1. ⇒ (MHT CET 2023 13th May Morning Shift )

In resonance tube, first and second resonance are obtained at depths $22.7 cm$ and $70.2 cm$ respectively. The third resonance will be obtained at a depth

A. 117.7 cm

B. 92.9 cm

C. 115.5 cm

D. 113.5 cm

Correct Option is (A)

First resonance will occur at $l_{1} + x = \frac{λ}{4}$ .... (i)

Second resonance will occur at $l_{2} + x = \frac{3 λ}{4}$ ..... (ii)

$\begin{aligned} l_{2} + x = 3 (l_{1} + x) \\ l_{2} + x = 3 l_{1} + 3 x \\ 2 x = l_{2} - 3 l_{1} \\ ∴ x = \frac{l_{2} - 3 l_{1}}{2} \\ = \frac{70.2 - 68.1}{2} = 1.05 cm \end{aligned}$

$∴$ Third resonance occurs at $l_{3} + x = \frac{5 λ}{4}$

$\begin{aligned} ∴ l_{3} & = 5 (l_{1} + x) - x \\ = 5 l_{1} + 4 x \\ = 113.5 + 4.2 \\ = 117.7 cm \end{aligned}$

2. ⇒ (MHT CET 2023 12th May Evening Shift )

End correction at open end for air column in a pipe of length ' $l$ ' is ' $e$ '. For its second overtone of an open pipe, the wavelength of the wave is

A. $\frac{2 (l + e)}{3}$

B. $\frac{2 (l + 2 e)}{3}$

C. $\frac{4 (l + e)}{5}$

D. $\frac{4 (l + 2 e)}{5}$

Correct Option is (B)

$\begin{aligned} ∴ & n_{3} = \frac{3 v}{2 l} = \frac{3 v}{2 (l + 2 e)} \\ λ_{3} = \frac{2 (l + 2 e)}{3} \dots . (∵ v = n_{3} λ) \end{aligned}$

3. ⇒ (MHT CET 2023 11th May Morning Shift )

The second overtone of an open pipe has the same frequency as the first overtone of a closed pipe of length ' $L$ '. The length of the open pipe will be

A. $\frac{L}{2}$

B. $L$

C. $2 L$

D. $4 L$

Correct Option is (C)

The length of closed pipe is denoted using L.

Let $l$ be the length of open pipe and $v$ be the velocity.

Frequency of second overtone of an open organ pipe is $n_{o} = \frac{3 v}{2 l}$

Frequency of first overtone of a closed pipe is $n_{c} = \frac{3 v}{4 L}$

Given: $n_{0} = n_{c}$

$\begin{aligned} \frac{3 v}{2 l} & = \frac{3 v}{4 L} \\ L & = \frac{l}{2} \\ ∴ l & = 2 L \end{aligned}$

4. ⇒ (MHT CET 2023 10th May Evening Shift )

The fundamental frequency of air column in pipe 'A' closed at one end is in unison with second overtone of an air column in pipe 'B' open at both ends. The ratio of length of air column in pipe ' $A$ ' to that of air column in pipe ' $B$ ' is

A. $1 : 6$

B. $3 : 8$

C. $2 : 3$

D. $3 : 4$

Correct Option is (A)

Fundamental frequency of a closed pipe, $n_{1} = \frac{v}{4 L_{1}}$

Frequency of the second overtone, $n_{2} = \frac{3 v}{2 L_{2}}$

Given $n_{1} = n_{2}$

$\Rightarrow \frac{v}{4 L_{1}} = \frac{3 v}{2 L_{2}}$

$∴ \frac{L_{1}}{L_{2}} = \frac{1}{6}$

5. ⇒ (MHT CET 2023 10th May Morning Shift )

An open organ pipe having fundamental frequency (n) is in unison with a vibrating string. If the tube is dipped in water so that $75 %$ of the length of the tube is inside the water then the ratio of fundamental frequency of the air column of dipped tube with that of string will be (Neglect end corrections)

A. $1 : 1$

B. $2 : 1$

C. $2 : 3$

D. $3 : 2$

Correct Option is (B)

$n_{open} = \frac{v}{2 L}$ ...... (i)

When dipped in water, pipe becomes closed at one end and open at the other.

Length available for resonance is

$\begin{aligned} l_{1} & = 25 % \times L \\ = L \times \frac{25}{100} \\ = L / 4 \end{aligned}$

$∴ n_{closed} = \frac{v}{4 l_{1}} = \frac{v}{4 \times \frac{L}{4}} = \frac{v}{L} ..... (ii)$

Comparing (i) and (ii),

$∴ \frac{n_{closed}}{n_{open}} = \frac{(\frac{v}{L})}{(\frac{v}{2 L})} = \frac{2}{1}$

⇒Superposition of Waves