First slide

Surface tension and surface energy

Question

A mercury drop of 1 cm radius is broken into 10⁶ small drops. The energy used will be (surface tension of mercury is $35 \times 10^{- 3} N / cm)$

Moderate

A

$4 .4 \times 10^{- 3} J$
B

$2 .2 \times 10^{- 4} J$
C

$8 .8 \times 10^{- 4} J$
D

$10^{4} J$

Solution

$\begin{array}{l} E = 4 {πR}^{2} T (n^{1 / 3} - 1) \\ = 4 \times 3 .14 \times 10^{- 4} \times 35 \times 10^{- 3} (10^{6 / 3} - 1) = 4 .4 \times 10^{- 3} J \end{array}$

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