A water drop of radius 1 cm is broken into eight equal droplets. Surface tension of water is 0.075 N m⁻¹. The gain in surface energy is ___________ × 10⁻⁷ J. (Take π = 3.14) Numerical answer type. Work it out, then open the answer.

Answer: 942. Use Δ E = 4π R² T(n1/3 - 1) with R = 0.01 m. n = 8 ⇒ n1/3 - 1 = 1 Δ E = 4(3.14)(10⁻⁴)(0.075) = 9.42 × 10⁻⁵ J = 942 × 10⁻⁷ J So the answer is 942.

A water drop of radius 1 cm is broken into eight equal droplets… | JEE Main 2022 (30 Jul, Shift 2)

Q45JEE Main 2022 · 30 JUL · SHIFT 2Medium

Splitting & Merging of Drops

A water drop of radius $1\text{ cm}$ is broken into eight equal droplets. Surface tension of water is $0.075\text{ N m}^{-1}$ . The gain in surface energy is ___________ $\times 10^{-7}\text{ J}$ . (Take $\pi = 3.14$ )

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 942

Use $\Delta E = 4\pi R^2 T\left(n^{1/3} - 1\right)$ with $R = 0.01\text{ m}$ .
$n = 8 \Rightarrow n^{1/3} - 1 = 1$
$\Delta E = 4(3.14)(10^{-4})(0.075) = 9.42 \times 10^{-5}\text{ J} = 942 \times 10^{-7}\text{ J}$
So the answer is 942.

Report a mistake

Splitting & Merging of Drops question from JEE Main 2022 (30 Jul, Shift 2)

More questions on Splitting & Merging of Drops