QBManager

Mechanical Properties of Fluids

142943 Due to capillary action, a liquid will rise in a tube if angle of contact is

1 acute

2 obtuse

3

90^{\circ}

4 zero

Explanation:

A Since, we know that, capillary rise of depression is given as-
$h = \frac{2 T \cos θ}{ρ gr}$
If $θ$ is less than $90^{\circ}$ , then $h$ will be positive liquid rises in a capillary tube, The angle of contact is acute.
Whereas liquid is depressed in a capillary tube, i.e. $θ$ is more than $90^{\circ}$ then the angle of contact is obtuse. 274. A vertical glass capillary tube of radius $r$ open at both ends contains some water (surface tension $T$ and density $ρ$ ). If $L$ be the length of the water column, then :

(a) $L = \frac{4 T}{r ρ g}$
(b) $L = \frac{2 T}{r ρ g}$
(c) $L = \frac{T}{4 r ρ g}$
(d) $L = \frac{T}{2 r ρ g}$
Ans: a :

Let, the radius of capillary tube is $r$ .

Net surface tension force in upward direction will be,
$F_{up} = 2 π r \cdot T \cos θ + 2 π r \cdot T \cos θ = 4 π r \cdot T \cos θ$
$F_{up} = 4 π r \cdot T \cos θ$
Due to the shape of meniscus at the upper and lower surfaces of the liquid, the vertical components of surface tension force will be added along vertical direction
So, net upward force-
$F_{up} = 4 T π r \cos θ$
$∵ θ = 0^{\circ} for glass \& water$
$F_{up} = 4 T \cdot π r$
Now, weight of liquid $mg$ can be balanced by surface tension force.
$∴ 4 T \cdot π r = mg$
$4 T \cdot π r = ρ Vg [∵ m = ρ V]$
$4 T \cdot π r = ρ A \cdot Lg [∵ V = A \times L]$
$4 T \cdot π r = ρ π r^{2} Lg [∵ A = π r^{2}]$
$L = \frac{4 T π r}{ρ π r^{2} \cdot g}$
$L = \frac{4 T}{ρ rg}$

BITSAT-2007

Mechanical Properties of Fluids

142944 A glass capillary tube of internal radius $r =$ $0.25 mm$ is immersed in water. The top end of the tube projected by $2 cm$ above the surface of the water. At what angle does the liquid meet the tube? Surface tension of water $= 0.7$ $N / m$ .

1

90^{\circ}

2

70^{\circ}

3

45^{\circ}

4

35^{\circ}

Explanation:

B Given that, $T = 0.7 N / m, r = 0.25 mm =$ $25 \times 10^{- 3} m$
We know that,
$h = \frac{2 T \cos θ}{r ρ g}$
$h = \frac{2 T}{r ρ g} [∵ For water and glass, θ = 0^{\circ}]$
$= \frac{2 \times 0.7}{0.25 \times 10^{- 3} \times 1000 \times 9.8}$
As the tube is only $2 cm$ above the water and so, water will rise by $2 cm$ and meet the tube at an angle such that,
$\cos θ = \frac{h^{'} r ρ g}{2 T}$
$= \frac{2 \times 10^{- 2} \times 0.25 \times 10^{- 3} \times 1000 \times 9.8}{2 \times 0.7}$
The liquid will meet the tube at an angle, $θ ≅ 70^{\circ}$

BITSAT-2015

Mechanical Properties of Fluids

142945 A capillary tube of radius $R$ is immersed in water and water rises in it to a height $H$ . Mass of water in the capillary tube is $M$ . If the radius of the tube is doubled, mass of water that will rise in the capillary tube will now be:

1

M

2

2 M

3

M / 2

4

4 M

Explanation:

B Given that, radius of capillary tube $= R$
Height of water $= H$
Mass of water in a capillary tube $= M$
Since, we know that hight of capillary rise in capillary tube having radius ' $r$ ' is given as-
$h = \frac{2 T \cos θ}{r ρ g}$

