\(\Delta L=10\log { _{ 10 }\left[ \frac { { I }_{ 1 } }{ { I }_{ 0 } } \right] =50dB } \)
\(\log { _{ 10 }\left[ \frac { { I }_{ 1 } }{ { I }_{ 0 } } \right] =5dB } \)
\(\frac { { I }_{ 1 } }{ { I }_{ 0 } } ={ 10 }^{ 5 }\Rightarrow { I }_{ 1 }={ 10 }^{ 5 }{ I }_{ 0 }\times { 10 }^{ -12 }{ Wm }^{ -2 }\)
I₁ = 10^-7 Wm^-2
Since three musical instruments are played,
therefore, I_total = 3I₁ = 3\(\times\)10^-7 Wm^-2.  

Given y₁ = 5 sin(240\(\pi\)t) and y₂ = 4 sin(244\(\pi\)t)
Comparing with y = A sin(2\(\pi\) f₁t), we get
2\(\pi\)f₁ = 240π \(\Rightarrow\) f₁ = 120Hz
2\(\pi\)f₂ = 244π \(\Rightarrow\) f₂ = 122Hz
The number of beats produced is | f₁ − f₂| = |120 − 122| = |− 2|=2 beats per sec

\(v=\sqrt { \frac { y }{ \rho } } =\sqrt { \frac { 2\times { 10 }^{ 11 } }{ 7800 } } =\sqrt { 0.2564\times { 10 }^{ 8 } } \)
= 0.506\(\times\)10⁴ms^-1= 5\(\times\)10³ms^-1
Therefore, longitudinal waves travel faster in a solid than in a liquid or a gas. Now you may understand why a shepherd checks before crossing railway track by keeping his ears on the rails to safeguard his cattle.

Observer is stationary
(i) Source (train) is moving towards an observer:
Apparent frequency due to train arriving station is
\({ f }_{ in }=\frac { f }{ \left( 1-\frac { { v }_{ s } }{ v } \right) } =\frac { 240 }{ \left( 1-\frac { 8 }{ 330 } \right) } =246Hz\)
(ii) Source (train) is moving away form an observer:
Apparent frequency due to train leaving station is
\({ f }_{ out }=\frac { f }{ \left( 1-\frac { { v }_{ s } }{ v } \right) } =\frac { 240 }{ \left( 1-\frac { 8 }{ 330 } \right) } =234Hz\)
So the number of beats = | f_in -f_out| = (246-234) = 12

Let l₂ be the length of the open organ pipe, with l₁ =30 cm the length of the closed organ pipe.
It is given that the third harmonic of closed organ pipe is equal to the fundamental frequency of open organ pipe.
The third harmonic of a closed organ pipe is 
\({ f }_{ 2 }=\frac { v }{ { \lambda }_{ 2 } } =\frac { 3v }{ { 4l }_{ 1 } } =3{ f }_{ 1 }\)
The fundamental frequency of open organ pipe is \({ f }_{ 1 }=\frac { v }{ { \lambda }_{ 1 } } =\frac { v }{ { 2l }_{ 2 } } \)
Therefore,
\(\frac { v }{ { 2l }_{ 2 } } =\frac { 3v }{ { 4l }_{ 1 } } \Rightarrow { l }_{ 2 }\frac { { 2l }_{ 1 } }{ 3 } =20 \ cm\)