Question

The difference in pressure across a soap bubble of diameter 0.005 m immersed in water
(surface tension is 0.072 Nm) is​

Answer 1

Answer:

Given : S=0.072 N/m

Diameter of water drop D=1.2mm=0.012 m

Thus radius of drop r=

2

0.012

=0.006 m

Excess pressure inside water drop ΔP=

r

2S

∴ ΔP=

0.006

2×0.072

=24 N/m

2

Answer 2

Answer:

Given :

S=

0.072

N/m

Diameter of water drop

D=

1.2mm=

0.012

m

Thus radius of drop

r=

2

0.012

=

0.006

m

Excess pressure inside water drop

ΔP=

r

2S

∴

ΔP=

0.006

2×0.072

= 24

N/m

2