The excess pressure inside a soap bubble A is twice that in another soap bubble B. The ratio of volumes of A and B is

Question

The excess pressure inside a soap bubble A is twice that in another soap bubble B. The ratio of volumes of A and B is (A) 1: 2 (B) 1: 4 (C) 1: 8 (D) 1: 16

Tardigrade Admin · Accepted Answer

Excess pressure inside the bubble is p=(4T/r) Here, pA=2pB (pA/pB)=(rB/rA) So, radius of bubble B is rB=2 Ratio of volumes (VA/VB)=( (rA/rB) )3=( (1/2) )3 (VA/VB)=(1/8)

Q. The excess pressure inside a soap bubble A is twice that in another soap bubble B. The ratio of volumes of A and B is

1 : 2

1 : 4

1 : 8

1 : 16

Excess pressure inside the bubble is $p = \frac{4 T}{r}$ Here, $p_{A} = 2 p_{B}$ $\frac{p _{A}}{p _{B}} = \frac{r _{B}}{r _{A}}$ So, radius of bubble $B$ is $r_{B} = 2$ Ratio of volumes $\frac{V _{A}}{V _{B}} = (\frac{r _{A}}{r _{B}})^{3} = (\frac{1}{2})^{3}$ $\frac{V _{A}}{V _{B}} = \frac{1}{8}$

Q. The excess pressure inside a soap bubble A is twice that in another soap bubble B. The ratio of volumes of A and B is

1 : 2

1 : 4

1 : 8

1 : 16

Excess pressure inside the bubble is p=r4T​ Here, pA​=2pB​ pB​pA​​=rA​rB​​ So, radius of bubble B is rB​=2 Ratio of volumes VB​VA​​=(rB​rA​​)3=(21​)3 VB​VA​​=81​

Excess pressure inside the bubble is $p = \frac{4 T}{r}$ Here, $p_{A} = 2 p_{B}$ $\frac{p _{A}}{p _{B}} = \frac{r _{B}}{r _{A}}$ So, radius of bubble $B$ is $r_{B} = 2$ Ratio of volumes $\frac{V _{A}}{V _{B}} = (\frac{r _{A}}{r _{B}})^{3} = (\frac{1}{2})^{3}$ $\frac{V _{A}}{V _{B}} = \frac{1}{8}$