datavorous/entrance-exam-dataset huggingface dataset:
An alternating current is given by the equation i=i1sinωt+i2cosωt. The rms current will be:

Options:
A 12i1+i22
B 12i1+i2
C 12i12+i2212
D 12i12+i2212

The correct answer is:
12i12+i2212

Solution:
i=i1sinωt+i2sinωt+90

i=i12+i22sinωt+ϕ

irms=i02=i12+i222

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (26 Feb Shift 1)
An alternating current of frequency $50 \mathrm{~Hz}$ has the peak value as $14 \cdot 14 \mathrm{~A}$. The time
taken by the alternating current in reaching from zero to maximum value and r.m.s. value of current will be respectively

Options:
A $0.025 \mathrm{~s}, 5 \mathrm{~A}$
B $0 \cdot 005 \mathrm{~s}, 5 \mathrm{~A}$
C $0 \cdot 005 \mathrm{~s}, 10 \mathrm{~A}$
D $0 \cdot 025 \mathrm{~s}, 10 \mathrm{~A}$

The correct answer is:
$0 \cdot 005 \mathrm{~s}, 10 \mathrm{~A}$

Solution:
(D)
$\mathrm{f}=50 \mathrm{~Hz}, \mathrm{~T}=\frac{1}{50} \mathrm{~s}=0.02 \mathrm{~s}$
Time taken to reach from zero to maximum value is
$\begin{array}{l}
t=\frac{T}{4}=\frac{0.02}{4}=0.005 \mathrm{~s} \\
I_{m s}=\frac{I_{0}}{\sqrt{2}}=\frac{14.14}{1.414}=10 \mathrm{~A}
\end{array}$

Topic: Physics, Alternating Current, JEE Main
An alternating e. \( \mathrm{m} \). f. of \( 100 \mathrm{~V} \) (rms) is applied to a series LCR circuit. At resonance, the potential difference across the inductance and across the capacitance is \( 400 \mathrm{~V} \) each. The potential difference across the resistance will be,

Options:
A \( 100 \mathrm{~V} \)
B \( 400 \mathrm{~V} \)
C \( 800 \mathrm{~V} \)
D zero

The correct answer is:
\( 100 \mathrm{~V} \)

Solution:
According to question, voltage of source, VS=100 V, voltage across inductor, VL=400 V and voltage across capacitance, VC=400 V. Let voltage across resistor be VR.

From the phasor diagram, we can write

VS=VL-VC2+VR2

Substituting the values in the formula, we get

100=400-4002+VR2⇒VR=100 V

Topic: Physics, Alternating Current, JEE Main
An alternating electric field of frequency f is applied across the radius R of a cyclotron to accelerate protons (mass m ). The operating magnetic field B used and K.E. of the proton beam produced by it are respectively ( e= charge on proton)

Options:
A 2Ïmfe,2Ï2mf2R2
B 2Ï2mfe2,4Ï2mf2R2
C Ï mfe ,Ï2mf2R2
D 2Ï2m2f2e ,2Ï2m2f2R2

The correct answer is:
2πmfe,2π2mf2R2

Solution:
T= 2πm eB
R= mf eB = P eB
B= 2πm e f
P=eBR=e× 2πmυ e R
K.E= P 2 2m = ( 2πmfR ) 2 2m
=2 π 2 m f 2 R 2

Topic: Physics, Magnetic Effects of Current, JEE Main
An alternating emf E=440sin100πt is applied to a circuit containing an inductance of 2πH. If an a.c. ammeter is connected in the circuit, its reading will be :

Options:
A 4.4 A
B 1.55 A
C 2.2 A
D 3.11 A

The correct answer is:
2.2 A

Solution:
Given that E=440sin100πt, L=2πH

Angular frequency of the source is ω=100π rad s-1.

