datavorous/entrance-exam-dataset huggingface dataset:
For a P-N junction diode which of the following statements is true about the order of magnitude of the various currents?

Options:
A Forward current is in mA and reverse current is in μA
B Forward current is in μA are reverse current is in mA
C Both forward and reverse currents are in μA
D Both forward and reverse currents are in mA

The correct answer is:
Forward current is in mA and reverse current is in μA

Solution:
For a P-N junction diode, the forward current is in mA and reverse current is in μA . In forward bias the majority charge carries drift in junction, while in reverse bias the majority charge carries drift away from the junction, only minority charge carries drift towards the junction.

Topic: Physics, Semiconductors, NEET
For a p-type semiconductor, which of the following statements is true?

Options:
A Electrons are the majority carriers and trivalent atoms are the dopants.
B Holes are the majority carriers and trivalent atoms are the dopants.
C Holes are the majority carriers and pentavalent atoms are the dopants.
D Electrons are the majority carriers and pentavalent atoms are the dopants.

The correct answer is:
Holes are the majority carriers and trivalent atoms are the dopants.

Solution:
Holes are the majority carriers and trivalent atoms are the dopants.

Topic: Physics, Semiconductors, NEET, NEET 2019 (All India)
For a parallel beam of monochromatic light of wavelength λ, diffraction is produced by a single slit whose width a is of the order of the wavelength of the light. If D is the distance of the screen from the slit, the width of the central maxima will be:

Options:
A 2Dλa
B Dλa
C Daλ
D 2Daλ

The correct answer is:
2Dλa

Solution:
Linear width of central maximum.
D 2θ=2Dθ=2Dλa

Topic: Physics, Wave Optics, NEET, NEET 2015 (Phase 1)
For a particle executing SHM, the displacement x is given by x=Acosâ¡Ït. Identify the graph which represents the variation of potential energy (PE) as a function of time t and displacement x.

Options:
A I, III
B II, IV
C II, III
D I, IV

The correct answer is:
I, III

Solution:
Potential energy as a function of displacement :
PE= 1 2 k y 2
â PE vs y graph will be an upwards parabola.
â graph III
Potential energy as a function of time:
PE= 1 2 k y 2 = 1 2 k ( A cos 2 Ït ) 2
âPE= 1 2 k A 2 cos 2 Ït= 1 2 k A 2 ( 1+cos2Ït ) 2
â PE vs time is a cosine wave with angular frequency 2Ïâgraph I.

Topic: Physics, Oscillations, NEET
For a particle moving along a circular path, the radial acceleration ar is proportional to time t. If at is the tangential acceleration, then which of the following will be independent of time t ?

Options:
A at

B arat

C arat
D ar(at)2

The correct answer is:
ar(at)2

Solution:
arât ; v2r=kt
v2=krt ; 2vdvdt=kr
at=kr2v=kr2rkt=12krt
at2ar=14krtÃkt=independent of t

Topic: Physics, Laws of Motion, NEET
For a particle moving in a straight line, the displacement of the particle at time t is given by S=t3-6t2+3t+7. The velocity of the particle when its acceleration zero is

Options:
A -9 ms-1
B -12 ms-1
C 3Â ms-1
D 42 ms-1

The correct answer is:
-9 ms-1

Solution:
v=3t2-12t+3;a=6t-12
When a=0,6t-12=0
or 6t=12 or t=2 s When t=2 s,
v=3Ã2Ã2-12Ã2+3=-9 ms-1

Topic: Physics, Motion In One Dimension, NEET
For a particle of a purely rotating body, v=rω, so correct relation will be

Options:
A ω∝1r
B ω∝v
C v∝1r
D ω is independent of r

The correct answer is:
ω is independent of r

Solution:
In case of purely rotating body ω and v are associated with this relation using r.

