Question

Work of 3.0×10 ^-4 J is required to be done in increasing the size of a soap film from 10cm×6cm to 10cm×11cm. The surface tension of the soap film is:

5×10^-2N/m

3×10^-2N/m

1.5×10^-2N/m

1.2×10^-2N/m​

Answer 1

Given :-

Area increased = (10 × 11 − 10 × 6) cm²

Number of sides the film have = 2

To Find :-

The surface tension of the film.

Solution :-

We know that,

• N/m = Newton meter
• J = Potential energy joule

Given that,

Area increased = (10 × 11 − 10 × 6) cm²

= 50 cm²

Number of sides the film have = 2

∴ Total increased area = 50 × 2

= 100 cm²

$\underline{\boxed{\sf Surface \ tension=\dfrac{Work \ done}{Increase \ in \ surface \ area} }}$

Substituting their values, we get

Surface tension = $\sf \dfrac{3 \times 10^{-4}}{100 \times 10^{-4}}$

$\sf =0.03 \ N/m$

$\sf =3 \times 10^{-2} \ N/m$

Therefore, the surface tension of the soap film is 3 × 10^-2 N/m