datavorous/entrance-exam-dataset huggingface dataset:
In the following reaction, we start with 2 mol of N2 and 5 mol of H2 exerting a total pressure of 7 atm at a given temperature in a closed vessel. When 50% of N2 is converted into NH3

N 2 +3H 2 →2 NH 3

Partial pressure of NH3 is :

Options:
A 2.8Â atm
B 2 atm
C 3.2Â atm
D 4Â atm

The correct answer is:
2 atm

Solution:
N 2 g + 3 H 2 g â 2 NH 3 g Initial 2 atm 5 atm 0 Partial pressures Reaction - 1 atm - 3 atm + 2 atm -------------- -------------- -------------- Final values 1 atm 2 atm 2 atm
Partial pressure of NH 3 =2 atm

Topic: Chemistry, States of Matter, NEET
$$
\text { In the following reactions, }
$$
$$
\text { the major products }(A) \text { and }(C) \text { are respectively }
$$

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, NEET, NEET 2011 (Screening)
In the following reactions, products (A) and (B) , respectively, are
NaOH hot and conc.  +  Cl 2 →(A) + side products
Ca(OH) 2 dry  +  Cl 2 → (B) + side products

Options:
A NaOCl and Ca(ClO3)2
B NaClO3 and Ca(OCl)2
C NaOCl and Ca(OCl)2
D NaClO3 and Ca(ClO3)2

The correct answer is:
NaClO3 and Ca(OCl)2

Solution:
Here the products formed (A) and (B), are NaClO3 and Ca(OCl)2 respectively
6NaOH+3Cl2→5NaCl+NaClO3+3H2O
2Ca(OH)2+Cl2→Ca(OCl)2+CaCl2+H2O

Topic: Chemistry, s Block Elements, NEET
In the following reactions, reactants A, B and C are:

C2H5NH2+Aâ¶C2H5N=CH-C6H5+H2O

Urea +Bâ¶H2N-NHCONH2+NH3

C 2 H 5 NH 2 +Câ C 2 H 5 Cl+ H 2 O+ N 2

Options:
A CH3CHO,NH2-NH2 and PCl5
B C6H5CHO, NH2-NH2 and SOCl2
C C6H5CHO, NH2-NH2 and NOCl
D CH3CHO, NH2-NH2 and PCl3

The correct answer is:
C6H5CHO, NH2-NH2 and NOCl

Solution:

Topic: Chemistry, Amines, NEET
In the following reactions, the major product W is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Amines, NEET
In the following sequence of reaction,

CH 3 CH 2 CH 2 CH 2 Br → alc. KOH A → HBr B → aq. KOH C ,

The product 'C' is ?

Options:
A Butan - 2 - ol
B Butan - 1 - ol
C Butyne
D Butene

The correct answer is:
Butan - 2 - ol

Solution:
C H 3 C H 2 C H 2 C H 2 Br â alc. KOH C H 3 âC H 2 âCH=C H 2 +HBr â HBr C H 3 C H 2 CH Br | âC H 3 â aq. KOH C H 3 âC H 2 â CH OH | âC H 3 ( A ) ( B ) ( C )

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
In the following sequence of reaction, what is D ?

Options:
A Primary amine
B An amide
C phenyl isocynate
D A chain lengthened hydrocarbon

The correct answer is:
phenyl isocynate

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, NEET
In the following sequence of reaction, what is D?

Options:
A Primary amine
B An amide
C Phenyl isocyanate
D A chain lengthened hydrocarbon

The correct answer is:
Phenyl isocyanate

Solution:

Topic: Chemistry, Hydrocarbons, NEET
In the following sequence of reactions,
$\mathrm{CH}_3-\mathrm{Br} \stackrel{\mathrm{KCN}}{\longrightarrow} A \stackrel{\mathrm{H}_3 \mathrm{O}^{+}}{\longrightarrow} B \xrightarrow[\mathrm{LiAlH}_4]{\text { ether }} C$
the end product $(C)$ is

Options:
A acetone
B methane
C acetaldehyde
D ethyl alcohol

The correct answer is:
ethyl alcohol

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, NEET, NEET 2012 (Screening)
In the following sequence of reactions, identify the final product.
Cyclopentane → Cl 2 /hv (I) → alc. KOH,  Δ (II) → (i)  O 3 (III)

Options:
A pentanal
B pentane-2-one
C pentan-3-one
D pentane-1,3-dial

The correct answer is:
pentane-1,3-dial

Solution:

Topic: Chemistry, Hydrocarbons, NEET
In the following sequence of reactions,
the compound Q formed will be

Options:
A aniline
B phenol
C benzaldehyde
D benzene sulphonic acid

The correct answer is:
phenol

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
In the following sequence of reactions, the product D is

CHâ¡CHâHBr AâHBr Bâalc KOH CâNaNH2 D

Options:
A Ethanol
B Ethyne
C Ethanal
D Ethene

The correct answer is:
Ethyne

Solution:

Topic: Chemistry, Hydrocarbons, NEET
In the following statements, one word per statement is underlined. This word can be changed to obtain the correct statement. In which one it should not be changed?

