JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

**Click Here for** **JEE main Previous Year Topic Wise Questions of Physics with Solutions **

**Download eSaral app for free study material and video tutorials.**

*Simulator*

**Previous Years JEE Advanced Questions**

**ab**and

**cd.**Then,

(A) $\mathrm{I}_{1}>\mathrm{I}_{2}$

(B) $\mathrm{I}_{1}<\mathrm{I}_{2}$

(C) $\mathrm{I}_{1}$ is in the direction ba and $\mathrm{I}_{2}$ is in the direction $\mathrm{cd}$

(D) $\mathrm{I}_{1}$ is in the direction ab and $\mathrm{I}_{2}$ is in the direction de

** [JEE-2009]**

**Sol.**(D)

As area of outer loop is bigger therefore emf induced in outer loop is dominant and therefore according to lenz law current in outer loop is Anticlockwise and inner loop is clockwise

** [JEE-2009]**

**Sol.**(B,D)

(a) when resistivity is low current induced will be more; therefore impulsive force on the ring will also be more and it jumps to higher levels. [But for this mass should be either less or equal to the other]

(A) IBL

(B) $\frac{\mathrm{IBL}}{\pi}$

(C) $\frac{\mathrm{IBL}}{2 \pi}$

(D) $\frac{\mathrm{IBL}}{4 \pi}$

**Sol.**(C)

$2 \mathrm{T} \sin \theta=\mathrm{IB}(2 \mathrm{R} \theta) .$ As $\theta$ is small so $\sin \theta \approx \theta$

$\Rightarrow 2 \mathrm{T}(\theta)=2 \mathrm{IBR} \theta \Rightarrow \mathrm{T}=\mathrm{BIR}=\frac{\mathrm{IBL}}{2 \pi} \quad[\because 2 \pi \mathrm{R}=\mathrm{L}]$

**[JEE 2011]**

**Sol.**(C)

Magnetic field lines and induced electric field lines always from closed loop.

**[JEE 2011]**

**Sol.**6

$\phi=\mathrm{B} \pi \mathrm{r}^{2}=\left(\frac{\mu_{0} I}{L}\right) \pi r^{2}=\mu_{0} I_{0} \frac{\pi r^{2}}{L} \cos 300 t \Rightarrow \varepsilon_{1}=\frac{d \phi}{d t}=\left(\frac{\mu_{0} I_{0} \pi r^{2}}{L}\right) 300 \sin 300 t$

$\mathrm{i}=\frac{\varepsilon}{R}=\left(\mu_{0} I_{0} \sin 300 t\right)\left[\frac{\pi r^{2}(300)}{L R}\right] \Rightarrow M=i \pi r^{2}=\left[\frac{\pi^{2} r^{4}(300)}{L R}\right] \mu_{0} I_{0} \sin 300 t$

**[JEE 2012]**

**Sol.**7

Magnetic field at a distance x along axis of a circular coil is given by $B(x)=\frac{\mu_{0} i R^{2}}{2\left(R^{2}+x^{2}\right)^{3 / 2}}$

$\phi(x)=\left[\frac{\mu_{0} i R^{2}}{2\left(R^{2}+x^{2}\right)^{3 / 2}}\right] N\left(a^{2}\right)\left(\cos 45^{\circ}\right)$

As $\mathrm{P}=\mathrm{Mi} \quad$ So $\mathrm{M}=\frac{\mu_{0} R^{2}(2)}{2\left[R^{2}+(\sqrt{3} R)^{2}\right]^{3 / 2}}\left(\frac{a^{2}}{\sqrt{2}}\right)=\frac{\mu_{0} a^{2}}{2^{7 / 2} R}$ thus $p=7$

(A) The emf induced in the loop is zero if the current is constant.

(B) The emf induced in the loop is infinite if the current is constant.

(C) The emf induced in the loop is zero if the current decreases at a steady rate.

(D) The emf induced in the loop is finite if the current decreases at a steady rate.

** [JEE 2012]**

**Sol.**(A,C)

Since flux of wire on the loop is zero therefore emf will not be induced.

**Paragraph for Questions 8 and 9**

A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity $\omega$. This can be considered as equivalent to a loop carrying a steady current $\frac{\mathrm{Q} \omega}{2 \pi}$. A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant $\gamma$.

$(\mathrm{A})-\gamma \mathrm{B} \mathrm{Q} \mathrm{R}^{2}$

$(\mathrm{B})-\gamma \frac{\mathrm{BQR}^{2}}{2}$

(C) $\gamma \frac{\mathrm{BQR}^{2}}{2}$

(D) $\gamma \mathrm{BQR}^{2}$

**[JEE Advance-2013]**

**Sol.**(B)

induce electric field $=\frac{\mathrm{R}}{2} \frac{\mathrm{d} \mathrm{B}}{\mathrm{dt}}=\frac{\mathrm{BR}}{2}$

torque on charge $=\frac{\mathrm{QBR}^{2}}{2}(-\hat{\mathrm{k}})$

by $\vec{\tau}=\frac{\mathrm{d} \overrightarrow{\mathrm{L}}}{\mathrm{dt}} \Rightarrow \int \mathrm{d} \overrightarrow{\mathrm{L}}=\int_{0}^{1} \vec{\tau} \mathrm{dt}$

