datavorous/entrance-exam-dataset huggingface dataset:
The heats of neutralisation of $\mathrm{CH}_{3} \mathrm{COOH}$, $\mathrm{HCOOH}, \mathrm{HCN}$ and $\mathrm{H}_{2} \mathrm{~S}$ are -13.2,-13.4,-2.9

and $-3.8 \mathrm{kCal}$ per equivalent respectively. Arrange the acids in increasing order of acidic strength.

Options:
A $\mathrm{HCOOH}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{H}_{2} \mathrm{~S}>\mathrm{HCN}$
B $\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{HCOOH}>\mathrm{H}_{2} \mathrm{~S}>\mathrm{HCN}$
C $\mathrm{H}_{2} \mathrm{~S}>\mathrm{HCOOH}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{HCN}$
D $\mathrm{HCOOH}>\mathrm{H}_{2} \mathrm{~S}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{HCN}$

The correct answer is:
$\mathrm{HCOOH}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{H}_{2} \mathrm{~S}>\mathrm{HCN}$

Solution:
The greater the (negative value) of heat of neutralisation, the more is the strength of the acid. Hence,

$$

\mathrm{HCOOH}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{H}_{2} \mathrm{~S}>\mathrm{HCN}

$$

Topic: Chemistry, Thermodynamics (C), JEE Main
The height at which the acceleration due to gravity becomes $\frac{\mathrm{g}}{9}$ (where $\mathrm{g}=$ the acceleration due to gravity on the surface of the earth) in terms of $R$, the radius of the earth is

Options:
A
$2 R$

B
$\frac{R}{\sqrt{2}}$

C
$\frac{R}{2}$

D
$\sqrt{2} \mathrm{R}$

The correct answer is:
$2 R$

Solution:
$g^{\prime}=\frac{G M}{(R+h)^2}$, acceleration due to gravity at height $h$
$$
\begin{aligned}
& \Rightarrow \frac{g}{9}=\frac{G M}{R^2} \cdot \frac{R^2}{(R+h)^2}=g\left(\frac{R}{R+h}\right)^2 \\
& \Rightarrow \frac{1}{9}=\left(\frac{R}{R+h}\right)^2 \Rightarrow \frac{R}{R+h}=\frac{1}{3} \\
& \Rightarrow 3 R=R+h \Rightarrow 2 R=h
\end{aligned}
$$

Topic: Physics, Gravitation, JEE Main, JEE Main 2009
The height at which the acceleration due to gravity becomes $\frac{g}{16}$ (where, $g=$ acceleration due to gravity on the surface of the earth) in terms of $R$ is, if $R$ is the radius of earth.

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

The correct answer is:
$3 R$

Solution:
As, acceleration due to gravity at height $h$,
$$
\begin{aligned}
& g^{\prime}=\frac{G M}{(R+h)^2}, \\
& \Rightarrow \quad \frac{g}{16}=\frac{G M}{R^2}\left(\frac{R^2}{(R+h)^2}\right)=g \frac{R^2}{(R+h)^2}\left(\because g^{\prime}=\frac{g}{16}\right) \\
& \Rightarrow \quad \frac{1}{4}=\frac{R}{R+h} \Rightarrow h=3 R \\
&
\end{aligned}
$$

Topic: Physics, Gravitation, JEE Main
The height at which the weight of a body becomes $1 / 16$ th, its weight on the surface of earth (radius $R$ ), is

Options:
A $5 R$
B $15 R$
C $3 R$
D $4 R$

The correct answer is:
$3 R$

Solution:
According to question $\frac{G M m}{(R+h)^2}=\frac{1}{16} \frac{G M m}{R^2}$
$\begin{array}{l}
\frac{1}{(R+h)^2} =\frac{1}{16 R^2} \\
\text {or } \frac{R}{R+h} =\frac{1}{4} \\
\text {or } \frac{R+h}{R} =4 \\
h =3 R
\end{array}$

Topic: Physics, Gravitation, JEE Main
The height ‘h’ at which the weight of a body will be the same as that at the same depth ‘h’ from the surface of the earth is (Radius of the earth is R and effect of the rotation of the earth is neglected)

