142964
Water rises in a capillary tube of radius $r$ up to a height ' $h$ '. The mass of water in a capillary is ' $m$ '. The mass of water that will rise in a capillary of radius $\frac{r}{4}$ will be
1 $\frac{4}{\mathrm{~m}}$
2 $4 \mathrm{~m}$
3 $\mathrm{m}$
4 $\frac{\mathrm{m}}{4}$
Explanation:
D Given that, Mass of the water in tube $\left(\mathrm{m}_{1}\right)=\mathrm{m}$ Radius $\left(\mathrm{r}_{1}\right)=\mathrm{r}$ In other tube, Mass of water in tube $\left(\mathrm{m}_{2}\right)=\mathrm{m}^{\prime}$ Radius $\left(\mathrm{r}_{2}\right)=\frac{\mathrm{r}}{4}$ The height of water column in a capillary tube is $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{rg}}$ $\mathrm{h} \propto \frac{1}{\mathrm{r}}$ We know that, Mass $=$ density $\times$ volume $\mathrm{m}=\rho \times \pi \mathrm{r}^{2} \mathrm{~h}$ $\mathrm{h}=\frac{\mathrm{m}}{\rho \pi \mathrm{r}^{2}}$ $\mathrm{h} \propto \frac{\mathrm{m}}{\mathrm{r}^{2}}$ From equating eq. (i) and (ii) we get $\frac{\mathrm{m}}{\mathrm{r}^{2}} \propto \frac{1}{\mathrm{r}}$ $\mathrm{m} \propto \mathrm{r}$ $\frac{\mathrm{m}}{\mathrm{m}^{\prime}}=\frac{\mathrm{r}}{\mathrm{r} / 4}$ $\mathrm{m}^{\prime}=\frac{\mathrm{m} \times \mathrm{r}}{4 \times \mathrm{r}}$ $\mathrm{m}^{\prime}=\frac{\mathrm{m}}{4}$
MHT-CET 2020
Mechanical Properties of Fluids
142966
A capillary tube is dipped into a liquid. If the levels of liquid inside and outside the tube are same, then the angle of contact is
1 $0^{\circ}$
2 $60^{\circ}$
3 $45^{\circ}$
4 $90^{\circ}$
Explanation:
D We know that, The height of liquid rise in a capillary tube is given as $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho g}$ When the level of liquids inside and outside are same then $\mathrm{h}=0$ Thus, $0=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}}$ $\cos \theta=0$ $\theta=90^{\circ}$
MHT-CET 2020
Mechanical Properties of Fluids
142969
Two capillary tubes of different diameters are dipped in water. The rise of water is :
1 the same in both tubes
2 greater in the tube of larger diameter
3 greater in the tube of smaller diameter
4 independent of the diameter of the tube
Explanation:
C We know that, Rise of water in capillary tube- $\mathrm{h}=\frac{4 \mathrm{~T} \cos \theta}{\rho \mathrm{dg}}$ $\mathrm{h} \propto \frac{1}{\mathrm{~d}}$ Where, $\mathrm{h}=$ Height of water capillary tube $\mathrm{d}=$ diameter of tube Hence, if the diameter of the tube will be greater, then its height will be less and vice versa.
