datavorous/entrance-exam-dataset huggingface dataset:
The bob of simple pendulum is hanging vertically down from a fixed identical bob by means of a sting of length $l$. If both bobs are charged with a charge $q$ each, time period of the pendulum is : (ignore the radii of the bobs)

Options:
A $2 \pi \sqrt{\frac{l}{g+\left(\frac{q^2}{l^2 m}\right)}}$
B $2 \pi \sqrt{\frac{l}{g-\left(\frac{q^2}{l^2 m}\right)}}$
C $2 \pi \sqrt{\frac{l}{g}}$
D $2 \pi \sqrt{\frac{l}{g-\left(\frac{q^2}{l}\right)}}$

The correct answer is:
$2 \pi \sqrt{\frac{l}{g}}$

Solution:
There will be a electrostatic repulsion between two charged bobs, but it does not affect the motion of pendulum. Thus, time period of pendulum remains same i.e., $T=2 \pi \sqrt{\frac{l}{g}}$.

Topic: Physics, Oscillations, TS EAMCET, TS EAMCET 2006
The bob of the simple pendulum is given a vertically downward velocity of magnitude v0=5 m s-1, as shown in the figure. The string can withstand a maximum tension equal to twice the weight of the bob and during motion the string breaks at an angle Î¸ with the vertical. If cosÎ¸=mn, where m and n are smallest integers, then, the value of m+n is

Options:

The correct answer is:
5

Solution:
Let us assume that the length of the string is l
T-mgcosâ¡Î¸=mv2l â¦â¦â¦i
mglcosâ¡Î¸=12mv2-12mv0 2 â¦â¦â¦ii
T=2 mg â¦â¦â¦iii

â´ cosâ¡Î¸=14

âm+n=5

Topic: Physics, Laws of Motion, JEE Main
The body cells in cockroach discharge their nitrogenous waste in the haemolymph mainly in the form of :

Options:
A Potassium urate
B Urea
C Calcium carbonate
D Ammonia

The correct answer is:
Potassium urate

Solution:
Malpighian tubules keep floating in haemolymph from where potassium waste diffuses into the tubule. Urate crystals are crystals of uric acid. Thus, cockroach is an uricotelic organism.

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2015 (Phase 2)
The body of the ovule is fused within the funicle at

Options:
A Micropyle
B Nucellus
C Chalaza
D Hilum

The correct answer is:
Hilum

Solution:
The structure of the ovule can be studied under the following headings:
a) Funicle: The stalk of the ovule by which it remains attached to the placenta.
b) Hilum: It is the junction between ovule and funicle or the point of attachment of funiculus to the body of ovule.
c) Integument: The one or more protective envelops surrounding the body of the ovule.
d) Micropyle: The pore or passage present at the tip of the ovule where the integument is absent.
e) Chalaza: Opposite to micropylar end representing the basal part of the ovule.
f) Nucellus: The parenchymatous mass of tissue enclosed within the integuments and forms the body of the ovule.
g) Embryo sac: It is also called the female gametophyte and is located in the nucellus. An ovule generally has a single embryo sac formed from a megaspore.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2020 (Phase 1)
The boiling point of a compound is raised by

Options:
A Intramolecular hydrogen bonding
B Intermolecular hydrogen bonding
C Covalent bonding
D Ionic covalent

The correct answer is:
Intermolecular hydrogen bonding

Solution:
Intermolecular hydrogen bonding compound have more boiling point. compare to intramolecular hydrogen bonding compound.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The boiling point of an azeotropic mixture of water and ethyl alcohol is less than that of theoretical value of water and alcohol mixture. Hence, the mixture shows

Options:
A that solution is highly saturated
B positive deviation from Raoult's law
C negative deviation from Raoult's law
D nothing can be said

The correct answer is:
positive deviation from Raoult's law

Solution:
The boiling point of an azeotropic mixture of water and ethyl alcohol is less than that of theoretical value of water and alcohol mixture. Hence, the mixture shows positive deviation from Raoult's law.
Positive deviations from Raoult's law are noticed when
(i) Exp. vaue of vapour pressure of mixture is more than calculated value.
(ii) Exp. value of boiling point of mixture is less than calculated value.
(iii) ΔHmixing=+ve
(iv) ΔVmixing=+ve .

