datavorous/entrance-exam-dataset huggingface dataset:
The hydrolysis of optically active \( 2 \)-bromobutane with aqueous \( \mathrm{NaOH} \) result in the formation of

Options:
A \( (+) \) butan-2-ol
B (-) butan-2-ol
C ( \( \pm \) ) butan-1-ol
D ( \( \pm \) ) butan-2-ol

The correct answer is:
( \( \pm \) ) butan-2-ol

Solution:
2-Bromobutane is a secondary alkyl halide, therefore, hydrolysis will proceed through \( S_{N} 1 \) mechanisms, hence, racemic
mixture will be formed, that is, (±) butan-2-ol.
(+) Butan-2-ol
(-) Butan-2-ol

Topic: Chemistry, Hydrocarbons, JEE Main
The hyperconjugative stabilities of tertiary − butyl carbocation and but-2−ene, respectively, are due to

Options:
A σ→p (empty) and σ→π* electron delocalisations.
B σ→σ* and σ→π electron delocalisations.
C σ→p (filled) and σ→π electron delocalisations.
D p (filled) →σ* and σ→π* electron delocalisations.

The correct answer is:
σ→p (empty) and σ→π* electron delocalisations.

Solution:
Hyperconjugation involves delocalisation of σ electron of α C-H with π empty p-orbital. It increases stability of carbocation, free radical and alkenes by decreasing energy.

CH3-CH=CH2σ-π* electron delocalization

Topic: Chemistry, General Organic Chemistry, JEE Main
The IUPAC name of the following compound is:

Options:
A 4-Methyl-3-hydroxypntanoic acid
B 4,4-Dimethyl-3-hydroxybutanoic acid
C 2-Methyl-3-hydroxypentan-5-oic acid
D 3-Hydroxy-4-methylpentanoic acid

The correct answer is:
3-Hydroxy-4-methylpentanoic acid

Solution:
3-Hydroxy-4-methyl pentanoic acid; -COOH group is principal functional group.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The $I$ - $V$ characteristics of a metal wire at two different temperatures $\left(T_{1}\right.$ and $T_{2}$ ) are given in the adjoining figure. Here, we can conclude that

Options:
A $T_{1}>T_{2}$
B $T_{1} < T_{2}$
C $T_{1}=T_{2}$
D $T_{1}=2 T_{2}$

The correct answer is:
$T_{1} < T_{2}$

Solution:
$R$ increases with temperature and slope of $V \cdot I$ graph gives resistance so $T_{1} < T_{2}$

Topic: Physics, Current Electricity, JEE Main
The $I-V$ characteristics of an electronic device shown in the figure. The device is:

Options:
A a diode which can be used as a rectifier
B a zener diode which can be used as a voltage regulator
C a transistor which can be used as an amplifier
D a solar cell

The correct answer is:
a zener diode which can be used as a voltage regulator

Solution:
Theory
Zener diode used as voltage regulator

Topic: Physics, Semiconductors, JEE Main, JEE Main 2024 (09 Apr Shift 2)
The I - V characteristics of an LED is:

Options:
A
B
C
D

The correct answer is:

Solution:
For same value of current higher value of voltage is required for higher frequency hence,
VTh∝ΔEg∝1λ
LED works in forward Bias. This graph is correct.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2013 (07 Apr)
The _________(i)________ vary in colour from olive green to various shades of brown depending upon the amount _________(ii)_______

Options:
A (i) red algae (ii) phycobilins
B (i) green algae (ii) chlorophyll
C (i) brown algae (ii) chlorophyll
D (i) brown algae (ii) carotenoids

The correct answer is:
(i) brown algae (ii) carotenoids

Solution:
The brown algae mainly present in seawater/marine water having chlorophyll a, c and fucoxanthin. Brown shades of brown algae depend upon the amount of xanthophyll (fucoxanthin) which is a type of carotenoid.

