at the instant t=0 a force F=kt (k is a constant) acts on a small body of mass m resting on a smooth horizontal surface.the time, when body leaves the surface is,

a)mgksinα b)ksinα/mg c)mgsinα/k d)mg/ksinα

From the question what I understand the force is applied at an angle $\alpha $ from the horizontal. The above figure represent the situation. The block will leave the surface when the vertical component of the applied force $\overrightarrow{F}$ will be equal to the blocks weight. Now vertical component of the force is $F\mathrm{sin}\alpha $. So we have

$F\mathrm{sin}\alpha =mg\phantom{\rule{0ex}{0ex}}\Rightarrow Kt\mathrm{sin}\alpha =mg\phantom{\rule{0ex}{0ex}}\Rightarrow t=\frac{mg}{K\mathrm{sin}\alpha}$

d) is the correct answer.

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