datavorous/entrance-exam-dataset huggingface dataset:
An ideal monoatomic gas initially at \( 300 \mathrm{~K} \) undergoes an isobaric expansion at a pressure of \( 2.5 \mathrm{kPa} \). If the volume increases from \( 1 \mathrm{~m}^{3} \) to \( 3 \mathrm{~m}^{3} \), then heat added to the gas and its final temperature respectively are

Options:
A \( 12500 \mathrm{~J}, 450 \mathrm{~K} \)
B \( 12500 \mathrm{~J}, 600 \mathrm{~K} \)
C \( 12500 \mathrm{~J}, 900 \mathrm{~K} \)
D \( 25000 \mathrm{~J}, 120 \mathrm{~K} \)

The correct answer is:
\( 12500 \mathrm{~J}, 900 \mathrm{~K} \)

Solution:
Given, T=300 K, P=2.5×103 Pa, Vi=1 m3 & Vf=3 m3, now heat added at constant pressure in terms of the molar specific heat capacity at constant pressure,

ΔQ=nCpT2-T1

=nγRγ-1T2-T1=nγγ-1RT2-RT1=γγ-1V2-V1P

This is a monoatomic gas so the ratio of molar specific heat at constant pressure and constant volume, γ=CPCV=53

∆Q=5353-1×2.5×103×3-1=12500 J

Now using idea gas equation at constant pressure,

V1V2=T1T2⇒13=300T2⇒T2=900 K

Topic: Physics, Kinetic Theory of Gases, JEE Main
An ideal monoatomic gas is confined in a cylinder by a spring loaded piston of cross section $8.0 \times 10^{-3} \mathrm{~m}^2$. Initially the gas is at $300 \mathrm{~K}$ and occupies a volume of $2.4 \times 10^{-3} \mathrm{~m}^3$ and the spring is in its relaxed state as shown in figure. The gas is heated by a small heater until the piston moves out slowly by $0.1 \mathrm{~m}$. The force constant of the spring is $8000 \mathrm{~N} / \mathrm{m}$ and the atmospheric pressure is $1.0 \times 10^5 \mathrm{~N} / \mathrm{m}^2$. The cylinder and the piston are thermally insulated. The piston and the spring are massless and there is no friction between the piston and the cylinder. The final temperature of the gas will be: (Neglect the heat loss through the lead wires of the heater. The heat capacity of the heater coil is also negligible).

Options:
A
$300 \mathrm{~K}$

B
$800 \mathrm{~K}$

C
$500 \mathrm{~K}$

D
$1000 \mathrm{~K}$

The correct answer is:
$500 \mathrm{~K}$

Solution:
No Solution Available

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2014 (11 Apr Online)
An ideal monoatomic gas is taken round the cycle. ABCDA as shown in figure. The work done during the cycle is

Options:
A -PV
B -2PV
C -12PV
D 0

The correct answer is:
-PV

Solution:
As net work done during a complete cycle is equal to area under the cycle so here work done is equal to -PV .

Topic: Chemistry, Thermodynamics (C), JEE Main
An ideal monoatomic gas undergoes an expansion from state A to state B following a process which is shown in the indicator diagram. If initially during the expansion, the gas was absorbing heat and later on it was rejecting heat, then what was the volume of the gas when it started rejecting the heat?

Options:
A V=32V0
B V=85V0
C V=158V0
D V=127V0

The correct answer is:
V=158V0

Solution:
From the graph we can find the pressure of the gas in terms of volume

P-P022V0-V=P02V0

P=3P02-P02V0V=nRTV

⇒nRT=3P02V-P02V0V2

nRdT=3P02dV-P0V0VdV

dW=PdV=3P02-P0V2V0dV

dU=32nRdT=9P04dV-3P02V0VdV

dQ=dU+dW=0

9P04-3P02V0VdV+3P02-P0V2V0dV=0

⇒154=2VV0

V=158V0

Topic: Physics, Thermodynamics, JEE Main
An ideal pendulum was oscillating with an angular amplitude θ = π / 3 inside a stationary elevator. When the pendulum was passing through equilibrium position, the lift suddenly starts moving upward with an acceleration a=g. Find the new angular amplitude of the oscillation.

