Q72 / 100 · BITSAT 2022 · Physics
In a resonance column first and second resonance are obtained at depths $24 \mathrm{~cm}$ and $78 \mathrm{~cm}$ the third resonance will be obtained at depth.
- A$160 \mathrm{~cm}$
- Correct answerB$132 \mathrm{~cm}$
- C$131 \mathrm{~cm}$
- D$152 \mathrm{~cm}$
Solution
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
$$
l_3=5(24)+4(3) \Rightarrow l_3=132 \mathrm{~cm} .
$$
Hence, the correct answer is option: 'B'.