datavorous/entrance-exam-dataset huggingface dataset:
The pH of a solution obtained by mixing equal volume of solutions having pH = 3 and pH = 4.
[log 5.5 = 0.7404]

Options:
A 3.26
B 3.5
C 4.0
D 3.42

The correct answer is:
3.26

Solution:
At pH =3,H+=10-3M
At pH =4,H+=10-4M
When in equal volume of the two solutions are mixed, the H+
=10-3+10-42=10-31+0.12=1.12×10-3
H+=5.5×10-4
-logH+=-log5.5-log10-4
pH =-0.7404+4=3.26

Topic: Chemistry, Ionic Equilibrium, JEE Main
The pH of ammonium phosphate solution, if pka of phosphoric acid and pkb of ammonium hydroxide are 5.23 and 4.75 respectively, is

Options:

The correct answer is:
7

Solution:
Since NH43PO4 is salt of weak acid H3PO4 and weak base NH4OH

pH=7+12pka-pkb

=7+125.23-4.75

=7.24≈7

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2021 (26 Feb Shift 2)
The pH of an aqueous solution containing 1M benzoic acid pKa=4.20 and 1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL of this buffer solution is __________mL.

Options:

The correct answer is:
100

Solution:
Benzoic acid+Sodium benzoateMolarity1M1MVolumeVamlVsmlmilli molesVa×1Vs×1

pH=4.5

pKa=4.20

pH=pKa+log[ salt ][ acid ]

4.5=4.2+logVsVaWhere,Vs- Volume of saltVa- Volume of acid

VsVa=2..........(i)

Vs+Va=300........(ii)

Va=100ml

Topic: Chemistry, Ionic Equilibrium, JEE Main
The pH of an aqueous solution containing 1M benzoic acid pKa=4.20 and 1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL of this buffer solution is __________mL.

Options:

The correct answer is:
100

Solution:
Benzoic acid+Sodium benzoateMolarity1M1MVolumeVamlVsmlmilli molesVa×1Vs×1

pH=4.5

pKa=4.20

pH=pKa+log[ salt ][ acid ]

4.5=4.2+logVsVaWhere,Vs- Volume of saltVa- Volume of acid

VsVa=2..........(i)

Vs+Va=300........(ii)

Va=100ml

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2024 (30 Jan Shift 2)
The $\mathrm{pH}$ of blood does not appreciably change by a small addition of an acid or a base because blood

Options:
A Contains serum protein which acts as buffer
B Contains iron as a part of the molecule
C Can be easily coagulated
D It is body fluid

The correct answer is:
Contains serum protein which acts as buffer

Solution:
Serum, the plasma component of blood which lacks coagulation factors, is similar to the interstitial fluid in which the correct composition of key ions acting as electrolytes is essential for normal functioning of muscles and nerves. Other components in the serum include proteins, which assist with maintaining pH and osmotic balance while giving viscosity to the blood; antibodies, or specialized proteins that are important for defense against viruses and bacteria; lipids, including cholesterol, which are transported in the serum; and various other substances including nutrients, hormones, metabolic waste, and external substances, such as drugs, viruses, and bacteria.

Topic: Chemistry, Ionic Equilibrium, JEE Main
The pH of blood stream is maintained by a proper balance of H2CO3 and NaHCO3 concentration. What volume of 5 M NaHCO3 solution should be mixed with a 10 mL sample of blood which is 2 M in H2CO3 in order to maintain its pH of 7.4?

[pKa for H2CO3=6.1]101.3=19.9

Options:
A 40 mL
B 38 mL
C 50 mL
D 79 mL

The correct answer is:
79 mL

Solution:
pH=pKa+logHCO3-H2CO3
Moles of H 2 CO 3 presenting blood sample
=2Ã0.01=0.02
Let V mL of NaHCO3 is required
=5ÃV=5V
So pH=pKa+log5V0.02
7.4=6.1+log5V0.02
V=79 mL

Topic: Chemistry, Ionic Equilibrium, JEE Main
The pH of blood stream is maintained by a proper balance of H 2 CO 3 and NaHCO3 concentration. What volume of 5M NaHCO3 solution should be mixed with 10 ml sample of blood which is 2 M in H2CO3 in order to maintain a pH of 7.4(Ka for H2CO3 in blood is 4.0×10-7) ?

