datavorous/entrance-exam-dataset huggingface dataset:
At a particular locus, the frequency of 'A' allele is 0.6 and that of 'a' is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium?

Options:
A 0.36
B 0.16
C 0.24
D 0.48

The correct answer is:
0.48

Solution:
The mathematical expression to calculate allele frequency can be given as p + q = 1. Where p represents the frequency of dominant allele and q represents the frequency of recessive allele. The bionomial expression of Hardy Weinberg's law to calculate genotype frequency is p2 + 2pq + q2 = 1
where p2 = homozygoys dominant
2pq = heterozygous dominant
q2 = homozygous recessive.
Frequency of heterozygous individuals
= 2 × 0.6 × 0.4 = 0.48.

Topic: Biology, Evolution, NEET
At a particular locus, the frequency of 'A' allele is 0.6 and that of 'a' is 0.4. What would be the frequency of heterozygotes in a randomly mating population of equilibrium?

Options:
A 0.16
B 0.36
C 0.48
D 0.24

The correct answer is:
0.48

Solution:
According to the Hardy-Weinberg equation,
P2+2PQ+Q2=1
Where, P= Dominant allele frequency
Q= Recessive allele frequency
P2= Homozygous dominant genotype
2PQ= Heterozygous genotype
Q2= Homozygous recessive genotype
Here, P = 0.6 and Q = 0.4
Therefore, 2PQ=2Ã0.6Ã0.4
=0.48.

Topic: Biology, Evolution, NEET
At a point A on the earth’s surface the angle of dip, δ=+25°. At a point B on the earth’s surface, the angle of dip, δ=-25°. We can interpret that:

Options:
A A and B are both located in the northern hemisphere.
B A is located in the southern hemisphere and B is located in the northern hemisphere.
C A is located in the northern hemisphere and B is located in the southern hemisphere.
D A and B are both located in the southern hemisphere.

The correct answer is:
A is located in the northern hemisphere and B is located in the southern hemisphere.

Solution:
As we know that the angle of dip is the angle between earth's resultant magnetic field from horizontal.

At the equator, dip is zero. At Northern Hemisphere, dip is positive. At Southern Hemisphere, dip is negative.

Since, the field lines originate from the South Pole, the angle of dip is negative in Southern Hemisphere. Similarly, as the field lines terminate at the North Pole, the angle of dip is positive in Northern Hemisphere.

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2019 (All India)
At a point in space, the electric field points towards north. In the region surrounding this point, the rate of change of potential will be zero along

Options:
A north
B south
C north-south
D east-west

The correct answer is:
east-west

Solution:
Eeast-west=0⇒dvdreast-west=0

Topic: Physics, Electrostatics, NEET
At a site, low grade copper ores are available and zinc and iron scraps are also available. Which of the two scraps would be more suitable for reducing the leached copper ore and why?

Options:
A Zinc scraps should be used because they are more readily available.
B Iron scraps should be used because zinc is more expensive.
C Zinc scraps should be used because they are more reactive.
D Iron scraps should be used because they are more stable in the reduction process.

The correct answer is:
Iron scraps should be used because zinc is more expensive.

Solution:
Correct Option is : (B)
Iron scraps should be used because zinc is more expensive.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
At a temperature of 30 Â°C, the susceptibility of a paramagnetic material is found to be X. Its susceptibility at 333 Â°C is

Options:
A 0.5 X
B 2Â X
C 11.1Â X
D 0.09 X

The correct answer is:
0.5 X

Solution:
Susceptibility of ferromagnetic material is inversely proportional to temperature
X∝1T
X1X2=T2T1
Given , T1=30 °C, T2=333 °C, X1=X
∴ X2=T1T2X=303 K606 KX=0.5 X

Topic: Physics, Magnetic Properties of Matter, NEET
At a temperature of 30Â°C , the susceptibility of a paramagnetic material is found to be X. Its susceptibility at 333Â°C is

