Longitudinal Stationary Waves (Organ Pipes) and Resonance Tube Questions in English

(C) Let the length of the open organ pipe be $L_o$ and the length of the closed organ pipe be $L_c$.
Given that $L_o = 2 L_c$.
The fundamental frequency of an open organ pipe is given by $f_o = \frac{v}{2 L_o} = 100 \ Hz$.
From this,we get $\frac{v}{L_o} = 200 \ Hz$.
Substituting $L_o = 2 L_c$,we get $\frac{v}{2 L_c} = 200 \ Hz$,which implies $\frac{v}{L_c} = 400 \ Hz$.
The fundamental frequency of a closed organ pipe is $f_c = \frac{v}{4 L_c} = \frac{400}{4} = 100 \ Hz$.
The harmonics of a closed organ pipe are odd multiples of the fundamental frequency $(f_c, 3f_c, 5f_c, \dots)$.
The third harmonic of the closed pipe is $3 f_c = 3 \times 100 = 300 \ Hz$.

(A) The frequency of the tuning fork is $f = 330 \, Hz$ and the speed of sound is $v = 330 \, m/s$.
The wavelength is $\lambda = \frac{v}{f} = \frac{330}{330} = 1 \, m = 100 \, cm$.
For a tube closed at one end,the resonance condition is $l_n = \frac{(2n-1)\lambda}{4}$,where $n = 1, 2, 3, ...$.
For $n=1$,$l_1 = \frac{\lambda}{4} = \frac{100}{4} = 25 \, cm$.
For $n=2$,$l_2 = \frac{3\lambda}{4} = \frac{300}{4} = 75 \, cm$.
For $n=3$,$l_3 = \frac{5\lambda}{4} = 125 \, cm$ (which is greater than the tube length $120 \, cm$).
The tube is initially resonant at $l_2 = 75 \, cm$ (since $l_1 = 25 \, cm$ is also possible,but the tube length is $120 \, cm$,so the air column length must be $75 \, cm$ to be resonant).
When filling with water,the air column length decreases. The next resonance occurs at $l_1 = 25 \, cm$.
The length of the water column required is $L - l_1 = 120 \, cm - 25 \, cm = 95 \, cm$ or $L - l_2 = 120 \, cm - 75 \, cm = 45 \, cm$.
The minimum length of the water column to get resonance again is $45 \, cm$.

(B) For an open pipe,the fundamental frequency is given by $n_0 = \frac{v}{2l}$.
Given $v = 360\ m/s$ and $l = 1.5\ m$,we have $n_0 = \frac{360}{2 \times 1.5} = \frac{360}{3} = 120\ Hz$.
For a closed pipe,the fundamental frequency is $n_1 = \frac{v'}{4l}$,where $v'$ is the speed of sound in the gas at $30\,^{\circ}C$.
Since both pipes resonate with the same tuning fork,$n_1 = n_0 = 120\ Hz$.
Thus,$120 = \frac{v'}{4 \times 1.5} \Rightarrow 120 = \frac{v'}{6} \Rightarrow v' = 720\ m/s$.
Using the relation $v \propto \sqrt{T}$,the speed of sound in the gas at $0\,^{\circ}C$ $(v_0)$ is related to the speed at $30\,^{\circ}C$ $(v')$ by $v_0 = v' \times \sqrt{\frac{T_0}{T_{30}}}$.
Here $T_0 = 273\ K$ and $T_{30} = 273 + 30 = 303\ K$.
$v_0 = 720 \times \sqrt{\frac{273}{303}} = 720 \times \sqrt{0.90099} \approx 720 \times 0.9492 \approx 683.4\ m/s$.
Rounding to the nearest integer,the speed is $683\ m/s$.

(B) When a high-pressure pulse (compression) reaches an open end,it reflects as a low-pressure pulse (rarefaction) because the pressure at the open end must remain constant (atmospheric pressure). Thus,statement $(II)$ is correct.
When a high-pressure pulse (compression) reaches a closed end,it reflects as a high-pressure pulse (compression) because the air particles cannot move past the boundary,causing a build-up of pressure. Thus,statement $(IV)$ is correct.
Therefore,the correct statements are $(II)$ and $(IV)$.

(A) Let $v$ be the speed of sound in air.
For a pipe $P_1$ closed at one end,the frequency of the first harmonic (fundamental frequency) is given by $f_1 = \frac{v}{4l_1}$.
For a pipe $P_2$ open at both ends,the frequency of the third harmonic is given by $f_2 = \frac{3v}{2l_2}$.
Since both pipes are in resonance with the same tuning fork,their frequencies are equal: $f_1 = f_2$.
Therefore,$\frac{v}{4l_1} = \frac{3v}{2l_2}$.
Simplifying the equation: $\frac{1}{4l_1} = \frac{3}{2l_2}$.
Rearranging for the ratio of lengths: $\frac{l_1}{l_2} = \frac{2}{12} = \frac{1}{6}$.
Thus,the ratio of the lengths of $P_1$ and $P_2$ is $1/6 \approx 0.167$.