Topic: Chemistry, Solutions, JEE Main
The boiling point of hydrogen halides varies in the order

Options:
A HF>HI>HBr>HCl
B HF>HCl>HBr>HI
C HI>HBr>HCl>HF
D HBr>HI>HCl>HF

The correct answer is:
HF>HI>HBr>HCl

Solution:
There is a presence of the dipole-dipole interactions i.e. hydrogen bonding in the hydrogen halides which is mainly responsible for the determining the order of boiling point, and the correct order will be

HF>HI>HBr>HCl

Here, the higher boiling point of hydrogen fluoride is due to the presence of strong hydrogen bonding between HF molecules which leads to the association of HF molecules and requires higher energy for boiling so it have higher melting point.

Topic: Chemistry, Chemical Bonding and Molecular Structure, TS EAMCET, TS EAMCET 2021 (04 Aug Shift 2)
The boiling point of the water is higher than liquid $\mathrm{HF}$. The reason is that

Options:
A Hydrogen bonds are stronger in water
B Hydrogen bonds are stronger in $\mathrm{HF}$.
C Hydrogen bonds are larger in number in $\mathrm{HF}$
D Hydrogen bonds are larger in number in water

The correct answer is:
Hydrogen bonds are larger in number in water

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The boiling points of the following hydrides follow the order of

Options:
A $\mathrm{NH}_3 \gt \mathrm{AsH}_3 \gt \mathrm{PH}_3 \gt \mathrm{SbH}_3$
B $\mathrm{SbH}_3 \gt \mathrm{AsH}_3 \gt \mathrm{PH}_3 \gt \mathrm{NH}_3$
C $\mathrm{SbH}_3 \gt \mathrm{NH}_3 \gt \mathrm{AsH}_3 \gt \mathrm{PH}_3$
D $\mathrm{NH}_3 \gt \mathrm{PH}_3 \gt \mathrm{AsH}_3 \gt \mathrm{SbH}_3$

The correct answer is:
$\mathrm{SbH}_3 \gt \mathrm{NH}_3 \gt \mathrm{AsH}_3 \gt \mathrm{PH}_3$

Solution:
Boiling points of $\mathrm{SbH}_3(254 \mathrm{~K}), \mathrm{NH}_3(238 \mathrm{~K}), \mathrm{AsH}_3(211 \mathrm{~K})$ and $\mathrm{PH}_3(185 \mathrm{~K})$ therefore boiling points are of the order $\mathrm{SbH}_3 \gt \mathrm{NH}_3 \gt \mathrm{AsH}_3 \gt \mathrm{PH}_3$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The bond angle H-X-H is the greatest in the compound :

Options:
A PH3
B CH4
C NH3
D H2O

The correct answer is:
CH4

Solution:
Molecule	Hybridization	Bond Angle

1.	PH3	No	93o

2.	CH4	sp3	109o 28′

3.	NH3	sp3	107o

4.	H2O	sp3	104.5o

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2016 (10 Apr Online)
The bond angle in ethylene is

Options:
A $180^{\circ}$
B $120^{\circ}$
C $109^{\circ}$
D $90^{\circ}$

The correct answer is:
$120^{\circ}$

Solution:
In ethylene both Carbon atoms are $s p^2$ - hybridised so $120^{\circ}$.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The bond angle in $\mathrm{NF}_{3}\left(102.3^{\circ}\right)$ is smaller than $\mathrm{NH}_{3}\left(107.2^{\circ}\right) .$ This is because of

Options:
A large size of F compared to H
B large size of N compared to F
C opposite polarity of $\mathrm{N}$ in the two molecules
D small size of $\mathrm{H}$ compared to $\mathrm{N}$

