In a resonance column first and second resonance are obtained at...

Given, first resonance column at depth, $l_1=24 \mathrm{ cm}$ Second resonance column at depth, $l_2=78 \mathrm{ cm}$ Third resonance column at depths, $l_3=$ ? As $$\begin{aligned}& l_1+x=\frac{\lambda}{4}=24 \quad \quad \ldots (i)\\ & l_2+x=\frac{3 \lambda}{4}=78 \quad \quad \ldots (ii)\\ & l_3+x=\frac{5 \lambda}{4} \quad \quad \ldots (iii) \end{aligned}$$ From $Eqs. (i) and (ii), x=\frac{l_2-3 l_1}{2}=\frac{78-3(24)}{2}=3 \quad \quad \ldots (iv) From Eqs. (iii) and (i), \frac{l_3+x}{l_1+x}=5$ l_3=5 l_1+5 x-x=5 l_1+4 x Substituting $l_1 and x$ value, we get l_3=5(24)+4(3) \Rightarrow l_3=132 $\mathrm{ cm}.$