datavorous/entrance-exam-dataset huggingface dataset:
The most stable ion is

Options:
A
$\left[\mathrm{Fe}(\mathrm{OH})_3\right]^{3-}$

B
$\left[\mathrm{Fe}(\mathrm{Cl})_6\right]^{3-}$

C
$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$

D
$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$

The correct answer is:
$\left[\mathrm{Fe}(\mathrm{Cl})_6\right]^{3-}$

Solution:
A more basic ligand forms stable bond with metal ion, $\mathrm{Cl}^{-}$ is most basic amongst all

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2002
The most stable ion is

Options:
A $\left[\mathrm{Fe}(\mathrm{OH})_3\right]^{3-}$
B $\left[\mathrm{Fe}(\mathrm{Cl})_6\right]^{3-}$
C $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$
D $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$

The correct answer is:
$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$

Solution:
$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$
The oxidation state of the $\mathrm{CN}$ ion is +3 in $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$. There the $\mathrm{CN}$ ion is the most stable ion.

Topic: Chemistry, Coordination Compounds, JEE Main
The most stable ion is

Options:
A \( \left[\mathrm{Fe}(\mathrm{OH})_{5}\right]^{3-} \)
B \( \left[\mathrm{FeCl}_{6}\right]^{3-} \)
C \( \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} \)
D \( \left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} \)

The correct answer is:
\( \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} \)

Solution:
CN-and OH- are strong nucleophiles.
FeOH53- is not formed.

Topic: Chemistry, Coordination Compounds, JEE Main
The most stable radical among the following is:

Options:
A

B
C
D

The correct answer is:

Solution:
Free radicals are electrophiles, hence electron donating groups stabilize the free radical. (4) has electron donating group (+ I), so it is stable

Topic: Chemistry, General Organic Chemistry, JEE Main
The most stable trihalide of nitrogen is.

Options:
A NF3
B NCl3
C NBr3
D NI3

The correct answer is:
NF3

Solution:
The size of halogen increases from F to I. The N-F bond is stronger than other N-X bonds due to the high electronegativity of fluorine. Larger the size of the halogen, less is the stability of nitrogen halide:

The stability order is NF3>NCl3>NBr3>NI3.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2022 (24 Jun Shift 1)
The most suitable method of separation of 1:1 mixture of ortho and para-nitrophenol is:

Options:
A Sublimation
B Chromatography
C Crystallisation
D Steam distillation

The correct answer is:
Steam distillation

Solution:
The ortho and para isomers can be separated by steam distillation. o-nitrophenol is more volatile due to intramolecular hydrogen bonding.

Topic: Chemistry, General Organic Chemistry, JEE Main
The most suitable method of the separation of a $1: 1$ mixture of ortho and para nitrophenols is

Options:
A Distillation
B Sublimation
C Crystallization
D Chromatography

The correct answer is:
Distillation

Solution:
$o-$ and $p$-nitrophenols are separated by steam distillation because o-nitrophenol is steam volatile while $p$-isomer is not.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The most suitable reagent for the conversion of R - CH 2 - OH → R - CHO is :

Options:
A KMnO4
B K2Cr2O7
C CrO3
D PCC (Pyridinium Chlorochromate)

The correct answer is:
PCC (Pyridinium Chlorochromate)

Solution:
• KMnO4 will oxidize the alcohol to carboxylic acid.
• K2Cr2O7 will also make carboxylic acids.
• âCrO3 in acetic acid is Jones reagent, it also converts alcohols to carboxylic acid.
•â PCC is a weak oxidising agent. It limits the oxidation to the aldehyde.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2014 (06 Apr)
The most suitable reagent for the given conversion is:

Options:
A B2H6
B NaBH4
C LiAlH4
D H2/Pd

The correct answer is:
B2H6

Solution:
B2H6 is very selective and usually used to reduce acid into alcohol.

Borane is commonly used for the reduction of carboxylic acids in the presence of esters, lactones, amides, halides and other functional groups. In addition, borane rapidly reduces aldehydes, ketones, and alkenes.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The most suitable reagent for the given conversion is:

Options:
A B2H6
B NaBH4
C LiAlH4
D H2/Pd

The correct answer is:
B2H6

Solution:
B2H6 is very selective and usually used to reduce acid into alcohol.

