 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2010

For a particle in uniform circular motion the acceleration $\overrightarrow a$at a point P(R, θ) on the circle of radius R is (here θ is measured from the x–axis)
A
$- {{{v^2}} \over R}\cos \theta \widehat i + {{{v^2}} \over R}\sin \theta \widehat j$
B
$- {{{v^2}} \over R}\sin \theta \widehat i + {{{v^2}} \over R}\cos \theta \widehat j$
C
$- {{{v^2}} \over R}\cos \theta \widehat i - {{{v^2}} \over R}\sin \theta \widehat j$
D
${{{v^2}} \over R}\widehat i + {{{v^2}} \over R}\widehat j$

Explanation

For a particle in uniform circular motion,

${a_c} = {{{v^2}} \over R}$ towards the center of the circle

From figure, $\overrightarrow a = {a_c}\cos \theta \left( { - \widehat i} \right) + {a_c}\sin \theta \left( { - \widehat j} \right)$

$= {{ - {v^2}} \over R}\cos \theta \widehat i - {{{v^2}} \over R}\sin \theta \widehat j$ 2

AIEEE 2009

Consider a rubber ball freely falling from a height $h=4.9$ $m$ onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic.

Then the velocity as a function of time and the height as a function of time will be :
A B C D Explanation

For downward motion $v=-gt$

The velocity of the rubber ball increases in downward direction and we get a straight line between $v$ and $t$ with a negative slope.

Also applying $y - {y_0} = ut + {1 \over 2}a{t^2}$

We get $y - h = - {1 \over 2}g{t^2} \Rightarrow y = h - {1 \over 2}g{t^2}$

The graph between $y$ and $t$ is a parabola with $y=h$ at $t=0.$ As time increases $y$ decreases.

For upward motion.
The ball suffer elastic collision with the horizontal elastic plate therefore the direction of velocity is reversed and the magnitude remains the same. Here $v=u-gt$ where $u$ is the velocity just after collision. As $t$ increases, $v$ decreases. We get a straight line between $v$ and $t$ with negative slope.

Also $y = ut - {1 \over 2}g{t^2}$

All these characteristics are represented by graph $(c).$
3

AIEEE 2009

A particle has an initial velocity $3\widehat i + 4\widehat j$ and an acceleration of $0.4\widehat i + 0.3\widehat j$. Its speed after 10 s is:
A
$7\sqrt 2$ units
B
7 units
C
8.5 units
D
10 units

Explanation

Given $\overrightarrow u = 3\widehat i + 4\widehat j,\,\,\overrightarrow a = 0.4\widehat i + 0.3\widehat j,\,\,t = 10s$

$\overrightarrow v = \overrightarrow u + \overrightarrow a t$

$= 3\widehat i + 4\widehat j + \left( {0.4\widehat i + 0.3\widehat j} \right) \times 10$

$= 7\widehat i + 7\widehat j$

We know speed is equal to magnitude of velocity.

$\therefore$ $\left| {\overrightarrow v } \right| = \sqrt {{7^2} + {7^2}} = 7\sqrt 2 \,\,\,$ units
4

AIEEE 2008

A body is at rest at $x=0.$ At $t=0,$ it starts moving in the positive $x$-direction with a constant acceleration. At the same instant another body passes through $x=0$ moving in the positive $x$ direction with a constant speed. The position of the first body is given by ${x_1}\left( t \right)$ after time $'t';$ and that of the second body by ${x_2}\left( t \right)$ after the same time interval. Which of the following graphs correctly describes $\left( {{x_1} - {x_2}} \right)$ as a function of time $'t'$ ?
A B C D Explanation For the body starting from rest

${x_1} = 0 + {1 \over 2}a{t^2} \Rightarrow {x_1} = {1 \over 2}a{t^2}$

For the body moving with constant speed

${x_2} = vt$

$\therefore$ ${x_1} - {x_2} = {1 \over 2}a{t^2} - vt$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {{d\left( {{x_1} - {x_2}} \right)} \over {dt}} = at - v$

at $t=0,$ $\,\,\,\,\,\,{x_1} - {x_2} = 0$, so graph should start from origin.

For $at < v;$ the slope is negative that means ${x_1} - {x_2}$ < 0 so initially velocity of 1st body is less than second body and velocity of 1st body is increasing gradually.

For $at = v;$ the slope is zero. So ${x_1} - {x_2}$ = 0 it means here velocity of both the bodies are same.

For $at > v;$ the slope is positive. So ${x_1} - {x_2}$ > 0 it means here velocity of first body is greater than second body.

We know the relation between distance and time is.

$S = ut + {1 \over 2}a{t^2}$, which is a equation parabola. So the graph should be a parabola.

These characteristics are represented by graph $(b).$