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PhysicsFluidSurface TensionEasy2 minQB

PhysicsEasysingle choice

Figure (a) and (b) show water drop and mercury drop in two identical conical glass pipes, then

Answer:

Solution:

1. Water Drop (Fig. a)

$P = P_{0} - \frac{2 T}{R}$ (where $P_{0}$ is atmospheric pressure).

2. Mercury Drop (Fig. b)

$P = P_{0} + \frac{2 T}{R}$ .

Water: Moves toward the narrow end.

Mercury: Moves toward the wide end.

Stream:JEESubject:PhysicsTopic:FluidSubtopic:Surface Tension

⏱ 2mℹ️ Source: QB

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