Options:
A 0.5 X
B 2Â X
C 11.1Â X
D 0.09 X

The correct answer is:
0.5 X

Solution:
Susceptibility of ferromagnetic material is inversely proportional to temperature
X∝1T
X1X2=T2T1
Given , T1=30 °C, T2=333 °C, X1=X
∴ X2=T1T2X=303 K606 KX=0.5 X

Topic: Physics, Magnetic Properties of Matter, NEET
At absolute zero, Si acts as

Options:
A non-metal
B metal
C insulator
D none of these

The correct answer is:
insulator

Solution:
Semiconductors, like Si , Ge , act as insulators at low temperature.

Topic: Physics, Semiconductors, NEET
At any instant of time $t$, the displacement of any particle is given by $2 t-1$ (SI unit) under the influence of force of 5 N . The value of instantaneous power is (in SI unit):

Options:
A 5
B 7
C 6
D 10

The correct answer is:
10

Solution:
$\begin{aligned} x & =2 t-1 \\ v & =\frac{d x}{d t}=2 \mathrm{~m} \mathrm{~s}^{-1} \\ P & =F . v \\ & =2 \times 5=10 \mathrm{~W}\end{aligned}$

Topic: Physics, Work Power Energy, NEET, NEET 2024
At any instant, two elements $X_1$ and $X_2$ have same number of radioactive atoms. If the decay constant of $X_1$ and $X_2$ are $10 \lambda$ and $\lambda$ respectively. Then the time when the ratio of their atoms becomes $\frac{1}{e}$ respectively will be:

Options:
A $\frac{1}{11 \lambda}$
B $\frac{1}{9 \lambda}$
C $\frac{1}{6 \lambda}$
D $\frac{1}{5 \lambda}$

The correct answer is:
$\frac{1}{9 \lambda}$

Solution:
By radioactivity law, we have,
$\mathrm{N}=\mathrm{N}_o e^{-\lambda t}$
(where, the symbols have their usual meanings) Hence, for two elements,
$\frac{\mathrm{N}_1}{\mathrm{~N}_2}=\frac{\mathrm{N}_o e^{-10 \lambda t}}{\mathrm{~N}_o e^{-\lambda t}}=\frac{1}{e}$
Hence, $e^{-9 \lambda t}=e$
$t=\frac{1}{9 \lambda}$

Topic: Physics, Nuclear Physics, NEET, NEET 2022 (Phase 2)
At any point x, 0, 0 the electric potential V is 1000x+1500x2+500x3 volt, then electric field intensity at x=1 m is

Options:
A 5500j^+k^ V m-1
B 5500i^ V m-1
C 55002j^+k^ V m-1
D 55002i^+k^ V m-1

The correct answer is:
5500i^ V m-1

Solution:
Electric potential
V=1000x+1500x2+500x3
Electric field
E=-dVdx=1000x2+21500x3+3500x4i^
At x=1,
E=5500i^ V m-1

Topic: Physics, Electrostatics, NEET
At certain Hill-station pure water boils at 99.72 Â°C . If Kb for water is 0.513 Â°Ckgmol-1 . The boiling point of 0.69 m solution of urea will be-

Options:
A 100.074 °C
B 103 °C
C 100.359 °C
D Un predictable

The correct answer is:
100.074 °C

Solution:
ΔTb=Kb×m
=0.513×0.69
∴∆Tb=0.354
So, Tb=0.354+99.72=100.074 °C

Topic: Chemistry, Solutions, NEET
At constant pressure, the ratio of increase in volume of an ideal gas, per degree rise in kelvin temperature, to its original volume is

Options:
A T2
B 1T
C T3
D T

The correct answer is:
1T

Solution:
According to ideal gas law,
pV=RTâV=RPT
VâT (At constant pressure)
Hence, V1V2=T1T2
â V2V1=T2T1
Where, V2 is the final volume
V2V1-1=T2T1-1
â V2-V1V1=T2-T1T1 âµ T2-T1=1K
â V2-V1V1=1T1=1T

