datavorous/entrance-exam-dataset huggingface dataset:
In the given circuit diagram, the currents, I1=-0.3 A, I4=0.8 A and I5=0.4 A, are flowing as shown. The currents I2, I3 and I6, respectively, are:

Options:
A 0.4 A, 1.1 A, 0.4 A
B 1.1 A, -0.4 A, 0.4 A
C 1.1 A, 0.4 A, 0.4 A
D -0.4 A, 0.4 A, 1.1 A

The correct answer is:
1.1 A, 0.4 A, 0.4 A

Solution:
Given I1=-0.3A,I4=0.8A,I5=0.4A,in series connection current will be same, hence I5=I6=0.4A,According to kirchoff's juncion rule, algebraic sum of the currents meeting at tthe juction is zero.i.e currents flowing into the junction is taken as positive and current flowing out of the junction is taken as negative.At the junction R, -I4+I2+I1=0⇒-0.8+I2-0.3=0⇒I2=1.1AAt junction Q, I6+I3-I2-I1=0⇒0.4+I3-1.1+0.3=0⇒I3=1.1-0.7=0.4AIt is problem of kirchhof's current law.
I6=0.4 A
I3=0.8 A-O.4A=0.4 A
I2=0.8+0.3A=1.1 A

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (12 Jan Shift 2)
In the given circuit diagram, when the current reaches a steady-state in the circuit, the charge on the capacitor of capacitance C will be:

Options:
A CEr1r1+r
B CE
C CEr1r2+r
D CEr2r+r2

The correct answer is:
CEr2r+r2

Solution:
In steady state, current through capacitor will be zero, so current through r1 will be zero.

Hence, the net current flowing in the circuit will be,
I=Er+r2
The potential difference across the capacitor will be,
V2=I×r2=Er2r+r2

Therefore, charge on the capacitor plate will be,
q=CV2=CEr2r+r2

Topic: Physics, Capacitance, JEE Main, JEE Main 2017 (02 Apr)
In the given circuit, initially switch S1 is closed and S2 and S3 are open. After charging of the capacitor, at t=0, S1 is open and S2 and S3 are closed. If the relation between inductance, capacitance and resistance is L = 4CR2, then find the time (in s) after which the current passing through the capacitor and inductor will be the same. [given R=ln2 mÎ©, L=2 mH]

Options:

The correct answer is:
1

Solution:
Charge on capacitor after complete charging=Cε
Now at t = 0 two circuits are formed,

(A) Discharging of capacitor

∴ q=Cεe-t/τL= Cεe-t/2RC
∴ i1= ε2Re-t/2RC
(B) Growth of current in L-R circuit
i2= ε2R1-e-t/τL
now i1= i2
ε2Re-t/τC = ε2R1-e-t/τL   …1
given L = 4CR2
∴ L2R= 2RC = 1ln2
⇒ t ln2 = ln2
⇒ t = 1 s

Topic: Physics, Electrostatics, JEE Main
In the given circuit of potentiometer, the potential difference E across AB(10 m length) is larger than E1 and E2 as well. For key K1 (closed), the jockey is adjusted to touch the wire at point J1 so that there is no deflection in the galvanometer. Now the first battery E1 is replaced by second battery E2 for working by making K1 open and K2 closed. The galvanometer gives then null deflection at J2. The value of E1E2 is a2, where a=_______ .

Options:

The correct answer is:
1

Solution:
Length of AB=10 m

For battery E1, balancing length is l1

l1=380 cm [from end A]

For battery E2, balancing length is l2

l2=760 cm [from end A]

Now, we know that E1E2=l1l2

⇒ E1E2=380760=12=ab

a=1

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (25 Feb Shift 1)
In the given circuit, rms value of current Irms through the resistor R is :

Options:
A 2 A
B 12 A
C 20 A
D 22 A

The correct answer is:
2 A

Solution:
Impedance of the given LCR circuit will be, Z=R2+XL-XC2⇒Z=1002+200-1002

=1002Ω

Therefore, RMS current can be written as,

Irms=VrmsZ=20021002

=2 A

Topic: Physics, Alternating Current, JEE Main, JEE Main 2023 (30 Jan Shift 2)
In the given circuit, Reading of ammeter is 1 A. If each of the 4 Î© resistor is replaced by 2 Î© resistor, the reading of ammeter will become nearly -

