143042
If the height of a mercury barometer is $75 \mathrm{~cm}$ at sea level and $50 \mathrm{~cm}$ at the top or a hill and the ratio of density of mercury to that of air is $10^{4}$ then the height of the hill is-
1 $1.25 \mathrm{~km}$
2 $2.5 \mathrm{~km}$
3 $250 \mathrm{~m}$
4 $750 \mathrm{~m}$
Explanation:
B Given that, Height $\left(\mathrm{h}_{1}\right)=75 \mathrm{~cm}$ Height $\left(\mathrm{h}_{2}\right)=50 \mathrm{~cm}$ Pressure difference between sea level and top of hill $\Delta \mathrm{P}=\left(\mathrm{h}_{1}-\mathrm{h}_{2}\right) \times \rho_{\mathrm{Hg}} \times \mathrm{g}$ $\Delta \mathrm{P}=(75-50) \times 10^{-2} \times \rho_{\mathrm{Hg}} \times \mathrm{g}$ $\Delta \mathrm{P}=25 \times 10^{-2} \rho_{\mathrm{Hg}} \times \mathrm{g}$ Pressure difference due to $h$ meter of air $\Delta \mathrm{P}=\mathrm{h} \times \rho_{\text {air }} \times \mathrm{g}$ On equating eq ${ }^{\mathrm{n}}$ (i) and (ii), we get- $25 \times 10^{-2} \rho_{\mathrm{Hg}} \times \mathrm{g}=\rho_{\text {air }} \times \mathrm{h} \times \mathrm{g}$ $\mathrm{h}=25 \times 10^{-2}\left(\frac{\rho_{\mathrm{Hg}}}{\rho_{\text {air }}}\right)$ $\mathrm{h}=25 \times 10^{-2} \times 10^{4}$ $\mathrm{~h}=2500 \mathrm{~m}$ Height of hill $=2.5 \mathrm{~km}$
BCECE-2012
Mechanical Properties of Fluids
143043
A soap bubble $A$ of radius $0.03 \mathrm{~m}$ and another bubble $B$ of radius $0.04 \mathrm{~m}$ are brought together so that the combined bubble has a common interface of radius $r$, then the value of $r$ is-
1 $0.24 \mathrm{~m}$
2 $0.48 \mathrm{~m}$
3 $0.12 \mathrm{~m}$
4 none of these
Explanation:
C Given that, Radius of bubble $\mathrm{A}\left(\mathrm{r}_{1}\right)=0.03 \mathrm{~m}$ Radius of another bubble B $\left(\mathrm{r}_{2}\right)=0.04 \mathrm{~m}$ Excess of pressure as compared to atmosphere inside $P_{1}=\frac{4 \sigma}{r_{1}}=\frac{4 \sigma}{0.03}$ $P_{2}=\frac{4 \sigma}{r_{2}}=\frac{4 \sigma}{0.04}$ In the double bubble the pressure difference between $\mathrm{A}$ and $\mathrm{B}$ on either side of the common surface is $\mathrm{P}_{1}-\mathrm{P}_{2}=\Delta \mathrm{P}$ $\frac{4 \sigma}{0.03}-\frac{4 \sigma}{0.04}=\frac{4 \sigma}{\mathrm{r}}$ $\Rightarrow \quad \frac{1}{0.03}-\frac{1}{0.04}=\frac{1}{\mathrm{r}}$ $\Rightarrow \quad \frac{0.04-0.03}{0.0012}=\frac{1}{\mathrm{r}}$ $\Rightarrow \quad \frac{0.01}{0.0012}=\frac{1}{\mathrm{r}}$ $\mathrm{r}=0.12 \mathrm{~m}$
BCECE-2007
Mechanical Properties of Fluids
143044
The excess pressure inside the first soap bubble of radius ' $R_{1}$ ' is two time, that inside the second soap bubble of radius ' $R_{2}$ '. The ratio of volumes of the first bubble to that of second bubble is
1 $1: 2$
2 $1: 8$
3 $1: 1$
4 $1: 4$
Explanation:
B Suppose that, $\mathrm{R} \rightarrow$ radius of soap bubble, $\sigma$ is surface tension Excess pressure of soap bubble, $\mathrm{P}=\frac{4 \sigma}{\mathrm{R}}$ $\Rightarrow \quad \mathrm{PR}=$ Constant $\therefore \quad \mathrm{P}_{1} \mathrm{R}_{1}=\mathrm{P}_{2} \mathrm{R}_{2}$ As given in the question $\mathrm{P}_{1}=2 \times \mathrm{P}_{2}$ Then, $\quad 2 \times \mathrm{P}_{2} \times \mathrm{R}_{1}=\mathrm{P}_{2} \times \mathrm{R}_{2}$ $\mathrm{R}_{2}=2 \times \mathrm{R}_{1}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1}{2}$ Ratio of volumes of the first bubble to that of second bubble $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\frac{4}{3} \pi \mathrm{R}_{1}^{3}}{\frac{4}{3} \pi \mathrm{R}_{2}^{3}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\left(\frac{1}{2}\right)^{3}=\frac{1}{8}$
MHT-CET 2020
Mechanical Properties of Fluids
143045
Due to surface tension, the excess pressure inside a smaller drop is 9 units. If 27 smaller drops combine, then the excess pressure inside the bigger drop is
1 4 units
2 1 units
3 2 units
4 3 units
Explanation:
D Express pressure inside a smaller drop $=9$ units $\mathrm{P}_{\text {In }}-\mathrm{P}_{\text {out }}=\frac{4 \mathrm{~T}}{\mathrm{r}}=9$ If 27 smaller drops combine then $27 \times \frac{4}{3} \pi \times r^{3}=\frac{4}{3} \pi \times r_{1}^{3}$ $27 \mathrm{r}^{3}=\mathrm{r}_{1}^{3}$ $\mathrm{r}_{1}=3 \mathrm{r}$ $\therefore$ Excess pressure $=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}=\frac{4 \mathrm{~T}}{3 \mathrm{r}}=\frac{9}{3}=3$ unit
143042
If the height of a mercury barometer is $75 \mathrm{~cm}$ at sea level and $50 \mathrm{~cm}$ at the top or a hill and the ratio of density of mercury to that of air is $10^{4}$ then the height of the hill is-
1 $1.25 \mathrm{~km}$
2 $2.5 \mathrm{~km}$
3 $250 \mathrm{~m}$
4 $750 \mathrm{~m}$
Explanation:
B Given that, Height $\left(\mathrm{h}_{1}\right)=75 \mathrm{~cm}$ Height $\left(\mathrm{h}_{2}\right)=50 \mathrm{~cm}$ Pressure difference between sea level and top of hill $\Delta \mathrm{P}=\left(\mathrm{h}_{1}-\mathrm{h}_{2}\right) \times \rho_{\mathrm{Hg}} \times \mathrm{g}$ $\Delta \mathrm{P}=(75-50) \times 10^{-2} \times \rho_{\mathrm{Hg}} \times \mathrm{g}$ $\Delta \mathrm{P}=25 \times 10^{-2} \rho_{\mathrm{Hg}} \times \mathrm{g}$ Pressure difference due to $h$ meter of air $\Delta \mathrm{P}=\mathrm{h} \times \rho_{\text {air }} \times \mathrm{g}$ On equating eq ${ }^{\mathrm{n}}$ (i) and (ii), we get- $25 \times 10^{-2} \rho_{\mathrm{Hg}} \times \mathrm{g}=\rho_{\text {air }} \times \mathrm{h} \times \mathrm{g}$ $\mathrm{h}=25 \times 10^{-2}\left(\frac{\rho_{\mathrm{Hg}}}{\rho_{\text {air }}}\right)$ $\mathrm{h}=25 \times 10^{-2} \times 10^{4}$ $\mathrm{~h}=2500 \mathrm{~m}$ Height of hill $=2.5 \mathrm{~km}$
BCECE-2012
Mechanical Properties of Fluids
143043
A soap bubble $A$ of radius $0.03 \mathrm{~m}$ and another bubble $B$ of radius $0.04 \mathrm{~m}$ are brought together so that the combined bubble has a common interface of radius $r$, then the value of $r$ is-