Options:
A 52R-R
B R2
C 5R-R2
D 3R-R2

The correct answer is:
5R-R2

Solution:
GM(R+h)2=GMR3(R-h)
R3=(R+h)2(R-h)
=R2+h2+2hR(R-h)
R3=R3+h2R+2hR2-R2h-h3-2h2R
h3+h2(2R-R)-R2h=0
h3+h2R-R2h=0
h2+hR-R2=0
h=-R±R2+4(1)R22
=-R+5R2
=(5-1)2R

Topic: Physics, Gravitation, JEE Main
The height ‘h’ at which the weight of a body will be the same as that at the same depth ‘h’ from the surface of the earth is (Radius of the earth is R and effect of the rotation of the earth is neglected)

Options:
A 52R-R
B R2
C 5R-R2
D 3R-R2

The correct answer is:
5R-R2

Solution:
GM(R+h)2=GMR3(R-h)
R3=(R+h)2(R-h)
=R2+h2+2hR(R-h)
R3=R3+h2R+2hR2-R2h-h3-2h2R
h3+h2(2R-R)-R2h=0
h3+h2R-R2h=0
h2+hR-R2=0
h=-R±R2+4(1)R22
=-R+5R2
=(5-1)2R

Topic: Physics, Gravitation, JEE Main, JEE Main 2020 (02 Sep Shift 2)
The height of a liquid column in a U tube is \( 0.3 \mathrm{~m} \). If the liquid in one of the limbs is depressed and then released, then the time period of the liquid column will be

Options:
A 1. \( 1 \mathrm{sec} \)
B \( 19 \mathrm{sec} \)
C \( 0.11 \mathrm{sec} \)
D \( 2 \mathrm{sec} \)

The correct answer is:
1. \( 1 \mathrm{sec} \)

Solution:
T=2πhg
T=2×3.14×0.39.8=1.1 sec

Topic: Physics, Oscillations, JEE Main
The height of a mercury barometer is $75 \mathrm{~cm}$ at sea level and $50 \mathrm{~cm}$ at the top of a hill. Ratio of density of mercury to that of air is $10^4$. The height of the hill is

Options:
A $250 \mathrm{~m}$
B $2.5 \mathrm{~km}$
C $2.5 \mathrm{~km}$
D $750 \mathrm{~m}$

The correct answer is:
$2.5 \mathrm{~km}$

Solution:
Difference of pressure between sea level and the top of hill
$\Delta P=\left(h_1-h_2\right) \times \rho_{H_g} \times g=(75-50) \times 10^{-2} \times \rho_{H_g} \times g$...(i)
and pressure difference due to $h$ meter of air
$\Delta P=h \times \rho_{\text {air }} \times g$...(ii)
By equating (i) and (ii) we get
$h \times \rho_{a b} \times g=(75-50) \times 10^{-2} \times \rho_{H_g} \times g$
$\therefore h=25 \times 10^{-2}\left(\frac{\rho_{H_g}}{\rho_{\text {aiv }}}\right)=25 \times 10^{-2} \times 10^4=2500 \mathrm{~m}$
$\therefore$ Height of the hill $=2.5 \mathrm{~km}$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The height of a right circular cylinder is decreasing while is diameter is increasing at a rate of $4 \mathrm{~cm} / \mathrm{see}$ so as to keep its volume unchanged. the rate of change in its lateral surface area (in $\mathrm{cm}^2 / \mathrm{sec}$ ) at the instant when its diameter is $8 \mathrm{~cm}$ and height is $12 \mathrm{~cm}$, is