Topic: Physics, Thermodynamics, NEET
At constant temperature on increasing the pressure of a gas by 10%, its volume will decrease by

Options:
A 9.09%
B 10%
C 5%
D 20%

The correct answer is:
9.09%

Solution:
Since, we know,
p∝1V
∴p1p2=V2V1
⇒p2=110100p1 (given)
∴V1V2=p2p1=110100
⇒V2=100110V1
∴ Change in volume, ΔV=V2-V1
=100110V1-V1=V1100-110110=-10V1110
Negative sign means volume is decreased.
∴ΔVV1×100=10110×100=9.09%

Topic: Physics, Kinetic Theory of Gases, NEET
At constant temperature, what is the relationship between the pressure of a gas and its concentration?

Options:
A p ∝ [gas]
B p = [gas]RT
C p = [gas]/RT
D p = [gas]/T

The correct answer is:
p ∝ [gas]

Solution:
At constant temperature, the pressure of the gas is proportional to its concentration, i.e., p ∝ [gas].

Topic: Chemistry, Chemical Equilibrium, NEET
At constant volume, 4 mol of an ideal gas when heated from 300 K to 500 K changes its internal energy by 5000 J. The molar heat capacity at constant volume is?

Options:
A 3.125
B 6.25
C 6.75
D -6.75

The correct answer is:
6.25

Solution:
Given ΔU=5000
n=4
ΔT=(500-300)
Using ΔU=nCvΔT
5000=4×Cv(500–300)
Cv=6.25 Jk–1mol–1

Topic: Chemistry, Thermodynamics (C), NEET
At Critical Micelle Concentration (CMC), the surfactant molecules,

Options:
A Dissociate
B Associate
C Become completely soluble
D Decompose

The correct answer is:
Associate

Solution:
At Critical Micelle Concentration (CMC), the surfactant molecules associate and form micelles.

Topic: Chemistry, Surface Chemistry, NEET
At dynamic equilibrium the reaction on both directions occur at the same rate and the mass on both sides of the equilibrium does not undergo any change. This condition can be achieved only when the value of ΔG is

Options:
A positive
B negative
C zero
D can be any of these values

The correct answer is:
zero

Solution:
At equilibrium all measurable properties become almost constant or unchanged i.e., ΔG=0

Topic: Chemistry, Chemical Equilibrium, NEET
At equilibrium, the concentrations of N2=3.0 × 10-3M, O2= 4.2 × 10-3M, and NO = 2.8 × 10-3M in a sealed vessel at 800 K. What will be Kc for the reaction N2(g) + O2(g) → 2NO(g)?

Options:
A 1.06 × 103
B 0.622
C 0.985
D 1.34

The correct answer is:
0.622

Solution:
Correct Option is : (B)
0.622

Topic: Chemistry, Ionic Equilibrium, NEET
At equilibrium, what is the ratio [Zn2+] / [Cu2+] equal to at T = 298K?

Options:
A Kc
B 0.059V/2
C 1.1V
D 2 × 1037

The correct answer is:
Kc

Solution:
At equilibrium,[Zn2+] / [Cu2+]is equal to Kc at T = 298K.

Topic: Chemistry, Electrochemistry, NEET
At G1 stage, which phenomenon takes place

Options:
A DNA synthesis
B RNA synthesis
C Reverse transcription
D All of these

The correct answer is:
RNA synthesis

Solution:
At the G1 phase, the RNA and protein are synthesized.
The G1 phase, or Gap 1 phase, is the first of four phases of the cell cycle that takes place in eukaryotic cell division. In this part of interphase, the cell synthesizes mRNA and proteins in preparation for subsequent steps leading to mitosis. G1 phase ends when the cell moves into the S phase of interphase.

