142938
If two capillary tubes of radii $r_{1}$ and $r_{2}$ in the ratio $1: 2$ are dipped vertically in water, then the ratio of capillary rises in the respective tubes is
1 $1: 4$
2 $4: 1$
3 $1: 2$
4 $2: 1$
5 $1: \sqrt{2}$
Explanation:
D Given that, Ratio of capillary tube radius are $r_{1}: r_{2}=1: 2$ Height of water, \(\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho g} \text { (capillary rise) }\) \(\therefore \mathrm{h} \propto \frac{1}{\mathrm{r}}\) \(\Rightarrow \frac{\mathrm{h}_1}{\mathrm{~h}_2} =\frac{\mathrm{r}_2}{\mathrm{r}_1}\) \(\frac{\mathrm{h}_1}{\mathrm{~h}_2}=\frac{2}{1}\) Hence, \(\mathrm{h}_1: \mathrm{h}_2=2: 1\)
MNR 1998
Mechanical Properties of Fluids
142940
The radius of the bore of a capillary tube is $\mathbf{r}$ and the angle of contact of the liquid is $\theta$. When the tube is dipped in the liquid, the radius of curvature of the meniscus of liquid rising in the tube is
1 $r \sin \theta$
2 $\frac{r}{\sin \theta}$
3 $r \cos \theta$
4 $\frac{r}{\cos \theta}$
Explanation:
D Let the radius of curvature of meniscus is ' $\mathrm{R}$ '. In the above figure, radius of capillary tube is ' $r$ ' and contact angle ' $\theta$ ' Now, in $\triangle \mathrm{AOB}-$ $\cos \theta=\frac{r}{R}$ $R=\frac{r}{\cos \theta}$
AP EAMCET (18.09.2020) Shift-I
Mechanical Properties of Fluids
142941
Two capillary tubes of same length each of 50 cm but of different radii $4 \mathrm{~mm}$ and $2 \mathrm{~mm}$ are connected in series. When water flows, the pressure difference between the ends of the first tube is
1 $\frac{p}{2}$
2 $\frac{\mathrm{p}}{17}$
3 $\frac{\mathrm{p}}{4}$
4 $\frac{\mathrm{p}}{8}$
Explanation:
B Given:- Radius of capillary tubes, $\mathrm{r}_{1}=4 \mathrm{~mm}, \mathrm{r}_{2}=2 \mathrm{~mm}$ Since, we know that water resistance according to Poiseuille's equation- $\mathrm{R} =\frac{8 \eta l}{\pi \mathrm{r}^{4}}$ $\mathrm{R} \propto \frac{1}{\mathrm{r}^{4}}$ $\Rightarrow \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{4}$ $\Rightarrow \quad \frac{R_{1}}{R_{2}}=\left(\frac{2}{4}\right)^{4}$ $R_{2}=16 R_{1}$ If $\mathrm{R}_{1}=\mathrm{R}$ then $\mathrm{R}_{2}=16 \mathrm{R}$ Pressure difference $(\Delta \mathrm{P})$ $\Delta \mathrm{P} =\frac{\mathrm{P}}{\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)} \times \mathrm{R}$ $\Delta \mathrm{P} =\frac{\mathrm{P}}{(\mathrm{R}+16 \mathrm{R})} \times \mathrm{R}$ $\therefore \quad \Delta \mathrm{P} =\frac{\mathrm{P}}{17 \mathrm{R}} \times \mathrm{R}$ $\Delta \mathrm{P} =\frac{\mathrm{P}}{17}$
AP EAMCET (21.04.2019) Shift-II
Mechanical Properties of Fluids
142942
With the increase in temperature, the angle of contact
1 Decreases
2 Increases
3 Remains constant
4 Sometimes increases and sometimes decreases
Explanation:
A We know that, height of capillary rise is given as- $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{gr}}$ Where $-\mathrm{T}=$ surface tension $\rho=\text { Density of fluid }$ $r=\text { Radius of capillary tube }$ $g=\text { acceleration due to gravity }$ When we increase the temperature of liquid like water then the intermolecular attraction between molecules decreases and they are moving in random order and separation between molecules increases due to which cohesive force decreases and surface tension also decreases. Hence, when we increase the temperature the value of contact angle will be decreases.
