datavorous/entrance-exam-dataset huggingface dataset:
The covalency of nitrogen in $\mathrm{HNO}_3$ is

Options:
A 0
B 3
C 4
D 5

The correct answer is:
5

Solution:
$(+1)+x+3(-2)=0 \Rightarrow 1+x-6=0 \Rightarrow x=6-1=5$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The covalent alkaline earth metal halide X=Cl,Br,I is:

Options:
A CaX2
B MgX2
C SrX2
D BeX2

The correct answer is:
BeX2

Solution:
Beryllium Halides are more covalent in nature, this is due to small size and high charge density of Be+2 ion. Hence it has high polarization power and more covalent character.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The covalent character of the following chlorides follow the order

Options:
A \(\mathrm{HgCl}_2 < \mathrm{CdCl}_2 < \mathrm{ZnCl}_2\)
B \(\mathrm{ZnCl}_2 < \mathrm{CdCl}_2 < \mathrm{HgCl}_2\)
C \(\mathrm{CdCl}_2 < \mathrm{ZnCl}_2 < \mathrm{HgCl}_2\)
D \(\mathrm{HgCl}_2 < \mathrm{ZnCl}_2 < \mathrm{CdCl}_2^2\)

The correct answer is:
\(\mathrm{ZnCl}_2 < \mathrm{CdCl}_2 < \mathrm{HgCl}_2\)

Solution:
\(\mathrm{ZnCl}_2\) and \(\mathrm{CdCl}_2\) are ionic, but \(\mathrm{HgCl}_2\) is covalent.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The critical angle for a denser-rarer interface is 45°. The speed of light in rarer medium is 3×108 m s-1. The speed of light in the denser medium is:

Options:
A 3.12×107 m s-1
B 5×107 m s-1
C 2.12×108 m s-1
D 2×108 m s-1

The correct answer is:
2.12×108 m s-1

Solution:
The critical angle and the refractive index of a media are related by the formula:

sinθc=1μ ...1

Substitute the value of the critical angle into equation (1) and solve to calculate the refractive index of the medium.

sin45°=1μ⇒12=1μ⇒μ=2

The refractive index of the medium can also be written as

μ=cv ...2

Substitute the values of the known parameters into equation (2) and solve to calculate the required speed of light within the medium.

2=3×108 m s-1v⇒v=3×108 m s-12=2.12×108 m s-1

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (11 Apr Shift 1)
The critical angle for a medium is $60^{\circ}$. The refractive index of the medium is

Options:
A $\frac{2}{\sqrt{3}}$
B $\frac{\sqrt{2}}{3}$
C $\sqrt{3}$
D $\frac{\sqrt{3}}{2}$

The correct answer is:
$\frac{2}{\sqrt{3}}$

Solution:
$\mu=\frac{1}{\sin C}=\frac{1}{\sin 60^{\circ}}=\frac{2}{\sqrt{3}}$

Topic: Physics, Ray Optics, JEE Main
The critical angle for diamond (refractive index $=2$ ) is

Options:
A About $20^{\circ}$
B $60^{\circ}$
C $45^{\circ}$
D $30^{\circ}$

The correct answer is:
$30^{\circ}$

Solution:
$\mu=\frac{1}{\sin C} \Rightarrow C=\sin ^{-1}\left(\frac{1}{2}\right)=30^{\circ}$

Topic: Physics, Ray Optics, JEE Main
The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3 and relative permeability 43 for this wavelength, will be:

Options:
A 15o
B 30o
C 45o
D 60o

The correct answer is:
30o

Solution:
V=1με
n=μrεr=2
sin ic=1n=12
ic=30°

Topic: Physics, Ray Optics, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The critical constants PC & TC for methane are 45 atm and 180 K . The correct statement is -

Options:
A VC=2.4Â L
B b=0.04 L/mol
C VC=0.8Â L
D b=0.8Â L/mol

The correct answer is:
b=0.04 L/mol

Solution:
The critical constants PC & TC for methane are 45 atm and 180 K .
Therefore, P C = a 27b 2 ,
T C = 8a 27 Rb
âPCTC=R8b
â b=0.04Lmol .