Options:
A π 3
B π 6
C cos -1 3 4
D π 2

The correct answer is:
cos -1 3 4

Solution:
1 - cos θ ′ = 1 2 1 - cos θ
cos θ ′ = 1 2 + 1 2 cos θ
⇒ θ ′ = cos -1 1 + cos θ 2
θ ′ = cos -1 1 + cos π / 3 2
= cos -1 1 + 1 / 2 2 = cos -1 3 4

Topic: Physics, Oscillations, JEE Main
An ideal refrigerator has a freezer at a temperature of $-13^{\circ} \mathrm{C}$. The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) will be

Options:
A $325^{\circ} \mathrm{C}$
B $325 K$
C $39^{\circ} \mathrm{C}$
D $320^{\circ} \mathrm{C}$

The correct answer is:
$39^{\circ} \mathrm{C}$

Solution:
Coefficient of performance
$\begin{gathered}K=\frac{T_2}{T_1-T_2} \Rightarrow 5=\frac{(273-13)}{T_1-(273-13)}=\frac{260}{T_1-260} \\ \Rightarrow 5 T_1-1300=260 \Rightarrow 5 T_1=1560 \\ \Rightarrow T_1=312 \mathrm{~K} \rightarrow 39^{\circ} \mathrm{C}\end{gathered}$

Topic: Physics, Thermodynamics, JEE Main
An ideal string is wrapped on a ring and the free end of the string is attached to the ceiling as shown in the figure. Initially, the system is held in equilibrium by an external agent and at some instant of time, the system is released from rest. If there is no slipping between the string and the ring, then the tension in the string is

Options:
A mg
B mg2
C mg3
D 2mg

The correct answer is:
mg2

Solution:
mg-T=ma
TR=mR2α
a=αR
mg-Tm=Tm
T=mg2

Topic: Physics, Rotational Motion, JEE Main
An ideal transformer with purely resistive load operates at 12 kV on the primary side. It supplies electrical energy to a number of nearby houses at 120 V. The average rate of energy consumption in the houses served by the transformer is 60 kW. The value of resistive load (Rs) required in the secondary circuit will be_________mΩ.

Options:

The correct answer is:
240

Solution:
The given data is

P=60×103 W

Vs=120 V

A transformer has a primary coil and a secondary coil.

The formula for the power delivered is given by

P=V2sRs

⇒60×103=120×120Rs⇒Rs=120×12060×103=240 mΩ

Topic: Physics, Alternating Current, JEE Main, JEE Main 2023 (06 Apr Shift 1)
An important reaction of acetone is auto-condensation in presence of concentrated sulphuric acid to give the aromatic compound

Options:
A mesitylene
B mesityl oxide
C trioxane
D phorone

The correct answer is:
mesitylene

Solution:
An important reaction of acetone is auto-condensation in presence of concentrated sulphuric acid to give the aromatic compound mesitylene.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
An impulse J is applied on a ring of mass m along a line passing through its centre O . The ring is placed on a rough horizontal surface. The linear velocity of centre of ring once it starts rolling without slipping is

Options:
A J m
B J 2m
C J 4m
D J 3m

The correct answer is:
J 2m

Solution:
Let v be the velocity of COM of the ring just after the impulse is applied and v' is the velocity when pure rolling starts.

Angular velocity ω of the ring at this instant will be ω = v' r

⇒Impulse = change in linear momentum
we have, J=mv
or v=J/m

Between the two positions shown in figure force of
friction on the ring acts backwards. Angular momentum
of the ring about bottommost point will remain conserved
∴        L i = L f
or mvr=mv'r+lω
= mv' r + mr 2 v' r = 2 mv' r
∴ v' = v 2 = J/2m

Topic: Physics, Rotational Motion, NEET
An inclined plane is bent in such a way that the vertical cross-section is given by y=x24 where y is in vertical and x in horizontal direction. If the upper surface of this curved plane is rough with coefficient of friction μ=0.5, the maximum height in cm at which a stationary block will not slip downward is________cm.