142938
If two capillary tubes of radii $r_{1}$ and $r_{2}$ in the ratio $1: 2$ are dipped vertically in water, then the ratio of capillary rises in the respective tubes is
1 $1: 4$
2 $4: 1$
3 $1: 2$
4 $2: 1$
5 $1: \sqrt{2}$
Explanation:
D Given that, Ratio of capillary tube radius are $r_{1}: r_{2}=1: 2$ Height of water, \(\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho g} \text { (capillary rise) }\) \(\therefore \mathrm{h} \propto \frac{1}{\mathrm{r}}\) \(\Rightarrow \frac{\mathrm{h}_1}{\mathrm{~h}_2} =\frac{\mathrm{r}_2}{\mathrm{r}_1}\) \(\frac{\mathrm{h}_1}{\mathrm{~h}_2}=\frac{2}{1}\) Hence, \(\mathrm{h}_1: \mathrm{h}_2=2: 1\)
MNR 1998
Mechanical Properties of Fluids
142940
The radius of the bore of a capillary tube is $\mathbf{r}$ and the angle of contact of the liquid is $\theta$. When the tube is dipped in the liquid, the radius of curvature of the meniscus of liquid rising in the tube is
1 $r \sin \theta$
2 $\frac{r}{\sin \theta}$
3 $r \cos \theta$
4 $\frac{r}{\cos \theta}$
Explanation:
D Let the radius of curvature of meniscus is ' $\mathrm{R}$ '. In the above figure, radius of capillary tube is ' $r$ ' and contact angle ' $\theta$ ' Now, in $\triangle \mathrm{AOB}-$ $\cos \theta=\frac{r}{R}$ $R=\frac{r}{\cos \theta}$
AP EAMCET (18.09.2020) Shift-I
Mechanical Properties of Fluids
142941
Two capillary tubes of same length each of 50 cm but of different radii $4 \mathrm{~mm}$ and $2 \mathrm{~mm}$ are connected in series. When water flows, the pressure difference between the ends of the first tube is
1 $\frac{p}{2}$
2 $\frac{\mathrm{p}}{17}$
3 $\frac{\mathrm{p}}{4}$
4 $\frac{\mathrm{p}}{8}$
Explanation:
B Given:- Radius of capillary tubes, $\mathrm{r}_{1}=4 \mathrm{~mm}, \mathrm{r}_{2}=2 \mathrm{~mm}$ Since, we know that water resistance according to Poiseuille's equation- $\mathrm{R} =\frac{8 \eta l}{\pi \mathrm{r}^{4}}$ $\mathrm{R} \propto \frac{1}{\mathrm{r}^{4}}$ $\Rightarrow \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{4}$ $\Rightarrow \quad \frac{R_{1}}{R_{2}}=\left(\frac{2}{4}\right)^{4}$ $R_{2}=16 R_{1}$ If $\mathrm{R}_{1}=\mathrm{R}$ then $\mathrm{R}_{2}=16 \mathrm{R}$ Pressure difference $(\Delta \mathrm{P})$ $\Delta \mathrm{P} =\frac{\mathrm{P}}{\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)} \times \mathrm{R}$ $\Delta \mathrm{P} =\frac{\mathrm{P}}{(\mathrm{R}+16 \mathrm{R})} \times \mathrm{R}$ $\therefore \quad \Delta \mathrm{P} =\frac{\mathrm{P}}{17 \mathrm{R}} \times \mathrm{R}$ $\Delta \mathrm{P} =\frac{\mathrm{P}}{17}$
AP EAMCET (21.04.2019) Shift-II
Mechanical Properties of Fluids
142942
With the increase in temperature, the angle of contact
1 Decreases
2 Increases
3 Remains constant
4 Sometimes increases and sometimes decreases
Explanation:
A We know that, height of capillary rise is given as- $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{gr}}$ Where $-\mathrm{T}=$ surface tension $\rho=\text { Density of fluid }$ $r=\text { Radius of capillary tube }$ $g=\text { acceleration due to gravity }$ When we increase the temperature of liquid like water then the intermolecular attraction between molecules decreases and they are moving in random order and separation between molecules increases due to which cohesive force decreases and surface tension also decreases. Hence, when we increase the temperature the value of contact angle will be decreases.