Options:
A $24 \pi$
B $-24 \pi$
C $48 \pi$
D $-48 \pi$

The correct answer is:
$48 \pi$

Solution:
Given, $\frac{d(2 r)}{d t}=4 \mathrm{~cm} / \mathrm{s}$
$\Rightarrow \quad \frac{d r}{d t}=2 \mathrm{~cm} / \mathrm{s}$
Volume of cylinder $=\pi r^2 h$
$$
\begin{aligned}
v & =\pi r^2 h \\
\Rightarrow \quad \frac{d v}{d t} & =\pi\left(r^2 \frac{d h}{d t}+2 r h \frac{d r}{d t}\right)
\end{aligned}
$$
Volume is unchanged and height of cylinder is decreasing
$$
\begin{aligned}
& \therefore & \pi r^2 h & =\pi\left(-r^2 \frac{d h}{d t}+2 r h(2)\right) \\
\Rightarrow & & r h & =-r \frac{d h}{d t}+4 h \\
\Rightarrow & & \frac{d h}{d t} & =\frac{4 h-r h}{r}
\end{aligned}
$$
Lateral surface area of cylinder $(S)=2 \pi r h$
$$
\frac{d S}{d t}=2 \pi\left(r \frac{d h}{d t}+h \frac{d r}{d t}\right)
$$
$$
\begin{aligned}
& \frac{d S}{d t}=2 \pi\left[\frac{r(4 h-r h)}{r}+h(2)\right] \\
& \frac{d S}{d t}=2 \pi(4 h-r h+2 h) \\
& \Rightarrow \quad \frac{d S}{d t}=2 \pi(6 h-r h) \\
& \frac{d S}{d t}=2 \pi(72-48) \quad[\because h=12, r=4] \\
& \frac{d S}{d t}=48 \pi \\
&
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main
The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is:

Options:
A 3
B 233
C 6
D 2 3

The correct answer is:
2 3

Solution:
From the diagram, we get radius of cylinder =R2-h22
⇒ Volume of cylinder V=πR2-h24h
For maximum volume: dVdh=0
⇒πR2-3h24=0
⇒h2=4R23=4×93=12
⇒h=23

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The height of a TV antenna is 200 m. The population density is 4000 km-2. Find the population benefited

Options:
A 3.2Ã108
B 3.2×107
C 3.2Ã106
D 3.2Ã105

The correct answer is:
3.2×107

Solution:
d=2hR;
Population converted=πd2× population density
=π(2Rh)×ρ
=227×2×6400×0.2×4000
=3.2×107

Topic: Physics, Communication System, JEE Main
The height of any point P above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point P will be : (Given g=acceleration due to gravity at the surface of earth).

Options:
A g2
B g4
C g3
D g9

The correct answer is:
g9

Solution:
Acceleration due to gravity at a height h from the surface of the Earth,
g'=g11+hR2
Given: h=2R
∴g'=g11+22
or g'=g9

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (25 Jun Shift 1)
The height of liquid column raised in a capillary tube of certain radius when dipped in liquid A vertically is, 5 cm. If the tube is dipped in a similar manner in another liquid B of surface tension and density double the values of liquid A, the height of liquid column raised in liquid B would be ______ m.

Options:
A 0.20
B 0.5
C 0.05
D 0.10

The correct answer is:
0.05

Solution:
The height h through which a liquid will rise in a capillary tube of radius r is given by,

h=2Tcosθrρg∝Tρ∝2T2ρ

As we can see in both cases, value of height will remain same. Therefore,

h'=h=5 cm= 0.05 m

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2023 (30 Jan Shift 1)
The height of mercury barometer is 75 cm at sea level and 50 cm at top of a hill. Ratio of density of mercury to that of air is 104. the height of the hill is

Options:
A 250 m
B 2.5 km
C 1.25 km
D 750 m

The correct answer is:
2.5 km

Solution:
Difference of pressure between sea level and the top of the hill
ΔP=h1-h2×ρHg×g=75-50×10-2×ρHg×g .......(i)
and pressure difference due to h metres of air ΔP=h×ρair×g ......(ii)
By equating (i) and (ii) we get h×ρair×g=(75-50)×10-2×ρHg×g
∴h=25×10-2ρHgρair=25×10-2×104=2500m
∴Height of the hill = 2.5 km.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The height of right circular cylinder of maximum volume inscribed in a sphere of diameter $2 a$ is

