datavorous/entrance-exam-dataset huggingface dataset:
Two circles with equal radii are intersecting at the points (0,1) and (0,-1) . The tangent at the point (0,1) to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is:

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

The correct answer is:
2

Solution:
$\because$ Two circles of equal radii intersect each other orthogonally. Then $R$ is mid point of $P Q$.

and $P R=O_{1} R=O_{2} R$

$P R=\frac{1}{2} \sqrt{(0-0)^{2}+(1+1)^{2}}=1$

$\therefore$ Distance between centres $=1+1=2$.

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Two circuits (a) and (b) have charged capacitors of capacitance C, 2C and 3C with open switches. Charges on each of the capacitor are as shown in the figures. On closing the switches
Circuit (a) Circuit (b)

Options:
A
No charge flows in (a) but charge flows from $R$ to $L$ in (b)

B
Charges flow from $L$ to $R$ in both (a) and (b)

C
Charges flow from $R$ to $L$ in (a) and from $L$ to $R$ in (b)

D
No charge flows in (a) but charge flows from $L$ to $R$ in (b)

The correct answer is:
Charges flow from $R$ to $L$ in (a) and from $L$ to $R$ in (b)

Solution:
Charge (or current) always flows from higher potential to lower potential.
Potential $=\frac{\text { Charge }}{\text { Capacitance }}$

Topic: Physics, Capacitance, JEE Main, JEE Main 2012 (07 May Online)
Two circular coils P and Q of 100 turns each have same radius of π cm. The currents in P and R are 1 A and 2 A respectively. P and Q are placed with their planes mutually perpendicular with their centers coincide. The resultant magnetic field induction at the center of the coils is x mT, where x= ______. [Use μ0=4π×10−7 T m A−1]

Options:

The correct answer is:
20

Solution:
Consider two coils as shown below:

Magnetic field due to P, BP=Nμ0i12r=100×μ0×12π×10-2=2×10−3 T

Magnetic field due to Q,

BQ=Nμ0i22r=100×μ0×22π×10-2=4×10−3 T

Therefore, Bnet=BP2+BQ2

=20 mT

Hence, x=20

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (31 Jan Shift 2)
Two circular concentric loops of radii r1=20 cm, r2=30 cm are placed in the X-Y plane as shown in the figure. A current I=7 A is flowing through them. The magnetic moment of this loop system is

Options:
A + 0.4 k^ Am2
B -1.5 k^ Am2
C +1.1 k^ Am2
D +1.3 j^ Am2

The correct answer is:
+1.1 k^ Am2

Solution:
Here, magnetic moment due to loop 1
M1=iA1
=iπr12
=7π0.202=0.28π
Similarly magnetic moment due to loop-2
M2=iA2=iπr22
=7π 0.302=0.63π
Net magnetic moment
=M1-M2
=0.63π-0.28π=1.1 A m2

Topic: Physics, Magnetic Effects of Current, JEE Main
Two circular loops L1 and L2 of wire carrying equal and opposite currents are placed parallel to each other with a common axis. The radius of loop L1 is R1 and that of L2 is R2. The distance between the centres of the loops is 3R1. The magnetic field at the centre of L2 shall be zero if

Options:
A R2=4R1
B R2=2R1
C R2=2R1
D R2=8R1

The correct answer is:
R2=8R1

Solution:
Net magnetic field at centre, Bnet=B1-B2

For magnetic field, Bnet to be zero, B1=B2

Magnetic field at the axis of coil 1 at a distance of 3R1

B1=μ0i2πR124πR12+3R123/2

B1=μ0i16R1 ...1

Magnetic field at the centre of coil 2.

