datavorous/entrance-exam-dataset huggingface dataset:
The force between the plates of a parallel plate capacitor of capacitance $C$ and distance of separation of the plates $d$ with a potential difference $V$ between the plates, is

Options:
A $\frac{C V^2}{2 d}$
B $\frac{c^2 V^2}{2 d^2}$
C $\frac{C^2 V^2}{d^2}$
D $\frac{V^2 d}{c}$

The correct answer is:
$\frac{C V^2}{2 d}$

Solution:
Total electric field between the plates of the capacitor:
$E_T=\frac{V}{d}$
Then, electric field due to only one plate:
$E_1=\frac{V}{2 d}$
$\therefore$ force of one plate on another:
$\mathrm{F}=\mathrm{E}_1 \times \mathrm{QF}=\mathrm{E}_1 \times \mathrm{CV}=\frac{\mathrm{V}}{2 \mathrm{~d}} \times \mathrm{CV}=\frac{\mathrm{CV}^2}{2 \mathrm{~d}}$

Topic: Physics, Capacitance, JEE Main
The force ‘F' acting on a particle of mass ‘m' is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from zero to 8 s is:

Options:
A 24 Ns
B 20 Ns
C 12 Ns
D 6 Ns

The correct answer is:
12 Ns

Solution:
As we know that:-

Area under force-time graph gives as momentum

∆P=ar of given curve

∆P=12×b×h + l1×b1 + l2×b2

∆P=1226-32+423
=6-6+12
=12

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
The force $\vec{F}=F \hat{i}$ on a particle of mass $2 \mathrm{~kg}$, moving along the $x$-axis is given in the figure as a function of its position $x$. The particle is moving with a velocity of $5 \mathrm{~m} / \mathrm{s}$ along the $x$-axis at $x=0$. What is the kinetic energy of the particle at $x=8 \mathrm{~m} ?$

Options:
A
$34 \mathrm{~J}$

B
$34.5 \mathrm{~J}$

C
$4.5 \mathrm{~J}$

D
$29.4 \mathrm{~J}$

The correct answer is:
$29.4 \mathrm{~J}$

Solution:
No Solution Available

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2012 (26 May Online)
The force F on a sphere of radius a, moving in a medium with velocity v , is given by F=6ÏÎ·av . The dimensions of Î· are

Options:
A ML−1T−1
B MT−1
C MLT−2
D ML−3

The correct answer is:
ML−1T−1

Solution:
η=Fav=[MLT−2][L][LT−1]=ML−1T−1

Topic: Physics, Units and Dimensions, JEE Main
The force is given in terms of time t and displacement x by the equation F=AcosBx+CsinDt

The dimensional formula of ADB is:

Options:
A M0 L T-1
B ML2 T-3
C M1 L1 T-2
D M2 L2 T-3

The correct answer is:
ML2 T-3

Solution:
[A]=MLT-2

[B]=L-1

[D]=T-1

ADB=MLT-2T-1L-1

ADB=ML2 T-3

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2021 (25 Jul Shift 2)
The force of attraction between \( +5 \mu \mathrm{C} \) and \( -5 \mu \mathrm{C} \) charges kept at some distance apart is \( 9 \mathrm{~N} \). When the two charges are made to contact and then separated again by the same distance the force acting between them becomes

Options:
A Infinite
B \( 9 \times 10^{9} \mathrm{~N} \)
C \( 1 \mathrm{~N} \)
D Zero

The correct answer is:
Zero

Solution:
When we touch two oppositely charged bodies, charge redistribution takes place and both the bodies become equally charged because they are identical. Since both bodies were equally positive and equally negative initially, after bringing them in contact with each other, they will have zero charge which results in zero force between them.

