datavorous/entrance-exam-dataset huggingface dataset:
In the Cannizzaro’s reaction given below :

2 Ph - CHO → KOH Ph - CH 2 OH + PhCO 2 -

The slowest step is

Options:
A The attack of —OH at the carbonyl group

B The transfer of hydride to the carbonyl group
C The abstraction of proton from the carboxylic acid

D The deprotonation of Ph—CH2OH

The correct answer is:
The transfer of hydride to the carbonyl group

Solution:
The slowest step in Cannizzaro reaction is transfer of hydride to the carbonyl group.

Topic: Chemistry, Aldehydes and Ketones, NEET
In the Carnot engine, efficiency is 40% at hot reservoir temperature T. For efficiency 50%, what will be the temperature of the hot reservoir?

Options:
A 2T5
B 6T
C 6T5
D T5

The correct answer is:
6T5

Solution:
The efficiency Î·=work doneheat input=WQ
Î·=1-T2T1
Where, T2= temperature of sink
And T1= temperature of hot reservoir
40100=1-T2T1
â T2T1=0.6
â T2=0.6T1
Now, 50100=1-T2T1â²âT2T1â²=0.5
0.6T1T1â²=0.5
â T1â²=0.60.5T1
â T1â²=65T1
â T1â²=65T âµ T1=T

Topic: Physics, Thermodynamics, NEET
In the case of a solid solution of CdCl2 and AgCl, what type of defect is observed?

Options:
A Vacancy defect
B Interstitial defect
C Frenkel defect
D Schottky defect

The correct answer is:
Vacancy defect

Solution:
In the case of a solid solution of CdCl2 and AgCl, a vacancy defect is observed.

Topic: Chemistry, Solid State, NEET
In the case of alkali metals, the covalent character decreases in the order

Options:
A $\mathrm{MCl}>\mathrm{MI}>\mathrm{MBr}>\mathrm{MF}$
B $\mathrm{MF}>\mathrm{MCl}>\mathrm{MBr}>\mathrm{MI}$
C $\mathrm{MF}>\mathrm{MCl}>\mathrm{MI}>\mathrm{MBr}$
D $\mathrm{MI}>\mathrm{MBr}>\mathrm{MCl}>\mathrm{MF}$

The correct answer is:
$\mathrm{MI}>\mathrm{MBr}>\mathrm{MCl}>\mathrm{MF}$

Solution:
Key Idea According to Fajan's rule, Covalent character $\propto \frac{1}{\text { size of cation }}$ $\propto$ In the given options, cation is same but anions are different. Among halogens the order of size is
$$
\mathrm{F} < \mathrm{Cl} < \mathrm{Br} < \mathrm{I}
$$
$\therefore$ Order of covalent character is
$$
\mathrm{MI}>\mathrm{MBr}>\mathrm{MCl}>\mathrm{MF}
$$

Topic: Chemistry, s Block Elements, NEET, NEET 2009 (Screening)
In the case of an ore containing ZnS and PbS, what depressant is used in the Froth Floatation method?

Options:
A NaOH
B NaCN
C HCl
D H2SO4

The correct answer is:
NaCN

Solution:
In the case of an ore containing ZnS and PbS, the depressant used in the Froth Floatation method is NaCN.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
In the case of aryl halides, what makes halogens (e.g., chlorine) ortho and para directing groups in electrophilic aromatic substitution reactions?

Options:
A Their inductive effect
B Their -I effect
C Their -M effect
D Their resonance effect

The correct answer is:
Their resonance effect

Solution:
Aryl halides like chlorobenzene have halogens as ortho and para directing groups due to the resonance effect. Although halogens exhibit a -I effect, the resonance effect increases electron density at the ortho and para positions, making them directing groups.