Topic: Physics, Rotational Motion, NEET
For a particle of mass m executing SHM with angular frequency Ï, the kinetic energy k is given by k=k0cos2â¡Ït. The equation of its displacement can be

Options:
A k0mω21/2sinâ¡ ωt
B 2k0mÏ21/2sinâ¡ Ït
C 2ω2m k01/2sinâ¡ ωt
D 2k0mω1/2sinâ¡ ωt

The correct answer is:
2k0mÏ21/2sinâ¡ Ït

Solution:
If m is the mass, r is the amplitude of oscillation, then maximum kinetic energy,
k0=12mÏ2r2 or r=2k0mÏ212
The displacement equation can be
y=rsinâ¡Ït=2k0mÏ212sinâ¡Ït

Topic: Physics, Oscillations, NEET
For a particle shows distance-time curve is given in this figure. The maximum instantaneous velocity of the particle is around the point

Options:
A D

B A

C B

D C

The correct answer is:
C

Solution:
Because the slope is highest at C, vâ¡ = dâ¡ sâ¡ dâ¡ t is maximum

Topic: Physics, Motion In One Dimension, NEET
For a photochemical reaction A→B,1.0×10-5 mole of B were formed on absorbing 1.2×1019 quanta each of λ=360 nm. The quantum efficiency is given by

Options:
A 0.50
B 1
C 0.1
D 0.2

The correct answer is:
0.50

Solution:
Number of mol of A reacted =1×10-5
Number of molecules of A =1×10-5×6.02×1023
Quantum efficiency =6.02×10181.2×1019=0.50

Topic: Chemistry, Chemical Kinetics, NEET
For a plane electromagnetic wave propagating in x-direction, which one of the following combinations gives the correct possible directions for the electric field (E) and magnetic field (B) respectively?

Options:
A j^+k^, j^+k^
B -j^+k^, -j^-k^
C j^+k^, -j^-k^
D -j^+k^, -j^+k^

The correct answer is:
-j^+k^, -j^-k^

Solution:
Direction =E→×B→

(i) (j^+k^)×(j^+k^)=0

(ii) (-j^+k^)×(-j^-k^)

=0+ı^-(-ı^)+0

=2ı^

(iii) (j^+k^)×(-j^-k^)=0

(iv) (-j^+k^)×(-j^+k^)

=0

Topic: Physics, Electromagnetic Waves, NEET, NEET 2021
For a prism of refracting angle A, the refractive index is cotâ¡A2. Then the minimum angle of deviation is

Options:
A 180°-A
B 180°-2A
C 90°-A
D A2

The correct answer is:
180°-2A

Solution:
cot⁡A2=sin⁡A+δm2sin⁡A2
cos⁡A2sin⁡A2=sin⁡A+δm2sin⁡A2
sin⁡90-A2=sin⁡A+δm2
Hence, δm=180°-2A

Topic: Physics, Ray Optics, NEET
For a prism of refractive index, 1.732, the angle of minimum deviation is equal to the angle of prism. Then the angle of the prism is

Options:
A 60°
B 70°
C 50°
D none of these

The correct answer is:
60°

Solution:
We know μ=sinδmin+A2sinA2
A=δmin
∴ 1.732=sin⁡Asin⁡A2
∴ A=60°

Topic: Physics, Ray Optics, NEET
For a radioactive sample, the initial activity of the material was 8 cps and after 3 h it becomes 1cps. The half-life of the sample is

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

The correct answer is:
1 h

Solution:
Here A0 = 8 counts, A=1 count and t=3 h.
AA0=12n
â 18=12n
or 123=12nân=3
So, T1/2 = tn = 33= 1 h

Topic: Physics, Nuclear Physics, NEET
For a reaction 12Aâ2B, the rate of disappearance of 'A' is related to the rate of appearance of 'B' by the expression-

Options:
A -dAdt=12d[B]dt
B -dAdt=14d[B]dt
C -dAdt=d[B]dt
D -dAdt=4d[B]dt

The correct answer is:
-dAdt=14d[B]dt

Solution:
12Aâ2B
Remember for a AâbB
-1adAdt=1bd[B]dt= Rate of reaction
For the given reaction
-2dAdt=12d[B]dt=rate of reaction
Rate of disappearance of A
=-dAdt=12Ã2d[B]dt
=14d[B]dt

Topic: Chemistry, Chemical Kinetics, NEET
For a reaction 2A+B→ Products, doubling the concentration of B alone doubles the rate and doubling the initial concentration of both the reactants increases the rate by a factor of 8 . The rate law for the reaction is

Options:
A Rate=k[A][B]2
B Rate=k[A]2[B]2
C Rate=k[A][B]
D Rate=k[A]2[B]

The correct answer is:
Rate=k[A]2[B]

