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Pressure inside two soap bubbles are `1.01` and `1.02` atmospheres. Ratio between their volumes isA. `2:1`B. `1:2`C. `8:1`D. `1:8`

Answer» Correct Answer - C
Excess pressure `P_(1) = 1.01 -1 = 0.01 ` atm
Excess pressure `P_(2) = 1.02 - 1 = 0.02 ` atm
As ` P prop 1/r " " :. (r_(1))/(r_(2)) = (P_(2))/(P_(1))= 0.02/0.01 = 2/1 `
` (v_(1))/(v_(2)) = (r_(1)^(3))/(r_(2)^(3)) = (2^(3))/(1^(3)) = 8/1 `


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