143052
If two soap bubbles of equal radii $r$ coalesce then the radius of curvature of interface between two bubbles will be
1 $\mathrm{r}$
2 0
3 infinity
4 $\frac{1}{2} \mathrm{r}$
Explanation:
C $P_{1}=P_{0}+\frac{4 T}{r}$ $\mathrm{P}_{2}=\mathrm{P}_{0}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}$ $\mathrm{P}_{1}>\mathrm{P}_{2}$ $\Delta \mathrm{P}=\mathrm{P}_{1}-\mathrm{P}_{2}=4 \mathrm{~T}\left[\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{r}_{2}}\right]$ $\frac{1}{\mathrm{r}}=\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{r}_{2}}$ $r=\frac{r_{1} r_{2}}{r_{2}-r_{1}}$ Both bubble of are equal radii, then $\mathrm{r}=\infty$
J and K CET- 2005
Mechanical Properties of Fluids
143054
Two equal drops of water are falling through air with a steady velocity of $5 \mathrm{~cm} / \mathrm{s}$. If the drops coalesce, the new steady velocity of the coalesced drop will be
1 $5 \mathrm{~cm} / \mathrm{s}$
2 $10 \mathrm{~cm} / \mathrm{s}$
3 $7.9 \mathrm{~cm} / \mathrm{s}$
4 $6 \mathrm{~cm} / \mathrm{s}$
Explanation:
C Let $\mathrm{R}$ is radius of bigger drop, $\mathrm{r}$ is radius of smaller drop $\frac{4}{3} \pi \mathrm{R}^{3}=2 \times \frac{4}{3} \times \pi \mathrm{r}^{3}$ $\mathrm{R}=(2)^{1 / 3} \cdot \mathrm{r}$ Steady velocity of bigger drop $\mathrm{v}_{1}=$ ? Steady velocity of smaller drop. $\mathrm{v}=5 \mathrm{~cm} / \mathrm{s}$ Terminal velocity, $\mathrm{v} \propto \mathrm{r}^{2}$ $\therefore \frac{\mathrm{v}_{1}}{\mathrm{v}}=\frac{\mathrm{R}^{2}}{\mathrm{r}^{2}}$ $\mathrm{v}_{1}=\frac{\left(2^{1 / 3}\right)^{2} \cdot r^{2}}{\mathrm{r}^{2}} \times 5$ $\mathrm{v}_{1}=5 \times 4^{1 / 3}=5 \times 1.59$ $\mathrm{v}_{1}=7.9 \mathrm{~cm} / \mathrm{s}$
J and K CET- 2000
Mechanical Properties of Fluids
143055
A number of water droplets of radius $r$ coalesce to form a drop of radius $R$. Assuming that the entire energy liberated due to coalesce goes into heating the drop, the rise in temperature $\mathrm{d} \theta$ is (surface tension of water $=T$ )
D Let, $r$ be the radius of small droplets and $R$ be the radius of forming droplet. A number of water droplets of radius $r$ coalesce to form a drop of radius $\mathrm{R}$. $n \times \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3}$ $n=\left(\frac{R}{r}\right)^{3} \ldots \ldots . .(i)$ The entire energy liberated due to coalesces goes into heating the drop. $\mathrm{H}=\frac{\Delta \mathrm{E}}{\mathrm{J}} \quad\{\because \mathrm{H}=\mathrm{ms} \Delta \theta\}$ $\mathrm{ms} \Delta \theta=\frac{\text { Surface tension } \times \text { Changein surface area }}{\mathrm{J}}$ $\mathrm{ms} \Delta \theta=\frac{\mathrm{T} \times 4 \pi\left(\mathrm{nr}^{2}-\mathrm{R}^{2}\right)}{\mathrm{J}}$ $\text { Here, } \mathrm{s}=1 \mathrm{cal} / \mathrm{gm}^{\circ} \mathrm{c}, \quad \mathrm{m}=\rho \mathrm{V}, \rho=1$ $\frac{4}{3} \pi \mathrm{R}^{3} \times 1 \times 1 \times \Delta \theta \times \mathrm{J}=\mathrm{T} \times 4 \pi\left(\mathrm{nr}^{2}-\mathrm{R}^{2}\right)$ From equation (i) we get, $\Rightarrow \mathrm{J} \times \Delta \theta \times \frac{\mathrm{R}^{3}}{3}=\mathrm{T} \times \mathrm{R}^{3}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$ $\Delta \theta=\frac{3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$