Topic: Biology, Cell Cycle and Cell Division, NEET
At high pressures, Z is

Options:
A >1
B <1
C =1
D Cannot say

The correct answer is:
>1

Solution:
Correct Option is : 1
>1

Topic: Chemistry, States of Matter, NEET
At how many places in Krebs cycle, FADH2 is/are formed?

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
In Krebs cycle, dehydrogenation reaction occurs four times; among which there are three points in the cycle where NAD+ is reduced to NADH+H+ and one point where FAD+ is reduced to FADH2 when succinate is converted to fumarate.

Topic: Biology, Respiration in Plants, NEET
At how many places in Kreb's cycle from pyruvic acid, the enzyme dehydrogenases are involved?

Options:
A 3
B 4
C 5
D 6

The correct answer is:
5

Solution:
In Kreb's cycle oxidation occurs by dehyrdrogenation in which the hydrogen released is accepted by hydrogen acceptors like NAD and FAD. A total of five steps involve the dehydrogenation reaction involving the enzyme dehydrogenase, in which there are four points in the cycle where NAD+ is reduced to NADH + H+ and one point FAD+ is reduced to FADH2.
The enzymes like pyruvic dehydrogenase, isocitric dehydrogenase, alpha-ketoglutaric acid dehydrogenase, succinate dehydrogenase and malate dehydrogenase are involved in this process.

Topic: Biology, Respiration in Plants, NEET
At its closest approach, the distance between the Mars and the Earth is found to be 60 million km. When the planets are at this closest distance, the time it would take to send a radio - message from a space probe of the Mars to the Earth is

Options:
A 5 s

B 200 s

C 0.2 s

D 500 s

The correct answer is:
200 s

Solution:
t = distance speed
= 6 0 Ã 1 0 6 Ã 1 0 0 0 m 3 Ã 1 0 8 m s -1 = 2 0 0 s

Topic: Physics, Electromagnetic Waves, NEET
At least _________ of sperms should have normal shape for normal fertility.

Options:
A 10%
B 30%
C 60%
D 80%

The correct answer is:
60%

Solution:
The human male ejaculates about 200 to 300 million sperms during coitus of which, for normal fertility, at least 60 per cent sperms must have normal shape and size and at least 40 per cent of them must show vigorous motility.

Topic: Biology, Reproductive Health, NEET
At low pressure and high temperature, the Van der Waals equation is finally reduced (simplified) to

Options:
A p+ a V m 2 V m âb =RT
B p V m âb =RT
C p+ a V m 2 V m =RT
D pV m =RT

The correct answer is:
pV m =RT

Solution:
At low pressure and high temperature, real gases behave like ideal gases. Accordingly, Van der Waals equation is modified as:
pV m =RT (for 1 mole)
âµ At low pressure and high temperature, the terms a V 2 and b becomes very small as compared to p and V.

Topic: Chemistry, States of Matter, NEET
At low pressure, the Vander Waal's equation is expressed as

Options:
A pV m =RT+1
B Z=pVmRT=1+bRTp
C Z=pVmRT=1-aRT
D Z= pV m RT =1− a RTV m

The correct answer is:
Z= pV m RT =1− a RTV m

Solution:
p+ a V m 2 V m âb =âRT
at low pressure bâ0
ââ P+ a V m 2 V m =RT
ââ PV m + a V m =RT
ââZâ=â1â a V m RT

Topic: Chemistry, States of Matter, NEET
At magnetic poles, the angle of dip is

Options:
A 45o
B 30o
C zero
D 90o

The correct answer is:
90o

Solution:
At magnetic poles, the magnetic needle becomes vertically downward the angle of dip becomes 90o

Topic: Physics, Magnetic Properties of Matter, NEET
At NTP water boils at 100 oC. Deep down a mine, water will boil at a temperature

Options:
A 100 oC
B >100 oC
C <100 oC
D will not boil at all

The correct answer is:
>100 oC

Solution:
Pressure inside the mines is greater than that of normal pressure or atmospheric pressure. Also we know that boiling point increases with increase in pressure. Therefore water will boil at a temperature greater than 100 oC.