Borane is commonly used for the reduction of carboxylic acids in the presence of esters, lactones, amides, halides and other functional groups. In addition, borane rapidly reduces aldehydes, ketones, and alkenes.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The most susceptible compound to nucleophilic attack at the carbonyl carbon among the given bellow group is

Options:
A MeCOCl
B MeCHO
C MeCOOMe
D MeCOOCOMe

The correct answer is:
MeCOCl

Solution:
Amongst aldehyde & the acid derivatives, acid chloride are the most susceptible to nucleophilic attack due to strong -I effect & weak +R effect of the Cl- atom as a result of which carbonyl carbon has the highest electron deficiency. The actual order is
MeCOCl>MeCOOCOMe >MeCOOMe >MeCHO .

Topic: Chemistry, General Organic Chemistry, JEE Main
The motion of a body falling from rest in a resistive medium is described by the equation dvdt=a-bv, where a and b are constants. The velocity at any time t is

Options:
A a(1-b2t)
B ab(1-e-bt)
C abe-t
D ab2(1-t)

The correct answer is:
ab(1-e-bt)

Solution:
dvdt=a-bv
or ∫0vdva-bv=∫0tdt
or -1bloga-bv0v=t
or a-bva=e-bt
or a-bav=e-bt
v=ab(1-e-bt)

Topic: Physics, Motion In One Dimension, JEE Main
The motion of a mass on a spring, with spring constant K is as shown in figure.

The equation of motion is given by, x(t)=Asinωt+Bcosωt with ω=Km.
Suppose that at time t=0, the position of mass is x(0) and velocity v(0), then its displacement can also be represented as x(t)=Ccos(ωt-ϕ), where C and ϕ are

Options:
A C=2v(0)2ω2+x(0)2, ϕ=tan-1v(0)x(0)ω
B C=2v(0)2ω2+x(0)2, ϕ=tan-1x(0)ω2v(0)
C C=v(0)2ω2+x(0)2, ϕ=tan-1x(0)ωv(0)
D C=v(0)2ω2+x(0)2, ϕ=tan-1v(0)x(0)ω

The correct answer is:
C=v(0)2ω2+x(0)2, ϕ=tan-1v(0)x(0)ω

Solution:
x=Asinωt+Bcosωt
v=dxdt=Aωcosωt-Bωsinωt
At t=0, x(0)=B
v(0)=Aω
x=Asinω+Bsinωt+90°

Anet=A2+B2
tanα=BA⇒cotα=AB
⇒x=A2+B2sin(ωt+α)
⇒x=A2+B2cos(ωt-(90-α))

x=Ccos(ωt-ϕ)

⇒C=A2+B2

C=[v(0)]2ω2+[x(0)]2
ϕ=90-α
tanα=cosα=AB

⇒tanϕ=v(0)x(0)·ω
ϕ=tan-1v(0)x(0)ω

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (22 Jul Shift 1)
The motion of a particle is described by the equation $u=a t$. The distance travelled by the particle in the first 4 seconds

Options:
A 4a
B 12a
C 6a
D 8a

The correct answer is:
8a

Solution:
$u=a t, x=\int u d t=\int a t d t=\frac{a t^2}{2}$
For $t=4 \mathrm{sec}, x=8 a$

Topic: Physics, Motion In One Dimension, JEE Main
The motion of a particle moving along the y-axis is represented as y=3t-2+5t-22. Identify the correct statement

Options:
A the initial t=0 velocity of the particle is 3 m s-1
B the acceleration of the particle is 5 m s-1
C the particle is at the origin at t=2 s
D all of the above

The correct answer is:
the particle is at the origin at t=2 s

Solution:
We know that vâ=dydt & aâ=dvâdt
v=dydt=3+10t-2
At t=0 velocity v=-17 m s-1
a=dvdt=10 m s-2
y=3t-2+5t-22
At the value of y will be y=32-2+52-22
therefore y=0

Topic: Physics, Motion In One Dimension, JEE Main
The motion of a particle varies with time according to the relation y=asin⁡ωt+bcos⁡ωt

