143067
Two soap bubbles of radii $r$ and $2 r$ are connected by a capillary tube-valve arrangement shown in the diagram. The valve is now opened. Then which one of the following will result.
1 The radii of the bubbles will remain unchanged
2 The bubbles will have equal radii
3 The radius of the smaller bubble will decrease and that of the bigger bubble will decrease
4 The radius of the smaller bubble will decrease and that of the bigger bubble will increase
Explanation:
D We know that, Pressure difference inside bubble $=\frac{4 \mathrm{~T}}{\mathrm{R}}$ For smaller soap, $\mathrm{P}_{\mathrm{o}}-\mathrm{P}_{\mathrm{i}}=\frac{4 \mathrm{~T}}{\mathrm{r}}$ For bigger soap, $\mathrm{P}_{\mathrm{o}}-\mathrm{P}_{\mathrm{i}}=\frac{4 \mathrm{~T}}{2 \mathrm{r}}$ We know that the pressure inside a bubble is inversely proportional to radii of the bubble. Hence pressure inside a small bubble is more than inside a bigger bubble. Thus, radius of the smaller bubble will decrease and that of the bigger bubble will increase
WB JEE 2013
Mechanical Properties of Fluids
143068
Two soap bubbles of radii $x$ and $y$ coalesce to constitutes a bubble of radius $z$. Then $z$ is equal to
1 $\sqrt{x^{2}+y^{2}}$
2 $\sqrt{x+y}$
3 $x+y$
4 $\frac{x+y}{2}$
Explanation:
A Given, radius of first soap bubble $=x$ Radius of second soap bubble $=\mathrm{y}$ Let $\mathrm{S}$ be the surface tension of the soap bubble. Excess pressure inside the soap bubble, $\mathrm{P}=\frac{4 \mathrm{~S}}{\mathrm{r}}$ When the two bubble, constitutes to form another bubble, then there is no loss of air inside the bubble. From the ideal gas equation, $\mathrm{PV}=\mathrm{n} R \mathrm{~T} \text {, }$ $\text { or } \quad \mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}}$ $\because \quad \mathrm{n}_{1}+\mathrm{n}_{2}=\mathrm{n}$ $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{RT}}+\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{RT}}=\frac{\mathrm{PV}}{\mathrm{RT}}$ $\mathrm{P}_{1} \mathrm{~V}_{1}+\mathrm{P}_{2} \mathrm{~V}_{2}=\mathrm{PV} \quad \text { (Here } \mathrm{T}=\text { constant ) }$ $\frac{4 S}{x} \times \frac{4}{3} \pi x^{3}+\frac{4 S}{y} \times \frac{4}{3} \pi y^{3}=\frac{4 S}{z} \times \frac{4}{3} \pi z^{3}$ $x^{2}+y^{2}=z^{2}$ $z=\sqrt{x^{2}+y^{2}}$
EAMCET 2003
Mechanical Properties of Fluids
143075
Two soap bubbles coalesce. It is noticed that, whilst joined together, the radii of the two bubbles are $a$ and $b$ where $a>b$. Then the radius of curvature of interface between the two bubble will be
1 $a-b$
2 $a+b$
3 $a b /(a-b)$
4 $a b /(a+b)$
Explanation:
C Let us consider the radius of curvature of the common internal film surface of the double bubble formed by two bubbles be $r$. Excess of pressure difference between $\mathrm{A}$ and $\mathrm{B}$ on either side of the common surface- $\left(\frac{4 \mathrm{~T}}{\mathrm{~b}}-\frac{4 \mathrm{~T}}{\mathrm{a}}\right)$ This will be equal to $\left(\frac{4 \mathrm{~T}}{\mathrm{r}}\right)$ $\therefore \frac{4 \mathrm{~T}}{\mathrm{r}}=\frac{4 \mathrm{~T}}{\mathrm{~b}} \frac{4 \mathrm{~T}}{\mathrm{a}}$ $\frac{1}{\mathrm{r}}=\left(\frac{1}{\mathrm{~b}}-\frac{1}{\mathrm{a}}\right)=\frac{(\mathrm{a}-\mathrm{b})}{\mathrm{ba}}$ $r=\frac{a b}{(a-b)}$
JIPMER-2017
Mechanical Properties of Fluids
143076
A soap bubble of radius $r$ is blown up to form a bubble of radius $2 \mathrm{r}$ under isothermal conditions. If $T$ is the surface tension of soap solution, the energy spent in the blowing
1 $3 \pi \mathrm{Tr}^{2}$
2 $6 \pi \mathrm{Tr}^{2}$
3 $12 \pi \mathrm{Tr}^{2}$
4 $24 \pi \mathrm{Tr}^{2}$
Explanation:
D Given, Radius of soap bubble $=\mathrm{r}$ Area of soap bubble of radius $r,\left(A_{1}\right)=4 \pi r^{2}$. Under isothermal condition radius became $2 \mathrm{r}$ then Area $\left(\mathrm{A}_{2}\right)=$ $4 \pi(2 r)^{2}=16 \pi r^{2}$ Energy spent, $\mathrm{W}=2 \mathrm{~T} \times \Delta \mathrm{A}$ $\mathrm{W}=\mathrm{T} \times 2\left(\mathrm{~A}_{2}-\mathrm{A}_{1}\right)$ $\mathrm{W}=\mathrm{T} \times 2\left(16 \pi \mathrm{r}^{2}-4 \pi \mathrm{r}^{2}\right)$ $\mathrm{W}=24 \pi \mathrm{Tr}^{2}$
143067
Two soap bubbles of radii $r$ and $2 r$ are connected by a capillary tube-valve arrangement shown in the diagram. The valve is now opened. Then which one of the following will result.