Topic: Physics, Thermal Properties of Matter, NEET
At rest, a well-trained athlete will have a heart rate _______ a normal person

Options:
A higher than
B lower than
C same as
D incomparable to

The correct answer is:
lower than

Solution:
In an adult, at rest, the resting heart rate ranges from 60-100 beats/ minute. The vigorous regular exercises which an athlete does helps the heart to function better and pump more blood at a lower rate as compared to the normal sedentary person. The aerobic exercises done by the athletes helps to strengthen the heart muscle and thus reducing the heart rate at rest which otherwise would be higher in a normal adult.

Topic: Biology, Body Fluids and Circulation, NEET
At room temperature, a diatomic gas is found to have an r.m.s speed of 1930 m s-1. The gas is :

Options:
A F2

B Cl2
C H2
D O2

The correct answer is:
H2

Solution:
vrms=3RTM â M=3RTvrms2=3Ã8.3Ã30019302=2 g mole-1
â H 2

Topic: Physics, Thermodynamics, NEET
At room temperature, copper has free electron density of 8.4 Ã 1028 m-3. The electron drift velocity in a copper conductor of cross-sectional area 10-6 m2 and carrying a current of 5.4 A, will be -

Options:
A 4Â m s-1
B 0.4Â m s-1
C 4Â cm s-1
D 0.4 mm s-1

The correct answer is:
0.4 mm s-1

Solution:
Use i=ne Avd
â´ vd=ineA
v d = 5.4 8.4Ã 10 28 Ã1.6Ã 10 â19 Ã 10 â6
v d = 5.4Ã 10 â3 8.4Ã1.6
v d =0.4â mmâ s â1

Topic: Physics, Current Electricity, NEET
At room temperature, the reaction between NO and O2 is given NO2 is fast, while that between CO and O2 is slower it is due to

Options:
A CO is smaller in size than that of NO
B CO is poisonous
C The activation energy for the reactions 2NO+O2→2NO2 is less then 2CO+O2→2CO2
D None of these

The correct answer is:
The activation energy for the reactions 2NO+O2→2NO2 is less then 2CO+O2→2CO2

Solution:
As The activation energy for the reactions 2NO+O2â2NO2 is less then 2CO+O2â2CO2 so NO reacts fast and CO reacts slowly.
Rate of reaction â1Activation energy.

Topic: Chemistry, Chemical Kinetics, NEET
At room temperature, what is the result when a primary alcohol reacts with Lucas reagent in the Lucas test?

Options:
A Immediate turbidity
B Turbidity after heating
C No turbidity at any temperature
D Formation of an alkyl halide

The correct answer is:
No turbidity at any temperature

Solution:
Primary alcohols do not produce turbidity at room temperature when they react with Lucas reagent.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
At room temperature, which metals are passive to dilute non-oxidizing acids?

Options:
A Titanium and copper
B Titanium and vanadium
C Vanadium and nickel
D Nickel and copper

The correct answer is:
Titanium and vanadium

Solution:
The paragraph states that titanium and vanadium are passive to dilute non-oxidizing acids at room temperature.

Topic: Chemistry, d and f Block Elements, NEET
At some instant, a radioactive sample S1 having an activity 5 Î¼Ci has twice the number of nuclei as another sample S2 which has an activity of 10 Î¼Ci . The ratio of half lives of S1 and S2 is

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

The correct answer is:
4:1

Solution:
Given: N 1 =2. N 2
Now, activity A 1 = Î» 1 N 1 =5Î¼ C i .....(1)
and A 2 = Î» 2 N 2 =10Î¼ C i .....(2)
Dividing ( 1 ) ( 2 ) we get
Î» 1 Î» 2 . N 1 N 2 = 5 10
â t 2 t 1 .2= 1 2 â t 1 t 2 = 4 1