J and K CET- 2000
Mechanical Properties of Fluids
143056
The excess of pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is $n$ times, the volume of the second, where $n$ is
1 0.125
2 1
3 2
4 4
Explanation:
A Radius of first soap bubble is $r_{1}$ and second soap bubble is $r_{2}$. Excess pressure inside a soap bubble, $\Delta \mathrm{P}_{1}=2 \Delta \mathrm{P}_{2}$ $\Rightarrow \frac{4 \mathrm{~T}}{\mathrm{r}_{1}}=2 \times \frac{4 \mathrm{~T}}{\mathrm{r}_{2}}$ $\Rightarrow \mathrm{r}_{2}=2 \mathrm{r}_{1}$ $\therefore$ Volume of bubble $\frac{4}{3} \pi \mathrm{r}_{1}^{3}=\mathrm{n} \times \frac{4}{3} \pi\left(2 \mathrm{r}_{1}\right)^{3}$ $\mathrm{r}_{1}^{3}=8 \mathrm{r}_{1}^{3} \times \mathrm{n}$ $\mathrm{n}=\frac{1}{8}$ $\mathrm{n}=0.125$
143052
If two soap bubbles of equal radii $r$ coalesce then the radius of curvature of interface between two bubbles will be
1 $\mathrm{r}$
2 0
3 infinity
4 $\frac{1}{2} \mathrm{r}$
Explanation:
C $P_{1}=P_{0}+\frac{4 T}{r}$ $\mathrm{P}_{2}=\mathrm{P}_{0}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}$ $\mathrm{P}_{1}>\mathrm{P}_{2}$ $\Delta \mathrm{P}=\mathrm{P}_{1}-\mathrm{P}_{2}=4 \mathrm{~T}\left[\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{r}_{2}}\right]$ $\frac{1}{\mathrm{r}}=\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{r}_{2}}$ $r=\frac{r_{1} r_{2}}{r_{2}-r_{1}}$ Both bubble of are equal radii, then $\mathrm{r}=\infty$
J and K CET- 2005
Mechanical Properties of Fluids
143054
Two equal drops of water are falling through air with a steady velocity of $5 \mathrm{~cm} / \mathrm{s}$. If the drops coalesce, the new steady velocity of the coalesced drop will be
1 $5 \mathrm{~cm} / \mathrm{s}$
2 $10 \mathrm{~cm} / \mathrm{s}$
3 $7.9 \mathrm{~cm} / \mathrm{s}$
4 $6 \mathrm{~cm} / \mathrm{s}$
Explanation:
C Let $\mathrm{R}$ is radius of bigger drop, $\mathrm{r}$ is radius of smaller drop $\frac{4}{3} \pi \mathrm{R}^{3}=2 \times \frac{4}{3} \times \pi \mathrm{r}^{3}$ $\mathrm{R}=(2)^{1 / 3} \cdot \mathrm{r}$ Steady velocity of bigger drop $\mathrm{v}_{1}=$ ? Steady velocity of smaller drop. $\mathrm{v}=5 \mathrm{~cm} / \mathrm{s}$ Terminal velocity, $\mathrm{v} \propto \mathrm{r}^{2}$ $\therefore \frac{\mathrm{v}_{1}}{\mathrm{v}}=\frac{\mathrm{R}^{2}}{\mathrm{r}^{2}}$ $\mathrm{v}_{1}=\frac{\left(2^{1 / 3}\right)^{2} \cdot r^{2}}{\mathrm{r}^{2}} \times 5$ $\mathrm{v}_{1}=5 \times 4^{1 / 3}=5 \times 1.59$ $\mathrm{v}_{1}=7.9 \mathrm{~cm} / \mathrm{s}$
J and K CET- 2000
Mechanical Properties of Fluids
143055
A number of water droplets of radius $r$ coalesce to form a drop of radius $R$. Assuming that the entire energy liberated due to coalesce goes into heating the drop, the rise in temperature $\mathrm{d} \theta$ is (surface tension of water $=T$ )
