A ring cut from a platinum tube of 8.5cm internal and 8.7cm external diameter .It is supported horizontally from a pan of a balance so that it comes in contact with the water in a glass vessel .The surface tension of water, if an extra 3.79g-wt is required to pull it away from water ( g = 980cm / s² )is found to be (71.77 - X ) dyne / cm,where X = ______

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answer is 3.

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Detailed Solution

$(2 π \times \frac{8 .7}{2} + 2 π \times \frac{8 .5}{2}) T = 3 .79 \times 980$

$\Rightarrow T = \frac{3 .79 \times 980}{2 π (\frac{8 .7 + 8 .5}{2})} = 68.74 d y n e / c m$

$\Rightarrow X = 71 .77 - 68.74 = 3.03$

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