Options:

The correct answer is:
25

Solution:
At maximum ht. block will experience maximum friction force. Therefore if at this height slope of the tangent is tanθ, then θ=Angle of repose

∴ tanθ=dydx=2x4=x2=0.5

⇒x=1 and therefore y=x24=0.25 m

=25 cm

∴ Answer is 25 cm

(Assuming that x & y in the equation are given in meter)

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (24 Feb Shift 1)
An inclined plane is located at angle Î±=53Â° to the horizontal. There is a hole at point B in the inclined plane as shown in the figure. A particle is projected along the plane with speed v0 at an angle Î²=37Â° to the horizontal in such a way so that it gets into the hole. Neglect any type of friction. Find the speed v0 (in m s-1) if h=1 m and l=8m.

Options:
A 9
B 3
C 6
D 12

The correct answer is:
9

Solution:
h=ltan⁡β-12gsin⁡αl2v02cos2⁡β
v0=gsin⁡αl22cos2⁡βltan⁡β-h
=10×45×8×82×45×458×34-1=9 m s-1

Topic: Physics, Motion In Two Dimensions, JEE Main
An inclined plane making an angle of 30° with the horizontal is placed in a uniform horizontal electric field 200NC as shown in the figure. A body of mass 1 kg and charge 5 mC is allowed to slide down from rest at a height of 1 m. If the coefficient of friction is 0.2, find the time taken by the body to reach the bottom.
g=9.8 m s-2; sin30°=12; cos30°=32

Options:
A 0.92 s
B 1.3 s
C 0.46 s
D 2.3 s

The correct answer is:
1.3 s

Solution:
here N=9.8cos30+1sin30≈9 N

soa=9.8sin30-1cos30-μN1

a=2.233 m s-2

By S=ut+12at2

=122.233t2

sin30°

t≈1.3 sec

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (26 Feb Shift 2)
An incompressible liquid flows through a horizontal tube as shown in the following fig. Then the velocity $v$ of the fluid is

Options:
A $3.0 \mathrm{~m} / \mathrm{s}$
B $1.5 \mathrm{~m} / \mathrm{s}$
C $1.0 \mathrm{~m} / \mathrm{s}$
D $2.25 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$1.0 \mathrm{~m} / \mathrm{s}$

Solution:
If the liquid is incompressible then mass of liquid entering through left end, should be equal to mass of liquid coming out from the right end.
$\therefore M=m_1+m_2 \Rightarrow A v_1=A v_2+1.5 A . v$
$\Rightarrow A \times 3=A \times 1.5+1.5 A \cdot v \Rightarrow v=1 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
An increase in bleeding time and delay in blood coagulation can possibly be due to the deficiency of which hormone?

Options:
A Thyroxine
B Adrenaline
C Noradrenaline
D Parathormone

The correct answer is:
Parathormone

Solution:
Parathormone or parathyroid hormone is produced and secreted by the parathyroid glands and their main function is to regulate serum calcium concentration by increasing the calcium levels in the blood. Calcium plays an important role in blood clotting and therefore if it is deficient there will be a delay in the clotting process. Thyroxine regulates metabolic processes of the body and adrenaline and nor-adrenaline are hormones of the sympathetic nervous system.

Topic: Biology, Chemical Coordination and Integration, NEET
An increase in equivalent conductance of a strong electrolyte with dilute dilution is mainly due to

Options:
A increase in number of ions
B increase in ionic mobility of ions
C 100 % ionization of electrolyte at normal dilution
D increase in both i.e., number of ions and ionic mobility of ions

The correct answer is:
increase in ionic mobility of ions

Solution:
As with dilution ionic mobility of ions increases which increases the equivalent conductance of a strong electrolyte.