Since $T, ρ, g$ and $θ$ are constant
$∴ h \propto \frac{1}{r}$
Since, radius $r = R^{1} = 2 R$
$\Rightarrow \frac{h_{1}}{h_{1}} = \frac{r^{1}}{r} = \frac{2 R}{R}$
$h^{1} = 1 / 2$
Mass $=$ volume $\times$ density
$= π (2 R)^{2} \frac{h}{2 ρ} = 2 (π R^{2} h ρ) [∵ π r^{2} h ρ = M]$
$= 2 M$

BITSAT-2017

Mechanical Properties of Fluids

142946 Figure shows a capillary rise $H$ . If the air is blown through the horizontal tube in the direction as shown, then rise in capillary tube will be

1

= H

2

> H

3

< H

4 zero

Explanation:

B
According to Bernoulli equation,
$P_{1} + \frac{1}{2} ρ v^{2} \times ρ gh = constant$
If velocity of air will increases the pressure at point (A) must be decreases to maintain equation (i).
According to question, air will start to blown through the horizontal tube so pressure at point (A) will decreases.
$∴$ The height of liquid rise in capillary tube will increases.

BITSAT-2018

Mechanical Properties of Fluids

142943 Due to capillary action, a liquid will rise in a tube if angle of contact is

1 acute

2 obtuse

3

90^{\circ}

4 zero

Explanation:

A Since, we know that, capillary rise of depression is given as-
$h = \frac{2 T \cos θ}{ρ gr}$
If $θ$ is less than $90^{\circ}$ , then $h$ will be positive liquid rises in a capillary tube, The angle of contact is acute.
Whereas liquid is depressed in a capillary tube, i.e. $θ$ is more than $90^{\circ}$ then the angle of contact is obtuse. 274. A vertical glass capillary tube of radius $r$ open at both ends contains some water (surface tension $T$ and density $ρ$ ). If $L$ be the length of the water column, then :

(a) $L = \frac{4 T}{r ρ g}$
(b) $L = \frac{2 T}{r ρ g}$
(c) $L = \frac{T}{4 r ρ g}$
(d) $L = \frac{T}{2 r ρ g}$
Ans: a :

Let, the radius of capillary tube is $r$ .

Net surface tension force in upward direction will be,
$F_{up} = 2 π r \cdot T \cos θ + 2 π r \cdot T \cos θ = 4 π r \cdot T \cos θ$
$F_{up} = 4 π r \cdot T \cos θ$
Due to the shape of meniscus at the upper and lower surfaces of the liquid, the vertical components of surface tension force will be added along vertical direction
So, net upward force-
$F_{up} = 4 T π r \cos θ$
$∵ θ = 0^{\circ} for glass \& water$
$F_{up} = 4 T \cdot π r$
Now, weight of liquid $mg$ can be balanced by surface tension force.
$∴ 4 T \cdot π r = mg$
$4 T \cdot π r = ρ Vg [∵ m = ρ V]$
$4 T \cdot π r = ρ A \cdot Lg [∵ V = A \times L]$
$4 T \cdot π r = ρ π r^{2} Lg [∵ A = π r^{2}]$
$L = \frac{4 T π r}{ρ π r^{2} \cdot g}$
$L = \frac{4 T}{ρ rg}$

BITSAT-2007

Mechanical Properties of Fluids

142944 A glass capillary tube of internal radius $r =$ $0.25 mm$ is immersed in water. The top end of the tube projected by $2 cm$ above the surface of the water. At what angle does the liquid meet the tube? Surface tension of water $= 0.7$ $N / m$ .

1

90^{\circ}

2

70^{\circ}

3

45^{\circ}

4

35^{\circ}

Explanation:

B Given that, $T = 0.7 N / m, r = 0.25 mm =$ $25 \times 10^{- 3} m$
We know that,
$h = \frac{2 T \cos θ}{r ρ g}$
$h = \frac{2 T}{r ρ g} [∵ For water and glass, θ = 0^{\circ}]$
$= \frac{2 \times 0.7}{0.25 \times 10^{- 3} \times 1000 \times 9.8}$
As the tube is only $2 cm$ above the water and so, water will rise by $2 cm$ and meet the tube at an angle such that,
$\cos θ = \frac{h^{'} r ρ g}{2 T}$
$= \frac{2 \times 10^{- 2} \times 0.25 \times 10^{- 3} \times 1000 \times 9.8}{2 \times 0.7}$
The liquid will meet the tube at an angle, $θ ≅ 70^{\circ}$