Solution:
2A+B→ Products
According to question : Rate of reaction is proportional to [B] as increase in rate is double when [B] is doubled.
Rate of reaction proportional to [A] [B] as increase in rate is 8 times when concentration of both reactant is doubled. It means that order of reaction is 3 and overall rate reaction should be r=K[A]2[B]

Topic: Chemistry, Chemical Kinetics, NEET
For a reaction $3 \mathrm{~A} \rightarrow 2 \mathrm{~B}$
The average rate of appearance of $B$ is given by $\frac{\Delta[B]}{\Delta t}$. The correct relation between the average rate of appearance of $B$ with the average rate of disappearance of $A$ is given in option

Options:
A $\frac{-2 \Delta[\mathrm{A}]}{3 \Delta \mathrm{t}}$
B $\frac{\Delta[\mathrm{A}]}{\Delta \mathrm{t}}$
C $\frac{-\Delta[\mathrm{A}]}{\Delta \mathrm{t}}$
D $\frac{-3 \Delta[\mathrm{A}]}{2 \Delta \mathrm{t}}$

The correct answer is:
$\frac{-2 \Delta[\mathrm{A}]}{3 \Delta \mathrm{t}}$

Solution:
For reaction, $3 \mathrm{~A} \longrightarrow 2 \mathrm{~B}$
Average rate of reaction $=-\frac{1}{3} \frac{\Delta[\mathrm{A}]}{\Delta \mathrm{t}}=\frac{1}{2} \frac{\Delta[\mathrm{B}]}{\Delta \mathrm{t}}$
So $\frac{\Delta[\mathrm{B}]}{\Delta \mathrm{t}}=-\frac{2}{3} \frac{\Delta[\mathrm{A}]}{\Delta \mathrm{t}}$

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2023 (Manipur)
For a reaction A+2Bâ2C , the following data were obtained. Initial concentration

[A]	[B]	Rate (moll-1min-1)

i.	1.0	1.0	0.15

ii.	2.0	1.0	0.30

iii.	3.0	1.0	0.45

iv.	1.0	2.0	0.15

v.	1.0	3.0	0.15

The rate law for this reaction

Options:
A R=k[A][B]0
B R=k[A][B]2
C R=k[A][B]
D R=k[A]2[B]

The correct answer is:
R=k[A][B]0

Solution:
For A from is i and ii
R2R1=c2c1n
or 0.300.15=21n
So n = 1
For B, from i and iv
R2R1=c2c1n
or 0.150.15=21n
or (2)°=(2)n
So n = 0
R=k[A][B]0

Topic: Chemistry, Chemical Kinetics, NEET
For a reaction A→B, enthalpy of reaction is -4.2 kJ mol-1 and enthalpy of activation is 9.6 kJ mol-1. The correct potential energy profile for the reaction is shown in the option.

Options:
A
B
C
D

The correct answer is:

Solution:
Only option (B) is exothermic. Rest all options are endothermic.

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2021
For a reaction, A + B → Product; the rate law is given by, r = k [A]1/2[B]2. What is the order of the reaction?

Options:
A Order of the reaction is 3.5
B Order of the reaction is 2.5
C Order of the reaction is 2
D Order of the reaction is 1.5

The correct answer is:
Order of the reaction is 2.5

Solution:
Correct Option is : (B)
Order of the reaction is 2.5

Topic: Chemistry, Chemical Kinetics, NEET
For a reaction A + B → Products, the rate law is Rate = k[A][B]3/2Can the reaction be an elementary reaction? Explain.

Options:
A During an elementary reaction, the number of atoms or ions colliding to react is referred to as molecularity. Had this been an elementary reaction the order of reaction with respect to B would have been 1, but in the given rate law it is (3/2) . This indicates that the reaction is not an elementary reaction.
B Yes,this reaction is an elementary reaction because order is fractional according to rate law
C No,this reaction is not an elementary reaction because the rate law indicates the order of the reaction is an integer.
D None of the above

The correct answer is:
During an elementary reaction, the number of atoms or ions colliding to react is referred to as molecularity. Had this been an elementary reaction the order of reaction with respect to B would have been 1, but in the given rate law it is (3/2) . This indicates that the reaction is not an elementary reaction.

