 1 the reactivity of alkyl halides follows the sequence

Options:
A R-I > R-Br > R-Cl > R-F
B R-Cl > R-F > R-Br > R-I
C R-F > R-Cl > R-Br > R-I
D R-I > R-F > R-Cl > R-Br

The correct answer is:
R-I > R-Br > R-Cl > R-F

Solution:
Nucleophilic substitution reaction depends on learning group ability
The correct order of reactivity is
RI > RBr > RCl > RF
It is due to fact, that weaker the base, better will be the leaving group.
Leaving group ability I â > Br â >Cl> F â

Topic: Chemistry, General Organic Chemistry, JEE Main
In the parabola $y^2=6 x$, the equation of the chord through vertex and negative end of latus rectum, is

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

The correct answer is:
$y+2 x=0$

Solution:
Vertex $\equiv(0,0)$. End points of latus rectum are $(a, \pm 2 a)$. Here $\quad a=\frac{6}{4}$
Therefore, $-v e$ end of latus rectum is $\left(\frac{3}{2},-3\right)$
Line through the point is $y=\frac{-3}{3 / 2} x$ or $y+2 x=0$.

Topic: Mathematics, Parabola, JEE Main
In the part of the circuit shown in the figure, the potential difference between points VA-VB=16 V. The current passing through the 2 Î© resistance will be

Options:
A 2.5 A
B 3.5 A
C 4.0 A
D zero

The correct answer is:
3.5 A

Solution:
VA-VB=16 V
â´4i1+2i1+i2-3+4i1=16 V ......(i)
Using Kirchhoff's second law in the closed loops we have
â9-i2-2i1+i2=0 ......(ii)
Solving equations (i) and (ii), we get
i1=1.5 A and i2=2 A
â´ Current through 2 Î© resistor =2+1.5=3.5 A

Topic: Physics, Current Electricity, JEE Main