B2=μ0i2R2 ...2

On equating 1 and 2

μ0i16R1=μ0i2R2

⇒R2=8R1

Topic: Physics, Magnetic Effects of Current, JEE Main
Two circular rings of identical radii and resistance 36 Î© are placed as shown in the figure. Conducting joints are made at points A and B and a cell of emf 20 V is connected between these two points. What is the power (in W) delivered by the cell? [C1 and C2 are the centres of the two rings]

Options:

The correct answer is:
100

Solution:
From the figure.
AC1 = AC2 = C1 C2 = radius
â´ â AC 1 B = 1 2 0 0

In between A and B, 4 resistances are in parallel

Hence the resistance of four sections are in parallel
Hence equivalent resistance R across AB is
1 R = 1 2 4 + 1 1 2 + 1 1 2 + 1 2 4
or R = 4 Î©
â´ Power V 2 R = 2 0 2 4 = 1 0 0 watt

Topic: Physics, Current Electricity, JEE Main
Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure pi and temperature T1 are connected through a narrow tube negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to T2 . The final pressure pf

Options:
A 2piT1T2T1+T2
B piT1T2T1+T2
C 2piT1T1+T2
D 2piT2T1+T2

The correct answer is:
2piT2T1+T2

Solution:
Initial moles and final moles are equal
nTi=nTf
P1VRT1+P1VRT1=PfVRT1+PfVRT2
2PiT1=PfT1+PfT2
Pf=2PiT2T1+T2

Topic: Chemistry, States of Matter, JEE Main
Two closed bulbs of equal volume V containing an ideal gas initially at pressure pi and temperature T1 are connected through a narrow tube of negligible volume, as shown in the figure below. The temperature of one of the bulbs is then raised to T2. The final pressure Pf is:

Options:
A 2piT2T1+T2
B 2pi T1T2T1+T2
C pi T1T2T1+T2
D 2piT1T1+T2

The correct answer is:
2piT2T1+T2

Solution:
Given, two closed bulbs of equal volumeV containing ideal gas initially of pressure pi and temperature T1, which are connected by a narrow tube of negligible volume.
To find the final pressure Pf, when one raised to T2:
Number of moles of gas does not change,
nTi=nTf
P i V RT 1 + P i V RT 1 = P f V RT 1 + P f V RT 2
2 P i T 1 = P f T 1 + P f T 2
2PiT1= Pf(T1 + T2)T1 × T2Pf = 2Pi (T2T1+ T2)

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2016 (03 Apr)
Two closed vessels of equal volume containing air at pressure $P_1$ and temperature $T_1$ are connected to each other through a narrow tube. If the temperature in one of the vessels is now maintained at $T_1$ and that in the other at $T_2$, what will be the pressure in the vessels

Options:
A $\frac{2 P_1 T_1}{T_1+T_2}$
B $\frac{T_1}{2 P_1 T_2}$
C $\frac{2 P_1 T_2}{T_1+T_2}$
D $\frac{2 P_1}{T_1+T_2}$

The correct answer is:
$\frac{2 P_1 T_2}{T_1+T_2}$

Solution:
$\frac{P_1}{T_1}+\frac{P_1}{T_1}=\frac{P}{T_1}+\frac{P}{T_2}$
$\frac{2 P_1}{T_1}=P\left(\frac{T_1+T_2}{T_1 T_2}\right)_{;} \quad \therefore P=\frac{2 P_1\left(T_1 T_2\right)}{T_1\left(T_1+T_2\right)}=\frac{2 P_1 T_2}{T_1+T_2}$

Topic: Chemistry, States of Matter, JEE Main
Two coaxial discs, having moments of inertia I1 and I12 , are rotating with respective angular velocities ω1 and ω12 , about their common axis. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If Ef and Ei are the final and initial total energies, then Ef-Ei is:

Options:
A I1ω126
B 38I1ω12
C -I1ω1212
D -I1ω1224

The correct answer is:
-I1ω1224

Solution:
As external torque is zero, angular momentum remains conserved.
I1ω1+I12ω12=I1+I12ω
∴ Common angular velocity ω=5ω16
∴ Ef-Ei=12×I1+I12ω2-12×I1ω1 2-12×Il2ω122
=12×3I1256ω12-916I1ω12
=75 I1ω12-81 I1ω12144
=-6I1ω12144
=-I1ω1224

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (10 Apr Shift 1)
Two coaxial long solenoids of equal lengths have currents i1 and i2, the number of turns per unit length n1 and n2 and radii r1 and r2 r2>r1 respectively. If n1i1=n2i2 and the two solenoids carry currents in the opposite sense, the magnetic energy stored per unit length is

