QJEEM19Phe9cd: JEE Main 2019 Physics - Class 11 JEEM19Phe9cd Solution: A resonance tube is old and has jagged end. It is still used in …

In first resonance, length of air column $= {\lambda \over 4}$ .

So, ${l_1} + e = {\lambda \over 4}$ or $11 \times 4 + 4e = \lambda$

So, speed of sound is

$\Rightarrow v = {f_1}\lambda = 512(44 + 4e)$ ...... (i)

And in second case,

$l{'_1} + e = {{\lambda '} \over 4}$ or $27 \times 4 + 4e = \lambda '$

$\Rightarrow v = {f_2}\lambda ' = 256(108 + 4e)$ ..... (ii)

Dividing both Eqs. (i) and (ii), we get

$1 = {{512(44 + 4e)} \over {256(108 + 4e)}} \Rightarrow e = 5$ cm

Substituting value of e in Eq. (i), we get

Speed of sound $v = 512(44 + 4e)$

$= 512(44 + 4 \times 5)$

$= 512 \times 64$ cm s ^{$-$ 1} = 327.68 ms ^{$-$ 1} $\approx$ 328 ms ^{$-$ 1}