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प्रश्न
A drop of mercury of radius 0.2 cm is broken into 8 droplets of the same size. Find the work done if the surface tension of mercury is 435.5 dyn/cm.
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उत्तर
Given:
R = 0.2 cm,
n = 8,
T = 435.5 dyn/cm
To find:
The work done
Solution:
As the volume of the liquid remains constant, volume of n droplets = volume of the drop
∴ `"n"xx4/3pi"r"^3 = 4/3pi"R"^3`
∴ `"r"^3 = "R"^3/"n"`
∴ r = `"R"/root(3)("n") = "R"/root(3)(8) = "R"/2`
Surface area of the drop = 4πR2
Surface area of n droplets = n × 4πr2
∴ The increase in the surface area = surface area of n droplets – surface area of drop
= n × 4πr2 - 4πR2
= 4π(nr2 - R2)
= `4π(8xx"R"^2/4 - "R"^2)`
= 4π(2 - 1)R2 = 4πR2
∴ The work done = surface tension × increase in surface area
= T × 4πR2 = 435.5 × 4 × 3.142 × (0.2)2
= 2.189 × 10-5 J
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