Question

A water drop of radius 1 cm is broken into 729 equal droplets. If surface tension of water is 75 dyne/cm, then the gain in surface energy upto first decimal place will be :
(Given \(\pi\) = 3.14)

• A. \(8.5 × 10^{–4} J\)
• B. \(8.2 × 10^{–4} J\)
• C. \(7.5 × 10^{–4} J\)
• D. \(5.3 × 10^{–4} J\)

Answer 1

The Correct Option is C

To solve the problem of determining the gain in surface energy when a water drop of radius 1 cm is broken into 729 equal droplets, let's follow these steps:

1. First, calculate the initial surface area of the large water drop.
2. The formula for the surface area of a sphere is given by: A = 4\pi r^2, where r is the radius of the sphere.
3. Initially, we have a single water drop of radius 1 cm. Therefore, the initial surface area is: A_1 = 4\pi \times (1)^2 = 4\pi \, \text{cm}^2.
4. Next, determine the radius of each of the 729 smaller droplets. Since the total volume must remain the same, we apply the formula for the volume of a sphere: V = \frac{4}{3}\pi r^3.
5. Set up the equality for volumes: \frac{4}{3}\pi (1)^3 = 729 \times \frac{4}{3}\pi r^3.
6. Simplifying, we get: (1)^3 = 729 \times r^3, which gives r^3 = \frac{1}{729}.
7. Thus, the radius of each small droplet is: r = \left(\frac{1}{729}\right)^{\frac{1}{3}} = \frac{1}{9} \, \text{cm}.
8. Now, calculate the total surface area of the 729 smaller droplets:
9. The surface area of one smaller droplet is: A_{\text{small}} = 4\pi \left(\frac{1}{9}\right)^2 = \frac{4\pi}{81} \, \text{cm}^2.
10. Therefore, the total surface area of all smaller droplets is: A_2 = 729 \times \frac{4\pi}{81} = 36\pi \, \text{cm}^2.
11. Calculate the gain in surface area: \Delta A = A_2 - A_1 = 36\pi - 4\pi = 32\pi \, \text{cm}^2.
12. The gain in surface energy can be calculated using: \Delta E = T \times \Delta A, where T is the surface tension of the water.
13. Given that the surface tension of water is 75 dyne/cm (or 75 \times 10^{-5} \, \text{N/cm}), we have: \Delta E = 75 \times 10^{-5} \times 32 \times 3.14 \, \text{J}.
14. Calculating this gives: \Delta E = 75 \times 10^{-5} \times 100.48 = 7.536 \times 10^{-3} \, \text{J}.
15. Therefore, the gain in surface energy up to the first decimal place is: 7.5 \times 10^{-4} \, \text{J}.

Hence, the correct answer is 7.5 \times 10^{-4} \, \text{J}.