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A drop of mercury of radius 2 mm is spli...

A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury `=0.465Jm^-2`

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The correct Answer is:
B, C, D
`r=2mm=2xx10^-3m,`
`T=0.465Jm^2`
`Volume of initial drop =8x(vol. of small drops)`
`=(4/3)piR^3=(4/3)pr^3xx8`
`=r=R/2=2`
Increase is surfac energy
`=TA\'-TA=8xx4pi^2T-4piR^2T`
`=4piT[98xx(R^2/4)-R^2]`
`=4piTR^2`
`=4xx(3.14)xx(0.465)xx(4xx10^-6)`
`=23.36 muJ`
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Knowledge Check

  • A drop of liquid of diameter 2.8 mm breaks up into 125 identical drops. The change in energy is nearly (s = 75 dyne/cm).

    A
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    B
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  • A drop of liquid of diameter 2.8 mm breaks up into 125 identical drops. The change in energy is nearly (s = 75 dyne/cm).

    A
    19 erg
    B
    zero
    C
    74 erg
    D
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