Options:
A 1 .11 A
B 1.25Â A
C 1.34Â A
D 1.68Â A

The correct answer is:
1 .11 A

Solution:
If rA=Ammeter Resistance
then ⇒ 1 A=10VReq=109+rA⇒ rA=1 Ω
When 4 Ω resistors are replaced by 2 Ω, then⇒ I=108+1=109 A=1.11 A

Topic: Physics, Current Electricity, JEE Main
In the given circuit the AC source has ω=100 rad s-1. Considering the inductor and capacitor to be ideal, what will be the current I flowing through the circuit?

Options:
A 5.9 A
B 6 A
C 0.94 A
D 3.16 A

The correct answer is:
3.16 A

Solution:
ZC=1ωC2+R2

=1100×100×10-62+1002

=1002+1002

=1002 Ω

ZL=(ωL)2+R2=(100×0.5)2+502=502 Ω

iL=200ZL=200502=42=22 A

iC=200ZC=2001002=2 A

cosϕ1=1001002=12, ϕ1=45°

cosϕ2=50502=12, ϕ2=45°

I=IC2+IL2; I=8+2; I=10=3.16 A

This question is bonus as the correct answer was not available for students to mark as option in original paper by NTA.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (26 Aug Shift 2)
In the given circuit the cells have zero internal resistance. The currents (in amperes) passing through resistance R1 and R2 respectively, are:

Options:
A 0,Â 1
B 2,Â 2
C 0.5 , 0
D 1,Â 2

The correct answer is:
0.5 , 0

Solution:
Assume VF as 0
as   V E = V B
⇒ Current in R 2 is zero
VA-VD=10 V
⇒I=1020=12 A from A to D

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (10 Jan Shift 1)
In the given circuit, the charge on 4 μF capacitor will be:

Options:
A 9.6 μC
B 5.4 μC
C 24 μC
D 13.4 μC

The correct answer is:
24 μC

Solution:
The given circuit can be redrawn as:
Charge on 4 μF capacitor:
Q=4×64+6×10
Q=24 μC

in series combination charge remains same.

so,

Q4μF=Q6μF=24 μC

Topic: Physics, Capacitance, JEE Main, JEE Main 2019 (12 Apr Shift 2)
In the given circuit, the current flowing through the resistance 20 Ω is 0.3 A, while the ammeter reads 0.9 A. The value of R1 is _____Ω.

Options:

The correct answer is:
30

Solution:
Given, i1=0.3 A, i1+i2+i3=0.9 A

So, VAB=i1×20 Ω=0.3×20 V=6 V

i2=6 V15Ω=25 A

Now,

i1+i2+i3=910 A

⇒310+25+i3=910

⇒710+i3=910

⇒i3=0.2 A

So, i3×R1=6 V

⇒(0.2)R1=6

⇒R1=60.2=30 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (29 Jan Shift 2)
In the given circuit, the current I through the battery will be

Options:
A 2.5 A
B 1 A
C 2 A
D 1.5 A

The correct answer is:
1.5 A

Solution:
In the first circuit diagram, the diodes D1 & D3 are forward biased and D2 is reverse biased(Hence, no current through it). So the circuit is simplified to the second diagram.

Hence, the equivalent resistance is

Req=20×1020+10

⇒Req=203 Ω

Using Ohm's law, the current is

I=10203=3020 A=1.5 A

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (15 Apr Shift 1)
In the given circuit the current in each resistance is

Options:
A \( 0.5 \mathrm{~A} \)
B \( 0 \mathrm{~A} \)
C \( 1 \mathrm{~A} \)
D \( 0.25 \mathrm{~A} \)

The correct answer is:
\( 0 \mathrm{~A} \)

Solution:
Let the potential at point C equals to 0 volt, then the potentials at other points will be as shown in the figure below.

Using ohm's law, V=IR.

If the potential difference is zero then, the current flowing through each resistor is zero.

Hence, current through each resistance is 0 A.

