143046
The excess pressure inside a soap bubble of volume ' $V$ ' is three times the excess pressure inside a second soap bubble of volume ' $V_{1}$ '. The value $\left(\frac{V_{1}}{V}\right)$ is
1 $3: 1$
2 $1: 9$
3 $1: 3$
4 $27: 1$
Explanation:
D Volume of first soap bubble, $V==\frac{4}{3} \pi r^{3}$ Volume of second soap bubble $\mathrm{V}_{1}=\frac{4}{3} \pi \mathrm{r}_{1}^{3}$ Excess pressure in first soap bubble, $\mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{r}}$ Excess pressure in second soap bubble $\mathrm{P}_{1}=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}$ According to question, $\mathrm{P}=3 \mathrm{P}_{1}$ $\frac{4 \mathrm{~T}}{\mathrm{r}}=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}} \times 3$ $\frac{\mathrm{r}}{\mathrm{r}_{1}}=\frac{1}{3}$ $\frac{\mathrm{V}}{\mathrm{V}_{1}}=\frac{\frac{4}{3} \pi \mathrm{r}^{3}}{\frac{4}{3} \pi \mathrm{r}_{1}^{3}}=\left(\frac{\mathrm{r}}{\mathrm{r}_{1}}\right)^{3}=\left(\frac{1}{3}\right)^{3}=\frac{1}{27}$ Ratio of $\frac{\mathrm{V}_{1}}{\mathrm{~V}}=\frac{27}{1}$ $\mathrm{V}_{1}: \mathrm{V}=27: 1$
MHT-CET 2020
Mechanical Properties of Fluids
143047
Water drops fall at regular intervals from a tap which is $\mathbf{5} \mathbf{~ m}$ above the ground. The third drops leaves the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant?
1 $2.5 \mathrm{~m}$
2 $3.75 \mathrm{~m}$
3 $4.0 \mathrm{~m}$
4 $1.25 \mathrm{~m}$
Explanation:
B Given that, For $1^{\text {st }}$ drop - $\text { Height of tap }(\mathrm{h})=5 \mathrm{~m}$ $\text { Initial velocity }(\mathrm{u})=0$ From, $\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^{2}$ $5=0+\frac{1}{2} \times 10 \times \mathrm{t}^{2} .$ $\mathrm{t}^{2}=1$ $\mathrm{t}=1 \mathrm{sec}$ It means that the third drop leaves after one second of the first drop. So the each drop leaves after every 0.5 sec. For $2^{\text {nd }}$ drop $u=0$ $t=0.5$ $v=u+g t$ $v=0+10 \times 0.5$ $v=5 m / s$ $\text { From, } v^{2}=u^{2}+2 g s$ $(5)^{2}=0+2 \times 10 \times \mathrm{s}$ $\mathrm{s}=\frac{25}{20}=1.25$ Position of the second drop from ground $=5-1.25=$ $3.75 \mathrm{~m}$
SCRA-2013
Mechanical Properties of Fluids
143048
The excess pressure inside a cylindrical drop of liquid or a cylindrical bubble of radius $r$ in a liquid of surface tension $T$ is
1 $\frac{\mathrm{T}}{4 \mathrm{r}}$
2 $\frac{\mathrm{T}}{\mathrm{r}}$
3 $\frac{2 \mathrm{~T}}{\mathrm{r}}$
4 $\frac{4 \mathrm{~T}}{\mathrm{r}}$
Explanation:
B Work done in increasing the radius from $\mathrm{r}$ to $\mathrm{r}+\mathrm{dr}$ $\mathrm{W}=\mathrm{Fdr}$ $\mathrm{W}=\mathrm{P}(2 \pi \mathrm{r} l) \mathrm{dr}$ Increase in area $=\Delta \mathrm{S}=2 \pi(\mathrm{r}+\mathrm{dr}) l-2 \pi \mathrm{r} l=2 \pi \mathrm{dr} l$ $\therefore \quad \mathrm{T}=\frac{\mathrm{W}}{\Delta \mathrm{S}}=\frac{\mathrm{P}(2 \pi \mathrm{r} l) \mathrm{dr}}{2 \pi \mathrm{dr} l}=\mathrm{Pr}$ $\Rightarrow P=\frac{T}{r}$