Options:
A 40 ml
B 35 ml
C 25 ml
D 38 ml

The correct answer is:
40 ml

Solution:
H2CO3 in blood = 2 M
Volume of blood = 10 ml
NaHCO3=5M
Let volume of NaHCO3 used = V ml
H2CO3 in mixture =2×10v+10
NaHCO3 in mixture =5×vv+10
pH = pKa + log salt acid
7.4 = -log (4×10−7) + log5 × vv + 102 × 10v + 10
v=40 ml

Topic: Chemistry, Ionic Equilibrium, JEE Main
The pH of pure water at 298 K and 308 K are 7 and 6 respectively. Calculate the heat of formation of water from H+ and OH– .

Options:
A ÎH=84.551Â kcal/mol
B ÎH=-84.551 kcal/mol
C ÎH=44.981Â kcal/mol
D ÎH=-44.981Â kcal/mol

The correct answer is:
ÎH=-84.551 kcal/mol

Solution:
At 298âK; [H+]=10â7
â´â K w 1 = 10 â14
At 308âK; H+=10â6
â´â K w 2 = 10 â12
Now using
2.303â log 10 Kw 2 Kw 1 = ÎH R T 2 â T 1 T 1 Ã T 2
2.303  log 10 10 −12 10 −14 = ΔH 2 308−298 298×308
â´ ÎH=84551.4 kcal/mol
=84.551 kcal/mol
Thus H2O âH++OH-; ÎH=84.551 kcal/mol
â´ H++OH-âH2O; ÎH-84.551 kcal/mol

Topic: Chemistry, Chemical Equilibrium, JEE Main
The pH of rain water is approximately:

Options:
A 5.6
B 6.5
C 7.5
D 7.0

The correct answer is:
5.6

Solution:
Normal, clean rain has a pH value of between 5.0 and 5.6, which is slightly acidic. However, when rain combines with sulfur dioxide or nitrogen oxides produced from power plants and automobiles the rain becomes much more acidic. Typical acid rain has a pH value of 4.0.

Rain water becomes acidic because gases present in environment are dissolved, so it's pH will be less than 7.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The phase (at a time $t$ ) of a particle in simple harmonic motion tells

Options:
A Only the position of the particle at time $t$
B Only the direction of motion of the particle at time $t$
C Both the position and direction of motion of the particle at time $t$
D Neither the position of the particle nor its direction of motion at time $t$

The correct answer is:
Both the position and direction of motion of the particle at time $t$

Solution:
The correct option is $\mathbf{3}$
Both the position and direction of motion of the particle at time $t$

Topic: Physics, Oscillations, JEE Main
The phase difference between the alternating current and emf is $\pi / 2$. Which of the following cannot be the constituent of the circuit?

Options:
A
C alone

B
R.L

C
L.C

D
L alone

The correct answer is:
R.L

Solution:
$0 < $ phase difference for $R-L$ circuit $ < \pi / 2$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2005
The phase difference between the current and voltage of $L C R$ circuit in series combination at resonance is

Options:
A 0
B $\pi / 2$
C $\pi$
D $-\pi$

The correct answer is:
0

Solution:
At resonance $L C R$ series circuit behaves as pure resistive circuit. For resistive circuit $=0$ degree

Topic: Physics, Alternating Current, JEE Main
The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is :

Options:
A $\pi$
B $0.70 \pi$
C zero
D $0.5 \pi$

The correct answer is:
$0.5 \pi$

Solution:
Let $y=A \sin \omega t=A \omega^2 \sin (\omega t+\pi)$
$\begin{aligned}
& \therefore \quad \frac{d y}{d t}=A \omega \cos \omega t \\
& \therefore \quad=A \omega \sin \left(\omega t+\frac{\pi}{2}\right) \\
& \text { Acceleration }=-A \omega^2 \cos \omega t=A \omega^2 \\
& \sin (\omega t+\pi) \\
& \therefore \text { Phase difference }=\pi-\frac{\pi}{2}
\end{aligned}$

Topic: Physics, Oscillations, JEE Main
The phase difference between the two particles situated on both the sides of a node is

Options:
A $0^{\circ}$
B $90^{\circ}$
C $180^{\circ}$
D $360^{\circ}$

The correct answer is:
$180^{\circ}$

Solution:
Both the sides of a node, two antinodes are present with separation $\frac{\lambda}{2}$
So phase difference between then
$\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}=\pi$

Topic: Physics, Waves and Sound, JEE Main
The phase difference between two points separated by 0.8 m in a wave of frequency 120 Hz is 0.5Ï. The wave velocity is