1 $0.24 \mathrm{~m}$
2 $0.48 \mathrm{~m}$
3 $0.12 \mathrm{~m}$
4 none of these
Explanation:
C Given that, Radius of bubble $\mathrm{A}\left(\mathrm{r}_{1}\right)=0.03 \mathrm{~m}$ Radius of another bubble B $\left(\mathrm{r}_{2}\right)=0.04 \mathrm{~m}$ Excess of pressure as compared to atmosphere inside $P_{1}=\frac{4 \sigma}{r_{1}}=\frac{4 \sigma}{0.03}$ $P_{2}=\frac{4 \sigma}{r_{2}}=\frac{4 \sigma}{0.04}$ In the double bubble the pressure difference between $\mathrm{A}$ and $\mathrm{B}$ on either side of the common surface is $\mathrm{P}_{1}-\mathrm{P}_{2}=\Delta \mathrm{P}$ $\frac{4 \sigma}{0.03}-\frac{4 \sigma}{0.04}=\frac{4 \sigma}{\mathrm{r}}$ $\Rightarrow \quad \frac{1}{0.03}-\frac{1}{0.04}=\frac{1}{\mathrm{r}}$ $\Rightarrow \quad \frac{0.04-0.03}{0.0012}=\frac{1}{\mathrm{r}}$ $\Rightarrow \quad \frac{0.01}{0.0012}=\frac{1}{\mathrm{r}}$ $\mathrm{r}=0.12 \mathrm{~m}$
BCECE-2007
Mechanical Properties of Fluids
143044
The excess pressure inside the first soap bubble of radius ' $R_{1}$ ' is two time, that inside the second soap bubble of radius ' $R_{2}$ '. The ratio of volumes of the first bubble to that of second bubble is
1 $1: 2$
2 $1: 8$
3 $1: 1$
4 $1: 4$
Explanation:
B Suppose that, $\mathrm{R} \rightarrow$ radius of soap bubble, $\sigma$ is surface tension Excess pressure of soap bubble, $\mathrm{P}=\frac{4 \sigma}{\mathrm{R}}$ $\Rightarrow \quad \mathrm{PR}=$ Constant $\therefore \quad \mathrm{P}_{1} \mathrm{R}_{1}=\mathrm{P}_{2} \mathrm{R}_{2}$ As given in the question $\mathrm{P}_{1}=2 \times \mathrm{P}_{2}$ Then, $\quad 2 \times \mathrm{P}_{2} \times \mathrm{R}_{1}=\mathrm{P}_{2} \times \mathrm{R}_{2}$ $\mathrm{R}_{2}=2 \times \mathrm{R}_{1}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1}{2}$ Ratio of volumes of the first bubble to that of second bubble $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\frac{4}{3} \pi \mathrm{R}_{1}^{3}}{\frac{4}{3} \pi \mathrm{R}_{2}^{3}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\left(\frac{1}{2}\right)^{3}=\frac{1}{8}$
MHT-CET 2020
Mechanical Properties of Fluids
143045
Due to surface tension, the excess pressure inside a smaller drop is 9 units. If 27 smaller drops combine, then the excess pressure inside the bigger drop is
1 4 units
2 1 units
3 2 units
4 3 units
Explanation:
D Express pressure inside a smaller drop $=9$ units $\mathrm{P}_{\text {In }}-\mathrm{P}_{\text {out }}=\frac{4 \mathrm{~T}}{\mathrm{r}}=9$ If 27 smaller drops combine then $27 \times \frac{4}{3} \pi \times r^{3}=\frac{4}{3} \pi \times r_{1}^{3}$ $27 \mathrm{r}^{3}=\mathrm{r}_{1}^{3}$ $\mathrm{r}_{1}=3 \mathrm{r}$ $\therefore$ Excess pressure $=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}=\frac{4 \mathrm{~T}}{3 \mathrm{r}}=\frac{9}{3}=3$ unit