Topic: Physics, Electrostatics, JEE Main
The force of attraction between two long parallel conductors \( 1 \mathrm{~m} \) apart carrying unit ampere current in same direction is

Options:
A \( 2 \times 10^{-7} \mathrm{Nm}^{-1} \)
B \( 10^{-7} \mathrm{~N} \mathrm{~m}^{-1} \)
C \( 10^{-5} \mathrm{~N} \mathrm{~m}^{-1} \)
D \( 0.5 \times 10^{-6} \mathrm{~N} \mathrm{~m}^{-1} \)

The correct answer is:
\( 2 \times 10^{-7} \mathrm{Nm}^{-1} \)

Solution:
Given: the distance between parallel conductors d=1 m, current in both parallel conductor is i1=i2=1 A.

The force per unit length of attraction between two parallel current-carrying conductors is given by,

f=μ0i1i22πd

f=2×10-7×1×11=2×10-7 N m-1.

Topic: Physics, Magnetic Effects of Current, JEE Main
The force of gravitation is

Options:
A Repulsive
B Electrostatic
C Conservative
D Non-conservative

The correct answer is:
Conservative

Solution:
Gravitation force is a conservative force.
A conservative force is a force with the property that the work done in moving a particle between two points is independent of the taken path.
Examples: gravitational force, spring force, magnetic force

Topic: Physics, Gravitation, JEE Main
The force of repulsion between two point charges is F, when these are at distance 0.5 m apart. Now the point charges are replaced by spheres of radii 5 cm each having the same charge as that of the respective point charges. The distance between their centres is again kept 0.5 m. Then the force of repulsion will

Options:
A increase
B decrease
C remain F
D become 10F9

The correct answer is:
remain F

Solution:
Charges may be assumed at the centre in case of uniformly charged sphere.

Topic: Physics, Electrostatics, NEET
The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. If the angle of inclination of the plane is $60^{\circ}$, then the coefficient of friction is

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

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

Solution:
According to the question, angle of inclination $\theta=60^{\circ}$.
Now, force of friction, $f=\mu N=\mu m g \cos \theta$ and net retarding force, $\left(F_1\right)=m g \sin \theta+f$ $\therefore$ Net accelerating force down the inclined plane is
$\therefore$ External force needed (up the inclined plane) to maintain sliding motion is (net retarding force)
From Eqs. (i),(ii) and (iii) we get
$m g \sin \theta+\mu m g \cos \theta=2(m g \sin \theta-\mu m g \cos \theta)$
or
$3 \mu m g \cos \theta=m g \sin \theta$
or
$$
\frac{\sin \theta}{\cos \theta}=3 \mu
$$
or
$$
\tan \theta=3 \mu \quad\left(\because \theta=60^{\circ}\right)
$$
or
$$
\tan 60^{\circ}=3 \mu
$$
$$
\begin{aligned}
\sqrt{3} & =3 \mu \\
\mu & =\frac{1}{\sqrt{3}}
\end{aligned}
$$
When inclination of plane is $60^{\circ}$ then the coefficient of friction, $\mu=\frac{1}{\sqrt{3}}$.

Topic: Physics, Laws of Motion, JEE Main
The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. If the angle of inclination of the plane is $60^{\circ}$, then the coefficient of friction is

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

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

Solution:
According to the question, angle of inclination $\theta=60^{\circ}$.
Now, force of friction, $f=\mu N=\mu m g \cos \theta$ and net retarding force, $\left(F_1\right)=m g \sin \theta+f$ $\therefore$ Net accelerating force down the inclined plane is
$\therefore$ External force needed (up the inclined plane) to maintain sliding motion is (net retarding force)
From Eqs. (i),(ii) and (iii) we get
$m g \sin \theta+\mu m g \cos \theta=2(m g \sin \theta-\mu m g \cos \theta)$
or
$3 \mu m g \cos \theta=m g \sin \theta$
or
$$
\frac{\sin \theta}{\cos \theta}=3 \mu
$$
or
$$
\tan \theta=3 \mu \quad\left(\because \theta=60^{\circ}\right)
$$
or
$$
\tan 60^{\circ}=3 \mu
$$
$$
\begin{aligned}
\sqrt{3} & =3 \mu \\
\mu & =\frac{1}{\sqrt{3}}
\end{aligned}
$$
When inclination of plane is $60^{\circ}$ then the coefficient of friction, $\mu=\frac{1}{\sqrt{3}}$.

