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# Work done in increasing the size of a soap bubble from a radius of 3cm to 5cm is nearly (Surface tension of soap solution = 0.03 Nm-1)(a) 0.2 πmJ(b) 2πmJ(c) 0.4πmJ(d) 4πmJ

Navjyot Kalra
askIITians Faculty 654 Points
7 years ago
(c) W = 2T∆V
W = 2T4π[(52) – (3)2] * 10-4
= 2 * 0.03 * 4π [25 – 9] * 10-4 J = 0.4π * 10-3 J
0.4πmJ
Sharva Potdar
26 Points
5 years ago
I couldn`t find the answer. The solution here given is wrong I guess,because the Surface Tension is given in N/m and the radius is not converted to metres. So the solution is not dimensionally correct,im getting the ans as 1 mJ
Mallika
16 Points
4 years ago
energy = 2TΔA where T is the tension of the soap bubble.2*0.3*4π(0.05^2-0.03^2) ( I have converted cm to m0.4πmJ
S Sharma
17 Points
3 years ago
Can't.
You see.              That there.              Is .
A 10^ -4 in.       The answer.     Which .
Is a proof that he
Converted cm into.m
See.        Correctly
Don't.      Make silly.      Mistakes
Yash Chourasiya
askIITians Faculty 256 Points
one year ago
Dear Student

W = (surface energy)final ​− (surface energy)initial​
W = T×4π[(5×10−2)2 − (3×10−2)2]×2
= 4π×0.03×16×10−4×2
= 4π×0.48×10−4×2
= 1.92π×10−4×2
= 3.94π×10−4
= 0.394πmJ ≈ 0.4πmJ