Topic: Chemistry, Electrochemistry, NEET
An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to

Options:
A increase in ionic mobility of ions
B $100 \%$ ionisation of electrolyte at normal dilution
C increase in both, ie, number of ions and ionic mobility of ions
D increase in number of ions

The correct answer is:
increase in ionic mobility of ions

Solution:
Key Idea $\lambda_{\text {eq }}=\kappa \times \mathrm{V}=\frac{\mathrm{k} \times 1000}{\text { normality }}$.
On dilution, the number of current carrying particles per $\mathrm{cm}^3$ decreases but the volume of solution increases. Consequently, the ionic mobility increases, which in turn increases the equivalent conductance of strong electrolyte.

Topic: Chemistry, Electrochemistry, NEET, NEET 2010 (Screening)
An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to:

Options:
A increase in both i.e. number of ions and ionic mobility of ions
B increase in number of ions
C increase in ionic mobility of ions
D \(100 \%\) ionization of electrolyte at normal dilution

The correct answer is:
increase in both i.e. number of ions and ionic mobility of ions

Solution:
On dilution, the number of charge carriers per \(\mathrm{m}^3\) decreases but the volume increases. Consequently, the ionic mobility increases which increase equivalent conductance.

Topic: Chemistry, Electrochemistry, JEE Main
An individual molecular structure within an antigen to which an individual antibody binds is as an:

Options:
A hinge region
B epitope
C Fab region
D Fc region

The correct answer is:
epitope

Solution:
Antigens are the exogenous substances (toxin components of the pathogen, foreign particle) that enters into the body and stimulates the production of antibodies. They are mainly proteins present on the surface of cells, microbes (bacteria, viruses). Combining sites of antigen, called antigenic determinant or epitopes react with the corresponding antigen-binding sites of antibodies called paratopes.

Topic: Biology, Human Health and Diseases, NEET
An inductance coil has a reactance of 100 Ω . When an AC signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45°. The self-inductance of the coil is

Options:
A 1.1×10-2 H
B 1.1×10-1 H
C 5.5×10-5 H
D 6.7×10-7 H

The correct answer is:
1.1×10-2 H

Solution:
tanθ=xLR=tan45°
xL=R
=100=xL2+R2
100=R2+R2
2R=100
R=502
∴xL=502
Lω=502
L=5022π×1000=252πmH
=1.1×10-2H

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2020 (02 Sep Shift 2)
An inductance of $1 \mathrm{mH}$ a condenser of $10 \mu \mathrm{F}$ and a resistance of $50 \Omega$ are connected in series. The reactances of inductor and condensers are same. The reactance of either of them will be

Options:
A $100 \Omega$
B $30 \Omega$
C $3.2 \Omega$
D $10 \Omega$

The correct answer is:
$10 \Omega$

Solution:
$\begin{aligned} & \text { Given } \omega L=\frac{1}{\omega C} \Rightarrow \omega^2=\frac{1}{L C} \\ & \text { or } \omega=\frac{1}{\sqrt{10^{-3} \times 10 \times 10^{-6}}}=\frac{1}{\sqrt{10^{-8}}}=10^4 \\ & X_L=\omega L=10^4 \times 10^{-3}=10 \Omega\end{aligned}$

Topic: Physics, Alternating Current, JEE Main
An inductive circuit contains resistance of $10 \Omega$ and an inductance of $20 \mathrm{H}$. If an ac voltage of $120 \mathrm{~V}$ and frequency $60 \mathrm{~Hz}$ is applied to this circuit, the current would be nearly

Options:
A $0.32 \mathrm{amp}$
B $0.016 \mathrm{amp}$
C $0.48 \mathrm{amp}$
D $0.80 \mathrm{amp}$

The correct answer is:
$0.016 \mathrm{amp}$

Solution:
$i=\frac{V}{\sqrt{R^2+\omega^2 L^2}}=\frac{120}{\sqrt{100+4 \pi^2 \times 60^2 \times 20^2}}=0.016 \mathrm{~A}$

