datavorous/entrance-exam-dataset huggingface dataset:
Ten points lie in a plane so that no three of them are collinear. The number of lines passing through exactly two of these points and dividing the plane into two regions each containing four of the remaining points is

Options:
A $1$
B $5$
C $10$
D dependent on the configuration of points

The correct answer is:
$5$

Solution:

Topic: Mathematics, Permutation Combination, KVPY, KVPY 2013 (SA)
Terpinen-4-ol is an active ingredient in tea tree oil has the following structure

The correct observations for terpinen-4-ol is/are

I. It rotates the plane of plane polarized light.

II. It reacts with Baeyer's reagent to form form a triol

III. On reaction with NaBr and H2SO4 , it gives a di bromo compound

IV. On ozonolysis it gives a compound with molecular formula C10H18O3

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

The correct answer is:
I, II, III and IV

Solution:
(I) Terpinen-4-ol is an optically active molecule because it is chiral molecule
(II)
(III)
(IV)

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Tertiary alkyl halide is practically to substitution by \( S_{N} 2 \) mechanism because of

Options:
A Insolubility
B Instability
C Inductive efffect
D Steric hindrance

The correct answer is:
Steric hindrance

Solution:
In \( S_{N} 2 \) reactions, nucleophile approach to the carbon bearing the leaving group, the presence of bulky substituents
inhibits the incoming nucleophile which decrease the rate of reaction, therefore, alkyl halides having bulkier substituents
or higher degree alkyl halides preferably react through \( S N_{1} \) path. The order of reactivity for \( S_{N} 1 \) path is \( 1^{\circ}>2^{\circ}>3 \)

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Tertiary alkyl halides are practically inert to substitution by $\mathrm{S}_{\mathrm{N}}{ }^2$ mechanism because of

Options:
A
insolubility

B
instability

C
inductive effect

D
steric hindrance

The correct answer is:
steric hindrance

Solution:
No Solution Available

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2005
Test cross in plants of in Drosophila involves crossing

Options:
A between two genotypes with recessive trait
B between two $\mathrm{F}_1$ hybrids
C the $F_1$ hybrid with a double recessive genotype
D between two genotypes with dominant trait

The correct answer is:
the $F_1$ hybrid with a double recessive genotype

Solution:
When $\mathrm{F}_1$ hybrid is crossed with recessive parent then it is called test cross. It is also a back cross. Test cross is done to test whether the dominant parent is heterozygous or homozygous.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2011 (Mains)
Tetraethyl lead is a

Options:
A Solvent
B Petroleum additive
C Oxidising agent
D Fire extinguisher

The correct answer is:
Petroleum additive

Solution:
Tetra ethyl lead C2H54Pb is used as an antiknock compound in gasoline.

Topic: Chemistry, Hydrocarbons, JEE Main
Tetragonal crystal system has the following unit cell dimensions

Options:
A $a=b=c$ and $\alpha=\beta=\gamma=90^{\circ}$
B $a=b \neq c$ and $\alpha=\beta=\gamma=90^{\circ}$
C $a \neq b \neq c$ and $\alpha=\beta=\gamma=90^{\circ}$
D $a=b \neq c$ and $\alpha=\beta=90^{\circ}, \gamma=120^{\circ}$

The correct answer is:
$a=b \neq c$ and $\alpha=\beta=\gamma=90^{\circ}$

Solution:
Tetragonal system has the unit cell dimension $a=b \neq c, \alpha=\beta=\gamma=90^{\circ}$.

Topic: Chemistry, Solid State, JEE Main
The _____1_____ is the main lymphoid organ where all blood cells _____2____ lymphocytes are produced. The_____3_____ is a lobed organ located near the heart and beneath the breastbone. The_____3_____is quite large at the time of birth but keeps reducing in size with age and by the time puberty is attained it reduces to a very small size. Both _____1_____and _____3_____provide micro-environments for the development and maturation of T-lymphocytes.

