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**

*t*= 4/3 s is

** [IIT JEE 2009]**

**Sol.**(D)

$\tau=8$

$\therefore \omega=\frac{\pi}{4}$

$\mathrm{x}=\sin \left(\frac{\pi}{4} \mathrm{t}\right)$

$\therefore a=-\frac{\pi^{2}}{16} \sin \left(\frac{\pi}{4} t\right)$

$-\frac{\pi^{2} \sqrt{3}}{32} \mathrm{cm} / \mathrm{s}^{2}$

**[IIT JEE 2009]**

**Sol.**(D)

$\frac{\mathrm{k}_{1} \mathrm{k}_{2}}{\mathrm{k}_{1}+\mathrm{k}_{2}} \mathrm{A}=\mathrm{k}_{1} \mathrm{x}=\mathrm{F}$

$\therefore \mathrm{x}=\frac{\mathrm{k}_{2} \mathrm{A}}{\mathrm{k}_{1}+\mathrm{k}_{2}}$

**[IIT JEE 2009]**

**Sol.**(C)

$\tau=2 \mathrm{k} \cdot \frac{\ell \theta}{2} \cdot \frac{\ell}{2}=\frac{\mathrm{m} \ell^{2}}{12} \cdot \omega^{2} \theta$

$\therefore \omega=\sqrt{\frac{6 \mathrm{k}}{\mathrm{m}}}$

$\therefore \mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{6 \mathrm{k}}{\mathrm{m}}}$

**Paragraph for Question No. 4 to 6**

When a particle of mass m moves on the x–axis in a potential of the form $\mathrm{V}(\mathrm{x})=\mathrm{k} \mathrm{x}^{2}$, it performs simple harmonic motion. The corresponding time period is proportional to $\sqrt{\frac{m}{k}}$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from $\mathrm{kx}^{2}$ and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x–axis. Its potential energy is $\mathrm{V}(\mathrm{x})=\alpha \mathrm{x}^{4}(\alpha>0)$ for $|\mathrm{x}|$ near the origin and becomes a constant equal to $\mathrm{V}_{0}$ for $|\mathrm{x}| \geq \mathrm{X}_{0}$ (see figure)

(A) $\mathrm{E}<0$

(B) E > 0

(C) $\mathrm{V}_{0}>\mathrm{E}>0$

(D) $\mathrm{E}>\mathrm{V}_{0}$

**[IIT-JEE 2010]**

**Sol.**(C)

Particle must be trapped.

$\therefore \mathrm{V}_{0}>\mathrm{E}>0$

*A*, the time period

*T*of this particle is proportional to :-

(A) $A \sqrt{\frac{m}{\alpha}}$

(B) $\frac{1}{A} \sqrt{\frac{m}{\alpha}}$

(C) $A \sqrt{\frac{\alpha}{m}}$

(D) $\frac{1}{A} \sqrt{\frac{\alpha}{m}}$

**[IIT-JEE 2010]**

**Sol.**(B)

Dimentions are

$\propto=\frac{\left[\mathrm{ML}^{2} \mathrm{T}^{-2}\right]}{\left[\mathrm{L}^{4}\right]}=\left[\mathrm{ML}^{-2} \mathrm{T}^{-2}\right]$

$\therefore \mathrm{T} \propto \frac{1}{\mathrm{A}} \sqrt{\frac{\mathrm{m}}{\alpha}}$

(A) proportional to $V_{0}$

(B) proportional to $\frac{V_{0}}{m X_{0}}$

(C) proportional to $\sqrt{\frac{V_{0}}{m X_{0}}}$

(D) Zero

**[IIT-JEE 2010]**

**Sol.**(D)

As $\mathrm{U}=\mathrm{V}_{0}=$ constant

$\therefore \mathrm{a}=0$

$x_{1}(t)=A \sin \omega t$ and $x_{2}(t)=A \sin \left(\omega t+\frac{2 \pi}{3}\right)$. Adding a third sinusoidal displacement

$x_{3}(t)=B \sin (\omega t+\phi)$ brings the mass to a complete rest. The values of $\mathrm{B}$ and $\phi$ are :-

** [IIT-JEE 2011]**

**Sol.**(B)

$\mathrm{x}^{\prime}=\mathrm{A} \sin (\omega \mathrm{t})+\mathrm{A} \sin \left(\omega \mathrm{t}+\frac{2 \pi}{3}\right)$

$=2 \mathrm{A} \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right) \cos \left(\frac{\pi}{3}\right)$

