A big drop is formed by coalescing 1000 small identical drops of water. If E1 be the total surface energy of 1000 small drops of water and E2 be the surface energy of single big drop of water, the E1 : E2 is x:1 where x=_____

The surface energy of a drop of water is directly proportional to its surface area. The surface area of a sphere (which is the shape of a drop of water) is given by the formula 4πr², where r is the radius of the sphere.

Step 1: Let's assume the radius of each small drop is r. Therefore, the total surface area of 1000 small drops is 1000 * 4πr² = 4000πr².

Step 2: When these 1000 drops coalesce to form a big drop, the volume remains the same (since volume is conserved). The volume of a sphere is given by the formula 4/3πr³. Therefore, the total volume of 1000 small drops is 1000 * 4/3πr³ = 4000/3πr³.

Step 3: The volume of the big drop is also 4000/3πR³, where R is the radius of the big drop. Since the volumes are equal, we can equate the two expressions and solve for R: r³ = R³ => R = r * (1000)^(1/3).

Step 4: The surface area of the big drop is 4πR² = 4π[r * (1000)^(1/3)]² = 4000πr²/1000^(1/3).

Step 5: The ratio of the total surface energy of 1000 small drops to the surface energy of the big drop is therefore (4000πr²) / (4000πr²/1000^(1/3)) = 1000^(1/3) = 10.

So, x = 10.