datavorous/entrance-exam-dataset huggingface dataset:
Liebermann test is given by which of the following

Options:
A CH3-CH2-COOH
B
C
D

The correct answer is:

Solution:
Only phenol gives Liebermann reaction. When phenol is dissolved in concentrated H2SO4 and few drops of aq. NaNO2 added, red colour is obtained which turns blue on addition of NaOH (aq).

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Liebermann’s test is used for which class of compounds -

Options:
A Alcohols
B Phenols
C Aldehydes
D Ketones

The correct answer is:
Phenols

Solution:
Liebermann’s test is given by phenol, phenol with NaNO2/H2SO4 give red coloured Indophenol which in alkaline medium change to blue.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Life cannot originate from inorganic materials at present because of

Options:
A A very high amount of oxygen in the atmosphere
B Very low atmospheric temperature
C Absence of raw materials
D High degree of environmental pollution

The correct answer is:
A very high amount of oxygen in the atmosphere

Solution:
The synthesis of organic molecules from simple molecules is no longer possible because of the oxidizing atmosphere.

Topic: Biology, Evolution, NEET
Life cycle of Ectocarpus and Fucus respectively are:

Options:
A Haplontic, Diplontic
B Diplontic, Haplodiplontic
C Haplodiplontic, Diplontic
D Haplodiplontic, Haplontic

The correct answer is:
Haplodiplontic, Diplontic

Solution:
Both Ectocarpus and Fucus belong to filamentous brown algae, which are found in the intertidal zones of rocky seashores almost throughout the world. Most algal genera are haplontic, some of them such as Ectocarpus, Polysiphonia, kelps are haplodiplontic, while Fucus is diplontic.

Topic: Biology, Plant Kingdom, NEET, NEET 2017
Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 1015, what is the energy of the source?

Options:
A 8.28 × 10–10J
B 9.65 × 10–10J
C 10.52× 10–10J
D 11.24× 10–10J

The correct answer is:
9.65 × 10–10J

Solution:
Correct Option is : (B)
9.65 × 10–10J

Topic: Chemistry, Structure of Atom, NEET
Ligase enzymes are used in genetic engineering to:

Options:
A Cut open plasmid DNA.
B Insert plasmids into bacteria
C Isolate the DNA making up a human gene
D Join human DNA to plasmid DNA.

The correct answer is:
Join human DNA to plasmid DNA.

Solution:
Ligase is an enzyme required to join the nucleotides together similar to the process in DNA replication of joining the Okazaki fragments. It does so by catalysing the joining of two molecules by making new chemical bond between them.

Topic: Biology, Biotechnology - Principles and Processes, NEET
Ligases is a class of enzymes responsible for catalysing the linking together of two compounds.
Which of the following bonds is not catalysed by it?

Options:
A $\mathrm{C}-\mathrm{C}$
B $\mathrm{P}-\mathrm{O}$
C $\mathrm{C}-\mathrm{O}$
D $\mathrm{C}-\mathrm{N}$

The correct answer is:
$\mathrm{C}-\mathrm{C}$

Solution:
Option (1) is the correct answer because, ligases are the enzymes that catalyse the linking together of 2 compounds, e.g., enzymes which catalyse joining of $\mathrm{C}-\mathrm{O}, \mathrm{C}-\mathrm{S}, \mathrm{C}-\mathrm{N}, \mathrm{P}-\mathrm{O}$ etc. bonds.
$\Rightarrow$ Options (2), (3) and (4) are wrong as ligase catalyse the joining of $\mathrm{P}-\mathrm{O}, \mathrm{C}-\mathrm{O} \& \mathrm{C}-\mathrm{N}$ bonds.

Topic: Biology, Biomolecules (B), NEET, NEET 2024 (Re-NEET)
Ligation of foreign DNA at which of the following site will result in loss of tetracycline resistance of pBR322?