(B) When a rod of length $L$ is clamped at its middle,the fundamental mode of longitudinal vibration has a node at the center and antinodes at the free ends.
This corresponds to a standing wave where the length of the rod is equal to half the wavelength: $L = \lambda / 2$.
Given $L = 100 \,cm = 1 \,m$,we have $\lambda = 2L = 2 \,m$.
The speed of sound $v$ is given by the relation $v = f \lambda$,where $f$ is the frequency.
Given $f = 2.53 \,kHz = 2.53 \times 10^3 \,Hz$.
Substituting the values: $v = (2.53 \times 10^3 \,Hz) \times (2 \,m) = 5.06 \times 10^3 \,m/s$.
Since $10^3 \,m = 1 \,km$,the speed is $5.06 \,km/s$.

(A) For an organ pipe closed at one end and open at the other,the resonance condition is given by $L = (2n - 1) \frac{\lambda}{4}$,where $n = 1, 2, 3, \dots$ represents the mode of vibration.
Rearranging for wavelength,we get $\lambda_n = \frac{4L}{2n - 1}$.
For the first three lowest frequencies (fundamental and first two overtones):
For $n = 1$: $\lambda_1 = \frac{4L}{2(1) - 1} = 4L$.
For $n = 2$: $\lambda_2 = \frac{4L}{2(2) - 1} = \frac{4L}{3}$.
For $n = 3$: $\lambda_3 = \frac{4L}{2(3) - 1} = \frac{4L}{5}$.
Thus,the wavelengths are $4L, 4L/3, 4L/5$.

(A) The fundamental frequency of a closed organ pipe is given by $n = \frac{v}{4(L + 0.6r)}$,where $v$ is the speed of sound,$L$ is the length,and $r$ is the radius.
Since $L \gg r$,the end correction $0.6r$ is negligible compared to $L$,so the frequency is approximately $n \approx \frac{v}{4L}$.
If the length $L$ is doubled $(L' = 2L)$ and the radius $r$ is halved $(r' = r/2)$,the new frequency $n'$ is:
$n' = \frac{v}{4(L' + 0.6r')} = \frac{v}{4(2L + 0.6(r/2))} = \frac{v}{4(2L + 0.3r)}$.
Since $L \gg r$,$0.3r$ is still negligible,so $n' \approx \frac{v}{4(2L)} = \frac{1}{2} \left( \frac{v}{4L} \right) = \frac{n}{2}$.
Thus,the frequency is halved.

(B) For a closed organ pipe,the frequency of the $n^{th}$ overtone is given by $f_n = (2n+1) \frac{v}{4L}$.
For the third overtone,$n=3$,so the frequency is $f_3 = (2(3)+1) \frac{v}{4L} = \frac{7v}{4L}$.
The wavelength $\lambda$ is given by $\lambda = \frac{v}{f_3} = \frac{4L}{7}$.
The displacement amplitude $A(x)$ at a distance $x$ from the closed end (which acts as a node for displacement) is given by $A(x) = A_{max} \sin(Kx)$,where $K = \frac{2\pi}{\lambda}$.
Here,$A_{max} = a$ and $K = \frac{2\pi}{4L/7} = \frac{7\pi}{2L}$.
Substituting $x = \frac{L}{7}$ into the equation:
$A(\frac{L}{7}) = a \sin\left(\frac{7\pi}{2L} \cdot \frac{L}{7}\right) = a \sin\left(\frac{\pi}{2}\right) = a(1) = a$.

(B) The fundamental frequency $f_0$ of a closed organ pipe of length $l$ is given by $f_0 = \frac{v}{4l}$,where $v$ is the speed of sound in air.
As water is poured into the pipe,the length of the air column $l$ decreases.
Since $f_0 \propto \frac{1}{l}$,as $l$ decreases,the fundamental frequency $f_0$ increases.
Eventually,the water level reaches a point where the rate of water inflow equals the rate of water outflow through the hole at the bottom.
At this point,the water level becomes constant,and consequently,the length of the air column $l$ becomes constant.
Therefore,the fundamental frequency $f_0$ first increases and then becomes constant.

(B) In a resonance tube experiment,the first resonance occurs when the length of the air column is $L_1 = \lambda/4 + e$,where $e$ is the end correction.
The second resonance occurs when the length of the air column is $L_2 = 3\lambda/4 + e$.
Subtracting the first equation from the second:
$L_2 - L_1 = (3\lambda/4 + e) - (\lambda/4 + e)$
$L_2 - L_1 = 2\lambda/4 = \lambda/2$.
Therefore,the wavelength $\lambda = 2(L_2 - L_1)$.

(A) The fundamental frequency of a pipe closed at one end $(COP)$ of length $L$ is given by $f_1 = \frac{v}{4L} = 412 \, Hz$.
This implies $\frac{v}{L} = 412 \times 4 = 1648 \, Hz$.
When the pipe is cut into two equal pieces of length $L' = \frac{L}{2}$,one piece remains a closed pipe $(COP)$ and the other becomes an open pipe $(OOP)$.
The fundamental frequency of the new closed pipe is $f_{COP} = \frac{v}{4L'} = \frac{v}{4(L/2)} = \frac{v}{2L} = 2 \times \frac{v}{4L} = 2 \times 412 = 824 \, Hz$.
The fundamental frequency of the new open pipe is $f_{OOP} = \frac{v}{2L'} = \frac{v}{2(L/2)} = \frac{v}{L} = 1648 \, Hz$.
Thus,the frequencies are $824 \, Hz$ and $1648 \, Hz$.

