datavorous/entrance-exam-dataset huggingface dataset:
In the context of water's dual role as an acid and a base, what determines whether water acts as an acid or a base?

Options:
A The temperature of the reaction
B The concentration of water in the solution
C The identity of the other reactants
D The pH of the solution

The correct answer is:
The identity of the other reactants

Solution:
The paragraph suggests that whether water acts as an acid or a base depends on the identity of the other reactants involved in the specific reaction.

Topic: Chemistry, Ionic Equilibrium, NEET
In the conversion of benzene sulphonic acid to sodium phenoxide, what type of reaction occurs?

Options:
A Acid-base reaction
B Redox reaction
C Nucleophilic substitution
D Elimination reaction

The correct answer is:
Acid-base reaction

Solution:
The conversion of benzene sulphonic acid to sodium phenoxide involves an acid-base reaction, neutralising the acidic sulfonic group with sodium hydroxide.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
In the conversion of carboxylic acids to esters, which reagent is typically used?

Options:
A LiAlH4
B H2O2
C Pd/C
D Alcohol and an acid catalyst

The correct answer is:
Alcohol and an acid catalyst

Solution:
The conversion of carboxylic acids to esters typically involves the use of alcohol and an acid catalyst.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
In the conversion of n-alkanes to benzene and its homologues, which of the following metals is often used as a catalyst support?

Options:
A Gold (Au)
B Aluminum (Al)
C Platinum (Pt)
D Iron (Fe)

The correct answer is:
Aluminum (Al)

Solution:
Aluminum is commonly used as a support material for catalysts in the aromatization or reforming reaction.

Topic: Chemistry, Hydrocarbons, NEET
In the coordination compound [PtCl4]2–, how many ligand atoms are directly attached to the central Pt ion?

Options:
A 2
B 4
C 6
D 8

The correct answer is:
4

Solution:
In the coordination compound [PtCl4]2–, four ligand atoms are directly attached to the central Pt ion.

Topic: Chemistry, Coordination Compounds, NEET
In the coordination compound [PtCl4]2–, what is the coordination polyhedron around the central Pt ion?

Options:
A Octahedral
B Square planar
C Tetrahedral
D Square bipyramidal

The correct answer is:
Square planar

Solution:
The coordination polyhedron around the central Pt ion in[PtCl4]2– is square planar.

Topic: Chemistry, Coordination Compounds, NEET
In the cubic close-packed (ccp) or face-centered cubic (fcc) structure, how many layers of spheres are required to complete one pattern cycle?

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
In the cubic close-packed (ccp) or face-centered cubic (fcc) structure, two layers of spheres are required to complete one pattern cycle.

Topic: Chemistry, Solid State, NEET
In the cubic close-packed (ccp) or face-centered cubic (fcc) structure, what pattern of layers is often written?

Options:
A ABCABC...
B AABBAA...
C ABAB...
D AAABBB...

The correct answer is:
ABCABC...

Solution:
In the cubic close-packed (ccp) or face-centered cubic (fcc) structure, the pattern of layers is often written as ABCABC...

Topic: Chemistry, Solid State, NEET
In the cyanide extraction process of silver from argentite ore, the oxidizing and reducing agents used are

Options:
A HNO3 and CO respectively
B HNO3 and Zn dust respectively
C O2 and Zn dust respectively
D O2 and CO respectively.

The correct answer is:
O2 and Zn dust respectively

Solution:
In the cyanide extraction process of silver from argentite ore, the oxidizing and reducing agents used are O2 and Zn dust respectively.
Ag 2 S+2 CN − + O 2 →2 Ag (CN) 2 − + SO 2
2 Ag (CN) 2 − +Zn→ Zn (CN) 4 2− +2Ag

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
In the cycle shown in Fig. 12.10, where a cycle followed by an engine (made of one mole of perfect gas in a cylinder with a piston) is depicted, with the following conditions:
A to B: Volume constant B to C: Adiabatic C to D: Volume constant D to A: Adiabatic VC=VD=2VA=2VBIn which part of the cycle is heat supplied to the engine from outside?

