A soap bubble of radius ‘ r ’ and surface tension ‘ T ’ is given a potential of ‘ V ’ volt . If the new radius ‘ R ’ of the bubble is related to its initial radius by equation, P 0 [ R 3 - r 3 ] + λ T [ R 2 - r 2 ] - ε 0 V 2 R/2 = 0 , where P 0 is the atmospheric pressure .Then find λ
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(4)
Sol. we know 
= V, Q = 
σ = 
= 
[ 
is excess pressure due to uniform charge distribution on the surface of a bubble pressure is larger
than outside]
Clearly P A × 
= 
⇒ P A = 
Now P A – P 0 = 
& P A – P 0 = 
......(3)
so, from (3) P A 
– P 0 = 
......(4)
P 0 – P 0 
= 
+ 
⇒ P 0 (R 3 – r 3 ) = 
{r 3 – rR 2 } + 
⇒ P 0 (R 3 – r 3 ) + 4T (R 2 – r 2 ) – 
= 0.
Hence provide Ans. λ = 4
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