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Mechanical Properties of Fluids

143009 A long glass capillary tube is dipped in water. It is known that water wets glass. The water level rises by $h$ in the tube. The tube is now pushed down so that only a length $h / 2$ is outside the water surface. The angle of contact at the water surface at the upper end of the tube will be

1

\tan_{2}^{- 1}

2

60^{\circ}

3

30^{\circ}

4

15^{\circ}

Explanation:

B Given that,
Rising the level of water in the tube $= h$
And, Length outside the water $= \frac{h}{2}$ So,
$h = \frac{2 s \cos 0^{\circ}}{r ρ g} (∵ \cos 0^{\circ} = 1)$
$h = \frac{2 s}{r ρ g}$
From the question,
$\frac{h}{2} = \frac{2 s \cos θ^{\circ}}{r ρ g}$
Dividing eq.(ii) by (i) we get -
$\frac{1}{2} = \cos θ$
$\cos 60^{\circ} = \cos θ$
$θ = 60^{\circ}$

VITEEE-2013

Mechanical Properties of Fluids

143010 A vessel, whose bottom has found holes with diameter of $1 mm$ is filled with water. Assuming that surface tension acts only at holes, then the maximum height to which the water can be filled in vessel without leakage is

1

3 cm

2

0.3 cm

3

3 mm

4

3 m

(Surface tension of water is

75 \times 10^{- 3} N / m

and

g

= 10 m / s^{2}

)

Explanation:

A Given that,
Diameter $= 1 mm$
Then, $r = 0.5 mm = 5 \times 10^{- 4} m$
The vertical force due to surface tension at the wheel is
$= (T \cos θ) L$
$= (T \cos θ) 2 π r$
$(∵ L = 2 π r)$
And it would balance the weight $mg$ of the drop
$m = π r^{2} ρ h$
So,
$T \cos θ (2 π r) = π r^{2} h ρ g$
$h = \frac{2 T \cos θ}{ρ rg}$
$h$ is maximum when $θ = 0^{\circ}$
$h = \frac{2 T \cos 0^{\circ}}{ρ rg} (∵ \cos 0^{\circ} = 1)$
$h = \frac{2 T}{ρ rg}$
$h_{max} = \frac{2 \times 75 \times 10^{- 3}}{10^{3} \times 5 \times 10^{- 4} \times 10}$
$h_{max} = 0.03 m$
$h_{max} = 3 cm$

J and K CET- 2004

Mechanical Properties of Fluids

143011 Three liquids of densities $ρ_{1}, ρ_{2}$ and $ρ_{3}$ (with $ρ_{1}$ $> ρ_{2} > ρ_{3}$ , having the same value of surface tension $T$ , rise to the same height in three identical capillaries. The angles of contact $θ_{1}, θ_{2}$ and $θ_{3}$ obey

1

\frac{π}{2} > θ_{1} > θ_{2} > θ_{3} \geq 0

2

0 \leq θ_{1} < θ_{2} < θ_{3} < \frac{π}{2}

3

\frac{π}{2} < θ_{1} < θ_{2} < θ_{3} < π

4

π > θ_{1} > θ_{2} > θ_{3} > \frac{π}{2}

Explanation:

B As we know that, liquid rise in a capillary tube
$h = \frac{2 T \cos θ}{ρ gr}$
According to question $r, h, T$ are same, therefore
$\cos θ = \frac{ρ grh}{2 T} (∵ \frac{grh}{2 T} = k)$
$\cos θ \propto ρ [\begin{matrix} θ = 0^{\circ} then \cos value maximum \\ \cos θ^{\circ} = \cos 0^{\circ} = 1 \\ and θ = \frac{π}{2}, \cos \frac{π}{2} = 0 minimum \end{matrix}]$
For rise -

Therefore, $0^{\circ} \leq θ_{1} < θ_{2} < θ_{3} < \frac{π}{2}$

NEET-2016

Mechanical Properties of Fluids

142939 Water rises in a glass capillary tube due to

1 surface tension of water

2 cohesive force of glass molecules

3 temperature of water

4 adhesive force between water molecules and the walls of the glass tube

UPSEE - 2011

Mechanical Properties of Fluids

142965 Which one of following statements about the angle of contact $(θ)$ , is wrong ?

1

θ > 0^{\circ}

for pure water - glass pair.