Solution:
Correct Option is : (A)
During an elementary reaction, the number of atoms or ions colliding to react is referred to as molecularity. Had this been an elementary reaction the order of reaction with respect to B would have been 1, but in the given rate law it is (3/2) . This indicates that the reaction is not an elementary reaction.

Topic: Chemistry, Chemical Kinetics, NEET
For a reaction, activation energy $\mathrm{E}_{\mathrm{a}}=0$ and the rate constant at $200 \mathrm{~K}$ is $1.6 \times 10^6 \mathrm{~s}^{-1}$. The rate constant at $400 \mathrm{~K}$ will be
[Given that gas constant $\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ ]

Options:
A $3.2 \times 10^4 \mathrm{~s}^{-1}$
B $1.6 \times 10^6 \mathrm{~s}^{-1}$
C $1.6 \times 10^3 \mathrm{~s}^{-1}$
D $3.2 \times 10^6 \mathrm{~s}^{-1}$

The correct answer is:
$1.6 \times 10^6 \mathrm{~s}^{-1}$

Solution:
Key Idea Rate constants at two different temperatures is given by Arrhenius equation as follows:
$$
\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right], T_2>T_1
$$
Given, $\mathrm{E}_{\mathrm{a}}=0, \mathrm{~T}_1=200 \mathrm{~K}, \mathrm{k}_1=1.6 \times 10^6 \mathrm{~s}^{-1}$
$$
\mathrm{T}_2=400 \mathrm{~K}, \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
$$
According to Arrhenius equation,
$$
\log \frac{\mathrm{k}_2}{\mathrm{k}_1}=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right]
$$
On substituting the given values in above equation, we get
$$
\begin{aligned}
& \log \frac{\mathrm{k}_2}{1.6 \times 10^6}=\frac{0}{2.303 \times 8.314}\left[\frac{\mathrm{l}}{200}-\frac{\mathrm{l}}{400}\right] \\
& \log \left(\frac{\mathrm{k}_2}{1.6 \times 10^6}\right)=0 \\
& \frac{\mathrm{k}_2}{1.6 \times 10^6}=10^0=1 \\
& \mathrm{k}_2=1.6 \times 10^6 \mathrm{~s}^{-1} \text { at } 400 \mathrm{~K}
\end{aligned}
$$
Thus, option (2) is correct.

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2019 (Odisha)
For a reaction between $A$ and $B$ the order with respect to $A$ is 2 and the other with respect to $B$ is 3 . The concentrations of both $\mathrm{A}$ and $\mathrm{B}$ are doubled, the rate will increase by a factor of:

Options:
A 12
B 16
C 32
D 10

The correct answer is:
32

Solution:
$$
\begin{aligned}
& \text { Rate }_1=k[\mathrm{~A}]^2[\mathrm{~B}]^3 \\
& \operatorname{Rate}_2=k[2 \mathrm{~A}]^2[2 \mathrm{~B}]^3 \\
& \therefore \quad \text { Rate }_2=32[\text { Rate }]_1 \\
&
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2013 (Karnataka)
For a reaction, I-+OCl-→IO-+Cl- in an aqueous medium, the rate of reaction is given by d[IO-]dt=k[I-][OCl-][OH-]
The overall order of reaction is

Options:
A -1
B 0
C 1
D 2

The correct answer is:
1

Solution:
d[IO-]dt=k[I-]1[OCl-]1[OH-]1
Here rate =K[I]1[OCl-]1[OH]1-
Order of reaction =1+1-1=1

Topic: Chemistry, Chemical Kinetics, NEET
For a reaction in which all reactants and products are liquids, which one of the following equations is most applicable?

Options:
A ∆H<∆E
B ∆H=∆S
C ∆H≈∆E
D Total W=0

The correct answer is:
∆H≈∆E

Solution:
As all reactant and product are liquid
âng=0
ΔH=ΔE− Δn g RT
ΔH=ΔE ∵ Δn g =0

Topic: Chemistry, Thermodynamics (C), NEET
For a reaction, temperature increase by 10oC, the equilibrium will be attained faster by

Options:
A 2 times
B Same
C 1/2 times
D 4 times

The correct answer is:
2 times

Solution:
Rate of reaction becomes almost twice with an increase of 10oC in temperature (from 298 K to 308 K).