Options:
A A to B
B B to C
C C to D
D D to A

The correct answer is:
A to B

Solution:
Correct Option is : (A)
A to B

Topic: Chemistry, Thermodynamics, NEET
In the cycle shown in Fig. 12.10, where a cycle followed by an engine (made of one mole of perfect gas in a cylinder with a piston) is depicted, with the following conditions:
A to B: Volume constant B to C: Adiabatic C to D: Volume constant D to A: Adiabatic VC=VD=2VA=2VBIn which part of the cycle heat is being given to the surrounding by the engine?

Options:
A A to B
B B to C
C C to D
D D to A

The correct answer is:
C to D

Solution:
Correct Option is : (C)
C to D

Topic: Chemistry, Thermodynamics, NEET
In the cyclic process shown in P-V diagram, the magnitude of the work done is

Options:
A πP2-P122
B πV2-V122
C π4(P2-P1)(V2-V1)
D π(V2-V1)2

The correct answer is:
π4(P2-P1)(V2-V1)

Solution:
Work done = Area bounded by the circle
=π(P2-P1)2(V2-V1)2
=π4(P2-P1)(V2-V1)

Topic: Chemistry, Thermodynamics (C), NEET
In the dark reaction of photosynthesis, assimilatory power is used for the reduction of

Options:
A PGA
B PGAL
C RuBP
D PEP

The correct answer is:
PGA

Solution:
The ATP and NADPH formed during light reaction is used for the assimilation of PGA to form carbohydrates in dark reaction.

Topic: Biology, Photosynthesis in Higher Plants, NEET
In the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by

Options:
A increasing the filament current
B decreasing the filament current
C decreasing the potential difference between the anode and filament
D increasing the potential difference between the anode and filament

The correct answer is:
increasing the potential difference between the anode and filament

Solution:
In the Davission and Germer experiment, the velocity of electron emitted from the electron gun can be increased by increasing the potential difference between the anode and filament.

Topic: Physics, Dual Nature of Matter, NEET, NEET 2011 (Screening)
In the Î²+ decay process, the following changes take place inside the nucleus

Options:
A AZX→A Z-1Y+e++γ
B AZX→A Z+1Y+e-+γ-
C AZX→YZA+e-+γ
D AZX→YZA+e-+γ-

The correct answer is:
AZX→A Z-1Y+e++γ

Solution:
Nuclides for which NZ is too small for stability can emit a positron, the electron's antiparticle which is identical to the electron, but with positive charge. The basic process called Î²+ decay.
XZA âYZ-1A+ e++Î³
(Î³=neutrino)

Topic: Physics, Nuclear Physics, NEET
In the decomposition of N2O5, what is the effect of higher temperature on the rate of decomposition?

Options:
A It decreases the rate of decomposition.
B It has no effect on the rate of decomposition.
C It increases the rate of decomposition.
D It stops the decomposition.

The correct answer is:
It increases the rate of decomposition.

Solution:
At a higher temperature (50°

Topic: Chemistry, Chemical Kinetics, NEET
In the dehalogenation of vicinal dihalides, which element is removed from the molecule to form an alkene?

Options:
A Carbon
B Hydrogen
C Halogen
D Zinc

The correct answer is:
Halogen

Solution:
In the dehalogenation of vicinal dihalides, halogen atoms are removed from the molecule to form an alkene.

Topic: Chemistry, Hydrocarbons, NEET
In the dehydrohalogenation reaction, the rate of the reaction is affected by the nature of the halogen atom. Which halogen has the lowest rate of reaction?

Options:
A Iodine
B Bromine
C Chlorine
D Fluorine

The correct answer is:
Fluorine

Solution:
Fluorine has the lowest rate of reaction in dehydrohalogenation among the halogens.

