c
(c) $\nu = \frac{P}{{2l}}{\left[ {\frac{F}{m}} \right]^{1/2}}$
$ \Rightarrow {\nu ^2} = \frac{{{P^2}}}{{4{l^2}}}\left[ {\frac{F}{m}} \right]$
$\therefore m \propto \frac{F}{{{l^2}{\nu ^2}}}$
$ \Rightarrow [m] = \left[ {\frac{{ML{T^{ - 2}}}}{{{L^2}{T^{ - 2}}}}} \right] = [M{L^{ - 1}}{T^0}]$