143018
A sphere of mass $M$ and radius $R$ is falling in a viscous fluid. The terminal velocity attained by the falling object will be proportional to:
1 $\mathrm{R}^{2}$
2 $\mathrm{R}$
3 $1 / \mathrm{R}$
4 $1 / R^{2}$
Explanation:
A The terminal velocity of any body falling in viscous fluid is given by- $\mathrm{v}_{\mathrm{T}}=\frac{2}{9 \eta} \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}$ Where, $\sigma$ is density of liquid, $\rho \text { is the density of sphere }$ $R=\text { Radius of body, } \eta=\text { viscosity of fluid. }$ So, From equation (i), Terminal velocity of body $\left(\mathrm{v}_{\mathrm{T}}\right)$ is directly proportional the $\mathrm{R}^{2}$. $\therefore \quad \mathrm{v}_{\mathrm{T}} \propto \mathrm{R}^{2}$
AIIMS-2004
Mechanical Properties of Fluids
143020
If the excess pressure inside air bubble of radius $r_{1}$ in air is equal to the excess pressure inside air bubble of radius $r_{2}$ inside the soap solution, then $r_{1}: r_{2}$ is
1 $2: 1$
2 $1: 2$
3 $1: 4$
4 $\sqrt{2}: 1$
5 $1: \sqrt{2}$
Explanation:
A Excess pressure inside air bubble having radius ' $r_{1}$ ' in air is given as- $\Delta \mathrm{P}_{1}=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}$ Now, again excess pressure inside air bubble having radius ' $r_{2}$ ' in soap solution is given by- $\Delta \mathrm{P}_{2}=\frac{2 \mathrm{~T}}{\mathrm{r}_{2}}$ Where, $\mathrm{T}=\text { Surface Tension }$ According to question, $\Delta \mathrm{P}_{1}=\Delta \mathrm{P}_{2}$ $\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}=\frac{2 \mathrm{~T}}{\mathrm{r}_{2}}$ $\frac{4 \mathrm{~T}}{2 \mathrm{~T}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ $\frac{2}{1}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ $\therefore \quad \mathrm{r}_{1}: \mathrm{r}_{2}=2: 1$
Kerala CEE 2012
Mechanical Properties of Fluids
143021
The excess pressure inside a spherical drop of water is four times that of another drop. then their respective mass ratio is
1 $1: 16$
2 $8: 1$
3 $1: 4$
4 $1: 64$
5 $32: 1$
Explanation:
D Excess pressure inside a spherical drop of water is given by $\Delta \mathrm{P}=\frac{2 \mathrm{~T}}{\mathrm{R}}$ Where $\mathrm{T}=$ Surface tension $\mathrm{R}=$ Radius of droplet $\Delta \mathrm{P} \propto \frac{1}{\mathrm{R}}$ $\therefore \quad \frac{\Delta \mathrm{P}_{1}}{\Delta \mathrm{P}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$ And given $\Delta \mathrm{P}_{2}=\mathrm{P}$, then $\Delta \mathrm{P}_{1}=4 \mathrm{P}$ Putting the above value in equation number (i) $\frac{4 \mathrm{P}}{\mathrm{P}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}} \Rightarrow \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1}{4}$ Now, If mass of water drop be $M$ then, $M=\rho V$ Where, $\rho=$ Density of water $\mathrm{V}=\text { Volume of water drop }$ $M=\rho \times \frac{4}{3} \pi R^{3}$ Since, $\pi$ and $\rho$ is constant $\therefore \quad \mathrm{M} \propto \mathrm{R}^{3}$ $\Rightarrow \quad \frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3} \quad\left(\because \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1}{4}\right)$ $\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\left(\frac{1}{4}\right)^{3} \Rightarrow \quad \frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\frac{1}{64}$ $\mathrm{M}_{1}: \mathrm{M}_{2}=1: 64$
Kerala CEE - 2009
Mechanical Properties of Fluids
143022
The excess pressure inside one soap bubble is three times that inside a second soap bubble, then the ratio of their surface areas is
1 $1: 9$
2 $1: 3$
3 $3: 1$
4 $1: 27$
5 $9: 1$
Explanation:
A The excess pressure inside a soap bubble is given as- $\Delta \mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{R}} \Rightarrow \Delta \mathrm{P} \propto \frac{1}{\mathrm{R}}$ Where, $T=$ Surface tension $\mathrm{R}=\text { Radius of soap bubble }$ $\Delta \mathrm{P}=\text { Excess pressure. }$ According to question- $\Delta \mathrm{P}_{1}=3 \Delta \mathrm{P}_{2}$ If $\Delta \mathrm{P}_{2}=\mathrm{P}$, then $\Delta \mathrm{P}_{1}=3 \mathrm{P}$ $\therefore \quad \frac{\Delta \mathrm{P}_{1}}{\Delta \mathrm{P}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{3}{1}$ Now, radius of surface area- $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{4 \pi \mathrm{R}_{1}^{2}}{4 \pi \mathrm{R}_{2}^{2}} \Rightarrow \frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{2}$ $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\left(\frac{1}{3}\right)^{2}$ $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{1}{9}$ $\mathrm{A}_{1}: \mathrm{A}_{2}=1: 9$
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Mechanical Properties of Fluids
143018
A sphere of mass $M$ and radius $R$ is falling in a viscous fluid. The terminal velocity attained by the falling object will be proportional to:
1 $\mathrm{R}^{2}$
2 $\mathrm{R}$
3 $1 / \mathrm{R}$
4 $1 / R^{2}$
Explanation:
A The terminal velocity of any body falling in viscous fluid is given by- $\mathrm{v}_{\mathrm{T}}=\frac{2}{9 \eta} \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}$ Where, $\sigma$ is density of liquid, $\rho \text { is the density of sphere }$ $R=\text { Radius of body, } \eta=\text { viscosity of fluid. }$ So, From equation (i), Terminal velocity of body $\left(\mathrm{v}_{\mathrm{T}}\right)$ is directly proportional the $\mathrm{R}^{2}$. $\therefore \quad \mathrm{v}_{\mathrm{T}} \propto \mathrm{R}^{2}$
AIIMS-2004
Mechanical Properties of Fluids
143020
If the excess pressure inside air bubble of radius $r_{1}$ in air is equal to the excess pressure inside air bubble of radius $r_{2}$ inside the soap solution, then $r_{1}: r_{2}$ is
1 $2: 1$
2 $1: 2$
3 $1: 4$
4 $\sqrt{2}: 1$
5 $1: \sqrt{2}$
Explanation:
A Excess pressure inside air bubble having radius ' $r_{1}$ ' in air is given as- $\Delta \mathrm{P}_{1}=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}$ Now, again excess pressure inside air bubble having radius ' $r_{2}$ ' in soap solution is given by- $\Delta \mathrm{P}_{2}=\frac{2 \mathrm{~T}}{\mathrm{r}_{2}}$ Where, $\mathrm{T}=\text { Surface Tension }$ According to question, $\Delta \mathrm{P}_{1}=\Delta \mathrm{P}_{2}$ $\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}=\frac{2 \mathrm{~T}}{\mathrm{r}_{2}}$ $\frac{4 \mathrm{~T}}{2 \mathrm{~T}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ $\frac{2}{1}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ $\therefore \quad \mathrm{r}_{1}: \mathrm{r}_{2}=2: 1$
Kerala CEE 2012
Mechanical Properties of Fluids
143021
The excess pressure inside a spherical drop of water is four times that of another drop. then their respective mass ratio is
1 $1: 16$
2 $8: 1$
3 $1: 4$
4 $1: 64$
5 $32: 1$
Explanation:
D Excess pressure inside a spherical drop of water is given by $\Delta \mathrm{P}=\frac{2 \mathrm{~T}}{\mathrm{R}}$ Where $\mathrm{T}=$ Surface tension $\mathrm{R}=$ Radius of droplet $\Delta \mathrm{P} \propto \frac{1}{\mathrm{R}}$ $\therefore \quad \frac{\Delta \mathrm{P}_{1}}{\Delta \mathrm{P}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$ And given $\Delta \mathrm{P}_{2}=\mathrm{P}$, then $\Delta \mathrm{P}_{1}=4 \mathrm{P}$ Putting the above value in equation number (i) $\frac{4 \mathrm{P}}{\mathrm{P}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}} \Rightarrow \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1}{4}$ Now, If mass of water drop be $M$ then, $M=\rho V$ Where, $\rho=$ Density of water $\mathrm{V}=\text { Volume of water drop }$ $M=\rho \times \frac{4}{3} \pi R^{3}$ Since, $\pi$ and $\rho$ is constant $\therefore \quad \mathrm{M} \propto \mathrm{R}^{3}$ $\Rightarrow \quad \frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3} \quad\left(\because \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1}{4}\right)$ $\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\left(\frac{1}{4}\right)^{3} \Rightarrow \quad \frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\frac{1}{64}$ $\mathrm{M}_{1}: \mathrm{M}_{2}=1: 64$
Kerala CEE - 2009
Mechanical Properties of Fluids
143022
The excess pressure inside one soap bubble is three times that inside a second soap bubble, then the ratio of their surface areas is
1 $1: 9$
2 $1: 3$
3 $3: 1$
4 $1: 27$
5 $9: 1$
Explanation:
A The excess pressure inside a soap bubble is given as- $\Delta \mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{R}} \Rightarrow \Delta \mathrm{P} \propto \frac{1}{\mathrm{R}}$ Where, $T=$ Surface tension $\mathrm{R}=\text { Radius of soap bubble }$ $\Delta \mathrm{P}=\text { Excess pressure. }$ According to question- $\Delta \mathrm{P}_{1}=3 \Delta \mathrm{P}_{2}$ If $\Delta \mathrm{P}_{2}=\mathrm{P}$, then $\Delta \mathrm{P}_{1}=3 \mathrm{P}$ $\therefore \quad \frac{\Delta \mathrm{P}_{1}}{\Delta \mathrm{P}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{3}{1}$ Now, radius of surface area- $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{4 \pi \mathrm{R}_{1}^{2}}{4 \pi \mathrm{R}_{2}^{2}} \Rightarrow \frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{2}$ $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\left(\frac{1}{3}\right)^{2}$ $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{1}{9}$ $\mathrm{A}_{1}: \mathrm{A}_{2}=1: 9$
143018
A sphere of mass $M$ and radius $R$ is falling in a viscous fluid. The terminal velocity attained by the falling object will be proportional to:
1 $\mathrm{R}^{2}$
2 $\mathrm{R}$
3 $1 / \mathrm{R}$
4 $1 / R^{2}$
Explanation:
A The terminal velocity of any body falling in viscous fluid is given by- $\mathrm{v}_{\mathrm{T}}=\frac{2}{9 \eta} \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}$ Where, $\sigma$ is density of liquid, $\rho \text { is the density of sphere }$ $R=\text { Radius of body, } \eta=\text { viscosity of fluid. }$ So, From equation (i), Terminal velocity of body $\left(\mathrm{v}_{\mathrm{T}}\right)$ is directly proportional the $\mathrm{R}^{2}$. $\therefore \quad \mathrm{v}_{\mathrm{T}} \propto \mathrm{R}^{2}$
AIIMS-2004
Mechanical Properties of Fluids
143020
If the excess pressure inside air bubble of radius $r_{1}$ in air is equal to the excess pressure inside air bubble of radius $r_{2}$ inside the soap solution, then $r_{1}: r_{2}$ is
1 $2: 1$
2 $1: 2$
3 $1: 4$
4 $\sqrt{2}: 1$
5 $1: \sqrt{2}$
Explanation:
A Excess pressure inside air bubble having radius ' $r_{1}$ ' in air is given as- $\Delta \mathrm{P}_{1}=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}$ Now, again excess pressure inside air bubble having radius ' $r_{2}$ ' in soap solution is given by- $\Delta \mathrm{P}_{2}=\frac{2 \mathrm{~T}}{\mathrm{r}_{2}}$ Where, $\mathrm{T}=\text { Surface Tension }$ According to question, $\Delta \mathrm{P}_{1}=\Delta \mathrm{P}_{2}$ $\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}=\frac{2 \mathrm{~T}}{\mathrm{r}_{2}}$ $\frac{4 \mathrm{~T}}{2 \mathrm{~T}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ $\frac{2}{1}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ $\therefore \quad \mathrm{r}_{1}: \mathrm{r}_{2}=2: 1$