Topic: Physics, Laws of Motion, AP EAMCET, AP EAMCET 2019 (20 Apr Shift 2)
The force required to stretch a wire of cross-section 1 cm2 to double its length will be :

(Given Yong's modulus of the wire =2×1011 N m-2)

Options:
A 1×107 N
B 1.5×107 N
C 2×107 N
D 2.5×107 N

The correct answer is:
2×107 N

Solution:
The wire is stretched to double its length. Therefore, if the length of the wore was l, change in the length ∆l=l.

Applying Hooke's law,

Y=stressstrain=FlA∆l⇒F=YA∆ll=YA⇒F=2×1011×1×10-4=2×107 N

=2×1011×10-42l-ll

=2×107 N

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2022 (28 Jul Shift 1)
The forces, which meet at one point but their lines of action do not lie in one plane, are called:-

Options:
A Non-coplanar and concurrent force.
B Coplanar and concurrent forces.
C Non-coplanar and non-concurrent forces.
D Coplanar and non-concurrent forces.

The correct answer is:
Non-coplanar and concurrent force.

Solution:
We know that when vectors lie on the same plane, then they are called coplanar vectors.

A concurrent vector system is a set of two or more vectors whose lines of action intersect at a point.

So, the forces, which meet at one point but their lines of action do not lie in one plane, are called non-coplanar and concurrent force.

Topic: Physics, Laws of Motion, JEE Main
The form of iron obtained from blast furnace is :

Options:
A Cast Iron
B Wrought Iron
C Steel
D Pig Iron

The correct answer is:
Pig Iron

Solution:
Iron obtained in molten form at bottom of blast furnace is pig Iron with composition
C - 3 to 4.3 %
Si - 1 to 2 %
P - 0.05 - 2 %
S - 0.05 - 1 %
Mn - 0.5 to 2 %

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2014 (09 Apr Online)
The formation of $\mathrm{CO}$ and $\mathrm{CO}_{2}$ illustrates the law of

Options:
A reciprocal proportion
B conservation of mass
C multiple proportion
D constant composition

The correct answer is:
multiple proportion

Solution:
Formation of $\mathrm{CO}$ and $\mathrm{CO}_{2}$ illustrates the law of multiple proportion that is constant mass of $\mathrm{C}$ reacts with different masses of oxygen. These masses here bears simple ratio of 1: 2 .

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The formation of cyanohydrin from acetone is which type of reaction?

Options:
A Electrophilic addition reaction
B Electrophilic substitution reaction
C Nucleophilic substitution reaction
D Nucleophilic addition reaction

The correct answer is:
Nucleophilic addition reaction

Solution:
Cyanohydrin formation:
This reaction is nucleophilic addition reaction.
Step (i): The nucleophilic carbon in the cyanide adds to the electrophilic carbon in the polar carbonyl group, electrons from the C = O move to the electronegative O creating an intermediate alkoxide.
Step (ii): An acid/base reaction protonation of the alkoxide oxygen creates the cyanohydrin product.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The formation of gas at the surface of tungsten due to adsorption is the reaction of order

Options:
A
0

B
1

C
2

D
insufficient data

The correct answer is:
0

Solution:
No Solution Available

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2002
The formation of molecular complex $\mathrm{BF}_3-\mathrm{NH}_3$ results in a change in hybridization of boron

Options:
A
from $s p^2$ to $d s p^2$

B
from $s p^2$ to $s p^3$

C
from $s p^3$ to $s p^2$

D
from $s p^3$ to $s p^3 d$

The correct answer is:
from $s p^2$ to $s p^3$

Solution:
$\operatorname{In~} \mathrm{BF}_3, \mathrm{~B}$ is $s p^2$ hybridized with one empty $p_{\mathrm{z}}$ orbital. The empty $p_{\mathrm{z}}$ orbital of $\mathrm{BF}_3$ can be filled by lone pair of molecules such as $\mathrm{NH}_3$. When this occurs a tetrahedral molecule or ion is formed which is $s p^3$ hybridized.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2012 (12 May Online)
The formation of nitric oxide by contact process $N_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}$. $\Delta \mathrm{H}=43.200 \mathrm{kcal}$ is favoured by

Options:
A Low temperature and low pressure
B Low temperature and high pressure
C High temperature and high pressure
D High temperature and excess reactants concentration

The correct answer is:
High temperature and excess reactants concentration

Solution:
High temperature and excess concentration of the reactant.