Topic: Physics, Current Electricity, JEE Main
In the given circuit, the current in resistance R3 is:

Options:
A 1 A
B 1.5 A
C 2 A
D 2.5 A

The correct answer is:
1 A

Solution:
Equivalent resistance, Req=2 Ω+4×44+4 Ω+1 Ω=5 Ω

Current through battery, i=VReq=105=2 A

Current in resistance R3=2×44+4

=2×48

=1 A

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (29 Jan Shift 2)
In the given circuit the current $l_1$ is

Options:
A $0.4 \mathrm{~A}$
B $-0.4 A$
C $0.8 \mathrm{~A}$
D $-0.8 \mathrm{~A}$

The correct answer is:
$-0.8 \mathrm{~A}$

Solution:
According to the principle of Wheatstone's bridge, the effective resistance between the given points is $4 \Omega$

Topic: Physics, Current Electricity, JEE Main
In the given circuit the current through Zener Diode is close to:

Options:
A $0.0 \mathrm{~mA}$
B $6.7 \mathrm{~mA}$
C $4.0 \mathrm{~mA}$
D $6.0 \mathrm{~mA}$

The correct answer is:
$0.0 \mathrm{~mA}$

Solution:
Since voltage across zener diode does not reach to breakdown voltage therefore its resistance will be infinite \& current through it is $0 .$

Topic: Physics, Semiconductors, JEE Main, JEE Main 2019 (11 Jan Shift 1)
In the given circuit the current through zener diode is:

Options:
A 3.3 mA
B 2.5 mA
C 5.5 mA
D 6.7 mA

The correct answer is:
3.3 mA

Solution:
VAB=10 V∴VBC=5 V, the current in R1= I1=5500
⇒I1=10 mA. Current in R2= I2=101500⇒ I2=203mA
Current in zener diode = I1- I2=10-203 mA=103 mA

Topic: Physics, Semiconductors, JEE Main, JEE Main 2018 (16 Apr Online)
In the given circuit, the equivalent resistance between the terminal A and B is ____ Ω

Options:

The correct answer is:
10

Solution:
In the given circuit, both 4 Ω resistors get shorted out, thus no current will flow from them. Removing the resistors from which no current will flow.

Now, the circuit can be redrawn as shown below.

Req=3+(2‖2)+6

⇒Req=3+2×22+2+6

⇒Req=10 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (25 Jan Shift 1)
In the given circuit the internal resistance of the 18V cell is negligible. If R1=400 Ω, R3=100 Ω and R4=500 Ω and the reading of an ideal voltmeter across R4 is 5 V, then the value of R2 will be:

Options:
A 550 Ω
B 300 Ω
C 450 Ω
D 230 Ω

The correct answer is:
300 Ω

Solution:
Given reading of an ideal voltmeter for resistance R4 is V4 = 5 V, then current (let i1)is given branch (as resistance R3 in series with resistance R4) is
i1=VR4=5500=0.01 A
Then for resistance R3, the potential difference will be

V3 = i1 R3 =0.01×100 =1V

So, for given branch of resistance potential difference is

V3 + V4 = 1 + 5 =6 V

From figure for resistance R1 , R3 and R4 a total applied potential difference is 18 V then,

V1+ V3+V4 = 18∴ V1 =18-V3+V4 =18-6 =12 V

So, the total current drawn from the source is

∴ Current through R2
i2=0·03-0·01= 0·02 A
and potential difference for R2 is

V2 =V3 + V4 =6 V ( R2 is in parallel arrangement with the resistances R3 and R4)

Now the value of resistance R2 is :
V2 = i2 R2⇒R2 = V2i2 =60·02 = 300 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (09 Jan Shift 2)
In the given circuit, the magnitude of VL and VC are twice that of VR. Given that f=50 Hz, the inductance of the coil is 1 Kπ mH. The value of K is _____ .

Options:

The correct answer is:
0

Solution:
Since the voltage across the inductor and capacitor is same, the circuit is in resonance. Therefore, the current in the circuit will be,

i=VR=2205=44 A.

The voltage across the inductor is double that across the resistor. Therefore, VL=2VR=440 V.