Options:
A The motion is oscillatory but not SHM
B The motion is SHM with amplitude a + b
C The motion is SHM with amplitude a2+b2
D The motion is SHM with amplitude a2+b2

The correct answer is:
The motion is SHM with amplitude a2+b2

Solution:
y=asin⁡ωt+bcos⁡ωt
Let a=rcos⁡θ and b=rsin⁡θ
y=rcos⁡θsin⁡ωt+rsin⁡θcos⁡ωt
=rsin⁡(ωt+θ)
∴ a2+b2=r2cos2⁡θ+sin2⁡θ=r2
or r=a2+b2

Topic: Physics, Oscillations, JEE Main
The motion of a simple pendulum executing S.H.M. is represented by the following equation y=Asinπt+ϕ, where time is measured in second. The length of pendulum is

Options:
A 97.23 cm
B 25.3 cm
C 99.4 cm
D 406.1 cm

The correct answer is:
99.4 cm

Solution:
Comparing the given equation y=Asinπt+ϕ with a standard equation y=Asinωt+ϕ. We have

ω=π rad s-1

Since time period is T=2πω=2 s, it is a second's pendulum.

T=2πLg⇒L=gπ2=99.4 cm

Topic: Physics, Oscillations, JEE Main, JEE Main 2022 (29 Jun Shift 2)
The mucosa of the stomach is lined with

Options:
A Non-glandular simple cuboidal epithelium
B Glandular simple cuboidal epithelium
C Non-glandular simple columnar epithelium
D Glandular simple columnar epithelium

The correct answer is:
Glandular simple columnar epithelium

Solution:
Simple columnar epithelia (single layer of tall and slender cells with basally placed nuclei):
Main function: Secretion and absorption.
If glands are present, it is of glandular type.
Location:
A. The lining of the stomach and colon.
B. The lining of the small intestine (with microvilli).

Topic: Biology, Digestion and Absorption, NEET
The mutual inductance between two coils is 1.25 henry. If the current in the primary changes at the rate of 80 ampere/second, then the induced e.m.f. in the secondary is

Options:
A $12.5 \mathrm{~V}$
B $64.0 \mathrm{~V}$
C $0.016 \mathrm{~V}$
D $100.0 \mathrm{~V}$

The correct answer is:
$100.0 \mathrm{~V}$

Solution:
$e=M \frac{d i}{d t}=1.25 \times 80=100 \mathrm{~V}$

Topic: Physics, Electromagnetic Induction, JEE Main
The $n$ rows each containing $m$ cells in series are joined in parallel. Maximum current is taken from this combination across an external resistance of $3 \Omega$ resistance. If the total number of cells used are 24 and internal resistance of each cell is $0.5 \Omega$ then

Options:
A $m=8, n=3$
B $m=6, n=4$
C $m=12, n=2$
D $m=2, n=12$

The correct answer is:
$m=12, n=2$

Solution:
Total cells $=m^* n=24$
For maximum current in the circuit $R=\frac{m r}{n}$
$\Rightarrow 3=\frac{m}{n} \times(0.5) \Rightarrow m=6 n$ .... (ii)
On solving equation (i) and (ii), we get $m=12, n=2$

Topic: Physics, Current Electricity, JEE Main
The $n^{\text {th }}$ roots of unity are in

Options:
A A.P.
B G.P.
C H.P.
D None of these

The correct answer is:
G.P.

Solution:
$x^n=1=\left(\cos 0+i \sin 0^{\circ}\right)=\cos 2 r \pi+i \sin 2 r \pi=e^{i 2 r \pi}$
$\Rightarrow \quad x=e^{i(2 r \pi / n)}, r=0,1,2$
Obviously the roots are $1, e^{2 \pi i / n}, e^{4 \pi / n}$ which are obviously in G.P. with common ratio $e^{2 \pi i / n}$

Topic: Mathematics, Complex Number, JEE Main
The nth roots of unity are in

Options:
A AP
B GP
C HP
D None of these

The correct answer is:
GP

Solution:
Let x=1

=cos0°+isin0°

⇒ x1/n=ei2πr/n; r=0, 1, 2,…

Then, the roots are 1, e2πin, e4πin,… which are clearly on GP with common ratio e2πi/n

