143038
A soap bubble in vacuum has a radius $3 \mathrm{~cm}$ and another soap bubble in vacuum has radius $4 \mathrm{~cm}$. If two bubbles coalesce under isothermal condition. Then the radius of the new bubble will be:
1 $7 \mathrm{~cm}$
2 $5 \mathrm{~cm}$
3 $4.5 \mathrm{~cm}$
4 $2.3 \mathrm{~cm}$
Explanation:
B Given that, The radius of bubble $\left(\mathrm{r}_{1}\right)=3 \mathrm{~cm}$ The radius of another bubble $\left(\mathrm{r}_{2}\right)=4 \mathrm{~cm}$ Isothermal condition- $\mathrm{PV}=\text { constant }$ $\mathrm{P}_{1} \mathrm{~V}_{1}+\mathrm{P}_{2} \mathrm{~V}_{2}=\mathrm{PV}$ $\Rightarrow \quad \frac{4 \sigma}{\mathrm{r}_{1}} \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}+\frac{4 \sigma}{\mathrm{r}_{2}} \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}=\frac{4 \sigma}{\mathrm{r}} \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}=\mathrm{r}^{2}$ Now putting the value of $r_{1}$ and $r_{2}$ we get - $3^{2}+4^{2}=r^{2}$ $r=5 \mathrm{~cm}$
AIIMS-2002
Mechanical Properties of Fluids
143039
A spherical drop of water has $1 \mathrm{~mm}$ radius. If the surface tension of water is $70 \times 10^{-3} \mathrm{~N} / \mathrm{m}$. Then the difference of pressures between inside and outside of the spherical drop is:
1 $140 \mathrm{~N} / \mathrm{m}^{2}$
2 $140 \mathrm{~N} / \mathrm{m}$
3 $35 \mathrm{Nm}^{2}$
4 none of these
Explanation:
A Given that, Radius of drop of water $(\mathrm{r})=1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m}$ Surface tension of water $(\sigma)=70 \times 10^{-3} \mathrm{~N} / \mathrm{m}$ Then, the difference of pressure between inside and outside spherical drop $\Delta \mathrm{P}=\frac{2 \sigma}{\mathrm{r}}=\frac{2 \times 70 \times 10^{-3}}{1 \times 10^{-3}}$ $\Delta \mathrm{P}=140 \mathrm{~N} / \mathrm{m}^{2}$
AIIMS-2001
Mechanical Properties of Fluids
143040
The excess pressure inside the first soap bubble is three times that inside second bubble then, the ratio of volume of the first to the second bubble will be:
1 $1: 27$
2 $3: 1$
3 $1: 3$
4 $1: 9$
Explanation:
A Given that, $\mathrm{P}_{1}=3 \mathrm{P}_{2}$ As we known, $\mathrm{P}_{1}=\frac{4 \sigma}{\mathrm{r}_{1}}$ And, $\mathrm{P}_{2}=\frac{4 \sigma}{\mathrm{r}_{2}}$ Put these value in equation (i), we get- $\frac{4 \sigma}{r_{1}}=3 \times \frac{4 \sigma}{r_{2}}$ $r_{2}=3 r_{1}$ $\frac{r_{1}}{r_{2}}=\frac{1}{3}$ So the ratios of volumes- $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\frac{4}{3} \pi \mathrm{r}_{1}^{3}}{\frac{4}{3} \pi \mathrm{r}_{2}^{3}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{3}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{1}{27}$ $\mathrm{V}_{1}: \mathrm{V}_{2}=1: 27$
AIIMS-1999
Mechanical Properties of Fluids
143041
Find the difference of air pressure between the inside and outside of a soap bubble $5 \mathrm{~mm}$ in diameter, if the surface tension is $1.6 \mathrm{~N} / \mathrm{m}$.