The correct answer is:
opposite polarity of $\mathrm{N}$ in the two molecules

Solution:
In $\mathrm{NF}_{3}$, $\mathrm{N}$ is less electronegative as compared to F but in $\mathrm{NH}_{3}$, it is more electronegative than $\mathrm{H}$. And in case of same central atom. as the electronegativity of other atoms increases, bond angle decreases, Thus, bond angle in $\mathrm{NF}_{3}$ is smaller than that in $\mathrm{NH}_{3}$ because of the difference in the polarity of $\mathrm{N}$ in these molecules.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The bond angles in NH3, NF3 and NCl3 are in the order :

Options:
A NCl 3 > NH 3 > NF 3
B NH 3 > NCl 3 > NF 3
C NF 3 > NH 3 > NCl 3
D NF 3 > NCl 3 > NH 3

The correct answer is:
NCl 3 > NH 3 > NF 3

Solution:
Comparison of NH3 and NF3 : Since N is more electronegative than hydrogen, shared pair of electrons of the N - H bond spent more time with nitrogen ( closer to nitrogen) giving a greater bond pair repulsion. On the other hand fluorine is more electronegative than nitrogen, shared pair of electrons of N - F bond spent more times with fluorine ( closer to fluorine ) giving less bond pair repulsion as compared to N - H bond. Although bond pair bond pair repulsion in both cases are less than lone pair - bond repulsion, contraction in bond angle is more in NF3 due to smaller bond pair repulsion in this case than in NH3. Hence bond angles are less than 109028' in both cases but it is further less in NF3

Comparison of NH3 and NCl3 : In NCl3, chlorine is much larger in size than hydrogen, they comes closer and their valence shell electrons start repelling one another. Therefore, in NCl3 the bond expand to some extent to compensate this repulsion:

Significant repulsion between the valence electrons of chlorine atom overcomes the lone pair bond - pair repulsion .
Hence, the overall order of bond angle is
NCl3 > NH3 > NF3

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The bond dissociation energy is highest for

Options:
A Cl2
B I2
C Br2
D F2

The correct answer is:
Cl2

Solution:
Fluorine (F2), Chlorine Cl2, Bromine Br2, Iodine I2 are called halogen compounds, they occur as diatomic molecules in which a covalent bond is formed after sharing 1 electron with each other. As it moves down from chlorine to iodine in the halogen group, the atomic size increases, and hence bond length also increases.

As bond length increases, the energy required to break bonds (dissociation energy) decreases.

So, chlorine will have the highest bond energy.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (29 Jan Shift 1)
The bond dissociation energy of B-F in BF3 is 646 kJ mol-1 whereas that of C-F in CF4 is 515 kJ mol-1 . The correct reason for higher B-F bond dissociation energy as compared to that of C-F is

Options:
A stronger σ bond between B and F in BF3 as compared to that between C and F in CF4
B significant pπ-pπ interaction between B and F in BF3 whereas there is no possibility of such interaction between C and F in CF4
C lower degree of pπ-pπ interaction between B and F BF3 than that between C and F in CF4
D smaller size of B- atom as compared to that of C- atom

The correct answer is:
significant pπ-pπ interaction between B and F in BF3 whereas there is no possibility of such interaction between C and F in CF4

Solution:
The bond dissociation energy of B-F in BF3 is 646 kJ mol-1 whereas that of C-F in CF4 is 515 kJ mol-1 due to significant pπ-pπ interaction between B and F in BF3 whereas there is no possibility of such interaction between C and F in CF4

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The Bond dissociation energy of B−F in BF 3 is 646 kJ  mol −1 whereas that of C−F in CF 4 is 515 kJ  mol −1 . The correct reason for higher B−F bond dissociation energy as compared to that of C−F is

Options:
A Smaller size of B atom as compared to that of C atom
B Stronger σ bond between B and F in BF3 as Compared to that between C and F in CF4
C Lower degree of pπ-pπ interaction between B and F in BF3 than that between C and F in CF4
D Significant pÏ-pÏ interaction between B and F in BF3 whereas there is no possibility of such interaction between C and F in CF4

The correct answer is:
Significant pÏ-pÏ interaction between B and F in BF3 whereas there is no possibility of such interaction between C and F in CF4

Solution:
In BF 3 , , B is sp 2 hybridised and has a vacant 2p -orbital which overlaps laterally with a filled 2p -orbital of F forming strong pÏ-pÏ bond. However in CF 4 , C does not have any vacant p -orbitals to undergo p -bonding. Thus BE B−F  >  BE C−F .