(B) For an open organ pipe,the frequency of the $n^{th}$ harmonic is given by the formula: $f = \frac{nv}{2l}$.
Given:
Length $l = 33 \, cm = 0.33 \, m$
Frequency $f = 1000 \, Hz$
Velocity of sound $v = 330 \, m/s$
Substituting the values into the formula:
$1000 = \frac{n \times 330}{2 \times 0.33}$
$1000 = \frac{n \times 330}{0.66}$
$1000 = n \times 500$
$n = \frac{1000}{500} = 2$
Since $n = 2$,the mode of vibration corresponds to the $2^{nd}$ harmonic.

(A) For a resonance column (closed at one end),the resonance lengths are given by:
$l_1 + e = \frac{\lambda}{4}$
$l_2 + e = \frac{3\lambda}{4}$
where $e$ is the end correction.
From the first equation,$\frac{\lambda}{4} = l_1 + e$.
Substituting this into the second equation:
$l_2 + e = 3(l_1 + e)$
$l_2 + e = 3l_1 + 3e$
$l_2 = 3l_1 + 2e$
Since the end correction $e$ is always positive $(e > 0)$,it follows that $l_2 > 3l_1$.

(B) For a pipe closed at one end,the frequency of the $n$-th overtone is given by $f_n = \frac{(2n+1)V}{4L_1}$,where $n$ is the overtone number. For the first overtone $(n=1)$,$f_1 = \frac{3V}{4L_1}$.
For a pipe open at both ends,the frequency of the $n$-th overtone is given by $f_n = \frac{(n+1)V}{2L_2}$. For the third overtone $(n=3)$,$f_3 = \frac{4V}{2L_2} = \frac{2V}{L_2}$.
Since both pipes are in resonance with the same tuning fork,their frequencies are equal: $\frac{3V}{4L_1} = \frac{2V}{L_2}$.
Rearranging for the ratio of lengths: $\frac{L_1}{L_2} = \frac{3}{4 \times 2} = \frac{3}{8} = 0.375$.

(B) The fundamental frequency of an open organ pipe of length $\ell$ is given by $n = \frac{V}{2\ell}$, where $V$ is the speed of sound.
For the two pipes, we have $n_1 = \frac{V}{2\ell_1}$ and $n_2 = \frac{V}{2\ell_2}$.
This implies $\ell_1 = \frac{V}{2n_1}$ and $\ell_2 = \frac{V}{2n_2}$.
When the two pipes are joined in series, the total length of the new pipe is $\ell = \ell_1 + \ell_2$.
The fundamental frequency $n$ of the new pipe is $n = \frac{V}{2\ell} = \frac{V}{2(\ell_1 + \ell_2)}$.
Substituting the values of $\ell_1$ and $\ell_2$, we get $n = \frac{V}{2(\frac{V}{2n_1} + \frac{V}{2n_2})} = \frac{V}{V(\frac{1}{n_1} + \frac{1}{n_2})} = \frac{1}{\frac{n_1 + n_2}{n_1 n_2}} = \frac{n_1 n_2}{n_1 + n_2}$.

(C) The given frequencies are $425 \, Hz$,$595 \, Hz$,and $765 \, Hz$.
Dividing these by their greatest common divisor $(85)$,we get the ratio: $425:595:765 = 5:7:9$.
Since the frequencies are in the ratio of odd integers $(5:7:9)$,the pipe must be a closed organ pipe.
The resonance frequencies for a closed pipe are given by $f_n = \frac{n V}{4l}$,where $n$ is an odd integer $(n = 1, 3, 5, ...)$.
For the first observed frequency,$n = 5$,so $\frac{5 V}{4l} = 425 \, Hz$.
Substituting $V = 340 \, m/s$:
$l = \frac{5 \times 340}{4 \times 425} = \frac{1700}{1700} = 1 \, m$.

(A) In a resonance tube,the first resonance occurs at length $\ell_1 = \frac{\lambda}{4} - e$ and the second resonance occurs at $\ell_2 = \frac{3\lambda}{4} - e$,where $e$ is the end correction.
Subtracting the two lengths: $\ell_2 - \ell_1 = \frac{3\lambda}{4} - \frac{\lambda}{4} = \frac{2\lambda}{4} = \frac{\lambda}{2}$.
Given $\ell_1 = 17 \, cm$ and $\ell_2 = 51 \, cm$,we have $\frac{\lambda}{2} = 51 - 17 = 34 \, cm$.
Therefore,the wavelength $\lambda = 2 \times 34 \, cm = 68 \, cm = 0.68 \, m$.
The speed of sound $v$ is given by $v = f \lambda$,where $f = 512 \, Hz$.
$v = 512 \times 0.68 = 348.16 \, m/s$.
Rounding to the nearest integer,the speed of sound is $348 \, m/s$.

(B) Given: Speed of sound,$v = 340 \, m \, s^{-1}$.
Length of pipe,$L = 17 \, cm = 0.17 \, m$.
Frequency of source,$f = 1.5 \, kHz = 1500 \, Hz$.
For a pipe closed at one end,the resonant frequencies are given by $f_n = \frac{nv}{4L}$,where $n = 1, 3, 5, 7, \dots$ represents the harmonic mode.
Setting $f_n = f$,we have $1500 = \frac{n \times 340}{4 \times 0.17}$.
$1500 = \frac{n \times 340}{0.68}$.
$1500 = n \times 500$.
$n = \frac{1500}{500} = 3$.
Thus,the $3^{\text{rd}}$ harmonic mode resonates with the source.

