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NEET Questions / Physics / Mechanical Properties Of Fluids / Force Due To Surface Tension
Two soap bubbles of radii and coalesce under isothermal conditions to form a bigger bubble of radius . If the atmospheric pressure is , the surface tension of the soap solution is:
T = P₀(R³ - r₁³ - r₂³) / 4(r₁² + r₂² - R²)
T = P₀(r₁³ + r₂³ - R³) / 4(r₁² + r₂² - R²)
T = P₀(R³ + r₁³ + r₂³) / 4(r₁² + r₂² - R²)
T = P₀(R² + r₁² + r₂²) / 4(r₁³ + r₂³ - R³)
A small air bubble of radius rises steadily through a liquid of density at a terminal velocity . If the coefficient of viscosity of the liquid is , the surface tension of the liquid is (neglecting the density of air):
\frac{2r\rho gv}{9\eta}
\frac{r\rho gv}{6\eta}
\frac{9\eta v}{2r\rho g}
Cannot be determined
A liquid rises to a height 'h' in a capillary tube of radius 'r'. If the radius of the capillary tube is doubled, the height to which the liquid rises in the tube will be:
h/4
h/2
h
2h
Two spherical bubbles of radii and in air combine to form a bigger bubble. The radius of the bigger bubble is:
A thin rectangular frame of wire is dipped vertically into water. When it is raised, a thin film of water is formed inside the frame. The force required to raise the frame against the surface tension of water (assuming negligible weight of the frame) is:
Sl
2Sl
Sl/2
4Sl
Two soap bubbles of radii and coalesce to form a single bubble of radius under isothermal conditions. If is the external pressure, then which of the following relations is correct?
P(r₁² + r₂² - R²) = 4T(r₁ + r₂ - R)
r₁² + r₂² = R²
r₁³ + r₂³ = R³
P(r₁³ + r₂³ - R³) = 4T(r₁² + r₂² - R²)
A small metal ring of radius 'r' is kept on a large soap bubble of radius 'R' (). If the surface tension of the soap solution is 'T', the force pulling the ring towards the center of the bubble is:
4πTr²/R
2πTr²/R
4πTR²/r
2πTR²/r