1
JEE Main 2018 (Online) 16th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
MCQ (Single Correct Answer)
+4
-1
Change Language
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
MCQ (Single Correct Answer)
+4
-1
Change Language
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
MCQ (Single Correct Answer)
+4
-1
Change Language
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
$$\theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1} + {\rho _2}} \over {{\rho _1} - {\rho _2}}}} \right)$$
JEE Main Subjects
EXAM MAP
Joint Entrance Examination
JEE MainJEE AdvancedWB JEE
Graduate Aptitude Test in Engineering
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
Medical
NEET