MCQ
A water drop of diameter $cm$ is broken into $64$ equal droplets. The surface tension of water is $0.075\,N / m$. In this process the gain in surface energy will be ...........$J$
  • $2.8 \times 10^{-4}$
  • B
    $1.5 \times 10^{-3}$
  • C
    $1.9 \times 10^{-4}$
  • D
    $9.4 \times 10^{-5}$

Answer

Correct option: A.
$2.8 \times 10^{-4}$
a
$d =2\,cm ; \quad r =1\,cm ; \quad T =0.075$

$\Delta SE = T \Delta A$

$=0.075\left( A _{ f }- A _{1}\right)$

$A _{ i }=4 \pi r ^{2}$

$A _{ t }=4 \pi r _{0}^{2} \times 64$

By volume conservation

$\frac{4}{3} \pi r ^{3}=64 \cdot \frac{4}{3} \pi r_{0}^{3}$

$I _{0}=\frac{ r }{4}$

$A _{ f }=4 \pi\left(\frac{ r }{4}\right)^{2} \cdot 64=16 \pi r ^{2}$

$\Delta SE =0.075\left(16 \pi r ^{2}-4 \pi r ^{2}\right)$

$=0.075\left(12 \pi(0.01)^{2}\right)$

$=2.8 \times 10^{-4}\,J$

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