(D) The tuning fork has a frequency $f = 256\, Hz$. It produces $1\, beat/s$ with the third normal mode of an open pipe. Therefore,the frequency of the third normal mode $f_3$ is $256 \pm 1\, Hz$,which is $255\, Hz$ or $257\, Hz$.
For an open pipe,the frequency of the $N^{th}$ normal mode is given by $f_N = \frac{N v}{2L}$,where $N=3$,$v = 340\, m/s$,and $L$ is the length of the pipe.
Case $1$: $255 = \frac{3 \times 340}{2L} \Rightarrow L = \frac{1020}{510} = 2\, m = 200\, cm$.
Case $2$: $257 = \frac{3 \times 340}{2L} \Rightarrow L = \frac{1020}{514} \approx 1.98\, m = 198\, cm$.
Given the options,$200\, cm$ is the correct value.

(A) For the first resonance,the effective length is $L_1 = \ell_1 + e$,where $\ell_1 = 10 \, cm$ and $e = 1 \, cm$.
So,$L_1 = 10 + 1 = 11 \, cm$.
Since $L_1 = \frac{\lambda}{4}$,we have $\lambda = 4 \times 11 = 44 \, cm$.
For the second resonance,the effective length is $L_2 = \ell_2 + e = \frac{3\lambda}{4}$.
Substituting the value of $\lambda$,we get $L_2 = 3 \times 11 = 33 \, cm$.
Therefore,$\ell_2 = 33 - e = 33 - 1 = 32 \, cm$.

(C) The frequency of a tuning fork is given by $n \propto \frac{1}{\sqrt{m}},$ where $m$ is the mass of the prongs.
When the arms of a tuning fork are filed,the mass $m$ decreases,which causes the frequency $n$ to increase.
In a resonance tube,the resonance condition is $n = \frac{v}{4(L+e)},$ where $L$ is the length of the air column and $e$ is the end correction.
Since $n$ increases,the term $(L+e)$ must decrease to maintain the resonance condition for a constant speed of sound $v.$
Therefore,the length $L$ of the air column should be decreased,not increased.
Thus,Statement $1$ is false and Statement $2$ is true.

(D) For a resonance tube,the resonance condition is $L + e = (2n-1) \frac{\lambda}{4}$,where $e$ is the end correction.
Given $L_1 = 16\,cm$ and $L_2 = 50\,cm$ for $f = 500\,Hz$.
$L_2 - L_1 = \frac{\lambda}{2} \implies 50 - 16 = 34\,cm = \frac{\lambda}{2} \implies \lambda = 68\,cm = 0.68\,m$.
Speed of sound $v = f \lambda = 500 \times 0.68 = 340\,m/s$.
Now,$L_1 + e = \frac{\lambda}{4} \implies 16 + e = \frac{68}{4} = 17 \implies e = 1\,cm = 0.01\,m$.
When the tube is taken out of water,it acts as an open organ pipe of length $L = 60.5\,cm + 2e = 60.5 + 2(1) = 62.5\,cm = 0.625\,m$.
The resonant frequencies of an open pipe are $f_n = \frac{n v}{2L}$.
For $n=1$,$f_1 = \frac{340}{2 \times 0.625} = \frac{340}{1.25} = 272\,Hz$.
For $n=2$,$f_2 = 2 \times f_1 = 544\,Hz$.

(B) The frequency of the tuning fork is $n = 264 \, Hz$. The velocity of sound is $v = 330 \, m/s$.
For a pipe closed at one end,the resonance frequencies are given by $n = \frac{(2k-1)v}{4L}$,where $k = 1, 2, 3, \dots$ is the harmonic number.
The fundamental length $(k=1)$ is $L_1 = \frac{v}{4n} = \frac{330}{4 \times 264} = 0.3125 \, m = 31.25 \, cm$.
The possible resonance lengths are $L = (2k-1) \times 31.25 \, cm$.
For $k=1$,$L = 31.25 \, cm$.
For $k=2$,$L = 3 \times 31.25 = 93.75 \, cm$.
Comparing with the given options,$93.75 \, cm$ is the correct value.

(A) For a closed organ pipe,the resonant frequencies are odd multiples of the fundamental frequency,given by $f_n = (2n + 1)f_0$,where $n = 0, 1, 2, \dots$ and $f_0 = 1500\, Hz$.
We need to find the number of overtones such that $f_n \leq 20,000\, Hz$.
$(2n + 1) \times 1500 \leq 20,000$
$2n + 1 \leq \frac{20,000}{1500} = 13.33$
$2n \leq 12.33 \implies n \leq 6.16$.
Since $n$ must be an integer,the possible values for $n$ are $0, 1, 2, 3, 4, 5, 6$.
Here,$n=0$ corresponds to the fundamental frequency.
The overtones correspond to $n = 1, 2, 3, 4, 5, 6$.
Thus,there are $6$ overtones that can be heard.