Options:
A 144 m s-1
B 384 m s-1
C 256 m s-1
D 720 m s-1

The correct answer is:
384 m s-1

Solution:
Here, f=120 Hz, x=0.8 m and Ï=0.5Ï.
From, Ï=2ÏÎ»x and Î»=2ÏxÏ=2ÏÃ0.80.5Ï=3.2 m
v=nÎ»=120Ã3.2=384 m s-1

Topic: Physics, Waves and Sound, JEE Main
The phase in which both the chromosome and DNA content becomes half is

Options:
A Anaphase
B Anaphase II
C Anaphase I
D Both (A) and (B)

The correct answer is:
Anaphase I

Solution:
During anaphase I the homologous chromosomes separate, while sister chromatids remain associated at their centromeres. As a result, this stage is called the reductional division stage.

Topic: Biology, Cell Cycle and Cell Division, NEET
The phenomenon of interaction between the member of two different species where one organism is being benefitted without affecting the other is

Options:
A Amensalism
B Commensalism
C Mutualism
D Predation

The correct answer is:
Commensalism

Solution:
Interaction between the population of two different species can be harmful, beneficial or neutral to one of the species or to both. Commensalism is the interaction between two species where one is benefitted whereas other is neither harmed nor benefitted. Mutualism refers to those species interactions where both the species are benefited from the association and are dependent on each other for survival. In Predation, only one species that is the predator is benefitted and the predator kills the prey. Amensalism is where one specie is harmed and the other remains unaffected

Topic: Biology, Organisms and Populations, NEET
The phenomenon of interference is based on

Options:
A Conservation of momentum
B Quantum nature of light
C Conservation of energy
D Conservation of charge

The correct answer is:
Conservation of energy

Solution:
In interference phenomenon, energy is distributed such that intensity of resultant wave is maximum at some points and minimum at another point, therefore phenomenon of interference is based on conservation of energy.

Topic: Physics, Wave Optics, JEE Main
The phenomenon of optical activity will be shown by

Options:
A
B
C
D

The correct answer is:

Solution:
Compounds which do not possess elements of symmetry are optically active.

Topic: Chemistry, Coordination Compounds, JEE Main
The phenomenon utilised in an optical fibre is

Options:
A Refraction
B Interference
C Polarization
D Total internal reflection

The correct answer is:
Total internal reflection

Solution:
When light rays falls on at an angle greater than the critical angle the light rays undergo total internal reflection which causes the light rays to reflect back to our eyes which cause the shining effect in the bubble. The picture below shows the different types of refraction caused in the air bubble in water.

Topic: Physics, Ray Optics, JEE Main
The photochemical smog does not generally contain

Options:
A NO
B NO2
C SO2
D HCHO

The correct answer is:
SO2

Solution:
The major components of photochemical smog are nitrogen oxides, ozone O3, PAN (peroxyacetylnitrate), and chemical compounds that contain the-CHO group (aldehydes). PAN and aldehydes can cause eye irritation and plant damage if their concentrations are sufficiently high.

Classical smog occurs in a cool, humid climate. Its components include smoke, fog and sulphur dioxide. It is reducing in nature.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2022 (25 Jul Shift 1)
The photodiode is used to detect the optical signals. These diodes are preferably operated in reverse biased mode because

Options:
A fractional change in majority carriers produce higher forward bias current
B fractional change in majority carriers produce higher reverse bias current
C fractional change in minority carriers produce higher forward bias current
D fractional change in minority carriers produce higher reverse bias current

The correct answer is:
fractional change in minority carriers produce higher reverse bias current

Solution:
In photo–diode, the diode is connected in reverse bias, so initially the current will be very small. If the light is incident on the diode, its photons produce more minority charge carriers, which are called photo-generated charge carriers, due to which the reverse current increases.

In reverse bias, the width of the depletion region is more and the current is already very small, So the fractional charge in the current due to the light will be large.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2022 (25 Jun Shift 1)
The photoelectric current from Na (work function, w0=2.3eV) is stopped by the output voltage of the cell

Pt(s)‖H2(g,1bar)|HCl(aq·,pH=1)|AgCl(s)∣Ag(s)
the pH of aq. HCl required to stop the photoelectric current from Kw0=2.25eV, all other conditions remaining the same, is ……….×10-2 (to the nearest integer).