Now the reactance of the inductor will be,

XL=ωL=100π2π=1002 Ω

Therefore, the peak current I0=E0XL=4401002=2.22 A

AC ammeter reads RMS value therefore reading will be Irms

Irms=I02=2.2 A

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (29 Jul Shift 1)
An alternating e.m.f. is applied to purely capacitive circuit. The phase relation between e.m.f. and current flowing in the circuit is
or
In a circuit containing capacitance only

Options:
A e.m.f. is ahead of current by $\pi / 2$
B Current is ahead of e.m.f. by $\pi / 2$
C Current lags behind e.m.f. by $\pi$
D Current is ahead of e.m.f. by $\pi$

The correct answer is:
Current is ahead of e.m.f. by $\pi / 2$

Solution:
For purely capacitive circuit $e=e_0 \sin \omega t$ $i=i_o \sin \left(\omega t+\frac{\pi}{2}\right)$ i.e. current is ahead of emf by $\frac{\pi}{2}$

Topic: Physics, Alternating Current, JEE Main
An alternating e.m.f. of frequency $v\left(=\frac{1}{2 \pi \sqrt{L C}}\right)$ is applied to a series $L C R$ circuit. For this frequency of the applied e.m.f.

Options:
A The circuit is at resonance and its impedance is made up only of a reactive part
B The current in the circuit is in phase with the applied e.m.f. and the voltage across $R$ equals this applied emf
C The sum of the p.d.'s across the inductance and capacitance equals the applied e.m.f. which is $180^{\circ}$ ahead of phase of the current in the circuit
D The quality factor of the circuit is $\omega L / R$ or $1 / \omega C R$ and this is a measure of the voltage magnification (produced by the circuit at resonance) as well as the sharpness of resonance of the circuit

The correct answer is:
The circuit is at resonance and its impedance is made up only of a reactive part

Solution:
As we know resonating angular frequency is
$\omega=\frac{1}{\sqrt{\mathrm{LC}}}$
so resonating frequency is
$\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$
so circuit is in resonance and impedence is made up of only resistance part

Topic: Physics, Alternating Current, JEE Main
An alternating volta ge $E=200 \sqrt{2} \sin (100 t)$ is connected to 1 microfarad capacitor through AC ammeter. The reading of ammeter shall be

Options:
A $10 \mathrm{~mA}$
B $20 \mathrm{~mA}$
C $40 \mathrm{~mA}$
D $80 \mathrm{~mA}$

The correct answer is:
$20 \mathrm{~mA}$

Solution:
Reading of ammeter $=I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_{C}}=\frac{V_{0} \omega C}{\sqrt{2}}$
$$
\begin{array}{l}
=\frac{200 \sqrt{2} \times 100 \times\left(1 \times 10^{-6}\right)}{\sqrt{2}}=2 \times 10^{-2} \mathrm{~A} \\
=20 \mathrm{~mA}
\end{array}
$$

Topic: Physics, Alternating Current, JEE Main
An alternating voltage e=2002sinâ¡100 t volt is connected to 1 Î¼F capacitor through AC ammeter. The reading of ammeter is

Options:
A 5 mA
B 10 mA
C 15 mA
D 20 mA

The correct answer is:
20 mA

Solution:
e= e 0 sinÏt
e 0 =200 2V
I rms = Ï rms X C = e 0 ÏC 2
= 200 2 Ã100Ã 10 â6 2
=2Ã 10 â2 =20 mA

Topic: Physics, Alternating Current, JEE Main
An alternating voltage is connected in series with a resistance $R$ and an inductance $L$ If the potential drop across the resistance is $200 \mathrm{~V}$ and across the inductance is $150 \mathrm{~V}$, then the applied voltage is

Options:
A $350 \mathrm{~V}$
B $250 \mathrm{~V}$
C $500 \mathrm{~V}$
D $300 \mathrm{~V}$

The correct answer is:
$250 \mathrm{~V}$

Solution:
The applied voltage is given by $V=\sqrt{V_R^2+V_L^2}$ $V=\sqrt{(200)^2+(150)^2}=250 \mathrm{volt}$

Topic: Physics, Alternating Current, JEE Main
An alternating voltage is given by E=100sin⁡ω+π6V . The voltage will be maximum for the first time when is (T = periodic time)

Options:
A T12
B T2
C T6
D T3

The correct answer is:
T6

Solution:
Given, alternating voltage is
E=100sin⁡ωt+π6V
The voltage will be maximum, when
sin⁡ωt+π6=1
⇒sin⁡ωt+π6=sinπ2
⇒ωt+π6=π2
⇒ωt=π3
⇒t=π3ω=π×T3×2π∵ω=2πT
=T6
Thus, the voltage will be maximum for the first time when t = T6 .