$\Delta \overrightarrow{\mathrm{L}}=\frac{\mathrm{QBR}^{2}}{2}(-\hat{\mathrm{k}})$

Change in magnetic dipole moment $=\gamma \Delta \overrightarrow{\mathrm{L}}$

$=\frac{\gamma \mathrm{QBR}^{2}}{2}(-\hat{\mathrm{k}})$

(A) $\frac{\mathrm{BR}}{4}$

(B) $\frac{\mathrm{BR}}{2}$

(C) BR

(D) 2BR

** [JEE Advance-2013]**

**Sol.**(B)

Magnitude of induced electric field $=\frac{\mathrm{R}}{2} \frac{\mathrm{dB}}{\mathrm{dt}}=\frac{\mathrm{BR}}{2}$

(A) The induced current in the wire is in opposite direction to the current along the hypotenuse.

(B) There is a repulsive force between the wire and the loop

(C) If the loop is rotated at a constant angular speed about the wire, an additional emf of $\left(\frac{\mu_{0}}{\pi}\right)$ volt is induced in the wire

(D) The magnitude of induced emf in the wire is $\left(\frac{\mu_{0}}{\pi}\right)$ volt.

** [JEE Advance-2013]**

**Sol.**(B,D)

by direction of induced electric field, current in wire is in same direction of current along the hypotenuse.

Flux through triangle if wire have current $i=\int_{0}^{0.1}\left(\frac{\mu_{0} i}{2 \pi \mathrm{x}}\right)(2 \mathrm{xd} \mathrm{x})=\frac{\mu_{0} \mathrm{i}}{10 \pi}$

$\Rightarrow$ Mutual inductance $=\frac{\mu_{0}}{10 \pi}$

Induced emf in wire $=\frac{\mu_{0}}{10 \pi} \frac{\mathrm{d} \mathrm{i}}{\mathrm{dt}}=\frac{\mu_{0}}{10 \pi} \times 10=\frac{\mu_{0}}{\pi}$

**[JEE Advance-2016]**

**Sol.**8

$\left.\mathrm{I}_{\max }=\frac{\varepsilon}{\mathrm{R}}=\frac{5}{12} \mathrm{A} \text { (Initially at } \mathrm{t}=0\right)$

$\mathrm{I}_{\min }=\frac{\varepsilon}{\mathrm{R}_{\mathrm{eq}}}=\varepsilon\left(\frac{1}{\mathrm{r}_{1}}+\frac{1}{\mathrm{r}_{2}}+\frac{1}{\mathrm{R}}\right) \quad$ (finally in steady state)

$=5\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{12}\right)$

$=\frac{10}{3} \mathrm{A}$

$\frac{\mathrm{I}_{\max }}{\mathrm{I}_{\min }}=8$

**[JEE Advance-2016]**

**Sol.**(C,D)

*At the point of crossing the wires remain electrically insulated from each other*. The entire loop lies in the plane (of the paper). A uniform magnetic field points into the plane of the paper. At t = 0, the loop starts rotating about the common diameter as axis with a constant angular velocity in the magnetic field. Which of the following options is/are correct?

(A) The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper

(B) The net emf induced due to both the loops is proportional to $\cos \omega t$

(C) The emf induced in the loop is proportional to the sum of the areas of the two loops

(D) The amplitude of the maximum net emf induced due to both the loops is equal to the

amplitude of maximum emf induced in the smaller loop alone

**[JEE Advance-2017]**

**Sol.**(A,D)

(A) The ratio of the currents through $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ is fixed at all times $(\mathrm{t}>0)$

(B) After a long time, the current through $L_{1}$ will be $\frac{V}{R} \frac{L_{2}}{L_{1}+L_{2}}$

(C) After a long time, the current through $\mathrm{L}_{2}$ will be $\frac{\mathrm{V}}{\mathrm{R}} \frac{\mathrm{L}_{1}}{\mathrm{L}_{1}+\mathrm{L}_{2}}$

(D) At $\mathrm{t}=0,$ the current through the resistance $\mathrm{R}$ is $\frac{\mathrm{V}}{\mathrm{R}}$

**[JEE Advance-2017]**

**Sol.**(A,B,C)

Since inductors are connected in parallel

$\mathrm{V}_{\mathrm{L}_{1}}=\mathrm{V}_{\mathrm{L}_{2}}$

$\mathrm{L}_{1} \frac{\mathrm{dI}_{1}}{\mathrm{dt}}=\mathrm{L}_{2} \frac{\mathrm{d} \mathrm{I}_{2}}{\mathrm{dt}}$

$\mathrm{L}_{1} \mathrm{I}_{1}=\mathrm{L}_{2} \mathrm{I}_{2}$

$\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{\mathrm{L}_{2}}{\mathrm{L}_{1}}$

Current through resistor at any time t is given by

$\left.\mathrm{I}=\mathrm{V} / \mathrm{R}_{(1-\mathrm{e}}^{-\frac{\mathrm{RT}}{\mathrm{L}}}\right)$ where $\mathrm{L}=\frac{\mathrm{L}_{1} \mathrm{L}_{2}}{\mathrm{L}_{1}+\mathrm{L}_{2}}$