Topic: Chemistry, Hydrocarbons, NEET
In the case of Bacillus thuringiensis, Bacillus itself is not killed by toxic protein crystals produced by it because

Options:
A Bt toxin protein is not produced in the bacteria
B Bt toxin protein is produced in very less amount in the bacillus
C Bt toxin exists as an inactive toxin
D the bacteria is resistant to the Bt toxin

The correct answer is:
Bt toxin exists as an inactive toxin

Solution:
The Bt toxin is produced by the soil bacterium Bacillus thuringiensis. This bacteria is unaffected by the toxin it produces because it is inactive and in crystalline form when inside the bacterium. This toxin needs to get activated to show its effects and therefore it kills only the insects which have ingested it as they get activated in the gut of the insect.

Topic: Biology, Biotechnology and Its Applications, NEET
In the case of peppered moth (Biston betutaria) the black-coloured form became dominant over the light-coloured form in England during industrial revolution. This is an example of

Options:
A natural selection whereby the darker forms were selected
B appearance of the darker coloured individuals due to very poor sunlight
C protective mimicry
D inheritance of darker colour character acquired due to the darker environment

The correct answer is:
natural selection whereby the darker forms were selected

Solution:
The given case is an example of natural selection. As a result of struggle for existence only those organisms could survive, which have favourable variations to adapt environmental conditions on with so many variations in populations of species the struggle for existence results in survival of the fittest. The survival of the fittest is the result of selection and proliferation of only those organisms, which were most suitably adapted to the environment and most successful in mating, ie, natural selection.

Topic: Biology, Evolution, NEET, NEET 2009 (Screening)
In the case of Poriferans, the spongocoel is lined with flagellated cells called

Options:
A Ostia
B Oscula
C Choanocytes
D Mesenchymal cells

The correct answer is:
Choanocytes

Solution:
Choanocytes (collar cells) form the lining of spongocoel in poriferans (sponges). Flagella in collar cells provide circulation to water in the water canal system. These distinctive cells line the interior body walls of sponges. These cells have a central flagellum that is surrounded by a collar of microvilli. Choanocytes are versatile cells. Their flagella beat to create the active pumping of water through the sponge, while the collars of the choanocytes are the primary areas where nutrients are absorbed into the sponge. Furthermore, in some sponges, the choanocytes develop into gametes.

Topic: Biology, Animal Kingdom, NEET, NEET 2017
In the case of the peppered moth, black coloured moth became prominent over light coloured moth in England during the industrial revolution. This is an example of:

Options:
A Stabilising natural selection
B Directional natural selection
C Disruptive natural selection
D Both (A) and (B)

The correct answer is:
Directional natural selection

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).
A study carried out in England for the population of peppered moth (Biston betularia) in the 1850s i.e. prior to industrialization revealed that the population of a white-winged moth on trees was larger than that of the black-winged moths, having more melanin. When the study was repeated in the 1920s i.e. after industrialization, it revealed that the population of black-winged moth was more compared to that of the white-winged moths. The explanation for this reversal was that the predators could easily spot a moth against a contrasting background.

Topic: Biology, Evolution, NEET
In the cell reaction Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s), how many electrons (ne-) are involved?

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
In the cell reaction,Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s), two electrons are involved.

Topic: Chemistry, Electrochemistry, NEET
In the cell reaction Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s), what is the role of Ni(s)?

Options:
A Ni(s) is the anode.
B Ni(s) is the cathode.
C Ni(s) is a spectator ion.
D Ni(s) is a catalyst.

The correct answer is:
Ni(s) is the cathode.

Solution:
In the cell reaction Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s), Ni(s) is the cathode where reduction occurs.

Topic: Chemistry, Electrochemistry, NEET
In the central nervous system, myelinated fibres form the ..., while the non-myelinated fibre form the ......

Options:
A Grey matter, white matter
B White matter, grey matter
C Ependymal cells, neurosecretory cells
D Neurosecretory cells, ependymal cells

The correct answer is:
White matter, grey matter

Solution:
The grey matter is composed of nerve cells, nerve fibres and neuroglia, which are non-myelinated, while white matter consists mostly of myelinated axons.

