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PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361181 If $h$ is the height of capillary rise and $r$ is the radius of the capillary tube, then which of the following relation will be correct?

1

h r =

constant

2

h / r^{2} =

constant

3

h r^{2} =

constant

4

h / r =

constant

Explanation:

$h = \frac{2 S \cos θ}{ρ r g} \Rightarrow h r = \frac{2 S \cos θ}{ρ g}$
$\Rightarrow h r = c o n s t a n t$
$[S . θ p, g a r e c o n s t a n t]$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361182 Find the pressure at point $A$ just below the meniscus is... $T$ is surface tension and $r$ is radius of the cavity. Assume that contact angle is $180^{\circ}$
supporting img

1

P_{atm}

2

P_{a t m} + \frac{2 T}{r}

3

P_{a t m} - \frac{2 T}{r}

4

P_{a t m} + \frac{4 T}{r}

Explanation:

The pressure difference across the surface is $P_{A} - P_{a t m} = \frac{2 T}{R}$
$R$ - radius of the meniscus
$\begin{aligned} R = r (For θ = 0^{\circ}) \\ P_{A} = P_{a t m} + \frac{2 T}{r} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361183 Two capillary tubes $P$ and $Q$ are dipped vertically in water. The height of water level in capillary tube $P$ is $\frac{2^{n d}}{3}$ of the height in capillary tube $Q$ . The ratio of their diameter is ____

1

2 : 3

2

3 : 2

3

3 : 4

4

4 : 3

Explanation:

$\begin{matrix} h = \frac{2 T \cos θ}{ρ r g} \\ \Rightarrow \frac{h_{p}}{h_{q}} = \frac{r_{q}}{r_{p}} \\ \frac{\frac{2}{3} h_{q}}{h_{q}} = \frac{r_{q}}{r_{p}} \\ \Rightarrow \frac{r_{q}}{r_{p}} = \frac{2}{3} \\ \Rightarrow \frac{d_{p}}{d_{q}} = \frac{3}{2} \end{matrix}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361184 A cube of side $a$ and mass $m$ just floats on the surface of water as shown in the figure. The surface tension and density of water are $T$ and $ρ$ , respectively. If angle of contact between cube and water surface is zero, find the distance $h$ (in metres) between the lower face of cube and surface of the water.
supporting img

1

\frac{m g + a T}{a^{2} ρ g}

2

\frac{m g + 4 a T}{a^{2} ρ g}

3

\frac{m g + 2 a T}{a^{2} ρ g}

4

\frac{4 a T}{a ρ g}

Explanation:

At equilibrium $m g + F_{S T} = F_{b}$
$m g + 4 a T = a^{2} h ρ g$
supporting img

$h = \frac{m g + 4 a T}{a^{2} ρ g}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361185 If a liquid rises to the same height in two capillaries of the same material at the same temperature, then

1 The weight of liquid in both capillaries are not equal

2 The radius of meniscus must be equal

3 The capillaries must be cylindrical and vertical

4 Given conditions are not possible

Explanation:

$h = \frac{2 T \cos θ}{r ρ g}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361181 If $h$ is the height of capillary rise and $r$ is the radius of the capillary tube, then which of the following relation will be correct?

1

h r =

constant

2

h / r^{2} =

constant

3

h r^{2} =

constant

4

h / r =

constant

Explanation:

$h = \frac{2 S \cos θ}{ρ r g} \Rightarrow h r = \frac{2 S \cos θ}{ρ g}$
$\Rightarrow h r = c o n s t a n t$
$[S . θ p, g a r e c o n s t a n t]$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361182 Find the pressure at point $A$ just below the meniscus is... $T$ is surface tension and $r$ is radius of the cavity. Assume that contact angle is $180^{\circ}$
supporting img

1

P_{atm}

2

P_{a t m} + \frac{2 T}{r}

3

P_{a t m} - \frac{2 T}{r}

4

P_{a t m} + \frac{4 T}{r}

Explanation:

The pressure difference across the surface is $P_{A} - P_{a t m} = \frac{2 T}{R}$
$R$ - radius of the meniscus
$\begin{aligned} R = r (For θ = 0^{\circ}) \\ P_{A} = P_{a t m} + \frac{2 T}{r} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361183 Two capillary tubes $P$ and $Q$ are dipped vertically in water. The height of water level in capillary tube $P$ is $\frac{2^{n d}}{3}$ of the height in capillary tube $Q$ . The ratio of their diameter is ____

1

2 : 3

2

3 : 2

3

3 : 4

4

4 : 3

Explanation:

$\begin{matrix} h = \frac{2 T \cos θ}{ρ r g} \\ \Rightarrow \frac{h_{p}}{h_{q}} = \frac{r_{q}}{r_{p}} \\ \frac{\frac{2}{3} h_{q}}{h_{q}} = \frac{r_{q}}{r_{p}} \\ \Rightarrow \frac{r_{q}}{r_{p}} = \frac{2}{3} \\ \Rightarrow \frac{d_{p}}{d_{q}} = \frac{3}{2} \end{matrix}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361184 A cube of side $a$ and mass $m$ just floats on the surface of water as shown in the figure. The surface tension and density of water are $T$ and $ρ$ , respectively. If angle of contact between cube and water surface is zero, find the distance $h$ (in metres) between the lower face of cube and surface of the water.
supporting img