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The bond dissociation energy of $\mathrm{B}-\mathrm{F}$ in $\mathrm{BF}_3$ is $646 \mathrm{~kJ}$ mol ${ }^{-1}$ whereas that of $\mathrm{C}-\mathrm{F}$ in $\mathrm{CF}_4$ is $515 \mathrm{~kJ}$ mol $^{-1}$. The correct reason for higher B-F bond dissociation energy as compared to that of $C$ - $F$ is :

Options:
A
smaller size of B-atom as compared to that of C-atom

B
stronger $\sigma$ bond between $\mathrm{B}$ and $\mathrm{F}$ in $\mathrm{BF}_3$ as compared to that between $\mathrm{C}$ and $\mathrm{F}$ in $\mathrm{CF}_4$

C
significant $\mathrm{p} \pi-\mathrm{p} \pi$ interaction between $\mathrm{B}$ and $\mathrm{F}$ in $\mathrm{BF}_3$ whereas there is no possibility of such interaction between $\mathrm{C}$ and $\mathrm{F}$ in $\mathrm{CF}_4$.

D
lower degree of $\mathrm{p} \pi-\mathrm{p} \pi$ interaction between $\mathrm{B}$ and $\mathrm{F}$ in $\mathrm{BF}_3$ than that between $\mathrm{C}$ and $\mathrm{F}$ in $\mathrm{CF}_4$.

The correct answer is:
significant $\mathrm{p} \pi-\mathrm{p} \pi$ interaction between $\mathrm{B}$ and $\mathrm{F}$ in $\mathrm{BF}_3$ whereas there is no possibility of such interaction between $\mathrm{C}$ and $\mathrm{F}$ in $\mathrm{CF}_4$.

Solution:
option itself is the reason

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The bond dissociation energy of $\mathrm{B}-\mathrm{F}$ in $\mathrm{BF}_3$ is $646 \mathrm{~kJ}$ mol ${ }^{-1}$ whereas that of $\mathrm{C}-\mathrm{F}$ in $\mathrm{CF}_4$ is $515 \mathrm{~kJ}$ mol $^{-1}$. The correct reason for higher B-F bond dissociation energy as compared to that of $C$ - $F$ is :

Options:
A
smaller size of B-atom as compared to that of C-atom

B
stronger $\sigma$ bond between $\mathrm{B}$ and $\mathrm{F}$ in $\mathrm{BF}_3$ as compared to that between $\mathrm{C}$ and $\mathrm{F}$ in $\mathrm{CF}_4$

C
significant $\mathrm{p} \pi-\mathrm{p} \pi$ interaction between $\mathrm{B}$ and $\mathrm{F}$ in $\mathrm{BF}_3$ whereas there is no possibility of such interaction between $\mathrm{C}$ and $\mathrm{F}$ in $\mathrm{CF}_4$.

D
lower degree of $\mathrm{p} \pi-\mathrm{p} \pi$ interaction between $\mathrm{B}$ and $\mathrm{F}$ in $\mathrm{BF}_3$ than that between $\mathrm{C}$ and $\mathrm{F}$ in $\mathrm{CF}_4$.

The correct answer is:
significant $\mathrm{p} \pi-\mathrm{p} \pi$ interaction between $\mathrm{B}$ and $\mathrm{F}$ in $\mathrm{BF}_3$ whereas there is no possibility of such interaction between $\mathrm{C}$ and $\mathrm{F}$ in $\mathrm{CF}_4$.