Options:
A 123bonemarrowincludingthymus
B 123spleenincludingthymus
C 123thymusexcludingbonemarrow
D 123bonemarrowexcludingspleen

The correct answer is:
123bonemarrowincludingthymus

Solution:
Lymphoid organs are the organs where the origin and/or maturation and proliferation of lymphocytes occur. The primary lymphoid organs are bone marrow and thymus where immature lymphocytes differentiate into antigen-sensitive lymphocytes. The bone marrow is the main lymphoid organ where all blood cells including lymphocytes are produced. The thymus is a lobed organ located near the heart and beneath the breastbone. The thymus is quite large at the time of birth but keeps reducing in size with age and by the time puberty is attained it reduces to a very small size. Both bone-marrow and thymus provide micro-environments for the development and maturation of T-lymphocytes.

Topic: Biology, Human Health and Diseases, NEET
The $3^{\prime}-5^{\prime}$ phosphodiester linkages inside a polynucleotide chain serve to join

Options:
A one DNA strand with the other DNA strand
B one nucleoside with another nucleoside
C one nucleotide with another nucleotide
D one nitrogenous base with pentose sugar

The correct answer is:
one nucleotide with another nucleotide

Solution:
In a polynucleotide chain, $3^{\prime}-5^{\prime}$ phospho diester bond is formed between carbon 3 of one nucleotide and carbon 5 of the other nucleotide, ie, it serves to join one nucleotide with another nucleotide in a polynucleotide chain.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2010 (Mains)
The 4th term of GP is 500 and its common ratio is 1m, m∈N. Let Sn denote the sum of the first n terms of this GP. If S6>S5+1 and S7<S6+12, then the number of possible values of m is ______

Options:

The correct answer is:
12

Solution:
Given:

T4=500

Let a= first term

r=common ratio =1m, m∈N

So,

ar3=500

⇒am3=500

Now,

Sn-Sn-1=arn-1

S6>S5+1

⇒S6-S5>1

⇒am5>1

⇒am3×1m2>1

⇒500m2>1

⇒m2<500 ...1

And,

S7-S6<12

⇒am6<12

⇒am3×2m3<1

⇒500×2m3<1

⇒m3>103

⇒m>10 ....2

From 1 and 2, we have

m=11, 12, 13………….., 22

So, number of possible values of m is 12.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (24 Jan Shift 1)
The 5th and the 31st terms of an arithmetic progression are, respectively 1 and –77. If the Kth term of the given arithmetic progression is –17, then the value of K is

Options:
A 12
B 10
C 11
D 13

The correct answer is:
11

Solution:
The fifth term is 1 . So, a+4d=1
The 31st term is -77 . So, a+30d =-77
⇒26 d =-78⇒d=-3  ⇒a=13
tk=-17⇒ a+k-1d=-17
⇒13+k-1-3=-17
⇒-3k=-33
⇒k=11

Topic: Mathematics, Sequences and Series, JEE Main
The 6 V Zener diode shown in the figure has negligible resistance and a knee current of 5 mA. The minimum value of R in Î© so that the voltage across it does not fall below 6 V is

Options:
A 40
B 60
C 72
D 80
E 120

The correct answer is:
80

Solution:
In Zener diode,
⇒ Vmax-VZR1≥Iknee+IR

R1 is the series resistance of 50 Ω and IR is the current through the resistance R
⇒ 10-650≥5m+IR
⇒ IR≤80 mA-5 mA
∵IRmax=75 mA (the maximum current through R)

So the minimum value of the resistance R is
Rmin=VIRmax=6 V75 mA=600075=80 Ω

Topic: Physics, Semiconductors, JEE Main
The 8th common term of the series
S1=3+7+11+15+19+…

S2=1+6+11+16+21+…. is

Options:

The correct answer is:
151

Solution:
Given,

Two A.P's as

3,7,11,15,...

