To solve this, we need to understand the principles of circular motion, especially as they relate to the condition for an object to just maintain contact with the circular path's highest point. When a cone filled with water is revolved in a vertical circle, the water will experience a centripetal force towards the centre of the circle (keeping it in circular motion) and the gravitational force downwards. At the top of the circle, these forces are in the same direction (towards the centre of the circle).

For the water to not fall out of the cone at the top of the circle, the centripetal force (which, in this case, comes from gravity itself) must be greater than or equal to the required centripetal force to keep the water moving in a circle. Mathematically, this condition can be expressed using the formula for centripetal force $$F = \frac{m v^2}{r}$$ and knowing that at the top of the circle, this force is provided by gravity $$F = m g$$ (where $m$ is the mass of the water, $g$ is the acceleration due to gravity (approximately $9.8 m/s^2$), $v$ is the velocity of the water, and $r$ is the radius of the circle).

By equating the gravitational force to the required centripetal force we get:

$$mg = \frac{m v^2}{r}$$

Since the mass $m$ cancels out, we don't need to know the quantity of water to solve this problem. Rearranging the equation for $v$, we find:

$$v^2 = rg$$

And solve for $v$:

$$v = \sqrt{rg}$$

The velocity of the object at the top of the circle can also be linked to the period of revolution $T$ (the time it takes to complete one revolution). The relationship between velocity, radius, and period is given by:

$$v = \frac{2 \pi r}{T}$$

Combining these equations, we can solve for $T$:

$$\sqrt{rg} = \frac{2 \pi r}{T}$$

Solving for $T$ gives us:

$$T = \frac{2 \pi r}{\sqrt{rg}}$$

Plugging in the values for $r = 4 m$ and $g = 9.8 m/s^2$:

$$T = \frac{2 \pi \times 4}{\sqrt{4 \times 9.8}}$$

$$T = \frac{8 \pi}{\sqrt{39.2}}$$

$$T = \frac{8 \pi}{6.26}$$

$$T \approx 4 s$$

Therefore, the correct answer is Option B: 4s.

Circular Motion

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