We know that VL=iXL=i×2πfL⇒L=44044×100π=110π H=100πmH

Therefore, K=1100=0.01≈0.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (28 Jun Shift 2)
In the given circuit the R.M.S. values of voltages across the capacitor C, inductor L and resistor R1 are 12 V, 10 V and 5 V respectively. Then the peak voltage across R2 is

Options:
A 72 V
B 69 V
C 138 V
D none of these

The correct answer is:
138 V

Solution:
In parallel combination each branch should have equal potential but in AC circuit in a branch,
Veff=VR2+(Vc-VL)2
VC2+VR12=VL2+VR22⇒122+52=102+VR22 ⇒(VR2)rms=69
V peak = 2 V rms
∴ VR2peak=138 V

Topic: Physics, Alternating Current, JEE Main
In the given circuit the R.M.S. values of voltages across the capacitor C, inductor L and resistor R1 are 12 V, 10 V and 5 V respectively. Then the peak voltage across R2 is

Options:
A 72 V
B 69 V
C 138 V
D none of these

The correct answer is:
138 V

Solution:
In a parallel combination, each branch should have equal potential. But in AC circuit, in a branch,
Veff=VR2+(Vc-VL)2
VC2+VR12=VL2+VR22⇒122+52=102+VR22 ⇒(VR2)rms=69
Vpeak=2Vrms
∴ VR2peak=138 V

Topic: Physics, Alternating Current, JEE Main
In the given circuit, the terminal potential difference of the cell is :

Options:
A $2 \mathrm{~V}$
B $3 \mathrm{~V}$
C $4 \mathrm{~V}$
D $1.5 \mathrm{~V}$

The correct answer is:
$2 \mathrm{~V}$

Solution:
$\begin{aligned} \mathrm{i} & =\frac{3}{1+2}=1 \mathrm{~A} \\ \mathrm{v} & =\mathrm{E}-\mathrm{ir} \\ & =3-1 \times 1=2 \mathrm{~V}\end{aligned}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (08 Apr Shift 1)
In the given circuit, the value of current IL will be _____ mA. (When RL=1 kΩ)

Options:

The correct answer is:
5

Solution:
Potential drop across the branch containing load resistance will be 5 V due to Zener diode. Therefore, potential drop across 800 Ω will be, 10 V-5 V=5 V.

Now, current through battery will be,

I=5800×1000 =6.25 mA

Current through load will be,

IL=51000 A

=5 mA

Topic: Physics, Semiconductors, JEE Main, JEE Main 2022 (24 Jun Shift 2)
In the given circuit the value of I1+I3I2 is:

Options:

The correct answer is:
2

Solution:
Potential distribution at each point in the circuit is shown in above diagram.

Since, both 10 Ω resistors are maintained at same potential difference, so current will be same, thus, using Ohm's law, I1=I2=20-1010=1 A

Now, current I3=1010=1 A

Hence, the value of I1+I3I2=1+11=2.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (01 Feb Shift 2)
In the given circuit, the voltage across load resistance (RL) is:

Options:
A 8.75 V
B 9.00 V
C 8.50 V
D 14.00 V

The correct answer is:
8.75 V

Solution:
Both diodes are in forward bias. In forward bias, potential drop across Ge and Si will be 0.3 V & 0.7 V respectively.

Therefore, current in the circuit i=15-0.3-0.74×1000=3.5 mA

Required potential drop across RL,
VL=iRL=3.5×10-3×2.5×1000 V

=8.75 V

Topic: Physics, Semiconductors, JEE Main, JEE Main 2024 (30 Jan Shift 2)
In the given circuit, the voltage across the load is maintained at $12 \mathrm{~V}$. The current in the zener diode varies from $0-50 \mathrm{~mA}$. What is the maximum wattage of the diode?

Options:
A $12 \mathrm{~W}$
B $6 \mathrm{~W}$
C $0.6 \mathrm{~W}$
D $1.2 \mathrm{~W}$

The correct answer is:
$0.6 \mathrm{~W}$

Solution:
In the circuit, the zener diode is used as
power regular therefore,
$P=V I$
$=12 \times 0.05=0.6 \mathrm{~W}$

Topic: Physics, Semiconductors, AIIMS, AIIMS 2012
In the given circuit, the voltmeter reading is 4.5 V . Assuming that the voltmeter is ideal, current through 12Î© resistance is -