Solution:
option itself is the reason

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2009
The bond dissociation enthalpy of gaseous \( \mathrm{H}_{2}, \mathrm{Cl}_{2} \) and \( \mathrm{HCl} \) are \( 435.243 \) and \( 431 \mathrm{~kJ} \mathrm{~mol}^{-1} \) respectively. In the same unit, enthalpy of formation of \( \mathrm{HCl} \) gas is

Options:
A \( -92 \)
B \( -184 \)
C \( +92 \)
D \( +184 \)

The correct answer is:
\( -92 \)

Solution:
The bond dissociation enthalpy of gaseous H2, Cl2 and HCl are 435, 243 and 431 kJ mol-1 respectively.

The enthalpy of formation of HCl gas is is given by

12 H2(g) + 12Cl2(g) →HClg ; ∆H

∆H =∑( Bond energy for reactants) -∑(Bond energy of products)

∆H =12435 + 243 - 431 ∆H =-92 kJ mol-1

Topic: Chemistry, Thermodynamics (C), JEE Main
The bond length between the hybridised carbon atom and other carbon atom is minimum in

Options:
A Butane
B Propyne
C Propene
D Propane

The correct answer is:
Propyne

Solution:
Bond length decreases with increase in number of bonds.

Topic: Chemistry, Hydrocarbons, JEE Main
The bond length of $\mathrm{HCl}$ molecule is $1.275 Å$ and its dipole moment is $1.03 \mathrm{D}$. The ionic character of the molecule (in percent) (charge of the electron $=4.8 \times 10^{-10} \mathrm{esu}$ ) is

Options:
A $100$
B $67.3$
C $33.6$
D $16.83$

The correct answer is:
$16.83$

Solution:
Given,
observed dipole moment $=1.03 \mathrm{D}$
Bond length of $\mathrm{HCl}$ molecule, $d=1.275 Å$
$$
=1.275 \times 10^{-8} \mathrm{~cm}
$$
Charge of electron, $\quad e^{-}=4.8 \times 10^{-10}$ esu
Percentage ionic character $=$ ?
Theoretical value of dipole moment $=e \times d$
$$
\begin{aligned}
& =4.8 \times 10^{-10} \times 1.275 \times 10^{-8} \mathrm{esu}-\mathrm{cm} \\
& =6.12 \times 10^{-18} \mathrm{esu}-\mathrm{cm} \\
& =6.12 \mathrm{D}
\end{aligned}
$$
Percentage ionic character
$$
\begin{aligned}
& =\frac{\text { observed dipole moment }}{\text { theoretical value of dipole moment }} \times 100 \\
& =\frac{1.03}{6.12} \times 100 \\
& =16.83 \%
\end{aligned}
$$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The bond length (pm) of $\mathrm{F}_2, \mathrm{H}_2, \mathrm{Cl}_2$ and $\mathrm{I}_2$, respectively is

Options:
A $144,74,199,267$
B $74,144,199,267$
C $74,267,199,144$
D $144,74,267,199$

The correct answer is:
$74,144,199,267$

Solution:
As the size of halogen atom increase, the bond length $X-X$ bond in $X_2$ molecule increase from $\mathrm{F}_2$ to $\mathrm{I}_2$ hence, the correct answer is (2).

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The bond order and magnetic behaviour of O2- ion are, respectively :

Options:
A 1.5 and diamagnetic.
B 1.5 and paramagnetic.
C 2 and diamagnetic.
D 1 and paramagnetic

The correct answer is:
1.5 and paramagnetic.

Solution:
O2-(Total electron =17){ Bond order =1.5, Paramagnetic }

σ1 s2,σ⋆1 s2,σ2 s2,σ⋆2 s2,σ2Pz2,π2Px2=π2Py2,π*2Px2=π*2Py1,σ⋆2Pz0

Number of unpaired electron =1

Bond order =nb-na2=10-72=1.5

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (26 Aug Shift 2)
The bond order and the magnetic characteristic of CN- are

Options:
A 212, diamagnetic
B 3, diamagnetic
C 3, paramagnetic
D 212, paramagnetic

The correct answer is:
3, diamagnetic

Solution:
According to MOT (If z is internuclear axis) The configuration of
CN−:σ1s2,σ*1s2,σ2s2,σ*2s2,π2px2=π2py2,σ2pz2