(D) For the first resonance,the length of the air column is $L = l + e$,where $l$ is the measured length and $e$ is the end correction.
For the first tuning fork: $f_1 = 512\,Hz$,$l_1 = 11\,cm$.
The condition for resonance is $L_1 = \frac{\lambda_1}{4} = \frac{v}{4f_1} \Rightarrow 11 + e = \frac{v}{4 \times 512} \quad (1)$
For the second tuning fork: $f_2 = 256\,Hz$,$l_2 = 27\,cm$.
The condition for resonance is $L_2 = \frac{\lambda_2}{4} = \frac{v}{4f_2} \Rightarrow 27 + e = \frac{v}{4 \times 256} \quad (2)$
Dividing equation $(1)$ by $(2)$:
$\frac{11 + e}{27 + e} = \frac{256}{512} = \frac{1}{2}$
$22 + 2e = 27 + e \Rightarrow e = 5\,cm$.
Substituting $e = 5$ into equation $(1)$:
$11 + 5 = \frac{v}{4 \times 512} \Rightarrow 16 = \frac{v}{2048}$
$v = 16 \times 2048 = 32768\,cm/s = 327.68\,m/s$.
Rounding to the nearest integer,the velocity is $328\,ms^{-1}$.

(C) In a resonance tube experiment,the speed of sound $\nu$ is related to the frequency $f$ and the difference between two successive resonance lengths $\ell_1$ and $\ell_2$ by the formula: $\nu = 2f(\ell_2 - \ell_1)$.
Given:
Frequency $f = 480\, Hz$
$\ell_1 = 30\, cm = 0.30\, m$
$\ell_2 = 70\, cm = 0.70\, m$
Substituting the values into the formula:
$\nu = 2 \times 480 \times (0.70 - 0.30)$
$\nu = 960 \times 0.40$
$\nu = 384\, m/s$.

(A) The condition for resonance in a closed pipe is given by $L_n = \frac{(2n-1)v}{4f}$,where $n$ is the resonance order,$v$ is the speed of sound,and $f$ is the frequency of the tuning fork.
For the first resonance in winter $(n=1)$: $L_1 = \frac{v_w}{4f} = 18\,cm$.
For the second resonance in summer $(n=2)$: $L_2 = \frac{3v_s}{4f} = x\,cm$.
Since the temperature in summer is higher than in winter,the speed of sound in summer $(v_s)$ is greater than in winter $(v_w)$,i.e.,$v_s > v_w$.
From the first resonance,we have $v_w = 4f \times 18 = 72f$.
Substituting $v_s > v_w$ into the second resonance equation: $x = \frac{3v_s}{4f} > \frac{3v_w}{4f} = 3 \times 18 = 54\,cm$.
Therefore,$x > 54$.

(D) For a pipe of length $L = 0.5\,m$ closed at one end,the resonant frequencies are given by $f_c = \frac{(2n-1)v}{4L}$,where $n = 1, 2, 3, \dots$
Substituting values: $f_c = \frac{(2n-1) \times 340}{4 \times 0.5} = (2n-1) \times 170\,Hz$.
This gives frequencies: $170\,Hz, 510\,Hz, 850\,Hz, \dots$
For a pipe of length $L = 0.5\,m$ open at both ends,the resonant frequencies are given by $f_o = \frac{mv}{2L}$,where $m = 1, 2, 3, \dots$
Substituting values: $f_o = \frac{m \times 340}{2 \times 0.5} = m \times 340\,Hz$.
This gives frequencies: $340\,Hz, 680\,Hz, 1020\,Hz, \dots$
Comparing the two sets of frequencies,there is no common frequency at which both pipes can resonate.

(A) For an open organ pipe of length $l$,the fundamental frequency $n$ is given by the formula:
$n = \frac{v}{2l}$
Given:
Velocity of sound $v = 350\, m/s$
Length of the pipe $l = 0.5\, m$
Substituting the values into the formula:
$n = \frac{350}{2 \times 0.5}$
$n = \frac{350}{1}$
$n = 350\, Hz$
Therefore,the fundamental frequency is $350\, Hz$.

(D) The resonance condition for a closed pipe is given by $L_n = (2n-1) \frac{\lambda}{4} + e$,where $e$ is the end correction.
For two successive resonance lengths $L_1$ and $L_2$,the difference is $\frac{\lambda}{2} = L_2 - L_1$.
Given $L_1 = 33.4 \, cm = 0.334 \, m$ and $L_2 = 101.8 \, cm = 1.018 \, m$.
$\frac{\lambda}{2} = 1.018 - 0.334 = 0.684 \, m$.
Therefore,$\lambda = 2 \times 0.684 = 1.368 \, m$.
The speed of sound $v$ is given by $v = f \lambda$.
$v = 256 \times 1.368 = 350.208 \, ms^{-1}$.
Rounding to the nearest value,$v \approx 350 \, ms^{-1}$.

(B) For an open organ pipe,the frequency of the first harmonic (fundamental frequency) is given by:
$f = \frac{v}{2L} = 480 \, Hz$
For a pipe closed at one end,the frequency of the first harmonic (fundamental frequency) is given by:
$f' = \frac{v}{4L'} = 480 \, Hz$
Since the tuning fork is the same,$f = f'$,therefore:
$\frac{v}{2L} = \frac{v}{4L'}$
By simplifying the equation:
$4L' = 2L$
$L' = \frac{2L}{4} = \frac{L}{2}$
Thus,the length of the pipe closed at one end should be $\frac{L}{2}$.