142938
If two capillary tubes of radii $r_{1}$ and $r_{2}$ in the ratio $1: 2$ are dipped vertically in water, then the ratio of capillary rises in the respective tubes is
1 $1: 4$
2 $4: 1$
3 $1: 2$
4 $2: 1$
5 $1: \sqrt{2}$
Explanation:
D Given that, Ratio of capillary tube radius are $r_{1}: r_{2}=1: 2$ Height of water, \(\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho g} \text { (capillary rise) }\) \(\therefore \mathrm{h} \propto \frac{1}{\mathrm{r}}\) \(\Rightarrow \frac{\mathrm{h}_1}{\mathrm{~h}_2} =\frac{\mathrm{r}_2}{\mathrm{r}_1}\) \(\frac{\mathrm{h}_1}{\mathrm{~h}_2}=\frac{2}{1}\) Hence, \(\mathrm{h}_1: \mathrm{h}_2=2: 1\)
MNR 1998
Mechanical Properties of Fluids
142940
The radius of the bore of a capillary tube is $\mathbf{r}$ and the angle of contact of the liquid is $\theta$. When the tube is dipped in the liquid, the radius of curvature of the meniscus of liquid rising in the tube is
1 $r \sin \theta$
2 $\frac{r}{\sin \theta}$
3 $r \cos \theta$
4 $\frac{r}{\cos \theta}$
Explanation:
D Let the radius of curvature of meniscus is ' $\mathrm{R}$ '. In the above figure, radius of capillary tube is ' $r$ ' and contact angle ' $\theta$ ' Now, in $\triangle \mathrm{AOB}-$ $\cos \theta=\frac{r}{R}$ $R=\frac{r}{\cos \theta}$
AP EAMCET (18.09.2020) Shift-I
Mechanical Properties of Fluids
142941
Two capillary tubes of same length each of 50 cm but of different radii $4 \mathrm{~mm}$ and $2 \mathrm{~mm}$ are connected in series. When water flows, the pressure difference between the ends of the first tube is
1 $\frac{p}{2}$
2 $\frac{\mathrm{p}}{17}$
3 $\frac{\mathrm{p}}{4}$
4 $\frac{\mathrm{p}}{8}$
Explanation:
B Given:- Radius of capillary tubes, $\mathrm{r}_{1}=4 \mathrm{~mm}, \mathrm{r}_{2}=2 \mathrm{~mm}$ Since, we know that water resistance according to Poiseuille's equation- $\mathrm{R} =\frac{8 \eta l}{\pi \mathrm{r}^{4}}$ $\mathrm{R} \propto \frac{1}{\mathrm{r}^{4}}$ $\Rightarrow \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{4}$ $\Rightarrow \quad \frac{R_{1}}{R_{2}}=\left(\frac{2}{4}\right)^{4}$ $R_{2}=16 R_{1}$ If $\mathrm{R}_{1}=\mathrm{R}$ then $\mathrm{R}_{2}=16 \mathrm{R}$ Pressure difference $(\Delta \mathrm{P})$ $\Delta \mathrm{P} =\frac{\mathrm{P}}{\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)} \times \mathrm{R}$ $\Delta \mathrm{P} =\frac{\mathrm{P}}{(\mathrm{R}+16 \mathrm{R})} \times \mathrm{R}$ $\therefore \quad \Delta \mathrm{P} =\frac{\mathrm{P}}{17 \mathrm{R}} \times \mathrm{R}$ $\Delta \mathrm{P} =\frac{\mathrm{P}}{17}$
AP EAMCET (21.04.2019) Shift-II
Mechanical Properties of Fluids
142942
With the increase in temperature, the angle of contact
1 Decreases
2 Increases
3 Remains constant
4 Sometimes increases and sometimes decreases
Explanation:
A We know that, height of capillary rise is given as- $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{gr}}$ Where $-\mathrm{T}=$ surface tension $\rho=\text { Density of fluid }$ $r=\text { Radius of capillary tube }$ $g=\text { acceleration due to gravity }$ When we increase the temperature of liquid like water then the intermolecular attraction between molecules decreases and they are moving in random order and separation between molecules increases due to which cohesive force decreases and surface tension also decreases. Hence, when we increase the temperature the value of contact angle will be decreases.