First terma1=3

Common differenced1=4

And,

1,6,11,16,...,

First terma2=1

Common difference d2=5

Now, common difference of the series of the common terms of the given A.P's is d=LCMd1,d2=LCM4,5=20

Now, series of common term is

11,31,51,71,...

So, 8th common term appearing in the both the series is

a8=11+8-1d

⇒a8=11+7×20=151

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (30 Jan Shift 2)
The 20th term from the end of the progression 20,1914,1812,1734,…,-12914 is :-

Options:
A -118
B -110
C -115
D -100

The correct answer is:
-115

Solution:
Given AP is 20,1914,1812,1734,...,-12914.

Common difference, d1=1914-20=-34

Reversing the AP such that first term becomes the last and last term becomes first.

⇒-12914,......,1914,20

This is also AP.

First term, a=-12914 and common difference, d=34.

⇒a20=-12914+(20-1)34

⇒a20=-5174+574

⇒a20=-4604

⇒a20=-115

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The 24 hour (diurnal) rhythm of our body such as the sleep-wake cycle is regulated by the hormone

Options:
A calcitonin
B prolactin
C adrenaline
D melatonin

The correct answer is:
melatonin

Solution:
Melatonin, also known as chemically N-acetyl 5-methoxytryptamine is a naturally occurring compound found in animals, plants and microbes. In animals, circulating levels of the hormone melatonin vary in a daily cycle, thereby allowing the entraiment of the circadian rythms of several biological functions.

Topic: Biology, Chemical Coordination and Integration, NEET, NEET 2011 (Mains)
The 50th term of the sequence 3+12+25+42+… is

Options:
A 5145
B 5148
C 5142
D 5195

The correct answer is:
5148

Solution:
Since in the sequence the second difference is constant. So,
Tn=an2+bn+c
3=a+b+c
12=4a+2b+c
25=9a+3b+c
⇒9=3a+b and 13=5a+b
⇒4=2a ⇒a=2
⇒b=3
⇒c=-2
⇒Tn=2n2+3n-2
T4=2×16+3×4-2=32+12-2=42
T50=2×2500+3×50-2
=5000+150-2=5148

Topic: Mathematics, Sequences and Series, JEE Main
The 71st electron of an element X with an atomic number of 71 enters the orbital:

Options:
A 5d
B 4f
C 6p
D 6s

The correct answer is:
5d

Solution:
71Lu :Xe 4f146s25d1 (The last electron enters 5d orbital)

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The above is a plot of binding energy per nucleon $E_b$, against the nuclear mass $M ; A, B, C, D, E, F$ correspond to different nuclei. Consider four reactions:
(i) $A+B \rightarrow C+\varepsilon$
(ii) $\mathrm{C} \rightarrow \mathrm{A}+\mathrm{B}+\varepsilon$
(iii) $\mathrm{D}+\mathrm{E} \rightarrow \mathrm{F}+\varepsilon$ and
(iv) $F \rightarrow D+E+\varepsilon$ where $\varepsilon$ is the energy released? In which reactions is $\varepsilon$ positive?

Options:
A
(i) and (iv)

B
(i) and (iii)

C
(ii) and (iv)

D
(ii) and (iii)

The correct answer is:
(i) and (iv)

Solution:
Reaction is fusion and $4^{\text {th }}$ reaction is fission.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2009
The above P-V diagram represents the thermodynamic cycle of an engine, operating with an ideal mono-atomic gas. The amount of heat, extracted from the source in a single cycle, is:

Options:
A 112P0V0
B 4P0V0
C P0V0
D 132P0V0

The correct answer is:
132P0V0

Solution:
The heat is extracted from the source during the process A→B and B→C

∆QAB+∆QBC
=n C v dT+n C p dT =32P0V0+522P0V0
= 1 3 2 P 0 V 0

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2013 (07 Apr)
The above reaction requires which of the following reaction conditions?