Options:
A EcoR I
B BamH I
C Pst I
D Pvu I

The correct answer is:
BamH I

Solution:
Option (2) is the answer because,
In pBR322, the restriction site for $B a m \mathrm{H} \mathrm{I}$ is present in tet $^{\mathrm{R}}$ gene. Therefore, ligation of a foreign DNA will results in loss of tetracycline resistance and recombinants will be tetracycline sensitive.
Option (1) is not the answer because, In pBR322, the restriction site of EcoR I is present outside the region of $a m p^R$ gene and tet ${ }^R$ gene.
Option (3) and (4) are not the answers because, in pBR322, the restriction sites for $P s t I$ and $P v u I$ are present in $\mathrm{amp}^{\mathrm{R}}$ gene. So, ligation of foreign DNA will result in loss of ampicillin resistance and recombinants will be ampicillin sensitive.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2023 (Manipur)
Light energy is converted into chemical energy during the light-dependent phase and chemical energy is used to

Options:
A help in the photolysis of water
B produce ATP and NADPH2
C reduce carbon dioxide to carbohydrate during the light-independent phase
D both a and b

The correct answer is:
reduce carbon dioxide to carbohydrate during the light-independent phase

Solution:
In light-dependent reactions, solar energy is utilized with the help of chlorophyll in plants and converted to chemical energy in the form of NADPH and ATP which are electron carrier molecules. This chemical energy produced during the light-dependent phase is then used to reduce the water and carbon dioxide to glucose in the light-independent phase.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Light from sodium lamp is made to pass through two polaroids placed one after the other in the path of light. Taking the intensity of the incident light as 100%, the intensity of the outcoming light can be varied in the range

Options:
A 0% to 100%
B 0% to 50%
C 0% to 25%
D 0% to 75%

The correct answer is:
0% to 50%

Solution:
Let I0 be the intensity of incident light. As the light coming from sodium lamp is unpolarized, so the intensity of light emerging from the first polaroid is
I1=I02
If θ is the angle between two polaroids, then the intensity of light emerging from the second polaroid is
I2=I1cos2θ=I02cos2θ
But I0=100% (given)
∴    I2=50%cos2θ
Since θ varies from 90° to 0°, so the intensity of the outcoming light can be varied from 0% to 50%.

Topic: Physics, Wave Optics, NEET
Light has a wavelength 600 nm in free space. It passes into the glass, which has an index of refraction of 1.50. What is the frequency of the light inside the glass?

Options:
A 3.3Ã1014Â Hz
B 5.0Ã1014 Hz
C 3.3Ã105Â Hz
D 5.0Ã105Â Hz

The correct answer is:
5.0Ã1014 Hz

Solution:
Given, the wavelength of light =600 nm
=600×10-9 m
n=1.5
As the frequency of light cannot change, c=νλ
Speed of light in free space =3×108 m s-1
∴3×108=ν×600×10-9
⇒ν=3×108600×10-9
=5×1014 Hz

Topic: Physics, Ray Optics, NEET
Light hydrocarbons can be converted to a mixture of gaseous hydrocarbons by

Options:
A Hydrolysis
B Oxidation
C Cracking
D Distillation under reduced pressure

The correct answer is:
Cracking

Solution:
Alkanes on heating in absence of air at high temperature undergo pyrolysis (cracking) to give a mixture of lower hydrocarbons, alkenes, and H2 .

Topic: Chemistry, Hydrocarbons, NEET
Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig.. A current in the coil produces a deflection of 3.5o of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

Options:
A 19.6cm
B 18.4cm
C 14.6cm
D 11.9cm

The correct answer is:
18.4cm

Solution:
Correct Option is : (B)
18.4cm

Topic: Chemistry, Ray Optics, NEET
Light is incident normally on the face AB of a prism as shown in the figure. A layer of liquid of refractive index Î¼ is applied on the face AC of the prism. If the prism is made of glass of refractive index 32, then the limit of Î¼ for which total internal reflection takes place on face AC is

