Q. Determine the energy stored in the surface of a soap bubble of radius 2.1 cm if its surface tension is $4.5 \times 10^{- 2}\ {Nm}^{- 1}$.
A) 8 mJ
B) 2.46 mJ
C) $4.93 \times 10^{- 4}\ J$
D) None of these
Explanation
(c) $Surface\ energy\ U = S \times 2 \times 4\pi R^{2}$
(As there are 2 surfaces in soap bubble )
$U = 4.5 \times 10^{- 2} \times 8\pi \times \left( 2.1 \times 10^{- 2} \right)^{2}$
$= 4.93 \times 10^{- 4}\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.