Topic: Biology, Plant Kingdom, NEET
The idea of froth flotation method came from a person X and this method is related to the process Y of ores. X and Y, respectivity, are:

Options:
A Washer man and reduction
B Fisher women and concentration
C Fisher man and reduction
D Washer woman and concentration

The correct answer is:
Washer woman and concentration

Solution:
Froth floatation is discovered by washer woman and it is a process of concentration of ores and in this process gangue is remove from sulphide ore.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The idea of matter waves was given by

Options:
A Davisson and Germer
B de-Broglie
C Einstein
D Planck

The correct answer is:
de-Broglie

Solution:
Electromagnetic waves behave like particles when they interact with matter. In 1924, Louis de Broglie proposed that electron and other particles can behave like waves called matter waves.

The wavelength of matter waves is λ=hp, called de Broglie wavelength, where p is the momentum of the particle.

Topic: Physics, Dual Nature of Matter, JEE Main
The idea of mutations was brought forth by

Options:
A Hugo de Vries, who worked on evening primrose
B Gregor Mendel, who worked on Pisum sativum
C Hardy Weinberg, who worked on allele frequencies in a population
D Charles Darwin, who observed a wide variety of organisms during sea voyage

The correct answer is:
Hugo de Vries, who worked on evening primrose

Solution:
Hugo de Vries worked on evening primrose (Oenothera lamarckiana) and put forward the idea of mutation.
Gregor Mendel worked on Pisum sativum and listed principles of genetics.
Charles Darwin observed a wide variety of organisms and proposed theory of natural selection.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2012 (Mains)
The idea of secondary wavelets for the propagation of a wave was first given by

Options:
A Newton
B Huygen
C Maxwell
D Fresnel

The correct answer is:
Huygen

Solution:
The idea of secondary wavelets is given by Huygen.

Topic: Physics, Wave Optics, JEE Main
The ideal gas equation for 1 mol of ideal gas is PV = nRT
The graph between P and T at constant volume, i.e. isochores are plotted as under

Which of the following order of volume is correct?

Options:
A V4>V3>V2>V1
B V1>V2>V3>V4
C V1=V2=V3=V4
D V2>V1>V3>V4

The correct answer is:
V1>V2>V3>V4

Solution:
At constant temperature T, V∝1P (for 1 mol gas).
Thus, correct sequence of volume will be
V1>V2>V3>V4

Topic: Chemistry, States of Matter, JEE Main
The identical spheres each of mass 2M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 4 m each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is 42x, where the value of x is ________.

Options:

The correct answer is:
3

Solution:
Position vector r→COM=m1r→1+m2r→2+m3r→3 m1+m2+m3

r→COM =2M×0+2M×4i^+2M×4j^6M

⇒r→=43i^+43j^

|r→|=432+432=423

Therefore, x=3.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2024 (01 Feb Shift 1)
The image formed by an objective of a compound microscope is

Options:
A Virtual and enlarged
B Virtual and diminished
C Real and diminished
D Real and enlarged

The correct answer is:
Real and enlarged

Solution:
Intermediate image means the image formed by objective, which is real, inverted and enlarged.

Topic: Physics, Ray Optics, JEE Main
The image formed by an objective of a compound microscope is

Options:
A
virtual and diminished

B
real and diminished

C
real and enlarged

D
virtual and enlarged

The correct answer is:
real and enlarged

Solution:
No Solution Available

Topic: Physics, Ray Optics, JEE Main, JEE Main 2003
The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 4 m away by means of a large convex lens. The maximum possible focal length of the lens required for this purpose will be

Options:
A 0.5 m
B 1.0 m
C 1.5 m
D 2.0 m

The correct answer is:
1.0 m

Solution:
Let a large convex lens is placed between two walls at a distance x from the wall on which an electric bulb is fixed.
Using
1f=1v-1u
=14-x-1-x ∵u=-x, v=4-x
=x+4-x4-xx
=4(4-x)(x)
or f=4-xx4 ……i
Now magnification m=vu=4-xx
⟹ 1=4-xx
⟹x=4-x
⟹2x=4
⟹x=2 m
From eq. (i)
f=(4-2)(2)4=2×24=1 m