Topic: Chemistry, States of Matter, JEE Main
The critical micelle concentration CMC of a cationic colloidal electrolyte is 10-3 M . If 1 mm3 contains 1013 micelles, the number of cations making one micelle is
(Given, NA=6.0Ã1023 mol-1 )

Options:

The correct answer is:
60.00

Solution:
At CMC, the number of cationic colloidal particles =10-3Ã6.0Ã1023= 6.0Ã1020 per L
Therefore, number of particles per mm3
= 6.0Ã1020106âµ1L=106mm3
= 6.0Ã1014 particles/ mm3
Number of micelles in 1 mm3= 1013 (given)
Number of cations forming one micelle
=6.0Ã10141013
=60 cations

Topic: Chemistry, Surface Chemistry, JEE Main
The crystal field splitting energy (CFSE) for CoCl64- is about 18000 cm-1 . What would be the CFSE value for [CoCl4]2- ?

Options:

The correct answer is:
8000

Solution:
CFSE Δ0 of [CoCl6]4-=18000cm-1
CFSE Δt of [CoCl4]2-=?
From formula, Δt=49×Δ0
=49×18000 cm-1
=8000 cm-1

Topic: Chemistry, Coordination Compounds, JEE Main
The crystal field stabilization energies (CFSE) of high spin and low spin $\mathrm{d}^{6}$ metal complexes in terms of $\Lambda_{0}$, respectively, are-

Options:
A $-0.4$ and $-2.4$
B $-2.4$ and $-0.4$
C $-0.4$ and $0.0$
D - $2.4$ and $0.0$

The correct answer is:
$-0.4$ and $-2.4$

Solution:

Topic: Chemistry, Coordination Compounds, JEE Main
The Crystal Field Stabilization Energy (CFSE) of CoFH2O33Δ0<P is :

Options:
A -0.8 Δ0+2P
B -0.4 Δ0
C -0.8 Δ0
D -0.4 Δ0+P

The correct answer is:
-0.4 Δ0

Solution:
CoH2O3F3 Co3+=3d04s0⇒t2g2v1,e91.1

CFSE=-0.4nt29+0.6nogΔ0+n(P)

=[-0.4×4+0.6×2]Δ∘+0

=-0.4∆0

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The crystal field stabilization energy (CFSE) of FeH2O6Cl2 and K2NiCl4, respectively, are:

Options:
A -0.4∆0 and -1.2∆t
B -2.4∆0 and -1.2∆t
C -0.4∆0 and -0.8∆t
D -0.6∆0 and -0.8∆t

The correct answer is:
-0.4∆0 and -0.8∆t

Solution:
FeH2O62+;Fe+2⇒3d6, Since H2O is
Weak field ligands,

CFSE =4×-0.4+0.6×2=-0.4â0
CFSE =4×0.4+-0.6×4=-0.8â0 and -0.8ât

Topic: Chemistry, Coordination Compounds, JEE Main
The crystal field stabilization energy (CFSE) of FeH2O6Cl2 and K2NiCl4, respectively, are:

Options:
A -0.4∆0 and -1.2∆t
B -2.4∆0 and -1.2∆t
C -0.4∆0 and -0.8∆t
D -0.6∆0 and -0.8∆t

The correct answer is:
-0.4∆0 and -0.8∆t

Solution:
FeH2O62+;Fe+2⇒3d6, Since H2O is
Weak field ligands,

CFSE =4×-0.4+0.6×2=-0.4â0
CFSE =4×0.4+-0.6×4=-0.8â0 and -0.8ât

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The crystal system of a compound with unit cell dimensions a=0.387, b=0.387 and c= 0.504 nm and α=β=90o and γ=120o is

Options:
A Hexagonal
B Cubic
C Rhombohedral
D Orthorhombic

The correct answer is:
Hexagonal

Solution:
For the given crystal,
a=b≠c,α=β=90o,γ=120o
These are the characteristic of a hexagonal system.