J and K CET- 2009
Mechanical Properties of Fluids
143050
A sphere of radius $R$ is gently dropped into liquid of viscosity $\eta$ is a vertical uniform tube. It attains a terminal velocity $v$. Another sphere of radius $2 R$ when dropped into the same liquid will attain its terminal velocity
1 $\mathrm{v}$
2 $2 \mathrm{v}$
3 $4 \mathrm{v}$
4 $9 \mathrm{v}$
Explanation:
C Terminal velocity of sphere into liquid is. $\mathrm{v}_{\mathrm{T}}=\frac{2}{9} \mathrm{R}^{2} \frac{(\rho-\sigma) \mathrm{g}}{\eta}$ $\mathrm{v} \propto \mathrm{R}^{2}$ $\frac{\mathrm{V}^{\prime}}{\mathrm{v}}=\frac{\mathrm{R}^{\prime^{2}}}{\mathrm{R}}$ $\Rightarrow \quad \frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\frac{4 \mathrm{R}^{2}}{\mathrm{R}^{2}}$ $\mathrm{v}^{\prime}=4 \mathrm{v}$
J and K CET- 2008
Mechanical Properties of Fluids
143051
Two liquid drops having diameters of $1 \mathrm{~cm}$ and $1.5 \mathrm{~cm}$. The ratio of excess pressures inside them is
1 $1: 1$
2 $5: 3$
3 $2: 3$
4 $3: 2$
Explanation:
D Diameter of $\mathrm{I}^{\text {st }}$ liquid drop $\left(\mathrm{D}_{1}\right)=1 \mathrm{~cm}$ Diameter of $\mathrm{II}^{\text {nd }}$ liquid drop $\left(\mathrm{D}_{2}\right)=1.5 \mathrm{~cm}$ For drop, Excess pressure, $\mathrm{P}=\frac{2 \mathrm{~T}}{\mathrm{R}}$ $\therefore \quad \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{\mathrm{D}_{2}}{\mathrm{D}_{1}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{1.5}{1}=\frac{15}{10}=3: 2$
143046
The excess pressure inside a soap bubble of volume ' $V$ ' is three times the excess pressure inside a second soap bubble of volume ' $V_{1}$ '. The value $\left(\frac{V_{1}}{V}\right)$ is
1 $3: 1$
2 $1: 9$
3 $1: 3$
4 $27: 1$
Explanation:
D Volume of first soap bubble, $V==\frac{4}{3} \pi r^{3}$ Volume of second soap bubble $\mathrm{V}_{1}=\frac{4}{3} \pi \mathrm{r}_{1}^{3}$ Excess pressure in first soap bubble, $\mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{r}}$ Excess pressure in second soap bubble $\mathrm{P}_{1}=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}$ According to question, $\mathrm{P}=3 \mathrm{P}_{1}$ $\frac{4 \mathrm{~T}}{\mathrm{r}}=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}} \times 3$ $\frac{\mathrm{r}}{\mathrm{r}_{1}}=\frac{1}{3}$ $\frac{\mathrm{V}}{\mathrm{V}_{1}}=\frac{\frac{4}{3} \pi \mathrm{r}^{3}}{\frac{4}{3} \pi \mathrm{r}_{1}^{3}}=\left(\frac{\mathrm{r}}{\mathrm{r}_{1}}\right)^{3}=\left(\frac{1}{3}\right)^{3}=\frac{1}{27}$ Ratio of $\frac{\mathrm{V}_{1}}{\mathrm{~V}}=\frac{27}{1}$ $\mathrm{V}_{1}: \mathrm{V}=27: 1$
MHT-CET 2020
Mechanical Properties of Fluids
143047
Water drops fall at regular intervals from a tap which is $\mathbf{5} \mathbf{~ m}$ above the ground. The third drops leaves the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant?
