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A $10 \mathrm{~cm}$ long wire is placed horizontally on the surface of water and is gently pulled up with a force of $2 \times 10^{-2} \mathrm{~N}$ to keep the wire in equilibrium. The surface tension, in $\mathrm{Nm}^{-1}$, of water is
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The correct answer is:
0.1
$T=\frac{F}{2 l}=\frac{2 \times 10^{-2}}{2 \times 10 \times 10^{-2}}=0.1 \mathrm{~N} / \mathrm{m}$
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