Topic: Chemistry, Solid State, JEE Main
The cube roots of unity when represented on the Argand plane form the vertices of an

Options:
A Equilateral triangle
B Isosceles triangle
C Right angled triangle
D None of these

The correct answer is:
Equilateral triangle

Solution:
The cube roots of unity are $1, \omega, \omega^2$. We know that if $z_1, z_2, z_3$ are the vertices of an equilateral triangle in the Argand plane. Then $z_1^2+z_2^2+z_3^2=z_1 z_2+z_2 z_3+z_3 z_1$
If we take $z_1=1, z_2=\omega, z_3=\omega^2$
Then $z_1^2+z_2^2+z_3^2=1+\omega^2+\omega^4=0$
and $z_1 z_2+z_2 z_3+z_3 z_1=1 . \omega+\omega . \omega^2+\omega^2 .1$
$=\omega+\omega^3+\omega^2=1+\omega+\omega^2=0$
Thus $z_1^2+z_2^2+z_3^2=z_1 z_2+z_2 z_3+z_3 z_1$
Therefore triangle is equilateral.
Note : Students should remember this question as a fact.

Topic: Mathematics, Complex Number, JEE Main
The current density in a cylindrical wire of radius r=4.0 mm is 1.0×106 A2 m2. The current through the outer portion of the wire between radial distances r2 and r is xπ A; where x is

Options:

The correct answer is:
12

Solution:
Current =J× Area

=J×πR2-R22=Jπ3R24=106×π×3×4×10-342

⇒π×3×164=12π A

Hence, the value of x=12.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (27 Jun Shift 2)
The current flowing in a coil of self inductance $0.4 \mathrm{mH}$ is increased by $250 \mathrm{~mA}$ in $0.1 \mathrm{sec}$. The e.m.f. induced will be

Options:
A $+1 \mathrm{~V}$
B $-1 \mathrm{~V}$
C $+1 \mathrm{mV}$
D $-1 \mathrm{mV}$

The correct answer is:
$-1 \mathrm{mV}$

Solution:
$e=-L \frac{d i}{d t}=$ $-0.4 \times 10^{-3} \times \frac{250 \times 10^{-3}}{0.1}=-1 \mathrm{mV}$

Topic: Physics, Electromagnetic Induction, JEE Main
The current flowing in two coaxial coils in the same direction. On increasing the distance between the two, the electric current will

Options:
A Increase
B Decrease
C Remain unchanged
D The information is incomplete

The correct answer is:
Increase

Solution:
Induced current in both the coils assist the main current so current through each coil increases

Topic: Physics, Electromagnetic Induction, JEE Main
The current flowing through a conductor connected across a source is 2 A and 1.2 A at 0oC and 100oC respectively. The current flowing through the conductor at 50oC will be _____ × 102 mA.

Options:

The correct answer is:
15

Solution:
By Ohm's law, V=IR.

Let the current and resistance at 0° C be i0 & R0 respectively.

Let the current and resistance at 100 °C be i100 & R100 respectively.

By the given condition,

i0R0=i100R100⇒2R0=1.2R0(1+100α)⇒1+100α=53⇒50α=13

Since the voltage remains constant,

i50R50=i0R0⇒i50=i0R0R50=2×R0R01+50α=21+13=1.5 A⇒i50=15×102 mA

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (12 Apr Shift 1)
The current flowing through an ac circuit is given by I=5sin120πt A. How long will the current take to reach the peak value starting from zero?

Options:
A 160 s
B 60 s
C 1120 s
D 1240 s

The correct answer is:
1240 s

Solution:
If initial value of current is zero, then it will take T4 time to reach the peak value, where T= time period.