BITSAT-2015

Mechanical Properties of Fluids

142945 A capillary tube of radius $R$ is immersed in water and water rises in it to a height $H$ . Mass of water in the capillary tube is $M$ . If the radius of the tube is doubled, mass of water that will rise in the capillary tube will now be:

1

M

2

2 M

3

M / 2

4

4 M

Explanation:

B Given that, radius of capillary tube $= R$
Height of water $= H$
Mass of water in a capillary tube $= M$
Since, we know that hight of capillary rise in capillary tube having radius ' $r$ ' is given as-
$h = \frac{2 T \cos θ}{r ρ g}$

Since $T, ρ, g$ and $θ$ are constant
$∴ h \propto \frac{1}{r}$
Since, radius $r = R^{1} = 2 R$
$\Rightarrow \frac{h_{1}}{h_{1}} = \frac{r^{1}}{r} = \frac{2 R}{R}$
$h^{1} = 1 / 2$
Mass $=$ volume $\times$ density
$= π (2 R)^{2} \frac{h}{2 ρ} = 2 (π R^{2} h ρ) [∵ π r^{2} h ρ = M]$
$= 2 M$

BITSAT-2017

Mechanical Properties of Fluids

142946 Figure shows a capillary rise $H$ . If the air is blown through the horizontal tube in the direction as shown, then rise in capillary tube will be

1

= H

2

> H

3

< H

4 zero

Explanation:

B
According to Bernoulli equation,
$P_{1} + \frac{1}{2} ρ v^{2} \times ρ gh = constant$
If velocity of air will increases the pressure at point (A) must be decreases to maintain equation (i).
According to question, air will start to blown through the horizontal tube so pressure at point (A) will decreases.
$∴$ The height of liquid rise in capillary tube will increases.

BITSAT-2018

Mechanical Properties of Fluids

142943 Due to capillary action, a liquid will rise in a tube if angle of contact is

1 acute

2 obtuse

3

90^{\circ}

4 zero

Explanation:

A Since, we know that, capillary rise of depression is given as-
$h = \frac{2 T \cos θ}{ρ gr}$
If $θ$ is less than $90^{\circ}$ , then $h$ will be positive liquid rises in a capillary tube, The angle of contact is acute.
Whereas liquid is depressed in a capillary tube, i.e. $θ$ is more than $90^{\circ}$ then the angle of contact is obtuse. 274. A vertical glass capillary tube of radius $r$ open at both ends contains some water (surface tension $T$ and density $ρ$ ). If $L$ be the length of the water column, then :

(a) $L = \frac{4 T}{r ρ g}$
(b) $L = \frac{2 T}{r ρ g}$
(c) $L = \frac{T}{4 r ρ g}$
(d) $L = \frac{T}{2 r ρ g}$
Ans: a :

Let, the radius of capillary tube is $r$ .

Net surface tension force in upward direction will be,
$F_{up} = 2 π r \cdot T \cos θ + 2 π r \cdot T \cos θ = 4 π r \cdot T \cos θ$
$F_{up} = 4 π r \cdot T \cos θ$
Due to the shape of meniscus at the upper and lower surfaces of the liquid, the vertical components of surface tension force will be added along vertical direction
So, net upward force-
$F_{up} = 4 T π r \cos θ$
$∵ θ = 0^{\circ} for glass \& water$
$F_{up} = 4 T \cdot π r$
Now, weight of liquid $mg$ can be balanced by surface tension force.
$∴ 4 T \cdot π r = mg$
$4 T \cdot π r = ρ Vg [∵ m = ρ V]$
$4 T \cdot π r = ρ A \cdot Lg [∵ V = A \times L]$
$4 T \cdot π r = ρ π r^{2} Lg [∵ A = π r^{2}]$
$L = \frac{4 T π r}{ρ π r^{2} \cdot g}$
$L = \frac{4 T}{ρ rg}$

BITSAT-2007

Mechanical Properties of Fluids

142944 A glass capillary tube of internal radius $r =$ $0.25 mm$ is immersed in water. The top end of the tube projected by $2 cm$ above the surface of the water. At what angle does the liquid meet the tube? Surface tension of water $= 0.7$ $N / m$ .