Topic: Chemistry, Chemical Equilibrium, JEE Main
The formation of the oxide ion $\mathrm{O}^{2-}(\mathrm{g})$ requires first an exothermic and then an endothermic step as shown below $\mathrm{O}(\mathrm{g})+\mathrm{e}^{-} \mathrm{O}^{-}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=-142 \mathrm{kJmol}^{-1}$
$\mathrm{O}^{-}(\mathrm{g})+\mathrm{e}^{-} \mathrm{O}^{2-}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=844 \mathrm{kJmol}^{-1}$

Options:
A
Oxygen is more electronegative

B
$\mathrm{O}^{-}$ion has comparatively larger size than oxygen atom

C
$\mathrm{O}^{-}$ion will tend to resist the addition of another electron

D
Oxygen has high electron affinity

The correct answer is:
$\mathrm{O}^{-}$ion will tend to resist the addition of another electron

Solution:
$$
\mathrm{O}^{-1}(\mathrm{~g})+\mathrm{e} \longrightarrow \mathrm{O}^{-2}(\mathrm{~g})
$$
Due to the electronic repulsion, amount of the energy is needed to add electron

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2004
The formation of which of the following polymers involves hydrolysis reaction?

Options:
A Bakelite
B Nylon- 6,6
C Terylene
D Nylon-6

The correct answer is:
Nylon-6

Solution:
Nylon - 6 formed from caprolactam first step of which involve ring opening by hydrolysis (base catalyzed).

Topic: Chemistry, Polymers, JEE Main, JEE Main 2017 (02 Apr)
The formula for chile salt petre and indian salt petre is

Options:
A NaNO3,KNO3
B Na2SO4.10H2O
C KNO3,NaNO3
D Na2S2O3.5H2O,KNO3

The correct answer is:
NaNO3,KNO3

Solution:
Chile salt petre is NaNO3 While KNO3 is Indian salt petre. Na2SO4.10H2O is Glouber salt and Na2S2O3.5H2O is known as Hypo.

Topic: Chemistry, s Block Elements, JEE Main
The formula for functional residual capacity is:

Options:
A TV+IRV
B ERV+RV
C VC+RV
D TV+IRV+ERV

The correct answer is:
ERV+RV

Solution:
Functional Residual Capacity (FRC) is the volume of air which remains in the lungs after a normal expiration. It is the sum of Expiratory Reserve Volume (ERV) and Residual Volume (RV) i.e. FRC=ERV+RV .
TV+IRV=IC (Inspiratory Capacity)
VC+RV=TLC (Total Lung Capacity)
TV+IRV+ERV=VC (Vital Capacity)

Topic: Biology, Breathing and Exchange of Gases, NEET
The formula for induced e.m.f. in a coil due to change in magnetic flux through the coil is (here $A=$ area of the coil, $B=$ magnetic field)

Options:
A $e=-A \cdot \frac{d B}{d t}$
B ${ }^{e=-B \cdot \frac{d A}{d t}}$
C $e-\frac{d}{d t}(A . B)$
D ${ }^{e=-\frac{d}{d t}(A \times B)}$

The correct answer is:
$e-\frac{d}{d t}(A . B)$

Solution:
$e=-\frac{d}{d t}(A . B)$
Induce e.m.f = Rate of change of flux link $=\frac{d \phi}{d t}$
$\phi=A \cdot B$
$e=-\frac{d}{d t}(A . B)$

Topic: Physics, Electromagnetic Induction, JEE Main
The formula for internal resistance of a cell will be, (Here \( l_{1} \& l_{2} \) are the balancing lengths of cell in open and closed circuit respectively.)