Topic: Mathematics, Complex Number, JEE Main
The $n^{\text {th }}$ term of an A.P. is $3 n-1$. Choose from the following the sum of its first five terms

Options:
A $14$
B $35$
C $80$
D $40$

The correct answer is:
$40$

Solution:
Here $T_n=3 n-1$, putting $n=1,2,3,4,5$ we get first five terms, $2,5,8,11,14$
Hence sum is $2+5+8+11+14=40$.
Aliter:
$S_n=\Sigma T_n=3 \Sigma n-\Sigma 1=\frac{3 n(n+1)}{2}-n$
$\Rightarrow \quad S_5=\frac{3.5 .6}{2}-5=40$

Topic: Mathematics, Sequences and Series, JEE Main
The $n^{\text {th }}$ term of series $\frac{1}{1}+\frac{1+2}{2}+\frac{1+2+3}{3}+\ldots \ldots .$. will be

Options:
A $\frac{n+1}{2}$
B $\frac{n-1}{2}$
C $\frac{n^2+1}{2}$
D $\frac{n^2-1}{2}$

The correct answer is:
$\frac{n+1}{2}$

Solution:
Given series
$\frac{1}{1}+\frac{1+2}{2}+\frac{1+2+3}{3}+\ldots \ldots$
So $n^{\text {th }}$ term of series is given by
$T_n=\frac{1+2+3+\ldots \ldots+n}{n}$ $=\frac{\frac{1}{2} n(n+1)}{n}=\frac{n+1}{2}$

Topic: Mathematics, Sequences and Series, JEE Main
The "N" which contribute least to the basicity for the compound is :

Options:
A N-9
B N-3
C N-1
D N-7

The correct answer is:
N-9

Solution:
N-9 lone pair is stabilized by resonance so it is not basic in nature.

Topic: Chemistry, General Organic Chemistry, JEE Main
The N which contribute least to the basicity of the compound is :

Options:
A N-9
B N-3
C N-1
D N-7

The correct answer is:
N-9

Solution:
N-9 lone pair is delocalized and participated in resonance, so it is not basic in nature.

Topic: Chemistry, Amines, JEE Main, JEE Main 2016 (10 Apr Online)
The narrow-spectrum antibiotic among the following is:

Options:
A Chloramphenicol
B Penicillin $G$
C Ampicillin
D Amoxicillin

The correct answer is:
Penicillin $G$

Solution:
Benzylpenicillin (Penicillin $\mathrm{G}$ ) is a narrow-spectrum antibiotic used to treat infections caused by susceptible bacteria.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
The natural domain of the function fx=sin-1â¡2x+Ï3 is

Options:
A -12,12
B -34,12
C 134,12
D -32,1

The correct answer is:
-34,12

Solution:
Here, 2x∈-1,1 and sin-1⁡2x+π3≥0
Hence, x∈-12,12 and x≥-34
So the domain is -34,12

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The natural domain of the real valued function defined by $f(x)=\sqrt{x^2-1}+\sqrt{x^2+1}$ is

Options:
A $1 \lt x \lt \infty$
B $-\infty \lt x \lt \infty$
C $-\infty \lt x \lt -1$
D $(-\infty, \infty)-(-1,1)$

The correct answer is:
$(-\infty, \infty)-(-1,1)$

Solution:
$f(x)=\sqrt{x^2-1}+\sqrt{x^2+1} \Rightarrow f(x)=y_1+y_2$
Domain of $y_1=\sqrt{x^2-1} \Rightarrow x^2-1 \geq 0 \Rightarrow x^2 \geq 1$
$x \in(-\infty, \infty)-(-1,1)$ and Domain of $y_2$ is real number, $\therefore$ Domain of $f(x)=(-\infty, \infty)-(-1,1)$.