Karnataka CET-2011
Mechanical Properties of Fluids
142971
The meniscus of mercury in a capillary glass tube is
1 concave
2 plane
3 cylindrical
4 convex
Explanation:
D The meniscus of mercury in a capillary glass tube is convex (obtuse angle). While the meniscus of water in a capillary glass tube is concave (acute angle). $\theta \lt 90^{\circ}$ (Acute angle) $\theta>90^{\circ}$ (obtuse angle)
J and K CET- 2007
Mechanical Properties of Fluids
142972
In a capillary tube of which the lower end dips in a liquid, the liquid rises to a height of $10 \mathrm{~cm}$. If a capillary of the same bore is taken, whose length is $5 \mathbf{c m}$ and dipped in liquid, then
1 a fountain of liquid will be obtain
2 the liquid will not rise in the tube at all
3 the liquid will rise upto the top and slowly ooze out of it
4 the liquid will rise to the top and will stay there
Explanation:
D Given that, Initially the height of the capillary tube $\mathrm{h}_{1}=10 \mathrm{~cm}$ And for the same bore (diameter) size of tube whose length is $\rightarrow 5 \mathrm{~cm}$ For the rise of capillary tube, $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho g r}$ $\mathrm{~h} \propto \frac{1}{\mathrm{r}}$ From the second condition when length of capillary tube is less $(5 \mathrm{~cm})$ then, It is understood from $h \propto \frac{1}{\mathrm{r}}$ that, when height above the surface is decreased, so there will not any motion as the water rises till height and stays. Overflow of water is not possible because there is not any external force experienced by the water molecules that makes it to overflow from the capillary tube. So, if the tube of insufficient height, the liquid will rise to the top of the tube and spread over the brim Thus, the radius of meniscus will adjust to a new value, so that rh remains constant, i.e. $\mathrm{hr}=$ constant
142964
Water rises in a capillary tube of radius $r$ up to a height ' $h$ '. The mass of water in a capillary is ' $m$ '. The mass of water that will rise in a capillary of radius $\frac{r}{4}$ will be
1 $\frac{4}{\mathrm{~m}}$
2 $4 \mathrm{~m}$
3 $\mathrm{m}$
4 $\frac{\mathrm{m}}{4}$
Explanation:
D Given that, Mass of the water in tube $\left(\mathrm{m}_{1}\right)=\mathrm{m}$ Radius $\left(\mathrm{r}_{1}\right)=\mathrm{r}$ In other tube, Mass of water in tube $\left(\mathrm{m}_{2}\right)=\mathrm{m}^{\prime}$ Radius $\left(\mathrm{r}_{2}\right)=\frac{\mathrm{r}}{4}$ The height of water column in a capillary tube is $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{rg}}$ $\mathrm{h} \propto \frac{1}{\mathrm{r}}$ We know that, Mass $=$ density $\times$ volume $\mathrm{m}=\rho \times \pi \mathrm{r}^{2} \mathrm{~h}$ $\mathrm{h}=\frac{\mathrm{m}}{\rho \pi \mathrm{r}^{2}}$ $\mathrm{h} \propto \frac{\mathrm{m}}{\mathrm{r}^{2}}$ From equating eq. (i) and (ii) we get $\frac{\mathrm{m}}{\mathrm{r}^{2}} \propto \frac{1}{\mathrm{r}}$ $\mathrm{m} \propto \mathrm{r}$ $\frac{\mathrm{m}}{\mathrm{m}^{\prime}}=\frac{\mathrm{r}}{\mathrm{r} / 4}$ $\mathrm{m}^{\prime}=\frac{\mathrm{m} \times \mathrm{r}}{4 \times \mathrm{r}}$ $\mathrm{m}^{\prime}=\frac{\mathrm{m}}{4}$
MHT-CET 2020
Mechanical Properties of Fluids
142966
A capillary tube is dipped into a liquid. If the levels of liquid inside and outside the tube are same, then the angle of contact is
1 $0^{\circ}$
2 $60^{\circ}$
3 $45^{\circ}$
4 $90^{\circ}$
Explanation:
D We know that, The height of liquid rise in a capillary tube is given as $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho g}$ When the level of liquids inside and outside are same then $\mathrm{h}=0$ Thus, $0=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}}$ $\cos \theta=0$ $\theta=90^{\circ}$
MHT-CET 2020
Mechanical Properties of Fluids
142969
Two capillary tubes of different diameters are dipped in water. The rise of water is :
1 the same in both tubes
2 greater in the tube of larger diameter
3 greater in the tube of smaller diameter
4 independent of the diameter of the tube
Explanation:
C We know that, Rise of water in capillary tube- $\mathrm{h}=\frac{4 \mathrm{~T} \cos \theta}{\rho \mathrm{dg}}$ $\mathrm{h} \propto \frac{1}{\mathrm{~d}}$ Where, $\mathrm{h}=$ Height of water capillary tube $\mathrm{d}=$ diameter of tube Hence, if the diameter of the tube will be greater, then its height will be less and vice versa.