Options:
A μ02n12i12πr22-r12
B μ0n12i12πr22-r12
C μ02n12i12π r12
D μ02n22i22π r22

The correct answer is:
μ02n12i12πr22-r12

Solution:
The magnetic field is non-zero only in the region between the two solenoids, where B=Âµ0 n2i2
â´ Energy stored per unit volume
=B22Î¼0=Î¼0n22 i222
The energy per unit length = energy per unit volume Ã area of cross section where B â 0
= Î¼0n22 i222Ï r22- r12
=Î¼0n12 i122Ïr22- r12 , since n1i1=n2i2

Topic: Physics, Magnetic Effects of Current, JEE Main
Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross sectional area $A=$ $10 \mathrm{~cm}^2$ and length $=20 \mathrm{~cm}$. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is $\left(\mu_0=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}\right)$

Options:
A
$2.4 \pi \times 10^{-5} \mathrm{H}$

B
$4.8 \pi \times 10^{-4} \mathrm{H}$

C
$4.8 \pi \times 10^{-5} \mathrm{H}$

D
$2.4 \pi \times 10^{-4} \mathrm{H}$

The correct answer is:
$2.4 \pi \times 10^{-4} \mathrm{H}$

Solution:
$$
M=\frac{\mu_0 N_1 N_2 A}{\ell}=2.4 \pi \times 10^{-4} \mathrm{H}
$$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2008
Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A = 10 cm 2 and length = 20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is Î¼ 0 = 4 Ï Ã 10 - 7 T m A -1

Options:
A 2.4 Ï Ã 10 - 4 H
B 2.4 π × 10 - 5 H
C 4.8 π × 10 - 4 H
D 4.8 π × 10 - 5 H

The correct answer is:
2.4 Ï Ã 10 - 4 H

Solution:
Mutual inductance
M= fluxâinâ( 2 ) currentâinâ( 1 )
= Ï 2 i 1
= B 1 . A 2 N 2 i 1
= Î¼ 0 N 1 i 1 l Ã A 2 N 2 i 1
= Î¼ 0 â N 1 . N 2 .A l
A1=A2=A
A = Ï r 1 2 = 10 cm 2 , l = 20 cm , N 1 = 300 , N 2 = 400 .
M = Î¼ 0 N 1 N 2 A l = 4 Ï Ã 10 - 7 Ã 300 Ã 400 Ã 10 Ã 10 - 4 0.20
= 2.4 Ï Ã 10 - 4 H .

Topic: Physics, Electromagnetic Induction, JEE Main
Two coaxial solenoids of different radii carry current I in the same direction. Let F1→ be the magnetic force on the inner solenoid due to the outer one and F2→ be the magnetic force on the outer solenoid due to the inner one. Then:

Options:
A F1→ is radially outwards and F2→=0
B F1â=F2â=0
C F1âÂ is radially inwards and F2âÂ is radially outwards
D F1â is radially inwards and F2â=0Â

The correct answer is:
F1â=F2â=0

Solution:
S2 is a solenoid with more radius than S1, the magnetic field of the solenoid S2 will be along the axis and constant, so the magnetic force on an element of solenoid S1 will be,

dF→1=il→×B→

dF→1=ilBj^×-i^=iBlk^

Similarly, for just next coil the magnetic field is in same direction, but the current element is in opposite direction, so,

dF→1=ilB-j^×-i^=-iBlk^.

From Symmetry, we can say that the overall force will be cancelled to the whole coil, hence, F→1=0

Similarly, from coil S1 to S2, The magnetic field of the inner solenoid is symmetric about its axis, but not exactly along the axis of the outer solenoid. As the outer solenoid is placed symmetrically the push of any coil will be equal and opposite to the pull of coil placed exactly the opposite distance from the centre. Hence, the net force on the outer coil will be zero, i.e., F→2=0.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2015 (04 Apr)
Two coherent light beams of intensity I and 4I are superposed. If the maximum and minimum possible intensities in the resulting beam are Imax and Imin, then ImaxâImin=nI. Find the value of n.

