 JEE Mains Previous Years Questions with Solutions

4.5
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1

AIEEE 2009

Three sound waves of equal amplitudes have frequencies $\left( {v - 1} \right),\,v,\,\left( {v + 1} \right).$ They superpose to give beats. The number of beats produced per second will be :
A
$3$
B
$2$
C
$1$
D
$4$

Explanation

Maximum number of beats $= \left( {v + 1} \right) - \left( {v - 1} \right) = 2$
2

AIEEE 2008

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 $\,\,\,\,\,$ $\beta = {{2\pi } \over 2} = \pi$
3

AIEEE 2007

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 $100.$
4

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 $= 105\,Hz.$