Options:
A 573 K, Cu, 300 atm
B 623 K, Cu, 300 atm
C 573 K, 300 atm
D 623 K, 300 atm

The correct answer is:
623 K, 300 atm

Solution:
Temperature =623 K

Pressure =300 atm

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The above splitting of d-orbitals takes place in the formation of complexes. The above splitting of d-orbitals takes place in the formation of complexes.

Options:
A tetrahedral
B square-planar
C octahedral
D both tetrahedral and square-planar

The correct answer is:
octahedral

Solution:
In an octahedral coordination entity with six ligands surrounding the metal atom/ion, there will be repulsion between the electrons in metal d orbitals and the electrons (or negative charges) of the ligands.

Such a repulsion is more when the metal d orbital is directed towards the ligand than when it is away from the ligand. Thus, the dx2-y2 and dz2 orbitals that point towards the axes along the direction of the ligand will experience more repulsion and will be raised in energy; and the dxy, dyz and dxz orbitals which are directed between the axes will be lowered in energy relative to the average energy in the spherical crystal field.

Thus, the given diagram explains d orbital splitting in an octahedral crystal field.

Topic: Chemistry, Coordination Compounds, JEE Main
The abscissa of A and B are the roots of the equation x2+2ax-b2=0 and their ordinates are the roots of the equation y2+2py-q2=0. The equation of the circle with AB as diameter is

Options:
A x2+y2+2ax+2py-b2-q2=0
B x2+y2+2ax+py-b2-q2=0
C x2+y2+2ax+2py+b2+q2=0
D None of these

The correct answer is:
x2+y2+2ax+2py-b2-q2=0

Solution:
Let, A â¡x1, y1 and B â¡x2, y2
x1, x2 are roots of x2+2ax-b2=0
â x-x1 x-x2=x2+2ax-b2
y1, y2 are roots of y2+2py-q2=0
â y-y1 y-y2=y2+2py-q2
Now, equation of circle with AB as diameter is
x-x1 x-x2+y-y1 y-y2=0
â x2+2ax-b2+y2+2py-q2=0
â x2+y2+2ax+2py-b2-q2=0

Topic: Mathematics, Circle, JEE Main
The abscissa(e) of the point(s), where the tangent to curve y=x3-3x2-9x+5 is parallel to the x-axis is(are)

Options:
A x=0
B x=1 and -1
C x=1 and -3
D x= -1 and 3

The correct answer is:
x= -1 and 3

Solution:
y=x3-3x2-9x+5
⇒ dydx=3x2-6x-9
We know that this equation gives the slope of the tangent to the curve. The tangent is parallel to x-axis.
Therefore, 3x2-6x-9=0
⇒ x=-1, 3 .

Topic: Mathematics, Application of Derivatives, JEE Main
The absolute configuration of

is

Options:
A (2R, 3R)
B (2R, 3S)
C (2S, 3R)
D (2S, 3S)

The correct answer is:
(2S, 3R)

Solution:
For 2nd carbon
the priority order a > b > c > d
a=-OH
b=-CH-|Â Â Â ClÂ Â CH3
c=-CO2H
d=-H
For 3rd carbon
the priority order a′>Â b′>Â c′>Â d′
a′=-Cl
b′=-CH-|Â Â Â OHCO2H
c′=-CH3
d′=-H

Topic: Chemistry, General Organic Chemistry, JEE Main
The absolute configuration of
is

Options:
A
$S, S$

B
R, R

C
$R, S$

D
$S, R$

The correct answer is:
R, R

Solution:
$$
\text { Both } C_1 \text { and } C_2 \text { have } R \text { - configuration. }
$$

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2008
The absolute configuration of

is:

Options:
A 2S,Â 3S
B 2R,Â 3R
C 2R,Â 3S
D 2S, 3R

The correct answer is:
2S, 3R

Solution:
at 2 Carbon

Rotation is clockwise, but lowest priority group is on horizontal line so configuration will be S
at 3 Carbon

Rotation is anticlockwise and lowest priority group is on horizontal line so, configuration will be R
∴ Configuration is 2S, 3R.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2016 (03 Apr)
The absolute configuration of the following compound is :