Topic: Biology, Neural Control and Coordination, NEET
In the certain process, 400 Cal of heat are supplied to a system and at the same time 105 J of mechanical work was done on the system. The increase in its internal energy is

Options:
A 20 cal
B 303 cal
C 404 cal
D 425 cal

The correct answer is:
425 cal

Solution:
dQ=400 cal, dW=-105 J
dU=dQ-dW
dU=400--25=425 cal
Note dW is negative because work is done on the system

Topic: Physics, Thermodynamics, NEET
In the chemical reaction
CH 3 CH 2 NH2+ CHCl 3 + 3 KOH → A + B + 3 H 2 O , the compound (A) and (B) are respectively

Options:
A C 2 H 5 NC and K 2 C O 3
B C H 3 C H 2 CON H 2 and 3KCl
C C 2 H 5 CN and 3KCl
D C 2 H 5 NC and 3KCl

The correct answer is:
C 2 H 5 NC and 3KCl

Solution:
Primary amines react with alcoholic alkali and chloroform to give an offensive odour compound i.e., isocyanide. This reaction is called carbylamines reaction.

This is carbylamines reaction.

C H 3 C H 2 N H 2 +CHC l 3 +3KOH → C 2 H 5 NC (ethyl isocyanide)         (A)   +   3KCl (Potassium    chloride)         (B)   +  3 H 2 O

Topic: Chemistry, Amines, NEET
In the chemical reactions,

Compounds A and B respectively are

Options:
A Fluorobenzene and phenol
B Benzene diazonium chloride and benzonitrile
C Nitrobenzene and chlorobenzene
D Phenol and bromobenzene

The correct answer is:
Benzene diazonium chloride and benzonitrile

Solution:
Formation of A is by diazotization and formation of B from A is by SN reaction.

Topic: Chemistry, Amines, NEET
In the circuit below, A and B represent two inputs and C represents the output. The circuit represents

Options:
A OR gate
B NOR gate
C AND gate
D NAND gate

The correct answer is:
OR gate

Solution:
It is OR gate. When either of input is high, the diode conducts and output is high.

Truth table for OR gate

A	B	C

0	0	0

0	1	1

1	0	1

1	1	1

Topic: Physics, Semiconductors, NEET
In the circuit below, A and B represent two inputs and C represents the output. The circuit represents

Options:
A OR gate
B NOR gate
C AND gate
D NAND gate

The correct answer is:
OR gate

Solution:
It is OR gate. When either of them conducts, the gate conducts.

Topic: Physics, Semiconductors, NEET
In the circuit below, the potential difference between A and B is

Options:
A 10 V
B 20 V
C 30 V
D 40 V

The correct answer is:
10 V

Solution:
â´ V1=1541+154Ã190
=1519Ã190=150â V
â V2=190-150 V=40 V
Now,
So, the potential difference between A and B will be,
VAB=11+3Ã40
=14Ã40=10 V

Topic: Physics, Electrostatics, NEET
In the circuit diagram shown below, the magnitude and direction of the flow of current, respectively, would be

Options:
A 73 A from a to b via e
B 73 A from b to a via e
C 1 A from b to a via e
D 1 A from a to b via e

The correct answer is:
1 A from a to b via e

Solution:
I=10–41+2+3=1 A (Clockwise so a to b via e )

Topic: Physics, Current Electricity, NEET
In the circuit diagram shown in Figure, the potentials of the points B, C and D are respectively-

Options:
A 12 V, 10 V, 6 V
B 11 V, 9 V, 6 V
C 11 V, 9 V, 0 V
D 12 V, 10 V, 0 V

The correct answer is:
11 V, 9 V, 6 V

Solution:
According to Kirchhoff's Voltage Law in loop ABCDA, Current flowing in the circuit is 1A

Topic: Physics, Current Electricity, NEET
In the circuit figure, the voltmeter reads 30 V. The resistance of the voltmeter is

Options:
A 1200 Î©
B 700Â Î©
C 400Â Î©
D 300Â Î©

The correct answer is:
1200 Î©

Solution:
The potential difference across 300 Î©=60-30=30V
Therefore the effective resistance of voltmeter resistance R and 400 Î© in parallel will be equal to 300 Î©, as 60 V is equally divided between two parts.
So, 300=RÃ400R+400
or 300R +120000=400R or R=1200 Î©

Topic: Physics, Current Electricity, NEET
In the circuit given below, a 100 W bulb B1 and two 60 W bulbs B2 and B3, are connected to a 250 V direct current source. If W1, W2 and W3 are the powers of the three bulbs then, which of the following statements is corect?