Karnataka CET-2011
Mechanical Properties of Fluids
142971
The meniscus of mercury in a capillary glass tube is
1 concave
2 plane
3 cylindrical
4 convex
Explanation:
D The meniscus of mercury in a capillary glass tube is convex (obtuse angle). While the meniscus of water in a capillary glass tube is concave (acute angle). $\theta \lt 90^{\circ}$ (Acute angle) $\theta>90^{\circ}$ (obtuse angle)
J and K CET- 2007
Mechanical Properties of Fluids
142972
In a capillary tube of which the lower end dips in a liquid, the liquid rises to a height of $10 \mathrm{~cm}$. If a capillary of the same bore is taken, whose length is $5 \mathbf{c m}$ and dipped in liquid, then
1 a fountain of liquid will be obtain
2 the liquid will not rise in the tube at all
3 the liquid will rise upto the top and slowly ooze out of it
4 the liquid will rise to the top and will stay there
Explanation:
D Given that, Initially the height of the capillary tube $\mathrm{h}_{1}=10 \mathrm{~cm}$ And for the same bore (diameter) size of tube whose length is $\rightarrow 5 \mathrm{~cm}$ For the rise of capillary tube, $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho g r}$ $\mathrm{~h} \propto \frac{1}{\mathrm{r}}$ From the second condition when length of capillary tube is less $(5 \mathrm{~cm})$ then, It is understood from $h \propto \frac{1}{\mathrm{r}}$ that, when height above the surface is decreased, so there will not any motion as the water rises till height and stays. Overflow of water is not possible because there is not any external force experienced by the water molecules that makes it to overflow from the capillary tube. So, if the tube of insufficient height, the liquid will rise to the top of the tube and spread over the brim Thus, the radius of meniscus will adjust to a new value, so that rh remains constant, i.e. $\mathrm{hr}=$ constant
142964
Water rises in a capillary tube of radius $r$ up to a height ' $h$ '. The mass of water in a capillary is ' $m$ '. The mass of water that will rise in a capillary of radius $\frac{r}{4}$ will be
1 $\frac{4}{\mathrm{~m}}$
2 $4 \mathrm{~m}$
3 $\mathrm{m}$
4 $\frac{\mathrm{m}}{4}$
Explanation:
D Given that, Mass of the water in tube $\left(\mathrm{m}_{1}\right)=\mathrm{m}$ Radius $\left(\mathrm{r}_{1}\right)=\mathrm{r}$ In other tube, Mass of water in tube $\left(\mathrm{m}_{2}\right)=\mathrm{m}^{\prime}$ Radius $\left(\mathrm{r}_{2}\right)=\frac{\mathrm{r}}{4}$ The height of water column in a capillary tube is $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{rg}}$ $\mathrm{h} \propto \frac{1}{\mathrm{r}}$ We know that, Mass $=$ density $\times$ volume $\mathrm{m}=\rho \times \pi \mathrm{r}^{2} \mathrm{~h}$ $\mathrm{h}=\frac{\mathrm{m}}{\rho \pi \mathrm{r}^{2}}$ $\mathrm{h} \propto \frac{\mathrm{m}}{\mathrm{r}^{2}}$ From equating eq. (i) and (ii) we get $\frac{\mathrm{m}}{\mathrm{r}^{2}} \propto \frac{1}{\mathrm{r}}$ $\mathrm{m} \propto \mathrm{r}$ $\frac{\mathrm{m}}{\mathrm{m}^{\prime}}=\frac{\mathrm{r}}{\mathrm{r} / 4}$ $\mathrm{m}^{\prime}=\frac{\mathrm{m} \times \mathrm{r}}{4 \times \mathrm{r}}$ $\mathrm{m}^{\prime}=\frac{\mathrm{m}}{4}$
MHT-CET 2020
Mechanical Properties of Fluids
142966
A capillary tube is dipped into a liquid. If the levels of liquid inside and outside the tube are same, then the angle of contact is
1 $0^{\circ}$
2 $60^{\circ}$
3 $45^{\circ}$
4 $90^{\circ}$
Explanation:
D We know that, The height of liquid rise in a capillary tube is given as $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho g}$ When the level of liquids inside and outside are same then $\mathrm{h}=0$ Thus, $0=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}}$ $\cos \theta=0$ $\theta=90^{\circ}$
MHT-CET 2020
Mechanical Properties of Fluids
142969
Two capillary tubes of different diameters are dipped in water. The rise of water is :
1 the same in both tubes
2 greater in the tube of larger diameter
3 greater in the tube of smaller diameter
4 independent of the diameter of the tube
Explanation:
C We know that, Rise of water in capillary tube- $\mathrm{h}=\frac{4 \mathrm{~T} \cos \theta}{\rho \mathrm{dg}}$ $\mathrm{h} \propto \frac{1}{\mathrm{~d}}$ Where, $\mathrm{h}=$ Height of water capillary tube $\mathrm{d}=$ diameter of tube Hence, if the diameter of the tube will be greater, then its height will be less and vice versa.