Topic: Physics, Ray Optics, JEE Main
The image of an illuminated square is obtained on a screen with the help of a converging lens. The distance of the square from the lens is 40 $\mathrm{cm}$. The area of the image is 9 times that of the square. The focal length of the lens is :

Options:
A
$36 \mathrm{~cm}$

B
$27 \mathrm{~cm}$

C
$60 \mathrm{~cm}$

D
$30 \mathrm{~cm}$

The correct answer is:
$30 \mathrm{~cm}$

Solution:
If side of object square $=\ell$ and side of image square $=\ell^{\prime}$
From question, $\frac{\ell^{\prime 2}}{\ell}=9$
or $\frac{\ell^{\prime}}{\ell}=3$
i.e., magnification $\mathrm{m}=3$
$\mathrm{u}=-40 \mathrm{~cm}$
$\mathrm{v}=3 \times 40=120 \mathrm{~cm}$
$$
\mathrm{f}=?
$$
From formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$$
\begin{aligned}
& \frac{1}{120}-\frac{1}{-40}=\frac{1}{\mathrm{f}} \\
& \text { or, } \frac{1}{\mathrm{f}}=\frac{1}{120}+\frac{1}{40}=\frac{1+3}{120} \therefore \mathrm{f}=30 \mathrm{~cm}
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2013 (22 Apr Online)
The image of \( P(a, b) \) in the line \( y=-x \) is \( Q \) and the image of \( Q \) in the \( y=x \) is \( R \), then the midpoint of \( P R \) is

Options:
A \( (a+b, b+a) \)
B \( \left(\frac{a+b}{2}, \frac{b+2}{2}\right) \)
C \( (a-b, b-a) \)
D \( (0,0) \)

The correct answer is:
\( (0,0) \)

Solution:
The image of Pa,b in the line y=-x is Q.

So, point Q is -b,-a.

Also, the image of Q in the line y=x is R.

So, point R is -a,-b.

By mid-point formula,

Mid-point of PR is a-a2,b-b2=0,0.

Topic: Mathematics, Straight Lines, JEE Main
The image of the line x-13=y-31=z-4-5 in the plane 2x-y+z+3=0 is the line

Options:
A x-33=y+51=z-2-5
B x-3-3=y+5-1=z-25
C x+33=y-51=z-2-5
D x+3-3=y-5-1=z+25

The correct answer is:
x+33=y-51=z-2-5

Solution:
We first find the intersection of line (L1) with the plane. Any general point on line can be written as 3k+1,k+3,-5k+4.

Substituting in equation of plane P we get

2(3k+1)-(k+3)+(-5k+4)+3=0
Which gives 6=0 which is not possible.

So, we conclude that the line is parallel to the plane.∵6-1-5=0

∴ The reflected line (L2) will also have the same direction ratios. i.e., (3,1,- 5).
Consider the normal perpendicular to the plane and line.
Its direction ratios are (2,-1,1)

∴ Equation of normal is

x-12=y-3-1=z-41

∴ Any general point can be written as (2r+1,-r+3,r+4) substituting in the equation of a plane,

2(2r+1)-(-r+3)+r +4+3=0

4r+2+ r-3+r+4+3 = 0

⇒6r+6=0

r=-1

â

∴ The point B is (-1,4,3)

B is the midpoint of segment AC.

C≡-3,5,2

∴ Equation of line L2 is x + 3 3 = y - 5 1 = z - 2 - 5

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2014 (06 Apr)
The image of the line x-22=y-3-3=z-41 in the plane x+y+z=6 can be

Options:
A r→=λ2i^-3j^+k^
B r→=2i^+j^+k^+α2i^-3j^+k^
C r→=j^+2k^+β2i^-3j^+k^
D r→=i^+2j^+u2i^-3j^+k^

The correct answer is:
r→=j^+2k^+β2i^-3j^+k^

Solution:
Note that the line is parallel to the plane, hence the image of the line will be a parallel line.