Topic: Physics, Alternating Current, JEE Main
An inductor 20 mH, a capacitor 50 μF and a resistor 40 Ω are connected in series across a source of emf V=10 sin ⁡340 t. The power loss in AC . circuit is:

Options:
A 0.51ââW
B 0.67  W
C 0.76  W
D 0.89  W

The correct answer is:
0.51  W

Solution:
Given : R=40Ω , L= 20 mH , C=50 μF

Comparing the given equation for voltage with general equation V=V0 sin ωt

ω=340 rad s-1 and peak voltage V0=10 V

Inductive reactance XLωL=340×20×10-3=68×10-1=6.8 Ω
Capacitive reactance XC
1ωC=1340×50×10-6=10434×5=234×103
=0.0588×103=58.82 Ω
Z=XL-XC2+R2⇒Z=ωL-1ωC2+R2
Z=58.8-6.82+402 ⇒ Z=2704+1600≈65.6 Ω
Irms=VrmsZ=1065.6×2                   (Vrms=V02)
Power =ErmsIrms cos ϕ=Irms2R

∵cos ϕ=RZ, Vrms=IrmsZ

=100×4065.62×2=200065.62
=0.51 W

Topic: Physics, Alternating Current, JEE Main
An inductor coil and a capacitor and a resistance of 5 Ω are connected in series to an a.c. source of rms voltage 30 V. When the frequency of the source is varied, a maximum r.m.s. current of 5 A is observed. If this inductor is connected in parallel with a resistance 5 Ω to a battery of emf 25 V and internal resistance 2.0 Ω , the current drawn from the battery is

Options:
A 1 5 0 1 7 A
B 2.5 A
C 1 2 5 3 3 A
D 1 2 5 9 9 A

The correct answer is:
1 5 0 1 7 A

Solution:
Let resistance of inductor be μ
I max = 3 0 5 + μ = 5 ⇒ μ = 1 Ω
Now, I = 2 5 5 6 + 2
= 1 5 0 1 7 A

Topic: Physics, Alternating Current, JEE Main
An inductor coil stores 64 J of magnetic field energy and dissipates energy at the rate of 640 W when a current of 8 A is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit (in s).

Options:
A 0.4
B 0.2
C 0.125
D 0.8

The correct answer is:
0.2

Solution:
Energy stored in an inductor, U=12LI2
⇒64=12L(8)2 ⇒L=2 H
and power dissipated in a resistor, P=I2R
⇒640=(8)2R ⇒R=10 Ω
Time constant of the LR DC circuit is,
τ=LR=210=0.2 s

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (26 Aug Shift 1)
An inductor ( L=0.03 H ) and a resistor (R=0.15  kΩ) are connected in series to a battery of 15 V E.M.F. in a circuit shown below. The key K1 has been kept closed for a long time. Then at t=0, K1 is opened and key K2 is closed simultaneously. At t=1 ms , the current in the circuit will be : Take, e5≈150

Options:
A 0.67 mA
B 100 mA
C 67 mA
D 6.7 mA

The correct answer is:
0.67 mA

Solution:
Case I: K1 is closed for long time,
for a long time, the inductor acts as a conducting wire, ⇒ current in the circuit =VR = 1 5 1 5 0 ⇒i0=0.1 A
Case II: K1 is open and K2 is closed
Current in the circuit, i=i0e-tτ; τ=LR
After t=1 ms=10-3 s
⇒i=i0e-10-3×1503×10-2 = 0.1 e - 1 5 3 =0.11e5=0.1150 A=0.67 mA.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2015 (04 Apr)
An inductor ( L=0.03 H ) and a resistor ( R=0.15 kΩ ) are connected in series to a battery of 15 V EMF in a circuit shown below. The key K1 has been kept closed for a long time. Then at t=0, K1 is opened and key K2 is closed simultaneously. At t=1 ms , the current in the circuit will be : e 5 ≅150