1 $2.5 \mathrm{~m}$
2 $3.75 \mathrm{~m}$
3 $4.0 \mathrm{~m}$
4 $1.25 \mathrm{~m}$
Explanation:
B Given that, For $1^{\text {st }}$ drop - $\text { Height of tap }(\mathrm{h})=5 \mathrm{~m}$ $\text { Initial velocity }(\mathrm{u})=0$ From, $\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^{2}$ $5=0+\frac{1}{2} \times 10 \times \mathrm{t}^{2} .$ $\mathrm{t}^{2}=1$ $\mathrm{t}=1 \mathrm{sec}$ It means that the third drop leaves after one second of the first drop. So the each drop leaves after every 0.5 sec. For $2^{\text {nd }}$ drop $u=0$ $t=0.5$ $v=u+g t$ $v=0+10 \times 0.5$ $v=5 m / s$ $\text { From, } v^{2}=u^{2}+2 g s$ $(5)^{2}=0+2 \times 10 \times \mathrm{s}$ $\mathrm{s}=\frac{25}{20}=1.25$ Position of the second drop from ground $=5-1.25=$ $3.75 \mathrm{~m}$
SCRA-2013
Mechanical Properties of Fluids
143048
The excess pressure inside a cylindrical drop of liquid or a cylindrical bubble of radius $r$ in a liquid of surface tension $T$ is
1 $\frac{\mathrm{T}}{4 \mathrm{r}}$
2 $\frac{\mathrm{T}}{\mathrm{r}}$
3 $\frac{2 \mathrm{~T}}{\mathrm{r}}$
4 $\frac{4 \mathrm{~T}}{\mathrm{r}}$
Explanation:
B Work done in increasing the radius from $\mathrm{r}$ to $\mathrm{r}+\mathrm{dr}$ $\mathrm{W}=\mathrm{Fdr}$ $\mathrm{W}=\mathrm{P}(2 \pi \mathrm{r} l) \mathrm{dr}$ Increase in area $=\Delta \mathrm{S}=2 \pi(\mathrm{r}+\mathrm{dr}) l-2 \pi \mathrm{r} l=2 \pi \mathrm{dr} l$ $\therefore \quad \mathrm{T}=\frac{\mathrm{W}}{\Delta \mathrm{S}}=\frac{\mathrm{P}(2 \pi \mathrm{r} l) \mathrm{dr}}{2 \pi \mathrm{dr} l}=\mathrm{Pr}$ $\Rightarrow P=\frac{T}{r}$
J and K CET- 2009
Mechanical Properties of Fluids
143050
A sphere of radius $R$ is gently dropped into liquid of viscosity $\eta$ is a vertical uniform tube. It attains a terminal velocity $v$. Another sphere of radius $2 R$ when dropped into the same liquid will attain its terminal velocity
1 $\mathrm{v}$
2 $2 \mathrm{v}$
3 $4 \mathrm{v}$
4 $9 \mathrm{v}$
Explanation:
C Terminal velocity of sphere into liquid is. $\mathrm{v}_{\mathrm{T}}=\frac{2}{9} \mathrm{R}^{2} \frac{(\rho-\sigma) \mathrm{g}}{\eta}$ $\mathrm{v} \propto \mathrm{R}^{2}$ $\frac{\mathrm{V}^{\prime}}{\mathrm{v}}=\frac{\mathrm{R}^{\prime^{2}}}{\mathrm{R}}$ $\Rightarrow \quad \frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\frac{4 \mathrm{R}^{2}}{\mathrm{R}^{2}}$ $\mathrm{v}^{\prime}=4 \mathrm{v}$
J and K CET- 2008
Mechanical Properties of Fluids
143051
Two liquid drops having diameters of $1 \mathrm{~cm}$ and $1.5 \mathrm{~cm}$. The ratio of excess pressures inside them is
1 $1: 1$
2 $5: 3$
3 $2: 3$
4 $3: 2$
Explanation:
D Diameter of $\mathrm{I}^{\text {st }}$ liquid drop $\left(\mathrm{D}_{1}\right)=1 \mathrm{~cm}$ Diameter of $\mathrm{II}^{\text {nd }}$ liquid drop $\left(\mathrm{D}_{2}\right)=1.5 \mathrm{~cm}$ For drop, Excess pressure, $\mathrm{P}=\frac{2 \mathrm{~T}}{\mathrm{R}}$ $\therefore \quad \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{\mathrm{D}_{2}}{\mathrm{D}_{1}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{1.5}{1}=\frac{15}{10}=3: 2$