(C) The given resonance frequencies are $f_1 = 425 \, Hz$,$f_2 = 595 \, Hz$,and $f_3 = 765 \, Hz$.
Taking the ratio of these frequencies: $425 : 595 : 765 = 5 : 7 : 9$.
Since the ratio of successive resonance frequencies is in the ratio of odd integers $(5:7:9)$,this indicates that the pipe is a closed organ pipe (closed at one end).
The resonance frequencies for a closed organ pipe are given by $f_n = \frac{n V}{4 \ell}$,where $n$ is an odd integer $(1, 3, 5, \dots)$.
For the first observed frequency $f_1 = 425 \, Hz$,we have $n = 5$ (since $425/85 = 5$).
Thus,$\frac{5 V}{4 \ell} = 425 \, Hz$.
Given $V = 340 \, m/s$,we substitute the value: $\frac{5 \times 340}{4 \ell} = 425$.
$\frac{1700}{4 \ell} = 425 \Rightarrow \frac{425}{\ell} = 425$.
Therefore,$\ell = 1 \, m$.

(B) The frequency of the $m^{th}$ overtone for a closed organ pipe is given by $f_{c} = \frac{(2m+1)V}{4L_{c}}$,where $m=1$ for the first overtone,so $f_{c} = \frac{3V}{4L_{c}}$.
The frequency of the $m^{th}$ overtone for an open organ pipe is given by $f_{o} = \frac{(m+1)V}{2L_{o}}$,where $m=1$ for the first overtone,so $f_{o} = \frac{2V}{2L_{o}} = \frac{V}{L_{o}}$.
Given that the frequencies are identical,$f_{c} = f_{o}$:
$\frac{3V}{4L_{c}} = \frac{V}{L_{o}}$
Rearranging the terms to find the ratio of lengths $L_{c}:L_{o}$:
$\frac{L_{c}}{L_{o}} = \frac{3}{4}$
Thus,the ratio of their lengths is $3:4$.

(A) The frequency of the tuning fork is $f = 330\,Hz$ and the speed of sound is $v = 330\,m/s$.
The wavelength is $\lambda = \frac{v}{f} = \frac{330}{330} = 1\,m = 100\,cm$.
For a closed organ pipe (resonance tube),the resonance occurs when the length of the air column $l$ satisfies $l = \frac{(2n-1)\lambda}{4}$ for $n = 1, 2, 3, \dots$.
Given the total length of the tube is $L = 120\,cm$,the length of the air column $l$ must be less than or equal to $120\,cm$.
For $n=1$,$l_1 = \frac{\lambda}{4} = \frac{100}{4} = 25\,cm$.
For $n=2$,$l_2 = \frac{3\lambda}{4} = \frac{300}{4} = 75\,cm$.
For $n=3$,$l_3 = \frac{5\lambda}{4} = \frac{500}{4} = 125\,cm$ (which is greater than $120\,cm$,so this is not possible).
The length of the water column $h$ is given by $h = L - l$.
To get the minimum length of the water column,we need the maximum possible length of the air column $l$ that is $\le 120\,cm$.
Thus,we take $l = 75\,cm$.
The minimum water column length is $h = 120\,cm - 75\,cm = 45\,cm$.

(B) Given:
Speed of sound,$v = 330 \, m/s$
Length of pipe,$L = 30 \, cm = 0.3 \, m$
Frequency of source,$f_n = 1.1 \, kHz = 1100 \, Hz$
For an open pipe (open at both ends),the frequency of the $n^{\text{th}}$ harmonic is given by:
$f_n = \frac{n v}{2L}$
Rearranging the formula to solve for $n$:
$n = \frac{2 L f_n}{v}$
Substituting the given values:
$n = \frac{2 \times 0.3 \times 1100}{330}$
$n = \frac{0.6 \times 1100}{330}$
$n = \frac{660}{330} = 2$
Thus,the $2^{\text{nd}}$ harmonic mode resonates with the source.

(C) The fundamental frequency of an open organ pipe is given by $f = \frac{v}{2L}$,where $v$ is the velocity of sound and $L$ is the length of the pipe.
Given lengths are $L_1 = 50 \, cm = 0.5 \, m$ and $L_2 = 50.5 \, cm = 0.505 \, m$.
The beat frequency is the difference between the fundamental frequencies: $f_1 - f_2 = 3 \, Hz$.
Substituting the formula: $\frac{v}{2 \times 0.5} - \frac{v}{2 \times 0.505} = 3$.
$v - \frac{v}{1.01} = 3$.
$v \left( 1 - \frac{1}{1.01} \right) = 3$.
$v \left( \frac{1.01 - 1}{1.01} \right) = 3$.
$v \left( \frac{0.01}{1.01} \right) = 3$.
$v = \frac{3 \times 1.01}{0.01} = 3 \times 101 = 303 \, m/s$.

(B) The fundamental frequency of a vibrating string is given by $f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \frac{m}{L}$.
Given: $T = 240 \, N$,$m = 3 \times 10^{-3} \, kg$,$L = 8 \, m$.
Linear mass density $\mu = \frac{3 \times 10^{-3}}{8} \, kg/m$.
Substituting the values: $f_n = \frac{n}{2 \times 8} \sqrt{\frac{240}{3 \times 10^{-3} / 8}} = \frac{n}{16} \sqrt{\frac{240 \times 8}{3 \times 10^{-3}}} = \frac{n}{16} \sqrt{640000} = \frac{n}{16} \times 800 = 50n \, Hz$.
The audible range is $120 \, Hz \leq 50n \leq 12020 \, Hz$.
Dividing by $50$: $2.4 \leq n \leq 240.4$.
Since $n$ must be an integer,$n$ can take values from $3$ to $240$.
The number of such frequencies is $240 - 3 + 1 = 238$.