Now, let the image of 2,3,4 in x+y+z=6 is x1,y1,z1

Then, x1-21=y1-31=z1-41=-22+3+4-61+1+1

i.e. x1,y1,z1≡0,1,2
So the required equation is r→=j^+2k^+β2i^-3j^+k^

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The image of the line x2=y-15=z+13 in the plane x+y+2z=3 meets the xz-plane at the point a,b,c, then the value of c is equal to

Options:
A 116
B 1296
C 1156
D 2323

The correct answer is:
1296

Solution:
Any general point on the line is 2λ,5λ+1,3λ-1
On satisfying this point on the plane, we get,
2λ+5λ+1+6λ-2=3
13λ=4⇒λ=413
So, coordinates of the point are 813,3313,-113
This point also lies on the image of the line
Image of point 0,1,-1 also lies on the image of the line
x-01=y-11=z+12=-2-46
x=43,y=73,z=53
Point is 43,73,53
Equation of image of the line is x-4328=y-73-8=z-5368
For xz-plane, putting y=0, we get,

x-4328=724=z-5368
⇒z=1296

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The image of the point (−1, 3, 4) in the plane x − 2y = 0 is

Options:
A
$\left(-\frac{17}{3},-\frac{19}{3}, 4\right)$

B
$(15,11,4)$

C
$\left(-\frac{17}{3},-\frac{19}{3}, 1\right)$

D
none of these

The correct answer is:
none of these

Solution:

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2006
The image of the point $(3,2,1)$ in the plane $2 x-y+3 z=7$ is

Options:
A $(1,2,3)$
B $(2,3,1)$
C $(3,2,1)$
D $(2,1,3)$

The correct answer is:
$(3,2,1)$

Solution:
We know that image $(x, y, z)$ of a point $\left(x_1, y_1, z_1\right)$ in a plane $a x+b y+c z+d=0$ is
$\begin{aligned}
\frac{x-x_1}{a} & =\frac{y-y_1}{b}=\frac{z-z_1}{c} \\
& =\frac{-2\left(a x_1+b y_1+c z_1+d\right)}{a^2+b^2+c^2}
\end{aligned}$
Here, point is $(3,2,1)$ and plane is
$\begin{array}{rlrl}
& 2 x-y+3 z=7 \\
& \therefore & \frac{x-3}{2} & =\frac{y-2}{-1}=\frac{z-1}{3} \\
& =\frac{-2[2(3)-(2)+3(1)-7]}{2^2+1^2+3^2} \\
\Rightarrow & \frac{x-3}{2} & =\frac{y-2}{-1}=\frac{z-1}{3}=-2(0) \\
\Rightarrow & & x & =3, y=2, z=1
\end{array}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The image of the point 3,5 in the line x-y+1=0, lies on :

Options:
A x-22+y-42=4
B x-42+y-42=8
C x-42+y+22=16
D x-22+y-22=12

The correct answer is:
x-22+y-42=4

Solution:
The slope of PP'=-1 ( As, PP' is perpendicular to the given line)

⇒k-5h-3=-1 ⇒k+h=8 ...1

Mid-point of PP' lies on the given line,

⇒h+32-k+52+1=0 ⇒ h+3-k-5+2=0 ⇒ h-k=0 ...2Solving equation 1 and 2 we get, h=4, k=4

Which satisfy the equation x-22+y-42=4.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2021 (25 Feb Shift 1)
The imaginary part of $\frac{(1+i)^2}{(2-i)}$ is

Options:
A $\frac{1}{5}$
B $\frac{3}{5}$
C $\frac{4}{5}$
D None of these

The correct answer is:
$\frac{4}{5}$

Solution:
We have $\frac{(1+i)^2}{2-i}=\frac{(2 i)(2+i)}{(2-i)(2+i)}=-\frac{2}{5}+i \frac{4}{5}$
Thus $(z)=\frac{4}{5}$

Topic: Mathematics, Complex Number, JEE Main
The imaginary part of 3+2-5412-3-2-5412,can be

Options:
A -6
B -26
C 6
D 6

The correct answer is:
-26

Solution:
|3+2-54|=9+216=15

3+2-541/2=±15+32+i15-32

=±3+i6

Also 3-2-541/2=±3-i6

Hence 3+2-541/2-3-2-541/2

=±26i or ±6

Hence, imaginary part =±26

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2020 (02 Sep Shift 2)
The immobile element in plants is

Options:
A calcium
B nitrogen
C phosphorus
D magnesium

The correct answer is:
calcium

Solution:
Nutrient mobility facilitates the identification of deficiencies in plant nutrients. Mobile nutrients such as nitrogen, potassium, magnesium, etc. can move from one part of the plant to the other part as per the requirement. While the immobile nutrients remain in their original place. They do not move to other parts that require them. Such nutrients include calcium, chlorine, copper, silicon etc.