$=\mathrm{A} \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right)$

$\therefore \mathrm{B}=\mathrm{A} \quad \phi=\frac{4 \pi}{3}$

(A) Restoring torque in case A = Restoring torque in case B

(B) Restoring torque in case A < Restoring torque in case B

(C) Angular frequency for case A> Angular frequency for case B

(D) Angular frequency for case A< Angular frequency for case B

**[IIT-JEE 2011]**

**Sol.**(A,D)

$\tau=\frac{\mathrm{mL}}{2} \mathrm{g} \cos \theta+\mathrm{mL} \mathrm{g} \cos \theta \mathrm{same}$

$\frac{\omega}{2 \pi}=\mathrm{f}$ for $\mathrm{A} \mathrm{mg} \mathrm{H}=\frac{1}{2} \mathrm{mv}^{2} \&$ for $\mathrm{B}=\operatorname{mg} \mathrm{x}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$

$\therefore \omega$

**[IIT-JEE 2012]**

**Sol.**(A)

Time of

$\therefore \mathrm{v}=\frac{\mathrm{g}}{\sqrt{2}}=\sqrt{50} \mathrm{m} / \mathrm{s}$

(A) the speed of the particle when it returns to its equilibrium position is $\mathrm{u}_{0}$

(B) the time at which the particle passes through the equilibrium position for the first

time is $t=\pi \sqrt{\frac{m}{k}}$

(C) the time at which the maximum compression of the spring occurs is $\mathrm{t}=\frac{4 \pi}{3} \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

(D) the time at which the particle passes through the equilibrium position for the second

time is $t=\frac{5 \pi}{3} \sqrt{\frac{m}{k}}$

**[JEE-Advanced-2013]**

**Sol.**(A,D)

Collision is elastic so energy is conserve in given question so when it return to equilibrium, its speed is $\mathrm{u}_{0}$

**[JEE-Advanced-2015]**

**Sol.**(B,D)

$\mathrm{P}_{1 \max }=\operatorname{mac}_{1}=\mathrm{b}$

$\mathrm{P}_{2 \max }=\mathrm{mR} \omega_{2}=\mathrm{R}$

$\frac{\omega_{1}}{\omega_{2}}=\frac{1}{n^{2}}$

$\frac{\omega_{2}}{\omega_{1}}=\mathrm{n}^{2}$

$\mathrm{E}_{1}=\frac{1}{2} \mathrm{m} \omega_{1}^{2} \mathrm{a}^{2}$

**[JEE-Advanced-2015]**

**Sol.**($(\mathrm{A})-\mathrm{P}, \mathrm{Q}, \mathrm{R}, \mathrm{T} ;(\mathrm{B})-\mathrm{Q}, \mathrm{S} ;(\mathrm{C})-\mathrm{P}, \mathrm{Q}, \mathrm{R}, \mathrm{S} ;(\mathrm{D})-\mathrm{P}, \mathrm{R}, \mathrm{T}$)

(A) The amplitude of oscillation in the first case changes by a factor of , whereas $\sqrt{\frac{\mathrm{M}}{\mathrm{m}+\mathrm{M}}}$ in the second case it remains unchanged

(B) The final time period of oscillation in both the cases is same

(C) The total energy decreases in both the cases

(D) The instantaneous speed at x0 of the combined masses decreases in both the cases.

** [JEE-Advanced-2016]**

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

$\mathrm{T}_{\mathrm{i}}=2 \pi \sqrt{\frac{\mathrm{M}}{\mathrm{K}}}, \mathrm{T}_{\mathrm{f}}=2 \pi \sqrt{\frac{\mathrm{M}+\mathrm{m}}{\mathrm{K}}}$

case (i) :

M $(\mathrm{A} \omega)=(\mathrm{M}+\mathrm{m}) \mathrm{V}$

$\therefore$ Velocity decreases at equilibrium position.

By energy conservation

$\mathrm{A}_{\mathrm{f}}=\mathrm{A}_{\mathrm{i}} \sqrt{\frac{\mathrm{M}}{\mathrm{M}+\mathrm{m}}}$

case ( ii ) :

No energy loss, amplitude remains same

At equilibrium $\left(\mathrm{x}_{0}\right)$ velocity $=\mathrm{A} \omega .$

In both cases $\omega$ decreases so velocity decreases in both cases

$\omega=\pi / 6 \mathrm{rad} \mathrm{s}^{-1} .$ If $|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{3}|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|$ at time $\mathrm{t}=\tau$ for the first time, the value of $\tau,$ in seconds, is

**[JEE-Advanced-2018]**

**Sol.**2.00 Sec