Given 2.303RTF=0.06V;EAgCl/Ag/Cl0=0.22V

Options:

The correct answer is:
142

Solution:
Sodium metal :

E=E0+KEmax ; Ecall0=0.22 V

Cell reaction

Cathode : AgCls+e- → Ags+Cl-aq

Anode : 12 H2g → H+aq+e-

Overall ; AgCls+12H2g → Ags+H+aq+Cl-aq

Ecell=Ecell0-0.061logH+ Cl-

Ecell=0.22-0.061log10-110-1=0.22+0.12=0.34 V

KEmax=Ecell=0.34 eV

So E=2.3+0.34=2.64 eV= Energy of photon incident

For potassium metal :

E=E0+KEmax

2.64=2.25+KEmax

KEmax=0.39=Ecell

Cell reaction

Cathode : AgCl(s)+e- → Ags+Cl-aq

Anode : 12H2g → H+aq+e-

Overall : AgCls+12H2g → Ags + H+aq+Cl-aq

Ecell=Ecell0-0.061log H+ Cl-

0.39=0.22-0.12 log H+

0.17=0.12×pH

pH=17/12=1.4166=1.42

Topic: Chemistry, Electrochemistry, JEE Main
The photoelectric current from Na (work function, w0=2.3eV) is stopped by the output voltage of the cell

Pt(s)‖H2(g,1bar)|HCl(aq·,pH=1)|AgCl(s)∣Ag(s)
the pH of aq. HCl required to stop the photoelectric current from Kw0=2.25eV, all other conditions remaining the same, is ……….×10-2 (to the nearest integer).

Given 2.303RTF=0.06V;EAgCl/Ag/Cl0=0.22V

Options:

The correct answer is:
142

Solution:
Sodium metal :

E=E0+KEmax ; Ecall0=0.22 V

Cell reaction

Cathode : AgCls+e- → Ags+Cl-aq

Anode : 12 H2g → H+aq+e-

Overall ; AgCls+12H2g → Ags+H+aq+Cl-aq

Ecell=Ecell0-0.061logH+ Cl-

Ecell=0.22-0.061log10-110-1=0.22+0.12=0.34 V

KEmax=Ecell=0.34 eV

So E=2.3+0.34=2.64 eV= Energy of photon incident

For potassium metal :

E=E0+KEmax

2.64=2.25+KEmax

KEmax=0.39=Ecell

Cell reaction

Cathode : AgCl(s)+e- → Ags+Cl-aq

Anode : 12H2g → H+aq+e-

Overall : AgCls+12H2g → Ags + H+aq+Cl-aq

Ecell=Ecell0-0.061log H+ Cl-

0.39=0.22-0.12 log H+

0.17=0.12×pH

pH=17/12=1.4166=1.42

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The photoelectric effect was discovered by

Options:
A Einstein
B Heinrich Rudolf Hertz
C Rutherford
D More than one of the above

The correct answer is:
Heinrich Rudolf Hertz

Solution:
The phenomenon of the photoelectric effect was discovered by Heinrich Rudolf Hertz.

Topic: Physics, Dual Nature of Matter, JEE Main
The photoelectric threshold of Tungsten is 2300 A∘. The energy of the electrons ejected from the surface by the ultraviolet light of wavelength 1800 A∘ is
(h=6.6×10-34J s, e=1.6×10-19 C)

Options:
A 0.15 eV

B 1.5 eV
C 15 eV

D 150 eV

The correct answer is:
1.5 eV

Solution:
Ek=hce1Î»-1Î»0 (in eV)
=6.6Ã10-34Ã3Ã1081.6Ã10-19 10101800-10102300=1.5 eV

Topic: Physics, Dual Nature of Matter, JEE Main
The photon of hard gamma radiation knocks a proton out of ${ }_{12}^{24} \mathrm{Mg}$ nucleus to form

Options:
A
the isotope of parent nucleus

B
the isobar of parent nucleus

C
the nuclide ${ }_{11}^{23} \mathrm{Na}$

D
the isobar of ${ }_{11}^{23} \mathrm{Na}$

The correct answer is:
the nuclide ${ }_{11}^{23} \mathrm{Na}$

Solution:
No Solution Available

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2005
The photosensitive metallic surface has work function $h v_0$ fall on this surface, the electrons come out with a maximum velocity of $4 \times 10^6 \mathrm{~m} / \mathrm{s}$. When the photon energy is increased $5 h v_0$. Then maximum velocity of photoelectrons will be:

Options:
A $2 \times 10^7 \mathrm{~m} / \mathrm{s}$
B $2 \times 10^6 \mathrm{~m} / \mathrm{s}$
C $8 \times 10^6 \mathrm{~m} / \mathrm{s}$
D $8 \times 10^5 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$8 \times 10^6 \mathrm{~m} / \mathrm{s}$