2

θ

is not constant for particular solid - liquid pair.

3

θ < 90^{\circ}

for kerosene - glass pair.

4

θ > 90^{\circ}

for mercury - glass pair.

Explanation:

B $θ$ is not constant for particular solid-liquid pair
The angle of contact lies between,
- If $θ < 90^{\circ}$ (acute angle) the liquid will have a meniscus concave upwards.
- If $θ > 90^{\circ}$ (obtuse) the mercury-glass surface will have a meniscus convex upwards.
- If $θ = 90^{\circ}$ the surface of liquid at the point of contact is plane.

MHT-CET 2020

Mechanical Properties of Fluids

143009 A long glass capillary tube is dipped in water. It is known that water wets glass. The water level rises by $h$ in the tube. The tube is now pushed down so that only a length $h / 2$ is outside the water surface. The angle of contact at the water surface at the upper end of the tube will be

1

\tan_{2}^{- 1}

2

60^{\circ}

3

30^{\circ}

4

15^{\circ}

Explanation:

B Given that,
Rising the level of water in the tube $= h$
And, Length outside the water $= \frac{h}{2}$ So,
$h = \frac{2 s \cos 0^{\circ}}{r ρ g} (∵ \cos 0^{\circ} = 1)$
$h = \frac{2 s}{r ρ g}$
From the question,
$\frac{h}{2} = \frac{2 s \cos θ^{\circ}}{r ρ g}$
Dividing eq.(ii) by (i) we get -
$\frac{1}{2} = \cos θ$
$\cos 60^{\circ} = \cos θ$
$θ = 60^{\circ}$

VITEEE-2013

Mechanical Properties of Fluids

143010 A vessel, whose bottom has found holes with diameter of $1 mm$ is filled with water. Assuming that surface tension acts only at holes, then the maximum height to which the water can be filled in vessel without leakage is

1

3 cm

2

0.3 cm

3

3 mm

4

3 m

(Surface tension of water is

75 \times 10^{- 3} N / m

and

g

= 10 m / s^{2}

)

Explanation:

A Given that,
Diameter $= 1 mm$
Then, $r = 0.5 mm = 5 \times 10^{- 4} m$
The vertical force due to surface tension at the wheel is
$= (T \cos θ) L$
$= (T \cos θ) 2 π r$
$(∵ L = 2 π r)$
And it would balance the weight $mg$ of the drop
$m = π r^{2} ρ h$
So,
$T \cos θ (2 π r) = π r^{2} h ρ g$
$h = \frac{2 T \cos θ}{ρ rg}$
$h$ is maximum when $θ = 0^{\circ}$
$h = \frac{2 T \cos 0^{\circ}}{ρ rg} (∵ \cos 0^{\circ} = 1)$
$h = \frac{2 T}{ρ rg}$
$h_{max} = \frac{2 \times 75 \times 10^{- 3}}{10^{3} \times 5 \times 10^{- 4} \times 10}$
$h_{max} = 0.03 m$
$h_{max} = 3 cm$

J and K CET- 2004

Mechanical Properties of Fluids

143011 Three liquids of densities $ρ_{1}, ρ_{2}$ and $ρ_{3}$ (with $ρ_{1}$ $> ρ_{2} > ρ_{3}$ , having the same value of surface tension $T$ , rise to the same height in three identical capillaries. The angles of contact $θ_{1}, θ_{2}$ and $θ_{3}$ obey

1

\frac{π}{2} > θ_{1} > θ_{2} > θ_{3} \geq 0

2

0 \leq θ_{1} < θ_{2} < θ_{3} < \frac{π}{2}

3

\frac{π}{2} < θ_{1} < θ_{2} < θ_{3} < π

4

π > θ_{1} > θ_{2} > θ_{3} > \frac{π}{2}

Explanation:

B As we know that, liquid rise in a capillary tube
$h = \frac{2 T \cos θ}{ρ gr}$
According to question $r, h, T$ are same, therefore
$\cos θ = \frac{ρ grh}{2 T} (∵ \frac{grh}{2 T} = k)$
$\cos θ \propto ρ [\begin{matrix} θ = 0^{\circ} then \cos value maximum \\ \cos θ^{\circ} = \cos 0^{\circ} = 1 \\ and θ = \frac{π}{2}, \cos \frac{π}{2} = 0 minimum \end{matrix}]$
For rise -

Therefore, $0^{\circ} \leq θ_{1} < θ_{2} < θ_{3} < \frac{π}{2}$

NEET-2016

Mechanical Properties of Fluids

142939 Water rises in a glass capillary tube due to

1 surface tension of water

2 cohesive force of glass molecules

3 temperature of water

4 adhesive force between water molecules and the walls of the glass tube

UPSEE - 2011

Mechanical Properties of Fluids

142965 Which one of following statements about the angle of contact $(θ)$ , is wrong ?

1

θ > 0^{\circ}

for pure water - glass pair.

2

θ

is not constant for particular solid - liquid pair.

3

θ < 90^{\circ}

for kerosene - glass pair.

4

θ > 90^{\circ}

for mercury - glass pair.

Explanation:

B $θ$ is not constant for particular solid-liquid pair
The angle of contact lies between,
- If $θ < 90^{\circ}$ (acute angle) the liquid will have a meniscus concave upwards.
- If $θ > 90^{\circ}$ (obtuse) the mercury-glass surface will have a meniscus convex upwards.
- If $θ = 90^{\circ}$ the surface of liquid at the point of contact is plane.

MHT-CET 2020

Mechanical Properties of Fluids

143009 A long glass capillary tube is dipped in water. It is known that water wets glass. The water level rises by $h$ in the tube. The tube is now pushed down so that only a length $h / 2$ is outside the water surface. The angle of contact at the water surface at the upper end of the tube will be

1

\tan_{2}^{- 1}

2

60^{\circ}

3

30^{\circ}

4

15^{\circ}

Explanation:

B Given that,
Rising the level of water in the tube $= h$
And, Length outside the water $= \frac{h}{2}$ So,
$h = \frac{2 s \cos 0^{\circ}}{r ρ g} (∵ \cos 0^{\circ} = 1)$
$h = \frac{2 s}{r ρ g}$
From the question,
$\frac{h}{2} = \frac{2 s \cos θ^{\circ}}{r ρ g}$
Dividing eq.(ii) by (i) we get -
$\frac{1}{2} = \cos θ$
$\cos 60^{\circ} = \cos θ$
$θ = 60^{\circ}$

VITEEE-2013

Mechanical Properties of Fluids

143010 A vessel, whose bottom has found holes with diameter of $1 mm$ is filled with water. Assuming that surface tension acts only at holes, then the maximum height to which the water can be filled in vessel without leakage is

1

3 cm

2

0.3 cm

3

3 mm

4

3 m

(Surface tension of water is

75 \times 10^{- 3} N / m

and

g

= 10 m / s^{2}

)

Explanation:

A Given that,
Diameter $= 1 mm$
Then, $r = 0.5 mm = 5 \times 10^{- 4} m$
The vertical force due to surface tension at the wheel is
$= (T \cos θ) L$
$= (T \cos θ) 2 π r$
$(∵ L = 2 π r)$
And it would balance the weight $mg$ of the drop
$m = π r^{2} ρ h$
So,
$T \cos θ (2 π r) = π r^{2} h ρ g$
$h = \frac{2 T \cos θ}{ρ rg}$
$h$ is maximum when $θ = 0^{\circ}$
$h = \frac{2 T \cos 0^{\circ}}{ρ rg} (∵ \cos 0^{\circ} = 1)$
$h = \frac{2 T}{ρ rg}$
$h_{max} = \frac{2 \times 75 \times 10^{- 3}}{10^{3} \times 5 \times 10^{- 4} \times 10}$
$h_{max} = 0.03 m$
$h_{max} = 3 cm$