Bond order=number of electrons in bonding orbitals-number of electrons in anti-bonding orbitals2

Bond order = 1 2 (10−4)=3
CN − is diamagnetic due to absence of unpaired electron.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2020 (07 Jan Shift 2)
The bond order in $N_2^{+}$ion is

Options:
A 1
B 2
C 2.5
D 3

The correct answer is:
2.5

Solution:
B.O. $=\frac{\text { No. of } N_b-\text { No. of } N_a}{2}=\frac{5}{2}=2.5$.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The bond order in $\mathrm{NO}$ is $2.5$ while that in $\mathrm{NO}^{+}$is 3 . Which of the following statements is true for these two species?

Options:
A
Bond length in $\mathrm{NO}^{+}$is greater than in $\mathrm{NO}$

B
Bond length is unpredictable

C
Bond length in $\mathrm{NO}^{+}$in equal to that in $\mathrm{NO}$

D
Bond length in $\mathrm{NO}$ is greater than in $\mathrm{NO}^{+}$

The correct answer is:
Bond length in $\mathrm{NO}$ is greater than in $\mathrm{NO}^{+}$

Solution:
Bond order of $\mathrm{NO}=2.5$
Bond order of $\mathrm{NO}^{+}=3$
Higher the bond order shorter is the bond length

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2004
The bond order in $\mathrm{O}_2^{+}$is

Options:
A 2
B 2.5
C 1.5
D 3

The correct answer is:
2.5

Solution:
B.O. $=\frac{N_b-N_a}{2}=\frac{8-3}{2}=\frac{5}{2}=2.5$.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The bond order of NO molecule is

Options:

The correct answer is:
2.5

Solution:
Bondâorderâ= 1 2 ( N b â N a )
N b â Total number of electrons in bonding molecular orbits.
N a â= Total number of electrons in anti-bonding molecular orbits.
NOâ Ï 1s 2 â Ï 1s *2 â Ï 1s 2 â Ï 2s *2 â Ï 2p z 2 â Ï 2p x 2 â=â Ï 2p y 2
π 2p x *1 =  π 2p y *0 ⇒ Bond order of NO =  1 2 10−5 =2.50

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The bond structure in n-type semiconductor is

Options:
A I
B II
C III
D IV

The correct answer is:
I

Solution:
The band structure in an n-type semiconductor is as
This is because a large energy difference (gap) exist between valence band and conduction band in case of a semiconductor and addition of an impurity (having higher valency) results in the n-type semiconductor.

Topic: Chemistry, Solid State, JEE Main
The bond that exists between $\mathrm{NH}_3$ and $\mathrm{BF}_3$ is called

Options:
A Electrovalent
B Covalent
C Coordinate
D Hydrogen

The correct answer is:
Coordinate

Solution:
$\mathrm{NH}_3$ has lone pair of electron while ${ }^{\mathrm{BF}_3}$ is electron deficient compound so they form a co-ordinate bond.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The $\pi$ - bonded organo metallic compound which has ethene as one of its component is

Options:
A Zeise's salt
B Ferrocene
C Dibenzene chromium
D Tetraethyl tin.

The correct answer is:
Zeise's salt

Solution:

Topic: Chemistry, Coordination Compounds, JEE Main
The Boolean expression ∼p⇒∼q is equivalent to

Options:
A (∼p)⇒q
B q⇒∼p
C p∨q
D p∧q

The correct answer is:
p∧q

Solution:
Recall p⇒q≡~p∨q

∼p⇒∼q=∼∼p∨∼q
=p∧q

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2019 (12 Apr Shift 2)
The Boolean expression ~p∨q∨~p∧q is equivalent to

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

The correct answer is:
~p

Solution:
Venn diagram of ~p∨q

Venn diagram of ~ p∧ q
Venn diagram of ~p∨q∨~p∧q
~p∨q=~p∧~q∨~p∧q
=~p∧~q∨q
=~p∧T=~p

