Question 122: Properties of Matter

Question

Two small drops of mercury each of radius R coalesce to form a single large drop. The ratio of total surface energy before and after the change is :

Accepted Answer

The volume of a sphere is given by $\frac{4}{3}\pi R^3$.
So the volume of the two small mercury drops each of radius $R$ is $2	imes \frac{4}{3}\pi R^3$.
When they coalesce to form a larger drop, the volume is conserved. So, the volume of the larger drop is also $2	imes \frac{4}{3}\pi R^3$.
Let's denote the radius of this large drop as $R'$.
Therefore, $2	imes \frac{4}{3}\pi R^3 = \frac{4}{3}\pi {R'}^3$.
Solving for $R'$, we get $R' = 2^{1/3}R$.
Now, the surface energy of a sphere is proportional to its surface area, and the surface area of a sphere is given by $4\pi R^2$.
So, the ratio of total surface energy before and after the change is:
$$\frac{{2	imes 4\pi R^2}}{{4\pi (2^{1/3}R)^2}} = \frac{{2}}{{2^{2/3}}} = {2^{1/3}}:1$$

Answer

$${2^{{1 \over 3}}}:1$$

Answer

$$1:{2^{{1 \over 3}}}$$

Answer

2 : 1

Answer

1 : 2

Properties of Matter

Two small drops of mercury each of radius R coalesce to form a single large drop. The ratio of total surface energy before and after the change is :

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