Topic: Physics, Alternating Current, JEE Main
An alternating voltage is represented as $E=20 \sin 300 \mathrm{t}$. The average value of voltage over one cycle will be

Options:
A Zero
B $10 \mathrm{volt}$
C $20 \sqrt{2}$ volt
D $\frac{20}{\sqrt{2}}$ volt

The correct answer is:
Zero

Solution:
$\mathrm{E}=20 \sin 300 \mathrm{t}$
Average value of any $\mathrm{AC}$ voltage over a cycle $=\mathrm{O}$ an it will have equal values for both positive and negative cycle, which will cancel out each other and resultant is zero

Topic: Physics, Alternating Current, JEE Main
An alternating voltage of amplitude $40 \mathrm{~V}$ and frequency $4 \mathrm{kHz}$ is applied directly across the capacitor of $12 \mu \mathrm{F}$. The maximum displacement current between the plates of the capacitor is nearly :

Options:
A $10 \mathrm{~A}$
B $12 \mathrm{~A}$
C $8 \mathrm{~A}$
D $13 \mathrm{~A}$

The correct answer is:
$12 \mathrm{~A}$

Solution:
Displacement current is same as conduction current in capacitor.
$\begin{aligned}
& X_C=\frac{1}{\omega \mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}} \\
& =\frac{1}{2 \pi \times 4 \times 10^3 \times 12 \times 10^{-6}}=3.317 \Omega
\end{aligned}$
$I=\frac{V}{X_C}=\frac{40}{3.317}=12 \mathrm{~A}$

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2024 (05 Apr Shift 1)
An alternating voltage source V=260sin(628t) is connected across a pure inductor of 5 mH. Inductive reactance in the circuit is:

Options:
A 3.14Ω
B 6.28Ω
C 0.5Ω
D 0.318Ω

The correct answer is:
3.14Ω

Solution:
Comparing the given equation of AC voltage with standard expression, V=V0sinωt, we get ω=628 rad s-1.

Inductive reactance is given by, XL=ωL.

=628×5×10-3

=3.14 Ω

Topic: Physics, Alternating Current, JEE Main, JEE Main 2023 (31 Jan Shift 2)
An alternating voltage V(t)=220sin100πt volt is applied to a purely resistive load of 50 Ω. The time taken for the current to rise from half of the peak value to the peak value is:

Options:
A 5 ms
B 3.3 ms
C 7.2 ms
D 2.2 ms

The correct answer is:
3.3 ms

Solution:
Time taken to change current value from half of the peak value to the peak value is,

t=T6

⇒t=2π6ω=π3ω=π300π=1300=3.33 ms

Topic: Physics, Alternating Current, JEE Main, JEE Main 2024 (30 Jan Shift 2)
An alternating voltage V(t)=220sin100πt volt is applied to a purely resistive load of 50 Ω . The time taken for the current to rise from half of the peak value to the peak value is:

Options:
A 7.21 ms
B 5.25 ms
C 2.24 ms
D 3.33 ms

The correct answer is:
3.33 ms

Solution:
i=VR=22050sin⁡100πt
i=imaxsin⁡100πt
For i=imax2:
imax2=imaxsin⁡100πt
sin⁡100πt=12 ⇒100πt=π6
t=1600 s
For i=imax:
imax=imaxsin⁡100πt
sin⁡100πt=1 ⇒100πt=π2
⇒t=1200 s

so time for imax2 to imax is
t(imax)-timax2=1200-1600=1300=3.33 ms

Topic: Physics, Alternating Current, JEE Main, JEE Main 2019 (08 Apr Shift 1)
An alternating voltage $V=V_{0}$ sin ot is appliedacross a circuit. As a result, a current $I=I_{0} \sin (\omega t-\pi / 2)$ flows in it. The power consumed per cycle is

Options:
A zero
B $0.5 \mathrm{~V}_{0} \mathrm{I}_{0}$
C $0.707 \mathrm{~V}_{0} \mathrm{I}_{0}$
D $1.414 \mathrm{~V}_{0} \mathrm{I}_{0}$

The correct answer is:
zero

Solution:
The phase angle between voltage $\mathrm{V}$ and current I is $\pi / 2$.