Options:
A W1 > W2 = W3
B W1 > W2 > W3
C W1 < W2 = W3
D W1 < W2 < W3

The correct answer is:
W1 < W2 < W3

Solution:
P = V 2 R so, R = V 2 P
∴ R 1 = V 2 100 and R 2 = R 3 = V 2 60
Now, W 1 = 250 2 R 1 + R 2 2 · R 1
W 2 = 250 2 R 1 + R 2 2 · R 2 and W 3 = 250 2 R 3
W1 : W2 : W3 = 15 : 25 : 64
or W1 < W2 < W3 .

Topic: Physics, Current Electricity, NEET
In the circuit of the figure, treat diode as ideal, current in the 4 Î© resistor is

Options:
A 2 A
B 3 A
C 127 A
D 3013 A

The correct answer is:
2 A

Solution:
The given circuit can be redrawn as,
This means, D2 is reversed based. Hence, this arm of the circuit will act as open arm i.e., the close circuit will have two resistors 2 Î© and 4 Î© which will be in series connection.
â Req=2 Î©+4 Î©
=6 Î©
So, current through 4 Î© resistor will be,
I=VR
=126
=2 A

Topic: Physics, Semiconductors, NEET
In the circuit shown, a potential difference of 60 V is applied across AB. The potential difference between the points M and N is

Options:
A 10 V
B 15 V
C 20 V
D 30 V

The correct answer is:
30 V

Solution:
Equivalent capacitor in circuit A'MNB' is C2
Charge on capacitor C between MN is 60×C2=30 C
∴VMN=30 CC=30 V

Topic: Physics, Electrostatics, NEET
In the circuit shown below, the charge on the 60 Î¼F capacitor is

Options:
A 150 μC
B 100 Î¼C
C 50 μC
D 75 μC

The correct answer is:
100 Î¼C

Solution:
The circuit shown in the figure is a balanced Wheatstone bridge, so the 80 μF doesn't have any effect on the circuit and it can be removed.

After removing the 80 μF, we see that the 30 μF and 60 μF are in series. Similarly, the 15 μF and 30 μF are also in series.

C1=30×6030+60=20 μF, C2=15×3015+30=10 μF

Hence, the equivalent capacitance of the entire circuit is

C=C1+C2=30 μF

Ceq=C×30C+30=15 μF

The charge that flows through the battery = 10×15=150 μC

This charge will be divided into two branches in the ratio of capacitances.

On solving we get that charge on 60 μF = 100 μC

Topic: Physics, Electrostatics, NEET
In the circuit shown below, the inductance $L$ is connected to an ac source. The current flowing in the circuit is $I=I_0 \sin \omega t$. The voltage drop $\left(V_L\right)$ across $L$ is

Options:
A $\omega L l_0 \sin \omega t$
B $\frac{I_0}{\omega L} \sin \omega t$
C $\frac{I_0}{\omega L} \cos \omega t$
D $\omega L I_0 \cos \omega t$

The correct answer is:
$\omega L I_0 \cos \omega t$

Solution:
$V_L$ leads current $/$ by $\frac{\pi}{2}$
$V_L=V_0 \sin \left(\omega t+\frac{\pi}{2}\right) \quad\left(\because I=I_0 \sin \omega t\right)$
$V_0=I_0 X_L$
$\Rightarrow V_L=10 X_L \cos (\omega t)=10 \omega L \cos (\omega t)$