Topic: Biology, Mineral Nutrition, NEET
The impact of immigration on population density is:

Options:
A Negative
B Both positive and negative
C Neutralized by natality
D Positive

The correct answer is:
Positive

Solution:
The population density of the population is mainly affected by four factors, i.e. natality, mortality, immigration and emigration.

Natality and immigration contribute to an increase in population density, while mortality and emigration contribute to decrease in population density.

Thus, the impact of natality and immigration is positive, while, mortality and emigration is negative.

The formula for calculating population density is: Nt=No[(B+I)-(D+E)]

Topic: Biology, Organisms and Populations, NEET, NEET 2020 (Phase 2)
The impedance of a circuit consists of $3 \mathrm{ohm}$ resistance and $4 \mathrm{ohm}$ reactance. The power factor of the circuit is

Options:
A 0.4
B 0.6
C 0.8
D 1.0

The correct answer is:
0.6

Solution:
$\begin{aligned} & Z=\sqrt{R^2+X^2}=\sqrt{4^2+3^2}=5 \\ & \therefore \cos \phi=\frac{R}{Z}=\frac{3}{5}=0.6\end{aligned}$

Topic: Physics, Alternating Current, JEE Main
The impedance of a circuit consists of $3 \Omega$ resistance and $4 \Omega$ reactance. The power factor of the circuit is

Options:
A 0.4
B 0.6
C 0.8
D 1

The correct answer is:
0.6

Solution:
$\tan \phi=\frac{\mathrm{X}}{\mathrm{R}} \times \frac{4}{3}$
Power factor $=\cos \phi=\frac{3}{5}=0.6$

Topic: Physics, Alternating Current, JEE Main
The important advantage(s) of Lintz and Donawitz (L.D.) process for the manufacture of steel is (are)

Options:
A the process is very quick
B operating costs are low
C better quality steel is obtained
D scrap iron can be used

The correct answers are:
the process is very quick, better quality steel is obtained, scrap iron can be used

Solution:
Lintz and Donawitz (L.D.) process is an important process for manufacturing of steel. In India, Rourkela steel plant is based upon this process.
This process is carried out in a converter similar to bessemer converter having lining (usually refractory made of $\mathrm{Mg} \mathrm{O}$ and $(\mathrm{CaO})$.
In this method scrap iron can be used along with flux in converter : On passing oxygen gas at $4-12$ atm pressure into converter, temperature of $2000-2500^{\circ} \mathrm{C}$ range produced which separates C. Si, Mn etc., impurities as slag from
iron quickly remaining pure iron is mixed with spiegeleisen (alloy Fe, $\mathrm{Mn}$, C) to produce better quality steel.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, WBJEE, WBJEE 2013
The important step in the extraction of metal from carbonate ore is

Options:
A Calcination
B Roasting
C Electro-reduction
D Cupellation

The correct answer is:
Calcination

Solution:
The important step in the extraction of metal from carbonate ore is calcination.
ZnCO3→ZnO+CO2

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
The impulse due to a force on a body is given by $\mathrm{I}=\int \mathrm{Fdt}$. If the force applied on a body is given as a function of time (t) as $\mathrm{F}=\left(3 \mathrm{t}^2+2 \mathrm{t}+5\right) \mathrm{N}$, then impulse on the body between $\mathrm{t}=3 \mathrm{sec}$ to $\mathrm{t}=5 \mathrm{sec}$ is:

Options:
A $175 \mathrm{~kg}-\mathrm{m} / \mathrm{sec}$
B $41 \mathrm{~kg}-\mathrm{m} / \mathrm{sec}$
C $216 \mathrm{~kg}-\mathrm{m} / \mathrm{sec}$
D $124 \mathrm{~kg}-\mathrm{m} / \mathrm{sec}$

The correct answer is:
$124 \mathrm{~kg}-\mathrm{m} / \mathrm{sec}$

Solution:
To find the impulse (I) on the bedy between $(\mathrm{t}=3)$ sec and $(\mathrm{t}=5)$ sece we need to calculate the integral of the force function ( $F(t)=3 t^2-2+2 t+$ 5) with respect to time over the glven interval.
Impulse (1) is given by:
$I=\int_8^5 F(t) d t$
Substitute ( $\mathbf{F}(t))$ :
$I=\int_3^5\left(3 t^2+2 t+5\right) d t$
We integrate each term separately:
$I=\int_3^5 3 t^2 d t+\int_3^5 2 t d t+\int_3^5 5 d t$
First, integrate $\left(3 t^2 2\right)$
$\int 3 t^2 d t=3 \cdot \frac{t_3}{3}=t^3$
Second, integrate ( $2 t$ ).
$\int 2 t d t=2 \cdot f_2^t=t^2$
Third, integrate ( 5 ):
$\int 5 d t=5 t$
Now, evaluate each integral from (3) to $(5)=$
For $\left(t^* 3\right)$
$\left[t^3\right]_3^5-5^3-3^3=125-27=98$
For $\left(t^* 2\right)=$
$\left[t^2\right]_3^5=5^2-3^2=25-9=16$
For $(5 t)$ :
$\begin{aligned} & (5 t]^5-5 \cdot 5-5 \cdot 3-25-15= \\ & 10\end{aligned}$
Sum these results:
$I=98+16+10=124$
So, the impulse on the body between $(t=3)$ sec and $(t=5)$ sec is ( 124 (kg-m/sec)).
Therefore, the correct answer is:
$124 \mathrm{~kg}-\mathrm{m} / \mathrm{sec}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
The inactivation rate of a viral preparation is proportional to the amount of virus. In the first minute after preparation, 10% of the virus is inactivated. The rate constant for viral inactivation is ___×10-3 min-1. (Nearest integer) [Use : ln10=2.303;log103=0.477 property of logarithm:logxy=ylogx

Options:

The correct answer is:
106

Solution:
As the unit of rate constant is min-1 so it must be a first order reaction
k×t=2.303logA0/At
in 1 min 10% is inactivated so taking
A0=100 At=90 in 1 min
So k×1=2.303×log10090

=2.303×(log10-2log3)
=2.303×(1-2×0.477)
=0.10593

=105.93×10-3
Answer is 106

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The incentre of the triangle with vertices $\mathrm{A}(1, \sqrt{3}), \mathrm{B}(0,0)$
and $\mathrm{C}(2,0)$ is

Options:
A $\left(1, \frac{\sqrt{3}}{2}\right)$
B $\left(\frac{2}{3}, \frac{1}{\sqrt{3}}\right)$
C $\left(\frac{2}{3}, \frac{\sqrt{3}}{2}\right)$
D $\left(1, \frac{1}{\sqrt{3}}\right)$

The correct answer is:
$\left(1, \frac{1}{\sqrt{3}}\right)$

Solution:
Vertices of triangle, $\mathrm{A}(1, \sqrt{3}), \mathrm{B}(0,0), \mathrm{C}(2,0)$
observe the figure, $\mathrm{BD}=1, \mathrm{DC}=1, \mathrm{AD}=\sqrt{3}$
So, $A B=2 . A C=2$ (Using Pythagoras theorem) So, given triangle is equilateral triangle In equilateral triangle, Incentre is controid only. $\therefore$ Incentre $=\left(\frac{1+0+2}{3}, \frac{\sqrt{3}+0+0}{3}\right)$
$=\left(\frac{3}{3}, \frac{\sqrt{3}}{3}\right)$
$=\left(1, \frac{1}{\sqrt{3}}\right)$