Options:
A \( \mathrm{r}=\left(\frac{\mathrm{l}_{1}-\mathrm{l}_{2}}{\mathrm{l}_{2}}\right) \mathrm{R} \)
B \( \mathrm{r}=\left(\frac{\mathrm{l}_{2}-\mathrm{l}_{1}}{\mathrm{l}_{2}}\right) \mathrm{R} \)
C \( \mathrm{r}=\left(\frac{\mathrm{l}_{1}-\mathrm{l}_{2}}{\mathrm{l}_{1}}\right) \mathrm{R} \)
D \( \mathrm{r}=\left(\frac{\mathrm{l}_{2}-\mathrm{l}_{1}}{\mathrm{l}_{1}}\right) \mathrm{R} \)

The correct answer is:
\( \mathrm{r}=\left(\frac{\mathrm{l}_{1}-\mathrm{l}_{2}}{\mathrm{l}_{2}}\right) \mathrm{R} \)

Solution:
Given that balancing lengths of potentiometer in open and closed circuit as l1,l2 and If R is the resistance of resistance box and x is the potential gradient then

Internal resistance of the cell will be,r=E–VVR=l1x–l2xl2xR=l1–l2l2R

Topic: Physics, Current Electricity, JEE Main
The formula of a gaseous hydrocarbon which requires 6 times of its own volume of O2 for complete oxidation and produces 4 times its own volume of CO2 is CxHy. The value of y is ________.

Options:

The correct answer is:
8

Solution:
Combustion reaction.

CxHy(g)+x+y4O2g→xCO2g+y2H2Oℓ

V 6 V -- - Vx=4V

⇒x=4

Since: I Vo2=6×VCxHy

⇒Vx+y4=6 V

⇒x+y4=6⇒4+y4=6

⇒y=8

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The formula of chromyl chloride is

Options:
A $\mathrm{CrCl}$
B $\mathrm{CrCl}_{3}$
C $\mathrm{CrOCl}_{2}$
D $\mathrm{CrO}_{2} \mathrm{Cl}_{2}$

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

Solution:
Chromyl chloride is inorganic compound with formula $\mathrm{CrO}_{2} \mathrm{Cl}_{2}$. It is reddish compound that is volatile at room temperature. It's structure is

Topic: Chemistry, d and f Block Elements, JEE Main
The formula of sodium nitroprusside is

Options:
A $\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]$
B $\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]$
C $\mathrm{NaFe}\left[\mathrm{Fe}(\mathrm{CN})_6\right]$
D $\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_6 \mathrm{NO}_2\right]$

The correct answer is:
$\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]$

Solution:
Sodium nitroprusside is known to have the formula $\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]$. It contains the nitrosyl ($\mathrm{NO}$) ligand, five cyanide ($\mathrm{CN}$) ligands, and an iron ($\mathrm{Fe}$) center. Therefore, the correct formula is $\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]$.

Topic: Chemistry, Coordination Compounds, JEE Main
The forward biased diode connection is :

Options:
A

B

C

D

The correct answer is:

Solution:
A junction diode is said to be forward-biased when the positive terminal of the external battery is connected to the p-region and the negative terminal to the n-region of the diode.

The potential of p-side is more positive that of n-side, hence diode is in forward biasing. In forward biasing it acts as short circuit, hence current flows.

In case of forward biased circuit P-junction should be at higher potential.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2014 (06 Apr)
The forward-bias voltage of a diode is changed from 0.6 V to 0.7 V, the current changes from 5 mA to 15 mA. What is the value of the forward bias resistance?

Options:

The correct answer is:
10

Solution:
The forward biased resistance of a diode is

R=∆V∆I=0.7-0.615-5×10-3
R=0.110×10-3=10 Ω

Topic: Physics, Semiconductors, JEE Main
The forward-biased diode connection among the following is:

Options:
A
B
C
D

The correct answer is:

Solution:
In case of forward biased circuit PN-junction, P should be at higher potential. Therefore, given below arrangement is the correct choice.

