Frequency is not affected by blowing of wind so long as source and listener are stationary.

From purely theoretical considerations, Newton came to the conclusion that velocity of longitudinal waves through any medium; solid, liquid or gas depends upon the elasticity and density of the medium. Newton gave the formula
$$v = \sqrt {\left( {\frac{E}{\rho }} \right)} $$
where
$$v =$$  velocity of sound in the medium
$$E =$$  coefficient of elasticity in the medium
$$\rho = $$  density of the (undisturbed) medium

In case of closed organ pipe frequency,
$$\eqalign{ & {f_n} = \left( {2n + 1} \right)\frac{v}{{4l}} \cr & {\text{for}} \cr & n = 0,{f_0} = 100\,Hz \cr & n = 1,{f_1} = 300\,Hz \cr & n = 2,{f_2} = 500\,Hz \cr & n = 3,{f_3} = 700\,Hz \cr & n = 4,{f_4} = 900\,Hz \cr & n = 5,{f_5} = 1100\,Hz \cr & n = 6,{f_6} = 1300\,Hz \cr} $$
Hence possible natural oscillation whose frequencies $$ < 1250\,Hz = 6\left( {n = 0,1,2,3,4,5} \right)$$

As we know that
Beat frequency $$ = {f_1} \sim {f_2} = n - \left( {n - 1} \right) = 1$$
and similarly for $$n$$ and $$n + 1$$
Beat frequency $$= n + 1 - n = 1$$

A pulse of a wave train when travels along a stretched string and reaches the fixed end of the string, then it will be reflected back to the same medium and the reflected ray suffers a phase change of $$\pi $$ with the incident wave and wave velocity after reflection will reverse.

KEY CONCEPT : The fundamental frequency for closed organ pipe is given by $${{\upsilon _c}} = \frac{v}{{4\,\ell }}$$   and
For open organ pipe is given by $${{\upsilon _0}} = \frac{v}{{2\,\ell }}$$
$$\therefore \frac{{{\upsilon _0}}}{{{\upsilon _c}}} = \frac{v}{{2\,\ell }} \times \frac{{4\,\ell }}{v} = \frac{2}{1}$$

Let the sound observed by the parachutist at $${t_0} = 12s$$   be produced at $${t_1}s.$$  Velocity of source at the instant of sound $$ = g{t_1}$$  and velocity of observer at the instant of observing same sound $$ = g{t_0}.$$  Hence the relation between apparent frequency $$f'$$ and original frequency $$f$$ will be $$f' = f\left( {\frac{{v + g{t_0}}}{{v - g{t_1}}}} \right).$$
$${\text{Here}}\,f = 800\,Hz,g = 10\,m/{s^2},$$

$$\eqalign{ & v = 330\,m/s,\,{t_0} = 12s\,{\text{and}} \cr & f' = 800 + 700 = 1500\,Hz \cr} $$
Putting these, we get, $${t_1} = 9s$$
Now the distance travelled by sound in $$\left( {{t_0} - {t_1}} \right)\sec $$   is
$$v\left( {{t_0} - {t_1}} \right) = \left( {h - \frac{1}{2}gt_0^2} \right) + \left( {h - \frac{1}{2}gt_1^2} \right)$$
Putting the values, we get, $$h = 1057.5\,m.$$

$$\eqalign{ & {\ell _1} + x = \frac{\lambda }{4}\,\,{\text{or, }}\lambda = 4\left( {{\ell _1} + x} \right) \cr & \left( {{\ell _2} + x} \right) = \frac{{3\lambda }}{4}\,\,{\text{or, }}\lambda = \frac{4}{3}\left( {{\ell _2} + x} \right) \cr & \therefore \,\,{\nu _1} = \frac{v}{{{\lambda _1}}} = \frac{v}{{4\left( {{\ell _1} + x} \right)}} \cr & \therefore \,\,{\nu _2} = \frac{v}{{{\lambda _2}}} = \frac{{3v}}{{4\left( {{\ell _2} + x} \right)}} \cr & {\text{Given, }}{\nu _1} = {\nu _2} \cr} $$

$$\eqalign{ & {\text{or, }}\frac{v}{{4\left( {{\ell _1} + x} \right)}} = \frac{{3\,v}}{{4\left( {{\ell _2} + x} \right)}} \cr & {\text{or, }}x = 0.025\,m \cr} $$

Frequency does not depend upon radius. As length is doubled, fundamental frequency becomes half.

Considering the end correction $$\left[ {e = 0.3\,D} \right],$$   we get
$$\eqalign{ & L + e = \frac{\lambda }{4} \cr & \Rightarrow \,\,L = \frac{\lambda }{4} - e \cr & = \frac{{336 \times 100}}{{512 \times 4}} - 0.3 \times 4 \cr & = 15.2\,cm\left[ {\because \,\,\lambda = \frac{v}{\upsilon }} \right] \cr} $$