Topic: Physics, Alternating Current, JEE Main
An alternating voltage $\mathrm{V}=\mathrm{V}_{0} \sin \omega \mathrm{t}$ is applied across a circuit. As a result, a current $\mathrm{I}=\mathrm{I}_{0} \sin \left(\omega \mathrm{t}-\frac{\pi}{2}\right)$ flows in it. The power consumed per cycle is

Options:
A zero
B $0.5 \mathrm{~V}_{0} \mathrm{I}_{0}$
C $0.707 \mathrm{~V}_{0} \mathrm{I}_{0}$
D $1.414 \mathrm{~V}_{0} \mathrm{I}_{0}$

The correct answer is:
zero

Solution:
The phase angle between voltage $V$ and current $I$ is $\frac{\pi}{2}$.

Therefore, power factor $\cos \phi=\cos \left(\frac{\pi}{2}\right)=0$.

Hence the power consumed is zero.

Topic: Physics, Alternating Current, JEE Main
An aluminium rod of length $1 \mathrm{~m}$ and a steel rod of length $2 \mathrm{~m}$ both having same cross-sectional area, are soldered together end-to-end. The thermal conductivity of aluminium rod and steel rod is $200 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ and $50 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ respectively. The temperatures of the free ends are maintained at $300 \mathrm{~K}$ and $500 \mathrm{~K}$. What is the temperature of the junction?

Options:
A $322 \mathrm{~K}$
B $350 \mathrm{~K}$
C $367 \mathrm{~K}$
D $400 \mathrm{~K}$

The correct answer is:
$322 \mathrm{~K}$

Solution:
Let junction temperature is $T K$, then in steady state heat flow rate is same through both rods.
So, using,
$\frac{Q}{t}=\frac{k A\left(T_2-T_1\right)}{l}$, we have,
$$
\frac{Q}{t}=\frac{k_s A_s(500-T)}{2}=\frac{k_{\mathrm{Al}} A_{\mathrm{Al}}(T-300)}{1}
$$
As areas of both rods is same, so
$$
\begin{array}{rlrl}
& & \frac{k_s(500-T)}{2} & =\frac{k_{\mathrm{Al}}(T-300)}{1} \\
\Rightarrow & & \frac{50(500-T)}{2} & =\frac{200(T-300)}{1} \\
\Rightarrow & 25 \times 500-25 T & =200 T-200 \times 300 \\
\Rightarrow & & 225 T & =12500+60000 \\
\Rightarrow & & T & =\frac{72500}{225}=322.2 \approx 322 \mathrm{~K}
\end{array}
$$

Topic: Physics, Thermal Properties of Matter, TS EAMCET, TS EAMCET 2018 (05 May Shift 1)
An aluminium rod with Young’s modulus Y=7.0 ×1010 N m-2 undergoes elastic strain of 0.04%. The energy per unit volume stored in the rod in SI unit

Options:
A 2800
B 11200
C 5600
D 8400

The correct answer is:
5600

Solution:
The data given is

Y=7×1010 N m-2 and∆ll=0.04100

Young's modulus is given by the ratio of stress and strain,

Y=FA∆ll=FlA∆l

The energy stored is given as

E=YA2l∆l2⇒E=YA2∆ll2×l ...(i)

The energy per unit volume can be written from equation (i)

EV=Y2×∆ll2 ...(ii)

Substituting the values in equation (ii)

EV=12×7×1010×0.04×0.04104=56×102

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2023 (08 Apr Shift 1)
An aluminum rod (Young's modulus $=7 \times 10^9 \mathrm{~N} / \mathrm{m}^2$ ) has a breaking strain of $0.2 \%$. The minimum cross-sectional area of the rod in order to support a load of $10^4 \mathrm{Newton } \mathrm{is}$

Options:
A $1 \times 10^{-2} \mathrm{~m}^2$
B $1.4 \times 10^{-3} \mathrm{~m}^2$
C $3.5 \times 10^{-3} \mathrm{~m}^2$
D $7.1 \times 10^{-4} \mathrm{~m}^2$

The correct answer is:
$7.1 \times 10^{-4} \mathrm{~m}^2$

Solution:
$Y=\frac{F / A}{\text { strain }} \Rightarrow A=\frac{F}{Y\times \text { strain }}=\frac{10^4}{7 \times 10^9 \times 0.002}$
$=\frac{1}{14} \times 10^{-2}=7.1 \times 10^{-4} \mathrm{~m}^2$

Topic: Physics, Mechanical Properties of Solids, JEE Main
An amine $\mathrm{C}_{3} \mathrm{H}_{9} \mathrm{N}$ reacts with benzene sulphonyl chloride to form a white precipitate which is insoluble in $\mathrm{aq} . \mathrm{NaOH}$. The amine is

Options:
A
B
C
D

The correct answer is:

Solution:
Since, the amine gives white precipitate with benzene sulphonyl chloride which is insoluble in aq. $\mathrm{Na} \mathrm{OH},$ so Ti must be a secondary amine lie. amine having RNHR group).
Among the given compound, only
is a secondary amine, so it represents the structure of given amine.