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

The correct answer is:
$\frac{2 a}{\sqrt{3}}$

Solution:
Let the radius and height of the cylinder are and $h$, respectively.
In $\triangle A O M$
$\therefore$
$$
\begin{array}{l}
r^{2}+\left(\frac{h^{2}}{4}\right)=a^{2} \\
h^{2}=4\left(a^{2}-r^{2}\right)
\end{array}
$$
Now, $V=\pi r^{2} h=\pi\left(a^{2} h-\frac{1}{4} h^{3}\right)$
For max or min,
$$
\begin{aligned}
& \frac{d V}{d h} &=\pi\left(a^{2}-\frac{3}{4} h^{2}\right)=0 \\
\Rightarrow & h &=\left(\frac{2}{\sqrt{3}}\right) a
\end{aligned}
$$
Now, $\quad \frac{d^{2} V}{d h^{2}}=-\frac{6 h}{4} < 0$
So, $V$ is maximum at $h=\frac{2 a}{\sqrt{3}}$.

Topic: Mathematics, Application of Derivatives, JEE Main
The height of transmitting antenna is 180 m and the height of the receiving antenna is 245 m. The maximum distance between them for satisfactory communication in line of sight will be: (given R=6400 km)

Options:
A 96 km
B 56 km
C 48 km
D 104 km

The correct answer is:
104 km

Solution:
The maximum distance for line of sight communication is

d=2hR+2h'R

The given data is

R=6400 kmh=180 m

h'=245 m

Thus, the value is

d=2×180×6400×103

+2×245×6400×103

=6×80×102+7×80×102

=104 km

Topic: Physics, Communication System, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The height of victoria's falls is 63 m. What is the difference in the temperature of water at the top and at the bottom of the fall? [Given 1 cal=4.2 J and specific heat of water =1 cal g-1 C∘-1

Options:
A 14.76 °C
B 1.476 °C
C 0.147 °C
D 0.014 °C

The correct answer is:
0.147 °C

Solution:
Here use the concept of conservation of energy, here potential energy is converted in to heat energy, ⇒mgh=msΔθ, here m, h & s, is the mass of the liquid, height of the fall, and specific heat capacity then, g×63=Δθ×4.2×103⇒Δθ=0.147°C.

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2021 (27 Aug Shift 2)
The height to which a cylindrical vessel be filled with a homogeneous liquid, to make the average force with which the liquid presses the side of the vessel equal to the force exerted by the liquid on the bottom of the vessel, is equal to

Options:
A Half of the radius of the vessel
B Radius of the vessel
C One-fourth of the radius of the vessel
D Three-fourth of the radius of the vessel

The correct answer is:
Radius of the vessel

Solution:
Pressure at the bottom $=h \rho g$
and pressure on the vertical surface $=\frac{1}{2} h \rho g$
Now, according to problem
Force at the bottom = Force on the vertical surface
$\Rightarrow{ }^{h \rho g \times \pi r^2=\frac{1}{2} h \rho g \times 2 \pi \cdot h} \Rightarrow h=r$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y=\left(8 t-5 t^2\right)$ meter and $x=6 t$ meter, where ${ }^t$ is in second. The velocity with which the projectile is projected is

Options:
A 8 m/sec
B 6 m/sec
C 10 m/sec
D Not obtainable from the data

The correct answer is:
10 m/sec

Solution:
$v_y=\frac{d y}{d t}=8-10 t \quad v_x=\frac{d x}{d t}=6$
at the time of projection i.e. $v_y=\frac{d y}{d t}=8$ and $v_x=6$
$\therefore v=\sqrt{v_x^2+v_y^2}=\sqrt{6^2+8^2}=10 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Motion In Two Dimensions, JEE Main
The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y=\left(8 t-5 t^2\right)$ meter and $x=6 t$ meter, where $t$ is in second. The velocity with which the projectile is projected is

Options:
A 8 m/sec
B 6 m/sec
C 10 m/sec
D Not obtainable from the data

The correct answer is:
10 m/sec

Solution:
$v_y=\frac{d y}{d t}=8-10 t \quad v_x=\frac{d x}{d t}=6$
at the time of projection i.e. $v_y=\frac{d y}{d t}=8$ and $v_x=6$
$\therefore v=\sqrt{v_x^2+v_y^2}=\sqrt{6^2+8^2}=10 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Motion In Two Dimensions, JEE Main
The helical structure of protein is stabilized by