1 The radii of the bubbles will remain unchanged
2 The bubbles will have equal radii
3 The radius of the smaller bubble will decrease and that of the bigger bubble will decrease
4 The radius of the smaller bubble will decrease and that of the bigger bubble will increase
Explanation:
D We know that, Pressure difference inside bubble $=\frac{4 \mathrm{~T}}{\mathrm{R}}$ For smaller soap, $\mathrm{P}_{\mathrm{o}}-\mathrm{P}_{\mathrm{i}}=\frac{4 \mathrm{~T}}{\mathrm{r}}$ For bigger soap, $\mathrm{P}_{\mathrm{o}}-\mathrm{P}_{\mathrm{i}}=\frac{4 \mathrm{~T}}{2 \mathrm{r}}$ We know that the pressure inside a bubble is inversely proportional to radii of the bubble. Hence pressure inside a small bubble is more than inside a bigger bubble. Thus, radius of the smaller bubble will decrease and that of the bigger bubble will increase
WB JEE 2013
Mechanical Properties of Fluids
143068
Two soap bubbles of radii $x$ and $y$ coalesce to constitutes a bubble of radius $z$. Then $z$ is equal to
1 $\sqrt{x^{2}+y^{2}}$
2 $\sqrt{x+y}$
3 $x+y$
4 $\frac{x+y}{2}$
Explanation:
A Given, radius of first soap bubble $=x$ Radius of second soap bubble $=\mathrm{y}$ Let $\mathrm{S}$ be the surface tension of the soap bubble. Excess pressure inside the soap bubble, $\mathrm{P}=\frac{4 \mathrm{~S}}{\mathrm{r}}$ When the two bubble, constitutes to form another bubble, then there is no loss of air inside the bubble. From the ideal gas equation, $\mathrm{PV}=\mathrm{n} R \mathrm{~T} \text {, }$ $\text { or } \quad \mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}}$ $\because \quad \mathrm{n}_{1}+\mathrm{n}_{2}=\mathrm{n}$ $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{RT}}+\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{RT}}=\frac{\mathrm{PV}}{\mathrm{RT}}$ $\mathrm{P}_{1} \mathrm{~V}_{1}+\mathrm{P}_{2} \mathrm{~V}_{2}=\mathrm{PV} \quad \text { (Here } \mathrm{T}=\text { constant ) }$ $\frac{4 S}{x} \times \frac{4}{3} \pi x^{3}+\frac{4 S}{y} \times \frac{4}{3} \pi y^{3}=\frac{4 S}{z} \times \frac{4}{3} \pi z^{3}$ $x^{2}+y^{2}=z^{2}$ $z=\sqrt{x^{2}+y^{2}}$
EAMCET 2003
Mechanical Properties of Fluids
143075
Two soap bubbles coalesce. It is noticed that, whilst joined together, the radii of the two bubbles are $a$ and $b$ where $a>b$. Then the radius of curvature of interface between the two bubble will be
1 $a-b$
2 $a+b$
3 $a b /(a-b)$
4 $a b /(a+b)$
Explanation:
C Let us consider the radius of curvature of the common internal film surface of the double bubble formed by two bubbles be $r$. Excess of pressure difference between $\mathrm{A}$ and $\mathrm{B}$ on either side of the common surface- $\left(\frac{4 \mathrm{~T}}{\mathrm{~b}}-\frac{4 \mathrm{~T}}{\mathrm{a}}\right)$ This will be equal to $\left(\frac{4 \mathrm{~T}}{\mathrm{r}}\right)$ $\therefore \frac{4 \mathrm{~T}}{\mathrm{r}}=\frac{4 \mathrm{~T}}{\mathrm{~b}} \frac{4 \mathrm{~T}}{\mathrm{a}}$ $\frac{1}{\mathrm{r}}=\left(\frac{1}{\mathrm{~b}}-\frac{1}{\mathrm{a}}\right)=\frac{(\mathrm{a}-\mathrm{b})}{\mathrm{ba}}$ $r=\frac{a b}{(a-b)}$