Options:
A 1Â A
B 0.5 A
C 0.25Â A
D 0.1Â A

The correct answer is:
0.5 A

Solution:
The voltmeter is ideal so it doesn't draw any current from the circuit.
Potential difference across 9 Ω resistance = 4.5 V (given)

So the current in the 9 Ω resistance = 4.59=0.5 A

i.e., I1′=0.5 A

The same current I1′ passes through 3 Ω. The resistors 9 Ω and 3 Ω are in series and their equivalent, i.e., 12 Ω is in parallel with 6 Ω between A and B . Dividing the current in the inverse ratio of resistances between A and B ,
I1′I1′′=612=12
I1′′=2I1′=2 × 0.5=1 A
and I1=I1′+I1′′=0.5+1=1.5 A
at junction C, I1 divides into three parts. Since the resistances 10Ω, 12Ω, 15 Ω are in parallel between C and D , current will distribute in the inverse ratio of resistances.
∴ I2′ : I2′′ : I2′′′=110 : 112 : 115
=6 : 5 : 4
I2′=515× 1.5=0.5 A
So I2′=6k, I2′′=5k, I2′′′=4k
( k being a constant of proportionality)
and I1=I2′+I2′′+I2′′′=15k
but I1=1.5 A
∴ 15 k=1.5
or k=0.1
so I2′′=5k=0.5 A
Thus current through 12 Ω resistance is 0.5 A

Topic: Physics, Current Electricity, JEE Main
In the given circuit, value of Y is:

Options:
A 0
B toggles between 0 and 1
C will not execute
D 1

The correct answer is:
0

Solution:
Y=AB-∙A¯
=AB̿+A¯
=AB+A¯
=0+0
=0

Topic: Physics, Semiconductors, JEE Main, JEE Main 2020 (08 Jan Shift 2)
In the given circuit, what is the current ( in A ) through the battery?

Options:

The correct answer is:
3

Solution:
In the given circuit diode D2 is reverse biased, so it will not conduct. Diode D1 and D3 are forward biased, so they will conduct, Thus the equivalent circuit is
â´Rnet=(4+4)Ã16(4+4)+16=8Ã1624=163 Î©
â´ current through the battery,
I=VRnet=16163=3 A

Topic: Physics, Semiconductors, JEE Main
In the given circuit, when $S_1$ is closed, the capacitor $C$ gets full charged. Then $S_1$ is kept open and $S_2$ is closed. Hence

Options:
A The current in the circuit is in the same direction.
B The instantaneous current in the circuit may be $V \sqrt{\frac{C}{L}}$.
C The energy stored in the circuit is purely in the form of magnetic energy.
D There is no exchange of energy between inductor $L$ and capacitor $C$.

The correct answer is:
The instantaneous current in the circuit may be $V \sqrt{\frac{C}{L}}$.

Solution:
When $S_1$ is closed, potential difference across capacitor is $V$.
Direction of current gets reversed when $S_2$ is closed.
When $S_1$ is opened and $S_2$ is closed,
$\begin{aligned}
& \frac{1}{2} C V^2=\frac{1}{2} L I^2 \\
& \Rightarrow I=V \sqrt{\frac{C}{L}}
\end{aligned}$
Total energy oscillates between capacitor and inductor.

Topic: Physics, Electromagnetic Induction, JEE Main
In the given circuits a and b, switches S1 and S2 are closed at t=0 and kept close for a long time. The variation of currents in the two circuits for t≥0 are shown in the options. (Figures are schematic and not drawn to scale.)

Options:
A
B
C
D

The correct answer is:

Solution:
In C-R series circuit,
q=q0 1-e-tτ
⟹q=CE 1-e-tRC
∴ I=dqdt
=CERC+e-tRC
I=ER e-tRC
⟹ the urrent decays exponentially. In L-R series circuit,
I=I0(1-e-tτ)
⇒I0=ER
⇒I=ER 1-e-RtL
⟹ current grows exponentially.
∴ For C-R circuit,
For L-R circuit,

Topic: Physics, Capacitance, JEE Main, JEE Main 2015 (10 Apr Online)
In the given compound here

If X and Y are the number of secondary and tertiary C-atoms respectively. Find the value of sum of X + Y here.