Options:
A 2S, 3R
B 2S, 3S
C 2R, 3S
D 2R, 3R

The correct answer is:
2S, 3S

Solution:
At C2:       At C3:
*4th group on Hz Clockwiseis  *4th group on Hz-line Clockwise
2-S          3-S

Topic: Chemistry, General Organic Chemistry, JEE Main
The absolute configurations of the following compounds respectively, are

Options:
A R and R
B S and S
C R and S
D S and R

The correct answer is:
S and R

Solution:
Using CIP (Cahn-Ingold-Prelog) ranking method, CH2SH>CH2OH>CH3>H is the ranking order.

∴

S - configuration

=R - configuration

Topic: Chemistry, General Organic Chemistry, JEE Main
The absolute difference of the coefficients of x10 and x7 in the expansion of 2x2+12x11 is equal to

Options:
A 133-13
B 113-11
C 103-10
D 123-12

The correct answer is:
123-12

Solution:
The given expansion is 2x2+12x11.

The general term in the binomial expansion of x+an is given by Tr+1=Crnxn-rar.

⇒Tr+1=Cr112x211-r12xr

=Cr11211-r-rx22-2r-r

Now for coefficient of x7⇒22-3r=7

⇒r=5

Coefficient of x7 is C511211-5×2

Now for coefficient of x10⇒22-3r=10

⇒r=4

Coefficient of x10 is C411211-2×4

We need the absolute difference, we get it by C511211-5×2-C411211-4×2.

=924-2640=1716

=123-12

Hence the absolute difference is =123-12.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The absolute enthalphy of neutralisation of the reaction \(M g O_{(s)}+2 H C l_{(a q)} \rightarrow M g C l_{2(a q)}+H_2 O_{(l)}\) will be

Options:
A Less than $-57.33 \mathrm{~kJ} \mathrm{~mol}^{-1}$
B $-57.33 \mathrm{~kJ} \mathrm{~mol}^{-1}$
C Greater than $-57.33 \mathrm{~kJ} \mathrm{~mol}^{-1}$
D $57.33 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
Less than $-57.33 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
Heat of neutralisation will be less than $-57.33 \mathrm{~kJ} /$ mole because some amount of this energy will be required for the dissociation of weak base $(M g O)$

Topic: Chemistry, Thermodynamics (C), JEE Main
The absolute value of the electron gain enthalpy of halogens satisfies:

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

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

Solution:
Order of electron gain enthalpy

(Absolute value)

Cl>F>Br>I

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The absolute zero is the temperature at which

Options:
A Water freezes
B All substances exist in solid state
C Molecular motion ceases
D None of the above

The correct answer is:
Water freezes

Solution:
water freezes

Topic: Physics, Thermal Properties of Matter, JEE Main
The absorbed water can rise to the highest point by

Options:
A root pressure
B imbibition force
C force of capillary
D transpiration pull

The correct answer is:
transpiration pull

Solution:
The transpirational pull is responsible for the transport of water to the highest point, from the roots to the leaves of the tree. This transpirational pull is due to the cohesion of water molecules (attraction between similar molecules), this force maintains a continuous channel for water.

Topic: Biology, Transport in Plants, NEET
The abstraction of proton will be fastest from which carbon in the following compound?

Options:
A x
B y
C z
D p

The correct answer is:
x

Solution:
Extended conjugation
Conjugate base is more stablized. So, abstraction of proton will be fastest at x position.

