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Physics Class 12th Mechanical Properties of Fluids Physics NCERT Exemplar Solutions Class 12th Chapter Ten

A water drop of radius 1cm is broken into 729 equal droplets. If surface tension of water is 75 dyne/cm, then the gain in surface energy upto first decimal place will be:

(Given π = 3.14)

Option 1 -

8.5 * 10^-4 J

Option 2 -

8.2 * 10^-4 J

Option 3 -

7.5 * 10^-4 J

Option 4 -

5.3 * 10^-4 J

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Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

3 months ago

Correct Option - 3

Detailed Solution:

$\frac{4}{3} π R^{3} = 729 \frac{4}{3} π r^{3}$

R³ = 729r³

$R = (729) \frac{1}{3} (r) (\frac{1}{3})$

R = 9r

$Δ u = T (4 π r^{2}) \times 729 - T \times 4 π R^{2}$

$= T \times 4 π \times 8 R^{2}$

$\begin{array}{l} = 75.39 \times 1 0^{- 5} J \\ Δ u = 7.5 \times 1 0^{- 4} J \end{array}$

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A water drop of radius 1cm is broken into 729 equal droplets. If surface tension of water is 75 dyne/cm, then the gain in surface energy upto first decimal place will be:

(Given π = 3.14)

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A water drop of radius 1cm is broken into 729 equal droplets. If surface tension of water is 75 dyne/cm, then the gain in surface energy upto first decimal place will be: (Given π = 3.14)

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A water drop of radius 1cm is broken into 729 equal droplets. If surface tension of water is 75 dyne/cm, then the gain in surface energy upto first decimal place will be:

(Given π = 3.14)