As we know, ω=120π

Therefore,

t=T4=2πω4 =π2ω=π2×120π=1240 s

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (27 Jun Shift 1)
The current gain Î² for a common emitter amplifier is 69. If the emitter current is 7 mA, the base current is

Options:
A 0.1 mA
B 1 mA
C 0.2 mA
D 2 mA

The correct answer is:
0.1 mA

Solution:
Here,
Î²=69
IE=7 mA
Since, Î²=Î±1-Î±
69=Î±1-Î±
69-69Î±=Î±
69=70Î±
Î±=6970=0.986
ICIE=0.986
â´IC=0.986 IE=0.986Ã7=6.9 mA
â´IB=IE-IC=7-6.9=0.1 mA

Topic: Physics, Semiconductors, JEE Main
The current gain in the common emitter amplifier mode of a transistor is 10. The input impedance is 20 kÎ© and load resistance is 100 kÎ©. The power gain is

Options:
A 300
B 500
C 200
D 100

The correct answer is:
500

Solution:
The power gain is defined as the ratio of change in output power and the change in input power.
Since, P=VI.

Therefore, Power gain = Current gain X Voltage gain
=β×βRoutRin=β2RoutRin

Given, β=10, Rin=20 kΩ and Rout=100 kΩ.
Hence, Power gain =10210020=100×5=500

Topic: Physics, Semiconductors, JEE Main
The current gain of a common emitter amplifier is 69. If the emitter current is 7.0 mA, collector current is :

Options:
A 69 mA
B 0.69 mA
C 6.9 mA
D 9.6 mA

The correct answer is:
6.9 mA

Solution:
β=69=ICIB
α=β1+β=6970=ICIE
IC=IE×6970
IC=6970×7
IC=6.9 mA.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2017 (09 Apr Online)
The current gain α of a transistor in common base mode is 0.995. Its gain β in the common-emitter mode is

Options:
A 200
B 99
C 199
D None of these

The correct answer is:
199

Solution:
The current gain α of a transistor in common base mode is 0.995
β is current gain in the common emitter mode
β=α1-α=0.9951-0.995=199

Topic: Physics, Semiconductors, JEE Main
The current I1 (in A) flowing through the 1 Ω resistor in the given circuit is

Options:
A 0.2
B 0.4
C 0.5
D 0.25

The correct answer is:
0.2

Solution:
I=12.5=0.4A
i=I2=0.2A

Topic: Physics, Current Electricity, JEE Main
The current (i) at time t=0 and t=∞ respectively for the given circuit is :

Options:
A 5E18,18E55
B 5E18,10E33
C 10E33,5E18
D 18E55,5E18

The correct answer is:
5E18,10E33

Solution:
At t=0, current through inductor is zero,

hence Req=5+1‖5+4=185

i1=E18/5=5E18

At t=∞, inductor becomes a simple wire and now the circuit will be as shown in figure

hence Req=5∥5+4∥1=3310; (∥⇒ parallel)

i2=E33/10=10E33

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (25 Feb Shift 1)
The current $\mathrm{I}$ drawn from the 5 volt source will be

Options:
A
$0.17 \mathrm{~A}$

B
$0.33 \mathrm{~A}$

C
$0.5 \mathrm{~A}$

D
$0.67 \mathrm{~A}$

The correct answer is:
$0.5 \mathrm{~A}$

Solution:
$i=\frac{5}{10}=0.5$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2006
The current i in a coil varies with time as shown in the figure. The variation of induced emf with time would be

Options:
A
B
C
D

The correct answer is:

Solution:
Induced emf, e = - L di dt
For 0 â¤ t â¤ T 4 ,
i-t graph is a straight line with positive constant slope.
â´ di dt = constant
â e = - ve and constant For 0 â¤ t â¤ T 4
For T 4 â¤ t â¤ T 2 ,
âµ i is constant â´ di dt = 0
â e = 0 For T 4 â¤ t â¤ T 2
For T 2 â¤ t â¤ 3 T 4 ,
i-t graph is a straight line with negative constant slope.
â´ di dt = constant
â e = + ve and constant For T 2 â¤ t â¤ 3 T 4
For 3 T 4 â¤ t â¤ T ,
i is zero â´ di dt = 0
â e = 0 For 3 T 4 â¤ t â¤ T
From this analysis, the variation of induced emf with time as shown in the figure below.

