 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2010

A particle is moving with velocity $\overrightarrow v = k\left( {y\widehat i + x\widehat j} \right)$, where K is a constant. The general equation for its path is
A
y = x2 + constant
B
y2 = x + constant
C
xy = constant
D
y2 = x2 + constant

Explanation

$\overrightarrow v = k\left( {y\widehat i + x\widehat j} \right)$ ........(1)

Also $\overrightarrow v = {v_x}\widehat i + {v_y}\widehat j$

$\overrightarrow v = {{dx} \over {dt}}\widehat i + {{dy} \over {dt}}\widehat j$ ........(2)

Equating (1) and (2), we get

${{dx} \over {dt}} = ky\,\,\,\,\,\,$ .......(3)

and $\,\,\,\,\,{{dy} \over {dt}} = kx$ ......(4)

Dividing (3) and (4), we get

${{dy} \over {dx}} = {x \over y}$

$\Rightarrow ydy = xdx$

Integrating both sides of above equation, we get

$\int {ydy} = \int {xdx}$

$\Rightarrow {y^2} = {x^2} +$ constant
2

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

$= {{ - {v^2}} \over R}\cos \theta \widehat i - {{{v^2}} \over R}\sin \theta \widehat j$ 3

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 $(c).$
4

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