143046
The excess pressure inside a soap bubble of volume ' $V$ ' is three times the excess pressure inside a second soap bubble of volume ' $V_{1}$ '. The value $\left(\frac{V_{1}}{V}\right)$ is
1 $3: 1$
2 $1: 9$
3 $1: 3$
4 $27: 1$
Explanation:
D Volume of first soap bubble, $V==\frac{4}{3} \pi r^{3}$ Volume of second soap bubble $\mathrm{V}_{1}=\frac{4}{3} \pi \mathrm{r}_{1}^{3}$ Excess pressure in first soap bubble, $\mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{r}}$ Excess pressure in second soap bubble $\mathrm{P}_{1}=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}$ According to question, $\mathrm{P}=3 \mathrm{P}_{1}$ $\frac{4 \mathrm{~T}}{\mathrm{r}}=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}} \times 3$ $\frac{\mathrm{r}}{\mathrm{r}_{1}}=\frac{1}{3}$ $\frac{\mathrm{V}}{\mathrm{V}_{1}}=\frac{\frac{4}{3} \pi \mathrm{r}^{3}}{\frac{4}{3} \pi \mathrm{r}_{1}^{3}}=\left(\frac{\mathrm{r}}{\mathrm{r}_{1}}\right)^{3}=\left(\frac{1}{3}\right)^{3}=\frac{1}{27}$ Ratio of $\frac{\mathrm{V}_{1}}{\mathrm{~V}}=\frac{27}{1}$ $\mathrm{V}_{1}: \mathrm{V}=27: 1$
MHT-CET 2020
Mechanical Properties of Fluids
143047
Water drops fall at regular intervals from a tap which is $\mathbf{5} \mathbf{~ m}$ above the ground. The third drops leaves the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant?
1 $2.5 \mathrm{~m}$
2 $3.75 \mathrm{~m}$
3 $4.0 \mathrm{~m}$
4 $1.25 \mathrm{~m}$
Explanation:
B Given that, For $1^{\text {st }}$ drop - $\text { Height of tap }(\mathrm{h})=5 \mathrm{~m}$ $\text { Initial velocity }(\mathrm{u})=0$ From, $\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^{2}$ $5=0+\frac{1}{2} \times 10 \times \mathrm{t}^{2} .$ $\mathrm{t}^{2}=1$ $\mathrm{t}=1 \mathrm{sec}$ It means that the third drop leaves after one second of the first drop. So the each drop leaves after every 0.5 sec. For $2^{\text {nd }}$ drop $u=0$ $t=0.5$ $v=u+g t$ $v=0+10 \times 0.5$ $v=5 m / s$ $\text { From, } v^{2}=u^{2}+2 g s$ $(5)^{2}=0+2 \times 10 \times \mathrm{s}$ $\mathrm{s}=\frac{25}{20}=1.25$ Position of the second drop from ground $=5-1.25=$ $3.75 \mathrm{~m}$
SCRA-2013
Mechanical Properties of Fluids
143048
The excess pressure inside a cylindrical drop of liquid or a cylindrical bubble of radius $r$ in a liquid of surface tension $T$ is
1 $\frac{\mathrm{T}}{4 \mathrm{r}}$
2 $\frac{\mathrm{T}}{\mathrm{r}}$
3 $\frac{2 \mathrm{~T}}{\mathrm{r}}$
4 $\frac{4 \mathrm{~T}}{\mathrm{r}}$
Explanation:
B Work done in increasing the radius from $\mathrm{r}$ to $\mathrm{r}+\mathrm{dr}$ $\mathrm{W}=\mathrm{Fdr}$ $\mathrm{W}=\mathrm{P}(2 \pi \mathrm{r} l) \mathrm{dr}$ Increase in area $=\Delta \mathrm{S}=2 \pi(\mathrm{r}+\mathrm{dr}) l-2 \pi \mathrm{r} l=2 \pi \mathrm{dr} l$ $\therefore \quad \mathrm{T}=\frac{\mathrm{W}}{\Delta \mathrm{S}}=\frac{\mathrm{P}(2 \pi \mathrm{r} l) \mathrm{dr}}{2 \pi \mathrm{dr} l}=\mathrm{Pr}$ $\Rightarrow P=\frac{T}{r}$