D Let, $r$ be the radius of small droplets and $R$ be the radius of forming droplet. A number of water droplets of radius $r$ coalesce to form a drop of radius $\mathrm{R}$. $n \times \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3}$ $n=\left(\frac{R}{r}\right)^{3} \ldots \ldots . .(i)$ The entire energy liberated due to coalesces goes into heating the drop. $\mathrm{H}=\frac{\Delta \mathrm{E}}{\mathrm{J}} \quad\{\because \mathrm{H}=\mathrm{ms} \Delta \theta\}$ $\mathrm{ms} \Delta \theta=\frac{\text { Surface tension } \times \text { Changein surface area }}{\mathrm{J}}$ $\mathrm{ms} \Delta \theta=\frac{\mathrm{T} \times 4 \pi\left(\mathrm{nr}^{2}-\mathrm{R}^{2}\right)}{\mathrm{J}}$ $\text { Here, } \mathrm{s}=1 \mathrm{cal} / \mathrm{gm}^{\circ} \mathrm{c}, \quad \mathrm{m}=\rho \mathrm{V}, \rho=1$ $\frac{4}{3} \pi \mathrm{R}^{3} \times 1 \times 1 \times \Delta \theta \times \mathrm{J}=\mathrm{T} \times 4 \pi\left(\mathrm{nr}^{2}-\mathrm{R}^{2}\right)$ From equation (i) we get, $\Rightarrow \mathrm{J} \times \Delta \theta \times \frac{\mathrm{R}^{3}}{3}=\mathrm{T} \times \mathrm{R}^{3}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$ $\Delta \theta=\frac{3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$
J and K CET- 2000
Mechanical Properties of Fluids
143056
The excess of pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is $n$ times, the volume of the second, where $n$ is
1 0.125
2 1
3 2
4 4
Explanation:
A Radius of first soap bubble is $r_{1}$ and second soap bubble is $r_{2}$. Excess pressure inside a soap bubble, $\Delta \mathrm{P}_{1}=2 \Delta \mathrm{P}_{2}$ $\Rightarrow \frac{4 \mathrm{~T}}{\mathrm{r}_{1}}=2 \times \frac{4 \mathrm{~T}}{\mathrm{r}_{2}}$ $\Rightarrow \mathrm{r}_{2}=2 \mathrm{r}_{1}$ $\therefore$ Volume of bubble $\frac{4}{3} \pi \mathrm{r}_{1}^{3}=\mathrm{n} \times \frac{4}{3} \pi\left(2 \mathrm{r}_{1}\right)^{3}$ $\mathrm{r}_{1}^{3}=8 \mathrm{r}_{1}^{3} \times \mathrm{n}$ $\mathrm{n}=\frac{1}{8}$ $\mathrm{n}=0.125$
143052
If two soap bubbles of equal radii $r$ coalesce then the radius of curvature of interface between two bubbles will be
1 $\mathrm{r}$
2 0
3 infinity
4 $\frac{1}{2} \mathrm{r}$
Explanation:
C $P_{1}=P_{0}+\frac{4 T}{r}$ $\mathrm{P}_{2}=\mathrm{P}_{0}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}$ $\mathrm{P}_{1}>\mathrm{P}_{2}$ $\Delta \mathrm{P}=\mathrm{P}_{1}-\mathrm{P}_{2}=4 \mathrm{~T}\left[\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{r}_{2}}\right]$ $\frac{1}{\mathrm{r}}=\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{r}_{2}}$ $r=\frac{r_{1} r_{2}}{r_{2}-r_{1}}$ Both bubble of are equal radii, then $\mathrm{r}=\infty$
J and K CET- 2005
Mechanical Properties of Fluids
143054
Two equal drops of water are falling through air with a steady velocity of $5 \mathrm{~cm} / \mathrm{s}$. If the drops coalesce, the new steady velocity of the coalesced drop will be
1 $5 \mathrm{~cm} / \mathrm{s}$
2 $10 \mathrm{~cm} / \mathrm{s}$
3 $7.9 \mathrm{~cm} / \mathrm{s}$
4 $6 \mathrm{~cm} / \mathrm{s}$
Explanation:
C Let $\mathrm{R}$ is radius of bigger drop, $\mathrm{r}$ is radius of smaller drop $\frac{4}{3} \pi \mathrm{R}^{3}=2 \times \frac{4}{3} \times \pi \mathrm{r}^{3}$ $\mathrm{R}=(2)^{1 / 3} \cdot \mathrm{r}$ Steady velocity of bigger drop $\mathrm{v}_{1}=$ ? Steady velocity of smaller drop. $\mathrm{v}=5 \mathrm{~cm} / \mathrm{s}$ Terminal velocity, $\mathrm{v} \propto \mathrm{r}^{2}$ $\therefore \frac{\mathrm{v}_{1}}{\mathrm{v}}=\frac{\mathrm{R}^{2}}{\mathrm{r}^{2}}$ $\mathrm{v}_{1}=\frac{\left(2^{1 / 3}\right)^{2} \cdot r^{2}}{\mathrm{r}^{2}} \times 5$ $\mathrm{v}_{1}=5 \times 4^{1 / 3}=5 \times 1.59$ $\mathrm{v}_{1}=7.9 \mathrm{~cm} / \mathrm{s}$
