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Question: A soap bubble, having radius of 1mm, is blown from a detergent solution having surface tension of \(...

A soap bubble, having radius of 1mm, is blown from a detergent solution having surface tension of 2.5×102N/m2.5\times 10^{-2}N/m . The pressure inside the bubble equals at a point Z0Z_{0} below the free surface of water in a container. Taking g=10m/s2g=10m/s^{2}, density of water=103kg/m3=10^{3}kg/m^{3}, the value of Z0Z_{0} is:
a.100cm100cm
b.10cm10cm
c.1cm1cm
d.0.5cm0.5cm

Explanation

Solution

Hint:
We know from surface tension that, the pressure outside the bubble must be equal to the pressure inside the bubble, it is given by the formulaP0+4TR=P0+ρgZ0P_{0}+\dfrac{4T}{R}= P_{0}+\rho g Z_{0}. We need to calculate pressure inside the bubble equals at a point .
Formula used: P0+4TR=P0+ρgZ0P_{0}+\dfrac{4T}{R}= P_{0}+\rho g Z_{0}

Complete step-by-step answer:
Surface tension is the property of liquid due to the force of attraction of particles on the surface layer, which tends to minimize the surface area. Its dimension is force per unit length. At any point in the medium, there is a force of attraction called the cohesive forces which act in every direction resulting in the net force zero. But at the surface, the molecules are pulled inwards towards the liquid, this causes the surface area of the liquid to minimize. Surfactants are used to minimize the surface tension and hence increase the surface area.
There are several effects of surface tension, such as:
Water droplets are spherical, as the sphere has minimum surface area to volume ratio.
A mixture of oil and water when mixed, doesn’t get dissolved as there is difference in surface tension among the dissimilar liquids.
Here it is given that, surface tensionT=2.5×102N/mT=2.5\times10^{-2 } N/m
Density of waterρ=103kg/m3\rho=10^{3}kg/m^{3}
Radius of bubbleR=103mR=10^{-3}m
g=10m/s2g=10m/s^{2}
We know from surface tension that;
Pressure inside the soap bubble P0+4TRP_{0}+\dfrac{4T}{R} must be equal to pressure at a point Z0Z_{0} below the surface P0+ρgZ0P_{0}+\rho g Z_{0}, where P0P_{0} is the atmospheric pressure.
Then, P0+4TR=P0+ρgZ0P_{0}+\dfrac{4T}{R}= P_{0}+\rho g Z_{0}
4TR=ρgZ0\dfrac{4T}{R}= \rho g Z_{0}
4TρgR=Z0\dfrac{4T}{\rho g R}= Z_{0}
Substituting, we get
Z0=4×2.5×102103×10×103=102m=1cmZ_{0}=\dfrac{4\times 2.5\times10^{-2}}{10^{3}\times 10\times 10^{-3}}=10^{-2}m=1cm

Hence the answer is c.1cm1cm

Note: Surface tension of the liquid leads it to occupy minimum surface area. The dimension of surface area is force per unit length. The pressure outside the bubble must be equal to the pressure inside the bubble.