So with \(v = n\lambda \), if \(v\) increases \(n\) will increase at
\({27^o}C,\,{v_1} = n\lambda \), at \({31^o}C\,,\,\,{v_2} = (n + x)\lambda \)
Now using \(v \propto \sqrt T \) \((\because v= \sqrt{\frac{{\gamma \,R\,T}}{{M}}})\)
\(\frac{{{v_2}}}{{{v_1}}} = \sqrt {\frac{{{T_2}}}{{{T_1}}}} = \frac{{n + x}}{n}\)
==> \(\frac{{300 + x}}{{300}} = \sqrt {\frac{{(273 + 31)}}{{(273 + 27)}}} = \sqrt {\frac{{304}}{{300}}} = \sqrt {\frac{{300 + 4}}{{300}}} \)
==> \(1 + \frac{x}{{300}} = {\left( {1 + \frac{4}{{300}}} \right)^{1/2}} = \left( {1 + \frac{1}{2} \times \frac{4}{{300}}} \right)\,\) \([\because {(\,1 + x)^n} = 1 + nx]\)
==> \(x = 2.\)
$(a)$ $\left(x^2-v t\right)^2$
$(b)$ $\log \left[\frac{(x+v t)}{x_0}\right]$
$(c)$ $e^{\left\{-\frac{(x+v t)}{x_0}\right\}^2}$
$(d)$ $\frac{1}{x+v t}$