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MHT CET 2021 (20 Sep, Shift 1) · Force due to Surface Tension

Force due to Surface Tension question from MHT CET 2021 (20 Sep, Shift 1)

Q14MHT CET 2021 · 20 SEP · SHIFT 1Easy

Force due to Surface Tension

The force required to take away a flat circular plate of radius 2 cm2\text{ cm} from the surface of water is [[Surface tension of water =70×10−3 N m−1,π=227]= 70 \times 10^{-3}\text{ N m}^{-1}, \pi = \frac{22}{7}]

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Answer: (b)

  1. Surface tension acts along the rim of the plate: F=T×2πrF = T \times 2\pi r.
  2. F=70×10−3×2×227×0.02F = 70\times10^{-3} \times 2 \times \frac{22}{7} \times 0.02.
  3. F=8.8×10−3 NF = 8.8 \times 10^{-3}\text{ N}. So the answer is (b).
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