Options:
A dipeptide bonds
B hydrogen bonds
C ether bonds
D peptide bonds

The correct answer is:
hydrogen bonds

Solution:
The $\alpha$-helix structure is formed when the chain of $\alpha$-amino acids coils as a right handed screw (called $\alpha$-helix) because of the formation of hydrogen bonds between amide groups of the same peptide chain, i.e., NH group in one unit is linked to carbonyl oxygen of the third unit by hydrogen bonding. This hydrogen bonding between different units is responsible for holding helix in a position.

Topic: Chemistry, Biomolecules, JEE Main
The helical structure of protein is stabilized by

Options:
A Dipeptide bonds
B Hydrogen bonds
C Ether bonds
D Peptide bonds

The correct answer is:
Hydrogen bonds

Solution:
The helical structure of protein is stabilized by hydrogen bonds between amide group of the same peptide chain. These bonds are formed by –NH- group of one unit and oxygen of carbonyl group of the third unit.

Topic: Chemistry, Biomolecules, JEE Main
The Henry's law constant for the solubility of $\mathrm{N}_2$ gas in water at $298 \mathrm{~K}$ is $1 \times 10^{+5} \mathrm{~atm}$. The mole fraction of air is 0.8 . The number of moles of $\mathrm{N}_2$ from air dissolved in 10 moles of water at $298 \mathrm{~K}$ and 5 atm pressure is

Options:
A $4 \times 10^{-5}$
B $4 \times 10^{-4}$
C $5 \times 10^{-4}$
D $4 \times 10^{-6}$

The correct answer is:
$4 \times 10^{-4}$

Solution:
At total pressure $5 \mathrm{~atm}$, the partial pressure of $\mathrm{N}_2=5 \times 0.8=4 \mathrm{~atm}$
According to Henry's Law,
$$
\mathrm{P}_{\mathrm{N}_2}=\mathrm{K}_{\mathrm{H}} \mathrm{x}_{\mathrm{N}_2}
$$
$\left(\mathrm{x}_{\mathrm{N}_2}\right)$ is the mole fraction of nitrogen gas dissolved in water.
or, $\quad \mathrm{x}_{\mathrm{N}_2}=\frac{4 \mathrm{~atm}}{1 \times 10^5 \mathrm{~atm}}$ or, $\mathrm{x}_{\mathrm{N}_2}=4 \times 10^{-5}$
$$
\begin{aligned}
& \frac{\mathrm{n}_{\mathrm{N}_2}}{\mathrm{n}_{\mathrm{N}_2}+\mathrm{n}_{\mathrm{H}_2 \mathrm{O}}} \simeq \frac{\mathrm{n}_{\mathrm{N}_2}}{\mathrm{n}_{\mathrm{H}_2 \mathrm{O}}}=4 \times 10^{-5}\left[\text { since } \mathrm{n}_{\mathrm{H}_2 \mathrm{O}} \gg \mathrm{n}_{\mathrm{N}_2}\right] \\
& \therefore \quad \mathrm{n}_{\mathrm{N}_2}=4 \times 10^{-5} \times 10=4 \times 10^{-4} \mathrm{~mol}
\end{aligned}
$$

Topic: Chemistry, Solutions, JEE Main
The Henryâs law constant for the solubility of N2 gas in water at 298 K is 1Ã10-5atm. The mole fraction of N2 in air is 0.8 . If the number of moles of N2 of air dissolved in 10 moles of water at 298 K and 5 atm is xÂ·10-4. Find the value of x.

