Liquid drops are falling slowly one-by-one from a vertical glass tube. Establish a relation between the weight of a drop W, the surface tension T and radius r of the bore of the tube. Assume the angle of contact to be zero

W = π r^{2} T

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W = 2 π r^{2} T

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W = (4 / 3) π r^{3} T

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W = 4 π r T

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Solution

The correct option is C $W = 4 π r T$
The weight of liquid drop is equal to surface tension force.
Therefore, $W = S . T = 2 π r . T$

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Liquid drops are falling slowly one-by-one from a vertical glass tube. Establish a relation between the weight of a drop W, the surface tension T and radius r of the bore of the tube. Assume the angle of contact to be zero

The correct option is C W=4πrTThe weight of liquid drop is equal to surface tension force. Therefore, W=S.T=2πr.T

The correct option is C $W = 4 π r T$
The weight of liquid drop is equal to surface tension force.
Therefore, $W = S . T = 2 π r . T$