Topic: Physics, Alternating Current, NEET, NEET 2024 (Re-NEET)
In the circuit shown below, the key K is closed at t=0. The current through the battery is

Options:
A 5 A at t=0 and 7 A at t=â
B 3 A at t=0 and 1 A at t=â
C 1 A at t=0 and 3 A at t=â
D 2 A at t=0 and 6 A at t=â

The correct answer is:
1 A at t=0 and 3 A at t=â

Solution:
At t=0, L behaves as open circuit and at t=â as short circuit.
So at t=0, i=1010=1 A [âµReq=10 Î©]
and at t=â, i=10103=3 A [âµReq=103 Î©]

Topic: Physics, Alternating Current, NEET
In the circuit shown below, the voltmeter is of large resistance. The E.M.F. of the cell is ε. The reading of the voltmeter is

Options:
A zero
B ε10
C ε5
D ε2

The correct answer is:
ε5

Solution:
From the figure, it is clear that
VA â VB =Îµ4+6 Ã 4=4Îµ10
VA â VD =Îµ6+4 Ã 6 =6Îµ10
VB â VD=Îµ5

Topic: Physics, Current Electricity, NEET
In the circuit shown, if a conducting wire is connected between points $\mathrm{A}$ and $\mathrm{B}$, the current in this wire will:

Options:
A flow from $B$ to $A$
B flow from $A$ to $B$
C flow in the direction which will be decided by the value of $V$
D be zero

The correct answer is:
flow from $B$ to $A$

Solution:
According to diagram
$\begin{array}{l}
V_A-V_B=\left[V-\left(\frac{V}{8} \times 4\right)\right]-\left[V-\left(\frac{V}{4} \times 1\right)\right] \\
=-\frac{V}{2}+\frac{V}{4}=-\frac{V}{4}
\end{array}$
$\Rightarrow V_B > V_A .$

Topic: Physics, Current Electricity, NEET, NEET 2006
In the circuit shown if the internal resistance of each cell is r, then the rate at which the chemical energy of ε1 is being consumed is

Options:
A ε25r
B 4ε25r
C 2ε25r
D 7ε25r

The correct answer is:
2ε25r

Solution:
i =ε1-ε22r+r+r+r=ε5r
P=2εi=2ε25r

Topic: Physics, Current Electricity, NEET
In the circuit shown in Fig. 2.7, initially K1 is closed and K2 is open. What are the charges on each capacitor? Then K1 was opened and K2 was closed (order is important). What will be the charge on capacitor C1 now? [C = 1µF]

Options:
A 5 µC
B 12 µC
C 18 µC
D 24 µC

The correct answer is:
18 µC

Solution:
Correct Option is : (C)
18 µC

Topic: Chemistry, Capacitance, NEET
In the circuit shown in Fig 3.6, two cells are connected in opposition to each other. Cell E1 has an emf of 6V and an internal resistance of 2Ω, and cell E2 has an emf of 4V and an internal resistance of 8Ω. What is the potential difference between points A and B?

Options:
A 3.6 V
B 4.4 V
C 5.6 V
D 6.8 V

The correct answer is:
5.6 V

Solution:
Correct Option is : (C)
5.6 V

Topic: Chemistry, Current Electricity, NEET
In the circuit shown in Fig, what is the value of RC?

Options:
A 0.56 kΩ
B 0.65 kΩ
C 0.42 kΩ
D 0.74 kΩ

The correct answer is:
0.56 kΩ

Solution:
Correct Option is : (A)
0.56 kΩ

Topic: Chemistry, Semiconductors, NEET
In the circuit shown in Fig., when the input voltage of the base resistance is 10V, Vbe is zero and Vce is also zero. What are the values of Ib?