1

\frac{m g + a T}{a^{2} ρ g}

2

\frac{m g + 4 a T}{a^{2} ρ g}

3

\frac{m g + 2 a T}{a^{2} ρ g}

4

\frac{4 a T}{a ρ g}

Explanation:

At equilibrium $m g + F_{S T} = F_{b}$
$m g + 4 a T = a^{2} h ρ g$
supporting img

$h = \frac{m g + 4 a T}{a^{2} ρ g}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361185 If a liquid rises to the same height in two capillaries of the same material at the same temperature, then

1 The weight of liquid in both capillaries are not equal

2 The radius of meniscus must be equal

3 The capillaries must be cylindrical and vertical

4 Given conditions are not possible

Explanation:

$h = \frac{2 T \cos θ}{r ρ g}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361181 If $h$ is the height of capillary rise and $r$ is the radius of the capillary tube, then which of the following relation will be correct?

1

h r =

constant

2

h / r^{2} =

constant

3

h r^{2} =

constant

4

h / r =

constant

Explanation:

$h = \frac{2 S \cos θ}{ρ r g} \Rightarrow h r = \frac{2 S \cos θ}{ρ g}$
$\Rightarrow h r = c o n s t a n t$
$[S . θ p, g a r e c o n s t a n t]$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361182 Find the pressure at point $A$ just below the meniscus is... $T$ is surface tension and $r$ is radius of the cavity. Assume that contact angle is $180^{\circ}$
supporting img

1

P_{atm}

2

P_{a t m} + \frac{2 T}{r}

3

P_{a t m} - \frac{2 T}{r}

4

P_{a t m} + \frac{4 T}{r}

Explanation:

The pressure difference across the surface is $P_{A} - P_{a t m} = \frac{2 T}{R}$
$R$ - radius of the meniscus
$\begin{aligned} R = r (For θ = 0^{\circ}) \\ P_{A} = P_{a t m} + \frac{2 T}{r} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361183 Two capillary tubes $P$ and $Q$ are dipped vertically in water. The height of water level in capillary tube $P$ is $\frac{2^{n d}}{3}$ of the height in capillary tube $Q$ . The ratio of their diameter is ____

1

2 : 3

2

3 : 2

3

3 : 4

4

4 : 3

Explanation:

$\begin{matrix} h = \frac{2 T \cos θ}{ρ r g} \\ \Rightarrow \frac{h_{p}}{h_{q}} = \frac{r_{q}}{r_{p}} \\ \frac{\frac{2}{3} h_{q}}{h_{q}} = \frac{r_{q}}{r_{p}} \\ \Rightarrow \frac{r_{q}}{r_{p}} = \frac{2}{3} \\ \Rightarrow \frac{d_{p}}{d_{q}} = \frac{3}{2} \end{matrix}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361184 A cube of side $a$ and mass $m$ just floats on the surface of water as shown in the figure. The surface tension and density of water are $T$ and $ρ$ , respectively. If angle of contact between cube and water surface is zero, find the distance $h$ (in metres) between the lower face of cube and surface of the water.
supporting img

1

\frac{m g + a T}{a^{2} ρ g}

2

\frac{m g + 4 a T}{a^{2} ρ g}

3

\frac{m g + 2 a T}{a^{2} ρ g}

4

\frac{4 a T}{a ρ g}

Explanation:

At equilibrium $m g + F_{S T} = F_{b}$
$m g + 4 a T = a^{2} h ρ g$
supporting img

$h = \frac{m g + 4 a T}{a^{2} ρ g}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361185 If a liquid rises to the same height in two capillaries of the same material at the same temperature, then

1 The weight of liquid in both capillaries are not equal

2 The radius of meniscus must be equal

3 The capillaries must be cylindrical and vertical

4 Given conditions are not possible

Explanation:

$h = \frac{2 T \cos θ}{r ρ g}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361181 If $h$ is the height of capillary rise and $r$ is the radius of the capillary tube, then which of the following relation will be correct?

1

h r =

constant

2

h / r^{2} =

constant

3

h r^{2} =

constant

4

h / r =

constant

Explanation:

$h = \frac{2 S \cos θ}{ρ r g} \Rightarrow h r = \frac{2 S \cos θ}{ρ g}$
$\Rightarrow h r = c o n s t a n t$
$[S . θ p, g a r e c o n s t a n t]$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361182 Find the pressure at point $A$ just below the meniscus is... $T$ is surface tension and $r$ is radius of the cavity. Assume that contact angle is $180^{\circ}$
supporting img

1

P_{atm}

2

P_{a t m} + \frac{2 T}{r}

3

P_{a t m} - \frac{2 T}{r}

4

P_{a t m} + \frac{4 T}{r}

Explanation:

The pressure difference across the surface is $P_{A} - P_{a t m} = \frac{2 T}{R}$
$R$ - radius of the meniscus
$\begin{aligned} R = r (For θ = 0^{\circ}) \\ P_{A} = P_{a t m} + \frac{2 T}{r} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361183 Two capillary tubes $P$ and $Q$ are dipped vertically in water. The height of water level in capillary tube $P$ is $\frac{2^{n d}}{3}$ of the height in capillary tube $Q$ . The ratio of their diameter is ____

1

2 : 3

2

3 : 2

3

3 : 4

4

4 : 3

Explanation:

$\begin{matrix} h = \frac{2 T \cos θ}{ρ r g} \\ \Rightarrow \frac{h_{p}}{h_{q}} = \frac{r_{q}}{r_{p}} \\ \frac{\frac{2}{3} h_{q}}{h_{q}} = \frac{r_{q}}{r_{p}} \\ \Rightarrow \frac{r_{q}}{r_{p}} = \frac{2}{3} \\ \Rightarrow \frac{d_{p}}{d_{q}} = \frac{3}{2} \end{matrix}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361184 A cube of side $a$ and mass $m$ just floats on the surface of water as shown in the figure. The surface tension and density of water are $T$ and $ρ$ , respectively. If angle of contact between cube and water surface is zero, find the distance $h$ (in metres) between the lower face of cube and surface of the water.
supporting img

1

\frac{m g + a T}{a^{2} ρ g}

2

\frac{m g + 4 a T}{a^{2} ρ g}

3

\frac{m g + 2 a T}{a^{2} ρ g}

4

\frac{4 a T}{a ρ g}

Explanation:

At equilibrium $m g + F_{S T} = F_{b}$
$m g + 4 a T = a^{2} h ρ g$
supporting img

$h = \frac{m g + 4 a T}{a^{2} ρ g}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361185 If a liquid rises to the same height in two capillaries of the same material at the same temperature, then

1 The weight of liquid in both capillaries are not equal

2 The radius of meniscus must be equal

3 The capillaries must be cylindrical and vertical

4 Given conditions are not possible

Explanation:

$h = \frac{2 T \cos θ}{r ρ g}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361181 If $h$ is the height of capillary rise and $r$ is the radius of the capillary tube, then which of the following relation will be correct?

1

h r =

constant

2

h / r^{2} =

constant

3

h r^{2} =

constant

4

h / r =

constant

Explanation:

$h = \frac{2 S \cos θ}{ρ r g} \Rightarrow h r = \frac{2 S \cos θ}{ρ g}$
$\Rightarrow h r = c o n s t a n t$
$[S . θ p, g a r e c o n s t a n t]$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361182 Find the pressure at point $A$ just below the meniscus is... $T$ is surface tension and $r$ is radius of the cavity. Assume that contact angle is $180^{\circ}$
supporting img

1

P_{atm}

2

P_{a t m} + \frac{2 T}{r}

3

P_{a t m} - \frac{2 T}{r}

4

P_{a t m} + \frac{4 T}{r}

Explanation:

The pressure difference across the surface is $P_{A} - P_{a t m} = \frac{2 T}{R}$
$R$ - radius of the meniscus
$\begin{aligned} R = r (For θ = 0^{\circ}) \\ P_{A} = P_{a t m} + \frac{2 T}{r} \end{aligned}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361183 Two capillary tubes $P$ and $Q$ are dipped vertically in water. The height of water level in capillary tube $P$ is $\frac{2^{n d}}{3}$ of the height in capillary tube $Q$ . The ratio of their diameter is ____

1

2 : 3

2

3 : 2

3

3 : 4

4

4 : 3

Explanation:

$\begin{matrix} h = \frac{2 T \cos θ}{ρ r g} \\ \Rightarrow \frac{h_{p}}{h_{q}} = \frac{r_{q}}{r_{p}} \\ \frac{\frac{2}{3} h_{q}}{h_{q}} = \frac{r_{q}}{r_{p}} \\ \Rightarrow \frac{r_{q}}{r_{p}} = \frac{2}{3} \\ \Rightarrow \frac{d_{p}}{d_{q}} = \frac{3}{2} \end{matrix}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361184 A cube of side $a$ and mass $m$ just floats on the surface of water as shown in the figure. The surface tension and density of water are $T$ and $ρ$ , respectively. If angle of contact between cube and water surface is zero, find the distance $h$ (in metres) between the lower face of cube and surface of the water.
supporting img

1

\frac{m g + a T}{a^{2} ρ g}

2

\frac{m g + 4 a T}{a^{2} ρ g}

3

\frac{m g + 2 a T}{a^{2} ρ g}

4

\frac{4 a T}{a ρ g}

Explanation:

At equilibrium $m g + F_{S T} = F_{b}$
$m g + 4 a T = a^{2} h ρ g$
supporting img

$h = \frac{m g + 4 a T}{a^{2} ρ g}$

PHXI10:MECHANICAL PROPERTIES OF FLUIDS

361185 If a liquid rises to the same height in two capillaries of the same material at the same temperature, then

1 The weight of liquid in both capillaries are not equal

2 The radius of meniscus must be equal

3 The capillaries must be cylindrical and vertical

4 Given conditions are not possible

Explanation:

$h = \frac{2 T \cos θ}{r ρ g}$