Options:

The correct answer is:
8

Solution:
As I∝a2 or a∝I, where a is the amplitude of the wave.
∴a1a2=I⁡4I⁡
⇒a1a2=12
I⁡maxI⁡min=a1+a2a1-a22
=1+21-22=91
∴I⁡max=9I⁡,I⁡min=I⁡
Now, Imax−Imin=8I
⇒n=8

Topic: Physics, Wave Optics, JEE Main
Two coherent light sources having intensity in the ratio 2x produce an interference pattern. The ratio Imax-IminImax+Imin will be

Options:
A 22xx+1
B 2x2x+1
C 22x2x+1
D 2xx+1

The correct answer is:
22x2x+1

Solution:
Given that, I1I2=2x.

We know,

Imax=I1+I22&Imin=I1-I22

∴ Imax-IminImax+Imin=2I1I2I1+I2=2I1/I2I1I2+1=22x2x+1

Topic: Physics, Wave Optics, JEE Main, JEE Main 2021 (25 Feb Shift 1)
Two coherent monochromatic light beams of intensities I and $4 \mathrm{I}$ are superimposed. The difference between maximum and minimum possible intensities in the resulting beam is $x \mathrm{I}$. The value of $x$ is $_________.

Options:

The correct answer is:
8

Solution:
$\begin{aligned} & I_{\max }=(\sqrt{\mathrm{I}}+\sqrt{4 \mathrm{I}})^2=9 \mathrm{I} \\ & \mathrm{I}_{\min }=(\sqrt{4 \mathrm{I}}-\sqrt{\mathrm{I}})^2=\mathrm{I} \\ & \therefore \mathrm{I}_{\max }-\mathrm{I}_{\min }=8 \mathrm{I}\end{aligned}$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Two coherent plane light waves of equal amplitude makes a small angle $\alpha(< < 1)$ with each other. They fall almost normally on a screen. If $\lambda$ is the wavelength of light waves, the fringe width $\Delta x$ of interference patterns of the two sets of waves on the screen is

Options:
A
$\frac{2 \lambda}{\alpha}$

B
$\frac{\lambda}{\alpha}$

C
$\frac{\lambda}{(2 \alpha)}$

D
$\frac{\lambda}{\sqrt{\alpha}}$

The correct answer is:
$\frac{\lambda}{(2 \alpha)}$

Solution:
$\Delta x=\frac{\lambda}{(2 \alpha)}$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2012 (19 May Online)
Two coherent point sources S1 and S2 are separated by a small distance d as shown in the figure. The fringes obtained on the screen will be

Options:
A semi-circles
B concentric circles
C points
D straight lines

The correct answer is:
concentric circles

Solution:
Path difference on circles around P is same. So, The fringes obtained on the screen in the given condition will be concentric circles.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2013 (07 Apr)
Two coherent point sources $S_1$ and $S_2$ vibrating in phase emit light of wavelength $\lambda$. The separation between them is $2 \lambda$ as shown in figure. The first bright fringe is formed at $P$ due to interference on a screen placed at a distance $D$ from $S_1(D>>\lambda)$, then $O P$ is

Options:
A $\sqrt{2} D$
B $1.5 \mathrm{D}$
C $\sqrt{3} D$
D $2 D$

The correct answer is:
$\sqrt{3} D$

Solution:
From the figure,
$\begin{aligned} & \left(S_1 P\right)^2=D^2+x^2 \\ & \text { and } \\ & \left(S_2 P\right)^2=(D-2 \lambda)^2+x^2 \\ & \text { So, }\left(S_1 P\right)^2-\left(S_2 P\right)^2=D^2+x^2 \\ & -\left[D^2+4 \lambda^2-4 D \lambda+x^2\right] \\ & =4 D \lambda-4 \lambda^2 \\ & \left(S_1 P+S_2 P\right)\left(S_1 P-S_2 P\right)=4 D \lambda-4 \lambda^2 \\ & {[D>>\lambda]} \\ & \text { Since, }[D>>\lambda] \\ & \therefore \quad S_1 P=S_2 P \\ & \end{aligned}$
and neglecting higher power of $\lambda$
$\begin{aligned} & \therefore \quad 2 S_1 P\left(S_1 P-S_2 P\right)=4 D \lambda \\ & 2\left(S_1 P\right)(\lambda)=4 D \lambda \\ & \text { or } \quad S_1 P=2 D \\ & \sqrt{D^2+x^2}=2 D \\ & x^2=3 D^2 \\ & x=\sqrt{3} D \\ & \end{aligned}$
Alternative
$\begin{gathered}2 \lambda \cos \theta=\lambda \\ \Rightarrow \quad \cos \theta=\frac{1}{2} \Rightarrow \theta=60^{\circ}, \tan \theta=\frac{x}{D} \\ \sqrt{3}=\frac{x}{D} \Rightarrow x=\sqrt{3} D\end{gathered}$