142938
If two capillary tubes of radii $r_{1}$ and $r_{2}$ in the ratio $1: 2$ are dipped vertically in water, then the ratio of capillary rises in the respective tubes is
1 $1: 4$
2 $4: 1$
3 $1: 2$
4 $2: 1$
5 $1: \sqrt{2}$
Explanation:
D Given that, Ratio of capillary tube radius are $r_{1}: r_{2}=1: 2$ Height of water, \(\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho g} \text { (capillary rise) }\) \(\therefore \mathrm{h} \propto \frac{1}{\mathrm{r}}\) \(\Rightarrow \frac{\mathrm{h}_1}{\mathrm{~h}_2} =\frac{\mathrm{r}_2}{\mathrm{r}_1}\) \(\frac{\mathrm{h}_1}{\mathrm{~h}_2}=\frac{2}{1}\) Hence, \(\mathrm{h}_1: \mathrm{h}_2=2: 1\)
MNR 1998
Mechanical Properties of Fluids
142940
The radius of the bore of a capillary tube is $\mathbf{r}$ and the angle of contact of the liquid is $\theta$. When the tube is dipped in the liquid, the radius of curvature of the meniscus of liquid rising in the tube is
1 $r \sin \theta$
2 $\frac{r}{\sin \theta}$
3 $r \cos \theta$
4 $\frac{r}{\cos \theta}$
Explanation:
D Let the radius of curvature of meniscus is ' $\mathrm{R}$ '. In the above figure, radius of capillary tube is ' $r$ ' and contact angle ' $\theta$ ' Now, in $\triangle \mathrm{AOB}-$ $\cos \theta=\frac{r}{R}$ $R=\frac{r}{\cos \theta}$
AP EAMCET (18.09.2020) Shift-I
Mechanical Properties of Fluids
142941
Two capillary tubes of same length each of 50 cm but of different radii $4 \mathrm{~mm}$ and $2 \mathrm{~mm}$ are connected in series. When water flows, the pressure difference between the ends of the first tube is
1 $\frac{p}{2}$
2 $\frac{\mathrm{p}}{17}$
3 $\frac{\mathrm{p}}{4}$
4 $\frac{\mathrm{p}}{8}$
Explanation:
B Given:- Radius of capillary tubes, $\mathrm{r}_{1}=4 \mathrm{~mm}, \mathrm{r}_{2}=2 \mathrm{~mm}$ Since, we know that water resistance according to Poiseuille's equation- $\mathrm{R} =\frac{8 \eta l}{\pi \mathrm{r}^{4}}$ $\mathrm{R} \propto \frac{1}{\mathrm{r}^{4}}$ $\Rightarrow \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{4}$ $\Rightarrow \quad \frac{R_{1}}{R_{2}}=\left(\frac{2}{4}\right)^{4}$ $R_{2}=16 R_{1}$ If $\mathrm{R}_{1}=\mathrm{R}$ then $\mathrm{R}_{2}=16 \mathrm{R}$ Pressure difference $(\Delta \mathrm{P})$ $\Delta \mathrm{P} =\frac{\mathrm{P}}{\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)} \times \mathrm{R}$ $\Delta \mathrm{P} =\frac{\mathrm{P}}{(\mathrm{R}+16 \mathrm{R})} \times \mathrm{R}$ $\therefore \quad \Delta \mathrm{P} =\frac{\mathrm{P}}{17 \mathrm{R}} \times \mathrm{R}$ $\Delta \mathrm{P} =\frac{\mathrm{P}}{17}$
AP EAMCET (21.04.2019) Shift-II
Mechanical Properties of Fluids
142942
With the increase in temperature, the angle of contact
1 Decreases
2 Increases
3 Remains constant
4 Sometimes increases and sometimes decreases
Explanation:
A We know that, height of capillary rise is given as- $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{gr}}$ Where $-\mathrm{T}=$ surface tension $\rho=\text { Density of fluid }$ $r=\text { Radius of capillary tube }$ $g=\text { acceleration due to gravity }$ When we increase the temperature of liquid like water then the intermolecular attraction between molecules decreases and they are moving in random order and separation between molecules increases due to which cohesive force decreases and surface tension also decreases. Hence, when we increase the temperature the value of contact angle will be decreases.
