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. The de-Broglie wavelength of an electron in the first Bohr orbit is
2. The angular momentum of an electron is J. The magnitude of the magnetic moment
3. The operation of a nuclear reactor is said to be critical, if the multiplication
4. A change of 0.04 V takes place between the base and the emitter when an input
5. The binding energy per nucleon of ₃⁷Li and ₂⁴He nuclei are 5.60 MeV and 7.06MeV

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