J and K CET- 2009
Mechanical Properties of Fluids
143050
A sphere of radius $R$ is gently dropped into liquid of viscosity $\eta$ is a vertical uniform tube. It attains a terminal velocity $v$. Another sphere of radius $2 R$ when dropped into the same liquid will attain its terminal velocity
1 $\mathrm{v}$
2 $2 \mathrm{v}$
3 $4 \mathrm{v}$
4 $9 \mathrm{v}$
Explanation:
C Terminal velocity of sphere into liquid is. $\mathrm{v}_{\mathrm{T}}=\frac{2}{9} \mathrm{R}^{2} \frac{(\rho-\sigma) \mathrm{g}}{\eta}$ $\mathrm{v} \propto \mathrm{R}^{2}$ $\frac{\mathrm{V}^{\prime}}{\mathrm{v}}=\frac{\mathrm{R}^{\prime^{2}}}{\mathrm{R}}$ $\Rightarrow \quad \frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\frac{4 \mathrm{R}^{2}}{\mathrm{R}^{2}}$ $\mathrm{v}^{\prime}=4 \mathrm{v}$
J and K CET- 2008
Mechanical Properties of Fluids
143051
Two liquid drops having diameters of $1 \mathrm{~cm}$ and $1.5 \mathrm{~cm}$. The ratio of excess pressures inside them is
1 $1: 1$
2 $5: 3$
3 $2: 3$
4 $3: 2$
Explanation:
D Diameter of $\mathrm{I}^{\text {st }}$ liquid drop $\left(\mathrm{D}_{1}\right)=1 \mathrm{~cm}$ Diameter of $\mathrm{II}^{\text {nd }}$ liquid drop $\left(\mathrm{D}_{2}\right)=1.5 \mathrm{~cm}$ For drop, Excess pressure, $\mathrm{P}=\frac{2 \mathrm{~T}}{\mathrm{R}}$ $\therefore \quad \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{\mathrm{D}_{2}}{\mathrm{D}_{1}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{1.5}{1}=\frac{15}{10}=3: 2$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Mechanical Properties of Fluids
143046
The excess pressure inside a soap bubble of volume ' $V$ ' is three times the excess pressure inside a second soap bubble of volume ' $V_{1}$ '. The value $\left(\frac{V_{1}}{V}\right)$ is
1 $3: 1$
2 $1: 9$
3 $1: 3$
4 $27: 1$
Explanation:
D Volume of first soap bubble, $V==\frac{4}{3} \pi r^{3}$ Volume of second soap bubble $\mathrm{V}_{1}=\frac{4}{3} \pi \mathrm{r}_{1}^{3}$ Excess pressure in first soap bubble, $\mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{r}}$ Excess pressure in second soap bubble $\mathrm{P}_{1}=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}$ According to question, $\mathrm{P}=3 \mathrm{P}_{1}$ $\frac{4 \mathrm{~T}}{\mathrm{r}}=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}} \times 3$ $\frac{\mathrm{r}}{\mathrm{r}_{1}}=\frac{1}{3}$ $\frac{\mathrm{V}}{\mathrm{V}_{1}}=\frac{\frac{4}{3} \pi \mathrm{r}^{3}}{\frac{4}{3} \pi \mathrm{r}_{1}^{3}}=\left(\frac{\mathrm{r}}{\mathrm{r}_{1}}\right)^{3}=\left(\frac{1}{3}\right)^{3}=\frac{1}{27}$ Ratio of $\frac{\mathrm{V}_{1}}{\mathrm{~V}}=\frac{27}{1}$ $\mathrm{V}_{1}: \mathrm{V}=27: 1$
MHT-CET 2020
Mechanical Properties of Fluids
143047
Water drops fall at regular intervals from a tap which is $\mathbf{5} \mathbf{~ m}$ above the ground. The third drops leaves the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant?