Topic: Physics, Electromagnetic Induction, JEE Main
The current I in an inductance coil varies with time according to the graph given in the figure.

Which one of the following graphs gives the variation of voltage with time across the inductor?

Options:
A
B
C
D

The correct answer is:

Solution:
The voltage across an inductor is

V=-Ldidt

So the voltage across the inductor will always be opposite to the rate of change of current.

When didt>0 then V<0 and vice versa

Also, the magnitude of V is proportional to the magnitude of didt.

When didtdecreases, V also decreases

Topic: Physics, Electromagnetic Induction, JEE Main
The current I in the given circuit will be

Options:
A 10A
B 20A
C 4A
D 40A

The correct answer is:
10A

Solution:
The circuit can be redrawn as :

We can see that the circuit is a balanced Wheatstone bridge

⇒RAB=4 Ω

Current through cell I=VRAB=404 A=10 A

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (26 Jul Shift 1)
The current $(I)$ in the inductance is varying with time according to the plot shown in figure.
Which one of the following is the correct variation of voltage with time in the coil?

Options:
A

B

C

D

The correct answer is:

Solution:
For inductor as we know induced voltage For $t=0$ to $t=T / 2$
$V=L \frac{d I}{d t}=L \frac{d}{d t}\left(\frac{2 I_0 t}{T}\right)=\text { constant, }$
For
$\begin{aligned}
t & =T / 2 \text { to } t=T \\
V & =\frac{L d I}{d t}\left(\frac{-2 I_0 t}{T}\right)=- \text { constant }
\end{aligned}$
Therefore, answer will represents by graph (d).

Topic: Physics, Electromagnetic Induction, JEE Main
The current i in the network is

Options:
A 0.2 A
B 0.6 A
C 0.3 A
D 0 A

The correct answer is:
0.3 A

Solution:
Both the diodes are in reverse biased.
I=930=310A=0.3 A

Topic: Physics, Semiconductors, JEE Main, JEE Main 2020 (09 Jan Shift 2)
The current $I$ is in the circuit shown is

Options:
A $1.33 \mathrm{A}$
B zero
C $2.00 \mathrm{A}$
D $1.00 \mathrm{A}$

The correct answer is:
$1.33 \mathrm{A}$

Solution:
The circuit diagram can be redrawn as

For the loop $A B C D A$,
$$
+2-2 I^{\prime}+2-2 I=0
$$
For the loop $A B F E A$,
$$
\begin{array}{r}
2-2 I+2-2\left(I-I^{\prime}\right)=0 \\
4-2 I-2 I+2 I^{\prime}=0
\end{array}
$$
$$
\begin{aligned}
4-4 I+2 I^{\prime} &=0 \\
2 &=2 I-I^{\prime}
\end{aligned}
$$
From Eqs (i) and (in), we get
$$
2=2 I-I^{\prime} \cdot 2=I+I^{\prime}
$$
$$
4=3 I, I=4 / 3=1.33 \mathrm{A}
$$

Topic: Physics, Current Electricity, JEE Main
The current in a coil of self inductance L=2.0 H is increasing according to the law i=2sint2. Find the amount of energy spent (in J) during the period when the current changes from 0 to 2 A.

Options:

The correct answer is:
4

Solution:
Let the current be 2 A at t=τ
Then 2=2sinτ2⇒τ=π/2
when the instantaneous current is i, the
self-induced emf is Ldidt. If dq amount of
charge is displaced in time dt, then elementary work done
=Ldidtdq=Ldidtidt= Lidi

∴ W=∫0t Lidi =∫0tL2sint2d2sint2

⇒W=∫0t8Lsint2cost2tdt=4L∫0tsin2t2dt

Let θ=2t2⇒dθ=4tdt
∴ the integral
=4L∫sinθdt/4L(-cosθ)=-Lcos2t2
∴ W=-cos2t20π/2=2L=4J

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2022 (25 Jun Shift 1)
The current in a conductor is expressed as I=3t2+4t3, where I is in Ampere and t is in second. The amount of electric charge that flows through a section of the conductor during t=1 s to t=2 s is ____________ C.

