Question

A mercury drop of radius 1 cm is sprayed into \(10^6\) drops of equal size. The energy expressed in joules is (surface tension of mercury is \(460 \times 10^{-3} \, \text{N/m}\)).

• A. \(0.057\)
• B. \(5.7\)
• C. \(5.7 \times 10^{-4}\)
• D. 5.7 × 10⁻6

Hint:

Surface energy change depends only on change in surface area: E = T Δ A Breaking droplets increases surface area significantly.

Answer 1

The Correct Option is B

Step 1: Surface energy: \(E = T \times \Delta A\)
Initial radius: \(R = 1 \, \text{cm} = 10^{-2} \, \text{m}\)
Step 2: Radius of each small drop:
\[ r = \frac{R}{100} = 10^{-4} \, \text{m} \]
Step 3: Increase in surface area:
\[ \Delta A = 10^6 (4 \pi r^2) - 4 \pi R^2 = 4 \pi (10^6 \times 10^{-8} - 10^{-4}) = 4 \pi (0.01 - 0.0001) \]
Step 4: Energy required:
\[ E = 460 \times 10^{-3} \times 4 \pi \times 0.0099 \approx 5.7 \, \text{J} \]