JIPMER-2017
Mechanical Properties of Fluids
143076
A soap bubble of radius $r$ is blown up to form a bubble of radius $2 \mathrm{r}$ under isothermal conditions. If $T$ is the surface tension of soap solution, the energy spent in the blowing
1 $3 \pi \mathrm{Tr}^{2}$
2 $6 \pi \mathrm{Tr}^{2}$
3 $12 \pi \mathrm{Tr}^{2}$
4 $24 \pi \mathrm{Tr}^{2}$
Explanation:
D Given, Radius of soap bubble $=\mathrm{r}$ Area of soap bubble of radius $r,\left(A_{1}\right)=4 \pi r^{2}$. Under isothermal condition radius became $2 \mathrm{r}$ then Area $\left(\mathrm{A}_{2}\right)=$ $4 \pi(2 r)^{2}=16 \pi r^{2}$ Energy spent, $\mathrm{W}=2 \mathrm{~T} \times \Delta \mathrm{A}$ $\mathrm{W}=\mathrm{T} \times 2\left(\mathrm{~A}_{2}-\mathrm{A}_{1}\right)$ $\mathrm{W}=\mathrm{T} \times 2\left(16 \pi \mathrm{r}^{2}-4 \pi \mathrm{r}^{2}\right)$ $\mathrm{W}=24 \pi \mathrm{Tr}^{2}$
143067
Two soap bubbles of radii $r$ and $2 r$ are connected by a capillary tube-valve arrangement shown in the diagram. The valve is now opened. Then which one of the following will result.
1 The radii of the bubbles will remain unchanged
2 The bubbles will have equal radii
3 The radius of the smaller bubble will decrease and that of the bigger bubble will decrease
4 The radius of the smaller bubble will decrease and that of the bigger bubble will increase
Explanation:
D We know that, Pressure difference inside bubble $=\frac{4 \mathrm{~T}}{\mathrm{R}}$ For smaller soap, $\mathrm{P}_{\mathrm{o}}-\mathrm{P}_{\mathrm{i}}=\frac{4 \mathrm{~T}}{\mathrm{r}}$ For bigger soap, $\mathrm{P}_{\mathrm{o}}-\mathrm{P}_{\mathrm{i}}=\frac{4 \mathrm{~T}}{2 \mathrm{r}}$ We know that the pressure inside a bubble is inversely proportional to radii of the bubble. Hence pressure inside a small bubble is more than inside a bigger bubble. Thus, radius of the smaller bubble will decrease and that of the bigger bubble will increase
WB JEE 2013
Mechanical Properties of Fluids
143068
Two soap bubbles of radii $x$ and $y$ coalesce to constitutes a bubble of radius $z$. Then $z$ is equal to
1 $\sqrt{x^{2}+y^{2}}$
2 $\sqrt{x+y}$
3 $x+y$
4 $\frac{x+y}{2}$
Explanation:
A Given, radius of first soap bubble $=x$ Radius of second soap bubble $=\mathrm{y}$ Let $\mathrm{S}$ be the surface tension of the soap bubble. Excess pressure inside the soap bubble, $\mathrm{P}=\frac{4 \mathrm{~S}}{\mathrm{r}}$ When the two bubble, constitutes to form another bubble, then there is no loss of air inside the bubble. From the ideal gas equation, $\mathrm{PV}=\mathrm{n} R \mathrm{~T} \text {, }$ $\text { or } \quad \mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}}$ $\because \quad \mathrm{n}_{1}+\mathrm{n}_{2}=\mathrm{n}$ $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{RT}}+\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{RT}}=\frac{\mathrm{PV}}{\mathrm{RT}}$ $\mathrm{P}_{1} \mathrm{~V}_{1}+\mathrm{P}_{2} \mathrm{~V}_{2}=\mathrm{PV} \quad \text { (Here } \mathrm{T}=\text { constant ) }$ $\frac{4 S}{x} \times \frac{4}{3} \pi x^{3}+\frac{4 S}{y} \times \frac{4}{3} \pi y^{3}=\frac{4 S}{z} \times \frac{4}{3} \pi z^{3}$ $x^{2}+y^{2}=z^{2}$ $z=\sqrt{x^{2}+y^{2}}$
EAMCET 2003
Mechanical Properties of Fluids
143075
Two soap bubbles coalesce. It is noticed that, whilst joined together, the radii of the two bubbles are $a$ and $b$ where $a>b$. Then the radius of curvature of interface between the two bubble will be
1 $a-b$
2 $a+b$
3 $a b /(a-b)$
4 $a b /(a+b)$
Explanation:
C Let us consider the radius of curvature of the common internal film surface of the double bubble formed by two bubbles be $r$. Excess of pressure difference between $\mathrm{A}$ and $\mathrm{B}$ on either side of the common surface- $\left(\frac{4 \mathrm{~T}}{\mathrm{~b}}-\frac{4 \mathrm{~T}}{\mathrm{a}}\right)$ This will be equal to $\left(\frac{4 \mathrm{~T}}{\mathrm{r}}\right)$ $\therefore \frac{4 \mathrm{~T}}{\mathrm{r}}=\frac{4 \mathrm{~T}}{\mathrm{~b}} \frac{4 \mathrm{~T}}{\mathrm{a}}$ $\frac{1}{\mathrm{r}}=\left(\frac{1}{\mathrm{~b}}-\frac{1}{\mathrm{a}}\right)=\frac{(\mathrm{a}-\mathrm{b})}{\mathrm{ba}}$ $r=\frac{a b}{(a-b)}$
JIPMER-2017
Mechanical Properties of Fluids
143076
A soap bubble of radius $r$ is blown up to form a bubble of radius $2 \mathrm{r}$ under isothermal conditions. If $T$ is the surface tension of soap solution, the energy spent in the blowing
1 $3 \pi \mathrm{Tr}^{2}$
2 $6 \pi \mathrm{Tr}^{2}$
3 $12 \pi \mathrm{Tr}^{2}$
4 $24 \pi \mathrm{Tr}^{2}$
Explanation:
D Given, Radius of soap bubble $=\mathrm{r}$ Area of soap bubble of radius $r,\left(A_{1}\right)=4 \pi r^{2}$. Under isothermal condition radius became $2 \mathrm{r}$ then Area $\left(\mathrm{A}_{2}\right)=$ $4 \pi(2 r)^{2}=16 \pi r^{2}$ Energy spent, $\mathrm{W}=2 \mathrm{~T} \times \Delta \mathrm{A}$ $\mathrm{W}=\mathrm{T} \times 2\left(\mathrm{~A}_{2}-\mathrm{A}_{1}\right)$ $\mathrm{W}=\mathrm{T} \times 2\left(16 \pi \mathrm{r}^{2}-4 \pi \mathrm{r}^{2}\right)$ $\mathrm{W}=24 \pi \mathrm{Tr}^{2}$
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Mechanical Properties of Fluids
143067
Two soap bubbles of radii $r$ and $2 r$ are connected by a capillary tube-valve arrangement shown in the diagram. The valve is now opened. Then which one of the following will result.