Options:
A In the living world

all

organisms have DNA as the genetic material which can be transmitted from one generation to another.
B Purine Pyrimidine ratio in Ribose Nucleic Acid can

never

be equal to 1.
C The 3rd carbon atom and the 5th carbon atom of

all

nucleotides in DNA are involved in the formation of phosphodiester bond.
D Monocyclic nitrogen bases

always

form weak hydrogen bonds with dicyclic nitrogen bases in B-DNA.

The correct answer is:
Monocyclic nitrogen bases

always

form weak hydrogen bonds with dicyclic nitrogen bases in B-DNA.

Solution:
In the living world, plant viruses and some animal viruses have RNA as the genetic material. Thus, in the living world,

most

organisms have DNA as the genetic material which can be transmitted from one generation to another. Even though RNA is single stranded, purine:pyrimidine ratio in ribose nucleic acid can

sometimes

be equal to 1. The 5' end of the DNA strand has a free phosphate group, and the 3' end of the DNA strand has a free -OH group. They do not form phosphodiester bond. Thus, the 3rd carbon atom and the 5th carbon atom of

some

nucleotides in DNA are involved in the formation of phosphodiester bond. Thymine (monocyclic nitrogen base: pyrimidine) always forms two weak hydrogen bonds with adenine (dicyclic nitrogen base: purine) and cytosine (monocyclic nitrogen base: pyrimidine) always forms three weak hydrogen bonds with guanine (dicyclic nitrogen base: purine) in B-DNA.

Topic: Biology, Molecular Basis of Inheritance, NEET
In the following table 'X' is formed by the division or branching of 'Y'. Which one is an exception?

âXâ	âYâ

A	Arteriole	Artery

B	Capillary	Arteriole

C	Venule	Vein

D	Renal Artery	Aorta

Options:
A A
B B
C C
D D

The correct answer is:
C

Solution:
Aorta is the largest artery that comes out of the heart. It gives rise to various blood vessels going to individual organs. The renal artery is a branch of the aorta. Arteries enter organs and divide to form arterioles. Arterioles further divide to form capillaries. Capillaries unite to form venules. Venules join to form veins. Thus, venules are not branches but tributaries of veins.

Topic: Biology, Body Fluids and Circulation, NEET
In the following the duration of intra-uterine life is matched with the events completed. Which of these is matched correctly?

i. End of the first month: Embryo heart is formed
ii. End of the second month: Foetus develops limbs and digits
iii. End of the fifth month: Hair on head
iv. End of the sixth month: Eyelashes are formed

Options:
A i, ii and iii
B ii and iv
C i and iii
D i, ii, iii and iv

The correct answer is:
i, ii, iii and iv

Solution:
Human pregnancy lasts for nine months and it officially begins after implantation i.e. 8th day after fertilization.The sign of developing foetus is first recorded at the end of 1st month as heart sounds are cleary audible. The foetus develops limbs and digits by the end of 60 days. Major systems also develop in the third month of pregnancy.Hence as the first trimester (3 months or 12 weeks) is completed the foetus has developed organ systems and limbs and digits. The first movements of the foetus and appearance of hair on the head are usually observed during the fifth month. By the end of 24 weeks (second trimester), the body is covered with fine hair, eye-lids separate, and eyelashes are formed.

Topic: Biology, Human Reproduction, NEET
In the following the most stable configuration of n-butane is

Options:
A
B
C
D

The correct answer is:

Solution:
The anit-conformation is the most stable conformation of n-butane. In this, the bulkyl methyl groups are as far apart as possible thereby keeping steric repulsion at a minimum.