Options:
A μ<334
B μ>52
C μ>2
D μ<3

The correct answer is:
μ<334

Solution:
For critical angle at the glass-liquid interface we have

Î¼1sinC=Î¼2sin90Â°

â32sinC=Î¼

âsinC=2Î¼3

For total internal reflection

60Â°>C

âsin60Â°>sinC

â32>2Î¼3

Î¼<334

Topic: Physics, Ray Optics, NEET
Light of energy 2.0 eV falls on a metal of work function 1.4 eV. The stopping potential is

Options:
A 0.6 V
B 2.0 V
C 3.4 V
D 1.4 V

The correct answer is:
0.6 V

Solution:
From Planck's quantum theory, the maximum kinetic energy (Ek) of photoelectron emitted from the metal is
Ek=hv-W
Where W is work function of metal and hv is the energy of the photon absorbed by the metal.
Given, hv=2 eV, W=1.4 eV
â´ Ek=2-1.4=0.6 eV
Hence, stopping potential is , Vs=Ee=0.6 eVe=0.6 V.

Topic: Physics, Dual Nature of Matter, NEET
Light of energy 2.0 eV falls on a metal of work function 1.4 eV. The stopping potential is

Options:
A 0.6 V

B 2.0 V

C 3.4 V

D 1.4 V

The correct answer is:
0.6 V

Solution:
From Planck's quantum theory, the maximum kinetic energy (Ek) of photoelectron emitted from the metal is
Ek=hf-W
Where W is the work function of metal and hf is the energy of the photon absorbed by the metal.
Given, hf=2 eV, W=1.4 eV
∴ Ek=2-1.4=0.6 eV
Hence, stopping potential is , Vs=Ee=0.6 eVe=0.6 V.

Topic: Physics, Dual Nature of Matter, NEET
Light of energy 2.0 eV falls on a metal of work function 1.4 eV. The stopping potential is

Options:
A 0.6 V
B 2.0 V
C 3.4 V
D 1.4 V

The correct answer is:
0.6 V

Solution:
From Planck's quantum theory, the maximum kinetic energy (Ek) of photoelectron emitted from the metal is
Ek=hν-W
Where W is the work function of metal and hν is the energy of the photon absorbed by the metal.
Given, hν=2 eV, W=1.4 eV
∴ Ek=2-1.4=0.6 eV
Hence, stopping potential is , Vs=Ee=0.6 eVe=0.6 V

Topic: Physics, Dual Nature of Matter, NEET
Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material, photoelectric current is emitted. If the frequency of light is halved and intensity is doubled, the photoelectric current becomes

Options:
A 4 times the orginal current
B 2 times the original current
C half the original current
D zero times the original current

The correct answer is:
zero times the original current

Solution:
v1=1.5v0
v2=v12=0.75v0<v0
⇒ photoelectric effect (P.E.E.) not possible.

Topic: Physics, Dual Nature of Matter, NEET
Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled?

Options:
A four times
B one-fourth
C zero
D doubled

The correct answer is:
zero

Solution:
K1=1.5 hv0−ϕ0=0.5 hv0

K2=1.5hv02−hv0=−0.25 hv0

∵ Kinetic energy can never be negative so, no emission and i=0

OR

In second case the incident frequency is halved
Incident frequency =1.52v0=0.75 v0
Now the incident frequency is less than threshold frequency, so no emission of electron take place therefore no current. i=0

Topic: Physics, Dual Nature of Matter, NEET, NEET 2020 (Phase 1)
Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. The ratio of maximum speeds of emitted electrons will be

Options:
A 1 : 4

B 1 : 2

C 1 : 1

D 1 : 5

The correct answer is:
1 : 2

Solution:
Here work function, Ï0 = 0.5 eV
According to Einstein's photoelectric equation
Maximum kinetic = Incident – Work function
the energy of the emitted photon
electrons energy
∴ Kmax1 = 1. eV – 0.5 eV = 0.5 eV ...(i)
and Kmax2 = 2.5 eV – 0.5 eV = 2 eV ...(ii)
Divide (i) by (ii), we get
K max 1 K max 2 = 0 · 5 eV 2 eV = 1 4
1 2 mv max 1 2 1 2 mv max 2 2 = 1 4 or v max 1 v max 2 = 1 4 = 1 2

