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

Work Of 30x10 Joule Is Required To Be Done In Physics Question

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 man goes 10 m towards north, then 20m towards east, the displacement is
  2. The temperature of a body falls from 50⁰C to 40⁰C in 10 minute. If the temperature
  3. Changing magnetic fields can set up current loops in nearby metal bodies
  4. Oil spreads over the surface of water where as water does not spread
  5. A coin of mass m and radius r having moment of inertia I about the axis passes

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.

NEETLab Mobile App

Share this page with your friends

Be the first to comment

Leave a Reply

Your email address will not be published.


*