Topic: Chemistry, General Organic Chemistry, NEET
In the following the most stable conformation of n- butane is

Options:
A
B
C
D

The correct answer is:

Solution:
Out of these conformations of n- butane the conformation that is ‘Anti’ is the most stable one as in it same atoms or groups are present at maximum distance (opposite or at 180o). The stability order can be given as follows

Topic: Chemistry, General Organic Chemistry, NEET
In the following the most stable conformation of n-butane is

Options:
A

B

C

D

The correct answer is:

Solution:
Key Idea The conformation in which the heavier groups are present at maximum possible distances, so that the forces of repulsion get weak, is more stable.
Among the given conformations of n-butane, the conformation shown in option (b) ie, anti conformation is most stable as in it the bulkier group (ie, $\mathrm{CH}_3$ group) are present at maximum possible distance.

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2010 (Screening)
In the following which one is the example of Bryophyta that has elaborate mechanism of spore dispersal?

Options:
A Polysiphonia
B Polytrichum
C Dictyota
D Marchantia

The correct answer is:
Polytrichum

Solution:
The sporophyte in mosses is more elaborate than that in liverworts. The capsule contains spores. Spores are formed after meiosis. The mosses have an elaborate mechanism of spore dispersal. Common examples of mosses are Funaria, Polytrichum, and Sphagnum. Polytrichum is a genus of mosses - commonly called haircap moss.

Topic: Biology, Plant Kingdom, NEET
In the following, which one of the diodes is reverse biased?

Options:
A
B
C
D

The correct answer is:

Solution:
A P-N junction is reverse biased if P terminal is at lower potential then the N terminal of the diode.

Topic: Physics, Semiconductors, NEET
In the following, which one of the diodes is reverse biased ?

Options:
A
B
C
D

The correct answer is:

Solution:
In a diode when potential of n is at high potential and potential at p is of low potential then it is called reversed biased.

Topic: Physics, Semiconductors, NEET
In the formation of calcium carbide (CaC2), what is the source of carbon?

Options:
A Quick lime
B Carbon dioxide
C Coke
D Calcium hydroxide

The correct answer is:
Coke

Solution:
Coke is the source of carbon in the formation of calcium carbide (CaC2). It reacts with quick lime to produce calcium carbide.

Topic: Chemistry, Hydrocarbons, NEET
In the formation of ionic halides, what type of bonding holds the metal cation and halogen anion together?

Options:
A Covalent bonding
B Metallic bonding
C Ionic bonding
D Hydrogen bonding

The correct answer is:
Ionic bonding

Solution:
Ionic halides are held together by ionic bonding, which is the electrostatic attraction between the metal cation and the halogen anion.

Topic: Chemistry, s Block Elements, NEET
In the formation of sodium chloride, what is the half reaction for the loss of electrons by sodium?

Options:
A 2 Na(s) → 2 Na+(g) + 2e-
B Cl2(g) + 2e- → 2 Cl-(g)
C 2 Na(s) + Cl2(g) → 2 Na+ Cl-(s)
D 2 Na(s) + 2e- → 2 Na+(g)

The correct answer is:
2 Na(s) → 2 Na+(g) + 2e-

Solution:
The passage provides the half reaction for the loss of electrons by sodium as 2 Na(s) → 2 Na+(g) + 2e-.

Topic: Chemistry, Redox Reactions, NEET
In the formula dn/dt = rN(K - N/K), K - N/K stands for

Options:
A Environmental resistance
B Reproductive potential
C Growth rate
D Carrying capacity

The correct answer is:
Environmental resistance

Solution:
In the formula dn dt = rN K - N K , K - N K stands for environmental resistance. Environmental resistance is the sum of the environmental factors (such as drought, mineral deficiencies and competition) that tend to restrict the biotic potential of an organism or kind of organism and impose a limit on numerical increase.

Topic: Biology, Organisms and Populations, NEET
In the formula to calculate the percentage of phosphorus when estimated as Mg2P2O7, what does the value 62 represent?

Options:
A Mass of the organic compound
B Mass of two phosphorus atoms present in Mg2P2O7
C Molar mass of Mg2P2O7
D Atomic mass of phosphorus

The correct answer is:
Mass of two phosphorus atoms present in Mg2P2O7

Solution:
In the formula to calculate the percentage of phosphorus when estimated as Mg2P2O7, the value 62 represents the mass of two phosphorus atoms present in Mg2P2O7.

Topic: Chemistry, General Organic Chemistry, NEET
In the Friedel-Crafts alkylation of benzene, why is it important to use excess benzene relative to the alkyl halide?

