Question

If 1000 droplets of water of surface tension 0.07 N/m, having same radius 1 mm each, combine to from a single drop In the process the released surface energy is - (Take π=\(\frac {22}{7}\))
 

• A. $8.8 \times 10^{-5} J$
• B. $7.92 \times 10^{-4} J$
• C. $7.92 \times 10^{-6} J$
• D. $9.68 \times 10^{-4} J$

Hint:

When droplets combine to form a single larger droplet, the total surface area decreases, and the surface energy is released in the process.

Answer 1

The Correct Option is B

To solve this problem, we need to find the released surface energy when 1000 small droplets combine to form a single large droplet. We are provided with the following information:

• Surface tension (\(T\)) = 0.07 N/m
• Radius of each small droplet (\(r\)) = 1 mm = 0.001 m
• Number of small droplets = 1000
• Value of \(\pi\) = \(\frac{22}{7}\)

The surface area of a single small droplet is given by the formula for the surface area of a sphere, \(A = 4\pi r^2\). Therefore, the total initial surface area of 1000 droplets is:

\(A_{\text{initial}} = 1000 \times 4 \pi r^2 = 1000 \times 4 \pi (0.001)^2\)

Calculate \(A_{\text{initial}}\):

\(A_{\text{initial}} = 1000 \times 4 \times \frac{22}{7} \times (0.001)^2 = 12.57 \times 10^{-3} \, \text{m}^2\)

When these droplets combine to form a single large droplet, the total volume remains the same. The volume of 1 droplet is:

\(V_{\text{single}} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (0.001)^3\)

Thus, the total volume of 1000 droplets:

\(V_{\text{total}} = 1000 \times \frac{4}{3} \pi (0.001)^3 = \frac{4}{3} \pi (0.01)^3\)

The new large droplet will also be spherical, and given by:

\(V_{\text{large}} = \frac{4}{3} \pi R^3\)

Equating the volumes:

\(\frac{4}{3} \pi R^3 = \frac{4}{3} \pi (0.01)^3\) \(R^3 = (0.01)^3\) \(R = 0.01 \, \text{m}\)

The surface area of the new large droplet is:

\(A_{\text{final}} = 4 \pi R^2 = 4 \pi (0.01)^2\)

Calculate \(A_{\text{final}}\):

\(A_{\text{final}} = 4 \times \frac{22}{7} \times (0.01)^2 = 1.257 \times 10^{-3} \, \text{m}^2\)

The change in surface area is:

\(\Delta A = A_{\text{final}} - A_{\text{initial}} = 1.257 \times 10^{-3} - 12.57 \times 10^{-3} = -11.31 \times 10^{-3}\)

The released surface energy (\(\Delta E\)) is given by:

\(\Delta E = T \times |\Delta A| = 0.07 \times 11.31 \times 10^{-3}\)

Calculate \(\Delta E\):

\(\Delta E = 0.07 \times 11.31 \times 10^{-3} = 7.92 \times 10^{-4} \, \text{J}\)

Therefore, the released surface energy is \(7.92 \times 10^{-4} \, \text{J}\), which corresponds to the correct answer. Thus, the correct option is:

$7.92 \times 10^{-4} J$