Transverse Stationary Waves and Sonometer Questions in English

(D) The fundamental frequency of a vibrating wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$, where $\mu$ is the mass per unit length.
Since the tension $T$ and the mass per unit length $\mu$ (as the wire is the same) remain constant, we have $n \propto \frac{1}{l}$.
Therefore, $\frac{n_1}{n_2} = \frac{l_2}{l_1}$.
Given $n_1 = 1 \ kHz$ and $n_2 = 1.001 \ kHz$, we have $\frac{l_2}{l_1} = \frac{1}{1.001}$.
Using the thermal expansion formula $l_2 = l_1(1 - \alpha \Delta t)$, where $\Delta t = 30^{\circ} C - 10^{\circ} C = 20^{\circ} C$:
$\frac{l_1}{1.001} = l_1(1 - \alpha \times 20)$.
$1 - 20\alpha = \frac{1}{1.001} \approx 1 - 0.001$.
$20\alpha = 0.001$.
$\alpha = \frac{0.001}{20} = 0.5 \times 10^{-4} /^{\circ} C$.

(B) Given, length of string, $l = 100 \,cm = 1 \,m$.
The resonant frequencies are given as $f_1 = 120 \,Hz, f_2 = 200 \,Hz, f_3 = 280 \,Hz$.
The resonant frequencies of a string fixed at both ends are given by $f_n = n f_0$, where $f_0$ is the fundamental frequency.
The fundamental frequency $f_0$ is the greatest common divisor $(GCD)$ of the given resonant frequencies.
$f_0 = \text{GCD}(120, 200, 280) = 40 \,Hz$.
The fundamental frequency for a string fixed at both ends is $f_0 = \frac{v}{2l}$.
Therefore, the speed of the wave is $v = 2 l f_0$.
Substituting the values, $v = 2 \times 1 \,m \times 40 \,Hz = 80 \,m/s$.

(B) The resonant frequencies are $160 \,Hz$, $240 \,Hz$, and $400 \,Hz$. The ratio of these frequencies is $160:240:400$, which simplifies to $2:3:5$.
Since the frequencies are harmonics of the fundamental frequency $f_0$, we can write $f_n = n f_0$, where $n$ is an integer.
Given the ratio $2:3:5$, the fundamental frequency $f_0$ must be $160/2 = 80 \,Hz$ (or $240/3 = 80 \,Hz$ or $400/5 = 80 \,Hz$).
The fundamental frequency of a string fixed at both ends is given by $f_0 = \frac{v}{2L}$.
Substituting the values $f_0 = 80 \,Hz$ and $v = 160 \,m/s$:
$80 = \frac{160}{2L}$
$80 = \frac{80}{L}$
$L = 1 \,m = 100 \,cm$.

(D) The fundamental frequency $f$ of a stretched string is given by the formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.
Given $f_1 = 300 \ Hz$, $L_1 = L$, and $T_1 = T_1$. Thus, $300 = \frac{1}{2L} \sqrt{\frac{T_1}{\mu}}$.
Given $f_2 = 100 \ Hz$, $L_2 = 2L$, and $T_2 = T_2$. Thus, $100 = \frac{1}{2(2L)} \sqrt{\frac{T_2}{\mu}} = \frac{1}{4L} \sqrt{\frac{T_2}{\mu}}$.
Dividing the two equations: $\frac{300}{100} = \frac{\frac{1}{2L} \sqrt{\frac{T_1}{\mu}}}{\frac{1}{4L} \sqrt{\frac{T_2}{\mu}}}$.
$3 = \frac{4L}{2L} \sqrt{\frac{T_1}{T_2}} = 2 \sqrt{\frac{T_1}{T_2}}$.
$1.5 = \sqrt{\frac{T_1}{T_2}}$.
Squaring both sides: $2.25 = \frac{T_1}{T_2}$, which is $\frac{9}{4} = \frac{T_1}{T_2}$.
Therefore, $\frac{T_2}{T_1} = \frac{4}{9}$.