Topic: Chemistry, Chemical Kinetics, NEET
For a real gas obeying van der waal's equation, a graph is plotted between PVm (yâaxis) and P (x-axis) where Vm is molar volume. y-intercept of the graph is

Options:
A RT
B P+aV2
C RTV-b
D Cannot be determined

The correct answer is:
RT

Solution:
P+aV2V-b=RT
Or PV+aV-Pb-abV2=RT
When, P=0, VM=V
PVM=Pb+RT
for 'r' intercept
PâO
â´Pb=O
â´PVm=RT

Topic: Chemistry, States of Matter, NEET
For a rigid diatomic molecule, the universal gas constant R=n C P, where, C P is the molar specific heat at constant pressure and n is a number. Hence, n is equal to

Options:
A 0.2257
B 0.4
C 0.2857
D 0.3557

The correct answer is:
0.2857

Solution:
For rigid diatomic molecule
CPCV=75
â´ CV=57CP
Also CP-CV=R
CP-57CP=R
27CP=R
â´ n=27=0.2857

Topic: Physics, Thermal Properties of Matter, NEET
For a sample of perfect gas, when its pressure is changed isothermally from pi to pf, the entropy change is given by

Options:
A ∆S=nRln⁡pfpi
B ∆S=nRln⁡pipf
C âS=nRTlnâ¡pfpi
D âS=RTlnâ¡pipf

The correct answer is:
∆S=nRln⁡pipf

Solution:
For a sample of perfect gas, when its pressure is changed isothermally from pi to pf, the entropy change is given by ∆S=nRln⁡pipf

Here, pi stands for initial pressure and pf stands for the final pressure.

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2016 (Phase 2)
For a satellite, escape velocity is 11 km s-1. If the satellite is launched at an angle of 60Â° with the vertical, then escape velocity will be

Options:
A 11 km s-1
B 113 km s-1
C 113 km s-1
D 33 km s-1

The correct answer is:
11 km s-1

Solution:
Since, escape velocity ve=2gRe is independent of angle of projection, so it will not change.

Topic: Physics, Gravitation, NEET
For a satellite in an orbit around the earth, the ratio of kinetic energy to potential energy is:

Options:
A $\frac{1}{2}$
B $\frac{1}{\sqrt{2}}$
C 2
D $\sqrt{2}$

The correct answer is:
$\frac{1}{2}$

Solution:
$-\frac{\mathrm{GMm}}{R^2}+m \omega^2 R=0$
$$
\begin{aligned}
\therefore \quad \frac{\mathrm{GMm}}{R^2} & =m \omega^2 R \\
\text { K.E. }=\frac{1}{2} I \omega^2 & =\frac{1}{2} m R^2 \omega^2 \\
& =\frac{\mathrm{GMm}}{2 R}
\end{aligned}
$$
$$
\begin{aligned}
& P . E .=-\frac{\mathrm{GMm}}{R} \therefore K . E=\frac{|P \cdot E|}{2} \\
& \therefore \quad \frac{K \cdot E}{P \cdot E}=\frac{1}{2}
\end{aligned}
$$

Topic: Physics, Gravitation, NEET, NEET 2005
For a series L-C-R circuit at resonance, the statement which is not true is

Options:
A Peak energy stored by a capacitor = peak energy stored by an inductor
B Average power = apparent power
C Wattles current is zero
D Power factor is zero

The correct answer is:
Power factor is zero

Solution:
For a series L - C - R
circuit at resonance
Phase difference, ϕ=0°
Power factor=cos⁡ϕ=1

Topic: Physics, Alternating Current, NEET
For a series LCR circuit the power loss at resonance is:

Options:
A $\frac{V^2}{\left(\omega L-\frac{1}{\omega C}\right)}$
B $r L \omega$
C $I^2 R$
D $\frac{V^2}{C \omega}$

The correct answer is:
$I^2 R$

Solution:
Power loss is given by
$$
P=V I \cos \phi
$$
At resonance current and voltage are in the same phase, that is $\phi=0$.
$$
\therefore P=V I \cos 0=V I=I^2 R
$$

Topic: Physics, Alternating Current, NEET, NEET 2002
For a solution at atmospheric pressure, water potential is

Options:
A Ψw=Ψs+Ψp
B Ψw=Ψs
C Ψw=Ψs-Ψp
D Ψw=0

The correct answer is:
Ψw=Ψs

Solution:
Î¨w is the water potential of cell, and it the kinetic energy present in water molecules. If some solutes is dissolved in pure water, the solution has fewer free water and the concentration of water decreases, reducing its water potential. For a solution at atmospheric pressure, Î¨w is equal to concentration of solutes present in it or Î¨s .

