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 current I flows along the length of an infinitely long, straight, thin walled pipe.
  2. The potential energy of a long spring when stretched by 2 cm is U.
  3. A coil of N turns and mean cross-sectional area A is rotating with uniform angular
  4. A uniform plank of Young’s modulus Y is moved over a smooth horizontal surface
  5. A closely wound solenoid of 2000 turns and area of cross-section 1.5 x 10⁻⁴ m² carries

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Your email address will not be published.


*