Topic: Physics, Semiconductors, JEE Main
The fossil bone has a ${ }^{14} \mathrm{C}:{ }^{12} \mathrm{C}$ ratio, which is $[1 / 16]$ of that in a living animal bone. If the halflife time of ${ }^{14} \mathrm{C}$ is 5730 years, then the age of the fossil bone is

Options:
A 11460 years
B 17190 years
C 22920 years
D 45840 years.

The correct answer is:
22920 years

Solution:
After 5730 years, the number of $\mathrm{C}^{14}$ remaining is $\frac{1}{2}$ th original. To have $(1 / 16)$ original value, it takes $(1 / 2)^4, 4$ half lives.
Therefore, the bone is $4 \times 5730$ years $=22920$ years old.

Topic: Physics, Nuclear Physics, AIIMS, AIIMS 2006
The four arithmetic means between 3 and 23 are

Options:
A $5,9,11,13$
B $7,11,15,19$
C $5,11,15,22$
D $7,15,19,21$

The correct answer is:
$7,11,15,19$

Solution:
Let four arithmetic means are $A_1, A_2, A_3$ and $A_4$.
So $3, A_1, A_2, A_3, A_4, 23$
$\Rightarrow T_6=23=a+5 d \Rightarrow d=4$
Thus $A_1=3+4=7, A_2=7+4=11$,
$A_3=11+4=15, A_4=15+4=19$

Topic: Mathematics, Sequences and Series, JEE Main
The four arms of a Wheatstone bridge (Fig.) have the following resistances: AB = 100Ω, BC = 10Ω, CD = 5Ω, and DA = 60Ω. A galvanometer of 15Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

Options:
A 3.32A
B 4.87A
C 6.45A
D 5.79A

The correct answer is:
4.87A

Solution:
Correct Option is : (B)
4.87A

Topic: Chemistry, Current Electricity, NEET
The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of 15Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

Options:
A 2.44μA
B 2.44 mA
C 4.87 mA
D 4.87μA

The correct answer is:
4.87 mA

Solution:
x-10100+x-y15+x-010=0

53x-20y=30 ……(1)

y-1060+y-x15+y-05=0

17y-4x=10 ……(2)

on solving (1) & (2)

x=0.865

y=0.792

ΔV=0.073R=15 Ω

⇒i=4.87 mA

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (17 Mar Shift 2)
The four quantum numbers of potassium's valence electron are

Options:
A \( 4,0,1, \frac{1}{2} \)
B \( 4,1,0, \frac{1}{2} \)
C \( 4,0,0, \frac{1}{2} \)
D \( 4,1,1, \frac{1}{2} \)

The correct answer is:
\( 4,0,0, \frac{1}{2} \)

Solution:
Valance shell of K=4s1

For which, n=4
l=0
m=0
s=+12 or 12

Topic: Chemistry, Structure of Atom, JEE Main
The four quantum numbers of the outermost orbital of $K$ (atomic no. $=19$ ) are

Options:
A $n=2 l=0, m=0, s=+\frac{1}{2}$
B $n=4, l=0, m=0, s=+\frac{1}{2}$
C $n=3, l=1, m=1, s=+\frac{1}{2}$
D $n=4, I=2 m=-1, s=+\frac{1}{2}$

The correct answer is:
$n=4, l=0, m=0, s=+\frac{1}{2}$

Solution:
$K_{19}=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6, 4 s^1$
for $4 s^1$ electrons.
$n=4, I=0, m=0$ and $s=+\frac{1}{2}$.