Karnataka CET-2011
Mechanical Properties of Fluids
142971
The meniscus of mercury in a capillary glass tube is
1 concave
2 plane
3 cylindrical
4 convex
Explanation:
D The meniscus of mercury in a capillary glass tube is convex (obtuse angle). While the meniscus of water in a capillary glass tube is concave (acute angle). $\theta \lt 90^{\circ}$ (Acute angle) $\theta>90^{\circ}$ (obtuse angle)
J and K CET- 2007
Mechanical Properties of Fluids
142972
In a capillary tube of which the lower end dips in a liquid, the liquid rises to a height of $10 \mathrm{~cm}$. If a capillary of the same bore is taken, whose length is $5 \mathbf{c m}$ and dipped in liquid, then
1 a fountain of liquid will be obtain
2 the liquid will not rise in the tube at all
3 the liquid will rise upto the top and slowly ooze out of it
4 the liquid will rise to the top and will stay there
Explanation:
D Given that, Initially the height of the capillary tube $\mathrm{h}_{1}=10 \mathrm{~cm}$ And for the same bore (diameter) size of tube whose length is $\rightarrow 5 \mathrm{~cm}$ For the rise of capillary tube, $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho g r}$ $\mathrm{~h} \propto \frac{1}{\mathrm{r}}$ From the second condition when length of capillary tube is less $(5 \mathrm{~cm})$ then, It is understood from $h \propto \frac{1}{\mathrm{r}}$ that, when height above the surface is decreased, so there will not any motion as the water rises till height and stays. Overflow of water is not possible because there is not any external force experienced by the water molecules that makes it to overflow from the capillary tube. So, if the tube of insufficient height, the liquid will rise to the top of the tube and spread over the brim Thus, the radius of meniscus will adjust to a new value, so that rh remains constant, i.e. $\mathrm{hr}=$ constant
142964
Water rises in a capillary tube of radius $r$ up to a height ' $h$ '. The mass of water in a capillary is ' $m$ '. The mass of water that will rise in a capillary of radius $\frac{r}{4}$ will be
1 $\frac{4}{\mathrm{~m}}$
2 $4 \mathrm{~m}$
3 $\mathrm{m}$
4 $\frac{\mathrm{m}}{4}$
Explanation:
D Given that, Mass of the water in tube $\left(\mathrm{m}_{1}\right)=\mathrm{m}$ Radius $\left(\mathrm{r}_{1}\right)=\mathrm{r}$ In other tube, Mass of water in tube $\left(\mathrm{m}_{2}\right)=\mathrm{m}^{\prime}$ Radius $\left(\mathrm{r}_{2}\right)=\frac{\mathrm{r}}{4}$ The height of water column in a capillary tube is $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{rg}}$ $\mathrm{h} \propto \frac{1}{\mathrm{r}}$ We know that, Mass $=$ density $\times$ volume $\mathrm{m}=\rho \times \pi \mathrm{r}^{2} \mathrm{~h}$ $\mathrm{h}=\frac{\mathrm{m}}{\rho \pi \mathrm{r}^{2}}$ $\mathrm{h} \propto \frac{\mathrm{m}}{\mathrm{r}^{2}}$ From equating eq. (i) and (ii) we get $\frac{\mathrm{m}}{\mathrm{r}^{2}} \propto \frac{1}{\mathrm{r}}$ $\mathrm{m} \propto \mathrm{r}$ $\frac{\mathrm{m}}{\mathrm{m}^{\prime}}=\frac{\mathrm{r}}{\mathrm{r} / 4}$ $\mathrm{m}^{\prime}=\frac{\mathrm{m} \times \mathrm{r}}{4 \times \mathrm{r}}$ $\mathrm{m}^{\prime}=\frac{\mathrm{m}}{4}$
MHT-CET 2020
Mechanical Properties of Fluids
142966
A capillary tube is dipped into a liquid. If the levels of liquid inside and outside the tube are same, then the angle of contact is
1 $0^{\circ}$
2 $60^{\circ}$
3 $45^{\circ}$
4 $90^{\circ}$
Explanation:
D We know that, The height of liquid rise in a capillary tube is given as $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho g}$ When the level of liquids inside and outside are same then $\mathrm{h}=0$ Thus, $0=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}}$ $\cos \theta=0$ $\theta=90^{\circ}$
MHT-CET 2020
Mechanical Properties of Fluids
142969
Two capillary tubes of different diameters are dipped in water. The rise of water is :
1 the same in both tubes
2 greater in the tube of larger diameter
3 greater in the tube of smaller diameter
4 independent of the diameter of the tube
Explanation:
C We know that, Rise of water in capillary tube- $\mathrm{h}=\frac{4 \mathrm{~T} \cos \theta}{\rho \mathrm{dg}}$ $\mathrm{h} \propto \frac{1}{\mathrm{~d}}$ Where, $\mathrm{h}=$ Height of water capillary tube $\mathrm{d}=$ diameter of tube Hence, if the diameter of the tube will be greater, then its height will be less and vice versa.
Karnataka CET-2011
Mechanical Properties of Fluids
142971
The meniscus of mercury in a capillary glass tube is
1 concave
2 plane
3 cylindrical
4 convex
Explanation:
D The meniscus of mercury in a capillary glass tube is convex (obtuse angle). While the meniscus of water in a capillary glass tube is concave (acute angle). $\theta \lt 90^{\circ}$ (Acute angle) $\theta>90^{\circ}$ (obtuse angle)
J and K CET- 2007
Mechanical Properties of Fluids
142972
In a capillary tube of which the lower end dips in a liquid, the liquid rises to a height of $10 \mathrm{~cm}$. If a capillary of the same bore is taken, whose length is $5 \mathbf{c m}$ and dipped in liquid, then
1 a fountain of liquid will be obtain
2 the liquid will not rise in the tube at all
3 the liquid will rise upto the top and slowly ooze out of it
4 the liquid will rise to the top and will stay there
Explanation:
D Given that, Initially the height of the capillary tube $\mathrm{h}_{1}=10 \mathrm{~cm}$ And for the same bore (diameter) size of tube whose length is $\rightarrow 5 \mathrm{~cm}$ For the rise of capillary tube, $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho g r}$ $\mathrm{~h} \propto \frac{1}{\mathrm{r}}$ From the second condition when length of capillary tube is less $(5 \mathrm{~cm})$ then, It is understood from $h \propto \frac{1}{\mathrm{r}}$ that, when height above the surface is decreased, so there will not any motion as the water rises till height and stays. Overflow of water is not possible because there is not any external force experienced by the water molecules that makes it to overflow from the capillary tube. So, if the tube of insufficient height, the liquid will rise to the top of the tube and spread over the brim Thus, the radius of meniscus will adjust to a new value, so that rh remains constant, i.e. $\mathrm{hr}=$ constant
142964
Water rises in a capillary tube of radius $r$ up to a height ' $h$ '. The mass of water in a capillary is ' $m$ '. The mass of water that will rise in a capillary of radius $\frac{r}{4}$ will be
1 $\frac{4}{\mathrm{~m}}$
2 $4 \mathrm{~m}$
3 $\mathrm{m}$
4 $\frac{\mathrm{m}}{4}$
Explanation:
D Given that, Mass of the water in tube $\left(\mathrm{m}_{1}\right)=\mathrm{m}$ Radius $\left(\mathrm{r}_{1}\right)=\mathrm{r}$ In other tube, Mass of water in tube $\left(\mathrm{m}_{2}\right)=\mathrm{m}^{\prime}$ Radius $\left(\mathrm{r}_{2}\right)=\frac{\mathrm{r}}{4}$ The height of water column in a capillary tube is $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{rg}}$ $\mathrm{h} \propto \frac{1}{\mathrm{r}}$ We know that, Mass $=$ density $\times$ volume $\mathrm{m}=\rho \times \pi \mathrm{r}^{2} \mathrm{~h}$ $\mathrm{h}=\frac{\mathrm{m}}{\rho \pi \mathrm{r}^{2}}$ $\mathrm{h} \propto \frac{\mathrm{m}}{\mathrm{r}^{2}}$ From equating eq. (i) and (ii) we get $\frac{\mathrm{m}}{\mathrm{r}^{2}} \propto \frac{1}{\mathrm{r}}$ $\mathrm{m} \propto \mathrm{r}$ $\frac{\mathrm{m}}{\mathrm{m}^{\prime}}=\frac{\mathrm{r}}{\mathrm{r} / 4}$ $\mathrm{m}^{\prime}=\frac{\mathrm{m} \times \mathrm{r}}{4 \times \mathrm{r}}$ $\mathrm{m}^{\prime}=\frac{\mathrm{m}}{4}$