Topic: Chemistry, General Organic Chemistry, JEE Main
The AC voltage across a resistance can be measured using a:

Options:
A Moving magnet galvanometer
B Moving coil galvanometer
C Hot wire voltmeter
D Potentiometer

The correct answer is:
Hot wire voltmeter

Solution:
The AC voltage can be measured by hot wire voltmeter. It uses the heating effect of the current.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2015 (11 Apr Online)
The acceleration ' $a$ ' (in $\mathrm{ms}^{-2}$ ) of a body, starting from rest varies with time $t$ (in $s$ ) following the equation $a=3 t+4$ The velocity of the body at time $t=2 \mathrm{~s}$ will be

Options:
A $10 \mathrm{~ms}^{-1}$
B $18 \mathrm{~ms}^{-1}$
C $14 \mathrm{~ms}^{-1}$
D $26 \mathrm{~ms}^{-1}$

The correct answer is:
$14 \mathrm{~ms}^{-1}$

Solution:
$a=3 t+4$
$\begin{aligned} & \frac{d V}{d t}=3 t+4 \\ & \int_0^V d V=\int_0^t(3 t+4) d t \\ & V=\frac{3 t^2}{2}+4 t=\frac{12}{2}+8=14 \mathrm{~m} / \mathrm{s}\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main
The acceleration due to gravity at a place is $\pi^2 \mathrm{~m} / \mathrm{sec}^2$. Then the time period of a simple pendulum of length one metre is

Options:
A $\frac{2}{\pi} \sec$
B $2 \pi \sec$
C $2 \mathrm{sec}$
D $\pi \mathrm{sec}$

The correct answer is:
$2 \pi \sec$

Solution:
$T=2 \pi \sqrt{l / g}=2 \pi \sqrt{\frac{1}{\pi^2}}=2 \mathrm{sec}$

Topic: Physics, Oscillations, JEE Main
The acceleration due to gravity at height h above the earth if h≪R (Radius of earth) is given by

Options:
A g'=g1-h22R2
B g'=g1-h2R
C g'=g1-2hR
D g'=g1-2h2R2

The correct answer is:
g'=g1-2hR

Solution:
The acceleration due to gravity g' at a height h from the surface of Earth is given by

g'=GM(R+h)2 ...1

Use the method of binomial expansion to obtain the approximate expression for the acceleration due to gravity at a particular height from the surface of the Earth.

g'=GMR21+hR2= GMR21+hR-2≈GMR21-2hR ...2

The formula to calculate the acceleration due to gravity at the surface of the Earth is given by

g=GMR2 ...(3)

Divide equation (2) by equation (3) and simplify to obtain the required relation.

g'g=GMR21-2hRGMR2= 1-2hR⇒g'=g1-2hR

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The acceleration due to gravity is found up to an accuracy of 4% on a planet. The energy supplied to a simple pendulum of known mass m to undertake oscillations of time period T is being estimated. If time period is measured to an accuracy of 3%, the accuracy to which E is known as ________%

Options:

The correct answer is:
14

Solution:
T=2πlg

Δll×100=Δgg×100+2ΔTT×100

⇒Δll×100=4+2×3=10%

E=12mω2A2=12mglA2Therefore, ΔEE×100=Δgg×100+Δll×100=4+10=14%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2021 (26 Aug Shift 2)
The acceleration due to gravity on the surface of earth is g. If the diameter of earth reduces to half of its original value and mass remains constant, then acceleration due to gravity on the surface of earth would be :

Options:
A g4
B 2g
C g2
D 4g

The correct answer is:
4g

Solution:
The acceleration due to gravity is related to the mass and the radius of the Earth by the formula:

g=GMR2⇒g∝1R2 ...1

Thus, from equation (1), it follows that

g2 g1=R12R22=R12R122=4⇒g2=4g

Topic: Physics, Gravitation, JEE Main, JEE Main 2024 (27 Jan Shift 1)
The acceleration experienced by a moving boat after is engine is cut-off, is given by a=-kv3, where k is a constant. If v0 is the magnitude of velocity at cut-off, then the magnitude of the velocity at time t after the cut off is

