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Question: If the excess pressure inside a soap bubble is balanced by an oil column of height 2 mm, then the su...

If the excess pressure inside a soap bubble is balanced by an oil column of height 2 mm, then the surface tension of soap solution will be (r = 1cm, density of oil = 0.8g/cm3)
A. 3.9N/m{\text{A}}{\text{. 3}}{\text{.9N/m}}
B. 3.9×102N/m{\text{B}}{\text{. 3}}{\text{.9}} \times {\text{1}}{{\text{0}}^{ - 2}}{\text{N/m}}
C. 3.9×103N/m{\text{C}}{\text{. 3}}{\text{.9}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{N/m}}
D. 3.9×101N/m{\text{D}}{\text{. 3}}{\text{.9}} \times {\text{1}}{{\text{0}}^{ - 1}}{\text{N/m}}

Explanation

Solution

Hint: The excess pressure inside the bubble is given by, P=4TrP = \dfrac{{4T}}{r} and the pressure to height column of oil is given as, P=ρhgP = \rho hg .

Complete step-by-step answer:

Formula used - P=4TrP = \dfrac{{4T}}{r} , P=ρhgP = \rho hg

Given that the excess pressure inside a soap bubble is balanced by an oil column of height 2 mm.
To find – surface tension of soap solution.

Now, as we know, the excess pressure inside the soap bubble is given by the formula, P=4TrP = \dfrac{{4T}}{r} and the pressure to height column of oil is given by the formula, P=ρhgP = \rho hg .

Now, since the excess pressure inside the bubble is balanced by the height of the column and so we can write that-

4Tr=ρhg\dfrac{{4T}}{r} = \rho hg
now, r=1cm=1×102mr = 1cm = 1 \times {10^{ - 2}}m
ρ=0.8g/cm3=0.8gcm3=0.8×103kg(102)3m3=0.8×103kg/m3\rho = 0.8g/c{m^3} = \dfrac{{0.8g}}{{c{m^3}}} = \dfrac{{0.8 \times {{10}^{ - 3}}kg}}{{{{({{10}^{ - 2}})}^3}{m^3}}} = 0.8 \times {10^3}kg/{m^3}
h=2mm=2×103mh = 2mm = 2 \times {10^{ - 3}}m
So, now using 4Tr=ρhg\dfrac{{4T}}{r} = \rho hg , we can write-

T=rρhg4T = \dfrac{{r\rho hg}}{4}
Putting the value, we get-

T=1×102×0.8×103×9.8×2×1034=0.8×102×9.8×24=7.84×1022 T=3.9×102N/m  T = \dfrac{{1 \times {{10}^{ - 2}} \times 0.8 \times {{10}^3} \times 9.8 \times 2 \times {{10}^{ - 3}}}}{4} = \dfrac{{0.8 \times {{10}^{ - 2}} \times 9.8 \times 2}}{4} = \dfrac{{7.84 \times {{10}^{ - 2}}}}{2} \\\ \Rightarrow T = 3.9 \times {10^{ - 2}}N/m \\\

So, the surface tension of soap solution is 3.9×102N/m{\text{3}}{\text{.9}} \times {\text{1}}{{\text{0}}^{ - 2}}{\text{N/m}} .

Hence, the correct option is B.

Note: Whenever such types of questions appear then always write down the things given in question. Then, as mentioned in the solution, that the excess pressure inside a soap bubble is balanced by an oil column, so we have written, 4Tr=ρhg\dfrac{{4T}}{r} = \rho hg and then after putting the values of density, r, h and g we found out the value of surface tension, T.