143042
If the height of a mercury barometer is $75 \mathrm{~cm}$ at sea level and $50 \mathrm{~cm}$ at the top or a hill and the ratio of density of mercury to that of air is $10^{4}$ then the height of the hill is-
1 $1.25 \mathrm{~km}$
2 $2.5 \mathrm{~km}$
3 $250 \mathrm{~m}$
4 $750 \mathrm{~m}$
Explanation:
B Given that, Height $\left(\mathrm{h}_{1}\right)=75 \mathrm{~cm}$ Height $\left(\mathrm{h}_{2}\right)=50 \mathrm{~cm}$ Pressure difference between sea level and top of hill $\Delta \mathrm{P}=\left(\mathrm{h}_{1}-\mathrm{h}_{2}\right) \times \rho_{\mathrm{Hg}} \times \mathrm{g}$ $\Delta \mathrm{P}=(75-50) \times 10^{-2} \times \rho_{\mathrm{Hg}} \times \mathrm{g}$ $\Delta \mathrm{P}=25 \times 10^{-2} \rho_{\mathrm{Hg}} \times \mathrm{g}$ Pressure difference due to $h$ meter of air $\Delta \mathrm{P}=\mathrm{h} \times \rho_{\text {air }} \times \mathrm{g}$ On equating eq ${ }^{\mathrm{n}}$ (i) and (ii), we get- $25 \times 10^{-2} \rho_{\mathrm{Hg}} \times \mathrm{g}=\rho_{\text {air }} \times \mathrm{h} \times \mathrm{g}$ $\mathrm{h}=25 \times 10^{-2}\left(\frac{\rho_{\mathrm{Hg}}}{\rho_{\text {air }}}\right)$ $\mathrm{h}=25 \times 10^{-2} \times 10^{4}$ $\mathrm{~h}=2500 \mathrm{~m}$ Height of hill $=2.5 \mathrm{~km}$
BCECE-2012
Mechanical Properties of Fluids
143043
A soap bubble $A$ of radius $0.03 \mathrm{~m}$ and another bubble $B$ of radius $0.04 \mathrm{~m}$ are brought together so that the combined bubble has a common interface of radius $r$, then the value of $r$ is-
1 $0.24 \mathrm{~m}$
2 $0.48 \mathrm{~m}$
3 $0.12 \mathrm{~m}$
4 none of these
Explanation:
C Given that, Radius of bubble $\mathrm{A}\left(\mathrm{r}_{1}\right)=0.03 \mathrm{~m}$ Radius of another bubble B $\left(\mathrm{r}_{2}\right)=0.04 \mathrm{~m}$ Excess of pressure as compared to atmosphere inside $P_{1}=\frac{4 \sigma}{r_{1}}=\frac{4 \sigma}{0.03}$ $P_{2}=\frac{4 \sigma}{r_{2}}=\frac{4 \sigma}{0.04}$ In the double bubble the pressure difference between $\mathrm{A}$ and $\mathrm{B}$ on either side of the common surface is $\mathrm{P}_{1}-\mathrm{P}_{2}=\Delta \mathrm{P}$ $\frac{4 \sigma}{0.03}-\frac{4 \sigma}{0.04}=\frac{4 \sigma}{\mathrm{r}}$ $\Rightarrow \quad \frac{1}{0.03}-\frac{1}{0.04}=\frac{1}{\mathrm{r}}$ $\Rightarrow \quad \frac{0.04-0.03}{0.0012}=\frac{1}{\mathrm{r}}$ $\Rightarrow \quad \frac{0.01}{0.0012}=\frac{1}{\mathrm{r}}$ $\mathrm{r}=0.12 \mathrm{~m}$
BCECE-2007
Mechanical Properties of Fluids
143044
The excess pressure inside the first soap bubble of radius ' $R_{1}$ ' is two time, that inside the second soap bubble of radius ' $R_{2}$ '. The ratio of volumes of the first bubble to that of second bubble is
1 $1: 2$
2 $1: 8$
3 $1: 1$
4 $1: 4$
Explanation:
B Suppose that, $\mathrm{R} \rightarrow$ radius of soap bubble, $\sigma$ is surface tension Excess pressure of soap bubble, $\mathrm{P}=\frac{4 \sigma}{\mathrm{R}}$ $\Rightarrow \quad \mathrm{PR}=$ Constant $\therefore \quad \mathrm{P}_{1} \mathrm{R}_{1}=\mathrm{P}_{2} \mathrm{R}_{2}$ As given in the question $\mathrm{P}_{1}=2 \times \mathrm{P}_{2}$ Then, $\quad 2 \times \mathrm{P}_{2} \times \mathrm{R}_{1}=\mathrm{P}_{2} \times \mathrm{R}_{2}$ $\mathrm{R}_{2}=2 \times \mathrm{R}_{1}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1}{2}$ Ratio of volumes of the first bubble to that of second bubble $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\frac{4}{3} \pi \mathrm{R}_{1}^{3}}{\frac{4}{3} \pi \mathrm{R}_{2}^{3}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\left(\frac{1}{2}\right)^{3}=\frac{1}{8}$
MHT-CET 2020
Mechanical Properties of Fluids
143045
Due to surface tension, the excess pressure inside a smaller drop is 9 units. If 27 smaller drops combine, then the excess pressure inside the bigger drop is
1 4 units
2 1 units
3 2 units
4 3 units
Explanation:
D Express pressure inside a smaller drop $=9$ units $\mathrm{P}_{\text {In }}-\mathrm{P}_{\text {out }}=\frac{4 \mathrm{~T}}{\mathrm{r}}=9$ If 27 smaller drops combine then $27 \times \frac{4}{3} \pi \times r^{3}=\frac{4}{3} \pi \times r_{1}^{3}$ $27 \mathrm{r}^{3}=\mathrm{r}_{1}^{3}$ $\mathrm{r}_{1}=3 \mathrm{r}$ $\therefore$ Excess pressure $=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}=\frac{4 \mathrm{~T}}{3 \mathrm{r}}=\frac{9}{3}=3$ unit
143042
If the height of a mercury barometer is $75 \mathrm{~cm}$ at sea level and $50 \mathrm{~cm}$ at the top or a hill and the ratio of density of mercury to that of air is $10^{4}$ then the height of the hill is-
1 $1.25 \mathrm{~km}$
2 $2.5 \mathrm{~km}$
3 $250 \mathrm{~m}$
4 $750 \mathrm{~m}$
Explanation:
B Given that, Height $\left(\mathrm{h}_{1}\right)=75 \mathrm{~cm}$ Height $\left(\mathrm{h}_{2}\right)=50 \mathrm{~cm}$ Pressure difference between sea level and top of hill $\Delta \mathrm{P}=\left(\mathrm{h}_{1}-\mathrm{h}_{2}\right) \times \rho_{\mathrm{Hg}} \times \mathrm{g}$ $\Delta \mathrm{P}=(75-50) \times 10^{-2} \times \rho_{\mathrm{Hg}} \times \mathrm{g}$ $\Delta \mathrm{P}=25 \times 10^{-2} \rho_{\mathrm{Hg}} \times \mathrm{g}$ Pressure difference due to $h$ meter of air $\Delta \mathrm{P}=\mathrm{h} \times \rho_{\text {air }} \times \mathrm{g}$ On equating eq ${ }^{\mathrm{n}}$ (i) and (ii), we get- $25 \times 10^{-2} \rho_{\mathrm{Hg}} \times \mathrm{g}=\rho_{\text {air }} \times \mathrm{h} \times \mathrm{g}$ $\mathrm{h}=25 \times 10^{-2}\left(\frac{\rho_{\mathrm{Hg}}}{\rho_{\text {air }}}\right)$ $\mathrm{h}=25 \times 10^{-2} \times 10^{4}$ $\mathrm{~h}=2500 \mathrm{~m}$ Height of hill $=2.5 \mathrm{~km}$
BCECE-2012
Mechanical Properties of Fluids
143043
A soap bubble $A$ of radius $0.03 \mathrm{~m}$ and another bubble $B$ of radius $0.04 \mathrm{~m}$ are brought together so that the combined bubble has a common interface of radius $r$, then the value of $r$ is-