Options:

The correct answer is:
8

Solution:
Number of s o C-atoms (X) = 4
Number of t o C-atoms (Y) = 4
Sum of X+Y=4+4=8.

Topic: Chemistry, Hydrocarbons, JEE Main
In the given compound, the number of \(2^{\circ}\) carbon atom \(/ \mathrm{s}\) is _______.

Options:
A Four
B Two
C One
D Three

The correct answer is:
One

Solution:
only one $2^{\circ}$ carbon is present in this compound.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (08 Apr Shift 1)
In the given diagram $A$ and $B$ represent
A B

Options:
A Mineralisation Demineralisation
B Ammonification Denitrification
C Denitrification Ammonification
D Denitrification Mineralisation

The correct answer is:
Denitrification Ammonification

Solution:
In given diagram (A) represents denitrification and (B) represents ammonification. Denitrification is conversion of nitrates into nitrogen gas by some microorganisms e.g., Pseudomonas denitrificans, Thiobacillus denitrificans, Micrococcus denitrificans. Ammonification is conversion of dead remains of living organisms into ammonia with the help of microorganisms like Bacillus ramosus, $B$. vulgaris, $B$. mesentericus etc.

Topic: Biology, Mineral Nutrition, AIIMS, AIIMS 2011
In the given diagram $\mathrm{C}(\mathrm{t})$ stands for the carrier wave and $\mathrm{m}(\mathrm{t})$ for the signal to be transmitted. What name do we give to the wave labeled as $C_m(t)$ in the diagram?

Options:
A Amplitude Modulated Wave
B Frequency Modulated Wave
C Phase Modulated Wave
D None of These

The correct answer is:
Frequency Modulated Wave

Solution:
$C_m(t)$ is the frequency modulated wave.

Topic: Physics, Communication System, JEE Main
In the given diagram of bacteriophage, A - D are labelled. Select the correct information?

Options:
A B - Collar, D - Tail fibres
B C - Sheath, A - Collar.
C A - Head, B - Tail fibres
D A - Head, D - Sheath

The correct answer is:
B - Collar, D - Tail fibres

Solution:
Bacteriophages are obligate parasites. These remain inactive outside the host (bacteria) but when they enter the host they use the cellular machinery of the host for their own multiplication and then destroy the host.

Topic: Biology, Biological Classification, NEET
In the given diagram, there is a conducting sphere of radius r1 which is surrounded by a dielectric of relative permittivity Îµr. If conducting sphere is given charge q, then the surface density of polarization charges, on the outer surface of the dielectric layer, is

Options:
A εrq4πr22
B q4πεrr22
C (εr-1)q4πr22
D (εr-1)q4πεrr22

The correct answer is:
(εr-1)q4πεrr22

Solution:
Induced charge =q1-1εr
Surface density =q1-1εr4πr22=qεr-14πεrr22

Topic: Physics, Electrostatics, JEE Main
In the given electrical circuit, if the switch $S$ is closed then the maximum energy stored in the inductors is:

Options:
A 3 J
B 9 J
C 12 J
D 6 J

The correct answer is:
3 J

Solution:
In figure below, if switch $S$ is open then total energy stored in the capacitors, is suppose $E_0$.
Energy in IF capacitor,
$$
E_1=\frac{1}{2} C V^2=\frac{1}{2} C\left(\frac{Q}{C}\right)^2=\frac{1}{2} \frac{Q^2}{C}=\frac{1}{2} \times \frac{4^2}{1}=8 \mathrm{~J}
$$
Similarly, $E_2=\frac{1}{2} \frac{Q^2}{C}=\frac{2^2}{2 \times 2}=1 \mathrm{~J}$
So, the total energy, $E_0=E_1+E_2=8+1=9 \mathrm{~J}$ Now, switch's is closed then the common potential of Capacitors,
$$
\begin{aligned}
V_{\text {common }} & =\frac{C_1 V_1+C_2 V_2}{C_1+C_2} \\
& =\frac{1 \times 4+2 \times 1}{1+2}=\frac{6}{3}=2 \mathrm{~V}
\end{aligned}
$$
Hence, now the new arrangement of energy,
and
$$
\begin{aligned}
& E_1=\frac{1}{2} C_1 V_{\text {common }}^2=\frac{1}{2} \times 1 \times 4=2 \mathrm{~J} \\
& E_2=\frac{1}{2} C_2 V_{\text {common }}^2=\frac{1}{2} \times 2 \times 4=4 \mathrm{~J}
\end{aligned}
$$
Now, from conservation of energy,
Energy stored in the inductor, $E_L \simeq E_0-\left(E_1+E_2\right)$
$$
=9-(2+4)=3 \mathrm{~J}
$$
Hence, the inductor has $3 \mathrm{~J}$ of energy.

