Question

A bubble of radius 2 cm is blown inside a cold drink using a straw. If the surface tension of the liquid is 60 dyne/cm. Find the work done (in ergs) in blowing the bubble.

Answer 1

Explanation:

Given A bubble of radius 2 cm is blown inside a cold drink using a straw. If the surface tension of the liquid is 60 dyne/cm. Find the work done (in ergs) in blowing the bubble.

Given r = 2 cm, T = 60 dyne / cm W = ?

We know that initial surface area of soap bubble = 0

Final surface area Δ A = 2 x 4 π r^2

Now increase in surface area = 2 x 4 π r^2

Work done W = T x Δ A

                       = 60 x 8 x 3.14 x (2)^2

                      = 6028.8 erg

The work done in blowing the soap bubble is 6028.8 ergs.

Answer 2

Explanation:

W=TΔA=T×2[4πR*2 ]=T×8πR*2

W=TΔA=T×2[4πR*2 ]=T×8πR*2

W=TΔA=T×2[4πR*2 ]=T×8πR*2 W=60×10*−3 ×8π×(0.2) 2

W=TΔA=T×2[4πR*2 ]=T×8πR*2 W=60×10*−3 ×8π×(0.2) 2 =1.92π10

W=TΔA=T×2[4πR*2 ]=T×8πR*2 W=60×10*−3 ×8π×(0.2) 2 =1.92π10 −2j

W=TΔA=T×2[4πR*2 ]=T×8πR*2 W=60×10*−3 ×8π×(0.2) 2 =1.92π10 −2j