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2018 (08 Apr)
The boolean expression ~p∧q∨q is equivalent to

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

The correct answer is:
p→p∨q

Solution:
Given ~p∧q∨q≡~p∨~q∨q≡~p∨t≡t

Also p→p∨q≡~p∨p∨q≡~p∨p∨q≡t∨q≡t

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2022 (27 Jun Shift 1)
The Boolean expression (p⇒q)∧(q⇒~p) is equivalent to :

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

The correct answer is:
~p

Solution:
(p⇒q)∧(q⇒~p)

=(~p∨q)∧(~q∨~p) {p⇒q=~p∨q}

=(~p∨q)∧(~p∨~q) {commutative property}

=~p∨(q∧~q) {distributive property}

=~p

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2021 (25 Jul Shift 1)
The Boolean expression p∧~q⇒q∨~p is equivalent to:

Options:
A q⇒p
B p⇒q
C ~q⇒p
D p⇒~q

The correct answer is:
p⇒q

Solution:
We have,

p	q	~p	~q	p∧~q	q∨~p	(p∧~q)
⇒(q∨~p)	p⇒q

T	F	F	T	T	F	F	F

F	T	T	F	F	T	T	T

T	T	F	F	F	T	T	T

F	F	T	T	F	T	T	T

∴ p∧~q⇒q∨~p

≡p⇒q

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The Boolean Expression p∧∼q ∨q∨∼p∧q is equivalent to

Options:
A p â¨q
B pÂ â¨â¼q
C â¼pâ§q
D pâ§q

The correct answer is:
p â¨q

Solution:
p â§ â¼q â¨qâ¨â¼pâ§q
Set equivalent
=A â©B- âªA- â©B âªB
=(A âªB-Aâ©B)âªB
=A âªB
Hence answer is p â¨q .

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2016 (03 Apr)
The Boolean expression (p∧q)⇒r∧q∧p is equivalent to:

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

The correct answer is:
(p∧q)⇒(r∧q)

Solution:
(p∧q)⇒((r∧q)∧p)≡(p∧q)⇒(r∧(p∧q))…….(i)

Case: I when p∧q=T and r=F

Statement-(i) is false as well as option (3) is also false
option (1), option (2) are True so given expression is not equivalent to option (1), option (2)

option (4) may be true if we consider q=T∴ option (4) is also not equivalent to option (1) Case : II p∧q=False

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2021 (27 Aug Shift 2)
The boolean expression \( (p \wedge \sim q) \vee(q \vee(\sim p \wedge q)) \) is:

Options:
A \( \sim p \wedge q \)
B \( p \wedge q \)
C \( p \vee q \)
D \( p \vee \sim q \)

The correct answer is:
\( p \vee q \)

Solution:
(p∧~q)∨q∨(~p∧q)

≡(p∧~q)∨q∨(~p∧q)

≡(p∨~q)∧t∨(~p∧q)

≡(p∨q)∨(~p∧q)

≡(p∨q∨~p)∧(p∨q∨q)≡q∨t∧p∨q≡t∧p∨q≡p∨q

Topic: Mathematics, Mathematical Reasoning, JEE Main
The Born-Haber cycle for KCl is evaluated with the following data:
ΔfHΘ for KCl=−436.7kJmol−1;ΔsubHΘ for K=89.2kJmol−1;

Δionization HΘ for K=419.0kJ mol −1;Δelectron gain HΘ for Cl(g)=−348.6kJmol−1

Δbond HΘ for Cl2=243.0kJmol−1
The magnitude of lattice enthalpy of KCl in kJmol-1 is (Nearest integer)

Options:

The correct answer is:
718

Solution:
So, ΔfHΘ=ΔfHΘKCl=Δsub HΘ+Δ1st ionisation HΘK+1/2ΔbondHΘCl2(g)+Δelectron gain HΘ+Δlattice HΘ-436.7=89.2+419+12243+-348.6+Δlattice HΘ

Δlattice HΘ=-717.8 kJ mole-1≈-718 kJ mole-1

Topic: Chemistry, Thermodynamics (C), JEE Main