\(\frac{2}{9}F\)
\(\frac{16}{9}F\)
\(\frac{8}{9}F\)
\(F\)

Let the masses are m and distance between them is l, then
F=\(\frac{Gm^2}{I^2}\)
When \(\frac{1}{3^{rd}}\) mass is transferred to the other then masses will be \(\frac{4m}{3}\) and \(\frac{2m}{3}\).
So new force will be
F′=\(G\frac{4m}{3}\)\(\frac{G\frac{4m}{3}×\frac{2m}{3}}{I^2}\)
=\(\frac{8}{9}\)\(\frac{Gm^2}{I^2}\)
=\(\frac{8}{9}F\)
\(\therefore ,\) The correct option is (C): \(\frac{8}{9}F\)
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