What is the work done in splitting a drop of water of $1 m m$ radius into $10^{6}$ droplets ? (Surface Tension of water $= 72 \times 10^{- 3} J / m^{2}$ )

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Solution

$L e t t h e r a d i u s o f l a r g e r d r o p = R a n d r a d i u s o f s m a l l e r d r o p s = r . S i n c e t h e t o t a l v o l u m e o f w a t e r d r o p l e t w i l l b e s a m e t h u s V o l u m e o f l a r g e r d r o p = v o l u m e o f s m a l l e r d r o p \frac{4}{3} π R^{3} = K \frac{4}{3} π r^{3} w h e r e K = c o n s t a n t N o w K = \frac{R^{3}}{r^{3}} a n d a r e a o f i n i t i a l d r o p = 4 π R^{2} a n d a r e a o f f i n a l d r o p = K 4 π r^{2} C h a n g e i n a r e a o f w a t e r d r o p = K 4 π r^{2} - 4 π R^{2} = 4 π (K r^{2} - R^{2}) A n d w o r k d o n e = F o r c e \times d i s p l a c e m e n t H e r e F o r c e = s u r f a c e t e n s i o n (T) d i s p l a c e m e n t = c h a n g e i n a r e a = 4 π (K r^{2} - R^{2}) W . D = 4 π T (K r^{2} - R^{2}); P u t t i n g t h e v a l u e o f r e s p e c t i v e q u a n t i t i e s w e g e t t h e e x p r e s s i o n = 4 π T R^{3} (\frac{1}{r} - \frac{1}{r}) = 4 π T R^{2} (K^{1 / 3} - 1) = 4 π \times 72 \times 10^{- 3} (10^{- 3})^{2} {10^{6 / 3} - 1} = 8.95 \times 10^{- 5} J$

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1 m m

64

72 \times 10^{- 3} J / m^{2}

10^{6}

10^{- 3} J / m^{2}

1 m m

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72 \times 10^{- 3} N / M

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

1 m m

10^{6}

72 \times 10^{- 3} N / m

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