J and K CET- 2004

Mechanical Properties of Fluids

143011 Three liquids of densities $ρ_{1}, ρ_{2}$ and $ρ_{3}$ (with $ρ_{1}$ $> ρ_{2} > ρ_{3}$ , having the same value of surface tension $T$ , rise to the same height in three identical capillaries. The angles of contact $θ_{1}, θ_{2}$ and $θ_{3}$ obey

1

\frac{π}{2} > θ_{1} > θ_{2} > θ_{3} \geq 0

2

0 \leq θ_{1} < θ_{2} < θ_{3} < \frac{π}{2}

3

\frac{π}{2} < θ_{1} < θ_{2} < θ_{3} < π

4

π > θ_{1} > θ_{2} > θ_{3} > \frac{π}{2}

Explanation:

B As we know that, liquid rise in a capillary tube
$h = \frac{2 T \cos θ}{ρ gr}$
According to question $r, h, T$ are same, therefore
$\cos θ = \frac{ρ grh}{2 T} (∵ \frac{grh}{2 T} = k)$
$\cos θ \propto ρ [\begin{matrix} θ = 0^{\circ} then \cos value maximum \\ \cos θ^{\circ} = \cos 0^{\circ} = 1 \\ and θ = \frac{π}{2}, \cos \frac{π}{2} = 0 minimum \end{matrix}]$
For rise -

Therefore, $0^{\circ} \leq θ_{1} < θ_{2} < θ_{3} < \frac{π}{2}$

NEET-2016

Mechanical Properties of Fluids

142939 Water rises in a glass capillary tube due to

1 surface tension of water

2 cohesive force of glass molecules

3 temperature of water

4 adhesive force between water molecules and the walls of the glass tube

UPSEE - 2011

Mechanical Properties of Fluids

142965 Which one of following statements about the angle of contact $(θ)$ , is wrong ?

1

θ > 0^{\circ}

for pure water - glass pair.

2

θ

is not constant for particular solid - liquid pair.

3

θ < 90^{\circ}

for kerosene - glass pair.

4

θ > 90^{\circ}

for mercury - glass pair.

Explanation:

B $θ$ is not constant for particular solid-liquid pair
The angle of contact lies between,
- If $θ < 90^{\circ}$ (acute angle) the liquid will have a meniscus concave upwards.
- If $θ > 90^{\circ}$ (obtuse) the mercury-glass surface will have a meniscus convex upwards.
- If $θ = 90^{\circ}$ the surface of liquid at the point of contact is plane.

MHT-CET 2020

Mechanical Properties of Fluids

143009 A long glass capillary tube is dipped in water. It is known that water wets glass. The water level rises by $h$ in the tube. The tube is now pushed down so that only a length $h / 2$ is outside the water surface. The angle of contact at the water surface at the upper end of the tube will be

1

\tan_{2}^{- 1}

2

60^{\circ}

3

30^{\circ}

4

15^{\circ}

Explanation:

B Given that,
Rising the level of water in the tube $= h$
And, Length outside the water $= \frac{h}{2}$ So,
$h = \frac{2 s \cos 0^{\circ}}{r ρ g} (∵ \cos 0^{\circ} = 1)$
$h = \frac{2 s}{r ρ g}$
From the question,
$\frac{h}{2} = \frac{2 s \cos θ^{\circ}}{r ρ g}$
Dividing eq.(ii) by (i) we get -
$\frac{1}{2} = \cos θ$
$\cos 60^{\circ} = \cos θ$
$θ = 60^{\circ}$

VITEEE-2013

Mechanical Properties of Fluids

143010 A vessel, whose bottom has found holes with diameter of $1 mm$ is filled with water. Assuming that surface tension acts only at holes, then the maximum height to which the water can be filled in vessel without leakage is

1

3 cm

2

0.3 cm

3

3 mm

4

3 m

(Surface tension of water is

75 \times 10^{- 3} N / m

and

g

= 10 m / s^{2}

)

Explanation:

A Given that,
Diameter $= 1 mm$
Then, $r = 0.5 mm = 5 \times 10^{- 4} m$
The vertical force due to surface tension at the wheel is
$= (T \cos θ) L$
$= (T \cos θ) 2 π r$
$(∵ L = 2 π r)$
And it would balance the weight $mg$ of the drop
$m = π r^{2} ρ h$
So,
$T \cos θ (2 π r) = π r^{2} h ρ g$
$h = \frac{2 T \cos θ}{ρ rg}$
$h$ is maximum when $θ = 0^{\circ}$
$h = \frac{2 T \cos 0^{\circ}}{ρ rg} (∵ \cos 0^{\circ} = 1)$
$h = \frac{2 T}{ρ rg}$
$h_{max} = \frac{2 \times 75 \times 10^{- 3}}{10^{3} \times 5 \times 10^{- 4} \times 10}$
$h_{max} = 0.03 m$
$h_{max} = 3 cm$

J and K CET- 2004

Mechanical Properties of Fluids

143011 Three liquids of densities $ρ_{1}, ρ_{2}$ and $ρ_{3}$ (with $ρ_{1}$ $> ρ_{2} > ρ_{3}$ , having the same value of surface tension $T$ , rise to the same height in three identical capillaries. The angles of contact $θ_{1}, θ_{2}$ and $θ_{3}$ obey

1

\frac{π}{2} > θ_{1} > θ_{2} > θ_{3} \geq 0

2

0 \leq θ_{1} < θ_{2} < θ_{3} < \frac{π}{2}

3

\frac{π}{2} < θ_{1} < θ_{2} < θ_{3} < π

4

π > θ_{1} > θ_{2} > θ_{3} > \frac{π}{2}

Explanation:

B As we know that, liquid rise in a capillary tube
$h = \frac{2 T \cos θ}{ρ gr}$
According to question $r, h, T$ are same, therefore
$\cos θ = \frac{ρ grh}{2 T} (∵ \frac{grh}{2 T} = k)$
$\cos θ \propto ρ [\begin{matrix} θ = 0^{\circ} then \cos value maximum \\ \cos θ^{\circ} = \cos 0^{\circ} = 1 \\ and θ = \frac{π}{2}, \cos \frac{π}{2} = 0 minimum \end{matrix}]$
For rise -

Therefore, $0^{\circ} \leq θ_{1} < θ_{2} < θ_{3} < \frac{π}{2}$

NEET-2016

Mechanical Properties of Fluids

142939 Water rises in a glass capillary tube due to

1 surface tension of water

2 cohesive force of glass molecules

3 temperature of water

4 adhesive force between water molecules and the walls of the glass tube

UPSEE - 2011

Mechanical Properties of Fluids

142965 Which one of following statements about the angle of contact $(θ)$ , is wrong ?

1

θ > 0^{\circ}

for pure water - glass pair.

2

θ

is not constant for particular solid - liquid pair.

3

θ < 90^{\circ}

for kerosene - glass pair.

4

θ > 90^{\circ}

for mercury - glass pair.

Explanation:

B $θ$ is not constant for particular solid-liquid pair
The angle of contact lies between,
- If $θ < 90^{\circ}$ (acute angle) the liquid will have a meniscus concave upwards.
- If $θ > 90^{\circ}$ (obtuse) the mercury-glass surface will have a meniscus convex upwards.
- If $θ = 90^{\circ}$ the surface of liquid at the point of contact is plane.

MHT-CET 2020

Mechanical Properties of Fluids

143009 A long glass capillary tube is dipped in water. It is known that water wets glass. The water level rises by $h$ in the tube. The tube is now pushed down so that only a length $h / 2$ is outside the water surface. The angle of contact at the water surface at the upper end of the tube will be

1

\tan_{2}^{- 1}

2

60^{\circ}

3

30^{\circ}

4

15^{\circ}

Explanation:

B Given that,
Rising the level of water in the tube $= h$
And, Length outside the water $= \frac{h}{2}$ So,
$h = \frac{2 s \cos 0^{\circ}}{r ρ g} (∵ \cos 0^{\circ} = 1)$
$h = \frac{2 s}{r ρ g}$
From the question,
$\frac{h}{2} = \frac{2 s \cos θ^{\circ}}{r ρ g}$
Dividing eq.(ii) by (i) we get -
$\frac{1}{2} = \cos θ$
$\cos 60^{\circ} = \cos θ$
$θ = 60^{\circ}$

