The excess of pressure inside two soap bubbles of diameters in the ratio 2 : 1 is
(a) 1 : 4
(b) 2 : 1
(c) 1 : 2
(d) 4 : 1
Answers
Answered by
18
The answer is option c) 1:2
Excess pressure inside a soap bubble, P = 4S / R ⇒ P is inversely proportional to R(radius)
Radius = Diameter / 2 ⇒ R₁ = 2 / 2 = 1 and R₂ = 1 / 2
So, P₁ / P₂ = R₂ / R₁ = (1/2) / 1 = 1 / 2 = 1 : 2
Hope this helps you..!
Excess pressure inside a soap bubble, P = 4S / R ⇒ P is inversely proportional to R(radius)
Radius = Diameter / 2 ⇒ R₁ = 2 / 2 = 1 and R₂ = 1 / 2
So, P₁ / P₂ = R₂ / R₁ = (1/2) / 1 = 1 / 2 = 1 : 2
Hope this helps you..!
Answered by
1
Answer:(c)1:2
Explanation:Formula of internal excess pressure inside a soap bubble which have two open surface is 4T/R...【where T=surface tension and R is internal radius of soap bubble ..
For first bubble it is 4T/2
For first bubble it is 4T/1
Then the ratio is 4T/2:4T/1
=>1/2:1
=>1:2
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