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If the radius and surface tension of a spherical soap bubble be `r` and `T` respectively. Find the charge uniformly distributed over the outer surface of the bubble, is required to double its radius. (Given that atmospheric pressure is `P_(0)` and inside temperature of the bubble during expansion remains constant.)A. `8pir[epsilon_(0)r(7P_(0)r+12T)]^(1//2)`B. `4pir[epsilon_(0)r(7P_(0)r+12T)]^(1//2)`C. `pir[epsilon_(0)r(7P_(0)r+12T)]^(1//2)`D. `8pir[epsilon_(0)r(7P_(0)r+12T)]^(1//3)` |
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Answer» Before charging the bubble pressure inside the bubble `P_(1)=P_(0)+4T//r` After charging electrostatic pressure will be `(sigma^(2))/(2epsilon_(0))` Which will pull the bubble outward. When radius becomes `2r` then `(P_(1))/(8)+(sigma^(2))/(2epsilon_(0))=P_(0)+(4T)/(2r)`, where `sigma=(q)/(4pi(2r)^(2))` |
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