1 $2560 \mathrm{~N} / \mathrm{m}^{2}$
2 $3720 \mathrm{~N} / \mathrm{m}^{2}$
3 $1208 \mathrm{~N} / \mathrm{m}^{2}$
4 $950 \mathrm{~N} / \mathrm{m}^{2}$
Explanation:
A Given that, Diameter $(\mathrm{d})=5 \mathrm{~mm}$ Radius $(\mathrm{r})=2.5 \mathrm{~mm}=2.5 \times 10^{-3} \mathrm{~m}$ Surface tension $(\sigma)=1.6 \mathrm{~N} / \mathrm{m}$ Difference of air pressure between inside and outside of soap bubble $\Delta \mathrm{P} =\frac{4 \sigma}{\mathrm{r}}=\frac{4 \times 1.6}{2.5 \times 10^{-3}}$ $\Delta \mathrm{P} =\frac{4 \times 10^{3} \times 16}{25}$ $\Delta \mathrm{P} =2560 \mathrm{~N} / \mathrm{m}^{2}$
143038
A soap bubble in vacuum has a radius $3 \mathrm{~cm}$ and another soap bubble in vacuum has radius $4 \mathrm{~cm}$. If two bubbles coalesce under isothermal condition. Then the radius of the new bubble will be:
1 $7 \mathrm{~cm}$
2 $5 \mathrm{~cm}$
3 $4.5 \mathrm{~cm}$
4 $2.3 \mathrm{~cm}$
Explanation:
B Given that, The radius of bubble $\left(\mathrm{r}_{1}\right)=3 \mathrm{~cm}$ The radius of another bubble $\left(\mathrm{r}_{2}\right)=4 \mathrm{~cm}$ Isothermal condition- $\mathrm{PV}=\text { constant }$ $\mathrm{P}_{1} \mathrm{~V}_{1}+\mathrm{P}_{2} \mathrm{~V}_{2}=\mathrm{PV}$ $\Rightarrow \quad \frac{4 \sigma}{\mathrm{r}_{1}} \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}+\frac{4 \sigma}{\mathrm{r}_{2}} \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}=\frac{4 \sigma}{\mathrm{r}} \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}=\mathrm{r}^{2}$ Now putting the value of $r_{1}$ and $r_{2}$ we get - $3^{2}+4^{2}=r^{2}$ $r=5 \mathrm{~cm}$
AIIMS-2002
Mechanical Properties of Fluids
143039
A spherical drop of water has $1 \mathrm{~mm}$ radius. If the surface tension of water is $70 \times 10^{-3} \mathrm{~N} / \mathrm{m}$. Then the difference of pressures between inside and outside of the spherical drop is:
1 $140 \mathrm{~N} / \mathrm{m}^{2}$
2 $140 \mathrm{~N} / \mathrm{m}$
3 $35 \mathrm{Nm}^{2}$
4 none of these
Explanation:
A Given that, Radius of drop of water $(\mathrm{r})=1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m}$ Surface tension of water $(\sigma)=70 \times 10^{-3} \mathrm{~N} / \mathrm{m}$ Then, the difference of pressure between inside and outside spherical drop $\Delta \mathrm{P}=\frac{2 \sigma}{\mathrm{r}}=\frac{2 \times 70 \times 10^{-3}}{1 \times 10^{-3}}$ $\Delta \mathrm{P}=140 \mathrm{~N} / \mathrm{m}^{2}$
AIIMS-2001
Mechanical Properties of Fluids
143040
The excess pressure inside the first soap bubble is three times that inside second bubble then, the ratio of volume of the first to the second bubble will be:
1 $1: 27$
2 $3: 1$
3 $1: 3$
4 $1: 9$
Explanation:
A Given that, $\mathrm{P}_{1}=3 \mathrm{P}_{2}$ As we known, $\mathrm{P}_{1}=\frac{4 \sigma}{\mathrm{r}_{1}}$ And, $\mathrm{P}_{2}=\frac{4 \sigma}{\mathrm{r}_{2}}$ Put these value in equation (i), we get- $\frac{4 \sigma}{r_{1}}=3 \times \frac{4 \sigma}{r_{2}}$ $r_{2}=3 r_{1}$ $\frac{r_{1}}{r_{2}}=\frac{1}{3}$ So the ratios of volumes- $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\frac{4}{3} \pi \mathrm{r}_{1}^{3}}{\frac{4}{3} \pi \mathrm{r}_{2}^{3}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{3}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{1}{27}$ $\mathrm{V}_{1}: \mathrm{V}_{2}=1: 27$
AIIMS-1999
Mechanical Properties of Fluids
143041
Find the difference of air pressure between the inside and outside of a soap bubble $5 \mathrm{~mm}$ in diameter, if the surface tension is $1.6 \mathrm{~N} / \mathrm{m}$.