Topic: Biology, Transport in Plants, NEET
For a son to be haemophilic,

Options:
A the mother must be haemophilic.
B the father must be haemophilic.
C the mother may be haemophilic or carrier for haemophilia.
D the father may be haemophilic or carrier for haemophilia.

The correct answer is:
the mother may be haemophilic or carrier for haemophilia.

Solution:
Haemophilia is an X linked recessive disease. A son is haemophilic if he has the genotype XhY. Son receives X chromosome from his mother only. Thus the mother must have atleast on Xh chromosome. Thus the genotype of the mother can be XhX or XhXh. Thus the mother may be haemophilic or carrier for haemophilia

Topic: Biology, Principles of Inheritance and Variation, NEET
For a sparingly soluble salt ApBq, the relationship of its solubility product (Ls) with its solubility (S) is given as:

Options:
A Ls=Sp + qÂ·ppÂ·qq
B Ls=Sp + q·pq·qp
C Ls=Spq·pp·qq
D Ls=Spq·p.qp + q

The correct answer is:
Ls=Sp + q·pp·qq

Solution:
A p B q (s) ⇌ p A aq +q +q B aq −p
K sp = A q + p B p - q
= pS p qS q
L s = K sp = p p q q S p + q in saturated solution.

Topic: Chemistry, Ionic Equilibrium, NEET
For a sphere made out of a certain material, the moment of inertia of the sphere is proportional to [radius of the sphere = R ]

Options:
A R2
B R3
C R4
D R5

The correct answer is:
R5

Solution:
Moment of inertia of a solid sphere is 25mR2
I=25mR2
I=25VρR2
I=25×43πR3ρR2
I=815πρR5
I∝R5

Topic: Physics, Rotational Motion, NEET
For a transistor, α=0.9 , the value of β is

Options:
A 1
B 0.09
C 0.9
D 9

The correct answer is:
9

Solution:
Using β=α1-α
we get β=0.90.1=9

Topic: Physics, Semiconductors, NEET
For a transistor, Î±dc and Î²dc are the current ratios, then the value of Î²dc-Î±dcÎ±dc â Î²dc is

Options:
A 1
B 1.5
C 2
D 2.5

The correct answer is:
1

Solution:
Î²dc=Î±dc1-Î±dc â 1-Î±dc=Î±dcÎ²dc
Î²dc-Î±dcÎ±dcâÎ²dc=Î²dc1-Î±dcÎ²dcÎ±dcâÎ²dc
=1-Î±dcÎ²dcÎ±dc=1-1-Î±dcÎ±dc
=1-1+Î±dcÎ±dc=1

Topic: Physics, Semiconductors, NEET
For a transistor $\frac{I_C}{I_E}=0.96$, then the current gain for common emitter configuration:

Options:
A 12
B 6
C 48
D 24

The correct answer is:
24

Solution:
For the common emitter configuration, the current transfer ratio is
$\begin{aligned}
& \alpha=\frac{I_C}{I_B}=0.96 \\
& \therefore \text { Current gain }=\beta=\frac{\alpha}{1-\alpha} \\
& =\frac{0.96}{1-0.96}=24
\end{aligned}$

Topic: Physics, Semiconductors, NEET, NEET 2002
For a transistor, the correct statement is

Options:
A Collector current increases with increase in emitter current
B Collector current decreases with increase in emitter current
C Collector current increases with decrease in emitter current
D Collector current decreases with decrease in emitter current

The correct answer is:
Collector current increases with increase in emitter current

Solution:
Collector current increases with the increase in emitter current.

