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Physics Question on mechanical properties of fluid

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

A

3.9 N/m3.9\,N/m

B

3.9×10−1N/m3.9\times 10^{-1}N/m

C

3.9×10−2N/m3.9\times 10^{-2}N/m

D

3.9 dyne/m3.9\,dyne/m

Answer

3.9×10−2N/m3.9\times 10^{-2}N/m

Explanation

Solution

Excess pressure inside a soap bubble of radius RR is
=4TR=\frac{4 T}{R} ... (i)
where TT is surface tension of liquid film. Pressure due to oil column
=hρg=h \rho g ... (ii)
where hh is height of column, pp the density and gg the gravity.
From Eqs. (i) and (ii), we get
4TR=hρg\frac{4 T}{R}=h \rho g
⇒T=hρgR4\Rightarrow T=\frac{h \rho g R}{4}
Given /t=2 mm=0.2 cm, g=980 cm s−2/ t=2\, mm =0.2\, cm,\, g=980\, cm\, s ^{-2}
ρ=0.8 g/cc,R=1 cm\rho=0.8\, g / cc, R=1\, cm
=T=0.2×0.8×9804=T=\frac{0.2 \times 0.8 \times 980}{4}
=3.92×10=3.92 \times 10 dyne cm−1cm^{-1}
In Nm−1, T=3.9×10×10−510−2Nm^{-1},\, T=3.9 \times 10 \times \frac{10^{-5}}{10^{-2}}
=3.9×10−2Nm−1=3.9 \times 10^{-2} Nm ^{-1}