(D) The fundamental frequency of a vibrating string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.
Initially, $f_1 = \frac{1}{2L_1} \sqrt{\frac{T_1}{\mu}}$.
Given that the length is decreased by $25 \%$, the new length $L_2 = L_1 - 0.25L_1 = 0.75L_1 = \frac{3}{4}L_1$.
The new frequency $f_2$ increases by $100 \%$, so $f_2 = f_1 + 1.00f_1 = 2f_1$.
Using the frequency formula for the new state: $f_2 = \frac{1}{2L_2} \sqrt{\frac{T_2}{\mu}}$.
Substituting $L_2$ and $f_2$: $2f_1 = \frac{1}{2(\frac{3}{4}L_1)} \sqrt{\frac{T_2}{\mu}} = \frac{4}{6L_1} \sqrt{\frac{T_2}{\mu}} = \frac{2}{3L_1} \sqrt{\frac{T_2}{\mu}}$.
Dividing the expression for $f_2$ by $f_1$: $\frac{2f_1}{f_1} = \frac{\frac{2}{3L_1} \sqrt{\frac{T_2}{\mu}}}{\frac{1}{2L_1} \sqrt{\frac{T_1}{\mu}}}$.
$2 = \frac{2}{3} \cdot 2 \cdot \sqrt{\frac{T_2}{T_1}} = \frac{4}{3} \sqrt{\frac{T_2}{T_1}}$.
$\sqrt{\frac{T_2}{T_1}} = 2 \cdot \frac{3}{4} = \frac{3}{2}$.
Squaring both sides, $\frac{T_2}{T_1} = (\frac{3}{2})^2 = \frac{9}{4}$.

(C) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.
Initially, $L_1 = 0.4 \,m$, $f_1 = 250 \,Hz$, and tension is $T_1 = T$.
So, $250 = \frac{1}{2 \times 0.4} \sqrt{\frac{T}{\mu}} \quad ...(i)$
Finally, $L_2 = 0.4 + 0.1 = 0.5 \,m$, and the new tension is $T_2 = T/4$.
So, $f_2 = \frac{1}{2 \times 0.5} \sqrt{\frac{T/4}{\mu}} \quad ...(ii)$
Dividing equation $(ii)$ by $(i)$:
$\frac{f_2}{250} = \frac{\frac{1}{2 \times 0.5} \sqrt{\frac{T}{4\mu}}}{\frac{1}{2 \times 0.4} \sqrt{\frac{T}{\mu}}} = \frac{0.4}{0.5} \times \sqrt{\frac{1}{4}} = \frac{4}{5} \times \frac{1}{2} = \frac{2}{5} = 0.4$
$f_2 = 250 \times 0.4 = 100 \,Hz$.

(C) The fundamental frequency of a vibrating wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.
Since the tension $T$ and mass per unit length $\mu$ are the same for all segments, the frequency $n$ is inversely proportional to the length $l$ of the segment $(n \propto \frac{1}{l})$.
Given the ratio of fundamental frequencies is $n_1 : n_2 : n_3 = 1 : 2 : 3$.
Therefore, the ratio of their lengths is $l_1 : l_2 : l_3 = \frac{1}{n_1} : \frac{1}{n_2} : \frac{1}{n_3} = \frac{1}{1} : \frac{1}{2} : \frac{1}{3} = 6 : 3 : 2$.
The total length of the wire is $L = l_1 + l_2 + l_3 = 99 \ cm$.
Sum of the ratio parts $= 6 + 3 + 2 = 11$.
Calculating individual lengths:
$l_1 = \frac{6}{11} \times 99 = 54 \ cm$
$l_2 = \frac{3}{11} \times 99 = 27 \ cm$
$l_3 = \frac{2}{11} \times 99 = 18 \ cm$
Thus, the lengths are $54 \ cm, 27 \ cm, 18 \ cm$.