Options:

The correct answer is:
4

Solution:
4.00
XN2=PN21KH=5Ã0Â·81Ã10-5
PN21=Pair mole fraction of N2
=4Ã10-5
∴ n N 2 n N 2 + n H 2 O =4× 10 –5 = n N 2 n N 2 +10 ≃ n N 2 10
â´n=4Ã10â4
Hence x=4

Topic: Chemistry, Solutions, JEE Main
The Henry's law constant for the solubility of $\mathrm{N}_2$ gas in water at $298 \mathrm{~K}$ is $1.0 \times 105 \mathrm{~atm}$. The mole fraction of $\mathrm{N}_2$ in air is $0.8$. The number of moles of $\mathrm{N}_2$ from air dissolved in 10 moles of water of $298 \mathrm{~K}$ and $5 \mathrm{~atm}$ pressure is

Options:
A
$4 \times 10^{-4}$

B
$4.0 \times 10^{-5}$

C
$5.0 \times 10^{-4}$

D
$4.0 \times 10^{-6}$

The correct answer is:
$4 \times 10^{-4}$

Solution:
$P_{\mathrm{N}_2}=K_{\mathrm{H}} \times$ mole-fraction $\left(\mathrm{N}_2\right)$ mole-fraction
$$
\mathrm{N}_2 \frac{1}{10^5} \times 0.8 \times 5=4 \times 10^{-5} \mathrm{~mol}^{-1}
$$
In 10 mole solubility in $4 \times 10^{-4}$.

Topic: Chemistry, Solutions, JEE Main
The high boiling point of ethanol $\left(78.2^{\circ} \mathrm{C}\right)$ compared to dimethyl ether $\left(-23.6^{\circ} \mathrm{C}\right)$, though both having the same molecular formulae $\mathrm{C}_6 \mathrm{H}_6 \mathrm{O}$, is due to

Options:
A Hydrogen bonding
B Ionic bonding
C Coordinate covalent bonding
D Resonance

The correct answer is:
Hydrogen bonding

Solution:
The high boiling point of ethanol $78.2 \mathrm{C}^{\circ}$ compared to dimethyl ether $-23.6 \mathrm{C}^{\circ}$, though they have the same molecular formula $[\mathrm{C} 2 \mathrm{H} 6 \mathrm{O}]$, is due to: Hydrogen bonding.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The higher concentration of which gas in air can cause stiffnes of flower buds?

Options:
A NO
B $\mathrm{CO}_{2}$
C SO.
D $\mathrm{CO}$

The correct answer is:
SO.

Solution:
High concentration of $\mathrm{SO}_{2}$ leads to stiffness of flower buds.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The higher stabilities of tert-butyl cation over isopropyl cation and trans-2- butene over propene, respectively, are due to orbital interactions involving

Options:
A σ→π and σ→π*
B σ→ vacant p and σ→π
C σ→σ* and σ→π
D σ→ vacant p and σ→π*

The correct answer is:
σ→ vacant p and σ→π*

Solution:
Tert-butyl cation is more stable than isopropyl cation, because of better hyper conjugation between σ-vacant and p-orbital.

whereas, trans-2-butene is more stable than propene because of hyper-conjugation σ-π*-orbitals.

Topic: Chemistry, General Organic Chemistry, JEE Main
The highest catenation ability is shown by

Options:
A Oxygen
B Carbon
C Sulphur
D Tellurium

The correct answer is:
Carbon

Solution:
Carbon shows Catenation to a very large extent due to its small size.
Hence, Option "2" is the correct answer.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The highest catenation ability is shown by

Options:
A Oxygen
B Carbon
C Sulphur
D Tellurium

The correct answer is:
Carbon

Solution:
Carbon shows Catenation to a very large extent due to its small size. Hence, Option "2" is the correct answer.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The highest electrical conductivity of the following aqueous solutions is of

Options:
A
0.1 M acetic acid

B
0.1 M chloroacetic acid

C
0.1 M fluoroacetic acid

D
0.1 M difluoroacetic acid

The correct answer is:
0.1 M difluoroacetic acid

Solution:
No Solution Available

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2005
The highest oxidation state in plutonium (At. No. = 94) is

Options:
A +6
B +4
C +5
D +7

The correct answer is:
+7

Solution:
The highest oxidation state in plutonium (Atomic number = 94) is + 7. It can show a greater range of oxidation states which (in part) is attributed to the fact 5f, 6d and 7s levels are of comparable energies. Plutonium (Pu) can show oxidation state of +3, +4, +5, +6 and +7 in its compound.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The highest oxidation state of $M n$ is shown by

Options:
A $\mathrm{K}_2 \mathrm{MnO}_4$
B $\mathrm{KMnO}_4$
C $\mathrm{MnO}_2$
D $\mathrm{Mn}_2 \mathrm{O}_2$

The correct answer is:
$\mathrm{KMnO}_4$

Solution:
$\mathrm{Mn}$ shows highest oxidation state in $\mathrm{KMnO} \mathrm{A}$.