Topic: Physics, Wave Optics, JEE Main
Two coherent sound sources A and B produce a sound of wavelength Î». If at t=0 a detector starts moving with constant velocity v0 from point O along y-direction, then the minimum time, after which it detects the sound of maximum intensity, is D >>Î»

Options:
A 7Â Dv0
B D3v0
C Dv0
D 7 D3v0

The correct answer is:
7 D3v0

Solution:
4Î» cos Î¸=3Î»
cos Î¸=34
âtan Î¸=yD
ây=7D3
t=7D3 v0

Topic: Physics, Waves and Sound, JEE Main
Two coherent sources of intensities $\mathrm{l}_{1}$ and $\mathrm{I}_{2}$ produce an interference pattern on screen. The maximum intensity in the interference pattern is

Options:
A $\left[\sqrt{\mathrm{I}_{1}}+\sqrt{\mathrm{I}_{2}}\right]^{2}$
B $\mathrm{I}_{1}+\mathrm{I}_{2}$
C $\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)^{2}$
D $\mathrm{I}_{1}^{2}+\mathrm{I}_{2}^{2}$

The correct answer is:
$\left[\sqrt{\mathrm{I}_{1}}+\sqrt{\mathrm{I}_{2}}\right]^{2}$

Solution:
Intensity is proportional to square of the amplitude.
$\mathrm{I} \propto \mathrm{a}^{2}$
$\therefore \sqrt{\mathrm{I}} \propto \mathrm{a}$
Maximum intensity is produced when the two amplitudes get added (phase difference is $2 n \pi$ ).
$\therefore \mathrm{I}_{\max } \propto\left(\mathrm{a}_{1}+\mathrm{a}_{2}\right)^{2} \quad$ or $\mathrm{I}_{\max } \propto\left(\sqrt{\mathrm{I}}_{1}+\sqrt{\mathrm{I}}_{2}\right)^{2}$

Topic: Physics, Wave Optics, JEE Main
Two coherent sources of intensities, $l_1$ and $I_2$ produce an interference pattern. The maximum intensity in the interference pattern will be

Options:
A $l_1+I_2$
B $I_1^2+I_2^2$
C $\left(I_1+I_2\right)^2$
D $\left(\sqrt{I_1}+\sqrt{I_2}\right)^2$

The correct answer is:
$\left(\sqrt{I_1}+\sqrt{I_2}\right)^2$

Solution:
Resultant intensity $I_R=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi$
For maximum $I_R, \quad \phi=0^{\circ}$
$\Rightarrow I_R=I_1+I_2+2 \sqrt{I_1 I_2}=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2$

Topic: Physics, Wave Optics, JEE Main
Two coherent sources of light interfere and produce fringe patterns on a screen. For central maximum, the phase difference between the two waves will be

Options:
A zero
B π
C 3π2
D π2

The correct answer is:
zero

Solution:
For central maximum, path difference is zero hence phase difference for coherent sources of light will be zero. And the relation between phase difference and path difference is, ∆ϕ=2πλ∆x, where Δx is path difference and ∆ϕ is phase difference. So due to zero path difference phase difference is also zero.

