 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2005

When two tuning forks (fork $1$ and fork $2$) are sounded simultaneously, $4$ beats per second are heated. Now, some tape is attached on the prong of the fork $2.$ When the tuning forks are sounded again, $6$ beats per second are heard. If the frequency of fork $1$ is $200$ $Hz$, then what was the original frequency of fork $2$ ?
A
$202$ $Hz$
B
$200$ $Hz$
C
$204$ $Hz$
D
$196$ $Hz$

Explanation

No. of beats heard when fork $2$ is sounded with fork $1$ $= \Delta n = 4$

Now we know that if on loading (attaching tape) an unknown fork, the beat frequency increases (from $4$ to $6$ in this case) then the frequency of the unknown fork $2$ is given by,

$n = {n_0} - \Delta n = 200 - 4 = 196Hz$
2

AIEEE 2004

The displacement $y$ of a particle in a medium can be expressed as, $y = {10^{ - 6}}\,\sin$ $\left( {100t + 20x + {\pi \over 4}} \right)$ $m$ where $t$ is in second and $x$ in meter. The speed of the wave is
A
$20\,\,m/s$
B
$5\,m/s$
C
$2000\,m/s$
D
$5\,\pi \,m/s$

Explanation

From equation given,

$\omega = 100$ and $k = 20,$ $v = {\omega \over k} = {{100} \over {20}} = 5m/s$
3

AIEEE 2003

A tuning fork of known frequency $256$ $Hz$ makes $5$ beats per second with the vibrating string of a piano. The beat frequency decreases to $2$ beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was
A
$256 + 2Hz$
B
$256 - 2Hz$
C
$256 - 5Hz$
D
$256 + 5Hz$

Explanation

A tuning fork of frequency $256$ $Hz$ makes $5$ beats/ second with the vibrating string of a piano. Therefore the frequency of the vibrating string of piano is $\left( {256 \pm 5} \right)$ $Hz$ ie either $261$$Hz or 251 Hz. When the tension in the piano string increases, its frequency will increases. Now since the beat frequency decreases, we can conclude that the frequency of piano string is 251 Hz 4 AIEEE 2003 MCQ (Single Correct Answer) The displacement y of a wave travelling in the x-direction is given by$$y = {10^{ - 4}}\,\sin \left( {600t - 2x + {\pi \over 3}} \right)\,\,metres$\$
where $x$ is expressed in metres and $t$ in seconds. The speed of the wave - motion, in $m{s^{ - 1}}$, is
A
$300$
B
$600$
C
$1200$
D
$200$

Explanation

$y = {10^{ - 4}}\sin \left( {600t - 2x + {\pi \over 3}} \right)$

But $y = A\sin \left( {\omega t - kx + \phi } \right)$

On comparing we get $\omega = 600;\,k = 2$

$v = {\omega \over k} = {{600} \over 2} = 300\,m{s^{ - 1}}$