1 $2.5 \mathrm{~m}$
2 $3.75 \mathrm{~m}$
3 $4.0 \mathrm{~m}$
4 $1.25 \mathrm{~m}$
Explanation:
B Given that, For $1^{\text {st }}$ drop - $\text { Height of tap }(\mathrm{h})=5 \mathrm{~m}$ $\text { Initial velocity }(\mathrm{u})=0$ From, $\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^{2}$ $5=0+\frac{1}{2} \times 10 \times \mathrm{t}^{2} .$ $\mathrm{t}^{2}=1$ $\mathrm{t}=1 \mathrm{sec}$ It means that the third drop leaves after one second of the first drop. So the each drop leaves after every 0.5 sec. For $2^{\text {nd }}$ drop $u=0$ $t=0.5$ $v=u+g t$ $v=0+10 \times 0.5$ $v=5 m / s$ $\text { From, } v^{2}=u^{2}+2 g s$ $(5)^{2}=0+2 \times 10 \times \mathrm{s}$ $\mathrm{s}=\frac{25}{20}=1.25$ Position of the second drop from ground $=5-1.25=$ $3.75 \mathrm{~m}$
SCRA-2013
Mechanical Properties of Fluids
143048
The excess pressure inside a cylindrical drop of liquid or a cylindrical bubble of radius $r$ in a liquid of surface tension $T$ is
1 $\frac{\mathrm{T}}{4 \mathrm{r}}$
2 $\frac{\mathrm{T}}{\mathrm{r}}$
3 $\frac{2 \mathrm{~T}}{\mathrm{r}}$
4 $\frac{4 \mathrm{~T}}{\mathrm{r}}$
Explanation:
B Work done in increasing the radius from $\mathrm{r}$ to $\mathrm{r}+\mathrm{dr}$ $\mathrm{W}=\mathrm{Fdr}$ $\mathrm{W}=\mathrm{P}(2 \pi \mathrm{r} l) \mathrm{dr}$ Increase in area $=\Delta \mathrm{S}=2 \pi(\mathrm{r}+\mathrm{dr}) l-2 \pi \mathrm{r} l=2 \pi \mathrm{dr} l$ $\therefore \quad \mathrm{T}=\frac{\mathrm{W}}{\Delta \mathrm{S}}=\frac{\mathrm{P}(2 \pi \mathrm{r} l) \mathrm{dr}}{2 \pi \mathrm{dr} l}=\mathrm{Pr}$ $\Rightarrow P=\frac{T}{r}$
J and K CET- 2009
Mechanical Properties of Fluids
143050
A sphere of radius $R$ is gently dropped into liquid of viscosity $\eta$ is a vertical uniform tube. It attains a terminal velocity $v$. Another sphere of radius $2 R$ when dropped into the same liquid will attain its terminal velocity
1 $\mathrm{v}$
2 $2 \mathrm{v}$
3 $4 \mathrm{v}$
4 $9 \mathrm{v}$
Explanation:
C Terminal velocity of sphere into liquid is. $\mathrm{v}_{\mathrm{T}}=\frac{2}{9} \mathrm{R}^{2} \frac{(\rho-\sigma) \mathrm{g}}{\eta}$ $\mathrm{v} \propto \mathrm{R}^{2}$ $\frac{\mathrm{V}^{\prime}}{\mathrm{v}}=\frac{\mathrm{R}^{\prime^{2}}}{\mathrm{R}}$ $\Rightarrow \quad \frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\frac{4 \mathrm{R}^{2}}{\mathrm{R}^{2}}$ $\mathrm{v}^{\prime}=4 \mathrm{v}$