Kerala CEE 2012
Mechanical Properties of Fluids
143021
The excess pressure inside a spherical drop of water is four times that of another drop. then their respective mass ratio is
1 $1: 16$
2 $8: 1$
3 $1: 4$
4 $1: 64$
5 $32: 1$
Explanation:
D Excess pressure inside a spherical drop of water is given by $\Delta \mathrm{P}=\frac{2 \mathrm{~T}}{\mathrm{R}}$ Where $\mathrm{T}=$ Surface tension $\mathrm{R}=$ Radius of droplet $\Delta \mathrm{P} \propto \frac{1}{\mathrm{R}}$ $\therefore \quad \frac{\Delta \mathrm{P}_{1}}{\Delta \mathrm{P}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$ And given $\Delta \mathrm{P}_{2}=\mathrm{P}$, then $\Delta \mathrm{P}_{1}=4 \mathrm{P}$ Putting the above value in equation number (i) $\frac{4 \mathrm{P}}{\mathrm{P}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}} \Rightarrow \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1}{4}$ Now, If mass of water drop be $M$ then, $M=\rho V$ Where, $\rho=$ Density of water $\mathrm{V}=\text { Volume of water drop }$ $M=\rho \times \frac{4}{3} \pi R^{3}$ Since, $\pi$ and $\rho$ is constant $\therefore \quad \mathrm{M} \propto \mathrm{R}^{3}$ $\Rightarrow \quad \frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3} \quad\left(\because \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1}{4}\right)$ $\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\left(\frac{1}{4}\right)^{3} \Rightarrow \quad \frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\frac{1}{64}$ $\mathrm{M}_{1}: \mathrm{M}_{2}=1: 64$
Kerala CEE - 2009
Mechanical Properties of Fluids
143022
The excess pressure inside one soap bubble is three times that inside a second soap bubble, then the ratio of their surface areas is
1 $1: 9$
2 $1: 3$
3 $3: 1$
4 $1: 27$
5 $9: 1$
Explanation:
A The excess pressure inside a soap bubble is given as- $\Delta \mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{R}} \Rightarrow \Delta \mathrm{P} \propto \frac{1}{\mathrm{R}}$ Where, $T=$ Surface tension $\mathrm{R}=\text { Radius of soap bubble }$ $\Delta \mathrm{P}=\text { Excess pressure. }$ According to question- $\Delta \mathrm{P}_{1}=3 \Delta \mathrm{P}_{2}$ If $\Delta \mathrm{P}_{2}=\mathrm{P}$, then $\Delta \mathrm{P}_{1}=3 \mathrm{P}$ $\therefore \quad \frac{\Delta \mathrm{P}_{1}}{\Delta \mathrm{P}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{3}{1}$ Now, radius of surface area- $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{4 \pi \mathrm{R}_{1}^{2}}{4 \pi \mathrm{R}_{2}^{2}} \Rightarrow \frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{2}$ $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\left(\frac{1}{3}\right)^{2}$ $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{1}{9}$ $\mathrm{A}_{1}: \mathrm{A}_{2}=1: 9$
143018
A sphere of mass $M$ and radius $R$ is falling in a viscous fluid. The terminal velocity attained by the falling object will be proportional to:
1 $\mathrm{R}^{2}$
2 $\mathrm{R}$
3 $1 / \mathrm{R}$
4 $1 / R^{2}$
Explanation:
A The terminal velocity of any body falling in viscous fluid is given by- $\mathrm{v}_{\mathrm{T}}=\frac{2}{9 \eta} \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}$ Where, $\sigma$ is density of liquid, $\rho \text { is the density of sphere }$ $R=\text { Radius of body, } \eta=\text { viscosity of fluid. }$ So, From equation (i), Terminal velocity of body $\left(\mathrm{v}_{\mathrm{T}}\right)$ is directly proportional the $\mathrm{R}^{2}$. $\therefore \quad \mathrm{v}_{\mathrm{T}} \propto \mathrm{R}^{2}$
AIIMS-2004
Mechanical Properties of Fluids
143020
