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A straw 6 cm long floats on water. The water film on one side has surface tension of 50 dyne/cm. On the other side, camphor reduces the surface tension to 40 dyne/cm. The resultant force acting on the straw is

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$60 dyne$

$10 d y n e$

$(\frac{50}{6} - \frac{40}{6}) dyne$

$90 d y n e$

answer is A.

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Detailed Solution

$F = (σ_{1} - σ_{2}) l = (50 \times 6 - 40 \times 6) = 60 dyn e$

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