Topic: Chemistry, Amines, JEE Main
An amine compound on reaction with benzenesulphonyl chloride gives an organic compound (P) which is highly acidic in nature. The compound P is

Options:
A
B
C
D

The correct answer is:

Solution:
We know that the reaction of the benzene sulfonyl chloride with primary amines gives a sulfonamide product that is soluble in alkali while with secondary amines it gives a sulfonamide product that is not soluble in alkali. Secondary amine is acidic in nature because amine contains one hydrogen with Nitrogen but in the case of primary amine it contains two hydrogen atoms with the nitrogen molecule that's why it is more acidic in nature

Hence, we conclude that primary amine is more acidic than secondary amine so the compound P will be:

Topic: Chemistry, Amines, JEE Main
An amine \((\mathrm{X})\) is prepared by ammonolysis of benzyl chloride. On adding p-toluenesulphonyl chloride to it the solution remains clear. Molar mass of the amine (X) formed is ______ \(\mathrm{g} \mathrm{mol}^{-1}\).
(Given molar mass in \(\mathrm{gmol}^{-1} \mathrm{C}: 12, \mathrm{H}: 1, \mathrm{O}: 16, \mathrm{~N}: 14\))

Options:

The correct answer is:
287

Solution:
Molar Mass of $(\mathrm{X})$ is $287 \mathrm{~g} \mathrm{~mol}^{-1}$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (06 Apr Shift 2)
An amine on reaction with benzenesulphonyl chloride produces a compound insoluble in alkaline solution. This amine can be prepared by ammonolysis of ethyl chloride. The correct structure of amine is:

Options:
A CH3CH2NH2
B CH3CH2CH2NHCH3
C
D

The correct answer is:

Solution:
An amine on reaction with benzenesulphonyl chloride produces a compound insoluble in alkaline solution. This amine can be prepared by ammonolysis of ethyl chloride.

It has to be 2° amine because on reaction with benzene sulphonylchloride it gives water in soluble product. As it is formed by ammonolysis of ethylchloride, so it has to be R-NH-Et type.

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (26 Feb Shift 1)
An amino acid having isoelectric point below 7 (at 25 o C ), when kept in a alkaline medium present in an electric field will show migration towards -

Options:
A Cathode
B Anode
C Either Cathode/Anode
D No migration

The correct answer is:
Anode

Solution:
In alkaline medium the zwitter ion will lose a proton & migrate towards anode due to net negative charge

Topic: Chemistry, Biomolecules, JEE Main
An ammeter has resistance R0 and range i. What resistance should be connected in parallel with it to increase its range by ni?

Options:
A R0/n-1
B R0/n+1
C R⁡ 0 / n⁡
D None of these

The correct answer is:
R⁡ 0 / n⁡

Solution:
Given, i⁡ g⁡ = i⁡ , G⁡ = R⁡ 0 ;
i1=n⁡i+i⁡=n⁡+1i⁡
∴S⁡=i⁡g⁡G⁡i1⁡-i⁡g⁡=i⁡R⁡0n⁡+1i⁡-i⁡=R⁡0n⁡

Topic: Physics, Current Electricity, JEE Main
An ammeter reads up to 1 ampere. Its internal resistance is $0.81 \mathrm{ohm}$. To increase the range to $10 \mathrm{~A}$ the value of the required shunt is

Options:
A
$0.03 \Omega$

B
$0.3 \Omega$

C
$0.9 \Omega$

D
$0.09 \Omega$

The correct answer is:
$0.09 \Omega$

Solution:
No Solution Available

Topic: Physics, Current Electricity, JEE Main, JEE Main 2003
An ammeter, voltmeter and a resistor are connected in series to a cell and the readings are noted as I and V. If another resistor R is connected in parallel with a voltmeter, then

Options:
A I and V increase
B I increases
C I and V will remain same
D I decreases

The correct answer is:
I increases

Solution:
When the resistor R is connected in parallel with voltmeter, the effective resistance in the circuit decreases because 1R=1R1+1R2
Therefore current, I=VR increases.
As I increases, potential across R increases and hence potential across voltmeter decreases. Hence reading of voltmeter decreases.