Topic: Chemistry, Redox Reactions, JEE Main
The highest possible oxidation states of uranium and plutonium, respectively, are

Options:
A 4 and 6
B 7 and 6
C 6 and 4
D 6 and 7

The correct answer is:
6 and 7

Solution:
The electronic configuration of Uranium is,
Rn 5f3 6d17s2
Hence, the maximum electrons it can lose is 3+1+2=6 (Inner electrons do not come out due to stability)
The electronic configuration of plutonium is, Rn 5f6 7s2
It exhibits a maximum oxidation state of +7.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The highest value of the calculated spin only magnetic moment (in BM) among all the transition metal complexes is

Options:
A 4.90 BM
B 3.87 BM
C 6.93 BM
D 5.92 BM

The correct answer is:
5.92 BM

Solution:
The maximum number of unpaired electrons for transition metals =5
or n=5
Spin only magnetic moment =nn+2BM
=55+2=35
=5.92 BM

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2019 (09 Jan Shift 1)
The hilum is a scar on the:

Options:
A Seed, where funicle was attached
B Fruit, where it was attached to pedicel
C Fruit, where style was present
D Seed, where micropyle was present

The correct answer is:
Seed, where funicle was attached

Solution:
Hilum is a scar on the surface of a seed marking its point of attachment to the seed stalk (funicle). Seed is developed from ovule and bears many structures that were present on the ovule.

Ovule is considered to be an integumented megasporangium. The ovule consists of the stalk and the body. The stalk is called funicle. One end of the funicle is attached to placenta and the other end to the body of the ovule. The point of attachment of funicle with the body is called hilum. Sometimes funicle gets fused with the body of the ovule one side and forms a ridge known as raphe. The body of the ovule shows two ends: the basal end, often called the chalazal end and the upper end is called micropylar end. The main body of the ovule is covered with one or two envelopes called integuments. These leave an opening at the top of the ovule called micropyle. The integuments enclose a large parenchymatous tissue known as nucellus.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2015 (Phase 1)
The Hinsberg's method is used for

Options:
A Preparation of primary amines
B Preparation of secondary amines
C Preparation of tertiary amines
D Separation of amine mixtures

The correct answer is:
Separation of amine mixtures

Solution:
The Hinsberg's method is used for separation of amine mixtures

Topic: Chemistry, Amines, JEE Main
The historic convention on Biological Diversity held in Rio de Janeiro in 1992 is known as

Options:
A
CITESConvention

B
The Earth Summit

C
G-16 Summit

D
MAB Programme

The correct answer is:
The Earth Summit

Solution:
The historic convention on biological diversity held in Rio de Janeiro (Brazil) in 1992 is known as Earth Summit (1st). CITES (Convention on International Trade in Endangered Species) of wild flora and fauna has helped in restricting poaching and loss of rare species. MAB stands for Man and Biosphere Programme that undertakes establishment and maintenance of biosphere reserves.

Topic: Biology, Biodiversity and its Conservation
The homoleptic and octahedral complex of Co2+ and H2O has ______ unpaired electron(s) in the t2g set of orbitals.

Options:

The correct answer is:
1

Solution:
The electronic configuration of Co2+ is 3d7. In an octahedral complex of Co2+ with H2O ligands, the d orbitals split into two sets of energy levels: t2g (lower energy) and eg (higher energy).

In this case, since the complex is octahedral, we have six ligands surrounding the central cobalt ion. Each ligand donates a pair of electrons to the cobalt ion, occupying the available d orbitals.

So, CoH2O62+=d7High spin=t2g5 eg2

In the t2g set of orbitals, there will be 1 unpaired electron

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (15 Apr Shift 1)