(4)

The fundamental frequency $\left(f\right)$ is related to the length of the pipe $\left(L\right)$ and speed of sound in air $\left(v\right)$ as

          $f=\frac{v}{\lambda }$

For a pipe open at both ends, the fundamental frequency corresponding to a wavelength $\lambda$ that is twice the length of the pipe is

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

When the pipe is dipped vertically in water so that half of its length is submerged, the new length of the air column becomes

        $L\text{'}=\frac{L}{2}$                                                           ...(ii)

For the new length of the air column, the wavelength $\lambda \text{'}$ is given by

          $\lambda \text{'}=4L\text{'}$

         $\lambda \text{'}=4\times \frac{L}{2}$                                  $\text{(Using (ii))}$

         $\lambda \text{'}=2L$

The new fundamental frequency,

          $f\text{'}=\frac{v}{\lambda \text{'}}=\frac{v}{2L}$                                                     ...(iii)

Comparing eq. (i) and (iii), we get

        $f\text{'}=f$

(3)

$4^{\text{th}}$ overtone of closed organ pipe $=9{\upsilon }_0=\frac{9\times \nu }{4L}$              ...(i)

3rd overtone of open organ pipe $=4{\upsilon }_0=\frac{4\times \nu }{2L\text{'}}$                  ...(ii)

As per question, $9{\upsilon }_0=4{\upsilon }_0$

From equations (i) and (ii), $\frac{9\times \nu }{4L}=\frac{4\times \nu }{2L\text{'}}$

$\Rightarrow \frac{L}{L\text{'}}=\frac{9}{8}$

(4)

Fundamental frequency for open organ pipe,

$f_o=\frac{v}{2L_o}$                                             ...(i)

Fundamental frequency for closed organ pipe,

$f_c=\frac{v}{4L_c}$                                            ...(ii)

$\frac{f_o}{f_c}=\frac{v}{2L_o}\times \frac{4L_c}{v}\Rightarrow \frac{f_o}{f_c}=\frac{2}{1}$         $(\because L_o=L_c\text{ (given)})$

(3)

Frequency $\left(\upsilon \right)$ = 800 Hz

As the pipe is closed at one end, so

$l_3-l_2=l_2-l_1=\frac{\lambda }{2}=21.5$$\text{cm}$

$\therefore$   $\lambda =43.0$ $\text{cm}$

As $\upsilon =\frac{v}{\lambda }\Rightarrow v=\upsilon \lambda$

$\therefore$   $v=\frac{800\times 43}{100}=344$$\text{m}$${\text{s}}^{-1}$