If the radius of a spherical liquid (of surface tension S) drop increases from r to r+ Δ r, the corresponding increase in the surface energy is

Question

If the radius of a spherical liquid (of surface tension S) drop increases from r to r+ Δ r, the corresponding increase in the surface energy is (A)  8π r Δ rS  (B)  4π r Δ rS  (C)  16π r Δ rS  (D)  2π r Δ rS

Tardigrade Admin · Accepted Answer

Work done in blowing a liquid drop or soap bubble W=S × 4 π[r22-r12] Here, W=S × 4 π[(r +Δ r)2-(r)2] =S × 4 π[r2+(Δ r)2+2 Δ r × r-r2] =8 π Δ r S r[(Δ r)2. is very small]

Q. If the radius of a spherical liquid (of surface tension $S$ ) drop increases from $r$ to $r + Δ r,$ the corresponding increase in the surface energy is

$8 π r Δ r S$

$4 π r Δ r S$

$16 π r Δ r S$

$2 π r Δ r S$

Work done in blowing a liquid drop or soap bubble
$W = S \times 4 π [r_{2}^{2} - r_{1}^{2}]$
Here, $W = S \times 4 π [(r + Δ r)^{2} - (r)^{2}]$
$= S \times 4 π [r^{2} + (Δ r)^{2} + 2Δ r \times r - r^{2}]$
$= 8 π Δ r S r [(Δ r)^{2}$ is very small]

Q. If the radius of a spherical liquid (of surface tension S) drop increases from r to r+Δr, the corresponding increase in the surface energy is

8πrΔrS

4πrΔrS

16πrΔrS

2πrΔrS