if all objects on the equator feels weightlessness then duration of the day will become___

We know that a faster rate of revolution generates a greater centrifugal force which reduces the weight of all things at or near the equator. At a particular limit of speed all objects located on the equator would be weightless.
For the given case, the acceleration due to gravity at the equator is given by,
$$\begin{gathered} \mathrm{g}\text{'}=\mathrm{g}-{\mathrm{R}}_{\mathrm{e}}{\mathrm{\omega }}^2 \\ \mathrm{or} \mathrm{\omega }=\sqrt{\frac{\mathrm{g}}{{\mathrm{R}}_{\mathrm{e}}}}=\sqrt{\frac{9.8}{6.4\times {10}^6}} \\ =1.24\times {10}^{-3}\mathrm{rad}/\mathrm{s} \\ \mathrm{In} \mathrm{this} \mathrm{situation}, \mathrm{the} \mathrm{period} \mathrm{of} \mathrm{rotation} \mathrm{of} \mathrm{the} \mathrm{earth} \mathrm{about} \mathrm{its} \mathrm{axis}, \mathrm{T} \mathrm{or} \mathrm{can} \mathrm{say} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{day} \mathrm{is} \\ \mathrm{T}=\frac{2\mathrm{\pi }}{\mathrm{\omega }} \\ =\frac{2\times 3.14}{1.24\times {10}^{-3}}=5075 \mathrm{seconds} \mathrm{or} 84.6 \min \end{gathered}$$