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**Previous Years JEE Advanced Questions**

**[IIT-JEE 2011]**

**Sol.**5

**[JEE Advanced 2014]**

**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

$\mathrm{t} \approx 2 \mathrm{s}$

**[JEE Advanced-2014]**

**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

$\mathrm{t}=\frac{500}{\mathrm{V}_{\mathrm{B}} \sin 30}=\frac{500}{100}=5$

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]**

**Sol.**(C)