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Q. A wire of length 10 cm is placed horizontal on the surface of water and is gently pulled up with a force of $ 1.8\times {{10}^{-2}}N $ to keep the wire in equilibrium. The surface tension of water will be

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Solution:

$ T=\frac{F}{2l} $ $ =\frac{1.8\times {{10}^{-2}}}{2\times 10\times {{10}^{-2}}} $ $ =0.09\,N\text{/}m $