Topic: Mathematics, Straight Lines, JEE Main
The incorrect expression among the following is :

Options:
A For isothermal process wreversible =-nRT lnVf Vi
B lnK=ΔH°-TΔS°RT
C ΔGSystem ΔSTotal =-T (at constant P)
D K=e-ΔG°/RT

The correct answer is:
lnK=ΔH°-TΔS°RT

Solution:
ΔG=ΔG°+RT lnQ at equilibrium Q=Keq and ΔG=0

ΔG°=-RT lnKeq and ΔG°=ΔH°-TΔS°

ΔH°-TΔS°=-RTlnKeq So, lnKeq=-ΔH°-TΔS°RT=-ΔH°+TΔS°RT

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2021 (31 Aug Shift 2)
The incorrect expression among the following is :

Options:
A
$\frac{\Delta \mathrm{G}_{\text {system }}}{\Delta \mathrm{S}_{\text {total }}}=-T$

B
In isothermal process $w_{\text {reversible }}=-n R T \ln \frac{V_f}{V_i}$

C
$\ln \mathrm{K}=\frac{\Delta \mathrm{H}^0-\mathrm{T} \Delta \mathrm{S}^0}{\mathrm{RT}}$

D
$\mathrm{K}=\mathrm{e}^{-\Delta \mathrm{G}^0 / \mathrm{RT}}$

The correct answer is:
$\ln \mathrm{K}=\frac{\Delta \mathrm{H}^0-\mathrm{T} \Delta \mathrm{S}^0}{\mathrm{RT}}$

Solution:
$\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{K}$ and $\Delta \mathrm{G}^0=\Delta \mathrm{H}^0-\mathrm{T} \Delta \mathrm{S}^0$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2012 (Offline)
The incorrect geometry is represented by:

Options:
A NF3 - trigonal planar
B BF3 - trigonal planar
C AsF5 - trigonal bipyramidal
D H2O - bent

The correct answer is:
NF3 - trigonal planar

Solution:
NF3
sp3 , Trigonal Pyramidal

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2018 (16 Apr Online)
The incorrect match in the following is:

Options:
A ΔG0<0,K>1
B ΔG0>0,K<1
C ΔG0=0,K=1
D ΔG0<0,K<1

The correct answer is:
ΔG0<0,K<1

Solution:
We know that ΔG°=-RTlnK

ΔG°=-2.303 RTlog⁡K

so for k=1 , log⁡K=0

hence ΔG°=0 ;

for k>1 , log⁡K>0

ΔG°=-ve

for k<1 , log⁡K<0

ΔG°>0

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2019 (12 Apr Shift 2)
The incorrect order of first ionisation energy is:

Options:
A Au > Cu > Ag
B Pt > Ni > Pd
C C > Pb > Sn
D B > Ga > Al

The correct answer is:
Pt > Ni > Pd

Solution:
First ionization energy of Pd > Nidue to its d 10 configuration.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The incorrect order of lattice energy is

Options:
A AlF3>MgF2
B Li3N>Li2O
C NaCl > LiF
D TiC > ScN

The correct answer is:
NaCl > LiF

Solution:
The incorrect order of lattice energy is NaCl > LiF as lattice energy of LiF is more than NaF
LE∝q+.q-r++r-
A. AlF3>MgF2 [Charge on cation]
B. Li3N>Li2O [Charge on anion]
C. NaCl<LiF [Size of cation and anion]
D. TiC > ScN [Charges on cation and anion].

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The incorrect postulates of the Dalton's atomic theory are :
(A) Atoms of different elements differ in mass.
(B) Matter consists of divisible atoms.
(C) Compounds are formed when atoms of different element combine in a fixed ratio.
(D) All the atoms of given element have different properties including mass.
(E) Chemical reactions involve reorganisation of atoms.
Choose the correct answer from the options given below :

Options:
A (C), (D), (E) only
B (B), (D) only
C (A), (B), (D) only
D (B), (D), (E) only

The correct answer is:
(B), (D) only

Solution:
$B, D$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2024 (05 Apr Shift 1)