(B) The frequency of a vibrating wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$. Since tension $T$ and mass per unit length $\mu$ are constant, we have $n \propto \frac{1}{l}$, which implies $nl = \text{constant}$.
Let $n$ be the frequency of the tuning fork.
At length $l_1 = 100 \,cm$, the frequency of the wire is $n_1 = n + 8$ (since beats are heard).
At length $l_2 = 101 \,cm$, the frequency of the wire is $n_2 = n - 8$ (as frequency decreases with increasing length).
Using $n_1 l_1 = n_2 l_2$:
$(n + 8) \times 100 = (n - 8) \times 101$
$100n + 800 = 101n - 808$
$101n - 100n = 800 + 808$
$n = 1608 \,Hz$.

(C) The frequency of the $n$-th harmonic for a string is given by $f_n = \frac{n}{2l} \sqrt{\frac{T}{\mu}}$, where $\mu = \pi r^2 \rho$.
For the first overtone of $A$ $(n=2)$: $f_{A} = \frac{2}{2l_A} \sqrt{\frac{T}{\pi r_A^2 \rho}} = \frac{1}{l_A r_A} \sqrt{\frac{T}{\pi \rho}}$.
For the second overtone of $B$ $(n=3)$: $f_{B} = \frac{3}{2l_B} \sqrt{\frac{T}{\pi r_B^2 \rho}} = \frac{3}{2l_B r_B} \sqrt{\frac{T}{\pi \rho}}$.
Given $f_A = f_B$ and $r_A = 2r_B$:
$\frac{1}{l_A (2r_B)} = \frac{3}{2l_B r_B}$.
Simplifying, $\frac{1}{2l_A} = \frac{3}{2l_B} \Rightarrow \frac{l_A}{l_B} = \frac{1}{3}$.
Thus, the ratio $l_A : l_B = 1 : 3$.

(C) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
Since $T$ (tension) and $m$ (mass per unit length) are constant for the same wire, $n \propto \frac{1}{l}$.
Given the ratio of frequencies $n_1 : n_2 : n_3 = 1 : 3 : 4$, the lengths of the segments must be in the inverse ratio:
$l_1 : l_2 : l_3 = \frac{1}{1} : \frac{1}{3} : \frac{1}{4}$.
To simplify this ratio, multiply by the least common multiple of $1, 3, 4$, which is $12$:
$l_1 : l_2 : l_3 = 12 : 4 : 3$.
The sum of the parts is $12 + 4 + 3 = 19$.
The total length of the wire is $114 \ cm$.
Therefore, the lengths are:
$l_1 = \frac{12}{19} \times 114 = 12 \times 6 = 72 \ cm$
$l_2 = \frac{4}{19} \times 114 = 4 \times 6 = 24 \ cm$
$l_3 = \frac{3}{19} \times 114 = 3 \times 6 = 18 \ cm$.
Thus, the lengths are $72 \ cm, 24 \ cm, 18 \ cm$.

(A) The fundamental frequency $f$ of a vibrating wire is given by the formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density.
Since $L$ and $\mu$ are constant for the same wire, we have $f \propto \sqrt{T}$.
Therefore, the ratio of frequencies is given by: $\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}}$.
Given $f_1 = 450 \,Hz$, $T_1 = 9 \,kg-wt$, and $f_2 = 900 \,Hz$.
Substituting the values: $\frac{900}{450} = \sqrt{\frac{T_2}{9}}$.
$2 = \sqrt{\frac{T_2}{9}}$.
Squaring both sides: $4 = \frac{T_2}{9}$.
$T_2 = 4 \times 9 = 36 \,kg-wt$.

(C) Given: $T_A = T_B$, $f_A = f_B$, $L_A = 80 \text{ cm}$, $L_B = x \text{ cm}$.
Ratio of densities: $\frac{d_A}{d_B} = 0.81$.
Ratio of diameters: $\frac{D_A}{D_B} = 2$.
Linear mass density $\mu = \text{Area} \times \text{density} = \frac{\pi D^2}{4} \times d$.
Therefore, $\frac{\mu_A}{\mu_B} = \left(\frac{D_A}{D_B}\right)^2 \times \frac{d_A}{d_B} = (2)^2 \times 0.81 = 4 \times 0.81 = 3.24$.
The fundamental frequency of a string is $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Since $f_A = f_B$ and $T_A = T_B$, we have $L_A \sqrt{\mu_A} = L_B \sqrt{\mu_B}$.
$\frac{L_B}{L_A} = \sqrt{\frac{\mu_A}{\mu_B}} = \sqrt{3.24} = 1.8$.
$x = 80 \times 1.8 = 144 \text{ cm}$.

