JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2009

MCQ (Single Correct Answer)
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
2

AIEEE 2008

MCQ (Single Correct Answer)
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 $(b).$
3

AIEEE 2007

MCQ (Single Correct Answer)
The velocity of a particle is v = v0 + gt + ft2. If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is
A
v0 + g/2 + f
B
v0 + 2g + 3f
C
v0 + g/2 + f/3
D
v0 + g + f

Explanation

Given that, v = v0 + gt + ft2

We know that, $v = {{dx} \over {dt}}$

$\Rightarrow dx = v\,dt$

Integrating, $\int\limits_0^x {dx} = \int\limits_0^t {v\,dt}$

or $\,\,\,\,\,x = \int\limits_0^t {\left( {{v_0} + gt + f{t^2}} \right)} dt$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {{v_0}t + {{g{t^2}} \over 2} + {{f{t^3}} \over 3}} \right]_0^t$

or $\,\,\,\,\,x = {v_0}t + {{g{t^2}} \over 2} + {{f{t^3}} \over 3}$

At $t = 1,\,\,\,\,\,x = {v_0} + {g \over 2} + {f \over 3}.$
4

AIEEE 2007

MCQ (Single Correct Answer)
A particle is projected at $60^\circ$ to the horizontal with a kinetic energy K. The kinetic energy at the highest point is
A
K/2
B
K
C
Zero
D
K/4

Explanation

Let $u$ be the velocity with which the particle is thrown and $m$ be the mass of the particle. Then

$KE = {1 \over 2}m{u^2}.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

At the highest point the velocity is $u$ $\cos \,{60^ \circ }$ (only the horizontal component remains, the vertical component being zero at the top-most point).

Therefore kinetic energy at the highest point,

${\left( {KE} \right)_H} = {1 \over 2}m{u^2}{\cos ^2}60^\circ$

$\,\,\,\,\,\,\,\,\,\,\,\, = {K \over 4}$ [ From eq $(1)$ ]

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