Topic: Mathematics, Functions, JEE Main
The natural frequency of an L-C circuit is 125000 cycles/s. When a dielectric medium of dielectric constant K is introduced between the plates of the capacitor, the frequency decreases by 25 kHz. The value of K is

Options:
A 3.05
B 2.18
C 1.56
D 1.74

The correct answer is:
1.56

Solution:
f=12πLC

When a dielectric is introduced the capacitance becomes

C⁡ ′ = K⁡ C⁡
⇒f⁡'f=C⁡C⁡'

125000-25000125000=C⁡KC

100125=1K

K=1251002=1.56

Topic: Physics, Alternating Current, JEE Main
The natural number m, for which the coefficient of x in the binomial expansion of xm+1x222 is 1540, is

Options:

The correct answer is:
13

Solution:
The general term in the binomial expansion,

Tr+1=Cr22xm22-rx-2r

⇒Tr+1=Cr22xm(22-r)-2r

For the coefficient of x

22 m-m r-2 r=1

⇒22 m-1=r(m+2)

⇒r=22m-1m+2

⇒r=22m+44-45m+2

r=22-3.3.5m+2⇒22-r=3·3·5m+2

Since, Cr22=C22-r22

So possible values of m=1,3,7,13,43, but C322=1540⇒22-r=3⇒m=13

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2020 (05 Sep Shift 1)
The naturally occurring amino acid that contains only one basic functional group in its chemical structure is

Options:
A asparagine
B histidine
C arginine
D lysine

The correct answer is:
asparagine

Solution:
Asparagine have only one basic functional Group Structure of Asparagine:

There are three amino acids that have basic side chains at neutral pH. These are arginine (Arg), lysine (Lys), and histidine (His). Their side chains contain nitrogen and resemble ammonia, which is a base. Their pKa's are high enough that they tend to bind protons, gaining a positive charge in the process.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2023 (13 Apr Shift 2)
The nature of a graph drawn for a freely falling body with time on the $x$-axis and speed on the $y$-axis is (Assuming initial speed to be zero.)

Options:
A a straight line with positive $y$-axis intercept.
B a straight line passing through the origin.
C a parabola.
D a straight line parallel to $y$-axis with positive $x$-axis intercept.

The correct answer is:
a straight line passing through the origin.

Solution:
As we know that, for free falling body the equation of velocity is
$$
v=u+g t
$$
Since, $u=0$ then
$$
\begin{aligned}
& v=g t \quad\left(\because g=10 \mathrm{~m} / \mathrm{s}^2\right) \\
& v=10 t
\end{aligned}
$$
Which is show that a straight line passing through the origin. So, the correct option is (2).

Topic: Physics, Motion In One Dimension, JEE Main
The nature of charge on resulting colloidal particles when FeCl3 is added to excess of hot water is:

Options:
A positive
B sometimes positive and sometimes negative
C neutral
D negative

The correct answer is:
positive

Solution:
If FeCl3 is added to hot water, a positively charged sol, hydrated ferric oxide is formed due to adsorption of Fe3+ ions.

Fe2O3·xH2O/Fe3+

Positively charged.

Thus, When FeCl3Ã¢â¬â¹ sol is treated with hot water, it forms Fe(OH)3​ forming Fe3+ ions(in excess amount) and these positive ions are adsorbed preferentially.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2021 (26 Feb Shift 2)
The nature of oxides V2O3 and CrO is indexed as 'X' and 'Y' type respectively. The correct set of X and Y is:

Options:
A X=acidic Y=acidic
B X=basic Y=amphoteric
C X=basic Y=basic
D X=amphoteric Y=basic

The correct answer is:
X=basic Y=basic

Solution:
Generally oxides of metals ae basic and oxides of non-metal are acidic.

Oxide	Nature

(i) V2O3	Basic

(ii) CrO	Basic

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2021 (27 Aug Shift 1)
The negation of (p∧(-q))∨(-p) is equivalent to

Options:
A p∧(-q)
B p∧q
C p∨(q∨(-p))
D p∧(q∧(-p))

The correct answer is:
p∧q

Solution:
The given statement is p∧~q∨~p

Let us take A∩Bc∪Ac.

Now let us apply distributive law to the above statement.

≡A∪Ac∩Bc∪Ac

≡U∩Bc∪Ac

≡Bc∪Ac

Now the given statement p∧~q∨~p will be equivalent to ~q∨~p or ~p∨~q.