(D) The frequency of a sonometer wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$. Since $f$ and $\mu$ are constant, we have $l \propto \sqrt{T}$.
Thus, $\frac{l_{\text{air}}}{l_{\text{water}}} = \sqrt{\frac{T_{\text{air}}}{T_{\text{water}}}}$.
In air, the tension $T_{\text{air}} = mg$. When immersed in water, the buoyant force $F_B$ acts upwards, so $T_{\text{water}} = mg - F_B$.
Given specific gravity $\sigma = 8$, the density of iron $\rho = 8 \rho_w$. The buoyant force $F_B = V \rho_w g = \frac{m}{\rho} \rho_w g = \frac{m}{8 \rho_w} \rho_w g = \frac{mg}{8}$.
Therefore, $T_{\text{water}} = mg - \frac{mg}{8} = \frac{7}{8} mg = \frac{7}{8} T_{\text{air}}$.
Substituting this into the ratio: $\frac{l_{\text{air}}}{l_{\text{water}}} = \sqrt{\frac{T_{\text{air}}}{\frac{7}{8} T_{\text{air}}}} = \sqrt{\frac{8}{7}}$.
Given $l_{\text{air}} = \frac{1}{\sqrt{7}} \times 1 \ m = \frac{1}{\sqrt{7}} \ m$.
Then $l_{\text{water}} = l_{\text{air}} \times \sqrt{\frac{7}{8}} = \frac{1}{\sqrt{7}} \times \sqrt{\frac{7}{8}} = \frac{1}{\sqrt{8}} \ m$.

(A) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$, where $l$ is the length, $T$ is the tension, and $m$ is the mass per unit length.
Initial frequency: $n_1 = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
New length: $l' = l - 0.40l = 0.6l$.
New tension: $T' = T + 0.44T = 1.44T$.
New frequency: $n_2 = \frac{1}{2l'} \sqrt{\frac{T'}{m}} = \frac{1}{2(0.6l)} \sqrt{\frac{1.44T}{m}}$.
Taking the ratio: $\frac{n_2}{n_1} = \frac{\frac{1}{2(0.6l)} \sqrt{\frac{1.44T}{m}}}{\frac{1}{2l} \sqrt{\frac{T}{m}}} = \frac{l}{0.6l} \times \sqrt{\frac{1.44T}{T}} = \frac{1}{0.6} \times \sqrt{1.44} = \frac{1.2}{0.6} = 2$.
Therefore, the ratio of the final to initial frequency is $2: 1$.

(B) The frequency of the $p$-th harmonic (or $(p-1)$-th overtone) for a string of length $l$, radius $r$, density $\rho$, and tension $T$ is given by $f = \frac{p}{2l} \sqrt{\frac{T}{\pi r^2 \rho}} = \frac{p}{2lr} \sqrt{\frac{T}{\pi \rho}}$.
For string $A$, the first overtone is the second harmonic $(p=2)$:
$f_A = \frac{2}{2l_A r_A} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{l_A r_A} \sqrt{\frac{T}{\pi \rho}}$.
For string $B$, the second overtone is the third harmonic $(p=3)$:
$f_B = \frac{3}{2l_B r_B} \sqrt{\frac{T}{\pi \rho}}$.
Given $f_A = f_B$ and $r_A = 2r_B$:
$\frac{1}{l_A r_A} = \frac{3}{2l_B r_B}$.
Substituting $r_A = 2r_B$:
$\frac{1}{l_A (2r_B)} = \frac{3}{2l_B r_B}$.
$\frac{1}{2l_A} = \frac{3}{2l_B}$.
$\frac{l_A}{l_B} = \frac{1}{3}$.
Thus, the ratio of the lengths $l_A : l_B = 1 : 3$.