Options:

The correct answer is:
22

Solution:
For variable current, integrate to get the total charge flown in a given interval.q=∫12idt=∫123t2+4t3dt

⇒q=t3+t412

⇒q=22 C

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (01 Feb Shift 1)
The current in an inductor is given by $\mathrm{I}=(3 \mathrm{t}+8)$ where $\mathrm{t}$ is in second. The magnitude of induced emf produced in the inductor is $12 \mathrm{mV}$. The self-inductance of the inductor _____ $\mathrm{mH}$.

Options:

The correct answer is:
4

Solution:
$\begin{aligned} & \mathrm{I}=3 \mathrm{t}+8 \\ & \varepsilon=12 \mathrm{mV} \\ & |\varepsilon|=\mathrm{L}\left|\frac{\mathrm{dI}}{\mathrm{dt}}\right| \\ & 12=\mathrm{L} \times 3 \\ & \mathrm{~L}=4 \mathrm{mH}\end{aligned}$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The current in the branch of a circuit shown below is

Options:
A 0 A
B 10-2A
C 1 A
D 0.10 A

The correct answer is:
0 A

Solution:
Since the diode is in reverse bias, there is no current in the branch.

Topic: Physics, Semiconductors, JEE Main
The current needed to reduce 26.6 g of nitrobenzene to aniline in acidic medium, is

Options:
A 0.4 F
B 0.6 F
C 0.8 F
D 1.2 F

The correct answer is:
1.2 F

Solution:
No. of gram equivalent = No. of Faraday
mole x n-factor = No. of Farady
0.2×6=1.2 F

Topic: Chemistry, Electrochemistry, JEE Main
The current of 5 A flows in a square loop of sides 1 m is placed in air. The magnetic field at the centre of the loop is X2×10-7 T. The value of X is _________.

Options:

The correct answer is:
40

Solution:
Magnetic field at the centre of the square loop is given by,

B=4μ0I4πdsin45°+sin45°, where d is perpendicular distance of the centre to one of its sides.

⇒B=4×μ0×54π×1212+12

=4×10-7×5×2×2

=402×10-7 T

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (30 Jan Shift 2)
The current of a conductor flowing through a conductor in terms of the drift speed of electrons is (the symbols have their usual meanings)

Options:
A

I= nAe v d

B

I=A ne v d

C

I= ne A v d

D

I=neA v d

Solution:

2385 Upvotes

Verified Answer

The correct answer is:

I=neA v d

Relation between current and drift velocity : Consider a conductor of length l and area of cross – section A having n electrons per unit length, as shown in the figure.

The volume of the conductor = Al

∴ Total number of electrons in the conductor

= Volume × electron density = Aln

If e is the charge of an electron, then

total charge contained in the conductor,

Q=enAl

Let a potential difference V is applied across the conductor. The resulting electric field in the conductor is given by

E= V l

Under the influence of this field E, free electrons begin to drift in a direction opposite to that of the field. Time taken by electrons to cross – over the conductor is

t= l v d

where, v d is the drift velocity of electrons.

Therefore, the current through the conductor is given by

I= Q t = enAl I/ v d

or I=neA v d

⇒Iα v d [ ∵n,e,A are all constant]

Hence, Current density is proportional to drift velocity.

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The correct answer is:

I=neA v d

Solution:
Relation between current and drift velocity : Consider a conductor of length l and area of cross – section A having n electrons per unit length, as shown in the figure.

The volume of the conductor = Al

∴ Total number of electrons in the conductor

= Volume × electron density = Aln

If e is the charge of an electron, then

total charge contained in the conductor,

Q=enAl

Let a potential difference V is applied across the conductor. The resulting electric field in the conductor is given by

E= V l

Under the influence of this field E, free electrons begin to drift in a direction opposite to that of the field. Time taken by electrons to cross – over the conductor is

t= l v d

where, v d is the drift velocity of electrons.

