Question

A drop of liquid of diameter $2.8\text{mm}$ break up into $216$ identical drops. What is the approximate change in energy of bigger drop?
(Given: Surface tension of liquid, $T=75\text{dyne/cm}$)

• A. $25\pi \text{erg}$
• B. $29.4\pi \text{erg}$
• C. $32\pi \text{erg}$
• D. $28\pi \text{erg}$

Answer 1

The correct option is B $29.4\pi \text{erg}$

$\Rightarrow$Change in energy (energy required) in splitting of bigger drop to $n$ smaller drops is,
$\Delta E=4\pi R^{2}T(n^{1/3}-1)...\left(i\right)$
Where, Radius of bigger drop
$R=1.4\times {10}^{-1}\text{cm}$
$n=$ Number of small drops
$T=$ Surface tension
From Eq. $\left(i\right)$ we get,
$\Delta E=4\times \pi \times (1.4\times {10}^{-1}{)}^2\times 75({216}^{1/3}-1)\text{dyne.cm}$
Since $1\text{erg}=1\text{dyne.cm}$
$\therefore \Delta E=29.4\pi \text{erg}$