(D) The frequency $n$ of a stretched string is given by the formula $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$, where $l$ is the length, $T$ is the tension, and $m$ is the mass per unit length.
From this, we have the relation $n \propto \frac{\sqrt{T}}{l}$.
Let the initial frequency be $n_1 = n$ and the final frequency be $n_2 = 2n$.
Let the initial length be $l_1 = l$ and the final length be $l_2 = \frac{3}{4}l$.
Using the proportionality, we get $\frac{n_1}{n_2} = \frac{l_2}{l_1} \sqrt{\frac{T_1}{T_2}}$.
Substituting the values: $\frac{n}{2n} = \frac{\frac{3}{4}l}{l} \sqrt{\frac{T_1}{T_2}}$.
$\frac{1}{2} = \frac{3}{4} \sqrt{\frac{T_1}{T_2}}$.
Squaring both sides: $\frac{1}{4} = \frac{9}{16} \frac{T_1}{T_2}$.
$\frac{T_2}{T_1} = \frac{9}{16} \times 4 = \frac{9}{4}$.
Thus, the tension must be changed by a factor of $\frac{9}{4}$.

(A) The fundamental frequency of a string fixed at both ends is given by $v = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$, where $n$ is the harmonic number, $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.
Linear mass density $\mu = \frac{M}{L}$.
Substituting $\mu$ into the formula:
$v = \frac{n}{2L} \sqrt{\frac{T}{M/L}} = \frac{n}{2L} \sqrt{\frac{TL}{M}}$.
Wait, let us re-evaluate: $v = \frac{n}{2L} \sqrt{\frac{T}{M/L}} = \frac{n}{2} \sqrt{\frac{T}{M/L \cdot L^2}} = \frac{n}{2} \sqrt{\frac{T}{ML}}$.
Thus, the correct formula is $v = \frac{n}{2} \sqrt{\frac{T}{ML}}$.

(B) The frequency of the $n^{th}$ harmonic for a stretched string is given by $f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.
Linear mass density $\mu = \rho \cdot A = \rho \cdot \pi r^2$, where $\rho$ is the density of the material and $r$ is the radius.
Thus, $f_n = \frac{n}{2L} \sqrt{\frac{T}{\rho \pi r^2}} = \frac{n}{2Lr} \sqrt{\frac{T}{\rho \pi}}$.
For the first wire: $n_1 = 2$ (first overtone is the second harmonic), $r_1 = 2r_2$, $L_1 = L_1$.
$f_{1, \text{overtone}} = \frac{2}{2L_1 r_1} \sqrt{\frac{T}{\rho \pi}} = \frac{1}{L_1 (2r_2)} \sqrt{\frac{T}{\rho \pi}}$.
For the second wire: $n_2 = 3$ (second overtone is the third harmonic), $r_2 = r_2$, $L_2 = L_2$.
$f_{2, \text{overtone}} = \frac{3}{2L_2 r_2} \sqrt{\frac{T}{\rho \pi}}$.
Given $f_{1, \text{overtone}} = f_{2, \text{overtone}}$:
$\frac{1}{2L_1 r_2} = \frac{3}{2L_2 r_2}$.
$\frac{1}{L_1} = \frac{3}{L_2} \text{ implies } \frac{L_1}{L_2} = \frac{1}{3}$.

(A) For a sonometer wire fixed at both ends, the frequency of the $n^{th}$ harmonic is given by $f_n = n f_1$, where $f_1$ is the fundamental frequency.
In a sonometer wire, the harmonics are $1^{st}$ (fundamental), $2^{nd}$, $3^{rd}$, $4^{th}$, etc.
The overtones are defined as frequencies higher than the fundamental frequency.
$1^{st}$ overtone = $2^{nd}$ harmonic $(n=2)$
$2^{nd}$ overtone = $3^{rd}$ harmonic $(n=3)$
$3^{rd}$ overtone = $4^{th}$ harmonic $(n=4)$
For the $n^{th}$ harmonic, the number of nodes is $(n+1)$ and the number of antinodes is $n$.
For the $4^{th}$ harmonic $(n=4)$:
Number of nodes = $4 + 1 = 5$
Number of antinodes = $4$
Therefore, there are $5$ nodes and $4$ antinodes.

