Question

A film of soap solution is trapped between a vertical frame and a light wire ab of
the length of 0.1 m. If g = 10 m/s^2, then the load mass that should be suspended from the
wire to keep it in equilibrium is : ( sigma= 25x10-3N / m)
A. 0.25gm
B. 0.3gm
C. 0.4gm
D. 0.5gm

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Answer 1

Answer:

(D) 0.5gm

Explanation:

in case of soap solutions two thin surfaces are formed so force can be calculated as

Force = 2×Surface tension×length

mg = 2×Surface tension×length

$m \times 10 = 2 \times 25 \times 10 {}^{- 3} \times 0.1$

$m \times 10 = 50 \times {10}^{ - 4}$

$m = 50 \times 10 {}^{ - 5} kg$

$m = 0.5gm$