Hence the given statement is equivalent to ~p∨~q,

Now negation of given statement by demorgan's law will be,

~~p∨~q≡p∧q

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The negation of ~p∧q∨p∧~q is

Options:
A p∨~q∨~p∨q
B p∨~q∧~p∨q
C p∧~q∧~p∨q
D p∧~q∧p∨~q

The correct answer is:
p∨~q∧~p∨q

Solution:
Let, s:~pâ§qâ¨pâ§~q
~s:~~pâ§qâ¨pâ§~q
~s:~~pâ§qâ§~pâ§~q
~s:pâ¨~qâ§~pâ¨q

Topic: Mathematics, Mathematical Reasoning, JEE Main
The negation of p∧( q→∼r ) is

Options:
A â¼pâ§ qâ§r
B  p∨( q∨r )
C  p∨( q∧r )
D â¼âpâ¨( qâ§r )

The correct answer is:
∼ p∨( q∧r )

Solution:
â¼â pâ§ qââ¼r =ââ¼pâ¨ â¼ qââ¼r
(âµâ~(pâ§q)=â~pâ¨~q & ~(pâq)=pâ§~q)
=â¼pâ¨( qâ§â¼( â¼r ) )
=â¼pâ¨( qâ§r )

Topic: Mathematics, Mathematical Reasoning, JEE Main
The negation of ∽s∨∽r∧s is equivalent to

Options:
A s ∧ r
B s∧~r
C s∧r∧~s
D s∨r∨~s

The correct answer is:
s ∧ r

Solution:
Given ∽∽s∨∽r∧s=s∧∽∽r∧s
∵~p∧q=∽p∨∽q
and ∽p∨q=∽p∧∽q
=s∧r∨∽s
=s∧r∨s∧∽s
=s∧r∨F
=s∧r

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2015 (04 Apr)
The negation of the Boolean expression p∨(~p∧q) is equivalent to :

Options:
A p∧~q
B ~p∧~q
C ~p ∨~q
D ~p∨q

The correct answer is:
~p∧~q

Solution:
Given statement is p∨(q∧~p)
∴ Negation is ~(p∨(q∧~p))
=(~p∧~q)∨(~p∧p)
=(~p∧~q)∨c
=~p∧~q

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2020 (06 Sep Shift 1)
The negation of the Boolean expression ~q∧p⇒~p∨q is logically equivalent to

Options:
A p⇒q
B q⇒p
C ~p⇒q
D ~q⇒p

The correct answer is:
~p⇒q

Solution:
To find negation of (~q∧p→~p∨q

Let ~q∧p=A & ~p∨q=B

~A→B =~~A∨B =A∧~B

Using above formula be get

~q∧p∧~~p∨q =~q∧p∧p∧~q

=A∧A =A=~q∧p =p∧~q or ~p→q

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2022 (25 Jun Shift 2)
The negation of the Boolean expression ~s∨~r∧s is equivalent to

Options:
A r
B s∧r
C s∨r
D ~s∧~r

The correct answer is:
s∧r

Solution:
Negation of given statement is

=~~s∨~r∧s

Using De Morgan's law ~A∨B=~A∧~B

=~~s∧~~ r ∧s

Using double negation law ~~A=A, and De Morgan's law

=s∧~~r∨~s

=s∧r∨~s

Using distributive law A∧B∨C=A∧B∨A∧C

=s∧r∨s∧~s

=s∧r∨F

=s∧r.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The negation of the Boolean expression x↔~y is equivalent to:

Options:
A (~x∧y)∨(~x∧~y)
B (x∧y)∨(~x∧~y)
C (x∧~y)∨(~x∧y)
D (x∧y)∧(~x∨~y)

The correct answer is:
(x∧y)∨(~x∧~y)

Solution:
Negation of x↔~y

≡~x↔~y

≡~x→~y∧~y→x

≡~x→~y∨~~y→x (using De'Morgan's laws)

≡x∧y∨~y∧~x

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2020 (05 Sep Shift 1)
The negation of the expression q∨((~q)∧p) is equivalent to

Options:
A (~p)∧(~q)
B p∧(~q)
C (~p)∨(~q)
D (~p)∨q

The correct answer is:
(~p)∧(~q)

Solution:
The negation of q∨((~q)∧p) is

~(q∨((~q)∧p))

≡~q∧~((~q)∧p)

≡~q∧(q∨~p)

≡(~q∧q)∨(~q∧~p)