Topic: Physics, Dual Nature of Matter, NEET
Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. The ratio of maximum speeds of both emitted electrons will be :

Options:
A 1 : 4

B 1 : 2

C 1 : 1

D 1 : 5

The correct answer is:
1 : 2

Solution:
K . E = ϕ - ϕ 0
K.E1=1eV-0.5eV=0.5eV
K.E2=2.5eV-0.5eV=2eV
K.E1K.E2=0.5eV2eV=14
v1v2=14=12

Topic: Physics, Dual Nature of Matter, NEET
Light of two different frequencies whose photons have energies $1 \mathrm{eV}$ and $2.5 \mathrm{eV}$ respectively illuminate a metallic surface whose work function is $0.5 \mathrm{eV}$ successively. Ratio of maximum speeds of emitted electrons will be

Options:
A $1: 2$
B $1: 1$
C $1: 5$
D $1: 4$

The correct answer is:
$1: 2$

Solution:
Kinetic energy
$$
\mathrm{KE}=\phi-\phi_0
$$
Here, $\mathrm{KE}_1=1-0.5=0.5 \mathrm{eV}$
$$
\begin{aligned}
& \mathrm{KE}_2=2.5-0.5=2 \mathrm{eV} \\
& \therefore \frac{\mathrm{KE}_1}{\mathrm{KE}_2}=\frac{0.5}{2}=\frac{1}{4} \\
& \text { or } \quad \frac{y_1^2}{y_2^2}=\frac{1}{4} \\
& \text { or } \quad \frac{y_1}{y_2}=\sqrt{\frac{1}{4}}=\frac{1}{2}
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2011 (Screening)
Light of wavelength 400 nm is incident continuously on a Caesium ball (work function 1.9 eV ). The maximum potential to which the ball will be charged is

Options:
A 3.1 V
B 1.2 V
C zero
D infinite

The correct answer is:
1.2 V

Solution:
eVs=124004000 eV-1.9 eV =3.1-1.9 eV
eVs=1.2 eV
Vs=1.2 V

Topic: Physics, Dual Nature of Matter, NEET
Light of wavelength 488 nm is produced by an Argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photo electrons is 0.38 V. Find the work function of the material form which the emitter is made.

Options:
A 1.38 eV
B 2.55 eV
C 2.17 eV
D 2.93 eV

The correct answer is:
2.17 eV

Solution:
E=hcλ=1242 eV488=2.55 eV
Kmax=eV0=e×0.38=0.38 eV
Kmax=hcλ-ϕ0=E-ϕ0
ϕ0=E-Kmax=2.55-0.38=2.17 eV

Topic: Physics, Dual Nature of Matter, NEET
Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is:

Options:
A <2.8×10-9 m
B ≥2.8×10-9 m
C ≤2.8×10-12 m
D <2.8×10-10 m

The correct answer is:
≥2.8×10-9 m

Solution:
KEmax=hcλ-Ψ
⇒KEmax=1240500-2.28
⇒KEmax=2.48-2.28=0.2 eV
λmin=h2mKEmax=203×10-342×9×10-31×0.2×1.6×10-19
λmin=259×10-9=2.80×10-9 m
So, λ≥2.8×10-9 m.