Options:
A 0.67 mA
B 100 mA
C 67 mA
D 6.7 mA

The correct answer is:
0.67 mA

Solution:
Case I: K 1 is closed for long tiime
for long time, inductor acts as a conducting wire.
â current in the circuit = V R = 1 5 1 5 0
i 0 = 0.1 A
Case II: K 1 is open and K 2 is closed
Current in the circuit
i = i 0 e - t Ï ; Ï = L R
After t = 1 ms = 1 0 - 3 s
i = i 0 e - 1 0 - 3 Ã 1 5 0 3 Ã 1 0 - 2
= 0.1 e - 1 5 3 = 0.1 1 e 5 = 0.1 1 5 0 A = 0.67 mA

Topic: Physics, Alternating Current, JEE Main
An inductor $(L=100 \mathrm{mH})$, a resistor $(R=100 \Omega)$ and a battery $(E=100 \mathrm{~V})$ are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points $A$ and $B$. The current in the circuit $1 \mathrm{~mm}$ after the circuit is

Options:
A
$1 \mathrm{~A}$

B
$1 / \mathrm{e} \mathrm{A}$

C
$e \mathrm{~A}$

D
$0.1 \mathrm{~A}$

The correct answer is:
$1 / \mathrm{e} \mathrm{A}$

Solution:
$\mathrm{I}=\mathrm{I}_0 \mathrm{e}^{-\mathrm{Rt} / \mathrm{L}}=\frac{1}{\mathrm{e}} \mathrm{A}$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2006
An inductor may store energy in

Options:
A its electric field
B its coil
C its magnetic field
D Both electric and magnetic fields

The correct answer is:
its magnetic field

Solution:
An inductor stored energy in its magnetic field.
Energy $1=\frac{1}{2} L i^{2}$, where $L i$ is magnetic flux.

Topic: Physics, Electromagnetic Induction, JEE Main
An inductor of 0.5 mH, a capacitor of 20 μF and resistance of 20 Ω are connected in series with a 220 V ac source. If the current is in phase with the emf, the amplitude of current of the circuit is x A. The value of x is-

Options:

The correct answer is:
242

Solution:
Here, emf and current are in phase, so the circuit is either pure resistive circuit or case of resonance,

XL=XC

Thus, the impedance of circuit is Z=R2+XL-XC2=R=20 Ω

Now, rms value of current is irms=22020=11 A

Using the relation between peak and rms value of current, we have peak value i0=irms2=112=242 A.

Hence, the value of x=242.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2023 (31 Jan Shift 1)
An inductor of 0.5 mH, a capacitor of 200 μF and a resistor of 2 Ω are connected in series with a 220 V ac source. If the current is in phase with the emf, the frequency of ac source will be ______×102 Hz.

Options:

The correct answer is:
5

Solution:
Given here: L=0.5×10-3 H

C=200×10-6 F

R=2 Ω.

If current is in phase with emf, so this is the condition of resonance.

f=fR=12πLC=12×3.14LC

f=16.280.5×10-3×200×10-6=503.55 Hz=5.03×102 Hz

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (29 Jun Shift 2)
An inductor of 10 mH is connected to a 20 V battery through a resistor of 10 kΩ and a switch. After a long time, when maximum current is set up in the circuit, the current is switched off. The current in the circuit after 1 μs is x100 mA. Then x is equal to ______ . (Take e-1=0.37)

Options:

The correct answer is:
74

Solution:
Imax=VR=20 V10 KΩ=2 mA

For LR-decay circuit,

I=Imaxe-Rt/L

I=2 mA e-10×103×1×10-610×10-3

I=2 mA e-1

I=2×0.37 mA

I=74100 mA

x=74

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (25 Jul Shift 1)
An inductor of inductance 2 μH is connected in series with a resistance, a variable capacitor and an AC source of frequency 7 kHz. The value of capacitance for which maximum current is drawn into the circuit is 1x F, where the value of x is ______.