(A) For a closed organ pipe,the fundamental frequency is given by $n = \frac{v}{4\ell}$,which implies $n \propto \frac{1}{\ell}$.
Since the pipe with the shorter length has a higher frequency,let $n_1$ be the frequency of the pipe with length $\ell_1 = 20\,cm$ and $n_2$ be the frequency of the pipe with length $\ell_2 = 20.5\,cm$.
Given $n_1 - n_2 = 5\,Hz$,we have $n_1 = n_2 + 5$.
Using the relation $\frac{n_1}{n_2} = \frac{\ell_2}{\ell_1}$,we substitute the values:
$\frac{n_2 + 5}{n_2} = \frac{20.5}{20}$
$20(n_2 + 5) = 20.5 n_2$
$20 n_2 + 100 = 20.5 n_2$
$0.5 n_2 = 100$
$n_2 = \frac{100}{0.5} = 200\,Hz$.
Therefore,$n_1 = 200 + 5 = 205\,Hz$.
The frequencies are $205\,Hz$ and $200\,Hz$.

(A) The fundamental frequency of a closed organ pipe is given by $n = \frac{v}{4(L + 0.6r)}$,where $v$ is the speed of sound,$L$ is the length,and $r$ is the radius.
Since $L \gg r$,the end correction $0.6r$ is often neglected,simplifying the formula to $n \approx \frac{v}{4L}$.
If the length $L$ is doubled $(L' = 2L)$ and the radius $r$ is halved $(r' = r/2)$,the new frequency $n'$ is given by:
$n' = \frac{v}{4(L' + 0.6r')} = \frac{v}{4(2L + 0.6(r/2))} = \frac{v}{4(2L + 0.3r)}$.
Since $L \gg r$,we can approximate $2L + 0.3r \approx 2L$.
Therefore,$n' \approx \frac{v}{4(2L)} = \frac{1}{2} \left( \frac{v}{4L} \right) = \frac{n}{2}$.
Thus,the frequency is halved.

(B) For a resonance tube closed at one end,the resonance lengths $L_1$ and $L_2$ are related to the wavelength $\lambda$ and end correction $e$ as follows:
$L_1 + e = \frac{\lambda}{4}$
$L_2 + e = \frac{3\lambda}{4}$
Dividing the second equation by the first,we get:
$\frac{L_2 + e}{L_1 + e} = 3$
$L_2 + e = 3L_1 + 3e$
$L_2 - 3L_1 = 2e$
$e = \frac{L_2 - 3L_1}{2}$
Given $L_1 = 10\, cm$ and $L_2 = 32\, cm$:
$e = \frac{32 - 3(10)}{2} = \frac{32 - 30}{2} = \frac{2}{2} = 1\, cm$.

(A) For an open organ pipe of length $L$,the fundamental frequency is given by $f = v / (2L)$,where $v$ is the speed of sound in air.
When the pipe is immersed in water up to half its length $(L/2)$,the water acts as a closed end at that point.
Thus,the effective length of the air column becomes $L' = L/2$.
Since the pipe is now closed at one end (by water) and open at the other,it behaves as a closed organ pipe of length $L' = L/2$.
The fundamental frequency of a closed organ pipe is given by $f' = v / (4L')$.
Substituting $L' = L/2$ into the formula,we get $f' = v / (4 * (L/2)) = v / (2L)$.
Comparing this with the original frequency $f = v / (2L)$,we find that $f' = f$.

(C) For an open organ pipe,the fundamental frequency $f$ is given by the formula:
$f = \frac{V}{2l}$
Given:
Velocity of sound $V = 340\, m/s$
Length of the pipe $l = 34\, cm = 0.34\, m$
Substituting the values into the formula:
$f = \frac{340}{2 \times 0.34}$
$f = \frac{340}{0.68}$
$f = 500\, Hz$
Therefore,the fundamental frequency is $500\, Hz$.

(A) The wavelength $\lambda$ of the sound wave is given by $\lambda = \frac{v}{f} = \frac{340\, m/s}{340\, Hz} = 1\, m = 100\, cm$.
The resonant lengths $l$ for a tube closed at one end are given by $l = \frac{(2n-1)\lambda}{4}$ for $n = 1, 2, 3, \dots$.
For $n=1$,$l_1 = \frac{\lambda}{4} = \frac{100\, cm}{4} = 25\, cm$.
For $n=2$,$l_2 = \frac{3\lambda}{4} = \frac{3 \times 100\, cm}{4} = 75\, cm$.
For $n=3$,$l_3 = \frac{5\lambda}{4} = \frac{5 \times 100\, cm}{4} = 125\, cm$.
Since the tube height is $120\, cm$,only $l_1 = 25\, cm$ and $l_2 = 75\, cm$ are possible.
To find the minimum height of water,we need the maximum possible air column length for resonance,which is $l_2 = 75\, cm$.
Minimum height of water $= \text{Total height} - l_2 = 120\, cm - 75\, cm = 45\, cm$.

