a
frequency of tuning fork $ = \frac{{\sqrt {\frac{{160}}{{\rm{m}}}} }}{{2l}}$
According to the question
$\frac{1}{{2l}}\sqrt {\frac{{160}}{{\rm{m}}}} - \frac{1}{{2l}}\sqrt {\frac{{129.6}}{{\rm{m}}}} = 10$
$\frac{1}{{2l}}\sqrt {\frac{{10}}{m}} [4 - 3.6] = 10 \Rightarrow \frac{1}{{2l}}\sqrt {\frac{{10}}{m}} = 25$
so ${{\rm{V}}_{{\rm{TF}}}} = \frac{1}{{2l}}\sqrt {\frac{{160}}{{\rm{m}}}} = 4\left[ {\frac{1}{{2l}}\sqrt {\frac{{10}}{{\rm{m}}}} } \right] = 100\,{\rm{Hz}}$