MHT-CET 2020
Mechanical Properties of Fluids
142966
A capillary tube is dipped into a liquid. If the levels of liquid inside and outside the tube are same, then the angle of contact is
1 $0^{\circ}$
2 $60^{\circ}$
3 $45^{\circ}$
4 $90^{\circ}$
Explanation:
D We know that, The height of liquid rise in a capillary tube is given as $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho g}$ When the level of liquids inside and outside are same then $\mathrm{h}=0$ Thus, $0=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}}$ $\cos \theta=0$ $\theta=90^{\circ}$
MHT-CET 2020
Mechanical Properties of Fluids
142969
Two capillary tubes of different diameters are dipped in water. The rise of water is :
1 the same in both tubes
2 greater in the tube of larger diameter
3 greater in the tube of smaller diameter
4 independent of the diameter of the tube
Explanation:
C We know that, Rise of water in capillary tube- $\mathrm{h}=\frac{4 \mathrm{~T} \cos \theta}{\rho \mathrm{dg}}$ $\mathrm{h} \propto \frac{1}{\mathrm{~d}}$ Where, $\mathrm{h}=$ Height of water capillary tube $\mathrm{d}=$ diameter of tube Hence, if the diameter of the tube will be greater, then its height will be less and vice versa.
Karnataka CET-2011
Mechanical Properties of Fluids
142971
The meniscus of mercury in a capillary glass tube is
1 concave
2 plane
3 cylindrical
4 convex
Explanation:
D The meniscus of mercury in a capillary glass tube is convex (obtuse angle). While the meniscus of water in a capillary glass tube is concave (acute angle). $\theta \lt 90^{\circ}$ (Acute angle) $\theta>90^{\circ}$ (obtuse angle)
J and K CET- 2007
Mechanical Properties of Fluids
142972
In a capillary tube of which the lower end dips in a liquid, the liquid rises to a height of $10 \mathrm{~cm}$. If a capillary of the same bore is taken, whose length is $5 \mathbf{c m}$ and dipped in liquid, then
1 a fountain of liquid will be obtain
2 the liquid will not rise in the tube at all
3 the liquid will rise upto the top and slowly ooze out of it
4 the liquid will rise to the top and will stay there
Explanation:
D Given that, Initially the height of the capillary tube $\mathrm{h}_{1}=10 \mathrm{~cm}$ And for the same bore (diameter) size of tube whose length is $\rightarrow 5 \mathrm{~cm}$ For the rise of capillary tube, $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho g r}$ $\mathrm{~h} \propto \frac{1}{\mathrm{r}}$ From the second condition when length of capillary tube is less $(5 \mathrm{~cm})$ then, It is understood from $h \propto \frac{1}{\mathrm{r}}$ that, when height above the surface is decreased, so there will not any motion as the water rises till height and stays. Overflow of water is not possible because there is not any external force experienced by the water molecules that makes it to overflow from the capillary tube. So, if the tube of insufficient height, the liquid will rise to the top of the tube and spread over the brim Thus, the radius of meniscus will adjust to a new value, so that rh remains constant, i.e. $\mathrm{hr}=$ constant