Therefore, the current through the conductor is given by

I= Q t = enAl I/ v d

or I=neA v d

⇒Iα v d [ ∵n,e,A are all constant]

Hence, Current density is proportional to drift velocity.

Topic: Physics, Current Electricity, NEET
The current sensitivity of a galvanometer can be increased by :
(A) decreasing the number of turns
(B) increasing the magnetic field
(C) decreasing the area of the coil
(D) decreasing the torsional constant of the spring

Choose the most appropriate answer from the options given below :

Options:
A (B) and (C) only
B (C) and (D) only
C (A) and (C) only
D (B) and (D) only

The correct answer is:
(B) and (D) only

Solution:
The current in the galvanometer is calculated by i=KNABθ.

Therefore, the sensitivity of the galvanometer,

dθdi=NABK

Hence, only options B and D are correct.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (28 Jul Shift 1)
The current sensitivity of moving coil galvanometer is increased by 25%. This increase is achieved only changing in the number of turns of coils and area of cross section of the wire while keeping the resistance of galvanometer coil constant. The percentage change in the voltage sensitivity will be:

Options:
A +25%
B -50%
C -25%
D Zero

The correct answer is:
+25%

Solution:
The current sensitivity is given as

Si=NBAC.

The voltage sensitivity is

Sv=NABCR, where R is the galvanometer resistance.

Since the resistance is constant,
ΔSi=ΔSv

Hence, the percentage change in voltage sensitivity is,

ΔSv=+25%.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (11 Apr Shift 1)
The current through an inductor of impedance 10 Î© lags behind the voltage by a phase of 60Â° when just the inductor is connected to the ac source. Now the inductor is connected to a 5Î© resistance in series, then the net impedance of the circuit is

Options:
A 15Â Î©
B 12Â Î©
C 13.2 Î©
D 18Â Î©

The correct answer is:
13.2 Î©

Solution:
10=r2+XL2 and XLr=tan⁡60°
10=r2+r32 or r=5 Ω, XL=53 Ω
Z=5+52+532=175=13.2 Ω

Topic: Physics, Electromagnetic Induction, JEE Main
The current through choke coil increases form zero to $6 A$ in 0.3 seconds and an induced e.m.f. of $30 \mathrm{~V}$ is produced. The inductance of the coil of choke is

Options:
A $5 \mathrm{H}$
B $2.5 \mathrm{H}$
C $1.5 \mathrm{H}$
D $2 \mathrm{H}$

The correct answer is:
$1.5 \mathrm{H}$

Solution:
$|e|=L \frac{d i}{d t} \Rightarrow 30=$ $=L \times \frac{(6-0)}{0.3} \Rightarrow L=1.5 \mathrm{H}$

Topic: Physics, Electromagnetic Induction, JEE Main
The current through the battery shown in the circuit is

Options:
A 1.33 A
B 1.71 A
C 2.00 A
D 2.31 A

The correct answer is:
2.00 A

Solution:
Since diode D 1 is reverse biased, therefore it will act like an open circuit.
Effective resistance of the circuit is R = 4 + 2 = 6 Ω .
Current in the circuit is I = E / R = 12 / 6 = 2 A.

Topic: Physics, Semiconductors, JEE Main
The current voltage relation of diode is given by I = (e1000 V/T - 1) mA, where the applied voltage V is in volts and the temperature T is in degree Kelvin. If a student makes an error measuring ± 0.01 V while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA ?

Options:
A 0.2 mA
B 0.02 mA
C 0.5 mA
D 0.05 mA

The correct answer is:
0.2 mA

Solution:
I = e 1000 V / T - 1 mA
When I = 5 mA , e 1000 V / T = 6 mA
Also, dI = e 1000 V / T × 1000 T · dV
= 6 mA × 1000 300 × 0.01
= 0.2 mA

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2014 (06 Apr)