l = 12m, M = 2.10 kg, T = 2.06\(\times \)10⁴N, v =?
\(\mu =\frac { M }{ l } =\frac { 2.10 }{ 12 } kg/m\)
\(\\ v=\sqrt { \frac { T }{ \mu } } =\sqrt { \frac { 2.06\times { 10 }^{ 4 } }{ 2.10/12 } } =3.43\times { 10 }^{ 2 }m/s\)

Frequency of A, v₀ = 512 Hz
Number of beats/s = 5
Frequency of B = 512\(\pm \) 5 = 517 or 517 Hz
On loading its frequency decreases from 571 to 507, so that number of beats/s remain 5.
Hence, frequency of B when not loaded = 517Hz.

Length of pipe(l) = 20 cm = 20 x 10^-2 m
\({ v }_{ funda }=\frac { v }{ 4L } =\frac { 330 }{ 4\times 20\times { 10 }^{ -2 } }\)
\( \\ { v }_{ funda }=\frac { 330\times 100 }{ 80 } =412.5\quad Hz\)
\(\\ \frac { { v }_{ given } }{ { v }_{ funda } } =\frac { 1237.5 }{ 412.5 } =3\)
Hence, 3rd harmonic mode of the pipe is resonantly excited by the source of given frequency.

On reflection from the denser medium, there will be a phase change of 180⁰
Net amplitude = \(\frac { 2 }{ 3 } \times 0.6=0.4\)
Hence, equation of reflected wave will be 
y = 0.4sin\(2\pi \left[ t+\frac { x }{ 2 } +\pi \right] \)
\( =0.4sin2\pi (t+{ x }/{ 2 })\)

We know that, speed,  \(v\infty \sqrt { T } \)
By formula v = \(\frac { xRT }{ p } \)
Where T is in kelvin 
\(\frac { v_{ t } }{ v_{ 0 } } =\sqrt { \frac { 273+t }{ 273+0 } } =3\)
\(\\ \Rightarrow \frac { 273+t }{ 273 } =9\Rightarrow \ t=9\times 273-273=2184^{ 0 }C\)