Topic: Chemistry, Structure of Atom, JEE Main
The four quantum numbers of the valence electron of potassium are

Options:
A $4,1,0$ and $\frac{1}{2}$
B $4,0,1$ and $\frac{1}{2}$
C $4,0,0$ and $+\frac{1}{2}$
D 4,1,1 and $\frac{1}{2}$

The correct answer is:
$4,0,0$ and $+\frac{1}{2}$

Solution:
Atomic number of potassium is 19 and hence electronic configurtion will be $1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 4 s^1$
Hence for $4 s^1$ electron value of Quantum number are
Prncipal quantum number $n=4$
Azimuthal quantum number $I=0$
Magnetic quantum number $m=0$
Spin quantum number $s=+1 / 2$

Topic: Chemistry, Structure of Atom, JEE Main
The four sketches (I, II, III and IV) given below, represent four different types of animal tissues. Which one of these is correctly identified in the options given, along with its correct location and function?
I.
II.
III.
IV.

Options:
A (a)
B (b)
C (c)
D (d)

The correct answer is:
(a)

Solution:
Glandular epithelium is located in intestine and deals with secretion.
Cardiac muscles are formed in heart and deals with contraction of heart.
Tendon attaches skeletal muscle to bones. Ciliated cubical epithelium is found in parts of uriniferous tubules (nephron) and helps in flow of urine.

Topic: Biology, Structural Organisation in Animals, NEET, NEET 2012 (Mains)
The fourth term in the expansion of $(1-2 x)^{3 / 2}$ will be

Options:
A $-\frac{3}{4} x^4$
B $\frac{x^3}{2}$
C $-\frac{x^3}{2}$
D $\frac{3}{4} x^4$

The correct answer is:
$\frac{x^3}{2}$

Solution:
Expansion of $(1-2 x)^{3 / 2}$
$=1+\frac{3}{2}(-2 x)+\frac{3}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}(-2 x)^2+\frac{3}{2} \cdot \frac{1}{2}\left(-\frac{1}{2}\right) \frac{1}{6}(-2 x)^3+\ldots .$.
Hence $4^{\text {th }}$ term is $\frac{x^3}{2}$

Topic: Mathematics, Binomial Theorem, JEE Main
The fourth term of an A.P. is three times of the first term and the seventh term exceeds the twice of the third term by one, then the common difference of the progression is

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

The correct answer is:
2

Solution:
Let the progression be $a$, $a+d$, a +2 d,

Then $x_{4}=3 x_{1} \Rightarrow a+3 d=3 a \Rightarrow 3 d=2 a \ldots$ (i)

Again $x_{7}=2 x_{3}+1$

$\Rightarrow a+6 d=2(a+2 d)+1 \Rightarrow 2 d=a+1 \quad \ldots$ (ii)

Solving (i) and (ii), we get $\mathrm{a}=3, \mathrm{~d}=2$

Topic: Mathematics, Sequences and Series, JEE Main
The fourth term of the arithmetic-geometric progression 6, 8, 8,.... is

Options:
A 8
B 12
C 323
D 649

The correct answer is:
649

Solution:
Let 6, (6+d)r, (6+2d)r2,....are in A.G.P.
⇒8=6+dr and 8=6+2dr2
⇒648=6+d2r26+2dr2
⇒36+d2+12d=48+16d
⇒d2-4d-12=0⇒d=6,-2
r=86+d⇒r=23,2
So, the next term of the A.G.P. is
6+3dr3=6+3×6233=24×827=649
or 6+3×-223=0

Topic: Mathematics, Sequences and Series, JEE Main
The fraction of a floating object of volume $V_0$ and density $d_0$ above the surface of a liquid of density $d$ will be

Options:
A $\frac{d_0}{d}$
B $\frac{d d_0}{d+d_0}$
C $\frac{d-d_0}{d}$
D $\frac{d d_0}{d-d_0}$

The correct answer is:
$\frac{d-d_0}{d}$

Solution:
For the floatation $V_0 d_0 g=V_{i n} d g \Rightarrow V_{i n}=V_0 \frac{d_0}{d}$
$\therefore \quad V_{a w t}=V_0-V_{m n}=V_0-V_0 \frac{d_0}{d}=V_0\left[\frac{d-d_0}{d}\right]$
$\Rightarrow \frac{V_{\text {out }}}{V_0}=\frac{d-d_0}{d}$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main