A nuclear pore allows free diffusion below about 40 kDa, which is why a larger protein needs a signal and energy to get in. A protein of 70 kDa needs that help. Set the ratio of that protein’s mass to the diffusion limit, to one decimal place.

1.75 (answers within ±0.2 count)

$70/40 = 1.75$ times the limit, so this protein needs a nuclear localisation signal and active transport. The cut-off is soft rather than sharp, but the principle is strict: large cargo requires a signal and energy.