J and K CET- 2000
Mechanical Properties of Fluids
143055
A number of water droplets of radius $r$ coalesce to form a drop of radius $R$. Assuming that the entire energy liberated due to coalesce goes into heating the drop, the rise in temperature $\mathrm{d} \theta$ is (surface tension of water $=T$ )
D Let, $r$ be the radius of small droplets and $R$ be the radius of forming droplet. A number of water droplets of radius $r$ coalesce to form a drop of radius $\mathrm{R}$. $n \times \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3}$ $n=\left(\frac{R}{r}\right)^{3} \ldots \ldots . .(i)$ The entire energy liberated due to coalesces goes into heating the drop. $\mathrm{H}=\frac{\Delta \mathrm{E}}{\mathrm{J}} \quad\{\because \mathrm{H}=\mathrm{ms} \Delta \theta\}$ $\mathrm{ms} \Delta \theta=\frac{\text { Surface tension } \times \text { Changein surface area }}{\mathrm{J}}$ $\mathrm{ms} \Delta \theta=\frac{\mathrm{T} \times 4 \pi\left(\mathrm{nr}^{2}-\mathrm{R}^{2}\right)}{\mathrm{J}}$ $\text { Here, } \mathrm{s}=1 \mathrm{cal} / \mathrm{gm}^{\circ} \mathrm{c}, \quad \mathrm{m}=\rho \mathrm{V}, \rho=1$ $\frac{4}{3} \pi \mathrm{R}^{3} \times 1 \times 1 \times \Delta \theta \times \mathrm{J}=\mathrm{T} \times 4 \pi\left(\mathrm{nr}^{2}-\mathrm{R}^{2}\right)$ From equation (i) we get, $\Rightarrow \mathrm{J} \times \Delta \theta \times \frac{\mathrm{R}^{3}}{3}=\mathrm{T} \times \mathrm{R}^{3}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$ $\Delta \theta=\frac{3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$
J and K CET- 2000
Mechanical Properties of Fluids
143056
The excess of pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is $n$ times, the volume of the second, where $n$ is
1 0.125
2 1
3 2
4 4
Explanation:
A Radius of first soap bubble is $r_{1}$ and second soap bubble is $r_{2}$. Excess pressure inside a soap bubble, $\Delta \mathrm{P}_{1}=2 \Delta \mathrm{P}_{2}$ $\Rightarrow \frac{4 \mathrm{~T}}{\mathrm{r}_{1}}=2 \times \frac{4 \mathrm{~T}}{\mathrm{r}_{2}}$ $\Rightarrow \mathrm{r}_{2}=2 \mathrm{r}_{1}$ $\therefore$ Volume of bubble $\frac{4}{3} \pi \mathrm{r}_{1}^{3}=\mathrm{n} \times \frac{4}{3} \pi\left(2 \mathrm{r}_{1}\right)^{3}$ $\mathrm{r}_{1}^{3}=8 \mathrm{r}_{1}^{3} \times \mathrm{n}$ $\mathrm{n}=\frac{1}{8}$ $\mathrm{n}=0.125$
143052
If two soap bubbles of equal radii $r$ coalesce then the radius of curvature of interface between two bubbles will be
1 $\mathrm{r}$
2 0
3 infinity
4 $\frac{1}{2} \mathrm{r}$
Explanation:
C $P_{1}=P_{0}+\frac{4 T}{r}$ $\mathrm{P}_{2}=\mathrm{P}_{0}+\frac{4 \mathrm{~T}}{\mathrm{r}_{2}}$ $\mathrm{P}_{1}>\mathrm{P}_{2}$ $\Delta \mathrm{P}=\mathrm{P}_{1}-\mathrm{P}_{2}=4 \mathrm{~T}\left[\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{r}_{2}}\right]$ $\frac{1}{\mathrm{r}}=\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{r}_{2}}$ $r=\frac{r_{1} r_{2}}{r_{2}-r_{1}}$ Both bubble of are equal radii, then $\mathrm{r}=\infty$
J and K CET- 2005
Mechanical Properties of Fluids
143054