VITEEE-2013

Mechanical Properties of Fluids

143010 A vessel, whose bottom has found holes with diameter of $1 mm$ is filled with water. Assuming that surface tension acts only at holes, then the maximum height to which the water can be filled in vessel without leakage is

1

3 cm

2

0.3 cm

3

3 mm

4

3 m

(Surface tension of water is

75 \times 10^{- 3} N / m

and

g

= 10 m / s^{2}

)

Explanation:

A Given that,
Diameter $= 1 mm$
Then, $r = 0.5 mm = 5 \times 10^{- 4} m$
The vertical force due to surface tension at the wheel is
$= (T \cos θ) L$
$= (T \cos θ) 2 π r$
$(∵ L = 2 π r)$
And it would balance the weight $mg$ of the drop
$m = π r^{2} ρ h$
So,
$T \cos θ (2 π r) = π r^{2} h ρ g$
$h = \frac{2 T \cos θ}{ρ rg}$
$h$ is maximum when $θ = 0^{\circ}$
$h = \frac{2 T \cos 0^{\circ}}{ρ rg} (∵ \cos 0^{\circ} = 1)$
$h = \frac{2 T}{ρ rg}$
$h_{max} = \frac{2 \times 75 \times 10^{- 3}}{10^{3} \times 5 \times 10^{- 4} \times 10}$
$h_{max} = 0.03 m$
$h_{max} = 3 cm$

J and K CET- 2004

Mechanical Properties of Fluids

143011 Three liquids of densities $ρ_{1}, ρ_{2}$ and $ρ_{3}$ (with $ρ_{1}$ $> ρ_{2} > ρ_{3}$ , having the same value of surface tension $T$ , rise to the same height in three identical capillaries. The angles of contact $θ_{1}, θ_{2}$ and $θ_{3}$ obey

1

\frac{π}{2} > θ_{1} > θ_{2} > θ_{3} \geq 0

2

0 \leq θ_{1} < θ_{2} < θ_{3} < \frac{π}{2}

3

\frac{π}{2} < θ_{1} < θ_{2} < θ_{3} < π

4

π > θ_{1} > θ_{2} > θ_{3} > \frac{π}{2}

Explanation:

B As we know that, liquid rise in a capillary tube
$h = \frac{2 T \cos θ}{ρ gr}$
According to question $r, h, T$ are same, therefore
$\cos θ = \frac{ρ grh}{2 T} (∵ \frac{grh}{2 T} = k)$
$\cos θ \propto ρ [\begin{matrix} θ = 0^{\circ} then \cos value maximum \\ \cos θ^{\circ} = \cos 0^{\circ} = 1 \\ and θ = \frac{π}{2}, \cos \frac{π}{2} = 0 minimum \end{matrix}]$
For rise -

Therefore, $0^{\circ} \leq θ_{1} < θ_{2} < θ_{3} < \frac{π}{2}$

NEET-2016

Mechanical Properties of Fluids

142939 Water rises in a glass capillary tube due to

1 surface tension of water

2 cohesive force of glass molecules

3 temperature of water

4 adhesive force between water molecules and the walls of the glass tube

UPSEE - 2011

Mechanical Properties of Fluids

142965 Which one of following statements about the angle of contact $(θ)$ , is wrong ?

1

θ > 0^{\circ}

for pure water - glass pair.

2

θ

is not constant for particular solid - liquid pair.

3

θ < 90^{\circ}

for kerosene - glass pair.

4

θ > 90^{\circ}

for mercury - glass pair.

Explanation:

B $θ$ is not constant for particular solid-liquid pair
The angle of contact lies between,
- If $θ < 90^{\circ}$ (acute angle) the liquid will have a meniscus concave upwards.
- If $θ > 90^{\circ}$ (obtuse) the mercury-glass surface will have a meniscus convex upwards.
- If $θ = 90^{\circ}$ the surface of liquid at the point of contact is plane.

MHT-CET 2020