(Take π=227)

Options:

The correct answer is:
3872

Solution:
For maximum current in series LCR circuit,

XC=XL⇒12πfC=2πfL

⇒C=14π2f2 L=14×π2×49×106×2×10-6

⇒C=13872 F

Therefore, x=3872.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2023 (29 Jan Shift 2)
An inductor of inductance 2.0  mH is connected across a charged capacitor of capacitance 5.0  μF and the resulting L-C circuit is set oscillating at its natural frequency. Let Q denote the instantaneous charge on the capacitor and I the current in the circuit. It is found that the maximum value of Q is 200  μC. When Q=100  μC, what is the value of dI / dt?

Options:
A 10000
B 1000
C 100000
D 100

The correct answer is:
10000

Solution:
This is a problem of L - C oscillations.
Charge stored in the capacitor oscillates simple harmonically as
Q= Q 0 sin( ÏtÂ±Ï )

Here, Q 0 = maximum value of Q=200ââÎ¼C=2Ã 10 â4 ââC

Ï = 1 LC = 1 2 Ã 10 - 3 5.0 Ã 1 0 - 6 = 10 4 s -1

At t=0,ââQ= Q 0 then

Q( t )= Q 0 cosÏt ... (i)

I t = dQ dt = - Q 0 Ï sin Ï t and ... (ii)

dI t dt = - Q 0 Ï 2 cos Ï t ... (iii)

Q=100ââÎ¼C or Q 0 2
cos Ï t = 1 2 or Ï t = Ï 3
dI dt = 2.0 Ã 1 0 - 4 C 1 0 4 s -1 2 1 2

dI dt = 1 0000 A / s

Topic: Physics, Alternating Current, JEE Main
An inductor of inductance L=400 mH and resistors of R1=2 Î© and R2=2 Î© are connected to a battery emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t=0. The potential drop across L as a function of time is

Options:
A 12e-5t V
B 12te3t V
C 61-e-t0.2 V
D 6 e-5t V

The correct answer is:
12e-5t V

Solution:
In the branch containing L and R2
i=ER21-e-R2tL
didt=ER2e-R2tL.R2L=ELe-R2tL
∴VL=Ldidt=Ee-R2tL
=12e-2t400×10-3=12  e-5t V

Topic: Physics, Alternating Current, JEE Main
An inductor of inductance $\mathrm{L}=400 \mathrm{mH}$ and resistors of resistances $\mathrm{R}_1$ $=2 \Omega$ and $R_2=2 \Omega$ are connected to a battery of emf $12 \mathrm{~V}$ as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at $t=0$. The potential drop across $L$ as a function of time is

Options:
A
$6 e^{-5 t} V$

B
$\frac{12}{t} e^{-3 t} V$

C
$6\left(1-e^{-t / 0.2}\right) V$

D
$12 e^{-5 t} V$

The correct answer is:
$12 e^{-5 t} V$

Solution:
$$
\begin{aligned}
& \mathrm{I}_1=\frac{\mathrm{F}}{\mathrm{R}_1}=\frac{12}{2}=6 \mathrm{~A} \\
& \mathrm{E}=\mathrm{L} \frac{\mathrm{dl_{2 }}}{\mathrm{dt}}+\mathrm{R}_2 \times \mathrm{I}_2 \\
& \mathrm{I}_2=\mathrm{I}_0\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{t}_{\mathrm{c}}}\right) \Rightarrow \mathrm{I}_{\mathrm{o}}=\frac{\mathrm{E}}{\mathrm{R}_2}=\frac{12}{2}=6 \mathrm{~A} \\
& \mathrm{t}_{\mathrm{c}}=\frac{\mathrm{L}}{\mathrm{R}}=\frac{400 \times 10^{-3}}{2}=0.2 \\
& \mathrm{I}_2=6\left(1-\mathrm{e}^{-\mathrm{t} / 0.2}\right)
\end{aligned}
$$
Potential drop across $L=E-R_2 l_2=12-2 \times 6\left(1-e^{-b t}\right)=12 e^{-5 t}$
Directions: Question numbers 28,29 and 30 are based on the following paragraph. Two moles of helium gas are taken over the cycle $A B C D A$, as shown in the $P-T$ diagram.

Topic: Physics, Electromagnetic Induction, JEE Main