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

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*Simulator***Previous Years JEE Advanced Questions**

Q. A train is moving along a straight line with a constant acceleration ‘a’. A boy standing in the train throws a ball forward with a speed of 10 m/s, at an angle of $60^{\circ}$ to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train, in $\mathrm{m} / \mathrm{s}^{2}$, is

**[IIT-JEE 2011]**
Q. A rocket is moving in a gravity free space with a constant acceleration of 2 $\mathrm{ms}^{-2}$ along + x direction (see figure). The length of a chamber inside the rocket is 4m. A ball is thrown from the left end of the chamber in + x direction with a speed of 0.3 $\mathrm{ms}^{-1}$ relative to the rocket. At the same time, another ball is thrown in –x direction with a speed of 0.2 $\mathrm{ms}^{-1}$ from its right end relative to the rocket. The time in seconds when the two balls hit each other is

**[JEE Advanced 2014]****Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...**

**Sol.**8 or 2 Assuming open chamber $\mathrm{V}_{\text {relative }}=0.5 \mathrm{m} / \mathrm{s}$ $\mathrm{S}_{\text {relative }}=4 \mathrm{m}$ time $=\frac{4}{0.5}=8 \mathrm{m} / \mathrm{s}$

**Alternate**Assuming closed chamber

**In the frame of chamber :**Maximum displacement of ball A from its left end is $\frac{\mathrm{u}_{\mathrm{A}}^{2}}{2 \mathrm{a}}=\frac{(0.3)^{2}}{2(2)}=0.0225 \mathrm{m}$ This is negligible with respect to the length of chamber i.e. 4m. So, the collision will be verym close to the left end. Hence, time taken by ball B to reach left end will be given by $\mathrm{S}=\mathrm{u}_{\mathrm{B}} \mathrm{t}+\frac{1}{2} \mathrm{at}^{2}$ $4=(0.2)(\mathrm{t})+\frac{1}{2}(2)(\mathrm{t})^{2}$ Solving this, we get $\mathrm{t} \approx 2 \mathrm{s}$

Q. Airplanes A and B are flying with constant velocity in the same vertical plane at angles $30^{\circ}$ and $60^{\circ}$ with respect to the horizontal respectively as shown in figure. The speed of A is $\mathrm{ms}^{-1}$. At time t = 0 s, an observer in A finds B at a distance of 500 m. This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t = $\mathrm{t}_{0}$, A just escapes being hit by B, $\mathrm{t}_{0}$ in seconds is

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**Sol.**5 As observed from A, B moves perpendicular to line of motion of A. It means velocity of B along A is equal to velocity of A $\mathrm{V}_{\mathrm{B}} \cos 30=100 \sqrt{3}$ $\mathrm{V}_{\mathrm{B}}=200$ If A is observer A remains stationary therefore $\mathrm{t}=\frac{500}{\mathrm{V}_{\mathrm{B}} \sin 30}=\frac{500}{100}=5$

Q. Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is that the instantaneous density $\rho$ remains uniform throughout the volume.
the rate of fractional change in density is $\left(\frac{1}{\rho} \frac{d \rho}{d t}\right)$ constant. the velocity v of any point on the surface of the expanding sphere is proportional to
(A) $R^{3}$
(B) $\frac{1}{R}$
(C) $\mathrm{R}$
(D) $R^{2 / 3}$

**[JEE Advanced-2017]**
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Where is solution

My memories I use to solve these questions. Physics was my favourite subject.

Easy peasy

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bhosdika

tumhari yyadd aa rahi ha

Please provide for vector also….

Nice

q no 3 is incorrect as the value for velocity of A is not given please fix it asap

too tough

Easy

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