Topic: Physics, Current Electricity, JEE Main
An ammeter with internal resistance $90 \Omega$ reads $1.85 \mathrm{~A}$ when connected in a circuit containing a battery and two resistors $700 \Omega$ and ${ }^{410 \Omega}$ in series. Actual current will be

Options:
A $1.85 \mathrm{~A}$
B Greater than $1.85 \mathrm{~A}$
C Less than $1.85 \mathrm{~A}$
D None of these

The correct answer is:
Greater than $1.85 \mathrm{~A}$

Solution:
In general, ammeter always reads less than the actual value because of its resistance.

Topic: Physics, Current Electricity, JEE Main
An ammoniacal metal salt solution gives a brilliant red precipitate on addition of dimethylglyoxime. The metal ion is :

Options:
A Cu2+
B Ni2+
C Fe2+
D Co2+

The correct answer is:
Ni2+

Solution:
Nickel cation reacts with dimethylglyoxime forms an insoluble red precipitate of nickel dimethylglyoxime.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2023 (24 Jan Shift 1)
An amount of solid $\mathrm{NH}_4 \mathrm{HS}$ is placed in a flask already containing ammonia gas at a certain temperature and $0.50 \mathrm{~atm}$. Pressure. Ammonium hydrogen sulphide decomposes to yield $\mathrm{NH}_3$ and $\mathrm{H}_2 \mathrm{~S}$ gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to $0.84 \mathrm{~atm}$. The equilibrium constant for $\mathrm{NH}_4 \mathrm{HS}$ decomposition at this temperature is

Options:
A
$0.30$

B
$0.18$

C
$0.17$

D
$0.11$

The correct answer is:
$0.11$

Solution:
Total pressure $=0.5+2 \mathrm{x}=0.84$
i.e., $\mathrm{x}=0.17$
$\mathrm{~K}_{\mathrm{p}}=\mathrm{p}_{\mathrm{NH}_3} \cdot \mathrm{p}_{\mathrm{H}_2 \mathrm{~s}}$
$=(0.67) .(0.17)$
$\quad=0.1139$.

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2005
An amplitude modulated signal is given by $\mathrm{V}(\mathrm{t})=10[1+0.3$ $\left.\cos \left(2.2 \times 10^{4} \mathrm{t}\right)\right] \sin \left(5.5 \times 10^{5} \mathrm{t}\right) .$ Here $\mathrm{t}$ is in seconds. The sideband frequencies (in $\mathrm{kHz}$ ) are, [Given $\pi=22 / 7$ ]

Options:
A 1785 and 1715
B $178.5$ and $171.5$
C $89.25$ and $85.75$
D $892.5$ and $857.5$

The correct answer is:
$89.25$ and $85.75$

Solution:
Equation given
$$
\begin{array}{l}
\mathrm{V}(\mathrm{t})=10\left[1+0.3 \cos \left(2.2 \times 10^{4}\right)\right] \\
\sin \left(5.5 \times 10^{5} \mathrm{t}\right) \\
=10+1.5\left[\sin \left(57.2 \times 10^{4} \mathrm{t}\right)+\sin \left(52.8 \times 10^{4} \mathrm{t}\right)\right] \\
\omega_{\mathrm{c}}+\omega_{\mathrm{w}}=57.2 \times 10^{4}=2 \pi \mathrm{f}_{1}
\end{array}
$$
$$
\mathrm{f}_{1}=\frac{57.2 \times 10^{4}}{2 \times\left(\frac{22}{7}\right)}=9.1 \times 10^{4} \simeq 91 \mathrm{KHz}
$$
$$
\omega_{\mathrm{c}}-\omega_{\mathrm{w}}=52.8 \times 10^{4}
$$
$$
\mathrm{f}_{2}=\frac{52.8 \times 10^{4}}{2 \times\left(\frac{22}{7}\right)} \simeq 84 \mathrm{KHz}
$$
Upper side band frequency $\left(f_{1}\right)$ is
$$
\mathrm{f}_{1}=\mathrm{f}_{\mathrm{c}}-\mathrm{f}_{\mathrm{w}}=\frac{52.8 \times 10^{4}}{2 \pi} \approx 85.00 \mathrm{kHz}
$$
Lower side band frequency $\left(\mathrm{f}_{2}\right)$ is
$$
\mathrm{f}_{2}=\mathrm{f}_{\mathrm{c}}+\mathrm{f}_{\mathrm{w}}=\frac{57.2 \times 10^{4}}{2 \pi} \approx 90.00 \mathrm{kHz}
$$