1 $2560 \mathrm{~N} / \mathrm{m}^{2}$
2 $3720 \mathrm{~N} / \mathrm{m}^{2}$
3 $1208 \mathrm{~N} / \mathrm{m}^{2}$
4 $950 \mathrm{~N} / \mathrm{m}^{2}$
Explanation:
A Given that, Diameter $(\mathrm{d})=5 \mathrm{~mm}$ Radius $(\mathrm{r})=2.5 \mathrm{~mm}=2.5 \times 10^{-3} \mathrm{~m}$ Surface tension $(\sigma)=1.6 \mathrm{~N} / \mathrm{m}$ Difference of air pressure between inside and outside of soap bubble $\Delta \mathrm{P} =\frac{4 \sigma}{\mathrm{r}}=\frac{4 \times 1.6}{2.5 \times 10^{-3}}$ $\Delta \mathrm{P} =\frac{4 \times 10^{3} \times 16}{25}$ $\Delta \mathrm{P} =2560 \mathrm{~N} / \mathrm{m}^{2}$
143038
A soap bubble in vacuum has a radius $3 \mathrm{~cm}$ and another soap bubble in vacuum has radius $4 \mathrm{~cm}$. If two bubbles coalesce under isothermal condition. Then the radius of the new bubble will be:
1 $7 \mathrm{~cm}$
2 $5 \mathrm{~cm}$
3 $4.5 \mathrm{~cm}$
4 $2.3 \mathrm{~cm}$
Explanation:
B Given that, The radius of bubble $\left(\mathrm{r}_{1}\right)=3 \mathrm{~cm}$ The radius of another bubble $\left(\mathrm{r}_{2}\right)=4 \mathrm{~cm}$ Isothermal condition- $\mathrm{PV}=\text { constant }$ $\mathrm{P}_{1} \mathrm{~V}_{1}+\mathrm{P}_{2} \mathrm{~V}_{2}=\mathrm{PV}$ $\Rightarrow \quad \frac{4 \sigma}{\mathrm{r}_{1}} \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}+\frac{4 \sigma}{\mathrm{r}_{2}} \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}=\frac{4 \sigma}{\mathrm{r}} \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}=\mathrm{r}^{2}$ Now putting the value of $r_{1}$ and $r_{2}$ we get - $3^{2}+4^{2}=r^{2}$ $r=5 \mathrm{~cm}$
AIIMS-2002
Mechanical Properties of Fluids
143039
A spherical drop of water has $1 \mathrm{~mm}$ radius. If the surface tension of water is $70 \times 10^{-3} \mathrm{~N} / \mathrm{m}$. Then the difference of pressures between inside and outside of the spherical drop is:
1 $140 \mathrm{~N} / \mathrm{m}^{2}$
2 $140 \mathrm{~N} / \mathrm{m}$
3 $35 \mathrm{Nm}^{2}$
4 none of these
Explanation:
A Given that, Radius of drop of water $(\mathrm{r})=1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m}$ Surface tension of water $(\sigma)=70 \times 10^{-3} \mathrm{~N} / \mathrm{m}$ Then, the difference of pressure between inside and outside spherical drop $\Delta \mathrm{P}=\frac{2 \sigma}{\mathrm{r}}=\frac{2 \times 70 \times 10^{-3}}{1 \times 10^{-3}}$ $\Delta \mathrm{P}=140 \mathrm{~N} / \mathrm{m}^{2}$
AIIMS-2001
Mechanical Properties of Fluids
143040
The excess pressure inside the first soap bubble is three times that inside second bubble then, the ratio of volume of the first to the second bubble will be:
1 $1: 27$
2 $3: 1$
3 $1: 3$
4 $1: 9$
Explanation:
A Given that, $\mathrm{P}_{1}=3 \mathrm{P}_{2}$ As we known, $\mathrm{P}_{1}=\frac{4 \sigma}{\mathrm{r}_{1}}$ And, $\mathrm{P}_{2}=\frac{4 \sigma}{\mathrm{r}_{2}}$ Put these value in equation (i), we get- $\frac{4 \sigma}{r_{1}}=3 \times \frac{4 \sigma}{r_{2}}$ $r_{2}=3 r_{1}$ $\frac{r_{1}}{r_{2}}=\frac{1}{3}$ So the ratios of volumes- $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\frac{4}{3} \pi \mathrm{r}_{1}^{3}}{\frac{4}{3} \pi \mathrm{r}_{2}^{3}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{3}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{1}{27}$ $\mathrm{V}_{1}: \mathrm{V}_{2}=1: 27$
AIIMS-1999
Mechanical Properties of Fluids
143041
Find the difference of air pressure between the inside and outside of a soap bubble $5 \mathrm{~mm}$ in diameter, if the surface tension is $1.6 \mathrm{~N} / \mathrm{m}$.
1 $2560 \mathrm{~N} / \mathrm{m}^{2}$
2 $3720 \mathrm{~N} / \mathrm{m}^{2}$
3 $1208 \mathrm{~N} / \mathrm{m}^{2}$
4 $950 \mathrm{~N} / \mathrm{m}^{2}$
Explanation:
A Given that, Diameter $(\mathrm{d})=5 \mathrm{~mm}$ Radius $(\mathrm{r})=2.5 \mathrm{~mm}=2.5 \times 10^{-3} \mathrm{~m}$ Surface tension $(\sigma)=1.6 \mathrm{~N} / \mathrm{m}$ Difference of air pressure between inside and outside of soap bubble $\Delta \mathrm{P} =\frac{4 \sigma}{\mathrm{r}}=\frac{4 \times 1.6}{2.5 \times 10^{-3}}$ $\Delta \mathrm{P} =\frac{4 \times 10^{3} \times 16}{25}$ $\Delta \mathrm{P} =2560 \mathrm{~N} / \mathrm{m}^{2}$
143038
A soap bubble in vacuum has a radius $3 \mathrm{~cm}$ and another soap bubble in vacuum has radius $4 \mathrm{~cm}$. If two bubbles coalesce under isothermal condition. Then the radius of the new bubble will be:
1 $7 \mathrm{~cm}$
2 $5 \mathrm{~cm}$
3 $4.5 \mathrm{~cm}$
4 $2.3 \mathrm{~cm}$
Explanation:
B Given that, The radius of bubble $\left(\mathrm{r}_{1}\right)=3 \mathrm{~cm}$ The radius of another bubble $\left(\mathrm{r}_{2}\right)=4 \mathrm{~cm}$ Isothermal condition- $\mathrm{PV}=\text { constant }$ $\mathrm{P}_{1} \mathrm{~V}_{1}+\mathrm{P}_{2} \mathrm{~V}_{2}=\mathrm{PV}$ $\Rightarrow \quad \frac{4 \sigma}{\mathrm{r}_{1}} \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}+\frac{4 \sigma}{\mathrm{r}_{2}} \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}=\frac{4 \sigma}{\mathrm{r}} \times \frac{4}{3} \pi \mathrm{r}^{3}$ $\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}=\mathrm{r}^{2}$ Now putting the value of $r_{1}$ and $r_{2}$ we get - $3^{2}+4^{2}=r^{2}$ $r=5 \mathrm{~cm}$
