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NEET PHYSICSMechanical Properties of FluidEasy

Question

Air is pushed carefully into a soap bubble of radius rr to double its radius. If the surface tension of the soap solution is TT, then the work done in the process is:

A

12πr2T12\pi r^2T

B

24πr2T24\pi r^2T

C

4πr2T4\pi r^2T

D

8πr2T8\pi r^2T

Step-by-Step Solution

The work done in increasing the size of a soap bubble is equal to the increase in its surface energy, which is the product of surface tension and the change in surface area.

  1. Identify Surface Area: A soap bubble has two free surfaces (inner and outer) in contact with air. Therefore, the effective surface area is 2×4πr2=8πr22 \times 4\pi r^2 = 8\pi r^2.

  2. Calculate Initial and Final Areas: Initial Radius (r1r_1) = rr Initial Surface Area (A1A_1) = 2×4πr2=8πr22 \times 4\pi r^2 = 8\pi r^2 Final Radius (r2r_2) = 2r2r (radius is doubled) Final Surface Area (A2A_2) = 2×4π(2r)2=2×4π(4r2)=32πr22 \times 4\pi (2r)^2 = 2 \times 4\pi (4r^2) = 32\pi r^2

  3. Calculate Change in Area (ΔA\Delta A): ΔA=A2A1=32πr28πr2=24πr2\Delta A = A_2 - A_1 = 32\pi r^2 - 8\pi r^2 = 24\pi r^2

  4. Calculate Work Done (WW): W=T×ΔA=T×24πr2=24πr2TW = T \times \Delta A = T \times 24\pi r^2 = 24\pi r^2 T

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Mechanical Properties of Fluid. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMechanical Properties of Fluidpushedcarefullybubbleradiusdouble

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