Question

Work done in splitting a drop of water of 1 mm radius into ${10}^6$ droplets is (Surface tension of water =$72\times {10}^{-3}J/m^2)$

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$=4\pi R^{2}T(n^{1/3}-1)$

$=4\pi \times ({10}^{-3}{)}^2\times 72\times {10}^{-3}\times ({10}^{6/3}-1)$

$=4\pi \times {10}^{-6}\times 72\times {10}^{-3}\times 99=8.95\times {10}^{-5}$J