If the excess pressure inside air bubble of radius $r_{1}$ in air is equal to the excess pressure inside air bubble of radius $r_{2}$ inside the soap solution, then $r_{1}: r_{2}$ is
1 $2: 1$
2 $1: 2$
3 $1: 4$
4 $\sqrt{2}: 1$
5 $1: \sqrt{2}$
Explanation:
A Excess pressure inside air bubble having radius ' $r_{1}$ ' in air is given as- $\Delta \mathrm{P}_{1}=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}$ Now, again excess pressure inside air bubble having radius ' $r_{2}$ ' in soap solution is given by- $\Delta \mathrm{P}_{2}=\frac{2 \mathrm{~T}}{\mathrm{r}_{2}}$ Where, $\mathrm{T}=\text { Surface Tension }$ According to question, $\Delta \mathrm{P}_{1}=\Delta \mathrm{P}_{2}$ $\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}=\frac{2 \mathrm{~T}}{\mathrm{r}_{2}}$ $\frac{4 \mathrm{~T}}{2 \mathrm{~T}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ $\frac{2}{1}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ $\therefore \quad \mathrm{r}_{1}: \mathrm{r}_{2}=2: 1$
Kerala CEE 2012
Mechanical Properties of Fluids
143021
The excess pressure inside a spherical drop of water is four times that of another drop. then their respective mass ratio is
1 $1: 16$
2 $8: 1$
3 $1: 4$
4 $1: 64$
5 $32: 1$
Explanation:
D Excess pressure inside a spherical drop of water is given by $\Delta \mathrm{P}=\frac{2 \mathrm{~T}}{\mathrm{R}}$ Where $\mathrm{T}=$ Surface tension $\mathrm{R}=$ Radius of droplet $\Delta \mathrm{P} \propto \frac{1}{\mathrm{R}}$ $\therefore \quad \frac{\Delta \mathrm{P}_{1}}{\Delta \mathrm{P}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$ And given $\Delta \mathrm{P}_{2}=\mathrm{P}$, then $\Delta \mathrm{P}_{1}=4 \mathrm{P}$ Putting the above value in equation number (i) $\frac{4 \mathrm{P}}{\mathrm{P}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}} \Rightarrow \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1}{4}$ Now, If mass of water drop be $M$ then, $M=\rho V$ Where, $\rho=$ Density of water $\mathrm{V}=\text { Volume of water drop }$ $M=\rho \times \frac{4}{3} \pi R^{3}$ Since, $\pi$ and $\rho$ is constant $\therefore \quad \mathrm{M} \propto \mathrm{R}^{3}$ $\Rightarrow \quad \frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3} \quad\left(\because \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1}{4}\right)$ $\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\left(\frac{1}{4}\right)^{3} \Rightarrow \quad \frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\frac{1}{64}$ $\mathrm{M}_{1}: \mathrm{M}_{2}=1: 64$
Kerala CEE - 2009
Mechanical Properties of Fluids
143022
The excess pressure inside one soap bubble is three times that inside a second soap bubble, then the ratio of their surface areas is
1 $1: 9$
2 $1: 3$
3 $3: 1$
4 $1: 27$
5 $9: 1$
Explanation:
A The excess pressure inside a soap bubble is given as- $\Delta \mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{R}} \Rightarrow \Delta \mathrm{P} \propto \frac{1}{\mathrm{R}}$ Where, $T=$ Surface tension $\mathrm{R}=\text { Radius of soap bubble }$ $\Delta \mathrm{P}=\text { Excess pressure. }$ According to question- $\Delta \mathrm{P}_{1}=3 \Delta \mathrm{P}_{2}$ If $\Delta \mathrm{P}_{2}=\mathrm{P}$, then $\Delta \mathrm{P}_{1}=3 \mathrm{P}$ $\therefore \quad \frac{\Delta \mathrm{P}_{1}}{\Delta \mathrm{P}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{3}{1}$ Now, radius of surface area- $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{4 \pi \mathrm{R}_{1}^{2}}{4 \pi \mathrm{R}_{2}^{2}} \Rightarrow \frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{2}$ $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\left(\frac{1}{3}\right)^{2}$ $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{1}{9}$ $\mathrm{A}_{1}: \mathrm{A}_{2}=1: 9$