Options:
A To increase the yield of alkylbenzene
B To reduce the reactivity of benzene
C To prevent the formation of side products
D To decrease the yield of alkylbenzene

The correct answer is:
To prevent the formation of side products

Solution:
Using excess benzene ensures that there is a sufficient amount of benzene to react with the electrophilic species and prevents the formation of side products by diluting the electrophile's reactivity.

Topic: Chemistry, Hydrocarbons, NEET
In the Friedel-Crafts alkylation reaction, what type of bond is formed between the alkyl group and the benzene ring?

Options:
A Covalent bond
B Ionic bond
C Metallic bond
D Hydrogen bond

The correct answer is:
Covalent bond

Solution:
In the Friedel-Crafts alkylation reaction, a covalent bond is formed between the alkyl group and the benzene ring through the electrophilic substitution mechanism.

Topic: Chemistry, Hydrocarbons, NEET
In the Friedel-Crafts alkylation reaction, what type of reaction mechanism is involved in the formation of alkylbenzenes?

Options:
A Substitution reaction
B Addition reaction
C Elimination reaction
D Redox reaction

The correct answer is:
Substitution reaction

Solution:
The formation of alkylbenzenes in the Friedel-Crafts alkylation reaction involves an electrophilic substitution mechanism where an electrophilic species replaces a hydrogen atom on the benzene ring.

Topic: Chemistry, Hydrocarbons, NEET
In the fully organized Polygonum type of embryo sac, what is the ratio of haploid, diploid and triploid nuclei?

Options:
A 3 : 1 : 3
B 6 : 0 : 1
C 6 : 1 : 0
D 3 : 2 : 3

The correct answer is:
6 : 1 : 0

Solution:
A fully organised Polygonum type of embryo sac in a 7-celled, 8-nucleate structure. The cells of egg appratus and antipodal are haploid, whereas the central cell is diploid, i.e., contains two polar nuclei. Triploid endosperm is formed after triple fusion, i.e, fusion of second male gamete with polar nuclei.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In the Gabriel synthesis, what reaction occurs when the potassium salt of phthalimide reacts with an alkyl halide?

Options:
A Reduction reaction
B Acid-base reaction
C Nucleophilic substitution reaction
D Addition reaction

The correct answer is:
Nucleophilic substitution reaction

Solution:
The reaction between the potassium salt of phthalimide and an alkyl halide in the Gabriel synthesis is a nucleophilic substitution reaction.

Topic: Chemistry, Amines, NEET
In the gas mask device, what is typically used as an adsorbent?

Options:
A Activated carbon
B Silica gel
C Aluminium oxide
D Charcoal and alumina mixture

The correct answer is:
Activated carbon

Solution:
The paragraph specifies that a gas mask typically consists of activated charcoal or a mixture of adsorbents.

Topic: Chemistry, Surface Chemistry, NEET
In the gas phase reaction, what does the term "x atm" represent?

Options:
A Initial pressure of reactant
B Increase in pressure of product
C Decrease in pressure of reactant
D Total pressure at time 't'

The correct answer is:
Decrease in pressure of reactant

Solution:
In the context of the gas phase reaction, "x atm" represents the decrease in pressure of reactant at time 't'.

Topic: Chemistry, Chemical Kinetics, NEET
In the genetic code dictionary, how many codons are used to in place of 10 code for all the 20 essential amino acids?

Options:
A 20
B 64
C 61
D 60

The correct answer is:
61

Solution:
The genetic codon is triplet and at each position of the triplet there are four possibilities due to presence of four nucleotides in nucleic acids, thereby meaning there are (\(4 \times 4 \times 4)\) or 64 possible codons. Out of these 64 possible codons, three codons UAA, UAG, UGA are termination codons. They do not code for any amino acids but result in termination of protein chain. Thus, 61 codons code for 20 essential amino acids.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2003
In the give reaction

The product is

Options:
A Optically active having (+) rotation
B Optically active having (-) rotation
C Optically inactive due to absence of stereocenter
D Optically inactive because product is meso

The correct answer is:
Optically inactive due to absence of stereocenter

Solution:
Here the product is 3-methylpentane which is optically inactive due to absence of stereocenter

Topic: Chemistry, Hydrocarbons, NEET
In the give sequence
HOCH2-CHOH-CH2OHâBr2 waterA+BâDil.NaOH(X)
'X' is

Options:
A Glyceraldehyde
B Dihydroxyacetone
C Fructose
D Glucose

The correct answer is:
Fructose

Solution:
Here X is fructose and it is formed as follows

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
In the given below, which ion cannot give iodometric titrations?