Options:
A 10µA
B 15µA
C 20µA
D 25µA

The correct answer is:
25µA

Solution:
Correct Option is : (D)
25µA

Topic: Chemistry, Semiconductors, NEET
In the circuit shown in figure, the base current IB is 10 Î¼A and the collector current is 5.2 mA. The voltage (VBE) across the base and emitter is

Options:
A 0.1 V
B 0.5 V
C 0.25 V
D 0.7 V

The correct answer is:
0.5 V

Solution:
VBE=VCC-IBRB

VBE=5.5-10×10-6×5×105=0.5 V

Topic: Physics, Semiconductors, NEET
In the circuit shown in figure, time constant and steady state current will be

Options:
A 0.25 s, 0.75 A
B 0.75 s, 0.25 A
C 0.25 s, 0.25 A
D 0.5 s, 0.5 A

The correct answer is:
0.25 s, 0.75 A

Solution:
Equivalent resistance is 8 Î©
Therefore time constant is,
Ï=LReq
Ï=28=0.25 s
And the steady-state current is,
i0=ÎµR
i0=68=0.75 A

Topic: Physics, Electromagnetic Induction, NEET
In the circuit shown in the figure, a conducting wire HE is moved with a constant speed v towards left. The complete circuit is placed in a uniform magnetic field Bâ perpendicular to the plane of circuit inwards. The current in HKDE is

Options:
A Clockwise
B Anticlockwise
C Alternating
D Zero

The correct answer is:
Zero

Solution:
The potential difference across capacitor V = Bvl = constant
Therefore, the charge stored in the capacitor is also constant. Thus, the current through the capacitor is zero.

Topic: Physics, Alternating Current, NEET
In the circuit shown in the figure below the cells E1 and E2 have e.m.f. 4 V and 8 V and internal resistance 0.5 Î© and 1.0 Î© respectively. Then the potential difference across the cell E1 and E2 will be-

Options:
A 3.75Â V,Â 7.5Â V
B 4.25 V, 7.5 V
C 3.75Â V,Â 3.75Â V
D 4.25Â V,Â 4.25Â V

The correct answer is:
4.25 V, 7.5 V

Solution:
Equivalent resistance of circuit is :
Req=8 Î©
So, I=8-48=0.5 A
V8V=E- Ir=8-0.51=7.5 V
V4V=E+Ir=4+0.50.5=4.25 V

Topic: Physics, Current Electricity, NEET
In the circuit shown in the figure, reading of voltmeter is V1 when only S1 is closed, reading of voltmeter is V2 when only S2 is closed and reading of voltmeter is V3 when both S1 and S2 are closed. Then

Options:
A V3>V2>V1
B V2>V1>V3
C V3>V1>V2
D V1>V2>V3

The correct answer is:
V2>V1>V3

Solution:
V2>V1>V3
When S1 closed

V1=3RR+3RE=34E
When S2 is closed

V2=6RR+6RE=67E
When both S1 and S2 are closed

V3=2RR+2RE=2E3
∴ V2>V1>V3

Topic: Physics, Current Electricity, NEET
In the circuit shown in the figure, reading of voltmeter is V1 when only S1 is closed, reading of voltmeter is V2 when only S2 is closed and reading of voltmeter is V3 when both S1 and S2 are closed. Then -

Options:
A V3>V2>V1
B V2>V1>V3
C V3>V1>V2
D V1>V2>V3

The correct answer is:
V2>V1>V3

Solution:
When only S1 is closed, i=ER+3R=34R
Reading V1=3R=3E4= 0.75 E
When only S2 is closed i=E7R
Reading V2=i6R= 6E7= 0.86 E
When S1 and S2 both are closed, Req=R+2R=3R
i =E3R, V3=i2R=2E3= 0.67 E, V2>V1>V3

Topic: Physics, Current Electricity, NEET
In the circuit shown in the figure, the input voltage Vi is 20 V, VBE=0 and VCE=0. The values of IB, IC and β are given by

Options:
A IB=20 μA, IC=5 mA, β=250
B IB=25 μA, IC=5 mA, β=200
C IB=40 μA, IC=10 mA, β=250
D IB=40 μA, IC=5 mA, β=125

