Soap bubble form in air .The excess pressure inside a soap bubble is twice the excess pressure in second bubble .The first bubble has n times the volume of the second bubble , finf n ….

Options

A. 1
B. 0.5
C. 0.25
D. 0.125

Answer

d

Explanation

Radius of first bubble `=R_(1)`
Radius of second bubble `=R_(2)`
Volume of first bubble =n (Volume of second bubble)
`(4)/(3)piR_(1)^(3)=n((4)/(3)piR_(2)^(3))`
`thereforeR_(1)^(3)=nR_(2)^(3)`
`therefore(R_(1))/(R_(2))=n^((1)/(3))` ...(1)
Excess pressure in bubble 1=n (Excess pressure in bubble 2)
`(P_(i)-P_(o))_(1)=2(P_(i)-P_(o))`
`(4T)/(R_(1))=2((4T)/(R_(2)))`
`therefore(R_(1))/(R_(2))=(1)/(2)` ....(2)
From equation (1) and (2),
`n^((1)/(3))=(1)/(2)`
`thereforen=(1)/(8)`
`thereforen=0.125`