 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2006

A string is stretched between fixed points separated by $75.0$ $cm.$ It is observed to have resonant frequencies of $420$ $Hz$ and $315$ $Hz$. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is
A
$105$ $Hz$
B
$1.05$ $Hz$
C
$1050$ $Hz$
D
$10.5$ $Hz$

Explanation

Given ${{nv} \over {2\ell }} = 315$ and $\left( {n + 1} \right){v \over {2\ell }} = 420$

$\Rightarrow {{n + 1} \over n} = {{420} \over {315}} \Rightarrow n = 3$

Hence $3 \times {v \over {2\ell }} = 315 \Rightarrow {v \over {2\ell }} = 105Hz$

Lowest resonant frequency is when $n=1$

Therefore lowest resonant frequency $= 105\,Hz.$
2

AIEEE 2006

A whistle producing sound waves of frequencies $9500$ $Hz$ and above is approaching a stationary person with speed $v$ $m{s^{ - 1}}.$ The velocity of sound in air is $300\,m{s^{ - 1}}.$ If the person can hear frequencies upto a maximum of $10,000$ $HZ,$ the maximum value of $v$ upto which he can hear whistle is
A
$15\sqrt 2 \,\,m{s^{ - 1}}$
B
${{15} \over {\sqrt 2 }}\,m{s^{ - 1}}$
C
$15\,\,m{s^{ - 1}}$
D
$30\,\,m{s^{ - 1}}$

Explanation

$v' = v\left[ {{v \over {v - {v_s}}}} \right] \Rightarrow 10000$

$= 9500\left[ {{{300} \over {300 - v}}} \right]$

$\Rightarrow 300 - v = 300 \times 0.95 \Rightarrow v$

$= 300 - 285 = 15\,m{s^{ - 1}}$
3

AIEEE 2005

An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency ?
A
$0.5\%$
B
zero
C
$20\%$
D
$5\%$

Explanation

$n' = n\left[ {{{v + {v_0}} \over v}} \right] = n\left[ {{{v + {v \over 5}} \over v}} \right] = n\left[ {{6 \over 5}} \right]$

${{n'} \over n} = {6 \over 5};{{n' - n} \over n}$

$= {{6 - 5} \over 5} \times 100 = 20\%$
4

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,

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