Solution:
According to the question
$$
\begin{gathered}
2 h v_0=h v_0+\frac{1}{2} m\left(4 \times 10^6\right)^2 \\
5 h v_0=h v_0+\frac{1}{2} m v^2 \\
\Rightarrow 4 \times \frac{1}{2} m\left(4 \times 10^6\right) \\
\Rightarrow \frac{1}{2} m v^2 \\
v=8 \times 10^6 \mathrm{~m} / \mathrm{s}
\end{gathered}
$$

Topic: Physics, Dual Nature of Matter, JEE Main
The photosensitive surface is receiving the light of wavelength 5000 A∘ at the rate of 10-8 J s-1. The number of photons received per second is (h=6.62×10-34 J s, c=3×108 m s-1)

Options:
A 2.5×1010
B 2.5×1011
C 2.5×1012
D 2.5×109

The correct answer is:
2.5×1010

Solution:
Energy of photon
E=hcÎ»
Given, Î»=5000 â« = 5Ã10-7 m
â´ E=6.6Ã10-34Ã3Ã1085Ã10-7
= 3.96Ã10-19 J
Energy received per second = 10-8 J s-1
â´ Number of photon's received per second
= Energy received per second Energy of one photon
= 10-83.96Ã10-19=2.5Ã1010

Topic: Physics, Dual Nature of Matter, JEE Main
The physical quantities not having same dimensions are

Options:
A
torque and work

B
momentum and Planck's constant

C
stress and Young's modulus

D
speed and $\left(\mu_0 \varepsilon_0\right)^{-1 / 2}$

The correct answer is:
momentum and Planck's constant

Solution:
$[$ momentum $]=[\mathrm{M}][\mathrm{L}]\left[\mathrm{T}^{-1}\right]=\left[\mathrm{MLT}^{-1}\right]$
(Planck's Constant) $=\frac{E}{\mathrm{u}}=\frac{[\mathrm{M}]\left[\mathrm{LT}^{-1}\right]^2}{\mathrm{~T}^{-1}}=\mathrm{ML}^2 \mathrm{~T}^{-1}$

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2003
The physical quantity which has the dimensional formula [M1T-3] is

Options:
A Surface tension
B Density
C Solar constant
D Compressibility

The correct answer is:
Solar constant

Solution:
MT-3=[ML2T-2]L2[T]=energyarea×time = dimensions of the solar constant.

Topic: Physics, Units and Dimensions, JEE Main
The physical sizes of the transmitter and receiver antenna in a communication system are:

Options:
A proportional to carrier frequency
B independent of both carrier and modulation frequency
C inversely proportional to modulation frequency
D inversely proportional to carrier frequency

The correct answer is:
inversely proportional to carrier frequency

Solution:
The physical sizes of transmitter and receiver antenna are directly proportional to carrier wave wavelength or inversely proportional to carrier wave frequency.

Topic: Physics, Communication System, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The phytohormone responsible for increasing the shelf life of cut flowers and vegetables also helps in

Options:
A Overcoming requirements of vernalisation
B Inducing parthenocarpy in tomato
C Developing epicotyl hook in seedlings
D Overcoming apical dominance and helps in sprouting of lateral buds

The correct answer is:
Overcoming apical dominance and helps in sprouting of lateral buds

Solution:
Apical dominance has been explained as due to the action of auxin present in the main apex. This view is amply supported by an experiment, where if the stem tip is cut off, the axillary buds found below sprout immediately. Instead, if an agar block containing auxin placed over the decapitated stem, the axillary buds remain suppressed. It is clear from the above experiment, that auxin present in the main apex somehow inhibits the growth of the axillary bud. The severity of apical dominance is greater on the axillary buds present nearer to the main apex. Nevertheless, apical dominance exerted by the auxins can be overcome by the application of cytokinin to axillary buds. This is because cytokinin induces cytokinesis

Topic: Biology, Plant - Growth and Development, NEET
The pieces which have been crushed into small pieces, further converted into powder form by which machine

Options:
A ball mill
B stamp mill
C crusher
D $A$ and $B$ both

The correct answer is:
$A$ and $B$ both

Solution:
Correct Option is 4 $A$ and $B$ both
pieces are reduced to fine powder with the help of a ball mill or stamp mill.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.
The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is:

Options:
A 5.755 mm
B 5.740Â mm
C 5.725 mm
D 5.950Â mm

The correct answer is:
5.725 mm

Solution:
least count =0.5 mm100=0.005 mm
Reading =5.5 mm+48-3×0.005 mm
=5.725 mm.