Topic: Chemistry, Hydrocarbons, NEET
In the density measurement of a cube, the mass and edge length are measured as 10.00Â±0.10 kg and 0.10Â±0.01 m, respectively. The error in the measurement of density is:

Options:
A 0.31
B 0.10
C 0.07
D 0.01

The correct answer is:
0.31

Solution:
Density ρ=massvolume
∆ρρ=∆mm+∆VV
=∆mm+3∆aa ∵V=a3
=0.110+30.010.1
=0.01+30.1
=0.31

Topic: Physics, Units and Dimensions, NEET
In the descending order of a group in this periodic table which of this following would be true?
1. All the atoms have the same number of valence electrons
2. Gram atomic volume increases
3. Electronegativity decreases
4. Metallic character decreases and the basic nature of their oxides decreases
Select the correct answer using the codes given below options

Options:
A 2, 3 and 4
B 1, 2 and 4
C 1, 3 and 4
D 1, 2 and 3

The correct answer is:
1, 2 and 3

Solution:
Fourth statement is not true as the correct statement is metallic character increases and the basic nature of their oxides increases. Rest other statements are correct.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
In the developing ovule of most of the Angiosperms, megaspore mother cell is situated within the nucellus

Options:
A towards the micropyle
B towards the chalaza
C In the centre, equidistant from micropyle and chalaza
D towards the raphe

The correct answer is:
towards the micropyle

Solution:
The developing ovule of most of the angiosperm originates from a single megaspore mother cell present in the nucellus that is situated towards the micropylar end.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In the developing ovule of most of the angiosperms, megaspore mother cell is situated within the nucellus

Options:
A towards the micropyle
B towards the chalaza
C in the centre, equidistant from micropyle and chalaza
D towards the raphe

The correct answer is:
towards the micropyle

Solution:
One cell of the nucellus towards the micropylar end enlarges and derives nutrition from other nucellar cells. This is called megaspore mother cell. It undergoes meiosis to form a linear tetrad of four haploid megaspores. Out of these, three megaspores towards the micropylar end disintegrate. The one towards the chalaza is the functional megaspore.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In the development of Angiosperms, generally

Options:
A The development of endosperm starts before the development of embryo.
B The development of endosperm starts after the development of an embryo.
C The development of endosperm starts along with the development of embryo.
D The endosperm disappears before the development of embryo starts.

The correct answer is:
The development of endosperm starts before the development of embryo.

Solution:
After double fertilization in Angiosperms, a triploid primary endosperm nucleus and a diploid zygote is formed. First the primary endosperm nucleus undergoes cell division to form a triploid endosperm. This will be used as a nutritive tissue for the developing embryo.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In the developmental history of mammalian heart, it is observed that it passes through a two-chambered fish-like heart, three-chambered frog-like heart and finally four-chambered stage. To which hypothesis can the above-cited statement be approximated?

Options:
A Biogenetic law
B Hardy-Weinberg law
C Lamarck's principle
D Mendelian principles

The correct answer is:
Biogenetic law

Solution:
The biogenetic law is a theory of development and evolution proposed by Ernst Haeckel in the 1860s. It is one of several recapitulation theories, which posit that the stages of development for an animal embryo are the same as other animals' adult stages or forms. Commonly stated as ontogeny recapitulates phylogeny, the biogenetic law theorizes that the stages an animal embryo undergoes during development are a chronological replay of that species' past evolutionary forms. The biogenetic law states that each embryo's developmental stage represents an adult form of an evolutionary ancestor. According to the law, by studying the stages of embryological development, one can understand the history and diversification of life on Earth.

Topic: Biology, Evolution, NEET
In the diagram, a nucleosome is given. Select the option with correct identification.

Options:
A

A	B	C
RNA	H1 histone	Non-histone protein

B

A	B	C

DNA	Linker histone	Histone octamer

C

A	B	C
DNA	Histone octamer	H1 histone

D

A	B	C

DNA	H1 histone	Histone tetramer

The correct answer is:

A	B	C

DNA	Linker histone	Histone octamer

Solution:
Histone octamer : It is made up of 8 molecules of histone proteins. Histone proteins are positively charged basic proteins. These proteins are rich in basic amino acids like lysines and arginines. Both of these amino acids possess positive charges in their side chains.
Nucleosome : DNA (negatively charged) wraps the positively charged histone octamer. In a nucleosome, about 200 bp are present in DNA.

Topic: Biology, Molecular Basis of Inheritance, NEET
In the diagram identify A, B and C.