Topic: Physics, Dual Nature of Matter, NEET, NEET 2015 (Phase 2)
Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is:

Options:
A <2.8Ã10-9 m
B â¥2.8Ã10-9 m
C â¤2.8Ã10-12 m
D <2.8Ã10-10 m

The correct answer is:
â¥2.8Ã10-9 m

Solution:
Given : work function ϕ of metal =2.28 eV
Wavelength of light λ=500 nm=500×10−9 m
KEmax=hcλ−ϕ
KEmax=6.6×10−34×3×1085×10−7×1.6×10-19−2.82
KEmax=2.48−2.28=0.2 eV
λmin=hp=h2m(KE)max
=203×10−342×9×10−31×0.2×1.6×10−19
λmin=259×10−9
=2.80×10−9 nm∴ λ≥2.8×10−9 m

Topic: Physics, Dual Nature of Matter, NEET
Light of wavelength 600 nm is incident normally on a slit of width 0.2 mm. The angular width of central maxima in the diffraction pattern is (measured from minimum to minimum)

Options:
A 6Ã10-3 rad
B 4Ã10-3 rad
C 2.4Ã10-3 rad
D 4.5Ã10-3 rad

The correct answer is:
6Ã10-3 rad

Solution:
Î»=600 nm=600Ã10-9m
d=0.2 mm=0.2Ã10-3m
âµ Linear width of central maximum,
x=2Î¸D
Where, 2Î¸: Angular width of central maximum.
We know that, x=2Î»Dd
Angular width 2Î¸.D=2Î»Dd
â 2Î¸=2Î»d=2Ã600Ã10-90.2Ã10-3
=6Ã10-7Ã104
=6Ã10-3 rad

Topic: Physics, Wave Optics, NEET
Light of wavelength 3000 â« is incident on a metal surface whose work function is 1eV. The maximum velocity of emitted photoelectrons will be

Options:
A 10 ms-1
B 103 ms-1
C 104 ms-1
D 106 ms-1

The correct answer is:
106 ms-1

Solution:
12mv2=hcÎ»-Ï0
=6.63Ã10-34Ã3Ã108(3Ã10-7)Ã1.6Ã10-19-1
=4.14-1=3.14eV
or v=2Ã3.14Ã1.6Ã10-199.1Ã10-31=106 ms-1

Topic: Physics, Dual Nature of Matter, NEET
Light of wavelength 5000 Aâ falls on a sensitive plate with the photoelectric work function of 1.9 eV. The maximum kinetic energy of the photoelectron emitted will be

Options:
A 1.16 eV

B 2.38 eV

C 0.58 eV

D 2.98 eV

The correct answer is:
0.58 eV

Solution:
As we know, energy of photo electrons,
E=hν=hcλ
E=6.6×10-343×1085000×10-10
=19.8×10-195 joule
=19.8×10-195×1.6×10-19 eV
= 19.8 8 = 2.48 eV
Now, hν=ϕ+12mvmax2
⇒ 12mvmax2=hν-ϕ
= 2.48 - 1.9 = 0.58 eV

Topic: Physics, Dual Nature of Matter, NEET
Light of wavelength Î»=5000 â« falls normally on a narrow slit. A screen placed at a distance of 1 m from the slit and perpendicular to the direction of light. First minima of the diffraction pattern is situated at 5 mm from the centre of central maximum. The width of the slit is

Options:
A 0.1 mm
B 1.0 mm
C 0.5 mm
D 0.2 mm

The correct answer is:
0.1 mm

Solution:
Position of nth minima
xn=nλDd
5×10-3=1×5000×10-10×1d
d=10-4 m=0.1 mm .

Topic: Physics, Wave Optics, NEET
Light of wavelength 5000 â« is incident over a slit of width 1 Î¼m. The angular width of central maxima will be

Options:
A 30°
B 60°
C 90°
D 120°

The correct answer is:
60°

Solution:
For first minima
sinθ=λa=5000×10-101×10-6=12
⇒ θ=30°

So the angular width of the central maxima is

2θ=60°

Topic: Physics, Wave Optics, NEET
Light of wavelength Î» which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with same velocity then stopping potential will

Options:
A increase
B decrease
C be zero
D become exactly half

The correct answer is:
increase

Solution:
According to photoelectric equation
hcÎ»-Ï=E where E=p22m(K.E)
If E is constant âthen 1Î»âÏ
If we decrease the wavelength Î» , then stopping potential Ï will increase such that hcÎ»-Ï= constant