Topic: Physics, Communication System, JEE Main, JEE Main 2019 (11 Jan Shift 1)
An amplitude modulated wave is as given in figure. Calculate the percentage modulation,

Options:
A 66.67 %
B 33.33 %
C 85.67 %
D None of These

The correct answer is:
66.67 %

Solution:
From the diagram,
Maximum voltage Vâ¡ max = 1 0 0 2 = 5 0 Vâ¡
Maximum voltage Vâ¡ min = 2 0 2 = 1 0 Vâ¡
Percentage modulation
Î¼ = Vâ¡ max - Vâ¡ min Vâ¡ max + Vâ¡ min Ã 1 0 0
= 5 0 - 1 0 5 0 + 1 0 Ã 1 0 0
= 4 0 6 0 Ã 1 0 0 = 6 6 Â· 6 7 %

Topic: Physics, Communication System, JEE Main
An amplitude modulated wave is represented by the equation $V_m=5(1+0.6 \cos 6280 t) \sin \left(211 \times 10^4 t\right) V$. The minimum and maximum amplitudes of the amplitude modulated wave are, respectively :

Options:
A $\frac{3}{2} V , 5 V$
B $2 V, 8 V$
C $5 V, 8 V$
D $3 V, 5 V$

The correct answer is:
$2 V, 8 V$

Solution:
$A_{\max }=A_c(1+\mu)$
The minimum amplitude is given by,
$A_{\text {min }}=A_c(1-\mu)$
From the given expression,
$V_m=5(1+0.6 \cos 6280 t) \sin \left(211 \times 10^4 t\right)$
Modulation index, $\mu=0.6$
$\begin{aligned} & A_c=5 \mathrm{~V} \\ & \Rightarrow A_{\max }=5(1+0.6)=8 \mathrm{~V} \\ & \Rightarrow A_{\min }=5(1-0.6)=2 \mathrm{~V}\end{aligned}$
Hence, option (2) is correct.

Topic: Physics, Communication System, JEE Main
An amplitude modulated waves is represented by expression vm=51+0.6 cos 6280t sin211×104t V. The minimum and maximum amplitudes of the amplitudes modulated wave are, respectively:

Options:
A 32V,5V
B 52V, 8V
C 5V,8V
D 3V,5V

The correct answer is:
52V, 8V

Solution:
From Given Equation

μ=0.6

Am=μAc

Amax+Amin2=Ac=5 V

Amax-Amin2=3 V

From Equation (1) + (2)

Amax=8 V

From Equation (1) – (2)

Amin=2 V

Topic: Physics, Communication System, JEE Main, JEE Main 2020 (02 Sep Shift 1)
An amplitude-modulated wave is represented by, Cm(t)=10(1+0.2cos12560t)sin111×104t V. The modulating frequency in kHz will be

Options:

The correct answer is:
2

Solution:
The function of the amplitude-modulated wave is represented by, Cm=AC1+AmACsin2πfmt, now compare it from the given equation,
Cm(t)=10(1+0.2cos12560t)sin111×104t V

⇒2πfm=12560 s-1⇒fm=2000 Hz=2 kHz.

Topic: Physics, Communication System, JEE Main, JEE Main 2021 (26 Aug Shift 1)
An analyst wants to convert 1 L HCl of pH=1 to a solution of HCl of pH=2. The volume of water needed to do this dilution is _____ mL. (Nearest integer)

Options:

The correct answer is:
9000

Solution:
The concentration of H+ or say HCl in both the given solution is 10–1 and 10–2 mole L–1 respectively.

For the given amount of HCl the concentration depends over dilution hence

M1V1=M2V2M1M2=V2V1

10-110-2=V21L

V2=10 L=10000 mL

∆V=10000-1000 mL=9000 mL

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2023 (12 Apr Shift 1)