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JEE Mains Previous Years Questions with Solutions

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1

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 $$100.$$
2

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 $$ = 105\,Hz.$$
3

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

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

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

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

$$ = 300 - 285 = 15\,m{s^{ - 1}}$$
4

AIEEE 2005

MCQ (Single Correct Answer)
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

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

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

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

Questions Asked from Waves

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