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

\({ v }_{ 1 }=\frac { 1 }{ { l }_{ 1 }{ D }_{ 1 } } \sqrt { \frac { { T }_{ 1 } }{ { \pi \rho }_{ 1 } } } \)
Where, D = diameter of wire
\( { v }_{ 2 }=\frac { 1 }{ { l }_{ 2 }{ D }_{ 2 } } \sqrt { \frac { { T }_{ 2 } }{ { \pi \rho }_{ 2 } } } \)  
\( { l }_{ 1 }={ l }_{ 2 },\ { \rho }_{ 2 }={ \rho }_{ 1 }\)
\( { T }_{ 2 }={ T }_{ 1 },\ { D }_{ 2 }={ 3D }_{ 1 }\)
\(\\ \Rightarrow { V }_{ 2 }=\frac { { V }_{ 1 } }{ 3 } \)
New frequency is \(\frac { 1 }{ 3 } \) rd of the original frequency.

Conceptual question based on fundamentals of characteristics of travelling wave.
The converse is not true means if the function can be represented in the form  y = f( x \(\pm \) vt ), it does not necessarily express a travelling wave. As the essential condition for a travelling wave is that the vibrating particle must have finite displacement value for all x and t.
(i) For x = 0
If t \(\rightarrow \)0, then (x - vt)²\(\rightarrow \)0 which is finite, hence, it is a wave as it passes the two tests.
(ii) log \(\left( \frac { x+vt }{ x_{ 0 } } \right)\)
l\( \\ At\ x=0\ and\ t=0,\)
\( f(x,t)=log\left( \frac { 0+0 }{ x_{ 0 } } \right) \)
= log 0 \(\rightarrow\) not defined
Hence, it is not a wave. 
(iii) \(\frac { 1 }{ x+vt } \)
\( \\ For\ x=0,\ t=0,\ f(x)\rightarrow \infty\)
Though the function is of (x\(\pm\) vt) type still at x = 0, it is infinite, hence, it is not a wave.

(a) Conceptual question based on fundamentals of characteristics of travelling wave.
The converse is not true means if the function can be represented in the form  y = f( x \(\pm \) vt ), it does not necessarily express a travelling wave. As the essential condition for a travelling wave is that the vibrating particle must have finite displacement value for all x and t.
\(\frac { 1 }{ x+vt } \)
\(\\ For\ x=0,\ t=0,\ f(x)\rightarrow \infty \)
Though the function is of (x±vt) type still at x=0, it is infinite, hence, it is not a wave.
(b) Conceptual question based on fundamentals of characteristics of travelling wave.
The converse is not true means if the function can be represented in the form  y = f( x ± vt ), it does not necessarily express a travelling wave. As the essential condition for a travelling wave is that the vibrating particle must have finite displacement value for all x and t.
For x = 0,
If t →0, then (x-vt)²→0 which is finite, hence, it is a wave as it passes the two tests.