Options:
A Ag+
B Fe3+
C Pb2+
D Cu2+

The correct answer is:
Pb2+

Solution:
Atom in highest oxidation state can oxidize iodide to liberate I2 which is volumetrically measured by iodometric titration using hypo solution.
2I-→I2
Pb+2→ Lowest oxidation state can not oxidise iodide to I2 .

Topic: Chemistry, Redox Reactions, NEET
In the given cell representation Zn | Zn2+ || Cu2+ | Cu, which is the negative electrode?

Options:
A Zn2+
B Cu2+
C Zn
D Cu

The correct answer is:
Zn

Solution:
In the cell Zn|Zn2+||Cu2+|Cu the negative electrode (anode) is Zn . In electrochemical cell representation anode is always written on left side while cathode on right side.

Topic: Chemistry, Electrochemistry, NEET
In the given chart of photophosphorylation, what does ‘A’ represent?

Options:
A PC
B FRS
C PQ
D Cyt -a3

The correct answer is:
PQ

Solution:
Plastoquinone transfer electron from photo system-II to photo system-I. Electrons released from P700 or photo system-I moves through Fe-S, plastoquinone, cytochrome and plastocyanin and are recycled to P700. In this process, only ATP is produced but no oxygen produced.

Topic: Biology, Photosynthesis in Higher Plants, NEET
In the given circuit diagram, current in 2 Î© resistor is 2 A , then the current in 6 Î© resistor will be

Options:
A 32 A
B 23 A
C 13 A
D 2 A

The correct answer is:
23 A

Solution:
As in the above branch potential across PQ is 6 V and equivalent resistance is 3 Î© , therefore current in this branch should be 2 A . Therefore voltage across 6 Î© is 4 V and hence current in it is 23 A .

Topic: Physics, Current Electricity, NEET
In the given circuit diagram. Find the heat generated on closing the switch S (Initially the capacitor of capacitance C is uncharged) -

Options:
A 32CV2
B CV2
C 12CV2
D 2 CV2

The correct answer is:
12CV2

Solution:
Before closing switch Ui=12 × 3C×2C3C+2C× V2
=35CV2
After closing switch Uf=12 CV2+35CV2
Energy supply by battery=CV × V=CV2
CV2=Uf- Ui+ΔH
CV2=12 CV2+ΔH
ΔH=12 CV2

Topic: Physics, Electrostatics, NEET
In the given circuit P and Q form the inputs. The output Y is

Options:
A Y= P ¯
B Y=P Q ¯
C Y=P+Q
D Y= Q ¯

The correct answer is:
Y= Q ¯

Solution:
From the given circuit, the output Y is
Y= ( P ¯ +Q )( P+Q ) ¯
Y= P ¯ ⋅P+ P ¯ Q+Q⋅P+Q⋅Q ¯
Y= Q+PQ+ P ¯ Q ¯ ( ∵ P ¯ ⋅P=0 )
Y= Q( 1+P+ P ¯ ) ¯ ( ∵1+A=1 )
Y= Q ¯

Topic: Physics, Semiconductors, NEET
In the given circuit switch K is open. The charge on the capacitor is C in steady-state is q1. Now the key is closed and steady-state charge on C is q2. The ratio of charges q1/q2 is -

Options:
A 3/2
B 3/1
C 1
D 1/2

The correct answer is:
3/2

Solution:
Initially when switch is open
Finally when the switch is closed
I=E3R; ∴ VA – VB=I 2R=E/3R × 2R=2E/3
∴ Charge on capacitor=CV ; q2=C2E3
∴q1q2=CE2/3CE=3 : 2

Topic: Physics, Electrostatics, NEET
In the given circuit switch K is open. The charge on the capacitor is C in steady-state is q1. Now the key is closed and steady-state charge on C is q2. The ratio of charges q1/q2 is -

Options:
A 3/2
B 3/1
C 1
D 1/2

The correct answer is:
3/2

Solution:
Initially when switch is open
Finally when the switch is closed
I=E3R; ∴ VA – VB=I×2R=E/3R × 2R=2E/3
∴ Charge on capacitor=CV ; q2=C2E3
∴q1q2=CE2/3CE=3 : 2

Topic: Physics, Electrostatics, NEET
In the given circuit the average power developed is:

Options:
A 502 watt
B 200 watt
C 1502 watt
D 2002 watt

The correct answer is:
200 watt

Solution:
Impedance of the circuit is given by
Z=R2+XL2=R2+(ωL)2 ,
Z=(50)2+(0.2×250)2
Z=2500+(50)2=502
Average power is given by
∴ P=VrmsIrmscos⁡ϕ