(B) The fundamental frequency of a sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Since $T$ and $\mu$ are constant, $f \propto \frac{1}{L}$.
Let the initial length be $L_1 = L$ and the initial frequency be $f_1 = 50 \text{ Hz}$.
The new length is $L_2 = L + 0.25L = 1.25L = \frac{5}{4}L$.
The new fundamental frequency $f_2$ is given by $\frac{f_2}{f_1} = \frac{L_1}{L_2} = \frac{L}{1.25L} = \frac{1}{1.25} = 0.8$.
So, $f_2 = 0.8 \times 50 \text{ Hz} = 40 \text{ Hz}$.
The frequency of the second harmonic is $f'_n = 2f_n$.
Initial second harmonic frequency $f'_1 = 2 \times 50 = 100 \text{ Hz}$.
New second harmonic frequency $f'_2 = 2 \times 40 = 80 \text{ Hz}$.
The change in frequency is $\Delta f' = 80 - 100 = -20 \text{ Hz}$.
The percentage change is $\frac{\Delta f'}{f'_1} \times 100 = \frac{-20}{100} \times 100 = -20\%$.
Thus, the frequency decreases by $20\%$.

(C) The frequency of a vibrating string is given by $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$, where $n$ is the number of loops (antinodes), $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.
Since the tuning fork is the same, the frequency $f$ remains constant.
Also, $L$ and $\mu$ are constant, so $n \propto \sqrt{T}$.
Given $T = Mg$, we have $n \propto \sqrt{M}$.
For the first case: $n_1 = 4$ and $M_1 = 1 \text{ kg}$.
For the second case: $n_2 = 2$ and $M_2 = M$.
Thus, $\frac{n_1}{n_2} = \sqrt{\frac{M_1}{M_2}}$.
Substituting the values: $\frac{4}{2} = \sqrt{\frac{1}{M}}$.
$2 = \sqrt{\frac{1}{M}}$.
Squaring both sides: $4 = \frac{1}{M}$.
Therefore, $M = \frac{1}{4} \text{ kg} = 0.25 \text{ kg}$.
Wait, re-evaluating the relationship: $n \propto \sqrt{T} \implies n^2 \propto T \implies n^2 \propto M$.
So, $n_1^2 M_2 = n_2^2 M_1$.
$4^2 \times M = 2^2 \times 1$.
$16M = 4$.
$M = \frac{4}{16} = 0.25 \text{ kg}$.
Given the options provided, there might be a misunderstanding in the question's premise or options. If the number of antinodes decreases, the tension must increase. Let's re-check: $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$. If $n$ decreases from $4$ to $2$, then $\sqrt{T}$ must increase by a factor of $2$, meaning $T$ must increase by a factor of $4$. Thus $M = 4 \times 1 \text{ kg} = 4 \text{ kg}$.

(C) The fundamental frequency $f$ of a stretched string is given by the formula $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density.
From this relation, we see that $f \propto \sqrt{T}$.
Let the initial frequency be $f_1$ and the final frequency be $f_2 = f_1 + 0.40f_1 = 1.4f_1$.
Let the initial tension be $T_1$ and the final tension be $T_2$.
Since $f \propto \sqrt{T}$, we have $\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{1.4f_1}{f_1} = \sqrt{\frac{T_2}{T_1}} \implies 1.4 = \sqrt{\frac{T_2}{T_1}}$.
Squaring both sides: $(1.4)^2 = \frac{T_2}{T_1} \implies 1.96 = \frac{T_2}{T_1}$.
Thus, $T_2 = 1.96T_1$.
The percentage increase in tension is $\frac{T_2 - T_1}{T_1} \times 100\% = (1.96 - 1) \times 100\% = 0.96 \times 100\% = 96\%$.

