A student observes that the absolute pressure inside a soap bubble of radius 2.0 mm in the atmosphere is exactly equal to the absolute pressure at a depth of 1.6 cm inside an oil tank. If the density of the oil is 800

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Quantrex Academy · Accepted Answer

Correct answer: D. 0.064 N m^ -1

The absolute pressure inside a soap bubble of radius

R

and surface tension

T

is given by:

P_{\text{bubble}} = P_0 + \dfrac{4T}{R}

The absolute pressure at a depth

h

in a liquid of density

\rho

is:

P_{\text{depth}} = P_0 + \rho g h

Equating the two pressures:

P_0 + \dfrac{4T}{R} = P_0 + \rho g h

\Rightarrow \dfrac{4T}{R} = \rho g h

\Rightarrow T = \dfrac{\rho g h R}{4}

Given values:

R = 2.0\text{ mm} = 2 \times 10^{-3}\text{ m}

h = 1.6\text{ cm} = 1.6 \times 10^{-2}\text{ m}

\rho = 800\text{ kg m}^{-3}

g = 10\text{ m s}^{-2}

Substituting the values:

T = \dfrac{800 \times 10 \times 1.6 \times 10^{-2} \times 2 \times 10^{-3}}{4}

T = \dfrac{128 \times 2 \times 10^{-3}}{4}

T = 64 \times 10^{-3}\text{ N m}^{-1} = 0.064\text{ N m}^{-1}

Answer:

0.064\text{ N m}^{-1}

A student observes that the absolute pressure inside a soap bubble...

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