Topic: Physics, Dual Nature of Matter, NEET
Light rays of wavelength 6000 â« and intensity 39.6 W m-2 is incident on a metal surface. If only one percent of photons incident on surface emit photoelectrons, then the number of electrons emitted per second per unit area from the surface will be

Options:
A 12×1018
B 10×1018
C 12×1017
D 12×1019

The correct answer is:
12×1017

Solution:
Î» = 6000Ã10-10 m
Î= 39.6 W m-2
1% of photons emit photoelectrons.
I=nhcAtÎ»
39.6 = nÃ6.6Ã10-34Ã3Ã1081Ã1Ã6000Ã10-10
39.6=nÃ1.1Ã10-24Ã3Ã105
np = 12Ã1019
No. of photoelectrons = 1% of photons
ne=1100Ã12Ã1019
ne=12Ã1017

Topic: Physics, Dual Nature of Matter, NEET
Light travels a distance x in time t1 in air and 10x in time t2 in another denser medium. What is the critical angle for this medium?

Options:
A sin-1(10t1t2)
B sin-1t2t1
C sin-110t2t1
D sin-1t110t2

The correct answer is:
sin-1(10t1t2)

Solution:
Applying Snell's law for refraction from denser to rarer medium, we get

μdenser×sinC=μair×sin90°⇒cvdensersinC=cvair×1⇒c10xt2sinC=cxt1⇒sinC=10t1t2⇒C=sin-110t1t2

Topic: Physics, Ray Optics, NEET, NEET 2023 (All India)
Light with an average flux of 20 W cm-2 falls on a non-reflecting surface at normal incidence having surface area 20 cm2. The energy received by the surface during time span of 1 min is:

Options:
A 12×103 J
B 24×103 J
C 48×103 J
D 10×103 J

The correct answer is:
24×103 J

Solution:
We know that, I=EAt

E=IAt

⇒E=2010−4×20×10−4×60

⇒E=24×103 J

Topic: Physics, Electromagnetic Waves, NEET, NEET 2020 (Phase 1)
Light with an energy flux of 25×104 W m-2 falls on a perfectly reflecting surface at normal incidence. If the surface area is 15 cm2, the average force exerted on the surface is:

Options:
A 1.25×10-6N
B 2.50×10-6N
C 1.20×10-6N
D 3.0×10-6N

The correct answer is:
2.50×10-6N

Solution:
Here, we have to calculate the average force calculated at the surface.

Given

I = 25×104

A = 15×10-4

We know that,

F=2IAC=225×10415×10-43×108=2.5×10-6N

Hence, force will be 2.5×10-6N

So Option (2) will be correct.

Topic: Physics, Electromagnetic Waves, NEET, NEET 2014
Lights of wavelengths Î»1=340 nm and Î»2=540 nm are incident on a metallic surface. If the ratio of the maximum speeds of the electrons ejected is 2, the work function of the metal is

Options:
A 2 eV
B 1.85 eV
C 1 eV
D 1.5 eV

The correct answer is:
1.85 eV

Solution:
λ1=350 nm, λ2=540 nm, V1V2=2⇒V1=2V and V2=V
hcλ1=ϕ+12m(2V)2
hcλ2=ϕ+12mV2
hcλ1-hcλ2×4=ϕ-4ϕ
hc(4540 nm-1340 nm)=3ϕ
ϕ=1.8 eV

Topic: Physics, Dual Nature of Matter, NEET
Likely bond angles of SF4 molecule are :

Options:
A 1200 , 1800
B 450 , 1180
C 1170 , 920
D 890,1170

The correct answer is:
890,1170

Solution:
SF4 molecule : Sp3d hybridization with one lone pair providing repulsion.
Therefore, bond angles change from 90 â¶ 89o (apical )
and from 120 â¶ 1170

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Lime stone is added in the extraction of lead from galena why?