J and K CET- 2008
Mechanical Properties of Fluids
143051
Two liquid drops having diameters of $1 \mathrm{~cm}$ and $1.5 \mathrm{~cm}$. The ratio of excess pressures inside them is
1 $1: 1$
2 $5: 3$
3 $2: 3$
4 $3: 2$
Explanation:
D Diameter of $\mathrm{I}^{\text {st }}$ liquid drop $\left(\mathrm{D}_{1}\right)=1 \mathrm{~cm}$ Diameter of $\mathrm{II}^{\text {nd }}$ liquid drop $\left(\mathrm{D}_{2}\right)=1.5 \mathrm{~cm}$ For drop, Excess pressure, $\mathrm{P}=\frac{2 \mathrm{~T}}{\mathrm{R}}$ $\therefore \quad \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{\mathrm{D}_{2}}{\mathrm{D}_{1}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{1.5}{1}=\frac{15}{10}=3: 2$
143046
The excess pressure inside a soap bubble of volume ' $V$ ' is three times the excess pressure inside a second soap bubble of volume ' $V_{1}$ '. The value $\left(\frac{V_{1}}{V}\right)$ is
1 $3: 1$
2 $1: 9$
3 $1: 3$
4 $27: 1$
Explanation:
D Volume of first soap bubble, $V==\frac{4}{3} \pi r^{3}$ Volume of second soap bubble $\mathrm{V}_{1}=\frac{4}{3} \pi \mathrm{r}_{1}^{3}$ Excess pressure in first soap bubble, $\mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{r}}$ Excess pressure in second soap bubble $\mathrm{P}_{1}=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}}$ According to question, $\mathrm{P}=3 \mathrm{P}_{1}$ $\frac{4 \mathrm{~T}}{\mathrm{r}}=\frac{4 \mathrm{~T}}{\mathrm{r}_{1}} \times 3$ $\frac{\mathrm{r}}{\mathrm{r}_{1}}=\frac{1}{3}$ $\frac{\mathrm{V}}{\mathrm{V}_{1}}=\frac{\frac{4}{3} \pi \mathrm{r}^{3}}{\frac{4}{3} \pi \mathrm{r}_{1}^{3}}=\left(\frac{\mathrm{r}}{\mathrm{r}_{1}}\right)^{3}=\left(\frac{1}{3}\right)^{3}=\frac{1}{27}$ Ratio of $\frac{\mathrm{V}_{1}}{\mathrm{~V}}=\frac{27}{1}$ $\mathrm{V}_{1}: \mathrm{V}=27: 1$
MHT-CET 2020
Mechanical Properties of Fluids
143047
Water drops fall at regular intervals from a tap which is $\mathbf{5} \mathbf{~ m}$ above the ground. The third drops leaves the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant?
