Velocity of sound $$ = \sqrt {\frac{{\gamma RT}}{M}} $$
When water vapour are present in air, average molecular weight of air decreases and hence velocity increases.

Maximum number of beats
$$ = \left( {v + 1} \right) - \left( {v - 1} \right) = 2$$

If $$T$$ is the time period, then time required for a point to move from maximum displacement to zero displacement is $$\frac{T}{4}.$$
$$\eqalign{ & {\text{So,}}\,\,\frac{T}{4} = 0.170\,\,{\text{or}}\,\,T = 0.170 \times 4 \cr & = 0.680\,s \cr} $$
Therefore, the frequency of wave is
$$n = \frac{1}{T} = \frac{1}{{0.680}} = 1.47\,Hz$$

Standard equation of plane progressive harmonic wave is
$$y = a\sin \frac{{2\,\pi }}{\lambda }\left( {vt - x} \right)\,......\left( {\text{i}} \right)$$
Given equation is
$$y = 0.5\sin \frac{{2\,\pi }}{{3.2}}\left( {64\,t - x} \right)\,......\left( {{\text{ii}}} \right)$$
Comparing Eq. (ii) with Eq. (i)
$$\eqalign{ & v = 64\,\,{\text{and}}\,\,\lambda = 3.2 \cr & \therefore {\text{Frequency }}\nu = \frac{v}{\lambda } = \frac{{64}}{{3.2}} \cr & = 20\,Hz \cr} $$

In this problem, the fundamental frequencies of each part could be find. The fundamental frequency of the complete wire could be find. One should check each option for the given values.

For I^st part, $${n_1} = \frac{v}{{2{l_1}}} \Rightarrow {l_1} = \frac{v}{{2{n_1}}}$$
For II^nd part, $${n_2} = \frac{v}{{2{l_2}}} \Rightarrow {l_2} = \frac{v}{{2{n_2}}}$$
For III^rd part, $${n_3} = \frac{v}{{2{l_3}}} \Rightarrow {l_3} = \frac{v}{{2{n_3}}}$$
For the complete wire, $$n = \frac{v}{{2l}} \Rightarrow l = \frac{v}{{2n}}$$
We have, $$l = {l_1} + {l_2} + {l_3}$$
$$\eqalign{ & \frac{v}{{2n}} = \frac{v}{{2{n_1}}} + \frac{v}{{2{n_2}}} + \frac{v}{{2{n_3}}} \cr & \frac{1}{n} = \frac{1}{{{n_1}}} + \frac{1}{{{n_2}}} + \frac{1}{{{n_3}}} \cr} $$

For first resonance
$${\ell _1} + e = \frac{\lambda }{4}$$

For second resonance
$${\ell _2} + e = \frac{{3\,\lambda }}{4}$$
$$\eqalign{ & {\text{But, }}\nu = \nu \lambda \cr & \therefore \,\,\nu = \nu \frac{4}{3}\left( {{\ell _2} + e} \right) \cr & \Rightarrow \,\,{\ell _2} + e = \frac{{3\nu }}{{4\nu }}\,\,\,.....\left( {\text{i}} \right) \cr & \therefore \,\,\nu = \nu 4\left( {{\ell _1} + e} \right) \cr & \Rightarrow \,\,{\ell _1} + e = \frac{\nu }{{4\nu }}\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Subtracting (i) and (ii),
$$\eqalign{ & \nu = 2\nu \left( {{\ell _2} - {\ell _1}} \right) \cr & \therefore \,\,\Delta \nu = 2\nu \left( {\Delta {\ell _2} + \Delta {\ell _1}} \right) \cr & = 2 \times 512 \times \left( {0.1 + 0.1} \right)\,cm/s \cr & = 204.8\,cm/s \cr} $$

$$\eqalign{ & T = \mu {v^2} = \mu \frac{{{\omega ^2}}}{{{k^2}}} = 0.04\frac{{{{\left( {\frac{{2\pi }}{{0.004}}} \right)}^2}}}{{{{\left( {\frac{{2\pi }}{{0.50}}} \right)}^2}}} \cr & = 6.25\,N \cr} $$

Frequency of first overtone in closed pipe,
$$v = \frac{{3v}}{{4{\ell _1}}}\sqrt {\frac{P}{{{\rho _1}}}} \,......\left( {\text{i}} \right)$$
Frequency of first overtone in open pipe,
$$v' = \frac{1}{{{\ell _2}}}\sqrt {\frac{P}{{{\rho _2}}}} \,......\left( {{\text{ii}}} \right)$$
From equation (i) and (ii)
$$ \Rightarrow {\ell _2} = \frac{4}{3}{\ell _1}\sqrt {\frac{{{\rho _1}}}{{{\rho _2}}}} $$

According to Hooke's law $${F_R} \propto x$$
[Restoring force $${F_R} = T,$$  tension of spring]
Velocity of sound by a stretched string $$v = \sqrt {\frac{T}{m}} $$   where $$m$$ is the mass per unit length
$$\eqalign{ & {\text{Hence}}\,\,v \propto \sqrt T \,\,{\text{or,}}\,\,\frac{v}{{v'}} = \sqrt {\frac{T}{{T'}}} \cr & {\text{or}}\,\,v' = v\sqrt {\frac{{T'}}{T}} = v\sqrt {\frac{{1.5x}}{x}} = 1.22\,v \cr} $$

Here, original frequency of sound, $${f_0} = 100\,Hz$$
Speed of source $${V_s} = 19.4\cos {60^ \circ } = 9.7$$

From Doppler's formula
$$\eqalign{ & f = {f_0}\left( {\frac{{V - {V_0}}}{{V - {V_s}}}} \right) \cr & f = 100\left( {\frac{{V - 0}}{{V - \left( { + 9.7} \right)}}} \right) \cr & f = 100\frac{V}{{V\left( {1 - \frac{{9.7}}{V}} \right)}} = \frac{{100}}{{\left( {1 - \frac{{9.7}}{{330}}} \right)}} = 103\,Hz \cr} $$
Apparent frequency $$f = 103\,Hz$$