Work of 3.0×10⁻⁴ joule is required to be done in increasing the size of a soap film

⇦

⇨

Work of 3.0×10⁻⁴ joule is required to be done in increasing the size of a soap film from 10cm x 6cm to 10cm x 11cm. The surface tension of the film is

Options

(a) 5 x10⁻² N/m
(b) 3 x10⁻² N/m
(c) 1.5 x10⁻² N/m
(d) 1.2 x10⁻² N/m

Correct Answer:

3 x10⁻² N/m

Explanation:

Surface tension, S = work done/increase in area
As the soap film has two surfaces.
S = (3.0 x 10⁻⁴ J)/[2 x (10 x 11 – 10 x 6 ) x 10⁻⁴ m²]
= 3 x 10⁻² N/m.

Related Questions:

1. A particle moves in xy plane according to the equation x = 4t² + 5t + 16 and y = 5t
2. The following four wires are made of the same material. Which of these will have
3. A force acts on a 3.0 g particle in such a way that the position of the particle
4. In Young’s double silt experiment, 12 fringes are observed to be formed
5. At the first minimum adjacent to the central maximum of a single-slit diffraction

Topics: Properties of Bulk Matter (130)
Subject: Physics (2479)

---

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

---

Share this page with your friends