≡(~q∧~p)

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2023 (01 Feb Shift 1)
The negation of the statement ((A∧(B∨C))⇒(A∨B))⇒A is

Options:
A equivalent to ~C
B equivalent to B∨~C
C a fallacy
D equivalent to ~A

The correct answer is:
equivalent to ~A

Solution:
Given,

((A∧(B∨C))⇒(A∨B))⇒A

We know that, X⇒Y≡~X∨Y,

Now using the above formula we get,

≡~(A∧(B∨C))∨(A∨B)⇒A

≡~(~(A∧(B∨C))∨(A∨B))∨A

≡(A∧(B∨C))∧~(A∨B)∨A

Now to solve (A∧(B∨C))∧~(A∨B) let us assume a universal set 1,2,3,4,5,6,7,8, set A=1,2,3, set B=3,4,5 and set C=5,6,7

So, A∧B∨C will be 3 and ~(A∨B) will be 6,7,8,

So, (A∧(B∨C))∧~(A∨B)=∅

≡F∨A

≡A

∴ Negation of statement=~A

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2023 (13 Apr Shift 1)
The negation of the statement "If I become a teacher, then I will open a school" is

Options:
A
I will become a teacher and I will not open a school

B
Either I will not become a teacher or I will not open a school

C
Neither I will become a teacher nor I will open a school

D
I will not become a teacher or I will open a school

The correct answer is:
I will become a teacher and I will not open a school

Solution:
$\sim(\sim p \vee q)=p \wedge \sim q$

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2012 (Offline)
The negation of the statement "If I will become famous then I will open a school" is

Options:
A I will become famous and I will not open a school
B Either I will not become famous or I will not open a school.
C Neither I will become famous nor I will open a school.
D I will not become famous or I will open a school.

The correct answer is:
I will become famous and I will not open a school

Solution:
a	b	a·b	a+b	a'	b'	a'·b	a·b'

1	0	0	1	0	1	0	1

1	1	1	1	0	0	0	0

0	1	0	1	1	0	1	0

0	0	0	0	1	1	0	0

From the table, we can say that X=a·b

Topic: Mathematics, Mathematical Reasoning, JEE Main
The negation of the statement “If I will become famous then I will open a school” is

Options:
A I will become famous and I will not open a school
B Either I will not become famous or I will not open a school.
C Neither I will become famous nor I will open a school.
D I will not become famous or I will open a school.

The correct answer is:
I will become famous and I will not open a school

Solution:
Let, p: I will become famous
q: I will open a school
Negation of p→q is ~p→q=p∧~q
i.e. I will become famous and I will not open a school.

Topic: Mathematics, Mathematical Reasoning, JEE Main
The negation of the statement "If I will become rich, then I will open a school", is

Options:
A I will become rich and I will not open a school
B Either I will not become rich or I will not open a school.
C Neither I will become rich nor I will open a school.
D I will not become rich or I will open a school.

The correct answer is:
I will become rich and I will not open a school

Solution:
Let, p: I will become rich
q: I will open a school
Negation of p→q is ~p→q=p∧~q
i.e., I will become rich and I will not open a school.

Topic: Mathematics, Mathematical Reasoning, JEE Main
The negation of the statement ~p∧p∨q is:

Options:
A ~p∨q
B ~p∧q
C p∨~q
D p∧~q

The correct answer is:
p∨~q

Solution:
~~p∧p∨q

p∨~p∧~q

p∨~p⏟t∧p∨~q

P∨~q

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The negation of the statement p∨q∧q∨~r is

Options:
A p∨r∧~q
B ~p)∨r∧~q
C ~p∨~q∨~r
D ~p∨~q∧~r

The correct answer is:
~p)∨r∧~q

Solution:
Given,

p∨q∧q∨~r

Now negation of above expression will be,

~p∨q∧q∨~r

=~p∨q∨~q∨~r as ~A∧B≡~A∨~B

=~p∧~q∨~q∧r

=~p∧~q∨~q∧~p∧~q∨r

=~q∧~p∨r∧~q∨r

=~q∧~p∨r∧~q∧~q∨r

=~q∧~p∨r∧~q

=~p∨r∧~q

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2023 (10 Apr Shift 1)