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 particle A suffers on oblique elastic collision with a particle B that
  2. The identical cells connected in series are needed to heat a wire of length one meter
  3. The ratio of the dimensions of planck’s constant and that of the moment of inertia is
  4. Potentiometer measures the potential difference more accurately than a voltmeter
  5. A radioactive element forms its own isotope after 3 consecutive disintegrations.

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Your email address will not be published.


*