Topic: Physics, Nuclear Physics, NEET
At some instant the ratio of the amount of two radioactive elements is 2:1. If their half-lives respectively are 12 hours and 16 hours, then after two days, the ratio of the amount of two substances is

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

The correct answer is:
1:1

Solution:
N1=124 N0=N016
and for the second element
N2=123 N0'=N0'8
N1N2=N0N0'816= N02N0'
Now given that N0 :N0'=2 :1
N1N2=22=1 :1

Topic: Physics, Nuclear Physics, NEET
At STP, 0.48 g of O2 diffused through a porous partition in 1200 seconds. What volume of CO2 will diffuse in the same time and under the same conditions ?

Options:
A 286.5 mL
B 346.7 mL
C 112.2 mL
D 224.8 mL

The correct answer is:
286.5 mL

Solution:
r 1 r 2 = V 1 / t V 2 / t = V 1 V 2 = M 2 M 1
Volume of 0.48 g of O2 at STP = 2 2 4 0 0 3 2 × 0 . 4 8 mL (32 g of O2 occupies 22400 mL at STP)
∴ V O 2 V CO 2 = M CO 2 M O 2 = 4 4 3 2 = 1 . 1 7 3
V CO 2 = V O 2 1 . 1 7 3 = 2 2 4 0 0 3 2 × 0 . 4 8 1 . 1 7 3 = 2 8 6 . 5 mL.

Topic: Chemistry, States of Matter, NEET
At temperatures above 1073K coke can be used to reduce FeO to Fe. How can you justify this reduction with the Ellingham diagram?

Options:
A As per Ellingham diagram at temperatures lower than 1073 K , ∆G(C,CO) < ∆G(Fe,FeO). Hence coke can reduce FeO to Fe.
B As per Ellingham diagram at temperatures greater than 1073 K , ∆G(C,CO) < ∆G(Fe,FeO). Hence coke can reduce FeO to Fe.
C As per Ellingham diagram at temperatures greater than 1073 K , ∆G(C,CO) > ∆G(Fe,FeO). Hence coke can reduce FeO to Fe.
D No,Ellingham diagram cannot provide any justification

The correct answer is:
As per Ellingham diagram at temperatures greater than 1073 K , ∆G(C,CO) < ∆G(Fe,FeO). Hence coke can reduce FeO to Fe.

Solution:
Correct Option is : (B)
As per Ellingham diagram at temperatures greater than 1073 K , ∆G(C,CO) < ∆G(Fe,FeO). Hence coke can reduce FeO to Fe.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
At the asymptote stage, the population is

Options:
A Stabilized
B Increasing
C Decreasing
D Extinct

The correct answer is:
Stabilized

Solution:
Asymptote stage is a stage at which population cannot increase due to the limited carrying capacity of earth.

Topic: Biology, Organisms and Populations, NEET
At the beginning of the final stage of mitosis,

Options:
A The chromosomes that have reached their respective poles decondense.
B Nuclear envelope assembles around the chromosome clusters.
C Chromosomes cluster at opposite spindle poles and their identity is lost as discrete elements.
D Nucleolus, golgi complex and ER reform.

The correct answer is:
The chromosomes that have reached their respective poles decondense.

Solution:
At the beginning of the final stage of mitosis, i.e., telophase, the chromosomes that have reached their respective poles decondense and lose their individuality. The individual chromosomes can no longer be seen and chromatin material tends to collect in a mass in the two poles.

Topic: Biology, Cell Cycle and Cell Division, NEET
At the critical micelle concentration, the surfactant molecules

Options:
A Become completely soluble
B Decompose
C Associate
D Dissociate

The correct answer is:
Associate

Solution:
AT (CMC) critical micelle concen-tration, the surfactant molecules associate to form miscelles. For soap CMC value is 10-3 mol/litre.

Topic: Chemistry, Surface Chemistry, NEET
At the critical point, what happens to the horizontal portion of the isotherm?