P= Vrms2Z2R
P=200225022×50=200 watt

Topic: Physics, Alternating Current, NEET
In the given circuit the cells have zero internal resistance. The currents (in amperes) passing through resistance R1 and R2 respectively, are:

Options:
A 0,Â 1
B 2,Â 2
C 0.5 , 0
D 1,Â 2

The correct answer is:
0.5 , 0

Solution:
Assume VF as 0
as   V E = V B
⇒ Current in R 2 is zero
VA-VD=10 V
⇒I=1020=12 A from A to D

Topic: Physics, Current Electricity, NEET
In the given circuit, the current flowing in the 1 Î© resistance will be

Options:
A 3 A
B 4 A
C 5 A
D 6 A

The correct answer is:
6 A

Solution:
All the three given cells are in parallel
â´ req=r3=33=1 Î©
Eeq=12 V
i=121+1=6 A

Topic: Physics, Current Electricity, NEET
In the given circuit,

the current through the battery is

Options:
A 1.5 A

B 2 A

C 3 A

D 5.33 A

The correct answer is:
3 A

Solution:
In the given circuit diode D2 is reverse biased, so it will not conduct. Diode D1 and D3 are forward biased, so they will conduct, Thus the equivalent circuit is
∴Rnet=(4+4)×16(4+4)+16=8×1624=163 Ω
∴ current through the battery,
I=VRnet=16163=3 A

Topic: Physics, Alternating Current, NEET
In the given circuit, the potential difference between points A and B is

Options:
A 6.7 V
B 3.7 V
C 4Â V
D 10Â V

The correct answer is:
6.7 V

Solution:
Let I1 is current flowing through first loop.
â´I1=23+2=25A
and I2 is current flowing through the second loop.
â´I2=43+5=12A
Let VA is potential of point A and VB is potential of point B.
Now, we apply Kirchhoff's law between points A and B.
VA+3I1+4+3I2=VB
VB-VA=1.2+4+1.5
=6.7 V
Hence, (1) is the correct option.

Topic: Physics, Current Electricity, NEET
In the given circuit the reading of voltmeter $V_1$ and $V_2$ are $300 \mathrm{~V}$ each. The reading to the voltmeter $\mathrm{V}_3$ and ammeter $A$ are respectively

Options:
A $150 \mathrm{~V}, 2.2 \mathrm{~A}$
B $220 \mathrm{~V}, 2.2 \mathrm{~A}$
C $220 \mathrm{~V}, 2.0 \mathrm{~A}$
D $100 \mathrm{~V}, 2.0 \mathrm{~A}$

The correct answer is:
$220 \mathrm{~V}, 2.2 \mathrm{~A}$

Solution:
For series $L C R$ circuit
voltage $\mathrm{V}=\sqrt{\mathrm{V}_{\mathrm{R}}^2+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^2}$
Since, $V_L=V_C$ hence $V=V_R=220 \mathrm{~V}$
also, current $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{220}{100}=2.2 \mathrm{~A}$

Topic: Physics, Alternating Current, NEET, NEET 2010 (Screening)
In the given circuit the reading of voltmeters V1 and V2 are 300 volt each. The reading of the voltmeter V3 and ammeter A are respectively

Options:
A 150  V and 2.2  A
B 220  V and 2.2  A
C 220  V and 2.0  A
D 100  V and 2.0  A

The correct answer is:
220  V and 2.2  A

Solution:
V= V R 2 + ( V L â V C ) 2
Since, V L = V C i.e. V 1 = V 2
â´V= V R =220 V
I= V R = 220 100 =2.2 A

Topic: Physics, Alternating Current, NEET
In the given circuits (a), (b) and (c), the potential drop across the two p-n junctions are equal in:

Options:
A Circuit (b) only
B Circuit (c) only
C Both circuits (a) and (c)
D Circuit (a) only

The correct answer is:
Both circuits (a) and (c)

Solution:
In circuit(a), both p-n junctions are in forward bias, hence current will flow and potential drop will be equal.

In circuit(c), both p-n junctions are in reverse bias and hence negligible current will flow through the circuit and potential drop will be equal.

In circuit(b), one p-n junctions is in reverse bias and one in forward bias. So, they will show different resistances. Also, negligible current will flow through the circuit. Hence, potential drop will be different.

Topic: Physics, Semiconductors, NEET, NEET 2022 (Phase 1)