If the excess pressure inside a soap bubble is balanced by oil column of height 2 mm, then the surface tension of soap solution will be ( r=1 cm and density d=0.8 g/ cc )

Question

If the excess pressure inside a soap bubble is balanced by oil column of height 2 mm, then the surface tension of soap solution will be ( r=1 cm and density d=0.8 g/ cc ) (A)  3.9 N/m  (B)  3.9× 10-1N/m  (C)  3.9× 10-2N/m  (D)  3.9 dyne/m

Tardigrade Admin · Accepted Answer

Excess pressure inside a soap bubble of radius

R

is

= \frac{4 T}{R}

... (i)
where

T

is surface tension of liquid film. Pressure due to oil column

= h ρ g

... (ii)
where

h

is height of column,

p

the density and

g

the gravity.
From Eqs. (i) and (ii), we get

\frac{4 T}{R} = h ρ g

\Rightarrow T = \frac{h ρ g R}{4}

Given

/ t = 2 mm = 0.2 c m, g = 980 c m s^{- 2}

ρ = 0.8 g / cc, R = 1 c m

= T = \frac{0.2 \times 0.8 \times 980}{4}

= 3.92 \times 10

dyne

c m^{- 1}

In

N m^{- 1}, T = 3.9 \times 10 \times \frac{1 0 ^{- 5}}{1 0 ^{- 2}}

= 3.9 \times 1 0^{- 2} N m^{- 1}

Q. If the excess pressure inside a soap bubble is balanced by oil column of height $2 mm,$ then the surface tension of soap solution will be ( $r = 1 c m$ and density $d = 0.8 g / cc$ )

$3.9 N / m$

$3.9 \times 1 0^{- 1} N / m$

$3.9 \times 1 0^{- 2} N / m$

$3.9 d y n e / m$

Q. If the excess pressure inside a soap bubble is balanced by oil column of height 2mm, then the surface tension of soap solution will be ( r=1cm and density d=0.8g/cc )

3.9N/m

3.9×10−1N/m

3.9×10−2N/m

3.9dyne/m

Q. If the excess pressure inside a soap bubble is balanced by oil column of height $2 mm,$ then the surface tension of soap solution will be ( $r = 1 c m$ and density $d = 0.8 g / cc$ )

$3.9 N / m$

$3.9 \times 1 0^{- 1} N / m$

$3.9 \times 1 0^{- 2} N / m$

$3.9 d y n e / m$