Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from ground, at the instant when the first drop strikes the ground is _____ m.
$$(g=10m/s^{2}$$

We need to find the height of the fourth drop when the first drop strikes the ground.

The drops are released from a height of 5 m at regular intervals, and when the first drop hits the ground the sixth drop begins to fall.

Since the first drop falls through 5 m, its fall time satisfies $$5 = \frac{1}{2}(10)t^2 \implies t = 1$$ s.

There are five intervals between the first and sixth drop, so the time interval is $$\tau = t/5 = 0.2$$ s.

At t = 1 s, the fourth drop has been falling for $$1 - 3\tau = 1 - 0.6 = 0.4$$ s.

The distance fallen by the fourth drop is $$d = \frac{1}{2}(10)(0.4)^2 = 5 \times 0.16 = 0.8$$ m.

Hence its height from the ground is $$5 - 0.8 = 4.2$$ m.

Therefore, the answer is Option 3: 4.2 m.