(A) In a sonometer wire fixed at both ends, the frequency of the $n^{th}$ harmonic is given by $f_n = n f_1$, where $f_1$ is the fundamental frequency.
The $n^{th}$ harmonic corresponds to the $(n-1)^{th}$ overtone.
Given that the wire vibrates in the third overtone, we have $n-1 = 3$, which implies $n = 4$.
For the $n^{th}$ harmonic, the number of loops formed is $n = 4$.
The number of antinodes is equal to the number of loops, which is $4$.
The number of nodes in a string fixed at both ends is $n+1 = 4+1 = 5$.
Therefore, the number of antinodes is $4$ and the number of nodes is $5$.

(A) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $T$ is the tension in the wire.
When the bob is hanging in air, the tension $T_1 = Mg$, so $n_1 \propto \sqrt{Mg}$.
When the bob is immersed in water, the buoyant force acts upwards, so the effective weight (tension) becomes $T_2 = Mg - V\rho_w g = Mg(1 - \frac{V\rho_w}{M}) = Mg(1 - \frac{\rho_w}{\rho_m})$, where $\rho_m$ is the density of the metal and $\rho_w$ is the density of water.
Thus, $n_2 \propto \sqrt{Mg(1 - \frac{\rho_w}{\rho_m})}$.
Taking the ratio: $\frac{n_1}{n_2} = \sqrt{\frac{1}{1 - \frac{\rho_w}{\rho_m}}} = \sqrt{\frac{\rho_m}{\rho_m - \rho_w}}$.
Squaring both sides: $\frac{n_1^2}{n_2^2} = \frac{\rho_m}{\rho_m - \rho_w}$.
Let the relative density $\sigma = \frac{\rho_m}{\rho_w}$. Then $\frac{n_1^2}{n_2^2} = \frac{\sigma}{\sigma - 1}$.
$n_1^2(\sigma - 1) = n_2^2 \sigma \implies \sigma(n_1^2 - n_2^2) = n_1^2$.
Therefore, $\sigma = \frac{n_1^2}{n_1^2 - n_2^2}$.

(B) The frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, which implies $n \propto \frac{1}{L}$.
Let the frequency of the tuning fork be $f$. The frequency of the wire is $n = \frac{k}{L}$.
Case $1$: $L_1 = 33 \text{ cm}$, beats = $5$. So, $n_1 = f \pm 5 = \frac{k}{33}$.
Case $2$: $L_2 = 32 \text{ cm}$, beats = $5$. So, $n_2 = f \pm 5 = \frac{k}{32}$.
Since $L$ decreases, $n$ increases. Thus, $n_2 > n_1$.
If $n_1 = f - 5$ and $n_2 = f + 5$, then $\frac{k}{32} = f + 5$ and $\frac{k}{33} = f - 5$.
Subtracting the two equations: $k(\frac{1}{32} - \frac{1}{33}) = 10 \Rightarrow k(\frac{1}{1056}) = 10 \Rightarrow k = 10560$.
Now, $f - 5 = \frac{10560}{33} = 320 \Rightarrow f = 325 \text{ Hz}$.

(C) The fundamental frequency of a stretched wire is given by the formula: $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $\mu$ is the linear mass density.
Linear mass density $\mu = \text{Area} \times \text{Density} = (\pi r^2) \rho = \pi (d/2)^2 \rho = \frac{\pi d^2 \rho}{4}$.
Substituting $\mu$ into the frequency formula: $n = \frac{1}{2L} \sqrt{\frac{T}{\frac{\pi d^2 \rho}{4}}} = \frac{1}{Ld} \sqrt{\frac{T}{\pi \rho}}$.
For the first wire: $n_A = \frac{1}{Ld} \sqrt{\frac{T}{\pi \rho}}$.
For the second wire: $n_B = \frac{1}{(2L)(3d)} \sqrt{\frac{2T}{\pi (2\rho)}} = \frac{1}{6Ld} \sqrt{\frac{T}{\pi \rho}}$.
Taking the ratio $n_B : n_A = \frac{\frac{1}{6Ld} \sqrt{\frac{T}{\pi \rho}}}{\frac{1}{Ld} \sqrt{\frac{T}{\pi \rho}}} = \frac{1}{6}$.
Thus, the ratio $n_B : n_A$ is $1 : 6$.