Topic: Physics, Wave Optics, JEE Main
Two coherent sources produce a dark fringe, when the phase difference between the interfering beams is

Options:
A zero
B \( 2 \pi \)
C \( \mathrm{n} \pi \)
D \( (2 n-1) \pi \)

The correct answer is:
\( (2 n-1) \pi \)

Solution:
We know that intensity in YDSE is given by

I=I1+I2+2I1I2cosϕ

There is a dark fringe, it means waves interfere destructively. So, intensity is minimum.

Now, intensity is minimum when

cosϕ=-1

⇒ϕ=2n-1π; n=1,2,3,.....

In this case minimum intensity is given by

Imin=I1-I22

Topic: Physics, Wave Optics, JEE Main
Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio:

Options:
A 25:9
B 16:9
C 5:3
D 4:1

The correct answer is:
25:9

Solution:
Imax=I1+I22
Imin=I1-I22
⇒ ImaxImin=(I1+I2)2(I1-I2)2
⇒ 161=(I1+I2)2(I1-I2)2
⇒ 41=I1+I2I1-I2
⇒I1I2=259

Topic: Physics, Wave Optics, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Two coils are fixed in space at some distance from each other such that there is a flux linkage between them. When coil 1 has no current and the current in the coil 2 increases at the rate 15.0 A s-1, the emf in the coil 1 is 25.0 mV. When coil 2 has a current of 3.6 A, the flux linkage in coil 2 (in mWb) is

Options:

The correct answer is:
6

Solution:
As e=Mdidt
â´ M=edi/dt=25Ã10-315.0=1.67Ã10-3H
As Ï =Mi
â´ Ï=1.67Ã10-3Ã3.6=6Ã10-3 Wb
= 6 mWb

Topic: Physics, Electromagnetic Induction, JEE Main
Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon

Options:
A
the rates at which currents are changing in the two coils

B
relative position and orientation of the two coils

C
the materials of the wires of the coils.

D
the currents in the two coils

The correct answer is:
relative position and orientation of the two coils

Solution:
No Solution Available

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2003
Two coils have a mutual inductance 0.005 H. The current changes in the first coil according to equation I=I0sinâ¡Ït, where I0=10 A and Ï=100Ï rad s-1. The maximum value of emf in the second coil is

Options:
A 2π
B 5π
C π
D 4π

The correct answer is:
5π

Solution:
emf, induced in second coil e=MdIdt
â e=MddtI0sinâ¡Ït=MI0Ïcosâ¡Ït
â e=0.005Ã10Ã100Ïcosâ¡Ït
â e=5Ïcosâ¡Ït, emax=5Ï
(emf is maximum for cosâ¡Ït=1)

Topic: Physics, Electromagnetic Induction, JEE Main
Two coils of self inductance $2 \mathrm{mH}$ and 8 $\mathrm{mH}$ are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is:

Options:
A $16 \mathrm{mH}$
B $10 \mathrm{mH}$
C $6 \mathrm{mH}$
D $4 \mathrm{mH}$

The correct answer is:
$4 \mathrm{mH}$

Solution:
Mutual inductance between coils is
$$
\begin{aligned}
M & =k \sqrt{I_1 I_2} \\
M & =1 \sqrt{2 \times 10^{-3} \times 8 \times 10^{-3}}\{x=1\} \\
& =4 \times 10^{-3}=4 \mathrm{mH}
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, JEE Main
Two coils of self inductance $L_1$ and $L_2$ are placed closer to each other so that total flux in one coil is completely linked with other. If $M$ is mutual inductance between them, then

Options:
A $M=L_1 L_2$
B $M=L_1 / L_2$
C $M=\sqrt{L_1 L_2}$
D $M=\left(L_1 L_2\right)^2$

The correct answer is:
$M=\sqrt{L_1 L_2}$

Solution:
$M=-\frac{e_2}{d i_1 / d t}=-\frac{e_1}{d i_2 / d t}$
Also, $e_1=-L_1 \frac{d i_1}{d t} \cdot e_2=-L_2 \frac{d i_2}{d t}$
$M^2=$ $=\frac{e_1 e_2}{\left(\frac{d i_1}{d t}\right)\left(\frac{d i_2}{d t}\right)}=L_1 L_2 \Rightarrow M=\sqrt{L_1 L_2}$