Options:
A A - Reissner's membrane; B - Organ of corti; C - Basilar membrane
B A - Reissner's membrane; B - Tectorial membrane; C - Basilar membrane
C A - Scala vestibuli; B - Scala media; C - Scala tympani
D A - Tectorial membrane; B - Reissner's membrane; C - Basilar membrane

The correct answer is:
A - Reissner's membrane; B - Organ of corti; C - Basilar membrane

Solution:

Topic: Biology, Neural Control and Coordination, NEET
In the diagram of the lateral view of the human brain, parts are indicated by alphabets. Select the answer in which these alphabets have been correctly matched with the parts which they indicate.

Options:
A A = Temporal lobe, B = Parietal lobe, C = Cerebellum, D = Medulla oblongata, E = Frontal lobe
B A = Frontal lobe, B = Temporal lobe, C = Cerebrum, D = Medulla oblongata, E = Occipital lobe
C A = Temporal lobe, B = Parietal lobe, C = Cerebrum, D = Medulla oblongata, E = Frontal lobe
D A = Frontal lobe, B = Temporal lobe, C = Cerebellum, D = Medulla oblongata, E = Parietal lobe

The correct answer is:
A = Frontal lobe, B = Temporal lobe, C = Cerebellum, D = Medulla oblongata, E = Parietal lobe

Solution:
A = Frontal lobe, B = Temporal lobe, C = Cerebellum, D = Medula oblongata, E = Parietal lobe

Topic: Biology, Neural Control and Coordination, NEET
In the diagram shown, the normal reaction force between $2 \mathrm{~kg}$ and $1 \mathrm{~kg}$ is (Given $g=10 \mathrm{~ms}^{-2}$) (Consider the surface, to be smooth):

Options:
A $25 \mathrm{~N}$
B $39 \mathrm{~N}$
C $6 \mathrm{~N}$
D $10 \mathrm{~N}$

The correct answer is:
$25 \mathrm{~N}$

Solution:
Total force along inclined plane,
$\begin{aligned}
f & =-\mathrm{F}_1+6 g \sin 30^{\circ}+\mathrm{F}_2 \\
f & =-60+6 g \sin 30^{\circ}+18 \\
f & =-60+6 \times 10 \times \frac{1}{2}+18 \\
f & =12 \mathrm{~N} \text { (downwards) } \\
\text {acceleration, } a & =\frac{f}{m}=\frac{12}{6}=2 \mathrm{~m} / \mathrm{s}^2
\end{aligned}$

As per question, the normal force between $2 \mathrm{~kg}$ and $1 \mathrm{~kg}$ is,
$\begin{aligned}
\mathrm{N}-18-10 \sin 30^{\circ} & =m a \\
\mathrm{~N}-18-10 \times \frac{1}{2} & =1 \times 2 \\
\mathrm{~N} & =2+23 \\
\mathrm{~N} & =25 \mathrm{~N}
\end{aligned}$

Topic: Physics, Laws of Motion, NEET, NEET 2022 (Phase 2)
In the diagram shown the potential energy U of a particle is plotted against its position x from the origin. Then which of the following statement is correct. A particle at

Options:
A x1 is in stable equilibrium
B x2 is in stable equilibrium
C x3 is in stable equilibrium
D None of these

The correct answer is:
None of these

Solution:
x = x1 and x = x3 are not equilibrium positions because dU dx ≠ 0 at these points.
x = x2 is unstable, as U is maximum at this point.

Topic: Physics, Work Power Energy, NEET
In the diagram, the graph between volume and pressure for a thermodynamic process is shown. If UA=0,UB=20 J and the energy given from B to C is 30 J, then at the stage of C, the internal energy of the system is

Options:
A 50 J
B 60Â J
C 30Â J
D 10Â J

The correct answer is:
50 J

Solution:
WBC=0
∴ ΔUBC=ΔQ
UC-UB=30
UC-20=30
UC=50 J

Topic: Physics, Thermodynamics, NEET
In the diagram, the intensity of waves arriving at D from two coherent sources s1 and s2 is I0. The wavelength of the wave is Î» = 4 m. Resultant intensity at D will be

Options:
A 4I0
B I0
C 2I0
D Zero

The correct answer is:
2I0

Solution:
Path difference
Δx=s2D-s1D=5-4=1 m
∴ The corresponding phase difference will be
ϕ = 2 π λ · Δ x = 2 π 4 1 = π 2
Using I=I1+I2+2I1 I2 cos ϕ
I=I0+I0+2I0 I0 cosπ2=2I0

Topic: Physics, Waves and Sound, NEET
In the diagram, the x-axis denotes the number of individuals with a certain phenotype. The dotted lines denote the phenotypes favoured by natural selection. Which of the following inferences can be drawn from the given diagram?

