A mercury drop of radius is broken into 125 equal size droplets. Surface tension of mercury is . The gain in surface energy is [2023]

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Solution

Q.
A mercury drop of radius $10^{- 3} m$ is broken into 125 equal size droplets. Surface tension of mercury is $0.45 {Nm}^{- 1}$ . The gain in surface energy is [2023]

1 $2.26 \times 10^{- 5} J$

2 $28 \times 10^{- 5} J$

3 $17.5 \times 10^{- 5} J$

4 $5 \times 10^{- 5} J$

Ans.
(1)

$Initial surface energy = 0.45 \times 4 π {(10^{- 3})}^{2}$

$\frac{4}{3} π {(10^{- 3})}^{2} = 125 \times \frac{4 π}{3} R_{new}^{2}$

$∴$ $10^{- 3} = 5 R_{new}$

$∴$ $R_{new} = \frac{10^{- 3}}{5} m$

$So, final surface energy = 0.45 \times 125 \times 4 π {(\frac{10^{- 3}}{5})}^{2}$

$Increase in energy = 0.45 \times 4 π \times {(10^{- 3})}^{2} [\frac{125}{25} - 1]$

$= 4 \times 0.45 \times 4 π \times 10^{- 6} = 2.26 \times 10^{- 5} J$

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