Correct Answer is 1

Given that the first three resonance frequencies are in the ratio of $1:3:5$, we can express them as follows:

$f_1=kf$ $f_3=3kf$ $f_5=5kf$

Where $k$ is a constant and $f_1,f_3$, and $f_5$ are the first, third, and fifth resonance frequencies, respectively. We are given that the frequency of the fifth harmonic is $405\mathrm{Hz}$, so we can write:

$f_5=5kf=405\mathrm{Hz}$

Now we can solve for the constant $k$:

$k=\frac{405}{5}=81\mathrm{Hz}$

We also know that the speed of sound in air is $v=324{\mathrm{ms}}^{-1}$. The relationship between the speed of sound, the frequency, and the wavelength of a standing wave in a closed pipe can be expressed as follows:

$v=f\lambda$

Where $\lambda$ is the wavelength of the wave. For the first harmonic in a closed pipe, the length of the pipe is equal to one-fourth of the wavelength:

$L=\frac{1}{4}\lambda$

We can now substitute the expression for the wavelength in terms of the length into the equation for the speed of sound:

$v=f_1\cdot 4L$

Now, we can substitute the value of $f_1=kf=81\mathrm{Hz}$ and the speed of sound $v=324{\mathrm{ms}}^{-1}$ into the equation:

$324=81\times 4L$

Now we can solve for the length of the organ pipe $L$:

$L=\frac{324}{81\times 4}=\frac{324}{324}=1\mathrm{m}$

The length of the organ pipe is $1\mathrm{m}$.

Correct Answer is 900

An organ pipe that is open at both ends resonates at all harmonics, including the fundamental (first harmonic), second harmonic, third harmonic, etc.

The frequency $f$ of the $n$-th harmonic for a pipe open at both ends is given by:

$f_n=\frac{nv}{2L}$,

where:

• $n$ is the number of the harmonic,
• $v$ is the speed of sound, and
• $L$ is the length of the pipe.

To find the frequency of the second harmonic ($n=2$), we can substitute the given values into the formula:

$f_2=\frac{2\times 360}{2\times 0.4}=900\text{Hz}$.

Therefore, the frequency of the second harmonic is $900\text{Hz}$.

Correct Answer is 104

$400=\frac{v}{4(L_1+e)}$ ..... (i)

$400=\frac{5v}{4(L_2+e)}$ ..... (ii)

$\Rightarrow L_1+e=\frac{\lambda }{4}=21$ cm

$L_2+e=\frac{5\lambda }{4}=105$ cm

$\Rightarrow$ e = 1 cm & L₂ = 104 cm

Correct Answer is 50

Given $340=\frac{n}{4\times 125}v$

$\Rightarrow n=5$

So $\lambda =100$ cm

So minimum height is $\frac{\lambda }{2}=50$ cm

Correct Answer is 80

texatom>$2\times \left(\frac{V}{2L_0}\right)=\left(\frac{V}{4L_c}\right)$

$\Rightarrow L_0=4L_c$

$=4\times 20$

$=80$ cm