Options:
A This natural selection will lead to stabilisation.
B In this pattern of natural selection, more individuals will acquire the mean character value.
C This natural selection will lead to a directional change.
D This natural selection will lead to disruption.

The correct answer is:
This natural selection will lead to disruption.

Solution:
Natural selection can lead to stabilisation (in which more individuals acquire mean character value), directional change (more individuals acquire value other than the mean character value) or disruption (more individuals acquire peripheral character value at both ends of the distribution curve). The characteristic two peaks of the graph clearly indicate a disruptive selection.

Topic: Biology, Evolution, NEET
In the dichromate ion Cr2O72-

Options:
A 4Cr-O bonds are equivalent
B 6Cr-O bonds are equivalent
C All Cr-O bonds are equivalent
D None of Cr-O bonds are equivalent

The correct answer is:
6Cr-O bonds are equivalent

Solution:
null

Topic: Chemistry, d and f Block Elements, NEET
In the diploblastic animals:

Options:
A Mesoderm and mesoglea both are absent
B Mesoderm is present but mesoglea is absent
C Mesoderm is absent but mesoglea is present
D Mesoderm and mesoglea both are present.

The correct answer is:
Mesoderm is absent but mesoglea is present

Solution:
Animals in which the cells are arranged in two embryonic layers, external ectoderm and an internal endoderm, are called diploblastic animals, e.g., coelenterates. An undifferentiated layer, mesoglea, is present in between the ectoderm and the endoderm.

Topic: Biology, Animal Kingdom, NEET
In the displacement versus time curve for a particle executing Simple Harmonic Motion (S.H.M.), as shown in Fig. 14.7, which points indicate the maximum speed of the oscillator?

Options:
A [A], [C], [E], [G]
B [B], [D], [F], [H]
C [A], [B], [G], [H]
D [C], [D], [E], [F]

The correct answer is:
[B], [D], [F], [H]

Solution:
Correct Option is : (B)
[B], [D], [F], [H]

Topic: Chemistry, Oscillations, NEET
In the DNA molecule

Options:
A the total amount of purine nucleotides and pyrimidine nucleotides is not always equal
B there are two strands, which run parallel in the $5^{\prime} \rightarrow 3$ direction
C the proportion of adenine in relation to thymine varies with the organism
D there are two strands, which run antiparallel-one in $5^{\prime} \rightarrow 3$ direction and other in $3 \rightarrow 5^{\prime}$

The correct answer is:
there are two strands, which run antiparallel-one in $5^{\prime} \rightarrow 3$ direction and other in $3 \rightarrow 5^{\prime}$

Solution:
In DNA molecule the adjacent deoxyribonucleotides are joined in a chain by phosphodiester bridges or bonds, which link the $5^{\prime}$ carbon of deoxyribose of one mononucleotide unit with 3 carbon of deoxyribose of next mononucleotide unit. According to Watson and Crick DNA molecule consists of two such polynucleotide chains wrapped helically around each other, with the sugar phosphate chain on the outside and purine and pyrimidine on the inside of helix. The two strands run antiparallely, ie, one strand has phosphodiester linkage in $3 \rightarrow 5^{\prime}$ direction while other strand has phosphodiester linkage in $5^{\prime} \rightarrow 3$ direction.
Chargaff (1950) suggested that despite wide compositional variations exhibited by different types of DNA the total amount of purines equaled the total amount of pyrimidines $(A+$ $\mathrm{G}=\mathrm{T}+\mathrm{C}$.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2008 (Screening)
In the earlier classification system, bacteria, fungi, ferns, gymnosperms and angiosperms are included under Plantae on the basis of

Options:
A mode of nutrition
B presence of similar type of DNA
C unicellular and multicellular characters
D presence of cell wall.

The correct answer is:
presence of cell wall.

