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A drop of radius 1mm is sprayed into 1000 droplets of same size. If surface tension of the liquid is 40 dyne (cm^-1), the amount of work done in the process is

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4.53erg

4.53J

4.53 \times 10^{- 6} J

9.06 \times 10^{- 6} J

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detailed solution

Correct option is C

Given,

$n = 1000, T = 40 dyne / cm$

$R = 1 mm = 10^{- 3} m & R = n^{\frac{1}{3}} r$

we know that,

$Work done = difference in surface area \times surface tension$

$W = 4 π ({nr}^{2} - R^{2}) T = 4 {πR}^{2} T (n^{\frac{1}{3}} - 1)$

$W = 4 (3.14) {(10 - 3)}^{2} (40) \times 10^{- 3} [{(1000)}^{\frac{1}{3}} - 1] = 4.53 \times 10^{- 6} J$

Hence the correct answer is $4.53 \times 10^{- 6} J .$

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A drop of radius 1mm is sprayed into 1000 droplets of same size. If surface tension of the liquid is 40 dyne (cm-1), the amount of work done in the process is

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A drop of radius 1mm is sprayed into 1000 droplets of same size. If surface tension of the liquid is 40 dyne (cm^-1), the amount of work done in the process is