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### JEE Main 2017 (Online) 9th April Morning Slot

A block of mass 0.1 kg is connected to an elastic spring of spring constant 640 Nm−1 and oscillates in a damping medium of damping constant 10−2 kg s−1 . The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to :
A
2 s
B
3.5 s
C
5 s
D
7 s
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### JEE Main 2018 (Online) 15th April Morning Slot

A thin uniform tube is bent into a circle of radius $r$ in the vertical plane. Equal volumes of two immiscible liquids, whose densities are ${\rho _1}$ and ${\rho _2}$ $\left( {{\rho _1} > {\rho _2}} \right),$ fill half the circle. The angle $\theta$ between the radius vector passing through the common interface and the vertical is :
A
$\theta = {\tan ^{ - 1}}\pi \left( {{{{\rho _1}} \over {{\rho _2}}}} \right)$
B
$\theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1}} \over {{\rho _2}}}} \right)$
C
$\theta = {\tan ^{ - 1}}\left( {{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}} \right)$
D
$\theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1} + {\rho _2}} \over {{\rho _1} - {\rho _2}}}} \right)$

## Explanation As system is in equilibrium so the pressuse at A from both side of the liquid must be equal .

(r cos $\theta$ + r sin $\theta$) $\rho$2g = (r cos $\theta$ $-$ r sin $\theta$) $\rho$1g

$\Rightarrow $$\,\,\, {{{\rho _1}} \over {{\rho _2}}} = {{\sin \theta + \cos \theta } \over {\cos \theta - \sin \theta }} = {{1 + \tan \theta } \over {1 - \tan \theta }} \Rightarrow$$\,\,\,$ $\rho$1 $-$ $\rho$1 tan$\theta$ = $\rho$2 + $\rho$2 tan$\theta$

$\Rightarrow $$\,\,\, (\rho 1 + \rho 2) tan\theta = \rho 1 - \rho 2 \Rightarrow$$\,\,\,$ tan$\theta$ = ${{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}$

$\Rightarrow$$\,\,\,$ $\theta$ = tan$-$1 $\left( {{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}} \right)$
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### JEE Main 2018 (Online) 15th April Evening Slot

Two simple harmonic motions, as shown below, are at right angles. They are combined to form Lissajous figures.
x(t) = A sin (at + $\delta$)
y(t) = B sin (bt)

Identify the correct match below.
A
Parameters   A $\ne$ B, a = b; $\delta$ = 0;
Curve    Parabola
B
Parameters    A = B, a = b; $\delta$ = ${\raise0.5ex\hbox{\pi } \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}}$
Curve    Line
C
Parameters    A $\ne$ B, a = b; $\delta$ = ${\raise0.5ex\hbox{\pi } \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}}$
Curve    Ellipse
D
Parameters    A = B, a = 2b; $\delta$ = ${\raise0.5ex\hbox{\pi } \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}}$
Curve    Circle
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### JEE Main 2018 (Online) 16th April Morning Slot

An oscillator of mass M is at rest in its equilibrium position in a potential
V = ${1 \over 2}$ k(x $-$ X)2. A particle of mass m comes from right with speed u and collides completely inelastically with M and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after 13 collisions is : (M = 10, m = 5, u = 1, k = 1)
A
${1 \over {\sqrt 3 }}$
B
${1 \over 2}$
C
${2 \over 3}$
D
${3 \over {\sqrt 5 }}$