1 $2.5 \mathrm{~m}$
2 $3.75 \mathrm{~m}$
3 $4.0 \mathrm{~m}$
4 $1.25 \mathrm{~m}$
Explanation:
B Given that, For $1^{\text {st }}$ drop - $\text { Height of tap }(\mathrm{h})=5 \mathrm{~m}$ $\text { Initial velocity }(\mathrm{u})=0$ From, $\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^{2}$ $5=0+\frac{1}{2} \times 10 \times \mathrm{t}^{2} .$ $\mathrm{t}^{2}=1$ $\mathrm{t}=1 \mathrm{sec}$ It means that the third drop leaves after one second of the first drop. So the each drop leaves after every 0.5 sec. For $2^{\text {nd }}$ drop $u=0$ $t=0.5$ $v=u+g t$ $v=0+10 \times 0.5$ $v=5 m / s$ $\text { From, } v^{2}=u^{2}+2 g s$ $(5)^{2}=0+2 \times 10 \times \mathrm{s}$ $\mathrm{s}=\frac{25}{20}=1.25$ Position of the second drop from ground $=5-1.25=$ $3.75 \mathrm{~m}$
SCRA-2013
Mechanical Properties of Fluids
143048
The excess pressure inside a cylindrical drop of liquid or a cylindrical bubble of radius $r$ in a liquid of surface tension $T$ is
1 $\frac{\mathrm{T}}{4 \mathrm{r}}$
2 $\frac{\mathrm{T}}{\mathrm{r}}$
3 $\frac{2 \mathrm{~T}}{\mathrm{r}}$
4 $\frac{4 \mathrm{~T}}{\mathrm{r}}$
Explanation:
B Work done in increasing the radius from $\mathrm{r}$ to $\mathrm{r}+\mathrm{dr}$ $\mathrm{W}=\mathrm{Fdr}$ $\mathrm{W}=\mathrm{P}(2 \pi \mathrm{r} l) \mathrm{dr}$ Increase in area $=\Delta \mathrm{S}=2 \pi(\mathrm{r}+\mathrm{dr}) l-2 \pi \mathrm{r} l=2 \pi \mathrm{dr} l$ $\therefore \quad \mathrm{T}=\frac{\mathrm{W}}{\Delta \mathrm{S}}=\frac{\mathrm{P}(2 \pi \mathrm{r} l) \mathrm{dr}}{2 \pi \mathrm{dr} l}=\mathrm{Pr}$ $\Rightarrow P=\frac{T}{r}$
J and K CET- 2009
Mechanical Properties of Fluids
143050
A sphere of radius $R$ is gently dropped into liquid of viscosity $\eta$ is a vertical uniform tube. It attains a terminal velocity $v$. Another sphere of radius $2 R$ when dropped into the same liquid will attain its terminal velocity
1 $\mathrm{v}$
2 $2 \mathrm{v}$
3 $4 \mathrm{v}$
4 $9 \mathrm{v}$
Explanation:
C Terminal velocity of sphere into liquid is. $\mathrm{v}_{\mathrm{T}}=\frac{2}{9} \mathrm{R}^{2} \frac{(\rho-\sigma) \mathrm{g}}{\eta}$ $\mathrm{v} \propto \mathrm{R}^{2}$ $\frac{\mathrm{V}^{\prime}}{\mathrm{v}}=\frac{\mathrm{R}^{\prime^{2}}}{\mathrm{R}}$ $\Rightarrow \quad \frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\frac{4 \mathrm{R}^{2}}{\mathrm{R}^{2}}$ $\mathrm{v}^{\prime}=4 \mathrm{v}$
J and K CET- 2008
Mechanical Properties of Fluids
143051
Two liquid drops having diameters of $1 \mathrm{~cm}$ and $1.5 \mathrm{~cm}$. The ratio of excess pressures inside them is
1 $1: 1$
2 $5: 3$
3 $2: 3$
4 $3: 2$
Explanation:
D Diameter of $\mathrm{I}^{\text {st }}$ liquid drop $\left(\mathrm{D}_{1}\right)=1 \mathrm{~cm}$ Diameter of $\mathrm{II}^{\text {nd }}$ liquid drop $\left(\mathrm{D}_{2}\right)=1.5 \mathrm{~cm}$ For drop, Excess pressure, $\mathrm{P}=\frac{2 \mathrm{~T}}{\mathrm{R}}$ $\therefore \quad \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{\mathrm{D}_{2}}{\mathrm{D}_{1}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{1.5}{1}=\frac{15}{10}=3: 2$
