The excess pressure inside a soap bubble is twice the excess pressure

1.	The excess pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is n times the volume of the second where n is(a) 4(b) 2(c) 1(d) 0125.
Answer» The correct answer is (d) 0125. Explanation: Let the radius of the first soap bubble = r, Excess pressure inside P = 4S/r Let the radius of the second soap bubble = r' Excess pressure inside P' = 4S/r' Since P = 2P' →4S/r =2* 4S/r' →1/r = 2/r' →r/r' = ½ The volume of first bubble = n* the volume of second bubble →4πr³/3 = n*4πr'³/3 →r³/r'³ = n →n = (r/r')³ = (½)³ =1/8 =0.125

The excess pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is n times the volume of the second where n is(a) 4(b) 2(c) 1(d) 0125.

Answer»

The correct answer is (d) 0125.

Explanation:

Let the radius of the first soap bubble = r,

Excess pressure inside P = 4S/r

Let the radius of the second soap bubble = r'

Excess pressure inside P' = 4S/r'

Since P = 2P'

→4S/r =2* 4S/r'

→1/r = 2/r'

→r/r' = ½

The volume of first bubble = n* the volume of second bubble

→4πr³/3 = n*4πr'³/3

→r³/r'³ = n

→n = (r/r')³ = (½)³ =1/8 =0.125