1

90^{\circ}

2

70^{\circ}

3

45^{\circ}

4

35^{\circ}

Explanation:

B Given that, $T = 0.7 N / m, r = 0.25 mm =$ $25 \times 10^{- 3} m$
We know that,
$h = \frac{2 T \cos θ}{r ρ g}$
$h = \frac{2 T}{r ρ g} [∵ For water and glass, θ = 0^{\circ}]$
$= \frac{2 \times 0.7}{0.25 \times 10^{- 3} \times 1000 \times 9.8}$
As the tube is only $2 cm$ above the water and so, water will rise by $2 cm$ and meet the tube at an angle such that,
$\cos θ = \frac{h^{'} r ρ g}{2 T}$
$= \frac{2 \times 10^{- 2} \times 0.25 \times 10^{- 3} \times 1000 \times 9.8}{2 \times 0.7}$
The liquid will meet the tube at an angle, $θ ≅ 70^{\circ}$

BITSAT-2015

Mechanical Properties of Fluids

142945 A capillary tube of radius $R$ is immersed in water and water rises in it to a height $H$ . Mass of water in the capillary tube is $M$ . If the radius of the tube is doubled, mass of water that will rise in the capillary tube will now be:

1

M

2

2 M

3

M / 2

4

4 M

Explanation:

B Given that, radius of capillary tube $= R$
Height of water $= H$
Mass of water in a capillary tube $= M$
Since, we know that hight of capillary rise in capillary tube having radius ' $r$ ' is given as-
$h = \frac{2 T \cos θ}{r ρ g}$

Since $T, ρ, g$ and $θ$ are constant
$∴ h \propto \frac{1}{r}$
Since, radius $r = R^{1} = 2 R$
$\Rightarrow \frac{h_{1}}{h_{1}} = \frac{r^{1}}{r} = \frac{2 R}{R}$
$h^{1} = 1 / 2$
Mass $=$ volume $\times$ density
$= π (2 R)^{2} \frac{h}{2 ρ} = 2 (π R^{2} h ρ) [∵ π r^{2} h ρ = M]$
$= 2 M$

BITSAT-2017

Mechanical Properties of Fluids

142946 Figure shows a capillary rise $H$ . If the air is blown through the horizontal tube in the direction as shown, then rise in capillary tube will be

1

= H

2

> H

3

< H

4 zero

Explanation:

B
According to Bernoulli equation,
$P_{1} + \frac{1}{2} ρ v^{2} \times ρ gh = constant$
If velocity of air will increases the pressure at point (A) must be decreases to maintain equation (i).
According to question, air will start to blown through the horizontal tube so pressure at point (A) will decreases.
$∴$ The height of liquid rise in capillary tube will increases.

BITSAT-2018

Mechanical Properties of Fluids

142943 Due to capillary action, a liquid will rise in a tube if angle of contact is

1 acute

2 obtuse

3

90^{\circ}

4 zero

Explanation:

A Since, we know that, capillary rise of depression is given as-
$h = \frac{2 T \cos θ}{ρ gr}$
If $θ$ is less than $90^{\circ}$ , then $h$ will be positive liquid rises in a capillary tube, The angle of contact is acute.
Whereas liquid is depressed in a capillary tube, i.e. $θ$ is more than $90^{\circ}$ then the angle of contact is obtuse. 274. A vertical glass capillary tube of radius $r$ open at both ends contains some water (surface tension $T$ and density $ρ$ ). If $L$ be the length of the water column, then :

(a) $L = \frac{4 T}{r ρ g}$
(b) $L = \frac{2 T}{r ρ g}$
(c) $L = \frac{T}{4 r ρ g}$
(d) $L = \frac{T}{2 r ρ g}$
Ans: a :

Let, the radius of capillary tube is $r$ .