Topic: Physics, Electromagnetic Induction, JEE Main
Two coils 'P' and 'Q' are separated by some distance. When a current of 3 A flows through coil 'P', a magnetic flux of 10-3 Wb passes through 'Q'. No current is passed through 'Q'. When no current passes through 'P' and a current of 2 A passes through 'Q', the flux through 'P' is:

Options:
A 3.67×10-3  Wb
B 6.67×10-4  Wb
C 3.67×10-4  Wb
D 6.67×10-3  Wb

The correct answer is:
6.67×10-4  Wb

Solution:
Since the coils are co-axial, the flux through the coils can be written as the product of their surface area and the magnetic field produced by the coil along its axis.

Consider the flux passing through Q

ϕQ=μ0i1R22R2+x232×πR2=10-3 Wb

And the flux passing through P:
ϕP=μ0i1r22R2+x232×πR2

Taking the ratio
ϕPϕQ=i2i1×R2+x232R2+x232⇒ϕP10-3=i2i1
⇒ϕP10-3=23
∴ϕP=6.67×10-4 Wb

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2019 (09 Apr Shift 2)
Two coils require 20 minutes and 60 minutes respectively to produce same amount of heat energy when connected separately to the same source. If they are connected in parallel arrangement to the same source; the time required to produce same amount of heat by the combination of coils, will be _____ min.

Options:

The correct answer is:
15

Solution:
If H is the heat required to heat the water, then

P1×60=H⇒P1=H60 and P2×20=H⇒P2=H20

When the coils are connected in parallel their powers will get added. If the required time now is t, then

P1+P2t=H⇒t=HH60+H20=15 min

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (29 Jun Shift 1)
Two coils, $\mathrm{X}$ and $\mathrm{Y}$, are kept in close vicinity of each other. When a varying current, $I(t)$, flows through coil $\mathrm{X}$, the induced $\operatorname{emf}(V(t))$ in coil $\mathrm{Y}$, varies in the manner shown here. The variation of $I(t)$, with time, can then be represented by the graph labelled as graph :
(A)
(B)
(C)
(D)

Options:
A
$\mathrm{A}$

B
$\mathrm{C}$

C
$\mathrm{B}$

D
$\mathrm{D}$

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

Solution:
Induced emf
$\varepsilon \propto \frac{-\mathrm{di}}{\mathrm{dt}}$

Topic: Physics, Electromagnetic Induction, JEE Main
Two coils, $\mathrm{X}$ and $\mathrm{Y}$, are kept in close vicinity of each other. When a varying current, $I(t)$, flows through coil $\mathrm{X}$, the induced $\operatorname{emf}(V(t))$ in coil $\mathrm{Y}$, varies in the manner shown here. The variation of $I(t)$, with time, can then be represented by the graph labelled as graph :
(A)
(B)
(C)
(D)

Options:
A
$\mathrm{A}$

B
$\mathrm{C}$

C
$\mathrm{B}$

D
$\mathrm{D}$

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

Solution:
Induced emf
$\varepsilon \propto \frac{-\mathrm{di}}{\mathrm{dt}}$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2013 (09 Apr Online)
Two common tangents to the circle $x^2+y^2=2 a^2$ and parabola $y^2=8 a x$ are

Options:
A
$x=\pm(y+2 a)$

B
$y=\pm(x+2 a)$

C
$x=\pm(y+a)$

D
$y=\pm(x+a)$

The correct answer is:
$y=\pm(x+2 a)$

Solution:
Any tangent to the parabola $y^2=8 a x$ is
$$
y=m x+\frac{2 a}{m}
$$
If (i) is a tangent to the circle, $x^2+y^2=2 a^2$ then, $\quad \sqrt{2} a=\pm \frac{2 a}{m \sqrt{m^2+1}}$
$$
\Rightarrow m^2\left(1+m^2\right)=2 \Rightarrow\left(m^2+2\right)\left(m^2-1\right)=0 ; \Rightarrow m=\pm 1
$$
So, $y=\pm(x+2 a)$

Topic: Mathematics, Circle, JEE Main, JEE Main 2002