Two equal drops of water are falling through air with a steady velocity of $5 \mathrm{~cm} / \mathrm{s}$. If the drops coalesce, the new steady velocity of the coalesced drop will be
1 $5 \mathrm{~cm} / \mathrm{s}$
2 $10 \mathrm{~cm} / \mathrm{s}$
3 $7.9 \mathrm{~cm} / \mathrm{s}$
4 $6 \mathrm{~cm} / \mathrm{s}$
Explanation:
C Let $\mathrm{R}$ is radius of bigger drop, $\mathrm{r}$ is radius of smaller drop $\frac{4}{3} \pi \mathrm{R}^{3}=2 \times \frac{4}{3} \times \pi \mathrm{r}^{3}$ $\mathrm{R}=(2)^{1 / 3} \cdot \mathrm{r}$ Steady velocity of bigger drop $\mathrm{v}_{1}=$ ? Steady velocity of smaller drop. $\mathrm{v}=5 \mathrm{~cm} / \mathrm{s}$ Terminal velocity, $\mathrm{v} \propto \mathrm{r}^{2}$ $\therefore \frac{\mathrm{v}_{1}}{\mathrm{v}}=\frac{\mathrm{R}^{2}}{\mathrm{r}^{2}}$ $\mathrm{v}_{1}=\frac{\left(2^{1 / 3}\right)^{2} \cdot r^{2}}{\mathrm{r}^{2}} \times 5$ $\mathrm{v}_{1}=5 \times 4^{1 / 3}=5 \times 1.59$ $\mathrm{v}_{1}=7.9 \mathrm{~cm} / \mathrm{s}$
J and K CET- 2000
Mechanical Properties of Fluids
143055
A number of water droplets of radius $r$ coalesce to form a drop of radius $R$. Assuming that the entire energy liberated due to coalesce goes into heating the drop, the rise in temperature $\mathrm{d} \theta$ is (surface tension of water $=T$ )
D Let, $r$ be the radius of small droplets and $R$ be the radius of forming droplet. A number of water droplets of radius $r$ coalesce to form a drop of radius $\mathrm{R}$. $n \times \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3}$ $n=\left(\frac{R}{r}\right)^{3} \ldots \ldots . .(i)$ The entire energy liberated due to coalesces goes into heating the drop. $\mathrm{H}=\frac{\Delta \mathrm{E}}{\mathrm{J}} \quad\{\because \mathrm{H}=\mathrm{ms} \Delta \theta\}$ $\mathrm{ms} \Delta \theta=\frac{\text { Surface tension } \times \text { Changein surface area }}{\mathrm{J}}$ $\mathrm{ms} \Delta \theta=\frac{\mathrm{T} \times 4 \pi\left(\mathrm{nr}^{2}-\mathrm{R}^{2}\right)}{\mathrm{J}}$ $\text { Here, } \mathrm{s}=1 \mathrm{cal} / \mathrm{gm}^{\circ} \mathrm{c}, \quad \mathrm{m}=\rho \mathrm{V}, \rho=1$ $\frac{4}{3} \pi \mathrm{R}^{3} \times 1 \times 1 \times \Delta \theta \times \mathrm{J}=\mathrm{T} \times 4 \pi\left(\mathrm{nr}^{2}-\mathrm{R}^{2}\right)$ From equation (i) we get, $\Rightarrow \mathrm{J} \times \Delta \theta \times \frac{\mathrm{R}^{3}}{3}=\mathrm{T} \times \mathrm{R}^{3}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$ $\Delta \theta=\frac{3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$
J and K CET- 2000
Mechanical Properties of Fluids
143056
The excess of pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is $n$ times, the volume of the second, where $n$ is
1 0.125
2 1
3 2
4 4
Explanation:
A Radius of first soap bubble is $r_{1}$ and second soap bubble is $r_{2}$. Excess pressure inside a soap bubble, $\Delta \mathrm{P}_{1}=2 \Delta \mathrm{P}_{2}$ $\Rightarrow \frac{4 \mathrm{~T}}{\mathrm{r}_{1}}=2 \times \frac{4 \mathrm{~T}}{\mathrm{r}_{2}}$ $\Rightarrow \mathrm{r}_{2}=2 \mathrm{r}_{1}$ $\therefore$ Volume of bubble $\frac{4}{3} \pi \mathrm{r}_{1}^{3}=\mathrm{n} \times \frac{4}{3} \pi\left(2 \mathrm{r}_{1}\right)^{3}$ $\mathrm{r}_{1}^{3}=8 \mathrm{r}_{1}^{3} \times \mathrm{n}$ $\mathrm{n}=\frac{1}{8}$ $\mathrm{n}=0.125$