(D) For an open organ pipe of length $\ell$,the frequency of the $p^{th}$ mode is given by $f_{open} = p \frac{v}{2\ell}$,where $v$ is the speed of sound.
For a closed organ pipe of length $\ell$,the frequency of the $p^{th}$ mode (where $p$ is the harmonic number) is given by $f_{closed} = (2p - 1) \frac{v}{4\ell}$.
To find the ratio of the $p^{th}$ mode frequency of the open pipe to the closed pipe,we divide the two expressions:
$\frac{f_{open}}{f_{closed}} = \frac{p \frac{v}{2\ell}}{(2p - 1) \frac{v}{4\ell}}$
Simplifying the expression:
$\frac{f_{open}}{f_{closed}} = \frac{p}{2\ell} \times \frac{4\ell}{2p - 1} = \frac{2p}{2p - 1}$
Thus,the ratio is $\frac{2p}{2p - 1}$.

(C) The frequency of a closed organ pipe is given by $n = \frac{v}{4L}$.
Given the initial frequency $n_1 = 256\, Hz$ and length $L_1 = 25.4\, cm = 0.254\, m$.
$256 = \frac{v}{4 \times 0.254} \implies v = 256 \times 4 \times 0.254 = 260.096\, m/s$.
When the length is increased by $2\, mm = 0.2\, cm$,the new length $L_2 = 25.4 + 0.2 = 25.6\, cm = 0.256\, m$.
The new frequency $n_2 = \frac{v}{4L_2} = \frac{260.096}{4 \times 0.256} = \frac{260.096}{1.024} = 254\, Hz$.
The number of beats per second is $|n_1 - n_2| = |256 - 254| = 2\, Hz$.

(A) Given lengths are $L_{1} = 50 \, cm = 0.5 \, m$ and $L_{2} = 50.5 \, cm = 0.505 \, m$.
For an open organ pipe,the fundamental frequency is given by $n = \frac{v}{2L}$.
Since $L_{2} > L_{1}$,the frequency $n_{1} > n_{2}$.
The beat frequency is $n_{1} - n_{2} = 3 \, Hz$.
Substituting the expressions for frequency: $\frac{v}{2L_{1}} - \frac{v}{2L_{2}} = 3$.
$\frac{v}{2} \left( \frac{1}{0.5} - \frac{1}{0.505} \right) = 3$.
$\frac{v}{2} \left( \frac{0.505 - 0.5}{0.5 \times 0.505} \right) = 3$.
$\frac{v}{2} \left( \frac{0.005}{0.2525} \right) = 3$.
$v \left( \frac{0.005}{0.505} \right) = 3$.
$v = \frac{3 \times 0.505}{0.005} = 3 \times 101 = 303 \, m/s$.

(A) The fundamental frequency $f$ of an open organ pipe of length $l$ is given by $f = \frac{v}{2l}$,where $v$ is the velocity of sound in the air column.
As the temperature $T$ increases,the velocity of sound $v$ increases according to the relation $v = \sqrt{\frac{\gamma RT}{M}}$.
While the length $l$ of the pipe may increase slightly due to thermal expansion,the increase in the velocity of sound $v$ is much more significant.
Since $f \propto v$ and $f \propto \frac{1}{l}$,the increase in $v$ dominates the effect of the slight increase in $l$,leading to an overall increase in the fundamental frequency $f$.
Therefore,both the Assertion and the Reason are correct,and the Reason provides a correct explanation for the Assertion.

(C) For a resonance column tube closed at one end,the resonance lengths $l_1, l_2, l_3, \dots$ are given by $l_n = (2n-1) \frac{\lambda}{4}$.
The difference between two successive resonance lengths is $\Delta l = l_{n+1} - l_n = \frac{\lambda}{2}$.
Given $l_1 = 9.75 \; cm$ and $l_2 = 31.25 \; cm$,the difference is $\Delta l = 31.25 \; cm - 9.75 \; cm = 21.50 \; cm = 0.215 \; m$.
Thus,$\frac{\lambda}{2} = 0.215 \; m$,which implies $\lambda = 0.43 \; m$.
The speed of sound $v$ is given by $v = f \lambda$.
Substituting the values $f = 800 \; Hz$ and $\lambda = 0.43 \; m$,we get $v = 800 \times 0.43 = 344 \; m/s$.

(B) The speed of sound in a gas is given by $v = \sqrt{\frac{\gamma P}{\rho}}$.
Assuming $\gamma$ and $P$ are the same for the gas and air,we have $\frac{v_{gas}}{v_{air}} = \sqrt{\frac{\rho_{air}}{\rho_{gas}}}$.
Given $\rho_{gas} = 2 \rho_{air}$,we get $\frac{v_{gas}}{300} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Thus,$v_{gas} = \frac{300}{\sqrt{2}} = 150 \sqrt{2} \; m/s$.
For an open organ pipe of length $L$,the fundamental frequency is $f_1 = \frac{v}{2L}$ and the second harmonic is $f_2 = \frac{2v}{2L} = \frac{v}{L}$.
The frequency difference is $\Delta f = f_2 - f_1 = \frac{v}{2L}$.
Substituting the values,$\Delta f = \frac{150 \sqrt{2}}{2(1)} = 75 \sqrt{2} \approx 75 \times 1.414 = 106.05 \; Hz$.
Rounding to the nearest integer,the answer is $106 \; Hz$.