Final answer:
To find the increase in surface energy when a drop of mercury is split into 8 identical droplets, we calculate the surface energy of the original drop and the surface energy of the individual droplets.
Step-by-step explanation:
To find the increase in surface energy when a drop of mercury is split into 8 identical droplets, we need to calculate the surface energy of the original drop and the surface energy of the individual droplets. The surface energy is given by the formula: Surface energy = Surface tension x Surface area. Since the droplets are identical, the surface area of each droplet will be one-eighth of the surface area of the original drop.
Given that the radius of the original drop is 2 mm, the surface area of the original drop is 4πr² = 4π(0.002)² = 0.05024 mm².
Therefore, the surface area of each droplet is 0.05024 mm² / 8 = 0.00628 mm².
Now, we can calculate the increase in surface energy. Since the surface area of each droplet is one-eighth of the surface area of the original drop, the increase in surface energy is also one-eighth of the surface energy of the original drop.
Given that the surface tension of mercury is 0.465 J/m², the increase in surface energy is 0.465 J/m² x 0.00628 mm² = 0.0029142 J.
The surface area of each droplet is one-eighth of the surface area of the original drop. The increase in surface energy is one-eighth of the surface energy of the original drop.