A water drop of radius r is divided into eight equal droplets. What is the work done in the process? (Given surface tension =σ)

Question

A water drop of radius r is divided into eight equal droplets. What is the work done in the process? (Given surface tension =σ) (A) 2 π pr 2 σ (B) 3 π r2 σ (C) 4 π r2 σ (D) 24 π r2 σ

Tardigrade Admin · Accepted Answer

(4/3) π r3=8((4/3) π r'3) ⇒ r'=(r/2) Δ A=8(4 π r2)-4 π r2=4 π[8((r/2))2-r2] =4 π(2 r2-r2)=4 π r2 Workdone =σ ⋅ Δ A=σ(4 π r2)

Q. A water drop of radius $r$ is divided into eight equal droplets. What is the work done in the process? (Given surface tension $= σ)$

$2 π p r^{2} σ$

$3 π r^{2} σ$

$4 π r^{2} σ$

$24 π r^{2} σ$

$\frac{4}{3} π r^{3} = 8 (\frac{4}{3} π r^{^{'} 3})$
$\Rightarrow r^{'} = \frac{r}{2}$
$Δ A = 8 (4 π r^{2}) - 4 π r^{2} = 4 π [8 (\frac{r}{2})^{2} - r^{2}]$
$= 4 π (2 r^{2} - r^{2}) = 4 π r^{2}$
Workdone $= σ \cdot Δ A = σ (4 π r^{2})$

Q. A water drop of radius r is divided into eight equal droplets. What is the work done in the process? (Given surface tension =σ)

2πpr2σ

3πr2σ

4πr2σ

24πr2σ