Topic: Physics, Semiconductors, NEET
For a transistor working as common base amplifier, the emitter current is 0.72 mA . The current gain is 0.96 . The collector current is

Options:
A 0.96×0.72 mA
B 0.960.72mA
C 0.96-0.72 mA
D 7.2A⁡-2×0.96 mA

The correct answer is:
0.96×0.72 mA

Solution:
α=iCiE
iC=αiE=0.96×0.72 mA

Topic: Physics, Semiconductors, NEET
For a transparent medium relative permeability and permittivity $\mu_{\mathrm{r}}$ and $\epsilon_{\mathrm{r}}$ are 1.0 and 1.44 respectively. The velocity of light in this medium would be

Options:
A $2.5 \times 10^8 \mathrm{~m} / \mathrm{s}$
B $3 \times 10^8 \mathrm{~m} / \mathrm{s}$
C $2.08 \times 10^8 \mathrm{~m} / \mathrm{s}$
D $4.32 \times 10^8 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$2.5 \times 10^8 \mathrm{~m} / \mathrm{s}$

Solution:
The speed of any electromagnetic wave (EM) like light in any medium of permeability, $\mu$ and permittivity, $\varepsilon$ is given by
$$
\mathrm{v}=\frac{1}{\sqrt{\mu \varepsilon}}
$$
Also, $\mu=\mu_0 \mu_r$ and $\varepsilon=\varepsilon_0 \varepsilon_r$ where, $\mu_0$ and $\varepsilon_0$ are permeability and permittivity in free space and $\mu_{\mathrm{r}}$ and $\varepsilon_{\mathrm{r}}$ are relative permeability and relative permittivity of medium respectively.
Here, $\mu_{\mathrm{r}}=1.0$
$$
\varepsilon_{\mathrm{r}}=1.44
$$
$$
\begin{aligned}
\therefore \quad \quad \quad \mathrm{v} & =\frac{1}{\sqrt{\mu_{\mathrm{r}} \varepsilon_{\mathrm{r}} \mu_0 \varepsilon_0}}=\frac{\mathrm{c}}{\sqrt{\mu_{\mathrm{r}} \varepsilon_{\mathrm{r}}}} \quad\left[\because \mathrm{c}=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\right] \\
& =\frac{3 \times 10^8}{\sqrt{1 \times 1.44}} \\
& =2.5 \times 10^8 \mathrm{~ms}^{-1}
\end{aligned}
$$

Topic: Physics, Electromagnetic Waves, NEET, NEET 2019 (Odisha)
For a weak acid $\mathrm{HA}$, the percentage of dissociation is nearly $1 \%$ at equilibrium. If the concentration of acid is $0.1 \mathrm{~mol} \mathrm{~L}^{-1}$, then the correct option for its $\mathrm{K}_{\mathrm{a}}$ at the same temperature is

Options:
A $1 \times 10^{-5}$
B $1 \times 10^{-3}$
C $1 \times 10^{-4}$
D $1 \times 10^{-6}$

The correct answer is:
$1 \times 10^{-5}$

Solution:
$\begin{aligned} & \mathrm{HA} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{-} \\ & \text {At } \mathrm{t}=0 \quad \mathrm{C} \quad- \\ & \text { At eq. } \mathrm{C}-\mathrm{C} \alpha \quad \mathrm{C} \alpha \quad \mathrm{C} \alpha \\ & \mathrm{K}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\frac{(\mathrm{C} \alpha)(\mathrm{C} \alpha)}{\mathrm{C}(1-\alpha)}=\frac{\mathrm{C} \alpha^2}{1-\alpha}=\mathrm{C}^2 \quad(\because \alpha \text { is very small }) \\ & =0.1(0.01)^2=1 \times 10^{-5} \\ & \end{aligned}$

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2023 (Manipur)
For a weak electrolyte α1 and α2 are in ratio of 1 : 2, for a given concentration ka1=2×10-4. What will be value of ka2?

Options:
A 8×10-4
B 2×10-4
C 4×10-4
D 1×10-4

The correct answer is:
8×10-4

Solution:
α1α2=ka1ka2
12=2×10-4ka2
14=2×10-4ka2
ka2=8×10-4

Topic: Chemistry, Ionic Equilibrium, NEET
For a zero order reaction
' R 0 is the initial concentration'

Options:
A t 1/2 ∝ R 0
B t 1/2 ∝ 1/R 0
C t 1/2 ∝ R 0 2
D t 1/2 ∝ 1/R 0 2

The correct answer is:
t 1/2 ∝ R 0

Solution:
For zero order reaction, t 1/2 ∝ R 0

Topic: Chemistry, Chemical Kinetics, NEET