A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. (Surface tension of mercury is 0.465 J m- 2 )

Question

A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. (Surface tension of mercury is 0.465 J m- 2 ) (A) 23.4μJ (B) 18.5μJ (C) 26.8μJ (D) 16.8μJ

Tardigrade Admin · Accepted Answer

Increase in surface energy or work done in splitting a big drop = 4 π R2 T (. n1 / 3 - 1 .) ⇒ W=4π × (. 2 × (10)- 3 .)2× 0.465(.81 / 3-1.)=23.4μJ

Q. A drop of mercury of radius $2 mm$ is split into $8$ identical droplets. Find the increase in surface energy. (Surface tension of mercury is $0.465 J m^{- 2}$ )

$23.4 μ J$

$18.5 μ J$

$26.8 μ J$

$16.8 μ J$

Increase in surface energy or work done in splitting a big drop $= 4 π R^{2} T (n^{1/3} - 1)$ $\Rightarrow W = 4 π \times (2 \times (10)^{- 3})^{2} \times 0.465 (8^{1/3} - 1) = 23.4 μ J$

Q. A drop of mercury of radius 2mm is split into 8 identical droplets. Find the increase in surface energy. (Surface tension of mercury is 0.465Jm−2 )

23.4μJ

18.5μJ

26.8μJ

16.8μJ