First slide

Surface tension

Question

The work done in splitting a drop of water of 1 mm radius into 10 6 droplets is [S.T. of water $72 \times 10^{- 3} J / m^{2})$

Moderate

A

$9 \cdot 98 \times 10^{- 5} joule$
B

$8 \cdot 95 \times 10^{- 5} joule$
C

$5 \cdot 89 \times 10^{- 5} joule$
D

$5 \cdot 98 \times 10^{- 5} joule$

Solution

. In splitting, the volume remains unchanged.
$∴ \frac{4}{3} π R^{3} = \frac{4}{3} π r^{3} \times n o r R = (n^{1 / 3}) r o r r = R / (n)^{1 / 3} w o r k d o n e W = s u r f a c e t e n s i o n x c h a n g e i n a r e a = 4 π r^{2} n - 4 π R^{2} = 4 π [\frac{n \times R^{2}}{n^{2 / 3}} - R^{2}] = 4 π R^{2} [n^{1 / 3} - 1] ∴ W = 4 π R^{2} [n^{1 / 3} - 1] T H e r o n = 10^{6}, R = 1 mm = 10^{- 3} m and T = 72 \times 10^{- 3} J / m^{2} Substituting these values, we get W = 8 \cdot 95 \times 10^{- 5} joule$

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