Kinematics 1D- JEE Advanced Previous Year Questions with Solutions
JEE Mains & Advanced
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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]
Ans. 5 
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]
[JEE Advanced 2014] Ans. 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 $\mathrm{t} \approx 2 \mathrm{s}$
$\mathrm{V}_{\text {relative }}=0.5 \mathrm{m} / \mathrm{s}$ $\mathrm{S}_{\text {relative }}=4 \mathrm{m}$ 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 
[JEE Advanced-2014]
[JEE Advanced-2014] Ans. 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 $\mathrm{t}=\frac{500}{\mathrm{V}_{\mathrm{B}} \sin 30}=\frac{500}{100}=5$
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]
Ans. (C)
Comments
Anshuman pandey
July 17, 2023, 6:35 a.m.
mai to class 9 me hu mai ye solve kar pa rha hu kya ye bat achi hai ya overconfidence wali bat hai?
Akash
March 29, 2022, 8:11 p.m.
Great
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Ashish chanchlani
Dec. 22, 2020, 2:26 p.m.
My memories I use to solve these questions. Physics was my favourite subject.