The correct answer is:
IB=40 μA, IC=5 mA, β=125

Solution:
Vi=IBRB+VBE (By Kirchoff's Voltage Law)
20=IB×500×103+0
IB=20500×103=40 μA
VCC=ICRC+VCE (By Kirchoff's Voltage Law)
20=IC×4×103+0
IC=5×10-3=5 mA
β=ICIB=5×10-340×10-6=125

Topic: Physics, Semiconductors, NEET, NEET 2018
In the circuit shown in the figure, V must be-

Options:
A 50Â V
B 100 V
C 75Â V
D 25Â V

The correct answer is:
100 V

Solution:
After redrawing the given circuit 100 Ω, 20 Ω and 25 Ω are in parallel combination.
Req=20 Ω
5A=VReq ⇒ V=100 V

Topic: Physics, Current Electricity, NEET
In the circuit shown in the figure, V must be

Options:
A 50Â V
B 100 V
C 75Â V
D 25Â V

The correct answer is:
100 V

Solution:
Req=20 Î©
5 A=VReq
âV=100 V

Topic: Physics, Current Electricity, NEET
In the circuit shown in the following figure, the value of Y is

Options:
A 0
B 1
C fluctuates between 0 and 1
D indeterminate as the circuit cannot be realized

The correct answer is:
0

Solution:
Lower NOT gate inverts input to zero. NOT gate from NAND gate inverts this output to 1 and upper NAND gate converts this input 1 and input 0 to 1.
Thus A=1 and B=1 become inputs of NAND gate giving final output as zero. Choice A is correct.

Topic: Physics, Semiconductors, NEET
In the circuit shown the cells $A$ and $B$ have negligible resistances. For $V_A=12 \mathrm{~V}$, $R_1=500 \Omega$ and $R=100 \Omega$ the galvanometer (G) shows no deflection. The value of $V_B$ is

Options:
A $4 \mathrm{~V}$
B $2 \mathrm{~V}$
C $12 \mathrm{~V}$
D $6 \mathrm{~V}$

The correct answer is:
$2 \mathrm{~V}$

Solution:
Applying Kirchhoff's law
$500 I+100 I=12$
So, $I=\frac{12 \times 10^{-2}}{6}=2 \times 10^{-2}$
Hence, $V_B=100\left(2 \times 10^{-2}\right)=2 \mathrm{~V}$

Topic: Physics, Current Electricity, NEET, NEET 2012 (Screening)
In the circuit shown, the current through the $4 \Omega$ resistor is $1 \mathrm{~A}$ when the points $P$ and $M$ are connected to a DC voltage source. The potential difference between the points $M$ and $N$ is

Options:
A $1.5 \mathrm{~V}$
B $1.0 \mathrm{~V}$
C $0.5 \mathrm{~V}$
D $3.2 \mathrm{~V}$

The correct answer is:
$3.2 \mathrm{~V}$

Solution:
Key Idea : In parallel resistances, potential difference across them is same.
Potential difference across PM
$V=4 \times 1=4 \text { volt (ie, across } 4 \Omega)$
Equivalent resistance of lower side arm,
$R_{e q}=1+\frac{0.5 \times 0.5}{0.5+0.5}=1.25 \Omega$
Now the circuit can be shown as

Let current I flows in lower branch, so
$\begin{aligned}
1.25 I & =4 \mathrm{~V} \\
I & =\frac{4}{1.25}=3.2 \mathrm{~A}
\end{aligned}$
Therefore, 3.2 A current flows in $1 \Omega$ resistance, hence potential difference between $M$ and $N$ is
$V^{\prime}=3.2 \times 1=3.2$ volt

Topic: Physics, Current Electricity, NEET, NEET 2008 (Screening)
In the circuit shown, the current through the ideal diode is

Options:
A 75 mA
B 20 mA
C 100 mA
D 25 mA

The correct answer is:
20 mA

Solution:
Ideal junction in forward bias behaves like a conducting wire.
∴ Current from the battery
i=280=25 mA
This current goes from diode as it behaves as short circuit.

Topic: Physics, Semiconductors, NEET