Solution:
In Linnaeus' time, a two-kingdom system of classification with Plantae and Animalia kingdoms was developed that included all plants and animals, respectively. This system did not distinguish between the eukaryotes and prokaryotes, unicellular and multicellular organisms and photosynthetic (green algae) and non-photosynthetic (fungi) organisms. Bacteria, blue-green algae, fungi, ferns, gymnosperms and angiosperms all possess a cell wall. Animals lack a cell wall.

Topic: Biology, Biological Classification, NEET
In the electrochemical cell
Zn|ZnSO4(0.01M)CuSO4(1.0M)|Cu, the emf of this Daniel cell is E1 . When the concentration of ZnSO4 is changed to 1.0M and that of CuSO4 changed to 0.01M, the emf changes to E2 . From the followings, which one is the relationship between E1 and E2 ?
(Given, RTF=0.059 )

Options:
A E1<E2
B E1>E2
C E2=0≠E1
D E1=E2

The correct answer is:
E1>E2

Solution:
E  =E o − 0.059 2 log 10 Zn 2+ Cu 2+
E 1 =EÂ°â 0.059 2 log 10 0.01 (1) 2
E 2 =EÂ°â 0.059 2 log 10 1 (0.01) 2
Hence E1>E2

Topic: Chemistry, Electrochemistry, NEET
In the electrochemical cell:
ZnZnSO40.01MCuSO41.0 MCu, the emf of this Daniel cell is E1 . When the concentration ZnSO4 is changed to 1.0M and that of CuSO4 changed to 0.01M, the emf changes to E2 . From the following, which one is the relationship between E1 and E2? (Given, RTF=0.059)

Options:
A E1=E2
B E1<E2
C E1>E2
D E2=0 ≠ E1

The correct answer is:
E1>E2

Solution:
ZnZnSO40.01 MCuSO41.0MCu
E=Ecello−2.303RT2Flog[Zn+2][Cu+2]
∴ E1=Ecello-2.303RT2×F×log⁡0.011
When concentrations are changed
∴ E2=Ecello-2.303RT2F×log⁡10.01
i.e., E1>E2

Topic: Chemistry, Electrochemistry, NEET, NEET 2017
In the electrolysis of silver nitrate, the mass of silver deposited is plotted against the charge

Slope of the line gives

Options:
A the equivalent mass of silver
B electrochemical equivalent of silver
C the value of faraday
D the current passed through the cell

The correct answer is:
electrochemical equivalent of silver

Solution:
W = ZQ
y = mx
Slope of the line (m) will give the value of electrochemical equivalent of silver.

Topic: Chemistry, Electrochemistry, NEET
In the electrolysis of water, 1 F of electrical energy would evolve-

Options:
A 1 mole of oxygen
B 1 g atom of oxygen
C 8 g of oxygen
D 22.4 L of oxygen

The correct answer is:
8 g of oxygen

Solution:
Hydrolysis of water can be represented by the following equations
2H2O â O2+ 4H++ 4e-
âµ 4 Faraday of charge liberate O2 = 32 g
∴ 1 Faraday of charge liberate O2 =324 g
= 8 g

Topic: Chemistry, Electrochemistry, NEET
In the electrolytic cell, flow of electrons is from

Options:
A cathode to anode in solution
B cathode to anode through external supply
C cathode to anode through internal supply
D anode to cathode through internal supply.

The correct answer is:
cathode to anode through external supply

Solution:
In an electrolytic cell, electrons do not flow themselves. It is the migration of ions towards oppositely charged electrodes that indirectly constitutes the flow of electrons from cathode to anode through external supply.

Topic: Chemistry, Electrochemistry, NEET
In the electrolytic process for the manufacture of chlorine, where is chlorine liberated?

Options:
A Cathode
B Anode
C Electrolyte
D Salt bridge

The correct answer is:
Anode

Solution:
In the electrolytic process for the manufacture of chlorine, chlorine is liberated at the anode.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
In the electrolytic refining process, what serves as the anode?

Options:
A A strip of pure metal
B A strip of impure metal
C A strip of metal oxide
D A strip of metal salt

The correct answer is:
A strip of impure metal

Solution:
The impure metal is made to act as the anode in the electrolytic refining process.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET