A ring of radius \(1.5\text{ cm}\) is floating horizontally on the surface of water. If this ring has to be raised up then how much additional force has to be applied to lift ring : (Surface tension of water \(73 \times 10^{-3}\text{ Newton/metre}\))
\(1.37 \times 10^{-2}\text{ N}\)
\(2.3 \times 10^{-2}\text{ N}\)
\(5.1 \times 10^{-3}\text{ N}\)
\(4 \times 10^{-2}\text{ N}\)
Solution:
The additional force required to lift the ring is \(F = 2 \times (2\pi r T) = 4\pi r T\). Substituting \(r = 1.5 \times 10^{-2}\text{ m}\) and \(T = 73 \times 10^{-3}\text{ N/m}\) gives \(F = 4 \times 3.14 \times 1.5 \times 10^{-2} \times 73 \times 10^{-3} \approx 1.37 \times 10^{-2}\text{ N}\).
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