After long time I $=\frac{\mathrm{V}}{\mathrm{R}}$

$\mathrm{I}_{1}+\mathrm{I}_{2}=\mathrm{I} \quad \ldots(\mathrm{i})$

$\mathrm{L}_{1} \mathrm{I}_{1}=\mathrm{L}_{2} \mathrm{I}_{2} \quad \ldots$ (ii)

From (i) & (ii) we get

$\mathrm{I}_{1}=\frac{\mathrm{V}}{\mathrm{R}} \frac{\mathrm{L}_{2}}{\mathrm{L}_{1}+\mathrm{L}_{2}}, \quad \mathrm{I}_{2}=\frac{\mathrm{V}}{\mathrm{R}} \frac{\mathrm{L}_{1}}{\mathrm{L}_{1}+\mathrm{L}_{2}}$

(D) value of current is zero at t = 0

value of current is $\mathrm{V} / \mathrm{R}$ at $\mathrm{t}=\infty$

Hence option (D) is incorrect.

(A) $\mathrm{I}_{\max }=\frac{\mathrm{V}}{2 \mathrm{R}}$

(B) $\mathrm{I}_{\max }=\frac{\mathrm{V}}{4 \mathrm{R}}$

(C) $\tau=\frac{\mathrm{L}}{\mathrm{R}} \ell \mathrm{n} 2$

(D) $\tau=\frac{2 \mathrm{L}}{\mathrm{R}} \ell \mathrm{n} 2$

**[JEE Advance-2017]**

**Sol.**(B,D)

$\mathrm{i}_{\max }=\left(\mathrm{i}_{2}-\mathrm{i}_{1}\right)_{\max }$

$\Delta \mathrm{i}=\left(\mathrm{i}_{2}-\mathrm{i}_{1}\right)=\frac{\mathrm{V}}{\mathrm{R}}\left[1-\mathrm{e}^{-\left(\frac{\mathrm{R}}{2 \mathrm{L}}\right) \mathrm{t}}\right]-\frac{\mathrm{V}}{\mathrm{R}}\left[1-\mathrm{e}^{\left(-\frac{\mathrm{R}}{\mathrm{L}}\right) \mathrm{t}}\right]$

$\frac{V}{R}\left[e^{-\left(\frac{R}{L}\right) t}-e^{-\left(\frac{R}{2 L}\right) t}\right]$

For $(\Delta \mathrm{i})_{\max } \frac{\mathrm{d}(\Delta \mathrm{i})}{\mathrm{dt}}=0$

$\frac{\mathrm{V}}{\mathrm{R}}\left[-\frac{\mathrm{R}}{\mathrm{L}} \mathrm{e}^{-\left(\frac{\mathrm{R}}{\mathrm{L}}\right) \mathrm{t}}-\left(-\frac{\mathrm{R}}{2 \mathrm{L}}\right) \mathrm{e}^{-\left(\frac{\mathrm{R}}{2 \mathrm{L}}\right) \mathrm{t}}\right]=0$

$\mathrm{e}^{-\left(\frac{\mathrm{R}}{\mathrm{L}}\right) \mathrm{t}}=\frac{1}{2} \mathrm{e}^{-\left(\frac{\mathrm{R}}{2 \mathrm{L}}\right) \mathrm{t}}$

$\mathrm{e}^{-\left(\frac{\mathrm{R}}{2 \mathrm{L}}\right) \mathrm{t}}=\frac{1}{2}$

$\left(\frac{\mathrm{R}}{2 \mathrm{L}}\right) \mathrm{t}=\ell \mathrm{n} 2$

$\mathrm{t}=\frac{2 \mathrm{L}}{\mathrm{R}} \ell \mathrm{n} 2 \rightarrow$ time when $\mathrm{i}$ is maximum.

$\dot{\mathbf{i}}_{\max }=\frac{V}{R}\left[e^{-\frac{R}{L}\left(\frac{2 L}{R} \ell_{n} 2\right)}-e^{-\left(\frac{R}{2 L}\right)\left(\frac{2 L}{R} \ln 2\right)}\right]$

$\left|\mathrm{i}_{\max }\right|=\frac{\mathrm{V}}{\mathrm{R}}\left|\left[\frac{1}{4}-\frac{1}{2}\right]\right|=\frac{1}{4} \frac{\mathrm{V}}{\mathrm{R}}$

$\mathrm{I}_{1}=\frac{\mathrm{V}}{\mathrm{R}} \frac{\mathrm{L}_{2}}{\mathrm{L}_{1}+\mathrm{L}_{2}}, \quad \mathrm{I}_{2}=\frac{\mathrm{V}}{\mathrm{R}} \frac{\mathrm{L}_{1}}{\mathrm{L}_{1}+\mathrm{L}_{2}}$

(D) value of current is zero at t = 0

value of current is $\mathrm{V} / \mathrm{R}$ at $\mathrm{t}=\infty$

Hence option (D) is incorrect.