Net surface tension force in upward direction will be,
$F_{up} = 2 π r \cdot T \cos θ + 2 π r \cdot T \cos θ = 4 π r \cdot T \cos θ$
$F_{up} = 4 π r \cdot T \cos θ$
Due to the shape of meniscus at the upper and lower surfaces of the liquid, the vertical components of surface tension force will be added along vertical direction
So, net upward force-
$F_{up} = 4 T π r \cos θ$
$∵ θ = 0^{\circ} for glass \& water$
$F_{up} = 4 T \cdot π r$
Now, weight of liquid $mg$ can be balanced by surface tension force.
$∴ 4 T \cdot π r = mg$
$4 T \cdot π r = ρ Vg [∵ m = ρ V]$
$4 T \cdot π r = ρ A \cdot Lg [∵ V = A \times L]$
$4 T \cdot π r = ρ π r^{2} Lg [∵ A = π r^{2}]$
$L = \frac{4 T π r}{ρ π r^{2} \cdot g}$
$L = \frac{4 T}{ρ rg}$

BITSAT-2007

Mechanical Properties of Fluids

142944 A glass capillary tube of internal radius $r =$ $0.25 mm$ is immersed in water. The top end of the tube projected by $2 cm$ above the surface of the water. At what angle does the liquid meet the tube? Surface tension of water $= 0.7$ $N / m$ .

1

90^{\circ}

2

70^{\circ}

3

45^{\circ}

4

35^{\circ}

Explanation:

B Given that, $T = 0.7 N / m, r = 0.25 mm =$ $25 \times 10^{- 3} m$
We know that,
$h = \frac{2 T \cos θ}{r ρ g}$
$h = \frac{2 T}{r ρ g} [∵ For water and glass, θ = 0^{\circ}]$
$= \frac{2 \times 0.7}{0.25 \times 10^{- 3} \times 1000 \times 9.8}$
As the tube is only $2 cm$ above the water and so, water will rise by $2 cm$ and meet the tube at an angle such that,
$\cos θ = \frac{h^{'} r ρ g}{2 T}$
$= \frac{2 \times 10^{- 2} \times 0.25 \times 10^{- 3} \times 1000 \times 9.8}{2 \times 0.7}$
The liquid will meet the tube at an angle, $θ ≅ 70^{\circ}$

BITSAT-2015

Mechanical Properties of Fluids

142945 A capillary tube of radius $R$ is immersed in water and water rises in it to a height $H$ . Mass of water in the capillary tube is $M$ . If the radius of the tube is doubled, mass of water that will rise in the capillary tube will now be:

1

M

2

2 M

3

M / 2

4

4 M

Explanation:

B Given that, radius of capillary tube $= R$
Height of water $= H$
Mass of water in a capillary tube $= M$
Since, we know that hight of capillary rise in capillary tube having radius ' $r$ ' is given as-
$h = \frac{2 T \cos θ}{r ρ g}$

Since $T, ρ, g$ and $θ$ are constant
$∴ h \propto \frac{1}{r}$
Since, radius $r = R^{1} = 2 R$
$\Rightarrow \frac{h_{1}}{h_{1}} = \frac{r^{1}}{r} = \frac{2 R}{R}$
$h^{1} = 1 / 2$
Mass $=$ volume $\times$ density
$= π (2 R)^{2} \frac{h}{2 ρ} = 2 (π R^{2} h ρ) [∵ π r^{2} h ρ = M]$
$= 2 M$

BITSAT-2017

Mechanical Properties of Fluids

142946 Figure shows a capillary rise $H$ . If the air is blown through the horizontal tube in the direction as shown, then rise in capillary tube will be

1

= H

2

> H

3

< H

4 zero

Explanation:

B
According to Bernoulli equation,
$P_{1} + \frac{1}{2} ρ v^{2} \times ρ gh = constant$
If velocity of air will increases the pressure at point (A) must be decreases to maintain equation (i).
According to question, air will start to blown through the horizontal tube so pressure at point (A) will decreases.
$∴$ The height of liquid rise in capillary tube will increases.

BITSAT-2018

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