Options:
A v02ktv02
B v01+2 ktv02
C v01-2 k v02
D v01+2 ktv02

The correct answer is:
v01+2 ktv02

Solution:
Given, acceleration a=-kv3
Initial velocity at cut-off, v1=v0
Initial time of cut-off, t=0
And final time after cut-off, t2=t
Again, a=dvdt=-kv3
Or dvv3=-kdt
Integrating both sides, with in the condition of motion.
∫v0vdvv3=-∫0tk dt
Or -12v2v0v=-kt0t
Or 12v2-12v02=kt
Or v=v01+2 kt v02

Topic: Physics, Motion In One Dimension, JEE Main
The acceleration of an electron in the first orbit of the hydrogen atom (n=1) is :

Options:
A h24πm2r3
B h2π2m2r3
C h28π2m2r3
D h24π2m2r3

The correct answer is:
h24π2m2r3

Solution:
For H-atom n=1
mv2r=14πϵoe2r2 ...1
mvr=h2π ...2
On squaring (2)
m2v2r2=h24π2
On dividing both sides by r3 ,
m2v2r=h24πr3 ⇒v2r=h24π2r3m2
Acceleration= h24π2r3m2.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2017 (09 Apr Online)
The acceleration of electron in the first orbit of a hydrogen atom is (where the symbols have their usual meanings)

Options:
A 4π2mh3
B h24π2mr
C h24π2m2r3
D m2h24π2r3

The correct answer is:
h24π2m2r3

Solution:
From mvr=nh2π,v=nh2πmr

For the first Bohr orbit, n=1
Acceleration, a=v2r=n2h24π2m2r2(r)=h24π2m2r3

Topic: Physics, Atomic Physics, JEE Main
The acceleration vs distance graph for a particle moving with initial velocity $5 \mathrm{~m} / \mathrm{s}$ is shown in the fiture. The velocity of the particle at $x=35 \mathrm{~m}$ will be

Options:
A $20.62 \mathrm{~m} / \mathrm{s}$
B $20 \mathrm{~m} / \mathrm{s}$
C $25 \mathrm{~m} / \mathrm{s}$
D $50 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$25 \mathrm{~m} / \mathrm{s}$

Solution:
Area of graph
$$
\begin{array}{l}
=\int \mathrm{adx}=\int \frac{\mathrm{vdv}}{\mathrm{dx}} \cdot \mathrm{d} \mathrm{x}==\int_{\mathrm{V}_{\mathrm{i}}}^{\mathrm{v}_{\mathrm{f}}} \mathrm{v} \mathrm{d} \mathrm{v} \\
=\left[\frac{\mathrm{v}^{2}}{2}\right]_{\mathrm{v}_{\mathrm{i}}}^{\mathrm{v}_{\mathrm{f}}}, 150+150=\frac{1}{2}\left[\mathrm{~V}_{\mathrm{f}}^{2}-\mathrm{V}_{\mathrm{i}}^{2}\right]
\end{array}
$$
$$
\begin{array}{l}
300 \times 2=V_{f}^{2}-(5)^{2} \\
V_{f}^{2}=600+25, V_{f}=25
\end{array}
$$

Topic: Physics, Motion In One Dimension, JEE Main
The acceptor level of a p-type semiconductor is $6 \mathrm{eV}$. The maximum wavelength of light which can create a hole would be : Given $\mathrm{hc}=1242 \mathrm{eV} \mathrm{nm}$.

Options:
A $414 \mathrm{~nm}$
B $103.5 \mathrm{~nm}$
C $207 \mathrm{~nm}$
D $407 \mathrm{~nm}$

The correct answer is:
$207 \mathrm{~nm}$

Solution:
$\begin{aligned} & \text { Energy }=\frac{\mathrm{hc}}{\lambda} \\ & \mathrm{E}=\frac{1240}{\lambda(\mathrm{nm})} \mathrm{eV} \\ & 6=\frac{1240}{\lambda(\mathrm{nm})} \\ & \lambda=\frac{1240}{6}=207 \mathrm{~nm}\end{aligned}$

Topic: Physics, Semiconductors, JEE Main, JEE Main 2024 (06 Apr Shift 2)