Options:
A It disappears.
B It merges into one point.
C It extends indefinitely.
D It becomes vertical.

The correct answer is:
It merges into one point.

Solution:
At the critical point, the horizontal portion of the isotherm merges into one point.

Topic: Chemistry, States of Matter, NEET
At the end of glycolysis, X is the net energy gain from one molecule of glucose via Y, but there is also energy stored in the form of Z. Identify X, Y and Z

Options:
A

X	Y	Z

1 ATP	Oxidative phosphorylation	NADH+H+

B

X	Y	Z

2 ATPs	Oxidative phosphorylation	NADH+H+

C

X	Y	Z

1 ATP	Substrate level phosphorylation	FADH+H+

D

X	Y	Z

2 ATPs	Substrate level phosphorylation	NADH+H+

The correct answer is:

X	Y	Z

2 ATPs	Substrate level phosphorylation	NADH+H+

Solution:
Glycolysis is a catabolic process where the substrate glucose is broken down into smaller molecules- pyruvic acid and occurs anaerobically. The site for glycolysis is the cytoplasm. It is a stepwise process involving a series of ten enzymes to catalyze the processes. In it, some of the intermediates may get involved in other metabolic processes than respiration. Like DHAP formed can be used in the lipid synthesis as it is a source of glycerol to join fatty acids. It is the connecting link or common pathway between both types of respiration. In aerobic respiration, the oxygen molecule is not involved in the glycolysis pathway.
ATP obtained from ADP and a phosphorylated intermediate instead of inorganic phosphate, Pi, is referred to as substrate-level phosphorylation. This process is observed in the glycolysis and the amount of ATP generated is relatively low. The net result of glycolysis is 2 ATP with 2 NADH2 which is used as an energy source later in ETC.

Topic: Biology, Respiration in Plants, NEET
At the first minimum adjacent to the central maximum of a single slit diffraction pattern, the phase difference between the Huygens' wavelet from the edge of the slit and the wavelet from the mid-point of the slit is

Options:
A π2 rad
B π rad
C π8 rad
D π4 rad

The correct answer is:
π rad

Solution:
According to the question for first minima, path difference ∆x,
AP-BP=λ
AP-MP=λ
So, phase difference,
ϕ=2πλ⋅∆x
=2πλ⋅λ2=π rad

Topic: Physics, Wave Optics, NEET
At the first minimum adjacent to the central maximum of a single-slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is:

Options:
A Ï2 radian
B π radian
C Ï8 radian
D Ï4 radian

The correct answer is:
π radian

Solution:
For first minima
AP-BP=λ
AP-MP=λ2
So phase difference =2πλ×λ2=π

Topic: Physics, Wave Optics, NEET, NEET 2015 (Phase 2)
At the macroscopic level, what aspects of a chemical reaction are of interest?

Options:
A Molecular structures of compounds
B Rate of reaction and reaction mechanisms
C Amounts reacted or formed and rates of consumption or formation
D Energy levels of reactants and products

The correct answer is:
Amounts reacted or formed and rates of consumption or formation

Solution:
At the macroscopic level, chemists are interested in amounts reacted or formed and the rates of their consumption or formation.

Topic: Chemistry, Chemical Kinetics, NEET
At the moment t=0, when the charge on the capacitor C1 is zero, the switch is closed. If I0 be the current through inductor at t=0, then for t>0 (initially C2 is uncharged)

Options:
A maximum current through inductor equals I02
B maximum current through inductor equals C1I0C1+C2
C maximum charge on C1=C1I0LC2C1+C2
D maximum charge on C1=C1I0LC1+C2

The correct answer is:
maximum charge on C1=C1I0LC1+C2

Solution:
The maximum current through the inductor is I0.
By conservation of energy,
12LI02=12C1+C2V2
V=LC1+C2I0
Q1=C1V=C1I0LC1+C2

Topic: Physics, Alternating Current, NEET