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. Which of the following physical quantities has neither dimensions nor unit
2. Which of the following are mechanical waves?
3. If r be the distance of a point on the axis of a magnetic dipole from its centre
4. X-rays when incident on a metal
5. The maximum velocity of a particle, executing simple harmonic motion with an amplitude

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