To solve this problem, we need to understand the relationship between the surface energy of a drop and its radius. The surface energy
E of a drop is proportional to its surface area
A. The surface area of a sphere (which is the shape of a water drop) can be calculated using the formula:
A=4Ï€r2where
r is the radius of the sphere. The surface energy is then given by:
E=σA=σ4πr2where
σ is the surface tension of the liquid. When 1000 small drops combine to form one large drop, the volume of the large drop is equal to the sum of the volumes of the small drops. The volume of a sphere is given by
V=‌πr3 If the radius of each small drop is
r, the total volume of the small drops is
1000×‌πr3. Let the radius of the large drop be
R. Then,
1000×‌πr3=‌πR3Canceling common terms and taking the cube root of both sides, we get
10001∕3×r=RSince
1000=103, the cube root of 1000 is 10 .
10×r=RSo the radius of the large drop is ten times the radius of each small drop.
Now we can compare the initial surface energy of all small drops with the final surface energy of the large drop. Let's denote the initial surface energy as
E‌initial ‌ and the final surface energy as
E‌final ‌.
E‌initial ‌=1000×σ4πr2E‌final ‌=σ4πR2Substituting
R=10r into the final energy equation, we get:
E‌final ‌=σ4π(10r)2E‌final ‌=σ4π(100r2)E‌final ‌=100σ4πr2
Now, we compare the final energy to the initial energy:
Hence, the ratio of the final surface energy to the total initial surface energy is
1:10 . Therefore, the correct answer is:
Option B 1 : 10