Options:
A It prevents the formation of PbSO4
B It remove the impurity of silica as fusible slag
C It converts lead silicate to lead oxide
D All of the above

The correct answer is:
All of the above

Solution:
CaO+SiO2âCaSiO3 Slag
PbO+SiO2âPbSiO3
CaO converts PbSiO3 to PbO,
PbSiO3+CaOâPbO+CaSiO3
and also prevents the formation of PbSO4.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
Limit of BOD prescribed by Central Pollution Control Board for the discharge of industrial and municipal waste waters into natural surface waters, is:

Options:
A $ < 30 \mathrm{ppm}$
B $ < 3.0 \mathrm{ppm}$
C $ < 10 \mathrm{ppm}$
D $ < 100 \mathrm{ppm}$

The correct answer is:
$ < 10 \mathrm{ppm}$

Solution:
The central pollution control board prescribed the BOD limit for the discharge of industrial and municipal waste water is less than $10 \mathrm{ppm}$.

Topic: Biology, Environmental Issues, NEET, NEET 2006
Limit of the environment to support a population is called its

Options:
A Biotic Potential
B Purifying capacity
C Carrying Capacity
D Environmental Resistance

The correct answer is:
Carrying Capacity

Solution:
In ecological terms, the carrying capacity of an ecosystem is the size of the community or population that can be supported indefinitely upon the available resources and services of that ecosystem. Living within the limits of an ecosystem depends on three factors: the number of resources available in the ecosystem, the size of the population or community and the number of resources each individual within the community is consuming.

Topic: Biology, Organisms and Populations, NEET
Limiting molar conductivity of $\mathrm{NH}_4 \mathrm{OH~} \left( \text{i.e., } \stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{OH}\right)}\right)$ is equal to

Options:
A $\stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}+\stackrel{\circ}{\Lambda}_{m(\mathrm{NaCl})}-\stackrel{\circ}{\Lambda}_{m(\mathrm{NaOH})}$
B $\stackrel{\circ}{\Lambda}_{m(\mathrm{NaOH})}+\stackrel{\circ}{\Lambda}_{m(\mathrm{NaCl})}-\stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}$
C $\stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{OH}\right)}+\stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}-\stackrel{\circ}{\Lambda}_{m(\mathrm{HCl})}$
D $\stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}+\stackrel{\circ}{\Lambda}_{m(\mathrm{NaOH})}-\stackrel{\circ}{\Lambda}_{m(\mathrm{NaCl})}$

The correct answer is:
$\stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}+\stackrel{\circ}{\Lambda}_{m(\mathrm{NaOH})}-\stackrel{\circ}{\Lambda}_{m(\mathrm{NaCl})}$

Solution:
According to Kohlrausch's law, limiting molar conductivity of $\mathrm{NH}_4 \mathrm{OH}$
$$
\Lambda_{m\left(\mathrm{NH}_4 \mathrm{OH}\right)}^{\circ}=\Lambda_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}^{\circ}+\Lambda_{m(\mathrm{NaOH})}^{\circ}-\Lambda_{m(\mathrm{NaCl})}^{\circ}
$$

Topic: Chemistry, Electrochemistry, NEET, NEET 2012 (Screening)
Linear polyenes on ozonolysis gives two moles of acetaldehyde and one mole of propanedinal. Linear polyene will be

Options:
A Alkadiene
B Alkatriene
C Alkatetraene
D Alkapentaene

The correct answer is:
Alkadiene

Solution:
Polyene is alkadiene i.e., hepta-2,5-di-ene and the reaction is as follows
CH3-CH=CH-CH2-CH=CH-CH3→CH3-CHO+OHC-CH2-CHO+CH3CHO

Topic: Chemistry, Hydrocarbons, NEET
Linkage isomerism in coordination compounds typically occurs with which type of ligands?

Options:
A Unidentate ligands
B Ambidentate ligands
C Polydentate ligands
D Monodentate ligands

The correct answer is:
Ambidentate ligands

Solution:
Linkage isomerism in coordination compounds arises when ambidentate ligands can bind to the metal center through different atoms.

Topic: Chemistry, Coordination Compounds, NEET