### JEE Mains Previous Years Questions with Solutions

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### AIEEE 2008

MCQ (Single Correct Answer)
A wave travelling along the $x$-axis is described by the equation $y(x, t)=0.005$ $\cos \,\left( {\alpha \,x - \beta t} \right).$ If the wavelength and the time period of the wave are $0.08$ $m$ and $2.0s$, respectively, then $\alpha$ and $\beta$ in appropriate units are
A
$\alpha = 25.00\pi ,\,\beta = \pi$
B
$\alpha = {{0.08} \over \pi },\,\beta = {{2.0} \over \pi }$
C
$\alpha = {{0.04} \over \pi },\,\beta = {{1.0} \over \pi }$
D
$\alpha = 12.50\pi ,\,\beta = {\pi \over {2.0}}$

## Explanation

$y\left( {x,t} \right) = 0.005\,\cos \left( {\alpha x - \beta t} \right)$ (Given)

Comparing it with the standard equation of wave

$y\left( {x,t} \right) = a\cos \left( {kx - \omega t} \right)$ we get

$k = \alpha$ $\,\,\,\,\,$ and $\,\,\,\,\,$ $\omega = \beta$

$\therefore$ ${{2\pi } \over \gamma } = \alpha$ $\,\,\,\,\,$ and $\,\,\,\,\,$ ${{2\pi } \over T} = \beta$

$\therefore$ $\alpha = {{2\pi } \over {0.08}} = 25\pi$ $\,\,\,\,\,$ and $\,\,\,\,\,$ $\beta = {{2\pi } \over 2} = \pi$
2

### AIEEE 2007

MCQ (Single Correct Answer)
A sound absorber attenuates the sound level by $20$ $dB$. The intensity decreases by a factor of
A
$100$
B
$1000$
C
$10000$
D
$10$

## Explanation

We have, ${L_1} = 10\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_0}}}} \right);$

${L_2} = 10\,\log \left( {{{{{\rm I}_2}} \over {{{\rm I}_0}}}} \right)$

$\therefore$ $\,\,{L_1} - {L_2} = 10\,\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_0}}}} \right) - 10\,\log \left( {{{{{\rm I}_2}} \over {{{\rm I}_0}}}} \right)$

or, $\Delta L = 10\,\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_0}}} \times {{{{\rm I}_0}} \over {{{\rm I}_2}}}} \right)$

or, $\Delta L = 10\,\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_2}}}} \right)$

or, $20 = 10\log \left( {{{{{\rm I}_1}} \over {{{\rm I}_2}}}} \right)$

or, $2 = \log \left( {{{{{\rm I}_1}} \over {{{\rm I}_2}}}} \right)$

or, ${{{{\rm I}_1}} \over {{{\rm I}_2}}} = {10^2}$

or, ${{\rm I}_2} = {{{{\rm I}_1}} \over {100}}.$

$\Rightarrow$ Intensity decreases by a factor $100.$
3

### AIEEE 2006

MCQ (Single Correct Answer)
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 $= 105\,Hz.$
4

### AIEEE 2006

MCQ (Single Correct Answer)
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

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

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

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

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

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