Question

The surface energy of a liquid drop is u. it is spreaed in 106 equal droplets. Then its surface energy becomes: एक द्रव बुंद की पृष्ठ ऊर्जा है। इससे 106 बराबर त्रिज्या की बूंदे बना दी जायें तो इसकी पृष्ठ ऊर्जा हो जायेगीः-

• A. u
• B. 100u
• C. 10000u
• D. 106u

Answer 1

Answer: (B)

100u

Answer 2

The problem involves the change in surface energy when a large liquid drop breaks into smaller droplets. The key principles are the conservation of volume and the definition of surface energy.

Let:

• $R$ be the radius of the initial large liquid drop.
• $r$ be the radius of each small droplet.
• $\sigma$ be the surface tension of the liquid.
• $N$ be the number of equal droplets.

Given:

• Initial surface energy of the large drop = $u$.
• Number of equal droplets = $10^6$.

1. Initial Surface Energy: The surface area of the initial large drop is $A_{initial} = 4\pi R^2$. The initial surface energy is given by: $u = \sigma \times A_{initial} = \sigma (4\pi R^2)$ --- (1)

2. Conservation of Volume: When the large drop breaks into $N$ smaller droplets, the total volume of the liquid remains constant. Volume of large drop = $N \times$ Volume of one small droplet $\frac{4}{3}\pi R^3 = N \times \frac{4}{3}\pi r^3$ $R^3 = N r^3$ From this, we can express $r$ in terms of $R$ and $N$: $r = \frac{R}{N^{1/3}}$ --- (2)

3. Final Surface Energy: The total surface area of the $N$ small droplets is $A_{final} = N \times (4\pi r^2)$. The final surface energy, $u_{final}$, is: $u_{final} = \sigma \times A_{final} = \sigma \times N (4\pi r^2)$ --- (3)

4. Substitute and Simplify: Substitute the expression for $r$ from equation (2) into equation (3): $u_{final} = \sigma \times N \left(4\pi \left(\frac{R}{N^{1/3}}\right)^2\right)$ $u_{final} = \sigma \times N \left(4\pi \frac{R^2}{N^{2/3}}\right)$ Rearrange the terms: $u_{final} = \sigma (4\pi R^2) \times \frac{N}{N^{2/3}}$ $u_{final} = \sigma (4\pi R^2) \times N^{(1 - 2/3)}$ $u_{final} = \sigma (4\pi R^2) \times N^{1/3}$

5. Relate to Initial Surface Energy: From equation (1), we know that $u = \sigma (4\pi R^2)$. Substitute this into the expression for $u_{final}$: $u_{final} = u \times N^{1/3}$

6. Calculate for $N = 10^6$: Given the options, we assume $N = 10^6$. $u_{final} = u \times (10^6)^{1/3}$ $u_{final} = u \times 10^{(6 \times 1/3)}$ $u_{final} = u \times 10^2$ $u_{final} = 100u$

Thus, if the liquid drop is spread into $10^6$ equal droplets, its surface energy becomes $100u$.