1 $0.24 \mathrm{~m}$
2 $0.48 \mathrm{~m}$
3 $0.12 \mathrm{~m}$
4 none of these
Explanation:
C Given that, Radius of bubble $\mathrm{A}\left(\mathrm{r}_{1}\right)=0.03 \mathrm{~m}$ Radius of another bubble B $\left(\mathrm{r}_{2}\right)=0.04 \mathrm{~m}$ Excess of pressure as compared to atmosphere inside $P_{1}=\frac{4 \sigma}{r_{1}}=\frac{4 \sigma}{0.03}$ $P_{2}=\frac{4 \sigma}{r_{2}}=\frac{4 \sigma}{0.04}$ In the double bubble the pressure difference between $\mathrm{A}$ and $\mathrm{B}$ on either side of the common surface is $\mathrm{P}_{1}-\mathrm{P}_{2}=\Delta \mathrm{P}$ $\frac{4 \sigma}{0.03}-\frac{4 \sigma}{0.04}=\frac{4 \sigma}{\mathrm{r}}$ $\Rightarrow \quad \frac{1}{0.03}-\frac{1}{0.04}=\frac{1}{\mathrm{r}}$ $\Rightarrow \quad \frac{0.04-0.03}{0.0012}=\frac{1}{\mathrm{r}}$ $\Rightarrow \quad \frac{0.01}{0.0012}=\frac{1}{\mathrm{r}}$ $\mathrm{r}=0.12 \mathrm{~m}$
BCECE-2007
Mechanical Properties of Fluids
143044
The excess pressure inside the first soap bubble of radius ' $R_{1}$ ' is two time, that inside the second soap bubble of radius ' $R_{2}$ '. The ratio of volumes of the first bubble to that of second bubble is
1 $1: 2$
2 $1: 8$
3 $1: 1$
4 $1: 4$
Explanation:
B Suppose that, $\mathrm{R} \rightarrow$ radius of soap bubble, $\sigma$ is surface tension Excess pressure of soap bubble, $\mathrm{P}=\frac{4 \sigma}{\mathrm{R}}$ $\Rightarrow \quad \mathrm{PR}=$ Constant $\therefore \quad \mathrm{P}_{1} \mathrm{R}_{1}=\mathrm{P}_{2} \mathrm{R}_{2}$ As given in the question $\mathrm{P}_{1}=2 \times \mathrm{P}_{2}$ Then, $\quad 2 \times \mathrm{P}_{2} \times \mathrm{R}_{1}=\mathrm{P}_{2} \times \mathrm{R}_{2}$ $\mathrm{R}_{2}=2 \times \mathrm{R}_{1}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1}{2}$ Ratio of volumes of the first bubble to that of second bubble $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\frac{4}{3} \pi \mathrm{R}_{1}^{3}}{\frac{4}{3} \pi \mathrm{R}_{2}^{3}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\left(\frac{1}{2}\right)^{3}=\frac{1}{8}$
MHT-CET 2020
Mechanical Properties of Fluids
143045
Due to surface tension, the excess pressure inside a smaller drop is 9 units. If 27 smaller drops combine, then the excess pressure inside the bigger drop is
1 4 units
2 1 units
3 2 units
4 3 units
Explanation:
D Express pressure inside a smaller drop $=9$ units $\mathrm{P}_{\text {In }}-\mathrm{P}_{\text {out }}=\frac{4 \mathrm{~T}}{\mathrm{r}}=9$ If 27 smaller drops combine then $27 \times \frac{4}{3} \pi \times r^{3}=\frac{4}{3} \pi \times r_{1}^{3}$ $27 \mathrm{r}^{3}=\mathrm{r}_{1}^{3}$ $\mathrm{r}_{1}=3 \mathrm{r}$ $\therefore$ Excess pressure $=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}=\frac{4 \mathrm{~T}}{3 \mathrm{r}}=\frac{9}{3}=3$ unit
