\(n = \frac{p}{{2l}}\sqrt {\frac{T}{m}} \)==> \(p\sqrt T = \)constant ==> \(\frac{{{p_1}}}{{{p_2}}} = \sqrt {\frac{{{T_2}}}{{{T_1}}}} \)
Hence \(\frac{4}{6} = \sqrt {\frac{{{T_2}}}{{(50 + 15)gm{\rm{ - }}force}}} \)==> \({T_2} = 28.8\,gm{\rm{ - }}f\)
Hence weight removed from the pan
\( = {T_1} - {T_2} = 65 - 28.8 = 3.62\,gm-force = 0.036 kg-f.\)