In a resonance column, first and second resonances are obtained at depths $22.7\, cm$ and $70.2\, cm .$ The third resonance will be obtained at a depth (in $cm$)
AIIMS 2019, Diffcult
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First resonance,
$l_{1}+x=\frac{\lambda}{4}=22.7 \ldots \ldots( I )$
Second resonance,
$l_{2}+x=\frac{3 \lambda}{4}=70.2 \ldots \ldots$ $(II)$
Third resonance,
$l_{3}+x=\frac{5 \lambda}{4} \ldots \ldots$ $(III)$
From equations $(I)$ and $(II),$
$x=\frac{l_{2}-3 l_{1}}{2}$
$x=\frac{70.2-68.1}{2}$
$x=1.05 cm$
Now from equations $(II)$ and $(III),$
$\frac{l_{3}+x}{l_{1}+x}=5$
$\frac{l_{3}+1.05}{68.1+1.05}=5$
After simplification,
$l_{3}=117.7 cm$
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