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72nd division on circular scale coincides with the reference line. The radius of the wire is

Options:
A 1.64 mm
B 0.82 mm
C 0.90 mm
D 1.80 mm

The correct answer is:
0.82 mm

Solution:
Least count =1 mm100=0.01 mm

zero error =+8×LC=+0.08 mm

True reading (Diameter)

=1 mm+72×LC- (Zero error)

=1 mm+72×0.01 mm-0.08 mm

=1.72 mm-0.08 mm

=1.64 mm

therefore, radius =1.642=0.82 mm.

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2021 (25 Feb Shift 1)
The $\mathrm{pK}_{\mathrm{a}}$ of a weak acid, $\mathrm{HA}$, is $4.80$. The $\mathrm{pK}_{\mathrm{b}}$ of a weak base, $\mathrm{BOH}$, is $4.78$. The $\mathrm{pH}$ of an aqueous solution of the corresponding salt, BA, will be

Options:
A
$9.58$

B
$4.79$

C
$7.01$

D
$9.22$

The correct answer is:
$7.01$

Solution:
It is a salt of weak acid and weak base
$$
\begin{aligned}
& {\left[\mathrm{H}^{+}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{w}} \times \mathrm{K}_{\mathrm{a}}}{\mathrm{K}_{\mathrm{b}}}}} \\
& \mathrm{pH}=7.01
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2008
The pK a values of ionisable groups in lysine are 2.18,8.95 and 10.79 respectively. Find isoelectric point of lysine.

Options:

The correct answer is:
9.87

Solution:
Isoelectric point= 8.95+10.79 2 =9.87

Topic: Chemistry, Biomolecules, JEE Main
The $\mathrm{pKa}$ of a weak acid $(\mathrm{HA})$ is $4.5$. The $\mathrm{pOH}$ of an aqueous buffered solution of $\mathrm{HA}$ in which $50 \%$ of the acid is ionized is

Options:
A
$4.5$

B
$2.5$

C
$9.5$

D
$7.0$

The correct answer is:
$9.5$

Solution:
For buffer solution
$\begin{aligned}
& \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { Salt }]}{[\text { Acid }]} \\
& =4.5+\log \frac{[\text { Salt }]}{[\text { Acid }]}
\end{aligned}$
as $\mathrm{HA}$ is $50 \%$ ionized so $[\text{Salt}]$=$[\text{Acid}]$
$\begin{aligned}
& \mathrm{pH}=4.5 \\
& \mathrm{pH}+\mathrm{pOH}=14 \\
& \Rightarrow \mathrm{pOH}=14-4.5=9.5
\end{aligned}$
Hence (C) is correct.

Topic: Chemistry, Ionic Equilibrium, JEE Main
The $\mathrm{pKa}$ of a weak acid $(\mathrm{HA})$ is $4.5$. The $\mathrm{pOH}$ of an aqueous buffered solution of $\mathrm{HA}$ in which $50 \%$ of the acid is ionized is

Options:
A
$4.5$

B
$2.5$

C
$9.5$

D
$7.0$

The correct answer is:
$9.5$

Solution:
For buffer solution
$\begin{aligned}
& \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { Salt }]}{[\text { Acid }]} \\
& =4.5+\log \frac{[\text { Salt }]}{[\text { Acid }]}
\end{aligned}$
as $\mathrm{HA}$ is $50 \%$ ionized so $[\text{Salt}]$=$[\text{Acid}]$
$\begin{aligned}
& \mathrm{pH}=4.5 \\
& \mathrm{pH}+\mathrm{pOH}=14 \\
& \Rightarrow \mathrm{pOH}=14-4.5=9.5
\end{aligned}$
Hence (C) is correct.

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2007
The plane 2x-2y+z=3 is rotated about its line of intersection with the xy-plane by an acute angle α . If the new position of the plane contains the point 3,1,1, then the value of cos⁡α is equal to

Options:
A 13
B 23
C 79
D 49

The correct answer is:
79

Solution:
Let the equation of the plane be

2x-2y+z-3+λz=0
∵ it contains the point 3,1,1
⇒6–2+1–3+λ=0
⇒λ=-2
Hence, the equation of the plane is
2x-2y-z-3=0
⇒cos⁡α=22-2-2+1-14+4+14+4+1=79

Topic: Mathematics, Three Dimensional Geometry, JEE Main