AIIMS-2002
Mechanical Properties of Fluids
143039
A spherical drop of water has $1 \mathrm{~mm}$ radius. If the surface tension of water is $70 \times 10^{-3} \mathrm{~N} / \mathrm{m}$. Then the difference of pressures between inside and outside of the spherical drop is:
1 $140 \mathrm{~N} / \mathrm{m}^{2}$
2 $140 \mathrm{~N} / \mathrm{m}$
3 $35 \mathrm{Nm}^{2}$
4 none of these
Explanation:
A Given that, Radius of drop of water $(\mathrm{r})=1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m}$ Surface tension of water $(\sigma)=70 \times 10^{-3} \mathrm{~N} / \mathrm{m}$ Then, the difference of pressure between inside and outside spherical drop $\Delta \mathrm{P}=\frac{2 \sigma}{\mathrm{r}}=\frac{2 \times 70 \times 10^{-3}}{1 \times 10^{-3}}$ $\Delta \mathrm{P}=140 \mathrm{~N} / \mathrm{m}^{2}$
AIIMS-2001
Mechanical Properties of Fluids
143040
The excess pressure inside the first soap bubble is three times that inside second bubble then, the ratio of volume of the first to the second bubble will be:
1 $1: 27$
2 $3: 1$
3 $1: 3$
4 $1: 9$
Explanation:
A Given that, $\mathrm{P}_{1}=3 \mathrm{P}_{2}$ As we known, $\mathrm{P}_{1}=\frac{4 \sigma}{\mathrm{r}_{1}}$ And, $\mathrm{P}_{2}=\frac{4 \sigma}{\mathrm{r}_{2}}$ Put these value in equation (i), we get- $\frac{4 \sigma}{r_{1}}=3 \times \frac{4 \sigma}{r_{2}}$ $r_{2}=3 r_{1}$ $\frac{r_{1}}{r_{2}}=\frac{1}{3}$ So the ratios of volumes- $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\frac{4}{3} \pi \mathrm{r}_{1}^{3}}{\frac{4}{3} \pi \mathrm{r}_{2}^{3}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{3}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{1}{27}$ $\mathrm{V}_{1}: \mathrm{V}_{2}=1: 27$
AIIMS-1999
Mechanical Properties of Fluids
143041
Find the difference of air pressure between the inside and outside of a soap bubble $5 \mathrm{~mm}$ in diameter, if the surface tension is $1.6 \mathrm{~N} / \mathrm{m}$.
1 $2560 \mathrm{~N} / \mathrm{m}^{2}$
2 $3720 \mathrm{~N} / \mathrm{m}^{2}$
3 $1208 \mathrm{~N} / \mathrm{m}^{2}$
4 $950 \mathrm{~N} / \mathrm{m}^{2}$
Explanation:
A Given that, Diameter $(\mathrm{d})=5 \mathrm{~mm}$ Radius $(\mathrm{r})=2.5 \mathrm{~mm}=2.5 \times 10^{-3} \mathrm{~m}$ Surface tension $(\sigma)=1.6 \mathrm{~N} / \mathrm{m}$ Difference of air pressure between inside and outside of soap bubble $\Delta \mathrm{P} =\frac{4 \sigma}{\mathrm{r}}=\frac{4 \times 1.6}{2.5 \times 10^{-3}}$ $\Delta \mathrm{P} =\frac{4 \times 10^{3} \times 16}{25}$ $\Delta \mathrm{P} =2560 \mathrm{~N} / \mathrm{m}^{2}$