1 The radii of the bubbles will remain unchanged
2 The bubbles will have equal radii
3 The radius of the smaller bubble will decrease and that of the bigger bubble will decrease
4 The radius of the smaller bubble will decrease and that of the bigger bubble will increase
Explanation:
D We know that, Pressure difference inside bubble $=\frac{4 \mathrm{~T}}{\mathrm{R}}$ For smaller soap, $\mathrm{P}_{\mathrm{o}}-\mathrm{P}_{\mathrm{i}}=\frac{4 \mathrm{~T}}{\mathrm{r}}$ For bigger soap, $\mathrm{P}_{\mathrm{o}}-\mathrm{P}_{\mathrm{i}}=\frac{4 \mathrm{~T}}{2 \mathrm{r}}$ We know that the pressure inside a bubble is inversely proportional to radii of the bubble. Hence pressure inside a small bubble is more than inside a bigger bubble. Thus, radius of the smaller bubble will decrease and that of the bigger bubble will increase
WB JEE 2013
Mechanical Properties of Fluids
143068
Two soap bubbles of radii $x$ and $y$ coalesce to constitutes a bubble of radius $z$. Then $z$ is equal to
1 $\sqrt{x^{2}+y^{2}}$
2 $\sqrt{x+y}$
3 $x+y$
4 $\frac{x+y}{2}$
Explanation:
A Given, radius of first soap bubble $=x$ Radius of second soap bubble $=\mathrm{y}$ Let $\mathrm{S}$ be the surface tension of the soap bubble. Excess pressure inside the soap bubble, $\mathrm{P}=\frac{4 \mathrm{~S}}{\mathrm{r}}$ When the two bubble, constitutes to form another bubble, then there is no loss of air inside the bubble. From the ideal gas equation, $\mathrm{PV}=\mathrm{n} R \mathrm{~T} \text {, }$ $\text { or } \quad \mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}}$ $\because \quad \mathrm{n}_{1}+\mathrm{n}_{2}=\mathrm{n}$ $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{RT}}+\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{RT}}=\frac{\mathrm{PV}}{\mathrm{RT}}$ $\mathrm{P}_{1} \mathrm{~V}_{1}+\mathrm{P}_{2} \mathrm{~V}_{2}=\mathrm{PV} \quad \text { (Here } \mathrm{T}=\text { constant ) }$ $\frac{4 S}{x} \times \frac{4}{3} \pi x^{3}+\frac{4 S}{y} \times \frac{4}{3} \pi y^{3}=\frac{4 S}{z} \times \frac{4}{3} \pi z^{3}$ $x^{2}+y^{2}=z^{2}$ $z=\sqrt{x^{2}+y^{2}}$
EAMCET 2003
Mechanical Properties of Fluids
143075
Two soap bubbles coalesce. It is noticed that, whilst joined together, the radii of the two bubbles are $a$ and $b$ where $a>b$. Then the radius of curvature of interface between the two bubble will be
1 $a-b$
2 $a+b$
3 $a b /(a-b)$
4 $a b /(a+b)$
Explanation:
C Let us consider the radius of curvature of the common internal film surface of the double bubble formed by two bubbles be $r$. Excess of pressure difference between $\mathrm{A}$ and $\mathrm{B}$ on either side of the common surface- $\left(\frac{4 \mathrm{~T}}{\mathrm{~b}}-\frac{4 \mathrm{~T}}{\mathrm{a}}\right)$ This will be equal to $\left(\frac{4 \mathrm{~T}}{\mathrm{r}}\right)$ $\therefore \frac{4 \mathrm{~T}}{\mathrm{r}}=\frac{4 \mathrm{~T}}{\mathrm{~b}} \frac{4 \mathrm{~T}}{\mathrm{a}}$ $\frac{1}{\mathrm{r}}=\left(\frac{1}{\mathrm{~b}}-\frac{1}{\mathrm{a}}\right)=\frac{(\mathrm{a}-\mathrm{b})}{\mathrm{ba}}$ $r=\frac{a b}{(a-b)}$
JIPMER-2017
Mechanical Properties of Fluids
143076
A soap bubble of radius $r$ is blown up to form a bubble of radius $2 \mathrm{r}$ under isothermal conditions. If $T$ is the surface tension of soap solution, the energy spent in the blowing
1 $3 \pi \mathrm{Tr}^{2}$
2 $6 \pi \mathrm{Tr}^{2}$
3 $12 \pi \mathrm{Tr}^{2}$
4 $24 \pi \mathrm{Tr}^{2}$
Explanation:
D Given, Radius of soap bubble $=\mathrm{r}$ Area of soap bubble of radius $r,\left(A_{1}\right)=4 \pi r^{2}$. Under isothermal condition radius became $2 \mathrm{r}$ then Area $\left(\mathrm{A}_{2}\right)=$ $4 \pi(2 r)^{2}=16 \pi r^{2}$ Energy spent, $\mathrm{W}=2 \mathrm{~T} \times \Delta \mathrm{A}$ $\mathrm{W}=\mathrm{T} \times 2\left(\mathrm{~A}_{2}-\mathrm{A}_{1}\right)$ $\mathrm{W}=\mathrm{T} \times 2\left(16 \pi \mathrm{r}^{2}-4 \pi \mathrm{r}^{2}\right)$ $\mathrm{W}=24 \pi \mathrm{Tr}^{2}$
