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In measuring the diameter of a drop of liquid the error is 1% and in measuring the surface tension of the liquid, the error is 2% . Then the maximum possible error in calculating surface energy of the drop will be

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3.5%

2.5%

answer is C.

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Detailed Solution

E = surface energy = $4 {πr}^{2} T$

$∴ (\frac{∆ E}{E}) x 100 % = 2 (\frac{∆ r}{r}) % + (\frac{∆ T}{T}) % = 2 x 1 % + 2 % = 4 %$

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