1
JEE Main 2018 (Online) 16th April Morning Slot
+4
-1
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
$${3 \over {\sqrt 5 }}$$
2
JEE Main 2018 (Online) 16th April Morning Slot
+4
-1
A particle executes simple harmonic motion and is located at x = a, b and c at times t0, 2t0 and 3t0 respectively. The freqquency of the oscillation is :
A
$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + c} \over {2b}}} \right)$$
B
$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + b} \over {2c}}} \right)$$
C
$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{2a + 3c} \over b}} \right)$$
D
$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + 2b} \over {3c}}} \right)$$
3
JEE Main 2018 (Online) 15th April Evening Slot
+4
-1
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; $$\delta$$ = $${\raise0.5ex\hbox{\scriptstyle \pi } \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}$$
Curve    Line
C
Parameters    A $$\ne$$ B, a = b; $$\delta$$ = $${\raise0.5ex\hbox{\scriptstyle \pi } \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}$$
Curve    Ellipse
D
Parameters    A = B, a = 2b; $$\delta$$ = $${\raise0.5ex\hbox{\scriptstyle \pi } \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}$$
Curve    Circle
4
JEE Main 2018 (Online) 15th April Morning Slot
+4
-1
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
$$\theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1} + {\rho _2}} \over {{\rho _1} - {\rho _2}}}} \right)$$
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