Topic: Physics, Electromagnetic Induction, JEE Main
In the given electromagnetic wave $\mathrm{E}_{\mathrm{y}}=600 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{Vm}^{-1}$, intensity of the associated light beam is (in $\mathrm{W} / \mathrm{m}^2$ : (Given $\epsilon_0=9 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$ )

Options:
A 243
B 729
C 972
D 486

The correct answer is:
486

Solution:
$\begin{aligned} \text { Intensity } & =\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2 \mathrm{c} \\ & =\frac{1}{2} \times 9 \times 10^{-12} \times(600)^2 \times 3 \times 10^8 \\ & =\frac{9}{2} \times 36 \times 3=486 \mathrm{w} / \mathrm{m}^2\end{aligned}$

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2024 (06 Apr Shift 2)
In the given figure, 1 represents isobaric, 2 represents isothermal and 3 represents adiabatic processes of an ideal gas. If $\Delta \mathrm{U}_{1}, \Delta \mathrm{U}_{2}, \Delta \mathrm{U}_{3}$ be the changes in internal energy in these processes respectively, then

Options:
A $\Delta \mathrm{U}_{1} < \Delta \mathrm{U}_{2} < \Delta \mathrm{U}_{3}$
B $\Delta \mathrm{U}_{1}>\Delta \mathrm{U}_{3} < \Delta \mathrm{U}_{2}$
C $\Delta \mathrm{U}_{1}=\Delta \mathrm{U}_{2}>\Delta \mathrm{U}_{3}$
D $\Delta \mathrm{U}_{1}>\Delta \mathrm{U}_{2}>\Delta \mathrm{U}_{3}$

The correct answer is:
$\Delta \mathrm{U}_{1}>\Delta \mathrm{U}_{2}>\Delta \mathrm{U}_{3}$

Solution:
$\Delta \mathrm{T}_{1}=\mathrm{T}_{0}$
$\Delta \mathrm{T}_{2}=0$
$\Delta \mathrm{T}_{3}=-\mathrm{ve}$
$\Rightarrow \Delta \mathrm{U}_{1}>\Delta \mathrm{U}_{2}>\Delta \mathrm{U}_{3}$

Topic: Physics, Thermodynamics, JEE Main
In the given figure a capacitor of plate area A is charged up to charge q. The mass of each plate is m2. The lower plate is rigidly fixed. Find the value of m1 so that the system remains in equilibrium -

Options:
A m2+q2ε0Ag
B m2
C q22Îµ0Ag+m2
D 2m2

The correct answer is:
q22ε0Ag+m2

Solution:
Now T=m2g+Felec â¦1
T=m1g .......2
m1g=m2g+q22AÎµ0
m1=m2+q22AÎµ0g

Topic: Physics, Electrostatics, NEET
In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillation will be:

Options:
A AM+mM
B AMM-m
C AM-mM
D AMM+m

The correct answer is:
AMM+m

Solution:
Momentum of system remains conserved.

pi=pf

MAω=m+MA'ω

MAkM=m+MA'km+M

A'=AMM+m

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (24 Feb Shift 1)
In the given figure an ammeter A consists of a $240 \Omega$ coil connected in parallel to a $10 \Omega$ shunt. The reading of the ammeter is $\qquad$ $\mathrm{mA}$

Options:

The correct answer is:
160

Solution:
Req
$\begin{aligned}
& =140.4+\frac{240 \times 10}{240+10} \\
& \text { Req }=140.4+\frac{2400}{250} \\
& \text { Req. }=150 \Omega
\end{aligned}$
$\begin{aligned} \therefore \text { Current in ammeter } & =\frac{24}{150} \\ & =160 \mathrm{~mA}\end{aligned}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (06 Apr Shift 2)